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CHBI 502
REACTION ENGINEERING
CHBI 502• HWs will be distributed after the chapters are covered, deadlines will be
posted– NO late submission
• 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)
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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 alland 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).
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• 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 Reactions One 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 EnginneringElements of Chemical Rxn EnginneringChemical kinetics is the study of chemical rxn rates and reaction Chemical kinetics is the study of chemical rxn rates and reaction mechanisms.mechanisms.Chemical reaction engineering (CRE) combines the study of chemical Chemical reaction engineering (CRE) combines the study of chemical kinetics with the reactors in which the reactions occur.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 carrying out desirable chemical reactions (what size and what type of equipment)(what size and what type of equipment)
Chemical Kinetics & Reactor DesignChemical Kinetics & Reactor Design
The reaction system thet operates in the safest and most efficient manner The reaction system thet operates in the safest and most efficient manner can be the key to the success of the plant.can be the key to the success of the plant.ca be t e ey to t e success o t e p a tca be t e ey to t e success o t e p a t
Modelling of;Modelling of;Chemical plantChemical plantPharmacokineticsPharmacokineticsMicroelectronicsMicroelectronicsDigestive system of an animalDigestive system of an animal
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Chapter1 MOLE BALANCESChapter1 MOLE BALANCES
1. Chemical Identity
A chemical species is said to have reacted when it has lost its h i l id tit Th id tit f h i l i i d t i dchemical 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:
1. Decomposition CH3CH3 → H2 + H2C = CH2
2 C bi ti N O 2NO2. Combination N2 + O2 → 2NO
3. Isomerization C2H5CH = CH2 → CH2 = C(CH3)2
2. Reaction Rate:
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 asidentity 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
rA = the rate of formation of species A per unit volume A p p
-rA = the rate of disappearance of species A per unit volume
rB = the rate of formation of species B per unit volume
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Example: A → B If B is being created 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
For catalytic reaction, we refer to –rA’, which is the rate of disappearance ofspecies 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 → dCA/dt = 0A
The rate law does not depend on reactor type!
-rA is the # of moles of A reacting (disappearing) per unit time per unit volume (mol/dm3 s)
In a reactor, two extreme conditions are considered:1. No mixing of streams2. Complete mixing (desirable)
Classification of reactions• 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)change with time (t)
FA0
FA
Batch CSTR)
Stirrer, rpm and design are important
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Thus,• Batch or continuous• Tank or tubular• Homogeneous or heterogeneous
Consider species j:
rj is the rate of formation of species j per unit volume [e.g. mol/dm3 s]
rj is a function of concentration, temperature, pressure, and the type of catalyst (if any)
rj is independent of the type of reaction system (batch, plug flow, etc.)
rj is 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 occurs [e.g. –rA = k(T) CA
2]
For example, the algebraic form of the rate law for –rA for A (products) may be;
AA Ckr ⋅=−a linear function of concentrations:
or can be determined by experiments:A
AA Ck
Ckr⋅+
⋅=−
2
1
1
2AA Ckr ⋅=−or, it may be some other algebraic function of conc’n as:
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3. General Mole Balance Equation:
IN – OUT + GENERATION = ACCUMULATION
]/[0 timemolesd
dNdVrFFV
AAAA ∫ =⋅+−
NA: # of moles of species A in][
00 dtAAA ∫ of species A in
the system at time t.
GA = rA V (if all system variables (T, CA, etc.) are spatially uniform throughout system volume)
[ ]volumevolumetime
molestime
moles
VrG AA
⋅⎥⎦⎤
⎢⎣⎡
⋅=⎥⎦
⎤⎢⎣⎡
⋅=
If the rate of formation of A varies with position:
∆V1
∆V2r1A1A
r2A
subvolumesMforVrGG
subvolumesallforetcVrGm
iiiA
m
iiAA
AA
∆⋅=∆=
∆⋅=∆
∑∑ 11
111 .
Basic equation for any species A entering, leaving, reacting ...dt
dNdVrFF
dVrG
AV
AAA
AA
ii
=⋅+−
⋅=
∫
∫
∑∑==
00
11
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Mole Balance on Different Reactor TypesMole 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.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
==
VdN
FF
A
AA 000
If perfect mixing
∫ ⋅=
⋅=
dVrdt
dN
Vrdt
jj
AA If perfect mixing
(no volume change throughout volume)
The # of moles changing (in A → B) is as follows:
NA0
NA1
NA NB
NB1
t1 t
NA1
tt1
⋅= AA Vr
dtdN
What time is the necessary to produce NA1
starting from NA0?
∫∫∫ ⋅=⇒
⋅=
⋅=
1
0
1
0
1
10
A
A
A
A
N
N A
AN
N A
At
A
A
VrdNt
VrdNdt
VrdNdt
dt
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Continuous Flow Reactors (CFR) operate at steady state.
Continuous Stirred Tank Reactor (CSTR)
Plug Flow Reactor (PFR)
Packed Bed Reactor (PBR)
CSTR
FA0
FA
usually used for liquid-phase rxns
usually operated at steady state
usually assumed to be perfectly mixed
T ≠ f(t,V)
General Mole Balance on System Volume V
IN – OUT + GENERATION = ACCUMULATION
∫ =⋅+−V
AAAA dt
dNdVrFF0
0
V
FA0
FA
Assumptions
AAA
AA
A
VrFF
VrdVr
dtdN
=⋅+−
⋅=⋅
=
∫0 0
0Steady State
Well mixed
A
AA
A
AA
rvCvCV
rFFV
−⋅−⋅
=
−−
=
00
0Design eq’n for CSTR
vCF AA ⋅= 00Molar flow rate concentration
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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)
∫ =⋅+−V
AAAA dt
dNdVrFF0
0
IN – OUT + GENERATION = ACCUMULATION
Reactants Products
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∆⋅=⋅=∆∆
∫ VrdVrGV
AAA
∆GAFA FjFA0
V V + ∆V
0=∆⋅+−
∆∆
∆+
∫VrFF
VrdVrG
AVVAVA
AAA
Divide by ∆V and rearrange:A
VAVVA rV
FF=
⎥⎥⎦
⎤
⎢⎢⎣
⎡
∆
−∆+
Taking the limit as ∆V→0:
∫=
=
A
A
F
F A
A
AA
rdFV
rdVdF
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)
General Balance on WGeneral Balance on W
∫ =⋅+−V
AAAA dt
dNdWrFF0
0 '
IN – OUT + GENERATION = ACCUMULATION
∆W∆W
W+∆WW
FA(W+∆W)FA(W)
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∫ =⋅+−V
AAAA
dNdVrFF 0 '
0=dt
dN A
Steady State
∫ AAA dt00
Differentiate with respect to W and rearrange
'AA r
dWdF
=
When pressure drop through the reactor and catalyst decay areWhen pressure drop through the reactor and catalyst decay are neglected, the integral eg’n can be used to find W:
∫∫ −==
0
0''
A
A
A
A
F
F A
AF
F A
A
rdF
rdFW
Batch Reactor Times
A → B
Calculate 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
-rA’ = k CA when k = 0.046 min-1rA k CA when k 0.046 min
AA
lawRatedt
dNVr
onAccumulatiGenerationOutInbalanceMol
=⋅+−
=+−
:
00
:
AA
AA
AAAAA
Nkdt
dNNkVr
VNkCkrCkr
lawRate
⋅−=⇒⋅−=⋅
⎟⎠⎞
⎜⎝⎛⋅−=⋅−=⇒⋅=−
:
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:
00⋅−
−=⋅
= ∫∫ VrdN
VrdNt
SolveN
N A
AN
N A
AA
A
A
A
ln1 00
⋅=⋅
=
⋅=⋅⋅=⋅⋅=⋅−
∫ NN
kNkdNt
NkVVNkVCkVr
A
AN
N A
A
AA
AA
A
A
)10ln(min046.0
110
0 ⋅=⇒= tNN AA
Therefore, t = 50 minutes
Summary of Chapter 1• Define the rate of chemical reaction. • Apply the mole balance equations to a batch reactor, CSTR, PFR,
and PBR. and PBR.
• Batch reactor: no in-out streams, no spatial variations in conc’n
• CSTR: no spatial variations in the tank, steady state
• PFR: spatial variations along the reactor steady statePFR: 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
dNdVrFF AV
AAA =⋅+− ∫0
0
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Reactor Differential Algebraic Integral
Batch
CSTR
PFR
PBR
general reaction, A->B
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Compartments for perfusion
StomachVG = 2.4 l
GastrointestinalVG = 2.4 ltG = 2.67 min
Liver
Alcohol
Perfusion interactions between compartments are shown by arrows.
VL = 2.4 ltL = 2.4 min
CentralVC = 15.3 ltC = 0.9 min
VG, VL, VC, and VM are -tissue water volumes for the gastrointestinal, liver, central and muscle compartments, respectively.
VS is the stomach contents volume.
Muscle & FatVM = 22.0 ltM = 27 min
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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
1
• Objectives:
CONVERSION AND REACTOR SIZINGCONVERSION 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 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)
2
CONVERSION AND REACTOR SIZINGCONVERSION AND REACTOR SIZING1. Conversion
Consider the general equation
dDCbBA dDcCbBaA +→+
We will choose A as our basis of calculation.
DadC
acB
abA +→+
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.
00 )()( AAAA FFXNNX −=
−=
Batch Flow
00 AA FX
NX ==
X = Moles of A reacted
Moles of A fed
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 = Xe.
3
Batch Reactor Design Equations:Batch Reactor Design Equations:
⎤⎡⎤⎡⎥⎤
⎢⎡ reactedAofMolesAofMolesAofMoles
2.2. Design EquationsDesign Equations
⎥⎦
⎤⎢⎣
⎡⋅⎥⎦⎤
⎢⎣⎡=
⎥⎥
⎦⎢⎢
⎣fedAofMoles
reactedAofMolesfed
AofMoles
consumedreacted
)(
[ ]0AN= [ ]X⋅ [1]
Now the # of moles of A that remain in the reactor after a time t, NA can be expressed in terms of NA0 and X;expressed in terms of NA0 and X;
[ ] [ ] [ ])1(0
00
XNN
XNNN
AA
AAA
−⋅=
⋅−=[2]
Vrdt
dN
mixingprefectVrdt
dN
AA
AA
⋅−=−
⋅= )( [3]
For batch reactors, we are interested in determining how long to leave the reactants in the reactor to achieve a certain conversion X.
dXdtdXN
dtdN
AA ⋅−= 00 (Since NA0 is constant) [4]
VrdtdXN
VrdtdXN
AA
AA
⋅−=⋅
⋅=⋅−
0
0
Batch reactor design eq’n (in differential form)
[5]
4
For a constant volume batch reactor: (V = V0)
==⋅ AAA
dtdC
dtVNd
dtdN
V0
0
)/(1 From [3]
∫⋅=
⋅−⋅=
=
X
A
AA
AA
VrdXNt
VrdXNdt
rdt
dC
0
0From [5]
Constant volume batch reactor
Batch time, t, required to achieve a ∫ ⋅− A Vr0 conversion X.
X
t
As t X
Flow Reactor Design Equations:Flow Reactor Design Equations:
reactedAofmolesfedAofmoles
For continuous-flow systems, time usually increases with increasing reactor volume.
AAA
A
FXFF
fedAofmolesreactedAofmoles
timefedAofmolesXF
=⋅−
⋅=⋅
00
0
inlet molar flow rate
Molar flow rate at which A is consumed within the system
Outlet flow rate
000
0 )1(vCF
XFF
AA
AA
⋅=−⋅=
moles /volume volume / time (volumetric flow rate, dm3/s)
5
For liquid systems, CA0 is usually given in terms of molarity (mol/dm3)
For gas systems, CA0 can be calculated using gas laws.
PP
Partial pressure
0
00
0
00 TR
PyTR
PC AAA ⋅
⋅=
⋅=
Py ⋅
Entering molar flow rate is
yA0 = entering mole fraction of AP t i t t l (kP )
0
000000 TR
PyvCvF AAA ⋅
⋅=⋅= P0 = entering total pressure (kPa)CA0 = entering conc’n (mol/dm3) R = 8.314 kPa dm3 / mol KT = T(K)
CSTR (Design Equation)CSTR (Design Equation)
DadC
acB
abA +→+For a rxn:
FF
A
AA
rFFV
−−
= 0
Substitute for FA
AAA
AAA
XFFFV
XFFF)( 000
00
⋅−−=
⋅−=
exitA
A
A
rXFV
r
)(0
−⋅
=
−
6
PFR (Design Equation)PFR (Design Equation)
rdVdF
AA −=−
dXFdFXFFF
AA
AAA
⋅−=⋅−=
0
00
Substitute back:
AAA r
dVdXF
dVdF
−=⋅=− 0
Seperate the variables V = 0 when X = 0
∫ −⋅=X
AA r
dXFV0
0
Applications of Design Equations for Applications of Design Equations for Continuous Flow ReactorsContinuous Flow Reactors
3.3. Reactor SizingReactor Sizing
Given –rA as 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 / -rA or 1 / -rA as a function of X. For FA0 / -rA vs. 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:
Levenspiel PlotsLevenspiel Plots
7
A particularly simple functional dependence is the first order dependence:
)1(0 XCkCkr AAA −⋅⋅=⋅=−
Specific rxn rate (function of T) initial conc’n( )
For this first order rxn, a plot of 1/-rA as a function of X yields :
⎟⎠⎞
⎜⎝⎛−
⋅⋅
=−XCkr AA 1
111
0
-1/rA
X
Example:Example: Let’s consider the isothermal gas-phase isomerization:
A → B
X -rA(mol/m3s)0 0 450 0.45
0.1 0.370.2 0.300.4 0.1950.60.7
0.1130.079
0 8 0 050.8 0.05
[T = 500 K]
[P = 830 kPa = 8.2 atm]
initial charge was pure A
8
Example:Example: Let’s consider the isothermal gas-phase isomerization:
A → B
X -rA(mol/m3s) 1 / -rA
0 0 45 2 220 0.45 2.220.1 0.37 2.700.2 0.30 3.330.4 0.195 5.130.60.7
0.1130.079
8.8512.7
0 8 0 05 20 00.8 0.05 20.0
[T = 500 K]
[P = 830 kPa = 8.2 atm]
initial charge was pure A
-1/rA
Draw Draw --1/r1/rAA vs X:vs X:We can use this figure to size flow reactors for
different entering molar flow rates.
Keep in mind :
1. if a rxn is carried out isothermally, the rate is
X
y,usually greatest at the start of the rxn, when the conc’n of reactant is greatest. (when x ≈ 0 → -1/rA is 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.
9
if FA0 = 0.4 mol/s, we can calculate [FA/-rA](m3)
Plot FA0/-rA vs X obtain Levenspiel Plot!
Example: Calculate volume to achieve 80 % conversion in CSTR.
3
8.0Xatrfind A
⎞⎛
=−
33
0
3
8.0
46802040
)(
201
mmmol
rXFV
smolm
r
exitA
A
A
=⋅⋅=
−⋅
=
⋅=⎟⎟
⎠
⎞⎜⎜⎝
⎛−
4.68.0204.0 msmols ⋅
1.5 m3
3.6 m3 CSTRs are usually used for liquid-phase rxns.
For instance:
4.4. Numerical Evaluation of IntergralsNumerical 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
10
Trapezoidal RuleTrapezoidal Rule
f(x1)
f(x)
2)]()([
)(
)]()([1)(
012
01
10
1
0
hxfxfA
hxfA
xfxfh
dxxfx
x
⋅−=
⋅=
+=⋅∫
A1
A2
x0 x1x
h
f(x0)
)]()([2
2)(
2)()(
2
10
010
21
2
xfxfh
xfxfxfh
AAA
+⋅=
⎥⎦⎤
⎢⎣⎡ −+⋅=
+=
Five Point Quadrature formula:Five Point Quadrature formula:
.....)233233(83)(
4)424(
3)(
6543210
0443210
0
4
0
fffffffhdxxf
xxhwherefffffhdxxf
Nx
x
x
x
++++++⋅=⋅
−=++++⋅=⋅
∫
∫For N+1 points, where N is an integer.
Example:Example: Consider the liquid phase reaction;
A → Products
which is to take place in a PFR. The following data was obtained in a batch reactor.
X 0 0.4 0.8-rA(mol/dm3s) 0.01 0.008 0.002
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?
11
Hint :FA0 = 2 mol/s, fed to a plug flow reactor
dXr
FVPFRX
AA ∫ −=∴
00
1Thus one needs (1/-rA) as a function of X.
00141: XFdXFVPFR A
X
A ⎥⎤
⎢⎡
++∆⋅== ∫
For Simpson’s three point formula we have:
[ ] 338.0
00
210
00
293500)125(410034.02:
)()()0(3:
dmmol
sdms
molr
dXFVPFR
XrXrXrrV
AA
AAAA
AA
=⎭⎬⎫
⎩⎨⎧
+⋅+⋅=−
=
⎥⎦
⎢⎣ −−=−−
∫
∫
To reach 80 % conversion your PFR must be 293 3 dm3To reach 80 % conversion, your PFR must be 293.3 dm3.
12
Sizing in PFRSizing in PFR
Example: Determine the volume in PFR to achieve a 80 % conversion.
FdX
rdVdXFPFRFor AA −=⋅
8080
0:
dXr
Fr
dXFVarrangingA
A
AA ⋅
−=
−⋅= ∫∫
8.0
0
08.0
00:Re
Let’s numerically evaluate the integral with trapezoidal rule
89.0)(0
0 =−
==XA
A
rFXf∫ ⇒⋅
−
8.0
0
0 dXr
F
A
A
0.8)(8.0
0 =−
==XA
A
rFXf
3556.34.089.8)0.889.0(28.0 mV =⋅=+⋅=
With five point quadrature V = 2.165 m3
Comparing CSTR & PFR SizingComparing CSTR & PFR Sizing
VCSTR > VPFR for the same conversion and rxn conditions.
The reason is that CSTR always operates at lowest rxn rate. PFR starts at a high y p grate, then gradually decreases to the exit rate.
13
Reactors in Series: The exit of one reactor is fed to the next one.
Given –rA as a function of conversion, one can design any sequence of reactors.
ipotoupreactedAofmolesX int=
reactorfirsttofedAofmolesX i =
Only valid if there are no side streams.
iAAAi XFFF ⋅−= 00
Example: Using Levenspiel plots to calculate conversion from known reactor volumes.
Pure A is fed at a volumetric flow rate 1000 dm3/h and at a concentration of 0.005 mol/dm3 to an existing CSTR, which is connected in series to an existing tubular reactor If the volume of the CSTR is 1200 dm3 and the tubular reactor volume isreactor. If the volume of the CSTR is 1200 dm3 and the tubular reactor volume is 600 dm3, 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.
14
Solution:
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
15
CSTRs in SeriesCSTRs in Series
Two CSTRs in series
FA0
Reactor 1:Reactor 1:
Mole Balance: FA0 – FA1 + rA1 V1 = 0 [1]
FA1 = FA0 – FA0 X1 [2]
Combining [1] & [2]:
-rA2
-rA1
FA1X1 = 0.4
FA2X1 = 0.8
Combining [1] & [2]:
V1 = FA0 (1 / -rA1) X1 [3]
Reactor 2:Reactor 2:
Mole Balance: FA1 – FA2 + rA2 V2 = 0 [4]
FA2 = FA0 – FA0 X2 [5]FA2 FA0 FA0 X2 [5]
Combining [4] & [5]:
)()()(12
2
0
2
200100
2
212 XX
rF
rXFFXFF
rFFV
A
A
A
AAAA
A
AA −−
=−
⋅−−⋅−=
−−
=
if we have the data:
31
4.01
01
3
4.01
0 82.005.24.0
FF
mXr
FVmr
FXA
A
XA
A
⎞⎜⎛⎞
⎜⎛
=⋅⎟⎟⎠
⎞⎜⎜⎝
⎛−
==⎟⎟⎠
⎞⎜⎜⎝
⎛−
==
312
2
02
3
8.02
0 2.3)(0.88.0 mXXr
FVmr
FXA
A
XA
A =−⋅⎟⎠
⎞⎜⎜⎝
⎛−
==⎟⎠
⎞⎜⎜⎝
⎛−
==
-FA0/rA
VCSTR,2 > VCSTR,1
0.4 0.8X
Total V = V1 + V2 = 4.02 m3 < 6.4 m3 volume necessary to get 80 % conversion with one CSTR
16
1 2 3 4
One can approximate a PFR by a large # of CSTRs in series:-FA0/rA
XX
FA0
-rA1
FA1X1 =0.4
PFRs in SeriesPFRs in Series∑=
=n
iiCSTRPFR VV
1,
VTOTAL = VPFR,1 + VPFR,2
The overall conversion of
∫∫∫ −⋅+
−⋅=
−⋅
2
1
12
00
00
0
X
X AA
X
AA
X
AA r
dXFr
dXFr
dXF
-rA2 FA2X1 = 0.8 two PFRs in series is the
same as one PFR with the same total volume
Reactors in Series: CSTR Reactors in Series: CSTR –– PFR PFR –– CSTRCSTR
Using the data in the table, calculate the reactor volumes V1, V2 and V3for the CSTR/PFR/CSTR reactors in series sequence along with the corresponding conversion.
17
Xvsr
FofplottheUseA
A .0⎟⎟⎠
⎞⎜⎜⎝
⎛−
(a) The CSTR design equation for Reactor 1 is:
⎞⎜⎛ ⋅0A XF
⎟⎠
⎞⎜⎜⎝
⎛−
=1
01
A
A
rXFV
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
0A
A
Fr
dVdX −
=
Rearranging and intergrating with limits
when V = 0 X = X1 = 0.4
when V = V2 X = X2 = 0.7
⎞⎛⎞⎛ 70X
∫∫ ⋅⎟⎟⎠
⎞⎜⎜⎝
⎛−
=⋅⎟⎟⎠
⎞⎜⎜⎝
⎛−
=7.0
4.0
002
1
dXr
FdXr
FVA
AX
X A
A
18
Choose three point quadrature formula with
15.02
4.07.02
12 =−
=−
=∆XXX
⎤⎡∆ 4 FFFX⎥⎦
⎤⎢⎣
⎡−
+−
⋅+
−∆
=)7.0()55.0(
4)4.0(3
0002
A
A
A
A
A
A
rF
rF
rFXV
Interpreting for (FA0/-rA) at X = 0.55 we obtain
30 370dmFA =⎞
⎜⎜⎛
55.0
370dmr XA
=⎠
⎜⎜⎝ − =
[ ] 33332 119600)370(4300
315.0 dmdmdmdmV =+⋅+=
The volume of the PFR is V2 = 119 dm3
(c) Reactor 3: CSTR
Balance
3332 0AAA VrFF
generationoutin
=⋅+−
+−
3
323
A
AA
rFFV
−−
=
303
202
)1(
)1(
XFF
XFF
AA
AA
−⋅=
−⋅=
333
233
03
180)4.07.0(600
)(
dmdmV
XXr
FVA
A
=−⋅=
−⋅−
=
The volume of last CSTR is 180 dm3
19
Summary:
CSTR X1 = 0.4 V1 = 120 dm3
PFR X2 = 0 7 V2 = 119 dm3PFR X2 0.7 V2 119 dm
CSTR 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?
or
V1 V2 V3 V4
The volumes are different!
V1 + V2 =? V3 + V4
For isothermal rxns
FA0
XAdiabatic rxns
The choice of reactors depend on ;
the Levenspiel plots
relative reactor sizes.
20
Space Time 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
0vV
=τ
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 106 tons/yr)
Tubular: 0 5s – 1h (50 to 5 x 106 tons/yr)Tubular: 0.5s 1h (50 to 5 x 10 tons/yr)
21
Space Velocity (SV) is defined as:
τ10 ==
VvSV
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.
Vv
GHSVV
vLHSV STPliq 00
==
22
• HW (due date: Feb 25):S l th bl i• Solve the prpblem in your own way.
• http://www.engin.umich.edu/~cre/web_mod/hippo/index.htm
• Suggested problems from the web:htt // i i h d / /• http://www.engin.umich.edu/~cre/
• Additional Homework Problems at each chapter
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 cell.
In Out
Corn Steep Liquor Penicillin
Penicillium chrysogenum
Assumption:
Penicillin is produced in the stationary state.
→ no cell growth & nutrients are used in making the product.
23
Mass balance for penicillin:
dNFdVrdVrG
flowinpenicilinnoFdt
dNGFF
onAccumulatiGenerationOutIn
pVV
in
ppoutin
=−⋅⇒⋅=
=
=++
=+−
∫∫
)_(0
dtFdVrdVrG outppp =⋅⇒⋅= ∫∫
Assuming steady state for the rate of production of penicilin in the cells stationary state,
p
out
p
outin
p
rFV
rFFV
dtdN
=⇒−−
=
= 0
Similarity, for Corn Steep Liquor with FC = 0
C
C
C
CC
rF
rFFV
−=
−−
= 00
No RNA is generated or destroyed.
Summary
00 )()( AAAA FFXNNX −=
−=
Batch Flow
00 AA FX
NX ==
X = Moles of A reacted
Moles of A fed
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 = Xe.
24
VrdtdXN AA ⋅−=⋅0
Batch reactor design eq’n (in differential form)
CSTRA XFV 0 ⋅ CSTR
PFR
exitArV
)(0
−=
∫ −⋅=X
AA r
dXFV0
0
RReactors in series
0vV
=τSpace time
1
Ch 3. Rate Laws and Stoichiometry
How do we obtain –rA = f(X)?
We do this in two steps
1. Rate Law – Find the rate as a function of concentration,
–rA = k fn (CA, CB …)
2. Stoichiometry – Find the concentration as a function of conversion
CA = g(X)
Part 1: Rate LawsBasic Definitions:A homogenous rxn is the one that involves only one phase.
A heterogeneous rxn involves more than one phase, rxn occurs at the interface
between the phases, i.e., solid gas phase.
An irreversible rxn proceeds in only one direction A → B
A reversible rxn can proceed in both directions A ↔ B (equilibrium)
Molecularity of a rxn is the number of atoms, ions, or molecules involved in a rxn
step.
The terms unimolecular, bimolecular, and termomolecular refer to rxns involving,
respectively, one, two or three atoms interacting in any one rxn step.
Unimolecular: radioactive decay
Bimolecular: rxns with free radicals Br. + X → XBr + Y.Termolecular: rxn pathway following a series of bimolecular rxns.
2
A rate law describes the behavior of a reaction. The rate of a reaction is afunction of temperature (through the rate constant) and concentration.
1. Relative Rates of Reaction
aA + bB → cC + dD
Rate of formation of C = c/a (Rate of disappearance of A)
The reaction :
2A + 3B → 5C
is carried out in a reactor. If at a particular point, the rate of disappearance of A is 10 mol/dm3s, what are the rates of B and C?
The rate of disappearance of A, -rA, is
or the rate of formation of species A is
The relative rates are
3
Species B
The rate of formation of species B (2A + 3B → 5C)
The rate of disappearance of B, -rb, is
Species C
The rate of formation of species C, rc, is
Rxn Order & Rate Law:
Algebraic equation that relates –rA to the concentrations of the reactants is called the “kinetic expression” or “rate law”.
Usually, rate can be written as the product of rxn rate constant and concentrations. In these expressions, usually the limiting reactant is chosen as basis for calculations.
-rA = kA(T) fn (CA, CB,…..)
In reality activities should be used
rA = -kA’ aAα aB
β
(ai = γi Ci)
rA = (-kA’ γAα γB
β) CAα CB
β
kA
4
Power Law and Elementary Rate LawsIn general
-rA = kA CAα CB
β
α : order in Aβ : order in Bn = α + β = overall rxn orderThe unit of –rA is always = concentration / timeFor a rxn with “n” order: {k} = (concentration)1-n / timeTherefore for a zero-, first-, second-, and third-order rxn
123
2
32
3
}{;)3(3
}{;)2(2
1}{;)1(1
}{;)0(
−⋅⎟⎟⎠
⎞⎜⎜⎝
⎛==−=
⋅==−=
==−=
==−=−
smoldmkCCkrnorder
smoldmkCkrnorder
skCkrnorder
sdmmolkkrnorderzero
BAAArd
AAAnd
AAAst
AA
Elementary Reactions
A reaction follows an elementary rate law if and only if the stoichiometric coefficients are the same as the individual reaction order of each species. For the reaction in the previous example (A + B → C + D), the rate law would be: -rA = k CA CB These rate laws can be derived from Collision Theory.
if 2NO + O2 → 2NO2 then –rNO = kNO (CNO)2 CO2 if elementary!!!
Question
What is the reaction rate law for the reaction A + ½ B → C if the reaction is elementary? What is rB? What is rC? Calculate the rates of A, B, and C in a CSTR where the concentrations are CA = 1.5 mol/dm3, CB = 9 mol/dm3 and kA = 2 (dm3/mol)(½) (1/s)
5
Solution:
Let’s calculate the rate if,
A + ½ B → C
Then,
For A + B ↔ C + D
⎥⎦
⎤⎢⎣
⎡ ⋅−⋅⋅=−
C
DCBAA K
CCCCkr Elementary
6
Non Elementary Rate Laws:
A large number of homogeneous or heterogeneous rxns do not follow simple rate laws.
If the rate law for the non-elementary reaction A + B → 2C + D is found to be –rA = k CA CB
2, then the rxn is said to be 2nd order in A and 1st order in B, and 3rd order overall.
For the homogeneous rxn
2
2
2
2
'122 222
2/322
O
ONON
ClCOCO
CkCk
rONON
CCkrCOClClCO
⋅+
⋅=−+→
⋅⋅=−→+
k and k’ strongly temperature dependent
In this case we can not state overall rxn order.
Here, we can speak of reaction orders under certain limiting conditionsas at very low conc’n of O2 (1 >> k’ CO2
)
2
22
2 ' O
ONONON Ck
Ckr
⋅
⋅=−
1st order (apparent rxn order)ONONON Ckr222
⋅=−
If 1 << k’ CO2–rN2O
0th order (apparently)
These types are very common for liquid & gas phase rxns on solidcatalysts
7
For the heterogeneous rxn
4662356 CHHCHCHHC catalyst +⎯⎯ →⎯+(toluene: T) (benzene: B) (methane: M)
In these types of rxns, partial pressures are used instead of conc’ns:
TTBB
THT PKPK
PPkr
⋅+⋅+
⋅⋅=−1
' 2
Per mass of catalystAdsorption constant{KT} = 1 / atm
Specific rxn rate {k} = molT / kg cat s atm2
if ideal gas law is applied → Pi = Ci R T
Example: Gas Phase Catalytic Reactions
When studying gas phase catalytic reactions the rate law is developedin terms of partial pressure,
To rewrite the rate law, just use ideal gas law to relate to concentrationsCA and CB
and then write concentration in terms of conversion
8
The net rate of formation of any species is equal to its rate of formation in the forward reaction plus its rate of formation in the reverse reaction: ratenet = rateforward + ratereverse
At equilibrium, ratenet ≈ 0 and the rate law must reduce to an equation that is thermodynamically consistent with the equilibrium constant for the reaction.
Example: Consider the exothermic, heterogeneous reaction; A + B → C
At low temperature, the rate law for the disappearance of A is
Recall PA = CA R T
At high temperature, the exothermic reaction is significantly reversible:
What is the corresponding rate law?
If the rate of formation of A for the forward reaction (A + B → C) is
then we need to assume a form of the rate law for the reverse reaction that satisfies the equilibrium condition. If we assume the rate law for the reverse reaction (C → A + B) is
Then: and:
9
Deriving –rA:
The forward rate is:
And the reverse rate law is:
The net rate for species A is the sum of the forward and reverse rate laws:
Substituting for rfor and rrev:
Solving for Kp:
Does this rate law satisfy our requirement at equilibrium.
For a rxn at equilibrium
At equilibrium, rnet ≈ 0, so;
Solving for Kp:
The conditions are satisfied.
10
Rate Law for Reversible Reactions
Example: Write the rate law for the elementary reaction
Here kfA and krA are the forward and reverse specific reaction rates bothdefined with respect to A.
(1)
(2)
At equilibrium
The equilibrium constant decreases as T increases (exothermic rxns)
Ke increases as T increases (endothermic rxns)
11
Reaction Rate Constant, k:
k is the specific rxn rate constant given by Arrhenius Equation:
k = A e–E/RT
where,
E: activation energy (cal/mol)
R: gas constant (cal/mol K)
T: temperature (K)
A: frequency factor (unit depends on rxn order)
T
kT → ∞ k → A
T → 0 k → 0
A ≈ 1013
EA (Activation Energy)
Molecules need some energy to distort/stretch their bonds so that they break them to form new bonds.
The steric & repulsion forces must be overcome as the reacting molecules come close together.
A A B C
2a b+c
+ A AB C
a+b a+c
12
A rxn coordinate is usually used:
A + BC ↔ A – B – C → AB + C
High energy complexEA-B-C
Energy Barrier
EAB + EC
Rxn Coordinate
EA + EBC
The rxn coordinate denotes the energy of the system as a function of progress along the reaction path.
More on Activation Energy
The activation energy can be thought of as a barrier to the reaction. One way to view the barrier to a reaction is through the reaction coordinates. These coordinates denote the energy of the system as a function of progress along the reaction path. For the reaction
the reaction coordinate is
for the reaction to occur, the reactants must overcome an energy barrier or activation energy EA.
13
Why is there an Activation Energy?
(1) the molecules need energy to distort or stretch their bonds in order to break them and to thus form new bonds
(2) as the reacting molecules come close together they must overcome both steric and electron repulsion forces in order to react
Energy Distribution of reacting molecules
In our development of collision theory we assumed all molecules had the same average energy. However, all the molecules don’t have the same energy, rather there is distribution of energies where some molecules have more energy than others. The distraction function f(E,T) describes this distribution of the energies of the molecules. The distribution function is read in conjunction with dE
f(E, T) dE = fraction of molecules with energies between E and E + dE
One such distribution of energies in the following figure
14
Increase Temperature
By increasing the temperature we increase the kinetic energy of the reactant molecules which can in turn be transferred to internal energy to increase the stretching and bending of the bonds causing them to be in an activated state, vulnerable to bond breaking and reaction.
As the temperature is increased we have greater number of molecules have energies EA and hence the reaction rate will be greater.
15
Activation EnergyThe activation energy is a measure of how temperature sensitive the reaction is. Reactions with large activation energies are very temperature sensitive.
If you know two points → EA is known!
REslope
T
kslope
−=
∆
∆=
)1(
ln
One can also write the specific reaction rate as:
16
Specific Reaction Rate Derivation
Taking the ratio:
This says if we know the specific rxn rate ko(To) at a temperature To, and we know E, then we can find reaction rate k(T) at any other temperature T for that rxn.
STOICHIOMETRY
We shall set up Stoichiometric Tables using A as our basis of calculation in the following reaction. We will use the stoichiometric tables to express the concentration as a function of conversion. We will combine Ci = f(X) with the appropriate rate law to obtain -rA = f(X).
17
where;and
1. Batch System Stoichiometric Reactor
Concentration -- Batch System:
Constant Volume Batch:Note:
if the reaction occurs in the liquid phase or
if a gas phase reaction occurs in a rigid (e.g., steel) batch reactorThen
if then
and we have -ra=f(x)
18
Example:
Write the rate law for the elementary liquid phase reaction
3A + 2B → 4C
solely in terms of conversion. The feed to the batch reactor is equal molar A and B with CA0 = 2 mol/dm3 and kA= .01 (dm3/mol)41/s.
1) Rate Law: -rA=kCA3CB
2
2) Stoichiometry:
Species A
Liquid phase, v = vo (no volume change)
Species B
What is ΘB?
Species A is the limiting reactant because the feed is equal molar in A and B, and two moles of B consumes 3 moles of A.
We now have -rA=f(X) and can size reactors or determine batch reaction times.
19
2. Flow System Stoichiometric Table
Where: and
Concentration -- Flow System:
Liquid Phase Flow System:
etc.
If the rate of reaction were -rA = kCACB then we would have
This gives us -rA = f(X). Consequently, we can use the methods discussed in Chapter 2 to size a large number of reactors, either alone or in series.
20
For Gas Phase Flow Systems: (Variable Volumetric Flow Rate)
Combining the compressibility factor equation of state with Z = Z0
with FT=CTV FTo=CToVo
we obtain
The total molar flow rate
Substituting for FT gives:
yA0 δ = εholds for both flow & batch reactors
Example: Calculate ε
For the gas phase reaction
2A + B C
the feed is equal molar in A and B. Calculate ε.
SolutionA is the limiting reactant
21
Gas Phase Flow Systems
etc.Again, these equations give us information about -rA = f(X), which we can use to size reactors. For example if the gas phase reaction has the rate law
then
with
The molar flow rate of A:
FA = FA0 + νA (FA0 X)
Stoichiometric coefficient
FA = FA0 (Θ + νA X)
22
Calculating the equilibrium conversion for gas phase reaction
Consider the following elementary reaction with KC and = 20 dm3/mol and CA0 = 0.2 mol/dm3. Pure A is fed. Calculate the equilibrium conversion, Xe, for both a batch reactor and a flow reactor. Assume constant volume.
Solution
At equilibrium
Stoichiometry
Constant Volume V = V0
23
Solving
24
Stoichiometric Table - Conversion
Let’s consider the production of ethyl benzene
The gas feed consists of 25% toluene and 75% ethylene. Set up a stoichiometric table to determine the concentrations of each of the reacting species and then to write the rate of reaction solely as a function of conversion. Assume the reaction is elementary with kT=250(dm6/mol2s). The entering pressure is 8.2 atm and the entering temperature is 227°C and the reaction takes place isothermally with no pressure drop.
25
Basis of Calculation
The stoichiometric ratio is one toluene to two ethylene (1/2). However, the feed is one toluene to three ethylene (1/3) and there is not sufficient toluene to consume all the ethylene. Therefore toluene is the limiting reactant and thus the basis of calculation.
Entering Concentrations of ethylene and toluene
Let A = toluene, B = ethylene, C = ethyl benzene and D = propylene
Ethylene
Since toluene, i.e. A, is the limiting reactant and has a stoichiometric coefficient of 1
Leaving FB
26
Complete the stoichiometric table including coolant flow rates
Write the volumetric flow rate in terms of conversion
P = P0 and T = T0
In terms of conversion
For a flow system at constant T and P
27
In terms of conversion
We now have –rA solely as a function of X
28
Summary
• We learned to write the rate expression for a given reaction• A B• or A ⇋B
• Meaning of rate constant, activation energy• k= A e {-E/RT}
Summary
where;and
1. Batch System Stoichiometric Reactor (similarly for CSTR and PFR!!)
1
ISOTHERMAL REACTOR DESIGN
In Chapter 1 & 2, we discussed balances on batch & flow reactors. In Chapter 3, we discussed rxns. Here, we will combine rxns and reactors.
TopicsPart 1: Mole Balances in Terms of Conversion
1. Algorithm for Isothermal Reactor Design 2. Applications/Examples of CRE Algorithm 3. Reversible Reactions 4. ODE Solutions to CRE Problems 5. General Guidelines for California Problems 6. PBR with Pressure Drop 7. Engineering Analysis
Part 2: Measures Other Than Conversion 1. Measures Other Than Conversion 2. Membrane Reactors 3. Semibatch Reactors
Part 1: Mole Balances in Terms of Conversion
1. Algorithm for Isothermal Reactor Design
The algorithm for the pathway of interest can be summarized as:
1. Mole Balance and Design Equation (choose reactor type)
2. Rate Law (choose rxn type; gas or liq. phase)
3. Stoichiometry
4. Combine
5. Evaluate
The Evaluate Step can be carried out
A. Graphically (Plots)
B. Numerically (Quadrature Formulas)
C. Analytically (Integral tables)
D. Using Software Packages
2
Elementary gas phase reaction in different reactor types
CSTRThe elementary gas phase reaction 2A + B → C
takes place in a CSTR at constant temperature (500 K) and constant pressure (16.4 atm). The feed is equal molar in A and B. x = 0.9, k= 10dm6/mol2s
VCSTR =?
Mole balance
Rate Law
Stoichiometry gas phase, isothermal (T = T0), no pressure drop (P = P0)
Derivation
3
Combine
Evaluate
PFR and Batch ReactorsElementary Gas Phase Reaction: 2A + B → CPFR Mole Balance
Rate Law
gas phase, isothermal (T = T0), no pressure drop (P = P0), CAo=CBo (Θ=1), v=vo(1+εX)
Stoichiometry
4
Combine
CAo=0.2, v=vo=25 dm3/s, k=10 dm6/mol2 s, ε=-0.5, X=0.9Parameter Evaluation
V=227 dm3
Remember that the reaction is:
For a gas phase system:
If the conditions are isothermal (T = T0) and isobaric (P = P0):
We must divide by the stoichiometric coefficient of our basis ofcalculation yielding:
Deriving CA and CB
5
This leaves us with CA as a function of conversion alone:
Similarly for CB:
And if the feed is equal molar, then:
Batch Reactor Constant Volume, V=Vo and the pressure changes.
Mole Balance
Rate Law
Stoichiometry
Combine
6
CAo=0.2, k=10 dm6/mol2 s,Parameter Evaluation
7
Scale – Up of Liquid – Phase Batch Reactor Data to Design of a CSTR:
Scale – Up a lab experiment to pilot – plant operation of full – scale production. Find k from experimental data and use it to design a full – scale flow reactor.
Batch Operation:
Liquid phase: (density change is small → V = V0)
Gas phase with constant volume V = V0.
AAAAA
AA
rdt
dCdt
VdNdt
dNVdt
dNV
rdt
dNV
===⋅=⎟⎠⎞
⎜⎝⎛⋅
=⎟⎠⎞
⎜⎝⎛⋅
0
0
/11
1
Since conc’n is a measured quantity in liquid – phase rxns:
AA r
dtdC
−=−
Let’s calculate the time necessary to achieve a given conversion X for the irreversible second order rxn:
A → B
)1(0
2
00
XCCCkr
VrdtdXN
AA
AA
AA
−⋅=⋅=−
⋅−=⋅ [1]
[2]
[3]
Combine [1], [2] and [3];2
0 )1( XCkdtdX
A −⋅⋅=
Rearrange: dtCkX
dXA ⋅⋅=
− 02)1(
@ t = 0; X = 0; T = T0 (isothermal) → k → constant
⎟⎠⎞
⎜⎝⎛−
⋅⋅
=⇒−⋅
= ∫∫ XX
Ckt
XdX
Ckt
A
X
A
t
11
)1(1
002
00
8
It is important to know the reaction time, tR, to achieve a certain conversion.
Flow reactors use characteristic rxn times, tR.
The time for a total cycle is much longer than tR, as one must account for the time to fill (tF), heat (th) and clean (tC).
t = tf + th + tC + tRDesign of Continuous Stirred Tank Reactors
CSTRs are usually used for liquid phase rxns:
A
A
exitA
A
rXCv
rXFV
−⋅⋅
=−
⋅= 000
)([1] (Design Eq’n for CSTR)
Divide by v0:
A
A
rXC
vV
−⋅
== 0
0
τ [2]
Volumetric flow rate
A single CSTR:
First order irreversible rxn:
kkX
XX
k
XCCCkr
AA
AA
⋅+⋅
=
⎟⎠⎞
⎜⎝⎛−
⋅=
−⋅=⋅=−
ττ
τ
1
11
)1(0
[3][4]
[5]
[6]
Combine [4] and [6]
Combine [2], [3] and [4]
kCC A
A ⋅+=
τ10
rxn Damköhler #,Da: a dimensionless # that for a first order rxn
says the degree of conversion that can be achieved in cont. flow reactor.
9
00
0 :=
⋅−=
tA
A
AoftransportconvectiveofrateAofrxnofrate
FVrDa
For a second order rxn:
000
20
0
0A
A
A
A
A CkCv
VCkF
VrDa ⋅⋅=⋅
⋅⋅=
⋅−= τ
90.01010.01.0
>≥<≤
XDaXDa
DaDaX+
=1
First order liq-phase rxn (Eq’n [6])
CSTRs in Series
CA0
CA1, X1
CA2, X2
-rA1
V1
-rA2
V2
For first order rxn (v = v0)
The effluent conc’n of reactant A from the first CSTR is
0
11
11
01 1 v
Vwherek
CC AA =
⋅+= τ
τ
From the mole balance on reactor 2:
22
210
2
212
)(
A
AA
A
AA
CkCCv
rFFV
⋅−
=−−
=
Solving for CA2:
)1)(1(1 1122
0
22
12 kk
Ck
CC AAA ⋅+⋅+
=⋅+
=τττ
10
If both reactors are of equal size, )( 21 τττ ==
and operate at the same T (k1 = k2 = k)
20
2 )1( kCC A
A ⋅+=
τfor n equal sized CSTR system
nA
nA
An DaC
kCC
)1()1(00
+=
⋅+=
τ
Substituting for CAn in terms of conversion
nn
nA
A
kDaX
DaCXC
)1(11
)1(11
)1()1( 0
0
⋅+−=
+−=
⇒+
=−⋅
τ
CSTRs in Parallel:Equal sized reactors are placed in parallel rather than in series:
FA01
X1
FA02 FA03 FA0n
X2 X3 Xn
Individual volume is given by
⎟⎟⎠
⎞⎜⎜⎝
⎛−
⋅=Ai
iiA r
XFV 0 [1]
Since Vi = Vj, then Xi = X; → -rA1 = -rA2 = ...... = -rA
nFF
nVV
AiA
i
00 =
= [2]
[3]
AA
Ai
iA
rXF
rX
nF
nV
−⋅=⎟⎟
⎠
⎞⎜⎜⎝
⎛−
= 00 .
Substitute [2] & [3] into [1]
[4]
11
Example: 2nd order rxn, v = v0, CA = CA0 (1-X); FA0 X = v0 CA0 X
Combine –rA = k CA2 and V = FA0 X / -rA
220
002
0
)1( XCkXCv
CkXFV
A
A
A
A
−⋅⋅⋅⋅
=⋅⋅
=
Divide by v0:
DaDaDaX
CkCkCkCk
X
XCkX
vV
A
AAA
A
⋅+−⋅+
=
⋅⋅⋅⋅⋅⋅−⋅⋅⋅+−⋅⋅⋅+
=
−⋅⋅==
241)21(
2)2()21()21(
)1(
0
20
200
200
ττττ
τ
Since DaCk A =⋅⋅ 0τfor a second order rxn.
Since X can not be greater than 1.0, ( - ) sign is chosen.
Example: The elementary liquid phase reaction
is carried out isothermally in a CSTR. Pure A enters at a volumetric flow rate of 25 dm3/s and at a concentration of 0.2 mol/dm3. What CSTR volume is necessary to achieve a 90% conversion when k = 10 dm3/(mol*s)?
Mole Balance
Rate Law
Stoichiometry liquid phase (v = vo)
12
Combine
Evaluate at X = 0.9,
V = 1125 dm3
Space Time
2. Applications/Examples of the CRE Algorithm
13
Applying the algorithm to the above reaction occuring in Batch, CSTR, PFR
14
3. Reversible Reaction
To determine the conversion or reactor volume for reversible reactions, one must first calculate the maximum conversion that can be achieved at the isothermal reaction temperature, which is theequilibrium conversion. (See Example 3-8 in the text for additional coverage of equilibrium conversion in isothermal reactor design.)
Equilibrium Conversion, Xe
Calculate Equilibrium Conversion (Xe) for a Constant Volume System
Example: Determine Xe for a PFR with no pressure drop, P = P0
Given that the system is gas phase and isothermal, determine the reactor volume when X = 0.8 Xe
First calculate Xe:Equilibrium constant Kc is;
Xe = 0.89
X = 0.8Xe = 0.711
( )
⎟⎠⎞
⎜⎝⎛=
−=
=
eABe
eAAe
Ae
BeC
XCC
XCCCCK
21
1
0
0
2
15
Deriving The Equilibrium Constant (KC) and Equilibrium Conversion (Xe) for a Non-Constant Volume System:
The reversible reaction:
which takes place in gas phase PFR. Since gas phase reactions almost always involve volume changes, we will have to account for volume changes in our calculations. The equilibrium constant, KC, for this reaction is:
where CAe and CBe are:
A is the limiting reactant A ⇔ B/2
Substituting for CAe and CBe gives us:
Substituting known values (CA0 = 0.2 mol/dm3 and KC = 100 dm3/mol), and realizing that:
Xe = 0.89Solving for the equilibrium conversion, Xe, yields:
16
Batch Reactor With a Reversible Reaction
The following reaction follows an elementary rate law
Initially 77% N2, 15% O2, 8% inerts are fed to a batch reactor where 80% of the equilibrium conversion (Xe = 0.02) is reached in 151 µ s. What is the specific reaction rate constant k1?
Additional Information
17
For 80% of equilibrium conversion X = 0.8 Xe = 0.016
Use Simpson's three point formula to integrate
with ∆X = 0.016/2 = 0.008
18
Tubular Reactors:
Gas-phase rxns are usually carried out in tubular reactors where the gas is generally turbulent.
Reactants Products
No radial variation in velocity, conc’n, T or –rA.
Design Eq’n: AA rdVdXF −=⋅0
In the absence of pressure drop or heat exchange, integral form of the design equation is used:
∫ −⋅=X
AA r
dXFV0
0
Pressure Drop in Reactors
In liquid-phase rxns, the conc’n of reactants is not effected by the changes in pressure.
However, in gas phase rxns, conc’n is propotional to total pressure.
Pressure Drop & Rate Law
Example: Analyze the following second order gas phase reaction that occurs isothermally in a PBR:
Mole Balance Must use the differential form of the mole balance to separate variables:
Rate Law
Second order in A and irreversible:
19
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
P = pressure (kPa)Dp = diameter of particle (m)Φ = porosity = volume of void / total bed volume1 – Φ = volume of solid / total VgC = 1 (metric)µ = viscosity of gas (kg / m s)z = length down the packed bed (m)u = superficial velocity = v0 (volumetric flow) / A2 (cross-area)
ρ = gas density (kg/m3)G = ρ u = superficial mass velocity (kg/m2 s)
vvststmm
⋅=⋅
−=••
ρρ 00
0)(
Ergun, S., 1952. Fluid flow through packed columns. Chemical Engineering Progress 48, pp. 89–94.
20
Variable Density
Catalyst Weight
Then
We will use this form for multiple reactions:
⎟⎟⎠
⎞⎜⎜⎝
⎛⋅=⋅=⋅+=
⎟⎟⎠
⎞⎜⎜⎝
⎛+=+=
δδεε
δδ
0
00
0
0
0000
1
1
T
AA
T
T
T
ATATT
FFyX
FF
XFFFXFFFSince
21
when ε<0; ∆P will be less than ε = 0.
when ε>0; ∆P will be greater than ε = 0.
Isothermal Operation
Recall that
Notice that
The two expressions are coupled ordinary differential equations. We can 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.
Analytical Solution , [e], PFR with
α−=⋅dWdyy2
WyWPPy
⋅−=
===
α10@)(1
20
CAUTION: Never use this form if
22
Combine
Solve
Could now solve for X given W, or for W given X.
For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop.
Effect of presure drop on P,rate, con’n, and x.
23
Pressure Drop in Pipes:
Pressure drop for gases flowing through pipes without packing can be neglected. For flow in pipes, ∆P is given by:
DgGf
dLduG
dLdP
⋅⋅⋅
−⋅−=22
Where
G = ρ u (g/cm3s) = mass velocity is constant along L.
u = average velocity of gas, cm/s.
f = fanning friction factor
f = fnc (Re, pipe roughness
2/1
00
2
0
)1(41 VDAP
VGfPP
pC
αρ
−=⎥⎦
⎤⎢⎣
⎡⋅⋅⋅⋅⋅⋅
−=
Example p 187
24
Optimum Particle Diameter
Laminar Flow, Fix P0, ρ0, Φρ0 = P0(MW)/RT0
ρ0P0 ∼ P02
Superficial mass velocity = ρ u
Cross-sectional area
25
Increasing the particle diameter descreases the pressure drop and increases the rate and conversion.
DP1 > DP2k1 > k2
Higher k, higher conversion
However, there is a competing effect. The specific reaction rate decreases as the particle size increases, therefore so deos the conversion.
k ∼ 1/Dp
26
The larger the particle, the more time it takes the reactant to get in and out of the catalyst particle. For a given catalyst weight, there is a greater external surgace area for smaller particles than larger particles. Therefore, there are more entry ways into the catalyst particle.
7. Engineering Analysis - Critical Thinking and Creative Thinking
We want to learn how the various parameters (particle diameter, porosity, etc.) affect the pressure drop and hence conversion. We need to know how to respond to "What if" questions, such as:"If we double the particle size, decrease the porosity by a factor of 3, and double the pipe size, what will happen to D P and X?"
To answer these questions we need to see how a varies with theseparameters
Turbulent Flow
27
Compare Case 1 and Case 2:For example, Case 1 might be our current situation and Case 2 might be the parameters we want to change to.
For constant mass flow through the system = constant
Laminar Flow
Part 2: Mole Balances in terms of Conc’n & Molar Flow Rates
In some cases, it is more convenient to deal with number of moles or molar flow rates rather than conversion.
Membrane reactors and multiple rxns taking place in gas phase are examples.
The main difference in molar flow rates, you have to write the mole balance on each and every species.
Membrane reactors can be used to increase conversion when the rxn is thermodynamically limited as well as to increase the selectivity when multiple rxns are occuring.
A. Membrane reactorsB. Multiple reactionLiquids: Use concentrations, i.e., CA
28
1. For the elementary liquid phase reaction carried out in a CSTR, where V, vo, CAo, k, and Kc are given and the feed is pure A, the combined mole balance, rate laws, and stoichiometry are:
There are two equations, two unknowns, CA and CBGases: Use Molar Flow Rates, I.E. FI
2. If the above reaction, ,carried out in the gas phase in a PFR, where V, vo,CAo,k, and Kc are given and the feed is pure A, the combined mole balance, rate laws, and stoichiometry yield, for isothermal operation (T=To) and no pressure drop (∆P=0) are:
29
• Microreactors
Are characterized by their high surface area to volume ratios (due to many micro-tubes and channels). Dchannel = 100µm, Lch= 2 cm
Control of heat and mass transfer resistance!!
They are used for highly exothermis rxns, for rxns with toxic or explosive intermediates. ALso,i for the productions of speciality chemicalsi combinatorial chemical screening, chemical sensors.
In modeling, we assume they are PFR. dFa/dV = rA
30
Membrane ReactorsMembrane reactors can be used to achieve conversions greater than the original equilibrium value. These higher conversions are the result of Le Chatelier's Principle; you can remove one of the reaction products and drive the reaction to the right. To accomplish this, a membrane that is permeable to that reaction product, but is impermeable to all other species, is placed around the reacting mixture.
Example: The following reaction is to be carried out isothermally in a membrane reactor with no pressure drop. The membrane is permeable to Product C, but it is impermeable to all other species.
For membrane reactors, we cannot use conversion. We have to work in terms of the molar flow rates FA, FB, FC.
Mole Balances
Rate Laws
31
StoichiometryIsothermal, no pressure drop
Polymath will combine for youCombine
Parameters
Polymath Solve
3. Semibatch Reactors Semibatch reactors can be very effective in maximizing selectivity in liquid phase reactions.
The reactant that starts in the reactor is always the limiting reactant
BU
AD
BAUU
BADD
k
k
CkCkUDySelectivit
CCkrCCkr
undesiredUBAproductDBA
U
D
⋅⋅
=
⋅⋅=
⋅⋅=
⎯→⎯+
⎯→⎯+
)/(
)()(
2
2
32
Three Forms of the Mole Balance Applied to Semibatch Reactors
If you have multiple reactions, use concentrations to make mole balances!!
ODE Solutions to CRE Problems
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
⎥⎦
⎤⎢⎣
⎡⋅+
⋅−⎥⎦
⎤⎢⎣⎡
⋅+−
⋅−=
−=
C
AAA
A
A
KXXC
XXCkr
Fr
dVdX
)1(21)1( 0
2
0
0
εε
[V, X] = ode45(@vdp1, [0 500], [0]) Matlab solution
33
Example: Elementary Irreversible Reaction
Consider the following irreversible elementary reaction
-rA = kCACB
The combined mole balance, rate law, and stoichiometry may be written in terms of number of moles, conversion, and/or concentrati
Polymath Equations
34
Equilibrium Conversion in Semibatch Reactors with Reversible Reactions
Consider the following reversible reaction:
Everything is the same as for the irreversible case, except for the rate law
Where:
At equilibrium, -rA=0, then
35
function f=volume(V,x)Ca0=0.2;Kc=100;Fa0=5;k=2;epsilon=-0.5;ra=k*(((Ca0*(1-x(1))/(1+epsilon*x(1)))^2)-(Ca0*x(1)/(2*Kc*(1+epsilon*x(1)))));f(1)=ra/Fa0;
>> [V x]=ode45('volume',[0 500],[0]);>> plot(V,x)>> grid>> xlabel('Volume')>> ylabel('conversion')
0 50 100 150 200 250 300 350 400 450 5000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Volume
conv
ersi
on
36
function f = packedbed(W,x)alpha = 0.0002;k = 10;epsilon = -0.5;FA0=2.5;CA0 = 0.2;f = zeros(2,1);CA = (CA0*(1-x(1))*x(2))/(1 + epsilon*x(1));ra = k*(CA.^2);f(1) = ra / FA0;f(2) = -alpha*(1 + epsilon*x(1))/(2*x(2));
>> [W x]=ode45('packedbed', [0 1000], [0;1]);>> plot(W,x(:,1))>> xlabel('W, kg')>> ylabel('X')>> plot(W,x(:,2))>> xlabel('W, kg')>> ylabel('y')
0 10 20 30 40 50 60 70 80 90 1000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
X
W, kg
0 10 20 30 40 50 60 70 80 90 1000.994
0.995
0.996
0.997
0.998
0.999
1
1.001
W, kg
y
37
Combining mole balances, rate laws and stoichiometry
38
Pressure Drop in Packed Bed Reactors
Ergun Equation
P = pressure (kPa)Dp = diameter of particle (m)Φ = porosity = volume of void / total bed volume1 – Φ = volume of solid / total VgC = 1 (metric)µ = viscosity of gas (kg / m s)z = length down the packed bed (m)u = superficial velocity = v0 (volumetric flow) / A2 (cross-area)
ρ = gas density (kg/m3)G = ρ u = superficial mass velocity (kg/m2 s)
vvststmm
⋅=⋅
−=••
ρρ 00
0)(
Ergun, S., 1952. Fluid flow through packed columns. Chemical Engineering Progress 48, pp. 89–94.
For gas phase reactions, as the pressure drop increases, the concentration decreases, resulting in a decreased rate of reaction, hence a lower conversion when compared to a reactor without a pressure drop.
Effect of presure drop on P,rate, con’n, and x.
39
Part 2: Mole Balances in terms of Conc’n & Molar Flow RatesIn some cases, it is more convenient to deal with number of moles or molar flow rates rather than conversion.
Membrane reactors and multiple rxns taking place in gas phase are examples.Microreactors with multiple reactions
1
There are some basic steps in Analysis of Rate Data
1. Postulate a rate law:
a) Power law model for homogeneous rxns:
b) Langmuir – Hinshelwood for heterogeneous rxns:
2. Select reactor type; i.e.; Batch, CSTR, PFR, PBR......
3. Process your data in term of measured variables (NA, CA or PA)
4. Do simplifications (assumptions), if –rA = k CACB and CA>>CB→ -rA ≈ k CB
βαBAA CCkr ⋅⋅=−
2)1('
BAA
BAA PPK
PPkr+⋅+
⋅⋅=−
Chapter 5Chapter 5
Collection & Analysis of Rate DataCollection & Analysis of Rate Data
5. For batch reactors, find reaction order
i. Find (- dCA / dt) from CA vs t data.
Take ln.
Find the order of rxn by ln (- dCA / dt) = ln k + α ln CA
Find the rate constant, k.
ii. Integral Form:
[ ]........1 10
α
α
−=
⋅=−
A
AA
Ck
t
Integrate
Ckdt
dC
6. For PBR, find –rA’ as a function of CA or PA.
2
SUMMARYSUMMARY
Differential Method:
α
αα
=
+=⎟⎠⎞
⎜⎝⎛−⋅=−
slope
Ckdt
dCCkdt
dCAA
AAA
A lnlnln
⎟⎠⎞
⎜⎝⎛−
dtdCAln
ACln
P
A
dtdC
⎟⎠⎞
⎜⎝⎛−ln
APCln
( )( )αAP
PAA C
dtdCk
/=
We should differentiate the concentration-time data either graphically or numerically.
1. Graphical differentiation
2. Numerical differentiation formula
3. Differentiation of a polynomial fit to the data.
1. Graphical
Tabulate (CAi, ti) observations and for each interval calculate
∆Cn = Cn – Cn-1 ∆tn = tn – tn-1
(∆CA/ ∆t)3
(∆CA/ ∆t)2
(∆CA/ ∆t)1
∆CA/ ∆t
C3 – C2t3 – t2C3t3
(dCA / dt)2C2t2
C2 – C1t2 – t1
(dCA / dt)1C1t1
C1 – C0t1 – t0
C0t0
(dCA / dt)∆CA∆tCiti
3
Plot these values as a histogram
Draw a smooth curve
Read estimates of (dCA/dt) at t1, t2, ......
∆x = xn – xn-1
∆y = yn – yn-1
Plot these
∫
∑
⋅=−
∆⋅∆∆
=−=
nx
xn
i
n
i in
dxdxdyyy
xxyyy
1
1
21
Try to find
Area(A) ≈ Area(B)
A + C ≈ B + D
AB
CD
Numerical MethodsNumerical MethodsIf the data points are equally spaced, i.e., t1 – t0 = t2 – t1 = ∆t
...........CA2CA1CA0Conc’n (mol/dm3)
...........t2t1t0time (min)
[ ]
[ ])()1()2(
)1()1(
210
421
21
243
0
nAnAnAt
A
iAiAt
A
AAA
t
A
CCCtdt
dC
CCtdt
dC
tCCC
dtdC
n
i
+−∆
=⎟⎠⎞
⎜⎝⎛
−∆
=⎟⎠⎞
⎜⎝⎛
∆−+−
=⎟⎠⎞
⎜⎝⎛
−−
−+
For initial point
interior point
End point
Polynomial fit:Polynomial fit:Fit the conc’n – time data to an nth order polynomial as:
nn
44
33
221oA t......a ta ta ta t a a C ++++=
{ } a........,,a ,a ,a n21oFind best values for1-n
n2
321A ta......n t3a t a 2 a
dtdC
+++=
4
Integral MethodIntegral Method
We first guess the rxn order and integrate the differential equation used to model the batch system. If the order is correct, the plot of conc’n time data should be linear.
This method is used when rxn order is known but EA and kA are unknown.
Non Non –– Linear RegressionLinear RegressionNon Non –– Linear LeastLinear Least--Square AnalysisSquare AnalysisWe want to find the parameter values (alpha, k, E) for which the sum of the squares of the differences, the measured rate (rm), and the calculated rate (rc) is a minimum.
# of parameters to be determined
# of runsThat is we want to be a minimum. For concentration-time data, we can integrate the mole balance equation for -rA = k CA
alpha to obtain
5
We find the values of alpha and k which minimize S2
α α
α
α
α
α
α
α
α
− −
−
−
−
⋅−−=
−−=
+−
=⋅
⋅=⋅
⋅=
1 10
10
1
)1(
1
11
tkCC
CConst
ConstCtk
CdCdtk
Ckdt
dC
AA
A
A
AA
AA
Vary α and k, obtain S2 (or use a search technique)
S(α’, k’) is a minimum (optimization methods)
Concentration vs Time Equations for the proposed rate equations can be done by differential or integration method.
Ex.:Ex.: Liquid phase rxn btw trimethylamine(A) and n-propyl bromide (B) was studied by Winkler & Hinshelwood. The results at 139.4oC are shown below. Initial sol’ns of A & B in benzene, 0.2 molal, were mixed and placed in constant temperature bath. After certain times, they were cooled to stop the rxn. Determine the first order and second order specificrates, k1 and k2, asssuming the rxn is irreversible. Use integration & differential methods.
55.21204
36.7593
25.7342
11.2131x (%)t, minRun
6
Sol’n: Sol’n: A + B → C+ + D-
Volume is constant
BAA
AA
CCkdt
dC
Ckdt
dC
⋅⋅==
⋅==
2And
1Ast
r-order 2
r-order 1
Integration MethodIntegration Method
[2]t C
1C1
Cdt
dC-
molal 0.1C C and equal are tscoefficien tricStoichiomeorder 2
[1] t kCCln-
order 1
2A0A
2A2
A
B0A0
nd
1A0
A
st
⋅=−
⋅=
==
⋅=
k
k
sgmolL
t
snt
xCCC
CCx AAA
AA
⋅×=
−⋅⋅⋅=⎟
⎠⎞
⎜⎝⎛ −⋅
⋅=
×=⋅
=⋅=
⋅=
−=−
=
−
−−
3
A02
14
A
A01
A
00
0
1063.1)112.01(1.0)6013(
112.01x-1
1C1 k
[2]in Substitute
1054.10888.0
1.0ln6013
1CCln1 k
[1]in Substitute0.888 0.1 Crun first For the
)1(
0.05521.711.12720040.03671.641.30354030.02571.701.46204020.01121.631.547801
CDk2 x 10-3 (L/mol s)k1 x 10-4 (s-1)t, secRun
If you repeat for four of the runs:
7
0 800 1600 2400 3200 4000 4800 5600 6400 72000
0.5
1
time, sec-ln
(CA
/CA
0)
0 800 1600 2400 3200 4000 4800 5600 6400 72000
5
10
15
time, sec
1/C
A-1
/CA
0
So, it is SECOND ORDER
Differential MethodDifferential Method
CD = CA0 – CA (moles of D produced = moles of A reacted)
CD = x CA0
0 1000 2000 3000 4000 5000 6000 7000 80000
0.01
0.02
0.03
0.04
0.05
0.06
time, sec
[D] m
oles
/L
A plot of CD vs t
curve) theof slope (the dt
dCdt
dCr DA =−=
8
Slopes determined from the curve are given as follows
0.450.050.050.640.060.040.790.070.031.140.080.021.380.090.011.580.100.0
r = dCD/dt (105 gmol/L)CACD
log rA = log k1 + log CA (from –rA = k1 CA)
log rA = log k2 + log CA2 = log k2 + 2 log CA (from –rA = k2 CA
2)
1st: log r vs log CA should yield a straight line with a slope of 1.0.
2nd: log r vs log CA should yield a straight line with a slope of 2.0.
log CA
log
r
slope = 2.0
log r = -2.76 + 2.0 log CA
The data suggest a slope of 2!
log k2 = -2.76
k2 = 1.73 x 10-3 L / mol s
9
Another Way (Batch Reactor Data)
B + A → products
-rA = kA CAαCB
β if α and β are both unknown, the rxn could be run in an excess of B so that CB remains essentially unchanged
CkCk 'k' whereC'k'r-
A) of (excess carried isrxn the, gdetermininIn CkCk k' whereCk'r-
A0AAABA
B0ABAAA
ααβ
ββα
α
≈=⋅=
≈=⋅=
Solving with search
[ ]( )24
1
1/110,
2 )1(∑=
−− ⋅−−−=i
iAiAm tkCCSαα α
0.04480.055272000.06330.036735400.07430.025720400.08880.0112780
CACDt
[ ]( )21
1/110,
2 )1(∑=
−− ⋅−−−=n
iiAiAm tkCabsCS
αα α fminsearch
10
Method of Initial Rates Method of Initial Rates
If there is a reverse rxn, it could render the differential method ineffective. In these cases, initial rates could be used for k & α.
Carry out a series of experiments with different CA0.
Determine –rA0 for each run.
-rA0 = k CA0 α → ln(-rA0) vs ln(CA0)
Instead of doing at different time steps, repeat the rxn.
Method of Half Lives:Method of Half Lives:
The half life of a rxn, t½, is defined as the ime it takes for the concentration of the reactant to fall to half of its initial value.
If you know t½, find k & α.
If there are more than 1 reactant, use method of excess.
1
0
1
/1
1
0
1
2/1
1
0
10
1
02/10
1)1(1
1)1(
12
1)1(
1
11)1(
1
21@0@
)(
−−
−−
−
−−
⎟⎟⎠
⎞⎜⎜⎝
⎛−−
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−−
=
⎟⎟
⎠
⎞
⎜⎜
⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛−
=
⎟⎟⎠
⎞⎜⎜⎝
⎛−
−=
====
⋅=−=−
→⋅=−
αα
αα
α
αα
α
α
α
α
α
α
An
A
A
A
AA
AAAA
AAA
AA
CkntSimilarly
Ckt
CC
kt
CCkt
CCttCCt
Ckrdt
dCproductsACkr
11
0
1
2/1 ln)1()1(12lnln ACk
t αα
α
−+−−
=−
2/1ln t
0ln AC
α−=1slope
Differential Reactors
A differential reactor is used to determine the rate of a rxn as a function of either concentration or partial pressure
FA0 FAe
∆L
FA0 FAe
∆L
∆W
Conversion is very small in bedCA is constant ≈ CA0 (gradientless)Rxn rate is uniform (almost isothermal)
Design Equation is similar to a CSTR:
wCv
wCCvr
wF
wxFr
wCvCvr
wFFr
wrFF
FF
PAeAA
PAA
AeAA
AeAA
AAeA
AeA
∆=
∆−
=
∆=
∆⋅
=
∆⋅−
=
∆−
=
=∆⋅+−
=+−
)()('
rate) flow of (in terms '
n)conc' of (in terms '
'
0'
0)cat of mass(cat of massrxn of Rate
000
0
00
0
0
0
12
Deviations from Ideal ReactorsDeviations from Ideal Reactors
When the mixing criteria of ideal reactors are not satisfied, mathematical expressions for the conservation equations become more difficult.
Deviations from ideal stirred – tank reactors
a) Stagnant regions
b) by – passing
Deviations from tubular-flow
a) Longtidunal mixing due to vortices and turbulence
b) Laminar flow (poor radial mixing)
c) By-passing in fixed-bed catalytic reactor.
1
Chapter 6 MULTIPLE REACTIONSChapter 6 MULTIPLE REACTIONS
1. Series Reactions
2. Parallel Reactions
3. Complex Reactions: Series and Parallel
4. Independent
If there are multiple rxns, use concentrations not conversions.
These might occur in combination or by themselves.
Consecutive rxns
intermediate
None of the products or reactants are common.
Desired & Undesired RxnsDesired & Undesired Rxns
We want to get the desired product, D and avoid the formation ofundesired one, U.
In the parallel rxn
UA
DAU
D
k
k
⎯→⎯
⎯→⎯
UDA UD kk ⎯→⎯⎯→⎯In series
Reactor SeperatorA
D
U Low HighEfficiency
Cost
Seperator Cost
Reactor CostTotal Cost
2
Selectivity and YieldSelectivity and Yield
Selectivity tells us how one product is favored over another when multiple rxns take place
Exit flow rate
For batch tendU
DDU
NNS =
~
Reaction yield = ratio of the rxn rate of a given product to the rxn rate of key reactant A.
Determine the instantaneous selectivity, SD/U, for the liquid phase reactions:
B3
A3U2
BA2U1
B2
A1D
CCkr UBA
CCkr UBA
CCkr DBA
2
1
=⎯→⎯+
=⎯→⎯+
=⎯→⎯+
Sketch the selectivity as a function of the concentration of A. Is there an optimum and if so what is it?
Ckk
CkCCkCCk
CCkrr
rS 2A32
A1
B3
A3BA2
B2
A1
UU
DUD/U
21
21 +=
+=
+=
[ ] [ ]
3
2*A
*A3
*A1
2*A321
A
kkC
C2kCkCkkk0dCdS
=
−+==Find the optimum concentration
3
Parallel Rxns:
For the following competing rxns
21
21
2U
1D
AU
D
U
DD/U
21
AUADUDA
AUUk
ADDk
Ckk
rrS
orders.rxn positive are and whereCkCkrrr-
A of ncedisappeara of Rate
Ckr )(undesired UA
Ckr (desired) DA
αα
αα
α
α
αα
−⋅==
+=+=
=⎯→⎯
=⎯→⎯
Maximizing the Desired Product for one ReactantMaximizing the Desired Product for one Reactant
Case1:
AD/UAU
DD/U21 C as S C
kkS If 21 αααα −⋅=⇒>
AD/UAU
DD/U21 C as S
C1
kkS If
12 αααα −⋅=⇒<Case2:
In Case 1: If gas rxn, we should run it without any inerts and at high pressures to keep CA high.
If liquid phase rxn, the diluents should be minimum. CSTR should not be used, batch or plug flow should be used.
In Case 2: Dilute the feed with inerts. CSTR should be used, because CAdecreases rapidly.
4
Effect of Temperature on Selectivity
It can be determined from
energy activation:E factor, freq:A where
eAA
kk~S ]/RT)E-([E-
U
D
U
DD/U
UD⋅=
⇒> EE UD
⇒< EE UD
SD/U
T
SD/U
T
Rxn should be operated at the highest possible T to max. SD/U
2. Series Reactions
This series reaction could also be written as Reaction (1) Reaction (2)
Species A:
Species B:
5
Using the integrating factor, i.f.:
at t = 0, CB = 0
When should you stop the reaction to obtain the maximum amount of B? Let's see
Then
And
6
The Integrating Factor The Integrating Factor
If you only had an expression of the form
things would be much easier, then you could integrate with respect to z and find y(z).
then you could manipulate equation (3):
where the term in brackets is the left hand side of equation (1). You need du/dz = u f(z). Recall
(5)
(4)
(3)
(2)
(1)
If you define and f(z) = dq/dz (i.e. , then
This satisfies the condition that du/dz=u f(z). Therefore, and substituting into equation (4),
where the term in brackets is the left hand side of equation (1).
(7)
(6)
CONCLUSION: If your problem is of the form
you can multiply both sides of the equation by the
7
(which you should be able to evaluate, since you know f(z)), to yield
or, substituting from equation (7)
DerivationDerivation
f(t)=k2, so
From equation (12), the solution is then
The constant can be obtained from the intitial condition that at t=0, CB=0;
8
3. Algorithm for Multiple Reactions
Reactor Selection & Operating Conditions:Reactor Selection & Operating Conditions:
Consider two simultaneous rxns
( ) ( )2121
222
111
BA2
1
U
DD/U
BA2Uk
BA1Dk
CCkk
rrS
CCkr UBA
CCkr DBA
ββαα
βα
βα
−−==
=⎯→⎯+
=⎯→⎯+
We usually want to max SD/U. So the reactors should be designed to max SD/U
A
B
A
B
A
B
A
B
D A
B C
D
∆H>>0
∆H>>0 Thermodinamically limited rxns A + B <-> C + D
9
Algorithm for Solution of Complex Reactions:Algorithm for Solution of Complex Reactions:
In combinations of parallel and series rxns, ODE solver packagesmake life easier.
• Write the rxns
• Write the mole balances
• Net rate laws
• Stoichiometry
• Combine & Solve
4. Applications of Algorithm 4. Applications of Algorithm
(1)
(2)
NOTE: The specific reaction rate k1A isdefined with respect to species A.
NOTE: The specific reaction rate k2C isdefined with respect to species C.
Case 1: PFRCase 1: PFR
Mole Balances
DD
CC
BB
AA
rdVdF :D
rdVdF :C
rdVdF :B
rdVdF :A
=
=
=
=
Rate Laws
10
Species A 2A1AA rrr +=
For reaction (1):
2BA1A1A
1A1C
1A1B
2BA1A1A
CC-kr
-rr2rr
CC-kr
=
=
=
=
For reaction (2):
2A
3C2C
2C2DD
2A
3C2C
2BA1AC
2BA1A1BB
2A
3C2C
2BA1AA
2C2D
2A
3C2C2C2A
2C2A
CCk31
3rrr
CCkCCkr
CCk2rr
CCk32CCkr
r31r
CCk32)-r(
32r
3r-
2r-
=−==
−=
−==
−−=
−=
−=−=
=
Species A
Species B
Species C
Species D
Writing the net rates
0
2A
3C2CD
0
DD
0
2A
3C2C
2BA1AC
0
CC
0
2BA1AB
0
BB
0
2A
3C2C
2BA1A
A
0
AA
CCk31
dVdC
vr
dVdC D Species
CCkCCkdVdC
vr
dVdC C Species
CCk2dVdC
vr
dVdC B Species
CCk32CCk
dVdC
vr
dVdC A Species
CombineCAv0FA
v0 v:liquid try Stoichiome
v
v
v
v
=
=
−=
=
−=
=
−−=
=
=
=
Evaluate solution using Polymath of MatLab
k1A = 0.5
k2C = 2.0
v0 =5.0
@ t=0; V=0, CA0=4, CB0=4, CC0=0, CD0=0
Vf = 5 dm3
11
2A
3C2C
D0
D
D
D
DD0
2A
3C2C
2BA1A
C0
C
C
C
CC0
2BA1A
BB00
B
BB0
2A
3C2C
2BA1A
AA00
A
AA0
CCk31
CvrF
r-)F-F(V D Species
CCkCCkCv
rF
r-)F-F(V C Species
CCk2)C-C(v
r-)F-F(V B Species
CCk32CCk
)C-C(vr-
)F-F(V A Species
===
−===
==
+==
Case 2: CSTRCase 2: CSTR
[ ][ ]
D02
A3
C2CD
C02
A3
C2C2
BA1AC
BB002
BA1AB
AA002
A3
C2C2
BA1AA
CvCCk31V)f(C
CvCCkCCkV )f(C
)C-C(vCCk2V -)f(C
)C-C(vCCk32CCkV -)f(C
−⎥⎦⎤
⎢⎣⎡⋅=
−−⋅=
+⋅=
+⎥⎦⎤
⎢⎣⎡ +⋅=
We will specify V, CA0, CB0 along with the specific reaction rates kij.
This formulation leaves us with four equations and four unknowns (CA, CB, CC and CD)
12
Case 3: Semibatch Liquid PhaseCase 3: Semibatch Liquid Phase
tv VV
VCCCk
31
dtdC D Species
VCCCkCCk
dtdC C Species
CCk2V
]C[Cdt
dC B Species
VCCCk
32CCk
dtdC A Species
00
D02A
3C2C
C
C02A
3C2C
2BA1A
C
2BA1A
BB00B
A02A
3C2C
2BA1A
A
+=
−=
−−=
−−
=
−−−=
v
v
v
v
hr 8 t0C 0C 0C 4C conditions Initial
2dmV dmmol4C
hrdm2.1v
hr moldm2k
hrmoldm0.5k
fDiCiBiAi
303B0
3
0
4
2
2C2
6
1A
=====
===
==Parameters
A
FB0
PolyMath Solutions
13
Mole balances
Net rates
Species ASpecies B
Species C
Species D
Rate Laws
Relative rates
Rxn (1) A + 2B ->C
Rxn (2) 2A + 3C ->D
THE ALGORITHMTHE ALGORITHM
2A
3C2C
2C2DD
2A
3C2C
2BA1AC
2BA1A1BB
2A
3C2C
2BA1AA
CCk31
3rrr
CCkCCkr
CCk2rr
CCk32CCkr
=−==
−=
−==
−−=Species A
Species B
Species C
Species D
StoichiometryTT
PP
FFC
TT
PP
FF
vFC 0
0T0
A0T0
0
0T0
A0
0
T0A ⋅⋅⋅=⋅⋅⋅=
14
200V 0,F 0,F 20,F 10,F 0,V Cond. Initial0.8C 1.3k 0.05,k Parameters
F F F FF
FF
FFCk
31
dVdF
FF
FFCk
31
FF
FFCk
dVdF
FF
FFCk2
dVdF
CCk32CCk
dVdF
FFC C
T T ,PP 0,P
fDCBA
T02C1A
ABCDT
2
T0
A
3
T0
C5T02C
D
2
T0
A
3
T0
C5T02C
2
T0
B
T0
A3T01A
C
2
T0
B
T0
A3T01A
B
2A
3C2C
2BA1A
A
T
iT0i
00
======
===
+++=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−⎟⎟
⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛=
⎟⎟⎠
⎞⎜⎜⎝
⎛⎟⎟⎠
⎞⎜⎜⎝
⎛−=
−−=
=
===∆Combine
Species A
Species B
Species C
Species D
A0
AA0
ABCDT
D
DD0
C
CC0
B
BB0
A
AA0
FF-FX
unknowns five equations Five
F F F FF :Totalr-
)F-F(V D Species
r-)F-F(V C Species
r-)F-F(V B Species
r-)F-F(V A Species
=
+++=
=
=
=
=For a CSTRFor a CSTR
15
ConcentrationConcentration--Time Trajectories Time Trajectories
1. The following concentration-time trajectories were observed in a batch reactor
Which of the following reaction pathways best describes the data
2. Sketch the concentration-time trajectory for the reaction
Choice B is the answer. Choices A and C are incorrect because they show species B eventually consumed, which is clearly not the case
Solution: Part 1
16
Solution: Part 2
(1) B virtually consumed so no more D can be produced in reaction 2.(2) Rates of Consumption of A and Ba are virtually the same.(3) Rate of consumption of B greater than that of A owing to Reaction 2
1
Ch. 7 Reaction Mechanisms, Pathways, Bioreactions and Bioreactors
Active Intermediates and Non-elementary Rate Laws:
Usually we have seen that rate can be expressed as:
-rA = k CAα where α = 1, 2, 3, 0
However, a large # of rxns, the orders are non-integers
Rate laws may also involve a # of elementary rxns and at least one active intermediate.
Active intermediate is a high energy molecule that reacts virtually as fast as it is formed.
It is found in small amounts and shown as A*.
A + M A* + M
The high E is stored in the chemical bonds where high-amplitude oscillations lead to bond ruptures, molecular rearrangement and decomposition
• Collision
• Free radicals, unpaired e-
• Ionic intermediates
• Enzyme-substrat complexes
In 1999, Ahmed Zewail got Nobel for fentosecond spectroscopy.
2
Pseudo-steady-state Hypothesis (PSSH)
The PSSH assumes that the net rate of species A* is zero.
rA* ≈ 0
The reaction
Why does the rate law decrease with increasing temperature?
has an elementary rate law
However
3
Mechanism: (1)
(2)
(3)
Assume that reactions (1) and (2) are elementary reactions, such that
The net reaction rate for is the sum of the individual reaction rates for :
The PSSH assumes that the net rate of is zero:
][O [NO] e A
A A][O [NO] k
k k r
C C kr since :nconc' NO lowat [NO]2
kk if
][Okk 2 *][NO k[NO]
2k if
][NO][Okk1 *][NO k[NO]
2k if
22RT
)E (E - E
2
312
2
2
31NO2
NONO332NO3
2
23
132
3
22
323
312 +
==
=>>
=>>
=<<
4
if E2 > (E1 + E3) as T increases r decreases.
Apparent rxn order at low conc’ns: 2nd order with NO
Apparent rxn order at high conc’ns: 1st order with NO
you can say
Rxns consist of
a) Activation step
b) Deactivation step
c) Decomposition step
rxn order 1 Ck r (inert)constant is C if
k C kC C k k r- r
*C k r :P offormation of rate thewriteP *A
M A M *A
P)A (for M *A M A
stApM
3M2
MA13Ap
A3p
k3
k2
k1
=
+==
=
⎯→⎯
+⎯→⎯+
→+⎯→⎯+
M
MAA*
+ +
If you have
5
Searching for a Mechanism
In many cases the rate data are correlated before a mechanism is found.
If you have
A
2A1
P
3A2
2A31
P
C k'1Ckr
kCkCkkr
+=
+=
Steps in Mechanism Search
1. Species having conc’ns appearing in the denominator of the rate law probably collide with the active intermediate
A + A* Products
2. If a constant appears in the denominator, one of the rxn steps is probably the spontaneous decomposition of the active intermediate
A* Decomposition products
3. Species having conc’ns in the numerator probably produce the active intermediate
Reactant A* + Other Products
1. Assume an active intermediate
2. Postulate a mechanism
3. Model each rxn as an elementary rxn
4. Write the rate laws for active intermediates
5. Use PSSH
6. Eliminate intermediate species
7. If the law obeys the experiment, OK.
Example: Chain rxns will have steps:
1. Initiation : Formation of active intermediate
2. Propagation : Interaction of A* with reactant
3. Termination : deactivation of A*
Active Intermediate C2H6 2CH3.
Reaction pathways help us to see the inter-connections for multiple rxns.
Metabolic pathways are examples.
Example 7-1 (pg 384)
6
Enzymatic Rxn Fundamentals
Enzymes (E) are proteins with catalytic properties. They act on substrates (S).
They increase the rate by 103 to 1017 times!
The rxn coordinate for enzyme catalysis:
S P
S E •
E P S E S E +→•↔+
Active intermediate (enzyme substrate complex)
Enzymes are highly specific (1 enzyme 1rxn), remain unchanged and can be used over and over.
A typical enzyme will catalyze the rxn of about 1000 substrate molecules every second.
Although conc’n are low, how do they bind specifically, motions are enormously fast at molecular level. It takes a small molecule 0.2 sec to diffuse a distance of 10 µm. Active site will be bombarded by about 500,000 random collisions with the substrate molecule per second.
They are named by the type of rxn they catalyze (-ase).
S
E
P
E E
P+
If the enzyme is exposed to extreme T or pH it will denature.
(unfolding will make the active site disappear)
Two models for SE interactions.
1. Lock & Key Model
PE E+
S
S E •
2. Induced fit model
+ S
E
EE
SP
7
Mechanisms
As an example, we will look at urea decomposition by urease.
E P E S OH2 +⎯⎯→⎯+
The corresponding mechanism:
E P W S E
S E S E
S E S E
3
2
1
k
k
k
+⎯→⎯+•
+⎯→⎯•
•⎯→⎯+
Corresponding rate laws are
(W) S) (E k rS) (E k r(S) (E) k- r
33P
22S
11S
•=
•=
=
Net rate of disappearance of S is: (-rS)
(*) S) (E k - (S) (E) k r- 21S •=
(1)
(2)
(3)
(from 1)
Net rate of formation of E-S complex
(W)k k (S) k(W) (S) )(E k k r-
......... k
.......... k kr-in back ngSubstituti
(S) k (W) k k(W)) k (k )(E E
(**) using (E)for Solving S) (E (E) )(E
:nconc' enzyme totalmeasurecan but we n,conc' enzyme unbound :(E) is therestillBut (W)k k
(W) (S) (E) k k r-
:gRearrangin(W)k k(S) (E) k k - (S) (E) k r-
(*) into Substitute
(**) (W) k k(S) (E) k S) (E
0 r PSSH, UsingS) (E (W) k - S) (E k - (S) (E) k r
321
t31S
2
31S
132
32t
t
32
31S
32
121S
32
1
ES
321ES
++=
+=
+++
=
•+=
+=
+=
+=•
=•= (combine 1 with 2 and 3)
8
Vmax
Vmax/2
Km
-rs
S
Michaelis – Menten Eq’n:
For a given enzyme conc’n, a sketch of rate of disappearance of the substrate is shown as a function of substrate conc’n:
Michaelis – Menten Plot
where Vmax is the maximum rate of rxn for a given total enzyme conc’n.Km is callad the Michaelis – Menten constant, it is the measure of attraction of the enzyme for its substrate (~affinity constant)The Km is equal to substrate conc’n at which the rate of rxn is equal to one-half of its max rate.
order zero isrxn he tV r-(S) K n,conc'high At
order 1 isrxn theK
(S) V r-
(S) K n,conc' substrate lowAt
(S) K(S) V r-
maxs
m
st
m
maxs
m
m
maxs
≅<<
=
>>
+=
Product – Enzyme Complex:
In many rxns, the Enzyme and Product Complex (E-P) is formed directly from the (E-S) complex as:
{ }Law Rate Haldane - Briggs C K K C
)K / C - (C V r-
:obtain weP), (E and S) (E toPSSH ApplyingP E P E S E S E
ppmaxs
cpsmaxs ++=
••
+↔•↔•↔+
Batch Reactor Calculations for Enzyme Rxns:
Let’s start writing the mole balance on urea in a batch reactor.
(2) -rdt
dC-constant is V and N} V {C phase liq.In
(1) V -rdt
dN- Urease CO 2NH Urease Urea
ureaurea
ureaurea
23
=
=
=
++→+
9
(6) V
X-CX-1
1lnVK t
X) - (1 C C :X of in terms (3)n Eq' Rewrite
(5) V
C-CCC
lnVK t
Integrate
(4) dCC VC K
r-dC t
:(2) into (3) Substitute
(3) C KC V r-
:Law Rate
max
urea,0
max
m
urease,0urease
max
ureaurea,0
urea
urea,0
max
m
C
Curea
ureamax
ureamC
C urea
urea
uream
ureamaxurea
urea,0
urea
urea,0
urea
+=
=
+=
+==
+=
∫∫
Determine Km and Vmax from batch reactor data using integral method. Divide both sides of (6) by (t Km / Vmax)
(7) tKX C
KV
X-11ln
t1
m
urea,0
m
max −= X-1
1ln t1 a where =
Draw a vs (X / t): slope = - Curease,0 / Km
intercept = Vmax / Km
substrate ofn conc' initial : S tKS - S -
KV
SSln
t1
S - S X S
S C and S X) - (1 S(7)) of (LHS Sby Multiply
0
m
0
m
max0
00
0urease,00
0
=
=
==
slope = - 1 / Km
intercept = Vmax / Km
S0 – S / t
SSln
t1 0
From a different approach, rewrite M-M Eq’n in terms of S:
10
Effect of temperature on Enzymatic Rxns
A very complex issue!
If the enzyme did not unfold, the rate would probably follow a Arrhenius temperature dependence.
However, at a certain temperature the enzyme will denature and its activity will be lost!
Therefore, we will first have an increase and then decrease:
Vmax
T
temp. İnactivation { thermal denaturation}
Inhibition of Enzyme Reactions:
Temperature, pH and the presence of an inhibitor influences the rates. Even the inhibition of a single (critical) enzyme may be fatal:
E.g.: Cyonide will cause the aerobic oxidation to stop when bound to cytochrome oxidase death (the whole pathway will be affected)
Asprin inhibits the enzyme used in synthesis of pain-producing reactants
E + I I • E (Inactive)
E • S + I I • E • S (Inactive)
E • S + I I • E • S (Inactive)
I • E + S I • E • S (Inactive)
Uncompetitive
Non-competitive
Competitive
Types of Enzyme Inhibition
Let’s look at them individually
11
Competitive Inhibition
Very important in drug design:
I competes with the substrate for the enzyme
I E I E [5]
(inactive) I E E I [4]
P E S E [3]
S E S E [2]
S E S E ]1[
I E I EPE S E S E
k5
k4
k3
k2
k1
+⎯→⎯•
•⎯→⎯+
+⎯→⎯•
+⎯→⎯•
•⎯→⎯+
•→+
•→•→+
(2) S) (E k S) (E k - (S) (E) k 0 rPSSH Applying
(1) S) (E k r
321SE,
3p
•+•==
•=
4
5I
1
32mt3max
mma
Im
maxsp
t
54I E
kk K and
kk k K and E k V
unchanged! are K andx V
KI 1 K S
S V r- r
:(4) and (3) (2), (1), Combining I) (E S) (E E E
:nconc' enzyme TheI) (E k- (I) (E) k 0 r
:Similarly
=+
==→
⎟⎟⎠
⎞⎜⎜⎝
⎛++
==
•+•+=
•==
max
mmI
mm
V 21it reach To
S. ofion decomposit for the needed is Slarger a means
constants) M-M(apparent K 'K KI1 K 'KSay
⇒
>⇒⎟⎟⎠
⎞⎜⎜⎝
⎛+=
12
decreases. r- and increases slope theincreases I As
Burk}r {Lineweave KI1
VK
S1
V1
r-1
:gRearrangin
S
Imax
m
maxS
⇒
⎟⎟⎠
⎞⎜⎜⎝
⎛++=
I2 > I1 Comp. inhibition I1
No I
Sr-1
maxV1
S1
Uncompetitive Inhibition
I has no affinity for the enzyme itself but for (E S) complex.
SE I SE I
(inactive) S E I S E I
E P S E
S E S E
S E S E
I S E k
I
PE SE SE
k5
k4
k3
k2
k1
2
•+⎯→⎯••
••⎯→⎯•+
+⎯→⎯
+⎯→⎯
⎯→⎯+
••
↓
+
+→•↔+
13
4
5I
Imaxmax
m
s
kk K where
KI 1
V1
VK
S1
r-1
:have we
=
⎟⎟⎠
⎞⎜⎜⎝
⎛++=
Uncomp Inh. I2 > I1Uncomp Inh. (I1)
No Inh.
Sr-1
S1
Slope remains constant, I increases, Intercept increases, -rS decreases.
Non – Competitive Inhibition (Mixed Inhibition)
I and S bind to different sites on E!
P S E(inactive) S E I E I S(inactive) S E I S E I
(inactive) S I I ES E S E
I S E S I E'K K
I I
P E S E S E
II
→•
••↔•+
••↔•+
•↔+
•↔+
••↔+•
++
+→•↔+
ββ
14
We have:
I1I2 > I1
No InhSr-
1
S1
⎟⎟⎠
⎞⎜⎜⎝
⎛++⎟⎟
⎠
⎞⎜⎜⎝
⎛+=
− Imax
m
Imaxs KI1
VK
S1
KI1
V1
r1
Both the slope and intercept increase as I increases.
Sr-1
S1
Non comp
Competitive Uncomp
No Inh.
Summary to compare three types of inhibition
Substrate Inhibition
In some cases, S can act as a inhibitor. If we have uncompetitive substrat inhibition:
inh.) uncomp normalin Swith I (replace
(1) /KS S K
S V r-
S E S S E S
I2
m
maxs ++=
••→•+
15
0)dSdr- (from K K S
SK V r- S K K / Sn conc'high at
KS V~ rs- K / S S Kn conc' lowat
sImmax
ImaxsmI
2
m
maxI
2m
==
=⇒+>>
⇒+>>
-rS
-rSmax
Smax
S
Enzyme Regeneration:
Glucose (SR) is oxidized to δ-gluconolactone (P) by glucose oxidase (E0)
P OH O S
OverallOH E O E
:regenerate torxns, catalysisfurther in used becannot EE P E P E S E S
22E
2R
2202R
R
RR0R0R
0 +⎯→⎯+
+→+
+↔•↔•↔+
δ-gluconolactone (P)
Glucose (SR) E0
ER
H2O2
O2
Enzyme CofactorsIn enzymatic rxns, a second substrate is needed to activate the enzyme. Cofactor / coenzyme bind to the enzyme
Unbound enzyme: apo-enzyme Bound enzyme: holo-enzyme
Multiple E and S Systems:We will look at multiple E and multiple S systems in which cyclic regeneration of the activated E occurs.
16
Bioreactors
A bioreactor is a special reactor that sustains and supports life for cells and tissue cultures. The use of living cells to produce marketable chemical products made the biotechnology a growing industry. Chemicals, agricultural products, food, etc.
Some organic chemicals are thought to be produced by living cells instead of petroleum.
The advantages are mild rxn conditions, high yields and stereospecific compounds.
Bacteria can be modified to be used as chemical factories.
In biosynthesis, the cells are called as “biomass”.
CellNutrients Products
The chemical rxns are simultaneous in the cell
We have
i) Nutrient degradation (fueling) rxns, ATP molecules are used to provide energy.
ii) Synthesis of small molecules (like aa, nucleotides)
iii) Synthesis of large molecules (like polymerization off aa to form proteins)
Cell Growth & Division
The four phases of cell division G1, S, G2 and M
The cell division eq’n can be written as
.substrates theof some are Phosphate andOxygen Nitrogen, Carbon,Product cells More Substrate cells +⎯⎯→⎯
17
Cell Growth
The number of living cells as a function of time can be shown in a batch reactor as:
In Phase I : lag phase, there is little increase in cell conc’n. Cells are adjusting to their new environment, synthesizing enzymes, getting ready to reproduce.
Lag cell conc’n
Phase II: exponential growth phase. Cell divide at maximum rate, all of the enzyme’s pathways are in place.
Phase III: Stationary phase: Cells reach a minimum biological place where the one or more of the nutrient are limiting.
Phase IV: Death phase: A decrease in live cell conc’n occurs.
Rate Laws
The most commonly used rate expression for
Cell + Substrate More cells + Product
is the Monod eq.’n for exp growth.
C K
CC r
C K
C
C r
ss
csmaxg
ss
smax
cg
+=
+=
=
µ
µµ
µ
Cell Growth Rate (g/dm3s)
Specific growth rate (s-1)
Cell conc’n (g/dm3)
µmax : max spec. growth rate
Ks : Monod constant (g/dm3)
Cs : Substrate (nutrient) conc’n (g/dm3)
For most of the bacteria Ks is small (~10-5 mol/dm3) rg = µmax CC
18
Growth rate, rg, depends on the nutrient conc’n (Cs)
Cs
rg
In many systems, product inhibits the growth rate. Wine production is an example, fermentation of glucose to produce ethanol is inhibited by the product ethanol (ethanol kills the yeast)
constant emprical :n)(g/dm stops metabolismat which n conc'product : *Cp
n}Eq.' {Monod *C
C -1 k where
C KC C k r
3
n
p
pobs
ss
scmaxobsg ⎟
⎟⎠
⎞⎜⎜⎝
⎛=
+=
µ
There are also Tessier & Moser eq’ns that fit experimental data better.
The cell death rate is given by rd = (kd + kt Ct) Cc
Conc’n of the substance toxic to the cell
Effect of Temperature can be given as
occurs.growth max at which temp.:Tmrategrowth max offraction : I'
wheree b1e T a I'
I' (Tm) (T)
/RTE-
/RTE-
2
1
+=
= µµrg
T
19
Stoichiometry for cell growth is complex, and chainges with microorganism/ nutrient system, and pH, T, etc.
Let’s look at the case where one nutrient is limiting:
Cell + Subs More cells + Products
S/CC/S
S
CC/S y
1y ; CC
consumed substrate of massformed cells new of mass y =
∆∆
==
Product formation can take place during different phases of the cell growth cycle if it occurs during the exp growth phase:
c
pP/C
ss
scmaxP/CcP/CgP/Cp
CC
formed cells new of massformedproduct of massy where
C KC C y C y r y r
∆∆−
==
+===
µµ
The stoichiometric yield coeff btw product & substrate is given by:
consumedsubstrateofmassformedproduct of mass yP/S =
Another term is the maintenance utilization term: (to maintain a cell’s daily activities)
Time cellsofmassemaintenancfor consumed substrate of mass m =
The rate of substrate consumption
rsm = m Cc
Neglecting the cell maintenance
Cc = yc/s [Cs0 – Cs]
20
Substrate Utilization
cPS/PgS/CS C m r y' r y' r
emaintenancfor consumed
Rate
product form toconsumed
Rate
cellsby consumed
Rate
nConsumptioSubstrate
of rateNet
++=−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
if there is production during growth, it is difficult to seperate amount of substrate used in production and growth
-rs = ys/c + m Cc
The corresponding rate law is
rp = rg yp/c growth associated product formation in the grwoth phase.
Stoichiometric coeff
Apparently, no net growth during stationary phase
Non-growth-associated product formation in stationary phase( )( )snsn
csnpp C+K
CCk=r
conc'n of a second nutrient which is used for maintenance and production (because the the nutrient is used up for growth)
21
Mass Balances
There are two ways that we could account for the growth of microorganisms. Either # of living cells or mass of the living cells.
For a CTSR: { called chemostats are the reactors that contain microorganisms}
CcCs
CsoGeneration+Out–=onAccumulatibalanceCell ln
( )Vrr+VCCV=dt
dCV dgcc −−∞0
Vr+CVCV=dt
dCV SSSSOS −0
00 =V=V
dgC
dgC rr=
dtdCVrVr=
dtdCV −⇒−
(divide by V)
( ) ( ) CgCSS
CgCSSS mCry=
dtdCVmCVry=Vr=
dtdCV −−⇒−− // (In growth phase)
(In most cases Cco = 0)
For a Batch System:
Cell balance
Substrate
Substrate Balance
( )Vry+VmC=Vr=dt
dCV ppsoss −/
for stationary phase (no growth)
22
Product Balance: ( )Vry=Vr=dt
dCV sspp
p −/
Design EquationsFor a CSTR
V=V0
CCO= 0
D=V 0
V
{D= 1/T }
is called the dilution rate (a parameter used in bioreactors)
CSTR Mass Balance: nGen'+Out=Acc −ln
( )dgCC rr+DC=
dtdC
−−0 { For Cell }
sSSOS r+DCDC=
dtdC
− { For Substrate }
( )( )SS
CSmaxCg C+K
CCµ=µC=r
Using Monod Eq'n; the growth rate is determined as:
For st–st operations
dgC rr=DC −
( ) sSSO r=CCD −
If we neglect death rate
µ=DVmuC=Vr=VC=F CgC ⇒0
divide by Cc V
You can control specific growth rate by D!
23
We know (Monod Eq’n)
Combining
If a single nutrient is limiting,- cell growth is the only process to substrate consumption.- cell maintanance is neglected.
Then;
ss
s
CKC+
= maxµµ
DDKC s
s −=
maxµ
cyrr s
gs =−
)(/ ssoscc CCyC −=
⎥⎦
⎤⎢⎣
⎡−
−=D
DKCyC ssoscc
max/ µ
wash-outAssume
if D>µ (at some time)
the dilution rate at which wash-out will occuris:
cc CD
dtdC )( −= µ
0=dr
cg Cr µ=
00 ==>< cc C
dtdC
)0( =cC
sos
so
CKCMD
+= max
max
24
0)(
)(max
//
=
−−=
dDDCd
DDKCDyDC
c
sosscc µ
••
= occ Cm υ timedm
dmmass
timemass 3
3=
cco DC
VC
=υ {Cell production per unit volume is the
mass flow rate of cells out of the reactor}
D for the max. cell production:
Subs
For max D!
cDCcC
sC
Physiologically Based Pharmacokinetic Models:(PBPK)
You can model runs in living systems. You can find conc’n-time profiles for medications, toxins, alcohol and drugs in the body!
Compartments of body are either a PFR or CSTR.Perfusion rate : Blood flow in/out of organs. unst-st models are used as CSTR liver : PFR(unst-st)
The interchange of material btw compartments is primarily through blood flow.
25
1υ
0AC0υ
1V 2V1υ
1AC 0υ 2AC
0υ
0AC
2υ
3AC2υ
1V 2V 3V
Blood
flow
Organ 1 Organ 2
For example, if organs are connected in series as shown
Then the balance eq’ns on species A in the tissue water volumes of the organs
, and are:
Organ 3
333023
3
222322112
2
111011
1
)(
)()(
)(
VrCCdt
dCV
VrCCCCdt
dCV
VrCCdt
dCV
AAAA
AAAAAA
AAAA
+−=
+−+−=
+−=
υ
υυ
υ
Air :matabolism rate of A in organs
1
Ch 8: Steady – State Non-isothermal Reactor DesignEnergy Balances, Rationale and Overview
Calculate the volume necessary to achieve a conversion, X, in a PFR for a first-order, exothermic reaction carried out adiabatically.
The combined mole balance, rate law, and stoichiometry yield:
For an adiabatic, exothermic reaction the temperature profile might look something like this:
To solve this equation we need to relate X and T. We will use the Energy Balance to relate X and T. For example, for an adiabatic reaction, , the energy balance can be written in the form
set X, then calculate T, -VA, and , increment X, then plot vs. X
2
1. Adiabatic (Q = 0); CSTR, PFR, Batch and PBR:
2. CSTR with heat exchanger, UA(Ta-T) and large coolant flow rate.
3 . PFR/PBR with heat exchange
3A. In terms of conversion, X
3B. In terms of molar flow rates, Fi
3
4. For Multiple Reactions
5. Coolant Balance
These are eq’ns we will use to solve rxn engineering problems with heat changes.
3. Energy Balance
Typical units for each term are J/s; i.e. Watts
4
Energy Balance
1st Law of Thermodynamics:
dE = dQ – dW (for a closed system)
Energy of the system
Heat flow to the system
Work done by the system on the surroundings
system}open an {for E F - E F WQdtdE
outout
flowmassby system
the toaddedenergy of rate
inin 321+−=••
Evaluation of the Work Term:
s1
i
~
i1
i
~
i
sf
WV P FV P F
WWW
+⋅⋅+⋅⋅−=
+=
∑∑==
•
n
i out
n
i in
flow shaft
∑∑==
••
⋅+⋅−+=n
iout
n
iins
1ii
1ii
sys H FH FWQdt
dE
Combining with i
~
ii VP U H +=
5
(1)
1. Replace Ei by Ei=Hi-PVi
2. Express Hi in terms of enthalpies of formation and heat capacities 3. Express Fi in terms of either conversion or rates of reaction4. Define ∆HRX
5. Define ∆CP
6. Manipulate so that the overall energy balance is either in terms of the Equations above 1.A, 1.B, 2, 3A, 3B, or 4 depending on the application
Step 1:Substitute
and,
into equation (1) to obtain the General Energy Balance Equation.
General Energy Balance:
For steady state operation:
6
We need to put the above equation into a form that we can easily use to relate X and T in order to size reactors. To achieve this goal, we write the molar flow rates in terms of conversion and the enthalpies as a function of temperature. We now will "dissect" both Fi and Hi.
Flow Rates, Fi
For the generalized reaction:
In general,
ad
ac
ab- 1-
FF where X) ( F F
DCBA
A0
i0iiiA0i
====
=Θ+Θ=
νννν
ν
(2)
(3)
[ ]
[ ]
∑∑∑
∑∑
∑∑
∆Θ=−
+=∆
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡⋅⋅Θ=
+Θ+=⋅⋅
= ∆=
==
X F (T)H - )H - (H F H F H F
H - H ab - H
ac H
ad H
X F )H ( - )H - (H F
........... )H - (H )H - (H F F H - F H
Then
A0RXii0iA0iii0i0
ABCDRX
n
1iA0
H
ii
n
1iiii0A0
n
1iBBB0AA0A0ii
n
1ii0i0
RX
434 21ν (4)
(5)
(6)
.Hin included be it will place, takeschange phase a If
0 X F (T)H -)H - (H F W - Q
:(6) and (1) combining ;0dtdEst -st for
RX
n
1iA0RXii0iA0
∆
=∆Θ+
⎭⎬⎫
⎩⎨⎧ =
∑=
•• (7)
7
Assuming no phase change:
QiRo
ii H )(TH H ∆+=
Enthalpy of formation at TR(ref. temp.)
Change in enthalpy when T is changed from TR to T.
If there is a phase change:
dT C )(TH H
change) phase no (if dT C H
dT C )(TH dT C )(TH H
T
TpiR
oii
T
TpiQi
T
Tipl,mmi
T
Tips,R
oii
R
2
1
R
M
R
∫
∫
∫∫
⋅+=
⋅=∆
+∆+⋅+=
(8)
(9)
(10)
(11)
2iiip T T C ⋅+⋅+= γβα
0 X F (T)H -]T -[T C F W - Q
:(7) into Substitute
]T -[T CdT C H H
n
1iA0RXi0piiA0
i0pi
T
Tpii0i
i0
=∆⋅⋅Θ+
⋅=⋅=−
∑
∫
=
••(13)
(12)
Combine eq’n (11) & (5)
pApBpCpDp
Ro
ARo
BRo
CRo
Do
RX
Rpo
RXRX
C - C ab - C
ac C
ad C
)(TH - )(TH ab - )(TH
ac )(TH
ad H
where
)T - (T C H H
+=∆
+=∆
∆+∆=∆ (14)
(15)
(16)
8
[ ] 0 X F )T - (T C H -]T -[T C F W - Qn
1iA0Rp
oRXi0piiA0 =⋅∆+∆⋅⋅Θ+ ∑
=
••(17)
Combine (13) and (14)
[ ] )T - (T C H-)T - (T C
X 0Q
0W
Rpo
RX
iopiis
∆+∆
Θ=
⎪⎭
⎪⎬⎫
=
= ∑•
•
Adiabatic Operation:
T
Xfor an exothermic, adiabatic rxn for an exo. rxn why does X
increase?
This is from E balance, not mole balance.
)T - (T CH Rpo
RX ∆>∆
so X vs T is linear!!!
Adiabatic Tubular Reactor
Rearrange (18):
[ ]∑∑
∆⋅+⋅Θ
⋅∆⋅+⋅Θ+⋅=
ppii
Rpopiio
RX
C X CT C X T CH-X
T (19)
(20) T) (X,r- dVdX F AA0 =
Combine with differential mole balance:
To obtain T, X and conc’n profiles along the reactor!
Use (19) to construct a table of T vs X.
Obtain k(T) as a function of X -rA as a function of X.
Look at Table 8-2A.
9
TFA0
To
FAe
Te
∑ ⋅ ii HF ∑ ⋅ ii HF
V V+∆V
0 Ws assume V T) - (Ta Ua T) - (TaA U Q =∆⋅⋅=∆⋅=∆••
Steady – State Tubular Reactor with Heat Exchange:
0H FH FQVViiVii =⋅−⋅+∆ ∑∑ ∆+
•
•
∆Q
(1)
The heat flow to the reactor
Overall heat transfer coefficient (U)
Heat exchange area (∆A)
Difference between ambient temperature (Ta) and reactor temperature (T)
T)-(Ta V Ua T) - (TaA UQ ⋅∆⋅=∆⋅=∆•
(2)
DVA 4= Diameter of reactor
If ∆V 0
dVdHF - H
dVdF
T)-(Ta Ua
(3)n Eq' diff.
)(-r r dVdF :balance mole
dV)H (F d
T)-(Ta Ua
iii
i
Aiii
ii
∑∑
∑
⋅=
==
⋅=
ν
(3)
(4)
(5)
dVdT C
dVdH
dT C H
pii
pii
⋅=
⋅=
(6)
Combine (5) & (4) & (6)
dVdTCF - )(-r H T)-(Ta Ua piiA
H
ii
RX
⋅⋅⋅= ∑∑∆
434 21ν (7)
10
Rearrange:
∑ ⋅−∆⋅
=pii
removedHeat generatedHeat
RXA
C FTa) -(T UaHr
dVdT
4 84 76484 76
(8)
X) ( F F iA0i ν+Θ⋅=
( ) { }PFR afor XC X F
Hr T) -(Ta Ua dVdT
piA0
RXA
∑ ⋅∆++Θ⋅∆⋅+
=νi
Substitute into (8)
for a PBR {dW = ρb dV}( )
∑ ⋅
∆⋅+⋅=
pii
RXAb
C F
H'r T) -(Ta Ua
dWdT ρ
These eq’ns will be coupled with mole balance eq’nsA0
A
Fr-
dVdX
=
Tao
FA0, To
V V + ∆V
R1R2
Reactants
Heat transfer fluid
The fluid will keep the rxn temperature constant for endo/exo – thermic rxns
You might have
A. Co – current Flow
B. Counter – current Flow
Balance on the Coolant Heat Transfer Fluid:
11
FA, T
V V + ∆V
FA, T
Ta: coolant temperature
mc, Hc mc, Hc
0 V Ta)-(T UaH m -H m
0 Q E -E
VVcc
Vcc
conductionVV
outV
in
=∆+
=+
∆+
••
•
∆+
••
Divide by ∆V and take limit as ∆V 0
A. Co – Current Flow
The Energy Balance
Cm
Ta) - (T Ua dVdT
dVdTC
dVdH
0 Ta) - (T Ua dVdHm
p
aap
c
c
⋅=⇒=
=+⋅
•
•
c
cexothermic
endothermic
Tao
Tao
V
TFAo, To
V V + ∆V
Ta2 Ta
V = 0 V = Vf
Cm
T) - (Ta Ua dVdT
pc
a
⋅= •
c
At the entrance X = 0; V = 0; Ta = Ta2
At the exit V = Vf; Ta = Tao
The sol’n to the counter-current flow problem to find {T X} is a trial & error procedure.
Assume a coolant temp at the entrance (Ta2)
Solve ODEs to calculate X, T and Ta as a function of V:
Find Ta(V = Vo)
a2
a2oaa2oa
Tanother assume Else
T)V(VT T-)V(VT If ==→<= ε
B. Counter – Current Flow
12
Equilibrium Conversion
As T X {For endothermic rxns}
As T X {For exothermic rxns}
Exothermic Rxns:
rxn}order {1st Kc 1
Kc Xe+
=
From Le Chaltlier’s Law
(Kc as T if ∆H<0) exothermic γ)(RTK
K pc =
To find the max X in an exothermic rxn carried adiabatically:
To1To
Xe
Xe1
Adabatic temperature
To1 >To
Energy balance
⎭⎬⎫
⎩⎨⎧
∆−
⋅⋅Θ=∑
X T as T tochanged is T if
)T(H)T-(T C
X
e
o1o
RXN
opiie
11R
)(THexp )(TK )(TK
0C if
T R)T-T(C )(TH
T R(T)H
dTdlnK
21
RRXNo
1p2p
~
2R
~
RRXNo
2RXNp
⎪⎪⎪⎪
⎭
⎪⎪⎪⎪
⎬
⎫
⎪⎪⎪⎪
⎩
⎪⎪⎪⎪
⎨
⎧
⎭⎬⎫
⎩⎨⎧
⎟⎟⎠
⎞⎜⎜⎝
⎛−
∆=
=∆
⋅∆+∆
=⋅
∆=
TT
p
p
13
Xe
Xeb
To increase the conversion in an exothermic rxn, use multiple reactors with interstate cooling:
For Endothermic Rxn (you need heating)
Xe
3
2
1
350 500 600
X
T, K
As T Xe but the (-rA) decreases!
(so the conversion is achieved at the end of the reactor)
So there must be an optimum temperature to achieve max X.
Curve A: Rxn rate slow, rxn dictated by rate of rxn and reactor volume
As T , r , X
Curve B: Rxn rate very rapid. Virtual equilibrium reached in X dictated by equilibrium conversion
Optimum Feed Temperature
Adiabatic Reactor of fixed size
Reversible & Exothermic Rxn T ; X as 0H if
XCH-T T
rx
pA
rxo
<∆
⋅∆
=
14
Adiabatic Rxn Algorithm
Suppose
1. Choose X
Calculate T
Calculate k
Calculate T/To
Calculate CA
Calculate CB
Calculate KC
Calculate -rA
2. Increment X and then repeat calculations.
3. When finished, plot vs. X or use some numerical technique to find V.
Levenspiel Plot for anexothermic, adiabatic reaction.
15
Consider:
For an exit conversion of 70%For an exit conversion of 40%PFR Shaded area is the volume.
For an exit conversion of 70%For an exit conversion of 40%
CSTR Shaded area is the reactor volume.
We see for 40% conversion very little volume is required.
CSTR+PFR
16
For an intermediate conversion of 40% and exit conversion of 70%
Looks like the best arrangement is a CSTR with a 40% conversion followed by a PFR up to 70% conversion.
Evaluating the Heat Exchanger Term
Energy transferred between the reactor and the coolant:
Assuming the temperature inside the CSTR, T, is spatially uniform:
17
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:
Since the coolant flow rate is high, Ta1 Ta2 Ta:
Multiple Steady States (MSS)
where Heat generated term
Heat removed term
18
R(T):
Varying Entering Temperature
Vary non-adiabatic parameter κ:
if you increase FA0 (molar flow rate) or decrease heat – exch area then κ will decrease
Increase To
Slope = Cpo (1+κ)
T
R(T)
κ decrease
T
R(T)κ=∞
κ= 0
Ta To
κκ
κ
+⋅+
=
⎪⎭
⎪⎬⎫
⎪⎩
⎪⎨⎧
=
1T T T
C FUa
a0c
p0A0
Low E
High EG(T)
T
increasing TG(T)
T
G(T)
For a first order rxn:
kkX⋅+
⋅=
ττ
1
19
Extinction temp
Ignition temp
Lower st-st
Upper st-st
Ignition – Extinction Curve:
The point of intersection of G(T) and R(T) give Tst-st.
By plotting Tst-st vs To, we obtain ignition – extinction curve.
As To Tst-st
Runaway Rxns in a CSTR: R(T),
G(T)
Tc T* Tangency pointκκ+⋅+
=1
T T T a0c
E
*2
c*
rcTR T-TT ⋅
==∆
Reactor T
if this diff. is exceeded, transition to the upper st – st will occur.
At this high temp, it is undesirable or even dangerous.
20
PFR
rxn) single a(for C F
(T))H)(-(-rT)-(T UadVdT
m
1ipii
RxnAa
∑=
∆+=
When q multiple rxns occur with m species:
[ ]species :j
rxn : i
C F
(T)H-)(-rT)-(T Ua
dVdT
m
1jpjj
1ijRxn,ija
∑
∑
=
=
∆+=
q
i
Ex:
[ ] [ ]
C FC FC F(T)H-)(-r(T)H-)(-rT)-(T Ua
dVdT
pCCpBBpAA
Rxn,2B2BRxn,1A1Aa
++∆+∆+
=
C B
B A
:1Rxn
2
1
k
k
⎯→⎯
⎯→⎯
CSTR
[ ] [ ] rxn) single a(for 0 Vr(T)H -]T -[T C F W - Qn
1iARXN0piiA0s =⋅∆⋅⋅Θ+ ∑
=
••
0 (T)HrV-]T -[T C F T)-UA(T
0 (T)HrV-]T -[T C F W - Q
n
1i 1ijRXN,ij0piiA0a
n
1i 1ijRXN,ij0piiA0s
=∆⋅⋅⋅Θ+
=∆⋅⋅⋅Θ+
∑ ∑
∑ ∑
= =
= =
••
q
i
q
i
When q multiple rxns take place0
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