Lecture 8 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.
Today’s 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 differential equations. We can only solve them simultaneously using an ODE solver such as Polymath. For the special case of isothermal operation and epsilon = 0, we can obtain an analytical 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 δ = 0 Gas Phase Reaction in PBR with δ = 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 α = 0.0099 kg-1 Find X at 100 kg
Example 1: Gas Phase Reaction in PBR for δ = 0 A + B 2C Case 1: Case 2:
Example 1: Gas Phase Reaction in PBR for δ = 0 1) Mole Balance: 2) Rate Law: 3) 4)
Example 1: Gas Phase Reaction in PBR for δ = 0 5)
Example 1: Gas Phase Reaction in PBR for δ = 0
Example A + B → 2C
Example A + B → 2C
Example 2: Gas Phase Reaction in PBR for δ ≠ 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 α = 0.02 kg-1
Example 2: Gas Phase Reaction in PBR for δ ≠ 0 A + 2B C 1) Mole Balance: 2) Rate Law: 3) Stoichiometry: Gas, Isothermal
Example 2: Gas Phase Reaction in PBR for δ ≠ 0 4) 5) 6) 7) Initial values: W=0, X=0, y=1 W=100 Combine with Polymath. If δ≠0, polymath must be used to solve.
Example 2: Gas Phase Reaction in PBR for δ ≠ 0
Example 2: Gas Phase Reaction in PBR for δ ≠ 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