introducing chemical engineering
TRANSCRIPT
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 1/80
Chapter 1
INTRODUCTION
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 2/80
1
INTRODUCTION
1. Introducing Chemical Engineering
Chemical Engineering covers a wide scope that is difficult to put in one definition.
However, a good definition is that chemical engineering is concerned with the design,
construction, and operation of chemical process plants.
In general, chemical plants perform the following :
Basically, there are 3 distinctive functions for a chemical plant.
- Crude Oil - Distillation of Steam reforming
dehydration crude oil of N.G → H2, CO
desalting
- Dehydration and - Separation of
The operation in the
chemical plant may involve
physical steps (separation)
and/or chemical steps (reactions)
Products (o u t p u t)
RawMaterials
Ener
( i n u t )
Purification of
raw materials
Separation into
products
Chemical
conversion
1 2 (3)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 3/80
2
sweetening of C1, C2, C3 and C4 - NH 3 synthesis
natural gas
- Fresh water from - Hydrotreating and
saline sources hydrocraking in
petroleum refining
Notice the difference between a physical separation process and a chemical
conversion step.
(Input) (Output) (Reactant) (Product)
2. Classification of Processes
Processes are classified according to different classifications as follows:
2.1. Physical Versus Chemical Processes
Physical processes include mixing, heat exchange (heating or cooling), condensation,
evaporation, separation and purification processes such as distillation, extraction, and
absorption. In these processes, there is no molecular change and thus no new species
formed or consumed. Chemical processes, on the other hand, involve chemical
reactions in which atoms are rearranged or redistributed to form new molecules or
products and some reactants are partially or totally consumed.
As shown in the diagram below, process industries usually consist of chemical
processes as the heart of the plant preceded and followed by physical processes :
M.S.F. NH 3
Synthesis(Reactor)
NH 3 N2 + H 2
FreshWater
Brine
Sea
Water
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 4/80
3
RawMaterial
2.2. Continuous Versus Batch Processes
Batch process - The feed is charged into the system at the beginning and
removed at one at the end .
Continuous processes - Inputs and outputs flow continuously throughout the
duration of the process.
Semi-batch processes - Also referred to as semi-continuous process, is any
process that is neither batch nor continuous.
Physical processtypically mixingor heat exchange
Chemical process
Physical processtypically heat exchange
or separationProducts
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 5/80
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 6/80
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 7/80
6
3.3. Functions of Chemical Engineers
(with the help of others)
Develop, design andengineer the complete
process (process design &the equipment)
Choose the properraw materials
Operate the plant efficiently,
safely and economically
Assure that the productsare in specifications
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 8/80
Chapter 2
DIMENSIONS AND UNITS
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 9/80
1
DIMENSIONS AND UNITS
Dimension is a property that can be measured, e.g. length (L), time (t), Mass (M), or
temperature (T). These are the basic dimensions relevant to chemical engineering.From these, other dimensions can be calculated by multiplying or dividing the basic
dimensions, such as velocity (L/t), volume (L 3), or density (M/L 3).
Units are specific values of dimensions that have been defined by convention, custom,
or law, such as grams for mass, seconds for time, and centimeters for length.
Treat the units as ALGEBRAIC symbols, i.e. you can add, subtract, or equate like
units but not unlike units.
3 kg + 5 kg = 8 kg 3x + 5x = 8x
15 cm - 12 cm = 3 cm 15x - 12x = 3x But
3 kg + 4 seconds = ? meaningless like 3x + 4y = ?
On the other hand, numerical values and their corresponding units may always be
combined by multiplication or division:
hkm
50h2km100
=
km360h4xh
km90 =
)quantityensionless(dim6xkm2km12
3 ft x 4 ft = 12 ft 2
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 10/80
2
1. System of Units
There are three systems of units. These are the CGS, SI, and the American
Engineering Systems of units. Each system of units has the following components
relevant to chemical engineering:
• Base units - units of length, mass, time, and temperature• Multiple units - multiples or fractions, e.g. cm, m, km.• Derived units - obtained by multiplying or dividing base units such as ft 3 and
3mkg
or defined equivalents, e.g. N = kg. 2s
m and 1 lb f = 32.2 2m
s
ft.lb
Table (1) shows a comparison of the various base and some selected derived units
used in the above mentioned systems of units. Table (2) gives more comprehensive
details.
Table (1)
Base Units
Quantity CGS SI American Eng.
Length centimeter (cm) meter (m) foot (ft)
Mass gram (g) kilogram
(kg)
pound mass (lb m)
Time seconds (s) seconds (s) seconds (s)
Temperature Kelvin (K) Kelvin (K) Rankine ( R )
Derived Units
Quantity CGS SI American Eng.
Volume liter ( l or 1) liter ( l or 1) cubic feet (ft 3)
Force dyne newton (N) pound force (lb f )
Pressure pascal (Pa) lb f /in2 (psi)
Energy, work erg gram-calorie joule (J)
Power watt (W) watt (W) horsepower (hp)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 11/80
3
2. Force and Weight
From Newton’s law, F = ma (where F is force, m mass, and a is acceleration).
Units of force 2s
mkg in SI units, 2s
cm.g in CGS units and 2m
s
ft.lb in
American Eng.
For simplicity, use derived force unit.
1 newton (N) ≡ 1 2
s
mkg
1 dyne ≡ 1 2s
cmg
In American Engineering, the derived force unit lb f (pound-force) is defined as the
product of a unit mass (l lb m) and the acceleration of gravity at sea level and 40 o
latitude which is 32.174 2s
ft .
1 lb f ≡ 32.174 lb m 2s
ft.
Since both units of mass and force in American Engineering system are called pounds,there is some confusion about this. So, always distinguish them in your mind, lb m ≡/
lb f .
Force conversion factor, g c = N
s/m.kg1
2
or =dyne
s/mcg1
2
or =f
.2m
lbs/ft.lb
174.32
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 12/80
4
Thus, the equation for force in defined units is:
cg
maF =
The weight of an object is the force exerted on the object by gravitational attraction.
Weight,cg
mgW = . g is the acceleration of gravity and is equal to :
kg N
8066.9g/gs
m8066.9g c2
⇒=
gdyn
66.980g/gs
cm66.980g c2
⇒=
m
f c2 lb
lb1g/g
s
ft174.32g ⇒=
Note that g is acceleration due to gravity which varies according to position on earth
and height while g c is conversion factor and is constant. This is a source of confusion
in American Engineering units.
3. The Mole Unit
The number of moles = ,weightmolecular
mass so, depending on the units of mass used,
the unit of moles is determined. When grams are used for mass, the resulting mole
unit is gram-moles, or simply moles, mass in kilograms will give mole unit in kg-mole, mass in pound-mass will give moles in lb-moles and so on. In the CGS and SI
systems of units, the mole unit is gram-mole (mole or g-mole) while in the American
Engineering, the pound-mole (lb-mole) is used.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 13/80
5
4. Conversion of Units
A measured quantity can be expressed in terms of any units having the appropriate
dimension. For example, velocity =.etc)hmin,s(
)mile,km,ft,cm(
time
length
Thus, the units of velocity may be ,h
km,
sft
etc.
Numerical value of velocity depends on units chosen.
BEWARE of errors due to improper use of units. Using units with your number help
guide you to the correct answer.
The Conversion Factor is a ratio used to convert a certain quantity with certain unit
to its equivalent in another unit.
e.g.( )
2
2
2
2
cm1mm100
cm1mm10
and,min1
s60,
cm100m1
=
For conversion to the new units, multiply the old unit by the conversion factor which
should be the equivalent ofunitoldunitnew
.
e.g. convert 500 g to kg
500 g x kg5.0g1000
kg1=
Alternatively, this may be done by using a dimensional equation as follows :
kg5.0g1000
kg1g500=
This should avoid mistakes, compare with
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 14/80
6
kgg
000,500kg1
g1000g500 2
=
of course this is meaningless, and certainly does not result in the desired units.
When converting a quantity having a compound unit
.etc,cm
g,
hkm
.g.e 3 , set up a
dimensional equation.
Table 3 gives the conversion factors you will need .
Example :
1. Convert a flow rate of 5 ft 3/s to its equivalent in m 3/day:
daym
233,12day1
h24h1min60
min1s60
ft1m028317.0
sft5 3
3
33
=
2. Convert a density of 1000 kg/m 3 to its equivalent in lb m/ft3 :
3m
3
3m
3 ft
lb4.62
ft3145.35m1
kg1lb20462.2
mkg1000 =
So, the dimensional equation has vertical lines set up to separate each ratio
(conversion factor), and these lines retain the same meaning as multiplication sign placed between each ratio. At any point in the dimensional equation, you can
determine the consolidated net units and see what conversions are still required.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 15/80
7
Dimensional Homogeneity:
As a rule, every valid equation must be dimensionally homogeneous; that is, all
additive terms on both sides of the equation must have the same dimensions.
e.g. )s(tsmg)s/m(V)s/m(V 2o
+=
dimensionally homogeneous
while V = V o + g is not homogeneous
This is also consistent in its unit.
If an equation is dimensionally homogeneous but its term are inconsistent in units,
these terms and hence equation can be made consistent in units by multiplying by the
conversion factor.
Thus, dimensional analysis can also be used to help identify the dimensions of terms
in an equation.
The opposite of the above rule is not necessarily true. An equation may be
dimensionally consistent but invalid.
e.g. If M mass of an object, then M = 2 M is dimensionally homogeneous but
incorrect.
Example :
Consider the equation
D(ft) = 3t(s) + 4
1. If the equation is valid, what are the dimensions of the constant 3 and 4?
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 16/80
8
2. If the equation is consistent in its units, what are the units of 3 and 4?
Solution
1. For the equation to be valid, it must be dimensionally homogeneous, so that each
term must have the dimension of length. The constant 3 must therefore have the
dimension length/time , and 4 must have the dimension length.
2. For consistency of units, the constants must be 3 ft/s and 4 ft.
Table 2.
SI units
Physical Quantity Name of UnitSymbol
for Unit * Definition of Unit
LengthMassTimeTemperatureAmount of substance
EnergyForcePowerDensityVelocityAccelerationPressureHeat Capacity
TimeTemperatureVolumeMass
Basic SI Units
metre, meterkilogramme, kilogramsecondkelvinmole
Derived SI Units
joulenewtonwattkilogram per cubic metermeter per secondmeter per second squarednewton per square meter, pascal
joule per (kilogram . kelvin)
Alternative Units
minute, hour, day, yeardegree Celsiuslitre, liter (dm 3)tonne, ton (Mg), gram
mkgsKmol
J NW
min, h, d, yoCLt, g
kg . m 2 . s -2 kg . m . s -2 → J . m -1 kg . m 2 . s 3 → J. s -1
kg . m -3 m . s -1 m . s -2
N . m -2, PaJ . kg -1 . K -1
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 17/80
9
American Engineering System Units
Physical Quantity Name of Unit Symbol
LengthMassForceTimeTemperature
EnergyPowerDensityVelocityAccelerationPressureHeat capacity
Basic Units
feet pound (mass) pound (force)second, hourdegree Rankine
Derived Units
British thermal unit, foot pound (force)horsepower
pound (mass) per cubic footfeet per secondfeet per second squared
pound (force) per square inchBtu per pound (mass) per degree F
ftlbm lb f s, hroR
Btu, (ft) (lb f )hplbm/ft3 ft/sft/s 2
lbf /in. 2 Btu/(lb m) (oF)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 18/80
Chapter 3
PROCESS VARIABLES
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 19/80
1
PROCESS VARIABLES
1. Liquid and Solid Densities
Density is the ratio of mass per unit volume, for example, kg / m 3, g / cm 3, lb m/ft 3. It
has both numerical values and units. To determine the density of a substance, you
must find both its mass and its volume. If the substance is solid, the common method
to determine its volume is to displace a measured quantity of inert liquid. For
example, a known weight of a material can be placed into a container of liquid of
known weight and volume, and the final weight and volume of the combination be
measured. The density (or specific gravity) of a liquid is commonly measured with a
hydrometer which consists of a body of known weight and volume which is dropped
into a liquid and the depth to which it penetrates into the liquid is noted.
Specific gravity is a dimensionless ratio of two densities - that of substance (A) of
interest to that of a reference substance.
Specific gravity =
lb ft
lb ft
g cm
g cm
kg m
kg m
m
A
m
ref
A
ref
A
ref
3
3
3
3
3
3
=
=
(1)
The reference substance of liquids and solids is normally water.
Gas densities are quite difficult to measure; one device used is the Edwards Balance
which compares the weight of a bulb filled with air to the same bulb filled with the
unknown gas. The specific gravity of a gas is frequently referred to air but may be
referred to other gases.
In petroleum industry the specific gravity of petroleum products is usually reported in
terms of hydrometer called °API, defined as,
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 20/80
2
o
o
o
API
SpGr
= −141 5
60
60
1315.
. (2)
or
SpGr API
60
60
141 5
1315
o
o o
=
+
.
. (3)
The density of liquids and solids does not change very much with pressure. For
common substances data is reported in literature. When a liquid or solid is heated, it
normally expands (i.e. its density decreases). In most process applications, however, it
can usually be assumed with little error that solid and liquid densities are independent
of temperature, provided that no phase change occur. Similarly changes in pressuredon not cause significant changes in the liquid or solid densities; these substances are
therefore termed incompressible .
2. Flow Rate
The flow rate of a process stream may be expressed as a mass flow rate (mass/time) oras a volumetric flow rate (volume/time). Suppose a fluid (gas or liquid) flows in the
cylindrical pipe shown below, where the shaded area represents a section
perpendicular to the direction of flow.
Figure1. Flow of a fluid in a pipe
If the mass flow rate of the fluid is ‘ m’ (kg/sec), then every second m kilograms of
fluid pass through the cross section. If the volumetric flow rate of the fluid at the
given cross-section is ‘ v’ (m 3 /sec), then every second, v cubic meters of fluid pass
m (kg fluid /sec)v (m 3 fluid /sec)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 21/80
3
through the cross-section. However, the mass flow rate m and volumetric flow rate v
are not independent quantities but are related through density.
3. Chemical Composition
3.1. 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 atomic weights of the
atoms that constitute a molecule of the compound.
A gram-mole( gmole or mole in SI units system) of a species is the amount of that
species whose mass in grams is numerically equal to its molecular weight. Other types
of moles (e.g. kg-mole or kmol, lb-mole, ton-mole etc.) are similarly defined. Carbon
monoxide (CO), for example, has a molecular weight of 28. One mole of CO therefore
contains 28 grams; 1 lb-mole contains 28 lbs.; 1 ton-mol contains 28 tons).
The molecular weight of a substance, M, can be expressed as M kg/kmol, M g/gmol.,
M lbs/lb-mol etc. The molecular weight may thus be used as a conversion factor that
relates mass and number of moles of the quantity of the substance. Therefore,
34 kg ammonia (NH 3) are equivalent to:
34117
233
33kg NH
kmol NH kg NH
kmol NH
= (4)
3.2. Mass and Mole Fractions
Process streams occasionally contain one substance but more often consist of mixture
of liquids or gases or solutions of one or more solutes in liquid solvent. Following
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 22/80
4
terms may be used to define the composition of a mixture of substances including a
species A.
Mass Fraction
x massof A
Total masskg of Atotal kg
g of Atotal g
or lb of Atotal lb A
m
m
=
, (5)
Mole Fraction
y molesof A
Total moleskmol of Atotal kmol
mol of Atotal mol
or lb mol of Atotal lb moles A = −
−
, (6)
For example, if a solution contains 15% A by mass (x A = 0.15) and 20% B (x B =
0.20), calculation of the mass of A in 175 kg of solution:
175015
26kg solution kg A
kg of sol kg A
..
= (7)
similarly, the mass flow rate of A in a stream of solution flowing at the rate of 53 lb m/
hr is given as,
Flow rate of A = 53015
8 0lbhr
lb Alb
lb Ahr
m m
m
m..
= (8)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 23/80
5
3.3. Conversion from a Composition by Mass to Molar Composition
Example : A mixture of gases has the following composition by mass:
O2 = 16%, CO = 4%, CO 2 = 17% N 2 = 63%Define the molar composition.
Solution:
Basis: 100 g of the mixture
n g total g O g total
mol O g O
mol O2100
016 132
052 2
2
=
=.
.
Similarly,
( )n mol CO =
=100 004128
0143. .
( )n mol CO2100 017
144
0386=
=. .
( )n mol N 2 100 0 631
282 25=
=. .
n n n n n mol T O CO CO N = + + + =2 2 2
3279.
Therefore,
yn
nmol O
total mol OO
T 2
2 053279
015 2= = =..
.
yn
n
mol CO
total mol COCO
T
= = =01433279
0044..
.
yn
n
mol CO
total mol CO
CO
T 2
2 0386
3279
012 2= = =.
.
.
yn
nmol N
total mol N N
T 2
2 2 253279
0 69 2= = =..
.
Σ yi = + + + =015 0 044 012 0 69 10. . . . .
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 24/80
6
3.4. Average Molecular Weight
The average molecular weight or mean molecular weight of a mixture, M (kg/kmol)
is the ratio of the mass of sample of the mixture (m total) to the moles (n total) of all thespecies in the mixture, i.e.
M m
n y M y M y M total
total = = + + +1 1 2 2 3 4
LLL (9)
or
M y M
where N total number of components
i ii
N =
=
=∑
1 (10)
Also,
1 1
1
2
2
3
3 1 M x M
x M
x M
x M
i
ii
N = + + + +
=∑LLL (11)
3.5. Concentration
The mass concentration of a component of a mixture or solution is the mass of this
component per unit volume of the mixture (g/cm 3, kg/m 3, lb m/ft 3). The concentration
can also be expressed as molar concentration which is the number of moles of the
component per unit volume of the mixture (gmol/cm 3, kmol/m 3, lbmol/ft 3). The
molarity of the solution is the value of the molar concentration of the solute expressed
in gram moles of solute per liter of solution. The density of a mixture of liquids can be
estimated by assuming that component volumes are additive: e.g. if 2 ml of liquid A
and 3 ml of liquid B are mixed, the resulting volume would be assumed to be 5 ml.
This leads to a simple averaging formula for the mixture density ( ρ);
1 1
1
2
2 ρ ρ ρ = +
x x (12)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 25/80
7
where ρ 1 and ρ 2 are the densities of component A and B respectively while x 1 and x 2
are their mass fractions respectively.
4. Pressure
Pressure is the ratio of a force to the area on which it acts. Therefore, the pressure
units are force units divided by area units (N / m 2, dynes / cm 2, or lb f / ft 2). The SI unit
of pressure is N / m 2 and is known as Pascal.
In most investigations we are concerned with absolute pressure. Most pressure
measuring devices read the difference between absolute pressure and atmospheric
pressure existing at the gauge and this is referred to as gauge pressure . This is shown
graphically in the figure below:
Figure 2. Illustration of terms used in pressure measurement
Pressure aboveAtmospheric Pressure Ordinary Pressure
gauge reading
Ordinary vacuum gauge readsdifference between atmosphericand absolute pressure
AtmosphericPressure
Absolute pressure thatis less than atmospheric
pressure
ZeroPressure
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 26/80
8
From the principles of hydrostatics, it is possible that pressure may be expressed as
head of a particular fluid - i.e. as a height of a hypothetical column of this fluid that
will exert the given pressure at its base if pressure at the top were zero. You can thus
speak of a pressure of 14.7 lb f /in 2 as equivalent of 33.9 ft of water (33.9 ft of H 2O) or
76 cm of mercury (76 cm Hg)
∆ ∆ P g h= ρ (13)
where,
ρ = density of fluid
g = acceleration due to gravity
∆h = head of the fluid.
If ∆ P = 2.0 ×10 5 Pa, the corresponding head of mercury can be calculated using the
above equation:
∆h N
m
mkg
kg N
mmm
mm= ×
= ×2 0 1013600 9 807
1015 105
2
3 33.
..
5. Temperature
The temperature is a measure of the degree of hotness or coldness of the body. There
are two commonly used scales for measuring temperature, namely the Fahrenheit and
Celsius scales. The Celsius scale was formerly called the Centigrade scale.
Until 1954 each of these scales was based on two fixed and easily duplicated points,the ice point and the steam point. The temperature of the ice point is defined as the
temperature of a mixture of ice and water which is in equilibrium with saturated air at
the pressure of 1 atm. The temperature of the steam point is the temperature of water
and steam which are in equilibrium at the pressure of 1 atm. On the Fahrenheit scale,
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 27/80
9
these two points are assigned the number 32 and 212 respectively. On the Celsius
scale the respective points are numbered 0 and 100. Letters ° F and ° C denote the
Fahrenheit and Celsius scales respectively.
In 1954 the Celsius scale was redefined in terms of a single fixed point and the
magnitude of the degree. The triple point of water (the state where the solid, liquid
and vapor phases of water exist together in equilibrium) assigned the value 0.01 ° C.
On the scale the steam point is experimentally found to be 100.0 ° C. Thus there is
essential agreement between the old and new temperature scales.
The absolute scale of temperature related to Celsius scale is referred to as the Kelvin
scale designated as K.
K C = +o
27315. (14)
The absolute scale related to the Fahrenheit scale is referred to as the Rankine scale
designated as R. The relation between the two scales is
R F = +o
459 67. (15)
The following relationships may be used to convert a temperature expressed in one
defined scale unit to its equivalent in another:
( )T K T C = +( ) .o
27315 (16)
T R T F ( ) ( ) .= +o
459 67 (17)
( )T R T K ( ) .= 18 (18)
( ) ( )T F T C o o
= +18 32. (19)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 28/80
Chapter 4
MATERIAL BALANCES AND APPLICATIONS
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 29/80
1
MATERIAL BALANCES AND APPLICATIONS
4.1. Introduction
Material balances are important first step when designing a new process or analyzing anexisting one. They are almost always prerequisite to all other calculations in the solution of
process engineering problems.
Material balances are nothing more than the application of the law of conservation of mass,which states that mass can neither be created nor destroyed. Thus, you cannot, for example,specify an input to a reactor of one ton of naphtha and an output of two tons of gasoline orgases or anything else. One ton of total material input will only give one ton of total output,i.e. total mass of input = total mass of output.
A material balance is an accounting for material. Thus, material balances are often comparedto the balancing of current accounts. They are used in industry to calculate mass flow rates ofdifferent streams entering or leaving chemical or physical processes.
4.2. The General Balance Equation
Suppose propane is a component of both the input and output streams of a continuous processunit shown below, these flow rates of the input and output are measured and found to bedifferent.
q in (kg propane/h) q out (kg propane/h)
If there are no leaks and the measurements are correct, then the other possibilities that canaccount for this difference are that propane is either being generated, consumed, oraccumulated within the unit.
A balance (or inventory) on a material in a system (a single process unit, a collection of units,or an entire process) may be written in the following general way:
Process unit
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 30/80
2
Input + generation output consumption = accumulation(entersthroughsystem
boundaries)
(producedwithinsystem
boundaries)
(leavesthroughsystem
boundaries)
(consumedwithinsystem)
(buildupwithinsystem)
This general balance equation may be written for any material that enters or leaves any process system; it can be applied to the total mass of this material or to any molecular oratomic species involved in the process.
The general balance equation may be simplified according to the process at hand. Forexample, by definition, the accumulation term for steady-state continuous process is zero.Thus the above equation becomes:
Input + generation = output + consumption
For physical process, since there is no chemical reaction, the generation and consumptionterms will become zero, and the balance equation for steady-state physical process will besimply reduced to:
Input = Output
4.3. Balances on Single and Multiple Physical Systems
4.3.1. Procedure for Material Balance Calculations
In material balance problems, you will usually be given a description of a process, the valuesof several process variables, and a list of quantities to be determined. In order to be trainedon using a systematic procedure to solve material balance problems, you are advised tofollow the steps summarized below:
1. Draw and label the process flow chart (block diagram). When labeling, write the values ofknown streams and assign symbols to unknown stream variables. Use the minimumnumber possible of symbols.
2. Select a basis of calculation. This is usually the given stream amounts or flow rates, if nogiven then assume an amount of a stream with known composition.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 31/80
3
3. Write material balance equations. Note in here the maximum number of independentequations you can write for each system is equal the number of species in the input andoutput streams of this system. Also note to first write balances that involve the fewestunknown variables.
4. Solve the equations derived in step 3 for the unknown quantities to be determined.
Notes
i. Minimize the symbols assigned to unknown quantities by utilizing all the given processspecifications and using laws of physics.
ii. After doing calculations on certain basis, you may scale up or scale down (convert to new basis) while keeping the process balanced. This is done by multiplying all streams (except
mass or mole fractions) by the scale factor which is equal to the ratio of the new streamamount or flow rate to the old one. You can only scale between mass amount or flow ratesregardless of units used but not from mass to molar quantity or flow rate.
The examples below will illustrate the procedure of balances on physical processes:
EXAMPLE: Balance on a mixing unit
An aqueous solution of sodium hydroxide contains 20% NaOH by mass. It is desired to produce an 8% NaOH solution by diluting a stream of the 20% solution with a stream of purewater.
1. Calculate the ratios (g H 2O/g feed solution) and (g product solution/g feed solution).
2. Determine the feed rates of 20% solution and diluting water needed to produce2310 lb m/min of the 8% solution.
Solution
We could take a basis of 2310 lb m product/min, but for illustrative purposes and to haveneater numbers to work with let us choose a different basis and scale the final results.
Basis: 100 g Feed Solution
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 32/80
4
Draw and label the flowchart, remembering that the amount of the product stream is nowunknown.
(Since the known stream amount is given in grams, it is convenient to label all unknownamounts with this unit.)
There are two unknowns - Q1 and Q2 - and since there are two substances - NaOH and H 2O -in the input and output streams, two balances may be written to solve for them. The totalmass balance and the water balance involve both unknowns, but the NaOH balance involvesonly one.
NaOH Balance
( ) ( )
( )( ) g2500.080g1000.20
out NaOHgin NaOHg
22 =⇒=
⇓
=
It is a good practice to write calculated variable values on the flowchart as soon as they areknown for ease of use in later calculations; at this point, 250 would therefore be written in
place of Q2 on the chart.
Total Mass Balance
OHg150
g250
g100
21
2
21
=
=
=+
⇓
⇓
Q
Q
100 g
0.20 g NaOH/g0.80 g H 2O/g
Q2 (g)
0.080 g NaOH/g0.920 g H 2O/g
Q1 (g H 2O)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 33/80
5
The desired ratios can now be calculated:
solutionfeedg
OHg1.5
solutionfeedg100
O)H(g 2150
211 =
⇒
solutionfeedg productg
2.5solutionfeedg100
O)H(g 25022
2 =
⇒
The scale factor is obtained as the true flow rate of the product stream divided by the ratecalculated on the assumed basis.
g/minlb
9.24 productg250n product/milb2310 mm
=
Feed Solution Flow Rate
minsolutionfeedlb
924g
/minlb9.24g100 mm =
Dilution Water Flow Rate
minOHlb
1386g
/minlb9.24g150 2mm =
Check: (924 + 1386) lb m/min = 2310 lb m/min
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 34/80
6
EXAMPLE: Scale up of a separation process flowchart
A 60 - 40 mixture (by moles) of A and B is separated into two fractions. A flowchart of the process is shown here.
It is desired to achieve the same separation with a continuous feed of 1250 lb-moles/h. Scalethe flowchart accordingly.
Solution
The scale factor is
molh/lbmoles
5.12mol100
h/lbmoles1250=
The masses of all streams in the batch process are converted to flow rates as follows:
Feed: specified)(ashmoles-lb
1250mol
moles/h-lb12.5mol100=
Top product stream: (50.0)(12.5) = 625 lb-moles/h
Bottom product stream: (12.5)(12.5) = 156 lb-moles A/h
(37.5)(12.5) = 469 lb-moles B/h
The units of the mole fractions in the top product stream may be changed from mol/mol to lb-mole/lb-mole, but their values remain the same. The flowchart for the scaled-up process isshown here.
100.0 mol0.60 mol A/mol0.40 mol B/mol
50.0 mol
0.95 mol A/mol0.05 mol B/mol
12.5 mol A37.5 mol B
625 lb-moles/hr0.95 lb-mole A/b-mole0.05 lb-mole B/b-mole1250 lb-moles/hr
0.60 lb-mol A/lb-mole0.40 lb-mol B/lb-mole
156 lb-moles A/hr469 lb-moles B/hr
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 35/80
7
EXAMPLE: Material balances on a distillation column
A mixture containing 45% benzene (B) and 55% toluene (T) by mass is fed to a distillationcolumn. An overhead stream of 95 wt% B is produced, and 8% of the benzene fed to thecolumn leaves in the bottom stream. The feed rate is 2000 kg/h. Determine the overhead flowrate and the mass flow rates of benzene and toluene in the bottom stream
Solution
Basis : Given Feed Rate
The labeled flowchart is as follows.
There are three unknowns on the chart - D , w B, and wT - and therefore three equations areneeded. We are entitled to write only two material balances since two species are involved inthe process; the third equation must therefore come from additional given information (theamount of benzene in the bottom stream.) The latter relation is.
chart)theonof placein72(WriteB/hkg72
kg/h]00)[(0.45)(200.080B/h)(kg
feed)in(B0.080stream bottominB
BB
B
ww
w
=
⇓
=⇓
=
Toluene and total mass balances each involve two unknowns, D and wT, but a benzene balance involves only one, D.
Benzene Balance
( )( )
chart)theonit(Writekg/h870D
B/hkg72
D0.95h
Bkg20000.45
B
B
=
⇓
=⇓
+=
w
w
2000 kg/h
0.45 kg B/kg0.55 kg T/kg
D kg/h
0.95 kg B/kg0.05 kg T/kg
Contains 8% of the Bin the feed
wB kg B/kgwT kg T/kg
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 36/80
8
Total Mass Balance (A toluene balance could be used equally well)
T/hkg1060
kg/h72
kg/h870D
Dh
kg2000
T
B
TB
=
⇓
=
=⇓
++=
w
w
ww
EXAMPLE: Two Unit Distillation Process
A labeled flowchart of a continuous steady-state two-unit distillation process is shown below.Each stream contains two components, A and B, in different proportions. Three streamswhose flow rates and/or compositions are not known are labeled 1, 2 and 3.
Calculate the unknown flow rates and compositions of streams 1, 2, and 3.
Solution
The systems about which balances might be taken are shown on the following representationof the flowchart.
40 kg/h0.900 kg A/kg0.100 kg B/kg
30 kg/h0.600 kg A/kg0.400 kg B/kg
30 kg/h0.300 kg A/kg0.700 kg B/kg
100 kg/h
0.500 kg A/kg0.500 kg B/kg
1 2 3
40 kg/h0.900 kg A/kg0.100 kg B/kg
30 kg/h0.600 kg A/kg0.400 kg B/kg
30 kg/h0.300 kg A/kg0.700 kg B/kg
100 kg/h
0.500 kg A/kg0.500 kg B/kg
Q1 (kg/h) Q2 (kg/h)
x1 (kg A/kg) x2 (kg A/kg)1- x1 (kg B/kg) 1- x2 (kg B/kg)
Q3 (kg/h)
x3 (kg A/kg)1- x3 (kg B/kg)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 37/80
9
The outer boundary encompasses the entire process. Two of the interior boundaries surroundthe individual process units, and the fourth boundary encloses a stream junction point.
Basis: Given Flow Rates
There are two unknowns in the streams that enter and leave the total process, Q3 and x3, andsince there are two independent components in these streams (A and B) we may write two
balances.
Overall Mass Balance
( ) ( )
kg/h60
hkg
3040hkg
30100
3
3
=
⇓
++=+
Q
Q
Overall Balance On A
( )( ) ( )( ) ( )( ) ( )( ) ( )( )
A/kgkg0.0833
60300.600400.900300.3001000.500
3
3
=
⇓
++=+
x
x
To determine the flow rate and composition of a connecting stream, we must write balanceson a subsystem whose boundary intersects this stream. Of the three such boundaries shown inthe flowchart, the middle one (about the stream junction) would not be a good one to use atthis point since its input and output streams contain four unknown quantities ( Q1, x1, Q2, x2),while the boundaries about the process units each intersect streams that contain twounknowns.
Let us choose the boundary about unit 1 for the next set of balances. There are twounknowns, Q1 and x1, in the streams that intersect this boundary, and up to two balances may
be written.
Total Mass Balance on Unit 1
h/kg60
hkg
40h/kg100
1
1
=
⇓
+=
Q
Q
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 38/80
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 39/80
11
EXAMPLE: 2SO 2+O 2→ 2SO 3. What is the stoichiometric coefficient of SO 2?
Solution The number that precede SO 2 is 2. Therefore, stoichiometric coefficient of SO 2 is2.
In a stoichiometric equation, the number of atoms in both sides of the equation must be balanced. In this example, the number of atoms of S and O are 2 and 6, respectively, in bothsides of equation.
Ratio of stoichiometric coefficients of two species is known as stoichiometric ratio .
EXAMPLE: 2SO 2+O 2→ 2SO 3. What is the stoichiometric ratio of SO 2 to SO 3?
Solution stoichiometric ratio of SO 2 to SO 3 =2 mole of SO reacted
2 mole of SO produced = 12
3
If proportion of chemical species fed to a reactor is same as the stoichiometric ratio, thenchemical species combine in stoichiometric proportion , otherwise one or more species will
be in excess of the other. The chemical compound which is present less than itsstoichiometric amount, will disappear first. This reactant will be the limiting reactant and allthe others will be excess reactants. Fractional and percentage excess are given by thefollowing formulas.
fractional excess =n n
nS
S
−
percentage excess = 100n nnS
S ×−
wheren = number of moles fed
ns = number of moles corresponding to stoichiometric amount
EXAMPLE: 100 moles of SO 2 and 100 moles O 2 are fed to a reactor and they reactaccording to 2SO 2+O 2→ 2SO 3. Find the limiting reactant, excess reactant,fractional excess and percentage excess?
Solution
ratio of SO 2 to O 2 fed = 100/100 = 1
stoichiometric ratio of SO 2/O2 = 2/1=2
Therefore, SO 2 is fed less than the stoichiometric proportion (or stoichiometric ratio). SO 2 isthe limiting reactant. The other reactant (O 2) will be the excess reactant .
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 40/80
12
n = number of moles of excess reactant (O 2) fed = 100
ns = stoichiometric amount of O 2 to react with 100 moles of the limiting reactant SO 2 = 50
Therefore, fractional excess = 0 .150
50100n
nn
S
S =−
=−
percentage excess = 100%100n
nnS
S =×−
4.4.2. Fractional Conversions, Extent of Reaction, Chemical Equilibrium
In many cases, chemical reactions do not go to completion and only a fraction will beconverted. Therefore, fractional and percentage conversions are used. They are defined asfollows,
fractional conversion (f) =mole reacted
mole fed to the reactor
percentage conversion = f ×100
fraction unreacted = (1-f)
EXAMPLE: 200 moles of SO 2 and 100 moles O 2 are fed to a reactor. Only 100 moles ofSO 2 react according to 2SO 2+O 2→2SO 3 Find fractional conversion, percentageconversion and fraction unreacted?
Solution
fractional conversion of SO 2 (f) =mole reacted
mole fed to the reactor =100/200=0.5
percentage conversion = f ×100=0.5 ×100=50%
fraction unreacted = (1-f)=1-0.5=0.5
When a reaction is not complete, remaining amount in the reactor will be given by
n ni io i= + β ζ
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 41/80
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 42/80
14
Solution
Basis : 100 mol Feed
The feed to the reactor contains
( )
( )
( )( )( )( ) excessinisO
1.51.5/1HC/O
1.6416.4/10.0HC/O
excessinis NH11/1H/C NH
1.2012.0/10.0H/C NH
mol16.4air mol
Omol0.210air mol78.0O
mol12.0 NH
mol10.0HC
2stoich632
in632
3stoich633
in633
2in2
in3
in63
⇒==
==
⇒==
==
==
=
=
⇓
Since propylene is fed in less than the stoichiometric proportion relative to the two otherreactants, propylene is the limiting reactant.
To determine the percentages by which ammonia and oxygen are in excess, we must firstdetermine the stoichiometric amounts of these reactants corresponding to the amount of
propylene in the feed (10 mol).
( ) 363
363stoich3 NHmol0.10
HCmol1 NHmol1HCmol0.10
NH ==
( ) 263
263stoich2 Omol0.15
HCmol1Omol5.1HCmol0.10
O ==
( ) ( )
( )[ ] 3
stoich3
stoich303 NH
NHexcess20%100%/101012
100%)(NH
)(NH NHexcess%
3
=×−=
×−
=
OHmoln
NHCmoln
Nmoln
Omoln
NHmolnHCmoln
2OH
33 NHC
2 N
2o
3 NH
63HC
2
33
2
2
3
63
100 mol
0.100 mol C 3H6/mol0.120 mol NH 3/mol0.780 mol air/mol
(0.21 mol O 2/mol air0.79 mol N 2/mol air)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 43/80
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 44/80
16
00.3n
1n
n
00.2n
00.1n
total
H
CO
OH
CO
2
2
2
m)equilibriuat presentCOof molsof (number
=
ξ=
ξ=
ξ−=
ξ−=
e
e
e
e
from which
/3.00ξy
/3.00ξy
)/3.00ξ(2.00y
)/3.00ξ(1.00y
eH
eCO
eOH
eCO
2
2
2
=
=
−=
−=
Substitution of these expressions in the equilibrium relation (with K = 1.00) yields
00.1)00.2)(00.1(
2
=ξ−ξ−
ξ
ee
e
This may be rewritten as a standard quadratic equation and solve to yield ξ e = 0.667. Thisquantity may in turn be substituted back into the expressions for y i(ξ e ) to yield
0.222y0.222,y0.444,y0.111,y222 HCOOHCO ====
The limiting reactant in this case is CO (verify). At equilibrium,
0.3330.6671.00n CO =−=
The fractional conversion of CO at equilibrium is therefore
f CO = (1.00 - 0.333) mol reacted/(1.00 mol fed) = 0.667
4.4.3. Multiple Reactions, Yield and Selectivity
In a chemical process, our objective is to produce a certain product (desired product), butthere may be several unwanted reactions which will produce undesirable by products.Therefore, we must maximize the production of a desired product in the process. Twoquantities, yield and selectivity , are used for this purpose and they are defined as follows,
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 45/80
17
yield =
completelyreactsreactantlimitingtheandreactionssideno werethereif formed productdesiredof moles
formed productdesiredof moles
selectivity =moles of desired product formed
moles of undesired product formed
When we have multiple reactions, the remaining amount or flow rate will be given by
jij j
ioi nn ξβ+= ∑
where i = compound i j = reaction jξ j = extent of reaction for the j th reactionβ ij = + ν i, stoichiometric coefficient of a product i in the j th reaction
= -ν i, stoichiometric coefficient of a reactant i in the j th reaction= 0, inert
EXAMPLE: Consider the following pair of reactions.
A → 2B (desired)
A → C (undesired)
100 moles of A are fed to a batch reactor and the final product contains 10 mol of A, 160 molof B and 10 mol of C. Calculate (1) percentage yield of B, (2) the selectivity of B relative toC and (3) the extents of the reactions.
Solution
Percentage Yield
moles of desired product (B) formed = 160
moles of desired product formed if there were no side reactions and the limiting reactant
reacts completely =reactedAof mole1
producedBof moles2 Aof moles100 × = 200 moles
yield (%) = 160/200*100 = 80%
Selectivity
moles of desired product (B) formed = 160
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 46/80
18
moles of undesired product (C) formed = 10
selectivity =moles of desired product formed
moles of undesired product formed = 160/10 = 16.
Extent of Reactions
j ij j
ioi nn ξβ+= ∑
for two reaction system as above, 2211ioi nn ξβ+ξβ+= ii
applying this equation for B, 120160 ξ+= , this gives ξ1 = 80
for C, 2010 ξ+= this gives ξ2 = 10.
check: for A,
2110010 ξ−ξ−= , 10 100 80 10 10= − − =
EXAMPLE: Yield and Selectivity in a Dehydrogenation Reactor
The reactions
4262
24262
CH2HHC
HHCHC
→++→
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 thefractional yield of ethylene is 0.471. Calculate the molar composition of the product gas andthe selectivity of ethylene to methane production.
Solution
Basis: 100 mol Feed
The outlet component amounts in terms of extents of reaction are as follows:
0.850 C 2H6 0.150 I
100 mol
I)(moln)CH(moln
)H(moln)HC(moln)HC(moln
5
44
23
422
621
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 47/80
19
n1 (mol C 2H6) = 85.0 - ξ1 - ξ2 n2 (mol C 2H4) = ξ1
n3 (mol H 2) = ξ1 - ξ2 n4 (mol CH 4) = 2 ξ2
n5 (mol I) = 15.0
Ethane Conversion
2162
62
62
621
85.0HCmol42.4
fedHCmol85.0fedHCmol
unreactedHCmol0.501)(1.000n
ξ−ξ−==
−=
Ethylene Yield
( ) 142422
62
4262
HCmol40.0HCmol85.00.471n
mol85.0HCmol1HCmol1fedHCmol85.0
ethylene possibleMaximum
ξ===⇓
==
Substituting 40.0 for ξ1 in (1) yield ξ2 = 2.6 mol. Then
I10.7%,CH3.7%,H26.7%,HC28.6%,HC30.3%
mol140.015.0)5.237.440.0(42.4n
Imol15.0n
CHmol5.22n
Hmol37.4n
424262
tot
5
424
2213
⇓
=++++=
=
=ξ=
=ξ−ξ=
Selectivity = (40.0 mol C 2H4)/(5.2 mol CH 4)4
42
CHmolHCmol
7.7=
4.4.4. Atomic and Molecular Balances
In a chemical process, molecules are either generated (produced) or consumed. Therefore,one should take into account the amounts (moles) generated, consumed, fed and remaining inmolecular balances .
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 48/80
20
EXAMPLE: In a steady state process, 100 moles ethane (C 2H6) react to produce ethylene(C 2H4) and hydrogen (H 2) according to C 2H6→ C2H4 + H 2. Product gas shows40 moles of hydrogen remaining. Perform molecular balances for all species.
Solution:
40 mol of H 2 q1 mol of C 2H6
100 mol C 2H6 q2 mol of C 2H4
Then, the molecular balance of H 2 is as follows,
input + generated (produced) = output + consumed (reacted)
input = 0.0
output = 40
consumed = 0
0 + generated (produced) = 40+ 0 ⇒ H 2 (generated) = 40 moles.
Molecular balance of C 2H6 is as follows,
input + generated (produced) = output + consumed (reacted)
input = 100generated = 0
output = q 1
100 + 0 = q 1 + consumed ⇒ C2H6 (consumed) = 100-q 1
H2 (generated) = 40 moles=C 2H6 (consumed) = 100-q 1 ⇒ q1 = 60 moles
Molecular balance of C 2H4 is as follows,
input + generated (produced) = output + consumed (reacted)
input = 0
output = q 2
consumed = 0
0 + generated = q 2 + 0 ⇒ C2H4 (generated) = q 2
C2H4 (generated) = q 2 = H 2 (generated) = 40 moles ⇒ q2 = 40 moles
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 49/80
21
According to conservation principle, atoms can neither be created (produced) nor destroyed(consumed). Therefore, in atomic balances there is no generation or consumption terms.Simply, input = output.
EXAMPLE: For the same problem as above, do atomic balances?
Solution
Atomic carbon balance: 100 × 2 = q 1× 2 + q 2× 2
Atomic hydrogen balance 100 × 6 = 40 × 2 + q 1× 6 + q 2× 4
Solving these two atomic balance equations, you will get the same answer as above.
EXAMPLE: Combustion of Methane
Methane is burned with oxygen to yield carbon dioxide and water. The feed contains 20mole% CH 4, 60% O 2, and 20% CO 2, and a 90% conversion of the limiting reactant isachieved. Calculate the molar composition of the product stream using (1) balances onmolecular species, (2) atomic balances, (3) the extent of reaction.
Solution
Basis : 100 mol Feed
OH2COO2CH 2224 +→+
Since a 2:1 ratio of O 2 to CH 4 would be stoichiometric and the actual ratio is 3:1, CH 4 is thelimiting reactant and O 2 is in excess.
Before the balances are written, the given process information should be used to determinethe unknown variables or relations between them. In this case, the methane conversion of90% tells us that 10% of the methane fed to the reactor emerges in the product, or
0.200 CH 4 0.600 O 2 0.200 CO 2
100 mol
O)H(moln)CO(moln
)O(moln)CH(moln
2OH
2CO
2O
4CH
2
2
2
4
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 50/80
22
44CH CHmol2.0fed)CHmol(20.00.100n4
=×=
Now all that remains are the balances. We will proceed by each of the indicated methods.
1. MolecularBalances
2CO
CO4
242
44
4
COmol38n
nCHmol1 producedCOmol1reactCHmol18
fedmolCOmol0.200mol100
output)generation(input
reactCHmol18fedmol
reactmol0.900fedCHmol20.0reactedCH
2
2
=
⇓
=+
=+
==
BalanceCO 2
OHmol36n
nreactCHmol1
producedOHmol2reactCHmol18
output)n(generatio
2OH
OH4
24
2
2
=
⇓
=
= BalanceO H 2
22O
4
24O
2
Omol24Omol36)(60n
reactCHmol1reactOmol2reactCHmol18
nmol
Omol0.600mol100
n)consumptiooutput(input
2
2
=−=
⇓
+=
+= BalanceO 2
In summary, the output quantities are 2 mol CH 4, 24 mol O 2, 38 mol CO 2, and 36 molH2O, for a total of 100 mol. (Since 3 moles of products are produced for every 3 moles ofreactants consumed, it should come as no surprise that total moles in = total moles out.)The mole fractions of the product gas components are thus:
0.02 mol CH 4/mol,0.24 mol O 2/mol,0.38 mol CO 2/mol,0.36 mol H 2O/mol
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 51/80
23
2. Atomic Balances
Referring to the flowchart, we see that a balance on atomic carbon involves only oneunknown (
2COn ), and a balance on atomic hydrogen also involves one unknown ( O2Hn ),
but a balance on atomic oxygen involves three unknowns. We should therefore write theC and H balances first, and then the O balance to determine the remaining unknownvariable. All atomic balances have the form input = output. (We will just determine thecomponent amounts; calculation of the mole fractions then follows as in part 1.)
C Balance
2CO
2
2CO
4
4
2
2
4
4
COmol38n
COmol1Cmol1COmoln
CHmol1Cmol1CHmol2.0
COmol1Cmol1COmol20.0
CHmol1Cmol1CHmol20.0
2
2
=⇓
+=
+
H Balance
OHmol36n
OHmol1Hmol2OHmoln
CHmol1Hmol4CHmol2
CHmol1Hmol4CHmol20
2OH
2
2OH
4
4
4
4
2
2
=⇓
+
=
O Balance
2O
222O
2
2
2
2
Omol24n
O)(1)Hmol(36(2))COmol(38(2))Omol(nOmol1Omol2COmol20
Omol1Omol2Omol60
2
2
=⇓
++=
+
confirming the results obtained using molecular balances.
3. Extent of Reaction
As discussed earlier, for all reactive species
ξβ+= iinout nn
For the species in this problem, we may write
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 52/80
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 53/80
25
4.5.2. Recycle and Purge
A material such as an inert gas or impurities which enter with the feed will remain in therecycle stream. This material will accumulate and the process will never reach steady state.To prevent this buildup, a portion of the recycle stream must be withdrawn as a purge
stream .
EXAMPLE: Dehydrogenation of Propane
Propane is dehydrogenated to from propylene in a catalytic reactor
26383 HHCHC +→
The process is to be designed for a 95% overall conversion of propane. The reaction productsare separated into two streams: the first, which contains H 2, C 3H6, and 0.555% of the propanethat leaves the reactor, is taken off as product; the second stream, which contains the balanceof the unreacted propane and 5% of the propylene in the product stream, is recycled to thereactor. Calculate the composition of the product, the ratio (moles recycled)/(moles freshfeed), and the single-pass conversion.
Solution
Basis : 100 mol Fresh Feed
Note: In labeling the feed stream to the reactor, we have implicitly used balances on propane and propylene about the stream junction.
In terms of the labeled variables, the quantities to be calculated are
Mole Fractions of Product Stream Components: ...,321
1
QQQQ
++etc.
Recycle Ratio:100
21 r r QQ +
Reactor Separation
Unit
Fresh feed100 mol C 3H8 100 + Q
r 1 mol C
3H
8Qr 2 mol C 3H6
Q r 1 mol C 3H8
Q r 2 mol C 3H6
P 1 mol C 3H8
P 2 mol C 3H6
P 3 mol H 2
Q1 mol C 3H8
Q2 mol C 3H6
Q2 mol H 2
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 54/80
26
Single-Pass Conversion: % )Q(
P )Q(
r
r 100100
100
1
11 ×+
−+
We must therefore calculate the values of Q1, Q2, Q3, Q r1 , Q r2 and P 1.
Let us first examine the overall process.
831
83
183
HCmol5
0.95HCmol100
HCmol10095%conversionOverall
=
⇓
=−
⇒=
Q
Q
Overall Propane Balance: input = output + consumption
reactedHCmol95C
)reactedHCmol(CHCmol5HCmol100
83
838383
=⇓
+=
Overall Propylene Balance: output = generation
632
83
63832
HCmol95
reactedHCmol1formedHCmol1reactedHCmol95
=
⇓
=
Q
Q
Similarly, overall H 2 balance yields
Q3 = 95 mol H 2
The analysis of the product is, therefore,
totalmol195
H%7.48Hmol95
HC%7.48HCmol95
HC%6.2HCmol5
22
6363
8383
⇒
We may now proceed to the analysis of the interior streams including the recycle stream.First, we summarize the information given in the problem statement.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 55/80
27
632r
6H3Cmol952
22r
831
8H3Cmol51
11
HCmol75.405.0
HCmol900P00555.0
=⇒=
=⇒=
=
=
QQQ
P Q
Q
Q
Next, write balances around the separation unit (which is nonreactive, so that input = outputfor all species.)
C 3 H 8 Balance About Separation Unit
831r
831
831
1r 11
HCmol895
HCmol5
HCmol900
=
=
=⇓
+=
Q
Q
P
QQ P
We now have all the variable values we need. The desired quantities are
feedfreshmolrecyclemol
00.9
mol75.4
mol895
feedfreshmol100
recyclemol)(ratiocycleRe
2r
1r
2r 1r
=
=⇓
+=
Q
Q
9.55%
HCmol900HCmol895
100%)(100
)(100
conversion passSingle
831
831r
1r
11r
==
⇓
×+
−+=−
P Q
Q
P Q
EXAMPLE: Recycle and Purge in the Synthesis of Ammonia
The fresh feed to an ammonia production process contains 24.75 mole % nitrogen, 74.25mole% hydrogen, and the balance inerts (I). The feed is combined with a recycle streamcontaining the same species, and the combined stream is fed to a reactor in which a 25%single-pass conversion of nitrogen is achieved. The products pass through a condenser inwhich essentially all of the ammonia is removed, and the remaining gases are recycled.However, to prevent buildup of the inerts in the system, a purge stream must be taken off.The recycle stream contains 12.5 mole% inerts. Calculate the overall conversion of nitrogen,the ratio (moles purge gas/mole of gas leaving the condenser), and the ratio (moles freshfeed/mole fed to the reactor).
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 56/80
28
Solution
Basis : 100 mol Fresh Feed .
N 2 + 3H 2 → 2NH 3
Strategy: Overall N, H, I balances ⇒ n6, n7, x ⇒ overall conversion of N 2
n6, 25% single pass conversion ⇒ n1, n4
n7, x, n 4, n2 balance around purge-recycle split point ⇒ n8
H2, I balances around recycle-fresh mixing point ⇒ n2, n3
Overall I balance: (0.0100)(100) = 0.125 n 7 n7 = 8.00 mol purge gas
/mol Nmol0.219
NHmol46.0n
3n(2))(0.8758.0025)(2)(100)(0.74
n(2))(8.0075)(2)(100)(0.24
2
36
6
6
==
⇒
+−=
+=
x x:balance H Overall
x:balance N Overall
92.9%100%fedmol24.75
effluentinmol00)(0.219)(8.fedmol24.75 =−
= xbalance N Overall 2
24
21
32
321
Nmol69.00)(0.75)(92.n
reactor tofeed Nmol92.0n
NHmol46.0 Nmol1
NHmol2
fedmol1
reactmol0.25fed Nmoln:
==
=⇓
=Conversion PassSingle25%
n7(mol purged)n8(mol recycled)
Reactor Condenser
0.125 I x N2
0.875 - x H2
n3 (mol. I)n4 (mol N 2)n5 (mol H 2)
0.125 I x N2
0.875 - x H2
n6 (mol NH 3)n4 (mol N 2)n5 (mol H 2)n6 (mol NH 3)
n3 (mol I)
n1 (mol N 2)n2 (mol H 2)n3 (mol I)
0.2475 N 2
0.7425 H 2 0.0100 I
100 mol
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 57/80
29
N 2 balance around purge-recycle split point: 69.0 = (0.219)n 8 + (0.219)(8.00)
⇓ n8 = 307 mol recycled
H 2 balance around mixing point: (100)(0.7425) + (307) (0.875 - 0.219) = n 2
⇓ n2 = 276 mol H 2
I balance around mixing point: (100)(0.01) + (307)(0.125) = n 3 n3 = 39.4 mol I
025.03078
8condenser leavingmoles
purgemoles =+
=
reactor tofedmolfeedfreshmoles
245.04.3927692100
=++
EXAMPLE: Recycle and Purge in the Synthesis of Methanol
Methanol may be produced by the reaction of carbon dioxide and hydrogen.
OHOHCHH3CO 2322 +→+
The fresh feed to the process contains hydrogen and carbon dioxide in stoichiometric proportion, and 0.5 mole% inerts (I). The reactor effluent passes to a condenser, whichremoves essentially all of the methanol and water formed, none of the reactants or inerts. Thelatter substances are recycled to the reactor. To avoid build-up of the inerts in the system, a
purge stream is withdrawn from the recycle. The feed to the reactor contains 2% inerts, andthe single-pass conversion is 60%. Calculate the molar flow rates of the fresh feed, the totalfeed to the reactor, and the purge stream for methanol production rate of 1000 mol/h.
Solution
Basis: 100 mol Combined Feed to the Reactor
As a general rule, the combined feed to the reactor is a convenient stream to use as a basis ofcalculation for recycle problems, provided that its composition is known. Since in this
process the reactants are fed in stoichiometric proportion and they are never separated fromeach other, they must be present in stoichiometric proportion throughout the process; that is,(CO 2/H2) = 1/3. The feed thus contains 2 mol I (2% of 100 mol), and 98 mol CO 2 + H 2, ofwhich 24.5 mol are CO 2 (1/4 of 98) and 73.5 mol are H 2.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 58/80
30
Take a moment to examine the chart labeling. In particular, notice that we have built in thefacts that CO 2 and H 2 are always present in a 1:3 ratio and that the compositions of thegaseous effluent from the condenser, the purge stream, and the recycle stream are allidentical. The more of this sort of information you can incorporate in the chart labeling, theeasier the subsequent calculations become. Let us outline the solution, referring to theflowchart.
1. Calculate n1, n2, and n3, from the feed to the reactor, the single-pass conversion and reactor balances.
2. Calculate n4, from a total mole balance about the condenser, and then x4 from a CO 2 balance about the condenser.
3. Calculate n0, and n6, from balances on total moles and I about the recycle-fresh-feedmixing point. (Two equations, two unknowns.) Then calculate n5, from a mole balanceabout the recycle-purge split point.
4. Scale up the calculated flows of fresh feed, combined reactor feed, and purge streams bythe factor (1000/ n2). The results will be the flow rates in mol/h corresponding to amethanol production of 1000/h.
60% Single-Pass Conversion
( )
OHmol7.14
OHCHmol7.14
generationoutput:OH,OHCHonBalances
reactCOmol7.145.24600.0
Hmol4.293COmol80.9)5.24(400.0
23
32
23
2
21
21
=
=
=⇓
=
=
==
n
n
nn
Reactor Condensern0 mol
n5 mol
x4 mol CO 2 /mol
3 x4 mol H 2 /mol
(1-4 x4) mol I/mol
n5 mol
x4 mol CO 2 /mol
3 x4 mol H 2 /mol
(1-4 x4) mol I/mol
n4 mol CO 2 /mol x4 mol CO 2 /mol3 x4 mol H 2 /mol
(1-4 x4) mol I/mol
100 mol
0.249 mol CO 2 /mol
0.746 mol H 2 /mol
0.005 mol I/mol
24.5 mol CO 2
75.5 mol H 2
2 mol I
n1 mol CO 2
3n1 mol H 2
n2 mol CH 3OHn3 mol H 2O2 mol I
n2 mol CH 3OH
n3 mol H 2O
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 59/80
31
Mole Balance About Condenser
mol2.41
for substitute
42
4
1
432321
=
⇓
++=++
n
n
nnnnnn
CO 2 Balance About Condenser
mol/COmol2379.0)2.41/80.9( 24
441
==
⇓
=
x
xnn
Mole Balance About Mixing Point
n0 + n6 =100
I Balance About Mixing Point
20484.000500.0
2379.0
2)41(00500.0
60
4
460
=+
=⇓
=−+
nn
x
xnn
Solving the above two equations simultaneously yields
no = 65.4 mol fresh feedn6 = 34.6 mol recycle
Mole balance About Purge Takeoff
purgemol6.66.342.415
654
=−=
⇓
+=
n
nnn
The required scale factor is
( )[ ] 17.142
2 h03.8mol/h/mol1000 −
=
⇒ 6 nn
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 60/80
32
The desired flow rates are therefore
65.4 × 68.06 = 4450 mol fresh feed/h100.0 × 68.03 = 6803 mol/h fed to reactor
6.6 × 68.03 = 449 mol/h purge
4.6. Combustion Reactions and Their Material Balances
4.6.1. Complete and Partial Combustion
Combustion is a rapid reaction of fuel with oxygen. Combustion products are CO 2, NO, CO,H2O, and SO 2. In a combustion reaction if CO is formed, then the reaction is incomplete andreferred as incomplete combustion or partial combustion. During a complete combustion ofa fuel, carbon will be oxidized to CO 2, hydrogen will be oxidized to H 2O, and sulfur will beoxidized to SO 2.
C + O 2 → CO 2 H2+½O 2 → H2O
C + ½O 2 → CO, S+O 2 → SO 2
Complete and incomplete combustion of C 3H8 are given by the following chemical reactions,complete C 3H8 +5O 2 → 3CO 2 + 4H 2O
incomplete C 3H8 + 7/2O 2 → 3CO + 4H 2O
4.6.2. Wet and Dry Basis
Product gas that leaves the combustion chamber is called stack or flue gas . Composition of aflue gas is given on a wet (including water ) or dry basis (excluding water).
EXAMPLE: Suppose a stack gas contains equimolar amounts of CO 2, N 2 and H 2O. Find thecomposition on wet and dry basis?
Solution
Suppose we have n moles of each, then
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 61/80
33
composition of CO 2 on wet basis = 100 basison wetmolestotal
COof moles 2 × = (n/3n) ×100 = 33.33%
composition of CO 2 on dry basis = 100
basisdryonmolestotal
COof moles 2 × = (n/2n) ×100 = 50%.
EXAMPLE: A stack gas contains 60 mole% N 2, 15% CO 2, 10% O 2, and the balance H 2O.Calculate the molar composition of the gas on a dry basis.
Solution
Basis: 100 mol Wet Gas
gasdrymolOmol118.0
8510
gasdrymolCOmol
176.08515
gasdrymol Nmol
706.08560
gasdrymol85Omol10
COmol15 Nmol60
2
2
2
2
2
2
=
=
=
⇓
EXAMPLE: Dry Basis ⇒ Wet Basis
An Orsat analysis (a technique for stack gas analysis) yields the following dry basiscomposition:
%10O
%11CO%14CO
%65 N
2
2
2
A humidity measurement shows that the mole fraction of H 2O in the stack gas is 0.07.Calculate the stack gas composition on a wet basis.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 62/80
34
Solution
Basis: 100 lb-moles Dry Gas
gasdrymolelbOHmolelb
0753.0gaswetmolelb/gasdrymolelb93.0
gaswetmolelb/OHmolelb07.0
gaswetmolelbgasdrymolelb
93.0gaswetmolelbOHmolelb
07.0
22
2
−−=
−−−−
⇓
−−
⇔−
−
gaswetmolelb5.107Omolelb00.10
)100.0)(100(
COmolelb0.11)110.0)(100(
COmolelb0.14)140.0)(100(
Nmolelb0.65gasdrymolelb Nmolelb650.0gasdrymolelb100
OHmoleslb53.7gaswetmolelb
OHmolelb0753.0gasdrymolelb100
2
2
22
22
−−=
−=−=
−=−
−−
−=−
−−
The mole fractions of each stack gas component may now easily be calculated:
etc...,.gaswetmoleslbOHmoleslb
0.070gaswetmoleslb107.5OHmoleslb7.53
y 22H 2 −
−=−−=O
4.6.3. Theoretical and Excess Air
Theoretical oxygen is the amount needed for complete combustion of reactants to form CO 2 and H 2O. Air that contains the theoretical amount of oxygen is called theoretical air .Theoretical air does not depend on how much of a reactant is converted. The difference
between the amount of air initial fed and the theoretical air is known as excess air. Therefore, percentage excess air is defined as,
% excess air = {[Air (fed) - Air (theoretical)]/Air (theoretical) }×100.
EXAMPLE: Theoretical and Excess Air
One hundred mol per hour of butane (C 4H10) and 5000 mol per hour of air are fed into acombustion reactor. Calculate the percent excess air.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 63/80
35
Solution
First, calculate the theoretical air from the feed rate of fuel and the stoichiometric equationfor complete combustion of butane.
hair mol
3094Omol
air mol76.4h
Omol650)air (
hOmol
650
HCmol
requiredOmol5.6
hHCmol100
)O(
OH5CO4O2
13
HC
2
2th
2
104
2104ltheoretica2
222104
==
=
=
+→+
Hence
%61.6100%309430945000
100%(air)(air)(air)
air excess%th
thfed=×
−
=×
−
=
If instead you had been given 61.6% excess air, you could have calculated
(air) fed = 1.616 (air) th = 1.616 (3094) = 5000 mol/h.
4.6.4. Combustion Material Balances
When doing material balances for combustion reactions, one should consider balances of thefollowings:
a. inert N 2 at both inlet and outlet
b. unreacted fuel
c. excess oxygen
d. combustion products CO 2, H 2), CO etc.
Suppose 15% excess of oxygen is fed,
oxygen fed = 1.15 × theoretical oxygen
nitrogen fed = (79/21) × oxygen fed
air fed = (100/21) × oxygen fed
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 64/80
36
EXAMPLE: Combustion of Ethane
Ethane (C 2H6) is burned with 50% excess air. The percentage conversion of the ethane is90%; of the ethane burned, 25% reacts to form CO and the balance to form CO 2. Calculatethe composition of the fuel gas and the ratio of water to dry fuel gas.
Solution
Basis: 100 mol Ethane Fed
OH3CO2O25
HC
OH3CO2O27
HC
2262
22262
+→+
+→+
Note:
1. Since no product stream mole fractions are known, subsequent calculations are easier ifindividual component amounts rather than a total amount and mole fractions are labeled.
2. Known information about the composition of air has been used to label the incoming N 2
stream
=
2179
76.3 .
3. If the ethane reacted completely, q1 would be omitted.
4. Since excess air is supplied, O 2 must appear in the product stream.
Theoretical O 2
mol350HCmol1Omol50.3HCmol100
62
262
=
O2 Fed ( )( ) fedOmol5253505.1 2
==Q
N 2 Fed (3.76)(525) = 1974 mol N 2 fed
Q mol O 2 3.76 Q mol N 2
100 mol C 2H6
OHmol
COmol
COmol
Nmol
Omol
HCmol
26
25
4
23
22
621
q
q
q
q
q
q
50% excess air
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 65/80
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 66/80
38
Atomic O Balance
22
2
2
2
2
2
22
2
2
Omol232
OHmol1Omol1OHmol270
COmol1Omol2COmol135
COmol1Omol1OCmol45
Omol1Omol2)Omol(
Omol1Omol2Omol525
OutputInput
=
⇓
+
+
+=
=
q
q
The analysis of the stack gas is now complete. Summarizing:
totalmol2666
OHmol270
gasdrymol2396
COmol135
COmol45
Nmol1974
Omol232
HCmol10
2
2
4
23
22
621
=+
=
=
=
=
=
6
5
q
q
q
q
q
q
Hence the flue gas composition on a dry basis is
molCOmol
0.0563gasdrymol2396
COmol135y
molCOmol
0.019gasdrymol2396
COmol45y
mol
Nmol0.824
gasdrymol2396
Nmol1974y
molOmol
0.0970gasdrymol2396
Omol232y
mol
HCmol0.00417
gasdrymol2396
HCmol10y
225
4
223
222
62621
==
==
==
==
==
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 67/80
39
and the mole ratio of water to dry flue gas is
gasfluedrymolOHmol
113.0gasfluedrymol2396
OHmol270 22 =
EXAMPLE: Combustion of a Natural Gas
A natural gas of an unknown composition is burned with air. An analysis of the product gasyields the following results
0.130 mole H 2O/mole wet gas
Orsat analysis of flue gas: 1.5% CO6.0% CO 2 8.2% O 2
84.3% N 2
Calculate the ratio of hydrogen to carbon in the gas, and speculate on what the gas might be.
Solution
Basis: 100 mol Dry Flue Gas
Since the composition of the fuel is unknown, label it as though carbon and hydrogen wereentering separately.
From the given mole fraction of water,
OHmol9.14gasdrymol870.0OHmol130.0gasdrymol100
22 ==wQ
qH mol HqC mol C
qO2 mol O 2
3.76 qO2 mol N 2
mol/ Nmol843.0
mol/Omol082.0
mol/COmol060.0
mol/COmol015.0
gasdrymol100
2
2
2
Qw mol H 2O
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 68/80
40
Atomic Carbon Balance
COmol1Cmol1
gasfluemolCOmol015.0gasfluemol100
c =q
( )( )( ) Cmol5.7106.0100
COinC 2
=+4 4 84 4 76
Atomic H Balance
CmolHmol
3.97Cmol7.5Hmol29.8
whichfrom
Hmol29.814.92OHmolHmol2O)H(mol
C
H
2
2wH
==
=×==
The natural gas may therefore be written with the formula (CH 3.97 )n. Since there is only onehydrocarbon for which the ratio of H to C is close to 3.97 - that is, CH 4 - we may conclude inthis case that the natural gas is essentially pure methane, with perhaps trace amounts of otherhydrocarbons. [If we had obtained, say, qH/qC ≈ 2, we could have gone no further than towrite the fuel as (CH 2)n: from the information given, there would have been no way todistinguish between C 2H4, C 3H6, a mixture of CH 4 and C 2H2, etc].
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 69/80
41
References
1. Felder, R. and Rousseau, R. W., Elementary Principles of Chemical Processes, 2 nd ed.,John Wily & Sons, 1986.
2. Himmelblau, D. M., Basic Principles and Calculations in Chemical Engineering, 4 th ed.,
Prentice Hall, New Jersey, 1982.
3. Perry, R. H. and Green, D., Perry’s Chemical Engineering Handbook, 6 th ed., McGrawHill, New York, 1984.
4. Al-Amer, M. J., Session 1 of a Short Course on “Introduction to Chemical Engineeringfor Non-Chemical Engineers”, KFUPM, Dhahran, 1998.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 70/80
Chapter 5
PVT CORRELATIONS
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 71/80
1
PVT CORRELATIONS
1. Ideal Gas
To solve most problems that involve gases, an expression is needed that relates
specific volume to temperature and pressure so that if any two of three properties are
known, the third can be calculated. An equation that relates the quantities (mass or
moles) and volume of a gas to its pressure and temperature is call Equation of State .
The simplest and most widely used equation of state is the ideal gas law which is
adequate for many engineering calculations over a wide range of conditions. At high
pressures and low temperatures, however, all gases deviate from ideal behavior andmore complex equations of state are needed to describe PVT relations.
1.1. Ideal Gas Law
The ideal gas law can be derived from the kinetic theory of gases by assuming that gas
molecules have negligible volume, exert no force on one another and collide
elastically with walls of their container. This law can be expressed as:
P V n R T = (1)
where,
P = Absolute pressure of the gas
V = Volume of the gas
n = Number of moles of the gas
R = The gas constantT = Absolute temperature of the gas
The above equation can be written for one mole of the gas, as
P v R T = (2)
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 72/80
2
The gas constant, R, has the units of (Pressure)(volume)/(mole)(temperature). Since
the product of pressure and volume has units of energy, R may also be expressed as
(energy)/(mole)(temperature). The units of R are listed on the last page of Felder and
Rousseau in various system of units.
1.2. Standard Conditions
The reference temperature of 0 °C (273 K, 492 R) and a pressure of one atmosphere
(1 atm) are commonly referred to as standard temperature and pressure. The gas law
gives,
P V n R T s s s s= (3)
or, R P
T V n
s
s
s
s
=
(4)
however, ( ) ( )V n
litersmol
STP ft
lbmol STP s
s
= =22 4 3593
.
The value of gas constant R can therefore be calculated from equation (4).
1.3. Ideal Gas Mixtures
Let us consider a mixture of ideal gases and assume that the mixture itself can be
considered an ideal gas. Let us further assume that the mixture consists of components
A and B and that initially the components are separated and exist at temperature and
pressure of the mixture as shown schematically in the figure below.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 73/80
3
Figure 3. The Amagat’s rule
We can then write the equation of state for the individual component on a molal basis
as follows:
P V n R T A A= (5)
PV n R T B B= (6)
At the same temperature and pressure, if the components are mixed together in one
container, then
P V n R T = (1)It therefore follows that
V V
nn
y A A A= = (7)
This equation simply means that volume fraction (V A/V) is equation to mole fraction.
Further,
n n n A B= + (8)
or PV RT
PV RT
PV RT
A
A
B
B
= + (9)
Gases A + BVolume V
T P
Gas AVolume V A
Gas BVolumeVB
T P P
T
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 74/80
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 75/80
5
Further,
n n n A B= + (8)
RT
V P
RT
V P
RT
PV B A+= (14)
i.e. A P P P += (15)
where each partial pressure is evaluated at the temperature and volume of the mixture.The above equation is called Dalton’s law of Partial Pressure .
2. Real Gas
As the temperature of the gas decreases and the pressure increases, the ideal gas law
discussed earlier provides increasingly poor description of the gas behavior. It
becomes essential to turn to other equations of state which may satisfactorily represent
the PVT behavior.
One approach to represent the PVT behavior of real gases is to use equations of state
that contain number of constants whose value differ for different gases. Another
approach is to correct the deviation from ideal behavior by including a term called the
compressibility factor in the ideal gas law. The resulting equation is called the
compressibility factor equation of state.
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 76/80
6
2.1. Compressibility Factor Equation of State
An equation of state that retains the simplicity of ideal gas law but describes the PVT
behavior of real gases over a much wider rage of conditions is of the following form:
PV Z n R T = (16)
P v Z R T = (17)
The coefficient Z is called the compressibility factor and the equation is called the
compressibility equation of state. A value of Z = 1 corresponds to ideal gas behavior.
The compressibility factor depends on the gas temperature and pressure and differs for
different gases at a given T and P.
2.2. Estimation of Compressibility Factors
2.2.1 Law of Corresponding States
It is possible to construct a single universal chart for the estimation of compressibility
factors for all gases if independent variables other than P and T are used. Figure 5
(Figure 5.4 in Felder and Rousseau)) shows such a plot called generalized
compressibility chart. The variables P and T and replaced by P r = (P/P c), and T r =
(T/T c) where P c and T c are critical pressure and temperature respectively of the gas.
The basis for the generalized compressibility chart is the experimentally observed fact
that the value of certain physical properties of gases, such as compressibility depend
on a greater extent on how near the gas is to its critical state. This suggests that a plot
of Z versus reduced temperature (T r ) and reduced pressure (P r ) should be
approximately be the same for all substances which indeed is the case. This assertion
is known as law of corresponding states .
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 77/80
7
2.3. Real Gas Mixtures
The techniques that is both simple and reasonably accurate is Kay’s Rule which usesthe compressibility factor equation of state:
PV Z n R T m= (18)
P v Z R T m= (19)
with Z m as the mean compressibility factor.
In order to determine Z m for the mixture of gases, A, B, C …… with the composition
of mole fraction y A, y B, y C ……the pseudocritical constants of the mixture as average
values of critical constants of the mixture components are calculated as follows:
Pseudocritical Temperature
T y T y T y T c A cA B cB C cC ' = + + + LLL (20)
Pseudocritical pressure
P y P y P y P c A cA B cB C cC ' = + + + LLL (21)
If the temperature and pressure of the mixture are known, pseudo reduced temperature
T T
T r
c
' '
= and pseudo reduced pressure P P
P r
c
' '
= are calculated and used to obtain
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 78/80
8
the value of Z m for the mixture using the generalized compressibility charts. This
value of Z m can be used to calculate V or v as desired from Eq. (18) or (19).
4. Selected Instruments For Process Variable
Measurement
4.1. Fluid Pressure Measurement
Several mechanical devices are used for the measurement of fluid pressure. The most
common of these devices is a Bourdon Gauge which is a hollow tube closed at one
end and bent into a C congifuration. The open end of the tube is exposed to the fluid
whose pressure is to be measured. As the pressure increases, the tube tends to
straighten, causing a pointer attached to the tube to rotate. The position of the pointer
on the calibrated dial gives the gauge pressure of the fluid. Figure 6 shows the
schematic diagram of the Bourdon gauge.
Figure 6. Bourdon Gauge
P (psi)
P
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 79/80
9
A manometer is a U-shaped tube partially filled with a liquid of known density. When
the ends of the tube are exposed to different pressures the fluid levels drops in the high
pressure arm and rises in the low pressure arm. The difference between the pressures
can be calculated from the measured difference between the liquid level in each arm.
Manometers are used in different ways as shown in Figure 7 (Figure 3.4-4 in Felder
and Rousseau). In each arm P 1 and greater than P 2.
Figure 7a shows an open ended manometer. One end is exposed to the fluid whose
pressure is to be measured and the other end to the atmosphere. Figure 7b shows a
differential manometer, used to measure the pressure difference between two points in
a process line. Figure 7c shows a sealed end manometer which has a vacuum enclosedat one end. If the open end of the sealed-end manometer is exposed to the atmosphere
(P1 = P atm) the device functions like a barometer.
4.2. Temperature Measurement
The temperature of a substance in a particular state of aggregation (solid, liquid or
gas) is a measure of the average kinetic energy possessed by the substance molecules.
Since the energy cannot be measured directly, the temperature must be determined
indirectly by measuring some physical property of the substance whose value depends
on the temperature in a known manner. Such properties and the temperature
measuring devices based on them include electrical resistance of the conductor
( Resistance Thermometer ) voltage at the junction of two dissimilar metals
(Thermocouple ), spectra of emitted radiation ( Pyrometer ), and volume of fixed massof fluid ( Thermometer ).
4.3. Flow Rate Measurement
8/13/2019 Introducing Chemical Engineering
http://slidepdf.com/reader/full/introducing-chemical-engineering 80/80
A flow meter is a device mounted in a process line that provides a continuous reading
of flow rate in the line. Two commonly used flow meters - The Rotameter and the
Orifice meter are shown in Figure 8 (Figure 3.2-1 in Felder and Rousseau).
The rotameter is a tapered vertical tube containing float, the larger the flow rate, the
higher the float rises in the tube. The orifice meter is an obstruction in flow channel
with a narrow opening through which the fluid passes. The fluid pressure drops from
the upstream side of the orifice to the downstream side. The pressure drop varies with
the flow rate; the greater the flow rate the larger is the pressure drop.