10 Simulink.pdf

8/11/2019 10 Simulink.pdf

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Simulink

Cheng-Liang Chen

PSELABORATORY

Department of Chemical EngineeringNational TAIWAN University


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MATrixLABoratory


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Simulink


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The SimulinkLibrary Browser


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Simulink Solution ofy= 10 sin(t)Check Results on Screen

dydt

= 10 sin(t) y(0) = 0, 0 t 13

y(t) =

130

(10 sin(t))dt + y(0)

Note: y(t) = 10(1 cos(t)) (exact solution)

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Simulink Solution ofy= 10 sin(t)Exporting toMATLAB

dydt

= 10 sin(t) y(0) = 0, 0 t 13

y(t) =

130

(10 sin(t))dt + y(0)

Note: y(t) = 10(1 cos(t)) (exact solution)

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Simulink Solution ofy= 10y+ 2 sin(4t)

dydt

= 10y+ 2 sin(4t) y(0) = 1, 0 t 3

y(t) =

30

(10y+ 2 sin(4t))dt + y(0)

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Linear State-Variable ModelsSimulink Model of A Two-Mass System

5x1+ 12x1+ 5x1 8x2 4x2 = 0

3x2+ 8x2+ 4x2 8x1 4x1 = f(t)

(m1= 5, M 2 = 3, c1= 4)

(c2= 8, k1= 1, k2= 4)

(x1(0) = 0.2; x1(0) = 0;)

(x2(0) = 0.5; x2(0) = 0;)

Let z1 x1, z3 x2

z1 = z2, z2 = (5z1 12z2+ 4z3+ 8z4)/5

z3 = z4, z4 = (4z1+ 8z2 4z3 8z4+ f(t))/3

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d

dt

z1

z2

z3

z4

=

0 1 0 0

1 12545

85

0 0 0 143

83

43

83

z1

z2

z3

z4

+

0

0

013

f(t)

x1

x2

=

1 0 0 0

0 0 1 0

z1

z2

z3

z4

+

0

0

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Process SimulationSimulation of A Gas Process

Consider the gas tank shown below. A fan blows air into a tank, and from thetank the air flows out through a valve. Suppose the air flow delivered by the fan isgiven by

fi(t) = 0.16mi(t)

where fi(t) is gas flow in scf/min, (scf is cubic feet at standard conditions of60oF

and1 atm); mi(t) is signal to fan, %. The flow through the valve is expressed by

fo(t) = 0.00506mo(t)

p(t)[p(t)p1(t)]

where fo(t) is gas flow, scf/min; mo(t) is signal to valve, %; p(t) is pressure intank, psia; p1(t) is downstream pressure from valve, psia.

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The volume of the tank is 20 ft3, and it can be assumed that the process occursisothermally at 60oF. The initial steady-state conditions are

fi= fo= 8 scfm; p= 40 psia; p1= 14.7 psia; mi= mo= 50 %

An unsteady-state mole balance around the control volume, defined as the fan,

tank, and outlet valve, is

dn(t)

dt =

V

RT

dp(t)

dt = fi(t) fo(t)

= 0.00263 lbmoles/scf is molar density of gas at standard conditions; R= 10.73psia-ft3/lbmoles-oR is ideal gas law constant; T= 520oR is gas temperature.

Please construct a Simulink model to simulate this process, and shows the

response of the pressure to a step change of5% in the signal to the inlet fan

(starts from time =5 min.)

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V

RT

dp(t)

dt = fi(t)fo(t)fi(t) = 0.16mi(t) (= 0.00263lbmole/scf, V = 20ft

3)

fo(t) = 0.00506mo(t)

p(t)[p(t)p1(t)]

mi(0) = mi= 50%, mo(0) = mo= 50%, p1(0) = p1= 14.7psia

fo(0) = fi(0) = 0.16mi(0) = (0.16)(50) = 8.0 scf/min

fo(0) = 0.00506mo(0)

p(0)[p(0)p1(0)]

p(0) = 39.8 psia

dp(t)dt

= RTV

[fi(t) fo(t)]

= (0.00263)(10.73)(520)

20 [fi(t) fo(t)]

= 0.734[fi(t) fo(t)] (now: mi= 55% at t= 5)

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fi(t) = 0.16mi(t), dp(t)

dt = 0.734[fi(t) fo(t)]

fo(t) = 0.00506mo(t)p(t)[p(t)p1(t)] po(0) = 39.8 mi: 50 55%

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subplot(2,1,1)

plot(dt,p,m,linewidth,2)

ylabel(\bf p(t),Fontsize,14);

title(\bf Gas pressure response to step fan change,Fontsize,14)subplot(2,1,2)

plot(dt,mi,b,linewidth,2)

ylabel(\bf m_i(t),Fontsize,14);

xlabel(\bf t (min),Fontsize,14);

set(gca,linewidth,3);

% set(gca,Fontsize,14);

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Process SimulationSimulation of A Stirred Tank Heater

The stirred tank is used to heat a process stream so that its premixed components

achieve a uniform composition. Temperature control is important in this process

because a high temperature tends to decompose the product while a lowtemperature results in incomplete mixing.

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The tank is heated by steam condensing inside a coil. Aproportional-integral-derivative (PID) controller is used to control the temperaturein the tank by manipulating the steam valve position.The feed has a density of68.0 lb/ft3, a heat capacity cp of0.80 Btu/lb-

oF. ThevolumeVof liquid in the reactor is maintained at 120ft3. The coil consists of205ft of4-in. schedule 40 steel pipe, weighting 10.8 lb/ft with a heat capacity of0.12

Btu/lb-o

F and an outside parameter of4.500 in. The overall heat transfercoefficientU, based on the outside area of the coil, has been estimated as 2.1Btu/min-ft2-oF. The steam available is saturated at a pressure of30 psia; it canbe assumed that its latent heat of condensation is constant at 966 Btu/lb. Itcan also be assumed that the inlet temperature Ti is constant.An energy balance on the liquid in the tank, assume negligible heat losses, perfect

mixing, and constant volume and physical properties, results in the equation

V cvdT(t)

dt =f(t)cpTi(t) + U A[Ts(t) T(t)] f(t)cpT(t)

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An energy balance on the coil, assuming that the coil metal is at the sametemperature as the condensing steam, results in (CM: heat capacitance of coilmetal, Btu/oF; w(t): steam rate, lb/min)

CMdTs(t)

dt =w(t) U A[Ts(t) T(t)]

The initial steady-state conditions are T(0) = 150oF andTs(0) = 230oF. Also the

initial design conditions are f(0) = 15 ft3/min, Ti(0) = 100oF, andw(0) = 42.2

lb/min.

Construct a Simulink diagram for the simulation of the heater. shows the

responses of the temperatures to a step changes in process flow.

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dT(t)

dt = 1

Vf(t)[T

i(t) T(t)] + UA

V cv[T

s(t) T(t)], T(0) = 150oF

dTs(t)dt =

1CM{w(t) U A[Ts(t) T(t)]} Ts(0) = 230

oF

Ti(0) = 100oF, f(0) = 15ft3/min, w(0) = 42.2lb/min

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Response of heater outlet temperature and steam chest temperature

to a step change in process flow

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Subsystem Block for The Stirred Tank Heater

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Process SimulationSimulation of A Batch Bioreactor

Many important specialty chemical products are produced in bioreactors byprocesses such as fermentation. Most of these processes are carried out in batchmode by filling a tank with a substrate solution and inoculating it with a smallamount of biomass. The biomass, feeding on the substrate, reproduces to producethe desired product, until the substrate is consumed. This example is presented

here to show some of the special characteristics of biochemical processes.A dynamic model of the growth of the biomass concentration x(t) and of theconsumption of the substrate concentration, s(t), is given on a per unit volumebsis as follows: dx(t)

dt = (t)x(t)

ds(t)dt

= 1y(t)

(t)x(t)

where y is the yield in biomass per unit mass of substrate and (t) is the biomassgrowth rate function (h1). This growth rate function is analogous to the kineticmodels used to model chemical reactors. It is designed to match experimental

data. Here we will use the Monod model with adaptability wich has the following

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form:d(t)

dt =

m

s(t)

k+ s(t) (t)

where is the adaptability parameter, and k andm are the parameters of the

model. Please use Simulink to simulate the model with the following data:= 15h1, k= 0.5g/liter, s(0) = 2.5g/liter, (0) =m= 1.2h

1, and

x(0) = 0.001g/liter.

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dx(t)

dt = (t)x(t)

ds(t)

dt = 1

y(t)x(t)

d(t)

dt =

m

s(t)k+s(t) (t)

= 15 h1, (0) =m= 1.2

k = 0.5 g/liter, s(0) = 2.5 g/liter, x(0) = 0.001 g/liter

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Process SimulationSimulation of A Pressure Tank

A stray bullet fired by a careless robber punctures the compressed air tank at a gas

station. The mass balance of air in the tank is

Vd(t)

dt =wi(t)Ao

2(t)[p(t)po]

where

(t) = MRT

p(t)

wi(t) kg/s, is the inlet flow from the air compressor, V = 1.5 m3, is the volume of

the tank, Ao= 0.785 cm2, is the area of the bullet hole, M= 29 kg/kmole, is the

molecular weight of air, R= 8.314 kPa-m3/kmole-K, is the ideal gas law

constant, and temperature T is assumed constant at 70oC, po= 500 kPa gauge.Use Simulink to simulate the process and plot the response of the pressure in the

tank.

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Process SimulationSimulation of A Mixing Tank

Computer-room Exercise

Consider the mixing process shown below. Assume

that the density of the input and output streams

are very similar and that the flow rates f1 and

f2 are constant. It is desired to understand

how each inlet concentration affects the outletconcentration. Develop the mathematical model.

Use Simulink to simulate the mixing process

and plot the response of the outlet concentration to a step change of5

gallon/minute (gpm) in flow f1. At the initial steady-state conditions the flow

from the tank is 100 gpm, and its concentration is 0.025 moles/cm3. The tank

volume is 200 gallons, and the feed compositions are 0.010 and0.05 moles/cm3.

Assume a tight level controller keeps the volume in the tank constant.

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Process SimulationSimulation of A Nonisothermal Chemical Reaction

Computer-room Exercise

Consider a stirred-tank reactor with reaction A B

as shown below. To remove the heat of reaction the

reactor is surrounded by a jacket through which a

cooling liquid flows. Let us assume that the heat

loss to the surroundings are negligible, and that

the thermodynamic properties, densities, and heat

capacities of the reactants and products are both

equal and constant. The heat of reaction is constant and is given by Hr inBtu/lbmole ofA reacted. Let us also assume that the level of liquid in the reactortank is constant; that is, the rate of mass into the tank is equal to the rate ofmass out of the tank. Finally, the rate of reaction is given by

rA(t) =koe

E/RT(t)

c

2

A(t)

lbmoles of A reacted

ft3-min

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where the frequency factor ko and energy of activation Eare constants. Thefollowing Table gives the steady-state values of the variables and other processspecifications. It is desired to find out how the outlet concentrations ofA andB,and the outlet temperature respond to changes in the inlet concentration ofA,

cAi(t); the inlet temperature of the reactant Ti(t); the inlet temperature ofcooling liquid Tci(t); and the flows f(t) andfc(t).

Process information

V = 13.26 ft3 ko= 8.33 108 ft3/(lbmole-min)

E= 27, 820 Btu/lbmole R= 1.987 Btu/(lbmole-

o

R)= 55 lbm/ft3 Cp= 0.88 Btu/(lbm-

oF)

Hr = 12, 000 Btu/lbmole U= 75 Btu/(h-ft2-oF)

A= 36 ft2 Cpc= 1.0 Btu/(lbm-oF)

Vc= 1.56 ft3

Steady-state values

CAi(t) = 0.5975 lbmole/ft3 Ti(t) = 635

oR

Tc= 602.7oR f= 1.3364 ft3/min

cA(t) = 0.2068 lbmole/ft3 T(t) = 678.9oR

Tci(t) = 540o

R fc(t) = 0.8771 ft3

/min

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Assume the reactor is initially at the design conditions. Use Simulink to simulatethe process and plot the response of the reactor temperature to a step change of0.25 ft3/min in process flow, and of0.1 ft3/min in coolant flow.

f(t)cAi(t) f(t)cA(t) V rA(t) = VdcA(t)dt

rA(t) = koeE/RT(t)c2A(t)

f(t)CpTi(t) U A[T(t) Tc(t)] f(t)CpT(t) V rA(Hr) = V CvdT(t)

dt

fc(t)cCpcTci(t) + U A[T(t) Tc(t)] fc(t)cCpcTc(t) = VccCvcdTc(t)

dt

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Thank You for Your AttentionQuestions Are Welcome

10 Simulink.pdf

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