Lecture 13 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors in which they take place. Lecture 13 – Tuesday 2/26/2013 Complex Reactions: A +2B C A + 3C D Example A: Liquid Phase PFR Example B: Liquid Phase CSTR Example C: Gas Phase PFR Example D: Gas Phase Membrane Reactors Sweep Gas Concentration Essentially Zero Sweep Gas Concentration Increases with Distance Example E: Semibatch Reactor 2 Gas Phase Multiple Reactions 3 New things for multiple reactions are: 1. Number Every Reaction 2. Mole Balance on every species 3. Rate Laws (a) Net Rates of Reaction for every species N rA r iA i 1 (b) Rate Laws for every reaction r1 A k 1 A C A C B 2 r2 C k 2 C C A C C 2 4 3 (c) Relative Rates of Reaction for every reaction For a given reaction i: (i) aiA+biB ciC+diD: riA riB riC riD ai bi ci di Reactor Mole Balance Summary Reactor Type Gas Phase Batch dN A dt dN Semibatch dt dN dt 5 B A Liquid Phase dC r AV dt dC A r AV rB V F B 0 dt dC B dt A rA rA rB 0C A V 0 C B 0 C B V Reactor Mole Balance Summary Reactor Type Gas Phase CSTR V PFR dF A dV PBR 6 dF A dW FA0 FA rA rA rA Liquid Phase C A 0 C A V 0 0 dC 0 dC rA A rA A r A dV dW Note: The reaction rates in the above mole balances are net rates. Batch CB V V0 CB V N T P0 T 0 NT0 P T N B N T 0 P T0 N T V 0 P0 T C B CT 0 7 NB N B P T0 N T P0 T Flow CB 0 CB FB FT P0 T 0 FT 0 P T F B FT 0 P T 0 FT 0 P0 T C B CT 0 F B P T0 FT P0 T Stoichiometry Concentration of Gas: F A T0 y C A C T 0 T F T FT F A F B FC F D Note: We could use the gas phase mole balances for liquids and then just express the concentration as: Flow: Batch: 8 CA CA FA 0 NA V0 Example A: Liquid Phase PFR The complex liquid phase reactions follow elementary rate laws: (1) A 2 B C r1 A k 1 A C A C 2 B NOTE: The specific reaction rate k1A is defined with respect to species A. ( 2 ) 3C 2 A D r2 C k 2 C C C C A 3 2 NOTE: The specific reaction rate k2C is defined with respect to species C. 9 Example A: Liquid Phase PFR Complex Reactions (1) A 2B C (2) A 3C D 1) Mole Balance on each and every species (1) dF A dV (3) 10 dF C dV rA rC (2) dF B dV (4) dF D dV rB rD Example A: Liquid Phase PFR 2) Rate Laws: Net Rates Rate Laws ( 5 ) rA r1 A r2 A ( 7 ) rB r1 B r2 B ( 6 ) rC r1C r2 C ( 8 ) rD 0 r2 D ( 9 ) r1 A k 1 A C A C B 2 (10 ) r2 C k 2 C C A C C 2 Relative Rates Reaction 1 r1 A 1 r1 B 2 (11) r1 B 2 r1 A 11 (12) r1 C r1 A 3 r1 C 1 Example A: Liquid Phase PFR Relative Rates Reaction 2 r2 A 2 r2 C 3 (13 ) r2 A r2 D 1 2 3 (14 ) r2 D r2 C r2 C 3 rA k 1 A C A C 2 B 2 3 2 rB 2 k 1 A C A C B 2 rC k 1 A C A C B k 2 C C A C C 2 12 rD k 2C 3 2 3 C AC C 3 k 2C C AC C 3 Example A: Liquid Phase PFR 3) Stoichiometry Liquid (15 ) C A F A 0 (16 ) C B F B 0 (17 ) C C FC 0 (18 ) C D F D 0 ~ (19 ) S C 13 D FC else 0 if V 0 . 00001 then F D Example A: Liquid Phase PFR Others FT Liquid – Not Needed (19 ) Liquid ( 20 ) C T 0 Liquid 4) Parameters ( 21 ) k 1 A 10 ( 22 ) k 2 C 20 ( 23 ) Liquid ( 24 ) C T 0 Liquid ( 25 ) V f 2500 ( 26 ) F A 0 200 ( 28 ) F B 0 200 14 ( 26 ) 0 100 – Not Needed – Not Needed Example B: Liquid Phase CSTR Same reactions, rate laws, and rate constants as Example A (1) A 2 B C r1 A k 1 A C A C B 2 NOTE: The specific reaction rate k1A is defined with respect to species A. ( 2 ) 3C 2 A D r2 C k 2 C C C C A 3 2 NOTE: The specific reaction rate k2C is defined with respect to species C. 15 Example B: Liquid Phase CSTR The complex liquid phase reactions take place in a 2,500 dm3 CSTR. The feed is equal molar in A and B with FA0=200 mol/min, the volumetric flow rate is 100 dm3/min and the reaction volume is 50 dm3. Find the concentrations of A, B, C and D existing in the reactor along with the existing selectivity. Plot FA, FB, FC, FD and SC/D as a function of V 16 Example B: Liquid Phase CSTR (1) A + 2B →C (2) 2A + 3C → D r1 A k 1 A C A C B 2 r2 C k 2 C C A C C 2 3 1) Mole Balance 17 (1) A 0 C A 0 0 C A r AV 0 (2) B 0 C B 0 0 C B rB V 0 (3) C 0 0 C C rC V 0 (4) D 0 0 C D rD V 0 Example B: Liquid Phase CSTR 2) Rate Laws: (5)-(14) same as PFR 3) Stoichiometry: (15)-(18) same as Liquid Phase PFR (19 ) S C / D FC F D 0 . 0001 0C C 0 C D 0 . 0001 4) Parameters: k1 A , k 2 C , C A 0 , C B 0 , V , 0 18 Example B: Liquid Phase CSTR In terms of molar flow rates (1) A + 2B →C (2) 2A + 3C → D r1 A k 1 A C A C B 2 r2 C k 2 C C A C C 2 1) Mole Balance (1–4) (1) f F A F A 0 F A r AV (=0) ( 2 ) f F B F B 0 F B rBV (=0) ( 3 ) f FC 0 FC rC V (=0) ( 4 ) f F D 0 F D rD V 19 (=0) 3 2) Rates (5–14) 3) Stoichiometry: (15–19) Same as Example A (15 ) C A FA 0 (16 ) C B FB 0 (17 ) C C FC 0 (18 ) C D FD 0 (19 ) SC D FC F D 0 . 00001 Example B: Liquid Phase CSTR In terms of concentration (1) A + 2B →C (2) 2A + 3C → D r1 A k 1 A C A C B 2 r2 C k 2 C C A C C 2 1) Mole Balance (1–4) (1) f C A 0 C A 0 0 C A rAV 2) Rates (5–14) (=0) ( 2 ) f C B 0 C B 0 0 C B rBV (=0) ( 3 ) f C C 0 0 C C rC V (=0) ( 4 ) f C D 0 0 C D rDV (=0) 20 3 Same as Example A 3) Stoichiometry: (15–19) (15 ) S C D FC F D 0 . 00001 Example C: Gas Phase PFR, No ΔP Same reactions, rate laws, and rate constants as Example A: (1) A 2 B C r1 A k 1 A C A C B 2 NOTE: The specific reaction rate k1A is defined with respect to species A. ( 2 ) 3C 2 A D r2 C k 2 C C C C A 3 2 NOTE: The specific reaction rate k2C is defined with respect to species C. 21 Example C: Gas Phase PFR, No ΔP 1) Mole Balance (1) dF A dV (2) dF B dV rA rB (3) dF C dV (4) dF D dV rC rD 2) Rate Laws: (5)-(14) same as CSTR 22 Example C: Gas Phase PFR, No ΔP 3) Stoichiometry: Gas: Isothermal T = T0 (15 ) C A C T 0 (17 ) C C C T 0 FA FT FC y y FT (16 ) C B C T 0 (18 ) C D C T 0 (19 ) FT F A F B FC F D Packed Bed with Pressure Drop dy 23 dW FT T FT 2 y FT 0 T 0 2 y FT 0 FB FT FD FT y y Example C: Gas Phase PFR, No ΔP 4) Selectivity S FC FD y 1 24 FC else 0 if V 0 . 00001 then FD 21 20 Example D: Membrane Reactor with ΔP Same reactions, rate laws, and rate constants as Example A: (1) A 2 B C r1 A k 1 A C A C B 2 NOTE: The specific reaction rate k1A is defined with respect to species A. ( 2 ) 3C 2 A D r2 C k 2 C C C C A 3 2 NOTE: The specific reaction rate k2C is defined with respect to species C. 25 Example D: Membrane Reactor with ΔP Because the smallest molecule, and the one with the lowest molecular weight, is the one diffusing out, we will neglect the changes in the mass flow rate down the reactor and will take as first approximation: m 0 m 1) Mole Balances A dF A dV B dF B dV 26 rA rB 1 2 C dF C dV D dF D dV rC R C rD 3 4 We also need to account for the molar rate of desired product C leaving in the sweep gas FCsg dF Csg RC dV Example D: Membrane Reactor with ΔP We need to reconsider our pressure drop equation. When mass diffuses out of a membrane reactor there will be a decrease in the superficial mass flow rate, G. To account for this decrease when calculating our pressure drop parameter, we will take the ratio of the superficial mass velocity at any point in the reactor to the superficial mass velocity at the entrance to the reactor. Fi MW i 0 0 G0 Fi 0 MW i G 27 Example D: Membrane Reactor with ΔP The superficial mass flow rates can be obtained by multiplying the species molar flow rates, Fi, by their respective molecular weights, Mwi, and then summing over all species: G G0 28 m AC 1 m 0 AC 1 F MW A F MW A i i0 i C1 i C1 F MW F MW i i0 i i Example D: Membrane Reactor with ΔP 2) Rate Laws: (5)-(14) same as Examples A, B, and C. 3) Stoichiometry: (15)-(20) same as Examples A and B (T=T0) dy dW FT dy 2 y FT 0 dV R C k C C C C CSweep 4) Sweep Gas Balance: FCsg V dF Csg 29 dV FCsg RC V V RC V 0 FT 2 y FT 0 21 Example E: Liquid Phase Semibatch Same reactions, rate laws, and rate constants as Example A: (1) A 2 B C r1 A k 1 A C A C B 2 NOTE: The specific reaction rate k1A is defined with respect to species A. ( 2 ) 3C 2 A D r2 C k 2 C C C C A 3 2 NOTE: The specific reaction rate k2C is defined with respect to species C. 30 Example E: Liquid Phase Semibatch The complex liquid phase reactions take place in a semibatch reactor where A is fed to B with FA0= 3 mol/min. The volumetric flow rate is 10 dm3/min and the initial reactor volume is 1,000 dm3. The maximum volume is 2,000 dm3 and CA0=0.3 mol/dm3 and CB0=0.2 mol/dm3. Plot CA, CB, CC, CD and SS/D as a function of time. 31 Example E: Liquid Phase Semibatch (1) A + 2B →C (2) 2A + 3C → D FA0 1) Mole Balances: dN A dt dN B dt dN C dt dN D 32 dt B rAV F A 0 N A0 0 rB V N B 0 C B 0V 0 2 . 000 rC V NC0 0 rD V N D0 0 Example E: Liquid Phase Semibatch 2) Rate Laws: (5)-(14) Net Rate, Rate Laws and relative rate – are the same as Liquid and Gas Phase PFR and Liquid Phase CSTR V V 0 v 0 t 15 CA NA CC NC V V 16 18 CB NB CD ND V V 17 19 3) Selectivity and Parameters: SC /D 33 NC if ( t 0 . 0001 ) then ND 0 10 dm 3 min V 0 100 dm 3 else ( 0 ) 20 FA 0 3 mol min End of Lecture 13 34