Download presentation

Presentation is loading. Please wait.

Published byRenee Seaton Modified about 1 year ago

1
Steady State Nonisothermal Reactor Design Dicky Dermawan ITK-330 Chemical Reaction Engineering

2
Rationale All reactions always accompanied by heat effect: exothermic reactions vs. endothermic reactions Unless heat transfer system is carefully designed, reaction mass temperature tend to change Design of heat transfer system itself requires the understanding of this heat effect Energy balance is also needed, together with performance equations derived from mass balance

3
Objectives Describe the algorithm for CSTRs, PFRs, and PBRs that are not operated isothermally. Size adiabatic and nonadiabatic CSTRs, PFRs, and PBRs. Use reactor staging to obtain high conversions for highly exothermic reversible reactions. Carry out an analysis to determine the Multiple Steady States (MSS) in a CSTR along with the ignition and extinction temperatures. Analyze multiple reactions carried out in CSTRs, PFRs, and PBRs which are not operated isothermally in order to determine the concentrations and temperature as a function of position (PFR/PBR) and operating variables

4
Why Energy Balance? Imagine that we are designing a nonisothermal PFR for a first order liquid phase exothermic reaction: Performance equation: Kinetics: The temperature will increase with conversion down the length of reactor Stoichiometry: Combine:

5
Energy Balance At steady state: Consider generalized reaction: Upon substitution:

6
Energy Balance (cont’) From thermodynamics, we know that: Thus:

7
Energy Balance (cont’) Upon substitution: Finally…. So what?

8
Energy Balance (cont’) For adiabatic reactions: The energy balance at steady state becomes: After rearrangement: When work is negligible: This is the X=X(T) we’ve been looking for!

9
Application to Adiabatic CSTR Design Case A: Sizing: X specified, calculate V (and T) Performance equation: Kinetics: Stoichiometry: Combine: Solve the energy balance for T Calculate k Calculate V using combining equation

10
Application to Adiabatic CSTR Design Case B (Rating): V specified, calculate X (and T) Performance equation: Kinetics: Stoichiometry: Mole balance: Energy balance: Find X & T that satisfy BOTH the material balance and energy balance, viz. plot X mb vs T and X eb vs T in the same graph: the intersection is the solution

11
Application to Adiabatic CSTR Design Example: P8-5A The elementary irreversible organic liquid-phase reaction: A + B C is carried out adiabatically in a CSTR. An equal molar feed in A and B enters at 27 o C, and the volumetric flow rate is 2 L/s. (a) Calculate the CSTR volume necessary to achieve 85% conversion (b) Calculate the conversion that can be achieved in one 500 L CSTR and in two 250 L CSTRs in series

12
Application to Adiabatic CSTR Design Case A: Sizing: X specified, calculate V (and T) Performance equation: Kinetics: Stoichiometry: Combine: Energy balance: Calculate k Calculate V using combining equation

13
Application to Adiabatic CSTR Design Case B (Rating): V specified, calculate X (and T) Performance equation: Kinetics: Stoichiometry: Mole balance: Energy balance:

14
Application to Adiabatic PFR/PBR Design Example for First Order Reaction Performance equation: Kinetics: Stoichiometry: Pressure drop: Energy balance: for PFR/small P: P/P 0 = 1 Gas liquid Combine: Thus The combination results in 2 simultaneous differential equations

15
Sample Problem for Adiabatic PFR Design P8-6A

16
Sample Problem for Adiabatic PBR Design

17
NINA = Diabatic Reactor Design Heat Transfer Rate to the Reactor Rate of energy transferred between the reactor and the coolant: The rate of heat transfer from the exchanger to the reactor: Combining:

18
NINA = Diabatic Reactor Design Heat Transfer Rate to the Reactor (cont’) At high coolant flow rates the exponential term will be small, so we can expand the exponential term as a Taylor Series, where the terms of second order or greater are neglected: Then: The energy balance becomes:

19
Sample Problem for Diabatic CSTR Design P8-4B

20

21
Application of Energy Balance to Diabatic Tubular Reactor Design Heat transfer in CSTR: In PFR, T varies along the reactor: Thus: For PBR: Thus:

22
Application of Energy Balance to Diabatic Tubular Reactor Design The steady state energy balance, neglecting work term: Differentiation with respect to the volume V: and recalling that Or: Inserting Coupled with Form 2 differential with 2 dependent variables X & T

23
Sample Problem for Diabatic Tubular Reactor Design

24
Design for Reversible Reactions Endotermik: X eq naik Endotermik: K naik dengan kenaikan T X eq naik reaksikan pada T max yang diperkenankan Eksotermik: X eq turun Eksotermik: K turun dengan kenaikan T X eq turun reaksikan pada T rendah Laju reaksi lambat pada T rendah! Ada trade off antara aspek termodinamika dan kinetika X eq = X eq (K) = X eq (T)

25
Design for Reversible Highly-Exothermic Reactions -r A = -r A (X,T) Generally:Higher X slower reaction rate Higher T faster rate At X = X eq : -r A = 0

26
Design for Equilibrium Highly-Exothermic Reactions #1 Starting with R-free solution, between 0 dan 100 o C determine the equilibrium conversion of A for the elementary aqueous reaction: A R A R The reported data is based on the following standard states of reactants and products: Assume ideal solution, in which case: In addition, assume specific heats of all solutions are equal to that of water

27
Design for Equilibrium Highly-Exothermic Reactions: Reaction Rate in X – T Diagram

28
Reaction Rate in The X – T Diagram at C A0 = 1 mol/L

29
Design for Equilibrium Highly-Exothermic Reactions: Optimum Temperature Progression in Tubular Reactor #3 a. Calculate the space time needed for 80% conversion of a feed starting with initial concentration of A of 1 mol/L b. Plot the temperature and conversion profile along the length of the reactor Let the maximum operating allowable temperature be 95 o C

30
Design for Reversible Reactions: Heat Effect

31
Design for Equilibrium Highly-Exothermic Reactions: CSTR Performance

32
#4 A concentrated aqueous A-solution of the previous examples, C A0 = 4 mol/L, F A0 = 1000 mol/min, is to be 80% converted in a mixed reactor. a. If feed enters at 25 o C, what size of reactor is needed? b. What is the optimum operating temperature for this purpose? c. What size of reactor is needed if feed enters at optimum temperature? d. What is the heat duty if feed enters at 25 o C to keep the reactor operation at its the optimum temperature?

33
Interstage Cooling

34
Review on Energy Balance in CSTR Operation Bila term kerja diabaikan dan H Rx konstan: Untuk CSTR: Pembagian kedua ruas dengan F A0 :

35
Multiple Steady State & Stability of CSTR Operation Review on Energy Balance in CSTR Operation

36
Multiple Steady State: Stability of CSTR Operation Temperature Ignition – Extinction Curve Finding Multiple Steady State: Varying T o Upper steady state Lower steady state Ignition temperature Extinction temperature Runaway Reaction

37
Sample Problem on Multiple Steady State in CSTR Operation P8-17B

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google