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ISOTHERMAL REACTOR DESIGN in Chapter 1 & 2, We Discussed ISOTHERMAL REACTOR DESIGN In Chapter 1 & 2, we discussed balances on batch & flow reactors. In Chapter 3, we discussed rxns. Here, we will combine rxns and reactors. Topics Part 1: Mole Balances in Terms of Conversion 1. Algorithm for Isothermal Reactor Design 2. Applications/Examples of CRE Algorithm 3. Reversible Reactions 4. ODE Solutions to CRE Problems 5. General Guidelines for California Problems 6. PBR with Pressure Drop 7. Engineering Analysis Part 2: Measures Other Than Conversion 1. Measures Other Than Conversion 2. Membrane Reactors 3. Semibatch Reactors Part 1: Mole Balances in Terms of Conversion 1. Algorithm for Isothermal Reactor Design The algorithm for the pathway of interest can be summarized as: 1. Mole Balance and Design Equation (choose reactor type) 2. Rate Law (choose rxn type; gas or liq. phase) 3. Stoichiometry 4. Combine 5. Evaluate The Evaluate Step can be carried out A. Graphically (Plots) B. Numerically (Quadrature Formulas) C. Analytically (Integral tables) D. Using Software Packages 1 Elementary gas phase reaction in different reactor types CSTR The elementary gas phase reaction 2A + B → C takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B. x = 0.9, k= 10dm6/mol2s VCSTR =? Mole balance Rate Law Stoichiometry gas phase, isothermal (T = T0), no pressure drop (P = P0) Derivation 2 Combine Evaluate PFR and Batch Reactors Elementary Gas Phase Reaction: 2A + B → C PFR Mole Balance Rate Law gas phase, isothermal (T = T ), no pressure Stoichiometry 0 drop (P = P0), CAo=CBo (Θ=1), v=vo(1+εX) 3 Combine C =0.2, v=v =25 dm3/s, k=10 dm6/mol2 s, ε=-0.5, Parameter Evaluation Ao o X=0.9 V=227 dm3 Deriving CA and CB Remember that the reaction is: For a gas phase system: If the conditions are isothermal (T = T0) and isobaric (P = P0): We must divide by the stoichiometric coefficient of our basis of calculation yielding: 4 And if the feed is equal molar, then: This leaves us with CA as a function of conversion alone: Similarly for CB: Batch Reactor Constant Volume, V=Vo and the pressure changes. Mole Balance Rate Law Stoichiometry Combine 5 6 2 Parameter Evaluation CAo=0.2, k=10 dm /mol s, 6 Scale – Up of Liquid – Phase Batch Reactor Data to Design of a CSTR: Scale – Up a lab experiment to pilot – plant operation of full – scale production. Find k from experimental data and use it to design a full – scale flow reactor. Batch Operation: Liquid phase: (density change is small → V = V0) Gas phase with constant volume V = V0. 1 ⎛ dN A ⎞ ⋅⎜ ⎟ = rA V ⎝ dt ⎠ 1 ⎛ dN A ⎞ 1 dN A dN A /V0 dCA ⋅⎜ ⎟ = ⋅ = = = rA V ⎝ dt ⎠ V0 dt dt dt Since conc’n is a measured quantity in liquid – phase rxns: dC − A = −r dt A Let’s calculate the time necessary to achieve a given conversion X for the irreversible second order rxn: A → B dX N ⋅ = −r ⋅V [1] A0 dt A 0 2 [2] − rA = k ⋅CA CA = CA0 ⋅(1− X ) [3] Combine [1], [2] and [3]; dX = k ⋅C ⋅(1− X )2 dt A0 Rearrange: dX = k ⋅C ⋅dt (1− X )2 A0 @ t = 0; X = 0; T = T0 (isothermal) → k → constant t 1 X dX 1 ⎛ X ⎞ t = ⇒ t = ⋅ ∫ ∫ 2 ⎜ ⎟ 0 k ⋅CA0 0 (1− X ) k ⋅CA0 ⎝1− X ⎠ 7 It is important to know the reaction time, tR, to achieve a certain conversion. Flow reactors use characteristic rxn times, tR. The time for a total cycle is much longer than tR, as one must account for the time to fill (tF), heat (th) and clean (tC). t = tf + th + tC + tR Design of Continuous Stirred Tank Reactors CSTRs are usually used for liquid phase rxns: F ⋅ X v ⋅C ⋅ X V = A0 = 0 A0 [1] (Design Eq’n for CSTR) (−rA )exit − rA Divide by v0: V C ⋅ X τ = = A0 [2] v0 − rA Volumetric flow rate A single CSTR: First order irreversible rxn: − rA = k ⋅CA [3] Combine [2], [3] and [4] [4] CA = CA0 ⋅(1− X ) 1 ⎛ X ⎞ τ = ⋅⎜ ⎟ [5] k ⎝1− X ⎠ τ ⋅k X = [6] 1+τ ⋅k C Combine [4] and [6] C = A0 A 1+τ ⋅k rxn Damköhler #, Da: a dimensionless # that for a first order rxn says the degree of conversion that can be achieved in cont. flow reactor. 8 − r ⋅V rate of rxn of A Da = A0 : F rate of convective transport of A A0 t=0 For a second order rxn: 2 − rA0 ⋅V k ⋅CA0 ⋅V Da = = =τ ⋅k ⋅CA0 FA0 v0 ⋅CA0 Da ≤ 0.1 X < 0.10 Da ≥10 X > 0.90 Da X = First order liq-phase rxn (Eq’n [6]) 1+ Da CSTRs in Series CA1, X1 C A0 For first order rxn (v = v0) CA2, X2 The effluent conc’n of reactant A from the first CSTR is -rA1 -rA2 C V V1 V2 A0 1 CA1 = where τ1 = 1+τ1 ⋅k1 v0 From the mole balance on reactor 2: FA1 − FA2 v0 (CA1 − CA2 ) V2 = = − rA2 k2 ⋅CA2 Solving for CA2: CA1 CA0 CA2 = = 1+τ 2 ⋅k2 (1+τ 2 ⋅k2 )(1+τ1 ⋅k1) 9 If both reactors are of equal size, (τ1 =τ 2 =τ ) and operate at the same T (k1 = k2 = k) C C = A0 A2 (1+τ ⋅k)2 for n equal sized CSTR system C C C = A0 = A0 An (1+τ ⋅k)n (1+ Da)n Substituting for CAn in terms of conversion C C ⋅(1− X ) = A0 ⇒ A0 (1+ Da)n 1 1 X =1− =1− (1+ Da)n (1+τ ⋅k)n CSTRs in Parallel: Equal sized reactors are placed in parallel rather than in series: FA01 FA02 FA03 FA0n X1 X2 X3 Xn Individual volume is given by ⎛ X ⎞ ⎜ i ⎟ [1] V = FA0i ⋅⎜ ⎟ ⎝ − rAi ⎠ Since Vi = Vj, then Xi = X; → -rA1 = -rA2 = ...... = -rA V V = [2] i n F [3] F = A0 A0i n Substitute [2] & [3] into [1] V F ⎛ X ⎞ X A0 ⎜ i ⎟ [4] = .⎜ ⎟ = FA0 ⋅ n n ⎝ − rAi ⎠ − rA 10 nd Example: 2 order rxn, v = v0, CA = CA0 (1-X); FA0 X = v0 CA0 X Combine –rA = k CA2 and V = FA0 X / -rA F ⋅ X v ⋅C ⋅ X V = A0 = 0 A0 2 2 2 k ⋅CA k ⋅CA0 ⋅(1− X ) Divide by v0: V X τ = = 2 v0 k ⋅CA0 ⋅(1− X ) (1+ 2⋅τ ⋅k ⋅C ) − (1+ 2⋅τ ⋅k ⋅C )2 − (2⋅τ ⋅k ⋅C )2 X = A0 A0 A0 2⋅τ ⋅k ⋅CA0 (1+ 2⋅ Da) − 1+ 4Da X = Since τ ⋅ k ⋅CA0 = Da 2⋅ Da for a second order rxn. Since X can not be greater than 1.0, ( - ) sign is chosen. Example: The elementary liquid phase reaction is carried out isothermally in a CSTR. Pure A enters at a volumetric flow rate of 25 dm3/s and at a concentration of 0.2 mol/dm3. What CSTR volume is necessary to achieve a 90% conversion when k = 10 dm3/(mol*s)? Mole Balance Rate Law Stoichiometry liquid phase (v = vo) 11 Combine Evaluate at X = 0.9, V = 1125 dm3 Space Time 2. Applications/Examples of the CRE Algorithm 12 Applying the algorithm to the above reaction occuring in Batch, CSTR, PFR 13 3. Reversible Reaction To determine the conversion or reactor volume for reversible reactions, one must first calculate the maximum conversion that can be achieved at the isothermal reaction temperature, which is the equilibrium conversion. (See Example 3-8 in the text for additional coverage of equilibrium conversion in isothermal reactor design.) Equilibrium Conversion, Xe Calculate Equilibrium Conversion (Xe) for a Constant Volume System Example: Determine Xe for a PFR with no pressure drop, P = P0 Given that the system is gas phase and isothermal, determine the reactor volume when X = 0.8 Xe First calculate Xe: CBe Equilibrium constant Kc is; KC = 2 CAe CAe = CA0 ()1− X e ⎛ 1 ⎞ CBe = CA0 ⎜ X e ⎟ ⎝ 2 ⎠ Xe = 0.89 X = 0.8Xe = 0.711 14 Deriving The Equilibrium Constant (KC) and Equilibrium Conversion (Xe) for a Non-Constant Volume System: The reversible reaction: which takes place in gas phase PFR. Since gas phase reactions almost always involve volume changes, we will have to account for volume changes in our calculations. The equilibrium constant, KC, for this reaction is: A is the limiting reactant A ⇔ B/2 where CAe and CBe are: Substituting for CAe and CBe gives us: 3 3 Substituting known values (CA0 = 0.2 mol/dm and KC = 100 dm /mol), and realizing that: Solving for the equilibrium conversion, Xe, yields: Xe = 0.89 15 Batch Reactor With a Reversible Reaction The following reaction follows an elementary rate law Initially 77% N2, 15% O2, 8% inerts are fed to a batch reactor where 80% of the equilibrium conversion (Xe = 0.02) is reached in 151 µ s. What is the specific reaction rate constant k1? Additional Information 16 For 80% of equilibrium conversion X = 0.8 Xe = 0.016 Use Simpson's three point formula to integrate with ∆X = 0.016/2 = 0.008 17 Tubular Reactors: Gas-phase rxns are usually carried out in tubular reactors where the gas is generally turbulent. Reactants Products No radial variation in velocity, conc’n, T or –rA. dX Design Eq’n: F ⋅ = −r A0 dV A In the absence of pressure drop or heat exchange, integral form of the design equation is used: X dX V = F ⋅ A0 ∫ 0 − rA Pressure Drop in Reactors In liquid-phase rxns, the conc’n of reactants is not effected by the changes in pressure.
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