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Simulink
Cheng-Liang Chen
PSELABORATORY
Department of Chemical EngineeringNational TAIWAN University
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Chen CL 1
MATrixLABoratory
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Simulink
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Chen CL 3
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)
C C
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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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Chen CL 6
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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Chen CL 7
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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Chen CL 9
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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Process SimulationSimulation of A Gas Process
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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Process SimulationSimulation of A Gas Process
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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Chen CL 14
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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Process SimulationSimulation of A Stirred Tank Heater
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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Process SimulationSimulation of A Stirred Tank Heater
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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Process SimulationSimulation of A Stirred Tank Heater
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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Process SimulationSimulation of A Stirred Tank Heater
Response of heater outlet temperature and steam chest temperature
to a step change in process flow
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Process SimulationSimulation of A Stirred Tank Heater
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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Process SimulationSimulation of A Batch Bioreactor
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 Batch Bioreactor
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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 Pressure Tank
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Process SimulationSimulation of A Pressure 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