Chemical Engineering Chapter 2 The point is to encourage the student to question

2-1

Solutions for Chapter 2 – Conversion and Reactor

Sizing

Synopsis

General: The overall goal of these problems is to help the student realize that if they

have –rA=f(X) they can “design” or size a large number of reaction systems. It sets the

stage for the algorithm developed in Chapter 4.

P2-1. This problem will keep students thinking about writing down what they learned

every chapter.

P2-5. This is a reasonably challenging problem that reinforces Levenspiels plots.

P2-6. Novel application of Levenspiel plots from an article by Professor Alice Gast at

Massachusetts Institute of Technology in CEE.

P2-7. Straight forward problem alternative to problems 8, 9, and 12.

P2-11. Great motivating problem. Students remember this problem long after the

course is over.

2-2

CDP2-C Similar to problems 2-8, 2-9, 2-12.

CDP2-D Similar to problems 2-8, 2-9, 2-12.

Summary

Assigned

Alternates

Difficulty

Time (min)

P2-1

O

15



P2-2

A

30

P2-9

AA

7,9,12

SF

45

P2-10

S

SF

15



P2-11

AA

SF

1

P2-12

AA

7,8,9

SF

60

Assigned

 = Always assigned, AA = Always assign one from the group of alternates,

O = Often, I = Infrequently, S = Seldom, G = Graduate level

Alternates

In problems that have a dot in conjunction with AA means that one of the

Time

Approximate time in minutes it would take a B/B+ student to solve the problem.

Difficulty

2-3

Summary Table Ch-2

Straight forward

1,2,3,4,10

Fairly straight forward

7,9,12

P2-1 Individualized solution.

P2-4 (a) Example 2-1 through 2-3

If flow rate FAO is cut in half.

v1 = v/2 , F1= FAO/2 and CAO will remain same.

P2-4 (b) Example 2-5

New Table: Divide each term

A

r

F

!

0

in Table 2-3 by 2.

Reactor 1 Reactor 2

P2-4 (c) Example 2-6

Now, FAO = 0.4/2 = 0.2 mol/s,

New Table: Divide each term

A

r

F

!

0

in Table 2-3 by 2.

X

0

0.1

0.2

0.4

0.6

0.7

0.8

[FAO/-rA](m3)

0.445

0.545

0.665

1.025

1.77

2.53

4

X1 = 0.603 for V1 = 0.551m3 X2 = 0.89 for V2 =

1.614m3

P2-4 (d) Example 2-7

(1)

For PFR,

0.2

For first CSTR,

X2 = 0.6,

3

1.32

AO

Fm

rA =

!

, V1 =

3

2 1

( ) .528

AO

F X X m

rA

!=

!

(2)

First CSTR remains unchanged

For PFR:

2-6

For CSTR,

(3) The worst arrangement is to put the PFR first, followed by the larger CSTR and finally the

smaller CSTR.

Conversion

Original Reactor Volumes

Worst Arrangement

X3 = 0.65

V3 = 0.10 (CSTR)

For PFR,

X1 = 0.2

For CSTR,

For X2 = 0.6,

3

1.32

AO

Fm

, V2 =

( )

2 1

AO

FX X

= 1.32(0.6 – 0.2) = 0.53 m3

rA =

!

P2-4 (e) Example 2-8

hrs5=

!

(1)

hrdmmol

XC

rA

A

3

0

/

5

8.05.2

!

==“

#

= 0.5 mol/dm3

P2-5

X

0

0.1

0.2

0.4

0.6

0.7

0.8

V = 1.0 m3

P2-5 (a) Two CSTRs in series

For first CSTR,

P2-5 (b)

Two PFRs in series

P2-5 (c)

Two CSTRs in parallel with the feed, FAO,

P2-5 (d)

Two PFRs in parallel with the feed equally divided between the two reactors.

2-8

P2-5 (e)

A CSTR and a PFR are in parallel with flow equally divided

P2-5 (f)

A PFR followed by a CSTR,

P2-5 (g)

A CSTR followed by a PFR,

P2-5 (h)

A 1 m3 PFR followed by two 0.5 m3 CSTRs,

P2-6 (a) Individualized Solution

P2-6 (b)

1) In order to find the age of the baby hippo, we need to know the volume of the stomach.

2-9

2

2) If Vmax and mao are both one half of the mother’s then

0

2 2

1

2mother

A

Ao

AM AM

m

r r

! “

# $

% &

=

‘ ‘

2-10

1

2

Ao

Ao Ao

m

m m

! “

! ” # $ ! “

P2-6 (c)

Vstomach = 0.2 m3

From the web module we see that if a polynomial is fit to the autocatalytic reaction we get:

Since the hippo needs 30% conversion to survive but only achieves 17.8%, the hippo cannot

survive.

2-11

P2-6 (d)

PFR CSTR

PFR:

Outlet conversion of PFR = 0.111

CSTR:

We must solve

P2-7

Exothermic reaction: A  B + C

X

r(mol/dm3.min)

1/-

r(dm3.min/mol)

0

1

0.20

1.67

0.6

P2-7 (a)

To solve this problem, first plot 1/-rA vs. X from the chart above. Second, use mole balance as

given below.

CSTR:

2-12

Mole balance:

( )( )

( )

min./5

4.0min/300

F

V 3

A0

dmmol

mol

r

X

A

CSTR =

!

=

=>

P2-7 (b)

For a feed stream that enters the reaction with a previous conversion of 0.40 and leaves at any

conversion up to 0.60, the volumes of the PFR and CSTR will be identical because of the rate is

P2-7 (c)

VCSTR = 105 dm3

Mole balance:

A

CSTR r

X

!

=A0

F

V

2-13

P2-7 (d)

From part (a) we know that X1 = 0.40.

Use trial and error to find X2.

Conversion = 0.64

P2-7 (e)

P2-7 (f)

See Polymath program P2-7-f.pol.

P2-8 (a)

S

r

XF

V

!

=0

P2-8 (b)

At a conversion of 80%,

g

hrdm

rS

3

8.0

1=

!

P2-8 (c)

!“

=

X

S

SPFR r

dX

FV

0

2-15

P2-8 (d)

To achieve 80% conversion with a CSTR followed by a CSTR, the optimum arrangement is to

have a CSTR with a volume to achieve a conversion of about 45%, or the conversion that

P2-8 (e)

SM

CS

sCK

CkC

r

+

=!

and

[ ] 001.01.0 0+!=SSC CCC

then

CskCk

CK

rS

SM

s2

2

1

+

=!

since CS < KM

P9-2

Irreversible gas phase reaction

2A + B  2C

See Polymath program P2-9.pol.

P2-9 (a)

PFR volume necessary to achieve 50% conversion

Mole Balance

P2-9 (b)

CSTR Volume to achieve 50% conversion

Mole Balance

P2-9 (c)

Volume of second CSTR added in series to achieve 80%

conversion

P2-9 (d)

Volume of PFR added in series to first CSTR to achieve

80% conversion

P2-9 (e)

For CSTR,

V = 60000 m3 (CSTR)

For PFR,

V = 60000 m3 (PFR)

Mole balance