CET260

Lecture 3:
Material balances with material balances
multiple units
Overview

■ Single pass conversion

■ Overall conversion
■ Degree of freedom table
■ Examples
■ Questions
Single pass conversion

■ Single pass conversion that takes place in the reactor or the unit
operation where conversion takes place
reactant input to reactor - reactant output from reactor
single pass conversion 
reactant input to reactor
 conversion related to reactor
Overall conversion

■ The overall conversion, on the other hand, has to do with the

fraction of reactant converted between the fresh feed and the final
product
reactant input to process - reactant output from process
overall conversion 
reactant input to process

 conversion related to overall process

Why is different?

■ Diagram of multiple units that includes a reactor

Combined
Feed (A)
2 Product (A, B)
3

Recycle Stream (A)

4

Fresh
Feed (A) Reactor Final
Separator
1 Mix Product (B)
5
Fractional excess

■ Another important definition is that of fractional excess or excess. The

fractional excess of a reactant j is the ratio of the excess to the
stoichiometric requirement of j:

fractional excess of j 
moles j fed   moles j stoichiometrically required 
moles j stoichiometrically required 
Degree of freedom
Reactor unit1 1 Separator 2 System Overall
No of Stream Variables x x x x x
No of Balance Equations x x x x x
No of Specified:-
Compositions x x x x x
Flows x x x x x
No of Subsidiary
Relations:- x x x x x
e.g., excess,
conversion etc

-x x -x x -x x -x x -x x
xx xx xx 0 xx
Example

Perchloric acid (HClO4) can be produced via the process shown in the
figure below. The reaction follows the stoichiometry:
Ba(ClO4)2 + H2SO4  BaSO4 + 2 HclO4
The H2SO4 fed to the reactor is 20% in excess of the stoichiometric
amount required for the reaction with the fresh feed of Ba(ClO4)2 and
1000 moles/h of stream 1 is fed. Determine the flowrate of the H2SO4
required and the flowrate of stream 5.
Question

■ Process flow diagram H2SO4

H2SO4 5

2 3 6
Reactor Separator 1 HClO4

H2SO4
1 Ba(ClO4)2 10%
BaSO4
Ba(ClO4)2 90% BaSO4
HClO4 7
HClO4 Ba(ClO4)2
4

Ba(ClO4) Separator 2
8
2
BaSO4
Ba(ClO4)2 2%
Solution

■ Block flow diagram

N2
2 N5
x 1 S
x 5S 1
Reactor N3 Separator
N1 3 1 N6
x S
1
x 0.9 3 N7 xH6 1
B x 0.1
B 7
x 1 3 xBS
H x BS
3 xB7
x P
N4 Separato
xB4 1 r N8
2 8
xBS 0.98
xB8
 Degree of freedom

Reactor Separator 1 Separator 2 System Overall

No of Stream Variables 9 8 5 15 8
No of Balance Equations 4 4 2 10 4
No of Specified:-
Compositions 2 1 1 3 2
Flows 1 0 0 1
No of Subsidiary
Relations:- 1 0 0 1 1
Excess H2SO4

-8 9 -5 8 -3 5 -15 15 -8 8
1 3 5 0 0
 Solution
Basis: Choose N1 = 1000
mol/h
. as the basis of calculation.

Subsidiary relation:
moles H2SO 4 fed  stoichiome tric moles H2SO 4 required
% excess H2SO 4  100
moles H2SO 4 fed
NS2  1000
20  100
1000

 NS2  1200 mol/h

Ba(ClO4)2 + H2SO4  BaSO4 + 2
HClO
Solution Ba(ClO4)2 + H2SO4  BaSO4 + 2 HClO4
Input + generation = output +
consumption
. H SO Balance:
2 4

 moles H 2 SO 4 entering   moles H 2 SO 4 leaving   moles H 2 SO 4 

     
 overall process   overall process    consumed 
 time   time   time 
     
     

NS2  N5S  νS r
x S2 N 2  x 5S N5  (1) r
N2  N5  r
1200  N5  r  1
Solution
Input + generation = output + consumption

Ba(ClO4)2 Balance:
.

 moles Ba(ClO 4 ) 2 entering   moles Ba(ClO 4 ) 2   moles Ba(ClO 4 ) 2 

     
 overall process   overall process    consumed 
 time   time   time 
     
     

N1B  NB8  ν B r
x 1B N1  x B8 N8  (1) r
0.91000  0.02 N8  r
900  0.02 N8  r  2
 Solution
Input + generation = output +
consumption
HClO4 Balance:
.
 moles HClO4 entering   moles HClO4   moles HClO4 leaving 
     
 overall process    generated   overall process 
 time   time   time 
     
     

N1H  νH r  NH6
x 1H N1  2r  x H6 N 6
0.11000  2r  N6
100  2r  N6  3 
 Solution
Input + generation = output + consumption

.
BaSO4 Balance:

 moles BaSO4 entering   moles BaSO 4   moles BaSO 4 leaving 

     
 overall process    generated   overall process 
 time   time   time 
     
     

8
0  ν BS r  NBS
8
r  x BS N8
r  0.98 N8  4 
Solution

(4) into (2):

900 = 0.02 N8 + 0.98 N8
N8 = 900 moles/h
r = (0.98)(900) = 882 mol/h

 from (1):
N5 = 1200 – 882 = 318 moles/h
Example 2:

The flow diagram for a combustion process is shown in the figure below. In this
process pure methane is combusted with air (consisting of 21% O 2 and 79% N2) in a
furnace. The reaction for this combustion is given by:
CH4 + 2 O2  CO2 + 2 H 2O
The CO2 found in the flue gas is scrubbed in an absorber using pure water as the
absorbing fluid (stream 5). Analysis of the resulting scrubbed gas (stream 4)
indicates the following composition: 80.8% N 2, 10.2% H2O, 2.05 CO2 and the rest
O2. If 1000 mol/h of methane is combusted with 50% excess oxygen, perform a
degree-of-freedom analysis to determine if the problem statement is well defined
and then determine the composition of the flue gas.
Process flow diagram

N2 80.8%
■ . H2O 10.2%
Flue Gas CO2 2.0%
N2 O2
CO2
4 5 H2O
H2O
CH4 O2
1 Furnace

3
2

Air
O2 21% 6
CO2
N2 79% H2O
Solution

Furnace Absorber System Overall

No of Stream Variables 8 11 15 11
No of Balance Equations 5 4 9 5
No of Specified:-
Compositions 1 3 4 4
Flows 1 0 1 1
No of Subsidiary Relations:-
Excess O2 1 0 1 1

-8 8 -7 11 -15 15 -11 11
0 4 0 0
Basis and equation

Basis: Choose N1 = 1000 mol/h as the basis of calculation.

CH4 + 2 O2  CO2 + 2 H 2O
M + 2O  C + 2W

O2 is in excess
Subsidiary relation
O2 fed  O2 stoichiome tric
. % excess  100
O2 stoichiome tric

NO2 2  21000 
50  100
21000 

 NO2 2  3000 mol/h

 NN2 2  11285.7
 N2  14285.7
Solution

O2 Balance:

■ Input + generation = output + consumption

NO2  0  N3O  νO r
3000  x 3O N3  2r  1
Solution

N2 Balance:

■ Input = output

NN2  NN3
11285.7  xN3 2 N3 2 
Solution
CH4 Balance:

■ Input + generation = output + consumption

N1M  0  NM3  νM r

1000 = 0 + r
 1000 = r . . . (3)
Solution

.
CO2 Balance:

Input + generation = output + consumption

0  ν C r  x 3CO 2 N3
r  x 3C N3 4 
Solution

H2O Balance:
.

Input + generation = output + consumption

From (3), r = 1000 mol/h

 from (1): xO3 N3 = 3000 – 2(1000) = 1000 mol/h

0  2 r  x 3W N3
2 r  xH3 2 O N3 5
Solution

.
and from (2): xN3 N3 = 11285.7
and from (4): xC3 N3 = 1000
and from (5): xW3 N3 = 2(1000) = 2000

15285.7 mol/h
Therefore the mole fractions can be calculated!
 Problem 1
Companies sometimes use two N2: 84.93% Stack gas Water (W)
burners as fuel supply. One is fired O2: 4.13% 100% H2O
with natural gas and the other with SO2: 0.10%
fuel oil. Each furnace has it’s own CO2: 10.84%
oxygen supply. For a stack gas
output of 6205 kmol/hr dry stack Air (A*)
gas, how much kmol/hr of natural Air (A)
Gas
furnace
Oil
furnace 4% CO2
gas was being consumed for the 21%O2 20% O2
analysis given? 79% N2 76% N2
Nat Gas Fuel Oil
96% CH4 50% C
2% C2H2 47% H2
2% CO2 3% S
Problem 2

The solvent ethyl ether, (C2H5)2O, is made

industrially by the dehydration of ethanol (C2H5OH)
according to the following reaction:
2 C2H5OH → (C2H5)2O + H2O
The flow process for the production and recovery of
ether is shown in the diagram below:
Problem 2- cont
Problem 2 cont

The feed rate to the process is 1000 mol/hr and the recycle
is half the feed rate to the process i.e. N2 = 500 mol/hr. In
the process shown a 90% overall conversion of C2H5OH is
obtained. Assuming the recycle to have the same
composition as the feed to the process, perform the
following: Fill in all relevant data on the diagram and degree-
of-freedom table supplied and from the results indicate
whether the problem statement is correctly defined and the
order of solving for all unknowns. Determine the composition
and flow-rate of the reactor product stream (N3).