ĐỘNG học và NHIỆT ĐỘNG học TRONG CÔNG NGHỆ lọc dầu

• Course Outline, Tentative schedule • • Review: Chemical Kinetics and Ch1-6, Two-three weeks • Chapter 1: Mole Balances • Chapter 2: Conversion and Reactor Sizing • Chapter 2: Conversi

• MT dates will be determined (mid of the semester)

• Final exam date will be determined by the registrar’s office

• You can contact me anytime through e-mail (okeskin@ )

• You are wellcome during office hours, you need to ask me if I will be at the

office other than these hours.

• No cheating (HWs, projects, exams)

POLICY ON COLLABORATION AND ORIGINALITY

Academic dishonesty in the form of cheating, plagiarism, or collusion are serious offenses and

are not tolerated at Koç University University Academic Regulations and the Regulations for

Student Disciplinary Matters clearly define the policy and the disciplinary action to be taken in

case of academic dishonesty

Failure in academic integrity may lead to suspension and expulsion from the University

Cheating includes, but is not limited to, copying from a classmate or providing answers or

information, either written or oral, to others Plagiarism is borrowing or using someone else’s

writing or ideas without giving written acknowledgment to the author This includes copying

from a fellow student’s paper or from a text (whether printed or electronic) without properly

citing the source Collusion is getting unauthorized help from another person or having

someone else write a paper or assignment You can discuss the lecture and reading material,

and the general nat re of the home ork problems ith an one Also o ma per se all

and the general nature of the homework problems, with anyone Also, you may peruse all

previous ChBI 502 material available anywhere, such as on the web, and in the library

accumulated over the years However, your final solutions should be your own original work

Jointly prepared solutions, and solutions closely resembling those available, are unacceptable.

• Textbook Elements of Chemical Reaction Engineering

(4th ed.), H.S Fogler Prentice Hall, Upper Saddle River, NJ (2005)

• Course Outline, Tentative schedule

•

• Review: Chemical Kinetics and Ch1-6, Two-three weeks

• Chapter 1: Mole Balances

• Chapter 2: Conversion and Reactor Sizing

• Chapter 2: Conversion and Reactor Sizing

• Chapter 3: Rate Law and Stoichiometry

• Chapter 4: Isothermal Reactor Design

• Chapter 5: Collection and Analysis of Rate Data

• Chapter 6: Multiple Reactions

• Chapter 7: Reaction Mechanisms, Pathways, Bioreactions and Bioreactors , Two weeks

• Chapter 8: Steady-State Nonisothermal Reactor Design, Two weeks

• Chapter 9: Unsteady-state Nonisothermal Reactor Design, One week

• Chapter 10: Catalysis and Catalytic Reactors, Two weeks

• Chapter 11: External Diffusion Effects on Heterogeneous ReactionsOne Week

• Chapter 11: External Diffusion Effects on Heterogeneous Reactions, One Week

• Chapter 12: Diffusion and Reaction in Porous Catalysts, Two Weeks

• Student presentations on projects, One week

Elements of Chemical Rxn Enginnering

Chemical kinetics is the study of chemical rxn rates and reaction

mechanisms.

Chemical reaction engineering (CRE) combines the study of chemical

kinetics with the reactors in which the reactions occur.

Objective of the course:

Objective of the course: Learn how to design equipment for Learn how to design equipment for

carrying out desirable chemical reactions

(what size and what type of equipment)

Chemical Kinetics & Reactor Design

The reaction system thet operates in the safest and most efficient manner

can be the key to the success of the plant.

chemical identity The identity of a chemical species is determinde

by the kind, number, and configuration of that species’ atoms

Three ways a chemical species can lose its chemical identity:

The reaction rate is the rate at which a species looses its chemical

identity per unit volume The rate of a reaction can be expressed as

identity per unit volume The rate of a reaction can be expressed as

the rate of disappearance of a reactant or as the rate of appearance

of a product Consider species A:

A → B

rAA= the rate of formation of species A per unit volume p p

-rA= the rate of disappearance of species A per unit volume

rB= the rate of formation of species B per unit volume

For catalytic reaction , we refer to –rA’, which is the rate of disappearance of

species A on a per mass of catalyst basis.

NOTE: dCA/dt is not the rate of reaction (This is only true for a batch system, we will see)

If continuous → dCAA/dt = 0

The rate law does not depend on reactor type!

-rAis the # of moles of A reacting (disappearing) per unit time per unit volume

• Ideal mixing (no axial mixing, complete radial mixing)

• Steady-state : conditions donot change with time at any point (PFR)

• Complete mixing

• non-steady-state: Uniform composition and temperture at any given instant,

change with time (t)

rjis the rate of formation of species j per unit volume [e.g mol/dm 3 s]

rjis a function of concentration, temperature, pressure, and the type of

catalyst (if any)

rjis independent of the type of reaction system (batch, plug flow, etc.)

rjis an algebraic equation, not a differential equation.

We use an algebraic equation to relate the rate of reaction, -rA, to the

concentration of reacting species and to the temperature at which the reaction

a linear function of concentrations:

A

C k r

⋅+

3 General Mole Balance Equation:

IN – OUT + GENERATION = ACCUMULATION

]/[

0 moles time

d

dN dV r F

F

V

A A

A

of species A in]

[0

0

dt

A A

the system at time t.

GA= rAV (if all system variables (T, CA, etc.) are spatially uniform

throughout system volume)

[ volume ]

volume time

moles time

moles

V r

for V r G

G

subvolumes all

for etc V

=

⋅ +

Mole Balance on Different Reactor Types

Batch Reactor is used for small-scale operations, for testing new

processes, for the manufacture of expensive products, and for the

processes that are not easy to convert to continuous

high conversion rates (time spend is longer)

high labor cost and & variability of products from batch-to-batch

=

=

V dN

F F

V r dt

j j

0

1

1 0

A

A A

A

N

A N

A t

A

A

V r

dN t

V r

Continuous Flow Reactors (CFR)operate at steady state

Continuous Stirred Tank Reactor (CSTR)

Plug Flow Reactor (PFR)

Packed Bed Reactor (PBR)

T ≠ f(t,V)

General Mole Balance on System Volume V

IN – OUT + GENERATION = ACCUMULATION

∫ ⋅ = +

A A A

dt

dN dV r F F

0 0

V

F A0

F A Assumptions

A A A

A A

A

V r F F

V r dV r dt dN

=

⋅ +

A

A A

A

A A

r

v C v C V

r

F F

0 Design eq’n for CSTR

v C

F A0= A0⋅

Tubular Reactors consists of a cylindirical pipe and is operated at

steady state Mostly used for gas phase rxns

PFR Derivation: uniform velocity in turbulent flow (no radial variation in

velocity, concentration, temperature, reaction rate)

∫ ⋅ = +

A A A

dt

dN dV r F F

0 0

IN – OUT + GENERATION = ACCUMULATION

∆

∆

∆ +

∫

V r F

F

V r dV r G

A V V A V

A

A A

A

Divide by ∆V and rearrange:

A V

A V V A

r V

F F

A A

r

dF V

r dV dF

0

Packed Bed Reactors (PBR) are not homogenous, the fluid-solid

heterogenous rxn take place on the surface of the catalyst Rate (r’) is

dependent on the mass of catalyst (W)

-rA’ = mol A reacted / (s) (g catalyst)

A A

dt

dN dW r F F

0

IN – OUT + GENERATION = ACCUMULATION

∆W W+∆W W

FA(W+∆W)

FA(W)

∫ ⋅ = +

A A A

dN dV r F

dt

0 0

Differentiate with respect to W and rearrange

'

A

A r dW

dF =

When pressure drop through the reactor and catalyst decay are

When pressure drop through the reactor and catalyst decay are

neglected, the integral eg’n can be used to find W:

A

F

F A A F

F A

A

r

dF r

dF W

Batch Reactor Times

A → BCalculate the time to reduce the number of moles by a factor of 10 (NA=

NA0/10) in a batch reactor for the above reaction with

-rrA’A= k C k CAAwhen k = 0.046 minwhen k 0.046 min-1

A A

law Rate

dt

dN V r

on Accumulati Generation

Out In

balance Mol

=

⋅ +

−

= +

−

:

0 0

:

A A

A A

A A

A A

A

N k dt

dN N

k V r

V

N k C k r C

k r

law Rate

N A

A

A

A A

N k V C k

V

r

A A N

A

A

A A

A

A

A

) 10 ln(

min 046 0

1 10

• Define the rate of chemical reaction

• Apply the mole balance equations to a batch reactor, CSTR, PFR,

and PBR

conc’n

• CSTR: no spatial variations in the tank, steady state

• PFR: PFR: spatial variations along the reactor steady state spatial variations along the reactor, steady state

• PBR: spatial variations along the reactor, steady state

Basic equation for any species A entering, leaving, reacting

dt

dN dV r

10/1/2010

Perfusion interactions between

compartments are shown by arrows

V L = 2.4 l

tL= 2.4 min Central

V C = 15.3 l

tC= 0.9 min

V G , V L , V C , and V Mare -tissue water

volumes for the gastrointestinal,

liver, central and muscle

compartments, respectively

V Sis the stomach contents volume

Muscle & Fat

VM= 22.0 l

t M = 27 min

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)

Dow, DuPont, Amoco, Chevron Pharmaceutical – Antivenom, Drug Delivery

Medicine –Pharmacokinetics, Drinking and

Driving

Microelectronics – CVD

• Objectives:

CONVERSION AND REACTOR SIZING

• Define conversion and space time

• Write the mole balances in terms of conversion for a

batch reactor, CSTR, PFR, and PBR

• Size reactors either alone or in series once given the

molar flow rate of A, and the rate of reaction, -rA, as

a function of conversion, X

• Conversion: Choose one of the reactants as the basis

of calculation and relate the other species involved in

the rxn to this basis.

• Space time: the time necessary to process one

reactor volume of fluid based on entrance conditions

(holding time or mean residence time)

CONVERSION AND REACTOR SIZING

1 Conversion

Consider the general equation

dD C

d C a

c B a

b

A + → +

The basis of calculation is most always the limiting reactant The

conversion of species A in a reaction is equal to the number of moles of A

reacted per mole of A fed

0

( A A FA FA

X N

For irreversible reactions, the maximum value of conversion, X, is

that for complete conversion, i.e X = 1.0

For reversible reactions, the maximum value of conversion, X, is the

equilibrium conversion, i.e X = X

Batch Reactor Design Equations:

of Moles fed

A of Moles consumed

reacted

) (

[ NA0]

Now the # of moles of A that remain in the reactor after a time t, NAcan be

expressed in terms of NA0and X;

expressed in terms of NA0and X;

) 1 (

0

0 0

X N

N

X N N

N

A A

A A

dN

mixing prefect

V r dt

dN

A A

A A

For batch reactors, we are interested in determining how long to leave the

reactants in the reactor to achieve a certain conversion X

dX

dt

dX N

dX

N

V r dt

dX

N

A A

A A

For a constant volume batch reactor: (V = V 0 )

V N d

dt

dN

V

0 0

)/(

A A

A A

V r

dX N

t

V r

dX N

of moles fed

A of moles

For continuous-flow systems, time usually increases with increasing reactor

volume

A A

A

A

F X F

F

fed A of moles

reacted A

of moles time

fed A of moles X

v C F

X F

F

A A

A A

For liquid systems, CA0is usually given in terms of molarity (mol/dm3)

For gas systems, CA0can be calculated using gas laws

P P

Partial pressure

0

0 0

0

0 0

T R

P y T R

Entering molar flow rate is

y A0 = entering mole fraction of A

0

0 0 0 0

0

0

T R

P y v C

v

A

A = ⋅ = ⋅ ⋅

P 0 = entering total pressure (kPa)

C A0 = entering conc’n (mol/dm 3 )

R = 8.314 kPa dm 3 / mol K

T = T(K)

CSTR (Design Equation)

D a

d C a

c B a

b

A + → +For a rxn:

F F

A

A A

r

F F

A

A A

A

X F F

F V

X F F

F

) ( 0 0

0

0 0

A

r

X F V

r

) (

PFR (Design Equation)

r dV

X F F F

A A

A A A

Substitute back:

A A

dV

dX F dV

V

0 0

Applications of Design Equations for

Continuous Flow Reactors

3.

3 Reactor Sizing Reactor Sizing

Given –rAas a function of conversion, -rA= f(X), one can size any type of

reactor We do this by constructing a Levenspiel Plot Here we plot either

FA0/ -rAor 1 / -rAas a function of X For FA0/ -rAvs X, the volume of a

CSTR and the volume of a PFR can be represented as the shaded areas

in the Levelspiel Plots shown below:

A particularly simple functional dependence is the first order dependence:

) 1 (

C k C k

k

rA A 1

1 1

1

0 -1/rA

0.1130.079

8.8512.7

0 → -1/rAis small)

2 As x → 1, –rA→ 0 thus 1/-rA→ ∞ & V → ∞

→ An infinite reactor volume is needed to reach complete conversion

For reversible reactions (A ↔ B), the max X is the equilibrium conversion

Xe At equilibrium, rA≈ 0

As X → Xe, –rA→ 0 thus 1/-rA→ ∞ & V → ∞

→ An infinite reactor volume is needed to obtain Xe

if FA0= 0.4 mol/s, we can calculate [FA/-rA](m3)

Plot FA0/-rAvs X obtain Levenspiel Plot!

Example: Calculate volume to achieve 80 % conversion in CSTR.

3

8 0

X at r

0

3

8 0

4 6 8 0 20

4 0

) (

20 1

m m

mol r

X F V

s mol

m r

exit A A A

4

s mol

4 Numerical Evaluation of Intergrals Numerical Evaluation of Intergrals

The integral to calculate the PFR volume can be evaluated using a

method such as Simpson’s One-Third Rule

NOTE:

NOTE: The intervals (∆X) shown in the

sketch are not drawn to scale They should be equal.

Simpson’s One-Third Rule is one of the most common numerical methods It

uses three data points One numerical methods for evaluating integrals are:

1 Trapezoidal Rule (uses two data points)

2 Simpson’s Three-Eighth’s Rule (uses four data points)

3 Five-Point Quadrature Formula

Trapezoidal Rule

f(x1)

f(x)

2)]

()([

)(

)]

()([1)(

0 1 2

0 1

1 0

1

0

h x f x f A

h x f A

x f x f h dx x f

2

)(2

)()(2

1 0

0 1 0

2 1 2

x f x f h

x f x f x f h

A A A

424(

3

)

(

6 5 4 3 2 1 0

0 4 4

3 2 1 0

0

4

0

f f f f f f f h

dx

x

f

x x h where f

f f f f

+++

If the molar feed of A to the PFR is 2 mol/s, what PFR volume is

necessary to achieve 80 % conversion under identical conditions as

those under which the batch data was obtained?

Hint :

FA0= 2 mol/s, fed to a plug flow reactor

dX r F

1

Thus one needs (1/-rA) as a function of X

0 0

14

0

0

0

2 1

0 0

0

293500

)125(41003

4.02

:

)()()0(3

:

dm mol

s dm s

mol r

dX F

V

PFR

X r X r X

r r

V

A A

A A

A A

A A

To reach 80 % conversion your PFR must be 293 3 dm3

To reach 80 % conversion, your PFR must be 293.3 dm3

Sizing in PFR

Example: Determine the volume in PFR to achieve a 80 % conversion

F dX

r dV

dX F PFR For A ⋅ = − A

8 8

0

:

dX r

F r

dX F

V arranging

A A A

0 0 0

: Re

Let’s numerically evaluate the integral with trapezoidal rule

89.0)

A

r

F X f

(

8 0

A

r

F X f

3

556.34.089.8)0.889

With five point quadrature V = 2.165 m3

Comparing CSTR & PFR Sizing

VCSTR> VPFRfor the same conversion and rxn conditions.

The reason is that CSTR always operates at lowest rxn rate PFR starts at a high y p g

rate, then gradually decreases to the exit rate.

Reactors in Series: The exit of one reactor is fed to the next one.

Given –rAas a function of conversion, one can design any sequence of

reactors

i po to up reacted A

of

moles

reactor first

to fed A of

moles

X i =

Only valid if there are no side streams

i A A

Ai F F X

F = 0− 0⋅

Example: Using Levenspiel plots to calculate conversion from known

reactor volumes

Pure A is fed at a volumetric flow rate 1000 dm 3 /h and at a concentration of 0.005

mol/dm 3 to an existing CSTR, which is connected in series to an existing tubular

reactor If the volume of the CSTR is 1200 dm 3 and the tubular reactor volume is

reactor If the volume of the CSTR is 1200 dm 3 and the tubular reactor volume is

600 dm 3 , what are the intermediate and final conversions that can be achieved

with the existing system? The reciprocal rate is plotted in the figure below as a

function of conversion for the conditions at which the reaction is to be carried out.

By trial and error, we find that a conversion of 0.6 gives the appropriate

CSTR volume of 1200 dm3

Therefore, the intermediate conversion is X = 0.6

Similarly for the PFR, through trial and error, we find that a conversion

of 0.8 gives the appropriate PFR volume of 600 dm3

Therefore, the final conversion is X = 0.8

)(

1 2 2 0 2

2 0 0 1 0 0 2

2 1

r

F r

X F F X F F r

F F V

A A A

A A A

A A

4 0 1

0 1

3 4

0 1

m X

r

F V m r

F X

A A X

A A

2 2

0 2

3 8

0 2

F X

A A X

VCSTR,2> VCSTR,1

0.4 0.8

X

Total V = V1+ V2= 4.02 m3< 6.4 m3 volume necessary to get 80 %

conversion with one CSTR

1 ,

VTOTAL= VPFR,1+ VPFR,2The overall conversion of

0 0

0 0

0

X

A X

A A X

A

A

r

dX F r

dX F r

dX

F

-rA2 FA2

same as one PFR with the same total volume

Reactors in Series: CSTR

Reactors in Series: CSTR –– PFR PFR – – CSTR CSTR

Using the data in the table, calculate the reactor volumes V1, V2and V3

for the CSTR/PFR/CSTR reactors in series sequence along with the

corresponding conversion

X vs r

F of

A

A

r

X F V

at X = X1= 0.4 the (FA0/ -rA1) = 300 dm3

V1= (300 dm3) (0.4) = 120 dm3

The volume of the first CSTR is 120 dm3

(b) Reactor 2: PFR The differential form of the PFR design is

4 0

0 0

2

1

dX r

F dX

r

F V

A A X

X A A

Choose three point quadrature formula with

15.02

4.07.02

−

∆

=

) 7 0 ( )

55 0 (

4 ) 4 0 ( 3

0 0

0 2

A A A

A A

A

r

F r

F r

F X V

Interpreting for (FA0/-rA) at X = 0.55 we obtain

3 0

15.0

dm dm

dm dm

A F r V F

generation out

in

=

⋅ +

A

A A

r

F F V

−

−

=

3 0

3

2 0

2

) 1 (

) 1 (

X F

F

X F

3

2 3 3

0

3

180 ) 4 0 7 0 ( 600

) (

dm dm

V

X X r

CSTR X1= 0.4 V1= 120 dm3

PFR X2= 0 7 V2= 119 dm3

PFR X2 0.7 V2 119 dmCSTR X3= 0.8 V3= 180 dm3

Total volume = 120 + 119 + 180 = 419 dm3

Reactor Sequencing

Is there any differences between having a CSTR – PFR system & PFR –

CSTR system? Which arrangement is best?

The choice of reactors depend on ;

the Levenspiel plots relative reactor sizes.

Space Time

The space time, tau, is obtained by dividing the reactor volume by

the volumetric flow rate entering the reactor:

V

Space time is the time necessary to process one volume of reactor

fluid at the entrance conditions This is the time it takes for the amount

of fluid that takes up the entire volume of the reactor to either

completely enter or completely exit the reactor It is also called holding

time or mean residence time

Example: v0= 0.01 m3/s and V = 0.2 m3 → τ = 0.2 m3/ 0.01 m3/s = 20 s

It would take 20 s for the fluid at the entrance to move to the exit

Typical space time for different reactors:

Batch : 15 min – 20 h (few kg/day – 100,000 tons/year ≈ 280 tons/day)

CSTR : 10 min – 4 h (10 to 3 x 106tons/yr)

Space Velocity (SV)is defined as:

instead of using volumetric flow rate at the entrance, you use liquid –

hourly & gas – hourly space velocities (LHSV, GHSV)

v0(for LHSV) is that of a liquid feed rate at 60°F or 75°F

v0(for GHSV) is that of the one that measured at STP

V

v GHSV V

v

• HW (due date: Feb 25):

Example: Consider cell as a reactor The nutrient corn steep liquor enters the cell of the

microorganism Penicillium chrysogenum and is decomposed to form such products as

amino acids, RNA and DNA Write an unsteady mass balance on (a) the corn steep liquor,

(b) RNA, and (c) pencillin Assume the cell is well mixed and that RNA remains inside the

Mass balance for penicillin:

dN F dV r dV

r G

flow in penicilin no

F

dt

dN G F F

on Accumulati Generation

Out In

p V

V in

p p out in

= +

−

∫

∫

) _ (

0

dt F dV r dV

r

G p=∫ p⋅ ⇒ ∫ p⋅ out =Assuming steady state for the rate of production of penicilin in the cells

stationary state,

p out p

out in p

r

F V r

F F V dt dN

C C

r

F r

F F V

For irreversible reactions, the maximum value of conversion, X, is

that for complete conversion, i.e X = 1.0

For reversible reactions, the maximum value of conversion, X, is the

equilibrium conversion, i.e X = X