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Pbr ejemplos páginas de modeling of chemical kinetics and reactor design   a. kayode coker
 

Pbr ejemplos páginas de modeling of chemical kinetics and reactor design a. kayode coker

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    Pbr ejemplos páginas de modeling of chemical kinetics and reactor design   a. kayode coker Pbr ejemplos páginas de modeling of chemical kinetics and reactor design a. kayode coker Document Transcript

    • Non-Isothermal Reactors 483Figure 6-12. Profiles of equilibrium conversion Xe versus temperature T forammonia synthesis. (Source: Schmidt, L. D., The Engineering of ChemicalReactions, Oxford University Press, New York, 1998.) Example 6-8 A tubular reactor is to be designed for the synthesis of methanolfrom a stoichiometric mixture of CO and H2. The reaction occurs inthe vapor phase using a solid catalyst in the form of porous spheres:CO + 2H2 = CH3OH. The average mixture physical and thermo-dynamic data at 500 K and 10 Mpa are Density = 30 kg/m3 Viscosity = 2 × 10–5 Pa • s Heat of reaction = –100 kJ/mol CH3OH Equilibrium constant = 0.3 × 104 Feed = Stoichiometric mixture at 473 K and 10 Mpa, gas velocity at reactor entrance is 0.15 m/sec Catalyst = Porous spheres 3 mm in diameter, packed in tubes 40 mm internal diameter Comment, from the reactor design and operation viewpoint, on the data.
    • 484 Modeling of Chemical Kinetics and Reactor DesignSolution k1   CO + 2 H 2 [ CH 3OH  ∆H R = −100 kJ  k2  mol • CH 3OH  The reaction is reversible and strongly exothermic. The equilibriumyield of CH3OH decreases as the temperature increases. Hence, a lowtemperature and increased pressure will be kept. The equilibrium constant Keq is expressed as k1 p CH 3OH K eq = = k 2 p CO • p 2 2 H Since the partial pressure is the mole fraction in the vapor phasemultiplied by the total pressure (i.e., pi = yiP), the equilibrium constantKeq is expressed as Keq = K y • P∆n, where ∆n = (1 – 1 – 2), thedifference between the gaseous moles of the products and the reactantsas in the ammonia synthesis reaction. p CH 3OH y CH 3OH 1  ∆G O  K eq = = • = exp − R  p CO • p 2 2 H y CO y 2 2 H P2  RT  The equilibrium constant Keq is determined at any temperature fromstandard state information on reactants and product. Considering thesynthesis of CH3OH, the equilibrium conversion Xe is determined fora stoichiometric feed of CO and H 2 at the total pressure. Theseconversions are determined by the number of moles of each speciesagainst conversion X by taking as a basis, 1 mole of CO.
    • Non-Isothermal Reactors 485Component Number of moles Ni Mole fraction yi 1− X CO 1 – X 3 − 2X 2(1 − X ) H2 2(1 – X) 3 − 2X X CH3OH X 3 − 2X Total ∑Ni = 3 – 2X ∑yi = 1.0 Substituting the expressions for the mole fractions of CO, H2, andCH3OH, respectively, for the equilibrium constant Keq yields X p CH 3OH y CH 3OH P K eq = = 1 • 2 = 3 − 2X  1 − X  P 2 − 2 X  P 2 2 p CO • p 2 2 H y CO y 2 2 H P  3 − 2X   3 − 2X  X (3 − 2 X ) 2 K eq = 4 (1 − X ) P 2 3 Figure 6-13 shows plots of equilibrium conversion versus tempera-ture. The plots indicate the conversion is low at operating temperatureT = 473 K (200°C), but ensures rapid reaction. The conversion perpass is low, therefore, it is important to maintain a high pressure toachieve a high conversion. Modern methanol plants operate at about250°C and 30–100 atm and give nearly equilibrium conversions usingCu/ZnO catalysts. The unreacted CO and H2 are recycled back intothe reactor.Reynolds Number Assuming a reasonable approach to plug flow in the reactor, andassuming the Reynolds number of the fluid in the reactor is d pρv Re p = µwhere Rep = Reynolds number of the fluid in the reactor dP = catalyst diameter
    • 486 Modeling of Chemical Kinetics and Reactor DesignFigure 6-12. Profiles of equilibrium conversion Xe versus temperature T formethanol synthesis. (Source: Schmidt, L. D., The Engineering of ChemicalReactions, Oxford University Press, New York, 1998.) ρ = fluid density µ = fluid viscosity v = fluid velocity The Reynolds number Rep is = (3 × 10 −3 )(30)(0.15) m • kg • m m • sec  Re p −5  •  2 × 10  m 3 sec kg  = 675The Reynolds number is 675, indicating that the fluid flow throughthe reactor is turbulent. Example 6-9 Design of Heterogeneous Catalytic Reactors A bench-scale study of the hydrogenation of nitrobenzene wasinvestigated by Wilson [2]. In this study, nitrobenzene and hydrogenwere fed at a rate of 65.9 gmol/hr to a 30 cm internal diameter (ID)reactor containing the granular catalyst. A thermocouple sheath, 0.9
    • Non-Isothermal Reactors 487cm in diameter, extended down the center of the tube. The voidfraction was 0.424 and the pressure atmospheric. The feed entered thereactor at 427.5 K, and the tube was immersed in an oil batch main-tained at the same temperature. The heat transfer coefficient from themean reaction temperature to the oil bath was determined experi-mentally to be 8.67 cal/hr • cm2 • °C. A large excess of hydrogenwas used so that the specific heat of the reaction mixture was equalto that of hydrogen. The change in total moles of the reaction wasneglected, and the heat of reaction was approximately constant andequal to –152,100 cal/gmol. The feed concentration of nitrobenzenewas 5 × 10–7 gmol/cm3. The global rate of reaction was representedby the expression 2, 958 − rp = 5.79 × 10 4 C 0.578 e Twhere rP = gmol nitrobenzene reacting/cm3hr, expressed in terms of void volume in the reactor C = concentration of nitrobenzene, gmol/cm3 T = temperature, K Calculate the reactor temperature and conversion as a function ofreactor length and comment on the results.Solution Concentration of nitrobenzene depends on both temperature andconversion. If u is the volumetric flowrate at a point in the reactorwhere the concentration is C, and uO is the value at the entrance, theconversion of nitrobenzene is u O C O − uC uC X= =1− (6-89) uOCO uOCO The volumetric flowrate depends on the temperature and changesas the temperature changes. Assuming a perfect gas behavior, CT X =1− (6-90) C O TO
    • 488 Modeling of Chemical Kinetics and Reactor Design Rearranging Equation 6-90 in terms of concentration gives C O TO C= (1 − X) (6-91) T = (5 × 10 −7 ) 427.5   T  (1 − X) (6-92) Substituting Equation 6-92 into the rate equation yields 0.578 − 2, 958 rp = 5.79 × 10 4 5 × 10 −7 × (1 − X) 427.5   e T  T  0.578 − 2, 958 1− X  = 438 e T (6-93)  T  From the differential mass balance (Equation 5-321) in terms of thevoid volume, FAO dX A = ( − rA ) dVp ′′′ (5-321)The feed rate of nitrobenzene is  273  ( FA = 65.9 (22, 400)  427.5  5.0 × 10 −7 ) = 1.16 gmol/hrHence, π 1.16 dX A = rp (0.424) (9 − 0.81) dl 4 1.16 dXA = 2.727(rP)dl dX A 2.727 dl = 1.16 rp ( ) 0.578 − 2, 958 1− X × 438  2.727 = e T 1.16  T 
    • Non-Isothermal Reactors 489 2, 958 1− X 0.578 − = 1, 038  dX A T e (6-94) dl  T The energy balance for a tubular reactor is: − ∆H R  ρuC p dT − u C AO  dX A = U πd t (TE − T )dl (6-95)  a In terms of the heat transfer coefficient, hO, − ∆H R  ρuC p dT − u C AO  dX A = h O πd t (TE − T )dl  a  or − ∆H R  ρuC p dT = u C AO  dX A − h O πd t (T − TE )dl  a  65.9 × 6.9dT = 1.16(152,00)dXA – 8.67(π)(3)(T – 427.5)dl ∆HR = –152,100 cal/gmol = 385.48 A  − 0.178(T − 427.5) dT dX  dl  (6-96) dl Substituting Equation 6-94 into Equation 6-96 gives   1− X 0.578 − 2, 958   = 385.481, 030  dT e T  − 0.178(T − 427.5) dl   T     2, 958 1− X 0.578 − = 397, 044.4  dT e T − 0.178 (T − 427.5) (6-97) dl  T  Equations 6-94 and 6-97 are first order differential equations, andit is possible to solve for both the conversion and temperature ofhydrogenation of nitrobenzene relative to the reactor length of 25 cm.A computer program PLUG61 has been developed employing theRunge-Kutta fourth order method to determine the temperature andconversion using a catalyst bed step size of 0.5 cm. Table 6-6 shows
    • 490 Modeling of Chemical Kinetics and Reactor Design Table 6-6 Longitudinal temperature profile and conversion in a reactor for the hydrogenation of nitrobenzene
    • Non-Isothermal Reactors 491the results of the program and Figure 6-14 illustrates the profile ofthe temperature against the catalyst-bed depth. The results show thatas the bed depth increases, the mean bulk temperature steadily increasesand reaches a maximum at about 13.0cm from the entrance to thereactor. This maximum temperature is referred to as the “hot spot” inFigure 6-14. Conversion and temperature profiles for the hydrogenationof nitrobenzene.
    • 492 Modeling of Chemical Kinetics and Reactor Designthe reactor. The conversion also increases and attains the maximumat about 24.0 cm from the entrance to the reactor. It is assumed thatthe radial temperature gradients in the plug flow reactor are negligible.Finally, these results are in agreement with the experimental resultsof Wilson, where the measured temperature was at the center ofthe reactor. TWO-DIMENSIONAL TUBULAR (PLUG FLOW) REACTOR Consider a reaction A → products in a tubular reactor with heatexchanged between the reactor and the surroundings. If the reactionis exothermic and heat is removed at the walls, a radial temperaturegradient occurs because the temperature at this point is greater thanat any other radial position. Reactants are readily consumed at thecenter resulting in a steep transverse concentration gradient. Thereactant diffuses toward the tube axis with a corresponding outwardflow of the products. The concentration and radial temperature gradi-ents in this system now make the one-dimensional tubular (plug flow)reactor inadequate. It is necessary to consider both the mass andenergy balance equations for the two dimensions l and r. With refer-ence to Figure 6-15, consider the effect of longitudinal dispersion andheat conduction. The following develops both material and energybalance equations for component A in an elementary annulus radiusδr and length δl. It is also assumed that equimolecular counter diffu-sion occurs. Other assumptions are: Figure 6-15. Differential section of two-dimensional tubular reactor.