Transcript Chemical Reaction

KINGDOM OF SAUDI ARABIA
Ministry of Higher Education
KING ABDULAZIZ UNIVERSITY
Faculty of Engineering Rabigh Branch
‫كلية الهندسة برابغ‬
Introduction to Chemical Engineering “CHEN 201”
Lecture 1 “Introduction”
Mohamed Ismail Bassyouni, Ph.D.
“Head of the Chemical Engineering Department”
1
Sep. 2015
Introduction to Materials Science & Engineering
Course Objective...
Introduce fundamental concepts in Chemical engineering calculations
You will learn about:
• Chemical Processes
• Mass Balance
• Energy Balance
This course will help you to:
• Carry out chemical engineering calculations
• realize Chemical engineering processes
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LECTURES
Instructor: Dr. Mohamed Ismail Bassyouni
Tutor: Eng. Ahmed Almalki
E-mail : [email protected]
Emal: [email protected]
Time:
Monday & Wednesday: 14-1650 am
Location: R 3
Activities:
• Present new material
• Announce reading and homework
• Take quizzes and midterms*
3
TEACHING
Name
Office
Tel.
Dr Mohamed Bassyouni
306
X-XXXX
E-mail
[email protected]
Teaching Assistant will
-Participate in recitation sessions
-Chemical Engineering (Calculations)
,
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COURSE MATERIALS (with text)
Textbook:
Hammelblau D. M. and Riggs J. B., Basic Principles and Calculations in
Chemical Engineering, 7th Edition, John Wiley, 2004
Reference:
Felder M., Rousaou W, Elementary Principles of Chemical Processes 5th Edition,
John Wiley, 1986
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GRADING
1- Weekly in-lecture quizzes
20%
Based on core homework problems
Your lowest quiz grade will be dropped
2- Midterm
20%
3- Assignment and homework 20%
5- Final
40%
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7
A chemist in your company's research and development division has discovered that if he mixes
two reactants in a certain proportion at an elevated temperature. He obtains a product
significantly more valuable than both reactants. The company contemplates manufacturing the
product using a process based on this reaction. At this point the matter becomes an engineering
problem or. more precisely. hundreds of engineering problems.
A+B C
1. What type of reactor should be used? A long pipe? A large tank? Several smaller tanks? An
extremely large test tube? How large? Made of what? Does it have to be heated? If so. how much
and how? With an electrical heater inside or outside the reactor? By passing a hot fluid through a
heating coil in the reactor'? By heating the reactants before they get into the reactor? Does the
reaction supply its own heat, so that heating is needed only for startup? If so, can the reactor "run
away" and possibly explode? Should control measures be introduced to prevent this? What kind?
2. Where should the reactants be obtained? Buy them, or make them? In what proportions should
they be fed to the reactor?
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3. Should the reactor effluent, which contains the product and unconsumed reactants. be sold
as is, or should the product be separated from the reactants and the latter be sent back to the
reactor? If separation is desirable. how can it be accomplished? Heat the mixture and draw off
and condense the vapor, which will be richer in the more volatile substances than the original
mixture? Add another substance that extracts the product and is immiscible with the reactants,
and then separate the two phases mechanically? If all of the process materials are gases at the
reaction temperature, can the mixture be cooled to a temperature at which the product
condenses but the reactants do not, or vice versa, or if they are liquids can the mixture be
cooled to a temperature at which the product crystallizes? If one of these alternatives is chosen,
what kind of equipment is needed'? What size? What materials? What are the heating or
cooling requirements? Are controls needed to keep the operation of the process within rigid
limits? What kind of controls? Should they be manual or automatic?
4. How should the reactant and product streams be moved to and from the reactor and any
heating, cooling, and separation equipment involved in the process? By gravity from a raised feed
tank? With pumps, or blowers, or compressors. or conveyor belts? What kinds'? How big? In
pipes made of what?
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5. Is enough known about the reaction system to be able to answer all of these questions. Or
should additional laboratory studies be carried out'? What studies? Can the laboratory data be
used directly to design the industrial plant, or should a smaller pilot plant be constructed first to
test the design? How much smaller?
6. What can possibly go wrong with the process, and what can be done if and when it does?
7. Are waste products produced by the process? In what quantities? Are they potentially harmful
if released untreated into the environment? If so, in what way? What should be done to reduce
pollution hazards? Chemically treat the wastes? Dump liquid and solid wastes into containers,
seal and cart them out to sea? Disperse gases in the tmosphere with a high stack? Precipitate
solids electrostatically from gas exhausts?
8. How much of the process should be automated. and how should the automation be done'?
9. How much will all of this cost? For how much can the product be sold, and to whom'? How
much money will the process net each year? Is it enough to make it worthwhile? If so, where
should the plant be built?
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10. Once the plant has been built, what procedure should be followed for startup?
11. Six months later when startup has been achieved, why is the product not coming out the
way it did in the laboratory? Is it an equipment malfunction or a change in conditions
somewhere between the laboratory and industrial process? How can we find out? What can be
done to correct the problem? Is it necessary to shut down the operation for modifications?
12. Is it significant or just a coincidental series of bad breaks that there have been three
explosions and four fires within six months in the reactor unit? In either case, how do we stop
them from recurring?
13. All sorts of other things are going wrong with the process operation. Why weren't they on the
list of things that could possibly go wrong? What can be done about them?
14. When the process finally starts working perfectly and the next day an order comes down to
change the product specifications, how can it be done without redesigning the entire process?
Why didn't they think of this before they built the plant?
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KINGDOM OF SAUDI ARABIA
Ministry of Higher Education
KING ABDULAZIZ UNIVERSITY
Faculty of Engineering Rabigh Branch
‫كلية الهندسة برابغ‬
Introduction to Chemical Engineering “ChE 201”
Lecture 2 “Unites and Dimensions”
Assis.Prof. Mohamed Ismail Bassyouni
“Head of the Chemical Engineering Department”
14
Sep 2011
Dimensions & Units
Dimensions: are our basic concepts of measurement such as
-Length
-Time
-Mass
-Temp….etc.
Units: are the means of expressing the diemntions such as:
-Feet
-Centimeters
-hours
Seconds,…etc.
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Dimensions & Units
• All measured property has a value and a unit
• A dimension is a property that can be measured, or calculated by
multiplying or dividing other dimensions.
• Measurable units are specific values of dimensions that have been
defined by convention, custom, or law.
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• In engineering one will come across many dimensionless
groups.
• It is important to note that the numerical value for a
dimensionless group is independent of the units chosen for the
primary quantities, provided the units are consistent.
• Dimensional Homogeneity ---- every valid equation must be
"dimensionally homogeneous" --- all additive terms must have
the same dimension.
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Examples
Dimension
: mass,
Unit: kg
5 Kilograms + 3 joules
 Meaningless
10 pounds + 5 grams
 Can be performed after the units are transformed to the same
10 sec +20 sec =30 sec
 Can be performed directly
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Systems of units
• Base Units
• Multiple Units
• Derived Units
• International System of Units: (SI) system
• Centimetre-gram-second system: CGS System
• American Engineering System
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Base units
for mass, length, time, temperature, electrical current,
and light intensity.
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Base Units
Quantity
Unit
Symbol
Length
Meter(SI)
Centimeter(CGS)
m
cm
Mass
Kilogram (SI)
Gram(CGS)
kg
g
Moles
Gram-mole
Time
Second
s
Temperature
Kelvin
K
Electric Current
Ampere
A
Light Intensity
Candela
cd
mol or gmol
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Multiple Unites
Which are defined as multiples or fractions of
base units such as minutes, hours, and
milliseconds, all of which are defined in terms of
the base unit of a second.
Multiple units are defined for convenience rather
than necessity: it is simply more convenient
to refer to 3 yr than to 94,608,000 s.
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Multiple Units
•
•
•
•
•
•
•
•
Tera (T) = 1012
Giga (G) = 109
Mega (M) = 10 6
Kilo (k)
= 103
Centi (c) = 10-2
Milli (m) = 10-3
Micro (μ) = 10-6
Nano (n) =10-9
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Derived units
Otained in one of two ways:
(a) By multiplying and dividing base or multiple units
(cm2, ftlmin, kg' m/s2 , etc.). Derived
units of this type are referred to as compound units.
(b) As defined equivalents of compound units (e.g., 1
erg (lg'cm/s\ 1lbf " 32.174 Ibm ·ft/s2).
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International System of Units: (SI) system
Name
Unit symbol
Quantity
metre
m
length
kilogram
second
ampere
kg
s
A
kelvin
K
mass
time
electric current
thermodynamic
temperature
Typical Symbol for
Variables
l (a lowercase L)
m
t
I (a capital i)
T
candela
cd
luminous intensity
Iv (a capital i with
lowercase nonitalicized v
subscript)
mole
mol
amount of
substance
n
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Centimeter-gram-second system: CGS System
Quantity
Symbol CGS unit
CGS unit
Definition
abbreviation
length
position
L, x
centimetre
cm
mass
m
gram
g
time
t
velocity
v
second
centimetre per
second
acceleration
a
force
energy
power
pressure
dynamic
viscosity
wavenumber
Equivalent
in SI units
1/100 of metre
= 10−2 m
= 10−3 kg
s
1/1000 of
kilogram
1 second
cm/s
cm/s
= 10−2 m/s
gal
Gal
cm / s2
= 10−2 m/s2
F
E
P
p
dyne
erg
erg per second
barye
dyn
erg
erg/s
Ba
g cm / s2
g cm2 / s2
g cm2 / s3
g / (cm s2)
= 10−5 N
= 10−7 J
= 10−7 W
= 10−1 Pa
μ
poise
P
g / (cm s)
= 10−1 Pa·s
k
kayser
cm−1
cm−1
−1
= 100 m27
=1s
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Convert an acceleration of 1 cm/s2 to its equivalent in km/yr2.
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KINGDOM OF SAUDI ARABIA
Ministry of Higher Education
KING ABDULAZIZ UNIVERSITY
Faculty of Engineering Rabigh Branch
‫كلية الهندسة برابغ‬
Introduction to Chemical Engineering “ChE 201”
Lecture 3 “Important terms”
Assis.Prof. Mohamed Ismail Bassyouni
“Head of the Chemical Engineering Department”
31
Sep. 2011
Dimensional consistency
A basic principle exists that equations must be dimensionally consistent
Example :
Van der waals equation
a will have the unit: atm (cm)6
b will have the unit: cm3
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Non dimensional groups
(cm)
(cm)
(s)
)g(
(cm) (s)
(cm) 3
(g)
All units cancel out
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Important terms
a) Precision:
refers to the degree of dispersion (or deviations) of measurements from their true values
b)Accuracy:
refers to how close measures value is to its true value
c) Mole and mole fraction
d) Density and specific gravity
e)Flow rate
f) Temperature
g) Pressure
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c) Define a kilogram mole, pound mole and gram mole
If a bucket holds 2 lb of NaOH (mol. Wt= 40), how many
(a) Pounds moles of NaOH does it contain?
(b) Gram moles of NaOH does it contains? ( Hint : 1lb=454 g)
= 0.05 lb mole NaOH
= 22.7 g mole NaOH
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How many pounds of NaOH are in 7.5 g mole of NaOH
Number of g mole = mass (g) / Molecular weight
7.5= mass (g) / 40
Mass (g)= 7.5 *40
Mass lb= 7.5 *40?454 = 0.661 lb
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Mass and Mole Fraction
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d) Density
Density
-The density of a substance is the mass per unit volume of the substance (kg/m3 , g/cm3, Ibm/ft3,
etc.).
-Densities of pure solids and liquids are essentially independent of pressure and vary relatively
slightly with temperature
specific gravity
The specific gravity of a substance is the ratio of the density p of the substance to the density
Pref of a reference substance at a specific condition:
SG = ρ/ ρref
Specific volume
The specific volume of a substance is the volume occupied by a unit mass of the substance; it is
the inverse of density
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e) Flow rate
Most processes involve the movement of material from one point to another-sometimes
between process units, sometimes between a production facility and a transportation
depot. Therate at which a material is transported through a process line is the flow rate
of that material.
The flow rate of a process stream may be expressed as a mass flow rate (mass/time)
or as a volumetric flow rate (volume/time).
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f) Temperature
Temperature and pressure do not have large influences on the densities of solids and
liquids. Nevertheless, the fact that mercury in a thermometer rises or falls with changing
temperature shows that the effect of temperature on liquid density is measurable.
the dependence of the volume of mercury on temperature as
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The process variables
The quantities used to describe a process are called process variables.
To design or analyze a process, we need to know the amounts, compositions, and
condition of materials entering, leaving, and within the process
The process variables of interest to chemical engineers are:
-Mass and Volume
-Composition
-Pressure
-Temperature
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1- Chemical composition
• Most materials are mixtures of various species.
• The physical properties of a mixture depend strongly on the mixture composition.
1.1 Moles and Molecular Weight
The atomic weight :
of an element is the mass of an atom on a scale that assigns 12C (the
isotope of carbon whose nucleus contains six protons and six neutrons) a mass of
exactly 12.
The molecular weight
of a compound is the sum of the atomic weights of the atoms that
constitute a molecule of the compound:
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Atomic weight of O = 16
Molecular weight of O2 = 32
Example:
Number of moles of 34 kg of ammonia (NH3 : M = 17)
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How many of each of the following are contained in 100.0 g of
CO2 (M = 44.01, atomic weight of Carbon = 12, Oxygen = 16)?
(1) mol CO2 ;
(2) lb-moles CO2 ; ( Hint : 1lb=454 g)
(3) mol C;
(4) mol O;
(5) mol O2;
(6) g O;
(7) g O2;
(8) molecules of CO2 .
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1.2 Mass and Mole Fractions and Average Molecular Weight
• Process streams occasionally contain one substance, but more often they consist of
mixtures of liquids or gases, or solutions of one or more solutes in a liquid solvent.
•Composition of a mixture of substances, including a species A.
•The percent by mass of A is 100xA,
• and the mole percent of A is 100yA
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A solution contains 15% A by mass (XA = 0.15) and 20 mole% B (YB = 0.20).
1. Calculate the mass of A in 175 kg of the solution.
2. Calculate the mass flow rate of A in a stream of solution flowing at a rate of 53
Ibm/h.
3. Calculate the molar flow rate of B in a stream flowing at a rate of 1000 mol/min.
4. Calculate the total solution flow rate that corresponds to a molar flow rate of 28
kmol B/s.
5. Calculate the mass of the solution that contains 300 Ibm of A.
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A mixture of gases has the following composition by mass:
Element
mass fraction
O2
16%
CO
4.0%
CO2
17%
N2
63%
What is the molar composition?
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The average molecular weight (or mean molecular weight) of a mixture
:
is the ratio of the mass of a sample of the mixture (mt) to the number of moles of all species (nt)
in the sample.
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Calculate the average molecular weight of air:
(1) from its approximate molar composition of 79% N2, 21% O2
(2) from its approximate composition by mass of 76.7% N2, 23.3% O2·
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1.3 Concentration
1-The Molarity of a solution
is the value of the molar concentration of the solute expressed in gram-moles solute/liter
solution (e.g., a 2-molar solution of A contains 2 mol A/liter solution).
2-Parts per Million and Parts per Billion
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2- Pressure
2.1 Fluid Pressure and Hydrostatic Head
• Pressure
is the ratio of a force to the area on which the force acts.
• Pressure units
are force units divided by area units (e.g., N/m2, dynes/cm2 , and Ibf/in2 or psi). The SI
pressure unit, N/m2, is called a Pascal (Pa).
• Hydrostatic pressure
• Head of fluid
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The fluid pressures referred to so far are all absolute pressures,
Many pressure-measuring devices give the gauge pressure
Psia : Pound /square inch absolute
Psig: Pound /square inch gauge
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The manometers was used to measure the pressure at
point 1 and 2 along a section of a piping. Please
determine the value of P1 and P2 in Psia.
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3- Temperature
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KINGDOM OF SAUDI ARABIA
Ministry of Higher Education
KING ABDULAZIZ UNIVERSITY
Faculty of Engineering Rabigh Branch
‫كلية الهندسة برابغ‬
Introduction to Chemical Engineering “Ch.E 201”
Lecture 5 “Material Balance”
Assis.Prof. Mohamed Ismail Bassyouni
“Head of the Chemical Engineering Department”
64
Oct. 2011
Law of conservation
The mass can neither be created nor destroyed
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Process Classification
1. Batch Process
The feed is charged (fed) into a vessel at the beginning
of the process and the vessel contents are removed
sometime later. No mass crosses the system boundaries
between the time the feed is charged and the time the
product is removed.
2. Continuous Process
The inputs and outputs flow continuously throughout
the duration of the process.
3. Semibatch process
Any process that is neither batch nor continuous
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Process Classification
Classification
Batch Process
Continuous Process
Semi batch Process
Example
Rapidly add reactants to a tank and
remove the products and unconsumed
reactants sometime later when the
system has come to equilibrium.
Pump a mixture of liquids into a
distillation column at a constant rate
and steadily withdraw product streams
from the top and bottom of the
column.
Allow the contents of a pressurized gas
container to escape to the atmosphere;
slowly blend several liquids in a tank
from which nothing is being withdrawn.
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Steady State and Unsteady State
(Transient Process)
i. Steady State Processes
• If the values of all the variables in a process (i.e., all
temperatures, pressures, volumes, flow rates) do not change
with time, except possibly for minor fluctuations about
constant mean values, the process is said to be operating at
steady state.
ii. Unsteady State (Transient Process)
• If any of the process variables change with time, transient or
unsteady-state operation is said to exist. Continuous
processes are usually run as close to steady state as possible.
• Unsteady-state (transient) conditions exist during the start-up
of a process and following changes-intentional or otherwise-in
process operation conditions
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Steady State and Unsteady State
(Transient Process)
•Batch Process
•Semibatch process
•Continuous Process


•Unsteady state
Unsteady state
or
Steady state
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Steady State and Transient Process
Example
1. A balloon filled with air at a steady rate of
2g/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.
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 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.
Results
1.
2.
3.
4.
Semibatch, Unsteady state
Batch, Unsteady state
Semibatch, Unsteady state
(a) Continuous, non-steady state
(b) Continuous , Steady State
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Steady State and Transient Process
Example
1. A balloon filled with air at a steady rate of
2g/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.
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 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.
Results
1.
2.
3.
4.
Semibatch
Batch
Semibatch
(a) Continuous, non-steady state
(b) Continuous , Steady State
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Open and Closed Systems
For Systems:
We must define the boundary of the system
Systems are of two types : closed and open
Closed system :
material is not crossing the boundary
Open system:
material is crossing the boundary
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Examples
D
F
Reaction
Water tank
Closed ?
Or open ?
Distillation
Column
Water
Tank
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The General Balance Equation
moin (CH4 Kg/hr)
Process Unit
moout (CH4 Kg/hr)
Suppose methane is a component of both the input and output
streams of a continuous process unit, and that in an effort to
determine whether the unit is performing as designed, the mass
flow rates of methane in both streams are measured and found to
be different
There are several possible explanations for the observed
difference between the measured flow rates:
1. Methane is being consumed as a reactant or generated as a
product within the unit.
2. Methane is accumulating in the unit-possibly adsorbing on the
walls.
3. Methane is leaking from the unit.
4. The measurements are wrong.
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Balance on Conserved Quantities
Differential balances, or balances that indicate what is happening in a system
at an instant in time. Each term of the 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, barrels/day). This is the type of balance usually
applied to a continuous process.
Integral balances, or balances that describe what happens between two
instants of time. Each term of the equation is an amount of the balanced
quantity and has the corresponding unit (people, g SO2, barrels). This type of
balance is 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.
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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
Let P denote people:
input + generation - output - consumption = accumulation
Each year the city's population decreases by 22,000 people.
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Material Balance Rules
The following rules may be used to simplify the material
balance equation:
1. If the balanced quantity is total mass, set generation = 0
and consumption = 0. Except in nuclear reactions, mass
can neither be created nor destroyed.
1. If the balanced substance is a nonreactive species
(neither a reactant nor a product), set generation = 0
and consumption = 0.
1. If a system is at steady state, set accumulation = 0,
regardless of what is being balanced. By definition, in a
steady-state system nothing can change with time,
including the amount of the balanced quantity.
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Balances on Continuous (Steady-State), and
Batch (Integral balance) Processes
For continuous processes at steady-state, the
accumulation term in the general balance equation
equals zero, and the equation simplifies to
Input + Generation = Output + Consumption
For integral balance on batch process
Accumulation = final output – initial input
= generation – consumption
Equating these two expression for the accumulation
field
Input + Generation = Output + Consumption
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Developing Flowchart to Solve the Problem
The catalytic dehydrogenation of propane is carried out in a
continuous packed bed reactor. One thousand kilograms per
hour of pure propane is preheated to a temperature of670oC
before it passes into the reactor. The reactor effluent gas, which
includes propane, propylene, methane, and hydrogen, is cooled
from 800°C to 110oC and fed to an absorption tower, where the
propane and propylene are dissolved in oil. The oil then goes to
a stripping tower in which it is heated, releasing the dissolved
gases; these gases are recompressed and sent to a distillation
column in which the propane and propylene are separated. The
propane stream is recycled back to join the feed to the reactor
preheater. The product stream from the distillation column
contains 98% propylene, and the recycle stream is 97% propane.
The stripped oil is recycled to the absorption tower.
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50 moles of C3H8
750 moles of O2
150 moles CO2
3760 moles N2
200 mole H2O
100 mol C3H8
Combustion
Chamber
Condensor
50 mol C3H8
750 mol O2
3760 mol N2
150 mol CO2
1000 mol O2
3760 mol N2
200 mol of H2O
Flowchart of Combustion-Condensation Process
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Flowchart scaling and basis of calculation
The procedure of changing the values of all stream amounts or
flow rates by a proportional amount while leaving the stream
compositions unchanged is referred to as scaling the flowchartscaling up if the final stream quantities are larger than the
original quantities, scaling down if they are smaller.
Example:
Suppose a kilogram of benzene is mixed with a kilogram of
toluene. The output from this simple process is obviously 2 kg of
a mixture that is 50% benzene by mass.
1 Kg C6H6
1 Kg C7H8
2 Kg
0.5 Kg C6H6/Kg
0.5 Kg C7H8/Kg
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1 Kg C6H6
Scaling Process
1 Kg C7H8
x 300
2 Kg
0.5 Kg C6H6/Kg
0.5 Kg C7H8/Kg
300 Kg C6H6
300 Kg C7H8
Kg to Kg/hr
600 Kg
0.5 Kg C6H6/Kg
0.5 Kg C7H8/Kg
Replace Kg with lbm
300 lbm/h C6H6
300 lbm/h C7H8
600 lbm/h
0.5 lbm C6H6/ lbm
0.5 lbm C7H8/ lbm
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Basis of Calculation
• A basis of calculation is an amount (mass or moles) or
flow rate (mass or molar) of one stream or stream
component in a process. The first step in balancing a
process is to choose a basis of calculation; all unknown
variables are then determined to be consistent with
this basis.
• If a stream amount or flow rate is given in a problem
statement, it is usually most convenient to use this
quantity as a basis of calculation.
• If no stream amounts or flow rates are known, assume
one, preferably that of a stream with a known
composition. If mass fractions are known, choose a
total mass or mass flow rate of that stream (e.g. 100 kg
or 100 kg/h) as a basis; if mole fractions are known,
choose a total number of moles or a molar flow rate.
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Balancing a 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.
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Degree-of-Freedom Analysis
• Before you do any lengthy calculations, you can use a
properly drawn and labeled flowchart to determine
whether you have enough information to solve a given
problem. The procedure for doing so is referred as
Degree-of-Freedom Analysis
• To perform a degree-of-freedom analysis, draw and
completely label a flowchart, count the unknown
variables on the chart, then count the independent
equations relating them, andsubtract the second
number from the first. The result is the number of
degrees of freedom of the process.
ndf = nunknowns – nindependent eqns
• There are three possibilities:
87
Degree-of-Freedom Analysis
1. If ndf = 0, there are n independent equations in n
unknowns and the problem can in principle be
solved.
2. If ndf > 0, there are more unknowns than
independent equations relating them, and at least
ndf additional variable values must be specified
before the remaining variable values can be
determined.
3. If ndf < 0, there are more independent equations
than unknowns. Either the flowchart is
incompletely labeled or the problem is
overspecified with redundant and possibly
inconsistent relations.
88
Degree-of-Freedom Analysis
Sources of equations relating unknown process stream
variables include the following
1.
2.
3.
4.
5.
6.
Material balances
An Energy Balance
Process Specifications
Physical properties and laws
Physical Constraints
Stoichiometric relations
89
General Procedure for Single-Unit Process
Material Balance Calculations
1. Choose as a basis of calculation an amount or flow
rate of one of the process streams.
2. Draw flowchart and fill in all known variable 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 or the other.
5. Do the degree-of-freedom analysis.
90
General Procedure for Single-Unit Process
Material Balance Calculations
6. If the number of unknowns equals the number of
equations relating them (i.e., if the system has zero
degrees 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.
91
Recycle and Bypass
There are several reasons for using recycle in a chemical
process
1.Recovery and reusing unconsumed reactants
2.Recovery of catalyst
3.Dilution of a process stream
4.Control of process variable
5.Circulation of a working fluid
A procedure that has several features in common with
recycle is bypass, in which a fraction of the feed to a
process unit is diverted around the unit and combined
with the output stream from the unit. By varying the
fraction of the feed that is bypassed, we can vary the
composition and properties of the product.
92
Examples of Recycle and Bypass
feed
PROCESS
Product
Bypass
93
Chemical Reaction Stoichiometry
• Stoichiometry is the theory of the proportions in
which chemical species combine with one another.
The stoichiometric equation of a chemical reaction is
a statement of the relative number of molecules or
moles of reactants and products that participate in
the reaction.
2SO2 + O2  2SO3
• For every two molecules (g-moles, Ib-moles) of S02
that react, one molecule (g-mole, lb-mole) of O2
reacts to produce two molecules (g-moles, lb-moles)
of S03. The numbers that precede the formulas for
each species are the stoichiometric coefficients of the
94
reaction components.
Chemical Reaction Stoichiometry
• A valid stoichiometric equation must be balanced; that
is, the number of atoms of each atomic species must
be the same on both sides of the equation, since atoms
can neither be created nor destroyed in chemical
reactions
• The stoichiometric ratio of two molecular species
participating in a reaction is the ratio of their
stoichiometric coefficients in the balanced reaction
equation. This ratio can be used as a 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 that
participated in the reaction
95
Limiting and Excess Reactants
• Two reactants, A and B, are said to be present in
stoichiometric proportion if the ratio (moles A
present)/(mole B present) equals the stoichiometric
ratio obtained from the balanced reaction
• The reactant that would run out if a reaction
proceeded to completion is called the limiting
reactant, and the other reactants are termed excess
reactants. A reactant is limiting if it is present in less
than its stoichiometric proportion relative to every
other reactant.
96
Chemical Reaction
The Limiting reactant
In chemical reaction one of the reactant will consume
first so it is called the limiting reagent
Other reactants remain in excess
The amount of product determined by the amount of
limiting reagent
Chemical Reaction
Determination of Limiting Reactant
1- Calculate the number of moles of each reactant (if
not given)
2- Divide that moles to there factors in chemical
reaction (ratio)
3- Compare the ratios, the smallest is the limiting
reagent
Chemical Reaction
Determination of Limiting Reactant
Example
Determine the limiting reactant for the following
reaction if 400 g Fe mixed with 300 g O2.
4 Fe (s) + 3 O2 (g)  2 Fe2O3 (s)
400/56 = 7.14
300/32 = 9.375
Stoichiometric amount of oxygen = 7.14*3/4= 5.25 moles < 9.375
So Fe is limiting reactant.
Chemical Reaction
The Reaction Yield
Theoretical Yield (T.Y.): the amount of product
produced when all limiting reagent reacted, or it is the
yield calculated from equation
Actual Yield (A.Y.): the amount of product obtained
actually
Due to many factors affecting on reaction, A.Y. is
always less than T.Y. so we can calculate the % yield
A.Y.
% Yield =
*100
T.Y.
Chemical Reaction
The Reaction Yield
Example
Calculate the % yield of urea, if 119 g of ammonia
reacts with 80 g of CO2 producing 100 g of urea
according to the following equation
2 NH3 (g) + CO2 (g)  CO(NH2 )2 (s) + H2O
% yield = actual/ theoritical = 100g/109.9
= 90.99%
Combustion Analysis
Determination of Empirical Formula
ascorbic acid contains 40.92% C and 4.58% H on
combustion (note that % O = 100 – 40.92 – 4.58 =
54.5 %)
The empirical formula can be determined as follows
MASS
MOLE
MOLE RATIO
C
1- Write the Mass % (or
Mass g)
H
40.92
4.58
2- Calculate No. of moles 40.92/12
3- Divide by smallest No.
4- Multiply by any No. (19) that makes all step 3
Nos are integer Nos
4.58/1
O
100-(40.92+4.58)
54.5/16
3.41/3.406 4.58/3.406 3.406/3.406
1.001
5- The No. of atoms are
6- The empirical formula is C3H4O3
1.3446
3
4
1
3
(*3)
Determine its empirical formula for 0.15 mole H , 0.05 mole C, and
0.025 mole O
3- Divide by smallest No.
4- Multiply by any No. (19) that makes all step 3
Nos are integer Nos
0.15/0.025 0.05/0.025 0.025/0.025
6
5- The No. of atoms are
6- The empirical formula is C2H6O
2
1
Combustion Analysis
Determination of Molecular formula
Relation between molecular formula and empirical
formula is
Molecular weight of unknown (g/mol)
Molecular Formula =
* Empirical formula
Molecular weight of Empirical formula (g/mol)
Combustion Analysis
Determination of Molecular formula
Example
What is the molecular formula of ascorbic acid if its
molecular weight is 176 g/mol
Molecular Formula =
176 (g/mol)
* C3H 4O3 = 2 * C3H 4O3 = C6 H8O6
(3*12  4*1  3*16)
Chemical Reaction
It is a process in which one or more substances
converted to other substances
This change involve a change in energy
Only change are happened no matter or energy
created or destroyed
Reactants  Product
The physical conditions can be written in the chemical
equation
(s) for Solid (l)
for Liquid (g) for Gas
(aq) for Aqueous solution
(ppt) for precipitation
Chemical Reaction
The chemical equation must be balanced
Balancing Chemical Equation
e. g. balance the following equation
C6 H12O6 + O2  CO2 + H2O
H.W. What is the coefficient of H2O when the equation
is balanced:
_ Al4C3 + _ H2O  _ Al(OH)3 + 3CH4
a. 13
b. 4
c. 6
D 12
H.W. What are the coefficients of Al4C3 ,H2O and
Al(OH)3, respectively, when the equation is balanced:
_ Al4C3 + _ H2O  _ Al(OH)3 + 3CH4
a. 4,1,5
B 1,12,4
a. 1,24, 4
b. 4,12,1
109
Chemical Reaction
Stoichiometry of Reaction
The following equation can be read as follows
C6 H12O6 + 6 O2  6 CO2 + 6 H2O
1 molecule of sugar + 6 molecules of O2 gas gives 6
molecules of CO2 and 6 molecules of water
1 mole of sugar + 6 moles of O2 gas gives 6 moles of
CO2 and 6 moles of water
180 g of sugar + 192 g of O2 gas gives 264 g of CO2
and 108 g of water
Chemical Reaction
Stoichiometry of Reaction
Not as Follows
1 g of sugar + 6 g of O2 gas gives 6 g of CO2 and 6 g
of water
So this numbers means the number of moles or
number of molecules not the number of grams,
if you want the number of grams you must
convert the number of moles to grams using the
molar mass (Mw or Aw) and vice versa
Chemical Reaction
The Amount of Reactants and Product
The amount of reactants or products can be
calculated from balanced chemical equation
The excess of one of reactants can’t do anything for
the amount of product produced
Example
How many grams of water produced when 7.00 g of
oxygen react with excess of hydrogen?
2H2 (g) + O2 (g)  2 H2O(g)
Chemical Reaction
2H2 (g) + O2 (g)  2 H2O(g)
Chemical Reaction
Examples
1-Calculate the number of moles of CO2 resulted from the
reaction of 4 moles of C2H6 with excess oxygen according to
the equation
2 C2 H6 + 7 O2  4 CO2 + 6 H2O
Chemical Reaction
Examples
2-Calculate the mass of chlorine that reacts with 1.220 g of
hydrogen to form hydrogen chloride according the following
equation:
H2 + Cl2  2 HCl
116
KINGDOM OF SAUDI ARABIA
Ministry of Higher Education
KING ABDULAZIZ UNIVERSITY
Faculty of Engineering Rabigh Branch
‫كلية الهندسة برابغ‬
Introduction to Chemical Engineering “Ch.E 201”
Lecture 5 Chemical Equilibria (Stoichiometry)
Assis.Prof. Mohamed Ismail Bassyouni
Syed Waheed ul Hasan
117
Dec. 2011
Chemical Equilibrium
Given a set of reactive species and reaction conditions:
(a) The field of chemical equilibrium thermodynamics
concerns itself with what will be the final (equilibrium)
composition of the reaction mixture
(b) Chemical kinetics deals with how long will the system
take to reach a specified state short of equilibrium?
Irreversible Reaction: The reaction that proceeds only in a
single direction (from reactants to products) and the
concentration of the limiting reactant eventually approaches
zero, is called an irreversible reaction.
118
Chemical Equilibrium
Reversible Reaction:
In a reversible reaction, reactants form products and
products undergo the reverse reactions to reform the
reactants. For example, consider the reaction in which
ethylene is hydrolyzed to ethanol:
C2H4 + H2O
C2H5OH
119
Calculation of an Equilibrium Composition
If the water-gas shift reaction:
CO(g)+H2O(g)
CO2(g)+H2(g)
proceeds to equilibrium at a temperature T(K), the mole
fractions of the four reactive species satisfy the relation:
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 of H2O, and no CO2 or H2, and the reaction
mixture comes to equilibrium at 1105 K. Calculate the
equilibrium composition and the fractional conversion of
the limiting reactant.
120
Solution
The strategy is to express all mole fractions in terms of a
single variable (ξe, the extent of reaction at equilibrium),
substitute in the equilibrium relation, solve for ξe, and
back-substitute to calculate the mole fractions and any
other desired quantity.
From the definition of extent of reaction, we can write
Number of gram mole of CO present at equilibrium:
nCO = 1mol – ξe
Similarly
nH2O = 2mol - ξe
nCO2 = ξe
nH2 = ξe
121
Solution
ntotal = 3 mol
From which the composition of the reaction mixture can
be given by:
yCO = (1mol – ξe) / 3 mol
Similarly
yH2O = (2mol - ξe ) / 3 mol
yCO2 = (ξe ) / 3 mol
yH2 = (ξe ) / 3 mol
Substitution of these expressions into the equilibrium
relation (with K = 1.00) yields
122
Solution
This equation may be rewritten as a standard quadratic
equation
ξe2= 2 - 3ξe + ξe2
2 - 3 ξe = 0
ξe = 2/3
ξe = 0.667 mol.
This quantity may in turn be substituted back into the
expression for yi to yield
yCO = 0.111
yH2O = 0.444
yCO2 = 0.222
yH2 = 0.222
123
Solution
Finding the limiting reactant:
Mole of CO provided = 1
Moles of CO/ stoichiometric coefficient = 1/1
Mole of H2O provided = 2
Moles of H2O/ stoichiometric coefficient = 2/1 = 2
Since the moles fed over stoichiometric coefficient of CO
is smaller than water, its our limiting reactant.
124
Solution
At equilibrium
nCO = 1mol – ξe
nCO = 1- 0.667 mol = 0.333 mol
The fractional conversion of CO at equilibrium:
fco = (1-0.333) mol CO reacted / (I mol CO fed)
fco = 0.667
125
Multiple Reactions
Reactants can usually combine in more than one way, and
the product once formed may react to yield something
less desirable.
This Truth results in
(a) Economic loss
(b) A greater quantity of raw materials must be fed to the
reactor to obtain a specified product yield.
126
Multiple Reactions
For
example,
ethylene
can
be
produced
by
the
dehydrogenation of ethane:
C2H6
C2H4+H2
Once some hydrogen is produced, it can react with ethane to
produce methane
C2H6+H2
2CH4
Moreover, ethylene can react with ethane to form propylene
and methane:
C2H4+C2H6
C3H6+CH4
127
Multiple Reactions
Since the object of the process is to produce ethylene,
only the first of these reactions may be regarded as
desirable; the second one consumes the reactant without
yielding the desired product and the third consumes both
the reactant and the desired product.
The engineer must consider not only how to maximize
the production of the desired product (C2H4), but also
how to minimize the production of undesired by-products
(CH4, C3H6)
128
Yield and Selectivity
The terms yield and selectivity are used to describe the
degree to which a desired reaction predominates over
competing side reactions.
129
Extent of Reactions
If a set of reactions takes place in a batch or continuous
steady-state reactor and vij is the stoichiometric
coefficient of substance i in reaction j (negative for
reactants, positive for products), we may then write
ni = nio + ∑vijξj
For a single reaction it reduces to
ni = nio + viξ
130
Extent of Reaction (example)
For example, consider the pair of reactions in which
ethylene is oxidized either to ethylene oxide (desired) or
to carbon dioxide (undesired):
C2H4+1/2 O2
C2H4+3 O2
C2H4O
2CO2 + 2H2O
The moles (or molar flow rates) of each of the five species
involved in these reactions can be expressed in terms of
the feed values and extents of reaction:
131
Extent of Reaction (example)
values of any two outlet amounts are given, the values of
ξ1 and ξ2 may be determined from the corresponding two
equations, and the remaining amounts may in turn be
calculated from the remaining three equations.
132
YIELD AND SELECTIVITY IN A DEHYDROGENATION REACTIOR
The reaction:
C2H6
C2H4 + H2
C2H6+ H2
2CH4
take place in a continuous reactor at steady state. The
feed contains 85.0 mole% ethane (C2H6) and the balance
inerts (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 the selectivity
of ethylene to methane production.
133
Solution
Basis: 100 mol of feed
134
YIELD AND SELECTIVITY IN A DEHYDROGENATION REACTIOR
Using the equation,
ni = nio + ∑vijξj
the outlet composition amounts in terms of extents of
reaction are as follows:
135
YIELD AND SELECTIVITY IN A DEHYDROGENATION REACTIOR
Etahne Conversion:
If the fractional conversion of ethane is 0.501, the fraction
unconverted (and hence leaving the reactor) must be (1-0.501)
= 42.2 mol C2H6 = 85 mol C2H6 – ξ1 - ξ2------(1)
Ethylene Yield:
136
YIELD AND SELECTIVITY IN A DEHYDROGENATION REACTIOR
n2 = 0.471 (85 mol C2H6) = 40 mol C2H4 = ξ1
Substituting 40 mol for ξ1 in equation 1, we get ξ2 as 2.6 mol.
Then
Once we know the number of moles, the percentage
composition of each species can be given as following:
137
YIELD AND SELECTIVITY IN A DEHYDROGENATION REACTIOR
Product
C2H6
C2H4
H2
CH4
I
Composition (%)
30.3
28.6
26.7
3.7
10.7
SELECTIVITY:
= (40 mol C2H4)/ (5.2 mol CH4)
= 7.7 mol C2H4/mol CH4
138
BALANCE ON REACTIVE PROCESSES
The reaction of dehydrogenation of ethane
C2H6
C2H4 + H2
take place in a continuous reactor at steady state. The
flowchart for the process is given as follows.
One hundred kmol/min of ethane is fed to the reactor.
The molar flow rate of H2 in the product stream is 40
kmol/min.
139
BALANCE ON REACTIVE PROCESSES
A number of different balances could be written on this
process.
1. balances on total mass,
2. C2H6
3. C2H4
4. H2
• The first has the simple form input = output
• The species balance must include a generation term (for
C2H4 and H2) or a consumption term (for C2H6).
• Notice, however, that balance equations may also be
written for atomic carbon and atomic hydrogen,
regardless of the molecular species in which the carbon
140
and hydrogen atoms happen to be found
BALANCE ON REACTIVE PROCESSES
• Balances on atomic species can be written
input=output, since atoms can neither be created
(generation = 0) nor destroyed (consumption = 0) in a
chemical reaction
• Molecular hydrogen balance is different from the atomic
hydrogen balance
141
BALANCE ON REACTIVE PROCESSES
142
BALANCE ON REACTIVE PROCESSES
From the atomic balance:
n.1+ n.2 = 100 --------- (1)
(Atomic C balance)
6n.1 + 4n.2 = 520 ------ (2) (Atomic H balance)
Multiplying (1) with 6 and subtracting (2) from the
resultant
n.2 = 40 kmol/min and by putting this in (1) we get
n.1= 60 kmol/min
143
BALANCE ON REACTIVE PROCESSES
In general, systems that involve chemical reactions may be
analyzed using
(a) molecular species balances (the approach always used for
nonreactive systems)
(b) atomic species balances
(c) extents of reaction
Each approach leads to the same results, but anyone of them
may be more convenient for a given calculation so
144
Bonus Question (+1)
In general, systems that involve chemical reactions may be
analyzed using
(a) molecular species balances (the approach always used for
nonreactive systems)
(b) atomic species balances
(c) extents of reaction
Each approach leads to the same results, but anyone of them
may be more convenient for a given calculation so
145
KINGDOM OF SAUDI ARABIA
Ministry of Higher Education
KING ABDULAZIZ UNIVERSITY
Faculty of Engineering Rabigh Branch
‫كلية الهندسة برابغ‬
Introduction to Chemical Engineering “Ch.E 201”
Lecture 6 Combustion
Assis.Prof. Mohamed Ismail Bassyouni
Syed Waheed ul Hasan
146
Dec. 2011
Combustion
Combustion:
The rapid reaction of a fuel with oxygen
Important Fuel:
Petroleum products, Coal, Gases like methane, LPG
Partial Combustion:
A combustion reaction in which CO is formed from a
hydrocarbon is referred to as partial combustion or
incomplete combustion of the hydrocarbon.
147
Composition of Stack Gases
Composition on wet Basis
The term composition on a wet basis is commonly used
to denote the component mole fractions of a gas that
contains water
Composition on dry Basis
Composition on a dry basis signifies the component mole
fractions of the same gas without the water
Gas analyzer for stack gases
The product gas that leaves a combustion furnace is
referred to as the stack gas or flue gas. Common
techniques for analyzing stack gases provide
compositions on a dry basis.
148
Flue or stack gas analyzer
149
Conversion from wet to dry basis
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.
Solution:
Basis: 100 mol wet gas
Moles of N2 = 60
Moles of CO2 = 15
Moles of O2 = 10
Total moles without considering water = 85 mol dry gas
Composition of dry gas:
fraction of nitrogen = 60 / 85 = 0.706
Fraction of carbon dioxide = 16 / 85 = 0.176
150
Fraction of oxygen = 10 / 85 = 0.118
Conversion from dry to wet basis
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 stack gas
composition on a wet basis.
151
SOLUTION
Basis: 100 lb mol of dry gas
Hence the gas in the assumed basis contains:
152
Solution
153
Theoretical and Excess Reactant
One hundred mol/h of butane (C4H10 ) and 5000 mol/h of
air are fed into a combustion reactor.
Calculate the percent excess air.
Solution:
First, calculate the theoretical air from the feed rate of
fuel and the stoichiometric equation for complete
combustion of butane
154
Solution
155
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.
156
SOLUTION
Basis: 100 mol of C2H6 fed
157
SOLUTION
158
SOLUTION
159
SOLUTION
160
SOLUTION
161
SOLUTION
162