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.
P1-4. 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.
P1-9 and P1-10. The results of these problems will appear in later chapters. Straight
forward application of chapter 1 principles.
1-2
P1-14. Many students like this straight forward problem because they see how CRE
P1-16. Open-ended problem.
P1-19 and P1-20. Help develop critical thinking and analysis.
CDP1-A Similar to problems 3, 4, 11, and 12.
Summary
Assigned
Alternates
Difficulty
Time (min)
P1-1
AA
SF
60
P1-2
I
SF
30
P1-3
O
SF
30
P1-4
O
SF
30
P1-5
AA
SF
30
AA
FSF
30
I
FSF
30
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)
____________
*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
P1-1 Individualized solution.
P1-2 Individualized solution.
1-4
P1-6
The general equation for a CSTR is:
A
AA
r
FF
V
!
!
=0
Here rA is the rate of a first order reaction given by:
))/5.0(1.0)(min23.0(
min)/10)(/5.0(1.0min)/10)(/5.0(
31
3333
00A0
dmmol
dmdmmoldmdmmol
kC
vCvC
V
A
A
!
!
=
!
=
V = 391.3 dm3
P1-7
1
0.23 mink!
=
NA0
at
0
!
=
, NAO = 100 mol and
! !
=
, NA = (0.01)NAO
1-5
P1-8 (a)
The assumptions made in deriving the design equation of a batch reactor are:
P1-8 (b)
The assumptions made in deriving the design equation of CSTR, are:
P1-8 (c)
The assumptions made in deriving the design equation of PFR are:
– No radial variation in properties of the system.
P1-8 (d)
The assumptions made in deriving the design equation of PBR are:
P1-8 (e)
For a reaction,
A B
-rA is the number of moles of A reacting (disappearing) per unit time per unit volume [=] moles/
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
!
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
dt =!+“
Assuming no accumulation and no spatial variation in rate, we get the same form as
above:
0
‘
j j
j
F F
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
1-7
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
dt
Assuming steady state for the rate of production of penicillin in the cells stationary state,
dNp
dt
= 0
And no variations
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
now then rank 9 in 1995 .
P1-12 (b)
Ranking of top 10 chemical companies in sales in year 2003 and 2002:
s
o
u
r
We have Chevron Phillips which jumped to 5 rank in 2003 from 8th rank in 2002 and Air
P1-12 (c)
Sulfuric acid is prime importance in manufacturing. It is used in some phase of the manufacture of nearly
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
1-9
P1-12 (d)
Annual Production rate of ethylene for year 2002 is 5.21x 1010 lb/year
P1-12 (e)
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
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.
1. High
Operating
cost.
2. Variable
product
quality.
PFR
One long reactor
or number of
CSTR’s in
1.
Primarily
gas Phase
1. Large Scale
pdn.
2. Fast reactions
1. High
conversion per
unit volume
1. Undesired
thermal
gradient.
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.
P1-14
Given
210
10*2 ftA =
RTSTP 69.491=
ftH 2000=
313
10*4 ftV =
T = 534.7
°
R PO = 1atm
P1-14 (a)
Total number of lb moles gas in the system:
P1-14 (b)
Molar flowrate of CO into L.A. Basin by cars.
N
P
0
V
⋅
R
T
⋅
:=
1-11
P1-14 (c)
Wind speed through corridor is v = 15mph
P1-14 (d)
Molar flowrate of CO into basin from Sant Ana wind.
P1-14 (e)
Rate of emission of CO by cars + Rate of CO by Wind – Rate of removal of CO =
CO
dN
P1-14 (f)
t = 0 ,
coOco CC =
P1-14 (g)
Time for concentration to reach 8 ppm.
8
03
2.04 10
CO
lbmol
Cft
!
=“
,
8
3
2.04 10
4
CO
lbmol
Cft
!
=“
From (f),
1-12
P1-14 (h)
(1) to = 0 tf = 72 hrs
co
C
= 2.00E-10 lbmol/ft3 a = 3.50E+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
ODE Report (RKF45)
Differential equations as entered by the user
[1] d(C)/d(t) = (a+b*sin(3.14*t/6)+F-v0*C)/V
Explicit equations as entered by the user
[1] v0 = 1.67*10^12
(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
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
(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)
– rA = k with k = 0.05 mol/h dm3
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
P1-15 (b)
– rA = kCA with k = 0.0001 s-1
CSTR:
PFR:
From above we already know that for a PFR
AA
AkCr
dV
vdC ==
0
Integrating
P1-15 (c)
– rA = kCA
2 with k = 3 dm3/mol.hr
CSTR:
V=CA0v0“CAv0
“r
=v0CA0(1“0.01)
kCA
2
!
!
!
P1-15 (d)
CA = .001CA0
t=dN
“r
AV
NA
NA0
#
Constant Volume V=V0
First order:
1-16
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
POLYMATH Results
Calculated values of the DEQ variables
Variable initial value minimal value maximal value final value
t 0 0 500 500
ODE Report (RKF45)
Explicit equations as entered by the user
1-17
When, tfinal = 800 and
30.00004 /( )k day rabbits=!
Plotting rabbits Vs. foxes
P1-17 (b)
POLYMATH Results
See Polymath program P1-17-b.pol.
1-18
POLYMATH Results
NLES Solution
Variable Value f(x) Ini Guess
NLES Report (safenewt)
P1-18 (a)
P1-18 (b)
An interpolation can be done on the logarithmic scale to find the desired values from the given data.
P1-18 (c)
We are given CA is 0.1% of initial concentration
CA = 0.001CA0
P1-18 (d) Safety of Plant.
P1-19 Enrico Fermi Problem – no definite solution
Cost vs Volume of reactor (log – log plot)
1-19
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:
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!=
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!=
( )( )
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:
CDP1-B
Given: Liquid phase reaction in a foam reactor, A
!
B
Consider a differential element,
V!
of the reactor:
By material balance
flow rate of A (g mol/sec.); V = volume of reactor