Industrial stoichiometry dlc 2105 29.10.2019 (1)

INDUSTRIAL STOICHIOMETRY
10/29/2019 M I L T O N A W E S U T A 1

• Stoichiometry (from the Greek stoikeion - element) is the
practical application of the law of multiple proportions.
• It is the study of the quantitative relationships between
the reactants and products formed by a chemical reaction
• Based on one of the basic laws of conservation i.e.
conservation of mass.
• The general conservation equation for any process system
is written as:
• Material out = Material in + Generation - Consumption –
Accumulation10/29/2019 M I L T O N A W E S U T A 2

• For a steady-state process the accumulation term will be zero.
• Except in nuclear processes, mass is neither created nor
consumed; but if a chemical reaction takes place a particular
chemical species may be formed or consumed in the process.
• If there is no chemical reaction the steady-state balance
reduces to:
• Material out = Material in
• Industrial Stoichiometry therefore is the industrial
application of this law
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Stoichiometry enables to us to account for the quantities
of material converted to products in a chemical change
 Need the knowledge of chemical equations
 Proper Balancing of these equations is key in Material
Balances at the macro/industrial level
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The Chemical reaction
• Chemical reactions stop when one of the reactants in used
up
• Mole ratio
• The ratio between the number of moles of two substances in
the balanced chemical equation is indicated by the
coefficients in equation
• So in stoichiometry:
• We can predict the amount of grams of product that will be
formed in the reaction
i.e. Knowing no. of grams of reactants one can predict the
grams of products that would be produced
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Balancing Chemical Equations
• With simple reactions it is usually possible to balance the
stoichiometric equation by inspection, or by trial and error
calculations.
• If difficulty is experienced in balancing complex equations,
the problem can always be solved by writing a balance for
each element present.
• E.gs:
1. Formation of water from hydrogen and oxygen
2. Combination of carbon with Oxygen to give carbon dioxide
3. Combustion of Ethane
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• H2 + O2 = H2O
• C + O2 = CO2
• C2H6 + O2 = CO2 + H2O
• C2H4 + Cl2 + O2 = C2H3Cl + H2O
• Solution:
• aC2H4 + bCl2 + cO2 = dC2H3Cl + eH2O
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Balance on(i) carbon
2a = 2d, a = d
(ii) hydrogen
4a = 3d + 2e
substituting d = a gives e =
a
2
(iii) chlorine
2b = d, hence b =
a
2
(iv) oxygen
2c = e, C =
e
2
=
a
4
putting a = 1, the equation becomes
C2H4 +
1
2
Cl2 +
1
4
O2 = C2H3Cl +
1
2
H2O
multiplying through by the largest denominator to remove fractions
4 C2H4 + 2Cl2 + O2 = 4C2H3Cl + 2 H2O

CONSERVATION OF MASS & ERNEGY
• Law of nature that states that Mass & energy cannot be
created or destroyed
• Mass & energy are very important when considering process
plants, operations analyzing plants and process units for
inventory & economic activities
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MATERIAL BALANCES
M I L T O N A W E S U T A

MATERIAL BALANCES
Defn
• The accounting of all mass in a chemical/pharmaceutical
process
• also referred to as mass balances.
• Two types of processes
a) Physical – No reaction
b) Chemical – Chemical change
• MB can be written in terms of:
i. Total Mass(number of moles)
ii. Mass(moles) of a chemical compound
iii. Mass(moles) on atomic particles
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CONSERVATION OF MASS
Distillation
Heat
Exchanger
Reactor
Separator
1
2
6
7
8
13
14
12
11
10
4
53
9
Mass is neither created nor destroyed

Application(s)
• ‘day to day’ operation of process for monitoring operating
efficiency
• Making calculations for
1. design and development of a process
2. Evaluating the existing process
i.e. quantities required, sizing equipment, number of items of
equipment
10/29/2019 13

Types of
balances
IntegralDifferential
– balances that indicate what is
happening in a system at an
instant time.
– balance equation is a rate (rate
of input, rate of generation,
etc.) and has units of the
balanced quantity unit divided
by a time unit (people/yr, g
SO2/s).
– usually applied to a
CONTINUOUS process.
– Balances that describe what happens
between two instants of time.
– balance equation is an amount of the
balanced quantity and has the
corresponding unit (people, g SO2).
– usually applied to a BATCH process,
with the two instants of time being
the moment after the input takes
place and the moment before the
product is withdrawn.
10/29/2019 M I L T O N A W E S U T A 14

The Universe
System
Surroundings
System & Universe
10/29/2019 M I L T O N A W E S U T A 15

Systems
• System – the PART of the universe that is under
consideration. It is separated from the rest of the
universe by it’s boundaries
– Open system  when matter CAN cross the boundary
– Closed system  when matter CANNOT cross the
boundary
– Isolated  Boundary seals matter and heat from
exchange with another system
open closed isolated↔↔matter
heat
heat
10/29/2019 M I L T O N A W E S U T A 16

10/29/2019 17
SYSTEMS
Systems
OPEN or CLOSED
Any arbitrary portion of or a whole process that you want to
consider for analysis
Reactor, the cell, mitochondria, human body, section of a pipe
Closed System
Material neither enters nor leaves the system
Changes can take place inside the system
Open System
Material can enter or leave the system across the boundaries

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STEADY/UNSTEADY-STATE PROCESSES
Steady-State
Nothing is changing with time
@ steady-state accumulation = 0
Rate of addition = Rate of removal
Unsteady-State (transient system)
{Input} ≠ {Output}
500 kg
H2O
100 kg/min
H2O
100 kg/min
H2O
CLASSICATION OF PROCESSES

Operation of
Continuous
Process
Unsteady
state
Steady
state
– All the variables (i.e.
temperatures,
pressure, volume,
flow rate, etc) do not
change with time
– Minor fluctuation can
be acceptable
– Process variable
change with
time, in
particular mass
flow rate.
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Batch Process
No material is transferred into/out of the system over the
period of interest.
E.g. heating a vessel of water
Continuous Process
Input and outputs flow continuously throughout the duration
of process
E.g. Pumping liquid into a distillation column & removing the
product stream from the top & bottom of the column.
Semibatch Process
Any process neither batch nor continuousslowly in a tank
E.g. blending 2 liquids

Process
Classification
SemibatchBatch
Continuous
– Feed is charge to the
process and product is
removed when the process
is completed
– No mass is fed or removed
from the process during
the operation
– Used for small scale
production (pharmaceutical
products)
– Input and output
is continuously
red and remove
from the process
– Used for large
scale production
– Neither batch nor
continuous
– During the process
a part of reactant
can be fed or a
part of product
can be removed.
10/29/2019 M I L T O N A W E S U T A 21

Define type and operation of process given below
• A balloon is filled with air at steady rate of 2 g/min
• A bottle of milk is taken from the refrigerator and left on
the kitchen
• Water is boiled in open flask
Answer
• Semibatch and unsteady state
• Batch and unsteady state
• Semibatch and unsteady state
10/29/2019 M I L T O N A W E S U T A 22

Classify the following:
1. A balloon is filled with air at a steady rate of 2g/minute
2. A bottle of soda is taken from a refrigerator and left on a table
3. Water is boiled in an open pan
4. Carbon monoxide and steam are fed into a tubular reactor at a
steady-rate and react to
form carbon dioxide and hydrogen. Products and unused reactants
are withdrawn at the other
end. The reactor contains air when the process is started up. The
temperature of the reactor is
also constant, and the composition and flow rate of the entering
reactant stream are also
independent of time. Classify the process (a) initially and (b) after a
long period of time has elapsed.
10/29/2019 M I L T O N A W E S U T A 23

• Suppose a cpd C is fed into a system at MCin & it is withdrawn
from the system at MCout & the amount of C in the system is
MCs.
•  the process unit has no losses
• If MCin ≠ MCout then,
i. either
a) MCin > MCout  C is consumed in the process unit as a reactant
b) MCin < MCout  C is generated as product
ii. Or
a) C is accumulated in the system and MCs increases; MCin < MCout
b) C (or Mcs) is reduced;MCin > MCout
10/29/2019 M I L T O N A W E S U T A 24

• The general balanced eqn is represented as
• Accumulation in the system is equal to inflow into system thru
the boundary plus generation within the system minus
consumption in the system minus outflow out of system
• INPUT – added materials to the system (FEED)
• OUTPUT – materials withdrawn from system (PRODUCTS)
• GENERATION / CONSUMPTION – products
formed/reactants consumed within a given system
• ACCUMULATION – increase in quantity of material within
the system with time i.e.
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{Input} + {Genn} - {Consumption} – {Output} = {Accumulation}

Examples
• Population in the Mbarara
• Each year a total of 100,000 people move into Mbarara,
75,000 people move out of the city, 55,000 are born while
45,000 die,
• Qn?
• Calculate the yearly population change in Mbarara?
• Soln
• 100,000 + 55,000 – 45,000 – 75,000 = 35,000
10/29/2019 M I L T O N A W E S U T A 26
{Input} + {Genn} - {Consumption} – {Output} = {Accumulation}

Key points on solving Material Balances
1. Draw a flow chart & indicate the input and output
streams/materials & compositions
2. Write a balanced equation if a chemical reaction is involved
3. List the terms to be determined, organize them in a table if
they are many.
4. Choose the basis of the calculation
5. Find the quantitative relationships when setting up a balanced
equation
6. Solve the unknowns by beginning with those of fewest/easiest
unknown variables
7. Use the right formula when necessary
8. Check the results10/29/2019 M I L T O N A W E S U T A 27

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Balances on Continuous Steady-state Processes
 Input + Generation = Output + Consumption
If the balance is on a nonreactive species, the generation and
consumption will be 0.
Thus, Input = Output
 Example
Distillation
1000 kg /h
Benzene + Toluene
%50 Benzene by mass
475 kg Toluene/h
M2 kg Benzene/h
m1 kg Toluene/h
450 kg Benzene/h
Input of 1000 kg/h of benzene+toluene containing 50% B by mass is
separated by distillation column into two fractions.
B: the mass flow rate of top stream=450 kg/h
T: the mass flow rate of bottom stream=475 kg/h

500 kg B/h
500 kg T/h
450 kg B/h
(kg T/h)
(kg B/h)
475 kg T/h
1m
2m
Steady state accumulation = 0
Since no chemical reactions occur generation & consumption = 0
INPUT = OUTPUT
Benzene Balance
500 kg B/h = 450 kg B/h + 2m
Toluene Balance
500 kg T/h = + 475 kg T/h1m
2m = 50 kg B/h
1m = 25 kg T/h
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10/29/2019 30
Solution of the example Input = Output
Benzene balance
1000 kg/h · 0.5 = 450 kg/h + m2
m2 = 50 kg/h Benzene
Toluene balance
1000 kg/h · 0.5 = 475 kg/h + m1
m1 = 25 kg/h Toluene
Balances on Continuous Steady-state Processes
.
.
.
.

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BALANCES ON BATCH PROCESSES
Initial Input + Generation = Final Output + Consumption
Objective: generate as many independent equations as the
number of unknowns in the problem
F
(W+A)
B
D F = B + D
F.xF = D.xD + B.xB
F.yF = D.yD + B.xB
x: mole fraction of W
y: mole fraction of A

10/29/2019 32
EXAMPLE (Batch Process)
 Centrifuges are used to seperate particles in the range of 0.1 to 100 µm
in diameter from a liquid using centrifugal force. Yeast cells are
recovered from a broth ( a mix with cells) using tubular centrifuge.
 Determine the amount of the cell-free discharge per hour if 1000 L/hr is
fed to the centrifuge, the feed contains 500 mg cells/L, and the product
stream contains 50 wt% cells. Assume that the feed has a density of 1
g/cm3.
Centrifuge
Feed (broth) 1000 L/hr
500 mg cells/L feed
( d= 1 g/cm3)
Concentrated cells P(g/hr)
50 % by weight cells
Cell-free discahrge D(g/hr)

10/29/2019 33
EXAMPLE (Batch Process)
Cell balance
Fluid balance
Input: (106 – 500) g/h fluid
Output 1: 1000g/h . 0.5 = 500 g/h fluid
Output 2: D(g/h) = (106 – 500)g/h – 500 g/h = (106 -103)g/h fluid
g/hr1000P
P[g/hr].
Pg1
cellsg0.5
mg1000
g1
.
feedL1
cellsmg500
.feedL1000


h
g
L
dm
dm
cm
cm
g
h
L 6
3
3
3
10
1
)
1
10
(
1
1000 

• When you are given process information and asked to
determine something about the process, ORGANIZE
the by drawing a flowchart
Represent
INPUTS
Represent
OUTPUTS
Represent
PROCESS UNIT
(Reactor, mixer,
separation units,
etc)
10/29/2019 M I L T O N A W E S U T A 34

• Write the values and units of all known stream variables at
the locations of the streams on the flowchart.
Example
• A stream containing 21 mole% O2 and 79% N2 at 320˚C and
1.4 atm flowing at a rate of 400 mol/h might be labeled as:
400 mol/h
0.21 mol O2/mol
0.79 mol N2/mol
T = 320˚C, P = 1.4 atm
Usually we write
of the stream!
,,, nmnm 
10/29/2019 M I L T O N A W E S U T A 35

• Process stream can be given in two ways:
60 kmol N2/min
40 kmol O2/min
0.6 kmol N2/kmol
0.4 kmol O2/kmol
100 kmol/min
3.0 lbm CH4
4.0 lbm C2H4
3.0 lbm C2H6
0.3 lbm CH4/lbm
0.4 lbm C2H4/lbm
0.3 lbm C2H6/lbm
10 lbm
As the total amount or
flow rate of the stream
and the fractions of each
component
Directly as the amount
or flow rate of each
component.
10/29/2019 M I L T O N A W E S U T A 36

• Assign algebraic symbols to unknown stream variables
[such as m (kg solution/min), x (lbm N2/lbm), and n
(kmol C3H8)] and write these variable names and their
associated units on the flowchart.
mol/h
0.21 mol O2/mol
0.79 mol N2/mol
T = 320˚C, P = 1.4 atm
n 400 mol/h
y mol O2/mol
(1-y) mol N2/mol
T = 320˚C, P = 1.4 atm
10/29/2019 M I L T O N A W E S U T A 37

If the mass of stream 1 is half that of stream 2, label the
masses of these streams as m and 2m rather than m1 and m2.
m
2m
m1
m2
10/29/2019 M I L T O N A W E S U T A 38

If you know that the mass fraction of nitrogen is 3 times than
oxygen, label the mass fraction as y g O2/g and 3y g N2/g rather
than y1 and y2.
y g O2/g
3y g N2/g
y1 O2/g
y2 g N2/g
When labeling component mass fraction or mole fraction, the
last one must be 1 minus the sum of the others
y mol O2/mol
(1-y) mol N2/mol
y1 mol O2/mol
y2 mol N2/mol
10/29/2019 M I L T O N A W E S U T A 39

gasinfractionmolesy
liquidinmoles)or(massfractioncomponentx
rateflowvolumeV
volumeV
rateflowmolarn
molesn
rateflowmassm
massm











10/29/2019 M I L T O N A W E S U T A 40

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COMBUSTION OF ETHANE
Ethane is burnt in a combustion reactor. The gas is fed into
the reactor which contains 7.5% ethane, 30% oxygen and
62.5% N2. If the C2H6 is completely burnt to CO2 and H2O
and the reactor is operating at a steady state, determine the
composition in mol% of the product gas exiting the reactor?
Combustion
Chamber
7.5% C2H6
30% O2
62.5% N2
Product gas
? Mole composition

Establish a balanced equation of combustion
• aC2H6 + bO2  cCO2 + dH2O
• Balance on (i) Carbon
•  2a = c  a = ……………………………………………………………(1)
(ii) oxygen
 2b = 2c + d …………………………………………………..(2)
 (iii) hydrogen
 6a =2d  3a = d  =d…………………………(3)
 Substituting for d in (2)
 2b = 2c + =  b =
• If c = 1 then a = ; b = & d =10/29/2019 M I L T O N A W E S U T A 42

10/29/2019 M I L T O N A W E S U T A 43
C2H6
O2
N2
CO2
H2O
Equation of combustion is:
C2H6 + O2  CO2 + H2O
2C2H6 + 7O2  4CO2 + 6H2O

Solution
Assuming the feed enters at 100% mol and also taken out at
steady state.
Quantities of materials in the reactor do not change
 {Accumulation} = 0
C2H6
O2
N2
CO2
H2O
10/29/2019 M I L T O N A W E S U T A 44
7.5 7.5
30
62.5
0
0
26.25
0
0
0
0
0
0
100
15
22.5
0
3.75
62.5
15
22.5
103.75
0
3.61
60.24
14.46
21.69
100

A Chemical plant produces an aqueous solution of sodium
hydroxide of 8% by mass by diluting a stream of 20%
solution by mass using a stream of pure water. What flow
rates of the pure water and 20% NaOH will produce
3,000Kgm/min of the 8% solution?
10/29/2019 M I L T O N A W E S U T A 45

1. Read the problem a few more times to clearly
understand it
2. On reading it is established that it is a mixing
problem
3. In the mixing we have two streams to be mixed.
4. Draw a mixer with all the streams
10/29/2019 M I L T O N A W E S U T A 46

Mixer
0.2KgmNaOH/KgT
0.8KgmH2O/KgT
H2O
0.08KgmNaOH/KgT
0.92KgmH2O/KgT
Rate = 3,000Kgm/min
S1Kgm/min
S2Kgm/min
Overall Mass balance:-
S1 + S2 = 3,000Kgm/min
10/29/2019 M I L T O N A W E S U T A 47

NaOH Balance
0.2 KgmNaOH
KgT
X S1
Kgm
min
= 0.08KgmNaOH
KgT
X 3,000Kgm
min
S1 = (0.08) X 3,000
(0.2)
= 1,200Kgm
min
But S1 + S2 = 3,000Kgm/min
Therefore S2 = 3,000- S1
= 3,000 – 1,200
= 1,800Kgm/min
10/29/2019 M I L T O N A W E S U T A 48

 An aqueous solution of NaOH contains 45% NaOH by mass. It is
desired to produce an 20 % NaOH solution by diluting a stream
of the 45 % solution with a stream of pure water.
 Calculate the ratios (liters H2O/kg feed solution) and (kg
product solution/ kg feed solution).
10/29/2019 M I L T O N A W E S U T A 49

Total Mass Balance
Nonreactive steady-state process input = output
= 225 kg NaOH
0.45 kg NaOH/kg
0.55 kg H2O/kg
0.20 kg NaOH/kg
0.80 kg H2O/kg
100 (kg) 2m (kg)
kg H2O
liters H2O
1m
1V
NaOH Balance
2m
(0.45 kg NaOH/kg)(100 kg)=(0.20 kg NaOH/kg) 2m
= 125 kg NaOH100 kg + = 2m
1m 1m
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1V
125 kg 1.00 liter
kg
= = 125 liters1V
100 kg
= 1.25 liters H2O/kg feed solution
1V
Diluted water volume
Ratios requested in problem statement
100 kg
=2.25 kg product solution/kg feed solution
2m
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An experiment on the growth rate of certain organism
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 rate of 20 cm3/min
B: Air (21% O2 and 79% N2)
C: Pure O2 with a molar flow rate one-fifth of the molar flow
rate of stream B
The output gas is analyzed and is found to contain 1.5 mole%
water. Draw and label the flowchart of the process, and
calculate all unknown stream variables.
10/29/2019 M I L T O N A W E S U T A 52

An experiment on the growth rate of certain organism 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 rate of 20 cm3/min
B: Air (21% O2 and 79% N2)
C: Pure O2 with a molar flow rate one-fifth of the molar flow rate of stream B
The output gas is analyzed and is found to contain 1.5 mole% water. Draw and label the flowchart of the process, and
calculate all unknown stream variables.
0.21 mol O2/mol
0.79 mol N2/mol
0.015 mol H2O/mol
y (mol O2/mol)
(0.985 – y)(mol N2/mol)
20.0 cm3 H2O (l)/min
(mol H2O/min)
1n0.2 (mol O2/min)
1n (mol air/min)
3n (mol/min)
2n
Evaporator
10/29/2019 M I L T O N A W E S U T A 53

0.21 mol O2/mol
0.79 mol N2/mol
0.015 mol H2O/mol
y (mol O2/mol)
(0.985 – y)(mol N2/mol)
(mol H2O/min)
1n (mol air/min)
3n (mol/min)
2n
2n
20.0 cm3 H2O 1.00 g H2O 1 mol
min cm3 18.02 g
=
H2O Balance
= 1.11 mol/min
Nonreactive steady-state process input = output
2n (mol/min) =
(mol) 0.015 mol H2O
(min) mol
3n
2n
3n = 74.1 mol/min
10/29/2019 M I L T O N A W E S U T A 54
Evaporator

0.21 mol O2/mol
0.79 mol N2/mol
0.015 mol H2O/mol
y (mol O2/mol)
(0.985 – y)(mol N2/mol)
(mol H2O/min)
1n (mol air/min)
3n (mol/min)
2n
N2 Balance
1n (mol) 0.79 mol N2
=
(mol) (0.985-y) (mol N2)
(min) mol (min) (mol)
3n y = 0.337 mol O2/mol
Total Mole Balance
0.2 + + =2n 3n1n 1n 3n = 60.8 mol/min
10/29/2019 M I L T O N A W E S U T A 55
Evaporator

Flowchart scaling
If final stream quantities
are larger than the
original quantities.
Scaling down What?
Procedure of changing the values of
all stream amounts or flow rates by
a proportional amount while leaving
the stream compositions unchanged.
Scaling up
if final stream quantities
are smaller than the
original quantities.
Suppose you have balanced a
process and the amount or flow
rate of one of the process
streams is n1.You can scale the
flow chart to make the amount
or flow rate of this stream n2 by
multiplying all stream amounts or
flow rate by the ratio n2/n1.
You cannot, however, scale
masses or mass flow rates to
molar quantities or vice versa by
simple multiplication; conversions
of this type must be carried out
using the methods as discussed in
mass fraction and mol fraction
section.
10/29/2019 M I L T O N A W E S U T A 56

1 kg C6H6
300 lbm/h
1 kg C7H8
300 kg C6H6
300 kg C7H8
300 lbm/h
2 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
600 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
600 lbm/h
0.5 lbm C6H6/lbm
0.5 lbm C7H8/lbm
x 300
kg kg/h
Replace kg with lbm
10/29/2019 M I L T O N A W E S U T A 57

3.0 kg/min of benzene and 1.0 kg/min of toluene are mixed
•Two unknown quantities – m and x, associated with process, so two equations are needed to
calculate them.
•For NONREACTIVE STEADY STATE process input = output.
•3 possible balance can be written – Balance on total mass, benzene, and toluene – any two of
which provide the equations needed to determine m and x.
For example,
Total Mass Balance:
3.0 kg/min + 1.0 kg/min = m kg/min = 4.0 kg/min
Benzene Balance:
3.0 kg C6H6/min = 4.0 kg/min (x kg C6H6/kg)
x = 0.75 kg C6H6/kg
m (kg/min)
x (kg C6H6/kg)
(1-x) (kg C7H8/kg)
3 kg C6H6/min
1 kg C7H8/min
10/29/2019 M I L T O N A W E S U T A 58

Rules of thumb for NONREACTIVE process
1. The maximum number of independent equations that can
be derived by writing balances on a nonreactive system
equals the number of chemical species in the input and
output streams.
2. Write balances first that involve the fewest unknown
variables.
10/29/2019 M I L T O N A W E S U T A 59

General Procedure for Single Unit Process
Material Balance Calculation
1. Choose as basis of calculation an amount or flow rate of one of the process streams.
2. Draw a flowchart and fill in all unknown variables values, including the basis of
calculation. Then label unknown stream variables on the chart.
3. Express what the problem statement asks you to determine in terms of the labeled
variables.
4. If you are given mixed mass and mole units for a stream (such as a total mass flow
rate and component mole fractions or vice versa), convert all quantities to one basis.
5. Do the degree-of-freedom analysis.
6. If the number of unknowns equals the number of equations relating them (i.e., if the
system has zero degree of freedom), write the equations in an efficient order
(minimizing simultaneous equations) and circle the variables for which you will solve.
7. Solve the equations.
8. Calculate the quantities requested in the problem statement if they have not
already been calculated.
9. If a stream quantity or flow rate ng was given in the problem statement and another
value nc was either chosen as a basis or calculated for this stream, scale the
balanced process by the ratio ng/nc to obtain the final result.
10/29/2019 M I L T O N A W E S U T A 60

light
an amount (mass or moles) of flow rate (mass
or molar) of one stream or stream component
in a process. All unknown variables are
determined to be consistent with the basis.
If a stream amount or flow rate is
given in problem, choose this
quantity as a basis
If no stream amount or flow rate are
known, assume one stream with known
composition. If mass fraction is
known, choose total mass or mass flow
rate as basis. If mole fraction is
known, choose a total moles or molar
flow rate as basis
Basis of
calculation
10/29/2019 M I L T O N A W E S U T A 61

light
A process used to determine if a material balance problem has
sufficient specifications to be solved.
It is an accounting of the number of unknowns in a problem and the
number of independent equations that can be written.
Procedure:-
1. draw and completely label the flowchart
2. count the unknown variables on the chart(n unknowns)
3. count the independent equations relating these variables
(n indep. eq.)
4. calculate degrees of freedom by subtracting step (3)
from step (2)(ndf)
Degree of
freedom
Independent equations Equations are independent if none of
them can be derived from the others. For example, not one of the set of
equations can be obtained by adding or subtracting multiples of the others.
ndf= n unknowns - n indep. eq.
10/29/2019 M I L T O N A W E S U T A 62

light
– n df = 0, there are n independent equations
and n unknowns. The problem can be solved.
– n df > 0, there are more unknowns that
independent equations. The problem is
underspecified. n df more independent
equations or specifications are needed to
solve the problem.
– n df < 0, there are more independent equations
than unknowns. The problem is over-specified
with redundant and possibly inconsistent
relations.
Possible
outcome
s of a
DFA:
10/29/2019 M I L T O N A W E S U T A 63

1.Material balances.
– For a nonreactive process, the number of independent
equation can be written is not more than number of molecules
species (nms) of the process
– If benzene and toluene is involved in stream, we can write
balance on benzene, toluene, total mass, atomic carbon and
etc., but only TWO INDEPENDENT balance equation exist
2.An energy balance.
– If the amount of energy exchanged between the system and
its surroundings is specified or if it is one of the unknown
process variables, an energy balance provides a relationship
between inlet and outlet material flows and temperatures.
– To be discussed later
10/29/2019 M I L T O N A W E S U T A 64

3. Process specifications
– The problem statement may specify how several process
are related.
– i.e: Outlet flow rate is two times than flow rate stream 1
or etc.
4. Physical properties and laws
– Two of the unknown variables may be the mass and
volume of a stream material,
– a tabulated specific gravity for liquids and solids or an
equation of state for gases would provide an equation
relating the variables.
10/29/2019 M I L T O N A W E S U T A 65

5. Physical constraints
– For example, if the mole fractions of the three
components A,B & C of a stream labeled are xA, xB, and
xC, then the relation among these variables is
– xA + xB + xC = 1.
– Instead label as xc, the last fraction should be 1-xA-xB
6. Stoichiometric relations
– If chemical reactions occur in a system, stoichiometric
equation provide a relationship between the quantities
of reactant and the product
– To be discussed later
10/29/2019 M I L T O N A W E S U T A 66

A stream of humid air enters a condenser in which 95 % of the
water vapor in the air is condensed. The flow rate 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 mole %
oxygen, with the balance nitrogen. The entering air contains 10.0
mole % water. Calculate the flow rate of the gas stream leaving
the condenser and the mole fractions of oxygen, nitrogen, and
water in this stream.
10/29/2019 M I L T O N A W E S U T A 67

Degree of freedom analysis:
 5 unknowns
 3 material balances ( since there are 3 molecular species in this nonreactive process)
 1 density relationship (relating the mole flow rate to the given volumetric flow rate of the
condensate
 1 the fractional condensation
0 degrees of freedom
10/29/2019 M I L T O N A W E S U T A 68

( mol O2/hr)
(mol N2/hr)
(mol H2O/hr)
225.0 L H2O(l) /min
(mol H2O(l)/min)
(95% of water in feed)
1n (mol/hr)
3n
2n
Condenser
10/29/2019 M I L T O N A W E S U T A 69
0.100 mol H2O/mol
0.900(mol dry air/mol)
0.21 mol O2/mol
0.79 mol N2/mol
Solvable since ndf = 0

( mol O2/hr)
(mol N2/hr)
(mol H2O/hr)
225.0 L H2O(l) /min
(mol H2O(l)/min)
(95% of water in feed)
1n
(mol/hr)
3n
2n
Condenser
10/29/2019 M I L T O N A W E S U T A 70
0.100 mol H2O/mol
0.900(mol dry air/mol)
0.21 mol O2/mol
0.79 mol N2/mol
Density relationship = (225
( )
)(1.00
( )
)( ⁄ ) = 12500 mol H2O/h = 12.5Kmol H2O/h2n
= (0.95)0.1 ;2n 1n

1. Absorption
– A phenomenon in which components in a mixture in one
phase contacts another bulky phase and some of the
components get dissolved
– This takes place in column or absorption tower
(absorber).
– A Scrubber is an absorption column designed to
remove undesirable components from a gas stream
– The bulky phase may be solid or liquid
– The two streams are better arranged to cause
countercurrent flow
10/29/2019 M I L T O N A W E S U T A 71

2. Water Boiler
– Process Unit in which a set of tubes is passed through
combustion fumes.
– The boiler feed water that is made to pass in the tubes is
heated by the hot combustion fumes to produce steam,
used predominantly in different heat exchange operations
in industry
– To be considered under Unit Operations
10/29/2019 M I L T O N A W E S U T A 72

• 3. Distillation
—An operation/process in which a liquid mixture of at least two
components is fed into a vertical column with a series of
vertically spaced horizontal plates of a solid packing through
which the mixtures flow in a counter current manner.
— The liquid feed mixture flows down the column while mixture
of the vapors formed flows upwards
—There is partial condensation of the rising vapors as there is
partial vaporization of the liquid phase
—As the vapor flow upwards there is progressive enrichment of
the vapor phase with the more volatile component of the feed
mixture
10/29/2019 M I L T O N A W E S U T A 73

– The liquid flowing down the column is progressively enriched
with the less volatile of the feed mixture
– The vapor leaving the top of the column is condensed
– Part of the condensate is taken of as the overhead product
and the rest is recycled in the column and refluxed
becoming the liquid stream
– The liquid leaving the bottom of the distillation unit is
partially vaporized. The vapors formed are recycled in the
column to form the vapor stream that flows upwards
– The residual is taken off at the bottom
– So the distillation unit comprises a column, Still and a
condenser
10/29/2019 M I L T O N A W E S U T A 74

4. Heat exchange(rs)
– Heat is made to flow from one region to another of
lower temperature through a specific medium
– Medium could be solid, Liquid or gas
– In most industrial processes, Liquid and/ or solid
media are commonly encountered
– Heat exchangers are the units in which heat
exchange takes place
– They are constructed in different designs
10/29/2019 M I L T O N A W E S U T A 75

5. Crystallization
– A process in which a liquid solution is cooled or solvent is
evaporated to an extent of formation of solid crystals
– The unit is called a crystallizer
– A slurry (suspension of solids in a liquid) leaves the
crystallizer
– The crystals may be separated from the slurry by filtration
or centrifugation
6. Evaporation/vaporization
– A unit operation in which a liquid or solid mixture is vaporize
to obtain a more concentrated phase
– The unit is called an evaporator- may be one or more, hence
single effect or Multieffect evaporation
10/29/2019 M I L T O N A W E S U T A 76

7. Extraction
– An operation in which a solid/Liquid mixture (solid/liquid +
feed carrier) is contacted with a third phase(commonly liquid
solvent) that is immiscible with the feed carrier but
preferentially dissolves the solid/liquid
– This is done in an extractor unit
– Soluble component is transferred to the solvent to form a
solution.
– The mixture of the solution and the exhausted inert feed
carrier are separated by for e.g. Settling by gravity in a
decanter or by other mechanical separation techniques
10/29/2019 M I L T O N A W E S U T A 77

8. Filtration
– A slurry of solid particles suspended in a liquid is passed
through a porous medium to separate the solids from the
liquid
– The liquid that passes through is the filtrate
– The solids and some entrained liquid remain as residue to form
a filter cake
– The unit is called a filter
– There are different types of filters depending the principle
applied and mode of operation
10/29/2019 M I L T O N A W E S U T A 78

9. Drying
– An operation in which liquid wetting a solid is made to
evaporate by either heating or a combination of heating and
cooling etc
– The vapor and gas evaporates into the outlet stream while the
solid and remaining liquid residue emerge as the second outlet
stream of the Drier
– The ideal aim of drying is to form a completely dry solid
10.Condensation
– An operation in which an incoming gas or compressed gas in to
a condenser is/are caused to liquefy
– The uncondensed gas and liquid leave the condenser as
separate streams
10/29/2019 M I L T O N A W E S U T A 79

11.Flash evaporation/distillation
– A special method of evaporation/ distillation in which a liquid
feed mixture at high pressure is suddenly exposed to lower
pressure causing evaporation
– The vapour product is richer in the more volatile component
and the residual liquid is rich in the less volatile component
12.Pumps
– Devices used to propel fluids from one location to another
through and enclosed channel
– May be pipes or tubes
13.Membranes
– Thin solid or liquid film through which one or more components
of a process stream can permeate10/29/2019 M I L T O N A W E S U T A 80

14. Mixing
In industrial process engineering, mixing is a unit
operation that involves manipulation of
a heterogeneous physical system with the intent to make it
more homogeneous.
Mixing is performed to allow heat or mass transfer
to occur between one or more streams, components
or phases.
The opposite of mixing is segregation.

• In real chemical industries, more than one unit processes
exist such as a separation unit after reactor and so on.
• Need to know term called SYSTEM in order to solve
material problem
• SYSTEM:
– Any portion of process that can be enclosed within a
hypothetical box (or boundary)
– It can be the entire process, an interconnected of
process unit, a single unit, a point which two or more
stream come together into one stream or etc.
– The inputs and outputs to a system are the process
streams that are intersect to the system boundary
10/29/2019 M I L T O N A W E S U T A 82

FEED 1
FEED 2
PRODUCT 1 PRODUCT 2 FEED 3
PRODUCT 3
UNIT 2UNIT 1
A
B
C
D
E
10/29/2019 M I L T O N A W E S U T A 83

• Solving material balances in multiple unit process is basically
the same as single unit processes
• In multiple unit, must isolate and write balance on several
subsystems to obtain enough equation to determine all
unknowns stream variables
• Always perform degree-of-freedom analysis before solving a
material balance of system.
10/29/2019 M I L T O N A W E S U T A 84

Recycling situation
• Normally in chemical reaction, some of unreacted reactant
also found in the product.
• This unreacted reactant can be separated and recycled back
to the reactor
Product
Recycle Stream
Fresh
Feed Reactor Separator
10/29/2019 M I L T O N A W E S U T A 85

Purpose of Recycle
1. Recovery of catalyst – catalyst is very expensive
2. Dilution of process stream – typically for slurry solution
3. Control of process variables – especially for the reaction that
release heat, heat can be reduce by lowering the feed
concentration
4. Circulation of working fluid - such as in refrigerator system
5. For chemical processes – improving the yield
10/29/2019 M I L T O N A W E S U T A 86

Bypass
• Fraction of the feed to a process unit is diverted around the unit
and combined with the output stream from the unit
• Used to control the composition of a final exit stream from a unit
by mixing the bypass stream & the unit exit stream in suitable
proportions to obtain desired final composition.
Product
Bypass Stream
Fresh
Feed
Process Unit
10/29/2019 M I L T O N A W E S U T A 87

Limiting Reactant, Excess Reactant and percentage Excess
To ensure complete conversion of a specific reactant, the
other reactant(s) is(are) fed in excess of the stoichiometric
quantities.
In stoichiometric reactions the reactant that would run out if
a reaction proceeded to completion is called the limiting
reactant, and the other reactants fed in excess are termed
excess reactants.
A reactant is limiting if it is present in less than its
stoichiometric proportion relative to every other reactant.
If all reactants are present in stoichiometric proportion, then
no reactant is limiting.
10/29/2019 M I L T O N A W E S U T A 88

Limiting Reactant & Excess Reactant
%100
n
n-n
ExcessPercentage
n
n-n
ExcessFractional
stoich
stoichfeed
stoich
stoichfeed


10/29/2019 M I L T O N A W E S U T A 89

Example
C2H2 + 2H2 ------> C2H6
Inlet condition: 20 kmol/h C2H2 and 50 kmol/h H2
What is limiting reactant and fractional excess?
(H2:C2H2) o = 2.5 : 1
(H2:C2H2) stoich = 2 : 1
H2 is excess reactant and C2H2 is limiting reactant
Fractional excess of H2 = (50-40)/40 = 0.25
10/29/2019 M I L T O N A W E S U T A 90

Fractional Conversion
• Fractional Conversion (f)
%100
fedmole
reactedmoles
f,ConversionPercentage
fedmole
reactedmoles
f,ConversionFractional


10/29/2019 M I L T O N A W E S U T A 91

Extent of Reaction
• Extent of Reaction, ξ
ξ = extent of reaction
ni = moles of species i present in the system after the
reaction occurred
nio = moles of species i in the system when the reaction starts
vi = stoichiometry coefficient for species i in the particular
chemical reaction equation


 iioi
iioi
vnn
vnn


or
10/29/2019 M I L T O N A W E S U T A 92

Extent of Reaction for Multiple Reaction
Concept of extent of reaction can also be applied
for multiple reaction
Only now each independent reaction has its own
extent.
ijj
jiioi vnn  
10/29/2019 M I L T O N A W E S U T A 93

Multiples Reaction, Yield & Selectivity
Some of the chemical reactions have side reactions which formed
undesired products- multiple reactions occur.
Effects of this side reaction might be:
1. Economic loss
2. Less of desired product is obtained for a given quantity of raw materials
3. Greater quantity of raw materials must be fed to the reactor to obtain a specified
product yield.
selectivity =
moles of desired product
moles of undesired product
10/29/2019 M I L T O N A W E S U T A 94

Yield
3 definitions of yield with different working definition
Yield =
Moles of desired product formed
Moles that would have been formed if there were no side
reaction and the limiting reactant had reacted completely
Yield =
Moles of reactant fed
Yield =
Moles of reactant consumed
10/29/2019 M I L T O N A W E S U T A 95

1. Forty-five hundred kilograms per hour of a solution that is one-third
K2CrO4 by mass is joined by a recycled 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 is cooled (causing crystals K2CrO4 to
come out solution) and then filtered. The filter cake consist 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.
Calculate the rate evaporation, the rate of production of crystalline
K2CrO4, the feed rates that evaporator and the crystallizer must be
designed to handle, and the recycle ratio (mass of recycle)/(mass of
fresh feed)
10/29/2019 M I L T O N A W E S U T A 96
EXERCISES

Limiting Reactant & Excess Reactant
10/29/2019 M I L T O N A W E S U T A 97
Consider the complete combustion of heptane by
the balanced reaction below
C7H16 + 11O2  7CO2 + 8H2O
• For the reactant side it is desired that heptane is
completely converted or used up
• This happens when Oxygen is fed in excess of the
stoichiometric quantities
•  Oxygen is the excess reactant and Heptane, the
limiting reactant
• The reaction stoichiometrically indicates that 11Kmol of
O2 is required to completely react with 1Kmol of heptane

2.Acrylonitrile is produced in the reaction of propylene,
ammonia, and oxygen
C3H6 + NH3 +3/2 O2 C3H3N + 3H2O
The feed contains 10.0 mole % 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 basis,
determine which reactant is limiting, the percentage by which
each of the other reactants is in excess, and the molar
amounts of all product gas constituents for a 30% conversion
of the limiting reactant (Assume basis 100 mol)
10/29/2019 M I L T O N A W E S U T A 98

Industrial stoichiometry dlc 2105 29.10.2019 (1)

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