Lecture 1 - University of Michigan

Lecture 1 - University of Michigan

Lecture 1 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. 1 Todays lecture Introduction Definitions General Mole Balance Equation Batch (BR) Continuously Stirred Tank Reactor

(CSTR) Plug Flow Reactor (PFR) Packed Bed Reactor (PBR) 2 Chemical Reaction Engineering Chemical reaction engineering is at the heart of virtually every chemical process. It separates the chemical engineer from other engineers. Industries that Draw Heavily on Chemical Reaction Engineering (CRE)

are: CPI (Chemical Process Industries) Examples like Dow, DuPont, Amoco, Chevron 3 4 Smog (Ch. 1) Wetlands (Ch. 7 DVDROM) Hippo Digestion (Ch. 2) Oil Recovery

(Ch. 7) 5 Lubricant Chemical Plant for Ethylene Glycol (Ch. Design (Ch. Cobra Bites (Ch. 6 DVDROM) Plant Safety Materials on the Web and CDROM

http://www.umich.edu/~ess en/ 6 Lets Begin CRE 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. 7

Chemical Identity A chemical species is said to have reacted when it has lost its chemical identity. The identity of a chemical species is determined by the kind, number, and configuration of that species atoms. 8 Chemical Identity A chemical species is said to have reacted

when it has lost its chemical identity. There are three ways for a species to loose its identity: 1. Decomposition H2C=CH2 CH3CH3 H2 + 2. Combination N2 + O2 2 NO 3. Isomerization CH2=C(CH3)2

9 C2H5CH=CH2 Reaction Rate The reaction rate is the rate at which a species looses its chemical identity per unit volume. The rate of a reaction (mol/dm3/s) can be expressed as either: The rate of Disappearance of reactant: rA

or as The rate of Formation (Generation) of product: rP 10 - Reaction Rate Consider the isomerization AB rA = the rate of formation of species A per unit volume -rA = the rate of a disappearance of species A per unit volume

rB = the rate of formation of species B per unit volume 11 Reaction Rate EXAMPLE: AB If Species B is being formed at a rate of 0.2 moles per decimeter cubed per second, ie, rB = 0.2 mole/dm3/s Then A is disappearing at the same rate: -rA= 0.2 mole/dm3/s The rate of formation (generation of A) is

rA= -0.2 mole/dm3/s 12 Reaction Rate For a catalytic reaction, we refer to -r A', which is the rate of disappearance of species A on a per mass of catalyst basis. (mol/gcat/s) NOTE: dCA/dt is not the rate of reaction 13

Reaction Rate 14 Consider species j: 1. rj is the rate of formation of species j per unit volume [e.g. mol/dm3s] 2. rj is a function of concentration, temperature, pressure, and the type of catalyst (if any) 3. rj is independent of the type of reaction system (batch, plug flow, etc.) 4. rj is an algebraic equation, not a differential equation

(e.g. = -rA = kCA or -rA = kCA2) General Mole Balance System Volume, V Fj0 Gj Molar Flow Rate of

Species j in Fj 0 15 mole time Fj

Molar Flow Molar Rate Molar Rate Rate of Generation Accumulation Species j out of Species j of Species j dN j Fj Gj

dt mole mole mole

time time time General Mole Balance If spatially uniform G j rjV If NOT spatially uniform V1 r j1

G j1 rj1V1 16 V2 rj 2 G j 2 rj 2 V2 General Mole Balance W

G j rji Vi i1 Take limit n G j rji Vi i1 lim V 0 n 17

rj dV General Mole Balance System Volume, V FA GA FA 0

General Mole Balance on System Volume V In Out Generation Accumulation dN A FA 0 FA rA dV dt 18 Batch Reactor Mole Balance Batch

dN A FA 0 FA rA dV dt FA 0 FA 0 Well Mixed 19 r dV r V A A

dN A rAV dt Batch Reactor Mole Balance Integrating dN A dt rAV when t = 0 NA=NA0 t=t NA=NA

NA dN A t rAV N A0 20 Time necessary to reduce number of moles of A from NA0 to NA.

Batch Reactor Mole Balance NA dN A t rAV N A0 NA 21 t

CSTR Mole Balance CSTR dN A FA 0 FA rA dV dt Steady State 22 dN A 0

dt CSTR Mole Balance Well Mixed r dV r V A A FA 0 FA rAV 0

23 FA 0 FA V rA CSTR volume necessary to reduce the molar flow rate from FA0 to FA. Plug Flow Reactor 24 Plug Flow Reactor Mole

Balance V FA 25 V FA

V V In Out Generation 0 at V at V V in V FA V FA V V rA V 0

Plug Flow Reactor Mole Balance Rearrange and take limit as VV0 lim V 0 26 FA V V FA V V

rA dFA rA dV This is the volume necessary to reduce the entering molar flow rate (mol/s) from FA0 to Plug Flow Reactor Mole Balance PFR

dN A FA0 FA rA dV dt Steady State dN A 0 dt FA0 FA rA dV 0 27 Alternative Derivation Plug Flow Reactor Mole Balance

Differientiate with respect to V dFA 0 rA dV dFA rA dV FA

dF A The integral form is: V r A FA 0 28 This is the volume necessary to reduce the entering molar flow rate (mol/s) from FA0 to the exit molar flow rate of F .

Packed Bed Reactor Mole Balance PBR FA W FA W W rAW dN A 0 dt Steady State lim

29 dN A dt W 0 FA W W FA W

W rA Packed Bed Reactor Mole Balance Rearrange: dFA rA dW The integral form to find the catalyst weight is:

FA dFA W rA F A0 30 PBR catalyst weight necessary to reduce

the entering molar flow rate FA0 to molar flow rate F . Reactor Mole Balance Summary React or Batch Differential 31

Integral NA dN A t rV N A0 A dN A rAV dt N

A t FA 0 FA V rA CSTR PFR Algebraic

dFA rA dV FA dFA drA FA 0 FA V

V 32 Fast Forward 10 weeks from now: Reactors with Heat Effects EXAMPLE: Production of Propylene Glycol in an Adiabatic CSTR Propylene glycol is produced by the

hydrolysis of propylene oxide: H 2 SO4 CH2 CH CH3 H 2O CH2 CH CH3 O OH OH

v0 Propylene Glycol What are the exit conversion X and exit temperature T? Solution Let the reaction be represented by 33 A+BC 34

35 36 37 38 Evaluate energy balance terms 39 40

41 Analysis We have applied our CRE algorithm to calculate the Conversion (X=0.84) and Temperature (T=614 R) in a 300 gallon CSTR operated adiabatically. T=535 R A+BC 42 X=0.84

T=614 R KEEPING UP 43 Separations Filtration Distillation Adsorption These topics do not build upon one

another 44 Reaction Engineering Mole Balance Rate Laws Stoichiometry These topics build upon one another

45 Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 46 Mole Balance 47

Rate Laws Heat Effects Isothermal Design Stoichiometry Rate Laws Mole Balance 48 End of Lecture 1 49

Additional Applications of Supplemental Slides CRE 50 Additional Applications of CRE 51 52

53 Compartments for perfusion Alcohol Stomach VG = 2.4 l Gastrointestinal VG = 2.4 l tG = 2.67 min Liver

VL = 2.4 l tL = 2.4 min Perfusion interactions between compartments are shown by arrows. VG, VL, VC, and VM are -tissue water volumes for the gastrointestinal, liver, central and muscle compartments, respectively. VS is the stomach contents volume. 54 Central

VC = 15.3 l tC = 0.9 min Muscle & Fat VM = 22.0 l tM = 27 min 55 56

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