Lecture 8 Chemical Reaction Engineering (CRE) is thefield that studies the rates and mechanisms ofchemical reactions and the design of the reactors inwhich they take place. Todays lecture Block 1: Mole Balances on PFRs and PBR
Must Use the Differential Form Block 2: Rate Laws Block 3: Stoichiometry Pressure Drop: Liquid Phase Reactions: Pressure Drop does not affect the concentrations in liquid phase rx. Gas Phase Reactions: Epsilon not Equal to Zero d(P)/d(W)=.. Polymath will combine with d(X)/f(W)=..for you Epsilon = 0 and Isothermal P=f(W) Combine then Separate Variables (X,W) and Integrate Reactor Mole Balances in terms of conversion
Differential Algebraic Integral Batch X t CSTR PFR PBR X W Concentration Flow System:
Gas Phase Flow System: Pressure Drop in Packed Bed Reactors
Note: Pressure drop does NOT affect liquid phase reactions Sample Question: Analyze the following second order gas phase reaction that occurs isothermally in a PBR: AB Mole Balance: Must use the differential form of the mole balance to separate variables: Second order in A and irreversible: Rate Law: Pressure Drop in Packed Bed Reactors
Stoichiometry: Isothermal, T=T0 Combine: Need to find (P/P0) as a function of W (or V if you have a PFR) Pressure Drop in Packed Bed Reactors
Ergun Equation: Constant mass flow: Pressure Drop in Packed Bed Reactors
Variable Density Let Pressure Drop in Packed Bed Reactors
Catalyst Weight Where Let Pressure Drop in Packed Bed Reactors
We will use this form for single reactions: Isothermal case Pressure Drop in Packed Bed Reactors
and or The two expressions are coupled ordinary differentialequations. We can only solve them simultaneously using anODE solver such as Polymath. For the special case ofisothermal operation and epsilon = 0, we can obtain ananalytical solution. Polymath will combine the mole balance, rate law and stoichiometry. PBR AB 1) Mole Balance: 2) Rate Law: PBR W P 1 CA 2 W -rA 3 W X 4 W 5 W 1.0 Example 1: Gas Phase Reaction in PBR for = 0
Gas Phase Reaction in PBR with = 0 (Polymath Solution) A + B 2C Repeat the previous one with equil molar feed of A and B and kA = 1.5dm9/mol2/kg/min = kg-1 Find X at 100 kg Example 1: Gas Phase Reaction in PBR for = 0
A + B 2C Case 1: Case 2: Example 1: Gas Phase Reaction in PBR for = 0
1) Mole Balance: 2) Rate Law: 3) 4) Example 1: Gas Phase Reaction in PBR for = 0
5) Example 1: Gas Phase Reaction in PBR for = 0 Example A + B 2C Example A + B 2C Example 2: Gas Phase Reaction in PBR for 0
Polymath Solution A + 2B C is carried out in a packed bed reactor in which there is pressure drop.The fed is stoichiometric in A and B. Plot the conversion and pressure ratio y = P/P0 as a function of catalyst weight upto 100 kg. Additional Information kA = 6dm9/mol2/kg/min = 0.02 kg-1 Example 2: Gas Phase Reaction in PBR for 0
A + 2B C 1) Mole Balance: 2) Rate Law: 3) Stoichiometry: Gas, Isothermal Example 2: Gas Phase Reaction in PBR for 0
4) 5) 6) 7) Initial values:W=0, X=0, y=1 W=100 Combine with Polymath. If 0, polymath must be used to solve. Example 2: Gas Phase Reaction in PBRfor 0 Example 2: Gas Phase Reaction in PBRfor 0 T = T0 Engineering Analysis Engineering Analysis Engineering Analysis Pressure Change Molar Flow Rate
Use for heat effects, multiple rxns Isothermal: T = T0 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance End of Lecture 8