Process Simulation

Process Simulation

Introduction

The model It is a representation of real object

(apparatus or part of the apparatus) It let us foresee behavior of physical objects

without experiments in the real world. However, usually the basis of the models are

experiments Approximation of constants/coefficients in model

equations Physical parameters of chemicals, etc.

Classification of the models Physical – mathematical

Physical – one physical quantity replaced with another (easier to measure) or use of scaled down objects (cars, planes in wind tunnels)

Mathematical - representation by the use of equations.

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 are found 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 model1. Balance dependences

Based upon basic nature laws of conservation of mass of conservation of energy of conservation of electric charge, etc.

Balance equation: Input – Output = Accumulation

Elements of the model2. Constitutive equations – apply to

unconvectional streams Newton eq. – for viscous friction Fourier eq. – for heat conduction Fick eq. – for mass diffusion

dtdx

dtxdm 2

2

S

STdtQ

2

2

xD

t

Elements of the model3. Phase equilibrium equations – important

for mass transfer 4. Physical properties equations – for

calculation parameters as functions of temperature, pressure and concentrations.

5. Geometrical dependences – involve influence of apparatus geometry on transfer coefficients – convectional streams.

Structure of the simulation model 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 vary with space – displaced parameters

Structure of the modelSteady state Unsteady state

Aggregated parameters

Algebraic eq. Ordinary differential eq.

Displaced parameters

Differential eq.1. Ordinary for 1-

dimensional case2. Partial for 2&3-

dimensional case (without time derivative, usually elliptic)

Partial differential eq.(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 Design (specification) problem Optimization problem Synthesis problem

by Rafiqul Gani

Flowsheeting problem Given:

All input information All operating condition All equipment parameters

To calculate: All outputs FLOWSHEET

SCHEMEINPUT

OPERATING CONDITIONS

EQUIPMENT PARAMETERS

PRODUCTS

R.Gani

Specyfication problem

Given: Some input&output information Some operating condition Some equipment parameters

To calculate: Rest of inputs&outputs Rest of operating condition Rest of equipment parameters

FLOWSHEETSCHEME

INPUT

OPERATING CONDITIONS

EQUIPMENT PARAMETERS

PRODUCTS

Specyfication problem

NOTE: degree of freedom is the same as in flowsheeting problem.

To gues: D, Qr

Solve the flowsheetingproblem

STOP

Is target product composition satisfied

?

Adjust D, Qr

Given: feed composition and flowrates, target product composition

Find: product flowrates, heating duties

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)

To gues: D, Qr


STOP


AND =min.

Adjust D, Qr


Find: product flowrate, heating dutie

Process synthesis/design problem the act of creation of a new process. Given:

inputs (some feeding streams can be added/changed)

Outputs (some byproducts may be unknown) To find:

flowsheet equipment parameters operations conditions

Process synthesis/design problem

flowsheetundefined

INPUT OUTPUT

To gues: D, Qr


STOP


AND =min.

Adjust D, QrAs well as

N, NF, R/D etc.


Find: product flowrate, heating dutie

Process synthesis/design problem

Separation method & equipment

methanolwater

methanol

water

Methods: distillation, membrane separation, flash, extractionEquipment: how many apparatus are needed

what is apparatus design and conditions

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 – flowsheeting programs. Equipped with:

Data base of apparatus models Data base of components 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. ASPEN One 2. HYSIM3. ChemCAD4. PRO/II5. ProMax

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 according to separately:

1. Material streams2. Apparatus

Apparatus-streams separation Why separate?

It’s make calculations easier Assumption:

All processes (chemical reaction, heat exchange etc.) taking places in the apparatus and streams are in the chemical and thermodynamical equilibrium state.

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=4Assumed: F1, F2, T, PCalculated:

•enthalpy•vapor fraction

Apparatus parameters 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 apparatus

mathematical model use) Detailed balance on the base of

apparatus model

Overall balance Apparatus is treated 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

Countercurrent, tube-shell heat exchangerGiven three streams data: 1, 2, 3 hence parameters of stream 4 can be easily calculated from the balance equation. 1

2

4

3

There is possibility that calculated temp. of stream 4 can be higher then inlet temp. of heating medium (stream 1).

DF=5

2134 ttcmttcm pBBpAA

Overall balance – ExampleGiven: 1. mA=10kg/s2. mB=20kg/s3. t1= 70°C4. t2=40°C5. t3=20°CcpA=cpB=f(t)

1, mB

2

4

3, mA

2134 ttmmtt

A

B

Ct 8040701020204

2134 ttcmttcm pBBpAA

Apparatus model involved Process is being described with use of

modeling equations (differential, dimensionless etc.)

Only physically possible 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 initial parameters of cut stream has to be

defined Calculations has to be repeated until cut

streams parameters are converted.

Loops and cut streams

Simulation of system with heat exchanger using Excel

I.Problem definitionSimulate 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

s6

s1

1

2

3 4

67

5

s2 s3 s4 s5

s7

s8

s9 s10

Stream s1

Ps1 =200kPa, ts1 = 85°C, f1s1 = 1000kg/h

Stream s6

Ps6 =200kPa, ts6 = 20°C, f2s6 = 1000kg/h

Numerical data:

Equipment parameters:1. L1=7m d1=0,025m2. L2=5m d2=0,16m, s=0,0016m, n=31...3. L3=6m, d3=0,025m4. 4=505. L5=7m d5=0,025m6. L6=10m, d6=0,025m7. 7=40

III. Stream summary table s1 s2 s3 s4 s5 s6 s7 s8 s9 s10

f1 f1s1 X X X X

f2 f2s6 X X X X

T Ts1 X X X X Ts6 X X X X

P Ps1 X x x X Ps6 X X X X

Uknown:Ts2, Ts3, Ts4, Ts5, Ts7, Ts8, Ts9, Ts10, Ps2, Ps3, Ps4, Ps7, Ps8, Ps9, f1s2,

f1s3, f1s4, f1s5, f2s7, f2s8, f2s9, f2s10

number of unknown variables: 24WE NEED 24 INDEPENDENT EQUATIONS.

f1s2= f1s1 f1s7= f1s6

f1s3= f1s2 f1s8= f1s7

f1s4= f1s3 f1s9= f1s8

f1s5= f1s4 f1s10= f1s9

12 TT

34 TT

45 TT

67 TT

89 TT

910 TT

Equations from equipment information

14 equations. Still do define 24-14=10 equations

Heat balance equations

8

7

2

3

T

TpS

T

TpT

QdTcm

QdTcm

QTTcf

QTTcf

pSs

pTs

786

321

2

1

New variable: QStill to define: 10+1-2=9 equations

Heat exchange equations

mm TkFQ

New variables: k, Tm: number of equations to find 9+2-2=9

101

65

10165

lnss

ss

ssssm

TTTT

TTTTT


Two new variables: T and S number of equations to find: 9+2-1=10

SstT

sk

11

1


Three new variables: NuT, NuS, deq,

number of equations to find: 10+3-3=10

2dNu TT

T

eq.dNu SS

S

22

22

22

ndDndDdeq


10000Re2300RelnRelnRelnReln

lnlnlnexp

5,2300Re62,1PrRe62,1

5,2300Re5,0PrRe5,0

10000RePrRe023,0

3/13/1

2

2

4,08,0

TTLTTuTLT

TuTLTLT

TTTT

TTHEX

HEXTT

TT

T

NuNuNu

GzGzld

GzGzld

Nu

T

s

TT d

fwd

2

12 14Re


10000Re2300RelnRelnRelnReln

lnlnlnexp

5,2300Re62,1PrRe62,1

5,2300Re5,0PrRe5,0

10000RePrRe023,0

3/13/1

4,08,0

SSLSTuSLS

TuSLSLS

SSHEX

HEXSS

SSHEX

HEXTS

SS

S

NuNuNu

GzGzld

GzGzld

Nu

Two new variables ReT and ReS,


SCSA

eqs

S

eqS F

dfwd

.6. 2Re

Pressure drop

Ps1-Ps2=P1

Ps2-Ps3=P2T

Ps3-Ps4=P3

Ps4-Ps5=P4

Ps6-Ps7=P5

Ps7-Ps8=P2S

Ps8-Ps9=P6

Ps9-Ps10=P7

Eight new variables: P1, P2T, P3, P4, P5, P2S, P6, P7, number of equations to find: 8+8-

8=8

Pressure drop

1

4

21

1

2

1116

2 dl

df

dlwP s

1

237,0

525,01

Re221,00032,0

10Re2300,Re

3164,0

2300Re,Re64

1

11

14Red

f s

Two new variables Re1 and 1, number of equations to find: 8+2-3=7

Pressure drop

Two new variables Re2t and 2T, number of equations to find: 7+2-3=6

THEX

HEXs

TT d

ldn

fdlwP

24

21

2

2

2116

2

T

T

2237,0

525,02

Re221,00032,0

10Re2300,Re

3164,0

2300Re,Re64

T

sT nd

f

2

12

14Re

Pressure drop

Two new variables Re3 and 3, number of equations to find: 6+2-3=5

34

21

3

2

3116

2 dl

df

dlwP s

3237,0

525,03

Re221,00032,0

10Re2300,Re

3164,0

2300Re,Re64

3

13

14Red

f s

Pressure drop

Number of equations to find: 5-1=4

44

21

4

2

4116

2 dfwP s

Pressure drop

Two new variables Re5 and 5,


5

4

22

5

2

5216

2 dl

df

dlwP s

5237,0

525,05

Re221,00032,0

10Re2300,Re

3164,0

2300Re,Re64

5

25

24Red

f s

Pressure drop

Two new variables Re2S and 2S, number of equations to find: 3+2-3=2

SeqCSA

s

SeqS d

lFf

dlwP

2.

22

2

2.

22

2216

2

S

S

2237,0

525,02

Re221,00032,0

10Re2300,Re

3164,0

2300Re,Re64

Seq

sS d

f

2.

22

24Re

Pressure drop

Two new variables Re6 and 6,


64

22

6

2

6216

2 dl

df

dlwP s

5237,0

525,06

Re221,00032,0

10Re2300,Re

3164,0

2300Re,Re64

5

26

24Red

f s

Pressure drop

Number of equations to find: 1-1=0 !!!!!!!!!!!!!!

74

22

7

2

7216

2 dfwP s

Agents parameters Temperatures are not constant Liquid properties are functions of temperature

•Specyfic heat cp

•Viscosity

•Density

•Thermal conductivity

•Prandtl number Pr

Agents parametersData are usually published in the tables

t cp Pr0,00 999,80 17,89 0,551 4237 13,76

10,00 999,60 13,04 0,575 4212 9,5520,00 998,20 10,00 0,599 4203 7,0230,00 995,60 8,014 0,618 4199 5,4540,00 992,20 6,531 0,634 4199 4,3350,00 988,00 5,495 0,648 4199 3,5660,00 983,20 4,709 0,659 4203 3,0070,00 977,70 4,059 0,668 4211 2,5680,00 971,80 3,559 0,675 4216 2,2290,00 965,30 3,147 0,68 4224 1,95

100,00 958,30 2,822 0,683 4229 1,75

Data in tables are difficult to use

Solution:

Approximate discrete data by the continuous functions.

Agents parameters

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

950

960

970

980

990

1000

1010

0 20 40 60 80 100

t [°C]

[k

g/m

3 ]

The end as of yet.

Process Simulation

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