Lecture notes - Chemical Engineering - Chapter 1-4 PDF

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Title:

Chemical Reaction Engineering

Course Codes:

CENG3003, BENG3008

Value:

½ Unit

Lecturers:

Prof. A Gavriilidis Dr N. Szita

Aims:

Development of the structure necessary for solving chemical reaction engineering problems. Ultimate goal is the design of chemical reactors.

Coursework: 4 sets Assessment: Written examination (80%) Coursework (20%)

1-1

SYNOPSIS Introduction: Brief survey of the scope of the subject together with a review of some of its foundations. Mole Balances: Definition of reaction rate. The general mole balance. The batch, plug flow and continuous stirred reactors. Industrial reactors. Conversion and Reactor Sizing: Definition of conversion. Design equations for batch and flow systems. Reactors in series. Space velocity and space time. Rate Laws and Stoichiometry: Concepts of reaction rate, reaction order, elementary reaction and molecularity. Stoichiometric table. Reactions with phase change. Isothermal Reactor Design: Design structure for isothermal batch, plug flow and continuous stirred reactors. Design of multiple reactor systems. Pressure drop in reactors. Reversible reactions. Catalysis and Catalytic Reactors: Catalyst definition and properties. Steps in catalytic reactions. Synthesising rate laws. Guidelines for design of reactors for gas-solid reactions. Heterogeneous data analysis for reactor design. Nonisothermal Reactor Design: The energy balance. Algorithms for nonisothermal plug flow and continuous stirred reactor design. Equilibrium conversion. Steady state multiplicity. Multiple Reactions: Conditions for maximising yield and selectivity in parallel and series reactions. Biochemical Reaction Engineering: Characteristics of enzyme catalysed reactions. Biocatalyst selection and production. Use of immobilised biocatalysts. Use of organic reaction media. Reactor selection and operation.

1-2

TEXTS Recommended: “Elements of Chemical Reaction Engineering” H.S. Fogler Prentice-Hall, 4th Ed., 2006 Other good texts: “Chemical Reaction Engineering” O. Levenspiel Wiley International, 3rd Ed., 1999 “Chemical Reactor Analysis and Design” G.F. Froment and K. B. Bischoff Wiley International, 2nd Ed., 1990 “Chemical Engineering Kinetics” J.M. Smith McGraw-Hill, 3rd Ed., 1981

1-3

INTRODUCTION

Raw Materials

Physical Treatment Steps

purification, heating, mixing, etc. Typical Chemical Process

Recycle

Chemical Treatment Steps

Physical Treatment Steps

distillation, extraction, filtration, etc.

Products

1-4

Knowledge and information from: Thermodynamics Chemical Kinetics Fluid Mechanics Heat/Mass Transfer Economics

Chemical Reactor Design

distinguishes the chemical engineer from other engineers

Most common industrial reactors BR:

Batch Reactor

CSTR:

Continuous-Stirred Tank Reactor

PFR:

Plug-Flow Reactor

1-5

1-6

CONTINUOUS STIRRED TANK REACTORS

1-7

PLUG FLOW REACTORS

1-8

1. MOLE BALANCES 1.1

Definition of the Rate of Reaction, -rA

Chemical Species: Chemical Compound or element with a given identity (determined by kind, number and configuration of atoms).

Chemical Reaction:

One or more chemical species have lost their chemical identity.

Types of Reactions Decompostition: CH(CH3)2

C3H6 cumene

Combination:

N2 + 3H2

benzene

propylene

2NH3

Izomerization: CH3 CH2 = C – CH2CH3

CH3 CH3C = CHCH3

1-9

Classification of Reactions Homogeneous:

reaction takes place in one phase alone

Heterogeneous:

requires the presence of at least two phases

Catalytic:

requires the presence of a catalyst

Noncatalytic:

does not require a catalyst

Homogeneous

Heterogeneous

Noncatalytic

Catalytic

Most gas-phase reactions

Most liquid-phase reactions

Burning gas flame

Enzyme reactions

Burning of coal

NH3 synthesis

Reduction of iron ore to iron

Cracking of crude oil

1-10

A

products

Reaction rate, -rA is defined as the number of moles of A reacting (disappearing) per unit volume (mol/m3 – s).

Note:

rA

dCA is NOT a definition for chemical reaction rate. dt

Rate law: function of T, C must be determined experimentally e.g.

rA

kC A ,

rA

k1C A 1 k 2C A

rA

kC A2

1-11

1.2

The General Mole Balance Equation moles

dN j

Fjo

Gj

Fj

In

Generation

Out

Accumulation

moles time

moles time

moles time

moles time

dt

system volume Fjo

Gj

Fj

If variables (C, T) uniform throughout system volume

Gj

rj V

moles time

moles volume time volume

If variables (C, T) NOT uniform in the system volume:

V1

r j1 rj2

V2

V

1-12

G j1 Gj

r j1 M i

V1 G

1

(subvolume V1) M

ji i

r

ji 1

V

(M subvolumes = total system volume)

i

By taking limits: M

,

V

0

V

Gj F jo

rj dV

Fj

Gj v

F jo

Fj

r j dV

dN i dt dN j dt

1-13

1.3 Batch Reactors no in flow,

no out flow,

perfectly mixed

Fjo = 0

Fj = 0

Gj = rj V

Mole balance becomes:

dN j dt

rj V

Design equation for BR

– Constant Volume BR Pressure gauge

CA

NA V

dN A rA V dt d CA V rA V dt dC A rA dt

V

dC A dt

rA V

1-14

– Constant Pressure BR - Movable piston

CA

NA V

dN A rA V dt d CA V rA V dt dC A dV V CA rA V dt dt dC A C A dV rA dt V dt dC A d ln V CA rA dt dt 1.41 Continuous-Stirred Tank Reactor

steady-state,

dN j

perfectly mixed

0

Gj = r j V

dt Recall mole balance:

Fjo – Fj + Gj =

V

F jo

dN j dt

Fj rj

Design equation for CSTR

1-15

Reactants well-mixed conditions the same everywhere conditions in the exit conditions in the tank. Products

Relationship between molar and volumetric flow rates: Fj = Cj·

moles time

=

moles volume

volume time

1.42 Tubular Reactor Flow field is modeled by plug flow, i.e. no radial variation of C, T. Therefore reactor is: plug-flow reactor (PFR). Concentration ( rj also) varies along the axial direction. Divide reactor into subvolumes V. Conditions (C, T) uniform in each V. y

Fjo

Fj, exit

y

Fj(y)

y+ y

V

Fj (y + y)

1-16

dN j

steady-state

0

dt

Mole balance for subvolume V:

Fj y V

Fj y A

y

rj V

0

y

Fj y

y

Fj y

Arj

y Take limit as y dF j

0

A rj

dy

Divide by –1, substitute Ady = dV dF j dV

rj

Design equation for PFR

Note that for a reactor in which the cross-section A varies along the reactor, the design equation remains the same.

Example 1-3 A

B

CA, exit = 0.1 CAO 3 O = 10 dm /min -rA = kCA k = 0.23 min-1 VPFR = ?

1-17

dF A dV

rA

d CA dV

O

dCA dV

O

O

k

dC dV

O

rA

kC A

dC A CA

dV

Integration limits:

V

0

CA

C AO

V

VPFR

CA

C A, exit

O

k V PFR

V PFR

C

,

dC A CA

A exit

C AO

O

k

ln

V 0

PFR

dV

C AO C A , exit

10 dm 3 / min 0 .23 / min

ln

C AO 0 .1 C AO

100 dm 3

1-18

BR

CSTR

PFR

Production Rate

Small

High

High

Capital Cost

Small

High

High

Labour Cost

High

Small

Small

Scale of Operation

Small

Large

Large

Reactant Conversion

High

Small

High

Temperature Control

Easy

Easy

Difficult

1-19

2. CONVERSION AND REACTOR SIZING 2.1

Definition of Conversion aA + bB

cC + dD

Basis of calculation: Limiting reactant, A . Arrange reaction on a “per mole of A” basis.

A

b B a

Conversion, XA:

2.2

XA

c C a

d D a

moles of A reacted moles of A fed

Batch Systems:

XA

N AO - N A N AO

Flow Systems:

XA

F AO - FA F AO

Design Equations

2.21 Batch Systems

2-1

X

N AO NA N AO

A

NA

N AO

N AO

dN A dt

0

recall :

dN A dt

N AO

N AO

XA dX A dt

rA V

dX A dt

rA V

design equation for BR (differential form)

If reactor volume is not constant then:

XA

t

V

f t :

Vdt

N AO

0

0

XA

V

f XA :

t

N AO 0

dX A rA

dX A rA V

2-2

2.2.2 Flow Systems

FAO FA FAO

XA FA

FAO

FAO

FAO X A

FA

FAO X A

Recall: FAO

FA

FAO X A

FAO

Design equation for CSTR

FAO X A

Recall: FAO

CSTR

rA V FAO X A rA exit

V FA

rA V

dFA dV

rA

dX A dV

rA

PFR Design equation for PFR

XA

V

FAO

dX A rA 0

2-3

2.3

Applications of the Design Equation

rA

f CA

CA

rA

f XA

f XA

rA

kC AO 1

1 rA

1 kC AO

XA 1 1 X

A

1 rA

XA 0

0.2

0.4

0.6

XA

0

r A : large

XA

1

rA

0

0.8

1.0

1 : small rA 1 rA

2-4

Example 2-1 (refer to previous figure) dm 3 s g mol

1 rA

FA0 = 20 mol/s 30 XA = 0.8 VCSTR = ? 20

10

V

FA 0 X A rA

XA 0

V V FA 0

1 rA

FA0 1 rA

0.2

0.4

0.6

0.8

1.0

XA

0. 8

From figure : X A V F A0 or

27.5 V F A0

V VCSTR

0.8

0. 8

22

1 rA

27.5

dm 3 s mol

area of shaded rectangle

mol dm3 s 20 22 s mol 440 dm 3

440 litres

2-5

Design Equation for PFR

FA0 dX A dV

rA

Integrate and rearrange XA

V

FA 0 0

dm 3 s g mol th

1 rA

dX A rA

30

20 0.8

10

dX A rA 0

I

0

0.2

0.4

0.6

0.8

1.0

From figure we can evaluate graphically 0 .8

0

dX A rA

V V PFR

dm 3 s 10 mol

mol 20 s

dm 3 s 10 mol

200 dm

3

2-6

Example 2-2 (refer to previous figure) Compare VPFR with VCSTR for XA = 0.6 FA0 = 5 mol/s

1 rA

CSTR

dm 3 s g mol 30

PFR

20

10 XA 0

VCSTR

0.2

1 rA

FA0

0.4

0.6

0.8

1.0

XA

5 mol 16 dm 3 s 0.6 s mol VCSTR

48 dm 3 VPFR < VCSTR 0.6

V PFR

F A0

dX A rA 0

5 mol 5.1 dm 3 s s mol VPFR

25.5 dm 3

(always true for iso-T, first order rxns)

2-7

2.4

Reactors in Series

Consider the system: XA = 0 FA0

XAI

V1

FA1

XA2 V2

XA3

V3

FA2

FA3

Define conversion

X A2

total number of A reacted up to point 2 moles of A fed to first reactor

valid when:

-

X A2

F A0

no sidestreams withdrawn feed enters only the first reactor

F A2

FA 2

FA0

FA 0

FA 0 X A 2

Mole balance for species A, for second reactor (CSTR)

in – out + generation = 0 FA1 – FA2 + rA2V2 = 0

F0A

F AO X 1A

V2

F0A

F0A X 2A

F A0 X A 2 rA2

r2A V2

0

X A1 evaluated at XA2 2-8

Example 2 – 3 Gas mixture of 50% A, 50% inerts PO TO R O

CAO FAO

yAO TO

C AO

10atm 149ºC 0.082 dm3.atm/mol.K 6 dm3/s (at 149oC) ? ?

PAO RTO

C AO

C AO

= = = = = =

= =

y AO PO RTO 0.5 (mole fraction of A) 149ºC = 422 K

0.5 10 atm dm 3 atm 0.082 mol K 0.144

422.2 K

mol dm 3

F AO

C AO

O

FAO

0.867

mol s

mol 0.144 dm3

dm 3 6.0 s

2-9

Laboratory data for examples 2 – 3 through 2 – 6 Isothermal, gaseous decomposition A T P yAO

O

= = =

49ºC 10atm 0.5 (rest inerts)

XA

-rA

3B

mol 3

dm s

1 / rA

0

0.0053

189

0.1

0.0052

192

0.2

0.0050

200

0.3

0.0045

222

0.4

0.0040

250

0.5

0.0033

303

0.6

0.0025

400

0.7

0.0018

556

0.8

0.00125

800

0.85

0.0010

1000

= 6.0 dm3/s

2-10

FAO XA1 = 0.4

FAe XA2 = 0.8

1 rA1

V1

F AO

X A1

V2

FAO X A 2 X A1 rA 2

Example 2 – 4 XA1 = 0.4 XA2 = 0.8 (i.e. FAe = 0.2·FAO) VCSTR1 = ? VCSTR2 = ? FAO = 0.867mol/s

1 rA 3

dm s g mol

800

CSTR2

CSTR1

400

XA 0

0.2

0.4

0.6

0.8

1.0

2-11

X

V1

1 rA1

0 .4

A1

1 r A1

F AO

X

1 rA 2

0 .8

A2

V2

1 rA

F AO

V2 V1

1 rA

F AO

V 2 CSTR

X2

0

FAO

X

dX rA

X

A2

A1

0 .8

0 .4

3

3

364 dm

(80%) in one CSTR X

0 . 867

A

800

0 .8

555 dm 3 iso

's

for PFRs V 1 PFR

However

0 .4

dm 3 s mol

800

V2

V V 1 CSTR

250

277 . 4 dm

same conversion V

0 . 867

dm 3 s 800 mol

mol 0 . 867 s

VT

dm 3 s mol

86 . 7 dm 3

V1 X

A1

250

X1

0

dX rA

T, 1st order V 2 PFRs because X2

X1

dX rA

XA1 = 0.4 FAe XA2 = 0.8 2-12

Example 2 – 5 XA1 = XA2 = VPFR1 = FAO =

0.4 0.8 ? VPFR2 = ? 0.867 mol/s

1 rA 3

dm s g mol

800

PFR1

PFR2

400 XA 0

0.2

0.4

0.6

0.8

1.0

Evaluate integrals numerically Simpson’s three point rule: X2 X

f x dx

0

x f x0 3

4 f x1

First reactor: X0 = 0, X1 = 0.2, X2 = 0.4,

V1

V1 V1

FAO

0.4 0

dX rA

mol 0.867 s

FAO

X 3

0.2 189 3

f x2

X = 0.2

1 rA 0 4 200

4

1 rA 0.2

1 rA 0.4

dm3 s 250 mol

71.6 dm 3

2-13

Second reactor: X0 = 0.4, X1 = 0.6, X2 = 0.8,

V2

V2

FAO

X 3

1 rA 0.4

mol 0.867 s

V2

153 dm3

VT

V1

V2

4

0.2 250 3

X = 0.2

1 rA 0.6 4 400

1 rA 0.8 dm3 s 800 mol

225 dm3

Scheme A: FAO

X1

X 2 = 0.8

Scheme B: FAO

X1

X 2 = 0.8

2-...