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Chemical EngineeringQ&A Library2.18 Perform a degrees of freedom analysis for the model inEqs. 2-64 through 2-68. Identify parameters, output variables,and inputs (manipulated and disturbance variables). tions in the inletMixture of A and Bmics of the pro-omatic control is9, CA, TV, p, T2.4 Dynareaction. In other words, the heat of mixing is neg-ligible compared to the heat of reaction.8. Shaft work and heat losses to the ambient can beneglected.decrease inlellllllddisplayed incentration asThe following form of the CSTR energy balanceis convenient for analysis and can be derived fromEqs. 2-62 and 2-63 and Assumptions 1-8 (Fogler, 2006;Russell and Denn, 1972),TR) have wide-embody manySTR models tendtypes of continu-and packed-bedodel provides aing principles forTable 2.3 NcCooling mediumat temperatureT.ParameterVpcdT= wC(T; - T) + (-AHR)Vkc,Figure 2.6 A nonisothermal continuous stirred-tank reactor.dt+ UA(T. – T)(2-68)For these assumptions, the unsteady-state mass balance {for the CSTR iswhere AHR is the heat of reaction per mole of A that isreacted.In summary, the dynamic model of the CSTR consistsof Eqs. 2-62 to 2-64, 2-66, 2-67, and 2-68. This model isnonlinear as a result of the many product terms and Tneexponential temperature dependence of k in Eq. 2-63.Consequently, it must be solved by numerical integra-tion techniques (Fogler, 2006). The CSTR model willbecome considerably more complex if-AHRversible chemicalets to form speciesB. We assume thatd(pV)(2-64)= pq, - pgdtBecause V and p are constant, Eq. 2-64 reduces to450respect to compo-(2-65)q = 9iThus, even though the inlet and outlet flow rates maychange due to upstream or downstream conditions,Eq. 2-65 must be satisfied at all times. In Fig. 2.6, bothflow rates are denoted by the symbol q.For the stated assumptions, the unsteady-state com-ponent balances for species A (in molar concentrationunits) is(2-62)e 400-1. More complicated rate expressions are considered.For example, a mass action kinetics model for asecond-order, irreversible reaction, 2A B, isgiven byper unit volume,units of reciprocaltion of species A.onstant is typicallyature given by the1 = kzc350 ---.(2-69)2. Additional species or chemical reactions areinvolved. If the reaction mechanism involved pro-duction of an intermediate species, 2A B → B,then unsteady-state component balances for bothA and B* would be necessary (to calculate c, andCR), or balances for both A and B could be written(to calculate c, and cB). Information concerningthe reaction mechanisms would also be required.y dea= q(CAi - CA) – VkcA(2-66)300dt(2-63)This balance is a special case of the general componentbalance in Eq. 2-7.Next,balance for the CSTR. But first we make five additionalFigure 2.7 Rea.changes in cooland from 300 toE is the activationThe expressions intheoretical consid-and E are usuallyconsideran unsteady-state energyweReactions involving multiple species are described byhigh-order, highly coupled, nonlinear reaction models,because several component balances must be written.assumptions:1.04. The thermal capacitances of the coolant and thecooling coil wall are negligible compared to thethermal capacitance of the liquid in the tank.5. All of the coolant is at a uniform temperature, T-(That is, the increase in coolant temperature as thecoolant passes through the coil is neglected.)0.9data. Thus, theseobe semi-empiricalin Section 2.2.CSTR is shown in0.8EXAMPLE 2.50.7To illustrate how the CSTR can exhibit nonlinear dynamicbehavior, we simulate the effect of a step change in thecoolant temperature T, in positive and negative directions.Table 2.3 shows the parameters and nominal operatingcondition for the CSTR based on Eqs. 2-66 and 2-68 for theexothermic, irreversible first-order reaction Atwo state variables of the ODES are the concentration ofA (c.) and the reactor temperature T. The manipulatedinput variable is the jacket water temperature, TTwo cases are simulated, one based on increased cool-ing by changing T, from 300 to 290 K and one reducing thecooling rate by increasing T, from 300 to 305 K.These model equations are solved in MATLAB with anumerical integrator (ode15s) over a 10-min horizon. The0.6of pure componentcooling coil is usedt the desired oper-cat that is released0.50.46. The rate of heat transfer from the reactor contentsto the coolant is given by0.30.2-0.1B. Thenitial CSTR model(2-67)Q = UA(T- T)mptions:where U is the overall heat transfer coefficient andA is the heat transfer area. Both of these modelparameters are assumed to be constant.7. The enthalpy change associated with the mixing ofthe feed and the liquid in the tank is negligible com-pared with the enthalpy change for the chemical0.Figure 2.8 React.changes in coolinand product streamsare denoted by p.ctor is kept constantReactant A concentration (mol/L)Reactor temperature (K):Question

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Step 1

The degree of freedom analysis of a process model ensures that its model equations are solvable. The expression to calculate the degree of freedom is:

Here, *N _{F}* is the number of degrees of freedom for the process model,

Step 2

The model given in equations 2-64 through 2-68 is:

Step 3

From the above equations, independent equations that are deduced for the model are:

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