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

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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
page-pf2
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
page-pf3
1-3
AA
FSF
30
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.
page-pf4
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
NA
1
NA
!
"
"
#
d+
at
0
!
=
, NAO = 100 mol and
! !
=
, NA = (0.01)NAO
t =
!
!
"
#
$
$
%
&
A
A
N
N
k
0
ln
1
( ) min100ln
23.0
1
=
t = 20 min
page-pf5
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.
page-pf6
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
0
Let
j
M
= molecular wt. of species j
page-pf7
1-7
dt
dm
dVrMww j
V
jjjj =+!"
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.
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,
page-pf8
P1-12 (a)
Ranking of 10 most produced chemicals in 1995 and 2002 are listed in table below:
Rank 2002
Rank 1995
Chemical
1
1
H2SO4
2
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
1
Dow Chemical
32632
2
2
DuPont
30249
3
3
ExxonMobil
20190
page-pf9
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
page-pfa
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
3
!!=
(See appendix B)
FA = 6.685 x 104 lb mol/hr
N
P
0
V
R
T
:=
page-pfb
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
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
page-pfc
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
page-pfd
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
page-pfe
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
page-pff
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
page-pf10
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
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
page-pf11
1-17
P1-17 (b)
POLYMATH Results
page-pf12
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)
page-pf13
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
3
73.1 0.37 /
200
Ao
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
A
dC
r
dt
=
Our first order rate law is:
A A
r kC!=
mole balance:
A
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
A
dC kC
dt
!="
=>
2
A
C
t
A
dC
k dt
C
!=
" "
=>
1 1
kt
C C
!=!+
page-pf14
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
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:

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