Chemical Engineering Chapter 1 Solutions For Mole Balances Synopsis General The Goal These Problems Are

1-1

Solutions for Chapter 1 – Mole Balances

Synopsis

General: The goal of these problems are to reinforce the definitions and provide an

understanding of the mole balances of the different types of reactors. It lays the

foundation for step 1 of the algorithm in Chapter 4.

P1-1. This problem helps the student understand the course goals and objectives.

P1-2. Part (d) gives hints on how to solve problems when they get stuck. Encourages

students to get in the habit of writing down what they learned from each

P1-3. Helps the student understand critical thinking and creative thinking, which are

two major goals of the course.

P1-4. Requires the student to at least look at the wide and wonderful resources

P1-5. The ICMs have been found to be a great motivation for this material.

P1-6. Uses Example 1-1 to calculate a CSTR volume. It is straight forward and gives

the student an idea of things to come in terms of sizing reactors in chapter 4. An

P1-7. Straight forward modification of Example 1-1.

P1-8. Helps the student review and member assumption for each design equation.

P1-9 and P1-10. The results of these problems will appear in later chapters. Straight

forward application of chapter 1 principles.

P1-11. Straight forward modification of the mole balance. Assigned for those who

emphasize bioreaction problems.

P1-12. Can be assigned to just be read and not necessarily to be worked. It will give

students a flavor of the top selling chemicals and top chemical companies.

P1-13. Will be useful when the table is completed and the students can refer back to it

in later chapters. Answers to this problem can be found on Professor Susan

1-2

P1-14. Many students like this straight forward problem because they see how CRE

principles can be applied to an everyday example. It is often assigned as an in-

P1-15. Shows a bit of things to come in terms of reactor sizing. Can be rotated from

P1-16. Open-ended problem.

P1-17. I always assign this problem so that the students will learn how to use

POLYMATH/MaLab before needing it for chemical reaction engineering

P1-18. Parts (a) and (b) are open-ended problem.

P1-19 and P1-20. Help develop critical thinking and analysis.

CDP1-A Similar to problems 3, 4, 11, and 12.

CDP1-B Points out difference in rate per unit liquid volume and rate per reactor

volume.

Summary

Assigned

Alternates

Difficulty

Time (min)



P1-1

AA

SF

60

P1-2

I

SF

30

P1-4

O

SF

30

P1-6

AA

1-15

SF

15



P1-8

S

SF

15

P1-10

S

SF

15

P1-12

I

– Read Only

SF

30

P1-14

O

FSF

30

P1-16

S

SF

15

P1-18

S

SF

30

P1-20

O

FSF

15

1-3

CDP1-A

AA

FSF

30

CDP1-B

I

FSF

30

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

problem, either the problem with a dot or any one of the alternates are always

Time

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

Difficulty

SF = Straight forward reinforcement of principles (plug and chug)

FSF = Fairly straight forward (requires some manipulation of equations or an

intermediate calculation).

____________

*Note the letter problems are found on the CD-ROM. For example A ≡ CDP1-A.

Summary Table Ch-1

Review of Definitions and Assumptions

1,5,6,7,8,9

Introduction to the CD-ROM

1,2,3,4

P1-1 Individualized solution.

P1-2 Individualized solution.

P1-3 Individualized solution.

P1-4 Individualized solution.

P1-5 Solution is in the decoding algorithm given with the modules.

1-4

P1-6

The general equation for a CSTR is:

AA

r

FF

V

!

=0

Given : CA = 0.1CA0 , k = 0.23 min-1, v0 = 10dm3 min-1, FA = 5.0 mol/hr

And we know that FA = CAv0 and FA0 = CA0v0

=> CA0 = FA0/ v0 = 0.5 mol/dm3

Substituting in the above equation we get:

P1-7

1

Combine:

dNA

dt

k!NA

“

0

!

=

0

t

t1

!

“

#

d1

k

$

%

&

‘

(

)

*

NA0

NA

1

NA

!

“

#

d+

at

0

!

=

, NAO = 100 mol and

! !

=

, NA = (0.01)NAO

 t =

!

“

#

$

%

&

A

N

k

0

ln

1

( ) min100ln

23.0

1

=

t = 20 min

1-5

P1-8 (a)

– Closed system: no streams carrying mass enter or leave the system.

P1-8 (c)

P1-8 (d)

P1-8 (e)

A B

-rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=] moles/

(dm3.s).

P 1-9

Rate of homogenous reaction rA is defined as the mole of A formed per unit volume of the reactor per

second. It is an Intensive property and the concentration, temperature and hence the rate varies with spatial

coordinates.

1-6

Also rj = ρb rj` and W = Vρb where ρb is the bulk density of the bed.

=>

‘

0

0 ( ) ( )

j j j b

F F r dV

!

=“+#

Hence the above equation becomes

0

‘

j j

j

Wr

=!

P1-10

Mole balance on species j is:

!=+“

V

j

jjj dt

dN

dVrFF

0

Let

j

M

= molecular wt. of species j

1-7

dt

dm

dVrMww j

V

jjjj =+!“

0

P1-11

Applying mole balance to Penicillin: Penicillin is produced in the cells stationary state (See Chapter 7), so

there is no cell growth and the nutrients are used in making product.

Let’s do part c first.

Therefore,

.

V

p

r dV

!

– Fp,out =

dNp

dt

Assuming steady state for the rate of production of penicillin in the cells stationary state,

And no variations

Or,

P1-12 (a)

Ranking of 10 most produced chemicals in 1995 and 2002 are listed in table below:

Rank 2002

Rank 1995

Chemical

1

H2SO4

2

N2

3

4

C2H4

s

o

u

r

c

all industrial products .It is used in production of every other strong acid. Because of its large number of

uses, it’s the most produced chemical. Sulfuric acid uses are:

It is consumed in production of fertilizers such as ammonium sulphate (NH4)2SO4 and

2003

2002

Company

Chemical Sales

($ million 2003)

1

Dow Chemical

32632

2

DuPont

30249

3

ExxonMobil

20190

1-9

P1-12 (d)

P1-12 (e)

P1-13

Type

Characteristics

Phases

Usage

Advantage

Disadvantage

Batc

h

All the reactants

fed into the

reactor. During

1. Liquid

phase

2. Gas

1. Small scale

pdn.

2. Used for lab

1. High

Conversion per

unit volume.

1. High

Operating

cost.

5. Simplicity of

construction.

6. Low

operating cost

7. Easy to clean

Input reqd.

PFR

One long reactor

or number of

CSTR’s in

series.

No radial

1.

Primarily

gas Phase

1. Large Scale

pdn.

2. Fast reactions

3. Homogenous

reactions

1. High

conversion per

unit volume

2. Easy to

maintain (No

1. Undesired

thermal

gradient.

2. Poor

temperature

1-10

PBR

Tubular reactor

that is packed

with solid

1. Gas

Phase

(Solid

1. Used primarily

in the

heterogeneous gas

1. High

conversion per

unit mass of

1. Undesired

thermal

gradient.

expensive.

P1-14

Given

210

10*2 ftA =

RTSTP 69.491=

ftH 2000=

313

10*4 ftV =

T = 534.7

°

R PO = 1atm

Rlbmol

ftatm

R

3

7302.0=

yA = 0.02

3

10

10*04.2

ft

lbmol

CS

!

=

C = 4*105 cars

FS = CO in Santa Ana wind FA = CO emission from autos

hr

ft

vA

3

3000=

per car at STP

P1-14 (a)

13 3

0.73 534.69

.

lbmol R

$ %

& ‘

P1-14 (b)

STP

A

ATAA TR

PCv

yFyF 0

!==

cars

ft

carhr

FT400000

359

3

!!=

(See appendix B)

FA = 6.685 x 104 lb mol/hr

N

P

0

V

⋅

R

T

⋅

:=

P1-14 (c)

W = 20 miles

The volumetric flowrate in the corridor is

FSv0CS

!:=

= 1.673 x 1013 ft3/hr

10

2.04 10!

” “

lbmol/ft3

Rate of emission of CO by cars + Rate of CO by Wind – Rate of removal of CO =

CO

dN

dt

dC

VCvFF co

cooSA =!+

(V=constant,

VCN coco =

)

P1-14 (f)

t = 0 ,

coOco CC =

0

co

coO

C

t

co

A S o co

C

dC

dt V

F F v C

=+!

” “

!

“

#

$

%

&

‘+

=

cooSA

coOoSA

oCvFF

CvFF

v

V

tln

P1-14 (g)

8

03

2.04 10

CO

lbmol

Cft

!

=“

,

8

3

2.04 10

4

CO

lbmol

Cft

!

=“

From (f),

0

3

4 3 13 8

33

3 3

13 4 3 13 8

3

.

ln

.

6.7 10 3.4 10 1.673 10 2.04 10

4ln

1.673 10 6.7 10 3.4 10 1.673 10 0.51 10

A S O CO

o A S O CO

F F v C

V

tv F F v C

lbmol lbmol ft lbmol

ft hr hr hr ft

ft lbmol lbmol ft lbmol

hr hr hr hr ft

!

” #

+!

=$ %

+!

& ‘

” #

(+( ! ( ( (

$ %

=$ %

( ( +( ! ( ( (

$ %

& ‘

t = 6.92 hr

1-12

P1-14 (h)

co

C

= 2.00E-10 lbmol/ft3 a = 3.50E+04 lbmol/hr

o

v

= 1.67E+12 ft3 /hr b = 3.00E+04 lbmol/hr

POLYMATH Results

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 72 72

C 2.0E-10 2.0E-10 2.134E-08 1.877E-08

v0 1.67E+12 1.67E+12 1.67E+12 1.67E+12

V 4.0E+13 4.0E+13 4.0E+13 4.0E+13

ODE Report (RKF45)

Explicit equations as entered by the user

[1] v0 = 1.67*10^12

[2] a = 35000

[3] b = 30000

(2) tf = 48 hrs

s

F

= 0

dt

dC

VCv

t

ba co

coo =!

“

#

$

%

&

‘

+6

sin

(

Now solving this equation using POLYMATH we get plot between Cco vs t

1-13

See Polymath program P1-14-h-2.pol.

POLYMATH Results

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 48 48

C 2.0E-10 2.0E-10 1.904E-08 1.693E-08

v0 1.67E+12 1.67E+12 1.67E+12 1.67E+12

ODE Report (RKF45)

Differential equations as entered by the user

[1] d(C)/d(t) = (a+b*sin(3.14*t/6)-v0*C)/V

Explicit equations as entered by the user

[1] v0 = 1.67*10^12

(3)

Changing a  Increasing ‘a’ reduces the amplitude of ripples in graph. It reduces the effect of the

sine function by adding to the baseline.

P1-15 (a)

CSTR: The general equation is

1-14

Substituting the values in the above equation we get,

05.0

10)5.0(01.010)5.0(

00A0 !

=

!

=k

vCvC

VA

 V = 99 dm3

PFR: The general equation is

kr

dF

A

A==

, Now FA = CAv0 and FA0 = CA0v0 =>

k

vdC A!=

0

A0

0

Hence V = 99 dm3

Volume of PFR is same as the volume for a CSTR since the rate is constant and independent of

P1-15 (b)

CSTR:

We have already derived that

PFR:

From above we already know that for a PFR

AA

AkCr

dV

vdC ==

0

Integrating

!

“=

!

V

C

AdV

dC

vA

0

1-15

P1-15 (c)

– rA = kCA

2 with k = 3 dm3/mol.hr

CSTR:

V=CA0v0“CAv0

“r

A

!

=v0CA0(1“0.01)

kCA

2

Substituting all the values we get

!

PFR:

dCAv0

dV =r

A=kCA

2

Integrating

!

=>

!

V=10dm3/hr

3dm3/mol.hr

(1

0.01CA0

“1

CA0

)

= 660 dm3

P1-15 (d)

CA = .001CA0

Constant Volume V=V0

!

Zero order:

t=1

k

CA0“0.001CA0

[ ]

=.999CAo

0.05 =9.99 h

First order:

!

Second order:

!

t=1

k

1

CA

“1

CA0

#

$

%

&

‘

( =1

3

1

0.0005 “1

0.5

#

$

%

&

‘

( =666 h

1-16

P1-16 Individualized Solution

P1-17 (a)

Initial number of foxes, y(0) = 200

Number of days = 500

1 2

dx k x k xy

dt =!

…………………………….(1)

3 4

dy k xy k y

dt =!

……………………………..(2)

Given,

1

2

3

1

4

0.02

0.00004 /( )

0.0004 /( )

0.04

k day

k day foxes

k day rabbits

k day

!

=

=“

=

See Polymath program P1-17-a.pol.

POLYMATH Results

Calculated values of the DEQ variables

Variable initial value minimal value maximal value final value

t 0 0 500 500

x 500 2.9626929 519.40024 4.2199691

y 200 1.1285722 4099.517 117.62928

ODE Report (RKF45)

Differential equations as entered by the user

Explicit equations as entered by the user

[1] k1 = 0.02

[2] k2 = 0.00004

1-17

P1-17 (b)

POLYMATH Results

1-18

POLYMATH Results

NLES Solution

Variable Value f(x) Ini Guess

NLES Report (safenewt)

Nonlinear equations

P1-18 (b)

Now we can interpolate to the get the cost at 6000 gallons and 15000 gallons

P1-18 (c)

We are given CA is 0.1% of initial concentration

 CA = 0.001CA0

k = 0.23 min-1 we get

3

300dmV =

which is three times the volume of reactor used

P1-18 (d) Safety of Plant.

P1-19 Enrico Fermi Problem – no definite solution

Cost vs Volume of reactor (log – log plot)

0

1

00.5 11.5 22.5 33.5 44.5

log volume (gallons)

1-19

CDP1-A (a)

If we assume ideal gas behavior, then calculating the moles of A initially present in the reactor is quite

simple. We insert our variables into the ideal gas equation:

8.3145 (500)

molK

# $

% &

Knowing the mole fraction of A (yAo) is 75%, we multiply the total number of moles (NTo) by the yA:

0.75 97.5 73.1

Ao

molesA N= = !=

The initial concentration of A (CAo) is just the moles of A divided by the volume:

3

73.1 0.37 /

200

Ao

N

moles moles

C moles dm

volume V dm

= = = =

CDP1-A (b)

Time (t) for a 1st order reaction to consume 99% of A.

A

dC

r

dt

=

Our first order rate law is:

A A

r kC!=

mole balance:

A

dC kC

dt =!

=>

A

C

t

A

dC

k dt

C

!=

” “

ln A

Ao

C

kt

C

! “

#=$ %

& ‘

, knowing CA=0.01 CAo and our rate constant (k=0.1 min-1), we can solve

for the time of the reaction:

1

1 4.61

ln(0.01) 46.1min

0.1min

t

k!

=!= =

CDP1-A (c)

rate law:

2

A A

r kC!=

mole balance:

2

A

dC kC

dt

!=“

=>

2

A

C

t

A

dC

k dt

C

!=

” “

=>

1 1

kt

C C

!=!+

1-20

( )( )

3 3

4 4 15.4 min .

0.7 / min 0.37 /

Ao

t

kC dm mol mol dm

= = =

To determine the pressure of the reactor following this reaction, we will again use the

ideal gas law. First, we determine the number of moles in the reactor:

0.25 0.2

T I A B C To Ao B C

N N N N N N N N N= + + + = + + +

0.8

B C Ao

N N N= =

(0.25) (0.2 0.8 0.8) 0.25(97.6) (1.8)(73.2) 156.1

T To Ao

N N N moles= + + + = + =

3

(156.1 ) 0.082 (500 )

32

200

T

dm atm

mole K

molK

N RT

P atm

V dm

! “

# $

% &

= = =

CDP1-B

Given: Liquid phase reaction in a foam reactor, A

!

B

Consider a differential element,

V!

of the reactor: