L7-1 Review: Liquid Phase Reaction in PFR Be able to do these 4 steps, integrate & solve for V for ANY ORDER RXN LIQUID PHASE: Ci ≠ f(P) → no pressure drop 2A → B -rA = kCA2 2nd order reaction rate Calculate volume required to get a conversion of XA in a PFR dX A rA Mole balance dV FA0 Rate law rA kCA 2 Stoichiometry (put CA in terms of X) CA CA0 (1 X A ) dX A dV Combine FA0 k CA0 2 XA 0 V dX A 1 XA 2 dV See Appendix A for integrals frequently used in reactor design 0 k CA02 1 X A FA0 k CA02 2 FA0 XA 1 X A V Liquid-phase 2nd order reaction in PFR Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-2 Review: Liquid Phase Reaction in PBR LIQUID PHASE: Ci ≠ f(P) → no pressure drop Be able to do these 4 steps, integrate & solve for V for ANY ORDER RXN 2A → B -r’A = kCA2 2nd order reaction rate Calculate catalyst weight required to get a conversion of XA in a PBR dX A r 'A Mole balance dW FA0 r 'A kCA 2 Rate law CA CA0 (1 X A ) Stoichiometry (put CA in terms of X) dX A dW Combine FA0 k CA0 2 XA dX A 0 1 XA W 2 dW 0 k CA02 1 X A FA0 k C A02 2 FA0 XA 1 X W A Liquid-phase 2nd order reaction in PBR Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-3 Review: Isobaric, Isothermal, Ideal Gas-Phase Rxns in Tubular Reactors Gas-phase reactions are usually carried out in tubular reactors (PFRs & PBRs) • Plug flow: no radial variations in concentration, temperature, & ∴ -rA • No stirring element, so flow must be turbulent FA0 FA C j0 jCA0 XA C j0 jCA0 XA P T0 Z0 Cj GAS PHASE: C j 1 XA 1 XA P0 T Z 1 1 1 Stoichiometry for basis species A: CA0 1 XA CA0 CA0 XA CA CA 1 XA 1 XA Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Review: Effect of on u and XA L7-4 NTf NT0 Change in total # moles at X A 1 NT0 total moles fed : expansion factor, the fraction of change in V per mol A reacted u0: volumetric flow rate u varies if gas phase & moles product ≠ u u 1 X Z T P0 0 A moles reactant, or if a DP, DT, or DZ occurs Z 0 T0 P No DP, DT, or DZ occurs, but moles product ≠ moles reactant → u u0 1 X A • = 0 (mol product = mol reactants): u u0: constant volumetric flow rate as XA ↑ < 0 (mol product < mol reactants): u < u0 volumetric flow rate ↓ as XA ↑ Q1: For an irreversible gas-phase reaction, how does the residence time and XA change when < 0? a)They don’t b)The residence time is longer & XA increases c)The residence time is longer & XA decreases d)The residence time is shorter, & XA decreases e)The residence time is shorter & XA increases Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Review: Effect of on u and XA L7-5 NTf NT0 Change in total # moles at X A 1 NT0 total moles fed : expansion factor, the fraction of change in V per mol A reacted u0: volumetric flow rate u varies if gas phase & moles product ≠ u u 1 X Z T P0 0 A moles reactant, or if a DP, DT, or DZ occurs Z 0 T0 P No DP, DT, or DZ occurs, but moles product ≠ moles reactant → u u0 1 X A • = 0 (mol product = mol reactants): u u0: constant volumetric flow rate as XA increases • < 0 (mol product < mol reactants): u < u0 volumetric flow rate decreases as XA increases • Longer residence time than when u u0 • Higher conversion per volume of reactor (weight of catalyst) than if u u0 • > 0 (mol product > mol reactants): u > u0 with increasing XA • Shorter residence time than when u u0 • Lower conversion per volume of reactor (weight of catalyst) than if u u0 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-6 Review: Isobaric, Isothermal, Ideal Rxn in PFR GAS PHASE: Be able to do these 5 steps, & solve for V for ANY ORDER RXN Ci = f() → no DP, DT, or DZ 2A → B -rA = kCA2 2nd order reaction rate Calculate PFR volume required to get a conversion of XA dX A rA Mole balance dV FA0 rA kCA 2 Rate law CA0 1 X A 1 XA Stoichiometry (put CA in terms of X) CA Combine dX A k CA0 1 X A dV 1 XA 2 FA0 2 V V FA0 k CA0 FA0 k CA02 2 XA 0 1 XA 2 dX A 2 1 XA 2 Integral A-7 in appendix 1 2 X A 2 1 ln 1 X A 2 X A 1 XA Gas-phase 2nd order rxn in PFR no DP, DT, or DZ Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-7 Review: Pressure Drop in PBRs GAS PHASE: A → B -r’A = kCA2 2nd order reaction rate Calculate dXA/dW for an isothermal ideal gas phase reaction with DP dX A FA0 r 'A Mole balance dW Rate law r 'A kCA 2 CA Stoichiometry (put CA in terms of X) Combine Relate P/P0 to W (Ergun equation) dX A dW CA0 1 X A P 1 X A P0 k CA02 FA 0 1 XA 2 P 2 1 XA 2 P0 dP T P0 1 X A dW 2 T0 P P0 Ergun Equation can be simplified by using y=P/P0 and T=T0: dy 1 X A dW 2y Simultaneously solve dXA/dW and dP/dW (or dy/dW) using Polymath Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-8 Review: Ergun Equation dP T P0 Calculates pressure drop in a packed bed. 1 X A This equation can be simplified to: dW 2 T0 P P0 Differential form of Ergun equation for pressure drop in PBR: y P P0 T dy 1 X A dW 2y T0 NTf NT0 y A0 NT0 1 : fraction of solid in bed = 20 A c c 1 P0 volume of solid total bed volume AC: cross-sectional area C: particle density : constant for each reactor, calculated using a complex equation that depends on properties of bed (gas density, particle size, gas viscosity, void volume in bed, etc) : constant dependant on the packing in the bed Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-9 L7: Unsteady-State Isothermal Reactor Operation: CSTR Start-Up and Semi-Batch Reactors CBu0 Semi-batch A A+B V0 V0 + u0t Vf start time t end • Time required to reach steady-state after CSTR start-up • Predicting concentration and conversion as a function of time Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-10 Start-Up of a Fixed-Volume CSTR Isothermal (unusual, but simple case), well-mixed CSTR Unsteady state: concentrations vary with time & accumulation is non-zero Goal: Determine the time necessary to reach steady-state operation u0CA CA0u0 In - Out + Generation = Accumulation moles A in CSTR dNA changes with FA0 F A rA V dt time until steady state is reached Use concentration rather than conversion in the balance eqs dNA CA0u0 C A u0 rA V dt C A0u0 C Au0 rA V dNA 1 V dt V Divide by V to convert dNA to dCA V u0 C A0 CA rA dC A dt Multiply by dC A C A0 CA rA dt Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-11 CSTR Start-Up: 1st Order Reaction dC A dC A Combine C C k C A0 A A dt rA kCA dt Integrate this eq to find CA (t) while 1st order rxn in CSTR is at unsteady-state: C A0 CA rA Bring variables to one side & factor dC A 1 C A0 CA 1 k dt CA0 dCA 1 1 k CA dt 1 k Put like variables with their integrals CA0 t 1 k CA dCA 1 k 1 k dt C ln 0 t A0 C 0 0 C A0 0 1 k A 1 k 1k t CA 1 e C t 1k C A0 A0 1 e CA 1 k 1 k CA Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-12 CSTR Start-Up: 1st Order Reaction dCA We integrated this eq to find CA (t) while CA0 CA kCA dt CSTR of 1st order rxn is in unsteady-state: At steady state, CA0 1 e t1k C C A0 C AS A 1 k t is large and: 1 k 0 Is this consistent with steady C C kC dCA No accumulation A0 A A state balance eq for CSTR? dt at steady state Yes, same! 0 C A0 C A0 C A kC A 0 C AS Goal: combine start-up and SS eqs 1 k to estimate time to reach SS (ts) In the unsteady state, CA0 CA 0 t s 1k 1 e 0.99 1 k when CA = 0.99CAS: 1 k Solve for ts to determine time to reach 99% of steady-state concentration t 1k t 1k t 1k 1 e s 0.99 0.01 e s ln 0.01 ln e s 1 k 4.6 t s 4.6 ts 1 k time to reach 99% of steady-state concentration in terms of k Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-13 CSTR Start-Up: 1st Order Reaction C 1 e C 1 k A0 t 1k A In the unsteady state, the time to reach CA = 0.99CAS is: When k is very small t 4.6 (slow rxn), 1>>k: s t 4.6 1 k When k is very big (fast rxn), 1<<k 63% of the steady-state concentration is achieved at: 1 k ts 4.6 k CA = 0.63CAS 99% of the steady-state 4.6 concentration is achieved at: 1 k CA 0.99CAS Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. Better Selectivity in a Semi-Batch Reactor L7-14 To enhance selectivity of desired product over side product 2 kP A B P rP kpCA CB Desired product P 2 kS A B S rS kSCA CB Undesired side product S Instantaneous selectivity, SP/S, is the ratio of the relative rates*: rP kPCA 2CB kP CA SP/S 2 k S CB rS k SCA CB Higher concentrations of A favor formation of the desired product P Higher concentrations of B favor formation of the undesired side product S To maximize the formation of the desired product: Slowly feed B into the reactor containing A Commonly used in bioreactors, when the enzyme is inhibited by excess substrate *We’ll look at this concept of instantaneous selectivity in more detail in L9 Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-15 Semi-Batch Reactor Design Equation CBu0 V0 + u0t Do a mole balance on A since it does not enter or leave the reactor (assume the reactor is well-mixed) In - Out + Generation = Accumulation FA0 0 dNA F A rA V dt dNA 0 rA V dt Use whatever units are most convenient (NA, CA, XA, etc) Convert NA to CA using: NA CA NA CA V V dC A dV rA V V CA dt dt dC A V rA V dt 2 parts: how CA changes with t and how V changes with t Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-16 Semi-Batch Reactor Design Equation CBu0 Do a mole balance on A since it does not enter or leave the reactor (assume the reactor is well-mixed) In - Out + Generation = 0 rA V V V0 + u0t 0 rA V t dC A dV CA dt dt Accumulation dNA dt 2 parts: how CA changes with t and how V changes with t Reactor volume at any time can be found with a mole balance In - Out + Generation = Accumulation d V u = u0 dV u0 0u0 0 0 V0 u0t V 0 dt dt Substitute: rA V V dC A C Au0 dt dC A rA V C Au0 V dt rA Rearrange to get in terms of dCA/dt CAu0 dCA V dt Balance on A Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-17 Semi-Batch Reactor Design Equation CBu0 Mole Balance on B In - Out + Generation = FB0 V0 + u0t 0 dNB rB V FB0 dt dNB rB V FB0 dt dVCB d CB V rB V FB0 dt dt dCB dV CB V rB V CB0u0 dt dt CBu0 V rB V Accumulation dNB dt dCB rB V CB0u0 dt u0 dV dt Substitute Rearrange to get in terms of dCB/dt u0 CB0 CB dCB Balance on B rB dt V Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-18 Semi-Batch Reactor Design Equation: in Terms of NA C u B 0 In - Out + Generation = 0 0 dNA rA V dt V0 + u0t rA V Accumulation dNA dt Substitute V V0 u0t Reactor design eq. provided that rA is a function of NA NA NB NA NB r k r k -rA = kACACB A A V V V u t 2 dNA rA V0 u0 t dt 0 0 dNA NA NB dNA k rA V dt V0 u0 t dt dNB NANB dNB k FB0 rA V FB0 NB comes from BMB: dt V u t dt 0 0 The design eq in terms of XA can be messy. Sometimes it gives a single equation when using Nj or Cj gives multiple reactor design equations. Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-19 Improving Yields of Reversible Rxns with Semi-Batch Reactors Semi-batch A+B V0 FD A+B⇌C+D V0 - u0t Vf To improve product yield in a reversible reaction: A l B l C l D g • Start with A(l) and B(l) in the reactor • D(g) bubbles out of the liquid phase, pushing the equilibrium to the right and forcing the reaction to go to completion Boil off water to produce high MW polymer Common industrial reaction: + n n + nylon H2O Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-20 Improving Yields of Reversible Rxns with Semi-Batch Reactors Semi-batch A+B V0 FD A+B⇌C+D V0 - u0t Vf To improve product yield in a reversible reaction: A l B l C l D g • Start with A(l) and B(l) in the reactor • D(g) bubbles out of the liquid phase, pushing the equilibrium to the right and forcing the reaction to go to completion How do we account for the loss of product D in the material balance? Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign. L7-21 Loss of Mass in Semi-Batch Reactor A l B l D(g) C l D g elementary rxn u = u0 0 Overall Mass balance: In - Out + Generation = Accumulation g dm m u0 m 0 gas leaving reactor dt time m m ↑want in terms of dV/dt V V u0 m dm 1 m dV V0 + u0t u Divide mass balance by 0 dt dt Relate ṁ to a rate: rA moles volume time From stoichiometry, rD = -rA Next, convert mass units to: time Conversion 2: Conversion 1: rA V massD molesD molesD MWD massD MWD r V MWD dV u0 A dt rA V MW moles time mass m time Substitute for ṁ One of the diff. eq. that are simultaneously solved (by Polymath) Slides courtesy of Prof M L Kraft, Chemical & Biomolecular Engr Dept, University of Illinois at Urbana-Champaign.