ChE 471

Lecture 6

October 2005

Ideal Reactors and Multiple Reactions Isothermal Operation

Selection of a proper flow pattern is the key factor in achieving desired selectivities and yields in multiple reactions. For every multiple reaction system of known stoichiometry it is possible to determine a priori which limiting flow pattern complete backmixing (CSTR) or no mixing (PFR) will yield superior yields or selectivities. The consideration of yields often is more important than reactor size in choosing the preferred reactor flow pattern. From Lecture 1 we know that all multiple reaction systems can be represented by a set of R independent reactions among the S chemical species present in the system:

j=1

ij

Aj=0

for

i = 1,2,3...R

(1*)

These stoichiometric relationships allow one to relate moles produced (or depleted) of each species to the molar extents of the R reactions:

& F j = F jo + ij X i
i =1

(1)

The rate of reaction of each species is given through the rates of the R independent reactions, ri, i = 1,2,R.
Rj = ij ri
i =1 R

(2)

CSTR Ideal Stirred Tank Continuous Flow Reactor

Fjo Qo V Fj Q

The design equation (i.e., the mass balance for species j) can be written for R species, j = 1,2,3R:
Fjo Fj + ij riV = 0
i =1 R

(3)

for j = 1, 2,...R

ChE 471

Lecture 6

October 2005

If the reaction rate ri for each independent reaction i can be represented by an n-th order form, of eq (4a) ri = k i C j ij

j =1 S

(4a)

then at P = const, T = const, the rate of the i-th reaction, ri, can be represented in terms of molar & of the reactions by: extents X i
R & + F ij X jo i S i =1 ij ri = k i C tot ,o S R j =1 & + F tot ,o ij X i = = j i 1 1

ij

(4b)

where ij is the reaction order of reaction i with respect to species j, Ftot,o is the total initial molar flow rate.

& . Substitution of equations (2) and (4b) into (3) results in set of R nonlinear equations in X i Three types of problems described below arise:
a) Given the feed flow rates, reactor size V and rate forms for all reactions one can calculate & ' s and from equation (1) get the composition of the outlet all the reaction extents X i stream.

In addition, from Lecture 1, at P = const, T = const:

R S & F + Q = Qo 1 X ij i tot , o i = 1 j = 1

(5)

The exit volumetric flow rate can be computed and effluent concentrations calculated
Cj = Fj Q

(6)

b) c)

Given the feed molar flow rates and composition, and the desired partial composition of the outflow, as well as the reaction rates, one can calculate the reactor size from equation (3) and the composition of other species in the outflow. Given molar feed rates and outflow molar flow rates for a given reactor size the rate of reaction for each species can be found from equation (3).

ChE 471

Lecture 6

October 2005

PFR Plug Flow Reactor

Fjo Qo V Fj Q

The design equation (i.e. the differential mass balance for species j) can be written for R species
dF j = ij ri
R

dV i =1 j = 1, 2, ...R

(7)

The initial conditions are

V=0

Fj = Fjo

(7a)

Using equations (1) and (4b) the above set of R first order differential equations can be & ' s as functions of V. integrated simultaneously and solved for X i a) b) Given the feed flow rate and composition, and the form of the reaction rates, one could determine what volume V is required to attain the desired product distribution. Given the feed and reactor volume and reaction rate forms, one can determine the exit product distribution.

Batch Reactor Autoclave of Constant Volume

The R species (for j=1, 2, 3..R) mass balances yield:

dn j dt = ij ri
i =1 R

(8)

Initial conditions are: t=0 nj = njo (8a)

Moles and extents are related by:

ChE 471

Lecture 6

October 2005

n j = n jo + ij X i
i =1

(9)

For j = 1, 2, 3S The rate form as a function of extents is given by

R s ij X i ri = k i C joij 1 + i =1 n jo j =1

ij

(10)

R ri = k i C jo + ij i i =1 j =1

ij

(10a)

where i =

Xi V

(10b)

One can solve the set of R first order differential equations to calculate the product distribution in time, or the desired time needed for a prescribed product distribution. The above approach, while well suited for the computer, does not provide us with the insight as to which flow pattern is better in a given process until we actually compute the answers for both limiting cases. In order to get better insight in the role of the flow pattern in product distribution in multiple reactions we will consider some simple systems and use the notions of yields and selectivity.
Classification of Multiple Reactions

A + B = R parallel C + D = P A+ B = R competitive A + 2B = S A + B = R consecutive (series reactions) R=S A + B = R mixed reactions R+ B = S

ChE 471

Lecture 6

October 2005

In Lecture 1 we have defined the various yields

ip ri Rp P i =1 {point (relative) yield y = = R A RA r iA i

i =1

Point (relative) yield measures the ratio of the production rate of a desired product P and the rate of disappearance of the key reactant A. Point yield is a function of composition and this varies along a PFR reactor, varies in time in a batch reactor, and is a constant number in a CSTR.
P F FPo {overall (relative) yield Y = P A FAo FA

Overall (relative) yield gives the ratio of the overall product P produced consumption of reactant A. In a CSTR the overall and point yield are identical.
P P Y ( ) = y A A In a PFR the overall yield is the integral average of the point yield: 1 P Y = A FAo FA
FAo

and the total

FA

P y dFA A

Overall operational yield is also often used, defined as the number of moles of the desired product produced per mole of key reactant fed to the system.
P FP FPo @ = A FAo

The relationship to overall relative yield is obvious P P @ = Y xA A A where xA is the overall conversion of A

ChE 471

Lecture 6

October 2005

xA =

FAo FA FAO

None of the above yields has been normalized, i.e., their maximum theoretical value may be more or less than one as dictated by stoichiometric coefficients. A normalized yield can be introduced by P y P A y = A y P max A P where y max is obtained by assuming that only the reactions leading from A to R occur. A Point selectivity and overall selectivity measure the ratio of formation of the desired product and one or more of the unwanted products, e.g. P Rp s = U Ru S= F p F po Fu Fuo

A general rule:
P dy A If > 0 PFR produces more P. dC A P dy A If < 0 CSTR produces more P dC A P If y is not a monotonic function of CA either reactor type may produce more P depending A on operating conditions. The case of monotonic point yield is illustrated below for the case with A = 0.
P y A

P y A

\ \ \ Area = Cp in PFR Area = Cp in CSTR

CA CAo CA

CA

CAo

CA

ChE 471

Lecture 6

October 2005

I.

Liquid Systems or Gases With

i =1 j =1

ij

=0

Competitive Reactions

a1A + b1B= p1P a2A + b2B = s2S Given the rate

1 B 1A r1 = k1C A CB 2B 2A r2 = k 2 C A CB

Point yield then is:

p1 Rp p1 r1 a1 P y = = = A R A a1 r1 + a 2 r2 1 + a 2 r2 a1 r1 p1 a1 P y = A 1 + a 2 k 2 C ( 2 A 1 A ) C ( 2 B 1 B ) A B a1 k1

P p y max = 1 A a1

1 P y = a k A 1 + 2 2 C ( 2 A 1 A ) C ( 2 B 1 B ) A B a1 k1 P We want y to be as high as possible. This implies: A i) ii) iii)

if 2 A < 1 A , 2 B < 1B , keep CA and CB as high as possible. PFR is better than CSTR. if 2 A = 1 A , 2 B < 1 B , keep CB as high as possible. PFR is better than CSTR. if 2 A = 1 A , 2 B > 1 B , keep CB as low as possible. CSTR is better than PFR. Try for yourself other combinations.

ChE 471

Lecture 6

October 2005

Example 1 A=P 2A = S RP = 1.0 CA (kmol/m3s) 2 (kmol/m3s) RS = 0.5 C A

Determine Cp in a) CSTR, b) PFR. The feed contains CAo = 1 (kmol/ms), Cpo = 0. Conversion of 98% is desired.

Rp Rp 1 P y = = = A R A R p + 2 Rs 1 + C A
P y max = 1 A P P y = y A A

To keep point yield as high as possible, it is necessary to keep CA low everywhere. CSTR will be better than PFR. Let us show this quantitatively.

a) CSTR By setting the overall yield equal to the point yield we can solve for the exit concentration of product P. Cp 1 P P = y = Y = A C Ao C A A 1+ CA
Cp = C Ao C A C Ao x A 1x0.98 = = = 0.961 kmol / m 3 1+ CA 1 + C Ao (1 x A ) 1 + 1 0.98

Overall yield

Cp P = 0.980 Y = A C Ao C A

Overall operational yield

P = 0.961 A

Required reactor size (space time)

C Ao C A C Ao x A = = 48.0( s ) 2 RA C Ao (1 x A ) + C Ao (1 x A ) 2

ChE 471

Lecture 6

October 2005

b) PFR Product concentration is obtained by integration of the point yield

Cp =
C Ao

CA

P y dC A = A

C Ao

CA

1+ C

dC A
A

1 + C Ao C p = l n 1+ C A

1+1 3 = ln 1 + 1 0.98 = 0.673(kmol / m )

P Overall yield Y = 0.687 A P Overall operational yield @ = 0.673 A

Required reactor size

C Ao

CA

dC A = RA

C Ao

CA

dC A C A (1 + C A )

= l n

C Ao (1 + C A ) 1 + 1 0.98 = ln = 3 .2 ( s ) (1 0.98)(1 + 1) C A (1 + C A )

Plug flow reactor is considerably smaller but CSTR gives a better yield and higher concentration of the desired product.

Example 2
A+B=P A+A=S Rp = 1.0 CACB (kmol/m3s) 2 (kmol/m3s) Rs = 0.5 C A

Given FAo = FBo = 1 (kmol/s; CAo = CBo = 1 (kmol/m3) CPo = CSo = 0 and desired P P P conversion xA = 0.98, determine C p , Y ,@ , S and required reactor space time for A A S a) CSTR, b) PFR.

R Rp CB 1 P y = p = = = A RA R p + 2 Rs CB + C A 1 + C A CB
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ChE 471

Lecture 6

October 2005

To maximize point yield one should keep the reactant concentration ratio CA/CB as low as possible everywhere. Eliminate CB in terms of Cp and CA using C j = C jo + ij i
C A = C Ao 1 2 2 C P = C Po + 1
i =1 R

i = 1, R

1 = C P C Po 2 =
Now:

1 [C Ao C A C P ] 2

C B = C Ao 1 = C Bo C p
C S = C So + 2 = 0 + = 1 C Ao C A C p 2

1 [C Ao C A C Bo + C B ] 2

a) CSTR

Cp C Ao C p P P = y = Y = A C Ao C A A C Ao C p + C A
Solve for Cp
2 Cp (C Ao _ C Bo )C p + C Bo (C Ao C A ) = 0

C Ao = C Bo = 1; C A = C Ao (1 x A ) = 1 0.98
2 Cp 2C p + 0.98 = 0

C p = 1 1 0.98= 0.859(kmol / m 3 C A = 0.02kmol / m 3 exit stream composition 3 C B = 0.141kmol / m 3 C S = 0.061kmol / m P Overall yield Y = 0.877 A

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ChE 471

Lecture 6

October 2005

P Overall operational yield @ = 0.859 A P Cp = 14.2 Overall selectivity S = S Cs Required Reactor Size

C Bo C B C Ao C B 1 0.141 = = = 306( s ) RB C AC B 0.02 x0.141

b) PFR dF p dFA = P = y RA A Rp

Since F j = QC j and Q = const dC p dC A C Bo C p C Bo C p + C A

at C A = C Ao , C p = 0 Rearrange: dC A CA + = 1 dC p C Bo C p d dC p at Cp = 0 CA C C p Bo 1 = C C Bo p

CA = CAo

Integrate from the indicated initial condition: C Bo C p C CA Ao = ln C C Ao C p C Bo Bo CA = 1 + ln(1 C p )(*) 1 Cp Substitute known quantities: CA = CAo (1-xA) = 1-0.98 = 0.02 (kmol/m3)
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ChE 471

Lecture 6

October 2005

Solve for Cp by trial and error:

(C p ) = C p 0.98 (1 C p )ln(1 C p ) = 0
D (C p ) = 2 + ln(1 C p )
n +1 p n (C p ) =C n D (C p ) n p

Newton-Raphson Algorithm

This yields:
C p = 0.613 kmol / m 3 C A = 0.02 kmol / m 3 Exit stream composition C B = 0.387 kmol / m 3 C s = 0.184 kmol / m 3

The last two concentrations above are evaluated using the stoichiometric relationship. P Y = 0.626 A P Overall operational yield @ = 0.613 A Overall yield P Overall selectivity S = 3.33 S Required reactor space time:

=
From (*)

C Bo

CB

Bo dC B dC B = R B C B =0.387 C A C B

=1

C A = (1 C p ) 1 + ln(1 C p ) From stoichiometry C B = C Bo C p = 1 C p Thus

C A = C B (1 + lnC B )

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ChE 471
C Bo =1 1+ ln C

Lecture 6

October 2005

Bo dC B e u = = e 2 u du C B = 0.387 C B (1 + lnC B ) 1+ ln C B

= e{E1 (1 + lnC B ) E1 (1 + lnC Bo )} = e{E1 (0.05) E1 (1)}

= 2.71{2.4679 0.2194} = 6.1( s )

where

E1 ( z ) =
z

e u du u

exponential integral

E1(z) values tabulated in M. Abramowitz & A. Stegun Handbook of Mathematical Functions, Dover Press, N.Y. 1964.
Comparison of CSTR & PFR

Reaction System: Decision variables:

A + B = P (desired) A+A=S xA = 0.98 F M B / A = Bo = 1 FAo r1 = 10 CACB (kmol/m3s) r2 = 0.5 CA2 (kmol/m3s) CSTR 0.859 14.2 306 (s) PFR 0.613 3.3 6.1 (s) Optimal Ideal Reactor 0.950 63.0 150 (s)

Rate Form:

Operational yield Overall selectivity Reactor space time

The last column of the above Table was computed based on an ideal reactor model shown below. We have B entering a plug flow reactor while FAo is distributed from the side stream into the reactor in such a manner that CA = 0.02 kmol/m3 everywhere in the reactor.
FAo

FBo

PFR

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ChE 471

Lecture 6

October 2005

From the expression for the point yield CB 1 P y = = A CB + C A 1 + C A CB it is clear that one needs to keep CA low and CB high. With the constraint of FAo = FBo the above ideal reactor accomplishes that requirement in an optimal manner. Could such a porous wall reactor with plug flow be constructed? It depends on the nature of the reaction mixture. However, we learn from the above that with our choice of decision variables maximum selectivity is 63, we can never do better than that! We also learn that a good reactor set up is a cascade of CSTRs.
FAo

FBo

The total number of reactors used will depend on economics With 2 reactors we get selectivity of over 20, with five we are close to optimum.

Examining the effect of decision variables we see that Cp increases with increased conversion of A. For conversions larger than 0.98 the reactor volume becomes excessive. If we took MB/A > 1 that would improve the yield and selectivity but at the expense of having to recycle more unreacted B. Let us ask the following question. How much excess B would we have to use in a PFR in order Cp = 14.2 at CAo = 1 to bring its overall selectivity to the level of a single CSTR i.e., S = Cs (kmol/m3) and xA = 0.98. So the goal is to choose C Bo in order to get at the exit of plug flow: Cp Cs = 14.2

From stoichiometry:

Cs =

1 1 1 C Ao C A C p = C Ao x A C p = 0.98 C p 2 2 2
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ChE 471

Lecture 6

October 2005

The prescribed desired selectivity is:

2C p 0.98 C p = 14.2C p = 0.859(kmol / m 3 )

C s = 0.061(kmol / m 3 ) The integrated equation for PFR is: Cp C CA Ao = ln 1 C C Bo C p C Bo Ao

The initial concentration of B is now the only unknown. Evaluate it by trial & error.
0.859 0.02 1 = ln 1 C gives C Bo 0.859 C Bo Bo C Bo = 3.79 kmol m3

Since CAo = l (kmol/m3), MB/A = 3.79 almost four times more B than A should be introduced in the feed to get the selectivity in a PFR to the level of a CSTR. A great excess of unreacted B has to be separated in the effluent: CB = CBo Cp = 3.790-0.859 = 2.931 (kmol/m3)
Consecutive Reactions

aA = p1P p2P = s S Two basic problems arise: a) conduct the reaction to completion, b) promote production of the intermediate. The first problem is trivial and can be reduced to a single reaction problem. Use the slowest reaction in the sequence to design the reactor. In order to maximize the production of intermediates PFR flow pattern is always superior to a CSTR flow pattern.

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ChE 471

Lecture 6

October 2005

Rp p r p 2 r2 p p r P y = = 11 = 1 2 2 a r1 a a r1 A RA P p y max = 1 A a
p r p k CP P y = 1 2 2 = 1 2 2 p1 r1 p1 k1C A A

One needs to keep CA high and Cp/CA low which is best accomplished in a PFR.
Example 1 A=P P=S

r1 = 1.0 CA (kmol/m3s) = 1 r2 = 0.5 CP (kmol/m3s) = 1

Starting with CAo = 1 (kmol/m3) and CPo = Cso = 0 find the maximum attainable CP in a) CSTR, b) PFR. We could continue to use the point yield approach. CSTR Stirred Tank Reactor
C 0.5C P P (C Ao C A ) C P = y (C Ao C A ) = A CA A

Solve for CP
CP = C A (C Ao C A ) 0.5(C Ao + C A )

Find optimal CA at which the CSTR should operate. dC P = 0 (C Ao 2C A )(C Ao + C A ) C A (C Ao C A ) = 0 dC A

2 2 CA + 2C Ao C A C Ao =0

C Aopt = C Ao C Pmax =

2 1 = 0.414(kmol / m 3 )

0.414(1 0.414) = 0.343(kmol / m 3 ) 0.5(1 + 0.414) C S = C Ao C A C P = 0.243kmol / m 3 Overall yield P Y = 0.585 A

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ChE 471

Lecture 6

October 2005

Operational yield Overall selectivity Required reactor space time:

P @ = 0.343 A P S = 1.4 S

C Ao C Aopt RA

C Ao C Aopt C Aopt

1 0.414 = 1.4( s) 0.414

b)

PFR Plug Flow Reactor dC P P 0.5C P C A = y = dC A CA A

at CA = CAo, CP = 0
dC P 0.5 C p = 1 dC A CA d CP 1 == dC A CA CA CP = 2 C Ao C A CA

]
]

CP = 2 C A

[C

Ao

CA

Find CA at the reactor exit by

dC P = 0 C Ao 2 C A = 0 dC A C Aopt = C Aopt C Ao 4 = 0.25(kmol / m 3 )

C Pmax = 2 0.25 1 0.25 = 0.5(kmol / m 3 )

C s = 1 0.5 0.25 = 0.25(kmol / m 3 )

Overall yield Operational yield

P 2 Y = = 0.667 A 3 P @ = 0.5 A

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ChE 471

Lecture 6

October 2005

Overall selectivity Required space time:

P S = 2.0 A

C Ao

C Aopt

C dC A = ln Ao = ln4 = 2ln2 = 1.39( s) CA C Aopt

The same results can be obtained by using the design equations (i.e. mass balance)for P & A. CSTR
C Ao C A CA CP C A 0.5C P

= =

CA = CP =

C Ao 1+ C Ao C A = 1 + 0.5 (1 + )(1 + 0.5 )

dC p d PFR

= 0 gives opt = 1.4( s ), etc.

dC A = C A C A = C Ao e d = 0, C A = C Ao = 1 dC P = C A 0.5C P d

=0

CP = 0

d 0.5 (e C P ) = C Ao e e 0.5 = C Ao e 0.5 d C P = 2(e 0.5 e ) dC P = 0 opt = 2ln 2, etc. d

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ChE 471

Lecture 6

October 2005

Mixed Reactions

This is the most frequently encountered type of multiple reactions which can be viewed as a combination of competitive and consecutive reactions. We can solve the problems involving these reactions either by setting R design equations for R components or by utilizing the concept of the point yield in simpler reaction schemes.
Example 1

2A + B = R 2B + R = S

-RA = 2k1 CA CB (kmol/m3min) Rs = k2 CB CR (kmol/m3min)

k1 = 10 k2 = 1 (m3/kmol min). R is the desired product. Find CR in a) CSTR, b) PFR, when CRo = Cso = 0. Decision variable CAo = CBo = 1 (kmol/m3). We can write two point yields: R k C C k2CBCR 1 k2 CR R = 1 y = R = 1 A B 2 k1 C A C B 2 k1 C A A RA k C A 2 CR R k C C k2CBCR k1 R y = R = 1 A B = B R B k1 C A C B + 2 k 2 C B C R C + 2 k 2 C A R k1 R The point yield y depends only on CR and CA and is simpler to use. A a) CSTR Stirred Tank Reactor k C A 2 CR k1 R (C Ao C A ) C R = y (C Ao C A ) = 2C A A

CR =

C A (C Ao C A ) C (C C A ) = A Ao 1.9C A + 0.1C Ao k2 k2 2 C + C A Ao k1 k1

Find optimum CA in a CSTR.

dC R k 2 2 =0 dC A k1 2 k2 2 k2 C A + 2 k C Ao C A k C Ao = 0 1 1

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ChE 471

Lecture 6

October 2005

C Aopt =

C Ao 2 k1 +1 k2 1

C Aopt =

2 x10 + 1

= 0.183(kmol / m 3 )

C Rmax =

0.183(1 0.183) = 0.334 (kmol / m 3 ) 1.9 x0.183 + 0.1

From stoichiometry 3 3 C B C Bo (C Ao C A ) + 2C R = 1 (1 0.183) + 2 x0.334 2 2 1 1 C S = (C Ao C A ) C R = (1 0183) 0.334 2 2 3 C B = 0.443(kmol / m ); C S = 0.0745(kmol / m 3 ) Overall yield Operational yield Overall selectivity R R Y = 0.409; Y = 0.600 A B R R @ = 0.334; @ = 0.334 A B R S = 4.48 = 4.5 S

Required reactor space time:

=
b)

C Ao C A 1 0.183 = = 5.0(min) 2k1C A C B 2 x1x0.183 x0.443

PFR

Plug Flow Reactor

dC R 1 k C R = y = + 2 R dC A 2 2 k1 C A A C A = C Ao ,C R = 0
1 k2 d k2 C A 2 k1 C R = C A 2 k1 2 dC A CR = C
k2 1 2 k 1 Ao

0.95 0.05 C A C Ao CA CA = 1 .9 k2 2 k1

k2 2 k1 A

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ChE 471

Lecture 6

October 2005

For optimal CA at reactor exit:

k2 dC R = 0 C Aopt = 2k dC A 1
1 k 1 2 2 k1

C Aopt = (0.05) C Rmax

1 0.95

= 0.0427(kmol / m 3 )

10.95 x0.0427 0.05 0.0427 = = 0.427(kmol / m 3 ) 1.9

3 C B = 1 (1 0.0427) + 2 x0.427 = 0.418(kmol / m 3 ) 2 CS = 1 (1 0.0427) 0.427 = 0.0517(kmol / m 3 ) 2 R R Y = 0.446; Y = 0.734 A B R R @ = 0.427; @ = 0.427 A B R S = 8.27 = 8.3 S

Overall yield Operational yield Overall selectivity Reactor space time:

C Ao

CA

dC A 2C A C B.

From stoichiometry C B = C Bo 3 (C Ao C A ) + 2C R 2

but all along the PFR

0.05 CA CA CR = 1.9

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Lecture 6

October 2005

3 0.05 (C Ao C A ) + 2 (C A CA ) 2 1.9 2 2 0.05 C B = 1.5 C A 0.5 C A + 1.9 1.9 0.05 C B = 0.447C A + 1.05C A 0.5 C B = C Bo

= 0 .5

dC A integrate numerically (2.759 min ) 1.05 0.5C A 0.0427 0.447C + 1.05C A

2 A

Caution must be exercised when using the point yield concept and finding maximum concentrations in mixed reactions. Sometimes formal answers will lie outside the physically permissible range if the other reactant is rate limiting.

For example in Example 1 if we take kmol CAo = 2 3 m kmol ; CBo = 1 3 m

We are feeding the reactants in stoichiometric ratio for reaction 1. Following the above described procedure in a CSTR we would find C A opt = C Rmax C Ao = 0.366(kmol / m 3 )

2 x10 + 1 0.366(2 0.366) = = 0.668(kmol / m 3 ) 1.9 x0.366 + 0.2

However these values are not attainable since from stoichiometry it follows that: 3 C B = 1 (2 0.366) + 2 x0.668 = 0.115 < 0 2 This indicates that B is not introduced in sufficient amount to allow the reactions to proceed to that point. If CBo = 1.115 (kmol/m3) then the above CR max can be obtained (theoretically) at CB = 0 and that would require an infinitely large reactor. Thus the maximum reactor size that is allowed would determine CBmin . (kmol/m3) (99% conversion of B). Calculate the resulting CA and CR from Say CBmin =0.01

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Lecture 6

October 2005

4 2 C A = C Ao C R (C Bo C Bmin ) 3 3 C A (C Ao C A ) CR = 1.9C A + 0.1C Ao C A = 0.461(kmol / m 3 ) That yields: C R = 0.660(kmol / m 3 ) C S = 0.110(kmol / m 3 ) Overall yield Operational yield Overall selectivity Reactor space time: R R Y = 0.429; Y = 0.667 A B R R @ = 0.330; @ = 0.660 B A R S = 6.0 S

=
II.

2 0.461 = 170min 2 x1x0.461x0.01

Systems with change in total volumetric flow rate ( A 0) gases

& & extents rather than concentrations. Same approach may be used but one needs to deal with X i Use relationships from Lecture 1.

Summary

PFR promotes more reactions of higher order with respect to reactions of lower order. CSTR favors reactions of lower order with respect to those of higher order. In consecutive reactions better yields are achieved always in PFR than in a CSTR for an intermediate product. Select judiciously the objective function to be optimized. Remember: Optimizing overall yield does not necessarily lead to the same result as maximizing the production rate (or concentration) of the desired product or as maximizing selectivity. Be aware of the relationship of the design equations and reaction stoichiometry.

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