Solution manual to chemical reactor analysis and design PDF

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CHAPTER 1 1.1. For the thermal cracking of ethane in a tubular reactor, the following data were obtained for the rate coefficient at different reference temperatures: T(°C) 702 725 734 754 773 789 803 810 827 837 -1 k(s ) 0.15 0.273 0.333 0.595 0.923 1.492 2.138 2.718 4.137 4.665 Solution The Arrhen...

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Int roduct ion t o Chemical Engineering Kinet ics and React or Design,2nd ed (2014) Vinicius Silva Alves [Charles G. Hill, T hat cher W. Root ] Int roduct ion t (z-lib.org) Camila Carolyne CHEG411: Chemical React ion Engineeirng. Fall 2010 Ahmed A Abdala

CHAPTER 1 1.1. For the thermal cracking of ethane in a tubular reactor, the following data were obtained for the rate coefficient at different reference temperatures: T(°C) -1 k(s )

702 0.15

725 0.273

734 0.333

754 0.595

773 0.923

789 1.492

803 2.138

810 2.718

827 4.137

Solution The Arrhenius expression k

E RT

A exp

is transformed logarithmically into: ln k

E RT

ln A

For each data point ln k and 1/T is calculated: x = 1/T∙10³

y = ln k

1.025 1.002 0.993 0.974 0.956 0.941 0.929 0.923 0.909 0.901

–1.897 –1.298 –1.100 –0.519 –0.080 0.400 0.760 1.000 1.420 1.540

The slope and the intercept (ln A) are calculated by linear regression:

E R

ln A

x

xy x

2

N x N

y m

x

N

with x = 1/T and y = ln k.

y 2

837 4.665

So: E R

28497. or E

56623 kcal/kmol or 2.37 10 5 kJ/kmol.

ln A = 27.245 or A = 6.800∙1011 s-1 Q

. S

B

Q

S

A

Solution

B

A

1.2. Derive the result given in Table 1.2.4.2-1 for the reaction

The continuity equation for species A reads: dC A dt

(1)

k CACB

To integrate (1), CB has to be expressed as a function of CA: CB

C B0

(C A 0

CA )

Hence, CA

C A0

dC A C A (C B 0 C A 0

t

CA )

k dt 0

or kt

CB0

CB 1 ln 0 CA 0 CA 0

CA CB

Expressing the concentrations CA and CB as a function of the conversion of the reactant, xA:

CA

CA 0 1 x A

CB

CB0

CB0

CA 0

CA

CB0

CA 0

CA 0 x A

Hence

kt

CB0

CB0 CA 0 1 x A 1 ln C A 0 C A 0 (C B 0 C A 0 x A )

CA 0 1 x A

or 1 M 1 xA ln M 1 M xA

C A 0 kt

with

M

C B0 C A0

1.3. Derive the solutions to the rate equation for the first order reversible reaction given in Section 1.2.3. Solution 1

For A

2

Q

dC A dt

rA

k 1C A

k1 C A

k 2 C A0

k 2CQ C Q0

CA

or dC A dt

k1

k 2 CA

k 2 C A0 C Q0

This is a standard form, with integrating factor exp [ (k1

k 2 ) dt ] exp (k1

k2 ) t

Thus

and

d k1 e dt

CA

e

k2 t

k1 k 2 t

k2 k1

CA

k2

e

k1 k 2 t

k 2 C A0

\ b k 2 (C A 0

CA 0

CQ 0

Ke

C Q0

CQ 0 e k 1

k2 t

dt

K)

k1 k 2 t

Now, at t = 0, CA = CA0, leading to CA

CA 0

CQ 0

k1C A 0

k2 k1

k2

k1

k 2CQ 0 k2

e

k1 k 2 t

which is the solution given in Section 1.2.3. The alternate approach in terms of conversions is somewhat simpler:

rA

k1

k 2 C A 0 x A eq

dC A dt

CA 0

CA 0

d xA dt

d xA dt

x A eq

xA

dx A dt

x A eq

Then k1

k2 xA

x A eq

and direct integration gives: ln

xA

x A eq

0 x A eq

k1

k2 t 0

k1

k2 t

or

ln 1

xA x A eq

1.4. A convenient laboratory technique for measuring the kinetics of ideal gas phase single reactions is to follow the change in total pressure in a constant volume and temperature container. The concentration of the various species can be calculated from the total pressure change. Consider the reaction aA bB ...

qQ sS ...

(a) Show that the extent can be found from:

ξ

V pt

p t0

RT Δ α

where Δ α q ... - a - b - ... (Note that the method can only be used for Δ α

0 .)

(b) Next show that partial pressure for the jth species can be found from pj

p j0

aj Δα

pt

p t0

(c) Use the method to determine the rate coefficient for the first-order decomposition of di-t-butyl peroxide

CH 3 3 COOC CH 3

2 CH 3 2 CO C2 H 6

3

The data given below are provided by J.H. Raley, R.E. Rust, and W.E. Vaughn [J.A.C.S., 70, 98 (1948)]. They were obtained at 154.6°C under a 4.2-mm Hg partial pressure of nitrogen which was used to feed the peroxide to the reactor. Determine the rate coefficient by means of the differential and integral method of kinetic analysis. t (min)

pt (mm Hg)

0 2 3 5 6 8 9 11 12 14 15 17 18 20 21 ∞

173.5 187.3 193.4 205.3 211.3 222.9 228.6 239.8 244.4 254.5 259.2 268.7 273.9 282.0 286.8 491.8

Solution a) For

0

α jA j j

Nj

N j0

α jξ

Also, RT N j, pt V

pj

pj j

Thus, ξ

Nj

αj j

j

V pt RT with Δ α

αj j

p t0

N j0

RT V

Nj j

ξ

V p t p t0 RT Δα

b) Use RT Nj V

pj

αj

p j0

Δα

pt

CH 3 3 COOC CH 3 rA

α jξ

V p t p t0 RT αj RT Δ α V

p j0

c)

RT N j0 V

1 dξ V dt kC A

p t0

3

2 CH 3 2 CO C 2 H 6

1 V 1 d pt V RT Δ α dt k

pA RT

k pA 0 RT

pt 0 ( 1) pt 2

pt 0

or d pt dt

pt 0

2k p A 0

1 pt 2

pt 0

Integrating yields:

ln

p A0

1 pt 2 p A0

p t0

kt

The integral method uses a plot of ln [ ] vs. t, and the slope gives the value of k. Alternatively, a linear or nonlinear regression method could be used: k = 0.0193 min-1. The differential method utilizes slopes from the p t

p t 0 data plotted vs.

1 p t p A 0 , the slope of which, or regression techniques, gives the rate 2 coefficient: k = 0.0196 min-1. p A0

0.5

0.4

- ln

1 pt 2 p A0

p A0

p t0

0.3

0.2

Slope = 0.0193 min-1

0.1

0 0

5

10

15

20

time(min) 1.5. The results of Problem 1.4 can be generalized for the measurement of any property of the reaction mixture that is linear in the concentration of each species: λj

K jC j

The λ j could be partial pressures (as in Problem 1.4) various spectral properties, ionic conductivity in dilute solutions, and so on. Then the total observed measurement for the mixture would be: λ

K jC j

λj j

j

a) For the general single reaction, α jA j

0

j

show that the relation between the extent of reaction and λ is

λ

α jK j

λ0 j

where λ0 j

K jC j0

ξ V

25

b) After a long (“infinite”) time, the extent ξ can be evaluated for irreversible reactions from the limiting reagent, and for reversible reactions from thermodynamics. Use this to formulate the desired relation containing only measured or determined variables (see Frost and Pearson [41]):

λ λ0 λ λ

ξ ξ

The total observed property for the mixture is given as K jC j

λ j

a) For the general reaction α jA j

0

j

the concentration at reaction extent ξ is Cj

Nj

N j0

V

V

αj

ξ V

Thus, Kj

λ

N j0 V

j

j

Kj

K jC j0

j

j

K jα j j

ξ V

ξ , constant volume V

b) For infinite time, either ξ can be found from the thermodynamic equilibrium constant KC j

C j0

ξ αj V

or from the limiting reactants: 0

C 0,LR

α LR

ξ V

In either case from part a):

αj

λ

α jK j

λ0

ξ V

Solving for the extents, ξ V

λ λ0 α jK j

ξ V

λ

and λ0 α jK j

Dividing the two gives the desired result:

λ λ0 λ λ0

ξ ξ

1.6. Show that the general expression for the concentration at which the autocatalytic reaction of Section 1.2.3.3 has a maximum rate is 0 A

Q

CC 1 12

C CQ

0 Q

x a m 0

Note that this agrees with the specific results in the example. Solution From Section 1.2.3.3, the rate written in terms of only CQ is:

r

k1 C0

CQ CQ

k 2 C Q2

The maximum rate will be at r C Q

r CQ

0

k 1C 0

2k 1C Q

0:

2k 2 C Q

or

CQ

max

k1 1 C0 k1 k 2 2

where K is the equilibrium constant. This can be written as

K 1 C0 2 1 K

CQ C Q0

C A0 1 K 1 21 K C Q0

max

C A0 1 1 2 C Q0 for the irreversible case with K

, as in the numerical calculations.

1.7. Derive the concentration as a function of time for the general three species first order reactions:

1

A

Q

2 4

3

5

6

S These should reduce to all the various results for first order reactions given in Sections 1.2 and 1.3. Also determine the equilibrium concentrations CAeq, CQeq, CSeq in terms of the equilibrium constants for the three reactions. Solution The general solution method is best explained in terms of the complex reaction network notation where the rate for the jth species is given by: dy j dt

k j1y1

N

k j2 y 2 ...

' k lj y j ...

(1)

l 1

A solution is assumed of the form:

yj

Yj0 e

(2)

λt

where Yj and λ are constants to be determined. Substituting (2) into (1) gives

Yj

λe

λt

k jl Y1e

λt

...

' k lj Yje

λt

...

or

k j1Y1

k j2 Y2 ...

' k lj

λ Yj ... 0

(3)

If equations (3) are written out for each of the j = 1, 2, … N species, they form a set of N equations for the N unknown, Yj. Note that they are homogeneous algebraic equations (RHS ≡ 0), and so they only possess solutions if the determinant of the coefficients is zero.

' k l1

λ

k12

k 21

' k l2

λ

k13

k1N

k 23

k 2N

(4)

0 k N1

k N2

k N3

' k lN

λ

When multiplied out; (14) will be an Nth order equation, with N-roots. For each of these roots, a set of the constants Yj can be found from (3): Yjr. Then the general solution will be N

yj

(5)

C r Yjr e λ rt

r 1

where Cr are constants of integration to be found from the initial conditions. For our specific problem, the determinant, (4), is:

k1

λ

k3

k1

k2

k2

λ

k4

k3 k6

k6

k5

λ

k4

0

(6)

k5

which when expanded gives: λ

λ2

λ k1

k2

k 2 k3

k4

k3 k5

k4

k5

k3 k5

k6

k1 k 4

k6

k 4k 6

k5

k6

0

or λ λ2

(7)

0

αλ β

Thus, the roots (eigenvalues) are

λ1 λ 2,3

0 α

α2 2

4β

(8)

(Note, Wei-Prater would number these λ 0 , λ1 , λ 2 .) Then the solutions will have the form CA

C1 YA1

C 2 YA2 e

λ2t

C 3 YA3 e

λ3t

(9)

CQ

C1 YQ1

C 2 YQ2 e

λ2t

C 3 YQ3 e

λ3t

(10)

CS

C1 YS1

C 2 YS2 e

where the Yjr are found from

λ2t

C 3 YS3 e

λ3t

(11)

λr

k1

k 3 Y1r

k 2 Y1r

λr

k2

k 4 Y1r

k 5 Y2r

k 1 Y2r

k 3 Y3r

0

k 6 Y2r

k 6 Y3r

0

k 5 Y3r

0

λr

k4

(12)

for each of the λr (actually, only two of the equations will be independent, and so two of the Yjr can be found in terms of the third). Finally, the integration constants, Cr, are found from the initial conditions: C A0

C1 YA1

C 2 YA2

C 3 YA3

C Q0

C1 YQ1

C 2 YQ2

C 3 YQ3

C S0

C1 YS1

C 2 YS2

(13)

C 3 YS3

Clearly, these manipulations are most easily done utilizing matrices. The equilibrium concentrations will be found from (9)-(11) for t C A,eq

C1 YA1

: (14)

where YA1 and C1 are known in terms of the kji and initial concentrations. From the principle of microscopic reversibility, it is also time that each reaction is individually at equilibrium, leading to:

CQ CA CA CS CS CQ

KI

eq

k1 k2

K II

eq

k4 k3

K III

eq

k6 k5

(15)

with (16)

K I K II K III 1 The total mass balance is CA

CQ

CS

Then, using (17) and (15):

C A0

C Q0

C S0

C0

(17)

1 C Ae K II

C Ae

K I C Ae

C Ae

C0 1 K I K II

C0

or

C1 YA1

Similarly

C Qe C Se

K II K II

C0 C1 YQ1 1 K III C0 C1 YS1 1 1 K III

(18)

1.8. For the complex reactions 1

aA bB q' Q b' B

qQ 2

sS

(a) Use (1.1.2-15) and (1.1.2-17) to express the time rates of change of NA, NB, NQ, and NS in terms of the two extents of reactions and stoichiometric coefficients a, b, b’, q, q’, and s; for example, dN A dt

a

dξ1 dt

(0)

dξ 2 dt

(b) In practical situations, it is often useful to express the changes in all the mole numbers in terms of the proper number of independent product mole number changes—in this case, two. Show that the extents in part (a) can be eliminated in terms of dNQ/dt and dNS/dt to give dN A dt

a dN Q q dt

dN B dt

b dN Q q dt

a q' dN S s q dt b q' s q

b' dN S s dt

This alternate formulation will be often used in the practical problems to be considered later in the book (c) For the general reaction N j 1

α ij A j i 1, 2, ..., M

The mole number changes in terms of the extents are dN j

dξ i dt

N

α ij

dt

j 1

or dN

αT

dt

dξ dt

where N is the N-vector of numbers of moles, ξ is the M-vector of extents, and α T is the transpose of the M x N stoichiometric coefficient matrix. Show that if an alternate basis of mole number change is defined as an M-vector dN b dt that the equivalent expressions for all the mole number changes are dN dt

αT αb

T

1

dN b dt

where α b is the M x M matrix of the basis species stoichiometric coefficients. Finally, show that these matrix manipulations lead to the same result as in part (b) if the basis species are chosen to be Q and S. Solution aA bB (a)

q' Q b' B Nj

qQ

ξ1

sS

ξ2

α ij ξi , N j i

N j0

α ijξi i

Thus, NA

a ξ1

(1)

NB

b ξ1 b' ξ2

(2)

NQ

q ξ1 q' ξ2

(3)

NS

(b)

s ξ2

Rearrange for ξ1 and ξ2 in terms of N Q , N S :

(4)

1 NS s 1 q' NQ ξ2 q q

ξ2

ξ1

(5) 1 q' 1 NQ NS q qs

(6)

Thus, a NQ q b NR r

NA NB

(c)

a q' NS sq b r' b' NS NS s r s

(7) (8)

For the general reaction N

0

α ijA j , i 1, 2, ..., M

(9)

j 1

The change in mole numbers is:

(10)

[α~ ]T ξ

N ~

~

If the new M-vector basis species are

N ~

(11)

the M x M matrix of basis species stoichiometric coefficients is part of the total stoichiometric matrix: β α

(12)

[α~ ]α 1,1,MM 11

α~

(The stoichiometric coefficient matrix can always be arranged to have the new basis species coefficients first.) Now the subset of the mole number change equations for the new basis species are: b

N~

b

[α~ ]T ξ

(13)

~

which is equiva...