Introduction
Process Simulation
Classification of the models

Black box – white box


Black box – know nothing about process in
apparatus, only dependences between
inputs and outputs are established.
Practical realisation of Black box is the
neural network
White box – process mechanism is well
<??> known and described by system of
equations
Classification of the models

Deterministic – Stochastic


Deterministic – for one given set of inputs
only one set of outputs is calculated with
probability equal 1.
Stochastic – random phenomenon affects
on process course (e.g. weather), output
set is given as distribution of random
variables
Classification of the models

Microscopic- macroscopic


Microscopic – includes part of process or
apparatus
Macroscopic – includes whole process or
apparatus
Elements of the model
Balance dependences
1.

Based upon basic nature laws





of
of
of
of
conservation
conservation
conservation
conservation
of
of
of
of
mass
energy
atoms number
electric charge, etc.
Balance equation (for mass):
(overall and for specific component without reaction)
Input – Output = Accumulation
or (for specific component if chemical reactions presents)
Input – Output +Source = Accumulation
Elements of the model
Constitutive equations
2.



Newton eq. – for viscous friction
Fourier eq. – for heat conduction
Fick eq. – for mass diffusion
2
m
d x
dt
Q
t

t
2
 
dx
dt


     Td S
S
 
2
 D
x
2
Elements of the model
3.
4.
5.
Phase equilibrium equations – important for
mass transfer
Physical properties equations – for
calculation parameters as functions of
temperature, pressure and concentrations.
Geometrical dependences – involve
influence of apparatus geometry on transfer
coefficients – convectional streams.
Structure of the simulation model


Structure corresponds to type of model
equations
Structure depends on:

Type of object work:



Continuous, steady running
Periodic, unsteady running
Distribution of parameters in space


Equal in every point of apparatus – aggregated
parameters (butch reactor with ideal mixing)
Parameters are space dependent– displaced parameters
Structure of the model
Steady state
Unsteady state
Aggregated
parameters
Algebraic eq.
Ordinary differential eq.
Displaced
parameters
Differential eq.
Partial differential eq.
1.
2.
Ordinary for 1dimensional case
Partial for 2&3dimensional case
(without time
derivative, usually
elliptic)
(with time derivative,
usually parabolic)
Process simulation

the act of representing some aspects of
the industry process (in the real world)
by numbers or symbols (in the virtual
world) which may be manipulated to
facilitate their study.
Process simulation (steady state)




Flowsheeting problem
Specification (design) problem
Optimization problem
Synthesis problem
by Rafiqul Gani
Flowsheeting problem

Given:




All of the input information
All of the operating condition
All of the equipment parameters
To calculate:

All of the outputs
INPUT
FLOWSHEET
SCHEME
OPERATING
CONDITIONS
PRODUCTS
EQUIPMENT
PARAMETERS
R.Gani
Specyfication problem
INPUT

Given:




OPERATING
CONDITIONS
Some input & some output information
Some operating condition
Some equipment parameters
To calculate:



FLOWSHEET
SCHEME
Undefined inputs&outputs
Undefined operating condition
Undefined equipment parameters
PRODUCTS
EQUIPMENT
PARAMETERS
Specyfication problem

NOTE: degree of freedom is the same as
in flowsheeting problem.
Given: feed composition and flowrates,
target product composition
Assume value to be guessed:
D, Qr
Solve the flowsheeting
problem
Is target product
composition satisfied
?
STOP
Find: product flowrates, heating duties
Adjust D, Qr
Process optimisation

the act of finding the best solution
(minimize capital costs, energy...
maximize yield) to manage the
process (by changing some
parameters, not apparatus)
Given: feed composition and flowrates,
target product composition
Assume value to be guessed:
D, Qr
Solve the flowsheeting
problem
Is target product
composition satisfied
AND =min.
STOP
Find: product flowrate, heating duty
Adjust D, Qr
Process synthesis/design problem


the act of creation of a new process.
Given:



inputs (some feeding streams can be
added/changed latter)
Outputs (some byproducts may be unknown)
To find:



Flowsheet (topology)
equipment parameters
operations conditions
Process synthesis/design problem
INPUT
flowsheet
undefined
OUTPUT
Given: feed composition and flowrates,
target product composition
Assume value to be guessed:
D, Qr, N, NF, R/D etc.
Solve the flowsheeting
problem
Is target product
composition satisfied
AND =min.
STOP
Find: product flowrate, heating duty,
column param. etc.
Adjust D, Qr
As well as
N, NF, R/D etc.
Process simulation - why?

COSTS



Material – easy to measure
Time – could be estimated
Risc – hard to measure and estimate
Software for process
simulation

Universal software:



Worksheets – Excel, Calc (Open Office)
Mathematical software – MathCAD, Matlab
Specialized software – process simulators.
Equipped with:




Data base of apparatus models
Data base of components and mixtures properties
Solver engine
User friendly interface
Software process simulators
(flawsheeting programs)
Started in early 70’
At the beginning dedicated to special
processes
Progress toward universality
Some actual process simulators:




1.
2.
3.
4.
5.
ASPEN Tech /HYSYS
ChemCAD
PRO/II
ProSim
Design II for Windows
Chemical plant system



The apparatus set connected with material
and energy streams.
Most contemporary systems are complex, i.e.
consists of many apparatus and streams.
Simulations can be use during:



Investigation works – new technology
Project step – new plants (technology exists),
Runtime problem identification/solving – existing
systems (technology and plant exists)
Chemical plant system

characteristic parameters can be
specified for every system separately
according to:
1.
2.
Material streams
Apparatus
Apparatus-streams separation

Assumption:


All processes (chemical reaction, heat
exchange etc.) taking places in the
apparatus and streams are in the chemical
and thermodynamical equilibrium state.
Why separate?

It’s make calculations easier
Streams parameters






Flow rate (mass, volume, mol per time
unit)
Composition (mass, volume, molar
fraction)
Temperature
Pressure
Vapor fraction
Enthalpy
Streams degrees of freedom
DFs=NC+2
e.g.: NC=2 -> DFs=4
Assumed: F1, F2, T, P
Calculated:
•enthalpy
•vapor fraction
Apparatus parameters & DF
Characteristics for each apparatus type.
E.g. heat exchanger :





Heat exchange area, A [m2]
Overall heat-transfer coefficient, U (k)
[Wm-2K-1]
Log Mean Temperature Difference, LMTD
[K]
degrees of freedom are unique to
equipment type
Calculation subject


Number of equations of mass and
energy balance for entire system
Can be solved in two ways:
Types of balance calculation


Overall balance (without use of
apparatus mathematical model)
Detailed balance on the base of
apparatus model
Overall balance

Apparatus is considered as a black box


Needs more stream data
User could not be informed about if the
process is physically possible to realize.
Overall balance – Example
3
2
Countercurrent, tube-shell heat exchanger
Given three streams data: 1, 2, 3 hence
parameters of stream 4 can be easily
calculated from the balance equation.
1
4
m A c pA t 4  t 3   m B c pB t1  t 2 
DF=5
There is possibility that calculated temp. of stream 4 can be
higher then inlet temp. of heating medium (stream 1).
Overall balance – Example
3, mA
2
1, mB
4
Given:
1. mA=10kg/s
2. mB=20kg/s
3. t1= 70°C
4. t2=40°C
5. t3=20°C
cpA=cpB=idem
m A c pA t 4  t 3   m B c pB t1  t 2 
t 4  t3 
t 4  20 
20
10
m B
m A
t 1  t 2 
70  40   80  C
Apparatus model involved




Process is being described with use of
modeling equations (differential,
dimensionless etc.)
Only physically acceptable processes taking
place
Less stream data required (smaller DF
number)
Heat exchange example: given data for two
streams, the others can be calculated from a
balance and heat exchange model equations
Loops and cut streams

Loops occur when:




some products are returned and mixed with input
streams
when output stream heating (cooling) inputs
some input (also internal) data are undefined
To solve:



one stream inside the loop has to be cut (tear
stream)
initial parameters of cut stream have to be defined
Calculations have to be repeated until cut streams
parameters are converted.
Loops and cut streams
Simulation of system with
heat exchanger using
MathCAD
I.Problem definition
Simulate system consists of: Shell-tube heat exchanger, four
pipes and two valves on output pipes. Parameters of input
streams are given as well as pipes, heat exchanger geometry
and valves resistance coefficients. Component 1 and 2 are
water. Pipe flow is adiabatic.
Find such a valves resistance to satisfy condition: both streams
output pressures equal 1bar.
II. Flawsheet
5
s6
s7
2
1
s1
4
3
s2
s3
s4
s8
s5
7
6
s9
s10
Numerical data:
Stream s1
Ps1 =200kPa, ts1 = 85°C, f1s1 = 10000kg/h
Stream s6
Ps6 =200kPa, ts6 = 20°C, f2s6 = 10000kg/h
Equipment parameters:
1.
L1=7m d1=0,025m
2.
L2=5m d2=0,16m, s=0,0016m, n=31...
3.
L3=6m, d3=0,05m
4.
z4=50
5.
L5=7m d5=0,05m
6.
L6=10m, d6=0,05m
7.
z7=40
III. Stream summary table
f1
s1
s2
s3
s4
s5
f1 s1
X
X
X
X
f2
s6
s7
s8
s9
s1 0
f2 s6
X
X
X
X
T
T s1
X
X
X
X
T s6
X
X
X
X
P
P s1
X
x
x
X
P s6
X
X
X
X
Uknown:Ts2, Ts3, Ts4, Ts5, Ts7, Ts8, Ts9, Ts10, Ps2, Ps3, Ps4, Ps5, Ps7, Ps8,
Ps9, Ps10, f1s2, f1s3, f1s4, f1s5, f2s7, f2s8, f2s9, f2s10
number of unknown variables: 26
WE NEED 26 INDEPENDENT EQUATIONS.
Equations from equipment information
T 2  T1
f1s2= f1s1
f1s7= f1s6
T 4  T3
f1s3= f1s2
f1s8= f1s7
T5  T 4
f1s4= f1s3
f1s9= f1s8
T7  T6
f1s5= f1s4
f1s10= f1s9
T9  T8
T10  T9
14 equations. Still do define 2614=12 equations
Heat balance equations
T2
 m
T
c p dT Q
T3
T8
 m
S
c p dT  Q
f 1 s 1 c pT T 2  T3   Q
f 2 s 6 c pS T8  T7   Q
T7
New variable: Q
Still to define: 12+1-2=11 equations
Heat exchange equations
Q  kFm  Tm
 Tm 
T s 5  T s 6   T s1  T s10 
T  T s 6 
ln s 5
T s1  T s10 
New variables: k, Tm: number of equations to find 11+2-2=11
Heat exchange equations
k 
1
1
aT

s
 st

1
aS
Two new variables: aT and aS
number of equations to find: 11+2-1=12
Heat exchange equations
aT 
Nu T  T
d2
aS 
Nu S  S
d eq.
D 2  nd 2
2
d eq 
2
D 2  nd 2
Three new variables: NuT, NuS, deq,
number of equations to find: 12+3-3=12
Heat exchange equations
Nu T
 0 , 023 Re T0 , 8 Pr T0 , 4  Re  10000

 0 ,5 Re Pr d HEX  0 , 5 Gz  Re  2300 , Gz  5
T
T
T
T

l HEX

1/ 3


d2 
 Re T Pr T
  1, 62 Gz T1 / 3  Re T  2300 , Gz  5
1
,
62


l 2 





ln Nu LT  ln Nu TuT
ln Re LT  ln Re T    2300  Re T  10000
 exp  ln Nu LT 
ln Re LT  ln Re TuT



Re T 
wd 2 


T
4 f 1s1
 d 2
T
Heat exchange equations
Nu S
 0 , 023 Re 0S , 8 Pr S0 , 4  Re  10000

 0 ,5 Re Pr d HEX  0 ,5 Gz  Re  2300 , Gz  5
S
T
S
S

l HEX

1/ 3


d HEX 
  1, 62 Gz 1S / 3  Re S  2300 , Gz  5
1, 62  Re S Pr S
l HEX 





ln Nu LS  ln Nu TuS
ln Re LS  ln Re S    2300  Re
 exp  ln Nu LS 
ln Re LS  ln Re TuS



Re
S

wd eq . 


S
f 2 s 6 d eq .
FCSA 
S
Two new variables ReT and ReS,
number of equations to find: 12+2-4=10
S
 10000
Pressure drop
P s1 -P s2 =  P 1
P s6 -P s7 =  P 5
P s2 -P s3 =  P 2 T
P s7 -P s8 =  P 2 S
P s3 -P s4 =  P 3
P s8 -P s9 =  P 6
P s4 -P s5 =  P 4
P s9 -P s1 0 =  P 7
E ig h t n ew v ariab les:  P 1 ,  P 2 T ,  P 3 ,  P 4 ,
 P 5 ,  P 2 S ,  P 6 ,  P 7 , n u m b er o f eq u atio n s to fin d :
1 0 + 8 -8 = 1 0
Pressure drop
w  l
2 2
2
 P1  
2
 64
 Re , Re  2300

 0 , 3164
5
1  
, 2300  Re  10
0 , 25
 Re
0 , 221

0
,
0032

0 , 237

Re

d
 
1
8 ff11ss11 ll
16
 d d  dd
2
44
11
1
1
Re 1 
4 f 1s1
d
Two new variables Re1 and 1,
number of equations to find: 10+2-3=9
Pressure drop
2
w  l
2
 P2 T  
 2T
2
d
 64
 Re , Re  2300

 0 ,3164
5

, 2300  Re  10
0 , 25
 Re
0 , 221

0
,
0032

0 , 237

Re

2T
2
16 8f 1fs 11 l HEX l
 2 2 4 4
n n d d indHEX d2inT
2
2T
One new variables and 2T,
number of equations to find: 9+1-3=7
Pressure drop
2
w  l
2
 P3  
2
d
 64
 Re , Re  2300

 0 ,3164
5
3  
, 2300  Re  10
0 , 25
 Re
0 , 221

0
,
0032

0 , 237

Re

 
3
2
168 ff11s 1 ll
2
4
dd  dd
4
33
Re 3 
4 f 1s1
d
3
Two new variables Re3 and 3,
number of equations to find: 7+2-3=6
3
Pressure drop
w 
2 2
2
 P4  z
2
 zz
4
168 f 1 s 1
2
4
dd 
4
44
Number of equations to find: 6-1=5
Pressure drop
w  l
22
2
 P5  
2
 64
 Re , Re  2300

 0 , 3164
5
5  
, 2300  Re  10
0 , 25
 Re
0 , 221

0
,
0032

0 , 237

Re

d
 
5
168 ff 22s 2
l
dd  d
2
44
Re 5 
5
4 f 2s2
d
5
Two new variables Re5 and 5,
number of equations to find: 6+2-3=4
5
Pressure drop
2 2
w  l2
16 ff 2
2s2
2
 P2 S  
2S
2
 64
 Re , Re  2300

 0 ,3164
5

, 2300  Re  10
0 , 25
 Re
0 , 221

0
,
0032

0 , 237

Re

d eq .
 
2S
ll2
2
 FF CSA
eq .

dd eq
CSA
. 2S
2s
2S
One new variables and 2S,
number of equations to find: 4+1-3=2
Pressure drop
w  l
22
2
 P6  
2
 64
 Re , Re  2300

 0 ,3164
5
6  
, 2300  Re  10
0 , 25
 Re
0 , 221

0
,
0032

0 , 237

Re

d


6
168 ff 22s 2
ll
2
44
dd  dd 66
Re 6 
4 f 2s2
d
5
5
Two new variables Re6 and 6,
number of equations to find: 2+2-3=1
Pressure drop
22
w 
2
 P7  z
2
 zz
7
168 ff 22s 2
dd 
2
44
7
7
Number of equations to find: 1-1=0 !!!!!!!!!!!!!!
Agents parameters
Temperatures are not constant
Liquid properties are functions of temperature
•Density 
•Viscosity 
•Thermal conductivity 
•Specyfic heat cp
•Prandtl number Pr
Agents parameters
Data are usually published in the tables
t
0,00
10,00
20,00
30,00
40,00
50,00
60,00
70,00
80,00
90,00
100,00

999,80
999,60
998,20
995,60
992,20
988,00
983,20
977,70
971,80
965,30
958,30

17,89
13,04
10,00
8,014
6,531
5,495
4,709
4,059
3,559
3,147
2,822

0,551
0,575
0,599
0,618
0,634
0,648
0,659
0,668
0,675
0,68
0,683
cp
4237
4212
4203
4199
4199
4199
4203
4211
4216
4224
4229
Pr
13,76
9,55
7,02
5,45
4,33
3,56
3,00
2,56
2,22
1,95
1,75
Agents parameters
Data in tables are difficult to use
Solution:
Approximate discrete data by the continuous functions.
Approximation

Approximating function



Polynomial
Approximation target: find optimal
parameters of approximating function
Approximation type

Mean-square – sum of square of
differences between discrete (from tables)
and calculated values is minimum.
Polynomial approximation
y = 1,540E-05x3 - 5,895E-03x2 + 2,041E-02x +
9,999E+02
1010
3
 [kg/m ]
1000
990
980
970
960
950
0
20
40
60
t [°C]
80
100
The end as of yet.