Reaction Models PDF

7 REACTION MODELS The selection of a reaction model is important in the design of reaction systems (eg. reactor and accompanying systems). A chosen model must closely correspond to what actually takes place and its rate expression must closely predict and describe the actual kinetics without too m...

7 REACTION MODELS The selection of a reaction model is important in the design of reaction systems (eg. reactor and accompanying systems). A chosen model must closely correspond to what actually takes place and its rate expression must closely predict and describe the actual kinetics without too many mathematical complexities. In the forgoing discussions on reaction models the reaction of single particles only shall be considered, even though it may not always represent actual conditions. Very dilute pulps, where the particles in suspension may be far enough from each other to interact with each other may be considered close to reactions of single particles. We shall also take a look at what happens in multiple particle systems. Since diffusional transport may differ for porous and nonporous particles these will be considered differently. 7.1 Reaction of a Single Nonporous Particle If the particle taking part in the reaction is nonporous then the fluid reactant is not able to penetrate into the solid, therefore the reaction takes place at the surface of the nonporous particle. Two variations of this are possible. In the first case the particle size decreases as the reaction proceeds with time, eg in gasification, dissolution reactions (Shrinking particle Model). In the second variation, a product is formed on the surface of the unreacted core and the unreacted core decreases as the reaction proceeds (Shrinking core Model). 7.1.1. CASE I: Shrinking Particle Model The reaction equation may be presented as: A(fluid) + bB(solid) = cR (fluid) + dD(removable solid) Examples of this reaction are the dissolution of metals/metal compounds, chlorination of metals, roasting of cinnabar (HgS), etc., reactions that give a soluble product, or gaseous product or a product that flakes off. The rate of consumption of fluid species A at the reaction site by the reaction is rA = - ks f(Cs) Cs is the concentration of A at the surface. At steady-state and neglecting accumulation in the boundary layer surrounding the solid, the rate of chemical reaction should be equal to the rate of mass transfer from bulk of fluid, ie. - ks f(Cs) = - km(Cb - Cs) (both processes are in series) Assuming first-order irreversible reaction, ksCs = km(Cb - Cs) Solving for Cs and substituting into rA = ksCs : 20 1 𝑟𝐴 = 𝐶𝑏 1 1 + 𝑘𝑚 𝑘𝑠 To obtain the overall conversion vs time relation, the rate of disappearance of A , rA, must be equated with the rate of consumption of solid B. From the stoichiometry of the reaction, 𝜌𝐵 𝑑𝑟𝑐 𝑟𝐴 = − B is the molar concentration of B 𝑏 𝑑𝑡 rC is particle radius at time t. From the above two equations, 𝑑𝑟𝑐 = 𝑑𝑡 𝑏 𝐶𝑏 /𝜌𝐵 1 1 + 𝑘𝑠 𝑘𝑚 All the parameters in this equation, except km, are independent of rc and therefore rearranging and integrating gives, 𝜌 𝑟𝑜 −𝑟𝑐 𝑡 = 𝑏 𝐶𝐵 [ 𝑏 𝑘𝑠 𝑟 + ∫𝑟 𝑐 𝑘 𝑜 𝑑𝑟𝑐 𝑚 (𝑟)𝑐 ] This expression gives the relationship between t and r C and assumes that both mass transfer rate and chemical reaction rate are equal. By introducing conversion (or fraction reacted), X, as related to r C , ie. 𝑋𝐵 = 𝑟𝑜3 −𝑟𝑐3 𝑟𝑜3 𝑟 = 1 − (𝑟𝑐 )3 𝑜 reaction time can be expressed as a function of conversion. 8.1.1.1 Chemical Reaction Controls: Then progress of reaction is unaffected by the changing size of the particle and the rate of reaction is proportional to the rate of disappearance of solid B during reaction. The differential equation for this is given by: − 𝑑𝑀 𝑑𝑟𝑐 = −𝜌𝐵 = 𝑏𝑘𝑠 𝐶𝑏 𝑑𝑡 𝑑𝑡 Rearranging and integrating gives, 𝑟 𝑡 𝜌 −𝜌𝐵 ∫𝑟 𝑐 𝑑𝑟𝑐 = 𝑏𝑘𝑠 𝐶𝑏 ∫0 𝑑𝑡 𝑡 = 𝑏 𝑘 𝐵𝐶 (𝑟𝑜 − 𝑟𝑐 ) 𝑜 𝑠 𝑏 In terms of conversion, 𝑡= 1 𝜌𝐵 𝑟𝑜 𝑟𝑐 𝜌𝐵 𝑟𝑜 (1 − ) = [1 − (1 − 𝑋𝐵 )3 ] 𝑏 𝑘𝑠 𝐶𝑏 𝑟𝑜 𝑏 𝑘𝑠 𝐶𝑏 At complete conversion, rc = 0 and t = τ 7.1.1.2 Film Diffusion Controls Film resistance at the surface of a particle depends on relative velocity between fluid and particle, fluid properties (eg. viscosity), particle size, among others. These have been correlated for various systems such as packed beds, fluidized beds and solids in free fall (eg. dilute solutions). For dilute solutions, mass transfer of a component of mole fraction, y, in fluid is: 𝑆ℎ = 1 1 𝑘𝑚 𝑑𝑝 𝑦 𝑑𝑝 𝜌 𝜈 1 𝜇 1 = 2 + 0.6 𝑅𝑒 2 𝑆𝑐 3 = 2 + 0.6 ( )2 ( )3 𝐷 𝜇 𝜌𝐷 As particle size changes during reaction so does km. Generally, km increases as fluid velocity increases and 21 for small dp and u (ie in the Stokes regime), km ≈ 1/dp for large dp and u km ≈ u1/2 / dp1/2 Solving conversion-time expression for small particles in the Stoke regime for time t and r = r C , 𝑑𝑛𝐵 = 𝜌𝐵 𝑑𝑉 = 4𝜋𝜌𝐵 𝑟𝑐2 𝑑𝑟𝑐 Also, 1 𝑑𝑛𝐵 𝑒𝑥 𝑑𝑡 −𝑆 1 = − 4𝜋𝑟 2 ∙ 4𝜋𝜌𝐵 𝑟𝑐2 𝑜 𝑑𝑟𝑐 𝑑𝑡 = −𝜌𝐵 𝑟𝑐2 𝑑𝑟𝑐 𝑟𝑜2 𝑑𝑡 = 𝑏 𝑘𝑚 𝐶𝑏 Sex is the external surface area In Stokes regime, the expression for mass transfer of component of mole fraction y in dilute fluid becomes: 𝑘𝑚 = 2𝐷 𝐷 = 𝑑𝑝 𝑦 𝑟𝑜 𝑦 Substituting and integrating gives, 𝑟 𝑏𝐶𝑏 𝑟𝑜2 𝐷 𝑐 𝜌𝐵 𝑦 − ∫𝑟 𝑜 𝑟𝑐2 𝑑𝑟𝑐 = 𝜌 𝑦𝑟𝑜2 𝑡 𝐵 𝑡 = 2𝑏𝐶 ∫0 𝑑𝑡 𝑏 𝑟 [1 − (𝑟𝑐 )2 ] 𝐷 0 Time taken for the complete disappearance of a particle is: 𝜌 𝑦𝑟𝑜2 𝐵  = 2𝑏𝐶 𝑏𝐷 Therefore: 𝑡 𝜏 𝑟 2 2 = 1 − (𝑟𝑐 ) = 1 − ( 1 − 𝑋𝐵 )3 𝑜 In terms of fraction converted, 𝑡= 2 𝜌𝐵 𝑦𝑟𝑜2 𝜌𝐵 𝑦𝑟𝑜2 𝑟𝑐 [1 − ( )2 ] = [1 − (1 − 𝑋𝐵 )3 ] 2𝑏𝐶𝑏 𝐷 𝑟𝑜 2𝑏𝐶𝑏 𝐷 7.1.2 CASE II: Shrinking Core Model In this case a product is formed on the surface of the particle and reaction takes place at the interface between the product layer and the unreacted core. Examples of this case are the reduction of metal oxides, oxidation of metals, roasting of sulphides and the leaching of minerals from ores. Such a reaction can be represented as: A(fluid) + bB(solid) = cR (fluid) + dD(solid) The overall process is made up of the following steps, (i) external mass transfer through the boundary layer (ii) diffusion through product layer (iii) chemical reaction at the interface between unreacted core and product layer. Any of the above factors may control the overall process. In treating this case we assume isothermal conditions, first-order irreversible reaction. 7.1.2.1. External Mass Transfer Controls The product layer offers no resistance to diffusion, the reaction at the interface is fast and there is no fluid reactant present at the surface (ie. Cs = 0). The concentration driving force (Cb - Cs) is constant at all times. 22 External mass transport rate across the boundary layer is given by: −𝑟𝐵 = 4𝜋𝑟𝑜2 𝑘𝑚 (𝐶𝑏 − 𝐶𝑠 ) or − 1 𝑑𝑛𝐵 1 𝑑𝑛𝐵 𝑏 𝑑𝑛𝐴 = = − = 𝑏 𝑘𝑚 (𝐶𝑏 − 𝐶𝑠 ) = 𝑐𝑜𝑛𝑠𝑡. 2 𝑆𝑒𝑥 𝑑𝑡 4𝜋𝑟𝑜 𝑑𝑡 4𝜋𝑟𝑜2 𝑑𝑡 The rate of disappearance of B, (-dnB), is proportional to the decrease in volume of unreacted core, (- dV), 4 −𝑑𝑛𝐵 = −𝑏 𝑑𝑛𝐴 = −𝜌𝐵 𝑑𝑉 = −𝜌𝐵 𝑑 ( 𝜋𝑟𝑐3 ) = −4𝜋𝜌𝐵 𝑟𝑐2 𝑑𝑟𝑐 3 1 𝑑𝑛𝐵 𝑒𝑥 𝑑𝑡 −𝑆 Therefore, 𝜌 1 = 4𝜋𝑟 2 ∙ 4𝜋𝜌𝐵 𝑟𝑐2 𝑑𝑟𝑐 𝑑𝑡 𝑜 𝑟 = 𝑏𝑘𝑚 𝐶𝑏 𝑡 − 𝑟𝐵2 ∫𝑟 𝑐 𝑟𝑐2 𝑑𝑟𝑐 = 𝑏𝑘𝑚 𝐶𝑏 ∫0 𝑑𝑡 𝑜 𝑜 Rearranging and integrating gives, 𝜌 𝑟𝑜 𝑡 = 3 𝑏 𝐵𝑘 𝑟 𝑚 𝐶𝑏 𝜌𝐵 𝑟𝑜 [1 − (𝑟𝑐 )3 ] = 3 𝑏 𝑘𝑚 𝐶𝑏 𝑜 𝑋𝐵 This expression is valid for both non-porous and porous products 7.1.2.2. Diffusion Through Product / Ash Layer Controls A pseudo-steady-state approximation is applied in solving the rate equation. The following assumptions are made: (i) the rate of reaction is given by the rate of diffusion to the reaction surface, (ii) the movement inwards of the reaction surface is slower than the rate of flow of fluid reactant towards the unreacted core (ie. the core is considered stationary) (iii) the flux relationship is written for a partially reacted particle and applied to all values of r C (ie. we integrate between 0 and rO). At steady-rate, the rate of reaction is constant: −𝑟𝐴 = − 𝑑𝑛𝐴 𝑑𝐶𝑏 = 4𝜋𝑟 2 𝐷𝑒 = 𝑐𝑜𝑛𝑠𝑡. 𝑑𝑡 𝑑𝑟 Integrating across product layer (from rO to rC) − 𝑑𝑛𝐴 𝑑𝑡 𝑟𝑐 𝑑𝑟 ∫𝑟𝑜 𝑟2 𝐶 =0 = 4𝜋𝐷𝑒 ∫𝐶 𝑏=𝐶 𝑑𝐶𝑏 𝑏 − or 𝑠 𝑑𝑛𝐴 𝑑𝑡 1 1 𝑐 𝑜 (𝑟 − 𝑟 ) = 4𝜋𝐷𝑒 𝐶𝑏 We then consider the change in the unreacted core with time. As reaction proceeds the product layer thickens and the rate of diffusion of A slows down. This means that as time changes, nA and rc also change (3 variables – time, nA and rc ). nA can be eliminated by expressing it in terms of rc (the rate of change of nA is proportional to rate of change of rc) , ie: −𝑏 𝑑𝑛𝐴 = −𝜌𝐵 𝑑𝑉 = −4𝜋𝜌𝐵 𝑟𝑐2 𝑑𝑟𝑐 Substituting this expression into the above and integrating gives, 𝑟 1 1 𝑜 𝑐 𝑜 𝜌 𝑟2 𝑡 −𝜌𝐵 ∫𝑟 𝑐 (𝑟 − 𝑟 ) 𝑟𝑐2 𝑑𝑟𝑐 = 𝑏𝐷𝑒 𝐶𝑏 ∫0 𝑑𝑡 𝑟 𝑟 𝑡 = 6𝑏𝐷𝐵 𝑜𝐶 [1 − 3(𝑟𝑐 )2 + 2(𝑟𝑐 )3 ] 𝑒 𝑏 𝑜 𝑜 or in terms of fraction converted, 𝜌 𝑟2 2 𝑡 = 6𝑏𝐷𝐵 𝑜𝐶 [1 − 3(1 − 𝑋𝐵 )3 + 2(1 − 𝑋𝐵 )] This expression is valid if there is no change in 𝑒 𝑏 particle size during reaction 23 If the volume of the particle changes, then rO also changes. Discussions on this can be found in literature, eg. Jander's or Crank, Ginstling and Brounshtein (CGB) simplified model or Valensi's exact solution can be found in Fathi Habashi: Principles of Extractive Metallurgy, Vol. I, General Principles, Gordon Breach Publishers. It is also important to note that the shape of the particle also affects the rate and therefore determines the rate expression. 7.1.2.3 Reaction is controlled by interfacial chemical reaction External mass transfer and diffusion in product or ash layer is fast (porous product layer). The reaction rate assuming first-order reaction is given by: − But 1 𝑑𝑛𝐵 𝑏 𝑑𝑛𝐴 =− = 𝑏𝑘𝑠 𝐶𝑏 𝑆𝑒𝑥 𝑑𝑡 𝑆𝑒𝑥 𝑑𝑡 −𝑑𝑛𝐵 = −4𝜋𝜌𝐵 𝑟𝑐2 𝑑𝑟𝑐 Therefore, 1 − 4𝜋𝑟 2 ∙ 𝜌𝐵 4𝜋𝑟𝑐2 𝑐 𝑑𝑟𝑐 𝑑𝑡 = −𝜌𝐵 𝑑𝑟𝑐 𝑑𝑡 = 𝑏𝑘𝑠 𝐶𝑏 Simplifying, rearranging and integrating rC between rO and rC gives, 𝑡= 7.2 1 𝜌𝐵 𝜌𝐵 𝑟𝑜 𝑟𝑐 𝜌𝐵 𝑟𝑜 (𝑟𝑜 − 𝑟𝑐 ) = (1 − ) = [1 − (1 − 𝑋𝐵 )3 ] 𝑏𝑘𝑠 𝐶𝑏 𝑏𝑘𝑠 𝐶𝑏 𝑟𝑜 𝑏𝑘𝑠 𝐶𝑏 Reaction of a Single Porous Particle In this case the fluid reactant reacts with solid during diffusion and therefore chemical reaction and diffusion occur in parallel over a diffuse zone rather than at a sharp boundary. 7.2.1 No solid product is formed When no solid product is formed (eg. in the combustion of porous carbon, dissolution of porous minerals and the Boudouard's solution loss reaction between porous carbon and carbon dioxide) and diffusion offers no resistance, the diffusion of fluid reactant within the pores of the solid creates a situation where the overall reaction is strongly influenced by pore diffusion and not controlled by it. Fluid reactant may diffuse deeply into the interior of the solid and reaction occurs throughout the solid with uniform concentration at its bulk value. However, under strong limitation of pore diffusion or external mass transfer control, as the reaction rate increases the fluid reactant reacts with solid as it penetrates and therefore reaction takes place in a narrow region near the external surface. The solid is consumed towards the centre. The reaction zone may be considered flat regardless of geometry and the mass balance written in one direction only is: 𝐷𝑒 𝑑2 𝐶𝑏 𝑑𝑥 2 − 𝑘𝑠 𝑆𝑣 𝐶𝑏𝑛 = 0 x is the distance normal to the external surface, SV is surface area per unit volume of solid, m2/m3 n is the reaction order The boundary conditions are: at x = 0 (outer surface of solid) C b = CS 24 at x → ∞, Cb = dCb/dx = 0 valid when it is recognised that the particle size is much larger than the thin layer where the reaction occurs. The solution to the above equation is given by: 1 𝑑𝐶𝑏 2 𝑘𝑠 𝑆𝑣 ) = −[( ∙ 𝐶𝑏𝑛+1 ]2 𝑑𝑥 𝑛 + 1 𝐷𝑒 At x = 0 (concentration gradient at the external surface), 𝑑𝐶𝑏 ) 𝑟𝐴 = −𝐷𝑒 ( 𝑑𝑥 𝑒𝑥𝑡. 1 𝑠𝑢𝑟𝑓. 𝑛+1 2 2 =( 𝑘 𝑆 𝐷 ) ∙ 𝐶𝑠 2 𝑛+1 𝑠 𝑣 𝑒 The above expression shows that: (i) rA ~ (k D) 1/2 the apparent activation energy is the arithmetic average of the activation energies of intrinsic reaction and diffusion. (ii) diffusion does not control the overall rate even when chemical reaction is fast (k S >> 0). Chemical reaction and diffusion occur in parallel 7.2.2 Solid product is formed The situation is similar to that discussed under nonporous particle. The only difference is that the initial reactant solid is porous and the reaction occurs in a diffuse zone. a. Pore diffusion in the product layer is fast, then concentration of fluid is uniform throughout solid and reaction occurs at a uniform rate. b. If chemical reaction is fast, reaction occurs in a narrow region between unreacted core and product layer and the situation is similar to diffusion controlled shrinking core reaction of a nonporous particle. 7.3 Grain Model This model is a more recent development in reaction modelling. It assumes that the porous solid is an aggregation of fine grains of various shapes (flat plates, long cylinders or spheres). Further assumptions in the treatment of the model are that: 1. the pseudo-steady-state approximation is valid for determining the concentration profile of the fluid reactant within the solid. 2. the resistance due to external mass transfer is negligible 3. intra-pellet diffusion is either equimolar counter diffusion or occurs at low concentration of the diffusing species. 4. diffusivities are constant throughout the solid 5. diffusion through the product layer around individual grain is fast. Mass balance for fluid reactant may be calculated by the application of Fick's 2nd Law: 𝐷𝑒 ∇2 𝐶𝑏 − 𝜈𝐴 = 0 ν is the local rate of consumption of fluid reactant A, moles/time . volume of solid Within each solid grain the mass balance for solid reactant B may be given as: −𝜌𝐵 𝑑𝑟𝑐 𝑑𝑡 = 𝑏𝑘𝑠 𝐶𝑏 25 By performing the necessary mathematical operations the solution giving the relationship between time and conversion is given by: 𝑡 ∗ ≃ 𝑔𝐹𝑔 (𝑥) + 𝜎 2 𝑝𝐹𝑝 (𝑥) t* is a dimensionless variable and σ is shrinking core reaction modulus 𝑏𝑘𝑠 𝐶𝑏 𝑡∗ = ( 𝜌𝐵 ∙ 𝐴𝑔 𝐹𝑔 ∙𝑉𝑔 )𝑡 A is the external surface area V is the external volume F is particle shape factor 1 𝑔𝐹𝑝 (𝑥) = 1 − (1 − 𝑋)𝐹𝑝 𝑝𝐹𝑔 (𝑥) ≡ 𝑥 2 if Fg = 1 (infinite slab) ≡ 𝑥 + (1 − 𝑥) ln(1 − 𝑥) if Fg = 2 (infinite cylinder) 2 ≡ 1 − 3(1 − 𝑋)3 + 2(1 − 𝑥) 8 if Fg = 3 (infinite sphere) REACTOR MODELS Metallurgical reactions (hydro-, pyro- and electrometallurgical) usually take place in reactors, and therefore knowledge of reactor design is a necessity. The object of design is to use the rate expression to predict the size of the reactor, which will be required to attain a given quantity of product from a given amount of feed material. In order to predict and optimize conversion rates, the relationship between concentration and other parameters such as mixing properties, temperature, pH, among others should be established. With the appropriate mathematical description (by modelling for example), the effluent reactant and product concentrations can be calculated as function of reactor volume (V), resident time (τ) and effluent flow rate. There are three basic types of reactors: 1. Batch Rector 2. Continuously Stirred Tank Reactor (CSTR), Continuous Flow Stirred Tank Reactor (CFSTR), Backmix reactor 26 3. 8.1 Plug Flow Reactor (Tubular reactor) Batch reactor Reactants are initially charged into a container, well mixed and allowed to react for a certain period. The resultant mixture is then discharged. This is an unsteady-state operation where composition changes with time, however, at any instant the composition throughout the reactor is uniform (because the reactor mixture is perfectly mixed). 8.2 CSTR It is a well-mixed stirred tank operated continuously and is normally run at steady-state (the effluent reactant and product concentrations are constant over time). Because of the perfect mixing the reactants entering the reactor assumed to be instantaneously distributed over the total reactor volume. As a result, there are no spatial variations in concentration, temperature or reaction rate. Thus, the concentration and temperature in the exit stream is the same. 8.3 Plug Flow Reactor (PFR) This consists of a pipe (tube), in which the reactants are continually consumed as they flow along the length of the reactor. The concentration varies continuously in the axial direction through the reactor, as a result of which the reaction rate, which is function of concentration, also varies axially. This is normally operated at steadystate. The necessary and sufficient condition for plug flow is for the residence time in the reactor to be the same for all fluid elements. Actually, the flow is highly turbulent and the flow field may be modelled by that of a plug flow. 8.4 Resident time - In a batch reactor each and every fluid element remains exactly the same reaction time in the reactor volume. - In a CSTR, the separate fluid elements have a wide range in residence times. In systems where mixing is highly non-ideal, the residence time distribution (RTD) is needed to characterise the mixing that occurs in the reactor. Real reactors never fully follow the ideal flow patterns. Deviations can be caused by channelling of fluid, recycling of fluid and creation of stagnant regions. The RTD of a reactor is a 27 characteristic of the mixing that occurs in a reactor. In ideal mixing, the space time (τ) (or mean residence time or holding time) is the proper performance measure of flow reactors. τ is the time required to process one reactor volume of feed based on entrance conditions. t= 8.5 reactor volume V = volumetric feed rate Fi Mass balance equations for the reactor models The mas balance is computed for any component A, which is the limiting component. Expressions are then derived to calculate the reactant concentration, reactor volume or the residence time. The macro mass balance equation is the basic equation for reaction engineering: V dVCA = Fi CAi - FoCA o + ò rA dV CA,i :is the input concentration dt o CA.o :is the output concentration V: reactor volume rA: reaction rate 8.5.1 Batch reactor V dVC A = ò rA dV dt 0 input = output = 0 If the reaction mixture is perfectly mixed, RA = const. dVCA = rA dV dt This expression at constant reactor volume reduces to the design equation, dC A n = rA the solution of which is rA = -knCA dt For Oth-order reactions CAo = CAi e- k1 t For first-order reactions, 8.5.2 CAo = CAi - ko t Continuously Stirred Tank Reactor At steady-state, dVC A =0 dt At perfect mixing, V òr A dV = rA V 0 rA = const. and The solution is given as: FCAi - FCAo = -rA V which when re-arranged gives the design equation: V= F (CAi - CA o ) -rA 28 For 1st - order reactions, V (CAi - CA o ) = F k1CA o Þ C Ao = C Ai 1 1+ k1 V F Since the CA,o is identical to the composition inside the reactor, the rate of reaction is evaluated at the exit conditions: rA = k1CAo For 0th -order reactions, V (CAi - CA o ) = F -ko 6.5.3 Þ CAo = CAi - ko V F Plug Flow Reactor (Tube Reactor) V dVC A =0 dt At steady-state, òr A dV = rA DV 0 The reaction rate (rA), which is a function of concentration, varies in axial direction. Therefore, the reactor is divided into ΔV volumes within which the reaction rate may be considered spatially uniform. The mass balance equation is: FCA z - FCA z+Dz + rA DV = 0 Using ΔV = AΔz, dividing through by Δz and taking the limit as Δz → 0 gives: lim Dz®0 FCA z - FCA z+Dz Dz = -rA DV dFCA = -A rA dz Using again dV = A dz , thus it is more convenient using reactor volume than reactor length and re-arranging : dFCA = rA dV Þ F dCA = rA dV For 1st -order reactions, at constant volumetric flow rate (F = const.), re-arranging and integrating gives: dC A k1 = dV CA F For 0th -order reactions ln and F CA o CAi dCA = -k0 dV k = 1 F Þ Þ CA o = CAi e CA o = CAi - k0 The above results show that: 29 V F - k1 V F 1. for 0th -order reactions, the CA,o are the same for all three reactors. 2. for a given duty, the same volume of batch or plug flow reactor is needed to do the job for all n. 3. for n>0, the PFR effluent reactant concentration (CA,o) is lower than that for CSTR, that is more reactant has been converted and the performance of PFR is better. 4. for any particular duty for n>0 the CSTR is always larger than the PFR. 5. when conversion is small, the rector performance is only slightly affected by flow type. increases very rapidly at high conversions. The ratio (GIVE EXAMPLES) 8.6 CSTR -s in series Many times reactors are connected in series, such that the exit stream of one reactor is the feed stream of the next reactor. Assumptions: 1. there are m CSTRs in series 2. all these reactors have the same volume 3. 1st -order reaction For 1 reactor, C0 1 1 = = Ci 1+ k V / m 1+ k t 1 1 F m V1 = V2 = ... = Vm = V/m where τ is the total residence time C o Co C m C C = x ... 2 1 Ci Ci Cm-1 C1 Cm However, This simplifies to C0 1 = Ci 1+ k t 1 m Þ t= m Ci m1 [( ) -1] k1 Co It can be shown that when m→ ∞ , Co will be the same as Co for a PFR. Thus tc = t p = m Ci ln[ ] k1 Co 30 V is the total volume 8.7 Comparison of PFR and CSTR required for the same job The required space time or residence time (and therefore the required volume) to reach the same output concentration (CA,o) of a PFR is lower than that for a CSTR for n>0. This can be shown graphically: CSTR PFR The design equation is: The design equation is CAi - C Ao V = tc = F -rA Plotting 1 -rA F dC A = rA dV Re-arranging and integrating gives: as a function of CA, CAo t= The area under the resulting curve equals the residence time (by definition) 31 C Ai V dC A dCA = ò =ò F CAi rA CAo -rA