Modeling and Simulation of Cstr

8/9/2019 Modeling and Simulation of Cstr

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MODELING AND SIMULATION OF CSTR

using

POLYMATH (Version 3.0.1)

Prepared by : Ranparia Hitesh M.Guided By : Parsana Vyomesh M.


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What is a MODEL ?

Aset of mathematical equations that allowsus to

predict the behavior ofchemical process.

CLASSIFICATIONOF MODELLinear / Non linearSteady state / Dynamic behavior Lumped parameter / Distributed parameter Discrete/Continuous

Deterministic/Stochastic

INTRODUCTION


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MODELING

The activities leading to the construction of themodel will be referred to as modeling.

Modelinginvolves:-

For the design ofcontrollers forchemical process,

modelingis a very critical step.

It givessimple description of how the process reactsIt givessimple description of how the process reactsto variousinputs.to variousinputs.

It is an abstraction which helps to avoid repetitiveIt is an abstraction which helps to avoid repetitive

experimentation and observations.experimentation and observations.

Much cheaper,safer and faster.Much cheaper,safer and faster.

Mass balance

Component balance Energy balance


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SIMULATION

Asimulationis a mathematical model of theprocess which attempts to predict how the

process would behave ifit wasconstructed.

It predicts:

the flow rate,composition, temp., pressureof the products.

how much raw material is beingused.

how much energy is beingconsumed.


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USE OF MATHMATICAL MODELUSE OF MATHMATICAL MODELRESEARCHANDDEVELOPMENT :RESEARCHANDDEVELOPMENT :--

Determiningchemical kinetic mechanisms and parameters from laboratory orDeterminingchemical kinetic mechanisms and parameters from laboratory orpilotpilot--plant reaction data.plant reaction data.

Exploring the effects of different operatingconditions for optimization and controlExploring the effects of different operatingconditions for optimization and controlstudiesstudies

AidinginscaleAidinginscale--up calculations.up calculations.

DESIGN:DESIGN:--

Exploring the sizing and arrangement of processing equipment for dynamicExploring the sizing and arrangement of processing equipment for dynamicperformance.performance.

Studying the interactions of various parts of the process, particularly whenStudying the interactions of various parts of the process, particularly whenmaterial recycle or heat integrationisused.material recycle or heat integrationisused.

Evaluating alternative process and control structures and strategies.Evaluating alternative process and control structures and strategies.

SimulatingstartSimulatingstart--up,shutdown, and emergency situations and procedures.up,shutdown, and emergency situations and procedures.

PLANTOPERATION:PLANTOPERATION:--

Troubleshootingcontrol and processing problems.Troubleshootingcontrol and processing problems.

AidinginstartAidinginstart--up and operator training.up and operator training.

Studying the effects of and the requirements for expansion projects.Studying the effects of and the requirements for expansion projects.

Optimizing plan operation.Optimizing plan operation.

It isusually much cheaper,safer, and faster to conduct the kinds ofstudies listedIt isusually much cheaper,safer, and faster to conduct the kinds ofstudies listedabove on a mathematical model than experimentally on an operatingunit.Thisisabove on a mathematical model than experimentally on an operatingunit.Thisis

not to say that plant tests are not needednot to say that plant tests are not needed


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CONTINUOUS STIRRED TANK REACTORMODEL

Dynamic behavior Lumped parameter Non isothermal Assume:-

A simple first order ,irreversible, exothermicreaction A B takes place in the reactor, whichis turn cooled by a coolant that flows through a

cooling jacket around the reactor.


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V , CA ,T

Tj , Vj

Fi, CAi,

Ti

FJ, TJi

FJ, TJ

F, CA, T

Temp.Controller

Level

Controller

CONTINUOUS STIRRED TANK REACTOR (non isothermal)


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ASSUMPTION

1.First order irreversible exothermic reaction.

2.Liquid density is constant, i =

3.Negligible heat losses

4.The volume of water in the jacket VJ is constant

5.Temperature everywhere in the jacket is TJ

6.Overall heat transfer coefficient U is constant

7. Fluid specific heat capacities are constant


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Fi*i - F* = d(*V) / dt

Rate of mass Rate of mass = Rate of change offow in flow out mass within system

but , i =

THE MASS BALANCE

Rate of flow of Rate of flow of Rate of disappearance = Rate of change of AA in A out of A due to reaction inside the tank

CAi*Fi - CA*F - r*V = d(CA*V) / dt

Assuming a constant amount of material in the reactor , we find that

Fi*i = F* ,but i = Fi = F & V is constant & k = k0*e

(-E / R*T)

F*(CAi CA) / V k0*e

(-E / R*T)

*CA = d(CA) / dt - - - - - 2

Fi - F = dV/dt - - - - - 1

THE COMPONENT BALANCE

2


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Rate of flow Rate of flow Rate at which energy Heat transfer = Rate of change ofof energy of energy is generated due to rate liquid energyinto out of chemical reaction

CSTR CSTR

[Fi*i*Cpi*(Ti Tref)] [ F**Cp*(T - Tref)]+[-k*CA*V*(H)]-Q = d(*Cp*V* T)/dt

The heat transfer between the process at temperature T and thecooling water at temperature Tj is described by

an overall heat transfer coefficient.Q = U*AH*(T Tj) - - - - -

Now, Fi =F & i = & Cp ,V = constant[F*(Ti T) / V] + [H*k0*e

(-E / R*T)*CA/*Cp] [ Q/*Cp*V] =d(T) / dt - - 3

Same for JACKETED energy balance

REACTOR ENERGY BALANCE

Fj*(Tji Tj) / V + U*AH*(T Tj) / j*Cj*Vj =d(Tj) / dt - - - 4

i


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A proportional level controller is assumed to change the outflowas the volume in the tank rises or falls: the higher the volume,the larger the outflow. The outflow is shut off completely whenthe volume drops to a minimum value Vmin.

F=Kv*(V-Vmin) ---- iiA second controller manipulates the flow rate of cooling water

to the jacket, Fj , in direct proportion to the temperature in the reactor.Fj=Fj -Kc*(T

set-T) ---- iii

Steadystate values: Initial Values:

F =40f t3/h V0 =48f t3CAi = 0.5 lb*mol A/ft

3 CA0 = 0.245 lb*mol A/ft3

Fj =49.9 ft3/h T0 = 600 R

DATA:- Ti =530 R Tj0 = 594.59 RParameter values:

Vj = 3.85 ft3

k0 = 7.08 * 1010

h-1

E = 30000 Btu/lb*mol R = 1.99 Btu/lb*mol*RU = 150 Btu/h*ft2*R AH = 250 ft

2

Tji = 530 R H = -30000 Btu/lb*molCp = 0.75 Btu/lbm*R Cj = 1.0 Btu/lbm*R = 5 0 l bm/ft

3 j = 62.3 lbm/ft3

kc =4( f t3/h)/ R Tset = 600 R


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By equations 1to4 and i to iii

PROGRAMING OF CSTRPOLYMATH (VER. -3.01)

%Disturbance is step change in CA at time zero from 0.5 to 0.55%

d(V)/d(t)=40-F

d(C)/d(t)=(F*(0.55-C))/V-k*C

d(T)/d(t)=(F*(530-T))/V+(30000*k*C)/37.5-Q/37.5/Vd(Tj)/d(t)=(Fj*(530-Tj))/3.85+Q/240

F=40-10*(48-V)

k=7.08*10^10*exp(-30000/1.99/T)

Fj=49.9-4*(600-T)Q=150*250*(T-Tj)

t(0)=0,V(0)=48,C(0)=0.245,T(0)=600,Tj(0)=594.59

t(f)= 6


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tt VV CC TT TjTj FF FjFj QQ

00 4848 0.2450.245 600.000600.000 594.590594.590 40.0040.00 49.90049.900 202875.00202875.00

0.150.15 4848 0.2500.250 600.197600.197 594.775594.775 40.0040.00 50.68750.687 203297.48203297.48

0.30.3 4848 0.2540.254 600.699600.699 594.957594.957 40.0040.00 52.69552.695 215309.24215309.24

0.450.45 4848 0.2560.256 601.249601.249 595.312595.312 40.0040.00 54.89854.898 222662.46222662.46

0.60.6 4848 0.2570.257 601.756601.756 595.517595.517 40.0040.00 56.92656.926 233972.47233972.47

0.750.75 4848 0.2580.258 602.134602.134 595.765595.765 40.0040.00 58.43758.437 238859.74238859.74

0.90.9 4848 0.2570.257 602.392602.392 595.855595.855 40.0040.00 59.46759.467 245122.52245122.52

1.051.05 4848 0.2570.257 602.523602.523 595.962595.962 40.0040.00 59.99359.993 246033.87246033.87

1.21.2 4848 0.2560.256 602.575602.575 595.957595.957 40.0040.00 60.20060.200 248157.91248157.91

1.351.35 4848 0.2560.256 602.564602.564 595.981595.981 40.0040.00 60.15660.156 246864.43246864.43

1.51.5 4848 0.2560.256 602.532602.532 595.940595.940 40.0040.00 60.02660.026 247173.40247173.40

1.651.65 4848 0.2560.256 602.486602.486 595.938595.938 40.0040.00 59.84559.845 245552.23245552.23

1.81.8 4848 0.2560.256 602.450602.450 595.901595.901 40.0040.00 59.70159.701 245581.50245581.50

1.951.95 4848 0.2560.256 602.420602.420 595.902595.902 40.0040.00 59.58159.581 244446.69244446.69

2.12.1 4848 0.2560.256 602.404602.404 595.880595.880 40.0040.00 59.51759.517 244671.09244671.09

R

ESU

LT

S


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tt VV CC TT TjTj FF FjFj QQ

2.252.25 4848 0.2560.256 602.394602.394 595.886595.886 40.0040.00 59.47759.477 244053.25244053.25

2.42.4 4848 0.2560.256 602.393602.393 595.875595.875 40.0040.00 59.47259.472 244418.25244418.25

2.552.55 4848 0.2560.256 602.394602.394 595.884595.884 40.0040.00 59.47459.474 244099.96244099.96

2.72.7 4848 0.2560.256 602.398602.398 595.879595.879 40.0040.00 59.49059.490 244458.74244458.74

2.852.85 4848 0.2560.256 602.400602.400 595.887595.887 40.0040.00 59.50159.501 244261.51244261.51

33 4848 0.2560.256 602.404602.404 595.883595.883 40.0040.00 59.51559.515 244537.67244537.67

3.153.15 4848 0.2560.256 602.405602.405 595.888595.888 40.0040.00 59.52159.521 244380.66244380.66

3.33.3 4848 0.2560.256 602.407602.407 595.885595.885 40.0040.00 59.52859.528 244569.34244569.34

3.453.45 4848 0.2560.256 602.407602.407 595.889595.889 40.0040.00 59.52859.528 244434.21244434.21

3.63.6 4848 0.2560.256 602.408602.408 595.886595.886 40.0040.00 59.53059.530 244560.25244560.25

3.753.75 4848 0.2560.256 602.407602.407 595.888595.888 40.0040.00 59.52859.528 244449.20244449.20

3.93.9 4848 0.2560.256 602.407602.407 595.886595.886 40.0040.00 59.52859.528 244537.41244537.41

4.054.05 4848 0.2560.256 602.407602.407 595.888595.888 40.0040.00 59.52659.526 244452.10244452.10


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Temp. V/S Time

594.500

595.000

595.500

596.000

596.500

0 2 4 6 8

Time

Tj


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Heat V/S Time

0.00

50000.00

100000.00

150000.00200000.00

250000.00

300000.00

0 2 4 6 8

Time

Q

Heat transferrate V/S Time


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Now for : +20%CA0, KC=2.5

0.22

0.23

0.24

0.25

0.26

0.27

0 2 4 6 8

Co

.C


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0

10

20

30

40

50

60

0 2 4 6 8

Time

Vol.(V)

598

600

602

604606

608

610

612

614

616

0 2 4 6 8

Tim e

Tem

p(T)


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594

596

598

600

602

604

606

0 2 4 6 8

Time

Temp(Tj)

0

10

20

30

40

50

0 2 4 6 8

Time

Flowrate(F)


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0

20

40

60

80

100

0 2 4 6 8

Time

Flowr

ate(Fj)

0

100000

200000

300000

400000

500000

0 2 4 6 8

Time

Hea

t(Q)


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Now for:- +20%F0, Kc=4

0.24

0.245

0.25

0.255

0.26

0.2650.27

0.275

0.28

0 2 4 6 8

o

c


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47.8

48

48.2

48.4

48.6

48.8

49

0 2 4 6 8

Time

Vol

.(V)

599

600

601

602

603

604

605

0 2 4 6 8

Time

Tem

p.(

T)


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594

594.5

595

595.5

596

596.5

597

0 2 4 6 8

Time

Tem

p.(

Tj)

0

10

20

30

40

50

60

0 2 4 6 8

Time

Flowr

ate(F)


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0

10

20

30

40

50

60

70

0 2 4 6 8

Time

Flowra

te(Fj)

0

50000

100000

150000

200000

250000

300000

0 2 4 6 8

Tim e

Heat(

Q)


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REFERENCES

1. Continuous stirred tank reactor modelsDr. M.J.WillisDept. of chemical & Process engineering,University of Newcastle.

2. Process Modeling, Simulation & Control for chemicalengineers.William L. Luyben. (second edition).

3. Chemical process controlGeorge Stephanopoulos.

4. Process dynamics Modeling, Analysis, and simulationB. Wayne Bequette.

Documents

Modeling and Simulation of Cstr