Python intro ch_e_comp

Introduction to Chemical Engineering Computing:

Extension to Python

Bruce A. Finlayson

Copyright, 2017

This pdf illustrates how to use the programming language Python to solve the problems
posed in the book Introduction to Chemical Engineering Computing, Bruce A. Finlayson,
Wiley (2006-2014). The material mirrors the use of MATLAB in the book, and solves the
examples in Chapters 2, 3, 4, and 8. In addition, a new Appendix F gives summaries of
many Python commands and examples that can be used independently of any
chemical engineering applications. Together with the book, chemical engineering
applications are illustrated using Microsoft Excel®, MATLAB®, Aspen Plus®, Comsol
Multiphysics®, and Python. This pdf is intended to be used in conjunction with the book,
and the treatment using Python mirrors that using MATLAB. Appendix F includes
information for setting up Python on your computer. This pdf is free for personal use
and is available at www.chemecomp.com, Supplement Using Python.

Introduction to Chemical Engineering Computing, Chapter 2 with Python– Copyright, Bruce A. Finlayson, 2017 2
CHAPTER 2. EQUATIONS OF STATE USING PYTHON
SOLVING EQUATIONS OF STATE (SINGLE EQUATION IN ONE UNKNOWN)
Nonlinear algebraic equations can be solved using Python, too. First, you have to define
the problem to solve by defining a function; then, you check it; finally, you issue a command to
solve it. See Appendix F for additional details.
Step 1 Import the fsolve program from SciPy and define the function (note the indentation after
the declaration def).
from scipy . optimize import fsolve
def f(x):
return x ** 2 - 2 * x - 8
Step 2 Check the function. Issue the command: print(f(1.1)) to get the result: –8.99. You can
easily calculate Eq. (2.8) to see that for x = 1.1, the function value is –8.99. Now you know the
function f is correct. Note that we used a value for x that meant that every term in the function
was important. If we had used x = 1e-5, then the x*x term would be negligible unless we
computed it to ten significant figures; hence we wouldn’t have checked the entire function. The
value x =1.0 is not a good choice either, since an incorrect function x-2*x-8 would return the
same value as x*x-2*x-8, hence the error would not be discovered. This is a trivial example, and
it is more important for more complicated problems.
Step 3 To find the value of x that makes f(x) = 0 in Python, use the ‘fsolve’ function. In the
command window, issue the following command.
x = fsolve (f, 0) # one root is at x = -2.0
print (' The root is %5.3f.' % x)
This command solves the following problem for x: starting from an initial guess of 0.
The answer is -2.0. You can test the result by saying:
print(f(x)))
which gives [ -2.55795385e-13]. Sometimes the function will have more than one solution, and
that can be determined only by using the command with a different initial guess.
To summarize the steps, step 1 defined the problem you wished to solve and evaluated it
for some x, step 2 checked your programming, and step 3 instructed Python to solve the problem.
It is tempting to skip the second step – checking your programming – but remember: if the
programming is wrong, you will solve the wrong problem. The last command gives a further
check that the zero of the function has been found.
When examining the command x = fsolve (f, x0), the f defines which problem to solve,
the x0 is your best guess of the solution and fsolve tells Python to vary x, starting from x0 until
the f is zero.
In all the commands, the f can be replaced by other things, say prob1. The answer can
€
f (x) = 0

Using Python in Chapter 2 – Copyright, Bruce A. Finlayson, 2017 3
also be put into another variable name: z = x.
z = x
print(z)
In the last example the result is put into the variable z. The options vector allows you to set
certain quantities, like the tolerance. Add to the fsolve command: xtol=1.5e-8. For the example
used above, you can find the other root by running the program with x0 = 3. Multiple roots can
be found only if you search for them starting with different guesses.
Example of a Chemical Engineering Problem Solved Using Python
Find the specific volume of n-butane at 500 ºK and 18 atm using the Redlich-Kwong
equation of state.
Step 1 First, you need to define the function that will calculate the f(x), here specvol(v), given
the temperature, pressure, and thermodynamic properties. The file is shown below.
def specvol(v):
# in K, atm, l/gmol
# for n-butane
Tc = 425.2
pc = 37.5
T = 393.3
p = 16.6
R = 0.08206
aRK = 0.42748 * (R * Tc) ** 2 / pc
aRK = aRK * (Tc / T) ** 0.5
bRK = 0.08664 * (R * Tc / pc)
return p * v ** 3 - R * T * v ** 2 + (aRK - p * bRK ** 2 - R * T * bRK) * v - aRK * bRK
This function, called specvol, defines the problem you wish to solve.
Step 2 To test the function specvol you issue the command:
print(specvol(2))
and get 25.98, which is correct. The specvol function causes Python to compute the value of the
function specvol when v = 2. You should check these results line by line, especially the
calculation of aRK, bRK, and y (just copy the code except for the return statement and calculate
aRK and bRK with a calculator.. Alternatively, you can use the spreadsheet you developed, put
in v = 1.506 and see what f(v) is; it should be the same as in MATLAB since the cubic function
and parameters are the same.
Step 3 Next you issue the command:
v = fsolve(specvol,2)

print(v)
and get 1.5064. In specvol the 2 is an initial guess of the answer. To check, you might evaluate
the function to find how close to zero f(v) is.
print (specvol(v))
and get 1.8e-15. Of course you expect this to be zero (or very close to zero) because you expect
Python to work properly. If Python can’t find a solution, it will tell you. If you use an initial
guess of 0.2, you might get the specific volume of the liquid rather than the gas. Python gives
0.18147.
Another Example of a Chemical Engineering Problem Solved Using Python
Next rearrange the Python code to compute the compressibility factor for a number of
pressure values. The compressibility factor is defined in Eq. (2.10).
(2.10)
For low pressures, where the gas is ideal, the compressibility factor will be close to 1.0. As the
pressure increases, it will change. Thus, the results will indicate the pressure region where the
ideal gas is no longer a good assumption. The following code solves for the Redlich-Kwong,
Redlich-Kwong-Soave, and Peng-Robinson equations of state and plots the compressibility
factor versus pressure as in Figure 2.3.
import numpy as np
import pylab as plt
# n-butane Redlich-Kwong, Eq. (2.5)
def specvolRK(v, p):
# in K, atm, l/gmol
# for n-butane
Tc = 425.2
pc = 37.5
T = 500
R = 0.08206
aRK = 0.42748 * (R * Tc) ** 2 / pc
aRK = aRK * (Tc / T) ** 0.5
bRK = 0.08664 * (R * Tc / pc)
return p * v ** 3 - R * T * v ** 2 + (aRK - p * bRK ** 2 - R * T * bRK) * v - aRK * bRK
€
Z =
pv
RT

# n-butane Redlich-Kwong-Soave, Eq. (2.5)
def specvolRKS(v, p):
# in K, atm, l/gmol
# for n-butane
Tc = 425.2
pc = 37.5
T = 500
R = 0.08206
acentric = 0.193
mRKS = 0.480 + (1.574 - 0.176*acentric)*acentric
alphaRKS = (1 + mRKS *(1-(T/Tc)**0.5)) ** 2
aRKS = 0.42748 * alphaRKS * (R * Tc) ** 2 / pc
bRKS = 0.08664 * (R * Tc / pc)
return p * v ** 3 - R * T * v ** 2 + (aRKS - p * bRKS ** 2 - R * T * bRKS) * v - aRKS * bRKS
# n-butane Peng-Robinson, Eq. (2.6)
def specvolPR(v, p):
# in K, atm, l/gmol
# for n-butane
Tc = 425.2
pc = 37.5
T = 500
R = 0.08206
acentric = 0.193
mPR = 0.37363 + (1.54226 - 0.26992*acentric)*acentric
alphaPR = (1 + mPR *(1-(T/Tc)**0.5)) ** 2
aPR = 0.45724 * alphaPR * (R * Tc) ** 2 / pc
bPR = 0.07780 * (R * Tc / pc)
return p*v**3+(bPR*p - R*T)*v**2+(aPR- *p*bPR**2- *R*T*bPR)*v +
(p*bPR**3 + R*T*bPR**2-aPR*bPR)
T = 500
R = 0.08206
pressure = np.arange(1, 27, 5)
print(pressure)
print(pressure[0])
print(pressure[5])
zcompRK = np.zeros(6,dtype=float)
zcompRKS = np.zeros(6,dtype=float)
zcompPR = np.zeros(6,dtype=float)
print(zcompRK)
for i in range(0, 6, 1):
p = pressure[i]
guess = R*T/p
v = fsolve(specvolRK, guess, p)
z = p * v / (R * T)
zcompRK[i] = z

v = fsolve(specvolRKS,v,p)
z = p * v / (R * T)
zcompRKS[i] = z
v = fsolve(specvolPR,v,p)
z = p * v / (R * T)
zcompPR[i] = z
print(zcompRK)
print(zcompRKS)
print(zcompPR)
plt . plot (pressure,zcompRK,'o-g',label='Redlich-Kwong')
plt . plot (pressure,zcompRKS,'x-b',label='Redlich-Kwong-Soave')
plt . plot (pressure,zcompPR,'s-r',label='Peng-Robinson')
plt . legend(loc='best)')
plt . xlabel('Pressure (atm)')
plt . ylabel('Z')
plt . title ('n-Butane')
plt . show()
Figure 2.3. Compressibility factor for n-butane, using Python
The first three commands bring in the needed routines – fsolve, numpy, and pylab (for plotting.
Then there are three definitions of functions that define the equation governing the specific
volume, Eq. (2.5) and (2.6). The main program sets the temperature, provides a vector of 6
pressures, equidistant from 1 to 27; pressure = [1, 6, 11, 16, 21, 26]. The index goes from 0 to
5. The vectors of compressibilites are also set with 6 values for each equation of state, starting
with 0, to be filled as the calculations proceed. Then a loop calculation is made for i from 0 to 5.

For each pressure in turn, the guess for the Redlich-Kwong equation of state is the result from
the ideal gas law. The result from the Redlich-Kwong equation of state is used for the guess
when solving the Redlich-Kwong-Soave equation of state, and that solution is used as the gues
for the Peng-Robinson equation of state. For each i, after the compressibility is found it is put in
a vector for that equation of state. Finally the results are plotted, with three curves on one plot.

CHAPTER 3. VAPOR-LIQUID EQUILIBRIA USING PYTHON
Example using Python
Before working this example, explore Appendix F so that you have all the program parts
needed here.
We want to solve Eq. (3.9) using Python, given the Ki and zi.
(3.9)
This is a nonlinear equation to solve for . Thus, you can apply here the same methods used
with Python in Chapter 2. Once the value of is known, you can calculate the value of the
liquid compositions, {xi}, and vapor compositions, {yi}, using Eq. (3.8) and (3.1). The mole
balance is then complete.
(3.8)
(3.1)
The program to do this is shown here.
# vapor-liquid equilibrium
# this is necessary to get fsolve, which solves the non-linear equations
# function definition, Eq. (3.9)
def vpequil(v):
z = [0.0, 0.1, 0.3, 0.4, 0.2] #note the first value in the matrix is not used
K = [0.0, 6.8, 2.2, 0.8, 0.052]
# print(z): z and K can be printed when checking the program
# print(K)
sum1 = 0.
for i in range(1,5): # note: in python the range stops one before the last value
# see the discussionin Appendix F on vectors
num = (K[i] - 1.0) * z[i]
denom = 1.0 + (K[i] - 1.0) * v
sum1 = sum1 + num / denom
#print(num) #these can be printed when checking the function
#print(denom)
#print(sum1)
return sum1
€
(Ki −1)zi
1+ (Ki −1)v'i=1
NCOMP
∑ = 0
€
v'
€
v'
€
xi =
zi
1+ (Ki −1)v'
€
yi = Kixi

Introduction to Chemical Engineering Computing, Chapter 3 with Python–Copyright, Bruce A. Finlayson, 2017 9
# check the function
print ('For testing %10.6f ' % vpequil(0.2))
# find the solution
v = fsolve (vpequil, 0.2)
print (v)
# find the composition
z = [0.0, 0.1, 0.3, 0.4, 0.2]
K = [0.0, 6.8, 2.2, 0.8, 0.052]
x = np.zeros(5,dtype=float) # these are necessary to introduce vectors
y = np.zeros(5,dtype=float)
for i in range(1,5): # remember that the indices go from 0 to 4
# we skip the first one, and the loop goes to one below 5, or 4
x[i] = z[i]/(1.0 + (K[i]-1.0)*v)
y[i] = K[i]*x[i]
print ('The liquid mole fractions are')
print (x)
print ('nThe vapor mole fractions are')
print (y)
The output is
For testing 0.241549
The vapor fraction is 0.425838
The liquid mole fractions are
[ 0. 0.0288196 0.19854325 0.43723857 0.33539858]
The vapor mole fractions are
[ 0. 0.19597326 0.43679516 0.34979086 0.01744073]
These agree with the results from Excel and MATLAB.

CHAPTER 4. CHEMICAL REACTION EQUILIBRIA USING PYTHON
Solution of Tables 4.2 and 4.3 Using Python
To solve Eq. (4.15) using Python, you define a function that will calculate the right-hand
side and use fsolve to find the value of x that makes it zero.
€
f (x) =148.4 −
x2
(1− x)2
(4.15)
The program that does this is shown here and explained step by step below.
def equil_eq(x):
return 148.4 - x*x/(1.0-x)**2
x = fsolve (equil_eq, 0.5)
print ('The root x is %10.5f' % x)
Step 1 Import fsolve from scipy.optimize.
Step 2 Construct a function evaluates the function, given x. The name is equil_eq and it is listed
below.
def equil_eq(x):
return 148.4 - x*x/(1.0-x)**2
Step 3 Call fsolve to find the value of x that makes the function equil_eq zero, using 0.5 as the
initial guess.
x = fsolve (equil_eq, 0.5)
Step 4 Print the result.
The result is:
The root x is 0.92414
The program is easily changed to allow different inlet mole fractions, rather than pure
carbon monoxide and water, i.e. Table 4.3. The data is entered in the function as a vector param
that is set outside the function.

Introduction to Chemical Engineering Computing, Chapter 4 with Python – Copyright, Bruce A. Finlayson, 2017 11
def equil_eq(x,param):
COin = param[1]
H2Oin = param[2]
CO2in = param[3]
H2in = param[4]
Kequil = 148.4
CO = COin - x
H2O = H2Oin - x
CO2 = CO2in + x
H2 = H2in + x
return Kequil-CO2*H2/(CO*H2O)
param = np.zeros(5,dtype=float)
param[1] = 1.
param[2] = 1.8
param[3] = 0.3
param[4] = 0.1
x = fsolve (equil_eq, 0.9, param)
The solution is the same as that found in Table 4.3.
The conversion x is 0.98836
Some times the solution cannot be found and you must try again with a different initial guess.
That was case here, and fsolve did not converge for an initial guess of 0.5.
Multiple Equations, Few Unknowns Using Python
Suppose you want to solve the following two equations:
€
10x + 3y2
= 3
x2
− exp(y) = 2
(4.16)
These can be solved using the fsolve program. Since the exponential is used it must be imported
from scipy, too.
from scipy import exp
import numpy as np
def prob2(p):
# vector components are transferred to the function
x = p[0]
y = p[1]
return 10*x + 3*y*y -3, x*x - exp(y) - 2

p = np.zeros(2,dtype=float)
p[0] = 1.5
p[1] = 2.5
p = fsolve(prob2,p)
print ('x = %8.5f' % p[0])
print ('y = %8.5f' % p[1])
The result is
x = -1.44555
y = -2.41216
The vector p could also be defined by the following command.
p = np.append(1.5, 2.5)
The function can also be calculated with the final solution to verify that it is correct.
x = p[0]
y = p[1]
z = 10*x + 3*y*y -3
print('The function is %10.5e ' % z)
z = x*x - exp(y) - 2
print(' %10.5e ' % z)
The result is
The function is -3.37508e-14
4.31655e-13
which is close to zero and indicates the solution is good.

CHAPTER 8. CHEMICAL REACTORS USING PYTHON
USING PYTHON TO SOLVE ORDINARY DIFFERENTIAL EQUATIONS
Simple Example
In Python, you define the problem by means of a function following the def command.
You then tell Python to solve the differential equation. This process is illustrated using a single,
simple differential equation:
(8.16)
Integrate this equation from t = 0 to t = 1. The exact solution can be found by quadrature and is
(8.17)
The program to do this is:
from scipy.integrate import odeint
import matplotlib.pyplot as plt
import numpy as np
def f(y, t):
# this is the rhs of the ODE to integrate , i.e. dy/dt=f(y,t)
return -10 * y
y0 = 1 # initial value
a = 0 # integration limits for t
b = 1
tspan = np. arange (a, b, 0.05) # values of t for which we require the solution y(t)
y = odeint (f, y0 , tspan) # actual computation of y(t)
# plot the solution
plt . plot (tspan, y)
plt . xlabel ('t'); plt . ylabel ('y(t)')
plt . show ()
# print the solution
print(tspan, y)
The import commands are necessary to get the programs that will be used. Then the function is
defined, f(y,t), which is the right-hand side of the differential equation. The time span is set and
values of the solution are to be obtained at intervals of 0.05. The odeint is called to solve the
differential equation, and it is then plotted, giving the same graph Figure 8.1.
€
dy
dt
= −10y, y(0) =1
y(t) = e − 10 t

Introduction to Chemical Engineering Computing, Chapter 8 with Python – Copyright, Bruce A. Finlayson,
2017
14
Passing Parameters
Still another way to introduce into the function is to use it as a parameter in the calling
argument. Modify the function and the calling arguments as follows. The red parts are the lines
that are changed.
def rhs(y,t,krate):
return - krate * y
b = 1
tspan = np. arange (a, b, 0.05) # values of t for which we require the solution y(t)
krate = 10.0 # set the rate constant
y = odeint (rhs, y0 , tspan, args=(krate,)) # actual computation of y(t)
# note how the krate has to be put into an args tuple
The solution is the same as before.
Example: Isothermal Plug Flow Reactor
The equations for all three species in the plug flow reactor are (p. 144)
(8.21)
At the inlet
(8.22)
and we take u = 0.5 m/s, = 0.3 m3
/kmol s, and the total reactor length as z = 2.4 m.
Step 1 The Python program requires a function that defines the right-hand side. The input
parameters to the function are the concentrations of all species. The function also needs the
velocity, u, and the rate constant, k. The distance from the inlet, z, takes the place of time and is
the independent variable. The code for the function follows. Note the fact that three values are
returned, the right-hand sides of the three derivatives.
def ydot(y, t):
# y(0) is CA, y(1) is CB, y(2) is CC
# k = 0.3 and u = 0.5
CA= y[0]
rate = 0.3*CA*CA
return (-2.*rate/0.5, +rate/0.5, 0.)
€
k
€
u
dCA
dz
= −2kCA
2
, u
dCB
dz
= +kCA
2
, u
dCC
dz
= 0
€
CA (0) = 2 kmol/m3
, CB (0) = 0, CC (0) = 2 kmol/m3
€
k

Step 2 You test this function by calling it with specific values for y to ensure that it is correct.
y0 = np.zeros(3,dtype=float)
y0 [0] = 0.2
y0 [1] = 0.3
y0 [2] = 0.4
a = 0.
VR = 2.6
tspan = np. arange (a, VR, 0.2)
print(ydot(y0,tspan))
gives the same answer as with MATLAB (page 145): -0.048, 0.024, 0. This is a very important
step, because this is where you add value. Python will integrate whatever equations you give it,
right or wrong, and only you can ensure that the program has solved the right equations.
Step 3 Next, write the code that serves as the driver. This code must (a) set any constants (here
they are just put into the function rate1 for simplicity), (b) set the initial conditions and total
reactor length, and (c) call the odeint solver.
y0[0] = 2.0
y0[1] = 0.0
y0[2] = 2.0
a = 0.
VR = 2.6
y = odeint(ydot,y0,tspan)
Step 4 The solution is then printedand plotted. The plot is the same as Figure 8.3.
print(tspan, y)
print(y[:,0])
# plot the solution
plt . plot(tspan, y[:,0],'*-')
plt . plot(tspan, y[:,1],'+-')
plt . plot(tspan, y[:,2],'x-')
plt . xlabel ('length (m)')
plt . ylabel ('concentrations (kgmol/m^3)')
plt .show()

2017
16
Example: Nonisothermal Plug Flow Reactor
Consider the model of a simple reactor oxidizing SO2 to form SO3 treated on pp. 146-9.
The equations are
(8.24)
where the reaction rate is
(8.25)
(8.26)
with the parameters: . The variable X is the concentration of
SO2 divided by the inlet concentration, 1–X is the fractional conversion, and T is the temperature
in K. The first equation is the mole balance on SO2, and the second is the energy balance. The
first term on the right-hand side of Eq. (8.24) represents cooling at the wall; the second term
there is the heat of reaction. The Python program to solve these equations follows. Note that
both the exp and sqrt functions need to be imported.
from scipy import exp, sqrt
import numpy as np
# define the functions or right-hand sides
def ydot(y, t):
# y(0) is X, y(1) is T
X = y[0]
T = y[1]
k1 = exp(-14.96 + 11070 / T)
k2 = exp(-1.331 + 2331 / T)
Keq = exp(-11.02 + 11570 / T)
term1 = X * sqrt(1 - 0.167 * (1 - X))
term2 = 2.2 * (1 - X) / Keq
denom = (k1 + k2 * (1 - X)) ** 2
rate = (term1 - term2) / denom
return (-50 * rate , -4.1 * (T - 673.2) + 1.02e4 * rate)
# set the initial conditions
y0[0] = 1.0
y0[1] = 673.2
€
dX
dz
= −50R',
dT
dz
= −4.1(T − Tsurr ) +1.02 104
R'
€
R'=
X[1− 0.167(1− X)]1/2
− 2.2(1− X)/Keq
[k1 + k2(1− X)]2
€
lnk1 = −14.96 +11070/T, lnk2 = −1.331+ 2331/T, lnKeq = −11.02 +11570/T
€
Tsurr = 673.2, T(0) = 673.2, X(0) =1

# set the integration limits and call odeint
a = 0.
L = 1.05
tspan = np. arange (a, L, 0.05)
print(tspan, y)
print(y[:,0])
# plot the solution in two plots
plt . plot(tspan, y[:,0],'*-')
plt . xlabel ('Dimensionless axial position (m)')
plt . show()
plt . plot(tspan, y[:,1],'+-')
plt . show()
The solution is the same as shown on page 149, Figure 8.3.
(a)

2017
18
(b)
Figure 8.3. (a) Fraction converted; (b) temperature, using Python
CHEMICAL REACTORS WITH MASS TRANSFER LIMITATIONS
When there are mass transfer limitations the reaction rate is calculated based on the
concentration on the solid catalyst, not the concentration in the fluid. As shown on p. 155, et.
seq. it is necessary to solver for the mass transfer as well as integrate the equation. The equations
are:
(8.39)
The mass transfer equation is:
(8.40)
Looking closely at Eq. (8.39)-(8.40) you can see that in order to solve the differential equations
in Eq. (8.39) you must solve Eq. (8.40) at every position z. Thus, this is a problem that combines
ordinary differential equations with nonlinear algebraic equations.
Python easily handles these kind of problems. We have to define a function to solve Eq.
(8.40) and call that function inside the function that defines the ordinary differential equations
(8.39). The function for Eq. (8.40) must appear in the code before the one for Eq. (8.39).
Basically you call a routine to integrate the ordinary differential equations (e.g., odeint). You
construct a right-hand side function (here called ydot) to evaluate the right-hand side. The input
€
u
dCA
dz
= −2ksCA,s
2
, u
dCB
dz
= +ksCA,s
2
, u
dCC
dz
= 0
€
kma(CA − CA,s) = ksCA,s
2

variables are z and the three concentrations, and the output variables are the three derivatives.
Take CA and solve Eq. (8.40) using the function mass_rxn to find CA,s at this location, z. Then
evaluate the rates of reaction in Eq. (8.39). The program follows.
def mass_rxn(CAs,CA): # define the mass balance equation
k = 0.3
km = 0.2
return km * (CA - CAs) - k * CAs * CAs
def ydot(y, t): # define the ordinary differential equations
CA= y[0]
CAguess = CA
CAs = fsolve(mass_rxn,CAguess,CA)
rate = 0.3*CAs*CAs
return (-2.*rate/0.5, +rate/0.5, 0.)
y0[0] = 2.0
y0[1] = 0.0
y0[2] = 2.0
# set the integration parameters
a = 0.
VR = 2.6
# integrate the ordinary differential equations
print(tspan, y) # print the solution
print(y[:,0])
# plot the solution
plt . plot(tspan, y[:,0],'*-',label='$A$')
plt . plot(tspan, y[:,1],'+-',label='$B$')
plt . plot(tspan, y[:,2],'x-',label='$C$')
plt . legend()
plt . xlabel ('Length (m)')
plt . ylabel ('Concentrations (kgmol/m^3)')
plt .show()
The solution is shown in Figure 8.10, which is the same as found with MATLAB. With mass
transfer resistance included, the outlet concentration of B is 0.61. When there was no mass
transfer limitation, the outlet concentration of B was 0.85. Thus, the reactor is not able to
produce as much product, and a bigger reactor is required.

2017
20
Figure 8.10. Solution to Eq. (8.39)-(8.40) when mass transfer is important, using Python
CONTINUOUS STIRRED TANK REACTORS
Eq. (8.15) gave the mass balance for a continuous stirred tank reactor (CSTR). A similar
equation can be written as an energy balance. This example considers a CSTR in which a first-
order reaction occurs, but the temperature also changes due to the heat of reaction. The
equations to be solved are:
(8.53)
The left-hand sides are the flow rate times a concentration or temperature difference between the
input and output, divided by the volume. The equations have been normalized by the inlet
concentration and temperature. The right-hand sides are the rate of reaction and the rate of
energy generation due to reaction, respectively.
The case described by Eq. (8.53) is for an adiabatic reactor. When the reactor is
adiabatic, the equations can be combined by multiplying the first equation by β and adding it to
the second equation; then the right-hand side vanishes.
(8.54)
This equation can be solved for T.
€
Q
VR
(1− c) = c exp[γ(1−1/T)]
Q
VR
(1− T) = −β c exp[γ(1−1/T)]
€
β(1− c) + (1− T) = 0

(8.55)
Now the mass balance can be considered a single equation in one unknown (c).
(8.56)
Solution using Python
The program to solve this equation just requires using fsolve to find the roots to a single
equation.
from scipy optimize import fsolve
import numpy as np
# define the function
def rate_T(c,param):
beta = param[1]
gamma = param[2]
flowvol = param[3]
T = 1 + beta*(1.0-c)
rate = c * exp(gamma * (1.0 - 1.0/T))
return flowvol * (1.0 - c) - rate
# set the parameters
param[1] = 0.15
param[2] = 30.0
param[3] = 8.7
print(param)
# print(rate_T(0.5,param)) #used for testing
# solve the problem
c = fsolve(rate_T,0.5,param)
print ('The root c is %10.4f' % c)
The answer is the same as before, 0.7311.
CSTR with multiple solutions
For a different set of parameters, the CSTR can have more than one solution. For this
problem, the solutions all lie between 0 and 1, because the concentration has been normalized by
the inlet value, where the normalized concentration is 1.0, and the reaction uses up the material.
Which solution you get depends upon the initial guess of c. Use Python to solve the problem
€
T =1+ β(1− c)
€
Q
VR
(1− c) = c exp[γ(1−1/{1+ β − βc})]

2017
22
when Q/VR = 25, β = 0.25, keeping γ = 30. Successive trials led to the results shown in Table
8.1, which are the same solutions obtained using Excel.
Initial guess of c Final result
0 0.0863
0.5 0.5577
1.0 0.9422
Table 8.1. Multiple solutions to Eq. (8.56) when Q/VR = 25, β = 0.25, γ = 30, using Python
Solutions to multiple equations using MATLAB.
When two or more variables must be found, as in Eq. (8.53), a solution can be found to
make both the equations zero without rearrangement like that to produce Eq. (8.56). The main
requirement is that the return statement has two parts to it (separated by commas).
def rate_T(y,param):
beta = param[1]
gamma = param[2]
flowvol = param[3]
c = y[0]
T = y[1]
rate = c * exp(gamma * (1.0 - 1.0/T))
return (flowvol * (1.0 - c) - rate, flowvol * (1.0 - T) + beta * rate)
param[1] = 0.15
param[2] = 30.0
param[3] = 8.7
y = fsolve(rate_T,[0.5, 1.1],param)
print ('The concentration is %10.4f ' % y[0])
print ('The temperature is %10.4f ' % y[1])
The answer is the same as before, 0.7311 and 1.0403.
TRANSIENT CONTINUOUS STIRRED TANK REACTORS
Reactors don’t always run at steady state. In fact, many pharmaceuticals are made in a
batch mode. Such problems are easily solved using the same techniques presented above
because the plug flow reactor equations are identical to the batch reactor equations. Even CSTRs
can be run in a transient mode, and it may be necessary to model a time-dependent CSTR to
study the stability of steady solutions. When there is more than one solution, one or more of
them will be unstable. Thus, this section considers a time-dependent CSTR as described by Eq.
(8.57).

(8.57)
The variables are
V = reactor volume, Q = volumetric flow rate
k0 = reaction rate constant, E = activiation energy
φ = void fraction, ρ = density, Cp = heat capacity per mass (8.58)
subscript f for fluid, s for solid
–ΔHrxn = heat of reaction, energy per mole
The non-dimensional form of these equations is
(8.59)
The parameters are defined as
(8.60)
The parameter Le is a Lewis number, and it includes the heat capacity of the system. The Da is a
Damköhler number and includes the rate of reaction. The parameters are taken as
(8.61)
import numpy as np
def ydot(y, tspan, param):
c = y[0]
T = y[1]
Damk = param[0]
Tin = param[1]
beta = param[2]
Lewis = param[3]
gamma = param[4]
€
V
dc'
dt'
= Q(c'−c'in ) −Vk0c'exp(−E /RT')
[φ(ρCp )f + (1− φ)(ρCp )s]V
dT'
dt'
= −(ρCp )f Q(T'−T'in ) + (−ΔHrxn )Vk0c'exp(−E /RT')
€
c'= concentration, T'= temperature, t'= time
€
dc
dt
= (1− c) − c • Da • exp[γ(1−1/T)]
Le
dT
dt
= (1− T) + β • c • Da • exp[γ(1−1/T)]
€
c =
c'
c'in
, T =
T'
T'in
, t =
Qt'
V
, Da =
V
Q
k0 exp(−
E
RT'in
)
Le =
φ(ρCp )f + (1− φ)(ρCp )s
(ρCp )f
, β =
(−ΔHrxn )c'in
(ρCp )f T'in
€
β = 0.15, γ = 30, Da = 0.115, Le =1080, c(0) = 0.7, T(0) =1

2017
24
rate = c*Damk*exp(gamma * (1.0 - 1.0/T))
return (1 - c - rate, (Tin - T - beta*rate)/ Lewis)
y0[0] = 0.7
y0[1] = 1.0
# set the integration conditions
tstart = 0.
tend = 2.1
tspan = np. arange (tstart, tend, 0.1)
# set the parameters
# Damk, Tin, beta, Lewis, gamma
param = (0.115, 1.0, 0.15, 1080.0, 30.0)
print(param)
# integrate the equations
y = odeint(ydot,y0,tspan,args=(param,))
print(tspan, y)
# plot the solution
plt . plot(tspan, y[:,0],'*-',label='$Concentration$')
plt . xlabel ('Time')
plt . ylim(0.65, 1.05)
plt . plot(tspan, y[:,1],'+-',label='$Temperature$')
plt . legend(loc=(0.1,0.5))
plt . xlabel ('Time')
plt . show()
Figure 8.15. Transient CSTR, up to t = 2, using Python

The steps in the program are the same as in other applications: define the function, set the initial
conditions, integration limits, parameters, integrate the differential equations, print the results,
and plot the graph. The only difference is that parameters are passed to the function, and these
need to be defined as a tuple, the tspan and param.
The result is shown in Figure 8.15. It looks like steady state is achieved by the time that t = 2.
This is not true, however. Integrate to t = 1000 and look at the results in Figure 8.16. It still has
not reached steady state. The reason is that the temperature responds much more slowly than
does the concentration. Thus, the concentration comes to a steady state value appropriate to the
current temperature, but then the temperature keeps changing and the concentration must change
to keep up. Notice the very rapid change of c from the initial value of 0.7 to about 0.89 in
Figure 8.16. This is because the value of c = 0.7 was not appropriate for a temperature of 1.0. In
mathematical terms, the time response of the two variables is very different (the eigenvalues of
the equation are widely separated), and the system is called stiff. See Appendix E for more
discussion about stiff equations.
Figure 8.16. Transient CSTR, up to t = 1000, using Python
Next we integrate to t = 40,000, as shown in Figure 8.17 with the following changes in the
program.
tstart = 0.
tend = 41250.0
This looks funny, but it is because the first data point for concentration (after the initial
condition) misses some of the detail shown in Figure 8.16. One way to improve the figure is to
just leave out the 1250.0 in tspan. Then the odeint will plot whatever points it has solved for.
Another solution is to integrate to 1000 and use that as the initial guess for another calculation to
40,000. The changes in the program are as follows.

2017
26
# set the integration conditions
tstart = 0.
tend = 1020.0
# integrate to t = 1000.
y = odeint(ydot,y0,tspan,args=(param,))
# set the integration conditions again
tstart = 1000.0
tend = 40000.0
# set the initial conditions to the ending conditions of the previous calculation
y0[0] = y[50,0]
y0[1] = y[50,1]
# continue the integration
y2 = odeint(ydot,y0,tspan,args=(param,))
# plot the extension
plt . plot(tspan, y2[:,0],'*-')
plt . plot(tspan, y2[:,1],'+-')
plt .show()
Figure 8.17. Transient CSTR, up to t = 40.000, using Python

Figure 8.17_revised. Transient CSTR, up to t = 40,000, using Python
Next change the parameter Le from 1080 to 0.1 and integrate to t = 100. The changes to
the code are:
param = (0.115, 1.0, 0.15, 0.1, 30.0)
and
tend= 100.025
tspan = np. 27rrange (tsta, tend, 0.025)
Figure 8.18 shows the limit cycle as the concentration and temperature never reach steady state.

2017
28
Figure 8.18. Transient CSTR, Le= 0.1, up to t = 100, using Python
Figure 8.19 shows the temperature plotted versus the concentration, and the limit cycle is clear.
Figure 8.19. Limit cycle display of Figure 8.18, using Python

Introduction to Chemical Engineering Computing, Appendix F – Copyright, Bruce A. Finlayson, 2017 29
APPENDIX F. HINTS WHEN USING PYTHON
This appendix provides hints and tips when using Python1
. Python is a programming
language first developed by Guido van Rossum in 1991, but it has been a good programming
language for scientific work only since about 2008. Since then mutually incompatible add-on
packages have been consolidated into a few compatible packages (see below).2
Python assumes
that you are a beginner in using Python, but not an absolute beginner in computer programming.
Most likely, you remember concepts from a computer programming class taken earlier. Included
in Appendix F are general features that are useful in all the applications solved with Python.
Other features are illustrated in the context of specific examples; a list of examples is provided at
the end of the appendix for handy reference. You’ll probably want to skim this appendix first,
then start working some of the problems that use Python to gather experience, and finally come
back and review this appendix in more detail. That way you won’t be burdened with details that
don’t make sense to you before you see where and how you need them.
Outline of Appendix F:
1. General features: loading Python, screen format, and creating a program, classes of
data
2. Programming options: input/output, functions, loops, conditional statements, timing,
matrices, matrix multiplication
3. Finding and fixing errors
4. Solving linear equations and finding eigenvalues of a matrix
5. Evaluate of an integral
6. Interpolation: splines and polynomials: spline interpolation, polynomial interpolation,
polynomials of degree n, fit a function to data and plot the result, fit a polynomial to
data
7. Solve algebraic equations
8. Integrate ordinary differential equations that are initial value problems: differential-
algebraic equations, checklist for using odeint
9. Plotting: simple plots, multiple plots, bold, italics, and subscripts, Greek letters,
contour plots, 3D plots
10. Python help
11. Applications of Python
F.1. GENERAL FEATURES
Loading Python, Screen Format, and Creating a Program
The web site https://www.python.org/ has options for downloading Python. The author
has chosen to use Pycharm, which is available at https://www.jetbrains.com/pycharm/; click
download. There is a Community version that is free, and that is adequate here. Documentation
1
Python is a programming language that is user-supported and available from the Python
Foundation: https://www.phthon.org The examples here use PyCharm ver. 3.6.
2
Python for Scientists, John M. Stewart, Cambridge University Press, Cambridge, England,
2014.

Appendix F – Hints to using Python, Copyright, Bruce A. Finlayson, 2017 30
is available at https://docs.python.org/3/library/index.html. Other sources include the
Introduction to Python for Computational Science and Engineering, by Hans Fangohr of the
University of Southhampton, available as a pdf from
https://www.southampton.ac.uk/~fangohr/training/python/pdfs/Python-for-Computational-
Science-and-Engineering.pdf and the books Chemical and Biomedical Engineering Calculations
Using Python, by Jeffrey J. Heys, Wiley, 2017 and Python For Scientists, by John M. Stewart,
Cambridge University Press, Cambridge, England, 2012.
The following websites may be useful.
http://docs.python.org/2/tutorial
http://matplotlib.org/api/pyplot_summary.html
http://matplotlib.org/gallery.html
http://www.scipy.org/NumPy_for_Matlab_Users
The following pdf documents are also useful, obtained from https://docs.scipy.org/doc/.
scipy-ref-0.18.1.pdf
numpy-user-1.11.0.pdf
The Python program includes many features, but not everything needed for scientific calculation.
Other modules can be added. The ones used here are math, numpy, scipy, and matplotlib. The
last three must be added to the same area where the Python program is stored. On a Macintosh,
open Terminal, back up one directory (cd ..) twice, then choose the directories in turn: cd usr, cd
local, cd bin. Then say python3.5 –m pip install numpy, etc. This need be done only once.
Then when you open Python you will be asked if you want to open an existing project or create a
new one. For a new one, provide a name. Then in the upper left choose the project name, then
File/New… in the pull-down menu. Several options are there; choose File and give it a name
ending with .py, such as screen.py. The upper right area is the editor where you type the program
screem.py, and the lower left area is where the output appears (unless you are writing to a file).
The screen will look like Figure F.1. Along the bottom there are several options. Python Consol
is where you can type in a Python command and immediately have it executed. That is useful for
testing commands and their syntax. Terminal is the area mentioned below for importing other
programs and can be accessed from this Python window. Run is where the output appears when
you run a program.

Figure F.1. Python Screen
However, when you type a command in the Python editor it will not execute until you say Run
(in the Run pull-down menu or the green triangle). If you just want to try some commands and
get the answer directly, there are several options. You can choose Python Console and enter the
command there. Or on the Macintosh open terminal independently of Python and say python.
The fact that you see >>> indicates that Python is waiting for input. The symbol >>> indicates a
computer command is expected and the result is printed immediately after you press return.
>>> 4+5
9
>>> type(4+5)
<class ‘int’>
In the terminal window you can say: import math, then dir(math) and see the functions that are
included. They include trigonometric functions, inverse trigonometric functions, hyperbolic
trigonometric functions, exponentials, error functions, as well as others. If you want to see what
modules you can import, go to the terminal window and say help(‘modules’). A long list is
provided which includes math, numpy, scipy3
, and matplotlib. To stop using the terminal as an
interpreter or consol say exit().
3
Some of the scipy programs require a FORTRAN compiler, but none of the examples here need
that. Legacy FORTRAN programs can be used; see Chapter 8 in Python For Scientists.

Once you are ready to create a program, type it in the editor window. Comment lines in
your code begin with a #. If you wish to have several lines as a paragraph of comments, then use
""" paragraph goes here """. In the examples below, the symbol >>> indicates a computer
command and the result is printed here immediately after. In actuality, you would put all the
computer commands together in the file and get all the output in the run window at the lower left
unless you were executing commands line by line in terminal.
The usual arithmetic symbols +, –, x, and / are used. Exponentiation is done using **, as in x**2.
In print commands with two or more variables (separated by commas) the comma itself provides
a space between the variables. The n indicates a carriage return.
There are also specialized programs submitted by others, but these require a compiler for another
language and are not treated here. You can see them at http://pypi.python.org/pypi.
Classes of Data
Variables can be integers, floating point numbers, strings, and Boolean variables. Setting
a = 4.0 provides a floating point number but setting b = 4 provides an integer.
>>>import numpy as np
>>>a = 4.0
>>>print(a)
4.0
>>>print(type(a))
<class 'float'>
b = 4
>>>print(b)
4
>>>print(type(b))
<class 'int'>
The floating point numbers involve 16 digits of precision, which is called double precision in C
and MATLAB. You can see all the digits in this way. Messages can be combined with the
numerical output for clarity.
b = 2.3456789
>>> print('long format %20.16e' % b)
long format 2.3456788999999998e+00
>>> print('long format %20.16f' % b)
long format 2.3456788999999998
Most of your work will involve numbers, but sometimes you will want to use text or words.
Strings are text.
>>> print("Chemical Engineering rules.")

Chemical Engineering rules.
>>>print(type(c))
<class 'str'>
Information read in by a file is read as a string and must be converted to float or int to use in
calculations.
A number can be converted to a character using the str command and back using the float
command.
>>>d = str(b)
>>>print(type(d))
<class 'str'>
>>>e = float(d)
>>>print(type(e)
<class 'float'>
Boolean variables are discussed below in conjunction with conditional statements.
Python is case sensitive so that t(i) and T(i) are different.
F.2. PROGRAMMING OPTIONS: INPUT/OUTPUT, FUNCTIONS, LOOPS,
CONDITIONAL STATEMENTS, TIMING, MATRICES
Input/Output
You can ask the user for input with the following command.
>>> viscosity = input('What is the viscosity (Pa s)?')
When you see ‘What is the viscosity (Pa s)?’ in the run-time window, type the value and press
return. Keep in mind that what comes in is treated as a string. Thus, if you typed 3.456 and
asked print(type(viscosity)) you would get <class 'str'>. You would have to use the
float(viscosity) command to convert it to another variable name as a floating point number. You
can test this with print(viscosity+viscosity), and you will get 3.4563.456. If you convert
viscosity to a float: viscosityf = float(viscosity) and printed viscosityf+viscosityf you would get
6.912. You can display in a specified format:
>>> a = 1.25456
>>>b = 5.4521
>>>print ('nThe values of a and b are %7.4f %5.3e' % (a, b))
gives
The values of a and b are 1.2546 5.452e+00
These are C-commands, and the notation means:

f – floating point
n is a carriage return
e – exponential format
5.3 means five characters, three after the decimal point.
To write a text file, you must open it, write it, and close it. The following commands write a text
file, closes the file, opens it and reads it.
# Open the file, write to it, and close it
>>>out_file = open ("test.txt", "w")
>>>out_file . write ("Writing text to file. This is the first line.n"+
"And the second line.")
>>>out_file . close ()
# Open the file, read it, and close it
>>>in_file = open ("test.txt", "r")
>>>text = in_file . read ()# read into a string variable ‘text’
>>>in_file . close ()
# Display the text
>>>print (text)
Writing text to file. This is the first line.
And the second line.
Numbers written to a file are written as text and must be converted to float or int when read in.
The first step is to put them into an array, then save the array as a text file. It is read as a text file
and is converted to a numbers.
>>> a = [2.345, 4.567, 6.789]
>>>np.savetxt('x.txt', a)
>>>xc=np.loadtxt('x.txt')
>>>print(xc)
[ 2.345 4.567 6.789]
>>>print (type(xc))
<class 'numpy.ndarray'>
>>>print (xc[1]*xc[2])
31.005363
Functions
A function is defined with the def command. The commands within the function must be
indented, and the last line returns the results. For example

>>>def f(x):
>>> return x ** 3 - 2 * x ** 2
will take the value of x and compute the number, returning it. The program following this
definition will not be indented – that is how the function definition is ended. The command
>>>print(f(3))
gives the value 9. There may be more than one argument, such as f(x,y,z), and numbers may be
assigned to them.
>>>def f(x=3,y=4,z)
In this case x will take the value 3, unless it is specified otherwise, and z must be specified before
the function is called. The function definition can be anywhere in the program as long as it is
defined before it is called. Examples with many input variables are given on pages 11, 21, 22, 44.
Loops
It is sometimes useful to execute a command over and over, putting each result in one element of
a vector. One option is to compute something when an index goes from a starting value to an
ending value.
>>>print ('loop using range')
>>>i = 0
# range(start, end, increment)
# does not include the last number, end
>>>for i in range(0,10,1):
>>> print(i)
>>> i = i + 1
The output is:
loop using range
0
1
2
…
8
9
Note that the loop does not include the ending value. The increment is optional. Another option
is to use the while command.
>>>print('nloop using while')

>>>i=0
# while(condition)
>>>while (i < 5):
>>> print(i)
>>> i=i+1
The output is:
loop using while
0
1
2
3
4
If you want the loop not to do the remaining calculations if a condition is true, the program will
continue the loop but not execute the rest of the commands, going back to the start of the loop.
>>>for i in range(…):
>>> <block1> # commands to be executed every time
>>> if <condition> #don’t execute the commands <block2>, but continue the loop
>>> continue
>>><block2>
If you want to stop before the end depending on a condition:
>>>for i in range(…):
>>> <block1> # commands to be executed every time
>>># don’t execute the commands <block2>, and get out of the loop
>>> if <condition>
>>> break
>>> <block2>
Conditional Statements
A Boolean number is either True or False. The following command produced 3.
>>>c = True
>>>if (c):
>>> print(3)
>>>else:
>>> print(2)
3
The following commands produced 2.
>>>c = False
>>>if not c:

>>> print(2)
>>>else:
>>> print(3)
2
Note the indentation of four spaces after the if/else commands. This is done automatically by
using a tab but is essential for telling where the section ends. Also note the : at the end of the
conditional statement.
Sometimes the program needs to execute different instructions depending upon a calculated
result. The following program prints negative if i is negative, 5 if it is between 0 and 5 (not
including 5) and 6 otherwise.
>>>if (i<0):
>>> print("negative")
>>>elif (i<5):
>>> print ("5")
>>>else:
>>> print ("6")
The other command are:
== which means the left-hand side is equal to the right-hand side,
not (var) == (condition) which means they are not equal,
(var) != (condition) which means they are not equal,
>= which means greater than or equal,
<= which means less than or equal.
One can also test two conditions with the <condition1> and <condition2> command. They both
have to be true for the expression following to be used. If one uses or then the expression
following is used if either condition1 or condition2 is true. The command not switches from true
to false.
Timing information
To measure the time a calculation takes, import time. Then say starttime = time.time() to
begin the clock. At a later point in the program say endtime = time.time(). Then the following
commands give you the time between those two calls.
>>>elapsedTime = endtime - starttime
>>>print('total time is %10.3f seconds' % elapsedTime
total time is 2.211 seconds
Matrices
Vectors are row (1x3)

>>>x = [2, 4, 6]
establishes a vector with three elements, numbered 0, 1, and 2; x(0) = 2,…x(2) = 6.
The matrix A
A =
2 6 6
8 10 12
1410 18
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
is set by
>>>A = ([2, 6, 6],[8, 10, 12], [14, 10, 18])
In Python these are called arrays, and each element of an array must be the same type. Python
has the capability of having lists, in which the elements need not be the same; some can be
numbers, some text, even pictures.
Two arrays can be added, multiplied, or divided, element by element.
>>>a = np.array([0,1,5])
>>>print(a)
[0 1 5]
>>>c = np.array([1,3,5])
>>>print(c)
[1 3 5]
>>>print(a+c)
[ 1 4 10]
>>>print(a*c)
[ 0 3 25]
>>>print(a/c)
[ 0. 0.33333333 1. ]
You can operate on one element of the array by using indices. In this example c[1] += 1 means
add 1 to c[1]. It could also be written c[1] = c[1] + 1.
>>>c[1] += 1
>>>print(c)
[1 4 5]

Matrix Multiplication
Take the matrix A =
2 6 6
8 10 12
1410 18
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
and the vector .
and compute the dot product.
A =
2 6 6
8 10 12
1410 18
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
2
3
4
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
=
46
94
130
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
(F.1)
import numpy as np
>>>AA=np.matrix([[2, 6, 6], [8, 10, 12], [14, 10, 18]])
>>>print(AA)
[[ 2 6 6]
[ 8 10 12]
[14 10 18]]
>>>Bb=np.array([[2], [3], [4]])
>>>print(Bb)
[[2]
[3]
[4]]
>>>ans = np.dot(AA,Bb)
>>>print(ans)
[[ 46]
[ 94]
[130]]
is [3x3][3x1] = [3x1] matrix. Note the brackets in Bb; Bb is a column vector, which is needed
for matrix multiplication. Note the difference between Bb and setting
>>>b=np.array([1, 3, 5])
>>>print(b)
[1 3 5]
F.3. FINDING AND FIXING ERRORS
One source of confusion is when you think you have defined a variable, but haven’t
really. Use print(x), Run to see the current value of x in the output window. The programs
illustrated in this book are fairly simple, and they are easy to debug. All programmers make
mistakes; good programmers learn to find them! Finding them is easy if you take the time. First
€
x'=
2
3
4
"
#
$
$
$
%
&
'
'
'

consider a line in a code. Set all the parameters used in the line and execute the line in the
console. Compare the answer with a calculation done with a calculator or one you did by hand.
Be sure that all the parameters you use are different, and don’t use zero or one, because some
mistakes won’t be picked up if the variable is zero or one. The parameters you use for testing
can be single digit numbers that are easy to calculate in your head, and they can be completely
unrelated to the problem you are solving. If you haven’t specified one of the parameters the
output will identify the missing parameter.
F.4. SOLVING LINEAR EQUATIONS AND FINDING EIGENVALUES OF A MATRIX
To solve the equation
A* x' = Bb , where A and Bb are
A =
2 6 6
8 10 12
1410 18
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
, Bb =
1
3
5
⎡
⎣
⎢
⎢
⎢
⎤
⎦
⎥
⎥
⎥
we say
>>>x=LA.solve(AA,Bb)
>>>print (x)
and get
[[ -2.50000000e-01]
[ -9.71445147e-17]
[ 4.16666667e-01]].
To get the inverse of A, say
>>>CC = LA.inv(A)
>>>print(CC)
[[-0.625 0.5 -0.125 ]
[-0.25 0.5 -0.25 ]
[ 0.625 -0.66666667 0.29166667]]
Multiplying A by CC gives a diagonal matrix.
>>>ans = np.dot(A,CC)
>>print(ans)
[[ 1.00000000e+00 0.00000000e+00 -8.32667268e-17]
[ 8.88178420e-16 1.00000000e+00 1.11022302e-16]
[ 1.77635684e-15 0.00000000e+00 1.00000000e+00]]

Eigenvalues are the polynomial roots to Eq. (F.2).
(F.2)
They are easy to calculate in Python using the eig command.
>>>import numpy
>>>import numpy . linalg as LA
>>>A = ([2, 4, 6],[8, 10, 12], [14, 16, 18])
>>>evalues , evectors = LA.eig (A)
>>>print ( evalues )
>>>print ( evectors )
gives the eigenvalues and eigenvectors.
[ 3.22336879e+01 -2.23368794e+00 -2.60735545e-15]
[[-0.23197069 -0.78583024 0.40824829]
[-0.52532209 -0.08675134 -0.81649658]
[-0.8186735 0.61232756 0.40824829]]
In this example, the matrix is nearly singular, and one of the eigenvalues is very small.
F.5. EVALUATE AN INTEGRAL
To evaluate the integral, Eq. F.3, import the function quad from scipy.integrate.
(F.3)
from scipy.integrate import quad
# function we want to integrate
def f(x):
return x * x
res, err = quad(f, 0, 2) # call quad to integrate f from 0 to 2
# print("The numerical result is {:f} (+-{g})".format(res,err))
print("The numerical result is {:f} ".format(res))
The output is
The numerical result is 2.666667
€
Aij − λδij = 0
€
area = x2
dx
0
2
∫

F.6. INTERPOLATION: SPLINES AND POLYNOMIALS
Spline Interpolation
Make a cubic B-spline pass through all the data points: [x(i), y(i), i = 1,...,m]. A cubic
spline is a cubic function of position, defined on small regions between data points. It is
constructed so the function and its first and second derivatives are continuous from one region to
another. It usually makes a nice smooth curve through the points. The following commands
create Figure F.2.
from scipy.interpolate import CubicSpline # import the functions
from math import cos
import numpy as np
x = np.arange (0, 11, 1) # use points from 0 to 10 with an interval of 1
y = np.cos(x) # calculate the cos of each x in the vector x
print (x)
print (y)
#Cubic Spline
f = CubicSpline(x, y) # compute the cubic spline
xnew = np.arange(0, 9, 0.1)# create a list of points closer together
ynew = f(xnew) # calculate the value at xnew using the interpolation
# function computed by Cubic Spline
plt.axis([0, 10, -1.5, 1]) # set the x- and y- axis
plt.plot(x, y, 'o', xnew, ynew, '-') # plot
plt.show()
Figure F.2. Spline fit of a set of data points
Polynomial Interpolation
To use a piecewise polynomial through the points, not a Cubic Spline, simply replace the
Cubic Spline code with the following.
0 2 4 6 8 10
−1.5
−1
−0.5
0
0.5
1

# Linear (default)
y = x * x
f = interpolate.interp1d (x, y)
xnew = np.arange(0, 9, 0.1)
ynew = f(xnew)
# use interpolation function returned by `interp1d`
plt.plot(x, y, '+', xnew, ynew, '-')
plt.show()
To use a quadratic polynomial, use the command
f = interpolate.interp1d (x, y, ‘quadratic’)
Polynomials of degree n
A polynomial of degree n can be created using poly1d.
(B.8)
p = poly1d([3,4,5])
In this case the polynomial is 3x2
+ 4x + 5 and that is printed out. The coefficients can be
obtained using
print (p.coeffs).
The polynomial can be differentiated
print(p.deriv())
to give 6x + 4 , and multiplied
print(p*p)
to give 9x4
+ 24x3
+ 46x2
+ 40x + 25 . To evaluate a polynomial use print(p(2)) to give 25.
Fit a Function to Data and Plot the Result
For data points x(i), yi(i),i =1,...,n fit a function that minimizes the least squares error.
import numpy as np
from scipy . optimize import curve_fit
import pylab as plt
€
y = p1xn
+ p2xn−1
+ ...+ pn x + pn+1

def f(x, a, b, c):
# Fit function y=f(x,p) with parameters p=(a,b,c).
return a * np. exp (- b * x) + c
# create fake data with n = 50 points
x = np. linspace (0, 4, 50)
y = f(x, a=2.5 , b=1.3 , c =0.5)
# add noise
yi = y + 0.2 * np. random . normal ( size =len (x))
# call curve fit function to get the best values of a, b, and c
popt , pcov = curve_fit (f, x, yi)
a, b, c = popt
print (" Optimal parameters are a=%g, b=%g, and c=%g" % (a, b, c))
# plotting
yfitted = f(x, * popt ) # equivalent to f(x, popt [0] , popt [1] , popt [2])
plt . plot (x, yi , 'o', label ='data $y_i$ ')
plt . plot (x, yfitted , '-', label ='fit $f(x_i)$')
plt . xlabel ('x')
plt . legend ()
plt . show ()
Figure F.3. Least Squares fit of data
Fit a polynomial to Data
To also get a polynomial giving the best fit to the same data, add the following code:
# fit with polynomial of order 3
pp = np.polyfit(x, yi, 3)

# print polynomial coefficient
print(pp)
# evaluate the polynomial at all x points
yfittedp = np.polyval(pp,x)
# output
print(x)
print(yfittedp)
plt . plot (x, yi, 'o', label ='data $y_i$ ')
plt . plot (x, yfittedp , '-', label ='fit $p(x_i)$')
plt . xlabel ('x')
plt . legend ()
plt . show ()
Figure F.4. Polynomial Least Squares Fit of Data
F.7. SOLVE ALGEBRAIC EQUATIONS
Solve for the vector . There are three steps.
Step 1 Import ‘fsolve’ from the scipy, the SCIentific Python package. Scipy must be imported to
your computer to add it to Pycharm.
Step2 Define the function.
€
fi ({y j }) = 0, i =1,...,n
€
{y j }

def prob2 (y):
u1, u2 = y
return (u1 ** 3 - 2 * u1 **2, u2 ** 2 - 2 * u2 + 1)
Step 3 Call ‘fsolve’ with an initial guess and print the results.
u1, u2 = fsolve (prob2, (3 , 1.5))
print ('nNow solving two equations with fsolven')
print (u1, u2)
The result is
Now solving two equations with fsolve
2.0 1.00000001516
You need to make several checks. The number of unknowns is set by the number of elements in
the initial guess, (in parentheses in the fsolve command). The defined function ‘prob2’
calculates , and there have to be as many elements in f as there are in y.
The function needs to be checked, of course. The only way you can make Python find the
solution to your problem is to make sure the function gives the correct set of f’s when given a set
of y’s. If this ‘fsolve’ does not work well, try making an initial value problem and solving it
using implicit methods, integrating to a long time.
Alternatively, try a different initial guess.
There are other options for fsolve, too. These can be found in the document scipy-ref-
0.18.pdf; search on scipy.optimize.fsolve.
F.8. INTEGRATE ORDINARY DIFFERENTIAL EQUATIONS THAT ARE INITIAL
VALUE PROBLEMS
To solve a single ordinary differential equation
we use the ‘odeint’ functions in Python. There are four steps.
Step 1 Import odeint, and matplotlib.pyplot and numpy
import numpy as N
€
fi, i =1,...,n, given {y j}
€
dyi
dt
= fi({y j}), yi(0) = initial guess
€
dy
dt
= f (t,y), y(0) = y0

Step 2 Define the function.
def f(y, t):
return -2 * y * t
Step 3 Set the initial condition, y0 and time span for the integration as well as the values of t for
which we want the solution from 0 to 10 at intervals of 0.1.
b = 10
t = N. arange (a, b, 0.1)
Step 4 Call odeint.
y = odeint(f, y0, t) or y = odeint(f, 3, (0,10)) # actual computation of y(t)
Since odeint will integrate whatever equation you give it, and the function ‘f’ is going to be used
many times, you must insure that it is correct: given a t and y it computes the correct f(t,y) and
returns the variable which is ydot. Once you have checked the program and run it, it is easy to
plot the solution.
plt . plot (t, y)
plt . xlabel ('t'); plt . ylabel ('y(t)')
plt . show ()
Numerical values can be obtained from
print(t, y)
to give
[ 0. 0.1 0.2 ……
[ 3.00000000e+00]
[ 2.45619230e+00]
[ 2.01096019e+00]…
To solve a set of ordinary differential equations
we do the same steps except that the initial condition is now a vector with N elements, the
function ‘rhs’ must compute N functions using the vector y with N elements, and we have to
return the result for each derivative.€
dyi
dt
= fi({y j}), yi(0) = yi0

# define the function evaluating the right-hand side
def ydot(y, t):
return -y[0], -y[1], -y[2]
y0[0] = 1.
y0[1] = 2.
y0[2] = 3.
# set the integration limits and intervals for obtaining the solution
a = 0.
b = 11.
tspan = np. arange (a, b, 0.2)
# call odeint to integrate the equations
print(tspan, y)
# plot the solution
plt . plot(tspan, y)
plt .show()
Examples with multiple parameters passed to the function are shown on pages 23 and 24. If your
system is stiff (the eigenvalues of the linearized system are significantly different) the integration
may taking an interminable amount of time. With MATLAB you had to use ode15s, which is
designed for stiff systems. With Python, though, the odeint is apparently a compiled program,
and it runs very fast so this usually won’t be a problem.
Figure F.5. Plot of Solution of Three ODEs

Differential-Algebraic Equations
Some problems may have ordinary differential equations and algebraic equations, too. There are
add-ons to Python that will do this. See, for example, Assimulo at
https://pypi.python.org/pypi/Assimulo.
Checklist for Using ‘odeint’
When using ‘odeint’, your function for the right-hand side must meet these conditions:
• The name in the calling command must be the same as the function name.
• The variable tspan (or whatever it is called) must have at least two values.
• The number of entries in the vector for the initial conditions must be the same as the
number of right-hand sides calculated in the function.
• The number of right-hand sides computed has to be equal to the number of differential
equations. They are written in the return statement, with commas between them.
Alternatively, the calculations can be put into a vector and the return statement can
refer to the vector.
• Variables can be used in the function. Then can be part of a vector of numbers, or specific
entries and are listed as arguments in the def function.
• The function will be called many times by the odeint function. However, you only have to
check the calculation once. Check the function by giving it t and all the y(i); compute
what you expect the right-hand sides to be and see that the computer gives those values.
This is the only way to ensure that Python is solving the problem you want solved.
F.9. PLOTTING
To plot in Python you need to have imported matplotlib, as discussed above. Many options are
available at http://matplotlib.org/gallery.html.
Simple Plots
import numpy as np
import pylab as plt
def f(x):
return 2. + 3.*x + 0.5*x*x
# create the data
x = np. linspace (0, 4, 50)
y = f(x)
# plotting
plt . plot (x,y,’o-g’)
plt . xlabel ('x')
plt . ylabel ('y')

plt . title ('Test Plot')
plt . show ()
plt.savefig ('Figure_F6.png')
Figure F.6. Simple Plot
In the call to plot, the ‘o-g’ means plot the data with a circle, use a line between data points, and
color it green. The plot appears on the screen. Press the image below the figure:
and you can save the figure to your computer.
More Complicated Plots
Possible symbols are (there are others):
+ plus ^ triangle (up)
o circle v triangle (down)
* star > triangle (right)
. point < triangle (left)
x cross p pentagram
s square h hexagram
none no marker (default)
The colors are in the default order for multiple plots (Pylab cycles through the first 5), but white
is not used.
Colors Line types
b blue – ___ solid
g green _ _ dashed
r red : dotted
c cyan _. dashdot
m magenta

y yellow Line width
k black lw = 2, etc.
w white
For a vector x with n entries, and a vector y with n entries, plot them in one of four ways:
1. plt(x,y) 2. plt.loglog(x,y) 3. plt.semilogx(x,y) 4. plt.semilogy(x,y)
To plot more than one variable:
plt . plot (x,y1,'o-g',x,y2,'x-r’) # if the y1 and y2 are known at the same values of x
plt(x1,y1,x2,y2) # if y1 is known at x1 and y2 is known at x2
You can also issue two plot commands, one for each line (y1 and y2), so that the line width can
be different for the two lines.
Limit the axes:
v = [0., 1., 0., 1.5]
plt . axis(v)
Add a legend:
plt . plot (tspan, y, label='First curve')
plt . legend(loc='best')
Add a grid with plt.grid().
Plot only one column of a matrix, here the second row:
plt.plot (x,mat[1,:],'o-g')
plt.show()
Multiple plots
The command: subplot(2,1,1) says there will be two plots and the next plot command will be
shown in the top half of the window. For the second figure, say subplot(2,1,2) and the next plot
will be in below the first one.
>>>plt.subplot(2,1,1)
>>>plt.plot (x,y,'o-g')
>>>plt.subplot(2,1,2)
>>>plt. plot (x,y2,'x-r')
>>>plt.show()

Figure B.7. Multiple Plots
The numbering system says that there are 2 rows, 1 figure across, and the last number refers to
the plots in the order they are defined. If you want four figures, two in the top row, then use
subplot(2,2,1), subplot(2,2,2), subplot(2,2,3), subplot(2,2,4). The first two numbers say that
there will be 2 x 2 plots. The last number represents the plot as defined in order.
Bold, Italics, and Subscripts
Other options are to change the fontstyle, fontsize, fontweight, linewidth, and Greek letters.
plt . xlabel ('t', fontstyle='italic',fontsize=24)
plt . ylabel ('y', fontweight='bold')
plt . plot (tspan, y, linewidth = 2)
To add a subscript use plt . title(r'$Q_1$') to make Q1.
To add a superscript use plt . title(r'$Q^1$') to make Q1
.
To use Greek letters use
plt . title(r'$alpha$')
The Greek letters are (These are TEX commands.)
α alpha Γ Gamma ∞ infty
β beta Δ Delta ≥ geq
γ gamma Θ Theta ≤ leq
. . . . ∂ partial
. . . . ± pm
ω omega Ω Omega Re
Im
Additional options can be found at
http://matplotlib.org/api/pyplot_api.html#matplotlib.pyplot.text.

Contour Plots
The following example comes from
http://matplotlib.org/examples/pylab_examples/contour_demo.html
import numpy as np
import matplotlib.mlab as mlab #MATLAB compatibility command
plt.rcParams['xtick.direction'] = 'out'
plt.rcParams['ytick.direction'] = 'out'
delta = 0.025
x = np.arange(-3.0, 3.0, delta) # sets the x-values
y = np.arange(-2.0, 2.0, delta) # sets the y-values
X, Y = np.meshgrid(x, y) # MATLAB command for generating the grid
Z1 = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0)
Z2 = mlab.bivariate_normal(X, Y, 1.5, 0.5, 1, 1)
# difference of Gaussians
Z = 10.0 * (Z2 - Z1)
plt.figure()
CS = plt.contour(X, Y, Z)
plt.clabel(CS, inline=1, fontsize=10)
plt.title('Simplest default with labels')
plt . show ( )
Figure F.8. Contour Plot of 2D Function

3D plots
To plot a function , create an grid (rectangular), evaluate the function at each grid
point, and plot. The program to generate Figure F.9 is:
import numpy as np
import matplotlib.mlab as mlab
import mpl_toolkits.mplot3d.axes3d as p3 # note the new import command
delta = 0.025
x = np.arange(-2.0, 2.0, delta) #create the x-grid from -2 to 2, increment 0.025
y = np.arange(-2.0, 2.0, delta) #create the y-grid from -2 to 2, increment 0.025
X, Y = np.meshgrid(x, y)
Z = X*X + Y*Y # evaluate the function
fig = plt.figure()
ax = p3.Axes3D(fig)
ax.plot_surface(X, Y, Z)
ax.set_xlabel('X')
ax.set_ylabel('Y')
ax.set_zlabel('Z')
plt.show()
Addition information can be found at
https://scipy.github.io/old-wiki/pages/Cookbook/Matplotlib/mplot3D.html
Figure F.9. 3D Plot of 2D Function using Python
€
z(x,y)
€
x − y

F.10. PYTHON HELP
For detailed information and a complete list of all commands in the SciPy file, go to
https://docs.python.org/3/library/index.html. You can also get the latest (and earlier) User Guides
and Reference Guides for Numpy and Scipy at https://docs.scipy.org/doc/ : numpy-user-
1.11.0.pdf and scipy-ref-0.18.1.pdf.
If you want to see what modules you can import, go to the terminal window and say
help(‘modules’). If you want to see what scipy contains, import scipy and say help(scipy). To see
the details in the math section, go to the console and say print(dir(math)).
A source with scientific examples is Introduction to Python for Computational Science and
Engineering, by Hans Fangohr of the University of Southhampton, available as a pdf from
https://www.southampton.ac.uk/~fangohr/training/python/pdfs/Python-for-Computational-
Science-and-Engineering.pdf
Additional useful websites include:
http://docs.python.org/2/tutorial
http://matplotlib.org/api/pyplot_summary.html
http://matplotlib.org/gallery.html
http://www.scipy.org/NumPy_for_Matlab_Users
There are also specialized programs submitted by others. See them at the Python Package Index:
http://pypi.python.org/pypi. These require a compiler for another language and are not treated
here, but they include methods to solve boundary value problems and partial differential
equations (like in Chapter 9). If you want to do serious numerical analysis using Python, these
programs should be considered. Modules there include methods using the finite element method,
methods for problems with strong convection, spectral methods, and collocation methods.
F.11. APPLICATIONS OF PYTHON
There are many chemical engineering examples in the book that can use Python.
• Solving a single nonlinear equation, Ch. 2, p. 15, 16-20; Ch. 3, p. 34-35, Ch. 4., p. 53-
56.
• Plotting, Ch. 2, p. 19.
• Multiple equations, few unknowns, Chapter 4, p. 58-59.
The Python programs applied to these problems are in the document Introduction to Chemical
Engineering Computing; Extension to Python.pdf

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