Example 8: CSTR with Multiple Solutions

Size: px

Start display at page:

Download "Example 8: CSTR with Multiple Solutions"

Sharlene Jennings
5 years ago
Views:

1 Example 8: CSTR with Multiple Solutions This model studies the multiple steady-states of exothermic reactions. The example is from Parulekar (27) which in turn was modified from one by Fogler (1999). The species and reactions do not represent a real system. Therefore, the physical properties have been specified to match the conditions given by Parulekar. Some conditions have been changed to give more realistic results, but the basic results regarding the multiple steady states have been retained. For this example, the intermediate compound, B, will be assumed the desired product. A Simple Reaction System Rg cal mol K A --> B B --> C gas law constant R R1 Thermophysical properties species molecular weights heats of formation free energies C, H, O atoms in each species A B C Inert Mw gm mol H f J mol F 1 3 cal mol Atom H f valueswere selected based on given H rxn, Parulekar (27) F will not be used for this problem E8: 1

2 The product of the molecular weight of A and the molar density, c, must equal the liquid density, which is assumed below to be that of water. The initial concentration of A was given as.3 mol/l by Parulekar. That would result in molecular weights of To be more realistic, the above molecular weights have been assumed, which makes the molar concentration 3 mol/l. These changes do not affect the solution because the two parameters, Mw and c, are always multiplied by each other, and the result is thus constant. NC rows (Mw) n NC 1 number of components component index Inlet conditions T 3 K reference temperature for dimensionless temperature (not the inlet temperature) Feed composition mole fractions 1 mfin A B C Inert mfin = 1 check sum M f ω mfin Mw molecular weight of feed mixture M f =.33 kg mol ( mfin Mw) 1 convert mole fractions to weight fractions M f ω = E8: 2

3 Fluid properties ρ 1 gm cm 3 c ρ = 3.33 mol M f L cp 3 3 J 3 mol K 3 Cp ( ω, Θ) cp ω Mw density of fluid molar density of feed heat capacities, molar basis heat capacities on a mass basis Cp is not a function of composition or temperature in this example, but the function is used to be consistent with other models and to make the model easily adaptable to other reaction systems. Stoichiometry 1 1 ν 1 1 rows are reactions, columns are components check stoichiometry and Mw = ν Mw kg mol This vector should be null if Mw and ν are correct. E8: 3

4 Heat of reaction H o ν H f standard state, at 298 K H o = J mol Cp ν cp = J mol K difference in heat capacities of products and reactants DHT (T) T This function provides the dependence of enthalpy on temperature. For cp constant with temperature, the function is simply the absolute temperature, T. H (T) H o + Cp ( DHT (T) DHT ( 298 K)) heat of reaction is constant for this example because Cp= Reaction Rate Expressions E activation energies cal mol pre-exponential rate constants E A 3.3 exp Rg 3 K 1 min A 1 E 4.58 exp 1 Rg 5 K 1 min The above rate constants result in a nominal space time of.1 min [Parulekar (27)]. To make the problem more typical for industry, the rate constants will be divided by 1. This will correspondingly increase the space times to the order of 1 min. These changes do not affect the multiple temperature solutions. E8: 4

5 A A 1 The rate expressions in terms of the state variables act 1 ω E Mw A exp θ T Rg ω Mw rxn ( ω, θ, p, act) c ω 1 E Mw A exp 1 1 θ T Rg 1 ω Mw The concentrations used in the rate expressions are in terms of mol/cu m. The concentrations have been written using the weight fractions, converting to mole fractions, and multiplying by the molar density, c. Pressure is not generally needed for liquid reaction rates, but it has been included in the argument list to be compatible with the other models. The catalyst activity, act, has been assumed to be 1. Net production rates T Rxn ( ω, θ, p, act) ν rxn ( ω, θ, p, act) mass The energy balance will use the difference in enthalpy of feed and effluent streams instead of a heat production term. E8: 5

6 CSTR with reactor temperature and space time as a parameters This is a design model for a non-adiabatic reactor. The adiabatic version will be solved in the next section..33 MwD diag (Mw) = kg mol a matrix version of the Mw vector is needed below MwDI MwD 1 the inverse is computed and assigned to a matrix to avoid repeat computations in the solve block Solve block The equations are written in vector/matrix form. Guess Values Constraints Solver ω ω ω ω + MwD Rxn ω, Tr, 1, act τ = T ρ 1 ω= T Atom MwDI ω ω = CSTR ( τ, Tr) find (ω) Initial guess for outlet weight fractions. Mass balances for each species. The last constraints are C, H atom balances which are not needed for this problem. They have been included so that the file may be easily modified for a realistic problem. The result, CSTR, is the outlet weight fraction vector as a function of τ and Tr. E8: 6

7 Define space time and temperature cases t i 8 time index τ_c t 1 (.1 t +.2) s space times 4 reactor temperature index Tout 2 K Tout Tout + 2 K outlet temperatures i + 1 i solve for each case ωout CSTR τ_c, Tout t, i t i outlet mass fractions for each case examine a case to check for errors tc 4 ic 15 select the space time and outlet temperature indexes of the case τ_c = 1 min Tout = 5K the selected case values for the independent variables tc ic ωout t, i Mw mfout t, i ωout t, i Mw outlet mole fractions, converted from outlet mass fractions E8: 7

8 inlet weight fractions outlet weight fractions inlet mole fractions outlet mole fractions 1 ω = ωout tc, ic = mfin = mfout tc, ic = ω = 1 ωout = 1 mfin = 1 mfout = 1 tc, ic check mass balances of elements tc, ic ω ωout T tc, ic Atom gm = Mw mol This result should be a null vector. Determine desired operating condition from plots conversion for each case ωout t, i conv 1 t, i ω intermediate B outlet mass fraction Bout ωout t, i, t i 1 E8: 8

9 Figure 8.1: Conversion Map Figure 8.2: B Effluent Weight Fraction conv Bout Conversion is sensitive to temperature but not space velocity for the ranges selected. There is a narrow window of reactor temperature that yields a high selectivity to the intermediate product. Select optimum (high conversion and selectivity to desired product) space time index and τopt 4 tout 15 τ opt = 1 min τ_c τopt T opt = 5K Tout tout values selected Tout index using above plots. Now that the desired reactor space time and temperature have been set, include the heat balance to see if operation at those conditions is feasible. E8: 9

10 Energy balance Thus far, the results have not been dependent on the inlet temperature. For an energy balance,the inlet temperature, needed. Tin, is Heat balance for a given (optimal) space time τ opt = 1 min Enthalpy out - Enthalpy in Del_H ( Tin, Tr, ω, ω1) H f ω1 + Cp ω1, Tr ( DHT (Tr) DHT ( 298 K)) H f ω + Cp ω, Tin ( DHT (Tin) DHT ( 298 K)) Mw T Mw T j 25 inlet temperature index Tin 25 K Tin Tin + 5 K inlet temperature vector j + 1 j DH Del_H Tin, Tout, ω, ωout j, i j i τopt, i matrix of enthalpy difference for the given space time, which is given implicitly via ωout τopt E8: 1

11 Figure 8.3: Enthalpy Change for Given Tau vertical axis: Tout index Look up the values that correspond to indexes below ⁶ ⁶ 1.5 1⁶ 3 1 1⁶ ⁵ ¹⁰ 5 1⁵ ¹⁰ ⁵ -1 1⁶ -1 1⁶ ⁶ -5 1⁵ 5 1⁵ 1 1⁶ 1.5 1⁶ 2 1⁶ 2.5 1⁶ 3 1⁶ horizontal axis: Tin index DH J kg Tin = K Tout = to see rest of vectors, click on the 3 dots and drag the view box K E8: 11

12 Figure E8.3 can be used to explore the possibility of operating at the desired outlet temperature in order to achieve the conversion and selectivity of the optimum. The chosen Tout index, tout = 15, may be found on the vertical axis, and then the heat exchange required at different inlet temperatures are examined by moving at constant Tout across the plot. The contours are constantheat exchanged, negative for cooling, positive for heating. Point 1on the plot is on the adiabatic curve for the selected outlet temperature. Point 2is on a curve with heat removal at the same outlet temperature. Thus, for a given τ, the approximate inlet and outlet temperatures and heat exchanged may be determined with the above plot. For the current example, it appears that multiple solutions will exist regardless of the chosen Tin and heat exchange pair. The designer may wish to discontinue this case and try using inerts to avoid the multiple solutions. For this example, the current case will be carried forward. The horizontal Tout line in Fig E8.3 intersects the adiabatic curve (see"1") at Tin index of about 1. This is Tin = 3K as 1 shown to the right of the graph. The slope of the heat exchange curve is positive at this intersection, indicating the operating point is stable. Instead of adiabatic operation, assume the inlet temperature is given as 36 K, or Tin index of 22. This is point "2" on the graph. The Tin and heat exchange chosen will be used for initial guesses for Fig E8.4. select inlet temperature and heat exchanged tin 22 index for Tin chosen from above plot Qm J kg given the outlet and inlet temperature indexes, estimate the heat exchange needed (the DH contours in Fig E8.3 are equal to Qm) Q ( x, y) y heat exchange curve (to plot the vertical line in Figure 8.4 at Qm) E8: 12

13 Example for Tin = 36K, Qm = J, Tout = 5K, τ = tin kg opt 6s tout Figure 8.4: Operating Curve and Heat Exchange 1 1³ Tout i (K) 6 Q Qm, Tout (K) i ⁶ ⁶ ⁶ -7 1⁵ ⁵ 3.5 1⁵ 7 1⁵ 1.5 1⁶ 1.4 1⁶ ⁶ 2.1 1⁶ DH tin, i J kg Qm J kg E8: 13

14 The red curve in Fig E8.4 is the heat produced at Tout, including the sensible heat of the inlet and outlet streams, but excluding any heat exchanged. The blue curve, Qm, is the amount of heat exchanged per kg of feed, a value chosen by the user. The solutions are where the heat exchanged equals the heat produced. Traditionally, as shown by Parulekar, the red curve would exclude the sensible heat and the second curve would be a sloping line representing the heat removedby the sensible heat difference. The intersections of the curves in the traditional plot are difficult to detect dueto the nearly equal slopes of the two curves. Parulekar therefore used an iterative calculation to find the roots. With the above plot, the intersections are readily determined using the Plots>Add Horizontal Marker feature. The horizontal line in the plot has been placed at the third highest Tout solution. In Fig E8.4, the first, third and fifth points are stable, and the second and fourth points are unstable. Stable points are on positive slopes of the red curve and unstable points are on the negative slopes. Even with cooling or heating, multiple solutions exist for the desired Tout. The vertical line in the above graph is moved to the left for cooling...to the right for heating (change Qm prior to the plot). Also, the red line can be moved to left orright by selecting a higher or lower inlet temperature (i.e. value of tin), respectively. The possibility of multiple solutions is a red flag for unstable operation even if the point is "stable". The dynamic CSTR model in Example 9 will be used to simulate the startup to the desired intermediate solution. A better option may be to include an inert in the feed. Operation at desired conditions The desired conditions have been selected from the simulations and plots E8.1-4, but the actual design condition has not yet been fully simulated with accuracy. The heat balance is now included in the model so that the reactor temperature becomes a dependent variable. The inlet temperature and space time are specified. Using this model, the conditions selected with the above procedure can be simulated to find the complete composition and outlet temperature at the given space time and inlet temperature. E8: 14

15 Desired conditions This area may be used to change conditions. previously set conditions If different conditions are desired, make the changes above this line. Change indexes for the temperatures and the space time, not the actual variable. This is required for the proper initial guesses and for specification of T. tout = 15 Tout index to set outlet temperature intial guess = 5K Tout tout tin = 22 Tin index to set inlet temperature T = 36K Tin tin τopt = 4 τ_c index to set space time τ = 6s τ_c τopt Qm = m s 2 heat exchanged E8: 15

16 Solve block Guess Values ω1 ωout Tout tout τopt, tout θ1 T initial guesses obtained from cases studied above Del_H T, θ1 T, ω, ω1 + Qm= The energy balance constraint is added. Note that in Prime, the equations may have units, as in the energy balance. Constraints ω ω1 + MwD Rxn ( ω1, θ1, 1, act) τ = ρ 1 ω1= T Atom MwDI ω ω1 = Solver x design θ design find ( ω1, θ1) E8: 16

17 results x design = mass fractions in reactor and in effluent Tr θ design T = K reactor temperature x design x design1 conv ad 1 =.946 conversion of A ω selectivity =.984 selectivity to B ω x design Summary Both a CSTR model and a design procedure were demonstrated. The procedure shown arrives at a desired design without trial and error. By first mapping the conversion and selectivity without the heat balance, multiple steady states are not an issue. After finding the desirable inlet and outlet temperatures and the heat exchange needed, those conditions and an outlet composition from a nearby mapped case provide the initial guesses needed to obtain the desired steady state solution. When compared to the combined analytical/numerical method used by Parulekar, the fully numerical method used above appears quite complicated. Parulekar's method achieves the goal of demonstrating the multiple steady state problem that can arise in CSTRs in a simple manner. The goal for the model presented above is a generic model that is easily changed for other systems. The system used for this example will be used for the dynamic CSTR model in Example 9. Exercise Explore the addition of inerts to obtain a system with a single stable operating point at high conversion and selectivity. An inert compound has been supplied. E8: 17

18 References Fogler, H.S., "Elements of Chemical Reaction Engineering", 3rd Ed., Prentice-Hall (1999). Parulekar, S.J., Illinois Inst. of Technology, Chemical Engineering Education, Winter (27) E8: 18

Chemical Kinetics and Reaction Engineering

Chemical Kinetics and Reaction Engineering MIDTERM EXAMINATION II Friday, April 9, 2010 The exam is 100 points total and 20% of the course grade. Please read through the questions carefully before giving