Lecture 5 Chemical Reaction Engineering (CRE) is the field that studies the rates and mechanisms of chemical reactions and the design of the reactors.

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Presentation transcript:

Lecture 5 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 5 – Thursday 1/24/2013 Block 1: Mole Balances Block 2: Rate Laws Block 3: Stoichiometry Stoichiometric Table: Flow Definitions of Concentration: Flow Gas Phase Volumetric Flow Rate Calculate the Equilibrium Conversion Xe

Reactor Mole Balances Summary Review Lecture 2 Reactor Mole Balances Summary in terms of conversion, X Reactor Differential Algebraic Integral CSTR PFR Batch X t PBR W

Algorithm How to find Review Lecture 3 Step 1: Rate Law Step 2: Stoichiometry Step 3: Combine to get

Reaction Engineering These topics build upon one another Review Lecture 3 Reaction Engineering Mole Balance Rate Laws Stoichiometry These topics build upon one another

Flow System Stoichiometric Table Species Symbol Reactor Feed Change Reactor Effluent A FA0 -FA0X FA=FA0(1-X) B FB0=FA0ΘB -b/aFA0X FB=FA0(ΘB-b/aX) C FC0=FA0ΘC +c/aFA0X FC=FA0(ΘC+c/aX) D FD0=FA0ΘD +d/aFA0X FD=FA0(ΘD+d/aX) Inert I FI0=FA0ΘI ---------- FI=FA0ΘI FT0 FT=FT0+δFA0X Where: and Concentration – Flow System

Stoichiometry Liquid Systems Concentration Flow System: Liquid Phase Flow System: Liquid Systems Flow Liquid Phase etc.

Liquid Systems If the rate of reaction were then we would have This gives us

Stoichiometry for Gas Phase Flow Systems Combining the compressibility factor equation of state with Z = Z0 Stoichiometry: We obtain:

Stoichiometry for Gas Phase Flow Systems Since , Using the same method,

Stoichiometry for Gas Phase Flow Systems The total molar flow rate is: Substituting FT gives: Where

For Gas Phase Flow Systems Concentration Flow System: Gas Phase Flow System:

For Gas Phase Flow Systems If –rA=kCACB This gives us FA0/-rA X

For Gas Phase Flow Systems

Example: Calculating the equilibrium conversion (Xef) for gas phase reaction in a flow reactor Consider the following elementary reaction where KC=20 dm3/mol and CA0=0.2 mol/dm3. Calculate Equilibrium Conversion or both a batch reactor (Xeb) and a flow reactor (Xef).

Gas Flow Example (Xef) Solution: Rate Law:

Gas Flow Example (Xef) Species Fed Change Remaining A FA0 -FA0X B +FA0X/2 FB=FA0X/2 FT0=FA0 FT=FA0-FA0X/2

Gas Flow Example (Xef) Gas isobaric P=P0 A FA0 -FA0X FA=FA0(1-X) B FA0X/2 FB=FA0X/2 Stoichiometry: Gas isothermal T=T0 Gas isobaric P=P0

Gas Flow Example (Xef) Pure A  yA0=1, CA0=yA0P0/RT0, CA0=P0/RT0 At equilibrium: -rA=0

Gas Flow Example (Xef) Flow: Recall Batch:

End of Lecture 5