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Process Simulation using Simulink

Cheng-Liang Chen

PSE

LABORATORY

Department of Chemical Engineering National TAIWAN University

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Process Simulation Simulation of A Gas Process Consider the gas tank shown below. A fan blows air into a tank, and from the tank the air flows out through a valve. Suppose the air flow delivered by the fan is given by fi(t) = 0.16mi(t) where fi(t) is gas flow in scf/min, (scf is cubic feet at standard conditions of 60oF and 1 atm); mi(t) is signal to fan, %. The flow through the valve is expressed by p 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 in tank, 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 occurs isothermally at 60oF. The initial steady-state conditions are f¯i = f¯o = 8 scfm; p¯ = 40 psia; p¯1 = 14.7 psia; m ¯i = m ¯ o = 50 % An unsteady-state mole balance around the control volume, defined as the fan, tank, and outlet valve, is V dp(t) dn(t) = = ρ¯fi(t) − ρ¯fo(t) dt RT dt ρ¯ = 0.00263 lbmoles/scf is molar density of gas at standard conditions; R = 10.73 psia-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 of 5% in the signal to the inlet fan (starts from time =5 min.)

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Process Simulation Simulation of A Gas Process V dp(t) = ρ¯fi(t) − ρ¯fo(t) RT dt fi(t) = 0.16mi(t) (¯ ρ = 0.00263 lbmole/scf, V = 20 ft3) p fo(t) = 0.00506mo(t) p(t)[p(t) − p1(t)] mi(0) = m ¯ i = 50%, mo(0) = m ¯ o = 50%, p1(0) = p¯1 = 14.7psia ⇒ fo(0) = fi(0) = 0.16mi(0) = (0.16)(50) = 8.0 scf/min p fo(0) = 0.00506mo(0) p(0)[p(0) − p1(0)] ⇒ p(0) = 39.8 psia dp(t) ρ¯RT ⇒ = [fi(t) − fo(t)] dt V (0.00263)(10.73)(520) = [fi(t) − fo(t)] 20 = 0.734[fi(t) − fo(t)] (now: mi = 55% at t = 5)

Chen CL

dp(t) = 0.734[fi(t) − fo(t)] pdt fo(t) = 0.00506mo(t) p(t)[p(t) − p1(t)] fo(0) = 39.8 mi : 50 → 55% fi(t) = 0.16mi(t),

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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 Simulation Simulation 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 low temperature results in incomplete mixing. The tank is heated by steam condensing inside a coil. A proportional-integral-derivative (PID) controller is used to control

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the temperature in the tank by manipulating the steam valve position. The feed has a density ρ of 68.0 lb/ft3, a heat capacity cp of 0.80 Btu/lb-oF. The volume V of liquid in the reactor is maintained at 120 ft3. The coil consists of 205 ft of 4-in. schedule 40 steel pipe, weighting 10.8 lb/ft with a heat capacity of 0.12 Btu/lb-oF and an outside parameter of 4.500 in. The overall heat transfer coefficient U , based on the outside area of the coil, has been estimated as 2.1 Btu/min-ft2-oF. The steam available is saturated at a pressure of 30 psia; it can be assumed that its latent heat of condensation λ is constant at 966 Btu/lb. It can 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 ρcv

dT (t) = f (t)ρcpTi(t) + U A[Ts(t) − T (t)] − f (t)ρcpT (t) dt

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

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

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

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initial design conditions are f (0) = 15 ft3/min, Ti(0) = 100oF, and w(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 Simulation Simulation of A Stirred Tank Heater dT (t) dt dTs (t) dt

= =

UA 1 f (t)[T (t) − T (t)] + i V V ρcv [Ts (t) 1 CM {λw(t) − U A[Ts (t) − T (t)]} o 3

− T (t)],

T (0) = 150oF

Ts(0) = 230oF

Ti(0) = 100 F, f (0) = 15ft /min, w(0) = 42.2lb/min

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Process Simulation Simulation 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 Simulation Simulation of A Stirred Tank Heater Subsystem Block for The Stirred Tank Heater

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Transfer Function Simulation Temperature Control of A Stirred Tank Heater (p.201)

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Transfer Function Simulation Temperature Control of A Stirred Tank Heater (p.201)

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Process Simulation Simulation of A Batch Bioreactor Many important specialty chemical products are produced in bioreactors by processes such as fermentation. Most of these processes are carried out in batch mode by filling a tank with a substrate solution and inoculating it with a small amount of biomass. The biomass, feeding on the substrate, reproduces to produce the 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 the consumption of the substrate concentration, s(t), is given on a per unit volume bsis as follows: dx(t) = µ(t)x(t) dt ds(t) 1 = − dt y(t) µ(t)x(t) where y is the yield in biomass per unit mass of substrate and µ(t) is the biomass growth rate function (h−1). This growth rate function is analogous to the kinetic models 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 form:   dµ(t) s(t) = α µm − µ(t) dt k + s(t)

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where α is the adaptability parameter, and k and µm are the parameters of the model. Please use Simulink to simulate the model with the following data: α = 15h−1, k = 0.5g/liter, s(0) = 2.5g/liter, µ(0) = µm = 1.2h−1, and x(0) = 0.001g/liter.

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Process Simulation Simulation of A Batch Bioreactor dx(t) dt dµ(t) dt

ds(t) 1 = µ(t)x(t) = − dt y µ(t)x(t) h i s(t) = α µm k+s(t) − µ(t) α = 15 h−1, µ(0) = µm = 1.2

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

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Process Simulation Simulation 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 p dρ(t) V = wi(t) − Ao 2ρ(t)[p(t) − po] dt where

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

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 Simulation Simulation of A Pressure Tank

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Process Simulation Simulation 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 outlet concentration. Develop the mathematical model. Use Simulink to simulate the mixing process and plot the response of the outlet concentration to a step change of 5 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 and 0.05 moles/cm3. Assume a tight level controller keeps the volume in the tank constant.

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Process Simulation Simulation 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 in Btu/lbmole of A reacted. Let us also assume that the level of liquid in the reactor tank is constant; that is, the rate of mass into the tank is equal to the rate of mass out of the tank. Finally, the rate of reaction is given by rA(t) = koe−E/RT (t)c2A(t)

lbmoles of A reacted ft3-min

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where the frequency factor ko and energy of activation E are constants. The following Table gives the steady-state values of the variables and other process specifications. It is desired to find out how the outlet concentrations of A and B, and the outlet temperature respond to changes in the inlet concentration of A, cAi(t); the inlet temperature of the reactant Ti(t); the inlet temperature of cooling liquid Tci(t); and the flows f (t) and fc(t). Process information V = E= ρ= ∆Hr = A= Vc =

13.26 ft3 27, 820 Btu/lbmole 55 lbm/ft3 −12, 000 Btu/lbmole 36 ft2 1.56 ft3

ko = R= Cp = U= Cpc =

8.33 × 108 ft3/(lbmole-min) 1.987 Btu/(lbmole-oR) 0.88 Btu/(lbm-oF) 75 Btu/(h-ft2-oF) 1.0 Btu/(lbm-oF)

Steady-state values CAi(t) = Tc = cA(t) = Tci(t) =

0.5975 lbmole/ft3 602.7oR 0.2068 lbmole/ft3 540oR

Ti(t) = f= T (t) = fc(t) =

635oR 1.3364 ft3/min 678.9oR 0.8771 ft3/min

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Assume the reactor is initially at the design conditions. Use Simulink to simulate the process and plot the response of the reactor temperature to a step change of 0.25 ft3/min in process flow, and of 0.1 ft3/min in coolant flow. f (t)cAi(t) − f (t)cA(t) − V rA(t) = rA(t) = f (t)ρCpTi(t) − U A[T (t) − Tc(t)] − f (t)ρCpT (t) − V rA(∆Hr ) = fc(t)ρcCpcTci(t) + U A[T (t) − Tc(t)] − fc(t)ρcCpcTc(t) =

dcA(t) V dt koe−E/RT (t)c2A(t) dT (t) V ρCv dt dTc(t) VcρcCvc dt