Download Solutions Manual for Elements of Chemical Reaction Engineering Fogler and more Exercises Chemical Principles in PDF only on Docsity!

Solutions Manual for

Elements of Chemical Reaction Engineering

Fourth Edition

Brian Vicente

Max Nori

H. Scott Fogler

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

chapter. It also gives tips on problem solving.

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

available on the CD-ROM and the Web.

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

alternative to P1- 15.

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

Montgomery’s equipment module on the CD-ROM. See P1- 17.

CDP1-A AA FSF 30

CDP 1 - 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

assigned.

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).

IC = Intermediate calculation required

M = More difficult

OE = Some parts open-ended.

____________

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,

Introduction to the CD-ROM 1,2,3,

Make a calculation 6

Open-ended 8,

P 1 - 1 Individualized solution.

P 1 - 2 Individualized solution.

P 1 - 3 Individualized solution.

P 1 - 4 Individualized solution.

P 1 - 5 Solution is in the decoding algorithm given with the modules.

P1- 6

The general equation for a CSTR is:

A

A A

r

F F

V

0

Here rA is the rate of a first order reaction given by: rA = - kCA

Given : CA = 0.1CA0 , k = 0.23 min

• 1 , v 0 = 10dm 3 min - 1 , FA = 5.0 mol/hr And we know that FA = CAv 0 and FA0 = CA0v 0 => CA0 = FA0/ v 0 = 0.5 mol/dm 3

Substituting in the above equation we get:

( 0. 23 min )( 0. 1 ( 0. 5 / ))

( 0. 5 / )( 10 /min) 0. 1 ( 0. 5 / )( 10 /min) 1 3

3 3 3 3 A0 0 0

mol dm

mol dm dm mol dm dm

kC

C v C v V

A

A !

V = 391.3 dm 3

P1- 7

1 k 0.23min

!

From mole balance:

dN A

dt

r A

!V

Rate law: r A ! k C A

r A ! k

N

A

V

Combine:

dN A

dt

!k N A

0

t

1 t

d

k

N A

NA

N

A

N

A

• d

at! = 0 , NAO = 100 mol and! =!, NA = (0.01)NAO

 t = !

A

A

N

N

k

0 ln

ln( 100 ) min

= t = 20 min

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

=>

'

0 = ( Fj 0 " Fj ) + # rj (! bdV )

Hence the above equation becomes

0 '

j j

j

F F

W

r

We can also just apply the general mole balance as

' ( 0 ) ( )

j j j j

dN F F r dW dt

Assuming no accumulation and no spatial variation in rate, we get the same form as

above:

0 '

j j

j

F F

W

r

P1- 10

Mole balance on species j is:

V j j j j dt

dN F F rdV

0

0

Let Mj = molecular wt. of species j

Then F (^) j 0 Mj = wj 0 = mass flow rate of j into reactor:

N (^) j Mj = mj = mass of species j in the reactor

Multiplying the mole balance on species j by Mj

V j j j j j j j j dt

dN F M FM M rdV M

0

0

Now Mj is constant:

( ) ( )

dt

dm

dt

dM N F M FM M rdV

j

V j j

j j!^ j j +" j j = =

0

0

dt

dm w w M rdV

j

V

j!^ j +" j j =

0

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.

[In flowrate (moles/time)] (^) penicillin + [generation rate (moles/time)]penicillin – [Out flowrate (moles/time)] (^) penicillin

= [rate of accumulation (moles/time)]penicillin

Fp,in + Gp – Fp,out =

dNp

dt Fp,in = 0 …………………………….(because no penicillin inflow)

Gp =.

V

! r dVp

Therefore,

V

! r dVp^ -^ Fp,out^ =

dNp

dt

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

dNp

dt

= 0

And no variations

p in , p out ,

p

F F

V

r

Or,

p out ,

p

F

V

r

Similarly, for Corn Steep Liquor with FC = 0

C

C

C

C C

r

F

r

F F

V

0 0

Assume RNA concentration does not change in the stationary state and no RNA is generated or destroyed.

P1- 12 (d)

Annual Production rate of ethylene for year 2002 is 5.21x 10 10 lb/year

Annual Production rate of benzene for year 2002 is 1.58 x 10 10 lb/year

Annual Production rate of ethylene oxide for year 2002 is 7.6 x 9 lb/year

P1- 12 (e)

Because the basic raw material ‘coal and petroleum’ for organic chemicals is very limited and their

production is not increasing as production of raw material for inorganic chemicals.

P1- 13

Type Characteristics Phases Usage Advantage Disadvantage

Batc

h

All the reactants

fed into the

reactor. During reaction nothing

is added or

removed. Easy

heating or cooling.

1. Liquid

phase

2. Gas phase
3. Liquid

Solid

1. Small scale

pdn.

2. Used for lab experimentation.
3. Pharmaceuticals
4. Fermentation
   1. High

Conversion per

unit volume.

2. Flexibility of

using for

multiple

reactions.

3. Easy to clean
   1. High

Operating

cost.

2. Variable

product

quality.

CST R

Continuous flow of reactants and

products.

Uniform composition

throughout.

1. Liquid phase
2. Gas –

liquid

3. Solid -

liquid

1. Used when agitation required.
2. Series

Configuration possible for

different

configuration streams

1. Continuous Operation.

2.Good

Temperature Control

3. Two phase

reactions possible.

4.Good Control

5. Simplicity of construction.
6. Low

operating cost

7. Easy to clean
   1. Lowest conversion

per unit

volume.

2. By passing

possible with

poor agitation.

3 High power

Input reqd.

PFR One long reactor

or number of CSTR’s in

series.

No radial variations. Conc.

changes along

the length.

Primarily gas Phase

1. Large Scale

pdn.

2. Fast reactions
3. Homogenous

reactions

4. Heterogeneous

reactions

5. Continuous pdn.
   1. High

conversion per unit volume

2. Easy to

maintain (No moving parts)

3. low

operating cost

4. continuous

operation

1. Undesired

thermal gradient.

2. Poor

temperature control

3. Shutdown

and cleaning expensive.

PBR Tubular reactor

that is packed

with solid

catalyst particles.

1. Gas

Phase

(Solid

Catalyst) 2.Gas –

solid

reactions.

1. Used primarily

in the

heterogeneous gas

phase reaction with solid catalyst

e.g Fischer

tropsch synthesis.

1. High

conversion per

unit mass of

catalyst

2. low

operating cost

3. Continuous operation
   1. Undesired

thermal

gradient.

2. Poor temperature

control

Channeling

4. Cleaning

expensive.

P1- 14

Given

10 2 A = 2 * 10 ft TSTP = 491. 69 R H = 2000 ft

13 3 V = 4 * 10 ft T = 534.7 ϒR PO = 1atm

lbmolR

atmft R

3

= 0. 7302 yA = 0.

3

10

2. 04 * 10 ft

lbmol C (^) S

!

C = 4* 5 cars

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

hr

ft v (^) A

3

= 3000 per car at STP

P1- 14 (a)

Total number of lb moles gas in the system:

N =

13 3

3

atm ft

atm ft R lbmol R

= 1.025 x 10 11 lb mol

P1- 14 (b)

Molar flowrate of CO into L.A. Basin by cars.

STP

A A A T A RT

v CP F y F y

0 = =!

cars ft

lbmol

hrcar

ft FT 400000 359

3

3 =!! (See appendix B)

FA = 6.685 x 10 4 lb mol/hr

N

P 0

⋅V

R⋅ T

:=

P1-14 (h)

(1) to = 0 tf = 72 hrs

Cco = 2.00E-10 lbmol/ft 3 a = 3.50E+04 lbmol/hr

vo = 1.67E+12 ft 3 /hr b = 3.00E+04 lbmol/hr

Fs = 341.23 lbmol/hr V = 4.0E+13 ft 3

dt

dC F vC V

t a b

co " + s! o co =

sin (

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

See Polymath program P 1 - 14 - h- 1 .pol.

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+ a 3.5E+04 3.5E+04 3.5E+04 3.5E+ b 3.0E+04 3.0E+04 3.0E+04 3.0E+ F 341.23 341.23 341.23 341. V 4.0E+13 4.0E+13 4.0E+13 4.0E+

ODE Report (RKF45)

Differential equations as entered by the user [1] d(C)/d(t) = (a+bsin(3.14t/6)+F-v0*C)/V

Explicit equations as entered by the user [1] v0 = 1.6710^ [2] a = 35000 [3] b = 30000 [4] F = 341. [5] V = 410^

(2) tf = 48 hrs Fs = 0 dt

dC vC V

t a b

co "! o co =

sin (

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

See Polymath program P 1 - 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+ a 3.5E+04 3.5E+04 3.5E+04 3.5E+ b 3.0E+04 3.0E+04 3.0E+04 3.0E+ V 4.0E+13 4.0E+13 4.0E+13 4.0E+

ODE Report (RKF45)

Differential equations as entered by the user [1] d(C)/d(t) = (a+bsin(3.14t/6)-v0*C)/V

Explicit equations as entered by the user [1] v0 = 1.6710^ [2] a = 35000 [3] b = 3000 0 [4] V = 410^

(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.

Changing b  The amplitude of ripples is directly proportional to ‘b’. As b decreases amplitude decreases and graph becomes smooth.

Changing v 0  As the value of v 0 is increased the graph changes to a “shifted sin-curve”. And as v 0 is decreased graph changes to a smooth increasing curve.

P1- 15 (a)

• rA = k with k = 0.05 mol/h dm 3

CSTR: The general equation is

A

A A

r

F F

V

0

Here CA = 0.01CA0 , v 0 = 10 dm 3 /min, FA = 5.0 mol/hr

Also we know that FA = CAv 0 and FA0 = CA0v 0 , CA0 = FA0/ v 0 = 0.5 mol/dm 3

P1- 15 (c)

• rA = kCA 2 with k = 3 dm 3 /mol.hr

CSTR:

V =

C A0 v 0 " C (^) A v 0

" rA

v 0 C (^) A 0 (1" 0.01)

kC (^) A

2

Substituting all the values we get

V =

(10 dm

3 / hr )(0.5 mol / dm

3 )(0.99)

(3 dm

3 / hr )(0.01* 0.5 mol / dm

3 )

2 => V = 66000 dm

3

PFR:

dC (^) A v 0

dV

= rA = kC (^) A

2

Integrating

v 0

k

dC (^) A

C A

2 C (^) A 0

C (^) A

" =^ #^ dV

0

V

"^ =>

v 0

k

C A

C A 0

) = V

=>

V =

10 dm

3 / hr

3 dm

3 / mol. hr

0.01 C A 0

C A 0

) = 660 dm 3

P1-15 (d)

CA = .001CA

t =

dN

N (^) A " rA V

N (^) A 0

Constant Volume V=V 0

t =

dC (^) A

C (^) A " rA

C (^) A 0

Zero order:

t =

k

[ C^ A 0 "^ 0.001 C^ A 0 ] =^

.999 C (^) Ao

= 9.99 h

First order:

t =

k

ln

C A 0

C A

' =^

ln

' = 6908 s

Second order:

t =

k

C A

C A 0

( =^

( =^666 h

P1- 16 Individualized Solution

P1- 17 (a)

Initial number of rabbits, x(0) = 500

Initial number of foxes, y(0) = 200

Number of days = 500

1 2

dx k x k xy dt

3 4

dy k xy k y dt

Given,

1 1 2 3 1 4

k day

k day foxes

k day rabbits

k day

!

!

See Polymath program P 1 - 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. y 200 1.1285722 4099.517 117. k1 0.02 0.02 0.02 0. k2 4.0E- 05 4.0E- 05 4.0E- 05 4.0E- 05 k3 4.0E- 04 4.0E- 04 4.0E- 04 4.0E- 04 k4 0.04 0.04 0.04 0.

ODE Report (RKF45)

Differential equations as entered by the user [1] d(x)/d(t) = (k1x)-(k2xy) [2] d(y)/d(t) = (k3xy)-(k4y)

Explicit equations as entered by the user [1] k1 = 0. [2] k2 = 0. [3] k3 = 0. [4] k4 = 0.

POLYMATH Results

NLES Solution

Variable Value f(x) Ini Guess x 2.3850387 2.53E- 11 2 y 3.7970279 1.72E- 12 2

NLES Report (safenewt)

Nonlinear equations [1] f(x) = x^3y-4y^2+3x-1 = 0 [2] f(y) = 6y^2-9xy-5 = 0

P 1 - 18 (a)

No preheating of the benzene feed will diminish the rate of reaction and thus lesser conversions will be

achieved.

P 1 - 18 (b)

An interpolation can be done on the logarithmic scale to find the desired values from the given data.

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

Cost of 6000 gal reactor = 1.905 x 10 5 $

Cost of 15000 gal reactor = 5.623 x 10 5 $

P 1 - 18 (c)

We are given CA is 0.1% of initial concentration

 CA = 0.001CA

Also from Example 1.3,

ln( )

0 0

A

A

C

C

k

v

V =

Substituting v0 = 10 dm 3 /min and

k = 0.23 min

• 1 we get

3 V = 300 dm

which is three times the volume of reactor used

in Example 1- 3.

P 1 - 18 (d) Safety of Plant.

P1- 19 Enrico Fermi Problem – no definite solution

P1- 20 Enrico Fermi Problem – no definite solution

P1- 21 Individualized solution.

Cost vs Volume of reactor (log - log plot)

0

1

2

3

4

5

6

0 0.5 1 1.5 2 2.5 3 3.5 4 4. log volume (gallons)

CDP1-A (a)

How many moles of A are in the reactor initially? What is the initial concentration of 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:

3

3

.^1

PV atm dm kPa n moles RT (^) kPa dm atm

molK

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

molesA = NAo = 0.75! 97.5 =73.

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

3 3

Ao Ao

moles N moles C moles dm volume V dm

CDP1-A (b)

Time (t) for a 1 st

order reaction to consume 99% of A.

A A

dC r dt

Our first order rate law is:! r (^) A = kCA

mole balance: A A

dC kC dt

0

A

Ao

t C A

C 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

ln(0.01) 46.1min 0.1min

t k

!

CDP1-A (c)

Time for 2nd order reaction to consume 80% of A and final pressure (P) at T = 127 C.

rate law:

2 A A ! r = kC

mole balance:

A^2 A

dC kC dt

2 0

A

Ao

t C A

C A

dC k dt C

A Ao

kt C C

We can solve for the time in terms of our rate constant (k = 0.7) and our initial

concentration (CAo):

Ao Ao

kt C C