biot numbers by david adrian - bssve.in numbers by david adrian ... consider the effect of biot...

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).

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

q r exterior = h(Ts − T∞ )n r

q r interior = −kT ∇T surface

The variables are defined as follows: q is the heat flux, h is the heat transfer coefficient in the fluid, T is the temperature, n r 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 /δT in systems without any interfacial reactions or phase changes. The variable δT is the thickness of the thermal boundary layer.)

Since our system is spherically symmetric: q r exterior = h(Ts −T∞ )e r r

r eq r interior = −kT

dT r

dr surface

dT− kT = h(Ts −T∞ )

dr surface

* I hope it is a review! TAs should be knowledgeable about this stuff if you have questions.

Cite as: David Adrian, 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].

www.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in

1www.onlineeducation.bharatsevaksamaj.net www.bssskillmission.in

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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 4πR2 3r

η = r / L = r = r =V (4 /3)πR3 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.

Using our definitions of dimensionless length and temperature, our Robin boundary condition becomes:

− kT

(T0 − T∞ ) dθ = h(Ts − T∞ )

L dη surface

dθ hL− = θ = N Bi θsurface surfacedη surface kT

hL L / kT "internal diffusion resistance"N Bi = = =

kT 1/ h "external convection resistance"By definition, our dimensionless temperature is at most 1. Consider the effect of Biot number on the problem.

Case 1: NBi<<1 dθ

− = NBiθ surfacedη surface

Since θ is at most one, in order for the equation to be true the surface gradient in surface

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 must be very small. However, since we scaled the derivative

surface

dθ properly, should be O(1). This means that θ is very small. Recall that if θ

surfacedη 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.

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).

The analogous case of a catalyst particle in a reactor fluid gives us the Robin boundary condition: W r

exterior = km (cs − c∞ )e r r

r dc r eWinterior = −De rdr surface

dc − D = km (cs − c∞ )e

dr surface

c − c∞θ = c0 − c∞

SA 4πR2 3rη = r / L = r = r =

V (4 /3)πR3 R

(c0 − c∞ ) dθ − D = km (cs − c∞ )e

L dη surface




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dθ k L − = m θ = N θ

surface Bi,m surfacedη Dsurface e

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.).

k LN Bi,m = m

De

D fluidkm = (stationary interface with no interface reaction) δm

N = D fluid L

Bi,m D δe 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 /δm is typically greater than one.

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.

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" NBi,m = = = >>1

De δm δm / D fluid "external diffusion resistance" 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.

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 = km (cb − cs ) ≈ kmcb because cs << cb when external limitations prevail.




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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.

FA

Fo

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 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)

[ ]

dNA = −F F +Gdt Ao A A

molesGA =sec

If homogeneous, G rA A= V

else, G rA A= ∫ dVwhere V is volume and rA is the rate of A created or destroyed (moles/sec/volume). The mass balance can also be written as:

d conv diffusionectionρA = ∇v Dρ ρ2A + ∇ A A+ r

dtwhere v is velocity, D is the diffusion coefficient, and ρA is the molar density of A (moles/volume). Example:

Cite as: William Green, Jr., 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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A+B→C+D

[ ][ ] [ ][ ] [ ][ ]mole A, ,literA Cr k A B r k A B A= − = + =

where k is a measurable rate constant that changes with respect to T and P but remains constant with respect to changing concentrations.

If [ ][ ]1r k A B= , then 1

,1 1A Ar rν=−

=

where ,1Aν is the stoichiometric coefficient for species A in reaction 1. Likewise, 1

,1 1C Cr rν=+

=

If there are n reactions involving species “A” then

,1

n

A Ai

r rν=

=∑ i i

[ ][ ]. - ( ( , , , )) ( , , , ) ( , , , )

( ) ( , , , )A AVol k T x y z t A x y z t B x y z t

G t r x y z t dxdydz= ∫∫∫

[ ] [ ][ ] *only true if there are no flows!

AN V

d Ak A B

dt= − →

A, assume r is true throughout volumeAA A

dN r dV r Vdt

= ≈∫

If the system is homogeneous and there are no flows:

AA

dN r Vdt

=

Therefore:

iff homogeneous, no flow, constant VAA

Nd rdt V⎛ ⎞ =⎜ ⎟⎝ ⎠

Extent of Reaction [ ]moles, extent of rxn.ξ =

[ ]moles , rate of extent of rxn.sec

ξ =

A B C D+ → +

10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 1 Prof. William H. Green Page 2 of 3 Cite as: William Green, Jr., 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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,initially

,initially ,

, if there is one reaction involving A

, if there are several reactionsA A

A A A n nn

N N

N N

ξ

ν ξ

= −

= −∑

, n is the reaction number

, A is the species

n n

A A

r dV

G r dV

ξ =

=

∫∫

Conversion

A B C D+ → + ,initial

,initial

(dimensionless)A AA

A

N NX

N−

=

,initial

( 1 since rxn is 1:1)CC

A

NXN

= ≈

A B C DA U

+ → +⎛ ⎞⎜ →⎝ ⎠

⎟ Selectivity is good if , bad if . A C→ A U→

*May worsen as rxn. goes on ( slows, keeps going) A C→ A U→




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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:

,AAo A A A A

dN F F G G r dVdt

= − + = ∫

Example:

CBA 2→+ *Reactions are reversible (often will neglect reverse)

AoA

Ao

N NXN

−= A Ao

AAo

F FXF

−= A

2C

CAo

FXF

=

FAo

FBo

C,A,B A,B

Figure 1. A reactor with reactants A and B constantly flowing in and product C and unused reactants A and B flowing out.

[ ] [ ]

[ ]

[ ][ ]

literssec

moles A flowing insec

moles A flowing outsec

1

Ao in oinput o

A ooutput

outoutput

inin

F A

F A

A vX

A v

= =

= =

= −

ut

v A v

v

=

Closed reactor, const. V Detailed balance- all steps in equilibrium, total system

[ ] [ ][ ]

[ ][ ][ ]

2

A

eq

d Ar k A B

dtC

KA B

= = −

=

when [ ][ ] [ ]2

eq

CA B

K= the rxn. stops -> rA=0




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[ ][ ] [ ]2

, , forward-reverse in one expression

same!

forA rev

eq eq

kCr k A B k

K K

⎛ ⎞⎜ ⎟= − − =⎜ ⎟⎝ ⎠

Catalysis

[ ][ ] [ ]( ),

may be includedi.e. when B is

very small

catalystA forward catr k A f B=

Rate limiting step determines the kinetics (slow step). The kinetics are insensitive to [B] because B is not part of this slow step.

[ ][ ][ ]

[ ] [ ] [ ][ ][ ]

2

, ,

2

, catalyst

A reverse A forwardeq

A net cateq

Cr r

K A B

Cr k A

K A B

=

⎛ ⎞⎜ ⎟= − −⎜ ⎟⎝ ⎠


Stoichiometric coefficient

, ,G RT o o

eq f products f reactantsK e G G GΔ Δ Δ Δ−= = −

A B C→ +

eqK [ ][ ][ ] [ ] moles

literB C

A= =

CK

[ ][ ]

[ ]

Partial Pressures

Ideal Gas:

B Ceq oo

A

B Cp p KPp P ART

n PpV nRTV 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.

, ,A A initial A i iN N ν ξ= + ∑

Standard state

Extent of rxn.

, ,1

rxnN

C C initial C i ii

N N ν ξ=

= + ∑



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column columnvector vector matrix vector

oN N ν ξ= + i

Do columns span space? Often no, limits on what is achievable.


( ) ( )d N F Ndt

=

Figure 2. A plot of the reaction trajectory. ξ is the extent of reaction.

Conservation Laws Conserve atoms

4 2 2 4 22CH O C H 2H O+ → +

constant moles

const.A A B B C CC N C N C N+ + =

4 2 4

4 2 4 2

,

,

:1 2

: 4 4 2CH C H C initial

CH C H H O H initial

C N N N

H N N N N

× + × =

× + × + =

ξ=Ν

NC

Αο

ξ

t=10

t=0.01

ξ1=0

NB

NA

Reaction trajectory- Plot state of system



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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: − =r fA A(C ,C B ,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

X

X

X

X X

CA

-rA

X=first order A Ar kC− =

=second order 2A Ar kC− =

=zero order Ar k− =

Figure 1. Rate versus concentration graphs for zero, first, and second order reactions.

Enzymes-Biological Catalysts S=Substrate E=Enzyme

Cite as: 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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Reaction Progress

Ener

gy

S

P

Activation energy decreases with E (enzyme)

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.

Time

CS Initial Rate

0

SS

t

dCrdt =

− = −

Figure 3. The slope of the concentration versus time curve at time = 0 can give the initial rate.

Michaelis Menten

X

CS

-rS

X

X X

X X

X X X X X

~1st order

~0th order

What gives change from 1st order to 0th order?

Figure 4. Michaelis-Menten kinetics. Hypothesized Mechanism: -Encounter complex ES formed -Irreversible reaction occurs -Rapid release of product from complex -Assume rapid equilibrium is reached in the formation of ES

In more complex cases, these are not always true

10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 3 Prof. K. Dane Wittrup Page 2 of 6 Cite as: 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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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

1

cat

k

k

k

E S ES

ES E P−

+

⎯⎯→ +

Steady state assumption on ES 0ESdCdt

→ =

Material Balance on ES:

0ESdC k C C k C k Cdt − (steady state) 1 1E S ES cat ES

= +CE CES

1 ,(CE E0 − C S )CS − CES (k−1 + kcat )

Free enzyme concentration ≠ C E ,0

because C CE S,0 ,0

C C≈

= − − ≈


,0

0E

k=

C

,0s S

Solve for CES (steady state solution)

,0

1

1

,0 max

1

1

max ,0

1

1

E SES

catS

p cat E S SS p cat ES

cat m SS

cat E

catm

C CC k k C

kdC k C C V Cr r k C k kdt K CC

kV k C

k kKk

−

−

−

=+

+

− = = = = =+ ++

= =

+= =

maximum reaction velocity

Michaelis constant

This model fits the data:

-As C K , S mmax

S Sm

Vr CK

− →

-As C K , − → S m maxSr V



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X

CS

-rS

X

X X

X X

X X X X The rate flattens out because all of the enzyme is bound.

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.

Growth Kinetics A population of single cells, growing without limitations N=#cells per volume

,dN Ndt

μ μ= = constant, specific growth rate

0tN N eμ= − exponential growth while true, "Malthusian growth"

Growth Stops -Run out of nutrients -Accumulate toxic byproducts

Nutrient Effect on μ (Monod)

max S

S S

CK Cμμ =

+

This expression is purely empirical (no mechanistic meaning). Even the fit is not that good.




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dN μmaxC= S Ndt KS S+ C

dCS −1

= μNdt Yx s

We will return to this when we look at bioreactors

μ

CI

I=inhibitor or toxin

( )

μ ∝ e−kCI

1μ ∝ 1− kCI

kμ ∝ I

k CI I+

All forms fit the data

Figure 6. Growth rate versus concentration of inhibitor. Many functional forms fit the curve.

CS

μ

X X

X X

X X

X X X X

KS

CS=concentration of growth limiting substrate, often glucose

Figure 7. Growth rate versus concentration of growth limiting substrate.




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Cell Growth as a Chemical Reaction Aerobic growth:

→2 2 2

Carbon source + O + Nitrogen Source Biomass + Byproducts + H O +CO

CH1.8O0.5N0.2

(for a typical microbial culture)

A yield coefficient can be defined:

A BAYB

ΔΔ

=

A=biomass, byproducts, CO2, heat (any of these) B=carbon source or oxygen

For glucose, 0.6 0.1x sY ≈ ±g biomass

g glucose

For oxygen, 2

1.9 0.7x OY ≈ ±2

g biomass

g O




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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

Figure 2. Two electrons in an orbital have opposite spin.

Bond Forming

H + +

0 e 2 e

empty orbital

H

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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(http://ocw.mit.edu)

Boltzmann Distribution

−E k T R ( ) ∼ Bp E e B k =

NA

E ≫ k T → very unlikelyB

E + E > EA B ActivationBarrier

A B

Small fraction willcollide correctlyand react

−E k T a B( ) ∼k T Ae

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

Example:

RC

O

OR + H2O RC

O

OH + ROH

(endothermic) O

k3k2k1 + ROH RC OR → + RO →OH + RC

O

OR k−1 R

O

OH R

O

O

OH (acid) minor (base catalyzed) species

minor (SS) species (SS)

+ H2O + OH

R

O

O R

O

OH

10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 4 Prof. William H. Green Page 2 of 4




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O

O O d RC OR

OH ≈ RC OR

≈ k1 OH

RC OR

0 OH k−1 + k2

dt

O

k RC OR

d [RO ]≈ 0 RO ≈

2 OH

dt k3 [acid]

rROH = k3 RO [acid] = k2

RC

O

OR = k k 1 2 OH

k−1 + k2 RC

O

OR OH

k eff

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.

RC

O

OR + H2O RC

O

OH + ROH

OH kk1 + R+

H+ + RC

O

OR RC OR 2 → k−1 + R

O

OH (acid) minor(acid catalyzed)

(SS) species (SS)

+ k3 +R + H O →ROH+H2

OH O d RC

+ OR

OH k1 H

+ RC OR

≈ 0 RC +

OR ≈

k−1 + k2dt

OH

d [R+ ] +

k2 RC +

OR

dt ≈ 0 R ≈

k [H O3 2 ]





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OH k k O

+ 1 2 + r = k R [H O ] = k

ROH 3 2 2 RC +

OR = k−1 + k2

H RC OR

keff

Ethylene(in plastics)

C H → H +C H 2 6 2 2 4

Ethyl radical C H + Hi → H + C H i 2 6 2 2 5

C H i → C H + Hi 2 5 2 4

−r = k Hi C H C H 1 [ 2 6 ]2 6

slow iC H inefficient, but important (radical creation) →2CH 2 6 3

CH i + C H → CH + C H i 3 2 6 4 2 5

2C H i → C H + C H (radical destruction) 2 5 2 6 2 4

(disproportionation)

C H + C H → 2C H i reverse disproportionation also happens 2 6 2 4 2 5





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Lecture 5: Continuous Stirred Tank Reactors (CSTRs)

This lecture covers: Reactions in a perfectly stirred tank. Steady State CSTR.

Continuous Stirred Tank Reactors (CSTRs) Continuous

Plug Flow Reactor CSTR

Open system Steady State Well mixed Open system

Steady State Continuous

Batch

Close system Well mixed Transient

In terms of conversion, XA?

Mole Balance on Component A In-Out+Production=Accumulation F F (steady state) Ao − +A rAV = 0 What volume do you need for a certain amount of conversion?

F XV = Ao 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: F C= ( =volumetric flow rate) Ao Aov0 0v

F FAo −V F= =AA FAo − X AF

−r AoA

Figure 2. A batch reactor. Figure 1. A plug flow reactor, and continuous stirred tank reactor.

F CA A= v 0

V C X= Ao A

v r0 − A

Vτ = ← average time a volume element of fluid stays in the reactor v0




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Ao A

A

C Xr

τ =−

Consider: 1st Order Reaction Kinetics concentration or conversion? A Ar kC− =

convert rate law from CA to XA

( )1A Ao AC C X= −

( )1A Aor kC X− = − A

( ) ( )1 1reactor size in terms of conversion and rate constant Ao A A

Ao A A

C X XkC X k X

τ⎫⎪= = ⎬− − ⎪⎭

rearrange to find how much conversion for a given reactor size

1AkX

kττ

=+

average reactor residence time τ ≡

1average time until reaction for a given molecule

k≡

We can now define a “Damköhler number”

reaction rateis the reaction rate law at the feed conditions

flow, Ao

AoAo

r VDa rF−

= =

For a liquid at constant pressure with 1st order kinetics: Da kτ=

1ADaX

Da⇒ =

+

therefore: As DaÆ, XA 1 As Da∞, XA 0 (molecule probably leaves before it can react) For a liquid at constant pressure with 2nd order kinetics:

( )

2

22 1A A

Ao A

r kC

kC X

− =

= −

AoAo A

A

CC Xr

τ = =− 2

A

Ao

X

kC ( ) ( )2 21 1A

A Ao

XX kC X

=− − A

solving for conversion:

( )1 2 1 42Ao Ao

AAo

kC kCX

kCτ τ

τ+ − +

=

10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture #5 Prof. William H. Green Page 2 of 4 Cite as: William Green, Jr., 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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2AokCDa =Ao

VC Ao

o

kCv

τ=

Thus, conversion can be put in terms of Da.

( )1 2 1 42A

Da DaX

Da+ − +

=

How long does it take for a CSTR to reach steady state? In-Out+Production=Accumulation

AAo A A

dNF F r Vdt

− + =

For a liquid at constant density this is:

AAo A A

dCC C rdt

τ τ− + =

non-dimensionalize

ˆ ÂA

Ao

C tC tC τ

= =

ˆ Âo Ao A Ao AC C C kC C ττ− − = AoC

τ

ˆÂdC

dt

ˆ ˆ1 1ˆ

DaA

AdC k Cdt

τ⎛ ⎞

+ + =⎜ ⎟⎜ ⎟⎝ ⎠

( )ˆ ˆ1 1ˆ

AA

dC Da Cdt

+ + =

with initial conditions: ˆ ˆ0, 0AC t= =

we have the solution:

( )ˆ(1 )1ˆ 11

Da tAC e

Da− += −

+

In nondimensional terms, it exponentially approaches a new steady state with a

characteristic time 1 Daτ+

.




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t

ˆAC

1

1 Da+

1

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 h

time: ln(2) τ

1+ Da

CSTRs in Series (Liquid and at constant pressure)

alf

CA0

CDa1 Da2

Figure 4. Two tanks in series. The output of the first tank is the input of the second tank.

1st order reaction kinetics CC A0

A1 = 1+ Da1

For the second reactor iterate

( )( )CC A0

A2 = 1 1+ +Da1 2Da

If the CSTRs are identical,

( )CCAn =

A0 1+ Da n

many CSTRs in series looks like a plug flow reactor.


A2



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Lecture 6: Concentration that Optimizes a Desired Rate

This lecture covers: Selectivity vs. conversion and combining reactors with separations.

A

P, A, U Separator Purer P (A, U)

recycle A (U, P)

Optimal

- What is the target? goal? Often it is to maximize profit or production

product byproducts Figure 1. Schematic of a reacting system with a recycle stream. const. (vol.) Simple Target: maximize F r p p= ⋅V r (concentrations, T) p Constraint: T T≤ max

High temperature maximum rate constant Simplest Case:

A→ P r kp = [A] [A A]⇒ [ ] feed

τ → 0 residence time in reaction

[ ] [ ]

[ ]

VF Pp = =V0 Pτ

(1− e−Da )

=Vk Afeed Da

Constraint: [P P] ≥ [ ] (purity constraint min )

X ≥ X conversion min

Fp = −k T( )max [A]0 m[P] i V =n

k T( )max [A] feedV (1− X )




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A+ →B C2 2nd order

r kc = 2 [A][B]

A, B

[ ][ ][ ]A B C

[A] = −[A X] feed(1 A)

[B] = −[B X] feed

(1 B )

.1 .1 r kc = −2 ([A] feed [B] feed

1 X )(1 X A − ).01

B

A, B B

[B] [A]

Separator A, B, C

(A)C

Figure 2. Schematic of a CSTR.

product may A→ B also react B → C A→U

F fB = ( )[A], ,[B] [C],[U ],τ ,T

6 variables fsolve (Matlab)

SS. 0 = −F Fin out+ r V A

A A

0 = −F Fin out+ rBV B B

0 = ... 0 = ...


undesirable

Figure 3. A reacting system with recycle stream.



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Simplest Case:

[ ]

k T( )

A B→1

k T

B →2 ( )

Ck T

A U→3 ( )

FA

A = in V k( )k 1

1 3+ + τ[U k] = 3 [A]τ

[C k] = 2τ [B] [ ][ ] k A

B = 1 1k2 + τ

( )F kAo 1τ

F TB τ , = ( 1k k1 3τ + +τ τ)(1+ k 2 )

optimize Da2 variables 1 Da2

Da contour plot

3

matlab fmincon (allows for constraints) Don’t use these – you may not find an actual optima.


FYield ≡ P

FA0

YieldA P→ →≡ X S A A P

SelectivityF

= P ( )F FA A−

0

∂ ∂F FB B= =0, 0∂ ∂T τ

SA P→

Figure 4. Sample contour plot for the 2- optimization. variable



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(especially important when A is expensive)

P U→

Y

τ

AP

→A P U→ →

Figure 5. Yield versus residence time. Intermediate P rises in concentration and then falls off as it is converted to U.




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Consider the unconstrained optimization of a CSTR with volume V.

A → Br = k[A]

q [A]0

[A] [B] V

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, [B] = k[A](V / q)

and FB = rV = k[A]V ;

thus production of B is maximized when [A] takes its maximum value, which is [A]0.

Continuing with the material balances, we find:

[A] = [A]0 =

[A]0

1+ (kV / q) 1+ kτ

When Da = kτ << 1, [A] goes to [A]0.

FB = rV = kV[A] = kV[A]o =

kV[A]o

1+ kτ 1+ kV / q

olim FB = lim⎜⎜⎛ kV[A]

⎟⎟⎞ = kV[A]0q→∞ q→∞⎝1+ kV / q ⎠

Unfortunately, in the limiting case of infinite flow rate, the concentration of B in the output solution is vanishingly small:

lim[B] = lim(k[A](V / q)) = lim⎜⎜⎛ k

[A]0 (V / q)⎟⎞⎟ = 0 .

q→∞ q→∞ q→∞⎝ 1+ (kV / q) ⎠

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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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?

CSTR Batch

Figure 1. Schematics of a batch reactor and a CSTR.

Small Amount of Material (small quantities) (does not tie up equipment continuously) Flexibility + - Expensive Reactants + - If product does not flow, + - Materials Handling (e.g. Polymers) Do not have to shut down and - + clean, less down time 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 In – Out + Product = Accumulation 0 0




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dNrA V = A dt

Constant V, dCr A

A = dt

In terms of conversion, dXC rA

Ao = dt A

Integrating, tCA dC

= ∫ A or CAo rA

t CX A dX

= AAo ∫

0 rA

1st Order Reaction A⎯k⎯→B − =r kA A AC = kC o (1− xA )

dx∫

xA A 1 1⎛ ⎞t C= ⎯Ao ⎯→ =t ln

0 ⎜ ⎟− −kCAo (1 xA ) k ⎝ ⎠1− xA

xA = −1 e−kt

90% conversion 1 1⎛ ⎞ 2.3t90.0% = =ln ⎜ ⎟ k k⎝ ⎠1 0− .9

(order of 1

) k

2nd Order Reaction A+ ⎯A Bk⎯→

− =r k 2 2A AC = kCAo (1− X A)2

t CX XA AdX A A1 dX

= =Ao ∫ ∫ 2 20 0− −kCAo (1 X A ) −kCAo (1− X A)2

1 Xt = A

kCAo 1− X A

kCX AotA =

1+ kCAot If k kfirstorder = second orderCA , which is faster? o

1 2 3 1st order 0.63 0.86 0.95 2nd x order 0.50 0.67 0.75 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 Lecture # Prof. K. Dane Wittrup Page 2 of 5 Cite as: 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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Batch Cycle Time

t

Charge Dischargerxn Clean Downtime, t d How long should t be? How high should X be? A

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 CA Ao V

Pr(Rate of Production) X CA Ao V

= t t+ d

What value of t will maximize Pr?

If there is a maximum of Pr vs. t, d Pr = 0 dt

Assume t constant. d =

dX( )t t Ad P optimum + −d X A

0 V= =r dtC dt Ao ( )t 2

optimum + td

dX( )t t Aoptimum + −d X

dt A = 0

Now specify kinetics. There may be no optimum.

1st order X e1 −kt A

dX A = ke−kt

dt( )t t − −ktoptimum ktoptimum

optimum + − d ke (1− e ) = 0Can numerically solve for t . optimum


= −



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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. slow

B

feed of

A

Figure 3. A fed-batch reactor with a slow feed of one reactant.

• 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.

2) Balances

A Balance

A

B

Figure 4. Fed-batch reactor with a feed of B.

In – Out + Product = Accumulation

d r( V)r tV( ) A

A = dt




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dCA d Vr tA AV( ) = +V C dt dt

Liquid V V= + 0 0v t

flow

dCA v= −r C0

dt A AV0

dilution

B Balance

dCB v v

= +r C0 0

dt B BV Vo − C B0 0

Addition Dilution




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Lecture 8: The Plug Flow Reactor

r k [ ]2A = − A

X A AF ro = − AV

[ ] ( )X FV A Ao (2nd order reaction)CSTR =

k A 2 2

01− X A

[ ] ( )Xt A

react. = k A

01− X A

V ABatch ( )[ ] =0

?

[ ]moles A V AF Batch 0

Ao = = time t treact + d

[ ] [ ] ( )F ⎡ ⎤

= +Ao XV t A Batch ⎢ ⎥A kd −

0 0⎢ ⎥⎣ ⎦A 1 X A

Assume XA=90% If treact>td then

[ ] [ ] ( )2FAo ⋅0.9VBatch =

A k0 0

A 1− 0.9

[ ] [ ]0.9FAo 1.8FV Ao

CSTR = ≤ k A 2 2k A

0 0

Figure 1. Three tanks in series.

[ ] [ ] VA A= +CSTR in

r A v0

If r kA = − [A] [ ][ ] [ ]A A

A = =in in out 1+ Da kV1+

v0

If n CSTRs are in series: V

each volume= n




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[ ][ ]A

A = in out ⎛ ⎞kV

n

1+ ⎜ ⎟⎝ ⎠nv0

improves productivity:

concentration of A in 1st one is higher than would be in one large CSTR

[ ] [ ]−kV

A A v0out =

0e

Batch

kV= ⇒3 95% conversions

v0

[ ][ ]A N X

A out = 0 A

CSTR 3 n 1 .75 series ⎛ ⎞

⎜ ⎟1+⎝ ⎠n

10 .93 100 .948

PFR

Figure 2. Diagram of a plug flow reactor. Plug Flow Reactor (behaves like an infinite number of infinitely small CSTRs) F F ( ) 0 CSTR Ain − +Aout rA ΔV =

⎛ ⎞F FAin −⎜ ⎟

Aout = −r⎝ ⎠ΔV A

dFA = −r design equation for PFR dV A

dFA = r C( )A B,C ,...dV

= −kC (for example)A BCF CA A= v 0

dFA FA BF= −k

dV v0 0v

This can be expressed as: dY

= F t( ,Y ) where t is replaced by V. dt

Example:

A+ B Ck

krev




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⎛ ⎞kFA BF k F⎜ ⎟− + rev C

v v2

⎛ ⎞F ⎜ ⎟0 0

d A ⎜ ⎟⎜ ⎟ kFA BF k FF = − + rev C⎜ ⎟B ⎜ ⎟

dV 2⎜ ⎟ ⎜ ⎟v0 0v⎝ ⎠FC ⎜ ⎟kFA BF k F

⎜ ⎟+ − rev CY ⎜ ⎟⎝ ⎠v v2

0 0

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

(v Aρ ) = const.

ρ =∑CW i i

For a liquid, dρ

= 0 dz

( )d vρA dv dA dρ= +ρ ρA v + Av = 0

dz dz dz dzRearrange:

dv ⎛ ⎞1 1dA dρ= −v ⎜ ⎟+

dz ⎝ ⎠A dz ρ dz

For a normal pipe dA

= 0 and for a liquid dz

dρ= 0

dz

Therefore: dv

= ⇒0 v v= dz 0

(We can’t assume this for gases!) For a PFR: dFA = rdV A

FA = v A[ ] ( )d vCA = rdV A

For liquids, v is constant so we can take it out of the differential.




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v dCr A , for liquidsA = area dz

dCA area= r

dz v A 0

Instead of treact we have zreact! area ⋅ lengthtpipe =

v0

[ ] ( )area ⋅ z Xt A

PFR = = v k0 A

01− X A

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

∑ PCi = RT

⎫P FC i ⎪ turns F's into concentrationsi = ⎬RT ∑

Fn ⎪

n ⎭ρ =∑CW , is molecular weight of i. i i Wi

( )←const.

mass flowratev = ρ z




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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 PFR

FV = Ao X −r A

A∫

XV A F= Ao dX A0 −rA

“Levenspiel Plot”

AX

Ao

A

Fr−

1st or 2nd

order reaction

CSTR Volume

• as X increases, decreases A CA

−r decreases, for 1st and 2nd order, A

so FAo increases −rA

PFR Volume

Figure 1. General Levenspiel Plot.

Ao

A

F A

r− o

A

Fr−

VCSTR

A

VPFR

X X A So PFR is always a smaller reactor for a given conversion when kinetics are positive order.

Figure 2. Levenspiel plots for a CSTR and a PFR for positive order reactions.




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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

x2

v1

v2

1AF

x1

2AF

Figure 3. Schematic of two CSTRs in series. FV Ao

1 1= X −rA1

2nd reactor: 0 In + Out + Prod = Acc F FA A− +

1 2rA2V2 = Steady state




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FFA A= − V (A0

2 0F X 2 2FA → =

0X 2 − X1) −rA2

Ao

A

Fr−

2V

2X

1V

X 1

X

Ao

A

Fr−

X

Multiple CSTRs begin to approximate a single PFR


Figure 5. Reactor volumes for multiple CSTRs in series.

Figure 4. Reactor volumes for 2 CSTRs in series.



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X

Ao

A

Fr−

Ao

A

Fr−

VCSTR

AX

VPFR VCSTR

VPFR

Designed final conversion

Final X X

Ao

A

Fr−

Ao

A

Fr−

VCSTR

AX

VPFR VCSTR

VPFR

Final X Designed final conversion

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. Choice of PFR vs CSTR depends on conversion. Choose the reactor that has the smallest volume reduce cost. Reactors:

X

FAo −rA

V CSTR

V PFR

Final X

CSTR

PFR

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 r k 1D = Cα D = Desired, U = Undesired d A

A→U r k 2U u= Cα A



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Define “selectivity” r kS CD d ( )α1 2−α

D U/ = = r k AU u

If α >α , as C increases, 1 2 A SD increases /U

-Favors PFR because C starts at C then drops whereas CSTR A Ao

concentrations are always at lower C . A

If α1 <α , as increases, 2 C decreases A SD /U

-CSTR favored

If α1 =α then 2kS d

D U/ = , no dependence on C k A

u

-Therefore no CSTR/PFR preference. Define a fractional yield

dC k Cα1

φ = =D d A − +dC 1 2

A dk CαA kuC

αA

Overall fractional yield All D producedΦ = All A consumed

For a CSTR: Φ =φ Εxit CA

ΔC CA = −A A0C

f

For a PFR: 1 C

Φ = At φdCΔC ∫ AC

A A0

If 1 2α α=

φ

Φ ACΔ

fAC

AC 0AC

Figure 8. Fractional yield versus concentration. Selectivity does not depend on CA.




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If α1 2>α

φ

AC

Figure 9. Fractional yield versus concentration when α1 > α2. CSTR PFR

φ

AC

φ

AC

fAC 0AC

fAC 0AC

CSTRΦ ACΔ PFRΦ ACΔ

Figure 10. Comparison of overall fractional yield for a CSTR and a PFR when α1 > α2. PFR is preferred because Φ >Φ , therefore the yield of D per mol A consumed PFR CSTR

is higher. If α <α 1 2

φ

AC

φ

AC fAC

0AC fAC

0AC

CSTRΦ ACΔ PFRΦ ACΔ

CSTR PFR Figure 11. Comparsion of overall fractional yield for a CSTR and a PFR when α1 > α2.

ΦPFR <Φ CSTR




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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

PFR CSTR

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.

Real mixed tank

stagnant

bypassing mixing

recirculation eddies

volumes

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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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

∞

= ∫ E ( ) dt = 1normalized t 0

∞

variance = σ 2 = t −τ 2E t dt (measures the broadness of the distribution) ∫ ( ) ( )

0

E

after t1

before t1

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.

Experimental Determination of E(t) Inflow should be something measurable Absorbance Fluorescence pH saltconductivity radioactivity

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. 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 10 Prof. K. Dane Wittrup Page 2 of 7




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A pulse input allows for easy interpretation because all materials enter the reactor at once.

t

Cin

input

detector t

Cin

curve

Figure 6. Schematic of a residencetime distribution experiment. The input curve enters the reactor; a detector detects concentration changes in the output stream.

out E ( ) t =

t

C (t )

∫Cout ( ) t dt 0

PFR (Ideal)

t

Cin

τ t

Cin

t0

Figure 7. Pulse input in ideal PFR. A pulse input in an ideal PFR becomes a pulse output.

E (t ) = δ (t −τ ) = 0 x ≠ 0

xδ ( ) = = ∞ x = 0

∞

∫ δ ( ) x dx = 1 −∞

∞

∫ f ( ) ( x δ x − a) dx = f ( ) a −∞

CSTR (Ideal) Transient material balance: InOut+Production=Accumulation

10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 10 Prof. K. Dane Wittrup Page 3 of 7




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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 C + 0 = V dC

0 dt

( ) 0

−t τ V

C t = C e , τ = ν0

C t t τ

t =

∫ ( ) τ

E ( ) ∞

( ) = e −

C t dt 0

CSTR

Figure 8. Pulse input in an ideal CSTR. In an ideal CSTR, a pulse input leads to a sharp peak with a tail.

∞ −t τ

mean residence time = ∫ te

dt =τ τ

0

CSTR (nonideal mixing) Bypassing: Divide input into 2 streams

0

Figure 9. A bypass is modeled by dividing the input stream into two streams, one of which does not enter the reactor.

V

0ν

Bν

SB ν ν





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Figure 11. Residencetime distribution for dead volumes. When a dead volume is present, a decreased amount of material is observed in the output stream.

measureable V=VSD+ VD

Vτ = SD SD <τ

ν ideal 0

PFR (Nonideal)

E

bypass portion

E

mixed

t t

combine

E

Perfect mixing τ = V

ν0

V Bypass τ = νt SB

Figure 10. Residencetime distribution determination for a bypass.

Dead volumes: Stagnant regions not getting mixed

VD

VSD

Eideal

dead volume t present

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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Figure 13. A pulse input can become an axially dispersed pulse output in a nonidealPFR.

There are two common models for dispersion in a tubular reactor:Tanks in a seriesTaylor dispersion model (based on the Peclet number)

To model the PFR as several tanks in a series, break the reactor volume, V, into nV

CSTRs of volume each. n

− τ(t τ

E t ) t= (

e i

,n −

τ i =1) !τ ni n

1

2

4

10 PFR

t

E(t)

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.

n−1

Axial Dispersion

1 2 3 n

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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2 τ 2 σ =

n τ 2

n = σ 2

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.

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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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 dN N streams N rxns

i = +∑ ∑F Vi m, ,cv υi l r dt l

m l= =1 1

stoichiometstoichiometric ric coefficient coefficient

r - depends on concentration l

- T - catalyst

dU total dV N s

∑treams

cv + =P Hcv conc ( )T F total +Q +W + (other energy termsdx dt m m m s )

m=1

flow work

extensive intensive heat do work Wshaft work s negative

work expansion work

If small control volume, pressure constant.

In 1cvV has a

fixed P

other control volumes

P2 P1 P1≠ P2

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 total and . cv r Tl ( )

dU total total N streams totalcv dU

= +cv dT ∑⎛ ⎞dU⎜ ⎟cv dNi

dt dT dt i=1 ⎝ ⎠dNi dt

heat capacity intensive substitute for of system contribution of

each species

dNi dt

Want dY

= F Y( ) dt

Assume ideal mixtures U Ntotal

cv ≈∑ iU i ( )Tcv

extensive intensive

dU totalcv =U Ti c( )

dN vi

If P=Constant (Isobaric)

dH total d ( )U + PV dU dP dV= = + V P+

dt dt dt dt dt0

dU totalcv dV

+ P cv

dt dt↓

dH total

dtAssume isobaric, all ideal mixtures, neglecting K.E., P.E., other energies

⎛ ⎞N species dT N streams N

⎜ ⎟∑ ∑N C cvi p, ,i = −∑

species

Fi m ( )Hi ( )Tm Hi (Tcv )⎝ ⎠i mdt i

N

∑ ∑streams N rxns

− +H Ti c( )v Vcvυi,l rl ( )Tcv Q +Wsi l

∑υ , ( ) i l H Ti cv ≡ ΔH stoichiometric coefficient rxn ( )Tcvi l




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N streams N rxns

− =∑ ∑ H Ti ( )cv Vcvυi,l rl ( )Tcv −∑V r ( )T ΔH cv l cv rxn ( )Tcvi l l l

Assume Q U≅ −A( )T T (conduction) a cv

area of contact


coolant heat reactor

transfer coefficient

Ws ≈ 0 (As a stirrer, heat negligible)

If designing engines W . s ≠ 0 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

dT N streams N species

cv = ≅0 (∑ ∑ F H( )T ) −H (T ) −∑V r dt i m, i m i cv cv lΔHrxn+UA(Ta −Tcv )

m i l

All depend on TCV

When we reach steady state, no more accumulation F F− + r V = 0 at steady-state A i, ,n A out A

See Fogler: 8.2.3 If just one reaction, one input stream, one output stream, and the system is at steady-state:

UA( )T T− += a i∑F

X , ,inputCp i (T T− in ) A F HAo ( )−Δ rxn



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In this special case, conversion and T linear 1 reaction making heat as product is made. When ΔH = (-) Exothermic, reactor is hotter than cooling reactor (heat transfer rxn

important) (+) Endothermic, reactor must be heated so that reaction will run G T( ) ≡ −( ΔHrxn )(−rA V FAo ) Generation

⎛ ⎞⎛ ⎞F ⎜ ⎟

R( )T C= +∑ i i, n UA⎜ ⎟p i, ⎜ ⎟1 ( )T −T c⎝ ⎠F FAo ⎜ ⎟∑ i, ,inC⎜ ⎟p i

⎝ ⎠K

Heat removal K = 0 Adiabatic K = Big Cooling

KTa i+TT nc =

1+ K R( )T linear with T G T( ) → constant at high T - not linear with T

•

•

•

( )G T

T

( )R T

rxnH−Δ

Three steady-state solutions

cool a lot

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.




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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:

•

•

•

( )G t

reactorT

( )R t removal of heat

Three steady-state solutions

generation

Figure 1. Three steady-state conditions shown on a G(T) versus T graph.

⎛ ⎞ξ1⎜ ⎟ξ2 = z⎜ ⎟ SS⎜ ⎟⎝ ⎠T

SS

dξ1 = 0dtdξ2 = 0 dt

dT= 0

dt

steady-state

⎧dξ⎪

1 ⎫= f Tdt 1 1( ,ξ ξ2 , ) ⎪

⎪ ⎪ vector notation ⎪ ⎪d dξ z

original eqns. ⎨ ⎬2 = →f2 1( ,ξ ξ = 2 ,T F) (z)⎪ ⎪dt dt⎪ ⎪dT

=⎪ ⎪f T3 1( ,ξ ξ2 , )⎩ dt ⎭


stability: we want any perturbation δ z from z to be self correcting SS



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i.e. d ( )δ z v= −( e)δ z dt

•

zδ 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 z z z SS

#

δ = −

d dz( )δ δz F= = ( z + z SS )dt dt

0 0 dF

≈ +δ δz F ( )z +O( z 2 )dz SS

Jacobian matrix

d ⎛ ⎞dF( )δ z z=∑⎜ ⎟n δdt dz m

⎝ ⎠m z SS

= J zδ

⎛ ⎞df1 1df df⎜ ⎟1

d dξ ξ⎜ ⎟1 2 dT⎜ ⎟df

= ⎜ ⎟2 2df dfJ 2 ⎜ ⎟d dξ ξ1 2 dT

Jacobian ⎜ ⎟df3 3df df⎜ ⎟3

⎝ ⎠d dξ ξ1 2 dT

Matrix d ( )δ z M= ( )δ z dt if eigenvalues of M<0 then stable



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CO + +H2 2O CO H2 +Q

toxic heat

1

X

original

w/ catalyst

equilibrium

rxn

doesn’t occur if equilibrium limited

T Figure 3. Conversion (X) versus Temperature (T). Data Collection:

- determining rate laws

Reactants Products, Unreacted Stuff, Byproducts

Reactor

,oTC τ

Figure 4. Schematic of a general reactor.

product conc. at output

[ ]0A

make plots - get empirical expression fit curve

Figure 5. Product concentration versus reactant concentration A. r(conc,T) ouput-input




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∫ r dxdydz → rV if homogeneous dV

“well-stirred” reactor (slow reactions) “no” conversion (really ~.1% conversion) C C= ± .1% can measure (output-input) 0

(r barely changes) *need very sensitive product detection “differential reactor” From data: guess mechanism vary ( k , k ) make a fit eq

1) Is mechanism consistent (error bars?) w/ data? 2) How to regress k ? (least squares method)




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Lecture 13: Biological Reactors- Chemostats

This lecture covers: theory of the chemostat, fed batch or semi-continuous fermentoroperations

Biological Reactors (Chemostat) Concentration/Combustion constant Biological CSTR

[S]0

V

F

F

[S], x

Figure 1. Diagram of a chemostat. F = Volumetric flow rate

biomassx =

volume [S]0 = Concentration of growth limiting substrate. (for growing cells) At steady-state, biomass balance In – Out + Prod = Acc Sterile feed: In=0 Steady state: Acc=0 − +Fx r V 0 at steady-state x =Cell growth kinetics r xx = μ

− +Fx μx V 0=

Solve Fμ = V

D=Dilution rateF 1

≡ = 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 at steady-state

1F S[ ]0 − −F S[ ] μx V = 0 Yx

s

Yield coefficient mass biomass created

mass substrate consumed

Divide by V

μxD S([ ]0 − =[S])

change in sugarYx

concentration s

At steady-state μ = D x = −Y Sx ([ ]0 [S])

s

What is the value of [S]? What more information do we need? μ = f ([S]) must choose a growth model to connect μ and [S] Monod growth model:

μmax[ ]Sμ = at steady-state K Ss + [ ]

μD max[ ]S=

K Ss + [ ]

K D[ ]S = s substitute in x equation μmax −D




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⎛ ⎞K Dx Y= −x ⎜ ⎟[ ]S s 0s ⎝ ⎠μmax −D

Specifying μ , max Ks , Yx , D , [ ]S , can predict 0 x , [S] .

s

x < 0 is non-physical but formally in solution μ −D can go to 0. If you turn knobs incorrectly: if D is too high, the cells cannot max

grow fast enough to reach steady-state. Washout will occur. so use x = 0 to find D max

μD max[ ]S 0

max = K Ss + [ ]0

For D D> “washout”, no steady-state. max

,x S

mass

volume

D maxD

S

x

washout

For real systems K S . Most cell growth systems reach maximum at fairly low s << [ ]0

concentrations; hence x is flat, then drops off sharply.

biomass is the product, is there a best operating condition?

hat should we consider?


Figure 2. Biomass/volume versus dilution rate. Beyond the maximum dilution rate, washout occurs.

If W dx

optimize x with respect to D? D=0 (no, because this would be batch reactor) dD

Define productivity as ( )( )

biomass= xD

reactor volume time



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d x( )D= 0 for optimum. (is a maximum)

dD

x

D

xD

D

optimumD

Figure 3. Left: Biomass/volume versus dilution rate. Right: Productivity versus dilution rate.

⎛ ⎞KD s optimum = −μmax ⎜ ⎟1⎜ ⎟K Ss +⎝ ⎠[ ]0

K Ss << [ ] 0

Doptimum ≈ μmax

≈ Dmax

Close to washout conditions. Operability would be difficult. We would not want to run too close to washout conditions. 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)

If you want


100 gxfinal = L

200

If 0g wtY ≈ 0.5 , [S] = ≈ 20% Toxic, sugar content cells will die x

s L volume 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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[S] F0

x(t) S(t), very small

Figure 4. Diagram of a fed-batch fermentor.

If μ is constant,


biomass=biomass tt 0 eμ =

There is a dilution term, because as we feed in fresh medium, volume will change. Volume often doubles.

x V V= x eμt 0 0

μx t0 0V eμ

Feed F S[ ] 0 =xsugar feed

Ys

sugar consumed

Assume all converted into biomass.

x0 0VF = μeμt [ ]S Y0 x

s

Exponential flow rate. Typically μ specified as “small.” Dilution: d V

= F dt

⎛ ⎞⎜ ⎟xV(t e) = +V 0

0 1 ( )μt −⎜ ⎟1 [ ]S Y⎜ ⎟0 x

⎝ ⎠s

( )biomass x eμt

x = = 0

V x1 1+ −0 eμt

Y Sx [ ]0 s

x eV μt

= 0 0

V



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ogistic equation”

“L

x

t

Figure 5. Graph of logistic growth.

If product is something cells are making: Product synthesis kinetics

1)


1 dP=αμ growth associated (e.g. ethanol)

x dtproductP ≡ volume

2) 1 dP

= β not growth associated (e.g. antibiotics, proteins, antibodies) x dtP x= β ∫

tdt integrate for amount of product.

0



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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 Complex

P + L C k

k

on

off

Figure 1. Protein-ligand binding. Association rate = k C on pCL

Dissociation rate = k C off c

@ equilibrium, k C = on pCL k Coff c

1

s

C Cp L k= =off K

C k dc on

L

mol s

In general, for protein-protein interactions, k m5 1− −1 on ≈10 ol s

half-time for complex dissociation ln 2τ1 2 = koff

Kd τ 1 2 types

mM milliseconds non-specific stickinessμM (micromolar) milliseconds-seconds cell surface, multi valent

nM minutes-hours antibodies, enzymespM hours-weeks growth factors

fM (femtomolar) weeks-months hycholase inhibitors

stronger interactions

Fractional saturation C CY = =c c

C Cp o, c +Cp




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C Cp L CK Y L

d = → =C Cc L + Kd

CL

Y

Kd log CL

Y

Figure 2. Left: Graph of fractional saturation versus ligand concentration. Right: Graph of fractional saturation versus the logarithm of ligand concentration. If Cp o, ≈C , then at equilibrium, L o, C CL L≠ ,o

C yL o, ,− C

Y = p o

C yL o, ,− +Cp o Kd

K Cd + + C 2

L, ,o Cp o − ( )Kd + +CL,o p o, − 4CY p o, C= L,o

2Cp o,

If instead C C , C ≈C L o, ,p o L L,o

C

Y = L o, C KL o, + d

How quickly is equilibrium reached?

dCc = −k Cdt on LCp koff C c

If C C “pseudo-1st order” L o, ,p o

k C on L = k Con L,o

C C p o, = +p Cc

c

(complexed)

C C p p= −,o C 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 14 Prof. K. Dane Wittrup Page 2 of 5 Cite as: 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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dCc = −k Con L, ,oCp koff Cc = k Con L o ( )Cp,o −Cc − kdt off Cc

= −k Con p, ,oCL o (konCL,o + koff Cc

C

⇒ =C t( ) C L o, (1− e−k tobsc p,o )

C KL o, + d

k k obs = +onCL,o koff

)

ln 2= half-time for reaching equilibrium

kobs

t

cC

ln 2

obsk

equlibrium

Figure 3. Concentration of complex versus time. Equilibrium is approached at long times.

Biosensor Surface plasmon resonance (label-free)

flow thin gold film

hν

amount of reflected light is a function of complex formation

Figure 4. Schematic of how surface plasmon resonance works.




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t

signal

flow signal

stop signal

on

off

kk

dK

offk

association dissociation equilibrium

Figure 5. Signal of detector versus time.

redundant estimates: k in both association & dissociation, off

kK off

d = in equilibriukon

phase best approach: fit one set of parameters to three phases of experiment. (global lesquares) Multivalency (Avidity)

cell cell

m

ast

surface cell

cell

multivalent label

Figure 6. Three examples of multiple protein-ligand binding. How does multivalency effect apparent interaction strength?




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free

bound

Keff

detected same

(high effective local concentration)

Figure 7. Multivalent binding equilibrium.




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Lecture 15: Gene Expression and Trafficking Dynamics

This lecture covers: Approach to steady state and receptor trafficking

Central dogma of molecular biology: DNA mRNA protein transcription translation

Material balance on one specific mRNA Accumulation = synthesis – degradation

moles mRNACmRNA ≡ cell volume

( )( )mol mRNAKr ≡ , transcription (function of gene dosage, inducers, etc.)

time cell volume

cell volumeVi ≡

vessel volume

( )d CmRNA Vi = −K Cr iV Vγ dt r mRNA i

γ r ≡ first order rate constant for mRNA degredation

V ≡ a function of time (cells grow, divide) i

can’t pull out of the derivative Do the chain rule:

d dVi mCC KV VRNAmRNA + =

dt i dt r i−γ rCmRNA V i

dCmRNA 1 d V= −K C i

r γ r mRNA −C dt mRNA Vi dt

simplify: 1 d Vi = μ (specific growth rate in exponential growth) Vi dt

dCmRNA = −K Cr γ μr mRNA − C dt mRNA

dilution by growth term (b/c concentration is on a per-cell volume basis)




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dCmRNA = −K Cr r( )γ μ+

dt mRNA

at steady-state:

KC rmRNA, SS =

( )γ r + μ transient case, analytical solution (just integrate)

KC er ⎛ ⎞1 − +( )μ γ r t

mRNA = −⎜ ⎟( )γ μr +

⎝ ⎠

independent of the transcription rate constant K r

t

mRNAC

1

rγ μ+

S.S.

Figure 1. Concentration of CmRNA versus time. At long times steady state is approached. Similar rate expression for the protein: (again, per-cell volume basis, analogous constants)


dCp = −K Cp mRNA ( )γ μp + C dt p

function of time, solved for above

dCp K

= −K er ( )1 (− +( )γ μr t

dt p p− γ μ+ )C ( )γ μr +

p

steady-state: d= 0 , t →∞

dtK K

C r pp S, S =

( )γ r p+ +μ γ( μ)




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Cp S, S K

= p Note: K , γ vary from protein to protein and condition C p p

mRNA, SS γ p + μto condition

Integrate dCp : dt⎛ ⎞( )γ μ+ −e e− +( )γ μp t

r p(γ + μ) − +( )γ μr t

C Cp p= +, SS ⎜ ⎟1 ⎜ ⎟γ γp r−⎝ ⎠

Usually, γ p rγ

in E. coli ln 2 ∼ 7 minutes on average. γ r

for most proteins, ln 2 ∼ hours to days. γ p

also, γ r μ

Apply assumptions to get:

K KC ep r 1 ( )p

p = −( ) ( )− +γ μ t

γ γr p + μ

Delays in synthesis time (seconds)

E. coli Yeast MammalsmRNA – 1 kb gene 10-20 30-50 30-50 Protein – 400 a.a. 20 20 60-400




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t

pC

Delay is generally small compared to 1

pγ μ+

Figure 2. Concentration of protein versus time. However, the delay can dramatically destabilize feedback loops.

Cellular compartmentalization C C→ where C C≡ for compartment 1, and p, 1 p,2 p, 1 p C Cp, 2 ≡ for compartment 2 p

rate = K C transport p,1

cell

out.

cyto.

prod. rec.

endocytosis

+ k

k

on

off

Figure 3. Diagram of protein-ligand binding on the cell surface.




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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

A B

Product

Figure 1. Bi-molecular reaction in a CSTR. Temperature drastically increases reaction rate.

Figure 2. Schematic of tube reactor. Catalyst dramatically increases reaction rate.

A catalyst

C2H6 Rxn occurs

Fire

Hot

Figure 3. Schematic of packed bed reactor. Catalyst: Accelerates rate of reaction but is not consumed




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E

rxn coordinate

products

transition state

~Ea reactants

Figure 4. Reaction diagram. rate constant:

k TB ⎡ ⎤( )G Gk exp ts −= − reactants⎢ ⎥

h R⎣ ⎦TG H= −TS

e e− −G R/ /T = H RTeS / R

E

rxn coordinate

no catalyst

with catalyst

Figure 5. Reaction diagram with and without catalyst. The reaction forms many intermediates. A catalyst lowers the energy of these intermediates. Acid/Base catalysis ROR + →H2O 2ROH

k HROR + H

1⊕ ROR

k ⊕−1


H k2

ROR→ +ROH R⊕ ⊕

k3

R⊕ ⊕+ →H O2 ROH + H



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HQSSA ROR , R⊕

⊕

⎡ ⎤Hd R⎢ ⎥OR

⊕ H⎣ ⎦O k1 1H + ⎡ ⎤≈ = ⎡ ⎤[ ]ROR −⎣ ⎦ ( )k

dt − + k2 ⎢ROR⊕

⎥ ⎣ ⎦

⎡ ⎤H k⎢ ⎥ROR = 1 ⎡ ⎤H + [ ]ROR

⊕ ⎣ ⎦ ⎣ ⎦QSSA

k k−1 2+

[ ]d ROH ⎡ ⎤H= 2k R2 ⎢ ⎥OR

dt ⎣ ⎦⊕QSSA

[ ]d ROH 2k k≈ =1 2 ⎡ ⎤H R+

⎣ ⎦[ ]OR r dt k−1 2+ k

r A A ∼ [ ]r ∼ [catalyst] (where ∼ denotes “proportional to”) A

H

⎡ ⎤+ +⎡ ⎤ NH R+ +⎢ ⎥O R ⎡R ⎤ = H + added = ⎡ ⎤⎣ ⎦ H +

⊕ ⎣ ⎦ ⎣ ⎦⎣ ⎦ V added

[ ] [ ][ ]

+ +⎛ ⎞k R⎡ ⎤ 2⎣ ⎦ ⎜ ⎟1+ +1 1OR k k RORH H= ⎡ ⎤⎜ ⎟ ⎣ ⎦

+ +⎝ ⎠k k k ( ) added1 2 3 k−1 2k H2O

[ ][ ]

k Reff OR ⎡ ⎤H +⎣ ⎦r = added

1+ k ROR[ ][ ][ ] [ ]

k Acatalystr =

1+ + +k AA Bk B ...

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


dNA =Vr dt A

f A( )[ ], ⎡ ⎤H +⎣ ⎦



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dNA = ( )area of metal r ′′ dt A

moles

area s

f ( )θ A

where N

θ A bound A = Ntotal sites

on surface




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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 radicals organometallics enzymes better mixing, uniformity transport limitations

immobilized enzymes metals solid acids, bases metal oxides, zeolites, clays, silica multiphase systems reuse catalyst easily product purity

1. New rate law on surface 2. Model the transport and mixing

( , , )fluid i

i x y zNC

Volume=

( , )

on surface

sites on surface

jsurface

j x y

NN

θ =

1j vacancyθ θ+ =∑

", ,

1 1( ) ( , )

rxn rxnN fluid N sufacefluid

i i n n i m mn m

Ar r C rV

Cν ν θ= =

⎛ ⎞= + ⎜ ⎟⎝ ⎠

∑ ∑

mols vol⋅

mol

s area⋅

QSSA for surface species: " ", ( , ) 0j j m mr r Cν θ= ≈∑

( )QSSA f Cθ = At surface

( )flowsjd

dtθ

=0 "

j avagadrosites

arear NN

⎛ ⎞+ ⎜ ⎟

⎝ ⎠

δ

A P A

Catalyst

surface

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: ( )A tot A A A PW C D y y W W= − ∇ + +

where yA is the mole fraction of A

totC const= diffusion convection

A AW D C C u= − ∇ + A

netA AF W= − ⋅∫ dn surface integral

A AF dxdydz= ∇∫∫∫ W⋅

( )( , , )AA A

dN dxdydz W r x y zdt

= ±∇ +∫∫∫

AA A

dC W rdt

= ∇ ⋅ +

*See Fogler 11-21

Continuity equation: 2 ( )AA A A

dC D C u C r Cdt

= ∇ − ⋅∇ +

Boundary Condition: ''into wall

A AW = −r at surface

1. Steady state

2. Gradients AdCdx

and AdCdy

are negligible

3. Velocity u towards the wall = 0 4. No reaction in the fluid

2

20 ACDz

∂=

∂

''A AW r= −

(''0

0

AzA

z

dCD rdz =

=

− = − )C (z=0 at the surface)

( 0)( ) ( 0)mainA A

A AC C zC z C z z

δ⎡ ⎤− =

= = + ⎢ ⎥⎣ ⎦

( )'', 0

0

( 0)mainA A A

A A zz

dC C C zD D r Cdz δ =

=

− == = −

Slow chemistry limit: ( 0) mainA AC z C= ≈

( )'' '' mainA A Ar r C≈

Fast chemistry limit: ( 0)AC z = ≈ 0

''mainA A

D C rδ

≈




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cD kδ

= “mass transfer coefficient”

c pk dSh

D= Sherwood number (dimensionless)

For spherical, catalyst particle with diameter dp: 1 2 1 32 0.6Re ScSh = +

ScDυ

= Re pudυ

= μυρ

=




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Lecture 18: External Mass-transfer Resistance

This lecture covers: Gas-liquid reactions in multiphase systems

fluid reactants products

sites rxn

intraparticle diffusion

external transport

adsorption desorption

mass transfer coefficient rate law

observed rxn rate =f(reactant concentration and product)

solid

Figure 1. Schematic of surface reaction kinetics. Analogies: noncovalent Langmuir biomolecular adsorption ↔interactions isotherms Michaelis- Langmuir-Hinshelwood Menton Haugen-Watson ↔enzyme kinetics kinetics (Briggs-Haldane, Henri) Logic:

• list rxns • hypothesize rate-limiting step • derive rate law • check for consistency w/ rate data

Single site, unimolecular decomposition

SA D

A S D

kk k

k k kA AS BS C S B C

− − −

+ + +

product site on catalyst S ≡




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rate of A adsorption dAr ≡

rate of rxn on surface sr ≡ rate of desorption Dr ≡


r k P C k C= −

dA A A S A AS−

1time

* units partial

pressureA

reactive sites

sites

mass catalyst

often C is given as fractional occupancy S θ

AA

A

kKk−

=

d

ASA A A S

A

Cr k P CK

⎛ ⎞= −⎜ ⎟

⎝ ⎠

Similarly, C BSS S AS

S

P Cr k CK

⎛ ⎞= −⎜ ⎟

⎝ ⎠

B SD D BS

D

P Cr k CK

⎛ ⎞= −⎜ ⎟

⎝ ⎠

1B

D

KK

= ( )D D BS B B Sr k C K P C= −

At steady-state, (or else you would accumulate molecules)

dA Sr r r= = D

For a rate-limiting step i , ,ji l

i j

rr rk k kl

Hypothesize, that adsorption is rate-limiting: , 0SA D

A S D

rr rk k k

≈

0S C BS BS CAS AS

S S

r P CC Ck K

≈ ⇒ − ⇒ ≈S

C PK

0DBS B B S BS B B S

D

r C K P C C K P Ck

≈ ⇒ − ⇒ ≈

B B S CAS

S

K P C PCK

≈



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d

ASA A A S

A

Cr k P CK

⎛ ⎞= −⎜ ⎟

⎝ ⎠

BA A S B C S

S A

Kk P C P P CK K

⎛ ⎞= −⎜ ⎟

⎝ ⎠

BA S A B C

S A

Kk C P P PK K

⎛ ⎞= −⎜ ⎟

⎝ ⎠

Material balance on (available sites): SC

0S S ASC C C CBS= + +

0

BS S B C S B B

S

KC C P P C K P CK

= + + S

1 BS B C B

S

KC P P KK

⎛ ⎞= + +⎜ ⎟

⎝ ⎠BP

Adsorption as rate-limiting step:

0

1

eq. driving force

BA S A B C

S A

AB

B C B BS

Kk C P P PK K

r K P P K PK

⎛ ⎞⎜ ⎟

−⎜ ⎟⎜ ⎟⎝− =

+ +

⎠ equilibrium driving force is 0 at equilibrium!

Surface reaction is rate-limiting:

0

1

B CS S A A

CA

B B A A

P Pk C K PK

rP K P K

⎛ ⎞−⎜ ⎟

⎝ ⎠− =+ +

Desorption is rate-limiting:

0

B CD S S A A

CA

C A A S A C A

P Pk C K K PK

rP P K K K P P

⎛ ⎞−⎜ ⎟

⎝ ⎠− =+ +

Initial rate expirements, approximate ( )A Ar f P− = , 0B CP P≈ ≈ , A SC

B

K KKK

= .

Adsorption limit:

0A A S Ar k C P− = o




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Ar−

AoP

Figure 2. Reaction rate vs. initial partial pressure of A for the absorption limiting case. Surface reaction limit:

0

1S S A Ao

AA Ao

k C K Pr

K P− =

+

Ar−

AoP

Figure 3. Reaction rate vs. initial partial pressure of A for the surface reaction limiting case. Desorption limit:

0A D Sr k C− =

Ar−

AoP

Figure 4. Reaction rate vs. initial partial pressure of A for the desorption limiting case.




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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

X

D.O.

μ

X

X X

X X

X X X X

critical value ≈ 0.01 mM

Figure 1. μ vs dissolved oxygen. D.O. = dissolved oxygen Equilibrium solubility of O2 1 mM ≈

O2

2 3

4

1

cell

bubble

Figure 2. Oxygen pathway.

1) Diffusion across stagnate gas film 2) Absorption 3) Stagnate liquid layer (rate-limiting step) 4) Diffusion and convection




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3) O2 flux = −k C( )* [=]l O2 2C O

mol

area time

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)

= α ⎛ ⎞P β

kl sa constantU ⎜ ⎟ where U⎝ ⎠V s is the superficial gas velocity

⎛ ⎞length ⎛ area ⎞k a[ ]= =⎜ ⎟⎜ ⎟ time−1 (s-1) l⎝ ⎠time ⎝ volume ⎠

lengthUS [ ]= (m/s) time

P power= (W/m3)

V volume

const. = 0.002 α = 0.2 β = 0.7

@ SS, O2 transport = O2 uptake by biomass

at equilibrium

mass transfer bulk liquid coefficient concentration

biomass growth rate


2 2

2

*( )X

O

l O OXk a C C

Yμ

− =

Crude limit:

g

dX< k a *

l OC X2Y

dt O2

or dX

dt

≈yield coefficient .4-.9

2

cell dry t.

w

g O



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O2 transport in tissues

o

o

o

o

o

capillary radius Rc

R0r

Figure 3. Krogh cylinder model. One-dimensional steady-state diffusion:

2 2

2

Fick's Law

O OO

D Cr V

r r r∂⎛ ⎞∂

=⎜ ⎟∂ ∂⎝ ⎠

metabolic consumption rate of oxygen, zero-order

(cylindrical coordinates) Boundary conditions: symmetry no-flux flux=0 @ r=R0

2

20O

O

CD

r∂

=∂

@ r=R0

2 2 ,O O plasmaC C= @ r=Rc

Integrate twice:

2

2

**2 *2

*,

1 2O

O plasma

C rr RC R

⎛ ⎞= +Φ − −⎜ ⎟

⎝ ⎠ln

where *0r r , R= *

0cR R R= , 2

2 2

2

,

14

char. rxn rate

char. transport rate

O

O plasma O

V RC D

Φ = =




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Figure 4. Dissolved oxygen vs. radius for various values of Φ .

r*

2

2 ,

O

O plasma

CC

1

Φ decreasing

O2 diffuses further before consumption as Φ decreases. When *R ≈ 0.05, 0 @

2OC = *r = 1 when Φ ≥ 0.2

O2

necrotic core

~50-100 mμ

o

Figure 5. Tumor micrometastases.



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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)

Figure 1. Sphere of Cells

Assume pseudo-homogeneous medium and Fick’s Law describes diffusion

[ ]d SFlux D

dr= − , where

#[ ] moleculesFluxarea time

=⋅

and 2

[ ] lengthDtime

=

n

s n sr k C− =

max ss

s s

V CrK C

− =+




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n=0 n=1

Cs

-rs

Figure 2. Rate of reaction versus species concentration

maxst1 order ss s s

s

V CC K rK

⇒ ≈

max0th orders s sC K r V⇒ ≈

Steady-state Shell Balance

In some cases, S still hasn’t penetrated to the center

t=0 t ∞

Figure 3. Time progression as species, S, enters the sphere

Thin shell r to (r+Δr) Sin by diffusion –Sout by diffusion –Scons by reaction=0

2 24 4 4nn sr r r

Flux r Flux r k C r rΔ

π π π+

⋅ − ⋅ − 2 0Δ =

Divide through by 4 rπΔ and take the limit as Δr 0

( )22 0n

n s

d Flux rk C r

dr

⋅− =

2 2 0nsn s

dCd D r k C rdr dr

⎛ ⎞− ⋅ − =⎜ ⎟⎝ ⎠

2

2

2 0 2nd order ODEns s ns

d C dC k Cdr r dr D

⎫+ − = ⎬

⎭

2 Boundary conditions:




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,0s sr RC C

==

Cs is finite everywhere, or

0

0s

r

dCdr =

=

Nondimensionalize

,0

s

s

Cr SR C

ρ = =

22

2

2 0nd S dS Sd d

φρ ρ ρ

+ − =

Boundary conditions: S=1 @ ρ=1 S is finite everywhere

( )( )

22 1,02

1,01

characteristic diffusion time

characteristic reaction time

nn s

nn s

R Dk R CD k C

φ−

−= = =

If diffusion is slow diffusion dominates If reaction is slow reaction dominates If φ2<<1 reaction limited regime If φ2>>1 diffusion limited regime

φ2<<1 φ2>>1

substrate throughout

substrate can’t make it to the center

Figure 4. Reaction and diffusion limited regimes

1 sinh( )sinh( )

S φρρ φ

=

sinh( )2

z ze ez−−

=




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φ ↑

1

1

S

ρ

0.1φ =

1φ =

10φ =

Figure 5. S versus ρ for various values of φ

Define “effectiveness factor” η

s S,0

overall rate of reaction

rate if C =C everywhereη =

overall reaction rate in sphere at steady state = [inward flux @ r=R (ρ=1)]*Area

24s

r R

dCD Rdr

π=

=

( ),0 ,01

4 4 cos sdSRDC RDCd ρ

π π φρ =

= = th 1φ −

( )2

3 coth 1η φ φφ

= −

η

φ

.01 0.1 1 10 100

1

0.1

0.01

1ηφ

∝

Figure 6. Log-log plot of effectiveness factor versus thiele modulus

Higher values of Thiele modulus effectiveness goes down *For a variety of reaction kinetics, geometries and rate laws, plots of η vs φ all look the same.




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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




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Lecture 21: Reaction and Diffusion in Porous Catalyst (cont’d)

This lecture covers: Packed bed reactors

A products

eff

AA

dF Ardz

=

Deff (actual is very complicated)

Figure 1. Packed Bed Reactor

Void Fraction φ ~.5

( , , )A j m lC x y z

1,30j = 1,30m =

1,30(points)

l =

(11-21) 2 0fluid

i i i iD C U C r∇ − ⋅∇ + = in the fluid i=1, fluidspeciesN

0ˆ

surfaceii i

surface

CD rn

∂+ =

∂ (boundary condition for the above)

Ergun’s Eq.:

(4-22) ( )

3

150 11 1.75c p p

dP G Gdz g D D

φ μφρ φ

⎡ ⎤−−= − +⎢ ⎥

⎢ ⎥⎣ ⎦

where TP Rρ= , zGUA

ρ =




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∂FA = Ar eff r reff = ( )effectiveness factor (

z A A ACAb)

∂ Ω(C z

b( )) (or η internal) A

Actual rate of reaction ( )r eff

Ω ≡ A Rate if C CA A= =( )z , and T T everywherebulk

area

total

z

bulk

eff idealii i

F Ar A rz

∂= = Ω

∂ [ ] mol

vol. sideal

ir =

wt. of catalyst in reactor

V reactorideal

i ir r⎛ ⎞′= ⎜ ⎟⎝ ⎠

(1 )i i cr r ρ φ′= −

(1 )surface area of cat.

wt. cat.i cr ρ φ⎛ ⎞′′= −⎜ ⎟⎝ ⎠

/c particlea m 2

[ ] ( )macroscopic surface area, visualam Sg

= +

F CA A= =ν AUC (some approximation) A

1 dF 2

A AdC d C= +U D− A

A dz dz a dz2

Figure 2. Flow over a sphere

Dispersion

hope this is 0!



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As A As pC C C r C Vη− =

Ask a ( ) ( ) ( )b

particlec c A

( )2

AC∂2 ( ) 0inside A AD r C r

r+ =

∂

0

0A

r

Cr =

∂=

∂ ( )b

Aeff inside c A AsD k C CC

r∂

= −∂

Ac A eff inside c A bulkr R

r R

Ck C D k Cr=

=

∂+ =

∂

Matlab:

1) Guess C ( surface) As CA

2) Use boundary conditions to get corresponding ∂CA ∂r r R=

3) Solve ODE (ode15s)

4) Vary guess C ) to make s∂C

( AA = 0 at center

∂r 1st oder irrev.

η

Ω = ηk S′′1 1 a bρ

kc ca

3η = 2 ( )φ1 coth(φi ) −1 φ

+ 1

φ ′1 1=… k ′



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Lecture 22: Combined Internal & External Transport Resistances

Packed Bed Reactor use PFR equation

effii

dF Ardz

=

z=0 z=L

Figure 1. Packed Bed Reactor

eff chem

i ir r≠ eff chemi ir r= Ω

if ? 1Ω ≈if ? 1Ω( ) –weird 1Ω >

porous catalyst particle

dp

BL: δ

Figure 2. Porous Catalyst Particle

Biot Number:

external heat transfer resistance

internal heat transfer resistanceiB ∼

Mass transfer Biot Number:




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m

c pi eff

inside

k dB

D= where fluid

c

Dk

δ∼

diameter

m

fluidi eff

inside

DB

D δ=

Mears’ Test:

if ( )

0.15observedA b p

c A bulk

r R nk C

ρ′< where n ≡ order of rxn

then (no external diffusion limitation) i.e. no changing

concentration across the boundary layer Ab AsC C≈

Similarly,

if 2

( )0.15

observed or theoryrxn A b p a

fluid

H r R Eh RT

ρ′′Δ< then bT Ts≈ (text: Eqn. 12-63)

no external diff. limit observedvs.A Ar r

[>] Weisz-Prater:

if 2( )

1observedA c p

effinside A bulk

r RD C

ρ′ (text: Eqn. 12-61)

then you can neglect internal diffusion limitations, i.e. ( 0As Ab AC C C r )≈ ≈ =

1000

100

10

1

.1

.1 1 10 100 1φ

η

0rxnHΔ >

0rxnHΔ =

or 0aE =

0.6β =

30a sE RT∼

00

temp. dep. & exothermicrxn

a

HEΔ ⎫

⎬⎭

Figure 3. (text: Figure 12-7)10.37 Chemical and Biological Reaction Engineering, Spring 2007 Lecture 22 Prof. William H. Green Page 2 of 3 Cite as: William Green, Jr., 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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rxn e As

t s

H D Ck T

β −Δ= where and eff

e insiD D≡ de tk ≡ heat conductivity

( ), ,eff i

c i b i s insider R

Ck C C Dr =

∂⎛ ⎞− = − ⎜ ⎟∂⎝ ⎠

rd= ∫ . i V

,eff i

c i inside c i bulkr Rr R

Ck C D k Cr=

=

∂⎛ ⎞− =⎜ ⎟∂⎝ ⎠

0

0i

r

Cr =

∂⎛ ⎞ =⎜ ⎟∂⎝ ⎠

if no significant external diffusion limit: ,i i bulkr R

C C =≈

C≈ −

A Ar k

aE RTe−

( )effA Ar f k≈ C

fit to: eff

aE RTeffA e− . (text: Table 12-1)




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Lecture 25: Course Review

Review

Fundamental Equations: Acc = Flow In – Flow Out + Reaction

, , ( , , , )mm o m out m

n F F r x y z t dxdydzt

∂= − +

∂ ∫∫∫

infinitesimal volume

mm m

C F rt

∂= ∇ ⋅ +

∂

rxnG RT

eqK e−Δ= ( eqK is unitless)

1

1

1

1

N products

reactants

n

j

m

meq

j

j

Pbar

KP

bar

ν

ν=

−

=

⎛ ⎞⎜ ⎟⎝ ⎠=⎛ ⎞⎜ ⎟⎝ ⎠

∏

∏

forwardc

reverse

kK

k=

2 2 2 2H O H O→ +

n PV RT=

2 2

2 2

1 1

1

H O

eqH O

P Pbar bar

KP

bar

⎛ ⎞⎛⎜ ⎟⎜⎝ ⎠⎝=

⎛ ⎞⎜ ⎟⎝ ⎠

⎞⎟⎠

[ ][ ][ ]

2 2

2 2c

H OK

H O=

units [ ]cmolKL

=

Convection dominated:




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Constant P, constant T, constant reactor V

m mF Cν≈ 2

m mW D C U C∇⋅ ≈ ∇ + ⋅∇ m

Pressure drop in Packed Bed:

Ergun Equation: ( )

3

150 11 1.75c p p

P G Gz pg D D

φ μφφ

⎡ ⎤−∂ −= − +⎢ ⎥

∂ ⎢ ⎥⎣ ⎦

totcv cv

j j sU VP F H Q

t t∂ ∂

+ = + +∂ ∂ ∑ W (U depends on T) tot

cv

( )in out k rxn

total

C T T Q r HTt C

ν⎧ ⎫− + + Δ∂ ⎪ ⎪= ⎨ ⎬∂ ⎪ ⎪⎩ ⎭

∑ where C is the heat capacity

Special Cases: Perfectly Homogeneous (“well stirred”, “perfectly mixed”)

no flows, “batch reactor”

( )( )mm

n r C t Vt

∂=

∂

CSTR, no t-dependence

( )0 in out outF F r C= − +

Homogeneous in x,y, not in z (no t-dependence) PFR (typically gives higher productivity than CSTR)

( )( )mm

F Ar C zz

∂=

∂

“sort of” PFR

( )( ) ( )avgmm

F Ar C z zz

∂=

∂Ω where ( )zΩ is the effectiveness factor

bubble

•

• 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 Lecture 25 Prof. William H. Green Page 2 of 3 Cite as: William Green, Jr., 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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( )2

2

2 ( ) 0m mm

C CD r Cr r r

∂ ∂⎛ ⎞+ +⎜ ⎟∂ ∂⎝ ⎠ r =

- nondimensionalize - some solutions in book - (18.03)

plug in to verify guess solution,matlab

Thiele modulus

φ 2r C

≡ m surface

D C 2solid m,s R

- if small (<1): reaction limited, ignore effectiveness factor Ω (internal diffusion

( )

fast) - if big: transport matters!

( ) from L interface bulkbubble

F Ak C C= −

Lk A correlations

C kc ∼ δ

erential equation into first-order ordinary differential

(sphere-packed bed)

( )into c bulk sparticle

F k A C= − D

Converting the second-order diffequations for MatLab solvers:

mm

C qr

∂=

∂

2 0m

m mqD q rr r

∂+ + =

∂

m m+ r MATLAB: ode15s

D

2m

qq rr

−∂=

∂



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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: ,A A bulk cF C UA= 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.

[ ]moles timeAF = [ ]moles fluid volume reactor volume timeAF ≠ ×

The correct relationship is:

,A A bulk cF C UAϕ=

[ ]moles time[ ]moles fluid volume fluid volume reactor volume×reactor volume timeAF = = × This is important for converting from the flowrate to the concentration correctly in a PBR. 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):

actualUV ϕ=

actualL

Vτ =

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 Q1 (NaOH) (mL/min) Q2 (Phen) (mL/min) 1 8.1 16.9 2 12 13 3 16.9 8.1 4 21 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.

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).

The four data sets have been fit to the model:

− ln I = A + B exp(− Ct)

I0 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.

Run A B C (sec-1) 1 2.76 -2.25 .0018 2 2.00 0.56 .0016 3 1.05 0.97 .0019 4 0.44 0.60 .0027

Cite as: Andreas Braumann and Michael Goodson, 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].www.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in


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1) Do the values of α, β, k1, and k2 that you obtained in the pre-lab provide semi-quantitative 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?

Cite as: Andreas Braumann and Michael Goodson, 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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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- (2) (pink) (colourless)

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-] (3)

r2 = k2[PHENOH3-] (4)

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.

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:

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

’)

1




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http://weblabs.cheng.cam.ac.uk/reactors_batchdata.html

ln( )I − ln(I t=∞ ) = exp[− (k ' + k )t] (6)ln(I t=0 ) − ln(I t=∞ ) 1 2

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

Q1

Q2

V Products NaOH

Phenolphthalein

Q

Figure 1: A schematic of the reactor

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.

βQ

Q αV(1–β)Q

Figure 2: A schematic of the bypass/dead volume reactor model

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http://weblabs.cheng.cam.ac.uk/reactors.html

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 , (7)

Nt /V

Bypass/Dead volume: c(t) = βδ ( )t τ + (1− β )2

exp − 1− β t , (8)

Nt /V α ατ

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 non-ideal 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.

Part D

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− ] , (9)

[ΡΗΕΝ 2− ] + [ΡΗΕΝΟΗ 3− ]

to avoid any ambiguities due to dilution of the feed streams.

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:

X Q(k2V + Q)Q1 = (10) 1− X k1V[ΟΗ− ]in

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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http://weblabs.cheng.cam.ac.uk/reactors_tracerdata.html

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).

Part F

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.

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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

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.




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HW6 Batch Reactor Balances

It is not legitimate to simply take the CSTR enthalpybalance 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 thissituation.

Given that there is negligible gas holdup in the reactor:NB (t) = NC (t) = 0 hence, dNB dNC= = 0 dt dt

Enthalpy balance:d H d (N AH A + NB H B + NC HC ) d (N AH A ) dN A dH A= = = H A + N Adt dt dt dt dt d H

= Q& − FB H B − FC HCdt dN dH

dtA H A +

dtA N A = Q& − FB H B − FC HC

Batch Reactor mole balances: dN A

dt = −kN A

dNB = 0 = −FB,out + kN Adt FB,out = kN A

dNC = 0 = −FC ,out + 2kN Adt FC ,out = 2kN A

Plugging in the three mole balance relationships into theenthalpy balance we find:

− kN AH A + dH A N A = Q& − kN AH B − 2kN AHCdt

dH A N A = Q& − kN A (H B + 2HC − H A ) = Q& − kN A∆Hrxn (T )dt

dH A dT = c

dt p,a dtdT

= Q& − kN A∆Hrxn (T )

= Q& − kmA∆H

rxn (T ) dt N Acp,a mAcp,a




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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)

and

XN2 = (moles of N2)/(total moles of N2 when all reactions are completed).

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.

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.

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 ReactionEngineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology.Downloaded on [DD Month YYYY].



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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:

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, k2~104 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.

Submit your Matlab program(s) to the 10.37 the course website.




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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.

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)

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)

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 ?




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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.




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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).

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 ξhas units of moles. Write algebraic equations for each X in terms of the ξ‘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.

By the convention used in this course, all the r’s have units of moles/second/liter. Write rN2, the rate ofproduction 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

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. Downloaded on [DD Month YYYY].



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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

a) (2 NaN3 Æ 2 Na + 3 N2)÷2 → NaN3 Æ Na + 3/2 N2

(10 Na + 2 NaNO3 Æ N2 + 6 Na2O)÷10 → Na + 1/5 NaNO3 Æ 1/10N2 + 6/10 Na2O (Na2O + 10 SiO2 Æ glass)x6/10 → 6/10Na2O + 6SiO2 Æ 6/10glass

Net reaction w/all Na safely sequestered as glass: NaN3 + 1/5NaNO3 + 6SiO2 Æ 8/5N2 + 3/5 glass 1mol nNaN3, = 150g = 2.3mol

o 65g 85gnNaNO , = nNaN , ⋅

1mol NaNO3 = 2.3 mol = 0.46mol or m = 0.46mol = 39gNaNO3 o 3 o 5mol NaN3 5 3 ,o 1mol

60gnSiO2 ,o = nNaN3,o ⋅ 6mol SiO2 = 6 ⋅ 2.3mol = 13.8mol or mSiO2 ,o = 13.8mol = 830g1mol NaN3 1mol

b) XNaN3 and XN2 are dimensionless. XNaN3 does not equal XN2 during the reaction.

X NaN3 =

2ξ1 X N2 = nN2 =

3ξ1 + ξ2

5 nNaN3 ,o nN2 , f

8 nNaN ,3 o

X NaN − X N = 2ξ1 −

3ξ1 + ξ2 =ξ1 − 5ξ2

3 2 nNaN3,o 85 nNaN3 ,o

8nNaN3 ,o

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).




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c) For no flows in/out and assuming a homogeneous reaction: dnA = ∑ν A,i

dξi = ∑ν A,i riV Can report dni/dt using extents or reaction ratesdt dti i

Species mol balance dni/dt (using extents) dni/dt (using reaction rates)

NaN3 nNaN3 = nNaN3,o - 2ξ1

dt d

dt dnNaN 123 ξ

= − rV dt

dnNaN 2 1 3 = −

NaNO3 nNaNO3 = nNaNO3,o - 2ξ2

dt d

dt dnNaNO 223 ξ

= − r Vdt

dn NaNO 2 2 3 −=

SiO2 nSiO2 = nSiO2,o - 10ξ3

dt d

dt dnSiO 3102 ξ

= − r Vdt

dn SiO 3102 −=

Na nNa = 2ξ1 - 10ξ2 dt d

dt d

dt dnNa 21 102 ξξ

−= ( )Vrrdt dnNa

21 102 −=

N2 nN2 = 3ξ1 + ξ2

dt d

dt d

dt dnN 2132 ξξ

+= ( )Vrr dt dnN

23 1 2 +=

Na2O nNa2O = 6ξ2 − ξ3

dt d

dt d

dt dnNa O 3262 ξξ

−= ( )Vrrdt

dnNa O 36 2

2 −=

Glass nglass = ξ3

dt d

dt dnglass ξ3= r V

dt dnglass

3 =

Rate of production of N2: rN = 3⋅ r1 + r22

d) See the Matlab solution provided. Plots showing XN2 vs. time, ξ3 vs. time, and volume vs. time are copied below.




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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 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)

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)

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 ?

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.




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Problem 2 Solution a) − rC2H3

= −d[C2 H3 ] Eq 5 dt

Substitute Eq 3 into Eq 5 −t τ

− rC H = −d[C2 H3 ] = −

d ([C2 H3 ]o e )= [C2 H3 ]o

1 e−t τ Eq 6

2 3 dt dt τ Substitute Eq 3 into Eq 2:

− rC2 H3 = (k0 + k[C2 H 4 ])[C2 H 3 ] = (k0 + k[C2 H 4 ])[C2 H 3 ]o e− τt Eq 7 Set Eq 6 = Eq 7 and simplify:

1τ =[C2 H 3 ]o

1 e−t τ = (k0 + k[C2 H 4 ]o )[C2 H 3 ]o e−t τ → Eq 8

k0 + k[C2 H4 ]oτ algebraic relationship between τ and k

Substitute Eq 3 into Eq 1: S( )t = b +m[C2H3 ]( )t = b +m[C2 H3 ]o e− Eq 9

Substitute Eq 8 into Eq 9: −t

1

S( )t = b +m[C2 H3 ]o e−t τ = b +m[C2 H3 ]o e k0 +k [C2 H4 ]o Eq 10

Compare Eq 10 and Eq 4 to define: 1Bn = b An = m[C2H3 ]o τ n =

k0 + k[C2 H4 ]o,n

Only τn depends on k and [C2H4]o,n.

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.

τt

k=1.11x107 L/mol-s

Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological ReaEngineering, Spring 2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of TechnoDownloaded on [DD Month YYYY].

ction logy. www.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in


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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 NH3 as the nitrogen source. The limiting substrate

is ethane, and Ysx

=22.8 gram dry weight per mole of ethane. Ysx

= Yield of biomass (x)

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 ash-

free biomass, CHaO

bN

c, and the formula weight per C-atom. Also determine Y

sx, in

the units of C-moles of biomass per C-mole of ethane.

b. Calculate the oxygen consumption Yxo

(moles of O2 per C-mole of biomass) when it is

assumed that CO2, H

2O and CH

aO

bN

c are the only metabolic products. Write the full

stoichiometric equation for the growth process, and determine the heat evolved per

kilogram dry weight. Assume ΔHethane

= 1560 kJ/mol, ΔHNH3

= 383 kJ/mol, ΔHbiomass

= 19 kJ/gram dry weight.

Problem 2. The gas phase homogeneous oxidation reaction

2 NO + O2 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 P + E (2)

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

concentration of [S], [ES], [E], and [P] in terms of k1, k2, and Keq,1, and

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.

c. The pseudo-steady approximation may be applied on the reactive intermediate

species [ES]. This approximation is: 0dt

d[ES]≈ . Using this pseudo-steady

approximation, verify that the rate of change of [S] is given by the expression:

[S]K[S]v

dtd[S]

M

max

+−= . What are and in terms of the other constants in this

problem: k

maxv MK

1, k2, Keq,1, [E]0, [S]0?

d. What does the rate dt

d[S] simplify to in the limit ? What about the limit

?

MK[S] >>

[S]KM >>

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 dt

d[S] regime these conditions lie within.

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.




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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) =

ethane mole-C 2ethane mol 1]

biomass molbiomass g 5.42[]

dry weight gbiomass g 3%)-(1[]

ethane moledry weight g22.8 [ ×÷×

=0.451 (C-mole biomass/C-mol ethane) b.

If assuming that CO2, H

2O and CH

aO

bN

c are the only metabolic products, then the overall

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

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)

Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].www.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in


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biomass mol-C 0.451ethane mol-C 1]

biomass mol-Cbiomass g 5.42[]

dry weight gbiomass g 3%)-(1[ ÷×÷

=-586 kJ/c-mol ethane. Then convert back again to per kilo dry weight Q=

1kg1000g

biomass mol-C 0.451ethane mol-C 1]

biomass mol-Cbiomass g 5.42[]

dry weight gbiomass g 3%)-(1[

ethane mol-ckJ586 ××÷×−

=-51.5 (MJ/ kg dry weight)

Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY].



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Problem 2.

For the reaction, , it is not possible for the apparent activation energy to

be negative, or equivalently, the rate constant decreases as T increases. We are asked to write down the elementary steps which include an NO

2k

2 NO2O2NO 2⎯→⎯+

3 species as an intermediate to explain this strange behavior. A possible mechanism:

3k

2 NOONO 1⎯→⎯+

2k

3 ONONO -1 +⎯→⎯

2k

3 2NONONO 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

0][NO][NOk-][NOk-][NO][Okdt

]d[NO3231-21

3 ==

From this we can obtain [NO]kk

][NO][Ok][NO21-

213 +=

Thus

[NO]kk][O[NO]k2k

[NO]kk][NO][Ok[NO])k-k(][NO][Ok

][NO][NOk-][NOk][NO][Okdt

]d[NOr

21-

22

21

21-

2121-21

3231-21NO

+−=

++−=

+−=−=

In order to have third-order reaction kinetics as the form , we

have to assume k

][O[NO]kr 22

forward effectiveNO −=

-1>> k2[NO], so that the overall reaction rate for NO is

1-

22

21NO k

][O[NO]k2kr −=

where 1-

21forward effective k

k2kk =

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.

a21- a,a11-

21overall a, EEE

kk2klnE +−∝∝ ~ΔH1, rxn+Ea2

If Ea1+Ea2-Ea, -1~ΔH1, rxn+Ea2<0, then we can have a negative apparent activation energy, for




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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.

3k

2 NOONO 1⎯→⎯+

2k

3 ONONO -1 +⎯→⎯

2k

3 2NONONO 2⎯→⎯+

NONO2NO 3k

2-2 +⎯→⎯

and still using PSSH on the intermediate NO3

0]NO[k][NO][NOk-][NOk-][NO][Okdt

]d[NO 222-3231-21

3 =+=

we can have the intermediate concentration:

[NO]kk]NO[k][NO][Ok][NO

21-

222-21

3 ++

=

Then in this case, the overall reaction rate is

[NO]kk])[O[NO]kk-][NOk2(k

][NOk[NO]kk

][NOk][NO][Ok[NO])k-k(][NO][Ok

][NOk][NO][NOk-][NOk][NO][Okdt

]d[NOr

21-

22

212

22-1-

222-

21-

222-21

21-21

222-3231-21NO

+=

++

++−=

++−=−=

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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Prob. 3 a. For the reactions

S-EES 1k⎯→⎯+

ESS-E -1k +⎯→⎯

EPS-E 2k +⎯→⎯

We can write down

S]-[Ek[S][E]kdt

]d[S1-1 +−=

S]-[EkS]-[Ek[S][E]kdt

]d[E21-1 ++−=


]S-d[E21-1 −−=

S]-[Ekdt

]d[P2=

With law of mass action on enzyme 0[E]S]-[E[E] =+ , 0[S]S]-[E[P][S] =++ , [P](t=0)=0,

[S](t=0)=[S]0, [E](t=0)=[E]0, and k-1=k1/Keq, 1

b. function [t,conc] = odehw2_prob3(k1, k2, keq1, tmax)

param = [k1,k2,keq1];

%initial concentrations

%conc0 = ([S],[ES],[E],[P])

conc0 = [0.01,0,1e-6,0];

%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;

dconcPdt = k2*concES;

%put derivative results back in column vector format for MATLAB

derivs = [dconcSdt; dconcESdt; dconcEdt; dconcPdt];

return;

c. Using this pseudo-steady approximation on intermediate species ES,


]S-d[E21-1 −−= =0

we know

21-

1

kkS][E][kS]-[E+

=

Using mass balance condition 0[E]S]-[E[E] =+

We know

S][kkk1

[E]S]-[E

1

21-

0

++

=

Therefore the reaction rate

m

max

1

21-022S KS][

S][

kkkS][

S][[E]kS]-[Ekdt

]d[Pdt

]d[Sr-+

−=+

+−=−=−==

V

where Km=(k-1+k2)/k1 and Vmax=k2[E]0. d. In the limit [S]>>Km, from




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maxm

max

KS][S][

dt]d[S VV

−≈+

−=

In the limit [S]<<Km, from

m

max

m

max

KS][

KS][S][

dt]d[S VV

−≈+

−=

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)

m

max

KS][

dt]d[S V

−=

This gives an exponential function for [S](t)

t)K

exp(][St)K

exp()0]([S(t)][Sm

max0

m

max VVt −=−==

While for [ES],

t)K

exp(][Sk

[E]S][k

[E]

S][k1

[E][ES](t)m

max0

M

0

M

0

M

0 V−=≈

+=

And for [P]

[P](t)=[S]0-[ES](t)-[S](t) t)]K

exp(1[][Sm

max0

V−−≈

Use matlab to solve the following non-linear ODE IVP:

[E])-([E]K

k[S][E]kdt

]d[S0

1 eq,

11 +−=

[E])-)([E]kK

k([S][E]kdt

]d[E02

1 eq,

11 ++−=

[E])-([E]kS]-[Ekdt

]d[Pr 022 ===

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.

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 106

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

t[sec]

[S][m

ol/L

]

[S](t)[M]

numericalanalytical

0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 106

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1x 10-8

t[sec]

[ES

] [M

]

[ES] vs. t

numericalanalytical




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0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5

x 106

0

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

t[sec]

[P][m

ol/L

]

[P](t)

numericalanalytical

0 0.5 1 1.5 2 2.5 3

x 10-8

0

1

2

3

4

5

6

7

8

9

10d[ES]/dt vs. t

t[sec]

d[E

S]/d

t [M

/sec

]




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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

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?

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.




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q0 FA0 FB0 FC0 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 FB2 = 0.75FB0

FC 2 = 0.75FC 0

q1 = FA1 + FB1 + FC1

q2 FA2 + FB2 + FC 2

A full flow diagram is shown below.

Determine the steady-state production of compound B in mol/h for options a) and b).




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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] .

First draw a diagram of the problem:

qA [A]0

qB [B]0

qout [A] [B]

The only unknown quantities in the above figure are [B] andqtot.

A volume balance on the carrier solvent gives the followingrelationship (constant fluid density):

qA + qB = qout = 1.16 L/h + 1.20 L/h = 2.36 L/h

Define the extent of reaction:

ξ& = rV[=] moltime

A material balance on component A yields the following:

dnA = FA0 − FA −ξ& = [A]0 qA − [A]qout −ξ& .dt

Setting the accumulation term to zero (steady stateoperation) and solving for the reaction rate we find:




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ξ& = [A]0 qA − [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 qB − [B]qout −ξ& .dt

Setting accumulation to zero, and solving for [B] we find:

[B] = [B]0 qB −ξ& .

qout

(38.9 mmol/L)( 1.20 L/h) − (4.23 mmol/h)[B] = = 18.0 mmol/L

(2.36 L/h)

Using the given rate law and knowing the extent ofreaction, we can now calculate the rate constant:

ξ& = rV = k[A][B]V

ξ&k = [A][B]V

(4.23 mmol/h) L −1 −1k = = 0.357 = 357M h(1.094 mmol/L)( 18.0 mmol/L)( 0.602 L) h mmol




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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:

ξ& = rV[=] moltime

Define the conversion in terms of the variables of the problem:

mol A reacted − rAV ξ& X A = = =

mol A fed q[A]0 q[A]0

A material balance on species A gives the followingequation:

dnA = 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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X A = [A]0 q − [A]q

= 1 − [A]

q[A]0 [A]0

[A] = (1− X A )[A]0

A material balance on species [B] gives the following:

dnB

dt = FB0 − FB + 0ξ& = [B]0 q − [B]q

At steady state:

[B] = [B]0 .

Plugging in the rate expression in the extent of reaction,

ξ& = rV = k[A][B]V ,

the steady state A balance becomes:

0 = [A]0 q − [A]q −ξ& = ([A]0 − [A])q − k[A][B]V .

Using the expressions derived for [B] and [A] in terms ofgiven values, we find:

0 = [A]0 X Aq − k[A]0 (1− X A )[B]0V

Solving the equation for V we find:

[A]0 X Aq X AqV = = k[A]0 (1− X A )[B]0 k(1− X A )[B]0

3(0.40)(0.5 m /ksec) 3V = = 43.7 m−3 3 3(1.15×10 m /(mol ⋅ ksec))(1 − 0.40)(6.63 mol/m )




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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 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 FB2 = 0.75FB0

FC 2 = 0.75FC 0

q1 = FA1 + FB1 + FC1

q2 FA2 + FB2 + FC 2

A full flow diagram is shown below.




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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 thesituation (the control volume).

q0 [A]0 [B]0=[C]0=0 [A] [B] [C]

V=15 L

Define extents of reaction:

ξ& 1 = r1V = k1[A]V

ξ& 2 = r2V = k−1[B][C]V

Steady state A, B, and C material balances give thefollowing equations:

0 = FA0 − FA −ξ& 1 + ξ& 2

0 = FB0 − FB + ξ& 1 −ξ& 2

0 = FC 0 − FC + ξ& 1 −ξ& 2

These three equations can be written in terms of only threevariables: [A], [B], and [C]. The system is fullyspecified. We must solve these equations simultaneously.The algebra is easier if we combine equations creatively tomake them simpler. When you combine two equations, youkeep the new equation and only one of the old ones, just asis done in linear algebra.

Notice that: FB0 = FC 0 = 0 .




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Hence, subtracting the B balance equation from the Cbalance 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 leftas a variable so that the same equations derived here willalso apply to the reactors in part b, where [B]0 will not necessarily equal zero.

Adding together the B and A balance gives the followingequation:

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])

Let this new equation replace the B mole balance.

Now we can write the A balance as a single equation with asingle variable, [B] using our two new equations:

0 = FA0 − FA −ξ& 1 + ξ& 2

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

20 = (k−1V )[B] + (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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b = (q0 + k1V )

c = (− q0[B]0 − k1V[A]0 − k1V[B]0 )

This quadratic equation in [B] can be solved with thequadratic formula:

2a

Since the coefficient b will always be POSITIVE, we knowthat we have to take the (+) root. The (-) root willalways be negative. Also notice that a is always positiveand c is always negative, so that the descriminant,

2(b − 4ac) > 0 ,

is always positive, and both roots are real. Hence, we maywrite:

2a

with no fear of getting a negative or imaginary result.

Converting all time to hours, we find that for part a

a = (k−1V ) = (4.49 ×10−4 M-1s−1 )⎛⎜ 3600s ⎞⎟(15 L) = 24.246

L2

⎝ h ⎠ mol ⋅ h

b = (q0 + k1V ) = (0.25 L/h)+ (1.82 ×10−4 s−1 )⎛⎜ 3600s ⎞⎟(15 L) = 10.078

L ⎝ h ⎠ h

c = (− q0[B]0 − k1V[A]0 − k1V[B]0 )

c = 0 − (1.82 ×10−4 s−1 )⎜⎛ 3600s ⎟⎞(15 L)( 0.25M)− 0 = −2.457

mol⎝ h ⎠ h

Plugging the values of the coefficients into the quadraticequation, we find:

[B] = 0.172 M

acbbB

4][

2 −±− =

acbbB

4][

2 −+− =




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Hence, the desired production rate of B is:

FB = q0[B] = ⎜⎛0.25

L ⎟⎞(0.172 M) = 0.0430 mol/h

⎝ h ⎠




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In order to do part b, it is helpful to notice thateverything 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 aroundthe entire system) once we knew the final outletconcentration of A.

current reactor output.

Reactor 1 q0=0.25L/h [A]0=0.25M [B]0=[C]0=0M

For simplicity of notation, let the zero subscript refersto the current reactor feed; no subscript refers to the

[A] [B] [C] V=5 L

Reactor 1 follows the same relationships as the singlereactor system, but with different volume. Hence the same material balances apply, and the final equation in terms of[B] is once again:

20 = (k−1V )[B] + (q0 + k1V )[B] + (− q0[B]0 − k1V[A]0 − k1V[B]0 )

0 = a[B]2 + b[B] + c

a = (k−1V )

b = (q0 + k1V )

c = (− q0[B]0 − k1V[A]0 − k1V[B]0 )

2a Plugging in the numbers, we find the quadratic coefficientsto be:

−4 -1 −1a = (k−1V ) = (4.49 ×10 M s )⎜⎛ 3600s ⎟⎞(5 L) = 8.082

L2

⎝ h ⎠ mol ⋅ h

−4 −1b = (q0 + k1V ) = (0.25 L/h) + (1.82 ×10 s )⎛⎜ 3600s ⎞⎟(5 L) = 3.526

L ⎝ h ⎠ h

acbbB

4][

2 −+− =




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c = (− q0[B]0 − k1V[A]0 − k1V[B]0 ) −4 −1c = 0 − (1.82 ×10 s )⎛⎜ 3600s ⎞

⎟(5 L)( 0.25M) − 0 = −0.819 mol

⎝ 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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Next we proceed to separator 1, carry forward our inputs,and rename our subscripted variables, so that they refer tothe current separator.

q0=0.25 L/h q1

[A]0 = 0.082 M Separator [A]1 [B]1[B]0=[C]0=0.168 M 1

q2 [A]2 [B]2

First we need to find the flow rate partitioning, then wecan 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 allstreams exiting the separator.) The separator equationsare:

FA2 = 0 FB2 = 0.75FB0

FC 2 = 0.75FC 0

q1 = FA1 + FB1 + FC1

q2 FA2 + FB2 + FC 2

Hence using the definition of molar flow rate, theknowledge that [B]0 = [C]0, and the above separatorrelationships, we find that:

q1 =(q0[A]0 )+ 0.25(q0[B]0 ) + 0.25(q0[C]0 ) =

[A]0 + 0.5[B]0

q2 0 + 0.75(q0[B]0 ) + 0.75(q0[C]0 ) 1.5[B]0

q1 = [A]0 + 0.5[B]0 =

(0.082)+ 0.5(0.168)= 0.659

q2 1.5[B]0 1.5(0.168)

q1 = 0.659q2

Assuming a constant density liquid phase, we have a volumebalance:

q0 = q1 + q2 = 1.659q2

0.25 L/hq2 = = 0.151L/h

1.659




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q1 = 0.659 × 0.151L/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 theconcentrations that are fed to the next reactor. Using themole balance on A:

FA0 = q0[A]0 = FA1 = q1[A]1

[A]1 = q0 [A]0 =

0.25 0.082

mol = 0.205 mol/L .

q1 0.100 L

Using the separator relationships for B and C:

FB1 = 0.25FB0

[B]1 = 0.25 q0 [B]0 = 0.25

0.25 0.168

mol = 0.105 mol/L

q1 0.100 L

[B]1 = [C]1




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Next we proceed to reactor 2 using the results fromseparator 1 and rename our variables so they refer to thecurrent reactor.

Reactor 2 q0=0.100L/h [A]0=0.205 M [B]0=[C]0=0.105 M

[A] [B] [C] V=5 L

All the same equations apply with different initialconcentrations and flow rates.

−4 -1 −1a = (k−1V ) = (4.49 ×10 M s )⎛⎜ 3600s ⎞⎟(5 L) = 8.082

L2

⎝ h ⎠ mol ⋅ h

−4 −1b = (q0 + k1V ) = (0.100 L/h)+ (1.82 ×10 s )⎛⎜ 3600s ⎞⎟(5 L) = 3.376

L ⎝ h ⎠ h

c = (− q0[B]0 − k1V[A]0 − k1V[B]0 ) −4 −1c =

⎛⎜− 0.100

L ⎞⎟(0.105M)− (1.82 ×10 s )⎛⎜ 3600s ⎞

⎟(5 L)[0.205 + 0.105]M ⎝ h ⎠ ⎝ h ⎠

mol c = −1.026

hThe quadratic solution gives:

[B] = 0.204 M

Solving for [A] from the [A]/[B] relationship we find:

[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 subscriptedvariables.

q0=0.100 L/h q1

[A]0 = 0.106 M Separator [A]1 [B]1[B]0=[C]0=0.204 M 2

q2 [A]2 [B]2

The flow rate partitioning is given by:

q =

[A] + 0.5[B] =(0.106) + 0.5(0.204)

= 0.6801 0 0

q2 1.5[B]0 1.5(0.204)

Hence, from the constant density volume balance:

q2 = q0 =

0.100 L/h = 0.0595 L/h

1.680 1.680

q1 = q0 − q1 = (0.100 − 0.0595) L = 0.0405 L/h

h

The amount of [B] recovered is:

FB2E = 0.75(q0[B]0 ) = 0.75(0.100L/h)(0.204 M) = 0.0153 mol/h

The concentrations of [A], [B], and [C] into the nextreactor are:

FA0 = q0[A]0 = FA1 = q1[A]1

[A]1 = q0 [A]0 =

0.100 0.106

mol = 0.262 mol/L .

q1 0.0405 L

FB1 = 0.25FB0

[B]1 = 0.25 q0 [B]0 = 0.25

0.100 0.204

mol = 0.126 mol/L

q1 0.0405 L

[B]1 = [C]1

Lastly, we rename our concentrations and proceed to reactor3.




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Lastly, we proceed to reactor 3 using the results fromseparator 2.

Reactor 3 q0=0.0405L/h [A]0=0.262 M [B]0=[C]0=0.126 M

[A] [B] [C] V=5 L

All the same equations apply with different initialconcentrations and flow rates.

a = (k−1V ) = (4.49 ×10−4 M-1s−1 )⎜⎛ 3600s ⎟⎞(5 L) = 8.082

L2

⎝ h ⎠ mol ⋅ h

b = (q0 + k1V ) = (0.0405 L/h) + (1.82 ×10−4 s−1 )⎛⎜ 3600s ⎞⎟(5 L) = 3.3165

L ⎝ h ⎠ h

c = (− q0[B]0 − k1V[A]0 − k1V[B]0 )

c = ⎛⎜− 0.100

L ⎞⎟(0.126M)− (1.82 ×10−4 s−1 )⎛⎜ 3600s ⎞

⎟(5 L)[0.262 + 0.126]M ⎝ h ⎠ ⎝ h ⎠

mol c = −1.284

h The quadratic solution gives:

[B] = 0.243 M

Solving for [A] from the A/B relationship equation we find:

[A] = ([A]0 + [B]0 − [B]) [A] = 0.262 M + 0.126 M − 0.243 M = 0.145 M

Since there is no separator, the amount of B recovered onthis step is just the amount leaving the reactor:

FB3E = [B]q0 = (0.243M)⎜⎛0.0405L ⎟⎞ = 0.0098mol/h

⎝ h ⎠ Thus the total amount of B recovered by this path is:

FB = FB1E + FB2E + FB3E = (0.0315 + 0.0153 + 0.0098) mol = 0.0566

mol h h

Another way to calculate it, and check our consistency, isto calculate the amount of B recovered from the overall A and B balances around the whole reaction system (any A thatdisappears must be present as B):




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0 = FAin − FAout + FBin − FBout

FBout = FAin − FAout = ⎜⎛0.25

L ⎟⎞(0.25M )− ⎜⎛0.0405

L ⎟⎞(0.145M ) = 0.0566

mol ⎝ h ⎠ ⎝ h ⎠ h

From the agreement of the numbers, it appears that mass wasconserved overall.

Notice that incorporating the selective separation processto remove product along the way, the amount of B recoveredwas improved with the same total volume of reactor.

Due to the truncation error and numerical rounding, acceptany answers within 2% of these values.

FB,3rctr = (0.0566 ± 0.0011) mol h

FB,1rctr = (0.0430 ± 0.0009) mol h

The modular setup of this problem (reactor units, separatorunits) makes a Matlab implementation straightforward. See included Matlab m-file hw3prob3.




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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 gas-phase 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.

(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?

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.

(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.




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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.




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Problem Set 4

Problem 1. (RC(O)OCH2)3 → (RCH2)3 + 3 CO2 r = k[(RC(O)OCH2)3] A → B + 3 C r = k[A]

Given: T0 = 150 C k(T0) = k0 = 5*10-3 (min-1) Ea = 85 kJ/mol FA,0 = 2.5 mol/min yA,0 = 1 T = 227 C X = 0.9 P = 10 atm

First, find k at the reaction temperature using Eq 3-21 from Fogler: E 1 1 85kJ / mol 1 1 R

T0 −

T −3 1 8.314⋅10−3 kJ /(mol ⋅K )

423.15K

− 500.15K

−1k(T ) = k(T0) ⋅ e = 5 ⋅10 ⋅ e = 0.206255min

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.

Species

A B C

Total

Feed Rate to Reactor (mol/min)

FA0

0 0

FA0

Change within Reactor (mol/min)

-FA0X FA0X 3FA0X

Effluent Rate fromReactor (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:

ν =νo

FT =ν0

FA0 (1 + 3X ) =ν0 (1 + 3X ) = FA0RT (1 + 3X )

FT 0 FA0 P

a) CSTR The design equation for CSTR volume in terms of conversion is (Eq 2-13 in Fogler):

FA0 X FA0 X FA0 X ν0 (1+ 3X )FA0 X FA0RT (1+ 3X )XVCSTR,a = (− rA )exit =

k[ ]A =

kFA

= kFA0 (1 − X ) =

Pk(1 − X ) ν

Plugging in numbers: mol .082 L ⋅atm (500.15K )(1 + 3 0.9 0.9 min mol ⋅K 3 2.5 ( ) )

VCSTR,a = =1655.36L ≈1.7 ⋅10 L1(10atm)

0.206255

min (1 − 0.9)




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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 X X X X

V = F dX = F dX

=F dX = F FA0RT (1+ 3X )dX =

FA0RT (1+ 3X )dXPFR,b A0 ∫ A0 ∫ [ ] A0 ∫ F A0 ∫ PkF ( − X ) Pk ∫ ( − X− r k A A 1 1 )0 A 0 0 k 0 A0 0

ν From integration by parts (or an integral table):

x2

∫(1 + mx)dx = (1 + m)ln

1− x1 + m(x1 − x2 )(1 − x) 1 − x2 x1

x (1 + mx) 1 When x1 = 0, this simplifies to: ∫ (1 − x) dx = (1+ m)ln1− x2

−mx2

0

Integrating and plugging in numbers gives: mol L ⋅atm

PFR,b FA0RT

( + )

1 − 3X

= 2.5

min 0.082

mol ⋅K (500.15K )

4ln

1 − 3 0.9

V = 1 3 ln ( )

Pk 1− X (10atm)0.206255 1 1− 0.9

min = 323.63L ≈ 3.2 ⋅102 L




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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.

FA0 FA1, FB1, FC1

X1

FA2, FB2, FC2

X2

FA0FA1, FB1, FC1

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 V =

FA0RT ( )3 ln 1

− 3 VPFR,bPk

− ( )1+ 1

+ 3X = 0 2 PFR,b Pk

1+

1− X1

X1

⇒ 2FA0RT

3 ln1− X1

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:

function [Xpfr]=partc;

X0=0.1; %initial guess for X_pfr

[Xpfr] = fsolve(@pfr_eqn,X0);

return

function F = pfr_eqn(X)

k = 0.206255; %1/minT = 227+273.15; %K FA0 = 2.5; %mol/minP = 10; %atm R = 0.082; %L*atm/mol/KV_pfr_b = 323.63; %L, from part b calculation

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:

f ( )xnxn+1 = 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: VCSTR,a kVCSTR,a kVCSTR,a FA2 0 = F − F + r = F 1− X − F 1− X − A = F X − X −A1 A2 A 2

A0 ( 1) A0 ( 2 ) 2

[ ] A0 ( 2 1) 2 ν

kVCSTR,a PFA0 (1− X 2 )

kVCSTR,aP (1− X 2 )0 = FA0 (X 2 − X1 )− 2

FA0RT (1+ 3X 2 )= FA0 (X 2 − X1)−

2RT (1+ 3X 2 )

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.

FA0FA0

FA2, FB2, FC2FA2, FB2, FC2FA1, FB1, FC1FA1, FB1, FC1

X1 X2X1 X2

CSTR equation will be the same as derived in part a, but with V = ½ VCSTR,a and X = X1:

1 FA0RT (1 + 3X1 )X1 1 FA0RT (1 + 3X1)X1

2 VCSTR,a =

Pk(1 − X1 )⇒

2 VCSTR,a −

Pk(1 − X1)= 0

Plugging in numbers and solving gives X1 ≈ 0.83

A mole balance on the PFR gives (see pg 15-16 in Fogler): 0 = FA V − FA V +∆V + rA∆V

Rearrange, divide by ∆V and take the limit as ∆V approaches zero to get: dFA = rA ⇒ dV =

dFA = d[FA0 (1 − X )]

= − FA0dX

= FA0 dX

dV rA rA rA − rA

Integrate with the limits V = 0 when X = XA1 and V = ½ VPFR,b when X = X2 to get:

1VPFR,b = FA0

X

∫ 2 dX

= FA0

X

∫ 2 dX

=FA0

X

∫ 2 dX

= FA0

X

∫ 2 FA0RT (1+ 3X )dX =

FA0RT X

∫ 2 (1 + 3X )dX

− rA [ ] k FA PkFA0 (1 − X Pk ( −2 k A ) 1 X )X1 X1 X1 ν

X1 X1

From integration by parts (or an integral table): x2 (1 + mx)dx = (1 + m)ln

1 − x1 + m(x1 − x2 ) ⇒ 1VPFR,b =

FA0RT (1 + 3)ln

1− X1 + 3(X1 − X 2 )

∫ (1 − x) 1 − x2 2 Pk 1− X 2 x1

FA0RT (1+ 3)ln

1− X1 + 3(X1 − X 2 )

−

VPFR,b = 0 Pk 1− X 2 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

Total

Initially (mol) NA0

0 0

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 = P ( + 3X ) = 2.7atm( + 3 0.9 = 9.99atm0 1 1 ( ) )

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 mol ⇒ N A0 = 2.5 mol

treacttreact min min

mol

Assuming ideal gas, Vbatch = N A0 RT

= 2.5

min treact RT

P0 P0

Find treact from design equation for a constant volume batch reactor (Eq 2-6 in Fogler):

[ ] k N A V kN (1 − XdX =− rAV

= k AV

= V = A0 )= k(1− X )

dt N A0 N A0 N A0 N A0

Rearrange and integrate: X t

∫dX

= k ∫dt ⇒ ln 1

= kt ⇒ treact =

1 ln 1

(1 − X ) 1 − x k 1 − X 0 0

Plug in numbers to get:

mol 1 mol 1 L ⋅atm 2.5 ln RT 2.5 ln 0.082 (500.15K ) min 1− X min 1− ( ) mol ⋅K 20.9

Vbatch = kP

= (0.206255min−1)(2.7atm)= 423.94L ≈ 4.2 ⋅10 L

0

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:

NH4NO3(liquid) Æ N2O(g) + 2 H2O(g)

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.

Also, FYI: Cp(steam) = 2 kJ/kg-degree F Cp(liquid NH4NO3) = 0.8 kJ/kg-degree F

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.




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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?

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.

(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?

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?

(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)

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.

Also, FYI: Cp(steam) = 2 kJ/kg-degree F Cp(liquid NH4NO3) = 0.8 kJ/kg-degree F

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.




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gaseous products at TR

liquid feed, including water and ammonium nitrate at 200oF

Pure liquid ammonium nitrate at TR

CSTR

This document will follow these symbolic conventions whenmultiple interpretations are possible:

An extensive property is given by an underbarA molar intensive property has no extra notational

information An intensive property that is per unit of mass is

given with a carrotExtents of reaction are in units of moles per time

Also, ideal mixing is assumed.

(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.

Input state: liquid water and liquid ammonium nitrate arefed at 200oF

Reaction State: temperature is TR, negligible gases arepresent

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 = N2O C = H2O

First, define extent of reaction ξ& in terms of reaction rate:




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ξ& = rV = k[A]V = kN A = (k / MWA )(N AMWA ) = (k / MWA )(mA )

Mole/mass balances on each component in the system yieldthe following relationships:Ammonium Nitrate Balance dmA = m& Ain − m& Aout −ξ& MWAdt Apply steady state, no liquid exits:0 = m& Ain − kmA

Equivalent mole balance:0 = FAin −ξ&

Nitrous oxide mass balance dmB = m& Bin − m& Bout +ξ& MWBdt Apply steady state, nothing enters

0 = −m& Bout + k mA MWB

MWA

Equivalent mole balance:0 = −FBout +ξ&

Water mass balance dmC ξ&= m& Cin − m& Cout + 2 MWCdt Apply steady state

0 = m& Cin − m& Cout + 2k mA MWC

MWA

Equivalent mole balance:0 = FCin − FCout + 2ξ&

Thus we can express the molar extent of reaction in termsof the molar flow rates:

(FCout − FCin )ξ& = 2

= FAin = FBout

Part (a) can be answered using the ss. mass balance oncomponent A:0 = m& Ain − kmA

Hence, we can find the holdup of ammonium nitrate in thereactor at 510 Fahrenheit to be:




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m& Ain = mA =(0.83)( 140kg/h)

= 378.5kgk(510°F ) 0.307h −1

(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 onthe open system.

Enthalpy Balance (assuming negligible gas holdup inreactor), mass and molar formats:

d⎜ c m T ⎟ d⎜ c N T ⎟d H

= ⎝

⎛∑ i

p,i i ⎠

⎞

= ⎝

⎛∑ i

p,i i ⎠

⎞

= d (mAcp, AT )

= d (N Acp,AT )

dt dt dt dt dtd H

=∑m& i,in H i −∑m& j ,out H

j + Q& −W& s =∑Fi,in Hi −∑Fj ,out H j + Q& −W& sdt i j i j

Focus on the molar format, as we will modify this toextract the delta H’s that we know. d (N Acp,AT )

=∑Fi,in Hi −∑Fj ,out H j + Q& −W& sdt i j

Apply steady state condition. Neglect shaft work.

0 = FA,in H A,in + FC ,in HC ,in − FC ,out HC ,out − FB ,out H B,out + Q&

Add and subtract the terms ± FA,in H A,out ± FC ,in HC ,out . We need to do this to get enthalpy differences that we know into theenthalpy balance, as absolute enthalpies have littlemeaning.

0 = FA,in (H A,in − H A,out )+ FC ,in (HC ,in − HC ,out )+K

+ FA,in H A,out − (FC ,out − FC ,in )HC ,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 ofreaction:

ξ& =(FCout − FCin ) = FAin = FBout2

& &0 = FA,in (H A,in − H A,out )+ FC ,in (HC ,in − HC ,out )+ξ[H A,out − H B,out − 2HC ,out ]+ Q

Recall the definition of the enthalpy of reaction




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∆H rxn (TR ) = ∑ν i Hi (TR ) − ∑ν i Hi (TR ) = H B (TR ) + 2HC (TR ) − H A (TR ) products reactants

Since the reaction enthalpy is approximately independent ofT near TR and the reaction occurs at the reactor outlet temperature (also the reactor temperature), the energybalance becomes:

0 = FA,in (H A,in − H A,out )+ FC ,in (HC ,in − HC ,out )−ξ& ∆H rxn + Q&

Putting the energy balance back in terms of mass (moreconvenient for this problem) we have:

0 = m& A,in (H A,in − H

A,out )+ m& C ,in (H C ,in − H

C ,out )− (k(TR ) mA ∆H rxn (TR ))+ Q&

Now we need to know how to calculate the enthalpydifferences in the inflow streams: we need to heat up theinflow to the reaction temperature, TR.

H A,in = molar enthalpy of liquid ammonium nitrate at 200 oF.

H A,out = molar enthalpy of liquid ammonium nitrate at TR

HC ,in = molar enthalpy of liquid water at 200 oF

HC ,out = molar enthalpy of steam at TR

(H C ,in − H

C ,out )= (H (water,200°F ) − H (steam,500°F ))+ (H (steam,(500°F )) − H (steam,TR )) ( − H )= − 2.2MJ/kg) ( − 500 F ) 2

kgF° ⎝1000kJ ⎠ ⎝−1.2 −

500F° ⎠MJ/kgH

C ,in ˆ

C ,out ( − TR ° ⎜⎜⎛ kJ

⎟⎟⎞⎜⎛ 1MJ

⎟⎞ = ⎜

⎛ TR ⎟⎞

⎝ ⎠

(H A,in − H

A,out )= −(TR − 200°F )⎛⎜⎜ 0.8kJ ⎞

⎟⎟ = −(0.8TR −160°F )kJ/kg = −(0.8TR −160°F )

MJ/kg ⎝ kgF° ⎠ 1000

Solving the energy balance for the heat required forisothermal operation at 510 Fahrenheit we find:

0 = m& A,in (H A,in − H

A,out )+ m& C ,in (H C ,in − H

C ,out )− (k(TR ) mA ∆H rxn (TR ))+ Q&

Q& = −m& A,in (H A,in − H

A,out )− m& C ,in (H C ,in − H

C ,out )+ (k(TR ) mA ∆H rxn (TR ))




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&Q = −(0.83)( 140kg/h)⎛⎜− (0.8(510°F )−160°F )

MJ/kg ⎞⎟ +L ⎝ 1000 ⎠

− (0.17)( 140kg/h)⎛⎜−1.2 −(510°F )⎞

⎟MJ/kg + ...⎝ 500F° ⎠

−1(0.307h )(378.5kg) ( − 0.740MJ/kg)

Q& = 28.8MJ/h + 52.8MJ/h − 86.0MJ/h = −4.4MJ/h If Ta = 100oF, then we may find UA (given that there isexcess cooling water, assume that the temperature change inthe cooling water is negligible).

Q& = UA(Ta −TR )

Q& /(Ta − TR ) = UA =− 4.4MJ / h

= 0.01073MJ / hF° = 0.00298kW / °F(100 − 510)F°

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.

Given two T values and two k values, calculate k(T). Note that in an Arrhenius expression, an ABSOLUTE temperaturescale must be used. Recall the temperature in Rankine (anabsolute scale) is just the Fahrenheit temperature plus459.67.

⎛ − b ⎞k(T / °F ) = Aexp⎜ ⎟ ⎝ T / °F + 459.67 ⎠

⎛ − b ⎞k(510) = Aexp⎜ ⎟ ⎝ 510 + 459.67 ⎠ ⎛ − b ⎞k(560) = Aexp⎜ ⎟ ⎝ 560 + 459.67 ⎠

− ln⎛⎜⎜

k(510) ⎞⎟⎟ = − ln⎛⎜

0.307 ⎞⎟ =

⎛⎜

b ⎞⎟ −

⎛⎜

b ⎞⎟

⎝ k(560) ⎠ ⎝ 2.91 ⎠ ⎝ 510 + 459.67 ⎠ ⎝ 560 + 459.67 ⎠ 2.24906 = (5.0692E − 05)b b = 44367

A = k(510) / exp ⎛⎜ − 44367 ⎞

⎟ = 2.28E19h−1

⎝ 510 + 459.67 ⎠ Note that k(T) is very sensitive to small changes in theactivation energy (or b in this case).




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ab = E (°Rankine)R

(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 thesteady state at 510oF and see if a perturbation could leadto a hotter, higher conversion steady state.Define G(T) and R(T) as follows:

− km ∆H G(T ) = A rxn =energy produced per unit mass of A fed. One

m& A,in

could also define G(T) as energy produced per mole fed, asit is in Fogler, but mass is a more convenient basis inthis problem.

G(T ) = R(T ) at steady state: R(T) is all the rest of the termsin the enthalpy balance (multiplied by -1, divided by themass flow rate in)

& R(T ) =

− m& A,in (H A,in − H

A,out )− m& C ,in (H C ,in − H

C ,out )− Q m& A,in

Near a steady state, there is no accumulation of A in thereactor and: m& A,in = kmA

Thus G(T) simplifies to: G(T ) = −∆H

rxn

Since there is no outflow, the conversion must always beunity. This makes the G(T) curve much simpler than it isin the standard CSTR in Fogler.

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)

T

G(T), R(T) G(T)

510oF

Since both G(T) and R(T) are lines, there is only onepossible intersection and one possible steady state. If you perturb the steady state to a slightly hottertemperature, R>G and the system will cool down. If youperturb the system to a slightly cooler temperature, G>Rand the system will warm up. Thus the steady state isSTABLE.

Stability analysis is only useful for small perturbationsfrom a steady state. Determining the magnitude of aperturbation that will cause the system to “blow up” isbeyond the scope of the problem.

The analysis in Fogler is to find perturbations that causemotion from one low-temperature steady state to a highertemperature steady state (ignition). As this transition is not possible given our G and R curves, the analysisrequired 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 mA (t = 0) = 378.5kg




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T (t = 0) = 510°F

Batch Reactor Balances: for subtle reasons, it is notpossible to just set the inflow terms in the CSTR energybalance to zero.

Given that there is negligible gas holdup in the reactor:N B (t) = NC (t) = 0 hence, dNB dNC= = 0 dt dt

Enthalpy:d H

= d (N AH A + NB H B + NC HC ) =

d (N AH A ) = dN A H A +

dH A N Adt dt dt dt dt d H

= Q& − FB ,out H B − FC ,out H Cdt dN A dH A &H A + N A = Q − FB ,out H B − FC ,out H Cdt dt

Batch Reactor mole balances dN A = −kN Adt dNB

dt = 0 = −FB ,out + kN A

FB,out = kN A

dNC = 0 = −FC ,out + 2kN Adt FC ,out = 2kN A

Plugging in the three mole balance relationships into theenthalpy balance we find:

− kN AH A + dH A N A = Q& − kN AH B − 2kN AHCdt

dH A N A = Q& − kN A (H B + 2HC − H A ) = Q& − kN A∆H rxn (T )dt

dH A dT = c

dt p ,a dtdT

= Q& − kN A∆H rxn (T )

= Q& − kmA∆H

rxn (T ) dt N Acp ,a mAc p,a




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k(T / °F ) = Aexp(−b /(T + 459.67)) ∆H rxn (T ) ≈ ∆H

rxn (510°F ) Now we have an ode set and initial conditions for a Matlab simulation: mA (t = 0) = 378.5kg T (t = 0) = 510°F dmA = − k(T )mAdt dT

= UA(Ta −T )

+− k(T )∆H

rxn

dt mAcp,a c 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 minutesthours = 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 NH4NO3T = x(2); %degFk = 2.28e19 *exp(-44367/(T+459.67));UA = 0.01073; %MJ/(h degF)Ta = 100; %degFcpa = 0.8/1000; %MJ/(kg NH4NO3 degF)dHrxn = -0.740; %MJ/kg NH4NO3dmdt = -k*m;%dTdt = (-k*dHrxn/cpa) + (UA*(Ta-T)/(m*cpa)) + (T*k); %slightly wrongdTdt = (-k*dHrxn/cpa) + (UA*(Ta-T)/(m*cpa)); %correctderivs = [dmdt;dTdt];return;




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Temperature Plot:

Perturbing the most sensitive problem parameter, theactivation energy term, b, by +1% gives similar lookingplots with explosion times shifted to 3-8 minutes.

(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?

The shutoff of the reactor turned a stable steady-statesystem of operation into an unsteady system, far from asteady state. Steady state in the batch reactor would meanthe reactor is empty at 100oF. In order for the batch reactor to reach this steady state, the reactor must cooldown. One can plot mass-temperature trajectories of thereactor operating in batch mode given different initialholdup masses and temperatures.




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Green circles represent initial conditions, red circlesrepresent final conditions after 4 hours of integration.Black lines represent trajectories and connect greencircles to red ones. Diverging trajectories lead to atemperature “explosion”. The absolutely final steady statewould be an empty reactor at the temperature of the coolingwater.

It is clear that above ~425oF it is not safe to shut off the reactor; this threshold “safety limit” temperatureincreases slightly as mass holdup decreases. Since the initial conditions are above this threshold, the suddenshutoff of the feed was a BAD IDEA.

In order to prevent the disaster, one must cool the reactorbelow the critical mass/temperature line before shuttingoff 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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Code for part e:function stabilitydiagram(tminutes);%stabilitydiagram(240); %integrate trajectories for at most 240 minutesthours = 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 stopplot(m(1),T(1),'og'); %green goplot(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;returnfunction derivs = odefun(t,x)m = x(1); %kg NH4NO3T = x(2); %degFk = 2.28e19 *exp(-44367/(T+459.67));UA = 0.01073; %MJ/(h degF)Ta = 100; %degFcpa = 0.8/1000; %MJ/(kg NH4NO3 degF)dHrxn = -0.740; %MJ/kg NH4NO3dmdt = -k*m;%dTdt = (-k*dHrxn/cpa) + (UA*(Ta-T)/(m*cpa)) + (T*k); %slightly wrongdTdt = (-k*dHrxn/cpa) + (UA*(Ta-T)/(m*cpa)); %correctderivs = [dmdt;dTdt];return;

(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 graduallyThe concentration of ammonium nitrate could be rampedup/down graduallyGreater cooling/heating control could be utilized atstartup and shutdown




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10.37Problem 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).

Problem 2.

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

Xbiomass productivity for a chemostat vs. a batch reactor i s approximately !n ?i;+frmnxttum.

Problem 3.

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?

Cite as: William Green, Ir., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring 2007. M I T OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month W].



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Problem 1. The binding of protein P with ligand L to form complex C is reversible, as told

CLP onk⎯→⎯+ and LPC offk +⎯⎯→⎯

We are given a table with various initial concentrations of L in order to estimate kon and koff and also Kd for the reaction.

CoffLPonC CkCCk

dtdC

−=

Also from material balances and stoichiometry, we have CP+CC=CP0 and CL+CC=CL0, therefore

CoffCL0CP0onC Ck)C)(CC-(Ck

dtdC

−−=

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

)t]kCexp[-(k1KC

CC)t]kCexp[-(k-1kCk

CCkC offL0ondL0

L0P0offL0on

offL0on

L0P0onC +−

+=+

+=

where on

offd k

kK ≡

Therefore, if we plot CC w.r.t time for each cases of CL0, we can fit according to an exponential

y=a[1-exp(-bt)], where b is konCL0+koff, and a is dL0

L0P0

KCCC+

. Values for a and b are shown in the

following table. L0(uM) a b

1 0.903 1.11135 1.0436 4.704715 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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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

time(t)

Con

cent

ratio

n of

C(n

M)

CL0=1uM

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time(t)

Con

cent

ratio

n of

C(n

M)

CL0=5uM




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0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

time(t)

CL0=15uM

Con

cent

ratio

n of

C(n

M)

0 5 10 15-2

0

2

4

6

8

10

12

14

16

CL0(uM)

b




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Problem 2. For a steady state chemostat, the material balance on cell mass yields

DVF

≡ =μ

the volumetric productivity is F(X- X0)=μV(X- X0)

For a batch reactor, the material balance on cell mass yields

XdtdX μ=

where the initial condition is X(t=0)=X0.

Therefore, we have )texp(XX 0 μ=

The volumetric productivity is

turn0

0

turn

0

tXXln1

)X-V(Xtt

)X-V(X

+=

+μ

Therefore, the ratio of the two

turn0

turn0

0t

XXln

tXXln1

)X-V(XX0) -V(X μ

μ

μ +=

+

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

turnmax0

tXXln μ+




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Problem 3. The expression that would be suitable to describe the change of protein expression is:

])(exp[1)(

kkC PPr

rPP tμγ

μγγ+−−

+=

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

])(exp[195.0 P tμγ +−−= or μγ +

−=P

05.0lnt

a) if cells are rapidly growing with a doubling time 30 min and stable protein with a degradation half-time one day, i.e.

30minln2=

μ and day 1ln2

P

=γ

So the half-time for switching is the time need to reach

])(exp[12195.0 P tμγ +−−=×

min9.27

12ln

min302ln

)210.95-ln(1-

)(

)210.95-ln(1-

P

≈+

×=

+

×=

day

tswitching μγ

b) if cells are not growing at all and the protein with a degradation half-time one hour, i.e.

hr 1ln2

P

=γ

So the half-time for switching is

hr

hr

tswitching 93.00

12ln

)210.95-ln(1-

)(

)210.95-ln(1-

P

=+

×=

+

×=

μγ

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.

1




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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)

Rate has units of mol/hr-m3 of porous material.

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:

CO + M = CO—M Keq1 at 700 K= 1 (atm)-1

NO + M = NO—M Keq2 at 700 K= 3 (atm)-1

The N—M formed in the rate-controlling step rapidly combine to make and release N2.

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.

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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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.

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?

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?

e) Compute how the NO conversion varies if you double the thickness of the porous layer.

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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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 internal-diffusion-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

e) Yes, the reaction is limited by internal diffusion.

f) Internal-diffusion limitations are seen at: 367K < T > 377K, FT0 > 1000 mol/hr

' mol actual _ rate(external _ lim) − rA

10 hr

,360K .25g) Ω =

ideal _ rate(rxn _ lim) =− rA

' 5000 mol ,360K ≈

.70 = 0.36

hr

' mol actual_rate(internal_lim) − rA 5000

hr ,367K

h) η = ideal_rate(rxn_lim) =

'

mol

− rA 5000 ,367K

hr Extrapolate the reaction-rate limited portion of the FT0 = 5000 mol/hr curve up to 367 K.

1.2η ≈ = 0.81.5

i) Fort a first order reaction with spherical pellets:

η =φ 3

2 (φ1 cothφ1 −1) = 0.8 1

Solve: φ1 = 2.0




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C sinh( )1 φ λΨ =

CAA

,S

= λ

sinhφ 1

1 with λ = r / R = 1/ 2

1 sinh( ) mol ( ) φ λ sinh 1 mol A CA,S λ sinhφ = 1

L ( ) sinh( )2

= 0.65 L

C = 1 2 1

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 diffusion limited

L

rnal diffusion limited

Rxn limited

ex

inteint

ternal diffusion limited

L

ernal diffusion limited

Rxnlimited

TT

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.

L

external diffusion limited

no external limitations

L

externaldiffusion limited

no external limitations

FT0FT0




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2. a)

rlim = kθNO−M θCO−M Area of noble metal

Volume of porous material

[i −M ]where θ i−M = [ ]oM

Use the adsorption steps (rapid equilibrium) to find θi−M .

[CO − ] θ M K P MKeq,1 =

M = CO−M [ ]o ⇒ θCO−M = eq,1 CO [ ]

PCO M M[ ]M PCO [ ] [ ]o

M [ ] [ ][NO − ] θ M K P MKeq,2 =

PNO

= NO−M

M o ⇒ θNO−M = eq,1

MNO

[ ]M PNO [ ] [ ]o

Use and overall site balance to find [M] (concentration of empty sites).

[ ] [ ] [ = M + CO −M ] [NO −M = M +θ [ ] + θ MM o + ] [ ] CO−M M o NO−M [ ]o

Substitute in the equations for θi−M :

[ ] [ ]M o Keq,1PCO [ ]M [ ] +

Keq,2 PNO [M ]M o = M + [ ]M o

M o [ ] [ ]

M o

Solve for [M] and use the simplifications given in the problem statement (Keq2PNO << Keq1PCO and PCO = PCO,S):

[ ]M = [ ]M

= [ ]Mo o

1+ Keq,1PCO + Keq,2 PNO 1+ Keq,1PCO ,S

Substitute θi−M and [M] into given rate equation:

P MKeq,1PCO ,S Keq,2 NO [ ]o 2 Area of noble metal

− rNO = rlim = k [ ]o [ ]o 1 + Keq,1PCO ,S


M M

Keq,1PCO ,SKeq,2 Area of noble metal − r = k P = k ' PNO (1 + Keq,1PCO ,S )2


NO NO

or, in terms of NO concentration:




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catalyst

− rNO = k ' RT [NO] = k"[NO]

where k" = k Keq,1Keq,2 PCO ,S

2 Area of noble metal

RT(1 + Keq,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 gaseBulk sgases

pipe wall

[NO]S

x

x = L catalyst

pipe wall

[NO]S

x

x = L

At steady state, diffusion of NO inside the catalyst is governed by:

2∂ [NO]Dpores + rNO = 0∂x2

2∂ [NO]Dpores 2 − k"[NO] = 0∂x

Non-dimensionalize:

ψ ≡ [NO] λ ≡

x ⇒

∂2 Ψ 2 −

k" L2 Ψ = 0

[NO]S L ∂λ Dpores

So, the Thiele modulus is defined as:

2 k" L2

φ ≡ Dpores

The equation: ∂∂

2

λΨ 2 −φ

2Ψ = 0

Has a solution of the form: ψ = Ae−λφ + Beλφ

The boundary conditions are:




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1. Ψ = 1 ([NO] = [NO]S at the surface)λ=1

∂Ψ2. = 0 (no flux at pipe wall)∂λ λ=0

From the second boundary condition:

∂ψ = −Aφe−λφ + Bφeλφ

∂λ

0 = − Aφe0 + Bφe0 ⇒ A = B

From the first boundary condition:

1 = Ae−φ + Aeφ ⇒ A = −φ

1 φe + e

Solution of the concentration profile of NO inside the catalyst (treated as a slab):

e−λφ + eλφ

ψ = e−φ + eφ

We want the internal effectiveness factor (η) as a function of the thiele modulus (φ).

actual rxn rateη = rxn rate is all catalyst was exposed to [NO]S

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.

actual rxn rate = (flux)(surface area ) = Dpores ∂[NO] Asurface = Dpores

[NO]S ∂Ψ Asurface∂x L ∂λ λ=1

actual rxn rate = Dpores [NO]S

−φe

−φ

−φ + φφ

eφ

AsurfaceL e + e

rxn rate if all catalyst was exposed to [NO]S = k"[NO]SVcatalyst




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η =

Dpores [NOL

]S −φ ee −φ

−φ

++

e φφ

eφ

Asurface

= Dpores − φe−φ + φeφ 1 eφ − e−φ

k"[NO]SVcatalyst k" L2 e−φ + eφ

= φ

eφ + e−φ

(note that the surface area exposed to the bulk gas divided by the volume of porous material is equal to 1/L)

c) Ω = actual rxn rate

=ηk"[NO]SVcatalyst = η[NO]S

rxn rate is all catalyst was exposed to [NO]bulk k"[NO]bulkVcatalyst [NO]bulk

Need to use flux balance to find [NO]S.

At steady state:

flux to the catalyst surface = rxn inside catalyst

kc Asurface ([NO]bulk − [NO]S ) =ηk"[NO]SVcatalyst

Solve for [NO]S:

[NO]S = kcAsurface [NO]bulk =

kc [NO]bulkηk"Vcatalyst + kc Asurface ηk" L + kc

Substitute into the omega equation:

ηkc [NO]bulk ηkcΩ = = [NO]bulk (ηk" L + kc ) ηk" L + kc

where kc = Dbulk/delta

d) Design equation for a PFR:

dX − rNO AcVporous _ material Ωk"[NO]AcVporous _ material Ωk"[NO]0 (1 − X NO )AcVporous _ materialNO = = =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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In order to calculate kc, need to first calculate delta using the equation given in the problem statement:

.001m ρUd pipeδ = where Re = Re0.5 µ

To calculate the superficial velocity (U) in each pipe, use the equation:

U = qo =

FNO ,o

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,totalFNO ,o =N pipes

The concentration of NO in the feed is found by multiplying the feed mole fraction by the total concentration:

FNO ,o P[NO]o = yNO ,oCtot = ∑Fi ,o RTi

See matlab code below:

function [X1f,X2f,X3f,X4f,X5f] = HW10_P2_allparts();%[X] = HW10_P2_allparts()

[z1,X1] = reactorsimulation(1E-6, 2000, 0.2);[z2,X2] = reactorsimulation(2E-6, 2000, 0.2);[z3,X3] = reactorsimulation(1E-6, 4000, 0.2);[z4,X4] = reactorsimulation(1E-6, 2000, 0.4);[z5,X5] = reactorsimulation(.4E-6, 5000, 0.2);

%[z,X] = reactorssimulation(catalyst thickness, number of pipes, length of%pipes, total mol/hr in feed)

X1f = X1(length(z1));X2f = X2(length(z2));X3f = X3(length(z3));X4f = X4(length(z4));X5f = 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)')

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)(parth)')

return;

function [z,X] = reactorsimulation(thicknessofcatalyst,numberofpipes,...lengthofpipes);

%a = NO%b = CO%Reactor/Catalyst Geometry and Chemistryparam.L = thicknessofcatalyst; %mN = numberofpipes;Lp = lengthofpipes; %mparam.T = 700; %in Kelvindpipe = 8/1000; %inner diameter of each pipe 8 mm, in mparam.Ac = pi*(dpipe/2)^2; %cross sectional area of each pipe, m^2

%Reactor/Feed ConditionsP = 1*101325; %pressure = 1atm, in Paparam.R = 8.314; %gas constant, J/(K mol)Ctot = P/param.R/param.T; %total concentration of gas, moles/m^3ya0 = 4/1624; %feed mole fraction of NOyb0 = 20/1624; %feed mole fraction of COFtot = (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/sFb0 = yb0*Ftot; %CO feed molar flow rate into each pipe, in mol/sparam.Ca0 = ya0 * Ctot; %feed concentration of NO, in mol/m^3Cb0 = yb0 * Ctot; %feed concentration of CO, in mol/m^3param.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/sU = q/param.Ac; %superficial velocity in each pipe, in m/s

%Physical Property DataDbulk = 1E-8; %in m^2/smu = 3E-5; %viscosity in Pa*s = kg/m/srho = 0.5; %density in kg/m^3Re = rho*U*dpipe/mu; %Reynolds numberdelta = 0.001/(Re^0.5); %thickness of boundary layer, in mparam.kc = Dbulk/delta; %m/sparam.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 parametersT = 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;

Dpores = 1E-9; %in m^2/sArea_metal_per_Vol_porous_material = 0.1*100^3; %m^2 platinum/m^3 porousmaterialKeq1 = 1/101325; %Pa^-1Keq2 = 3/101325; %Pa^-1k = 400/60/60; %reaction rate constant, in mol/s/m^2(of noble metal)

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^-1return




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d) Original conditions (does not meet emission specifications).

e) Double thickness of porous layer - more conversion (more catalyst).




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f) Double number of parallel pipes – more conversion (lower flow rate).

g) Double length of pipes – higher conversion.




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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.




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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:

rdf r

where time is the independent variable, f is the (unknown) vector of dependent

r g( f , t)= ,dt

r

variable. Given an initial condition for each dependent variable, f0 , the equations can be

rvariables, and g is the vector of functions giving the rate-of-change of each dependent r

rintegrated 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.

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 r 2 A→ B →C

r1 = k1[A]

r2 = k2[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

Cite as: David Adrian, 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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d[B] = k [A] − k [B]

dt 1 2

d[C] = k2[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.

The derivative function has the form:

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.

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:

1

2

dt

subject to the following initial condition:

r f0

f1 (t) 15exp( t)

f 2 (t)

Feel free to use the structure of the example file, odesintegrate.m as a template to build from.

f f

1 2

0 5

df1

dt df

10 10

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:

⎡⎢⎣

−5exp( 5t)−−=

−10exp( 5t) . =

⎤⎡⎤⎡⎤−⎡⎥⎦

⎢⎣⎥⎦−⎢

⎣=

⎥⎥⎥2

⎢⎢⎢

⎦⎣

⎤⎥⎦

=




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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:

function [value,isterminal,direction] = eventfunctionname(t,f,param);

value = f(1); %every zero in the value vector is an event

isterminal = 1; %isterminal =1/0: stop/keep integrating

direction = 0; %0/+1 or -1: direction doesn’t matter/does matter

return;

See learntouseevents.m for a more sophisticated example of the use of event functions, including recovery of the log of events.

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.

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,…);

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):

x = linspace(0,10,101); y1 = sin(x); y2 = cos(x);

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…);

Also, a grid can be overlaid on the plot with

grid on;

This should cover most plotting needs you will have. Advanced options are available. Look up the Matlab “help plot” for details.


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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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:

initial_guess=[1.00,1.00,0,0];

Note that we guessed our values as those that you should get if there were no noise in the data. Finishing up, call:

options=fitoptions(model);set(options,'StartPoint',initial_guess);fresult = fit(x,f,model,options);

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

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:

0 = g r(x r) .

This could be interpreted as directly finding the steady state of the equation

r rdf r

= 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.

The calling sequence and use of fsolve is

x = fsolve(@gfun,x0,options,param);

Where x0 is the initial guess, param are additional constants needed by the g function, and gfun is the gfunction:

[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.

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:

J ij = ∂∂

xg

j

i .

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.


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:

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.

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.

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.

Cite as: David Adrian, 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].www.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in


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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

0 0A A AF F r V− + =

[ ]0 00AF A ν=

[ ]AF A ν=

For a liquid phase with constant density: 0ν ν=

For A B→ the reaction moles are the same, so 0ν ν=

For 2A B→ , 0ν ν≠

[ ]ξ = moles (extent of reaction) (-) for a reactant and (+) for a product

0 ,1

rxnsN

i i i nn

N N nυ ξ=

= + ∑

Suppose A B C→ +

0

0

A AA

A

N NXN

−=

( )0 1A A AN N X= −

Thermodynamics

Suppose 1

12k

kA B C

−+

G RTeK e Δ−=

, ,f products f reactantsG G GΔ Δ Δ° °= −

[ ][ ][ ]c

B CK

A= has units. You need to use standard states, such as 1M, to make it

dimensionless.

Enzyme Catalysis

Reaction Progress

Ener

gy

S

P

with enzyme

without enzyme

Figure 1. Energy diagram for a reaction with and without enzyme.




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E S+ ES ES E→ + P Pseudo steady state approximation:

[ ]d ES= 0

dt[ES ] = f ( )other species

Cell Growth # cellsN =

volumeN N= t

0eμ

Monod kinetics:

[ ][ ]

μmax Sμ =

K SS +ΔAYA =

B ΔB

Rate Constants k T( ) = Ae−E Ra T Given k1 and k2, you can calculate k and a different temperature.

CSTRs F XV = A0 A

−rA

If the reaction is 1st order and it consists of liq ds with constant density:

Incorporates changing volumetric flow rate

uiXτ = A

k X(1− A)V volumeτ = = ν volumetric flow rate0

τ kX A = 1+τ k

Da = τ k =Damköhler number: ratio of kinetic effect to volumetric effect or ratio of reaction rate to dilution rate 2nd order reaction:

10.37 Chemical and Biological Reaction Engineering, Spring 2007 Exam1 Review Page 2 of 10 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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( )Xτ = A

kC 1 2A A0 − X

Da =τ kCA 0

1 2+ −Da 1 4+ DaX A = 2Da

For constant density and 1st order reaction:

dNF F AA A0 − + rAV =

dtdCC C A

A A0 − + rAτ τ= dt

Let:

ˆ CA tC tA = =ˆ CA0 τ

Nondimesionalize:

dCA + +(1 Da)C 1dt A =

Solve given C tˆ 0 0atA = =ˆ

1C eˆ ( )1 − +(1 Da t) ˆA = − 1+ Da

Tanks in series: 1st order reaction

( )C

C A,0A n, =

1+ Da n

Reactor Design Equations

CSTR: F XV = A0 A

−rA

Batch: dNr VA = A dt

N V= [A] If V changes, then V must remain in the differential A

PFR: dX A −r

= A Adz FA0

PBR: Pressure drop consideration If ( ) ( )2g gA B→

Use 0 0 (conservation of mass) ν ρ νρ=




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Introduce ideal gas law T

PmF RT

ρ = TmnRT P P F RTνρ

⎛ ⎞= = =⎜ ⎟

⎝ ⎠

( )0 1T TF F Xε= +

Reactor volume:

Ao

A

Fr−

VPFR (area under curve)

AX

VCSTR positive order reactions

Figure 2. Levenspiel plot for a CSTR and a PFR for positive order reactions.

Selectivity

instφ

AC fAC 0AC

overallΦ ACΔ

CSTR

Figure 3. Fractional yield versus concentration. Overall yield times concentration difference shown for a CSTR. CSTR

1 2k kA P C⎯⎯→ ⎯⎯→ 3kA U⎯⎯→

[ ] [ ]( ) [ ]1 3 0Ak A k A V F A ν+ = −

Non-ideal reactors Residence time distribution E(t) 10.37 Chemical and Biological Reaction Engineering, Spring 2007 Exam1 Review Page 4 of 10 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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t( ) ( )reactort Cδ ⎯⎯⎯→ E(t) must have a pulse trace

0

0

( )( ) ( ) 1( )

C tE t E tC t dt

∞

∞= =∫∫

0

( )mt tE t∞

= ∫ dt

Mean residence time, tm, for an: Ideal CSTR: τ Ideal PFR: τ

( )22

0

( )mt t E t dtσ∞

= −∫

Variance, σ2, for an: Ideal CSTR: τ2

Ideal PFR: 0 ( ) VE t tδν

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

000

( ) ( ) ( )f t t t dt f tδ∞

− =∫ property of a dirac delta function

For a CSTR, ( )teE t

τ

τ

−

=

Example 1

L z

FA0 FA, FB, FC

Figure 4. Schematic of a PFR with inflow of A and outflow of A, B, and C.

1 2r rA B C⎯⎯→ ⎯⎯→

1 1 2 2A Br k C r k C= =

0

( )moles of B produced

moles of A inB

BA

F LYF

==

Mole balance on B

1 2 1 21 B

B Axs

dF r r r k C k CA dz

= = − = − B

0B BF C ν=

0 0A AF C 0ν=



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01 2

BA B

xs

dC k C k CA dzν

= −

01

AA

xs

dC k CA dzν

= −

1

0

xsA

A

k AdC dzC ν

−=

1

0

ln xsA

k AC z φν

−= +

Initial condition at z=0 gives:

0ln 0AC φ= +

10

0

exp xsA A

k AC C zν

⎡ ⎤−= ⎢ ⎥

⎣ ⎦

2 1 10

0 0 0

expxs xs xsBB A

k A k A k AdC C Cdz ν ν ν

⎡ ⎤−+ = ⎢ ⎥

⎣ ⎦z

[ ]0

time, call it xsA z τν

=

This is the time it takes for something to flow to the end of the reactor (of length z).

0 velocityxs

VAν

=

12 1 0

kBB A

dC k C k C ed

τ

τ−+ =

Integrating factor: 2ke τ

( )2 121 0

k kkB A

d C e k C ed

ττ

τ−⎡ ⎤ =⎣ ⎦

( )2 11 1 0

2 1

k kk AB

k CC e ek k

ττ φ−= +−

Initial condition: z=0, CB=0 B

1 0

2 1

0 Ak Ck k

φ= +−

1 21 0

2 1

( ) k kAB

k CC ek k

eτ ττ − −⎡ ⎤= −⎣ ⎦−

21 C

xs

dF rA dz

= +

11 A

xs

dF rA dz

= −

1 21 B

xs

dF r rA dz

= −




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1 0CA B

xs

dFdF dFA dz dz dz

⎡ ⎤+ + =⎢ ⎥⎣ ⎦

0A A BF F F F= + + C

B

0C A AF F F F= − +

F

τ

A B C

find

τ∗

Figure 5. Graphs of flow rates of A, B, and C as a function of residence time.

( )1 2* *1 01 2

2 1

0k kAB k CdC k e k ed k k

τ τ

τ− −= − +

−=

1 2* *1 2

k kk e k eτ τ− −=

1 1 2 2ln * ln *k k k kτ τ− = −

( )11 2

2

ln *k k kk

τ= −

1

2

1 2

ln*

kk

k kτ =

−

L’Hopital’s rule: * *

( )( ) 0lim lim( ) 0 ( )

x x x x

dA xA x dxdo dyB x dB x

dxdx

→ →=

Find τ* for k1=k2

1 2 1 2

1

2

11 1lim * lim

1k k k k

kk k

τ→ →

= = =

0 0

* xs xsA L Aτ τ zν ν

= =




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0* length of reactorxs

LA

τ ν= =

Example 2 If 2A B→ (Assume negligible pressure drop)

Ar kC=

L z

FA0

Figure 6. Schematic of a PFR.

AA A

dF r kCdV

= − = −

0ν ν↔

( ,totalPV nRT F n Vν= = )=

1total total total

RTF FP

Cν −= =

total A BF F= + F

A A A totalF C y Fν= =

A AA total

A B

dF F Pky C kdV F F RT

= − = −+

2 2B AA total

A B

dF F Pky C kdV F F RT

= + = ++

If P or T changes, you need other equations.

Derivation of E(t) for a CSTR

0 ( )N tδ

Figure 7. Schematic of a CSTR. No reaction:




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0

0

( )

( )

dN N t Cdt

NN tV

δ ν

δ ν

= −

= −

0 ( )dN N N tdt V

ν δ+ =

Integrating factor: exp tVν⎛ ⎞

⎜ ⎟⎝ ⎠

0exp exp ( )d N t t Ndt V V

ν ν tδ⎛ ⎞⎛ ⎞ ⎛ ⎞=⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

0 0exp exp 0N t N NV Vν ν φ⎛ ⎞ ⎛ ⎞= ⋅ = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Initial condition: t=0, N=N0 φ=0

0 expN N tVν⎛ ⎞= −⎜ ⎟

⎝ ⎠

1Vν

τ=

0 exptC C

τ−⎛ ⎞= ⎜ ⎟

⎝ ⎠

0

( ) CE tCdt

∞=

∫

( )0 0 000 0

e 0 1t t

Cdt C e dt C C Cτ τ0τ τ τ

∞ ∞ ∞− −⎡ ⎤ ⎡ ⎤= = − = − − =⎣ ⎦⎢ ⎥⎣ ⎦∫ ∫

0

0

( )t t

C e eE tC

τ τ

τ τ

− −

= =

Long-chain approximation 1 2

1

3

4 Deactivate

k kk

k

k

E S ES P

tRNA E

E

E

−+ ⎯⎯→ +

⎯⎯→

⎯⎯→

Enzyme propagates a long time before it is destroyed. LCA: [ ] [ ]3 4k tRNA k E=

(assume 1st order) If there are other steps, add them into the equation




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5 destructionkES ⎯⎯→

[ ] [ ] [ ]3 4 5k tRNA k E k ES= +

Suppose there is a production term 6kC E⎯⎯→

Add another term

[ ] [ ] [ ] [ ]6 3 4 5k C k tRNA k E k ES+ = +

0 ,AA A

A

FdX r kCAdz r

= − =−

This is a single differential equation in terms of X. Use for PFR with gas flow.

0 0ρ ν ρ= ν

( )0 0

0

11A

A

C X T PCX PTε

−=

+




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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?

Small CSTR

Big CSTRFeed Effluent

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 (Eqn 3) a = 2x104 liter/mole-second b=108 liter/mole.

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.

(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.

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.

(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 = RUeq (1− e−kobs t ).

[L]o (nM) kobs (s-1) RUeq 0.14 0.00177 1.63 0.84 0.00186 9.34 2.1 0.00201 21.5 5.60 0.00245 47.1 14.00 0.00350 82.5 35.0 0.00612 118 84.00 0.0122 141 210.00 0.0280 155 560.00 0.0718 161 1400.00 0.177 163

A) (60 points) Determine Kd, kon, and koff. Are these data self-consistent?

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?

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) a=0.0113 exp(10000/T)

C(s) + H2O(g) Æ H2(g) + CO(g) (Reaction 2) ∆Hrxn2 = +130 kJ/mole

r2 = k2[O2][H2O]/([H2O] + a[H2]) 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-stateapproximation 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

Adiabatic PFR, Volume V1 Only reaction 1

Isothermal CSTR Volume V2 T=1100 K Reaction 1 & 2

Feed 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.

(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?

(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?

(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: A + H2O⎯ → ⎯ ⎯ 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?

c) (10 points) What is the ratio of the steady-state productivity (moles/time) of B for

case b) relative to case a)? Problem 2. (30 points) 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) 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 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. 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: dY/dz = f(Y) 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).




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Solutions. Problem 1. a) For first-order reaction kinetics

sec/3.010000min102.51

sec/3.010000min102.5

k1k

Da1DaX

13-

13-

A

LL

LL

××+

××=

+=

+=

−

−

ττ

=0.58

b) Consider the non-steady state design equation for CSTR, where we can have

dtdNrFF A

AAA0 =+− V

Since in liquid phase with constant density, we have

dtdCkC-CC A

AAA0 ττ =−

or equivalently

τττ A0

AA CCk1

dtdC

=+

+

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

)k1exp(C]k1exp[1k1

C)(C 0 tA,A0

A tttττ

ττ

τ+

−++

−−+

= =

Therefore the conversion at time t is

]k1exp[C

C]k1exp[1

k111)(X

A0

0 tA,A ttt

ττ

ττ

τ+

−−+

−−+

−= =

After 60 min of changing flow rate,

598.0]k1exp[C

C]k1exp[1

k111min)60(X

A0

0 tA,A =

+−−

+−−

+−= = tt

ττ

ττ

τ

The new steady state fractional conversion is

7.0sec/3.010000min102.51

7.0sec/3.010000min102.5

k1k

Da1DaX

13-

13-

A

×××+

×××

=+

=+

=−

−

LL

LL

ττ

=0.665

c) The steady state productivity (moles/time) of B is FB, and the ratio of that in b) vs. that in a) is B

803.058.0L/sec3.0

665.07.0L/sec3.0FF

a B,

b B, =×××

=

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

vVk1

vVk

XA

+=

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.

vVk1

vVk

X1

1

A1

+=

We are told V1=600L-400L=200L, XA1=0.75, therefore we can calculate

vk

=)X-(1V

X

A11

A1= =0.015.

Then for the well-stirred reactor, V=600L

vVk1

vVk

XA

+= =0.9




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Problem 3. For a PFR reactor, the design equation is

in celluose,

celluosecelluose

Fr-S

dzdX

=

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

watercellulosein total, MW

20%)1(gram/sec 60MW

20%gram/sec 60n −×+

×= =

mol/sec 683.2gram/mol 18

20%)1(gram/sec 60gram/mol 720

20%gram/sec 60 =−×

+×

=

Using ideal gas law to get the volumetric flow rate:

PRTn

v in total,in = = liter/sec 61.6

atm 20K600J/mol/K 8.314mol/sec 2.683=

××

Also

30cellulose,in mol/m 52.2

600KK) l8.314J/(moatm 02

)18gram/mol

%201gram/mol720

%20(

gram/mol 720%20

RTPy

][cellulose =×−

+==

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.

a[CO]1][cellulosekr- celluose +

=

where the concentrations should be expressed (constant pressure and temperature are assumed)

X)(1X)-(1][cellulose][cellulose in ε+

=

and

X)(124X][cellulose][H[CO] in2 ε+

==

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.


A surface-catalyzed reaction follows Rideal-Eley kinetics as follows:

A

A

k

kA S A

−

⎯⎯⎯→+ ←⎯⎯⎯ S

12

kAS A A S+ ⎯⎯→ + 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 A

2 as a function of the concentration of

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

An equimolar feed of liquid X-OH and Y-H at 25o

C are fed to a CSTR. At 25o

C, where 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

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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)

a) If the reaction is carried out with the reactor at steady-state at the inlet temperature of 25o

C, is T greater than, less than, or equal to Ta? (5 points)

b) When running the reactor at T = 25o

C 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 = 105o

C. At this temperature, all of the H2O formed evaporates, but the

other species are still liquids. The heat of vaporization of H2O at 105o

C is +40 kJ/mole.

When T=105o

C, the reaction runs to 50% conversion 10x faster than it did at 25o

C, so we

increase the flowrates until the reactor is making 10x as much product as it did at 25o

C (still

at 50% conversion). When we achieve the new steady-state high-productivity operation at

105o

C, will the magnitude of Q (i.e. |Q|) be larger, smaller, or the same as it was when we

were operating at 25o

C? At this steady-state condition, is T greater than, less than, or equal

to Ta? (20 points)

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 185o

C

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 = 300o

C.) 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)

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)




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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

k = 2x104 M−1s-1. The two species are mixed at an initial protein concentration of 3 nM on

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 koff [P]eq [L]eqK d = = kon [C]eq

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

0 = [C]eq 2 − ([L]0 + [P]0 + Kd )[C]eq + [P]0[L]0 = 0

[C] =([L]0 + [P]0 + Kd )± ([L]0 + [P]0 + Kd )2 − 4[P]0[L]0 = 0eq 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] = [L]0[P]0 0.462nMeq Kd + [P]0

[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 = 0.0588 Kd + [L]0

[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]eq

2).)

Start with the full dynamic equation: d[C]

= kon [L][P] − koff [C] = kon ([L]0 −[C])( [P]0 − [C])− koff [C]dt

Make an appropriate approximation: ([P]0 −[C]) ≈ [P]0

d[C] ≈ kon [L]0[P]0 − kon [P]0[C] − koff [C]

dt Rearrange and solve using the integrating factor: d[C]

+ [C](kon [P]0 + koff )= kon [L]0[P]0dtd ([C]exp[(kon [P]0 + koff )t])= kon [L]0[P]0 exp[(kon [P]0 + koff )t]dt

on 0 0 on 0 off([C ] exp [(kon [P]0 + koff ) t ]) = k [L] [P] exp ((k [P] + k )t )

+ I .C .(kon [P]0 + koff ) [C] =

([[ PL ]]

0

0[ +

PK ]0

d )+ I.C.exp[− (kon [P]0 + koff ) t]

Using the initial condition, [C](t = 0) = 0, we find the integration constant to be: [L]0[P]0I .C. = −

([P]0 + Kd ) Hence,




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[C](t) = [L]0[P]0 1− exp[− (kon [P]0 + koff )t]

([P]0 + Kd )

The equilibrium value is clearly the value when t gets large. [C](t) = [C]eq 1− exp(− (kon [P]0 + koff ) 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*) = 0.95 = 1− exp[− (kon [P]0 + koff )t *]

[C]eq

0.05 = exp[− (kon [P]0 + koff ) t *]

− ln(0.05) − ln(0.05) 3.00 t* = = = =11500s

−5 −1 −1(kon [P]0 + koff ) (kon [P]0 + kon Kd ) 2x10 nM s [3nM +10nM ]

t* = 11500s ≈ 3.2h


A surface-catalyzed reaction follows Rideal-Eley kinetics as follows:

kA→A + S← AS k− A

k1AS + A → A2 + S

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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r = k C C −

CAS where K =

k A Ad A A v A

K A k− A

rs = k1CAS CA

Adsorption is at rapid equilibrium, so rAd ≈ 0k A

CACv = CAS ⇒ CAS = K ACACvK A

Overall site balance in terms of So:

SSo = Cv + CAS = Cv + K ACACv ⇒ Cv = o

1+ K ACA

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:

k K S C 2

rA '2 = rs = k1CASCA = k1K ACvCA

2 = 1 A o A

1 + K ACA


It is desired to make a product X-Y via this reaction:

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

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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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

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)

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.

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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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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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.

k = 1.7 x 10-4

s-1

at 298 K.

Ea = 41.87 x 103 kJ/kmol

ΔHR(298) = -167.5 x 103 kJ/kmol

CA0 = 2.0 kmol/m3 (Feed is pure A)

V = 0.5 m3

ρ = 1050 kg/m3

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. Problem 2. Consider an organism which follows Monod equation of growth with μmax = 0.5 h-1 and Ks = 2 g/L.

a) 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?

b) 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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Cite as: William Green, Jr., and K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on [DD Month YYYY]. www.bsscommunitycollege.in www.bssnewgeneration.in www.bsslifeskillscollege.in


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biot numbers by david adrian - bssve.in numbers by david adrian ... consider the effect of biot...

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