Chapter 4
Fundamentals of
Material Balances
Material Balance-Part 1
Process Classifications
3 type of chemical processes:
- Concept of boundary of the process
1. Batch process
– 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
– Operate in unsteady state
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Shoukat Choudhury
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Process Classifications
2. Continuous process
– Input and output is continuously fed and
remove from the process
– Operate in steady state
– Used for large scale production
3. Semibatch process
– Neither batch nor continuous
– During the process a part of reactant can be
fed or a part of product can be removed.
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Shoukat Choudhury
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2 type of process operations:
1. Steady state
– All the variables (i.e. temperatures, pressure,
volume, flow rate, etc) do not change with
time
2. Unsteady state or transient
– Process variable change with time
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Shoukat Choudhury
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Try This…
Define type and operation of processes
given:
• A balloon is being filled with air at steady
rate of 2 g/min
Semibatch and unsteady state
• A bottle of milk is taken from the
refrigerator and left in the kitchen
Batch and unsteady state
• Water is boiled in an open flask
Semibatch and unsteady state
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Shoukat Choudhury
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General Balance Equation:
INPUT + GENERATION – OUTPUT – CONSUMPTION = ACCUMULATION
Balances on Continuous Steady-State Process
• Steady state;
accumulation = 0
INPUT + GENERATION = OUTPUT + CONSUMPTION
• If balance on steady-state non-reactive processes;
generation = 0, consumption = 0
INPUT = OUTPUT
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Shoukat Choudhury
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Example 4.2-­‐2 One thousands kilogram per hour of a mixture of Benzene (B) and Toluene (T) containing 50% Benzene by mass is separated by a disEllaEon into two fracEons. The mass flow rate of Benzene in the top stream is 450 kg/h and that of Toluene in the boIom stream is 475 kg/h. The operaEon is at steady state. Calculate the unknown components flow rates in the output streams. n3/29/15
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Shoukat Choudhury
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Flowchart n3/29/15
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Shoukat Choudhury
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Steps for Material Balance Calculations:
1. Learn how to organize information about
process variables
- Flow Chart drawing and Labeling
2. Choose a Basis
3. Set up material balance equations
4. solve the equations for unknown variables.
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Shoukat Choudhury
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Flowcharts
• When you are given process information and
asked to determine something about the
process, it is essential to organize the
information in a way that is convenient for
subsequent calculations.
• The best way to do this is to draw a flowchart
– using boxes or other symbols to represent process
units (reactors, mixers, separation units, etc.)
– lines with arrows to represent inputs and outputs.
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Shoukat Choudhury
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Flowcharts……
• The flowchart of a process can help get material balance
calculations started and keep them moving.
• Flowchart must be fully labeled when it is first drawn,
with values of known process variables and symbols for
unknown variables being written for each input and
output stream.
• Flowchart will functions as a scoreboard for the problem
solution: as each unknown variable is determined its
value is filled in, so that the flowchart provides a
continuous record of where the solution stands and what
must still be done.
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Shoukat Choudhury
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Step 1.2: Labeling a flowchart
2 suggestions for labeling flowchart:
1. Write the values and units of all known stream
variables at the locations of the streams on the
flowchart.
For 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
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Shoukat Choudhury
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Labeling a flowchart-continue
Process stream can be given in two ways:
a) As the total amount or flow rate of the stream
and the fractions of each component
b) Or directly as the amount or flow rate of each
component.
10 lbm
0.3 lbm CH4/lbm
0.4 lbm C2H4/lbm
0.3 lbm C2H6/lbm
3.0 lbm CH4
4.0 lbm C2H4
3.0 lbm C2H6
100 kmol/min
0.6 kmol N2/kmol
0.4 kmol O2/kmol
60 kmol N2/min
40 kmol O2/min
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Shoukat Choudhury
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Labeling a flowchart(continued)
2. 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 units
on the flowchart.
n! mol/h
400 mol/h
0.21 mol O2/mol
0.79 mol N2/mol
T = 320˚C, P = 1.4 atm
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nProf.
y mol O2/mol
(1-y) mol N2/mol
T = 320˚C, P = 1.4 atm
Shoukat Choudhury
n14
Labeling a flowchart-continue
• If that 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.
• If you know that mass fraction of nitrogen is 3 times
than oxygen, label mass fractions as
y g O2/g and 3y g N2/g rather than y1 and y2.
• When labeling component mass fraction or mole fraction,
the last one must be 1 minus the sum of the others.
• If volumetric flow rate of a stream is given, it is generally
useful to label the mass or molar flow rate of this stream
or to calculate it directly, since balance are not written
on volumetric qualities.
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Shoukat Choudhury
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Consistent on Notation
m = mass
! = mass flow rate
m
n = moles
! = molar flow rate
n
V = volume
! = volume flow rate
V
x = component fraction (mass or moles) in liquid
y = moles fraction in gas
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Shoukat Choudhury
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Example 4.2-­‐3 Two methanol water mixtures are contained in separate flasks. The first mixture contains 40 wt% methanol, and the second contains 70% methanol. If 200 g of the first mixture is combined 150 g of the second, what are the mass and composiEon of the product? n3/29/15
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Shoukat Choudhury
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Try This..
nExample
4.2.3
Two methanol-water mixture are contained
in separate flasks. The first mixture contains
40wt% methanol and the second flask
contains 70wt% methanol. If 20 Kg of the
first mixture are going to be mixed with
15000 g of the second in a mixing unit,
what are the mass and composition of the
product of the mixing unit?
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Shoukat Choudhury
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nExample
Try This..
4.3.1
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.
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Shoukat Choudhury
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Solution
0.200n!1 mol O2/min
n!1 mol air/min
n! 3 mol/min
Evaporation
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 cm3 H2O (l)/min
n! 2 mol H2O/min
Ø
Ø
Ø
Ø
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convert water flow-rate to mol/min
Water balance
Total mol
Nitrogen / oxygen balance
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Shoukat Choudhury
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Flowchart Scaling & Basis of Calculation
• Flowchart scaling – procedure of changing
the values of all stream amounts or flow
rates by a proportional amount while leaving
the stream compositions unchanged. The
process would still be balance.
• Scaling-up – if final stream quantities are
larger than the original quantities.
• Scaling down – if final stream quantities are
smaller than the original quantities.
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Shoukat Choudhury
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Flowchart Scaling & Basis of Calculation
1 kg C6H6
2 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
1 kg C7H8
x 300
300 kg C6H6
600 kg
0.5 kg C6H6/kg
0.5 kg C7H8/kg
300 kg C7H8
kg
kg/h
Replace kg with lbm
300 lbm/h
600 lbm/h
0.5 lbm C6H6/lbm
0.5 lbm C7H8/lbm
300 lbm/h
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Shoukat Choudhury
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Flowchart Scaling & Basis of Calculation
• 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.
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Shoukat Choudhury
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Basis of Calculation
• A basis of calculation is 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.
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Flowchart n3/29/15
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Shoukat Choudhury
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Shoukat Choudhury
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Product Separation
Reactants
Products
Product
Separation
Unit
Reactor
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Shoukat Choudhury
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K-salt Recovery
• In a steady state process crystalline potassium
chromate (K2CrO4) is recovered from an aqueous
solution of this salt.
• 10 ton per hour of a solution that is 30% K2CrO4 by
mass is fed into an evaporator. The concentrated
stream leaving the evaporator contains 50% 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 contain 35% K2CrO4
by mass; the crystals account for 95% of the total
mass of the filter cake. The filtrate drains out of the
system also contains 35% K2CrO4.
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Shoukat Choudhury
n41
10 ton per hour of a solution that is 30% K2CrO4 by mass is fed into an
evaporator. The concentrated stream leaving the evaporator contains 50%
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 contain 35% K2CrO4 by mass; the
crystals account for 95% of the total mass of the filter cake. The filtrate drains
out of the system also contains 35% K2CrO4.
.
m1 kg W/h
Filter Cake
Fresh Feed
10,000 kg/h
0.30 kg K/kg
0.70 kg W/kg
.
.
m2 kg/h
Evaporator
0.50 kg K/kg
0.50 kg W/kg
Crystallizer
& Filter
m3 kg/h
.
m4 kg/h
0.35 kg K/kg
0.65 kg W/kg
Filtrate
.
0.35 kg K/kg
m 5 kg/h 0.65 kg W/kg
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Shoukat Choudhury
n42
m1 = 4000 kg W/h
m2 = 6000 kg K-soln/h
m3 = 1385 kg K(S)/h
m4 = 73 kg K-soln/h
m5 = 4542 kg K-soln/h
• Product rate, m3 = 1385 Kg/h
• Filtrate soln, m5 = 4542 Kg/h
• If filtrate is discarded, a huge loss in raw
material and money
• Therefore, recycle is the solution
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Shoukat Choudhury
n43
K-salt Recovery
•
•
•
•
•
In a steady state process crystalline potassium chromate (K2CrO4) is
recovered from an aqueous solution of this salt.
10 ton per hour of a solution that is 30% K2CrO4 by mass is joined
by a recycle stream containing 35% K2CrO4, and combined stream
is fed into an evaporator.
The concentrated stream leaving the evaporator contains 50%
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 K2CrO4crystals and a solution that contain
35% K2CrO4 by mass; the crystals account for 95% of the total mass
of the filter cake. The solution that passes through the filter also
35% K2CrO4, is the recycle stream.
Calculate the rate of evaporation, the rate of production of
crystalline K2CrO4, the feed rates that the evaporator and the
crystallizer must be designed to handle and the recycle ratio
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Shoukat Choudhury
n44
Product Separation & Recycling
.
m 2 kg W/h
Fresh Feed
10,000 kg/h
0.30 kg K/kg
0.70 kg W/kg
.
.
m1 kg/h
MP
Evaporator
m3 kg/h
0.50 kg K/kg
0.50 kg W/kg
Recycle
Crystallizer
& Filter
Filter Cake
.
m4 kg/h
.
m 5 kg/h
0.35 kg K/kg
0.65 kg W/kg
.
m 6 kg/h
0.35 kg K/kg
0.65 kg W/kg
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Shoukat Choudhury
n45
Product Separation
& Recycling
m1 = 19662 kg mixed feed /h
nWithout
Recycle
m2 = 6900 kg Water evaporated/h
m2 = 4000 kg W/h
m3 = 12762 kg fed to crystallizer/h
m3 = 6000 kg K-soln/h
m4 = 2945 kg crystals/h
m4 = 1385 kg K(S)/h
m5 = 155 kg K-soln/h
m5 = 73 kg K-soln/h
m6 = 9662 kg recycle/h
m6 = 4542 kg K-soln/h
Recycle Ratio= 0.9662 Kg recycle/Kg
Fresh feed
Comment:
Crystal production increases 113% due
to recycling
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Shoukat Choudhury
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Recycle and Bypass
nWhat
is wrong here?
• Reuse of reactants/feed
• Recovery of catalyst
• Dilution of process streams (improve filter
operation)
• Control of a process variable (Reduce
reactant concentration)
• Circulation of a working fluid
(Refrigeration)
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Shoukat Choudhury
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Product Separation and Recycle
• Normally, reactions are not complete
– Separation and recycle
– Improved yield, conversion ,…
Reactants
Products
Product
Separation
Unit
Reactor
Recycle
üProcess feed
üOverall conversion
üFresh feed
üOnce-through conversion
üGross product
üNet product
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Shoukat Choudhury
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Balances on Reactive Systems
• Stoichiometry
The theory of proportions in which chemical
species combine with one another.
– Example: 2 SO2 + O2 à 2 SO3
• Stoichiometric Ratio
• Ratio of stoichiometric coefficients
– Example
2 mol SO3 produced
1 mol O2 reacted
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Shoukat Choudhury
2 mol SO2 reacted
2 mol SO3 produced
n49
Terminology
• Limiting reactants
Exist less than stoichiometric proportion
• Excess reactants
Exist more than stoichiometric proportion
• Example
2SO2 + O2 à 2 SO3
(30 mol) (10 mol)
Excess
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Limiting
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Terminology
n − ns
• Fractional excess n
s
• Percent excess n − ns
× 100
ns
• Example
– H2 + Br2 à HBr
– H2 : 25 mol /hr
– Br2 : 20 mol /hr
– Fractional Excess H2 = (25 – 20 ) /20 = 0.25
– Percent Excess = 25 %
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Terminology
• Fractional conversion
– Chemical reactions are not always completed.
– Factional conversion
• f = (moles reacted) / (moles fed)
– When fresh feed consists of more than one
material the conversion must be stated for a
single component, usually the limiting
reactant.
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Shoukat Choudhury
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Terminology
Overall Conversions =
reactant input to process - reactant output from process
reactant input to process
Single-Pass Conversions =
reactant input to reactor - reactant output from reactor
reactant input to reactor
• In general, high overall conversions can be
achieved in two ways:
– Design the reactor to yield a high singlepass conversion, or
– Design the reactor to yield a low singlepass conversion and follow it with a
separation unit to recover and recycle
unconsumed reactant.
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Problem
• Consider the reaction
6 NaClO3 + 6 H2SO4 + CH3OH → 6 ClO2 + 6 NaHSO4 + CO2
• if the reactor feed has the composition
(mol%) of 36% NaClO3, 54% H2SO4, and
the rest CH3OH, which is the limiting
reactant?
• Calculate the reactant flows required to
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Shoukat
Choudhury
produce 10 ton Prof.
per
hour
of
ClO2 assuming54
n
n
n
Balances of Atomic and
Molecular Species
•
Methods for solving mass balances with reactions
– Using balances on molecular species
– Using balances of atoms
– Using the extent of reaction
For multiple reactions, sometimes it is more convenient to use
atomic balances
• Atomic species balances generally lead to the most straightforward
solution procedure, especially when more than one reaction is
involved.
• Extents of 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.
•
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Shoukat Choudhury
n55
A simple Problem
The oxidation of ethylene to produce
ethylene oxide proceeds as: 2 C2H4 + O2
= 2 C2H4O
The feed to the reactor contains 100 Kmol
C2H4 and 200 Kmol O2. What is the
limiting reactant? What is the percentage
excess of the other reactant? If the
reaction proceeds to a point where the
fractional conversion of the limiting
reactant is 50%,Prof.how
much
of each
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n
n
n
Multiple Reaction, Yield,
Selectivity
• Multiple reaction : one or more reaction
– Side Reaction : undesired reaction
– Example: Production of ethylene
C2H6 à C2H4 + H2
Side Reactions
C2H6 + H2 à 2CH4
C2H4 + C2H6 à C3H6 + CH4
– Design Objective
• Maximize desired products (C2H4)
• Minimize undesired products (CH4, C3H6)
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Shoukat Choudhury
n57
Terminology
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
moles of desired product formed
moles of reactant fed
moles of desired product formed
nYield =
moles of reactant consumed
nYield
=
moles of desired product formed
SELECTIVITY =
moles of undesired product formed
– Yield and Selectivity are used to describe the degree to
which a desired reaction predominates over competing side
reactions.
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Shoukat Choudhury
n58
Determination of yield and
Reaction: selectivity
C2H6 + 2.5O2 → 2CO +
3H2O
C2H6 + 3.5O2 → 2CO2
+3H2O
Undesired product: CO
Products
REACTOR conversion
80%
reactor
100 mol C H
n
2
500 mol O2
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6
n20
mol C2H6
n120 mol CO2
n40 mol CO
n240 mol O2
n240 mol H2O
nProf.
Yield = 0.6 or 1.2 or 1.5
Selectivity = 3.0
Shoukat Choudhury
n59
Problem
Ethane is burned with air in a continuous
steady-state combustion reactor to yield a
mixture of carbon monoxide, carbon
dioxide, and water. The feed to the reactor
contains 10% C2H6. The percentage
conversion of ethane is 80%, and gas
leaving the reactor contains 8 mol CO2 per
mol of CO.
Determine molar composition
nReaction:
of productnCgas.
2H6 + 5/2 O2 → 2CO + 3H2O
nC2H6
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+ 7/2 O2 → 2CO2 +3H2O
nProf.
Shoukat Choudhury
n60
Concept
of
Purgewhy
needed?
• Production of Ethylene Oxide
•
•
•
Reaction: 2C2H4 + O2 → 2C2H4O
Mixture of Ethylene and air stream is charged to the reactor
Reactor effluent is charged to absorber and gas stream
containing N2, O2 and unreacted ethylene is charged back to
reactor
Recycle
Fresh Feed
Solvent
Reactor
Absorber
Products
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n61
Production of Ethylene Oxide
• Problem: accumulation of N2
• Solution: allow purging of inert species
Purge stream
Recycle
50 mol C2H4/s
10 mol C2H4/s
25 mol O2/s
5 mol O2/s
565 mol N2/s
113 mol N2/s
Solvent
Fresh Feed
60 mol C2H4/s
30 mol O2/s
113 mol N2/s
100 mol C2H4/s
Reactor
50 mol C2H4/s
50 mol C2H4O/s
25 mol O2/s
50 mol O2/s
565 mol N2/s
565 mol N2/s
Absorber
Products
50 mol C2H4O/s
solvent
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n62
Purging
Getting rid of undesired materials in recycle stream.
Reactants
Product
Separation
Unit
Reactor
Products
Recycle
Purging
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Definition
• A Recycle Stream is a term denoting a
process stream that returns material from
downstream of a process unit back to the
process unit.
• A Bypass stream - one that skips one or
more stages of the process and goes
directly to another downstream stage.
• A Purge stream – a stream bled off to
remove an accumulation of inert or
unwanted material that might otherwise
build up in the recycle
stream.
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Example 4.7-3, page 139
Recycle and Purge in the Synthesis of Methanol
Methanol is produced in the reaction of carbon dioxide and
hydrogen: CO2 + 3H2 → CH3OH + H2O
The fresh feed to the process contains hydrogen, carbon
dioxide and 0.40 mole percent inerts(I). The reactor effluent
passes to a condenser that removes essentially all of the
methanol and water formed and none of the reactants or
inerts. The latter substances are recycled to the reactor. To
avoid buildup of the inerts in the system, a purge stream is
withdrawn from recycle. The feed to the reactor contains
28.0 mole% CO2, 70 mole% H2 and 2% inerts. The single
pass conversion of H2 is 60%. Calculate the molar flow rates
and molar compositions of the fresh feed, the total feed to
the reactor, the recycle stream, and the purge stream for a
methanol production rate of 155 kmol methnol/hr.
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Reaction: CO2 + 3H2 → CH3OH + H2O
Basis: 155 kmol CH3OH/h
Recycle
Purge
N8 kmol/h
N7 kmol/h
x6C mol CO2/mol
x6C mol CO2/mol
x6H mol H2/mol
x6H mol H2/mol
(1-x6C-x6H) mol I/mol
(1-x6C-x6H) mol I/mol
N6 kmol/h
x6C mol CO2/mol
x6H mol H2/mol
(1-x6C-x6H)mol I/mol
n1 kmol/h
n2 kmol/h
x1C mol CO2/mol
0.28 mol CO2/mol
REACTOR
(0.996-x1C)mol H2/mol 0.70 mol H2/mol
0.02 mol I/mol
0.004 mol I/mol
CONDENSER
n3 kmol CO2/mol
n4 kmol H2/mol
n5 kmol I/mol
155 kmol CH3OH/h
155 kmol H2O/h
155 kmol CH3OH/h
155 kmol H2O/h
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Combustion Reaction
• Combustion
– A rapid reaction of a fuel with oxygen
– Fuels : coal, fuel oil, gas fuel, solid fuel, …
– Complete combustion / incomplete
combustion
– Wet basis composition / dry basis composition
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Remember:
Orsat analysis yields dry basis composition
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Terminology
• Theoretical oxygen : Amount of oxygen
needed for complete combustion
– all carbon in the fuel is oxidized to CO2 and
– all the hydrogen is oxidized to H2O
• Theoretical air : The quantity of air that
contains theoretical oxygen
Air(theo) = 4.76 x O2(theo)
• Excess air : The amount by which the
air fed to reactor exceeds the
(moles air)fed − (moles air)theroretical
×100%
theoretical air (moles air )theoretical
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• Percent excessProf.air
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Composition of Flue or stack gas
• Wet Basis = => Dry Basis
- Basis: 1 mole wet gas
- basic idea, subtract the water and express the rest in %
• Dry Basis = => Wet Basis
- Basis: 1 mole wet gas
- need one extra information: How much water is there in
one mole of wet gas? (Say, y mole fraction out of 1 mole
wet gas)
- subtract water from one mole of wet gas and get the
mole of dry gas. (dry gas=1-y)
- Now, dry gas fraction is (1-y)*yi, where yi is mole
fraction of ith dry gas components
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•
a)
b)
c)
d)
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A problem with purge and
recycle
The fresh feed to an ammonia production process contains
nitrogen and hydrogen in stoichiometric proportion, along with
2 mole% inert gas. The feed is combined with a recycle stream
containing the same three species and the combined stream is
fed to a reactor in which a single pass conversion of 20% is
achieved. The reactor effluent flows to a condenser. A liquid
stream containing essentially all of the ammonia formed in the
reactor and a gas stream containing all the inerts and the
unreacted nitrogen and hydrogen leave the condenser. 10% of
the gas stream leaving the condenser is removed as purge and
the rest constitutes the recycle stream.
Draw a complete flow chart of the process.
Completely label the flow-chart
Find the overall conversion of N2
Find the total feed flow rates to the reactor
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Questions
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Shoukat Choudhury
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Chemical Reaction
• What is final composition ?
– Chemical
ˆνi
K
=
∏
f
i
equilibrium
thermodynamics
K = ∏ aνi i
d ln K ΔH r
=
dT
RT
• How long it will
to reach
equilibrium ?
r = ktake
(T ) f (compositio
n)
i
=k T e
– Chemical kinetics
m − E / RT
o
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C AnCbl ...
Shoukat Choudhury
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Sources of equations of unknown process
variables:
1.
2.
3.
4.
5.
6.
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Material balances
An energy balance
Process specifications
Physical properties and laws
Physical constraints
Stoichiometric relations
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Variables in process design or analysis
–
–
–
–
–
Temperature
Pressure
Flow rate
Chemical composition
Physical properties
Physical properties
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–
–
–
–
–
–
Chemical composition
–
Specific gravity
–
Specific volume
–
Density
Specific heat
–
Enthalpy
Heat of reaction,nProf.
etc.Shoukat Choudhury
Mass fraction
Mole fraction
Mass and molar
composition
Concentration: mass
conc., molar conc.,
molality
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Topic Outcomes
• Make your conception clear about the following terms:
– Batch, semibatch, continuous, transient, and steady-state
processes
– Recycle, bypass and purge
– Degrees-of-Freedom
– Fractional conversion of a limiting reactant
– Percentage excess of a reactant
– Yield and selectivity
– Dry-basis composition of a mixture containing water
– Theoretical air and percent excess air in a combustion reaction
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Topic Outcomes
• Given a process description:
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– Draw and fully label a flowchart
– Choose a convenient basis of calculation
– For a multiple-unit process, identify the
subsystems for which balances might be
written
– Perform DoF analysis for the overall system
and each possible subsystems
– Write in order the equations you would use
to calculate specified process variables
– Perform the calculations
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Topic Outcomes
• Do these computations
– for single-unit and multiple-unit processes and
– for processes involving recycle, bypass, or
purge streams
• If the system involves reactions, you
should be able to use
– molecular species balances,
– atomic species balances, or
– extents of reaction for both the DoF analysis
calculations
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Shoukat Choudhury
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Topic Outcomes
• Given a combustion reactor and
information about the fuel composition
– calculate the feed rate of air from a given
percent excess or vice versa
• Given additional information about the
conversion of the fuel and the absence or
presence of CO in the product gas
– calculate the flow rate and composition of the
product gas
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