L7b-1 Copyright © 2014, Prof. M. L. Kraft All rights reserved. Review: Fixed-Volume CSTR Start-Up Isothermal (unusual, but simple case), well-mixed CSTR Unsteady state: concentrations vary with time & accumulation is non-zero Goal: Determine the time required to reach steady-state operation and C A as a function of time moles A in CSTR  wrt time while in unsteady state InOut - + Generation = Accumulation C A0  0 0CA0CA Use concentration rather than conversion in the balance eqs Integrate to find C A (t) while CSTR of 1 st order rxn is in unsteady-state:

L7b-2 Copyright © 2014, Prof. M. L. Kraft All rights reserved. Review: Time to Reach Steady-State At steady state, t is large and: 0 In the unsteady state, when C A = 0.99C AS : time to reach 99% (C A = 0.99C AS ) of steady-state concentration in terms of  k 99% of the steady-state concentration is achieved at: When k is very small (slow rxn), 1>>  k: When k is very big (fast rxn), 1<<  k 63% of the steady-state concentration is achieved at: C A = 0.63C AS

L7b-3 Copyright © 2014, Prof. M. L. Kraft All rights reserved. Semi-batch FBFB Review: Enhanced Yield in Semi- Batch Reactor V0V0 VfVf FDFD V 0 -  0 t Scenario 2: Improve the product yield obtained from a reversible reaction Allowing D(g) to bubble out of solution pushes equilibrium towards completion A+B ⇌ C+D A+B Scenario 1: Enhance selectivity of desired product over undesired side product Higher concentrations of A favor formation of the desired product Higher concentrations of B favor formation of the undesired side product A A+B →P V 0 +  0 t Scenario 1 shown in blue. Scenario 2 shown in red.

L7b-4 Copyright © 2014, Prof. M. L. Kraft All rights reserved. Review: Mole Balance on A for Semi-Batch Reactor CB0CB0 V 0 +  0 t InOut - + Generation = Accumulation Use whatever units are most convenient (N A, C A, X A, etc) Convert N A to C A using: InOut - + Generation = Accumulation Reactor volume balance:  =  0   Rearrange to get in terms of dC A /dt Goal: Find how C A  with time (assume reactor is well-mixed)

L7b-5 Copyright © 2014, Prof. M. L. Kraft All rights reserved. Review: Mole Balance on B in Semi-Batch Reactor CB0CB0 V 0 +  0 t Mole balance on B: InOut - + Generation = Accumulation Substitute Balance on B Rearrange to get in terms of dC B /dt Goal: Find how C B  with time (assume reactor is well-mixed)

L7b-6 Copyright © 2014, Prof. M. L. Kraft All rights reserved. Review: Semi-Batch Mole Balances in Terms of N A CB0CB0 V 0 +  0 t InOut - + Generation = Accumulation N B comes from basic mole balance: The design eq in terms of X A can be messy. Sometimes it gives a single equation when using N j or C j gives multiple reactor designs Substitute: -r A = k A C A C B and Goal: Find how N A & N B  with time (reactor is well-mixed)

L7b-7 Copyright © 2014, Prof. M. L. Kraft All rights reserved. X=0 P 0 = 20 atm PBR, 1000 kg cat X=? P = ? atm What conversion and P is measured at the outlet of the PBR? The rxn is isothermal at 300 K, assume ideal gas behavior, and the feed contains pure A (g). 2A→B -r A = kC A 2 α = /kg k=0.1 dm 6 /mol∙min∙kg cat at 300 K F A0 = 10 mol min 1. Mole balance 2. Rate law 3.Stoichiometry (put C A in terms of X) 4.Combine 5.Relate P/P 0 to W 1

L7b-8 Copyright © 2014, Prof. M. L. Kraft All rights reserved. X=0 P 0 = 20 atm PBR, 1000 kg cat X=? P = ? atm What conversion and P is measured at the outlet of the PBR? The rxn is isothermal at 300 K, assume ideal gas behavior, and the feed contains pure A (g). 2A→B -r A = kC A 2 α = /kg k=0.1 dm 6 /mol∙min∙kg cat at 300 K F A0 = 10 mol min 4.Combine 5.Relate P/P 0 to W Simultaneously solve dX A /dW and dP/dW (or dy/dW) using Polymath First, need to determine  C A0, &  0. What is C A0 ?

L7b-9 Copyright © 2014, Prof. M. L. Kraft All rights reserved. X=0 P 0 = 20 atm PBR, 1000 kg cat X=? P = ? atm What conversion and P is measured at the outlet of the PBR? The rxn is isothermal at 300 K, assume ideal gas behavior, and the feed contains pure A (g). 2A→B -r A = kC A 2 α = /kg k=0.1 dm 6 /mol∙min∙kg cat at 300 K F A0 = 10 mol min 4.Combine 5.Relate P/P 0 to W Simultaneously solve dX A /dW and dP/dW (or dy/dW) using Polymath First, need to determine  C A0, &  0. What is C A0 ?

L7b-10 Copyright © 2014, Prof. M. L. Kraft All rights reserved. X=0 P 0 = 20 atm PBR, 1000 kg cat X=? P = ? atm What conversion and P is measured at the outlet of the PBR? The rxn is isothermal at 300 K, assume ideal gas behavior, and the feed contains pure A (g). 2A→B -r A = kC A 2 α = /kg k=0.1 dm 6 /mol∙min∙kg cat at 300 K F A0 = 10 mol min Simultaneously solve dX A /dW and dP/dW (or dy/dW) using Polymath

L7b-11 Copyright © 2014, Prof. M. L. Kraft All rights reserved. X=0 P 0 = 20 atm PBR, 1000 kg cat X=? P = ? atm What conversion and P is measured at the outlet of the PBR? The rxn is isothermal at 300 K, assume ideal gas behavior, and the feed contains pure A (g). 2A→B -r A = kC A 2 α = /kg k=0.1 dm 6 /mol∙min∙kg cat at 300 K F A0 = 10 mol min Simultaneously solve dX A /dW and dP/dW (or dy/dW) using Polymath

L7b-12 Copyright © 2014, Prof. M. L. Kraft All rights reserved. X = 0.93 What conversion and P is measured at the outlet of the PBR? The rxn is isothermal at 300 K, assume ideal gas behavior, and the feed contains pure A (g). P = atm

L7b-13 Copyright © 2014, Prof. M. L. Kraft All rights reserved. What conversion can be achieved in a fluidized CSTR with the same catalyst weight and P 0 = P (ideal gas behavior, pure A feed)? 1 0 Use info from PBR to determine F A0, C A0 & k Isothermal and  =0. Ergun eq for P/P 0 becomes: Plug into C A : Do not plug in P and P 0 that occurred in PBR yet! Use Ergun eq to get P/P 0 as a function of W, plug into design eq & integrate over W! Use PBR expt parameters to solve for α

L7b-14 Copyright © 2014, Prof. M. L. Kraft All rights reserved. What conversion can be achieved in a fluidized CSTR with the same catalyst weight and P 0 = P (ideal gas behavior, pure A feed)? Use info from PBR to determine F A0, C A0 & k Plug C A into PBR design eq: Rearrange Integrate so that we can get values of unknowns

L7b-15 Copyright © 2014, Prof. M. L. Kraft All rights reserved. What conversion can be achieved in a fluidized CSTR with the same catalyst weight and P 0 = P (ideal gas behavior, pure A feed)?

L7b-16 Copyright © 2014, Prof. M. L. Kraft All rights reserved. What conversion can be achieved in a fluidized CSTR with the same catalyst weight and P 0 = P (ideal gas behavior, pure A feed)? Plug this value into the CSTR eq

L7b-17 Copyright © 2014, Prof. M. L. Kraft All rights reserved. What conversion can be achieved in a fluidized CSTR with the same catalyst weight and P 0 = P (ideal gas behavior, pure A feed)? Conversion in fluidized CSTR, no pressure drop

L7b-18 Copyright © 2014, Prof. M. L. Kraft All rights reserved. A + 2B → CElementary rxn, feed is a stoichiometric mixture Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction P 0 = 6 atm; T = 443K &  0 = 50 dm 3 /min How many kg of catalyst is required to achieve X A = 0.8? 1. What is C A0 ? Known:  0 and X A Unknown: C A0 & -r’ A Fluidized CSTR design eq: Feed is a stoichiometric mixture → 1 part A, 2 parts B

L7b-19 Copyright © 2014, Prof. M. L. Kraft All rights reserved. A + 2B → C Elementary rxn, feed is a stoichiometric mixture Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction P 0 = 6 atm; T = 443K &  0 = 50 dm 3 /min How many kg of catalyst is required to achieve X A = 0.8? 2. What is –r’ A ? Known:  0, X A, & C A0 (0.055 mol/dm 3 ) Unknown: -r’ A Express rate law in terms of partial pressure, not C j 2a. What is P A ? For ideal, isobaric, isothermal rxn: Substitute for C j & C j0

L7b-20 Copyright © 2014, Prof. M. L. Kraft All rights reserved. A + 2B → C Elementary rxn, feed is a stoichiometric mixture Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction P 0 = 6 atm; T = 443K &  0 = 50 dm 3 /min How many kg of catalyst is required to achieve X A = 0.8? 2. What is –r’ A ? Known:  0, X A, & C A0 (0.055 mol/dm 3 ) Unknown: -r’ A Units on k necessitate expressing rate law in terms of partial pressure, not C j 2a. What is P A ? A =-1  A =1

L7b-21 Copyright © 2014, Prof. M. L. Kraft All rights reserved. A + 2B → C Elementary rxn, feed is a stoichiometric mixture Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction P 0 = 6 atm; T = 443K &  0 = 50 dm 3 /min How many kg of catalyst is required to achieve X A = 0.8? 2. What is –r’ A ? Known:  0, X A, & C A0 (0.055 mol/dm 3 ) Unknown: -r’ A 2b. What is P B ? B =-2

L7b-22 Copyright © 2014, Prof. M. L. Kraft All rights reserved. A + 2B → C Elementary rxn, feed is a stoichiometric mixture Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction P 0 = 6 atm; T = 443K &  0 = 50 dm 3 /min How many kg of catalyst is required to achieve X A = 0.8? 2. What is –r’ A ? Known:  0, X A, & C A0 (0.055 mol/dm 3 ) Unknown: -r’ A 2c. What is k at 443K?

L7b-23 Copyright © 2014, Prof. M. L. Kraft All rights reserved. 2. What is –r’ A ? Known:  0, X A, & C A0 (0.055 mol/dm 3 ) Unknown: -r’ A A + 2B → C Elementary rxn, feed is a stoichiometric mixture Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction P 0 = 6 atm; T = 443K &  0 = 50 dm 3 /min How many kg of catalyst is required to achieve X A = 0.8?

L7b-24 Copyright © 2014, Prof. M. L. Kraft All rights reserved. How many kg of catalyst is required to achieve X A = 0.8? C A0 =0.055 mol/dm 3 A + 2B → C Elementary rxn, feed is a stoichiometric mixture Fluidized CSTR, isothermal, isobaric, ideal, gas-phase reaction P 0 = 6 atm; T = 443K &  0 = 50 dm 3 /min