Introduction to Chemical Engineering
CHE-201
INSTRUCTOR:
Dr. Nabeel Salim Abo-Ghander
Chapter 4
Fundamentals of Material Balance
Introduction
Feed
Product
Feed
Product
Feed
Material enters the process
Process
Outlet
Input
A Process is an operation or series of operations in which certain objectives are
achieved.
Product
Material leaves the process
2
4.1
Process classification
Let’s Consider the following situation:
100 g gold
Goldsmith
Shop
50 golden pieces
Each of which is 3 g
Is it possible?!!
Conservation of Mass:
Mass can neither be created nor destroyed.
3
4.1
Process classification
Why do we need
To classify chemical
Processes?
4
4.1
Process classification
Classification of Chemical Processes:
1. Batch Processes.
2. Continuous Processes.
3. Semibatch Process.
5
Batch Process
Feed
Batch
Process
Product
Feed is charged at the beginning of the process and
the product is collected some time later
6
Continuous Process
Feed
Continuous
Process
Product
Feed and product flow continuously throughout
the duration of the process
7
Semibatch Process
Feed
Semibatch
Process
Product
Any process which are not batch or continuous
(Inputs without outputs OR outputs without inputs)
Feed
Semibatch
Process
Product
8
Process classification
Chemical Processes can be also classified into two main categories:
1. Steady-State:
Process Variables
Process variables don’t vary with time, i.e. y  f t 
time
2. Unsteady-State (Transient):
Process variables vary with time, i.e. y  f t 
Process Variables
4.1
time
9
4.1
Process classification
Summary:
All batch processes are considered to be unsteady-state if the changes
between the initial and final time is required to determine .
Continuous processes are operated in the unsteady state at the startup, then it is operated almost at steady state mode.
10
Example 4.1-1:
Classification of processes:
1.
A balloon is filled with air at a steady rate of 2 g/min.
2.
A bottle of milk is taken from the refrigerator and left on the kitchen table.
3.
Water is boiled in an open flask.
11
4.1
Process classification
Let’s Consider the following situation:
min CH4/s
Process
mout CH4/s
m in  mout
1. A chemical reaction is taking place.
2. Leakage exist.
3. Wrong measurement.
12
4.2
Balances
enters through
system boundaries
Input
-
Output
produced within system
+
leaves through
system boundaries
generation
-
Buildup within system
consumption =
Accumulation
consumed within system
This equation can be applied for any conserved quantity such as total mass,
mass of a specific species, energy, momentum.
13
4.2
Balances
Example 4.2-2
The general Balance Equation
Each year 50,000 people move into a city, 75,000 people move out, 22,000 are born,
and 19,000 die. Write a balance on the population of the city.
14
4.2
Balances
Two types of balances may be written:
Aspects
Differential balance
Integral Balance
Indication
Indicates what happens at a certain
moment of time
Indicates what happens between
two instants of time
Balanced quantity
rate
quantity
Unit of balanced
quantity
Quantity/time
Quantity
Process
Continuous
Batch with two instants of time
Mathematical
Model
Algebraic equations
Differential equations
15
4.2
Input
Balances
-
Output
+
generation
-
consumption =
Accumulation
Special Cases:
1.
If the balanced quantity is the total mass, then:
generation
=
0
consumption
=
0
Input
2.
Output
=
Accumulation
If the balanced substance is a nonreactive species, then:
Input
3.
-
generation
=
0
consumption
=
0
-
Output
=
Accumulation
If the system is at steady state, then:
accumulation
Input
-
Output
+
=
generation
0
-
consumption
16
4.2b
Input
-
Balances on continuous steady-state processes:
Output
+
generation
-
consumption =
Accumulation
For continuous steady-state processes:
Accumulation
Input
-
Output
=
=
consumption
0
-
generation
If the balance is on a nonreactive species:
generation
=
0
consumption
=
0
Input
=
Output
17
4.2b
Balances on continuous steady-state processes:
Example 4.2-2: Material Balance on a Continuous Distillation Process
One thousand kilograms per hour of a mixture of benzene (B) and toluene (T) containing
50% benzene by mass is separated by distillation into two fractions. The mass flow rate of
benzene in the top stream is 450 kg/h and that of toluene in the bottom stream is 475
kgT/h. The operation is at steady state. Write balances on benzene and toluene to calculate
the unknown component flow rates in the output streams.
450 kg B/h
m1 (kg T/h)
500 kg B/h
500 kg T/h
Distillation
Process
m2 (kg B/h)
475 kg T/h
18
4.2c
Integral Balance on Batch Processes:
Input
-
Output

Input
=
0
Output
=
0
Accumulation
final output
+
generation
-
consumption =
Accumulation
=
-
initial input
=
generation
-
consumption
If the balance is on a nonreactive species:
generation
=
0
consumption
=
0
initial input
=
final output
19
4.2c
Integral Balance on Batch Processes:
Example 4.2-3: Balances on a Batch Mixing Processes
Two methanol-water mixtures are contained in separate flasks. The first mixture contains
40.0 wt% methanol, and the second contains 70.0 wt% methanol. If 200 g of the first
mixture is combined with 150 g of the second, what are the mass and composition of the
product?
200 g
0.400 g CH3OH/ g
0.600 g H2O/ g
Mixer
m (g)
x (g CH3OH/ g)
(1-x) (g H2O/ g)
150 g
0.700 g CH3OH/ g
0.300 g H2O / g
Example 4.2-4: Integral Balance on a Semibatch Process is not included in the material
20
4.3
Flow Chart
1. Represent each process unit by a simple box or any other simple.
Process
Unit
Process
Unit
Process
Unit
2. Feeds and Products are represented by lines and arrows.
Total mass (molar)
flow rate
Mass (molar)
composition
Other variables like
temperature and pressure
OR
Component
Mass (molar)
Flow rates
Other variables like
temperature and pressure
21
4.3
Flow Chart
Example:
100 kmol/min
0.6 kmol N2/kmol
0.4 kmol O2/kmol
T = 320oC, P = 1.4 atm
10 lbm mixture
0.3 lbm CH4/lbm
0.4 lbm C2H4/lbm
0.3 lbm C2H6/lbm
T = 320oC, P = 1.4 atm
OR
OR
60 kmol N2
40 kmol O2
T = 320oC, P = 1.4 atm
3.0 lbm CH4
4.0 lbm C2H4
3.0 lbm C2H6
T = 320oC, P = 1.4 atm
22
4.3
Flow Chart
3. Assign algebraic symbols to unknown stream variables associating them with
units in parenthesis.
n (mol/min)
0.21 mol O2/mol
0.79 mol N2/mol
AND
400 mol/min
y (mol O2/mol)
(1-y)(mol N2/mol)
4. The following variables are normally used to describe:
n
Molar quantity (mole unit)
V
m
Volume (volume unit)
n
V
m
Molar flow rate (mol/time)
T
P
Temperature (temperature unit)
Mass (mass unit)
Volumetric flow rate (volume/time)
Mass flow rate (mass/time)
Pressure (pressure unit)
23
4.3a
Flow Chart
Example 4.3-1:
Flowchart of an Air Humidification and Oxygenation Process
An experiment on the growth rate of a certain organisms requires an environment of
humid air enriched in oxygen. Three input streams are fed into an evaporation chamber
to produce an output stream with the desired composition.
A:
liquid water, fed at a rate of 20.0 cm3/min
B:
Air (21 mole % O2, 79 mole% N2)
C:
Pure oxygen, with a molar flow rate one-fifth of the molar flowrate of stream
B
The output gas is analyzed and is found to contain 1.5 mole% water. Draw and label a
flowchart of the process, and calculate all unknown stream variables.
24
4.3a
Flow Chart
Wet air containing 30.0 mole% water, flowing at a rate of 1 mole/s passes through a
condenser. Air leaving the condenser is 15.0 mole% water. Draw a fully labeled flowchart.
25
4.3a
Flow Chart
A feed stock available at the rate of 1000 mol/h and consisting of (all in mol %):
20% Propane (C3)
30% Isobutane (i-C4)
20% Isopentane (i-C5)
30% Normal pentane (C5)
is to be separated into two fractions by distillation. The distillate (top product) is to
contain all of the propane feed to the unit and 80% of the isopentane fed to the
unit and is to consist of 40 mole% isobutane. The bottom stream is to contain all
the normal pentane fed to the unit. Draw the flowchart of the process
26
4.3a
Flow Chart
Acetone can be recovered from a carrier gas by dissolving it in a pure water
stream in a unit called an absorber. In this process unit, 200 lbm/h of a stream
containing 20% acetone and 80% gas is treated with 1000 lbm/h of a pure
water stream to yield an acetone free overhead gas and an acetone-water
solution. Assume no carrier gas dissolves in water.
Draw a flowchart for the process.
27
4.3b
Flowchart Scaling and Basis of Calculation
Consider the following:
1 kg C6H6
2 kg
1 kg C7H8
0.5 kg C6H6/kg
0.5 kg C7H8/kg
×10
Scaling Up
The final quantities are larger than the original quantities
10 kg C6H6
20 kg
10 kg C7H8
0.5 kg C6H6/kg
0.5 kg C7H8/kg
28
4.3
Flowchart Scaling and Basis of Calculation
1 kg C6H6
2 kg
1 kg C7H8
0.5 kg C6H6/kg
0.5 kg C7H8/kg
×1/2
Scaling Down
The final quantities are smaller than the original quantities
0.5 kg C6H6
1.0 kg
0.5 kg C7H8
0.5 kg C6H6/kg
0.5 kg C7H8/kg
29
4.3
Flowchart Scaling and Basis of Calculation
Summary:
1. Balanced processes can be always scaled up or down by multiplying all steams of the
old process by a factor while maintaining stream compositions unchanged.
2.
the scale factor is defined as :
n
amount (flowrate) of new stream
amount (flowrate) of the correspond ing old stream
3. When balanced processes are scaled up or down, compositions of all streams must
remain unchanged.
4. Changing balanced processes from batch to continuous by dividing quantities by unit
time will not change the original process.
30
4.3b
Flowchart Scaling and Basis of Calculation
1 kg C6H6
2 kg
1 kg C7H8
0.5 kg C6H6/kg
0.5 kg C7H8/kg
1 kg C6H6/hr
2 kg/hr
1 kg C7H8/hr
0.5 kg C6H6/kg
0.5 kg C7H8/kg
31
4.3
Flowchart Scaling and Basis of Calculation
Example 4.3-2:
Scale-up of a Separation Process Flowchart
A 60-40 mixture (by moles) of A and B is separated into two fractions. A flowchart of
the process is shown here:
50.0 mol
0.95 mol A/mole
0.05 mol B/mole
100.0 mol
0.6 mol A/mole
0.4 mol B/mole
12.5 mol A
37.5 mol B
It is desired to achieve the same separation with a continuous feed of 1250 lb-moles/h.
scale the flowchart accordingly.
32
4.3c
Balancing a Process
Humid Air
0.30 mole H2O/mole
0.70 mole air/mole
Condenser
Humid Air
0.15 mole H2O/mole
0.85 mole air/mole
H2O (l)
Can we perform mass balance calculation and why?
33
4.3b
Flowchart Scaling and Basis of Calculation
Calculation can be made for any basis of any convenient set of stream
amounts or flow rates and the results can afterward be scaled to any desired
extent.
Basis of calculation is an amount (mass or moles) of flow rate (mass or
molar) of one stream or stream component in a process.
Choosing basis of calculation is considered to be the first step in balancing a
process.
If no stream amounts or flow rates are known, assume one, preferably that of
known compositions.
34
4.3c
Balancing Processes
Consider the following system of equations:
2 x  5 y  10
x  3y  3
The solution set is (15, -4)
The equations in the above system is recognized as independent equations, i.e. can’t be
derived algebraically from each other. This results in a unique solution set
Consider the following system of equations:
2 x  5 y  10
4 x  10 y  20
The equations in the above system is recognized as dependent equations, i.e. Eq(2) =
2×Eq(1). This results in infinitely many solutions!
35
4.3c
Balancing Processes
3.0 kg C6H6/min
m (kg/min )
x (kg C6H6/kg)
(1-x) (kg C7H8/kg)
1.0 kg C7H8/min
Total mass balance:
3.0  1.0  m
(1)
3.0  x  m
(2)
1.0  1  x   m
(3)
C6H6 balance:
C7H8 balance:
Conclusion:
For two components, we can write two balance equations and for “n” components we can
write “n” balance equations plus one total mass balance equation = (n+1) equations.
36
4.3c
Balancing Processes
Dependant and Independent Equations:
Addition of (2) and (3) yields:
3.0  1.0  xm  1  x m
Total mass balance equation
Substation of (2) and (1) yields:
3.0  1.0  3.0  m  xm
Toluene balance equation
Substation of (3) and (1) yields:
3.0  1.0  1.0  m  1  x m
Benzene balance equation
Conclusion:
Although we can write three balance equations, we can only solve for two variables as
they are not linearly independent.
37
4.3c
Balancing Processes
Summary:
Total number of balance equation which can be written on physical systems equals
number of components +1
Number of independent balance equations which can be written on physical systems
equals the number of the chemical components.
Try always to write balances which involve the fewest number of unknowns
variables.
38
4.3c
Balancing Processes
Example 4.3-3:
Balances on a Mixing Unit
An aqueous solution of sodium hydroxide contains 20.0% NaOH by mass. It is desired
to produce an 8.0% NaOH solution by diluting a stream of the 20% solution with a
stream of pure water. Calculate the ratios (liters H2O/kg feed solution) and (kg product
solution/kg feed solution)
39
4.3d
Degree-of-freedom Analysis
Is there a way which can save
my time and tell me if the
mass balance problem is
solvable or not?
40
4.3d
Degree-of-freedom Analysis
Solvable problems mean that enough information is available so that all variable
can be uniquely determined.
Degree of Freedom Analysis:
is an analysis which checks if enough information is available so that mass balance
problems are solvable or not before attempting to derive any equations.
DF  nunknowns  nindep. equations
DF = 0
Correctly specified system
DF > 0
Underspecified system, i.e. more unknowns are there than number of
equations.
DF < 0
Overspecified system, i.e. more information is available than required.
41
4.3d
Degree-of-freedom Analysis
Sources of Equations:
1. Material balances
2. An energy balance
3. Process specifications
4. Physical properties and laws.
5. Physical constraints.
42
4.3d
Degree-of-freedom Analysis
Example 4.3-4
Degree-of-Freedom Analysis
A stream of humid air enters a condenser in which 95% of the water vapor in the air is
condensed. The flowrate of the condensate (the liquid leaving the condenser) is
measured and found to be 225 L/h. dry air may be taken to contain 21.0% oxygen, with
the balance nitrogen. Calculate the flowrate of the gas stream leaving the condenser and
the mole fractions of oxygen, nitrogen, and water in this stream.
43
4.3d
Degree-of-freedom Analysis
m1 (kg/s)
0.30 mole H2O/mole
0.70 mole air/mole
Condenser
m3 (kg/s)
0.15 mole H2O/mole
0.85 mole air/mole
m2 (kg/s)
H2O (l)
Perform Degree of Freedom on the above system?
44
4.3d
General Procedure for Single-Unit Process Material
Balance Calculations
Overhead Product
m2 (kg/h)
2000 L/h
m1 (kg/h)
0.45 kg B/kg
0.55 kg T/kg
0.95 mol B/mole
0.05 mol T/mole
Bottom Product
mB3 (kg/h) (8% of B in feed)
mT3 (kg/h)
45
4.4
Balances on Multiple-Unit Processes
Flowchart of a two-unit process
Feed 2
A
E
C
B
Feed 1
D
Unit
1
Product 1
Unit
2
Product 2
Product 3
Feed 3
46
4.4
Balances on Multiple-Unit Processes
Mixing Points:
stream 2
stream 2
stream 1
stream 3
stream 1
stream 3
 The resulting stream after mixing has different flowrate and compositions than the
constituting streams.
 The total balance equations which can be written is number of chemical species +1.
 The number of independent equations equals the number of chemical species.
47
4.4
Balances on Multiple-Unit Processes
Splitting Points:
stream 1
stream 3
stream 1
stream 2
stream 3
stream 2
 The resulting streams after mixing have different flowrate but the same compositions as
the main stream.
 The total balance equations which can be written is only ONE.
48
4.4
Balances on Multiple-Unit Processes
Procedure to solve multiple unit systems:
1. Draw a fully labeled flowchart.
2. Divide the multiple systems into subsystems.
3. Perform degree of freedom on each subsystems.
4. Solve systems of zero degree of freedom.
49
4.4
Balances on Multiple-Unit Processes
Example 4.4-1
Two-Unit Process
A labeled flowchart of a continuous steady-state two-unit process is shown below. Each
stream contains two components, A, B, in different proportions. Three streams whose
flow rates and/or compositions are not known are labeled 1,2, and 3. Calculate the
unknown flowrates and composition of streams 1,2, and 3?
40.0 kg/h
0.900 kg A/kg
0.100 kg B/kg
100.0 kg/h
0.500 kg A/kg
0.500 kg B/kg
Unit
I
1
30.0 kg/h
0.600 kg A/kg
0.400 kg B/kg
2
Unit
II
3
30.0 kg/h
0.300 kg A/kg
0.700 kg B/kg
50
4.4
Balances on Multiple-Unit Processes
100 kg M
75 kg M
Extractor
Extractor
100 kg
0.5 kg W/kg
0.5 kg A/kg
mA2 (kg A)
mM2 (kg M)
mW2 (kg W)
43.1 kg
0.053 kg A/kg
0.016 kg M/kg
0.931 kg W/kg
m3 (kg)
0.09 (kg A/kg)
0.88 (kg M/kg)
0.03 (kg W/kg)
m1 (kg)
0.275 (kg A/kg)
xM1 (kg M/kg)
(0.725-xM1 )(kg W/kg)
mA4 (kg A)
mM4 (kg M)
mW4 (kg W)
Distillation
Unit
m5 (kg)
0.97 kg A/kg
0.02 kg M/kg
0.01 kg W/kg
mA6 (kg A)
mM6 (kg M)
mW6 (kg W)
51
4.5
Recycle and by-pass
Consider the following:
Impractical as large
number of units
are sometimes needed
to be built
Feed
Feed
Process
Unit
Process
Unit
Product 1
Product
Process
Unit
Product 2
52
4.5
Recycle and by-pass
Fresh feed
Feed
Process
Unit
Product
Recycle stream
A recycle stream is simply a stream which is split off from the outlet of a unit
and sent back as the input of an upstream unit
53
4.5
Recycle and by-pass
Example 4.5-1
Material balance on an Air Conditioner.
Fresh air containing 4.00 % water vapor is to be cooled and dehumidified to a water
content of 1.70 mole % H2O. A stream of fresh air is combined with a recycle stream of
previously dehumidified air and passed through the cooler. The blended stream entering
the unit contains 2.30 mole % H2O. In the air conditioner, some of the water in the feed
stream is condensed and removed as liquid. A fraction of the dehumidified air leaving the
cooler is recycled and the remainder is delivered to a room. Taking 100 mol of
dehumidified air delivered to the room as a basis of calculation. Calculate the moles of
fresh feed, moles of water condensed, and moles of dehumidified air recycled.
54
4.5
Recycle and by-pass
n5 (mol)
0.983 mol DA/mole
0.017 mol W/mole
n1(mol)
n2 (mol)
0.960 mol DA/mole 0.977 mol DA/mole
0.040 mol W/mole
0.023 mol W/mole
AIR
COND.
n4(mol)
100 mol
0.983 mol DA/mole 0.983 mol DA/mole
0.017 mol W/mole 0.017 mol W/mole
DA = Dry Air
W = Water
n3 [mol W(l)]
55
4.5
Recycle and by-pass
The flowchart of a steady-state process to recover crystalline chromate (K2CrO4) from an aqueous solution of this salt is
shown below:
H2O
4500 kg/h
33.3% K2CrO4
EVAPORAT
OR
49.4% K2CrO4
Filtrate
36.4% K2CrO4 solution
CRYSTALIZER
AND FILTER
Filter cake
K2CrO4 solid
36.4 % K2CrO4 solution
(the crystals constitute 95% by
mass of the filter cake)
Forty-five hundred kilograms per hour of a solution that is one-third K2CrO4 by mass is joined by a recycle stream containing
36.4% K2CrO4, and the combined stream is fed into an evaporator. The concentrated stream leaving the evaporator contains
49.4% K2CrO4; this stream is fed into a crystallizer in which it is cooled (causing crystals of K2CrO4 to come out of solution)
and then filtered. The filter cake consists of K2CrO4 crystals and a solution that contains 36.4 % K2CrO4 by mass; the crystal
account for 95% of the total mass of the filter cake. The solution that passes through the filter, also 36.4% K2CrO4, is the
recycle stream.
1.
Calculate the rate of evaporation, the rate of production of crystalline K2CrO4, the feed rates that the evaporator and
crystallizer must be designed to handle, and the recycle ratio (mass of recycle)/(mass of fresh feed).
2.
Suppose that the filtrate were discarded instead of being recycled. Calculate the production rate of crystals. What are
the benefits and costs of the recycling?
56
4.5
Recycle and by-pass
m2 (kg W(v)/h)
Fresh feed
m1 (kg /h)
4500 kg/h
x1 (kg K/kg)
(1-x1) (kg W/kg)
EVAPORATOR
m3 (kg /h)
0.494 kg K/kg
0.506 kg W/kg
CRYSTALIZER
AND FILTER
Filter cake
m4 (kg K(s)/h)
m5 (kg soln/h)
0.364 kg K/kg
0.636 kg W/kg
Filtrate (recycle)
m6 (kg /h)
0.364 kg K/kg
0.636 kg W/kg
57
4.5
Recycle and by-pass
m1 (kg W(v)/h)
Fresh feed
4500 kg/h
EVAPORATOR
m2 (kg /h)
0.494 kg K/kg
0.506 kg W/kg
CRYSTALIZER
AND FILTER
m3 (kg K(s)/h)
m4 (kg soln/h)
0.364 kg K/kg
0.636 kg W/kg
m5 (kg /h)
0.364 kg K/kg
0.636 kg W/kg
58
4.5
Recycle and by-pass
Fresh feed
Feed
Process
Unit
Product
Recycle stream
Fresh feed
Feed
Process
Unit
Product
Bypass stream
59
4.5
Recycle and by-pass
Reasons for installing recycle lines:
1.
Recovery of catalyst.
2.
Dilution of a process stream.
3.
Control of a process variable
4.
Circulation of working fluid.
60
4.6
Chemical Reaction Stoichiometry
Stoichiometry:
is the theory of the proportions in which chemical species combine with one another.
Stoichiometric equation:
is a statement of the relative number of molecules or moles of reactants and products
that participate in the reaction.
2SO2  O2  2SO3
It tells the reactive species and relative amounts in mole, k-mole, or lb-mole.
It must be balanced to be used.
Atoms can’t be generated nor created in chemical reactions.
61
4.6
Chemical Reaction Stoichiometry
Stoichiometric Ratio:
is defined for two species participating in a reaction in the ratio of their stoichiometric
coefficients in the balanced reaction equations.
2SO2  O2  2SO3
2 mol SO3 generated
1 mol O2 consumed
2 lb  moles SO2 consumed
2 lb  moles SO3 generated
It can be used as conversion factor to calculate the amount of a particular reactant (or
product) that was consumed or (produced), given a quantity of another reactant of product
the participated in the reaction.
Example:
Calculate the amount of oxygen required to produce 1600 kg/h of SO3?
62
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Consider:
2SO2  O2  2SO3
SO2
O2
SO3
4 moles
2 moles
4 moles
Case 1:
Stoichiometric proportion:
For any two chemical species A and B, they are at their stoichiometric proportion if their
feed ratio equals the stoichiometic ratio obtained from the balanced reactions equation.
63
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Consider:
2SO2  O2  2SO3
SO2
O2
SO3
4 moles
1 moles
2 moles
Case 2:
Comment:
If the reaction goes to completion, only 2 moles of SO3 are produced, 2 moles of SO2
consumed and all oxygen are consumed.
We say that:
O2 is the limiting reactant
SO2 is excess reactant
Limiting reactant:
The reactant that would run out if a reaction proceed to completion. It presents in less
than its stoichiometric proportion relative to every other reactant.
64
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Consider:
2SO2  O2  2SO3
SO2
O2
SO3
4 moles
3 moles
2 moles
Case 2:
Comment:
If the reaction goes to completion, only 2 moles of SO3 are produced, 4 moles of SO2
consumed and 2 moles of oxygen are consumed.
We say that:
O2 is excess reactant
SO2 is the limiting reactant
Limiting reactant:
The reactant that would run out if a reaction proceed to completion. It presents in less
than its stoichiometric proportion relative to every other reactant.
65
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Fractional Excess:
n A feed  n A stoich
fractional excess of A 
n A stoich
Fractional Conversion:
f
moles reacted
moles fed
66
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Reactor Flowchart:
Feed
Reactor
Product
Product stream of a reactor is composed of a matrix of:
1.
Unconverted reactants.
2.
Products formed.
3.
Inert materials
67
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Reactor Flowchart:
A B C  D
Feed
Reactor
Feed
Reactor
n (mole/s)
yA (mole A/mole)
yB (mole B/mole)
yC (mole C/mole)
(1-yA -yB- yC) (mole D/mole)
nA (mole A/s)
nB (mole B/s)
nC (mole C/s)
nD (mole D/s)
68
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Techniques to balance reactive processes:
1.
Extent of chemical reaction.
2.
Molecular balance.
3.
Atomic balance.
69
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Extent of chemical reactions:
Input
-
Output
+
generation
-
consumption =
Accumulation
Assumptions:
 Steady state, i.e. accumulation =0
Generation = consumption   i
This can be reduced to:
ni  nio  i
nio : feed molar flowrate of component i.
ni : product molar flowrate of component i.
 i : stoichiometric number s obtained from the balanced stoichiometric equation and
defined to be negative for reactants and positive for products.
70
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Acrylonitrile is produced in the reaction of propylene, ammonia, and oxygen:
C3 H 6  NH 3 
3
O2  C3 H 3 N  3H 2O
2
The feed contains 10.0% propylene, 12.0% ammonia, and 78.0% air. A fractional conversion
of 30.0% of the limiting reactant is achieved. Taking 100 mol of feed as a basis, determine
which reactant is limiting, the percentage by which each of the other reactants is excess, and
the molar amounts of all product gas constituents for a 30.0% conversion of the limiting
reactant.
71
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
C3 H 6  NH 3 
3
O2  C3 H 3 N  3H 2O
2
100 mole
Reactor
0.10 mole C3H6 /mole
0.12 mole NH3 /mole
0.78 mole Air /mole
0.21 mole O2 / mole Air
0.79 mole N2 / mole Air
n1 (mole C3H6)
n2(mole NH3)
n3(mole O2)
n4 (mole C3H3N)
n5 (mole H2O)
n6 (mole N2)
72
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
Consider the ammonia formation reaction:
N 2  3H 2  2 NH 3
Suppose the feed to a continuous reactor consists of 100 mol/s of nitrogen, 300 mol/s of
hydrogen, and 1 mol of argon (an inert gas), calculate the molar flowrates of all
components knowing that the fractional conversion of hydrogen is 0.6?
73
4.6b
Limiting and Excess Reactants, Fractional Conversion,
and Extent of Reaction
N 2  3H 2  2 NH 3
Fractional conversion of hydrogen is 0.60
n1 (mole N2/s)
100 mole N2/s
300 mole H2/s
Reactor
n2(mole H2/s)
n3(mole NH3/s)
n4 (mole Ar/s)
1.0 mole Ar /s
n1  100  
n2  300  3
n3  0.0  2
n4  1.0  0
74
75
Chemical Species
Chemical Equilibrium:
Chemical Species
4.6c
Progress of Reaction
Progress of Reaction
Irreversible reaction
Reversible reaction
aA  bB  cC  dD
aA  bB  cC  dD
Objective:
We are interested in knowing the molar quantities (compositions) at equilibrium.
76
4.6c
Chemical Equilibrium:
Example 4.6-2:
Calculation of an Equilibrium Composition
If water-gas shift reaction:
COg   H 2Og   CO2 g   H 2 g 
Proceed to equilibrium at a temperature T(K), the mole fraction of the four reactive species satisfy
the relation:
yCO2 y H 2
yCO y H 2O
 K T 
Where K(T) is the reaction equilibrium constant. At T=1105 K, K=1.00. Suppose the feed to a
reactor contains 1.00 mol of CO, 2.00 mol H2O, and no CO2 or H2, and the reaction mixture comes
to equilibrium at 1105K. Calculate the equilibrium composition and the fractional conversion of the
limiting reactant.
77
4.6c
Chemical Equilibrium:
COg   H 2Og   CO2 g   H 2 g 
yCO2 y H 2
yCO y H 2O
 K T 
n1 (mole CO)
1.0 mole CO
2.0 moles H2O
Reactor
n2(mole H2O)
n3(mole CO2)
n4 (mole H2)
78
4.6d
Multiple Reactions, Yield, and Selectivity:
Consider:
Process: Dehydrogenation of Ethane to produce ethylene
C 2 H 6  C2 H 4  H 2
C 2 H 6  H 2  2CH 4
C2 H 4 C 2 H 6  C3 H 6  CH 4
Main reaction
Side reactions
Two major mathematical quantities are needed to be defined as they measure the
performance of the production of the process.
79
4.6d
Multiple Reactions, Yield, and Selectivity:
Yield 
moles of desired product formed
moles that would have been formed if
there were no side reactions and the
limiting reactant had reacted completely
Selectivity 
moles of desired product formed
moles of undersired product formed
High values of selectivity and yield signify that the undesired side reactions have been
successfully suppressed relative to the desired reaction.
Extent of Chemical Reaction:
R
ni  nio    ij j
j 1
80
4.6d
Multiple Reactions, Yield, and Selectivity:
Example 4.6-3
Yield and Selectivity in a Dehydrogenation Reactor
The reactions:
C2 H 6  C2 H 4  H 2
C2 H 6  H 2  2CH 4
Take place in a continuous reactor at steady state. The feed contains 85.0 mole%, ethane (C2H6) and
the balance inert (I). The fractional conversion of ethane is 0.501, and the fractional yield of ethylene
is 0.471. calculate the molar composition of the product gas and selectivity of ethylene to methane
production.
81
4.6d
Multiple Reactions, Yield, and Selectivity:
C2 H 6  C2 H 4  H 2
C2 H 6  H 2  2CH 4
f = 0.501
n1 (mole C2H6)
0.85 mole C2H6/mol
Reactor
n2(mole C2H4)
n3(mole H2)
n4 (mole CH4)
0.15 mol I/ mol
Y = 0.471
n5 (mol I)
82
4.7
BALANCES ON REACTIVE PROCESSES:
Molecular Balance:
For each chemical species participating in a chemical reaction, a generation or
consumption term is defined.
Atomic Balance:
Atoms can’t be destroyed nor created in a chemical reaction.
83
4.7
BALANCES ON REACTIVE PROCESSES:
Example:
Ethane is dehydrogenated according to the following stoichiometric equation:
C2 H 6  C2 H 4  H 2
In a steady-state continuous reactor, one hundred kmol/min of ethane is fed to the
reactor. If the conversion on the reactor is obtained to be 0.45, calculate the molar
flowrate of all components constituting the product stream using the molecular balance
and atomic balance techniques?
84
4.7a
BALANCES ON REACTIVE PROCESSES:
n1 (kmole C2H6/min)
100 kmol C2H6/min
Reactor
n2 (kmole C2H4/min)
n3 (kmole H2/min)
Fractional conversion is 0.45
85
4.7b
Independent Equations, Independent Species and Independent
Reactions:
n1 (mol O2/s)
n3 (mol O2/s)
3.67n1 (mol N2/s)
3.67n3 (mol N2/s)
n4 (mol CCl4(v)/s)
n2 (mol CCl4(l)/s)
n5 (mol CCl4(l)/s)
O2-balance:
n1  n3
N2-balance:
3.67n1  3.67n3
CCl4-balance:
n2  n4  n5
If two molecular species are in the same ratio to each other wherever they appear in a
process and this ratio is incorporated in the flowchart labeling, balances on those
species will not be independent equations.
86
4.7b
Independent Equations, Independent Species and Independent
Reactions:
Chemical reaction equations:
A  2B
1
B C
2
A  2C
3
Chemical reactions are independent if the stoichiometric equation of any one of them
cannot be obtained by adding and subtracting multiples of the stoichiometic equations
of the others.
87
4.7c
Molecular Species Balances:
Degree of Freedom is obtained by:
No. of unknown labeled variables
+ No. of independent chemical equations.
- No. of independent molecular species balances.
- No. of other equations relating unknown variables
= No. of degrees of freedom
88
4.7c
Molecular Species Balances:
n1 (kmole C2H6/min)
100 kmol/min
Reactor
n2 (kmole C2H4/min)
n3 (kmole H2/min)
Fractional conversion is 0.45
89
4.7d
Atomic Species Balances:
Degree of Freedom is obtained by:
No. of unknown labeled variables
- No. of independent atomic species balances.
- No. molecular balances on independent nonreactive species
- No. of other equations relating unknown variables
= No. of degrees of freedom
90
4.7d
Atomic Species Balances:
n1 (kmole C2H6/min)
100 kmol/min
Reactor
n2 (kmole C2H4/min)
n3 (kmole H2/min)
Fractional conversion is 0.45
91
4.7e
Extent of Reaction:
Degree of Freedom is obtained by:
No. of unknown labeled variables
+ No. of independent chemical equations.
- No. of independent reactive species
- No. of independent nonreactive species.
- No. of other equations relating unknown variables
= No. of degrees of freedom
92
4.7e
Extent of Reaction:
n1 (kmole C2H6/min)
100 kmol/min
Reactor
n2 (kmole C2H4/min)
n3 (kmole H2/min)
Fractional conversion is 0.45
93
Summary of Balances on Reactive Systems:
Atomic species balances
generally lead to the most straightforward solution
procedure, especially when more than one reaction is involved.
Extent of chemical reaction are convenient for chemical equilibrium problems and
when equation-solving software is to be used.
Molecular species balances require more complex calculations than either of the
other two approaches and should be used only for simple systems involving one
reaction.
94
4.7e
Extent of Reaction:
Example 4.7-1
Incomplete combustion of Methane
Methane is burned with air in a continuous steady-state combustion reactor to yield a
mixture of carbon monoxide, carbon dioxide, and water. The reactions taking place are:
CH 4 
3
O2  CO  2 H 2O
2
CH 4  2O2  CO2  2 H 2O
The feed to the reactor contains 7.80% CH4, 19.4% O2, 72.8% N2. the percentage
conversion of methane is 90.0%, and the gas leaving the reactor contains 8.0 mol
CO2/mol CO. Carry out a degree-of-freedom analysis on the process. Then calculate the
molar composition of the product stream using molecular species balances, atomic
species balances, and extent of reaction.
95
4.7f
75 mol
A/ min
Product Separation and Recycle:
100 mol
A/ min
25 mol A/ min
REACTOR
75 mol B/ min
PRODUCT
SEPARATION
UNIT
75 mol
B/ min
25 mol A/ min
96
4.7f
Product Separation and Recycle:
Two types of conversions are defined:
Overall Conversion 
reactant input to process  reacrant output from process
reactant input to process
Single  Pass Conversion 
reactant input to reactor  reacrant output from reactor
reactant input to reactor
97
4.7f
75 mol
A/ min
Product Separation and Recycle:
100 mol
A/ min
25 mol A/ min
REACTOR
75 mol B/ min
PRODUCT
SEPARATION
UNIT
75 mol
B/ min
25 mol A/ min
98
4.7f
Product Separation and Recycle:
Example 4.7-2
Dehydrogenation of Propane
Propane is dehydrogenated to form propylene in a catalytic reactor:
C3 H 8  C3 H 6  H 2
The process is designed for a 95% overall conversion of propane. The reaction products are
separated are separated into two streams: the first, which contains H2, C3H6, and 0.555% of
the propane that leaves the reactor, is taken off as product: the second stream, which
contains the balance of the unreacted propane and 5% of the propylene in the first stream, is
recycled to the reactor. Calculate the composition of the product, the ratio (mole
recycled)/(mole fresh feed), and the single-pass conversion.
99
4.7f
Product Separation and Recycle:
100 mol
C3H8
REACTOR
n1 (mol C3H8)
n2 (mol C3H6)
n3 (mol C3H8)
n4 (mol C3H6)
n5 (mol H2)
SEPARATOR
Product
n6 (mol C3H8)
n7 (mol C3H6)
n8 (mol H2)
Recycle
n9 (mol C3H8)
n10 (mol C3H6)
100
4.7g
Purging
Recycle
Purge stream
40 mol C2H4/s
20 mol O2/s
452 mol N2/s
10 mol C2H4/s
5 mol O2/s
113 mol N2/s
Solvent
Fresh feed
60 mol C2H4/s
30 mol O2/s
113 mol N2/s
100 mol C2H4/s
50 mol O2/s
565 mol N2/s
REACTOR
50 mol C2H4/s
25 mol O2/s
565 mol N2/s
50 mol C2H4O/s
C2 H 4  O2  2C2 H 4O
ABSORBER
Product
50 mol C2H4O/s
Solvent
101
4.7g
Purging
Example 4.7-3
Recycle and Purge in the Synthesis of Methanol
Methanol is produced in the reaction of carbon dioxide and hydrogen:
CO2  3H 2  CH 3OH  H 2O
The fresh feed to the process contains hydrogen, carbon dioxide, and 0.400 mole% inert (I). The
reactor effluent passes through a condenser to remove essentially all of the methanol and water
formed and none of the reactants or inert.. The later substances are recycled to the reactor. To
avoid buildup of the inert in the system, a purge stream is withdrawn from the recycle.
The feed to the reactor (not the fresh feed to the process) contains 28.0 mole% CO2, 70.0 mole%
H2, and 2.00 mole% inert. The single-pass conversion of hydrogen is 60.0%. Calculate the molar
flow rates and molar composition of the fresh feed, the total feed to the reactor, the recycle stream,
and purge stream for a methanol production rate of 155 kmol CH3OH/h.
102
4.7g
Purging
nr (mol)
np (mol)
x5C (mol CO2/mol)
x5H (mol H2/mol)
(1- x5C- x5H)
x5C (mol CO2/mol)
x5H (mol H2/mol)
(1- x5C- x5H)
n5 (mol)
no
x0C (mol CO2/mol) 0.280 mol CO2/mol
(0.996- x0C)mol H2/mol 0.700 mol H2/mol
0.00400 mol I/mol
0.020 mol I/mol
REACTOR
n1 (mole CO2)
n2 (mole H2)
2.0 mol I
n3 (mole
CH3OH)
n4 (mole H2O)
CO2  3H 2  CH 3OH  H 2O
x5C (mol CO2/mol)
x5H (mol H2/mol)
(1- x5C- x5H)
CONDENSER
Product
103
4.8
Combustion Reactions
Combustion: the rapid reaction of fuel with oxygen.
fuel  oxygen  matrix of gases
They are run to produce tremendous amount of energy used to run turbine and
boilers.
Fuel can be:
1. Coal (carbon, some hydrogen and sulfur and various noncombustible materials.
2. Fuel oil (high molecular weight of hydrocarbons, some sulfur).
3. Gaseous fuel (natural gas such as methane).
4. Liquefied petroleum gas (propane and/or butane).
Elements of the matrix of gas product are CO2, CO, H2O, SO2 and at high temperature
NO is produced.
104
4.8
Combustion Reactions
Combustion reactions can be divided into two main categories:
1. Complete combustion.
2. Incomplete combustion (Partial).
Complete Combustion
Incomplete combustion
Gas Matrix
(Products)
Production of CO2 and
steam
Production of CO and
steam
Energy
High energy
Low energy
Oxygen required
More
Less
105
4.8
Combustion Reactions
Example 4.8-1: Classify each of the following reactions as complete or incomplete:
C  O2  CO2
1
C  O2  CO
2
C3 H8  5O2  3CO2  4 H 2O
7
C3 H 8  O2  3CO  4 H 2O
2
CS 2  3O2  CO2  2S2O
106
4.8
Combustion Reactions
Flowchart of a combustion reactor:
Flue gas
Stack gas
Gaseous products
Fuel
Air
Combustion
Reactor
0.21 mol O2/mol
0.79 mol N2/mol
Unburned fuel
CO
CO2
SO2
H2O
Noncombustible material
Compositions of gaseous products can be given in two ways:
1.
Composition on a dry basis.
2.
Composition on a wet basis.
107
4.8
Combustion Reactions
Example 4.8-1: Composition on Wet and Dry Bases
a. A stack gas contains 60.0 mole% N2, 15.0% CO2, 10.0% O2, and the balance
H2O. Calculate the molar composition of the gas on a dry basis.
b. An orsat analysis (a technique for stack analysis) yields the following dry basis
composition:
N2
65%
CO2
14%
CO
11%
O2
10%
A humidity measurement shows that the mole fraction of H2O in the stack gas is
0.0700. Calculate the sack gas composition on a wet basis.
108
4.8b
Theoretical and Excess Air
Theoretical Oxygen:
The moles (batch) or molar flowrate (continuous) of O2 needed for complete
combustion of all the fuel fed to the reactor, assuming that all carbon in the fuel is
oxidized to CO2, hydrogen to H2O, and sulfur to SO2.
Theoretical Air:
The amount of air containing the theoretical oxygen.
Excess Air:
The amount by which the air fed to the reactor exceeds the theoretical air.
moles air fed  moles air theoretica l
Percent Excess Air :
 100%
moles air theoretica l
moles air fed  1  fraction of excess air  moles air theoretica l
109
4.8b
Theoretical and Excess Air
Example 4.8-2 Theoretical and Excess Air
One hundred mol/h of butane (C4H10) and 5000 mol/h of air are fed to a combustion
reactor. Calculate the percent excess air.
110
4.8c
Material Balances on Combustion Reactors
Example 4.8-3
combustion of Ethane
Ethane is burned with 50% excess air. The percentage conversion of the ethane is 90%;
of the ethane burned, 25% reacts to form CO and the balance reacts to form CO2.
calculate the molar composition of the stack gas on a dry basis and the mole ratio of
water to dry stack gas.
111
4.8c
Material Balances on Combustion Reactors
Example 4.8-4 Combustion of a Hydrocarbon of Unknown Composition
A hydrocarbon gas is burned with air. The dry-basis product gas composition is 1.5
mole% CO, 6.0% CO2, 8.2% O2, and 84.3% N2. there is no atomic oxygen in the
fuel. Calculate the ratio of hydrogen to carbon in the fuel gas and speculate on what
the fuel might be. The calculate the percent excess air fed to the reactor.
112
113