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Biot Numbers by David Adrian
This document is a review* of some of the concepts of heat and mass transfer,
particularly focusing on the dynamics at the interface between two disparate materials,
such as the boundary of a solid particle submerged in a fluid. In the cases considered,
the interface is stationary and there is no phase change or chemical reaction at the
interface.
The Biot number is a dimensionless group that compares the relative transport
resistances, external and internal. It arises when formulating and non-dimensionalizing
the boundary conditions for the typical conservation of species/energy equation for
heat/mass transfer problems.
If your problem consists of an object suspended in a well mixed fluid, commonly you
only need to calculate the dynamics of the object (such as the temperature as a function of
position and time). If we focus on the fluid/object interface, the convective flux from the
bulk fluid to the object must equal the diffusive flux from the surface to the interior of the
object. This is typically formulated as a Robin boundary condition at the interface. For
example, consider the unsteady heat transfer in a solid sphere at initial temperature T0
submerged in a fluid of temperature T∞ (this is also the “bulk” temperature, and could be
given the symbol Tb).
N
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V
S
At the fluid-solid interface, the flux of heat into the sphere from the fluid must equal the
flux of heat from the surface of the sphere to the interior.
r
r
qexterior = qinterior
r
r
qexterior = h(Ts − T∞ )n
r
qinterior = −kT ∇T surface
S
.B
W
The variables are defined as follows: q is the heat flux, h is the heat transfer coefficient in
r
the fluid, T is the temperature, n is the outwardly pointing normal from the solid, and kT
is the thermal conductivity of the solid. (Also recall that the heat transfer coefficient can
be obtained from correlations, and is basically just kT , fluid / δ T in systems without any
W
W
interfacial reactions or phase changes. The variable δT is the thickness of the thermal
boundary layer.)
Since our system is spherically symmetric:
r
r
qexterior = h(Ts − T∞ )er
dT
r
qinterior = −kT
dr
− kT
*
dT
dr
surface
r
er
surface
= h(Ts − T∞ )
I hope it is a review! TAs should be knowledgeable about this stuff if you have questions.
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We may non-dimensionalize the system by changing variables. We must pick a
characteristic length scale and temperature scale with the hope that our dimensionless
temperature scale will be O(1)†, i.e. it goes from 0 to 1. Also, we want the derivatives of
that function with respect to our dimensionless length to also be O(1). This is getting a
good scaling for the problem.
In this problem, a good dimensionless temperature is:
T − T∞
θ =
T0 − T∞
This temperature variable starts out at 1 and decays to zero once the solid and fluid temperatures are equal. A good dimensionless length is the inverse of the surface area to volume ratio: SA
4πR 2
3r
=r
=
η =r/L =r
V
(4 / 3)πR 3 R
L = R/3= O(R) in the case of a sphere so “L = R” would also be an okay choice. L is the
“typical” length scale that heat in the solid particle must diffuse to get to the surface.
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Using our definitions of dimensionless length and temperature, our Robin boundary
condition becomes:
(T0 − T∞ ) dθ
− kT
= h(Ts − T∞ )
L
dη
surface
−
dθ
dη
N Bi =
=
hL
θ
surface
kT
hL
L / kT
=
surface
= N Bi θ
V
S
S
.B
surface
W
"internal diffusion resistance"
=
W
1/ h
"external convection resistance"
kT
By definition, our dimensionless temperature is at most 1. Consider the effect of Biot
number on the problem.
W
Case 1: NBi<<1
dθ
−
= N Bi θ
dη surface
Since θ
surface
surface
is at most one, in order for the equation to be true the surface gradient in
our dimensionless temperature is also small (as small as the Biot number). This means
that a good approximation to the dynamics can be found from a uniform temperature
throughout the sphere (a lumped system model). This means that the “external
convection resistance” dominates the problem, and the “internal diffusion resistance” is
†
“Big O Notation” is used to state in rough terms the magnitude of terms relative to each other, usually
only considering the order of magnitude or scaling
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small relative to the convection resistance so that it can be neglected when determining
the total resistance. (Think back to the “resistors in series” analogy in heat transfer.)
For example, a small heat transfer Biot numbers can arise in the case of a small
conductive metal sphere in a stagnant fluid such as air. The convection speed is very
slow (the resistance is large and h is small because the fluid is stagnant) and the
conduction speed is very high (the thermal resistance is small, or the thermal conductivity
is large, or the distance that heat has to diffuse in the object is very small).
Case 2: NBi>>1
On the other hand, if our Biot number is very large, the gradient must either be very large
at the surface or θ surface must be very small. However, since we scaled the derivative
properly,
dθ
dη
should be O(1). This means that θ
surface
is very small. Recall that if θ
surface
is zero, this corresponds to the surface being in equilibrium with the bulk temperature (it
takes the bulk value). This simplifies our boundary condition for the problem so that we
can just use the condition that T (r = R) = Ts = T∞ and still get good results.
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Mass Transfer Biot Numbers
In the case of mass transfer, the definition of the Biot number could get a little more
complicated because the partition coefficient between phases is involved. In this class,
however, we aren’t worrying about partition coefficients, probably because we can’t
easily measure the internal concentrations anyway so we lump their effect into other
unknown constants (such as the surface reaction rate constant).
V
S
S
.B
The analogous case of a catalyst particle in a reactor fluid gives us the Robin boundary
condition:
r
r
Wexterior = k m (c s − c∞ )er
r
dc
r
Winterior = −De
er
dr surface
W
W
− De
θ=
W
dc
dr surface
c − c∞
= k m (c s − c∞ )
c0 − c∞
η =r/L =r
− De
SA
V
=r
4πR 2
(4 / 3)πR 3
(c0 − c∞ ) dθ
L
dη
=
3r
R
= k m (c s − c∞ )
surface
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−
dθ
dη
=
surface
km L
De
θ
surface
= N Bi,m θ
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4
surface
In this case, we should take a practical look at the terms in the Biot number and see if it is
stuck in a particular range (always large, always small, etc.).
N Bi,m =
km =
km L
De
D fluid
δm
N Bi,m =
(stationary interface with no interface reaction)
D fluid L
De δ m
The boundary layer is the largest it can be in a stagnant system, and in that case it is
approximately the same length scale as L. (Consider the example of the Sherwood or
Nusselt number for spheres in a stagnant medium.) Thus the ratio L / δ m is typically
greater than one.
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S
Also, the diffusivity of the species in the fluid is typically much greater than the effective
diffusivity in the particle, where the species has to navigate the tortuous pore space. Thus
D fluid / De is also typically much greater than one.
S
.B
Let’s consider the mass transfer Biot number definition again and consider what it means
in words:
D fluid L
L / De
"internal diffusion resistance"
N Bi ,m =
=
=
>> 1
De δ m δ m / D fluid "external diffusion resistance"
W
W
So the consequence of this ratio is that whenever we are looking at a reactor where we have bulk kinetic data, if we find out that we have an external diffusion limitation, we know that we must also have an internal diffusion limitation. W
However, we might not be all that concerned, because when you have an external limitation, you may be able to predict the reaction rate purely by considering the surface flux: − r'' = W = k m (cb − c s ) ≈ k m cb because c s << cb when external limitations prevail. www.bsscommunitycollege.in
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Adrian, course materials for 10.37
Chemical and Biological Reaction
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 1: Preliminaries & Remembrance of Things Past
This lecture covers: reaction stoichiometry, lumped stoichiometries in complex
systems such as bioconversions and cell growth (yields), extent of reaction,
independence of reactions, measures of concentrations, single reactions and reaction
networks, and bioreaction pathways.
Fo
FA
Figure 1. A schematic of a control volume with inflow of Fo and outflow of species A,
FA.
F = total molar flow rate (moles/sec)
Fo = total molar flow rate entering control volume
FA = molar flow rate of species A
FAo = molar flow rate of species A entering control volume
NA = Moles of species A
N
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V
S
S
.B
Mass balance: (change in “A” inside control volume)= (amount of “A” that entered)(amount of “A” that exited) + (amount of “A” created inside the control volume) –
(amount of “A” destroyed inside the control volume)
dN A
= FAo − FA + GA
dt
moles
GA [ = ]
sec
If homogeneous, GA = rAV
W
W
W
else, GA = ∫ rA dV
where V is volume and rA is the rate of A created or destroyed (moles/sec/volume).
The mass balance can also be written as:
diffusion
convection
d ρA P
2
= v∇ρ A + D∇ ρ A + rA
dt
where v is velocity, D is the diffusion coefficient, and ρA is the molar density of A
(moles/volume).
Example:
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A+B→C+D
rA = − k [ A][ B ] , rC = + k [ A][ B ] ,
[ A][ = ]
mole A
liter
where k is a measurable rate constant that changes with respect to T and P but
remains constant with respect to changing concentrations.
r1 = k [ A][ B ] , then
=−1
P
rA = ν A,1 r1
If
ν A,1 is the stoichiometric coefficient for species A in reaction 1.
=+1
P
rC = ν C ,1 r1
where
Likewise,
If there are n reactions involving species “A” then
n
rA = ∑ν A,i ri
N
I
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E
i =1
GA (t ) = ∫∫∫
rA ( x, y, z, t )
dxdydz
Vol . - k (T ( x , y , z ,t )) A( x , y , z ,t ) B ( x , y , z ,t )
[
][
]
V
S
P
d [ A]
NA V
= − k [ A][ B ] → *only true if there are no flows!
dt
dN A
= ∫ rA dV ≈ rAV , assume rA is true throughout volume
dt
S
.B
W
If the system is homogeneous and there are no flows:
dN A
= rAV
dt
Therefore:
W
W
d ⎛ NA ⎞
⎜
⎟ = rA iff homogeneous, no flow, constant V
dt ⎝ V ⎠
Extent of Reaction
ξ [ =] moles, extent of rxn.
moles
, rate of extent of rxn.
sec
A+ B → C + D
ξ [ = ]
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 1
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N A = N A,initially − ξ , if there is one reaction involving A
N A = N A,initially − ∑ν A,nξ n , if there are several reactions
n
ξn = ∫ rn dV , n is the reaction number
GA = ∫ rA dV , A is the species
Conversion
A+ B → C + D
N
− NA
(dimensionless)
X A = A,initial
N A,initial
NC
( ≈ 1 since rxn is 1:1)
XC =
N A,initial
⎛ A+ B → C + D⎞
⎟ Selectivity is good if A → C , bad if A → U .
⎜
A →U
⎝
⎠
*May worsen as rxn. goes on ( A → C slows, A → U keeps going)
N
I
.
E
V
S
S
.B
W
W
W
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Prof. William H. Green
Lecture 1
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 2: The Reaction Rate & Reaction Mechanisms
The lecture covers: Definitions in terms of reacting compounds and reaction extent,
rate laws, Arrhenius equation, elementary, reversible, non-elementary, catalytic
reactions.
From previous lecture:
dN A
= FAo − FA + GA , GA = ∫ rA dV
dt
Example:
A + B → 2C
*Reactions are reversible (often will neglect reverse)
XA =
N Ao − N A
N Ao
XA =
FAo − FA
FAo
A,B
XC =
FC
2 FAo
N
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C,A,B
FAo
FBo
V
S
Figure 1. A reactor with reactants A and B constantly flowing in and product C and
unused reactants A and B flowing out.
FAo =
S
.B
moles A flowing in
= [ A]input vin = [ A]o vo
sec
liters
sec
W
moles A flowing out
FA =
= [ A]output vout
sec
[ A]output vout
X = 1−
[ A]in vin
W
W
Closed reactor, const. V
Detailed balance- all steps in equilibrium, total system
d [ A]
dt
= rA = − k [ A][ B ]
[C ]
K eq =
[ A][ B ]
2
when
[ A][ B ]
[C ]
=
2
K eq
the rxn. stops -> rA=0
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2
⎛
C]
[
⎜
rA = − k [ A][ B ] −
⎜
K eq
⎝
9
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⎞
k
⎟ , krev = for , forward-reverse in one expression
⎟
K eq
⎠
same!
Catalysis
( [ B ])
rA, forward = kcat [ catalyst ][ A] f
may be included
i.e. when B is
very small
Rate limiting step determines the kinetics (slow step). The kinetics are insensitive to
[B] because B is not part of this slow step.
[C ]
K eq [ A][ B ]
2
rA,reverse = rA, forward
rA,net
2
⎛
⎞
C]
[
⎟
= − kcat [ catalyst ] ⎜ [ A] −
⎜
K eq [ A][ B ] ⎟
⎝
⎠
K eq = e − ΔG RT
ΔG = ΔG of , products − ΔG of ,reactants
A→ B+C
[ B ][C ] = moles
K eq =
[ ]
liter
[ A]
pB pC
pA Po
= K eq =
V
S
S
.B
KC
Partial Pressures
N
I
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E
Standard state
[ B ][C ]
W
W
Po
[ A] RT
W
Ideal Gas: pV = nRT ⇒
n
P
=
V RT
Po is the standard state pressure (1 atm), this makes the units cancel. Using partial
pressures is accurate within 10%, more error with liquids.
Stoichiometric coefficient
N A = N A,initial + ∑ν A,iξi
Extent of rxn.
N rxn
N C = N C ,initial + ∑ν C ,iξi
i =1
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 2
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column
vector
column
vector
10
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matrix vector
N = No + ν i ξ
Do columns span space? Often no, limits on what is achievable.
NC
Reaction trajectoryPlot state of system
ξ=ΝΑο
t=10
ξ
d
(N ) = F (N)
dt
t=0.01
NB
ξ1=0
N
I
.
E
V
S
S
.B
NA
Figure 2. A plot of the reaction trajectory. ξ is the extent of reaction.
W
Conservation Laws
W
Conserve atoms
2CH 4 + O 2 → C2 H 4 + 2H 2 O
C A N A + CB N B + CC N C = const.
constant moles
W
C :1× NCH 4 + 2 × NC2 H 4 = N C ,initial
H : 4 × NCH 4 + 4 × NC2 H 4 + 2 N H 2O = N H ,initial
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 2
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 3: Kinetics of Cell Growth and Enzymes
This lecture covers: cell growth kinetics, substrate uptake and product formation in
microbial growth, enzyme kinetics, and the Michaelis-Menten rate form.
Biological Rate Laws- Enzymes and Cell Growth
Rate Law: − rA = f (C A ,CB ,CP ,T , pH ,...)
Why would you need a rate law?
-Predictive description of a production process
-Design tool for forming a desired product
-Consistency or inconsistency with alternative mechanism
For a hypothesized mechanism, we can often derive an exact, closed form analytical
solution.
If not, it’s used to approximate:
-some reactions go rapidly to equilibrium
-the concentrations of some species rapidly reach their steady state values
-the rate-limiting step
Or, as a last resort- numerical solution
N
I
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V
S
S
.B
X=first order
-rA
X
W
X
X
=second order
−rA = kC A 2
=zero order
−rA = k
W
X
X
− rA = kC A
W
CA
Figure 1. Rate versus concentration graphs for zero, first, and second order
reactions.
Enzymes-Biological Catalysts
S=Substrate
E=Enzyme
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Energy
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Activation energy
decreases with E
(enzyme)
S
P
Reaction Progress
Figure 2. An energy diagram for a reaction with and without an enzyme. The
activation energy is lower with an enzyme, so the reaction proceeds faster.
CS
Initial Rate
− rS = −
dCS
dt
t =0
N
I
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E
Time
Figure 3. The slope of the concentration versus time curve at time = 0 can give the
initial rate.
V
S
S
.B
Michaelis Menten
~1st order
W
~0th order
-rS
X
X
X
X
X
X
W
X X X
W
X
X
What gives change
from 1st order to 0th
order?
CS
Figure 4. Michaelis-Menten kinetics.
Hypothesized Mechanism:
-Encounter complex ES formed
-Irreversible reaction occurs
In more complex cases,
-Rapid release of product from complex
these are not always true
-Assume rapid equilibrium is reached in the formation of ES
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 3
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Later, Briggs-Haldane derived a law with steady state assumption for ES and they
got the same rate law as Michaelis-Menten. However, the Briggs-Haldane method is
more generally applicable.
1
ZZZ
X
E + S YZZ
Z ES
k
k
−1
ES ⎯⎯→ E + P
kcat
Steady state assumption on ES
→
dCES
=0
dt
Material Balance on ES:
dCES
= k1CE CS − k−1CES − kcat CES ≈ 0 (steady state)
dt
Free enzyme concentration ≠ CE ,0
because CE ,0
CE ,0 = CE + CES
0 = k1 (CE ,0 − CES )CS − CES (k−1 + kcat )
Cs ≈ CS ,0
CES
− rS = rp =
dC p
W
W
= kcat CES
W
dt
V
S
S
.B
Solve for CES (steady state solution)
CE ,0CS
=
k−1 + kcat
+ CS
k1
N
I
.
E
CS ,0
kcat CE ,0CS
V C
=
= max S
k−1 + kcat
+ C S K m + CS
k1
Vmax = kcat CE ,0 = maximum reaction velocity
Km =
k−1 + kcat
= Michaelis constant
k1
This model fits the data:
Vmax
CS
Km
-As CS K m , − rS → Vmax
-As CS K m , − rS →
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 3
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-rS
X
X
X
X
X X X
X
The rate flattens out
because all of the
enzyme is bound.
X
X
CS
Figure 5. Michaelis-Menten kinetics.
-This rate law reappears in heterogeneous catalysis (Langmuir-Hinshelwood)
-In the actual Michaelis-Menten derivation Kmax was an equilibrium constant
-Briggs and Haldane showed that this rate law still works for steady state
-Steady state approximation is more general
-Steady state approximation is consistent with the data
-Steady state is defined over a limited period of time. Because there is an
irreversible step, eventually all of the intermediate will be consumed.
-Equilibrium assumption is not exactly correct. The irreversible step prevents
true equilibrium.
N
I
.
E
V
S
Growth Kinetics
S
.B
A population of single cells, growing without limitations
N=#cells per volume
dN
= μ N , μ = constant, specific growth rate
dt
N = N 0e μ t − exponential growth while true, "Malthusian growth"
W
W
Growth Stops
-Run out of nutrients
-Accumulate toxic byproducts
W
Nutrient Effect on μ (Monod)
μ C
μ = max S
K S + CS
This expression is purely empirical (no mechanistic meaning). Even the fit is not that
good.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 3
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μ C
dN
= max S N
dt K S + CS
15
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We will return to this when we
look at bioreactors
dCS
−1
=
μN
dt
Yx s
I=inhibitor or toxin
μ ∝ e − kC
μ
I
μ∝
1
(1 − kCI )
μ∝
kI
k I + CI
All forms fit the data
CI
N
I
.
E
Figure 6. Growth rate versus concentration of inhibitor. Many functional forms fit
the curve.
V
S
μ
X
X
X
X
X
X
X X X
X
S
.B
W
CS=concentration of growth limiting
substrate, often glucose
W
W
KS
CS
Figure 7. Growth rate versus concentration of growth limiting substrate.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 3
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Cell Growth as a Chemical Reaction
Aerobic growth:
Carbon source + O2 + Nitrogen Source → Biomass + Byproducts + H2O +CO2
CH1.8O0.5N0.2
(for a typical microbial culture)
A yield coefficient can be defined:
YA B =
ΔA
ΔB
A=biomass, byproducts, CO2, heat (any of these)
B=carbon source or oxygen
For glucose, Yx s ≈ 0.6 ± 0.1
For oxygen,
Yx O2 ≈ 1.9 ± 0.7
g biomass
g glucose
g biomass
N
I
.
E
g O2
V
S
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 3
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 4: Reaction Mechanisms and Rate Laws
Fundamentals of Chemical Reactions
­PSSA (SS, QSSA, PSSH)
­long chain approximation
­rate­limiting step
A+B
Stable molecules: neutral, closed shells
(­)
(e­)
nucleus
(+)
nucleus
bond
Figure 1. Stable molecules.
Pauli Exclusion Principle
­You can’t put 2 identical e­ in the same exact spot
N
I
.
E
V
S
S
.B
Figure 2. Two electrons in an orbital have opposite spin.
W
Bond Forming
0 e­
+
2 e­
W
W
H
H
+
empty
orbital
H
H
Figure 3. Bond formation. On the left, an empty orbital receives two electrons from
another orbital. On the right, half­filled orbitals on the H atom mix to form a filled
bonding orbital with two electrons.
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Boltzmann Distribution
p ( E ) ∼ e − E k BT
kB =
R
NA
E ≫ k BT → very unlikely
E A + EB > E Activation
Barrier
B
A
Small fraction will
collide correctly
and react
k (T ) ∼ Ae − Ea
k BT
A is the prefactor, proportional to the number of ways the molecules get together
with sufficient energy to react.
Reactive Intermediates
­charged
acid/base chemistry
­empty orbital
metal catalyst
­single e­ orbital
free radical
N
I
.
E
V
S
Example:
O
O
RC
OR + H2O ⇌ RC
O
S
.B
OH + ROH
(endo­
thermic)
O­
W
k2
k1
⇀
RC OR → R
OH + RC OR ↽
k−1
­
(base catalyzed)
W
OH
W
OH
(acid)
+ RO­
k3

→
O
R
O­
+ ROH
minor
species (SS)
O
O
R
minor
species (SS)
O
­
O
+ H2O ⇌
R
OH
+ OH­
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 4
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­

 O
d RC OR
 OH 
≈0
dt
­
O


O
 k1 OH­  
RC OR
RC OR  ≈


k−1 + k2

 OH
­
 O



RC
OR
k2



RO­  ≈  OH
k3 [acid]
d [RO­ ]
≈0
dt
­
rROH
O

O

kk
­
= k3 RO  [ acid] = k2 RC OR = 1 2 OH­  

RC OR

 k−1 + k2

 OH 
keff
Rate Limiting Step
­Only 1 rate constant of keff is really relevant
­What do you have most of in a reaction mix? This is the material preceding the rate
limiting step.
O
O
RC
N
I
.
E
OR + H2O
S
.B
OH
O
k2
⇀
→
RC OR 
H + RC OR ↽
k−1
R
+
k1
+
W
(acid catalyzed)
(SS)
R
+
W
+ H2O 
→ ROH+H
+
k3

d RC
+
dt
OH
W

OR
 ≈0
d [R + ]
≈0
dt
V
S
⇌ RC OH + ROH
O
OH
(acid)
+ R+
minor
species (SS)
O
+ 



k
H
1
 OH 
  RC OR


RC OR ≈
k−1 + k2
 +

 OH 
k2 RC OR
 +


R +  ≈ 
k3 [H2O
]
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 4
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 OH 
 O

kk
rROH = k3 R+  [H2O] = k2 RC OR = 1 2 H+  

 +
 k −1 + k 2
RC OR


keff
Ethylene
(in plastics)
C2H6 → H2 +C2H4
Ethyl radical
C2H6 + H → H2 + C2H5
i
i
C2H5 → C2H4 + H
i
i
−rC2H6 = k1 Hi  [C2H6 ]
C2H6
i
CH3
slow

→ 2CH3i
inefficient, but important (radical creation)
+ C2H6 → CH4 + C2H5i
i
2C2H5
→ C2H6 + C2H4
N
I
.
E
(radical destruction)
(disproportionation)
C2H6 + C2H4 → 2C2H5 reverse disproportionation also happens
i
V
S
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 4
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 5: Continuous Stirred Tank Reactors (CSTRs)
This lecture covers: Reactions in a perfectly stirred tank. Steady State CSTR.
Continuous Stirred Tank Reactors (CSTRs)
Batch
Continuous
Plug Flow Reactor
CSTR
Open system
Steady State
Well mixed
Open system
Steady State
Continuous
Close system
Well mixed
Transient
N
I
.
E
Figure 2. A batch reactor. Figure 1. A plug flow reactor, and continuous stirred
tank reactor.
Mole Balance on Component A
In-Out+Production=Accumulation
FAo − FA + rAV = 0 (steady state)
V=
FAo − FA
−rA
S
.B
FA = FAo − X A FAo
In terms of
conversion, XA?
V
S
W
W
W
What volume do you need for a certain amount of conversion?
V=
FAo X A
−rA
where rA is evaluated at the reactor concentration. This is the same as the exit
concentration because the system is well mixed.
For a liquid phase with constant P:
FAo = C Ao v0 (v0 =volumetric flow rate)
FA = C Av0
V C Ao X A
=
v0
−rA
V
τ = ← average time a volume element of fluid stays in the reactor
v0
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τ=
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C Ao X A
−rA
Consider: 1st Order Reaction Kinetics
− rA = kC A concentration or conversion?
Æ convert rate law from CA to XA
C A = C Ao (1 − X A )
− rA = kC Ao (1 − X A )
τ=
C Ao X A
XA
⎪⎫
=
⎬ reactor size in terms of conversion and rate constant
kC Ao (1 − X A ) k (1 − X A ) ⎪⎭
Æ rearrange to find how much conversion for a given reactor size
XA =
τ≡
1
≡
k
τk
1+τ k
average reactor residence time
N
I
.
E
average time until reaction for a given molecule
We can now define a “Damköhler number”
reaction rate − rAoV
, rAo is the reaction rate law at the feed conditions
Da =
=
flow
FAo
V
S
S
.B
st
For a liquid at constant pressure with 1 order kinetics:
Da = kτ
⇒ XA =
Da
1 + Da
W
W
therefore:
As DaÆ, XAÆ1
As Da∞, XAÆ0 (molecule probably leaves before it can react)
W
For a liquid at constant pressure with 2nd order kinetics:
− rA = kC A2
= kC Ao 2 (1 − X A )
τ=
2
C Ao X A
C Ao X A
XA
=
=
2
2
−rA
kC Ao (1 − X A )
kC Ao 2 (1 − X A )
solving for conversion:
XA =
(1 + 2τ kC Ao ) −
1 + 4τ kC Ao
2τ kC Ao
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture #5
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Da =
23
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kC Ao 2 V
= τ kC Ao
C Ao vo
Thus, conversion can be put in terms of Da.
XA =
(1 + 2 Da ) −
1 + 4 Da
2 Da
How long does it take for a CSTR to reach steady state?
In-Out+Production=Accumulation
FAo − FA + rAV =
dN A
dt
For a liquid at constant density this is:
C Ao − C A + rAτ = τ
dC A
dt
Ænon-dimensionalize
C
Cˆ A = A
C Ao
t
tˆ =
τ
τ C Ao dCˆ A
C Ao − C Ao Cˆ A − kC Ao Cˆ Aτ =
dtˆ
τ
N
I
.
E
Da
dCˆ A ⎛ P ⎞ ˆ
+ ⎜ 1 + kτ ⎟ C A = 1
⎟
dtˆ ⎜⎝
⎠
dCˆ A
+ (1 + Da ) Cˆ A = 1
ˆ
dt
V
S
S
.B
with initial conditions:
Cˆ A = 0, tˆ = 0
we have the solution:
Cˆ A =
(
1
ˆ
1 − e− (1+ Da )t
1 + Da
W
)
W
In nondimensional terms, it exponentially approaches a new steady state with a
W
characteristic time
τ
1 + Da
.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture #5
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Cˆ A
1
1 + Da
1
tˆ
1 + Da
Figure 3. Approach to steady state in a continuous stirred tank reactor (CSTR).
The time at which ½ of the steady state concentration of CA is achieved is the half
time:
ln(2)
τ
1 + Da
N
I
.
E
CSTRs in Series
(Liquid and at constant pressure)
V
S
CA0
W
S
.B
Da2
Da1
CA2
W
Figure 4. Two tanks in series. The output of the first tank is the input of the second
tank.
W
1st order reaction kinetics
C A1 =
C A0
1 + Da1
For the second reactorÆ iterate
C A2 =
C A0
(1 + Da1 )(1 + Da2 )
If the CSTRs are identical,
C An =
C A0
(1+ Da )
n
Æ many CSTRs in series looks like a plug flow reactor.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture #5
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 6: Concentration that Optimizes a Desired Rate
This lecture covers: Selectivity vs. conversion and combining reactors with
separations.
Optimal
A
- What is the target? goal? Æ Often it is to maximize profit or
production
recycle A (U, P)
product
Purer P
(A, U)
Separator
P, A, U
byproducts
N
I
.
E
Figure 1. Schematic of a reacting system with a recycle stream.
const. (vol.)
Fp = rp ⋅V
Simple Target: maximize
V
S
S
.B
rp (concentrations, T)
Constraint: T ≤ Tmax
High temperature Æ maximum rate constant
Simplest Case:
W
W
rp = k [ A]
A→ P
W
Fp = [ P ]V0 = [ P ]
[ A] ⇒ [ A] feed
τ →0
residence
time in reaction
V
τ
(1− e − Da )
= Vk [ A] feed
Da
Constraint: [ P ] ≥ [ P ]min
(purity constraint )
Æ X ≥ X min conversion
{
}
Fp = k (Tmax ) [ A]0 − [ P ]min V = k (Tmax ) [ A] feed V (1− X )
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A + B → 2C
rc = 2k [ A][ B ]
26
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2nd order
A, B
[ A] = [ A] feed (1 − X A )
[ B ] = [ B ] feed (1 − X B )
.1
[ A][ B ][C ]
.01
Figure 2. Schematic
of a CSTR.
A, B
B
[ B ] [ A]
A, B, C
N
I
.
E
Separator
Figure 3. A reacting system with
recycle stream.
(A)C
A→ B
B→C
A →U
.1
rc = 2k [ A] feed [ B ] feed (1 − X A )(1 − X B )
S
.B
product may
also react
undesirable
V
S
W
W
FB = f ([ A] , [ B ] , [C ] , [U ] ,τ , T )
W
6 variables
Æ fsolve (Matlab)
SS.
0 = Fin − Fout + rAV
A
A
0 = Fin − Fout + rBV
B
B
0 = ...
0 = ...
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture #6
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Simplest Case:
k1 (T )
A→B
k2 (T )
B→C
k3 (T )
FA
A→U
in
A
=
[ ]
V (k1 + k3 + 1 )
τ
[U ] = k3 [ A]τ
[C ] = k2τ [ B ]
k [ A]
[ B] = 1 1
k2 +
τ
FB (τ , T ) =
FAo k1τ
N
N
I
.
E
(k1τ + k3τ + 1)(1+ k2τ )
N
N N
optimize
Da1
2 variables
Æ contour plot
Da2
V
S
Da3
S
.B
Figure 4. Sample contour plot for the 2variable optimization.
W
W
matlabÆfmincon
(allows for constraints)
W
∂FB
∂FB
= 0,
=0
∂T
∂τ Don’t use these – you may not find an actual optima.
Yield ≡
FP
FA0
Yield A→ P
≡ X A S A→ P
Selectivity =
S A→ P
FP
( FA0 − FA )
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture #6
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(especially important when A is expensive)
Y
A
→
P
A → P →U
P →U
τ
Figure 5. Yield versus residence time. Intermediate P rises in
concentration and then falls off as it is converted to U.
N
I
.
E
V
S
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture #6
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Consider the unconstrained optimization of a CSTR with volume V.
A → B
r = k[ A]
q [A]0
V
[A] [B]
The goal is to maximize FB with respect to changes in the volumetric flow rate, q.
FB = q[B]
Steady state material balances on species A and B give:
0 = FA0 − FA − rV = q([ A] 0 − [ A]) − kV[ A]
0 = FB0 − FB + rV = −q[B] + kV[ A]
Hence,
N
I
.
E
V
S
[B] = k[ A](V / q )
and
S
.B
FB = rV = k[ A]V ;
thus production of B is maximized when [A] takes its maximum value, which is [A]0.
W
Continuing with the material balances, we find:
[ A] =
W
[ A]0
[ A]0
=
1+ (kV / q ) 1+ kτ
W
When Da = kτ << 1, [A] goes to [A]0.
FB = rV = kV[ A] =
kV[ A]o
1+ kτ
=
kV[ A]o
1+ kV / q
⎛ kV[ A]o ⎞
⎟ = kV[ A]0
lim FB = lim⎜⎜
⎟
q→∞
q→∞
⎝ 1+ kV / q ⎠
Unfortunately, in the limiting case of infinite flow rate, the concentration of B in the
output solution is vanishingly small:
⎛
⎞
[ A]0
(
V / q )⎟⎟ = 0 .
lim[B] = lim(k[ A](V / q )) = lim⎜⎜ k
q→∞
q→∞
q→∞
⎝ 1 + (kV / q )
⎠
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and Biological Reaction Engineering,
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 7: Batch Reactors
This lecture covers batch reactor equations, reactor sizing for constant volume and
variable volume processes.
Batch Reactors
Run at non-steady state conditions
Which to choose? Batch vs. CSTR?
Batch
CSTR
N
I
.
E
Figure 1. Schematics of a batch reactor and a
CSTR.
Small Amount of Material
(small quantities)
Flexibility
+
V
S
S
.B
Expensive Reactants
+
(does not tie up
equipment continuously)
-
-
If product does not flow,
Materials Handling (e.g. Polymers)
+
-
Do not have to shut down and
clean, less down time
-
+
W
W
W
Captial costs? For size of reactor,
for given conversion
+
(concentration stays
higher longer)
Operability & Control (T, P, p4)
e.g. Exothermic reaction
-
-
+
(Manipulate only one
setpoint, steady state.
You can control additional
variables. Such as flow rates.)
Material Balance
0
In – Out + Product = Accumulation
0
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rA V =
31
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dN A
dt
Constant V,
dC A
dt
rA =
In terms of conversion,
Integrating,
C Ao
dX A
= rA
dt
X A dX
dC A
A
or t = C Ao ∫
0
C Ao r
r
A
A
t=∫
CA
A ⎯k⎯
→B
1st Order Reaction
− rA = kC A = kC Ao (1− x A )
t = C Ao ∫
xA
0
dx A
− kC Ao (1 − x A )
→
⎯⎯
t=
1 ⎛ 1 ⎞
ln ⎜
⎟
k ⎝ 1− x A ⎠
xA = 1 − e −kt
(order of
1 ⎛ 1 ⎞ 2.3
ln ⎜
⎟=
k ⎝ 1 − 0.9 ⎠ k
t90.0% =
90% conversion
1
)
k
W
−rA = kC A2 = kC Ao 2 (1− X A )2
XA
dX A
1
=
t = C Ao ∫
2
2
0
−kC Ao
−kC Ao (1 − X A )
W
t=
If
W
XA
1
kC Ao 1− X A
XA =
V
S
S
.B
A + A ⎯k⎯
→B
2nd Order Reaction
N
I
.
E
∫
XA
0
dX A
(1− X A ) 2
kC Aot
1+ kC Aot
kfirst order = ksecond order C Ao , which is faster?
1st order
2nd order
1
0.63
0.50
2
0.86
0.67
3
0.95
0.75
}x
A
For a given Damkohler number, 1st order is faster. The second order reaction has
greater concentration dependence. Exponential approach (1st order) is faster.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture #
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Batch Cycle
Time
t
Charge
rxn
Discharge
Clean
Downtime,
td
How long should t be?
How high should X A be?
Economic calculation:
Compare economics of further conversion to a different use
of equipment
Chemical consideration: Will product degrade? Assume product stable.
Product produced in one cycle = X AC Ao V
Pr(Rate of Production)
=
N
I
.
E
X AC Ao V
t + td
What value of t will maximize Pr?
If there is a maximum of Pr vs. t,
Assume td = constant.
(toptimum + td )
0=
d Pr
=0
dt
V
S
dX A
− XA
dt
(toptimum + td ) 2
S
.B
d Pr
= C Ao V
dt
dX A
− XA = 0
dt
(toptimum + td )
W
W
Now specify kinetics. There may be no optimum.
W
X A = 1 − e− kt
dX A
= ke− kt
dt
− ktoptimum
− kt
(toptimum + td )ke
− (1− e optimum ) = 0
1st order
Can numerically solve for
toptimum .
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture #
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Semi-batch Reactor
or
Figure 2. Schematics of two types of fed-batch reactors.
1) Why?
• To remove “poisonous” product
• Make room in reactor (expansion of product)
• If a reactant has a negative order effect on rate, add in small quantities
• Selectivity
A + B → Desired
(control)
A + A → Byproduct
Start with B, slowly feed A.
N
I
.
E
slow
feed of
A
V
S
Figure 3. A fed-batch
reactor with a slow feed of
one reactant.
S
.B
B
•
•
•
To shift equilibrium, strip off product
To control evolution of heat
In biological cases
Fed-batch
- Feed in carbon source slowly to avoid overflow metabolism
- (glucose)
- O2 sparingly soluble, must feed.
W
W
W
2) Balances
B
A Balance
Figure 4. Fed-batch
reactor with a feed of B.
A
In – Out + Product = Accumulation
rA V(t ) =
d (rA V)
dt
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture #
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dC A
dV
+ CA
dt
dt
Liquid V = V0 + v0t
rA V(t ) = V
flow
dilution
v
dC A
= rA − 0 C A
dt
V0
B Balance
v
v
dCB
= rB + 0 CBo − 0 CB
dt
V0
V0
N
Addition
Dilution
N
I
.
E
V
S
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture #
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 8: The Plug Flow Reactor
rA = −k [ A]
2
X A FAo = −rAV
X A FAo
nd
VCSTR =
(2 order reaction)
2
2
k [ A]0 (1− X A )
treact . =
VBatch
XA
k [ A]0 (1− X A )
( [ A] ) = ?
FAo =
0
moles A
VBatch =
time
=
VBatch [ A]0
treact + td
⎡
⎤
XA
⎢t d +
⎥
k [ A]0 (1 − X A ) ⎦⎥
⎣⎢
FAo
[ A]0
N
I
.
E
Assume XA=90%
If treact>td then
VBatch =
2FAo ⋅ 0.9
[ A]0 k [ A]0 (1− 0.9 )
VCSTR =
0.9FAo
k [ A]0
2
≤
V
S
S
.B
1.8FAo
k [ A]0
2
W
W
W
Figure 1. Three tanks in series.
[ A]CSTR = [ A]in + rA
[ ]
[ A]in
If rA = − k A
[ A]out =
1+ Da
V
v0
[ A]in
=
1+
kV
v0
If n CSTRs are in series:
each volume =
V
n
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[ A]out =
36
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[ A]in
⎛ kV ⎞
1+ ⎜
⎟
⎝ nv0 ⎠
n
Æimproves productivity:
concentration of A in 1st one is higher than would be in one large CSTR
= [ A]0 e
[ A]out
Batch
− kV
v0
kV
= 3 ⇒ 95% conversions
v0
[ A]0
out
=
[ A]CSTR
⎛ 3⎞
⎜1+ ⎟
⎝ n⎠
series
N
1
10
100
n
XA
.75
.93
.948
PFR
N
I
.
E
V
S
Figure 2. Diagram of a plug flow reactor.
Plug Flow Reactor (behaves like an infinite number of infinitely small CSTRs)
FAin − FAout + rA ( ΔV ) = 0 CSTR
S
.B
⎛ FAin − FAout ⎞
⎜
⎟ = −rA
ΔV
⎝
⎠
dFA
= −rA design equation for PFR
dV
dFA
= r ( C A ,CB ,...)
dV
= −kC ACB (for example)
FA = C Av0
dFA
F F
= −k A B
dV
v0 v0
dY
= F ( t ,Y ) where t is replaced by V.
This can be expressed as:
dt
W
W
W
Example:
ZZZ
A + B YZ
ZX
ZC
k
k
rev
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 8
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⎛ kFA FB krev FC ⎞
+
⎜−
⎟
v02
v0 ⎟
⎜
⎛ FA ⎞
d ⎜ ⎟ ⎜ kFA FB krev FC ⎟
FB = ⎜ −
+
⎟
dV ⎜⎜ ⎟⎟ ⎜
v0 ⎟
v02
FC ⎠
⎝N
⎜ kF F k F ⎟
Y
⎜⎜ + A2 B − rev C ⎟⎟
v0
v0 ⎠
⎝
F ( t ,Y )
z
Figure 3. Diagram of a plug flow reactor showing flow in the z-direction.
dV = area ⋅ dz
Mass flow rate is constant
N
I
.
E
( v ρ A ) = const.
ρ = ∑ CiWi
dρ
=0
dz
d ( vρ A)
dv
dA
dρ
+ Av
=0
= ρ A + ρv
dz
dz
dz
dz
For a liquid,
V
S
S
.B
Rearrange:
⎛ 1 dA 1 d ρ ⎞
dv
= −v ⎜
+
⎟
dz
⎝ A dz ρ dz ⎠
dA
dρ
=0
= 0 and for a liquid
For a normal pipe
dz
dz
dv
= 0 ⇒ v = v0
Therefore:
dz
W
W
W
(We can’t assume this for gases!)
For a PFR:
dFA
= rA
dV
FA = v [ A]
d ( vC A )
dV
= rA
For liquids, v is constant so we can take it out of the differential.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 8
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v dC A
, for liquids
area dz
dC A area
=
rA
dz
v0
rA =
Instead of treact we have zreact!
area ⋅length
v0
XA
area ⋅ z
=
=
v0
k [ A]0 (1− X A )
t pipe =
t PFR
Flow is driven by the pressure drop across the pipe.
P0
Pfinal
Figure 4. Diagram of a pipe showing pressure upstream and downstream.
PV = NRT
P
∑ Ci = RT
⎫
Fi ⎪
⎬ turns F's into concentrations
∑n Fn ⎪⎭
ρ = ∑ CiWi , Wi is molecular weight of i.
P
Ci =
RT
v=
mass flowrate
N
I
.
E
V
S
S
.B
← const.
W
ρ (z)
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 8
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 9: Reactor Size Comparisons for PFR and CSTR
This lecture covers reactors in series and in parallel, and how the choice of reactor
affects selectivity versus conversion.
PFR vs. CSTR: Size and Selectivity
Material balance:
CSTR
V=
PFR
FAo
XA
−rA
V=∫
XA
0
FAo
dX A
−rA
“Levenspiel Plot”
• as X A increases, C A decreases
FAo
− rA
1st or 2nd
order
reaction
V
S
Figure 1. General Levenspiel Plot.
S
.B
XA
CSTR Volume
N
I
.
E
−rA decreases, for 1st and 2nd order,
F
so Ao increases
−rA
W
PFR Volume
W
FAo
− rA
W
FAo
− rA
VPFR
VCSTR
XA
XA
Figure 2. Levenspiel plots for a CSTR and a PFR for positive order reactions.
So PFR is always a smaller reactor for a given conversion when kinetics are positive
order.
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Non-monotonically positive order kinetics arise:
• Autocatalytic reactions (e.g. cell growth)
• Adiabatic or non-isothermal exothermic reactions
• Product inhibited reactions (some enzymes)
Series of Reactors
Example: 2 CSTRs
FAo
v1
x1
N
I
.
E
FA1
v2
x2
V
S
S
.B
FA2
Figure 3. Schematic of two CSTRs in series.
F
V1 = Ao X 1
−rA1
W
W
0
2nd reactor:
In + Out + Prod = Acc
W
FA1 − FA2 + rA2 V2 = Steady state
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 9
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FA2 = FA0 − X 2 FA0
FAo
− rA
→
V2 =
FA0
−rA2
V1
( X 2 − X1 )
Figure 4. Reactor volumes for 2 CSTRs
in series.
V2
X1
41
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X2
X
N
I
.
E
Multiple CSTRs begin to
approximate a single PFR
FAo
− rA
V
S
X
S
.B
W
W
Figure 5. Reactor volumes for multiple CSTRs in series.
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 9
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FAo
− rA
FAo
− rA
VPFR
42
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VCSTR
VPFR
VCSTR
X
XA
Designed final
conversion
Final X
Figure 6. Levenspiel plots comparing CSTR and PFR volumes for changing kinetics.
Left: The CSTR has the smaller volume. Right: The PFR eventually has the smaller
volume.
N
I
.
E
Choice of PFR vs CSTR depends on conversion. Choose the reactor that has the
smallest volume Æreduce cost.
Reactors:
FAo
−rA
V
S
VCSTR
S
.B
CSTR
W
VPFR
W
W
PFR
X
Final X
Figure 7. To achieve the desired conversion with smaller reactor volumes, use a
combination. In this case, use a CSTR then a PFR. By doing so, the reactor volume is
less than the area underneath the curve.
For competing parallel reactions, selectivity for desired product can dominate the
choice.
Example
A→ D
rD = kd C Aα1
A →U
α2
D = Desired, U = Undesired
rU = ku C A
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 9
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If
α1 > α 2 , as C A
rD kd (α1 −α 2 )
= CA
rU ku
SD /U =
Define “selectivity”
43
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increases, S D /U increases
-Favors PFR because C A starts at C Ao then drops whereas CSTR
concentrations are always at lower C A .
If
α1 < α 2 , as C A
increases, S D /U decreases
-CSTR favored
If
α1 = α 2
then
S D /U =
kd
, no dependence on C A
ku
-Therefore no CSTR/PFR preference.
Define a fractional yield
φ=
kd C αA1
dCD
=
− dC A kd C αA1 + ku C αA 2
Overall fractional yield
For a CSTR:
Φ=
Φ = φ Εxit C
A
ΔC A = C A − C A
0
Φ=
For a PFR:
If
1
ΔC A
α1 = α 2
φ
∫
C At
CA 0
N
I
.
E
All D produced
All A consumed
V
S
S
.B
f
φ dC A
W
W
W
C Af
ΦΔC A
C A0
CA
Figure 8. Fractional yield versus concentration. Selectivity does not depend on CA.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 9
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If
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α1 > α 2
φ
CA
Figure 9. Fractional yield versus concentration when α1 > α2.
CSTR
PFR
φ
φ
Φ PFR ΔC A
N
I
.
E
Φ CSTR ΔC A
C Af
C Af
C A0
V
S
CA
C A0
CA
S
.B
Figure 10. Comparison of overall fractional yield for a CSTR and a PFR when α1 >
α2.
PFR is preferred because Φ PFR >Φ CSTR , therefore the yield of D per mol A consumed
W
is higher.
If
W
α1 < α 2
φ
W
C Af
Φ CSTR ΔC A
C A0
Φ PFR ΔC A
φ
C Af
CA
C A0 C A
CSTR
PFR
Figure 11. Comparsion of overall fractional yield for a CSTR and a PFR when α1 >
α2.
Φ PFR <Φ CSTR
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 9
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 10: Non­ideal Reactor Mixing Patterns
This lecture covers residence time distribution (RTD), the tanks in series model, and
combinations of ideal reactors.
Non­Ideal Mixing
CSTR
PFR
Figure 1. Ideal PFR with pulse input. A
pulse input will yield an output profile
that is a pulse input.
Figure 2. Ideal CSTR with pulse input. A pulse input will yield an output profile that
is a sharp peak with a tail.
N
I
.
E
Real mixed tank
bypassing
mixing
V
S
S
.B
recirculation
eddies
W
W
stagnant
volumes
W
Figure 3. A real mixed tank. In a real mixed tank there are portions that are not
well mixed due to stagnant volumes, recirculation eddies, and mixing bypasses.
In a real PFR there is back­mixing and axial dispersion. In a packed bed reactor
(PBR) channeling can occur. This is where the fluid channels through the solid
medium.
Residence Time Distribution
A useful diagnostic tool is the residence time distribution (RTD). The residence time
is how long a particle stays in the reactor once entering.
E ( t ) dt ≡ Probability that a fluid element entering the vessel at t=0 exits between
time t and t+dt.
Probability density function for exit time, t, as a random variable.
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2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on
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t
∫ E ( t ) dt
Probability that fluid element exits before time t.
0
∞
∫ E ( t ) dt
Probability of exiting at time later than t.
t
∞
mean t = tE ( t ) dt = τ
∫
0
∞
normalized =
∫ E ( t ) dt = 1
0
∞
variance = σ =
2
∫ ( t − τ ) E ( t ) dt
2
(measures the broadness of the distribution)
0
E
before t1
N
I
.
E
after t1
V
S
S
.B
t1
t
Figure 4. E(t) versus t. At a given time point, some material has exited and some
material will still exit at a later time.
W
Experimental Determination of E(t)
W
Inflow should be something measurable
­Absorbance
­Fluorescence
­pH
­salt­conductivity
­radioactivity
W
Use one of two types of input concentration curves:
Pulse
Cin
Step
Cin
t
t
Figure 5. Two types of input. A pulse input is a spike of infinite height but zero
width, ideally. A step input is a constant concentration over a period of time.
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Lecture 10
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A pulse input allows for easy interpretation because all materials enter the reactor at
once.
Cin
Cin
t
t
detector
input
curve
Figure 6. Schematic of a residence­time distribution experiment. The input curve
enters the reactor; a detector detects concentration changes in the output stream.
E (t ) =
Cout ( t )
t
∫ C ( t ) dt
out
0
N
I
.
E
PFR (Ideal)
Cin
V
S
Cin
S
.B
t
t0
t
W
τ
Figure 7. Pulse input in ideal PFR. A pulse input in an ideal PFR becomes a pulse
output.
E (t ) = δ (t −τ )
W
W
= 0
δ ( x) = 
= ∞
x≠0
x=0
∞
∫ δ ( x ) dx = 1
−∞
∞
∫ f ( x ) δ ( x − a ) dx = f ( a )
−∞
CSTR (Ideal)
Transient material balance:
In­Out+Production=Accumulation
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Lecture 10
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Since all the material is added at once, In=0. The tracer used is non­reactive.
Therefore there is no production. This gives:
0 −ν 0 C + 0 = V
C ( t ) = C0 e
E ( t ) =
∞
−t
τ
dC
dt
, τ=
C (t )
∫ C ( t ) dt
=
V
ν0
e
−t
τ
τ
0
CSTR
N
I
.
E
Figure 8. Pulse input in an ideal CSTR. In an ideal CSTR, a pulse input leads to a
sharp peak with a tail.
∞
mean residence time =
∫
0
te
−t
τ
τ
dt = τ
V
S
S
.B
CSTR (non­ideal mixing)
Bypassing: Divide input into 2 streams
ν0
W
νB
W
V
W
ν SB
ν0
Figure 9. A bypass is modeled by dividing the input stream into two streams, one of
which does not enter the reactor.
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Prof. K. Dane Wittrup
Lecture 10
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bypass portion
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mixed
E
E
t
t
combine
E
Perfect mixing
τ=
V
ν0
Bypass τ = V
ν SB
t
Figure 10. Residence­time distribution determination for a bypass.
Dead volumes: Stagnant regions not getting mixed
N
I
.
E
E
ideal
VSD
dead volume
present
VD
t
V
S
S
.B
Figure 11. Residence­time distribution for dead volumes. When a dead volume is
present, a decreased amount of material is observed in the output stream.
W
measureable V=VSD+ VD
τ SD =
VSD
ν0
W
< τ ideal
W
PFR (Non­ideal)
Channeling
channeling
bed
channel
PFR­like
Figure 12. Channeling. In channeling, the residence­time distribution will show
peaks for each channel as well as the one for the main portion of the reactor.
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Prof. K. Dane Wittrup
Lecture 10
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Axial Dispersion
Figure 13. A pulse input can become an axially dispersed pulse output in a non­ideal
PFR.
There are two common models for dispersion in a tubular reactor:
­Tanks in a series
­Taylor dispersion model (based on the Peclet number)
To model the PFR as several tanks in a series, break the reactor volume, V, into n
CSTRs of volume
V
each.
n
1
N
I
.
E
n
3
2
Figure 14. n tanks in series. The output of tank 1 is the input to tank 2. The output
is sampled at tank n for dispersion.
E (t ) =
E(t)
−t
t n −1
τ
e τi , τ i =
n
n
( n − 1)!τ i
W
1
2
S
.B
PFR
10
V
S
W
W
4
t
Figure 15. E(t) plots for 1, 2, 4, and 10 tanks and a PFR. Notice how the E(t) curve
approaches the PFR pulse as more tanks are used.
The numbers above represent numbers of CSTRs. Without enough CSTRs, the peak
is not a good approximation to the narrow peak for a PFR when there is a pulse
input.
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Prof. K. Dane Wittrup
Lecture 10
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σ2 =
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τ2
n
τ
σ2
2
n=
We can physically measure τ and we can determine σ from experimentally measuring
E(t).
RTD (residence time distribution) are useful for diagnosis, but not for reactor design.
To calculate conversion, the most straightforward tactic is to model the non­ideal
system as compartmental combinations of ideal reactors.
Figure 16. Recirculation. Recirculation can be modeled by a PFR followed by a CSTR
with a recycle stream.
Figure 17. Partially dead volumes. Dead
volumes can be modeled as separate
CSTRs that exchange material with each
other.
N
I
.
E
V
S
S
.B
W
W
W
Figure 18. Bypass. A
bypass can be modeled
as a CSTR along one
route with a PFR along
the bypass route.
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Prof. K. Dane Wittrup
Lecture 10
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 11: Non-isothermal Reactors, equilibrium limitations, and stability
This lecture covers: Derivation of energy balances for ideal reactors; equilibrium
conversion, adiabatic and non-adiabatic reactor operation.
Non-isothermal Reactors
N rxns
dN i N streams
= ∑ Fi ,m + Vcv ∑ υi ,l rl
dt
m =1
l =1
stoichiomet
stoichiometric
ric
coefficient
rl - depends on concentration
-T
- catalyst
N streams
dU cvtotal
dV
+ P cv = ∑ H mconc (Tm ) Fmtotal + Q +Ws + (other
dx
dt
m=1
flow work
intensive
extensive
work
expansion
work
heat
shaft
work
N
I
.
E
energy terms)
V
S
do work Æ
Ws negative
S
.B
W
If small control volume, pressure constant.
W
P1
W
In Vcv1 has a
P2
P1 ≠ P2
other control
volumes
fixed P
Figure 1. Schematic of a PFR with small control volumes, each with a fixed P.
PFR has many small control volumes, each with its own constant P.
For isothermal –
–
–
–
Q adjusted to keep T constant
Practical – have big cooling bath
or just operate at a particular temperature found after reactor
built
⇒ not a good strategy, for design we want to know ahead of
time
before assumed uniform T, actually have hot spots
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Where is T? In U cv
and
total
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rl (T ) .
dU cvtotal dU cvtotal dT N streams ⎛ dU cvtotal ⎞ dN i
=
+ ∑ ⎜
⎟
dt
dT dt
i=1 ⎝ dN i ⎠ dt
heat capacity
of system
intensive
contribution of
each species
substitute for
dN i
dt
dY
= F (Y )
dt
Want
Assume ideal mixtures
U cvtotal ≈ ∑ N iU i (Tcv )
extensive
N
I
.
E
intensive
total
cv
dU
= U i (Tcv )
dN i
If P=Constant (Isobaric)
S
.B
dH total d (U + PV ) dU dP
dV
=
=
+
V +P
dt
dt
dt N
dt
dt
W
0
total
cv
dU
dV
+ P cv
dt dt
↓
total
dH
dt
V
S
W
W
Assume isobaric, all ideal mixtures, neglecting K.E., P.E., other energies
⎛ N species
⎞ dTcv N streams N species
N
C
= ∑ ∑ Fi ,m ( H i (Tm ) − H i (Tcv ) )
⎜ ∑ i p ,i ⎟
m
i
⎝ i
⎠ dt
−
N streams N rxns
∑ ∑ H (T
i
i
∑υ
i
i ,l
H i (Tcv ) ≡ Δ H rxn (Tcv )
cv
)Vcvυi ,l rl (Tcv ) + Q +Ws
l
stoichiometric coefficient
l
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 11
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−
N streams N rxns
∑ ∑ H (T
i
i
l
cv
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)Vcvυi ,l rl (Tcv ) = −∑ Vcv rl (Tcv )Δ H rxn (Tcv )
l
l
Assume
Q ≅ UA(Ta − Tcv )
heat
transfer
coefficient
(conduction)
coolant
reactor
area of
contact
W s ≈ 0
(As a stirrer, heat negligible)
If designing engines W s ≠ 0 .
N
I
.
E
Now just put into MATLAB and solve
Chapter 8 in Fogler
– lots of special case equations
– be careful of assumptions
Special case: Start up CSTR to a steady state
want to know ultimate T
V
S
S
.B
N streams N species
dTcv
= 0 ≅ ∑ ∑ Fi ,m ( H i (Tm ) − H i (Tcv ) ) − ∑ Vcv rl Δ H rxn +UA(Ta − Tcv )
dt
l
m
i
W
W
W
All depend on TCV
When we reach steady state, no more accumulation
FA, in − FA, out + rAV = 0 at steady-state
See Fogler: 8.2.3
If just one reaction, one input stream, one output stream, and the system is at
steady-state:
XA =
UA(T − Ta ) + ∑ Fi , input C p , i (T − Tin )
FAo (−ΔH rxn )
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 11
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In this special case, conversion and T linear
1 reaction making heat as product is made.
When ΔH rxn = (-) Exothermic, reactor is hotter than cooling reactor (heat transfer
important)
(+) Endothermic, reactor must be heated so that reaction will run
G (T ) ≡ (−ΔH rxn )(−rA V FAo )
Generation
⎛
⎞
⎜
⎟
⎛ Fi , in
⎞
UA
R(T ) = ⎜ ∑
C p, i ⎟ ⎜1 +
⎟ (T − Tc )
FAo
Fi, inC p , i ⎟
∑
⎝
⎠ ⎜ ⎟
⎜
K
⎝
⎠
Heat removal
K =0
K = Big
Tc =
Adiabatic
Cooling
N
I
.
E
KTa + Tin
1+ K
V
S
R(T ) linear with T
S
.B
G (T ) → constant at high T
- not linear with T
W
cool
a lot
G (T )
W
W
Three steady-state solutions
R(T )
•
−ΔH rxn
•
•
T
Figure 2. Graph of G(T) versus T. Three steady-state points are shown where R(T)
intersects with the heat of reaction.
With multiple steady states must consider stability.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 11
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 12: Data collection and analysis
This lecture covers: Experimental methods for the determination of kinetic
parameters of chemical and enzymatic reactions; determination of cell growth
parameters; statistical analysis and model discrimination
Continuing the stability and multiple steady-state discussion from Lecture 11:
removal of heat
R (t )
•
G (t )
•
Three steady-state solutions
generation
•
N
I
.
E
Treactor
Figure 1. Three steady-state conditions shown on a G(T) versus T graph.
V
S
⎛ ξ1 ⎞
⎜ ⎟
⎜ ξ 2 ⎟ = z SS
⎜T ⎟
⎝ ⎠ SS
dξ1
=0
dt
dξ 2
=0
dt
dT
=0
dt
S
.B
W
steady-state
W
W
⎧ dξ1
⎫
⎪ dt = f1 (ξ1 , ξ 2 , T ) ⎪
⎪
⎪
⎪ dξ 2
⎪ dz
original eqns. ⎨
= f 2 (ξ1 , ξ 2 ,T ) ⎬ →
= F ( z)
dt
dt
⎪
⎪
⎪ dT
⎪
⎪ dt = f 3 (ξ1 , ξ 2 , T ) ⎪
⎭
⎩
stability: we want any perturbation
δz
from
vector notation
z SS to be self correcting
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i.e.
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d
(δ z ) = (−ve)δ z
dt
δz
what does perturbation cause?
- back to steady-state or off elsewhere?
•
Figure 2. A small perturbation moves the system away from steady state. Does the
system move back or does it move to elsewhere?
dz
= F( z)
dt
#
z = z SS + δ z
δ z = z − z SS
d
dz
(δ z ) =
= F (δ z + z SS )
dt
dt
0
0
dF
2
δ z + F ( z SS ) + O(δ z )
≈
dz
N
I
.
E
Jacobian matrix
V
S
⎛ dF ⎞
d
(δ z ) = ∑ ⎜ n ⎟ δ zm
dt
⎝ dzm ⎠ z
S
.B
SS
= Jδz
⎛ df1
⎜
⎜ dξ1
⎜ df
J =⎜ 2
⎜ dξ1
Jacobian ⎜ df
⎜ 3
⎝ dξ1
df1
dξ 2
W
df1 ⎞
⎟
dT ⎟
df 2 ⎟
⎟
dT ⎟
df 3 ⎟
⎟
dT ⎠
W
df 2
dξ 2
W
df 3
dξ 2
Matrix
d
(δ z ) = M (δ z )
dt
if eigenvalues of M<0 then stable
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 12
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CO + H 2O R CO2 + H 2 + Q
toxic
heat
1
doesn’t occur if equilibrium limited
w/ catalyst
X
equilibrium
original
rxn
T
Figure 3. Conversion (X) versus Temperature (T).
Data Collection:
-
N
I
.
E
determining rate laws
Products,
Unreacted Stuff,
Byproducts
Reactor
Reactants
CoT ,τ
S
.B
Figure 4. Schematic of a general reactor.
W
product
conc.
at
output
V
S
W
W
make plots
- get empirical expression
Æ fit curve
[ A]0
Figure 5. Product concentration versus reactant concentration A.
r(conc,T)
ouput-input
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 12
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→ rV
∫ r dxdydz
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if homogeneous
dV
“well-stirred” reactor
(slow reactions)
“no” conversion
(really ~.1% conversion)
C = C0 ± .1% Æ can measure (output-input)
(r barely changes)
*need very sensitive product detection
“differential reactor”
From data:
guess mechanism
vary ( k , k eq )Æmake a fit
1) Is mechanism consistent (error bars?) w/ data?
2) How to regress k ? (least squares method)
N
I
.
E
V
S
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 12
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13: Biological Reactors- Chemostats
This lecture covers: theory of the chemostat, fed batch or semi-continuous fermentor
operations
Biological Reactors (Chemostat)
Concentration/Combustion constant
Biological CSTR
[S]0
F
[S], x
V
F
N
I
.
E
Figure 1. Diagram of a chemostat.
F = Volumetric flow rate
x=
biomass
V
S
volume
S
.B
[S]0 = Concentration of growth limiting substrate. (for growing cells)
At steady-state, biomass balance
W
In – Out + Prod = Acc
W
Sterile feed: In=0
Steady state: Acc=0
W
− Fx + rx V = 0 at steady-state
Cell growth kinetics rx = μ x
− Fx + μ x V = 0
F
Solve μ =
V
F 1
=
D=Dilution rate ≡
V τ
μ=D
Biological
Mechanical
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When at steady-state, can control cell mass.
Allows precisely reproducible cell states.
Not easy to run at steady-state.
Material balance on [S] (sugar concentration)
In – Out + Prod = Acc
0
F [ S ]0 − F [ S ] −
at steady-state
1
μx V = 0
Yx
s
mass biomass created
Yield coefficient
mass substrate consumed
N
I
.
E
Divide by V
μx
D ([ S ]0 − [S]) =
Yx
change in sugar
concentration
V
S
s
At steady-state
x = Yx ([ S ]0 − [S])
W
s
W
What is the value of
μ = f ([S])
S
.B
μ=D
[S] ? What more information do we need?
W
Å must choose a growth model to connect
μ
and
[S]
Monod growth model:
μ=
μmax [ S ]
K s + [S ]
[S ] =
Æ at steady-state Æ
Ks D
μmax − D
D=
μmax [ S ]
K s + [S ]
substitute in x equation
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13
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⎛
Ks D ⎞
x = Yx ⎜ [ S ]0 −
⎟
μmax − D ⎠
s ⎝
Specifying
μmax , K s , Yx , D , [ S ]0 , can predict x , [S] .
s
x < 0 is non-physical but formally in solution
μmax − D
can go to 0. If you turn knobs incorrectly: if D is too high, the cells cannot
grow fast enough to reach steady-state. Washout will occur.
so use x = 0 to find Dmax
Dmax =
μmax [ S ]0
K s + [ S ]0
For D > Dmax “washout”, no steady-state.
S
x, S
x
mass
V
S
S
.B
volume
D
W
W
N
I
.
E
washout
Dmax
Figure 2. Biomass/volume versus dilution rate. Beyond the maximum dilution rate,
washout occurs.
For real systems K s << [ S ]0 . Most cell growth systems reach maximum at fairly low
W
concentrations; hence x is flat, then drops off sharply.
If biomass is the product, is there a best operating condition?
What should we consider?
dx
optimize x with respect to D? D=0 (no, because this would be batch reactor)
dD
Define productivity as
biomass
(reactor
volume )( time )
= xD
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13
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d ( xD)
= 0 for optimum. (is a maximum)
dD
x
xD
D
D
Doptimum
Figure 3. Left: Biomass/volume versus dilution rate. Right: Productivity versus
dilution rate.
⎛
Ks
Doptimum = μmax ⎜⎜1 −
K s + [ S ]0
⎝
⎞
⎟⎟
⎠
N
I
.
E
K s << [ S ]0
Doptimum ≈ μ max
≈ Dmax
V
S
S
.B
Close to washout conditions.
Operability would be difficult. We would not want to run too close to washout
conditions.
W
W
Fed-batch fermentor (microbes or mammalian cells)
-used to achieve very high cell densities (e.g. hundreds of grams cell dry weight
(c.d.w)/liter)
W
If you want x final =
100 g
L
If Yx ≈ 0.5 , [S]0 =
wt
200 g
≈ 20%
Å Toxic, sugar content cells will die
L
volume
s
Why do we not feed all at once? Cells will die.
Calculate medium feed rate in order to hold
μ
constant.
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Prof. K. Dane Wittrup
Lecture 13
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[S]0
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F
x(t)
S(t), very small
Figure 4. Diagram of a fed-batch fermentor.
If
μ
is constant, biomass = biomass t =0 e
μt
There is a dilution term, because as we feed in fresh medium, volume will change.
Volume often doubles.
x V = x0 V0 e μt
μ x0 V0 e μt
F[S ] =
0
Feed
sugar feed
Yx
s sugar consumed
N
I
.
E
Assume all converted into biomass.
F=
x0 V0
μ e μt
[ S ]0 Yx
s
Exponential flow rate. Typically
μ
S
.B
Dilution:
dV
=F
dt
V
S
specified as “small.”
W
W
⎛
⎞
x0
⎜
⎟
V(t ) = V0 ⎜1 +
e μt − 1) ⎟
(
⎜ [ S ]0 Yx
⎟
s
⎝
⎠
W
x=
biomass
V
x0 e μt
=
1+
x0
e μt − 1)
(
Yx [ S ]0
s
=
x0 V0 e
V
μt
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Prof. K. Dane Wittrup
Lecture 13
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“Logistic equation”
x
Figure 5. Graph of logistic growth.
t
If product is something cells are making:
Product synthesis kinetics
1)
1 dP
= αμ
x dt
P≡
2)
N
I
.
E
growth associated (e.g. ethanol)
product
volume
V
S
1 dP
=β
x dt
not growth associated (e.g. antibiotics, proteins, antibodies)
S
.B
t
P = β ∫ xdt
integrate for amount of product.
0
W
W
W
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Prof. K. Dane Wittrup
Lecture 13
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 14: Kinetics of Non-Covalent Biomolecular Interactions
This lecture covers : Significance, typical values and diffusion limit, approach to
equilibriu, and multivalency
Noncovalent Interactions
Protein
Ligand
P
+
L
Complex
kon
U
C
koff
Figure 1. Protein-ligand binding.
Association rate =
konC p CL
Dissociation rate =
koff Cc
@ equilibrium,
N
I
.
E
konC p CL = koff Cc
V
S
1
s
C p CL
Cc
=
koff
kon
S
.B
= Kd
W
L
mol s
W
W
In general, for protein-protein interactions,
half-time for complex dissociation
τ1 2 =
nM
pM
fM (femtomolar)
stronger
interactions
Fractional saturation Y =
ln 2
koff
τ1 2
types
milliseconds
milliseconds-seconds
non-specific stickiness
cell surface, multi valent
minutes-hours
hours-weeks
weeks-months
antibodies, enzymes
growth factors
hycholase inhibitors
Kd
mM
μM (micromolar)
kon ≈ 105 mol −1s −1
Cc
Cc
=
C p , o Cc + C p
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Kd =
C p CL
Cc
→Y =
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CL
CL + K d
Y
Y
Kd
CL
log CL
Figure 2. Left: Graph of fractional saturation versus ligand concentration. Right:
Graph of fractional saturation versus the logarithm of ligand concentration.
If
C p ,o ≈ CL ,o , then at equilibrium, CL ≠ CL ,o
Y=
Y=
If instead
CL ,o − yC p ,o + K d
K d + CL,o + C p ,o − ( K d + CL,o + C p ,o ) 2 − 4C p ,o CL,o
2C p ,o
CL ,o
V
S
S
.B
CL ,o C p ,o , CL ≈ CL,o
Y=
N
I
.
E
CL ,o − yC p ,o
W
CL ,o + K d
W
How quickly is equilibrium reached?
W
dCc
= konCLC p − koff Cc
dt
If
CL , o C p , o
“pseudo-1st order”
konCL = konCL ,o
C p ,o = C p + Cc
(complexed)
C p = C p ,o − Cc
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Prof. K. Dane Wittrup
Lecture 14
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dCc
= konCL,oC p − koff Cc = konCL ,o (C p,o − Cc ) − koff Cc
dt
= kon C p,oCL ,o − (konCL,o + koff )Cc
⇒ Cc (t ) = C p,o
CL ,o
CL ,o + K d
(1− e −kobs t )
kobs = konCL,o + koff
ln 2
= half-time for reaching equilibrium
kobs
Cc
equlibrium
ln 2
kobs
N
I
.
E
t
V
S
Figure 3. Concentration of complex versus time. Equilibrium is approached at long
times.
S
.B
Biosensor
W
Surface plasmon resonance (label-free)
W
flow
W
hν
thin gold film
amount of reflected light is a
function of complex formation
Figure 4. Schematic of how surface plasmon resonance works.
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Prof. K. Dane Wittrup
Lecture 14
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Kd
signal
kon
koff
koff
t
flow
signal
association
stop
signal
equilibrium
dissociation
Figure 5. Signal of detector versus time.
redundant estimates:
N
I
.
E
koff in both association & dissociation, K d =
V
S
phase
koff
kon
in equilibrium
best approach: fit one set of parameters to three phases of experiment. (global least
squares)
S
.B
Multivalency (Avidity)
W
cell
cell
W
W
cell
surface
cell
multivalent
label
Figure 6. Three examples of multiple protein-ligand binding.
How does multivalency effect apparent interaction strength?
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 14
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free
(high effective local
concentration)
U
K eff
U
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bound
detected same
Figure 7. Multivalent binding equilibrium.
N
I
.
E
V
S
S
.B
W
W
W
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Prof. K. Dane Wittrup
Lecture 14
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 15: Gene Expression and Trafficking Dynamics
This lecture covers: Approach to steady state and receptor trafficking
Central dogma of molecular biology:
DNA
Æ
mRNA
Æ
protein
transcription
translation
Material balance on one specific mRNA
Accumulation = synthesis – degradation
CmRNA ≡
cell volume
mol mRNA
Kr ≡
Vi ≡
moles mRNA
( time )( cell volume )
N
I
.
E
, transcription (function of gene dosage, inducers, etc.)
V
S
cell volume
vessel volume
d ( CmRNA Vi )
= K r Vi − γ r CmRNA Vi
dt
S
.B
W
γ r ≡ first order rate constant for mRNA degredation
W
Vi ≡ a function of time (cells grow, divide)
Æ can’t pull out of the derivative
W
Do the chain rule:
d Vi
dC
+ Vi mRNA = K r Vi − γ r CmRNA Vi
dt
dt
dCmRNA
1 d Vi
= K r − γ r CmRNA − CmRNA
dt
Vi dt
CmRNA
simplify:
1 d Vi
=μ
Vi dt
(specific growth rate in exponential growth)
dCmRNA
= K r − γ r CmRNA − μ CmRNA
dt
dilution by growth term
(b/c concentration is on a per-cell volume basis)
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dCmRNA
= K r − (γ r + μ )CmRNA
dt
at steady-state:
CmRNA, SS =
Kr
(γ r + μ )
transient case, analytical solution (just integrate)
Kr ⎛
− ( μ +γ r ) t ⎞
⎜1 − e
⎟
(γ r + μ ) ⎝
⎠
CmRNA =
independent of the transcription rate constant K r
N
I
.
E
S.S.
CmRNA
1
γr + μ
V
S
S
.B
t
Figure 1. Concentration of CmRNA versus time. At long times steady state is
approached.
W
Similar rate expression for the protein:
(again, per-cell volume basis, analogous constants)
dC p
dt
W
W
= K p CmRNA − (γ p + μ )C p
function of time, solved for above
dC p
dt
= Kp
Kr
(1 − e−(γ r + μ )t ) − (γ p + μ )C p
(γ r + μ )
d
=0, t →∞
dt
Kr K p
steady-state:
C p , SS =
(γ r + μ )(γ p + μ )
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Prof. K. Dane Wittrup
Lecture 15
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C p , SS
Kp
=
CmRNA, SS
Note:
γp +μ
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K p , γ p vary from protein to protein and condition
to condition
Integrate
dC p
:
dt
⎛ (γ r + μ )e − (γ p + μ )t − (γ p + μ )e − (γ r + μ )t
C p = C p , SS ⎜ 1 +
⎜
γ p −γr
⎝
Usually,
in E. coli
γp
γr
ln 2
γr
∼ 7 minutes on average.
for most proteins,
also,
γr
⎞
⎟
⎟
⎠
ln 2
γp
∼ hours to days.
μ
N
I
.
E
Apply assumptions to get:
Cp =
(1 − e
+ μ)
K p Kr
γ r (γ p
− (γ p + μ )t
)
Delays in synthesis
V
S
mRNA – 1 kb gene
Protein – 400 a.a.
W
W
W
S
.B
E. coli
10-20
20
time (seconds)
Yeast
30-50
20
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Mammals
30-50
60-400
Lecture 15
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Cp
t
Delay is generally small compared to
1
γp +μ
Figure 2. Concentration of protein versus time.
However, the delay can dramatically destabilize feedback loops.
N
I
.
E
Cellular compartmentalization
C p, 1 → C p,2
rate =
where
C p, 1 ≡ C p for compartment 1, and C p, 2 ≡ C p for compartment 2
V
S
K transport C p,1
+
koff
o
c y ut.
to
.
prod.
S
.B
kon
rec.
W
W
W
endocytosis
cell
Figure 3. Diagram of protein-ligand binding on the cell surface.
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Prof. K. Dane Wittrup
Lecture 15
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 16: Catalysis
This lecture covers: Inorganic and enzyme catalysis and their properties; kinetics of
heterogeneous catalytic reactions; adsorption isotherms, derivation of rate laws; and
Langumuir-Hinshelwood kinetics
What initiates the reaction?
A + B → starts upon mixing
B
A
Product
N
I
.
E
Figure 1. Bi-molecular reaction in a CSTR.
Temperature drastically increases reaction rate.
V
S
Hot
C2H6
Rxn occurs
S
.B
Fire
Figure 2. Schematic of tube reactor.
W
Catalyst dramatically increases reaction rate.
W
A
W
catalyst
Figure 3. Schematic of packed bed reactor.
Catalyst: Accelerates rate of reaction but is not consumed
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transition state
E
products
~Ea
reactants
rxn coordinate
Figure 4. Reaction diagram.
rate constant:
k BT
⎡ (G − Greactants ) ⎤
exp ⎢ − ts
⎥
h
RT
⎣
⎦
G = H − TS
k=
N
I
.
E
e − G / RT = e− H / RT e S / R
no catalyst
V
S
E
S
.B
with catalyst
W
rxn coordinate
W
Figure 5. Reaction diagram with and without catalyst.
W
The reaction forms many intermediates. A catalyst lowers the energy of these
intermediates.
Acid/Base catalysis
ROR + H 2O → 2ROH
k1
ROR + H ⊕
k−1
H
H
ROR
⊕
k2
ROR → ROH + R ⊕
⊕
k3
R ⊕ + H 2O → ROH + H ⊕
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 16
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H
ROR , R ⊕
QSSA
⊕
⎡ H ⎤
d ⎢ ROR ⎥
H
⊕
⎦ = k ⎡ H + ⎤ ROR − (k + k ) ⎡ ROR ⎤
O≈ ⎣
[
]
⎢
⎥
−
1⎣
1
2
⎦
dt
⎣ ⊕ ⎦
⎡ H ⎤
k1
⎡⎣ H + ⎤⎦ [ ROR ]
=
⎢ ROR
⎥
⊕
k
k
+
−1
2
⎣
⎦ QSSA
d [ ROH ]
dt
⎡ H ⎤
= 2k2 ⎢ ROR ⎥
⎣ ⊕ ⎦ QSSA
d [ ROH ]
2k1k2
⎡⎣ H + ⎤⎦ [ ROR ] = r
≈
dt
k−1 + k2
rA ∼ [ A]
rA ∼ [ catalyst ]
N
I
.
E
(where ∼ denotes “proportional to”)
N +
⎡ H ⎤
⎡⎣ H + ⎤⎦ + ⎢ R O R ⎥ + ⎡⎣ R + ⎤⎦ = H added = ⎡⎣ H + ⎤⎦
added
⊕
V
⎣
⎦
⎛ k [ ROR ]
⎞
k1k2 [ ROR ]
⎡⎣ H + ⎦⎤ ⎜1 + 1
= ⎣⎡ H + ⎦⎤
+
⎟
⎜
⎟
added
k1 + k2
k3 (k−1 + k2 ) [ H 2O ] ⎠
⎝
V
S
r=
r=
keff [ ROR ] ⎡⎣ H + ⎤⎦
1+ k [ ROR ]
S
.B
W
added
W
k [ catalyst ] [ A]
1+ k A [ A] + k B [ B ] + ...
W
All the things that the
catalyst binds to
Langmuir-Hinshelwood: all reagents bind to catalyst, bound forms react
Eley-Rideal: one reagent binds, 2nd reagent reacts with bound form
dN A
= VrA
dt
f
([ A] , ⎡⎣ H ⎤⎦ )
+
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 16
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moles
dN A
= ( area of
dt
metal
) rA′′
area s
f (θ A )
where
θA =
N A bound
N total sites
on surface
N
I
.
E
V
S
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 16
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 17: Mass Transfer Resistances
This lecture covers: External diffusion effects, non-porous packed beds and
monoliths, and immobilized cells
Table 1. Homogeneous vs. Heterogeneous Catalysis
Homogeneous
vs.
Heterogeneous Catalysis
acids,bases
immobilized enzymes
radicals
metals
organometallics
solid acids, bases
enzymes
metal oxides, zeolites, clays, silica
better mixing, uniformity
multiphase systems
transport limitations
reuse catalyst easily
product purity
1. New rate law on surface
2. Model the transport and mixing
Ci fluid
( x, y,z ) =
θ
surface
j ( x, y )
ri
fluid
=
N
I
.
E
Ni
Volume
N j on surface
N sites on surface
∑ θ j + θvacancy = 1
=
N rxn fluid
∑
n =1
⎛ A⎞
ν i ,n rn (C ) + ⎜ ⎟
⎝V ⎠
mol
s ⋅ vol
∑
W
W
"
dθ j
W
ν r (C ,θ )
"
i ,m m
m =1
QSSA for surface species: rj =
V
S
S
.B
N rxn suface
mol
s ⋅ area
∑ν
r (C ,θ ) ≈ 0
"
j ,m m
θQSSA = f (C )
At surface
0
⎛ area ⎞
= ( flows ) + rj" ⎜
⎟ N avagadro
dt
⎝ N sites ⎠
A
A
P
δ
surface
Catalyst
Figure 1. Schematic of boundary layer at catalyst surface for a
turbulent, well mixed system where CA is a function of x.
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(
Flux of A to the surface: WA = −Ctot D∇ y A + y A WA + WP
)
where yA is the mole fraction of A
Ctot = const
diffusion
convection
P
WA = − D∇C A + C A u
FAnet = − v∫ WA ⋅ dn
Åsurface integral
JK
FA = ∫∫∫ dxdydz ∇ ⋅ WA
JK
dN A
= ∫∫∫ dxdydz ±∇WA + rA ( x, y, z )
dt
dC A JK
= ∇ ⋅ WA + rA
dt
(
)
*See Fogler 11-21
Continuity equation:
dC A
= D∇ 2C A − u ⋅∇C A + rA (C )
dt
N
I
.
E
Boundary Condition:
WAinto wall = −rA'' at surface
1. Steady state
2. Gradients
dC A
dC A
and
are negligible
dx
dy
S
.B
3. Velocity u towards the wall = 0
4. No reaction in the fluid
∂ 2C A
0=D
∂z 2
WA = −rA''
−D
dC A
dz
V
S
W
W
= − rA'' ( C z =0 ) (z=0 at the surface)
z =0
W
⎡ C main − C A ( z = 0) ⎤
C A ( z ) = C A ( z = 0) + ⎢ A
z⎥
δ
⎣
⎦
main
dC
C
− C A ( z = 0)
=D A
= −rA'' ( C A, z =0 )
D A
dz z =0
δ
Slow chemistry limit: C A ( z
= 0) ≈ C Amain
(
rA'' ≈ rA'' C Amain
Fast chemistry limit: C A ( z
D
δ
)
= 0) ≈ 0
C Amain ≈ rA''
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 17
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D
δ
81
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= kc “mass transfer coefficient”
Sh =
kc d p
D
ÅSherwood number (dimensionless)
For spherical, catalyst particle with diameter dp:
Sh = 2 + 0.6 Re1 2 Sc1 3
ud
υ
μ
Sc =
Re = p υ =
D
υ
ρ
N
I
.
E
V
S
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 17
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 18: External Mass-transfer Resistance
This lecture covers: Gas-liquid reactions in multiphase systems
fluid
reactants
products
external
transport
mass
transfer
coefficient
rate law
observed rxn rate
=f(reactant concentration
and product)
intraparticle
diffusion
adsorption desorption
solid
sites
rxn
N
I
.
E
Figure 1. Schematic of surface reaction kinetics.
V
S
Analogies:
noncovalent
biomolecular
interactions
↔
Michaelis↔
Menton
enzyme kinetics
(Briggs-Haldane, Henri)
W
W
Langmuir-Hinshelwood
Haugen-Watson
kinetics
W
Logic:
•
•
•
•
S
.B
Langmuir
adsorption
isotherms
list rxns
hypothesize rate-limiting step
derive rate law
check for consistency w/ rate data
Single site, unimolecular decomposition
kA
kS
kD
k− A
k− S
k− D
A R AS R BS + C R S + B + C
S ≡ site on catalyst
product
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rAd ≡ rate of A adsorption
rs ≡ rate of rxn on surface
rD ≡ rate of desorption
rAd = k A PACS − k− AC AS
* units
partial
reactive
pressureA sites
1
sites
time
mass catalyst
θ
often CS is given as fractional occupancy
KA =
kA
k− A
⎛
C ⎞
rAd = k A ⎜ PACS − AS ⎟
KA ⎠
⎝
⎛
PC ⎞
Similarly, rS = k S ⎜ C AS − C BS ⎟
KS ⎠
⎝
⎛
PC ⎞
1
Æ
rD = k D ⎜ CBS − B S ⎟ Æ K B =
KD
KD ⎠
⎝
At steady-state,
N
I
.
E
V
S
rD = k D ( CBS − K B PB CS )
S
.B
rAd = rS = rD (or else you would accumulate molecules)
For a rate-limiting step
i,
W
r r
ri
j, l
ki
k j kl
W
W
Hypothesize, that adsorption is rate-limiting:
r
rA
r
S , D ≈0
kA
kS k D
rS
PC
C P
≈ 0 ⇒ C AS − C BS ⇒ C AS ≈ BS C
KS
kS
KS
rD
≈ 0 ⇒ CBS − K B PB CS ⇒ CBS ≈ K B PB CS
kD
K PC P
Æ C AS ≈ B B S C
KS
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 18
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⎛
C ⎞
rAd = k A ⎜ PACS − AS ⎟
KA ⎠
⎝
⎛
⎞
KB
= k A ⎜ PACS −
PB PC CS ⎟
KS K A
⎝
⎠
⎛
⎞
KB
PB PC ⎟
= k ACS ⎜ PA −
KS K A
⎝
⎠
CS0 = CS + C AS + CBS
Material balance on CS (available sites):
C S0 = C S +
KB
PB PC CS + K B PB CS
KS
⎛ K
⎞
= CS ⎜1 + B PB PC + K B PB ⎟
⎝ KS
⎠
Adsorption as rate-limiting step:
N
I
.
E
eq. driving force
⎞
⎛
⎜
⎟
KB
k ACS0 ⎜ PA −
PB PC ⎟
KS K A
⎜
⎟
⎝
⎠
− rA =
KB
PB PC + K B PB
1+
KS
S
.B
Surface reaction is rate-limiting:
⎛
PP ⎞
k S CS0 K A ⎜ PA − B C ⎟
KC ⎠
⎝
− rA =
1 + PB K B + PA K A
V
S
equilibrium driving force is 0 at equilibrium!
W
W
W
Desorption is rate-limiting:
⎛
PP ⎞
k D CS0 K S K A ⎜ PA − B C ⎟
KC ⎠
⎝
− rA =
PC + PA K A K S + K A PC PA
Initial rate expirements, approximate − rA = f ( PA ) , PB ≈ PC ≈ 0 , K C =
K AKS
.
KB
Adsorption limit:
− rA = k ACS0 PAo
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 18
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− rA
PAo
Figure 2. Reaction rate vs. initial partial pressure of A for the absorption limiting
case.
Surface reaction limit:
− rA =
kS CS0 K A PAo
1 + K A PAo
N
I
.
E
− rA
V
S
PAo
S
.B
Figure 3. Reaction rate vs. initial partial pressure of A for the surface reaction
limiting case.
W
Desorption limit:
− rA = k D CS0
− rA
W
W
PAo
Figure 4. Reaction rate vs. initial partial pressure of A for the desorption limiting
case.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 18
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 19: Oxygen transfer in fermentors
This lecture covers: Applications of gas-liquid transport with reaction
Gas-liquid mass transfer in bioreactors
Microbial cells often grown aerobically in stirred tank reactors
-oxygen supply is often limiting
μ
X
X
X
X
X X X
X
critical value
X
X
D.O.
V
S
Figure 1. μ vs dissolved oxygen.
S
.B
D.O. = dissolved oxygen
Equilibrium solubility of O2
bubble
O2
≈ 1 mM
W
W
W
N
I
.
E
≈ 0.01 mM
4
cell
1 2 3
Figure 2. Oxygen pathway.
1)
2)
3)
4)
Diffusion across stagnate gas film
Absorption
Stagnate liquid layer (rate-limiting step)
Diffusion and convection
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at equilibrium
3) O2 flux = kl (CO2 − CO2 )
[=]
*
mass transfer
coefficient
mol
area time
bulk liquid
concentration
What is the value for the interfacial area?
Important
-
system parameters:
liquid physical properties (surface tension, viscosity)
power input/volume (stirring, propeller size)
superficial gas velocity
empirical correlations (TIB 1:113 ’83)
β
⎛P⎞
kl a = constantU s ⎜ ⎟ where Us is the superficial gas velocity
⎝V ⎠
⎛ length ⎞ ⎛ area ⎞
−1
kl a[=] ⎜
(s-1)
⎟⎜
⎟ = time
⎝ time ⎠ ⎝ volume ⎠
N
I
.
E
α
U S [ =]
V
S
length
time
power
S
.B
P
=
V volume
(m/s)
(W/m3)
W
const. = 0.002
α = 0.2
β = 0.7
W
W
@ SS, O2 transport = O2 uptake by biomass
kl a (C − CO2 ) =
*
O2
biomass growth rate
μX
YX O
2
dX
< kl aCO* 2 YX O
Crude limit:
2
dt
or
dX
dt
yield coefficient ≈ .4-.9
g cell dry wt.
g O2
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 19
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O2 transport in tissues
o
o
r
o
R0
o
o
capillary radius Rc
Figure 3. Krogh cylinder model.
One-dimensional steady-state diffusion:
DO2 ∂ ⎛ ∂CO2 ⎞
⎜r
⎟ = VO2
r ∂r ⎝ ∂r ⎠
metabolic consumption rate of
oxygen, zero-order
Fick's Law
N
I
.
E
(cylindrical coordinates)
V
S
Boundary conditions:
S
.B
symmetry
no-flux
flux=0 @ r=R0
∂CO2
= 0 @ r=R0
∂r
CO2 = CO2 , plasma @ r=Rc
DO2
W
W
W
Integrate twice:
CO2
CO2 , plasma
⎛
r* ⎞
= 1 + Φ ⎜ r *2 − R*2 − 2 ln * ⎟
R ⎠
⎝
where
r * = r R0 , R* = Rc R0 , Φ =
1 VO2
char. rxn rate
R2
=
4 CO2 , plasma DO2 char. transport rate
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 19
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1
CO2
Φ decreasing
CO2 , plasma
r*
Φ.
Figure 4. Dissolved oxygen vs. radius for various values of
O2 diffuses further before consumption as
When R ≈ 0.05,
*
decreases.
CO2 = 0 @ r * = 1 when Φ ≥ 0.2
~50-100
o
Φ
N
I
.
E
μm
V
S
O2
S
.B
necrotic core
Figure 5. Tumor micrometastases.
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 19
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 20: Reaction and Diffusion in Porous Catalyst
This lecture covers: Effective diffusivity, internal and overall effectiveness factor,
Thiele modulus, and apparent reaction rates
Reaction & Diffusion
-Diffusion in a porous solid phase
Ex. Precious metals on ceramic supports or drug/nutrient delivery through tissues
-Derive steady state material balance accounting for diffusion and reaction in a
spherical geometry
-Thiele modulus (φ)
R
r
[S]0= surface concentration of a growth
substrate (ex. glucose and O2)
N
I
.
E
V
S
S
.B
Figure 1. Sphere of Cells
Assume pseudo-homogeneous medium and Fick’s Law describes diffusion
Flux = − D
d [S ]
dr
− rs = knCsn
V C
− rs = max s
K s + Cs
, where
# molecules
length 2
and D [=]
time
area ⋅ time
W
Flux [=]
W
W
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-rs
n=1
n=0
Cs
Figure 2. Rate of reaction versus species concentration
Cs K s ⇒ 1st order rs ≈
Vmax Cs
Ks
Cs K s ⇒ 0th order rs ≈ Vmax
N
I
.
E
Steady-state Shell Balance
In some cases, S still hasn’t
penetrated to the center
V
S
S
.B
t=0 Æ
W
tÆ ∞
W
Figure 3. Time progression as species, S, enters the sphere
W
Thin shell r to (r+Δr)
Sin by diffusion –Sout by diffusion –Scons by reaction=0
Flux ⋅ 4π r 2
r + Δr
− Flux ⋅ 4π r 2 − kn Csn 4π r 2 Δr = 0
r
Divide through by 4πΔr and take the limit as ΔrÆ0
(
d Flux ⋅ r 2
dr
) −k C r
n
n 2
s
=0
dC
d ⎛
⎞
− D s ⋅ r 2 ⎟ − knCsn r 2 = 0
⎜
dr ⎝
dr
⎠
⎫
d 2Cs 2 dCs kn n
+
− Cs = 0 ⎬ 2nd order ODE
2
dr
r dr D
⎭
2 Boundary conditions:
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 20
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Cs
r=R
92
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= Cs ,0
Cs is finite everywhere, or
dCs
dr
=0
r =0
Nondimensionalize
C
r
S= s
R
Cs ,0
ρ=
d 2 S 2 dS
+
− φ 2S n = 0
dρ2 ρ dρ
Boundary conditions: S=1 @ ρ=1
S is finite everywhere
φ =
2
kn R 2Csn,0−1
D
=
(R
2
D
)
(1 k C )
n
n −1
s ,0
characteristic diffusion time
=
characteristic reaction time
N
I
.
E
If diffusion is slow Æ diffusion dominates
If reaction is slow Æ reaction dominates
If φ2<<1 Æ reaction limited regime
If φ2>>1 Æ diffusion limited regime
2
φ <<1
φ >>1
substrate
throughout
V
S
S
.B
2
W
W
substrate can’t make
it to the center
W
Figure 4. Reaction and diffusion limited regimes
S=
1 sinh(φρ )
ρ sinh(φ )
sinh( z ) =
e z − e− z
2
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 20
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φ = 0.1
S
φ =1
φ↑
φ = 10
ρ
1
Figure 5. S versus ρ for various values of φ
Define “effectiveness factor” η
overall rate of reaction
η=
rate if Cs =CS,0 everywhere
overall reaction rate in sphere at steady state = [inward flux @ r=R (ρ=1)]*Area
=D
dCs
dr
4π R 2
r=R
= 4π RDCs ,0
η=
3
φ2
dS
dρ
N
I
.
E
= 4π RDCs ,0 (φ coth φ − 1)
ρ =1
(φ coth φ − 1)
V
S
S
.B
1
η
W
0.1
0.01
.01
W
W
0.1
1
10
η∝
1
φ
100
φ
Figure 6. Log-log plot of effectiveness factor versus thiele modulus
Higher values of Thiele modulus Æ effectiveness goes down
*For a variety of reaction kinetics, geometries and rate laws, plots of η vs φ all look
the same.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 20
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Shrinking Core Model
In cases with noncatalytic and irreversible reaction, diffusion limit is describable by
the “shrinking core model”.
Figure 7. Shrinking core model
Rapid, irreversible reaction limited by rate of diffusion of a reactant from the surface
The following must be written down for the shell balance:
1. Rate of reaction
2. Rate of diffusion
3. Rate of movement of the core
N
I
.
E
V
S
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 20
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2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 21: Reaction and Diffusion in Porous Catalyst (cont’d)
This lecture covers: Packed bed reactors
dFA
= ArAeff
dz
A
products
N
I
.
E
V
S
Deff (actual is very complicated)
S
.B
Figure 1. Packed Bed Reactor
Void Fraction
φ ~.5
W
C A ( x j , ym , zl )
j = 1,30
m = 1,30
l = 1,30
(points)
W
W
(11-21)
Di ∇ 2Ci − U ⋅∇Ci + ri fluid = 0 Å in the fluid
Di
∂Ci
∂nˆ
+ ri surface = 0
i=1, N species
fluid
(boundary condition for the above)
surface
Ergun’s Eq.:
(4-22)
dP
G 1−φ
=−
dz
ρ gc Dp φ 3
where P =
ρ RT
,
ρU z =
⎡150 (1 − φ ) μ
⎤
+ 1.75G ⎥
⎢
Dp
⎢⎣
⎥⎦
G
A
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∂FA
= ArAeff
∂z
96
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rAeff = (effectiveness factor )rAC Ab ( z )
(
)
Ω C Ab ( z ) (or η internal)
area
total
Ω≡
∂Fi
= Ari eff = AΩri ideal
∂z
Actual rate of reaction
Rate if
ri ideal [=]
(rAeff )
C A = C Abulk ( z ), and T = Tbulk everywhere
mol
vol. s
⎛ wt. of catalyst in reactor ⎞
ri ideal = ri′ ⎜
⎟
V reactor
⎝
⎠
N
I
.
E
ri = ri′ ρc (1 − φ )
⎛ surface area of cat. ⎞
= ri′′ ⎜
⎟ ρ c (1 − φ )
wt. cat.
⎝
⎠
ac / m particle
V
S
S
.B
m2
[=] Sa + (macroscopic surface area, visual)
g
FA = ν C A = AUC A (some approximation)
W
W
W
Dispersion
dC A
d 2C A
1 dFA
= +U
− Da
A dz
dz
dz 2
hope this is 0!
Figure 2. Flow over a sphere
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 21
Page 2 of 3
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(
97
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)
kc ac C Ab − C As = η ( C As ) rA particle ( C As ) V p
Dinside
∂ 2C A
+ rA ( C A (r ) ) = 0
∂r 2
∂C A
=0
∂r r =0
kc C A r = R + Deff inside
∂C A
∂r
Deff inside
∂C A
= kc C Ab − C As
∂r
(
)
= kc C A bulk
r=R
Matlab:
1) Guess C As ( C A surface)
2) Use boundary conditions to get corresponding
3) Solve ODE (ode15s)
4) Vary guess ( C As ) to make
∂C A
∂r
r=R
∂C A
= 0 at center
∂r
N
I
.
E
st
1 oder irrev.
η
Ω=
1+
η k1′′ Sa ρb
η=
3
φ12
V
S
(φ1 coth(φi ) − 1)
S
.B
kc ac
φ1 = … k1′′
W
W
W
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Prof. William H. Green
Lecture 21
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 22: Combined Internal & External Transport Resistances
Packed Bed Reactor Æ use PFR equation
dFi
= Ari eff
dz
z=0
z=L
Figure 1. Packed Bed Reactor
ri eff ≠ ri chem
ri eff = Ωri chem
Æ
if Ω ≈ 1 ?
if Ω 1 ?
( Ω > 1 ) –weird
BL:
δ
N
I
.
E
porous
catalyst
particle
V
S
S
.B
W
W
W
dp
Figure 2. Porous Catalyst Particle
Biot Number:
Bi ∼
external heat transfer resistance
internal heat transfer resistance
Mass transfer Biot Number:
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Bim =
kc d p
where kc ∼
99
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D fluid
eff
Dinside
D
diameter
Bim = efffluid
Dinside
δ
δ
Mears’ Test:
rA′ (observed) ρb R p n
if
kc C A bulk
< 0.15
where n ≡ order of rxn
then C Ab ≈ C As (no external diffusion limitation) Æ i.e. no changing
concentration across the boundary layer
Similarly,
if
ΔH rxn rA′′ (observed or theory) ρb R p Ea
h fluid RT 2
rAno external diff. limit vs. rAobserved
V
S
[>]
S
.B
Weisz-Prater:
if
rA′ (observed) ρc R p
N
I
.
E
< 0.15 then Tb ≈ Ts (text: Eqn. 12-63)
2
W
eff
Dinside
C A bulk
(text: Eqn. 12-61)
1
W
then you can neglect internal diffusion limitations, i.e. C As ≈ C Ab ≈ C A ( r = 0)
1000
η
W
Ea ∼ 30 RTs
β = 0.6
100
ΔH rxn
Ea
10
1
.1
1
10
.1
ΔH rxn > 0
100
0⎫
⎬ temp. dep. & exothermic
0 ⎭
φ1
ΔH rxn = 0
or Ea = 0
Figure 3. (text: Figure 12-7)
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 22
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β=
100
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−ΔH rxn DeC As
eff
where De ≡ Dinside and kt ≡ heat conductivity
ktTs
⎛ ∂Ci ⎞
eff
kc ( Ci ,b − Ci , s ) = − Dinside
⎜
⎟
⎝ ∂r ⎠ r = R
= ∫ ri dV .
kc Ci
r=R
⎛ ∂Ci ⎞
eff
− Dinside
= kc Ci , bulk
⎜ r ⎟
⎝ ∂ ⎠ r =R
⎛ ∂Ci ⎞
⎜ r ⎟ =0
⎝ ∂ ⎠ r =0
if no significant external diffusion limit: Ci
r =R
≈ Ci , bulk
rA ≈ −k C A
e − Ea
V
S
rAeff ≈ f (k ) C A
fit to:
N
I
.
E
RT
Aeff e− Ea
eff
RT
.
(text: Table 12-1)
S
.B
W
W
W
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 22
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 25: Course Review
Review
Fundamental Equations:
Acc = Flow In – Flow Out + Reaction
∂nm
= Fm ,o − Fm ,out + ∫∫∫ rm ( x, y, z , t ) dxdydz
∂t
infinitesimal volume
∂Cm
= ∇ ⋅ Fm + rm
∂t
K eq = e−ΔGrxn
RT
νn
N products
⎛ Pm
⎞
⎜ 1 bar ⎟
∏
⎝
⎠
K eq = m =1
−ν j
reactants
⎛ Pj
⎞
∏
⎜ 1 bar ⎟
⎠
j =1 ⎝
Kc =
N
I
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E
( K eq is unitless)
k forward
V
S
S
.B
W
kreverse
W
H 2O2 → H 2 + O2
n
P
=
V RT
⎛ PH 2
⎞⎛ PO2
⎞
⎜ 1 bar ⎟⎜ 1 bar ⎟
⎠⎝
⎠
K eq = ⎝
⎛ PH 2O2
⎞
⎜
1 bar ⎟⎠
⎝
W
Kc =
[ H 2 ][O2 ]
[ H 2O2 ]
units K c [ =]
mol
L
Convection dominated:
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Constant P, constant T, constant reactor V
Fm ≈ ν Cm
∇ ⋅ Wm ≈ D∇ 2Cm + U ⋅∇Cm
Pressure drop in Packed Bed:
Ergun Equation:
⎤
G 1 − φ ⎡150 (1 − φ ) μ
∂P
=−
+ 1.75G ⎥
3 ⎢
pg c D p φ ⎣⎢
Dp
∂z
⎦⎥
∂U cvtot
∂V
+ P cv = ∑ Fj H j + Q + Ws
∂t
∂t
( U cv depends on T)
∂T ⎪⎧ν C (Tin − Tout ) + Q + ∑ rk ΔH rxn ⎪⎫
=⎨
⎬
∂t ⎪⎩
Ctotal
⎪⎭
where
tot
C is the heat capacity
N
I
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Special Cases:
Perfectly Homogeneous (“well stirred”, “perfectly mixed”)
V
S
Æ no flows, “batch reactor”
∂nm
= rm ( C (t ) ) V
∂t
S
.B
Æ CSTR, no t-dependence
0 = Fin − Fout + r ( Cout )
W
Homogeneous in x,y, not in z (no t-dependence)
W
PFR (typically gives higher productivity than CSTR)
∂Fm
= Arm ( C ( z ) )
∂z
W
“sort of” PFR
∂Fm
= Arm ( C avg ( z ) ) Ω( z )
∂z
•
bubble •
where Ω ( z ) is the effectiveness factor
cat
Figure 1. a) mass transfer from gas to liquid b) mass transfer into catalyst particle
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 25
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D
103
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∂ 2Cm ⎛ 2 ⎞ ∂Cm
+⎜ ⎟
+ rm ( C ( r ) ) = 0
∂r 2 ⎝ r ⎠ ∂r
-
nondimensionalize
some solutions in book
(18.03)
guess solution, plug in to verify
matlab
Thiele modulus
φ2 ≡
rm ( Csurface )
Dsolid Cm , s R 2
- if small (<1): reaction limited, ignore effectiveness factor Ω (internal diffusion
fast)
- if big: transport matters!
N
I
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E
Ffrom = Ak L (Cinterface − Cbulk )
bubble
kL A Æ correlations
V
S
(sphere-packed bed)
S
.B
= kc A(Cbulk − Cs )
Finto
particle
kc ∼
D
δ
Converting the second-order differential equation into first-order ordinary differential
equations for MatLab solvers:
∂Cm
= qm
∂r
D
W
W
W
∂qm 2
+ qm + rm = 0
∂r r
2
−
qm + rm
∂qm
r
=
Æ
Æ MATLAB: ode15s
∂r
D
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. William H. Green
Lecture 25
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Error in Fogler: Superficial Velocity or Actual Velocity
See page 781 of the 4th edition for the external transport limited PBR derivation (no axial
diffusion or radial dispersion).
Fogler states that: FA = C A,bulkUAc
where U is the superficial molar average velocity. This is defined as the velocity that the
fluid would have if no catalyst were present. It can also be thought of as the velocity that
would be measured immediately upstream or downstream of the packed bed.
This equation is incorrect if you consider the units, there should be an additional factor of
phi, the void volume, in it. The bulk concentration should be measured by taking a fluid
sample, not measuring the total amount of a component per unit volume of the entire
reactor.
FA[=]moles time
FA[≠ ]moles fluid volume × reactor volume time
N
I
.
E
The correct relationship is:
V
S
FA = C A,bulkϕUAc
S
.B
FA[=]moles time[=]moles fluid volume × fluid volume reactor volume×reactor volume time
W
This is important for converting from the flowrate to the concentration correctly in a
PBR.
W
If you want to know the residence time in your PBR, the important velocity is the
ACTUAL average axial velocity of the fluid flowing in the void space, not the superficial
velocity. The reduction in reactor volume by incorporating catalyst causes the residence
time to be shorter than predictions using the superficial velocity. In cases where the fluid
is water (constant volumetric flow rate):
W
Vactual = U
ϕ
τ = LV
actual
A similar mistake with the superficial velocity is made on page 843. A simple remedy is
to replace U with Uφ in these equations.
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Department of Chemical Engineering University of Cambridge – Weblab Exercise (due April 6, 2007 at 1 pm) Part 2 – Post Lab Write-Up Four runs were completed during the two recitations on Monday, March 19, 2007.
The settings for the four runs are given below:
Run
1
2
3
4
Q1 (NaOH) (mL/min)
8.1
12
16.9
21
Q2 (Phen) (mL/min)
16.9
13
8.1
4
The baseline intensity before we started run 1 was I0 = 2977 ± 4, the stirrer was
kept at full power during all four runs, and the temperature of the feed streams and
reactor during the four runs was approximately 18oC.
N
I
.
E
The data from the four experiments are tabulated in .txt files and are posted on the
10.37 Stellar webpage. In each file, the first column is the time (sec) and the
second column is -ln(I/I0).
V
S
S
.B
The four data sets have been fit to the model:
 I 
− ln  = A + B exp(− Ct )
 I 0 
W
W
The parameter values for A, B and C are tabulated below. In addition, figures
showing the experimental data and the fit model for each run are included at the
end of this assignment.
W
Run
1
2
3
4
A
2.76
2.00
1.05
0.44
B
-2.25
0.56
0.97
0.60
C (sec-1)
.0018
.0016
.0019
.0027
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1) Do the values of α, β, k1, and k2 that you obtained in the pre-lab provide semiquantitative agreement with the data measured during recitation? How far off are
the predictions that you would make using the model and parameters suggested
from the pre-lab information? What could you do to get better predicted
parameters?
2) Can you determine the conversion in the experiment we did on Monday from
the dataset? Would it be helpful to know the value of "b" that relates ln(I/I0) to the
concentrations? Explain.
3) From the noise level in the baseline, how close would one expect the agreement
to be if the model were perfect and the parameters α, β, k1, and k2, and V and flow
rates were all known perfectly?
N
I
.
E
V
S
S
.B
W
W
W
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N
I
.
E
V
S
S
.B
W
W
W
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N
I
.
E
V
S
S
.B
W
W
W
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Department of Chemical Engineering University of Cambridge – Weblab Exercise (due April 6, 2007 at 1 pm) Note: Parts A – D should be completed before the Weblab on Monday, March 19, 2007,
but don’t have to be turned in until April 6, 2007.
Introduction
This exercise concerns the reaction of phenolphthalein (PHEN) in an aqueous solution of
sodium hydroxide:
PHEN + 2OH- → PHEN2- + 2H2O
(1)
PHEN2- + OH- ↔ PHENOH3(pink)
(colourless)
(2)
Reaction (1) can be assumed to occur rapidly, so the rate determining step is reaction (2).
The rate laws for reaction (2) are:
r1 = k1[PHEN2-][OH-]
N
I
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E
(3)
3-
r2 = k2[PHENOH ]
(4)
V
S
The purpose of this exercise is to determine the reaction constants for this reaction, and to
predict the operating conditions required to achieve a given conversion of PHEN2- to
PHENOH3- in a non-ideal reactor. You will test your predictions by running experiments
on the reactor in question.
S
.B
W
PART A
http://weblabs.cheng.cam.ac.uk/reactors_batchdata.html contains experimental data from
reactions (1) and (2) carried out in a batch reactor. To track the progress of the reaction, a
spectrophotometer has been used to measure the intensity of light (550 nm) transmitted
through the reaction solution. This intensity, I, is related to the concentration, c, by the
Beer-Lambert law:
W
W
c = –b * ln (I/I0)
(5)
Note: As concentration increases (more pink stuff), the intensity of light passing through
the sample decreases (c up, I down). b is positive, but knowing the value is unnecessary
because it will cancel out.
Show that the time variation of the intensity, I, through the experiment can be expressed
as: (give an expression for the pseudo first order rate constant k1’)
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ln (I ) − ln (I t=∞ )
= exp − (k1' + k 2 )t
ln (I t=0 ) − ln (I t=∞ )
[
]
(6)
Use the experimental data to determine the rate constants k1 and k2 for the first set
of data (collected at 20.0 °C). The above data was taken in an IDEAL BATCH
REACTOR, unlike Part B and beyond.
Part B
The reaction is to be carried out in a continuous reactor (CSTR - Figure 1). More
information on the reactor setup can be found at
http://weblabs.cheng.cam.ac.uk/reactors.html. The available reactant concentrations are:
NaOH:
[OH-]in = 0.20 mol/l
Phenolphthalein:
[PHEN]in = 7.2 x 10-5 mol/l
N
I
.
E
Q1
NaOH
Q
V
Phenolphthalein
Products
V
S
Q2
S
.B
Figure 1: A schematic of the reactor
W
In most cases, modeling a reactor as a stirred tank is not sufficiently accurate. One model
for a non-ideal reactor is the bypass/dead volume model, in which a fraction β of the flow
bypasses the reaction zone, and only a fraction α of the reactor volume is utilized. This is
shown schematically in figure 2.
W
W
Q
βQ
(1–β)Q
αV
Figure 2: A schematic of the bypass/dead volume reactor model
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Reactor characteristics can be determined by examining the residence time distributions
under continuous operation. Using Laplace Transforms show that for a pulse input of Nt
moles of tracer at t = 0, the outlet concentrations for the two reactor models are:
CSTR:
c(t )
= e −t / τ ,
N t /V
Bypass/Dead volume:
(1 − β ) exp − 1 − β t  ,
c(t )
= βδ (t )τ +


N t /V
α
 ατ 
(7)
2
(8)
where τ = V / Q .
Part C
Use the experimental data at http://weblabs.cheng.cam.ac.uk/reactors_tracerdata.html to
determine α and β for the bypass/dead volume model when applied to the reactor in nonideal setup. Note: Data is given for the reactor when run as close to being a perfect
CSTR as possible, as well as when it is run as a non-ideal CSTR. The perfect CSTR data
will allow you to relate I (t=0) to I0, which is just a reference value. I (t=0) is the intensity
directly AFTER the pulse is given, when the concentration is Nt / V.
N
I
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E
V
S
Part D
S
.B
It is desired to run the reactor at a product flowrate, Q, of 25 ml/min and a conversion, X,
of 30 %. In this case, conversion is defined as:
X =
[ΡΗΕΝΟΗ 3− ]
,
[ΡΗΕΝ 2− ] + [ΡΗΕΝΟΗ 3− ]
W
(9)
W
to avoid any ambiguities due to dilution of the feed streams.
W
Assume that this reactor can be modeled as an ideal continuous stirred-tank reactor
(CSTR) of volume V = 250 ml. Use steady state material balances to show that to achieve
a product flowrate, Q, at conversion, X, the NaOH flowrate must be:
Q1 =
X Q(k 2V + Q )
1 − X k1V[ΟΗ − ]in
(10)
Now assume that the reactor behaves according to the bypass/dead volume model. What
value of Q1 is required to give a product flowrate of 25 ml/min at a conversion of 30%?
In order to avoid too much unnecessary algebra, it may be useful to set up an Excel
spreadsheet (or similar) so that you can determine the conversions at different flowrates
with your calculated values of the parameters α and β.
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Part E
Experimental Procedure – this will be done as a group in recitation on Monday,
March 19, 2007
The aims of the experiment are:
1. To test the assumption that the reactor behaves as a CSTR.
2. To test the non-ideal model and parameters you have derived in your preliminary
analysis.
To determine a baseline reading for the intensity, set the flowrate of NaOH to the value
calculated from equation (10). Once a steady value for I0 is achieved, set the flowrate of
phenolphthalein to give a total flowrate of 25 ml/min.
Let the reactor equilibrate and record the steady state intensity.
Adjust the flowrates of the two reactants to the values calculated for the non-ideal reactor
setup. Again, let the reactor equilibrate and record the steady state intensity.
The plan is that each recitation will perform two runs, and you will analyze the data from
both recitations (up to four sets of data).
N
I
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E
Part F
V
S
Experimental Data Analysis
Data analysis requirements will be posted on the 10.37 Stellar website after the Weblab is
completed, and will be due with the rest of the write-up on April 6, 2007.
S
.B
W
W
W
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For Fogler 2-5 and 6-6
In the first problem, Fogler 2-5, the textbook asks to use the figure to calculate the
conversions. However, the figure itself has problems such as not matching with the table
above it (less data points and one misrepresented point) and difficult to get accurate
results. You are required to use the values in the table instead of the figure and Matlab to
solve this problem. The use of matlab involves numerical interpolation (by command
interp1, together with a spline scheme to get a smooth interpolation), numerical
integration (quad), and numerical solution to equations (fzero). These will be also used in
the RTD problem. Of course, any method getting the accurate results will be accepted.
A sample code (showing you how to use these functions) is as follows:
x0=fzero(@equation,1)
y=quad(@curve,0,1);%to integrate function curve(x) with x from 0 to 1
function y=curve(x)
data=[1 2 3 4 5;1.1 2.1 3.01 4.03 5.2]';
y=interp1(data(:,1),data(:,2),x,'spline');
return
function y=equation(x)
t=0.3;y=curve(x)*x-t;
return
N
I
.
E
V
S
S
.B
In the problem of Fogler 6-6, plot the selectivities vs. concentrations using Matlab, not by
crude sketching. This will help you in understanding the obscure problem
statement better.
W
W
W
Cite as: William Green,
Jr., and K. Dane Wittrup, course
materials for 10.37 Chemicalwww.bsslifeskillscollege.in
and Biological Reaction Engineering,
www.bsscommunitycollege.in
www.bssnewgeneration.in
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
www.onlineeducation.bharatsevaksamaj.net
www.bssskillmission.in
114
HW6 Batch Reactor Balances
It is not legitimate to simply take the CSTR enthalpy
balance and set the flow terms to zero. This is an
incorrect enthalpy balance on the reactor in batch mode.
See below for a review of how to use thermodynamics in this
situation.
Given that there is negligible gas holdup in the reactor:
N B (t) = N C (t) = 0
hence,
dN B dN C
=
=0
dt
dt
Enthalpy balance:
d H d (N A H A + N B H B + N C H C ) d (N A H A ) dN A
dH A
=
=
=
HA +
NA
dt
dt
dt
dt
dt
dH &
= Q − FB H B − FC H C
dt
dN A
dH A
HA +
N A = Q& − FB H B − FC H C
dt
dt
Batch Reactor mole balances:
dN A
= −kN A
dt
dN B
= 0 = −FB,out + kN A
dt
FB,out = kN A
N
I
.
E
V
S
S
.B
dN C
W
W
= 0 = −FC ,out + 2kN A
dt
FC ,out = 2kN A
Plugging in the three mole balance relationships into the
enthalpy balance we find:
dH A
− kN A H A +
N A = Q& − kN A H B − 2kN A H C
dt
dH A
N A = Q& − kN A (H B + 2H C − H A ) = Q& − kN A ∆H rxn (T )
dt
dH A
dT
= c
p,a
dt
dt
&
dT Q − kN A ∆H rxn (T ) Q& − km A ∆Ĥ rxn (T )
=
=
dt
N A c p,a
m A ĉ p,a
W
Cite as: David Adrian,
course materials for 10.37 Chemical
and Biological Reaction Engineering,
Spring 2007. MIT OpenCourseWare
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
(http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].��
www.onlineeducation.bharatsevaksamaj.net
www.bssskillmission.in
10.37 Spring 2007 Homework 1 Due noon Wednesday, Feb. 14. Problem 1. Airbags contain a mixture of NaN3, NaNO3, and SiO2. When the vehicle is in
a crash, the following reactions are initiated:
2 NaN3 Æ 2 Na + 3 N2
10 Na + 2 NaNO3 Æ N2 + 6 Na2O Na2O + 10 SiO2 Æ glass a) If 150 g of NaN3 are used in an airbag, how many grams of NaNO3 and SiO2 must be
included so that all of the sodium in the system can be safely sequestered as glass? Note
the sodium-containing compounds NaN3, Na, and Na2O are all dangerous and toxic.
b) The most important species for airbag performance in a crash are NaN3 and N2, so
there are two obvious definitions of conversion:
XNaN3 = (moles NaN3 reacted)/(initial moles NaN3)
N
I
.
E
and
XN2 = (moles of N2)/(total moles of N2 when all reactions are completed).
V
S
What units do XNaN3 and XN2 have? Does XNaN3 equal XN2? If not, how different could
they be?
S
.B
There are three other related quantities, ξ1, ξ2, and ξ3, the extents of reactions 1,2, and 3.
Note that each ξ has units of moles. Write algebraic equations for each X in terms of the
ξ‘s.
W
W
c) Suppose that reaction 1 has a rate expression r1=k1/V (this reaction proceeds at a steady rate as a reaction front moves through the solid NaN3), reaction 2 has a rate expression r2=k2[Na][NaNO3], and reaction 3 has a rate expression r3=k3[Na2O]/V. W
By the convention used in this course, all the r’s have units of moles/second/liter. Write
rN2, the rate of production of N2 per unit volume, in terms of r1, r2, and r3.
Write the equations for rate of change of the number of moles, dni/dt, for all the chemical
species (i=N2, NaN3, Na, NaNO3, Na2O, SiO2, glass).
d) Of course the volume of the airbag, V, is dramatically changing during the course of
the reaction due to the creation of a gas, N2, inside the bag. If the bag can expand fast
enough to so that the pressure inside the bag is similar to the pressure outside the bag, by
the ideal gas law one would expect:
V = Vo + VN * nN2
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction
Engineering,
Spring 2007. MIT OpenCourseWare
(http://ocw.mit.edu), Massachusetts
Institute of Technology.
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
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and under this condition the bag would expand depending on the rate at which gas is
created:
dV/dt = VN * dnN2/dt
where VN is the molar volume of a gas at atmospheric pressure (~22 liter/mole) and nN2 is
the number of moles of N2 in the airbag. The initial volume of the airbag Vo ~70 cm3.
However there is a physical limit on how fast the airbag can expand. When an airbag is
expanded by gas pressure, the radius of the bag cannot grow faster than the speed of
pressure fronts in the gas, approximately the speed of sound:
dR/dt < csound
csound ~ 300 m/s in air.
so there is an upper bound on how fast the airbag can grow; for a spherical airbag:
N
I
.
E
dV/dt = 4π R2 dR/dt
so dV/dt < 4π (3V/4π)2/3 csound = (36πV2)1/3csound
V
S
So a reasonable approach to model this numerically is
if (V<Vo + VN*nN2)
dV/dt = (36πV2)1/3csound
else
dV/dt = VN * dnN2/dt
endif
S
.B
W
W
Using a numerical ODE solver in Matlab, solve the coupled system of differential
equations for the n’s and V. Take k1~103 moles/s, k2~104 liter/mole-s, k3~105 liter/s.
Make and turn in a plot of XN2 vs. time, ξ3 vs. time and volume vs. time for the first 10
milliseconds of operation. Also, make and turn in a plot of volume vs. time for just the
first 0.1 milliseconds of operation. Does the volume vs. time behavior make physical
sense? If not, go back and modify your Matlab program to fix the non-physical dV/dt
behavior.
W
Submit your Matlab program(s) to the 10.37 the course website.
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Engineering,
Spring 2007. MIT OpenCourseWare
(http://ocw.mit.edu), Massachusetts
Institute of Technology.
Downloaded on [DD Month YYYY].
116
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Problem 2. One of the students in this class recently measured the reaction of vinyl
radical (C2H3) with ethene (C2H4), a reaction important in flames, pyrolysis, and
polymerization reactors. Vinyl radical absorbs purple light, so the amount of light
absorbed is proportional to the concentration of the vinyl radical. In each experiment the
student measured the time variation in the amount of purple light passing through his
constant volume sample using a photodetector. The voltage signal from the photodetector
is linearly related to the absorbance, which is proportional to [C2H3], so
Signal(t) = b + m[C2H3](t)
Eq. (1)
where b is an uninteresting number related to how well the electronics baseline was
zeroed out before each experiment.
He performed similar experiments many times, each time with different initial
concentrations of ethene in the sample. From these experiments, he extracted the rate
constant “k” at various temperatures and pressures.
N
I
.
E
The reaction of interest is:
C2H3 + C2H4 Æ products
V
S
This reaction is very exothermic, so the reaction is essentially irreversible (i.e. when
equilibrium is achieved the vinyl concentration is too small to detect). You expect this
reaction to follow elementary-step kinetics, i.e.
-rC2H3=(k0 + k[C2H4])[C2H3]
S
.B
(Eq. 2)
W
k0 accounts for all other first-order loss processes of C2H3 in the experiment (e.g.
unimolecular reaction). Because the initial concentration of C2H3 is much smaller than
the concentration of C2H4, it is reasonable to assume that the concentration of C2H4 does
not vary significantly during each experiment. Therefore one expects a simple
exponential decay of [C2H3]:
W
W
[C2H3] = [C2H3]o e-t/τ
(Eq. 3)
a) Write out the algebraic relationship between τ and k. Fit the measured signal for the
nth experiment Sn to this form:
Sn(t) = Bn + An exp(-t/τn)
(Eq. 4)
Give expressions for An, Bn, and τn in terms of b, m, [C2H3]o, k0, k, and [C2H4]0,n. Which
of the three fit parameters An, Bn, and τn depends on k and [C2H4]0 ?
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Engineering,
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(http://ocw.mit.edu), Massachusetts
Institute of Technology.
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b) Use Matlab to plot 1/τn vs. [C2H4]n, where τn is the exponential decay time constant
determined by fitting the data from the nth experiment. How can you use this plot to
determine the rate constant “k”?
Your assignment is to compute the rate constant “k” for the reaction of interest from the
student’s data, contained in files vinylethene1, vinylethene2, and vinylethene3 on the
10.37 course website. In each file the first column is the time in seconds, and the second
column is the measured signal Sn. The first dataset is for [C2H4]=6.7x10-4 M, the second
for [C2H4]=4x10-4 M and the third for [C2H4]=1.33x10-4 M.
Turn in the value of “k” you derived from modeling the student’s experimental data
(don’t forget to specify the units of “k”!), and also turn in plots comparing your model
predictions using this “k” with the experimental data.
Submit your Matlab program(s) to the 10.37 course website.
N.B. Notice that in this type of “pseudo-first-order” experiment, one can determine “k”
without knowing [C2H3]o, the calibration constant “m” relating the signal to [C2H3], what
the products of the reaction are, nor what the competing reactions are (that contribute to
k0). Because of these simplifications, this type of experiment is very widely used to
determine rate constants.
N
I
.
E
V
S
S
.B
W
W
W
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction
Engineering,
Spring 2007. MIT OpenCourseWare
(http://ocw.mit.edu), Massachusetts
Institute of Technology.
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Downloaded on [DD Month YYYY].
118
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www.bssskillmission.in
10.37 Spring 2007 Homework 1 Due Wednesday, Feb. 14. Problem 1. Airbags contain a mixture of NaN3, NaNO3, and SiO2. When the vehicle is in a crash, the following reactions are initiated: 2 NaN3 Æ 2 Na + 3 N2
10 Na + 2 NaNO3 Æ N2 + 6 Na2O Na2O + 10 SiO2 Æ glass a) If 150 g of NaN3 are used in an airbag, how many grams of NaNO3 and SiO2 must be included so that all
of the sodium in the system can be safely sequestered as glass? Note the sodium-containing compounds
NaN3, Na, and Na2O are all dangerous and toxic.
b) The most important species for airbag performance in a crash are NaN3 and N2, so there are two obvious
definitions of conversion:
XNaN3 = (moles NaN3 reacted)/(initial moles NaN3) and XN2 = (moles of N2)/(total moles of N2 when all reactions are completed). N
I
.
E
What units do XNaN3 and XN2 have? Does XNaN3 equal XN2? If not, how different could they be?
There are three other related quantities, ξ1, ξ2, and ξ3, the extents of reactions 1,2, and 3. Note that each ξ
has units of moles. Write algebraic equations for each X in terms of the ξ‘s. V
S
c) Suppose that reaction 1 has a rate expression r1=k1/V (this reaction proceeds at a steady rate as a reaction
front moves through the solid NaN3), reaction 2 has a rate expression r2=k2[Na][NaNO3], and reaction 3 has
a rate expression r3=k3[Na2O]/V.
S
.B
By the convention used in this course, all the r’s have units of moles/second/liter. Write rN2, the rate of
production of N2 per unit volume, in terms of r1, r2, and r3. W
Write the equations for rate of change of the number of moles, dni/dt, for all the chemical species (i=N2, NaN3, Na, NaNO3, Na2O, SiO2, glass). W
W
d) Of course the volume of the airbag, V, is dramatically changing during the course of the reaction due to
the creation of a gas, N2, inside the bag. If the bag can expand fast enough to so that the pressure inside the
bag is similar to the pressure outside the bag, by the ideal gas law one would expect:
V = Vo + VN * nN2
and under this condition the bag would expand depending on the rate at which gas is created: dV/dt = VN * dnN2/dt where VN is the molar volume of a gas at atmospheric pressure (~22 liter/mole) and nN2 is the number of moles of N2 in the airbag. The initial volume of the airbag Vo ~70 cm3. However there is a physical limit on how fast the airbag can expand. When an airbag is expanded by gas pressure, the radius of the bag cannot grow faster than the speed of pressure fronts in the gas, approximately the speed of sound: dR/dt < csound
csound ~ 300 m/s in air. Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction
Engineering,
Spring 2007. MIT OpenCourseWare
(http://ocw.mit.edu), Massachusetts
Institute of Technology.
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Downloaded on [DD Month YYYY].
119
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www.bssskillmission.in
so there is an upper bound on how fast the airbag can grow; for a spherical airbag:
dV/dt = 4π R2 dR/dt
so dV/dt < 4π (3V/4π)2/3 csound = (36πV2)1/3csound
So a reasonable approach to model this numerically is
if (V<Vo + VN*nN2)
dV/dt = (36πV2)1/3csound
else
dV/dt = VN * dnN2/dt
endif
Using a numerical ODE solver in Matlab, solve the coupled system of differential equations for the n’s and
V. Take k1~103 moles/s, k2~106 liter/mole-s, k3~105 liter/s. Make and turn in a plot of XN2 vs. time, ξ3 vs.
time, and volume vs. time for the first 10 milliseconds of operation. Also, make and turn in a plot of
volume vs. time for just the first 0.1 milliseconds of operation. Does the volume vs. time behavior make
physical sense? If not, go back and modify your Matlab program to fix the non-physical dV/dt behavior.
Submit your Matlab program(s) to the 10.37 course website.
Problem 1 Solution
N
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E
a)
(2 NaN3 Æ 2 Na + 3 N2)÷2
(10 Na + 2 NaNO3 Æ N2 + 6 Na2O)÷10
(Na2O + 10 SiO2 Æ glass)x6/10
→ NaN3 Æ Na + 3/2 N2
→ Na + 1/5 NaNO3 Æ 1/10N2 + 6/10 Na2O
→ 6/10Na2O + 6SiO2 Æ 6/10glass
V
S
S
.B
Net reaction w/all Na safely sequestered as glass:
1
NaN3 + /5NaNO3 + 6SiO2 Æ 8/5N2 + 3/5 glass
1mol
= 2.3mol
65g
85g
1mol NaNO3 2.3
n NaNO3 ,o = n NaN 3 ,o ⋅
=
mol = 0.46mol or m NaNO3 ,o = 0.46mol
= 39g
5mol NaN 3
5
1mol
6mol SiO2
60g
= 830g
n SiO2 ,o = n NaN 3 ,o ⋅
= 6 ⋅ 2.3mol = 13.8mol or mSiO2 ,o = 13.8mol
1mol NaN 3
1mol
n NaN3 ,o = 150g
W
W
W
b) XNaN3 and XN2 are dimensionless. XNaN3 does not equal XN2 during the reaction.
X NaN 3 =
2ξ1
X N2 =
n NaN 3 ,o
X NaN 3 − X N 2 =
2ξ1
n NaN 3 ,o
−
nN2
nN2 , f
ξ − 5ξ 2
3ξ1 + ξ 2
= 1
8
n NaN 3 ,o 8n NaN 3 ,o
5
=
3ξ1 + ξ 2
8
n NaN 3 ,
o
5
The difference between XNaN3 and XN2 will be determined by the kinetics of r1 and r2 (i.e.
the rate of change of ξ1 compared to ξ2).
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Engineering,
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Institute of Technology.
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c) For no flows in/out and assuming a homogeneous reaction:
dn A
=
dt
∑ν
i
A,i
dξi
=
dt
∑ν
Can report dni/dt using extents or reaction rates
A,i riV
i
Species
mol balance
dni/dt (using extents)
NaN3
nNaN3 = nNaN3,o - 2ξ1
dn NaN 3
dn NaNO3
nNaNO3 = nNaNO3,o - 2ξ2
dt
SiO2
dn SiO2
nSiO2 = nSiO2,o - 10ξ3
dt
Na
nNa = 2ξ1 - 10ξ2
N2
nN2 = 3ξ1 + ξ2
Na2O
nNa2O = 6ξ2 − ξ3
Glass
nglass = ξ3
dn NaN 3
dξ1
dt
dξ
= −2 2
dt
= −2
dt
NaNO3
dni/dt (using reaction rates)
= −10
dt
dn
NaNO
dξ 3
dt
dt
dn SiO 2
dn Na
dξ
dξ
= 2 1 − 10 2
dt
dt
dt
dn N 2
dξ1 dξ 2
=3
+
dt
dt
dt
dn Na2O
dξ 2 dξ 3
=6
−
dt
dt
dt
dn glass dξ 3
=
dt
dt
dt
3
= −2r1V
= − 2 r2V
= − 10 r3V
dn Na
= (2r1 − 10r2 )V
dt
dn N 2
= (3r1 + r2 )V
dt
dn Na2O
= (6r2 − r3 )V
dt
dn glass
= r3V
dt
N
I
.
E
Rate of production of N2: rN 2 = 3⋅ r1 + r2
V
S
d) See the Matlab solution provided. Plots showing XN2 vs. time, ξ3 vs. time, and
volume vs. time are copied below.
S
.B
W
W
W
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Engineering,
Spring 2007. MIT OpenCourseWare
(http://ocw.mit.edu), Massachusetts
Institute of Technology.
Downloaded on [DD Month YYYY].
www.onlineeducation.bharatsevaksamaj.net
www.bssskillmission.in
Problem 2. One of the students in this class recently measured the reaction of vinyl radical (C2H3) with
ethene (C2H4), a reaction important in flames, pyrolysis, and polymerization reactors. Vinyl radical absorbs
purple light, so the amount of light absorbed is proportional to the concentration of the vinyl radical. In
each experiment the student measured the time variation in the amount of purple light passing through his
constant volume sample using a photodetector. The voltage signal from the photodetector is linearly related
to the absorbance, which is proportional to [C2H3], so
Signal(t) = b + m[C2H3](t)
Eq. (1)
where b is an uninteresting number related to how well the electronics baseline was zeroed out before each
experiment.
He performed similar experiments many times, each time with different initial concentrations of ethene in
the sample. From these experiments, he extracted the rate constant “k” at various temps and pressures.
The reaction of interest is:
C2H3 + C2H4 Æ products
This reaction is very exothermic, so the reaction is essentially irreversible (i.e. when equilibrium is
achieved the vinyl concentration is too small to detect). You expect this reaction to follow elementary-step
kinetics, i.e.
-rC2H3=(k0 + k[C2H4])[C2H3]
(Eq. 2)
N
I
.
E
k0 accounts for all other first-order loss processes of C2H3 in the experiment (e.g. unimolecular reaction).
Because the initial concentration of C2H3 is much smaller than the concentration of C2H4, it is reasonable to
assume that the concentration of C2H4 does not vary significantly during each experiment. Therefore one
expects a simple exponential decay of [C2H3]:
[C2H3] = [C2H3]o e-t/τ
(Eq. 3)
V
S
S
.B
a) Write out the algebraic relationship between τ and k. Fit the measured signal for the nth experiment Sn
to this form:
W
Sn(t) = Bn + An exp(-t/τn)
(Eq. 4)
W
Give expressions for An, Bn, and τn in terms of b, m, [C2H3]o, k0, k, and [C2H4]0,n. Which of the three fit
parameters An, Bn, and τn depends on k and [C2H4]0 ?
W
b) Use Matlab to plot 1/τn vs. [C2H4]n, where τn is the exponential decay time constant determined by
fitting the data from the nth experiment. How can you use this plot to determine the rate constant “k”?
Your assignment is to compute the rate constant “k” for the reaction of interest from the student’s data,
contained in files vinylethene1, vinylethene2, and vinylethene3 on the 10.37 course website. In each file the
first column is the time in seconds, and the second column is the measured signal Sn. The first dataset is for
[C2H4]=6.7x10-4 M, the second for [C2H4]=4x10-4 M and the third for [C2H4]=1.33x10-4 M.
Turn in the value of “k” you derived from modeling the student’s experimental data (don’t forget to specify
the units of “k”!), and also turn in plots comparing your model predictions using this “k” with the
experimental data. Submit your Matlab program(s) to the 10.37 course website.
N.B. Notice that in this type of “pseudo-first-order” experiment, one can determine “k” without knowing
[C2H3]o, the calibration constant “m” relating the signal to [C2H3], what the products of the reaction are, nor
what the competing reactions are (that contribute to r0). Because of these simplifications, this type of
experiment is very widely use’d to determine rate constants.
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Engineering,
Spring 2007. MIT OpenCourseWare
(http://ocw.mit.edu), Massachusetts
Institute of Technology.
Downloaded on [DD Month YYYY].
122
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Problem 2 Solution
a) − rC2 H 3 = −
d [C2 H 3 ]
dt
Eq 5
Substitute Eq 3 into Eq 5
− rC2 H 3 = −
(
)
d [C 2 H 3 ]o e −t τ
d [C 2 H 3 ]
1
=−
= [C 2 H 3 ]o  e −t τ
dt
dt
τ 
Eq 6
Substitute Eq 3 into Eq 2:
− rC2 H 3 = (k 0 + k [C 2 H 4 ])[C 2 H 3 ] = (k 0 + k [C 2 H 4 ])[C 2 H 3 ]o e − t τ
Eq 7
Set Eq 6 = Eq 7 and simplify: [C 2 H 3 ]o  1 e −t τ = (k 0 + k [C 2 H 4 ]o )[C 2 H 3 ]o e −t τ →
τ 
τ=
1
k 0 + k [C 2 H 4 ]o
Eq 8
algebraic relationship between τ and k
Substitute Eq 3 into Eq 1:
S (t ) = b + m[C 2 H 3 ](t ) = b + m[C 2 H 3 ]o e − t τ
Eq 9
Substitute Eq 8 into Eq 9: S (t ) = b + m[C 2 H 3 ]o e
= b + m[C 2 H 3 ]o
−t τ
−t

1

 k + k [C H ]
0
2 4 o

e




N
I
.
E
Eq 10
Compare Eq 10 and Eq 4 to define: Bn = b
An = m[C2 H 3 ]o
τn =
1
k 0 + k [C2 H 4 ]o,n
Only τn depends on k and [C2H4]o,n.
V
S
S
.B
b) See the Matlab solution provided. To determine k, fit 1/τn vs. [C2H4]o,n to a straight
line. Slope = k (L/mol-s), y-intercept = k0 (1/s). Figures from Matlab are copied below.
k=1.11x107 L/mol-s
W
W
W
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.B
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10.37 Spring 2007
Problem Set 2 due Wednesday, Feb. 21.
Problem 1. The microorganism Mycobacterium vaccae is able to grow with ethane as the
sole source of carbon and energy and NH as the nitrogen source. The limiting substrate
3
is ethane, and Y =22.8 gram dry weight per mole of ethane. {Y = Yield of biomass (x)
sx
sx
from ethane (s = substrate)}.
a. Except for small amounts of S and P, an analysis of dry cell mass is C, 47.60 wt%; N,
7.30 wt%; H, 7.33 wt%; ash, 3.00 wt%. The remainder is taken to be oxygen, which
can not be detected in the analysis. Determine the elemental composition for the ashfree biomass, CH O N , and the formula weight per C-atom. Also determine Y , in
a
b
N
I
.
E
c
the units of C-moles of biomass per C-mole of ethane.
sx
V
S
b. Calculate the oxygen consumption Y (moles of O per C-mole of biomass) when it is
xo
2
S
.B
assumed that CO , H O and CH O N are the only metabolic products. Write the full
2
2
a
b
c
stoichiometric equation for the growth process, and determine the heat evolved per
W
kilogram dry weight. Assume ΔH
ethane
= 19 kJ/gram dry weight.
= 1560 kJ/mol, ΔH
NH3
= 383 kJ/mol, ΔH
biomass
W
W
Problem 2. The gas phase homogeneous oxidation reaction
2 NO + O2 Æ 2 NO2
is known to have a third-order rate law:
rNO = -2k[NO]2[O2]
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at least under atmospheric conditions. However, the rate constant decreases as T
increases, contrary to what happens in all direct elementary step termolecular
reactions. So this reaction must actually go through more than one elementary step.
Provide a mechanism that explains this strange behavior, that includes an NO3
species as an intermediate. Under what conditions would you expect rNO to deviate
significantly from the normal third-order expression above? What is the PSSH rate
law for the reverse reaction?
Problem 3. The Michaelis-Menton reaction mechanism usually assumed for enzymatic
reactions is:
S + E = E-S
(1)
E-S Æ P + E
(2)
N
I
.
E
a. Consider a well-mixed batch reactor with initial enzyme concentration [E]0 and
initial substrate concentration [S]0. Write expressions for the rate of change of
V
S
concentration of [S], [ES], [E], and [P] in terms of k1, k2, and Keq,1, and
S
.B
concentration variables.
b. Write a Matlab function that solves this set of differential equations for the
concentration of all species in time given inputs k1, k2, Keq,1, [E]0, [S]0.
W
c. The pseudo-steady approximation may be applied on the reactive intermediate
W
species [ES]. This approximation is:
W
d[ES]
≈ 0 . Using this pseudo-steady
dt
approximation, verify that the rate of change of [S] is given by the expression:
v [S]
d[S]
. What are v max and K M in terms of the other constants in this
= − max
dt
K M + [S]
problem: k1, k2, Keq,1, [E]0, [S]0?
d. What does the rate
d[S]
simplify to in the limit [S] >> K M ? What about the limit
dt
K M >> [S] ?
e. Consider the following conditions: k1=109 liter/mole-s, k2=1 s-1, Keq,1= 1
liter/mole, [E]0=10-6 M, [S]0=0.01 M. Find analytical solutions for [S](t), [ES](t),
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and [P](t). Hint: determine what
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127
d[S]
regime these conditions lie within.
dt
Compare the analytical solution with the full numerical solution by plotting them
together: plot [S](t), [ES](t) and [P](t) for both the numerical and analytical
solutions (three plots). Use a solid line for the analytical solutions and open
symbols for the numerical solutions. Run the simulation at least until the
conversion XP = [P]/[S]0 ≥ 99%. After approximately how much time does the
pseudo-steady approximation become valid? Hint: look at the short-time
behavior of [ES](t) in your numerical solution to find the answer.
N
I
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V
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S
.B
W
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128
Problem 1.
a.
C: H: O: N=47.60%/12: 7.33%/1: (1-47.60%-7.30%-7.33%-3.00%)/16: 7.30%/14=1: 1.85: 0.55:
0.13
Therefore, the elemental composition for the ash-free biomass is CH1.85O0.55N0.13.
Thus, the formula weight per C-atom is:
1*12 (g/mol) +1.85*1(g/mol)+0.55*16 (g/mol)+0.13*14(g/mol)=24.5 g/mol.
Since ethane is the sole carbon source, from the conservation of C-atom, we know
Ysx=moles of biomass(x)/moles of ethane(s)
=
[ 22.8
g dry weight
g biomass
g biomass
1 mol ethane
] × [(1 - 3%)
] ÷ [24.5
]×
mole ethane
g dry weight
mol biomass 2 C - mole ethane
=0.451 (C-mole biomass/C-mol ethane)
N
I
.
E
b.
If assuming that CO , H O and CH O N are the only metabolic products, then the overall
2
2
a
b
V
S
c
metabolic reaction is
0.5 C2H6+Yso O2+Ysn NH3 → Ysx CH1.85O0.55N0.13 + YscCO2 + Ysw H2O
From a), we already got Ysx=0.451.
Use mass balance conditions on each atom:
C: 0.5*2=Ysx+Ysc
N: Ysn=Ysx*0.13
H: 0.5*6+Ysn*3=Ysx*1.85+Ysw*2
O: Yso*2=Ysx*0.55+Ysc*2+Ysw
S
.B
W
W
W
After solving this set of linear equations, we finally get
Ysc=0.549 (mol CO2/C-mol ethane),
Ysn= 0.0589 (mol NH3/C-mol ethane),
Ysw=1.17 (mol H2O/C-mol ethane),
Yso=1.26 (mol O2/C-mol ethane)
Therefore, the full stoichiometric equation for the growth process
0.5 C2H6+1.26 O2+0.0589 NH3 → 0.451 CH1.85O0.55N0.13 + 0.549CO2 + 1.17 H2O
The oxygen consumption is
Yxo=Yso/Ysx=1.26/0.451=2.79 (mol O2/C-mol biomass)
Then we can determine the heat evolved per kilogram dry weight from the enthalpy of combustion
data:
Q=0.5*ΔHcomb(ethane)+0.0589*ΔHcomb(NH3)-0.451*ΔHcomb(biomass)
=-(0.5*1560 kJ/mol+0.0589*383 kJ/mol-19 (kJ/g dry weight)
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÷ [(1 - 3%)
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129
g biomass
g biomass
1 C - mol ethane
] × [ 24. 5
]÷
g dry weight
C - mol biomass 0.451 C - mol biomass
=-586 kJ/c-mol ethane.
Then convert back again to per kilo dry weight
Q=
1 C - mol ethane
1000g
g biomass
g biomass
kJ
]×
] ÷ [24.5
− 586
× [(1 - 3%)
×
0.451
C
mol
biomass
1kg
C
mol
biomass
g dry weight
c - mol ethane
=-51.5 (MJ/ kg dry weight)
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S
.B
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Problem 2.
2
For the reaction, 2NO + O 2 ⎯⎯→
2 NO2 , it is not possible for the apparent activation energy to
k
be negative, or equivalently, the rate constant decreases as T increases. We are asked to write
down the elementary steps which include an NO3 species as an intermediate to explain this strange
behavior.
A possible mechanism:
k1
NO + O 2 ⎯⎯→
NO3
k -1
NO3 ⎯⎯→
NO + O 2
k2
NO3 + NO ⎯⎯→
2NO 2
So the reaction rates: r1=k1[NO][O2], r-1=k-1[NO3], r2=k2[NO3][NO]
If using PSSH for the intermediate NO3, we have
d[NO3 ]
= k1[NO][O2 ] - k -1[NO3 ] - k 2 [NO3 ][NO] = 0
dt
From this we can obtain [NO3 ] =
k1[NO][O2 ]
k -1 + k 2 [NO]
V
S
Thus
rNO
N
I
.
E
S
.B
d[NO]
=−
= − k1[NO][O2 ] + k -1[NO3 ] - k 2 [NO3 ][NO]
dt
k [NO][O2 ]
= −k1[NO][O2 ] + (k -1 - k 2 [NO]) 1
k -1 + k 2 [NO]
W
2k1k 2 [NO]2 [O 2 ]
k -1 + k 2 [NO]
W
=−
W
In order to have third-order reaction kinetics as the form rNO = −k effective forward [NO] [O 2 ] , we
2
have to assume k-1>> k2[NO], so that the overall reaction rate for NO is
rNO = −
where k effective forward =
2k1k 2 [NO]2 [O 2 ]
k -1
2k1k 2
k -1
It is therefore under the condition when k-1~ k2[NO] or k-1<< k2[NO] for rNO to deviate
significantly from the normal third-order expression above.
Also let’s see what happens to the activation energy.
E a, overall ∝ ln
If Ea1+Ea2-Ea, -1~ΔH1,
rxn+Ea2<0,
2k1k 2
∝ E a1 − E a, -1 + E a2 ~ΔH1, rxn+Ea2
k -1
then we can have a negative apparent activation energy, for
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131
example, if elementary step 1 has a significantly negative ΔH1, rxn, so as long as Ea2 is not too high
the overall process will have a negative Ea, overall.
For the reverse reaction of the overall reaction, including the reverse reaction for the second
elementary step, i.e.
k1
NO + O 2 ⎯⎯→
NO3
k -1
NO3 ⎯⎯→
NO + O 2
k2
NO3 + NO ⎯⎯→
2NO 2
k -2
2NO 2 ⎯⎯→
NO3 + NO
and still using PSSH on the intermediate NO3
d[NO3 ]
= k1[NO][O2 ] - k -1[NO3 ] - k 2 [NO3 ][NO] + k - 2 [ NO 2 ]2 = 0
dt
N
I
.
E
we can have the intermediate concentration:
[NO3 ] =
k1[NO][O2 ] + k -2 [ NO 2 ]2
k -1 + k 2 [NO]
V
S
Then in this case, the overall reaction rate is
rNO
S
.B
d[NO]
=−
= −k1[NO][O2 ] + k -1[NO3 ] - k 2 [NO3 ][NO] + k - 2 [NO 2 ]2
dt
k [NO][O2 ] + k - 2 [NO 2 ]2
= −k1[NO][O2 ] + (k -1 - k 2 [NO]) 1
+ k - 2 [NO 2 ]2
k -1 + k 2 [NO]
=
W
2(k -1k - 2 [NO 2 ]2 - k1k 2 [NO]2 [O 2 ])
k -1 + k 2 [NO]
W
W
Again, if k2[NO]<<k-1, then
rNO = 2 k-2[NO2]2 – 2 k1k2/k-1 [NO]2[O2]
Notice that the second term is what we got before for the forward reaction, i.e.
kforward=2k1k2/k-1
The first term gives the effective rate constant for the reverse process:
kreverse=2k-2
Note that kforward/kreverse = k1k2/k-1k-2 = Kc1Kc2 = Kc,overall where Kc’s are equilibrium constants.
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132
Prob. 3
a. For the reactions
k1
S + E ⎯⎯→
E -S
k -1
E - S ⎯⎯→
S+ E
k2
E - S ⎯⎯→
P+E
We can write down
d[S]
= −k1[S][E] + k -1[E - S]
dt
d[E]
= − k1[S][E] + k -1[E - S] + k 2 [E - S]
dt
d[E - S]
= k1[S][E] − k -1[E - S] − k 2 [E - S]
dt
d[P]
= k 2 [E - S]
dt
N
I
.
E
With law of mass action on enzyme [E] + [E - S] = [E]0 , [S] + [P] + [E - S] = [S]0 , [P](t=0)=0,
V
S
[S](t=0)=[S]0, [E](t=0)=[E]0, and k-1=k1/Keq, 1
b.
S
.B
function [t,conc] = odehw2_prob3(k1, k2, keq1, tmax)
param = [k1,k2,keq1];
W
%initial concentrations
%conc0 = ([S],[ES],[E],[P])
W
conc0 = [0.01,0,1e-6,0];
W
%use ode15s at the function derivhw2
%t is the time vector output
%conc is the 4 column matrix solution containing the concentrations of
%[S],[ES],[E],[P]
options = odeset('AbsTol', 1e-9, 'RelTol', 1e-6);
[t,conc] = ode15s(@derivhw2_prob3,[0;tmax],conc0,options,param);
%this is the function inputed into ode15s
function derivs = derivhw2_prob3(t,conc,param)
%extract constants
k1 = param(1);
k2 = param(2);
keq1 = param(3);
%This is the order of the variables in the concentration vector
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%concS = [S] in M
%concES = [ES] in M
%concE = [E] in M
%concP = [P] in M
%switch from list of f's to actual names for ease of formulation of
concS = conc(1);
concES = conc(2);
concE = conc(3);
concP = conc(4);
%defining the rate equations
dconcSdt = -k1*concS*concE + (k1/keq1)*concES;
dconcEdt = -k1*concE*concS + k2*concES + (k1/keq1)*concES;
dconcESdt = k1*concE*concS - k2*concES - (k1/keq1)*concES;
N
I
.
E
dconcPdt = k2*concES;
%put derivative results back in column vector format for MATLAB
V
S
derivs = [dconcSdt; dconcESdt; dconcEdt; dconcPdt];
return;
S
.B
c. Using this pseudo-steady approximation on intermediate species ES,
d[E - S]
= k1[S][E] − k -1[E - S] − k 2 [E - S] =0
dt
W
we know
W
[E - S] =
W
k1[S][E]
k -1 + k 2
Using mass balance condition [E] + [E - S] = [E]0
We know
[E - S] =
[E]0
k + k2
1 + -1
k1[S]
Therefore the reaction rate
- rS =
d[S]
d[P]
[S]
V [S]
=−
= −k 2 [E - S] = −k 2 [E]0
= − max
k + k2
dt
dt
[S] + K m
[S] + -1
k1
where Km=(k-1+k2)/k1 and Vmax=k2[E]0.
d. In the limit [S]>>Km, from
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134
d[S]
V [S]
= − max
≈ −Vmax
dt
[S] + K m
In the limit [S]<<Km, from
d[S]
V [S]
V [S]
= − max
≈ − max
dt
[S] + K m
Km
e.
Consider the conditions: k1=109 liter/mole-s, k2=1 s-1, Keq,1= 1 liter/mole, [E]0=10-6 M, [S]0=0.01
M.
We know that now Km=(k-1+k2)/k1=k2/k1+1/Keq,1~1 M, [S]0=0.01 M,
therefore, [S]0<<Km
since [S] is decreasing monotonically, [S]<<Km is always correct.
So now we can use the result from d)
d[S]
V [S]
= − max
dt
Km
N
I
.
E
This gives an exponential function for [S](t)
[S](t) = [S](t = 0) exp(−
V
S
S
.B
While for [ES],
[ES](t) =
Vmax
V
t) = [S]0 exp(− max t)
Km
Km
[E]0
[E]
[E]
V
≈ 0 [S] = 0 [S]0 exp(− max t)
k
kM
kM
Km
1+ M
[S]
W
W
And for [P]
W
[P](t)=[S]0-[ES](t)-[S](t) ≈ [S]0 [1 − exp(−
Vmax
t)]
Km
Use matlab to solve the following non-linear ODE IVP:
d[S]
k
= −k1[S][E] + 1 ([E]0 - [E])
dt
K eq, 1
d[E]
k
= − k1[S][E] + ( 1 + k 2 )([E]0 - [E])
dt
K eq, 1
r=
d[P]
= k 2 [E - S] = k 2 ([E]0 - [E])
dt
with the initial conditions [S]t=0=[S]0=0.01M, k1=109 liter/mole-s, k2=1 s-1, Keq,1= 1 liter/mole, and
[E]0=10-6 M. From d[ES]/dt=0, we can determine the time to reach pseudo steady state is
approximately 7×10-9 sec, which is really really short. For this specific condition, pseudo steady
state works very well. This may also be seen from a direct comparison of the analytical/full
numerical solutions [ES](t) plots on a short time scale. [E-S] should rapidly rise from zero to the
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PSSA value (on the order of 10-8 seconds). The only observable difference between the
numerical and analytical solutions is this initial jump in [E-S] on the short time scale.
[S](t)[M]
0.01
numerical
analytical
0.009
0.008
0.007
[S][mol/L]
0.006
0.005
0.004
N
I
.
E
0.003
0.002
V
S
0.001
0
0
0.5
1
1.5
W
-8
1
S
.B
2
x 10
2.5
t[sec]
3
3.5
4
x 10
[ES] vs. t
numerical
analytical
W
0.8
0.7
5
6
W
0.9
4.5
[ES] [M]
0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
t[sec]
3
3.5
4
4.5
5
6
x 10
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[P](t)
0.01
0.009
0.008
numerical
analytical
0.007
[P][mol/L]
0.006
0.005
0.004
0.003
0.002
N
I
.
E
0.001
0
0
0.5
1
1.5
2
2.5
t[sec]
d[ES]/dt vs. t
3.5
4
4.5
5
6
V
S
x 10
S
.B
10
9
W
8
W
7
d[ES]/dt [M/sec]
3
W
6
5
4
3
2
1
0
0
0.5
1
1.5
t[sec]
2
2.5
3
-8
x 10
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N
I
.
E
V
S
S
.B
W
W
W
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10.37 HW 3
Spring 2007
Problem 1. A CSTR of volume 0.602 liters (constant density liquid phase) is operated in
which the following reaction occurs:
A+ B→C+ D
The feed rate of A is 1.16 liters/hr of a solution at concentration 5.87 mmol/L. The feed rate of B is 1.20 liters/hr at a concentration of 38.9 mmol/L. The outlet concentration of species A is 1.094 mmol/L. Calculate the rate constant for this reaction assuming a mass-action rate law of the form: r = k[ A][B] .
Problem 2. Consider the catalyzed reaction:
A+ B → B+C
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I
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E
with the second-order rate constant 1.15 x 10-3 m3/mol/ksec. The rate law is
r = k[ A][B] .
V
S
What volume of CSTR would be necessary to give 40% conversion of species A if the
feed concentration of A is 96.5 mol/m3, the feed concentration of B is 6.63 mol/m3, and
the flow rate is 0.5 m3/ksec?
S
.B
W
Problem 3. Two configurations of CSTRs are contemplated for performing reversible
hydrolysis of compound A to produce compounds B and C. The forward reaction is
pseudo-first-order with respect to A, with rate constant k1= 1.82 x 10-4 s-1. The reverse
reaction is second-order with rate constant k-1 = 4.49 x 10-4 M-1s-1 .
(1) A → B + C; r1 = k1[ A]
(2) B + C → A; r2 = k −1[B][C]
The feed is a dilute aqueous solution of A (concentration 0.25 mol/L) at a rate of 0.25
liters per hour.
W
W
Consider the following two configurations:
a) a single 15 liter CSTR.
b) three 5-liter CSTRs in series, with 75% of product species B & C selectively removed
between stages 1 and 2 and between stages 2 and 3, with appropriate adjustment in flow
rate; the volumetric flow rates in the two streams leaving a separator are proportional to
the total number of moles of A, B, and C in each stream. See the separator diagram
below.
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q0 FA0 FB0 FC0
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139
q1 FA1 FB1 FC1
Separator
q2 FA2 FB2 FC2
If q is the volumetric flow rate and F is the molar flow rate, then the separator follows the
relationships:
FA2 = 0
FB 2 = 0.75 FB0
FC 2 = 0.75FC 0
F + FB1 + FC1
q1
= A1
q2 FA2 + FB2 + FC 2
A full flow diagram is shown below.
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V
S
S
.B
W
W
W
Determine the steady-state production of compound B in mol/h for options a) and b).
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140
10.37 HW 3
Spring 2007
Problem 1. A CSTR of volume 0.602 liters (constant density liquid phase) is operated in
which the following reaction occurs:
A+ B→C+ D
The feed rate of A is 1.16 liters/hr of a solution at concentration 5.87 mmol/L. The feed rate of B is 1.20 liters/hr at a concentration of 38.9 mmol/L. The outlet concentration of species A is 1.094 mmol/L. Calculate the rate constant for this reaction assuming a mass-action rate law of the form: r = k[ A][B] .
First draw a diagram of the problem:
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qA [A]0
qout [A] [B]
qB [B]0
V
S
The only unknown quantities in the above figure are [B] and
qtot.
S
.B
A volume balance on the carrier solvent gives the following
relationship (constant fluid density):
W
q A + q B = qout = 1.16 L/h + 1.20 L/h = 2.36 L/h
W
W
Define the extent of reaction:
mol
ξ& = rV [=]
time
A material balance on component A yields the following:
dn A
= FA0 − FA − ξ& = [ A]0 q A − [ A]qout − ξ& .
dt
Setting the accumulation term to zero (steady state
operation) and solving for the reaction rate we find:
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141
ξ& = [ A]0 q A − [ A]qout
ξ& = (5.87mmol/L)(1.16L/h ) − (1.094mmol/L )(2.36L/h ) = 4.23mmol/h
A material balance on component B yields the following:
dnB
= FB0 − FB − ξ& = [B]0 q B − [B]qout − ξ& .
dt
Setting accumulation to zero, and solving for [B] we find:
[B] =
[B]0 q B − ξ&
. qout
[B] =
(38.9 mmol/L)(1.20 L/h ) − (4.23 mmol/h )
= 18.0 mmol/L
(2.36 L/h )
N
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Using the given rate law and knowing the extent of
reaction, we can now calculate the rate constant:
ξ& = rV = k[A][B]V
ξ&
k=
k=
V
S
S
.B
[ A][B]V
(4.23 mmol/h)
L
= 0.357
= 357M −1h −1
(1.094 mmol/L)(18.0 mmol/L)(0.602 L )
h mmol
W
W
W
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142
Problem 2. Consider the catalyzed reaction:
A+ B → B+C
with the second-order rate constant 1.15 x 10-3 m3/mol/ksec. The rate law is
r = k[ A][B] .
What volume of CSTR would be necessary to give 40% conversion of species A if the
feed concentration of A is 96.5 mol/m3, the feed concentration of B is 6.63 mol/m3, and
the flow rate is 0.5 m3/ksec?
First draw a diagram of the problem:
q [A]0 [B]0
[A] [B] [C]
Define the extent of reaction:
mol
ξ& = rV[=]
time
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V
S
S
.B
Define the conversion in terms of the variables of the
problem:
XA =
mol A reacted
W
W
mol A fed
W
=
− rAV
q[ A]0
=
ξ&
q[ A]0
A material balance on species A gives the following
equation:
dn A
= FA0 − FA − ξ& = [A]0 q − [ A]q − ξ&
dt
At steady state:
0 = [ A]0 q − [ A]q − ξ&
ξ& = [ A]0 q − [A]q
Thus our conversion definition is equivalent to:
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XA =
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143
[ A]0 q − [ A]q
[ A]
=1−
q[ A]0
[ A]0
[ A] = (1 − X A )[ A]0
A material balance on species [B] gives the following:
dnB
= FB0 − FB + 0ξ& = [B]0 q − [B]q
dt
At steady state:
[B] = [B]0 .
Plugging in the rate expression in the extent of reaction,
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ξ& = rV = k[ A][B]V ,
the steady state A balance becomes:
V
S
0 = [ A]0 q − [ A]q − ξ& = ([ A]0 − [ A])q − k[ A][B]V .
S
.B
Using the expressions derived for [B] and [A] in terms of
given values, we find:
W
0 = [ A]0 X A q − k[ A]0 (1 − X A )[B]0 V
W
Solving the equation for V we find:
V =
V
W
[ A]0 X A q
X Aq
=
k[ A]0 (1 − X A )[B]0 k (1 − X A )[B]0
(0.40)(0.5 m 3 /ksec)
=
= 43.7 m 3
3
3
−3
(1.15 ×10 m /(mol ⋅ ksec))(1 − 0.40)(6.63 mol/m )
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144
Problem 3. Two configurations of CSTRs are contemplated for performing reversible
hydrolysis of compound A to produce compounds B and C. The forward reaction is
pseudo-first-order with respect to A, with rate constant k1= 1.82 x 10-4 s-1. The reverse
reaction is second-order with rate constant k-1 = 4.49 x 10-4 M-1s-1 .
(1) A → B + C; r1 = k1[ A]
(2) B + C → A; r2 = k −1[B][C]
The feed is a dilute aqueous solution of A (concentration 0.25 mol/L) at a rate of 0.25
liters per hour.
Consider the following two configurations:
a) a single 15 liter CSTR.
b) three 5-liter CSTRs in series, with 75% of product species B & C selectively removed
between stages 1 and 2 and between stages 2 and 3, with appropriate adjustment in flow
rate; the volumetric flow rates in the two streams leaving a separator are proportional to
the total number of moles of A, B, and C in each stream. See the separator diagram
below.
q0 FA0 FB0 FC0
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q1 FA1 FB1 FC1
Separator
V
S
q2 FA2 FB2 FC2
S
.B
W
If q is the volumetric flow rate and F is the molar flow rate, then the separator follows the
relationships:
FA 2 = 0
FB 2 = 0.75FB 0
FC 2 = 0.75FC 0
F + FB1 + FC1
q1
= A1
q2 FA2 + FB2 + FC 2
A full flow diagram is shown below.
W
W
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145
Determine the steady-state production of compound B in mol/h for options a) and b).
First, consider the 1-reactor case. Draw a diagram of the
situation (the control volume).
q0 [A]0
[B]0=[C]0=0
V=15 L
[A] [B] [C]
V
S
Define extents of reaction:
ξ&1 = r1V = k1[ A]V
ξ&2 = r2V = k −1[ B][C ]V
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S
.B
W
Steady state A, B, and C material balances give the
following equations:
W
0 = FA0 − FA − ξ&1 + ξ&2
0 = F − F + ξ& − ξ&
W
B0
B
1
2
0 = FC 0 − FC + ξ&1 − ξ&2
These three equations can be written in terms of only three
variables: [A], [B], and [C]. The system is fully
specified. We must solve these equations simultaneously.
The algebra is easier if we combine equations creatively to
make them simpler. When you combine two equations, you
keep the new equation and only one of the old ones, just as
is done in linear algebra.
Notice that:
FB0 = FC 0 = 0 .
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146
Hence, subtracting the B balance equation from the C
balance equation gives the result:
FB = FC
q0 [ B] = q0 [C ]
[ B] = [C ]
Let this new equation replace the C mole balance.
The equality of concentrations of C and B is true whenever
[C]0 and [B]0 are equal. [B]0 has been intentionally left
as a variable so that the same equations derived here will
also apply to the reactors in part b, where [B]0 will not
necessarily equal zero.
Adding together the B and A balance gives the following
equation:
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0 = FA0 − FA + FB0 − FB = q([ A]0 − [ A] + [ B]0 − [ B])
Dividing by q0 and solving for [A] we get:
[ A] = ([ A]0 + [ B]0 − [ B])
V
S
S
.B
Let this new equation replace the B mole balance.
Now we can write the A balance as a single equation with a
single variable, [B] using our two new equations:
W
W
0 = FA0 − FA − ξ&1 + ξ&2
W
0 = q0 ([ A]0 − [ A]) − k1V [ A] + k −1V [ B][C ]
0 = q0 ([ A]0 − ([ A]0 + [ B]0 − [ B ])) − k1V ([ A]0 + [ B]0 − [ B]) + k −1V [ B][ B]
0 = q0 [ B] − q0 [ B]0 − k1V [ A]0 − k1V [ B]0 + k1V [ B] + k −1V [ B]2
0 = (k −1V )[ B]2 + (q0 + k1V )[ B] + (− q0 [ B]0 − k1V [ A]0 − k1V [ B ]0 )
0 = a[ B ]2 + b[ B ] + c
a = (k −1V )
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147
b = (q0 + k1V )
c = (− q0 [ B]0 − k1V [ A]0 − k1V [ B]0 )
This quadratic equation in [B] can be solved with the
quadratic formula:
[ B] =
− b ± b 2 − 4ac
2a
Since the coefficient b will always be POSITIVE, we know
that we have to take the (+) root. The (-) root will
always be negative. Also notice that a is always positive
and c is always negative, so that the descriminant,
(b
2
− 4ac ) > 0 ,
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is always positive, and both roots are real.
write:
[ B] =
Hence, we may
V
S
− b + b 2 − 4ac
2a
S
.B
with no fear of getting a negative or imaginary result.
Converting all time to hours, we find that for part a
W
L2
⎛ 3600s ⎞
a = (k −1V ) = (4.49 ×10 M s )⎜
⎟(15 L ) = 24.246
mol ⋅ h
⎝ h ⎠
W
−4
-1 −1
W
L
⎛ 3600s ⎞
b = (q0 + k1V ) = (0.25 L/h ) + (1.82 ×10 −4 s −1 )⎜
⎟(15 L ) = 10.078
h
⎝ h ⎠
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
mol
⎛ 3600s ⎞
c = 0 − (1.82 ×10 −4 s −1 )⎜
⎟(15 L )(0.25M ) − 0 = −2.457
h
⎝ h ⎠
Plugging the values of the coefficients into the quadratic
equation, we find:
[B] = 0.172 M
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148
Hence, the desired production rate of B is:
L⎞
⎛
FB = q0 [B] = ⎜ 0.25 ⎟(0.172 M ) = 0.0430 mol/h
h⎠
⎝
N
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V
S
S
.B
W
W
W
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149
In order to do part b, it is helpful to notice that
everything flows from left to right, with no recycle loops.
Hence we can solve each reactor and separator in sequence,
and add up the total B recovery at the end of the process.
Also, we could get it from an overall balance (a box around
the entire system) once we knew the final outlet
concentration of A.
For simplicity of notation, let the zero subscript refers
to the current reactor feed; no subscript refers to the
current reactor output.
Reactor 1 q0=0.25L/h [A]0=0.25M [B]0=[C]0=0M V=5 L
[A] [B] [C]
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Reactor 1 follows the same relationships as the single
reactor system, but with different volume. Hence the same
material balances apply, and the final equation in terms of
[B] is once again:
V
S
S
.B
0 = (k −1V )[B]2 + (q0 + k1V )[B] + (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
0 = a[B]2 + b[B] + c
a = (k −1V )
W
W
b = (q0 + k1V )
W
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
[ B] =
− b + b 2 − 4ac
2a
Plugging in the numbers, we find the quadratic coefficients
to be:
L2
⎛ 3600s ⎞
a = (k −1V ) = 4.49 ×10 −4 M -1s−1 ⎜
⎟(5 L ) = 8.082
mol ⋅ h
⎝ h ⎠
(
)
L
⎛ 3600s ⎞
b = (q0 + k1V ) = (0.25 L/h ) + (1.82 ×10 −4 s−1 )⎜
⎟(5 L ) = 3.526
h
⎝ h ⎠
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150
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
mol
⎛ 3600s ⎞
c = 0 − (1.82 ×10 −4 s−1 )⎜
⎟(5 L )(0.25M ) − 0 = −0.819
h
⎝ h ⎠
Plugging the quadratic coefficients into the formula for
[B] we find:
[B] = 0.168 M
Using our formula for [A] in terms of [B], we find:
[ A] = ([ A]0 + [B]0 − [B])
[ A] = 0.25 M + 0 M − 0.168 M = 0.082 M
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V
S
S
.B
W
W
W
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151
Next we proceed to separator 1, carry forward our inputs,
and rename our subscripted variables, so that they refer to
the current separator.
q0=0.25 L/h
[A]0 = 0.082 M
[B]0=[C]0=0.168 M
q1
[A]1
[B]1
Separator
1
q2
[A]2
[B]2
First we need to find the flow rate partitioning, then we
can calculate the new concentrations from an A and B
balance. (Since the separator treats B and C the same,
these concentrations will continue to be equal in all
streams exiting the separator.) The separator equations
are:
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FA 2 = 0
FB 2 = 0.75FB 0
V
S
FC 2 = 0.75FC 0
F + FB1 + FC1
q1
= A1
q2 FA 2 + FB 2 + FC 2
S
.B
Hence using the definition of molar flow rate, the
knowledge that [B]0 = [C]0, and the above separator
relationships, we find that:
W
q1
q2
q1
q2
W
(q0 [ A]0 ) + 0.25(q0 [B]0 ) + 0.25(q0 [C]0 ) [ A]0 + 0.5[B]0
=
0 + 0.75(q0 [B]0 ) + 0.75(q0 [C]0 )
1.5[B]0
W
=
=
[ A]0 + 0.5[B]0
1.5[B]0
=
(0.082) + 0.5(0.168)
= 0.659
1.5(0.168)
q1 = 0.659q2
Assuming a constant density liquid phase, we have a volume
balance:
q0 = q1 + q2 = 1.659q2
q2 =
0.25 L/h
1.659
= 0.151 L/h
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152
q1 = 0.659 × 0.151 L/h = 0.100 L/h
Thus we have recovery of product B at this stage of:
FB1E = 0.75(q0 [B]0 ) = 0.75(0.25 L/h )(0.168 M ) = 0.0315 mol/h
The last things to calculate at this step are the
concentrations that are fed to the next reactor. Using the
mole balance on A:
FA0 = q0 [ A]0 = FA1 = q1[ A]1
[ A]1 =
q0
q1
[ A]0 =
0.25
0.100
0.082
mol
L
= 0.205 mol/L . Using the separator relationships for B and C:
N
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FB1 = 0.25FB 0
[B]1 = 0.25
q0
q1
[B]0 = 0.25
0.25
0.100
0.168
mol
L
= 0.105 mol/L
V
S
S
.B
[ B ]1 = [C ]1
W
W
W
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Next we proceed to reactor 2 using the results from
separator 1 and rename our variables so they refer to the
current reactor.
Reactor 2
q0=0.100L/h
[A]0=0.205 M
[B]0=[C]0=0.105 M
V=5 L
[A] [B] [C]
All the same equations apply with different initial
concentrations and flow rates.
L2
⎛ 3600s ⎞
(
)
a = (k −1V ) = 4.49 ×10 −4 M -1s−1 ⎜
5
L
=
8.082
⎟
mol ⋅ h
⎝ h ⎠
(
)
L
⎛ 3600s ⎞
b = (q0 + k1V ) = (0.100 L/h ) + (1.82 ×10 −4 s−1 )⎜
⎟(5 L ) = 3.376
h
⎝ h ⎠
N
I
.
E
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
(
)
L⎞
⎛
⎛ 3600s ⎞
c = ⎜ − 0.100 ⎟(0.105M ) − 1.82 ×10 −4 s −1 ⎜
⎟(5 L )[0.205 + 0.105]M
h⎠
⎝
⎝ h ⎠
mol
c = −1.026
h
The quadratic solution gives:
V
S
S
.B
W
[B] = 0.204 M
W
Solving for [A] from the [A]/[B] relationship we find:
W
[ A] = ([ A]0 + [B]0 − [B])
[ A] = 0.205 M + 0.105 M − 0.204 M = 0.106 M
These concentrations will be fed to separator 2.
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Next we proceed to separator 2 and rename our subscripted
variables.
q0=0.100 L/h
[A]0 = 0.106 M
[B]0=[C]0=0.204 M
q1
[A]1
[B]1
Separator
2
q2
[A]2
[B]2
The flow rate partitioning is given by:
q1
q2
=
[ A]0 + 0.5[B]0
1.5[B]0
(0.106) + 0.5(0.204)
= 0.680
1.5(0.204)
=
Hence, from the constant density volume balance:
q2 =
q0
1.680
0.100 L/h
=
1.680
N
I
.
E
= 0.0595 L/h
q1 = q0 − q1 = (0.100 − 0.0595)
L
V
S
= 0.0405 L/h
S
.B
h
The amount of [B] recovered is:
W
FB 2 E = 0.75(q0 [B]0 ) = 0.75(0.100L/h )(0.204 M ) = 0.0153 mol/h
W
The concentrations of [A], [B], and [C] into the next
reactor are:
W
FA0 = q0 [ A]0 = FA1 = q1[ A]1
0.100
q
mol
[ A]1 = 0 [ A]0 =
0.106
= 0.262 mol/L . 0.0405
q1
L
FB1 = 0.25FB 0
0.100
q
mol
[B]1 = 0.25 0 [B]0 = 0.25
0.204
= 0.126 mol/L
0.0405
q1
L
[ B ]1 = [C ]1
Lastly, we rename our concentrations and proceed to reactor
3.
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155
Lastly, we proceed to reactor 3 using the results from
separator 2.
Reactor 3
q0=0.0405L/h
[A]0=0.262 M
[B]0=[C]0=0.126 M
V=5 L
[A] [B] [C]
All the same equations apply with different initial
concentrations and flow rates.
L2
⎛ 3600s ⎞
a = (k −1V ) = 4.49 ×10 −4 M -1s −1 ⎜
⎟(5 L ) = 8.082
mol ⋅ h
⎝ h ⎠
L
⎛ 3600s ⎞
b = (q0 + k1V ) = (0.0405 L/h ) + 1.82 ×10 −4 s −1 ⎜
⎟(5 L ) = 3.3165
h
⎝ h ⎠
c = (− q0 [B]0 − k1V[ A]0 − k1V[B]0 )
(
)
(
(
)
N
I
.
E
)
L⎞
⎛
⎛ 3600s ⎞
c = ⎜ − 0.100 ⎟(0.126M ) − 1.82 ×10 −4 s −1 ⎜
⎟(5 L )[0.262 + 0.126]M
h⎠
⎝
⎝ h ⎠
mol
c = −1.284
h
The quadratic solution gives:
V
S
S
.B
[B] = 0.243 M
Solving for [A] from the A/B relationship equation we find:
W
[ A] = ([ A]0 + [B]0 − [B])
[ A] = 0.262 M + 0.126 M − 0.243 M = 0.145 M
W
W
Since there is no separator, the amount of B recovered on
this step is just the amount leaving the reactor:
L⎞
⎛
FB3E = [B]q0 = (0.243M)⎜ 0.0405 ⎟ = 0.0098mol/h
h⎠
⎝
Thus the total amount of B recovered by this path is:
FB = FB1E + FB 2 E + FB 3 E = (0.0315 + 0.0153 + 0.0098)
mol
h
= 0.0566
mol
h
Another way to calculate it, and check our consistency, is
to calculate the amount of B recovered from the overall A
and B balances around the whole reaction system (any A that
disappears must be present as B):
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156
0 = FAin − FAout + FBin − FBout
L⎞
L⎞
mol
⎛
⎛
FBout = FAin − FAout = ⎜ 0.25 ⎟(0.25M ) − ⎜ 0.0405 ⎟(0.145M ) = 0.0566
h⎠
h⎠
h
⎝
⎝
From the agreement of the numbers, it appears that mass was
conserved overall.
Notice that incorporating the selective separation process
to remove product along the way, the amount of B recovered
was improved with the same total volume of reactor.
Due to the truncation error and numerical rounding, accept
any answers within 2% of these values.
FB ,3rctr = (0.0566 ± 0.0011)
FB ,1rctr = (0.0430 ± 0.0009 )
mol
h
mol
N
I
.
E
h
The modular setup of this problem (reactor units, separator
units) makes a Matlab implementation straightforward. See
included Matlab m-file hw3prob3.
V
S
S
.B
W
W
W
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157
10.37 Problem Set 4 Due March 7, 2007
Problem 1:
Biodiesel@MIT is working to convert the waste cooking oils from local dining halls and
restaurants into diesel fuel to run the MIT shuttle buses.
Waste cooking oils can be converted to biodiesel in several ways. One way is by gasphase pyrolysis:
(RC(O)OCH2)3 Æ (RCH2)3 + 3 CO2
This can be modeled as first-order irreversible reaction with a measured k=5x10-3 min-1 at
150 C and a measured Ea=85 kJ/mole.
Pure cooking oil is injected into a hot (T=227 C) reactor at a rate of 2.5 mole/min. At this
temperature all the species are in the vapor phase. The steady-state pressure in the reactor
is 10 atm.
N
I
.
E
(a) If the reactor is a CSTR, what reactor volume is required to achieve 90% conversion? (b) If the reactor is a PFR and the pressure drop is negligible, what reactor volume is
required to achieve 90% conversion?
(c) If you had a PFR half the volume you computed in part (b), and then fed its output
into a CSTR half the volume you computed in part (a), what would the conversion
be? What if you hooked them up the other way round: the half-size CSTR first
and the half-size PFR afterwards?
V
S
S
.B
Suppose the reaction is carried out in a batch reactor, by filling it with enough
cooking oil and heating rapidly to 227 C, so that when the oil all vaporizes, but before
any significant reaction has occurred, the initial pressure in the reactor will be 2.7
atm.
W
W
W
(d) What will the pressure be in the isothermal batch reactor when the reaction has
run to 90% conversion?
(e) What would the batch reactor volume have to be if we were to process 3600
moles/day (= 2.5 moles/minute) of cooking oil this way? Assume that the batch
reactor can be emptied and refilled very rapidly, and that it is not necessary to
clean the reactor between batches.
(f) Which reactor (CSTR, PFR, batch) would you recommend be used for this process? Explain briefly. www.bsscommunitycollege.in
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Problem 2: Do Fogler problem 4-19 parts (a) through (e). This problem is about a
microreactor constructed by our new Department Head and some of his former graduate
students.
For part (c), does the pressure ratio profile for the new particle diameter make physical
sense? Why or why not? Decrease G (superficial mass velocity) to 3.5 kg/(m2*s) and re
plot the molar flow rates, X (conversion) and y (pressure ratio). Does the pressure ratio
make physical sense? Turn in plots for both cases (original G and new G).
For part (d), use Kc = 0.03 m3/mol.
Print out hard copies of all Matlab programs, and all figures, and staple them to your
handwritten solutions. Submit your Matlab programs to the 10.37 course website.
N
I
.
E
V
S
S
.B
W
W
W
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159
Problem Set 4
Problem 1. (RC(O)OCH2)3 → (RCH2)3 + 3 CO2
A
→ B
+3C
r = k[(RC(O)OCH2)3]
r = k[A]
Given:
T0 = 150 C
FA,0 = 2.5 mol/min
X = 0.9
Ea = 85 kJ/mol
T = 227 C
k(T0) = k0 = 5*10-3 (min-1)
yA,0 = 1
P = 10 atm
First, find k at the reaction temperature using Eq 3-21 from Fogler:
k(T ) = k(T0 ) ⋅ e
E 1 1 
 − 
R  T0 T 
85kJ / mol

1
1

1  8.314⋅10 −3 kJ /(mol⋅K )  423.15K − 500.15K 

=  5 ⋅10−3
= 0.206255min −1
⋅e
min 

Next, make a stoichiometric table for the flow system (see Table 3-4 in Fogler). This
table applies to both a PFR and CSTR reactor.
A
B
C
Feed Rate to Reactor
(mol/min)
FA0
0
0
Total
FA0
Species
Change within Reactor
(mol/min)
-FA0X
FA0X
3FA0X
V
S
N
I
.
E
S
.B
Effluent Rate from
Reactor (mol/min) FA = FA0(1 – X) FB = FA0X FC = 3FA0X FT = FA0(1 + 3X)
Since this is a gas-phase reaction, with a change in the total number of moles, the
volumetric flow rate (ν) will not be constant. Simplify Eq 3-41 in Fogler for the steady
state (constant P and T) ideal gas case to:
W
W
 F 
 F (1 + 3X ) 
F RT
 = ν 0 (1 + 3X ) = A0
(1 + 3X )
ν = ν o  T  = ν 0  A0
F
F
P
A0
 T0 


W
a) CSTR The design equation for CSTR volume in terms of conversion is (Eq 2-13 in Fogler): VCSTR,a =
ν (1 + 3X )FA0 X FA0 RT (1 + 3X )X
FA0 X
F X
F X
=
= A0 = A0 = 0
(− rA )exit k [A] k  FA 
kFA0 (1 − X )
Pk (1 − X )


ν


Plugging in numbers:
VCSTR,a
mol 
L ⋅ atm 

(500.15K )(1 + 3(0.9 ))0.9
 .082
 2.5
mol
min
⋅K 


= 1655.36L ≈ 1.7 ⋅103 L
=
1 

(10atm ) 0.206255 (1 − 0.9 )
min 

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b) PFR
Neglect pressure drop, so ν equation is the same as above. The design equation for a PFR
in terms of conversion is (Eq 2-16 in Fogler):
X
VPFR,b = FA0
∫
0
dX
dX
dX
FA0 RT (1+ 3X )
F RT
= FA0
=FA0
= FA0
dX = A0
FA
(
)
k [A]
− rA
PkF
X
−
Pk
1
A0
0
0 k
0
X
X
∫
X
∫
∫
ν
X
(1+ 3X )
∫ (1 − X ) dX
0
From integration by parts (or an integral table):
x2
∫
x1
(1 + mx ) dx = (1 + m )ln  1 − x1  + m(x − x )
1
2


(1 − x )
 1 − x2 
(1 + mx ) dx = (1 + m )ln  1  − mx
2


(1 − x )
 1 − x2 
0
x
When x1 = 0, this simplifies to:
∫
Integrating and plugging in numbers gives:
VPFR,b
F RT
= A0
Pk

 1
(1 + 3)ln 1− X


mol 
L ⋅ atm 

 2.5
 0.082
(500.15K )
 
  1 

min 
mol ⋅ K 

−
=
3X


4ln 1− 0.9  − 3(0.9 )
1








(10atm ) 0.206255 
min 

N
I
.
E
= 323.63L ≈ 3.2 ⋅102 L
V
S
S
.B
W
W
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c) Now find X in a CSTR/PFR combination for a given V.
First: PFR then CSTR, where the volume of each is ½ the volume calculated in parts a
and b.
FA1, FB1, FC1
FA0
X1
FA2, FB2, FC2
X2
PFR equation will be the same as derived in part b, but with V = ½ VPFR,b and X = X1:

 1 
1
F RT 
 − 3 X 1 
VPFR,b = A0
(1+ 3)ln
1−
2
Pk 
X

1

VPFR,b Pk
⇒
2FA0 RT
 1 
 + 3X 1 = 0
− (1 + 3)ln
 1− X 1 
There are many methods that can be used to find the roots of the above equation (solver
function in Excel, fsolve in Matlab, etc). For example, using fsolve in Matlab:
N
I
.
E
function [Xpfr]=partc;
X0=0.1; %initial guess for X_pfr
V
S
[Xpfr] = fsolve(@pfr_eqn,X0);
return
S
.B
function F = pfr_eqn(X)
k = 0.206255; %1/min
T = 227+273.15; %K
FA0 = 2.5; %mol/min
P = 10; %atm
R = 0.082; %L*atm/mol/K
V_pfr_b = 323.63; %L, from part b calculation
W
W
W
F = V_pfr_b*P*k/(2*FA0*R*T)-(1+3)*log(1/(1-X))+3*X;
return
Can also solve for X1 by hand using an iterative method. For example, using the
Newton-Raphson Method:
xn +1 = xn −
f ( xn )
f ' ( xn )
where f’(xn) is the derivative of f, evaluated at x = xn.
Keep iterating until xn+1 ≈ xn.
Regardless of the method used, should find that X1 ≈ 0.75
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A mole balance on the CSTR gives:
kV
kV
V

F 
0 = FA1 − FA2 + rA  CSTR,a  = FA0 (1− X 1 ) − FA0 (1− X 2 ) − CSTR,a [A] = FA0 ( X 2 − X 1 ) − CSTR,a  A2 
2
2
2
 ν 


0 = FA0 ( X 2 − X 1 ) −
kV
P (1− X 2 )
kVCSTR,a  PFA0 (1− X 2 ) 

 = FA0 ( X 2 − X 1 ) − CSTR,a
(
)
(1+ 3X 2 )
2
1+
3X
2RT
F
RT
2 
 A0
Again, solve the above equation either by hand or with a program to find that X2 ≈ 0.95
Now: CSTR followed by PFR, where the volume of each is ½ the volume calculated in
parts a and b.
FA0
N
I
.
E
FA2, FB2, FC2
FA1, FB1, FC1
X1
X2
CSTR equation will be the same as derived in part a, but with V = ½ VCSTR,a and X = X1:
1
F RT (1 + 3X 1 )X 1
VCSTR,a = A0
2
Pk (1 − X 1 )
V
S
1
F RT (1 + 3X 1 )X 1
=0
VCSTR,a − A0
2
Pk (1 − X 1 )
⇒
S
.B
Plugging in numbers and solving gives X1 ≈ 0.83
W
A mole balance on the PFR gives (see pg 15-16 in Fogler):
0 = FA V − FA V +∆V + rA ∆V
W
Rearrange, divide by ∆V and take the limit as ∆V approaches zero to get:
dX
dFA
dF
d [FA0 (1 − X )] − FA0dX
= rA ⇒ dV = A =
=
= FA0
dV
rA
rA
rA
− rA
W
Integrate with the limits V = 0 when X = XA1 and V = ½ VPFR,b when X = X2 to get:
1
VPFR,b = FA0
2
X 2
∫
X1
dX
= FA0
− rA
X2
∫
X1
dX
=FA0
k [A]
X2
∫
X1
X2
F RT
FA0 RT (1 + 3X )
dX
dX = A0
= FA0
F
Pk
PkFA0 (1 − X )
k A
X 1
∫
ν
X2
∫
X1
(1 + 3X ) dX
(1 − X )
From integration by parts (or an integral table):
x2
∫
x1
(1 + mx ) dx = (1 + m )ln  1 − x1  + m(x − x ) ⇒
1
2


(1 − x )
 1 − x2 
FA0 RT
Pk
1
F RT
VPFR,b = A0
2
Pk


 1 − X1 
(1 + 3)ln 
 + 3( X 1 − X 2 )


1 − X 2 

 VPFR ,b
 1− X 1 
=0
(1+ 3)ln 
 + 3( X 1 − X 2 ) −
2


 1− X 2 
Plugging in numbers and solving gives X2 ≈ 0.93
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Now, consider an isothermal batch reactor system with P0 = 2.7 atm. Make a
stoichiometric table for the batch system (see table 3-3 in Fogler)
Species
A
B
C
Initially (mol)
NA0
0
0
Total
NA0
Change (mol)
-NA0X
NA0X
3NA0X
Remaining (mol)
NA = NA0(1 – X) NB = NA0X NC = 3NA0X NT = NA0(1 + 3X)
d) The volume of the batch reactor is fixed, so the change in moles as the reaction
proceeds will cause an increase in P. For an isothermal, constant volume batch reactor
(Eq 3-38 in Fogler, rearranged):
P = P0 (1 + 3X ) = 2.7atm (1 + 3(0.9 )) = 9.99atm
e) Want to process 2.5 mol/min of cooking oil in the batch reactor (assume that the down
time between batches is negligible).
⇒ N A0 =  2.5
N A0
mol
= 2.5
treact
min

Assuming ideal gas, V batch
N
I
.
E
mol 
treact
min 
mol 

 2.5
 t react RT
N A0 RT
min 

=
=
P0
P0
V
S
S
.B
Find treact from design equation for a constant volume batch reactor (Eq 2-6 in Fogler):
N
k AV
dX − rAV k [A]V
kN A0 (1 − X )
=
=
= V
=
= k (1 − X )
dt
N A0
N A0
N A0
N A0
X
∫
0
t
dX
= k dt
(1 − X )
∫
W
W
Rearrange and integrate:
W
0
⇒ ln
1 
 = kt
1− x 
⇒
treact =
1  1 
ln

k 1− X 
Plug in numbers to get:
Vbatch
mol   1 
L ⋅ atm 

mol   1 

 0.082
 ln
(500.15K )
 RT  2.5
 2.5
 ln
min   1 − (0.9 ) 
mol ⋅ K 
min   1 − X 


=
= 423.94L ≈ 4.2 ⋅102 L
=
−1
kP0
0.206255min (2.7atm )
(
)
e) Since the characteristic reaction time is on the order of minutes, a flow reactor is
recommended for this process. If minimizing volume is the most important design
criteria, a single PFR is the best choice. If maximizing conversion is the most important
design criteria, the half-size PFR followed by a half-size CSTR is the best choice.
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Homework 6 10.37 Spring 2007 Due Wednesday, March 21 Problem 1: Quoting from the Wilmington, DE Morning News, Aug. 3, 1977:
“Investigators sift through the debris from blast in quest for the cause [why the new chemical plant exploded]. A company spokesman said it appears… likely that the [fatal] blast was caused by [rapid decomposition of] … ammonium nitrate [NH4NO3] used to produce nitrous oxide [N2O].” In the process, a T=200oF aqueous solution, 83 wt% ammonium nitrate, is fed into a
CSTR. When the process is running normally at steady state, about 140 kg/hr of the
aqueous solution is injected, and the temperature in the reactor, TR, is 510oF. At this
temperature, the water evaporates rapidly, but the molten ammonium nitrate remains in
the CSTR, slowly decomposing by this reaction:
N
I
.
E
NH4NO3(liquid) Æ N2O(g) + 2 H2O(g)
k(T=510oF) = 0.307/hour ∆Hrxn(T=510oF) = -740 kJ/kg of ammonium nitrate
V
S
Note that it takes about 2.2 MJ to convert a kg of liquid water at 200oF to a kg of steam at
500oF.
S
.B
Also, FYI:
Cp(steam) = 2 kJ/kg-degree F
Cp(liquid NH4NO3) = 0.8 kJ/kg-degree F
W
W
Neglect non-ideal mixing effects and assume that NH4NO3 enters as a liquid. Assume
that ∆Hrxn is approximately constant in the reactor temperature ranges of the problem. A
diagram is shown below.
W
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gaseous products at TR
liquid feed, including water and
ammonium nitrate at 200oF
Pure liquid
ammonium
nitrate at TR
CSTR
(a) During normal steady-state operation, what mass (kg) of ammonium nitrate resides in
the reactor? Note that there is negligible hold-up of any gases within the reactor.
(b) How much cooling capacity (in kW) is required for this reactor when it is running in
steady state? If this is provided by excess cooling water with an average temperature
Ta=100oF, what is the product of the heat transfer coefficient and the area, UA, in kW/Fo?
N
I
.
E
V
S
During the investigation, it was hypothesized that the temperature increased, accelerating
the reaction. The rate constant of the decomposition reaction increases with temperature,
e.g. at T=560oF, k=2.91 hr -1. The reaction follows the Arrhenius T-dependence.
S
.B
(c) Using the stability criteria explained in the vicinity of Eq. 8-75, should the reactor
operate stably at 510oF? What is the critical temperature above which runaway reaction
could occur?
W
W
It is believed that pressure fluctuations were detected in the feed stream and it was shut
off by a plant operator about 4 minutes before the explosion occurred.
W
(d) Write and (using Matlab) solve a set of differential equations describing what
happened in the reactor after the feed was shut off. Plot the temperature in the reactor vs.
time. Do you predict an explosion?
(e) Was the operator wise to quickly shut off the feed of aqueous ammonium nitrate
solution when he feared something was going wrong in the reactor? Is there something
else he should have done to prevent the disaster?
(f) Propose a procedure for safely starting-up and shutting-down a process like this (a
qualitative description will suffice).
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HOMEWORK 6 KEY
Problem 1: Quoting from the Wilmington, DE Morning News, Aug. 3, 1977:
“Investigators sift through the debris from blast in quest for the cause [why the new chemical plant exploded]. A company spokesman said it appears… likely that the [fatal] blast was caused by [rapid decomposition of] … ammonium nitrate [NH4NO3] used to produce nitrous oxide [N2O].” In the process, a T=200oF aqueous solution, 83 wt% ammonium nitrate, is fed into a
CSTR. When the process is running normally at steady state, about 140 kg/hr of the
aqueous solution is injected, and the temperature in the reactor, TR, is 510oF. At this
temperature, the water evaporates rapidly, but the molten ammonium nitrate remains in
the CSTR, slowly decomposing by this reaction:
NH4NO3(liquid) Æ N2O(g) + 2 H2O(g)
N
I
.
E
k(T=510oF) = 0.307/hour ∆Hrxn(T=510oF) = -740 kJ/kg of ammonium nitrate
Note that it takes about 2.2 MJ to convert a kg of liquid water at 200oF to a kg of steam at
500oF.
V
S
S
.B
Also, FYI:
Cp(steam) = 2 kJ/kg-degree F
Cp(liquid NH4NO3) = 0.8 kJ/kg-degree F
W
Neglect non-ideal mixing effects and assume that NH4NO3 enters as a liquid. Assume
that ∆Hrxn is approximately constant in the reactor temperature ranges of the problem. A
diagram is shown below.
W
W
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gaseous products at TR
Pure liquid
ammonium
nitrate at TR
liquid feed, including water and
ammonium nitrate at 200oF
CSTR
This document will follow these symbolic conventions when
multiple interpretations are possible:
An extensive property is given by an underbar
A molar intensive property has no extra notational
information
An intensive property that is per unit of mass is
given with a carrot
Extents of reaction are in units of moles per time
N
I
.
E
Also, ideal mixing is assumed.
V
S
S
.B
(a) During normal steady-state operation, what mass (kg) of ammonium nitrate resides in
the reactor? Note that there is negligible hold-up of any gases within the reactor.
W
W
Input state: liquid water and liquid ammonium nitrate are
fed at 200oF
W
Reaction State: temperature is TR, negligible gases are
present
Output state: only gases may exit, gases exit at TR
The components are renamed A, B, and C as in the following:
A(l) => B(g) + 2C(g)
A = ammonium nitrate
B = N2 O
C = H2 O
First, define extent of reaction ξ& in terms of reaction
rate:
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ξ& = rV = k[ A]V = kN A = (k / MW A )( N A MW A ) = (k / MW A )(m A )
Mole/mass balances on each component in the system yield
the following relationships:
Ammonium Nitrate Balance
dm A
= m& Ain − m& Aout − ξ& MW A
dt
Apply steady state, no liquid exits:
0 = m& Ain − km A
Equivalent mole balance:
0 = F − ξ&
Ain
Nitrous oxide mass balance
dmB
= m& Bin − m& Bout + ξ& MWB
dt
Apply steady state, nothing enters
MWB
0 = −m& Bout + k m A
MW A
Equivalent mole balance:
0 = −FBout + ξ&
Water mass balance
dmC
= m& Cin − m& Cout + 2ξ&
MWC
dt
Apply steady state
MWC
0 = m& Cin − m& Cout + 2k m A
MW A
Equivalent mole balance:
0 = FCin − FCout + 2ξ&
N
I
.
E
V
S
S
.B
W
W
W
Thus we can express the molar extent of reaction in terms
of the molar flow rates:
(F − FCin )
ξ& = Cout
= FAin = FBout
2
Part (a) can be answered using the ss. mass balance on
component A:
0 = m& Ain − km A
Hence, we can find the holdup of ammonium nitrate in the
reactor at 510 Fahrenheit to be:
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m& Ain
k(510°F )
= mA =
(0.83)(140kg/h )
0.307h −1
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169
= 378.5kg
(b) How much cooling capacity (in kW) is required for this reactor when it is running in
steady state? If this is provided by excess cooling water with an average temperature
Ta=100oF, what is the product of the heat transfer coefficient and the area, UA, in kW/Fo?
To answer part (b), we need an enthalpy (energy) balance on
the open system.
Enthalpy Balance (assuming negligible gas holdup in
reactor), mass and molar formats:
⎛
⎞
⎛
⎞
d⎜ ∑ ĉ p ,i miT ⎟ d⎜ ∑ c p ,i N iT ⎟
dH
⎠ = d (m A ĉ p , AT ) = d (N A c p , AT )
⎠= ⎝ i
= ⎝ i
dt
dt
dt
dt
dt
d H
= ∑ m& i,in Ĥ i − ∑ m& j,out Ĥ j + Q& −W& s = ∑ Fi,in H i − ∑ F j,out H j + Q& −W& s
dt
i
j
i
j
N
I
.
E
V
S
Focus on the molar format, as we will modify this to
extract the delta H’s that we know.
d (N A c p, AT )
= ∑ Fi,in H i − ∑ F j,out H j + Q& −W& s
dt
i
j
Apply steady state condition. Neglect shaft work.
S
.B
W
0 = FA,in H A,in + FC ,in H C ,in − FC ,out H C ,out − FB,out H B,out + Q&
W
Add and subtract the terms ± FA,in H A,out ± FC ,in H C ,out . We need to do
this to get enthalpy differences that we know into the
enthalpy balance, as absolute enthalpies have little
meaning.
W
0 = FA,in (H A,in − H A,out ) + FC ,in (H C ,in − H C ,out ) + K
+ FA,in H A,out − (FC ,out − FC ,in )H C ,out − FB,out H B,out + Q&
Use the extent of reaction definition from the mass
balances to simplify the expression and extract the heat of
reaction:
(F − FCin )
ξ& = Cout
= FAin = FBout
2
0 = FA,in (H A,in − H A,out ) + FC ,in (H C ,in − H C ,out ) + ξ&[H A,out − H B,out − 2H C ,out ] + Q&
Recall the definition of the enthalpy of reaction
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∆H rxn (TR ) =
∑ν
i
∑ν
H i (TR ) −
products
i
170
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H i (TR ) = H B (TR ) + 2H C (TR ) − H A (TR )
reactants
Since the reaction enthalpy is approximately independent of
T near TR and the reaction occurs at the reactor outlet
temperature (also the reactor temperature), the energy
balance becomes:
0 = FA,in (H A,in − H A,out ) + FC ,in (H C ,in − H C ,out ) − ξ& ∆H rxn + Q&
Putting the energy balance back in terms of mass (more
convenient for this problem) we have:
(
)
(
) (
)
0 = m& A,in Ĥ A,in − Ĥ A,out + m& C ,in Ĥ C ,in − Ĥ C ,out − k(TR ) m A ∆Ĥ rxn (TR ) + Q&
Now we need to know how to calculate the enthalpy
differences in the inflow streams: we need to heat up the
inflow to the reaction temperature, TR.
N
I
.
E
H A,in = molar enthalpy of liquid ammonium nitrate at 200
o
F.
H A,out = molar enthalpy of liquid ammonium nitrate at TR
V
S
H C ,in = molar enthalpy of liquid water at 200
o
F
S
.B ) (
H C ,out = molar enthalpy of steam at TR
(Ĥ
(Ĥ
(Ĥ
) (
) = (− 2.2MJ/kg ) − (T
C ,in
− Ĥ C ,out = Ĥ (water,200°F ) − Ĥ (steam,500°F ) + Ĥ (steam,(500°F )) − Ĥ (steam,TR )
C ,in
− Hˆ C ,out
A,in
− Ĥ A,out
W
)
)
⎛ kJ ⎞⎛ 1MJ ⎞ ⎛
TR ⎞
⎟⎜
− 500°F )⎜⎜ 2
=
−1.2
−
⎟
⎜
⎟MJ/kg
⎟
500F° ⎠
⎝ kgF° ⎠
⎝ 1000kJ ⎠ ⎝
⎛ 0.8kJ ⎞
(0.8TR −160°F )
⎟ = −(0.8TR −160°F )kJ/kg = −
MJ/kg
= −(TR − 200°F )⎜⎜
⎟
1000
⎝ kgF° ⎠
W
R
W
Solving the energy balance for the heat required for
isothermal operation at 510 Fahrenheit we find:
(
)
(
)
0 = m& A,in Ĥ A,in − Ĥ A,out + m& C ,in Ĥ C ,in − Ĥ C ,out − (k(TR ) m A ∆H rxn (TR ) ) + Q&
Q& = −m& A,in Ĥ A,in − Ĥ A,out − m& C ,in Ĥ C ,in − Ĥ C ,out + (k(TR ) m A ∆H rxn (TR ) )
(
)
(
)
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171
⎛ (0.8(510°F ) −160°F )
⎞
Q& = −(0.83)(140kg/h )⎜ −
MJ/kg ⎟ + L
1000
⎝
⎠
(510°F ) ⎞
⎛
− (0.17 )(140kg/h )⎜ −1.2 −
⎟MJ/kg + ...
500F° ⎠
⎝
(0.307h )(378.5kg ) (− 0.740MJ/kg )
−1
Q& = 28.8MJ/h + 52.8MJ/h − 86.0MJ/h = −4.4MJ/h
If Ta = 100oF, then we may find UA (given that there is
excess cooling water, assume that the temperature change in
the cooling water is negligible).
Q& = UA(Ta − TR )
− 4.4MJ / h
Q& / (Ta − TR ) = UA =
= 0.01073MJ / hF° = 0.00298kW / °F
(100 − 510)F°
N
I
.
E
During the investigation, it was hypothesized that the temperature increased, accelerating
the reaction. The rate constant of the decomposition reaction increases with temperature,
e.g. at T=560oF, k=2.91 hr -1. The reaction follows the Arrhenius T-dependence.
V
S
Given two T values and two k values, calculate k(T). Note
that in an Arrhenius expression, an ABSOLUTE temperature
scale must be used. Recall the temperature in Rankine (an
absolute scale) is just the Fahrenheit temperature plus
459.67.
S
.B
W
− b
⎛
⎞
k(T / °F ) = Aexp⎜
⎟
⎝ T / °F + 459.67 ⎠
− b
⎛
⎞
k(510) = Aexp⎜
⎟
⎝ 510 + 459.67 ⎠
− b
⎛
⎞
k(560) = Aexp⎜
⎟
⎝ 560 + 459.67 ⎠
⎛ k(510) ⎞
0.307 ⎞ ⎛
b
b
⎞ ⎛
⎞
⎟ = − ln⎛⎜
− ln⎜⎜
⎟=⎜
⎟−⎜
⎟
⎟
⎝ 2.91 ⎠ ⎝ 510 + 459.67 ⎠ ⎝ 560 + 459.67 ⎠
⎝ k(560) ⎠
2.24906 = (5.0692E − 05)b
b = 44367
⎛ − 44367 ⎞
A = k(510) / exp⎜
⎟ = 2.28E19h −1
⎝ 510 + 459.67 ⎠
Note that k(T) is very sensitive to small changes in the
activation energy (or b in this case).
W
W
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b=
Ea
R
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172
(°Rankine)
(c) Using the stability criteria explained in the vicinity of Eq. 8-75, should the reactor
operate stably at 510oF? What is the critical temperature above which runaway reaction
could occur?
This question is very poorly worded as to its intent. The
intent is to ask you to do a stability analysis on the
steady state at 510oF and see if a perturbation could lead
to a hotter, higher conversion steady state.
Define G(T) and R(T) as follows:
− km A ∆Ĥ rxn
G(T ) =
=energy produced per unit mass of A fed. One
m& A,in
could also define G(T) as energy produced per mole fed, as
it is in Fogler, but mass is a more convenient basis in
this problem.
N
I
.
E
G(T ) = R(T ) at steady state: R(T) is all the rest of the terms
in the enthalpy balance (multiplied by -1, divided by the
mass flow rate in)
− m& A,in Ĥ A,in − Ĥ A,out − m& C ,in Ĥ C ,in − Ĥ C ,out − Q&
R(T ) =
m& A,in
Near a steady state, there is no accumulation of A in the
reactor and:
m& A,in = km A
Thus G(T) simplifies to:
G(T ) = −∆Ĥ rxn
Since there is no outflow, the conversion must always be
unity. This makes the G(T) curve much simpler than it is
in the standard CSTR in Fogler.
(
)
(
V
S
)
S
.B
W
W
W
Also notice that R(T) is a simple linear function of T.
Thus if we sketch R(T) and G(T), the following results:
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R(T) G(T), R(T)
G(T)
510oF
T
N
I
.
E
Since both G(T) and R(T) are lines, there is only one
possible intersection and one possible steady state. If
you perturb the steady state to a slightly hotter
temperature, R>G and the system will cool down. If you
perturb the system to a slightly cooler temperature, G>R
and the system will warm up. Thus the steady state is
STABLE.
V
S
S
.B
Stability analysis is only useful for small perturbations
from a steady state. Determining the magnitude of a
perturbation that will cause the system to “blow up” is
beyond the scope of the problem.
W
W
W
The analysis in Fogler is to find perturbations that cause
motion from one low-temperature steady state to a higher
temperature steady state (ignition). As this transition is
not possible given our G and R curves, the analysis
required is much simpler.
It is believed that pressure fluctuations were detected in the feed stream and it was shut
off by a plant operator about 4 minutes before the explosion occurred.
(d) Write and (using Matlab) solve a set of differential equations describing what
happened in the reactor after the feed was shut off. Plot the temperature in the reactor vs.
time. Do you predict an explosion?
Unsteady Problem
m A (t = 0) = 378.5kg
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174
T (t = 0) = 510°F
Batch Reactor Balances: for subtle reasons, it is not
possible to just set the inflow terms in the CSTR energy
balance to zero.
Given that there is negligible gas holdup in the reactor:
N B (t) = N C (t) = 0
hence,
dN B dN C
=
=0
dt
dt
Enthalpy:
d H d (N A H A + N B H B + N C H C ) d (N A H A ) dN A
dH A
=
=
=
HA +
NA
dt
dt
dt
dt
dt
dH &
= Q − FB ,out H B − FC ,out H C
dt
dN A
dH A
HA +
N A = Q& − FB ,out H B − FC ,out H C
dt
dt
Batch Reactor mole balances
dN A
= −kN A
dt
dN B
= 0 = −FB ,out + kN A
dt
FB ,out = kN A
N
I
.
E
V
S
S
.B
dN C
W
W
= 0 = −FC ,out + 2kN A
dt
FC ,out = 2kN A
W
Plugging in the three mole balance relationships into the
enthalpy balance we find:
dH A
− kN A H A +
N A = Q& − kN A H B − 2kN A H C
dt
dH A
N A = Q& − kN A (H B + 2H C − H A ) = Q& − kN A ∆H rxn (T )
dt
dH A
dT
= c
p ,a
dt
dt
dT Q& − kN A ∆H rxn (T ) Q& − km A ∆Ĥ rxn (T )
=
=
dt
N A c p ,a
m A ĉ p ,a
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175
k(T / °F ) = Aexp(−b / (T + 459.67 ))
∆H rxn (T ) ≈ ∆Ĥ rxn (510°F )
Now we have an ode set and initial conditions for a Matlab
simulation:
m A (t = 0) = 378.5kg
T (t = 0) = 510°F
dm A
= − k(T )m A
dt
dT UA(Ta − T ) − k(T )∆Ĥ rxn
=
+
dt
m A ĉ p ,a
ĉ p ,a
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Code:
function [m,T]=hw6p1(tminutes);
%[m,T]=hw6p1(5); %run simulation for 5 minutes, crashes at ~4.5 minutes
thours = tminutes/60;
x0 = [378.5;510];
options = odeset('RelTol',10^-6,'AbsTol',10^-9);
[t,x] = ode15s(@odefun,[0,thours],x0,options);
m = x(:,1);
T = x(:,2);
figure(1);
plot(t*60,m);
title('NH_4NO_3 holdup');
xlabel('time/minutes since feed shutoff')
ylabel('m')
figure(2);
plot(t*60,T);
title('Temperature in Reactor');
xlabel('time/minutes since feed shutoff')
ylabel('T/degF')
return;
function derivs = odefun(t,x)
m = x(1); %kg NH4NO3
T = x(2); %degF
k = 2.28e19 *exp(-44367/(T+459.67));
UA = 0.01073; %MJ/(h degF)
Ta = 100; %degF
cpa = 0.8/1000; %MJ/(kg NH4NO3 degF)
dHrxn = -0.740; %MJ/kg NH4NO3
dmdt = -k*m;
%dTdt = (-k*dHrxn/cpa) + (UA*(Ta-T)/(m*cpa)) + (T*k); %slightly wrong
dTdt = (-k*dHrxn/cpa) + (UA*(Ta-T)/(m*cpa)); %correct
derivs = [dmdt;dTdt];
return;
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177
Temperature Plot:
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Perturbing the most sensitive problem parameter, the
activation energy term, b, by +1% gives similar looking
plots with explosion times shifted to 3-8 minutes.
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(e) Was the operator wise to quickly shut off the feed of aqueous ammonium nitrate
solution when he feared something was going wrong in the reactor? Is there something
else he should have done to prevent the disaster?
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The shutoff of the reactor turned a stable steady-state
system of operation into an unsteady system, far from a
steady state. Steady state in the batch reactor would mean
the reactor is empty at 100oF. In order for the batch
reactor to reach this steady state, the reactor must cool
down. One can plot mass-temperature trajectories of the
reactor operating in batch mode given different initial
holdup masses and temperatures.
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178
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Green circles represent initial conditions, red circles
represent final conditions after 4 hours of integration.
Black lines represent trajectories and connect green
circles to red ones. Diverging trajectories lead to a
temperature “explosion”. The absolutely final steady state
would be an empty reactor at the temperature of the cooling
water.
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It is clear that above ~425oF it is not safe to shut off the
reactor; this threshold “safety limit” temperature
increases slightly as mass holdup decreases. Since the
initial conditions are above this threshold, the sudden
shutoff of the feed was a BAD IDEA.
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In order to prevent the disaster, one must cool the reactor
below the critical mass/temperature line before shutting
off the feed. This could be accomplished by injecting more
water into the system, gradually decreasing the flow rate,
increasing the cooling rate, etc.
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179
Code for part e:
function stabilitydiagram(tminutes);
%stabilitydiagram(240); %integrate trajectories for at most 240 minutes
thours = tminutes/60;
figure(3); clf; hold on;
options = odeset('RelTol',10^-6,'AbsTol',10^-9);
mlist = linspace(200,500,21);
Tlist = linspace(350,560,21);
for(i = 1:length(mlist))
for(j=1:length(Tlist))
[t,x] = ode15s(@odefun,[0,thours],[mlist(i),Tlist(j)],options);
m = x(:,1);
T = x(:,2);
plot(m(end),T(end),'or'); %red stop
plot(m(1),T(1),'og'); %green go
plot(m,T,'-k'); %black trajectory
end
end
title('stability diagram at flow shutoff')
xlabel('NH_4NO_3 holdup mass, kg')
ylabel('reactor temperature, degF')
xlim([0,500])
ylim([0,600])
grid on;
hold off;
return
function derivs = odefun(t,x)
m = x(1); %kg NH4NO3
T = x(2); %degF
k = 2.28e19 *exp(-44367/(T+459.67));
UA = 0.01073; %MJ/(h degF)
Ta = 100; %degF
cpa = 0.8/1000; %MJ/(kg NH4NO3 degF)
dHrxn = -0.740; %MJ/kg NH4NO3
dmdt = -k*m;
%dTdt = (-k*dHrxn/cpa) + (UA*(Ta-T)/(m*cpa)) + (T*k); %slightly wrong
dTdt = (-k*dHrxn/cpa) + (UA*(Ta-T)/(m*cpa)); %correct
derivs = [dmdt;dTdt];
return;
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(f) Propose a procedure for safely starting-up and shutting-down a process like this (a
qualitative description will suffice).
Examples of ideas:
The flowrate could be turned on/shut off gradually
The concentration of ammonium nitrate could be ramped
up/down gradually
Greater cooling/heating control could be utilized at
startup and shutdown
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10.37 Problem set 7
Due 4/11/07
Problem 1.
A protein P reversibly binds a ligand L to f m l a complex C.The table below lists complex
concentration measured as a function of time t with varying initial ligancl concentrations ( [L],
= 1. 5 . or 15 pM). The initial protein concentration [PI, was always 1 nM. Estimate k,,, kd,
and Kd of the reaction. (Conaributed by 19 Bmmford).
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Problem 2.
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The objective of this exercise is to compare the volumetric productivity of a steady-state
chenlostat to that of a batch reactor. The batch operating time is the time for exponential
biomass growth from X, to X plus a turnaround time F~,,,,.Show that the ratio of volumetric
X
frmnxttum.
biomass productivity for a chemostat vs. a batch reactor i s approximately !n ?i;
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+
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Problem 3.
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The notion of computers with circuits built from cells has been proposed previously. If
the switches in such a computer involve changes in the level of expressed proteins, what
expression would describe the time to change from an "off state" (no expression) to an
"on state" (95% of the new steady-state level)? What would the half-time for switching
be in the following two cases: a) cells rapidly growing (doubling time 30 minutes) and a
stable protein (degradation half-tune one day); or b) cell not growing at all (infinite
doubling time) and a protein with a degradation half-time of 1 hour? How do these
switchmg times compare to those for silicon logic circuits? Would you invest in a
company developing such cellular computers?
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181
Problem 1.
The binding of protein P with ligand L to form complex C is reversible, as told
k on
off
P + L ⎯⎯→
C and C ⎯k⎯
→P + L
We are given a table with various initial concentrations of L in order to estimate kon and koff and
also Kd for the reaction.
dCC
= k on C P C L − k off CC
dt
Also from material balances and stoichiometry, we have CP+CC=CP0 and CL+CC=CL0, therefore
dCC
= k on (C P0 - C C )(CL0 − CC ) − k off CC
dt
In this problem, we can safely assume that CL0-CC≈CL0 since CL0>> CP0 in all three cases of
different CL0.
Thus, the integrated analytic expression for CC becomes
CC =
k on C P0 C L0
C C
{1 - exp[-(k on C L0 + k off )t]} = P0 L0 {1 − exp[-(k on C L0 + k off )t]}
k on C L0 + k off
C L0 + K d
where K d ≡
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k off
k on
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Therefore, if we plot CC w.r.t time for each cases of CL0, we can fit according to an exponential
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C P0 C L0
. Values for a and b are shown in the
C L0 + K d
y=a[1-exp(-bt)], where b is konCL0+koff, and a is
following table.
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L0(uM)
a
b
1
0.903
1.1113
5
1.0436
4.7047
15
0.9932
15.1079
One important observation in this table is that parameter a does not change much when initial
ligand concentration is changed, indicating CL0=1μM is already above the saturating value.
Therefore, value for koff can not be obtained accurately from this design of experiments. We can
only conclusively obtain the value for kon.
So we fit a linear express of b vs. CL0 to get kon.
kon=0.0010 nM-1 sec-1
An estimate on koff would be
0<=koff<<konCL0,min =0.0010 nM-1 sec-1*1uM=1 sec-1
Similarly an estimate on Kd is
0<= Kd<< CL0,min=1μM.
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CL0=1uM
0.4
0.35
Concentration of C(nM)
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.05
0.1
0.15
0.2
0.25
time(t)
CL0=5uM
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0.35
0.4
0.45
0.5
0.4
0.45
0.5
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1
0.9
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0.8
Concentration of C(nM)
0.3
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0.7
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0.6
0.5
0.4
0.3
0.2
0.1
0
0
0.05
0.1
0.15
0.2
0.25
time(t)
0.3
0.35
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CL0=15uM
1
0.9
Concentration of C(nM)
0.8
0.7
0.6
0.5
0.4
0.3
0.2
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0.1
0
0
0.05
0.1
0.15
0.2
0.25
time(t)
0.3
0.35
0.4
0.45
0.5
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16
14
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b
8
6
4
2
0
-2
0
5
10
15
CL0(uM)
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184
Problem 2.
For a steady state chemostat, the material balance on cell mass yields
D≡
F
=μ
V
the volumetric productivity is
F(X- X0)=μV(X- X0)
For a batch reactor, the material balance on cell mass yields
dX
= μX
dt
where the initial condition is X(t=0)=X0.
Therefore, we have X = X 0 exp(μt )
The volumetric productivity is
V(X - X 0 )
V(X - X 0 )
=
1
X
t + t turn
ln
+ t turn
X0
μ
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Therefore, the ratio of the two
μV(X - X0)
V(X - X 0 )
1
X
ln
+t
μ X 0 turn
= ln
X
+ μt turn
X0
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In practice, for chemostat, in order to maximize the productivity of biomass (DX), the operating
condition for μ is close μmax. Therefore the ratio above is approximately
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ln
X
+ μ max t turn
X0
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185
Problem 3.
The expression that would be suitable to describe the change of protein expression is:
k Pk r
{1 − exp[−(γ P + μ )t ]}
γ r (γ P + μ )
CP =
where the meaning of each symbol is in accord with what we did in class.
The time required to change from an “off state” to an “on state” (95% of the steady-state value) is
0.95 = 1 − exp[−(γ P + μ )t ] or t = −
ln 0.05
γP + μ
a) if cells are rapidly growing with a doubling time 30 min and stable protein with a degradation
half-time one day, i.e.
ln2
μ
= 30min and
ln2
γP
= 1 day
So the half-time for switching is the time need to reach
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1
= 1 − exp[−(γ P + μ )t ]
2
1
1
- ln(1 - 0.95 × ) - ln(1 - 0.95 × )
2 =
2 ≈ 27.9 min
=
ln 2
ln 2
(γ P + μ )
+
30 min 1day
0.95 ×
t switching
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b) if cells are not growing at all and the protein with a degradation half-time one hour, i.e.
ln2
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γP
= 1 hr
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So the half-time for switching is
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t switching
1
1
- ln(1 - 0.95 × ) - ln(1 - 0.95 × )
2 =
2 = 0.93hr
=
ln
2
(γ P + μ )
+0
1hr
So in both cases, the switching times are much much longer than that of the current electronic
circuits, which is on the order of ~ns-μs. Thus, it would not be promising in realizing a computer
for practical uses.
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10.37 Spring 2007
Problem Set 10
Due Wednesday, May 9
Figure 1.
1) The catalytic reaction (A Æ B) takes place within a fixed bed containing spherical
porous catalyst X22. Figure 1 shows the overall rates of reaction at a point in the
reactor as a function of temperature for various entering total molar flow rates,
FT0.
a) Is the reaction limited by external diffusion?
b) If your answer to part (a) was “yes,” under what conditions [of those shown (i.e. T, FT0)] is the reaction limited by external diffusion?
c) Is the reaction “reaction-rate-limited”?
d) If your answer to part (c) was “yes,” under what conditions [of those
shown (i.e. T, FT0)] is the reaction limited by the rate of the surf
e) Is the reaction limited by internal diffusion?
f) If your answer to part (e) was “yes,” under what conditions [of those
shown (i.e. T, FT0)] is the reaction limited by the rate of internal diffusion?
g) For a flow rate of 10 mol/hr, determine (if possible) the overall
effectiveness factor, Ω, at 360 K.
h) Estimate (if possible) the internal effectiveness factor, η, at 367 K.
i) If the concentration at the external catalyst surface is 1 mol/L, calculate (if
possible) the concentration at r = R/2 inside the porous catalyst at 367 K.
j) If the reactor must achieve a conversion X=0.05, qualitatively sketch how
the length of the reactor will have to vary with T and with FT0. Assume
the reactor is isothermal.
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10.37 Spring 2007
Problem Set 10
Due Wednesday, May 9
2) A large fraction of all the platinum-group metals that have ever been mined
currently reside in catalytic converters on automobiles. As noble metals prices rise
(prices remain near record high levels) and the number of vehicles in the world
increases, it is imperative to minimize the amount of catalyst employed, while still
meeting the emissions regulations. Your object in this problem is to first build a
Matlab simulation of a catalytic converter, and then explore the geometrical
tradeoffs involved that affect the total amount of catalyst employed.
The input to the converter is the car exhaust stream; when it is running fuel-rich (e.g.
during acceleration) it is 1200 moles/hr of N2, 200 moles/hr of CO2, 200 moles/hr of
H2O(steam), 20 moles/hr CO, and 4 moles/hr NO all at T=700 K and a total pressure
of 1 atm. The most difficult requirement is that output needs to be < 0.4 moles/hr of
NO. Assume that the rate-limiting (and essentially irreversible) elementary step is
CO—M + NO—M Æ CO2 + N—M + M
Rate at 700 K=(400 mole/hr-m2) θNO—M θCO—M(Area of noble metal/ volume
porous material)
3
Rate has units of mol/hr-m of porous material.
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Where M is a noble metal site and the fraction of the surface covered by CO--M and
NO--M is given by Langmuir-Hinshelwood equilibria:
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Keq1 at 700 K= 1 (atm)-1
Keq2 at 700 K= 3 (atm)-1
CO + M = CO—M
NO + M = NO—M
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The N—M formed in the rate-controlling step rapidly combine to make and release
N2.
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For the purposes of this problem you can neglect the binding by the other gas-phase
species at this temperature. (In the real system there are several other
complications…). To make the numerical solution easier, we suggest that you assume
that the partial pressure of CO is constant inside any pore, i.e.
[CO(inside pore)]=[CO,s]
and that you also assume that Keq2*PNO is negligible compared to Keq1*PCO. This
is approximately correct even if the chemistry is fast because the CO is present in
excess, and in any case Keq2*PNO is always << 1. If you are a careful engineer, you
can check the validity of these approximations after you have a solution in hand.
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The converters are monolithic reactors, you can approximate the geometry as a lot of
8 mm i.d. pipes in parallel, with each pipe’s walls lined with a thin layer of porous
material (0.1 m2 platinum/ ml porous material).
Assume the reactor is isobaric and isothermal.
Diffusivity of all species in bulk, D=1e-8 m^2/s
Diffusivity of all species in pores, De=1e-9 m^2/s
2
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10.37 Spring 2007
Problem Set 10
Due Wednesday, May 9
Assume gas density and viscosity constant along reactor at
Density = 0.5 kg/m^3
Viscosity = 3e-5 Pa*s
Use the boundary layer approximation:
Delta = 0.001 m / (Re^0.5)
Re= density*(superficial velocity)*diameter/viscosity
Use mass transfer approximation:
kc=D/delta
a) Rewrite the rate law so it is first order with respect to the NO concentration in the
gas within the catalyst pores. It should have the form:
Rate= k’ RT * [NO]
In which k’ is derived from equilibrium balances and given k and
area/volume term.
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b) Derive the Thiele modulus assuming the thin porous layer can be modeled as a
slab. Derive the concentration profile of the NO inside the pore. What is the
effectiveness factor?
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c) Use a flux balance to find the overall effectiveness factor, omega, for the reaction.
d) Using your overall effectiveness factor, which allows you to write the reaction
rate using the bulk NO and CO concentrations in the pipe, run a simulation for
this base case: the porous layer is 1 µm thick, the pipes are 20 cm long, and there
are 2000 identical pipes in parallel. Does this achieve the emission target for NO?
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e) Compute how the NO conversion varies if you double the thickness of the porous
layer.
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f) Compute how the NO conversion varies if you double the number of parallel
pipes.
g) Compute how the NO conversion varies if you double the length of the pipes.
h) Make a guess at a design that would achieve the same or better NO conversion as
the base case but uses less noble metal. Run a simulation for this case to see if
your estimate is right. Would the pressure drop be a lot different in your design
than in the base case? (You do not need to simulate the pressure drop.)
3
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189
10.37 Spring 2007
Problem Set 10 Solutions
1. a) Yes, the reaction is limited by external diffusion.
b) Table 12-1 shows that external diffusion-limited ra varies with U1/2 and nearly linearly with
T. This is the case for:
FT0 = 10 mol/hr, all T
FT0 = 100 mol/hr, about 362K < T < 375K
c) Yes, the reaction is “reaction-rate-limited”.
d) According to Fogler, rA varies exponentially with both reaction-rate-limitations and internaldiffusion-limitations, and is nearly independent of FT0. However, rA varies more with T more
strongly in the reaction-rate limited regime. So, the reaction-rate-limitation is overcome before
the internal-diffusion-limitation as T is increased. This is the case for:
T < 362K, FT0 = 100 mol/hr T < 365K, FT0 > 1000 mol/hr N
I
.
E
e) Yes, the reaction is limited by internal diffusion.
f) Internal-diffusion limitations are seen at:
367K < T > 377K, FT0 > 1000 mol/hr
V
S
S
.B
 mol

− rA' 10
,360K 
actual _ rate(external _ lim )
hr

 ≈ .25 = 0.36
g) Ω =
=
mol
ideal _ rate(rxn _ lim )

 .70
− rA'  5000
,360K 
hr


W
W
mol


− rA'  5000
,367K 
actual_rate(internal_lim )
hr


=
h) η =
mol
ideal_rate(rxn_lim )


− rA'  5000
,367K 
hr


Extrapolate the reaction-rate limited portion of the FT0 = 5000 mol/hr curve up to 367 K.
W
η≈
1.2
= 0.8
1.5
i) Fort a first order reaction with spherical pellets:
3
η = 2 (φ1 coth φ1 − 1) = 0.8
φ1
Solve: φ1 = 2.0
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Ψ=
CA
1  sinh (φ1λ ) 

= 
C A,S λ  sinh φ1 
C A = C A,S
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190
with λ = r / R = 1 / 2
1  sinh (φ1λ ) 
mol
mol  sinh (1) 

 = 1
(2
)
 = 0.65
λ  sinh φ1 
L
L
 sinh (2 ) 
j) For a given FT0, the reaction rate will increase either exponentially or linearly with T (meaning
the needed pipe length to achieve a certain conversion will decrease either exponentially or
linearly with T).
external
ex
ternal diffusion limited
L
Rxn
limited
int
inte
ernal
rnal diffusion limited
N
I
.
E
T
V
S
For a given T, increasing the flowrate will increase the pipe length needed for a given
conversion. However, while in the external diffusion limited region, an increase in FT0 also
increases the reaction rate (i.e. decreases the pipe length). The higher rA will offset somewhat
the effect of the higher flow rate, and therefore the needed pipe length will increase slowly with
FT0 while in the external-diffusion limited region. When the process is internal-diffusion or rxn
rate limited, rA no longer increases with FT0, so the pipe length needed for a given conversion
will increase sharply.
S
.B
W
L
W
W
no external
limitations
external
diffusion limited
FT0
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191
2. a) 

Area of noble metal
rlim = kθ NO − M θ CO − M 

 Volume of porous material 
where θ i−M =
[i − M ]
[M ]o
Use the adsorption steps (rapid equilibrium) to find θ i−M .
K eq,1 =
[CO − M ] = θ CO−M [M ]o
PCO [M ]
PCO [M ]
⇒
θ CO−M =
K eq,2 =
[NO − M ] = θ NO−M [M ]o
PNO [M ]
PNO [M ]
⇒
θ NO−M =
K eq,1 PCO [M ]
[M ]o
K eq,1 PNO [M ]
[M ]o
N
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Use and overall site balance to find [M] (concentration of empty sites).
[M ]o = [M ] + [CO − M ] + [NO − M ] = [M ] + θ CO−M [M ]o + θ NO−M [M ]o
Substitute in the equations for θ i−M :
[M ]o = [M ] +
K eq,1 PCO [M ]
[M ]o
S
.B
K eq,2 PNO [M ]
[M ]o +
V
S
[M ]o
[M ]o
W
Solve for [M] and use the simplifications given in the problem statement (Keq2PNO << Keq1PCO
and PCO = PCO,S):
[M ] =
W
W
[M ]o
1+ K eq,1 PCO + K eq,2 PNO
=
[M ]o
1+ K eq,1PCO,S
Substitute θ i − M and [M] into given rate equation:
− rNO = rlim
− rNO = k
K P
K P 
[M ]o
= k eq,1 CO,S eq,2 NO 

[M ]o [M ]o  1 + K eq,1 PCO,S




2


Area of noble metal


 Volume of porous material 
K eq,1 PCO,S K eq,2 

Area of noble metal
 P = k' PNO

2 
(1 + K eq,1 PCO,S )  Volume of porous material  NO
or, in terms of NO concentration:
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192
− rNO = k ' RT [NO ] = k " [NO ]
where k" = k
K eq,1 K eq,2 PCO,S 

Area of noble metal
 RT

2 
(1 + K eq,1PCO,S )  Volume of porous material 
b) We now have a rate expression that is first order in [NO]. The thickness of the catalyst on the
surface is small, so the curvature can be ignored and the catalyst is modeled as a slab:
Bulk gases
gases
[NO]S
x=L
x
catalyst
N
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E
pipe wall
At steady state, diffusion of NO inside the catalyst is governed by:
D pores
∂ 2 [NO ]
+ rNO = 0
∂x 2
D pores
∂ 2 [NO ]
− k "[ NO ] = 0
∂x 2
W
x
λ≡
L
S
.B
W
W
Non-dimensionalize:
[NO]
ψ ≡
[NO]S
V
S
⇒
∂ 2 Ψ  k" L2 
−
Ψ=0
∂λ2  D pores 
So, the Thiele modulus is defined as:
φ2 ≡
k" L2
D pores
The equation:
∂2Ψ
− φ 2Ψ = 0
2
∂λ
Has a solution of the form: ψ = Ae − λφ + Be λφ
The boundary conditions are:
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193
1. Ψ λ =1 = 1 ([NO] = [NO]S at the surface)
2.
∂Ψ
= 0 (no flux at pipe wall)
∂λ λ =0
From the second boundary condition:
∂ψ
= − Aφe −λφ + Bφe λφ
∂λ
0 = − Aφe 0 + Bφe 0
⇒
A=B
From the first boundary condition:
1 = Ae −φ + Aeφ
⇒
A=
1
e + eφ
N
I
.
E
−φ
Solution of the concentration profile of NO inside the catalyst (treated as a slab):
ψ =
V
S
e − λφ + e λφ
e −φ + eφ
S
.B
We want the internal effectiveness factor (η) as a function of the thiele modulus (φ).
η=
actual rxn rate
rxn rate is all catalyst was exposed to [NO]S
W
W
The actual reaction rate is the integral of the concentration times the first order rate constant over
the entire slab thickness. However, an easier way to find this is to use the fact that the flux at the
surface of the catalyst has to be equal to the total reaction occurring inside.
W
actual rxn rate = (flux )(surface area ) = D pores
actual rxn rate = D pores
[ NO ]S
L
 − φe −φ + φeφ

−φ
φ
 e +e
[ NO ]S ∂Ψ
∂[ NO ]
Asurface
Asurface = D pores
∂x
L ∂λ λ =1

 Asurface

rxn rate if all catalyst was exposed to [NO]S = k"[NO]S Vcatalyst
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D pores
η=
[NO ]S  − φe −φ + φeφ  A
L  e −φ + eφ
k"[NO]S Vcatalyst


surface
=
194
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D pores  − φe −φ + φeφ

k" L2  e −φ + eφ
 1  eφ − e −φ
 =  φ
−φ
 φ e +e



(note that the surface area exposed to the bulk gas divided by the volume of porous material is
equal to 1/L)
c) Ω =
ηk"[NO]S Vcatalyst η[NO]S
actual rxn rate
=
=
rxn rate is all catalyst was exposed to [NO]bulk k"[NO]bulk Vcatalyst [NO]bulk
Need to use flux balance to find [NO]S.
At steady state:
flux to the catalyst surface = rxn inside catalyst
N
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E
k c Asurface ([NO]bulk − [NO]S ) = ηk"[NO]S Vcatalyst
Solve for [NO]S:
[NO]S =
k c Asurface [NO]bulk
ηk"Vcatalyst + k c Asurface
Substitute into the omega equation:
Ω=
=
k c [NO]bulk
ηk" L + k c
V
S
S
.B
W
ηk c [NO]bulk
ηk c
=
[NO]bulk (ηk" L + k c ) ηk" L + k c
W
W
where kc = Dbulk/delta
d) Design equation for a PFR:
dX NO − rNO AcV porous _ material Ωk"[NO]AcV porous _ material Ωk"[NO]0 (1 − X NO )AcV porous _ material
=
=
=
dz
FNO,0Vreactor
FNO,0Vreactor
FNO,0Vreactor
Note that the change in the number of moles can be ignored because epsilon ≈ 0 (mole fraction
of NO in the feed is extremely small).
The Vporous_material/Vreactor term is needed because the rate expression derived in part a) is per
volume of porous material, but the rate expression in the PFR design equation needs to be on a
per volume of reactor basis.
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195
In order to calculate kc, need to first calculate delta using the equation given in the problem
statement:
δ=
.001m
Re 0.5
where Re =
ρUd pipe
µ
To calculate the superficial velocity (U) in each pipe, use the equation:
U=
FNO ,o
qo
=
Ac [NO]o Ac
The feed molar flow rate of NO in each pipe is found by dividing the total feed molar flow rate
of NO by the number of pipes:
FNO ,o =
FNO ,o ,total
N pipes
N
I
.
E
The concentration of NO in the feed is found by multiplying the feed mole fraction by the total
concentration:
[NO]o = y NO ,o Ctot =
FNO ,o P
∑ Fi ,o RT
S
.B
i
See matlab code below:
V
S
W
function [X1f,X2f,X3f,X4f,X5f] = HW10_P2_allparts();
%[X] = HW10_P2_allparts()
[z1,X1]
[z2,X2]
[z3,X3]
[z4,X4]
[z5,X5]
=
=
=
=
=
W
W
reactorsimulation(1E-6, 2000, 0.2);
reactorsimulation(2E-6, 2000, 0.2);
reactorsimulation(1E-6, 4000, 0.2);
reactorsimulation(1E-6, 2000, 0.4);
reactorsimulation(.4E-6, 5000, 0.2);
%[z,X] = reactorssimulation(catalyst thickness, number of pipes, length of
%pipes, total mol/hr in feed)
X1f
X2f
X3f
X4f
X5f
=
=
=
=
=
X1(length(z1));
X2(length(z2));
X3(length(z3));
X4(length(z4));
X5(length(z5));
figure(1);
plot(z1,X1);
xlabel(['position along length of pipe (m)'])
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ylabel('NO conversion')
title('Original Conditions (part d)')
figure(2);
plot(z2,X2);
xlabel(['position along length of pipe (m)'])
ylabel('NO conversion')
title('Double Thickness of Porous Layer (part e)')
figure(3);
plot(z3,X3);
xlabel(['position along length of pipe (m)'])
ylabel('NO conversion')
title('Double Number of Parallel Pipes (part f)')
figure(4);
plot(z4,X4);
xlabel(['position along length of pipe (m)'])
ylabel('NO conversion')
title('Double Length of Pipes (part g)')
N
I
.
E
figure(5);
plot(z5,X5);
xlabel(['position along length of pipe (m)'])
ylabel('NO conversion')
title('Thinner Catalyst Layer, More Pipes (same amount of noble metal)(part
h)')
V
S
S
.B
return;
function [z,X] = reactorsimulation(thicknessofcatalyst,numberofpipes,...
lengthofpipes);
W
%a = NO
%b = CO
%Reactor/Catalyst Geometry and Chemistry
param.L = thicknessofcatalyst; %m
N = numberofpipes;
Lp = lengthofpipes; %m
param.T = 700; %in Kelvin
dpipe = 8/1000; %inner diameter of each pipe 8 mm, in m
param.Ac = pi*(dpipe/2)^2; %cross sectional area of each pipe, m^2
W
W
%Reactor/Feed Conditions
P = 1*101325; %pressure = 1atm, in Pa
param.R = 8.314; %gas constant, J/(K mol)
Ctot = P/param.R/param.T; %total concentration of gas, moles/m^3
ya0 = 4/1624; %feed mole fraction of NO
yb0 = 20/1624; %feed mole fraction of CO
Ftot = (1200+200+200+20+4)/60/60/N; %total feed flow rate per pipe (mol/s)
param.Fa0 = ya0*Ftot; %NO feed molar flow rate into each pipe, in mol/s
Fb0 = yb0*Ftot; %CO feed molar flow rate into each pipe, in mol/s
param.Ca0 = ya0 * Ctot; %feed concentration of NO, in mol/m^3
Cb0 = yb0 * Ctot; %feed concentration of CO, in mol/m^3
param.theta_b = yb0/ya0; %feed ratio for CO
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q=param.Fa0/param.Ca0; %volumetric flow rate into each pipe, in m^3/s
U = q/param.Ac; %superficial velocity in each pipe, in m/s
%Physical Property Data
Dbulk = 1E-8; %in m^2/s
mu = 3E-5; %viscosity in Pa*s = kg/m/s
rho = 0.5; %density in kg/m^3
Re = rho*U*dpipe/mu; %Reynolds number
delta = 0.001/(Re^0.5); %thickness of boundary layer, in m
param.kc = Dbulk/delta; %m/s
param.Vporous_mat = pi*dpipe*param.L*Lp; %volume of catalyst (m^3)
param.Vreactor = param.Ac*Lp; %volume of reactor (m^3)
odeoptions = odeset('AbsTol',1e-9,'RelTol',1e-12);
[z,X]=ode15s(@diffeqs,[0 Lp],0,odeoptions,param);
return
function dXdz = diffeqs(z,X,param);
%unpack the parameters
T = param.T;
R = param.R;
kc = param.kc;
Ac = param.Ac;
L = param.L;
theta_b = param.theta_b;
Vporous_mat = param.Vporous_mat;
Vreactor = param.Vreactor;
Ca0 = param.Ca0;
Fa0 = param.Fa0;
N
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V
S
S
.B
Dpores = 1E-9; %in m^2/s
Area_metal_per_Vol_porous_material = 0.1*100^3; %m^2 platinum/m^3 porous
material
Keq1 = 1/101325; %Pa^-1
Keq2 = 3/101325; %Pa^-1
k = 400/60/60; %reaction rate constant, in mol/s/m^2(of noble metal)
W
W
W
Ca = Ca0*(1-X);
Cb = Ca0*(theta_b-X);
Pbs = Cb*R*T; %CO pressure in pores (Pa)
k_double_prime =
k*R*T*Keq1*Keq2*Pbs/(1+Keq1*Pbs)^2*Area_metal_per_Vol_porous_material;
%mol/s/m^3(catalyst)*1/Pa
phi = (k_double_prime/Dpores)^.5*L;
eta = 1/phi*(exp(phi)-exp(-phi))/(exp(phi)+exp(-phi));
omega = eta*kc/(eta*k_double_prime*L+kc);
r = k_double_prime*Ca*omega; %mol/s/m^3(catalyst)
dXdz = r/Fa0*Vporous_mat/Vreactor*Ac; %m^-1
return
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198
d) Original conditions (does not meet emission specifications).
N
I
.
E
e) Double thickness of porous layer - more conversion (more catalyst).
V
S
S
.B
W
W
W
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199
f) Double number of parallel pipes – more conversion (lower flow rate).
N
I
.
E
V
S
g) Double length of pipes – higher conversion.
S
.B
W
W
W
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200
h) Open ended.
Improve reaction rate by increasing T or changing the catalyst to improve surface area to volume
ratio.
Improve external mass transfer by making more turbulent flow.
Improve internal mass transfer by making a larger number of shallower pores.
Many other answers possible. Equation 4-36 in Fogler gives an equation for calculating pressure
drop in pipes. Whether pressure drop changes significantly or not will depend on the parameters
changed.
The figure below is for a thinner catalyst layer (0.4 um) and more pipes, but with the same total
amount of noble metal. The conversion is slightly higher (71%) than the base case (64%). For
the case shown below, an increase in the total number of pipes (i.e. decrease in the superficial
velocity through each individual pipe) would decrease the pressure drop compared to the base
case. The pressure drop in the base case is assumed to be negligible, so an increase in the total
number of pipes should have no significant effect on the pressure drop.
N
I
.
E
V
S
S
.B
W
W
W
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201
MATLAB REVIEW
The focus of this document is to review common, useful, higher-level Matlab operations
that will be employed on your assignments, such as: ordinary differential equation time
integrators (i.e. ode45 and ode15s, including event functions), curve fitting to a model,
plotting options for x-y plots, and solving sets of nonlinear equations (fsolve). Additional
examples will be made available for future topics. None of these exercises are required
work, but are intended to be a great review and resource for future questions.
Task 1: Time integration of ODE’s
Typically, a set of unsteady chemical engineering equations can be cast into the following
form:
accumulation = input – output + generation
which leads to an unsteady set of equations of the form:
N
I
.
E
r
r r
df
=
g( f ,t) ,
dt
r
where time is the independent variable, f is the (unknown) vector of dependent
r
variables, and g is the vector of functions giving the rate-of-change of each dependent
r
variable. Given an initial condition for each dependent variable, f 0 , the equations can be
r
integrated in time to find f (t ) . However, while an analytical expression may not be
available for the solution, a numerical integration technique can always find a solution to
this type of problem.
V
S
S
.B
W
W
Open odesintegrate.m. This file solves for the concentrations of two chemical species in
a solution a batch reactor (perfect mixing, isothermal, constant volume) using Matlab’s
ode solvers. The system is:
r1
r2
W
A→ B →C
r1 = k1 [ A]
r2 = k 2 [B]
and applying the general chemical engineering balance equation on the reactor, we find
that the ode’s governing species are:
d[ A]
= −k1 [ A]
dt
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d[B]
= k1 [ A] − k 2 [B]
dt
d[C]
= k 2 [B]
dt
Since [C] can be expressed in terms of [A] and [B] using mass balances, the third ode is
redundant and need not be followed explicitly. Consider the following initial condition:
[A0] = 1 M, [B0]=[C0] = 0 M. Through two reaction steps, the [A] is converted to [C].
Look at the example file odesintegrate.m. In particular, look at the calling sequence of
the ode solver, which is generally:
[t,f]=ode_solver(@derivative_function_name,[t0,tf],f0,options,param);
where the ode solver can be ode45 or ode15s, for example. The returned values are a
column vector t of time steps, and a matrix f, where each column corresponds to a
variable.
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The derivative function has the form:
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dfdt = derivative_function_name(t,f,param);
Basically, you need to pick a solver, set the function that calculates the time derivatives
r
df
, set options (such as error tolerances), and pass additional necessary constants.
dt
Options and param may each be [] when using default error tolerances and no additional
constants are needed in the derivative function. Note that only the current value of t
(scalar) and the column vector of the current unknowns, f are passed to the derivative
function.
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IMPORTANT: always use column vectors for dfdt and f. This is the format Matlab
expects and use of row vectors leads to errors that you will have to debug later.
Try running each set in the EXAMPLE sequence of odesintegrate.m. (Type “help
odesintegrate” at the Matlab prompt and run each set of parameters in sequence.) This
example sequence illustrates a difference between the performance of ode45 and ode15s.
As k1 is ramped up with a constant k2, the [A] eventually disappears “instantaneously”
relative to the rate of reaction of [B] into [C]. This essentially looks like [A0] =[C0]= 0
and [B0] = 1 case. See Figure 1. Note that the number of time steps that the ode solver
requires to get an accurate result increases as k1 gets much larger than k2. Whenever a
process has two or more vastly differing characteristic rates, the problem can be called
“stiff”.
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Figure 1. Top: [A0] = 1, [B0] = 0; Bottom: [A0] = 0, [B0]=1; Since k1 >> k2, these
dynamic results are very similar when viewed over the timescale of decay of [B]. Results
obtained with integrator ode45.
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While ode45 is a more accurate time integrator in general, and is a good overall tool,
ode15s handles stiff problems much more quickly; ode15s has better stability properties
and thus can take larger time steps. (This is because ode45 is explicit and ode15s is
implicit. If you would like to more about the inner workings of the time integrators, feel
free to ask.)
Programming Exercise 1:
Consider the following pair of coupled linear odes:
⎡
df 1 ⎤
⎢ dt ⎥ ⎡
−
1 2
⎤
⎡
f1 ⎤
⎢ df ⎥ =
⎢
⎥⎢ ⎥
⎢
2 ⎥
⎣
0
−
5
⎦
⎣
f 2 ⎦
⎣
dt ⎦
subject to the following initial condition:
r ⎡10⎤
f 0 =
⎢ ⎥
⎣10⎦
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Your task is to solve this set of equations numerically in Matlab using ode45. Ensure you
have the correct solution by plotting your numerical expressions for f1 and f2 along with
the analytical expressions:
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f1 (t) =
15exp(−
t)
−
5exp(−5t)
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f 2 (t) =
10 exp(−5t) .
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Feel free to use the structure of the example file, odesintegrate.m as a template to build
from.
W
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Task 2: Using events to terminate ode solvers early:
A priori, you won’t always know how long to run a time integration. Matlab incorporates
“event function” functionality into the various ode solvers to allow for this capability.
The best way to understand how this works is to see an example. Look at
learntouseevents.m.
A set of event functions is attatched to an ode solver with the following options:
options = odeset('Events',@eventfunctionname);
Each event function in the set triggers an event when the function “value” returned is
zero. The event function has the following format:
[value,isterminal,direction] = eventfunctionname(t,f,param);
The event function takes the current value of time, the unknown vector f, and any
additional problem parameters and returns any number of event function values. Each of
the three returned variables, value, isterminal, and direction are column vectors with one
entry per event function. For example, if you wanted to have a single event function
terminate integration as soon as f(1) became zero, you would write the code:
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function [value,isterminal,direction] = eventfunctionname(t,f,param);
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value = f(1); %every zero in the value vector is an event
isterminal = 1; %isterminal =1/0: stop/keep integrating
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direction = 0; %0/+1 or -1: direction doesn’t matter/does matter
return;
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See learntouseevents.m for a more sophisticated example of the use of event functions,
including recovery of the log of events.
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Programming exercise 2:
Using the same solver and code as exercise 1, add an event function that terminates
integration when f2 = 1.
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Task 3: Plot Management
Trying to put lots of things on the same graph can be a pain, so here is a summary of the
figure creation process.
A plotting command replaces whatever is currently plotted on the active figure by
default. A new figure can be created with the command:
figure;
or
figure(n); (n is a positive integer, the figure number)
In the second case, the figure will always have the number n and if figure n exists it will
be replaced unless the hold is on (more on that later). Figure n becomes the “active
figure” on which all plot commands will be placed.
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X-Y plots are made by calling the plot command, which may take a variable number of
arguments:
plot(x1,y1,formatstring1,x2,y2,formatstring2,x3,y3,formatstring3,…);
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See Matlab documentation under “LineSpec” for examples of format strings (or “help
plot”). If several data sets are given in a single plot call, Matlab automatically cycles
through different colors. If you want to have separate plots in separate statements on the
same figure, the “hold on” command must be used, and formats SHOULD be explicitly
specified so you can tell which line is which. For example, try the following series of
commands (copy and paste):
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x = linspace(0,10,101); y1 = sin(x); y2 = cos(x);
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figure(1); plot(x,y1); plot(x,y2);
figure(2); plot(x,y1); hold on; plot(x,y2); hold off;
figure(3); plot(x,y1,'ob'); hold on; plot(x,y2,'or'); hold off;
figure(4); plot(x,y1,x,y2);
In figure 1, the second plot command over-wrote the first. In figure 2, the second dataset
plotted is by default the same color as the first; in figure 3, this is explicitly corrected and
open circles are used and no line is used. In figure 4, the Matlab defaults automatically
cycle through colors. I recommend you explicitly give figure numbers, so that you don’t
have to close your old figures when doing several runs in a row.
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Plotting ranges can be set with the following options:
xlim([xmin,xmax]);
ylim([ymin,ymax]);
The title can be specified with:
title(titlestring);
and the legend can be specified with: legend(series1string,series2string…); or legend({series1string,series2string,series3string…}); N
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Also, a grid can be overlaid on the plot with grid on;
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This should cover most plotting needs you will have. Advanced options are available.
Look up the Matlab “help plot” for details.
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Programming exercise 3:
Using the results of exercise 1 or 2, plot f1 with red upward facing triangles (no line) and
f2 with a blue dashed line (no symbol). Ensure the minimum x and y values on the plot
are zero, and all the data are visible in the plotted range.
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W
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Task 4: Model fitting
The Matlab “fit” function is a versatile tool for fitting parameters to user-specified
models. This example will show you how to fit some noisy data to a sine function.
First, you need a noisy data set f(x) (a column vector);
x = linspace(0,2*pi,101)';
f = sin(x+0.1*randn(size(x)))+0.1*randn(size(x));
plot(x,f);
Next set the model type to be fit. The call:
model=fittype('A*sin(B*x+C)+D','ind','x');
declares a 4-parameter model based on the sine function and declares x to be the
independent variable. Next you should guess the parameters in the order they appear
from left to right:
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initial_guess=[1.00,1.00,0,0];
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Note that we guessed our values as those that you should get if there were no noise in the
data. Finishing up, call:
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options=fitoptions(model);
set(options,'StartPoint',initial_guess);
fresult = fit(x,f,model,options);
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The values of A, B, C, and D are stored in the struct fresult. The resulting parameters can
be viewed as a whole by typing:
fresult
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and the two plots can be compared by using the model directly as if it were a function.
Try typing:
plot(x,f,x,fresult(x));
Individual parameters from the fit can be extracted with struct “dot” notation. For
example, to see parameter A type
fresult.A
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Programming Exercise 4:
Collect and save the f1 output you get from exercise 1. Add some noise to it, using the
command:
f1 = f1 + (0.1*randn(size(f1)));
Fit a 4-parameter model to fit f1(t) = Aexp(Bt) + Cexp(Dt). Do you recover the
coefficients A, B, C, and D that you expect?
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Task 5: Solving Nonlinear Systems of Equations
Matlab has the ability to solve nonlinear systems of equations using fsolve, where equations are of the form: r r
0 = g ( x) .
This could be interpreted as directly finding the steady state of the equation
r
r r
df
= g( f , t) .
dt
The calling sequence of fsolve is very similar to the use of ode integrators. The main
difference is that instead of supplying a function to calculate the derivatives, you supply a
function to calculate the right-hand-side of the equation, g. When g is not zero, this is
also called the residual.
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The calling sequence and use of fsolve is
x = fsolve(@gfun,x0,options,param);
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Where x0 is the initial guess, param are additional constants needed by the g function,
and gfun is the gfunction:
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[y,Jacobian] = gfun(x,param);
gfun returns the value of each of the g(x) functions in y. When all the entries in the
column vector y are zero, the system is solved with the current iterate of x.
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The Jacobian matrix is an expression that contains the rate of change of each of the g
functions with respect to each of the parameters x (as in Newton iteration). Entries are
given by:
∂g
J ij = i .
∂x j
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The Jacobian matrix is an optional returned value in gfun. In general, Matlab will use a
numerical approximation of the analytical Jacobian. The user specifically mentions that
an specified analytical Jacobian will be given with the statement:
options = optimset('Jacobian','on');
The use of fsolve is demonstrated in the example file multiplefsolve.m. Given a perimeter
and area, this function finds the length and width of an appropriate rectangle if one exists:
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g1 (L,W ) = 2L + 2W − P
and
g 2 (L,W ) = LW − A ,
where L = length, W = width, P = perimeter, and A = area. Notice that the g functions
must be rearranged so that they are zero when satisfied by appropriate L and W.
Programming exercise 5:
Following the example of multiplefsolve.m, make a program that finds the height and
radius of a cylinder with a given area and volume. Multiple solutions may exist! Verify
that your program gives you R = 1 and H = 1 when Area = 4*pi and Volume = pi with an
appropriate initial guess.
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Good programming practices:
Use meaningful names for variables. If you have a list of 10 different chemical species
stored in x(1), x(2)…x(10), you shouldn’t be using them in that form in your reaction
equations. Consider renaming them as soon as you enter the function or declare
descriptive names equal to your integer indices, e.g.
Cl = 1;
Na = 2;
NaCl = 3;
Then you can say:
x(Cl)*x(Na);
which is much easier to understand and debug than:
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x(1) * x(2);
Go for simplicity and clarity before going for elegance. Colon notation is an effective
way to make Matlab programs more compact, but for loops are much more intuitive. Use
for loops if colons give you trouble. Furthermore, in general, several simple steps are
easier to understand than one long, complicated step.
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Watch out for the differences between matrix and element-wise operations. This
leads to common mistakes. Most functions, such as sin() and exp() can accept a matrix of
any size rather than a single value. They operate on each matrix element individually,
returning the sin() or exp() of each element, respectively. However, whenever you use
the operations (*,/) and the objects being multiplied/divided are vectors or matrices, be
sure to know the difference between (*,/) and (.*,./). The first uses matrix multiplication
and the second does element-wise operations, operating on corresponding entries. If this
gives trouble, feel free to use for loops to operate on one value at a time.
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Learn to use the Matlab debugger. Programs usually fail on the first try. Matlab has a
useful debugger, where you can put breakpoints in the code, examine variable values
during execution, and even execute any additional operations from the command line
(such as make plots). As an alternative to placing a breakpoint, the “keyboard;”
command will do the same thing. To exit debug mode at the command line, type
“return;”. On the command line, “who” is a useful command that displays the names of
all the variables that are known at this point in the code. Each one can be inspected or
operated upon.
Use semicolons. If semicolons are omitted at the end of a line, more output is sent to the
screen. In the case of large matrices, it can be a startlingly large amount.
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Use functions rather than scripts. A function takes input and gives output; this makes
it easy to build a code as a group of working parts with well-defined interfaces. If a
function can’t change its input to mess up the calling program. A script works at the
same level as the regular Matlab prompt. In a script, all variables are effectively global,
and thus from run to run things might change depending on what is stored in all current
values of the variables. Also, if you run two different scripts, they could use the same
variable names and operations and results can get mixed up.
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10.37 Chemical and Biological Reaction Engineering, Spring 2007
Exam 1 Review
In-Out+Production=Accumulation
Accumulation=0 at steady state
FA0 − FA + rAV = 0
FA0 = [ A]0 ν 0
FA = [ A]ν
For a liquid phase with constant density:
ν0 =ν
For A → B the reaction moles are the same, so
For A → 2 B ,
ξ [ =]
ν0 =ν
ν0 ≠ν
moles (extent of reaction)
(-) for a reactant and (+) for a product
N rxns
N i = N i 0 + ∑ υi ,nξ n
n =1
Suppose A → B + C
XA =
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N A0 − N A
N A0
N A = N A0 (1 − X A )
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Thermodynamics
Suppose
Ke = e
−1
− ΔG RT
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ΔG = ΔG °f , products − ΔG °f ,reactants
Kc =
S
.B
ZZZX
A + B YZZ
Z 2C
k
k1
[ B ][C ]
[ A]
W
has units. You need to use standard states, such as 1M, to make it
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dimensionless.
Enzyme Catalysis
Energy
without enzyme
with enzyme
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P
Reaction Progress
Figure 1. Energy diagram for a reaction with and without enzyme.
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E + S U ES
ES → E + P
Pseudo steady state approximation:
d [ ES ]
=0
dt
[ ES ] = f (other species)
Cell Growth
# cells
N=
volume
N = N0e μ t
Monod kinetics:
μ=
μmax [ S ]
KS + [S ]
YA =
B
ΔA
ΔB
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Rate Constants
k (T ) = Ae − Ea
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RT
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.B
Given k1 and k2, you can calculate k and a different temperature.
CSTRs
V=
W
FA0 X A
−rA
W
W
Incorporates changing
volumetric flow rate
If the reaction is 1st order and it consists of liquids with constant density:
XA
k (1 − X A )
V
volume
τ= =
τ=
ν0
volumetric flow rate
τk
1+ τ k
Da = τ k =Damköhler number: ratio of kinetic effect to volumetric effect or
XA =
2nd order reaction:
ratio of reaction rate to dilution rate
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Exam1 Review
Page 2 of 10
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τ=
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XA
kC A0 (1 − X A )
2
Da = τ kC A0
XA =
1 + 2 Da − 1 + 4 Da
2 Da
For constant density and 1st order reaction:
dN A
dt
dC A
C A0 − C A + rAτ = τ
dt
FA0 − FA + rAV =
Let:
C
Cˆ A = A
C A0
t
tˆ =
τ
Nondimesionalize:
dĈ A
+ (1 + Da)Ĉ A = 1
dtˆ
Solve given Cˆ A = 0 at tˆ = 0
1
ˆ
1 − e− (1+ Da )t
Cˆ A =
1 + Da
(
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)
Tanks in series: 1st order reaction
C A, n =
(1+ Da )
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C A,0
n
W
Reactor Design Equations
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FA0 X A
−rA
dN A
Batch: rAV =
dt
N A = V [ A] If V changes, then V must remain in the differential
CSTR:
PFR:
V=
W
dX A −rA
=
Adz FA0
PBR: Pressure drop consideration
If
A( g ) → 2 B( g )
Use
ν 0 ρ0 = νρ
(conservation of mass)
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Exam1 Review
Page 3 of 10
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Introduce ideal gas law
⎛
⎞
m
= Pν = P = FT RT ⎟
⎜ nRT
ρ
⎝
⎠
Pm
FT RT
ρ=
FT = FT 0 (1 + ε X )
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Reactor volume:
positive order
reactions
VCSTR
FAo
− rA
VPFR (area under curve)
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XA
Figure 2. Levenspiel plot for a CSTR and a PFR for positive order reactions.
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Selectivity
CSTR
S
.B
φinst
W
W
W
C Af
Φ overall ΔC A
C A0
CA
Figure 3. Fractional yield versus concentration. Overall yield times concentration
difference shown for a CSTR.
CSTR
k1
k2
A ⎯⎯
→ P ⎯⎯
→C
k3
A ⎯⎯
→U
( k1 [ A] + k3 [ A])V = FA0 − [ A]ν
Non-ideal reactors
Residence time distribution E(t)
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Exam1 Review
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reactor
δ (t ) ⎯⎯⎯
→ C (t )
E(t) must have a pulse trace
E (t ) =
C (t )
∞
∫ C (t )dt
0
∫ E (t ) = 1
∞
0
∞
tm = ∫ tE (t )dt
0
Mean residence time, tm, for an:
Ideal CSTR: τ
Ideal PFR: τ
∞
σ 2 = ∫ ( t − tm ) E (t )dt
2
0
Variance, σ2, for an:
Ideal CSTR: τ2
Ideal PFR: 0
⎛
⎛ V
⎜ E (t ) = δ ⎜ t −
⎝ ν
⎝
⎞⎞
⎟⎟
⎠⎠
N
I
.
E
∞
∫ f (t )δ (t − t )dt = f (t )
Åproperty of a dirac delta function
e−t τ
V
S
0
0
0
For a CSTR, E (t ) =
τ
S
.B
Example 1
z
W
L
W
FA0
FA, FB, FC
W
Figure 4. Schematic of a PFR with inflow of A and outflow of A, B, and C.
r1
r2
A ⎯⎯
→ B ⎯⎯
→C
r1 = k1C A r2 = k2CB
YB =
moles of B produced
moles of A in
=
FB ( L)
FA0
Mole balance on B
1 dFB
= rB = r1 − r2 = k1C A − k2CB
Axs dz
FB = CBν 0
FA0 = C A0ν 0
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ν 0 dCB
= k1C A − k2CB
Axs dz
ν 0 dC A
= −k1C A
Axs dz
dC A −k1 Axs
=
dz
CA
ν0
−k A
ln C A = 1 xs z + φ
ν0
Initial condition at z=0 gives:
ln C A0 = 0 + φ
⎡ −k A ⎤
C A = C A0 exp ⎢ 1 xs z ⎥
⎣ ν0
⎦
⎡ −k A
kA
dCB k2 Axs
+
CB = 1 xs C A0 exp ⎢ 1 xs
dz
ν0
ν0
⎣ ν0
Axs
ν0
⎤
z⎥
⎦
N
I
.
E
z [ = ] time, call it τ
This is the time it takes for something to flow to the end of the reactor (of length z).
ν0
G
= V velocity
Axs
dCB
+ k2CB = k1C A0 e− k1τ
dτ
kτ
Integrating factor: e 2
d
⎡⎣CB ek2τ ⎤⎦ = k1C A0 e( k2 − k1 )τ
dτ
kC
k −k τ
CB ek1τ = 1 A0 e( 2 1 ) + φ
k2 − k1
V
S
S
.B
W
W
W
Initial condition: z=0, CB=0
B
kC
0 = 1 A0 + φ
k2 − k1
kC
CB (τ ) = 1 A0 ⎡⎣e− k1τ − e− k2τ ⎤⎦
k2 − k1
1 dFC
= + r2
Axs dz
1 dFA
= −r1
Axs dz
1 dFB
= r1 − r2
Axs dz
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1 ⎡ dFA dFB dFC ⎤
+
+
=0
Axs ⎢⎣ dz
dz
dz ⎥⎦
FA0 = FA + FB + FC
FC = FA0 − FA + FB
F
find
A
C
B
τ∗
τ
N
I
.
E
Figure 5. Graphs of flow rates of A, B, and C as a function of residence time.
(
)
dCB k1C A0
=
−k1e− k1τ * + k2 e− k2τ * = 0
dτ
k2 − k1
k1e− k1τ * = k2 e− k2τ *
ln k1 − k1τ * = ln k2 − k2τ *
k
ln 1 = ( k1 − k2 )τ *
k2
k
ln 1
k2
τ* =
k1 − k2
V
S
S
.B
W
W
dA( x)
A( x) 0
dx
L’Hopital’s rule: lim
= do lim
x → x* B ( x )
x → x*
dy
0
dB ( x)
dx
dx
W
Find τ* for k1=k2
1
k1 1 1
= =
k1 → k2
k1 → k2 1
k2 k
A L
A z
τ * = xs
τ = xs
lim τ * = lim
ν0
ν0
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L=
τ *ν 0
Axs
221
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= length of reactor
Example 2
If A → 2 B (Assume negligible pressure drop)
r = kC A
z
L
FA0
Figure 6. Schematic of a PFR.
dFA
= −rA = − kC A
dV
ν0 ↔ν
PV = nRT
( Ftotal = n , ν = V )
RT
−1
ν = Ftotal
= Ftotal Ctotal
P
Ftotal = FA + FB
FA = C Aν = y A Ftotal
dFA
FA
P
= − ky ACtotal = − k
dV
FA + FB RT
dFB
FA
P
= +2ky ACtotal = +2k
dV
FA + FB RT
N
I
.
E
V
S
S
.B
W
W
If P or T changes, you need other equations.
W
Derivation of E(t) for a CSTR
N 0δ (t )
Figure 7. Schematic of a CSTR.
No reaction:
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dN
= N 0δ (t ) −ν C
dt
N
= N 0δ (t ) −ν
V
dN ν
+ N = N 0δ (t )
dt V
⎛ν ⎞
t⎟
⎝V ⎠
d ⎛
⎛ν ⎞⎞
⎛ν ⎞
⎜ N exp ⎜ t ⎟ ⎟ = exp ⎜ t ⎟ N 0δ (t )
dt ⎝
⎝V ⎠⎠
⎝V ⎠
⎛ν ⎞
⎛ν ⎞
N exp ⎜ t ⎟ = N 0 ⋅ exp ⎜ 0 ⎟ = N 0 + φ
⎝V ⎠
⎝V ⎠
Integrating factor: exp ⎜
Initial condition: t=0, N=N0 Æ φ=0
⎛ ν ⎞
N = N 0 exp ⎜ − t ⎟
⎝ V ⎠
ν 1
=
V τ
⎛ −t ⎞
C = C0 exp ⎜ ⎟
⎝τ ⎠
C
E (t ) = ∞
∫ Cdt
N
I
.
E
V
S
S
.B
0
∞
∞
∫ Cdt = ∫ C e
−t
0
0
τ
0
−t
dt = C0 ⎡ −τ e
⎣⎢
−t
Ce τ e τ
=
E (t ) = 0
τ
C0τ
−t
τ
∞
⎤ = C ⎡ −τ ( 0 − 1) ⎤ = C τ
0 ⎣
0
⎦
⎦⎥ 0
W
W
W
Long-chain approximation
k2
1
ZZZ
X
E + S YZZ
→P+ E
Z ES ⎯⎯
k
k
−1
tRNA ⎯⎯
→E
k3
k4
E ⎯⎯
→ Deactivate
Enzyme propagates a long time before it is destroyed.
LCA: k3 tRNA = k 4 E
[
]
[ ]
st
(assume 1 order)
If there are other steps, add them into the equation
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k5
ES ⎯⎯
→ destruction
k3 [tRNA] = k4 [ E ] + k5 [ ES ]
Suppose there is a production term
k6
C ⎯⎯
→E
Add another term
k6 [C ] + k3 [tRNA] = k4 [ E ] + k5 [ ES ]
F
dX
= A0 , − rA = kC A
Adz −rA
This is a single differential equation in terms of X. Use for PFR with gas flow.
ρ0ν 0 = ρν
CA =
C A0 (1 − X ) T0 P
1+ ε X
P0T
N
I
.
E
V
S
S
.B
W
W
W
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10.37 Final Exam
Spring 2007
There are 4 problems. Pick any 3 of these problems to do and turn in. (If you turn in
solutions to 4 problems, we will not count the problem where you earned the lowest
score.) Please turn in your solution for each problem separately. Write your name on
every blue book or sheet of paper you turn in.
Problem 1. (100 points) A reactant stream is split to feed, in parallel, two CSTRs, one of
which is twice the volume of the other. The effluents from the two CSTRs are combined.
The reaction of interest follows a first-order rate law.
a) (50 points). How should the feed stream be split in order to maximize the total
reactant conversion in the combined effluent from the two CSTRs?
N
I
.
E
Small
CSTR
Feed
S
.B
V
S
Big CSTR
Effluent
W
W
W
b) (50 points). Using the same feed, what overall conversion would be obtained by
placing these same two CSTRs in series (i.e. the effluent of the first is the feed of
the second)? (Consider both possible topologies – one where the large CSTR
feeds the smaller, and one where the small CSTR feeds the larger.) Is this
conversion superior or inferior to the best case calculated in a)?
1
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Problem 2. (100 points) An appealing partial solution to the greenhouse gas problem is
to convert biomass into liquid fuels. Most of biomass is composed of linked C6 sugars
(C6H12O6) and C5 sugars (C5H10O5), which can be broken down by an enzyme secreted
by fungi. Ideally, the sugars could then be converted into ethanol or other liquids in
subsequent reactions.
This mechanism is proposed for the enzymatic breakdown:
C11H22O11 + Enzyme Æ Complex
(reaction 1)
Complex Æ Enzyme + C6H12O6 + C5H10O5
(reaction 2)
Reaction 1 is expected to be reversible under some conditions, but reaction 2 is
expected to be irreversible.
Experimental rate data on this enzyme from low-conversion batch-reactor
experiments can be fit to this expression:
d[C5H10O5]/dt = r = a[C11H22O11]0[Enzyme]0/{1 + b*[C11H22O11]0}
a = 2x104 liter/mole-second
b=108 liter/mole.
(Eqn 3)
where [C11H22O11]0 and [Enzyme]0 are the initial concentrations added to the mixture, i.e.
(moles added / volume of solution), not necessarily the actual concentrations of these
species in the beaker when the reaction is running, since some of the enzyme will exist in
the form of the complex.
N
I
.
E
V
S
(a) (20 points) Is the observed rate law (Eqn 3) consistent with the mechanism shown
above? If so, give an expression for b in terms of k1, k-1, and k2. If not consistent, explain.
S
.B
Suppose we could tether 10-9 mole of enzyme within a 4 mm diameter porous particle
without affecting the rate law (i.e. r is given by Eqn. 3). The diffusivity inside the porous
particles is 10-10 m2/s. In the bulk fluid D = 7x10-10 m2/s. Suppose we then filled a packed
bed reactor (internal diameter 2 cm, length 30 cm) with many particles like this, and
flowed an aqueous solution of C11H22O11 through the reactor at rate of 1 liter/minute. The
void fraction of the packed bed φ=0.4. For concentrations of C11H22O11 below 0.5 M, the
viscosity and density of the solution is essentially the same as that of water.
W
W
W
(b) (30 points) Write an equation for the Thiele modulus for this system, as a function of
[C11H22O11]. Over what range of [C11H22O11] is it reasonable to neglect diffusive
transport limitations?
(c) (30 points) If [C11H22O11] is always in the range where transport limitations are
negligible, what differential equation(s) should be solved to compute the conversion?
What Matlab program would you use to solve the equation(s) numerically? Write the
differential equation(s) in the form dY/dt = F(Y) required by the Matlab solvers.
(d) (20 points) Write (but do not attempt to solve) the differential equation(s) with
boundary conditions that would have to be solved to compute the effectiveness factor Ω
if [C11H22O11]bulk had a value outside the range specified in part (b). Are you missing any
data needed to calculate Ω?
2
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Problem 3. (100 points) A biosensor experiment is performed with a small amount of
immobilized protein and flowing soluble ligand. The observed rate constant during the
association phase, and the signal output at equilibrium, are given as a function of ligand
concentration in the table below. The equilibrium signal RUeq is proportional to the
concentration of protein/ligand complex at the surface. During the association phase the
signal follows the following function: RU = RU eq (1− e−kobs t ).
kobs (s-1)
0.00177
0.00186
0.00201
0.00245
0.00350
0.00612
0.0122
0.0280
0.0718
0.177
[L]o (nM)
0.14
0.84
2.1
5.60
14.00
35.0
84.00
210.00
560.00
1400.00
RUeq
1.63
9.34
21.5
47.1
82.5
118
141
155
161
163
N
I
.
E
A) (60 points) Determine Kd, kon, and koff. Are these data self-consistent?
V
S
B) (40 points) In a separate experiment, soluble protein (2 nM) and ligand (0.1 nM)
are mixed. At equilibrium, what fraction of ligand is complexed with the protein?
At what time following mixing will 95% of this equilibrium value be attained?
S
.B
W
W
W
Problem 4. (100 points total) “Clean Coal” technology is based on first converting the
coal into syngas (an H2 + CO mixture). The syngas can then be purified and used to make
clean synthetic fuels, or to generate electricity (with CO2 sequestration). The largest and
most expensive reactor in a “clean coal” plant is the gasifier. The main reactions are:
C(s) + ½ O2(g) Æ CO(g) (Reaction 1)
∆Hrxn1 = -110 kJ/mole
r1 = k1(T)[O2]2[H2]/([H2O] + a[H2])
k1(T)=(107 liter/mole-s)exp(-2000/T)
C(s) + H2O(g) Æ H2(g) + CO(g) (Reaction 2)
r2 = k2[O2][H2O]/([H2O] + a[H2])
a=0.0113 exp(10000/T)
∆Hrxn2 = +130 kJ/mole
k2 = 5x104 s-1 at T=1100 K
a=100
at T=1100 K
3
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Problem 4 (continued) The peculiar-looking rate laws come from a quasi-steady-state­
approximation treatment of the radicals which are the reactive intermediates in these
reactions. Reaction 1 is much faster than reaction 2 at low T, so the gasifier system can
be modeled as an adiabatic PFR (volume V1) where most of the O2 is consumed but the
H2 formation is negligible, followed by an isothermal CSTR (volume V2) where both
reactions occur.
Gasifier
Feed
Adiabatic PFR,
Volume V1
Only reaction 1
Isothermal
CSTR
Volume V2
T=1100 K
Reaction 1 & 2
syngas
The final 1100 K output stream coming out of the gasifier has no carbon and a negligible
concentration of O2(g), but large concentrations of CO, H2, and steam. Assume that the
heat capacity of the feed stream is 3 MJ/ton-Kelvin and that the heat capacity per ton
does not change significantly with temperature or the change in composition through the
reactions. The process is carried out at Ptotal=40 bar. The input stream is at 700 K, and
consists of 120 tons per hour of C(s), 120 tons per hour of steam, 64 tons per hour of O2,
and 0.02 tons per hour of H2.
N
I
.
E
V
S
S
.B
(a) (20 points) What is the molar flow rate of CO, H2, and H2O at the output? What is
the volumetric flow rate of the output?
W
(b) (25 points) How much heat must be transferred per second to maintain the output
at 1100 K? Is the heat flowing into or out of the reactor? How could you adjust
the composition of the feed to reduce the amount of heat transfer required, while
still maintaining the same Carbon feed rate and output temperature?
W
W
(c) (30 points) Write the equation(s) that should be solved to compute the conversion
of O2 in the adiabatic PFR, V1=1 m3, in a form that can be solved numerically by
Matlab. What Matlab program would you use to solve these equations?
(d) (25 points) What CSTR reactor size V2 is required so that 99% of the initial feed
O2 and 99% of the initial feed carbon will be consumed by the time the mixture
leaves the isothermal (1100 K) CSTR? Hint: How much carbon was consumed by
reacting with O2? So how much carbon must be consumed by reaction 2?
4
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10.37 first midterm
March 23, 2007
3 problems total
Problem 1. (40 points total)
The following liquid-phase hydration reaction occurs in a 10,000 L CSTR:
H 2O
A ⎯+⎯
⎯→ B
With a first-order rate constant of 2.5 x 10-3 min-1.
a) (20 points) What is the steady-state fractional conversion of A if the feed rate is
0.3 L/sec and the feed concentration CA,o = 0.12 mol/L?
b) (10 points) If the feed rate suddenly drops to 70% of its original value and is
maintained there, what is the fractional conversion of A after 60 minutes, and
what is the new steady state fractional conversion?
N
I
.
E
c) (10 points) What is the ratio of the steady-state productivity (moles/time) of B for
case b) relative to case a)?
V
S
S
.B
Problem 2. (30 points)
W
W
W
A 600 L tank reactor gives 75% conversion for a first order irreversible reaction.
However, the paddle turbine motor is underpowered and so the tank is not well stirred. In
fact, a pulse-tracer experiment to determine the residence time indicates that
approximately 400 L of the tank may be considered as a dead volume that does not
interact appreciably with the input or output streams. If you replace the stirrer with one
sufficiently strong to obtain complete mixing throughout the reactor volume, what
conversion will be obtained?
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Problem 3. (30 points)
In a biorefinery, a steady-state PFR operating at 600 K is used to convert cellulose
oligomers CnH2nOn suspended in water into “syngas” (a mixture of CO and H2 that can be
easily converted into many different fuels or chemicals):
C24H48O24(g) Æ 24 CO(g) + 24 H2(g)
The cellulose oligomers are introduced as a 20 wt% slurry in liquid water; 60 grams of
the slurry are introduced every second. At this high temperature, it is a good
approximation to assume all the species in the reactor are in the gas phase. The reaction
rate law has been measured at 600 K under these conditions to be:
r = k [CnH2nOn(g)] / {1 + a [CO(g)]}
where k= 0.01 s-1 and a = 10 liter/mole
N
I
.
E
The pressure in the reactor is 20 atm. There is no significant pressure drop in the reactor.
The cross-sectional area inside the reactor is 10 cm2, and the reactor is 3 m long.
V
S
Write the differential equation(s) that would have to be solved to predict the
moles/second of H2 coming out of the reactor. Write the equation(s) in the standard ODE
format:
S
.B
dY/dz = f(Y)
W
where Y(z) are the unknown(s) you want to compute, and there are no other
unknowns on the right hand side of the equation(s). Please do not attempt to solve the
equation(s).
W
W
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230
Solutions.
Problem 1.
a) For first-order reaction kinetics
10000 L
Da
kτ
0.3L / sec =0.58
XA =
=
=
1 + Da 1 + kτ 1 + 2.5 × 10-3 min −1× 10000 L
0.3L / sec
2.5 × 10-3 min −1×
b) Consider the non-steady state design equation for CSTR, where we can have
FA0 − FA + rAV =
dN A
dt
Since in liquid phase with constant density, we have
C A0 − C A - kC Aτ = τ
dC A
dt
or equivalently
dCA 1 + kτ
C
CA = A0
+
dt
τ
τ
N
I
.
E
with the initial conditions CA,t=0=(1-0.58)×0.12 mol/L=0.05 mol/L, new τ=10000L/(0.7×0.3*60
L/min)=793.7 min
Therefore, integrate this equation we can have
C A (t ) =
C A0
1 + kτ
1 + kτ
{1 − exp[−
t ]} + C A, t = 0 exp(−
t)
1 + kτ
τ
τ
X A (t ) = 1 −
V
S
S
.B
Therefore the conversion at time t is
C
1 + kτ
1
1 + kτ
t ]} − A, t = 0 exp[−
t]
{1 − exp[−
τ
τ
C A0
1 + kτ
W
After 60 min of changing flow rate,
W
X A (60 min) = 1 −
W
C
1 + kτ
1
1 + kτ
t ]} − A, t = 0 exp[−
t ] = 0.598
{1 − exp[−
τ
τ
C A0
1 + kτ
The new steady state fractional conversion is
10000 L
2.5 × 10-3 min −1×
Da
kτ
0.3L / sec× 0.7 =0.665
XA =
=
=
1 + Da 1 + kτ 1 + 2.5 × 10-3 min −1× 10000 L
0.3L / sec× 0.7
c) The steady state productivity (moles/time) of B is FB, and the ratio of that in b) vs. that in a) is
B
FB, b 0.3L/sec × 0.7 × 0.665
=
= 0.803
FB, a
0.3L/sec × 0.58
Therefore, the productivity will be decreased by lowering the flow rate even though a higher
conversion is to be achieved.
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Problem 2.
We know for first order reaction, conversion XA has the following relation with rate constant k,
CSTR reactor volume V and volumetric flow rate v
V
v
XA =
V
1+ k
v
k
When we have a dead volume (denoted as subscript 1) which does not interact with the input and
output streams, we can deduce this amount of volume from the overall reactor volume.
V1
v
X A1 =
V
1+ k 1
v
k
We
are
told
V1=600L-400L=200L,
XA1=0.75,
therefore
X A1
k
==
=0.015.
v
V1 (1 - X A1 )
we
can
calculate
Then for the well-stirred reactor, V=600L
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V
S
V
v =0.9
XA =
V
1+ k
v
k
S
.B
W
W
W
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232
Problem 3.
For a PFR reactor, the design equation is
dX celluose
-r
= S celluose
dz
Fcelluose, in
Now S=10cm2, length=2m, Fcelluose, in=vin[cellulose]in, in order to know v, we have to convert 60
grams of slurry into total moles of cellulose and water per second, that is
n total, in =
=
60 gram/sec × 20% 60 gram/sec × (1 − 20%)
+
=
MWcellulose
MWwater
60 gram/sec × 20% 60 gram/sec × (1 − 20%)
= 2.683 mol/sec
+
720 gram/mol
18 gram/mol
Using ideal gas law to get the volumetric flow rate:
v in =
n total, in RT 2.683 mol/sec × 8.314 J/mol/K × 600K
=
= 6.61 liter/sec
20 atm
P
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Also
20%
y
P
20 atm
720 gram/mol
[cellulose]in = cellulose,0 =
= 2.52 mol/m3
20
%
1
−
20
%
RT
(
+
) 8.314J/(mol K) × 600K
720gram/mol 18gram/mol
V
S
S
.B
where ycellulose,0 is the molar fraction of cellulose in the inlet.
In this problem, since the gas phase reaction creates more molecules, the volumetric flow rate is
not a constant.
W
W
W
- rcelluose = k
[cellulose]
1 + a[CO]
where the concentrations should be expressed (constant pressure and temperature are assumed)
[cellulose] = [cellulose]in
(1 - X)
(1 + εX)
[CO] = [H 2 ] = [cellulose]in
24X
(1 + εX)
and
Here ε=(24+24-1)ycellulose,0=47*0.0062=0.292.
Now the final molar flow rate of H2 is
vfinal[H2]=24vin[cellulose]inXf
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10.37 Exam 2 25 April, 2007
100 points
Problem 1: 35 points
A protein and ligand bind reversibly with Kd = 10 nM . The association rate constant
kon= 2x104 M-1s-1. The two species are mixed at an initial protein concentration of 3 nM
and an initial ligand concentration of 0.2 nM.
a) At equilibrium, what fraction of the ligand will be complexed with protein? (15 points)
b) At what time will the fraction of ligand in complex reach 95% of the equilibrium
value? (20 points)
Justify any assumptions you make to simplify equations.
Problem 2: 30 points
A surface-catalyzed reaction follows Rideal-Eley kinetics as follows:
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kA
⎯⎯⎯
→ AS
A + S ←⎯⎯
⎯
k− A
k1
AS + A ⎯⎯
→ A2 + S
Where A and A2 are in the gas phase, S is a reactive site on the surface, and AS is a
V
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molecule of A adsorbed to a reactive site.
S
.B
Assuming that:
adsorption of A is at rapid equilibrium
reaction of AS with A is rate-limiting
desorption of A2 is very rapid
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W
Derive the steady-state rate law for production of A as a function of the concentration of
2
W
A and the total initial reactive site density So .
Problem 3: 35 points
It is desired to make a product X-Y via this reaction:
X-OH + Y-H → X-Y + H2O
o
o
An equimolar feed of liquid X-OH and Y-H at 25 C are fed to a CSTR. At 25 C, where all 4
material species are liquids, the heat of reaction ∆Hrxn=-200 kJ/mole, and the heat capacity of
o
each liquid-phase species is 4 kJ/(kg C ). The molecular weight of X-OH is 150 g/mole, and the
molecular weight of Y-H is 100 g/mole. The temperature inside the reactor (T) is controlled by
putting the reactor in thermal contact with a fluid flowing over the outside of the reactor at
temperature Ta. To a good approximation, the heat transfer rate (Q, in watts) from the fluid
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flowing over the outside the reactor to the contents of the reactor is given by the linear
expression: Q = UA(Ta-T)
o
a) If the reaction is carried out with the reactor at steady-state at the inlet temperature of 25 C, is
T greater than, less than, or equal to Ta? (5 points)
o
b) When running the reactor at T = 25 C to 50% conversion, the productivity is unacceptably
low. To try to accelerate the reaction, it is decided to increase the steady-state reactor
o
temperature to T = 105 C. At this temperature, all of the H2O formed evaporates, but the
o
other species are still liquids. The heat of vaporization of H2O at 105 C is +40 kJ/mole.
o
o
When T=105 C, the reaction runs to 50% conversion 10x faster than it did at 25 C, so we
o
increase the flowrates until the reactor is making 10x as much product as it did at 25 C (still
at 50% conversion). When we achieve the new steady-state high-productivity operation at
o
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105 C, will the magnitude of Q (i.e. |Q|) be larger, smaller, or the same as it was when we
o
were operating at 25 C? At this steady-state condition, is T greater than, less than, or equal
to Ta? (20 points)
V
S
S
.B
c) Since operating hot improved our productivity, but conversion is still pretty low, the operator
o
tries to improve things by cranking up the temperature, preheating the inlet streams to 185 C
and increasing Ta. For good measure the operator simultaneously cranks up the reactor
W
pressure from 1 bar to 100 bar; at this high pressure all the species remain as liquids. (The
W
o
reactor is safe at this condition, and even up T = 300 C.) Curiously, the conversion and
W
productivity of the reactor do not increase under these severe conditions, instead they
decrease. Propose an explanation for this experimental observation. (5 points)
d) Your manager gives the operator who turned up the temperature (without doing any
calculations first) a formal reprimand, saying the operator is probably lucky
that the conversion went down instead of increasing. Why do you think the
manager was happy that conversion was low instead of increasing a lot? (5 points)
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological
Reaction Engineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts
Institute of Technology. Downloaded on [DD Month YYYY].
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235
10.37 Exam 2
25 April, 2007
100 points
Problem 1: 35 points
A protein and ligand bind reversibly with K d = 10 nM . The association rate constant
kon = 2x10 4 M−1s-1 . The two species are mixed at an initial protein concentration of 3 nM
and an initial ligand concentration of 0.2 nM.
a) At equilibrium, what fraction of the ligand will be complexed with protein? (15 points)
b) At what time will the fraction of ligand in complex reach 95% of the equilibrium
value? (20 points)
Justify any assumptions you make to simplify equations.
P + L ←→ C
k off [P] eq [L] eq
Kd =
=
k on
[C] eq
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Using a batch reactor mole balance and looking at the reaction stoichiometry, it is easy to
see that given the initial conditions any unit of complex formed takes away a unit of
protein and ligand:
[P] + [C] = [P] 0
[L] + [C] = [L] 0
Using these in the equilibrium equation we can get a quadratic equation in [C]eq.
([P]0 − [C]eq )([L]0 − [C]eq )
Kd =
[C]eq
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.B
W
0 = [C]eq − ([L]0 + [P]0 + K d )[C]eq + [P]0 [L]0 = 0
2
[C]eq =
W
([L]0 + [P]0 + K d ) ± ([L]0 + [P]0 + K d )2 − 4[P]0 [L]0
W
=0
2
Reject the positive root, it is too large (larger than the initial amount of ligand and
protein).
[C]eq = 0.0456nM
[C]eq
= 0.228 = 23%
[L]0
At this point, it is interesting to look at different approximations to the expression.
Good Approximation: [P]0 >> [C]
This leads to
([P]0 )([L]0 − [C]eq )
Kd ≈
[C]eq
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[C]eq =
[L]0 [P]0
K d + [P]0
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236
0.462nM
[C]eq
= 0.231 = 23%
[L]0
Bad Approximation: [L]0 >> [C]
([P]0 −[C]eq )([L]0 )
Kd ≈
[C]eq
[C]eq =
[L]0 [P]0
K d + [L]0
= 0.0588
[C]eq
= 0.294 = 29%
[L]0
The error in [C]eq of the bad approximation is about 30% of the true answer, whereas the
good approximation is only off by about 1%.
By noticing that the “good” approximation is a good approximation, the dynamic
equation becomes easier to solve. (As an aside, an even better approximation would be
just to neglect the second order term that is O([C]eq2).)
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Start with the full dynamic equation:
d[C]
= k on [L][P] − k off [C] = k on ([L]0 −[C])([P]0 − [C]) − k off [C]
dt
Make an appropriate approximation:
([P]0 −[C]) ≈ [P]0
d[C]
≈ k on [L]0 [P]0 − k on [P]0 [C] − k off [C]
dt
Rearrange and solve using the integrating factor:
d[C]
+ [C](k on [P]0 + k off ) = k on [L]0 [P]0
dt
d
([C]exp (kon [P]0 + koff ) t ) = kon [L]0 [P]0 exp (kon [P]0 + koff ) t
dt
([C ] exp [(k on [P]0 + k off ) t ]) = k on [L]0 [P]0 exp ((k on [P]0 + k off )t ) + I .C.
(k on [P]0 + k off )
V
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W
W
[
[C] =
W
[ L ]0 [ P ] 0
([ P]0 + K d )
]
[
[
]
+ I.C.exp − (k on [P]0 + k off ) t
]
Using the initial condition, [C](t = 0) = 0, we find the integration constant to be:
[L]0 [P]0
I.C. = −
([P]0 + K d )
Hence,
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[C](t) =
[L]0 [P]0
([P]0 + K d )
{1− exp[− (k
on
237
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]
[P]0 + k off ) t }
The equilibrium value is clearly the value when t gets large.
[C](t) = [C]eq {1− exp(− (k on [P]0 + k off ) t )}
In order to find the point at 95% of the equilibrium value, rearrange and solve for the
time when [C]/[C]eq=0.95:
[C](t*)
[C]eq
[
]
= 0.95 = {1− exp − (k on [P]0 + k off ) t * }
[
]
0.05 = exp − (k on [P]0 + k off ) t *
t* =
(k
− ln(0.05)
on [P]0 + k off )
=
− ln(0.05)
(kon [P]0 + kon K d )
=
3.00
2x10 nM s [3nM +10nM ]
−5
= 11500s
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−1 −1
t* = 11500s ≈ 3.2h
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S
.B
Problem 2: 30 points
A surface-catalyzed reaction follows Rideal-Eley kinetics as follows:
kA
→
AS
A + S ←
k− A
W
W
k1
AS + A 
→ A2 + S
W
Where A and A2 are in the gas phase, S is a reactive site on the surface, and AS is a
molecule of A adsorbed to a reactive site.
Assuming that:
• adsorption of A is at rapid equilibrium
• reaction of AS with A is rate-limiting
• desorption of A2 is very rapid
Derive the steady-state rate law for production of A2 as a function of the concentration of
A and the total initial reactive site density So .
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
C 
rAd = k A  C A C v − AS 
KA 

where K A =
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238
kA
k−A
rs = k1C AS C A
rAd
≈0
kA
Adsorption is at rapid equilibrium, so
C ACv =
C AS
KA
⇒
C AS = K A C A C v
Overall site balance in terms of So:
S o = C v + C AS = C v + K A C A C v
⇒
Cv =
So
1 + K AC A
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Given that the surface reaction is the rate limiting step, and the stoichiometric coefficient
is +1 for A2, the rate of production of A2 is:
rA' 2 = rs = k1C AS C A = k1 K AC v C A2 =
k1 K A S o C A2
1 + K AC A
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S
S
.B
Problem 3: 35 points
W
It is desired to make a product X-Y via this reaction:
W
X-OH + Y-H → X-Y + H2O
An equimolar feed of liquid X-OH and Y-H at 25oC are fed to a CSTR. At 25oC, where
W
all 4 material species are liquids, the heat of reaction ∆Hrxn=-200 kJ/mole, and the heat
capacity of each liquid-phase species is 4 kJ/(kg Co). The molecular weight of X-OH is
150 g/mole, and the molecular weight of Y-H is 100 g/mole. The temperature inside the
reactor (T) is controlled by putting the reactor in thermal contact with a fluid flowing
over the outside of the reactor at temperature Ta. To a good approximation, the heat
transfer rate (Q, in watts) from the fluid flowing over the outside the reactor to the
contents of the reactor is given by the linear expression:
Q = UA(Ta-T)
a) If the reaction is carried out with the reactor at steady-state at the inlet
temperature of 25oC, is T greater than, less than, or equal to Ta? (5 points)
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239
For an exothermic reaction, to maintain the products at the same temperature as the
reactants one must remove heat. So T must be greater than Ta, i.e. Ta must be below
room temperature.
b) When running the reactor at T = 25oC to 50% conversion, the productivity is
unacceptably low. To try to accelerate the reaction, it is decided to increase the
steady-state reactor temperature to T = 105oC. At this temperature, all of the H2O
formed evaporates, but the other species are still liquids. The heat of vaporization
of H2O at 105oC is +40 kJ/mole. When T=105oC, the reaction runs to 50%
conversion 10x faster than it did at 25oC, so we increase the flowrates until the
reactor is making 10x as much product as it did at 25oC (still at 50% conversion).
When we achieve the new steady-state high-productivity operation at 105oC, will
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the magnitude of Q (i.e. |Q|) be larger, smaller, or the same as it was when we
were operating at 25oC? At this steady-state condition, is T greater than, less than,
or equal to Ta? (20 points)
V
S
S
.B
If the reaction ran 100% to completion, it would release (200 kJ/mole)(1 mole/0.25
kg entering the reactor)=800kJ/kg entering the reactor. Since we only have 50%
conversion, the chemical heat release is half as much, 400 kJ/kg. Heating the feed
from 25 C to 105 C requires: (80 degrees)(4 kJ/kg) = 320 kJ/kg entering the reactor.
Evaporating the water formed by the reaction requires an additional (40 kJ/mole
H2O)(0.5 mole H20/ 0.25 kg entering reactor) = 80 kJ/kg. Since the heat release and
the heat required to warm up the mixture and evaporate the water balance, Q=0 now.
In contrast, when we ran the reactor at 298 K, we had to remove heat at a rate of 200
kJ/kg. So the magnitude of Q is much lower now than before, and T should
approximately equal Ta.
W
W
W
c) Since operating hot improved our productivity, but conversion is still pretty low,
the operator tries to improve things by cranking up the temperature, preheating
the inlet streams to 185oC and increasing Ta. For good measure the operator
simultaneously cranks up the reactor pressure from 1 bar to 100 bar; at this high
pressure all the species remain as liquids. (The reactor is safe at this condition,
and even up T = 300oC.) Curiously, the conversion and productivity of the reactor
do not increase under these severe conditions, instead they decrease. Propose an
explanation for this experimental observation. (5 points)
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240
It is possible that we are running into equilibrium limitations on the reaction. For
exothermic reactions, Keq decreases with increasing temperature. So if we are
equilibrium limited at high T, we would expect the conversion to decrease as T
increases. (Note that by increasing the pressure so dramatically, the operator
prevented most of the H2O from evaporating, hence the concentration of H2O in the
liquid phase is probably much higher now than it was in the 105 C case where the
water evaporated; this contributes to the equilibrium limitation).
d) Your manager gives the operator who turned up the temperature (without doing
any calculations first) a formal reprimand, saying the operator is probably lucky
that the conversion went down instead of increasing. Why do you think the
manager was happy that conversion was low instead of increasing a lot? (5 points)
Exothermic reactions can “run away” if T goes high enough, i.e. past a certain point,
the steady-state conversion will suddenly jump from a low number to a very high
conversion, releasing essentially the whole exothermicity, and jumping the T to a
temperature above the safety limits of the reaction vessel. In the present case, if the
reverse reaction were negligible the reaction would release 800 kJ/kg, enough to
increase the temperature inside the reactor by up to 200 degrees, so it could have
exceeded the T=300 C safety limit. The results could have been fatal to the operators
or anyone else nearby. But fortunately in this case it was lucky that high conversion
was not achievable due to the small value of the equilibrium constant.
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V
S
S
.B
W
W
W
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241
Four major areas (√: covered in the following example problems):
Non-isothermal reactors (√)
Biological reactors (√)
Ligand/receptor binding kinetics
Surface reactions/catalysis kinetics
Go over the first problem in PS7 to review the ligand/receptor binding kinetics and
the first problem in PS8 to review the surface reactions/catalysis kinetics.
Problem 1.
The irreversible liquid phase reaction A →R + S is carried out in a CSTR. The reaction is
first order in A. The feed stream is available at a temperature of 298 K.
-4 -1
k = 1.7 x 10 s at 298 K.
3
Ea = 41.87 x 10 kJ/kmol
3
ΔHR(298) = -167.5 x 10 kJ/kmol
3
CA0 = 2.0 kmol/m (Feed is pure A)
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3
V = 0.5 m
3
ρ = 1050 kg/m
Cp = 4.19 kJ/kg/K
These values can be considered to be constant over the used interval of concentration and
temperature. The CSTR is made of carbon steel and weighs 800 kg.
Cp,steel = 502.4 J/kg/K
Calculate:
a) Conversion and heat duty for an isothermal reactor operating at 298 K.
b) Conversion and reactor temperature for an adiabatic reactor with inlet temperature
of 298 K.
c) Conversion and preheating temperature for an adiabatic reactor with a reactor
temperature of 363 K.
d) Conversion and heat duty if the reactor is operated non-adiabatically without
preheating and at a temperature of 363 K.
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Problem 2.
Consider an organism which follows Monod equation of growth with μmax = 0.5 h-1 and Ks = 2 g/L.
a)
b)
In a continuous perfectly mixed vessel at steady state with no cell death, if the substrate
concentration in the feed is Sfeed = 50 g/L, the yield Y = 1 (g cells / g substrate), what
dilution rate D gives the maximum volumetric productivity?
For the same dilution rate as part (a) using tanks of the same size in series, how many
vessels will be required to reduce the substrate concentration to less than 1 g/L?
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242
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.B
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W
W
Cite as: William Green,
Jr., and K. Dane Wittrup, course
materials for 10.37 Chemical
and Biological Reaction Engineering,
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
www.onlineeducation.bharatsevaksamaj.net
www.bssskillmission.in
243
N
I
.
E
V
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S
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W
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Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering,
www.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
www.onlineeducation.bharatsevaksamaj.net
www.bssskillmission.in
244
N
I
.
E
V
S
S
.B
W
W
W
Cite as: William Green,
Jr., and K. Dane Wittrup, course
materials for 10.37 Chemical
and Biological Reaction Engineering,
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].
www.onlineeducation.bharatsevaksamaj.net
www.bssskillmission.in
245
N
I
.
E
V
S
S
.B
W
W
W
Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
www.bsscommunitycollege.in
www.bssnewgeneration.in
www.bsslifeskillscollege.in
2007. MIT OpenCourseWare
(http://ocw.mit.edu), Massachusetts
Institute of Technology.
Downloaded on [DD Month YYYY].