UNIT ONE
INDUSTRIAL PROCESSES
Industrial processes are procedures involving chemical, physical, electrical or mechanical steps to aid
in the manufacturing of an item or items, usually carried out on a very large scale.
Industrial processes are the key components of heavy industry. The parameters or quantities that we
wish to control at the correct limit are called Process Variables. A variable is something that can vary
or change. Because process variables can and do change, instrumentation systems measure the variable
then control the variable to keep the variable within the given limits.
A process variable, process value, process quantity or process parameter is the current measured
value of a particular part of a process which is being monitored or controlled at the correct limit. An
example of this would be the temperature of a furnace. Four commonly measured variables which
affect chemical and physical processes are: pressure, temperature, level and flow.
Process- any operation that causes a physical or chemical change in a substance and can consist of
several process units. Process variable is a condition of the process fluid (a liquid or gas) that can
change the manufacturing process in some way. Common process variables include: pressure, density,
flow rate, chemical composition, temperature and mass.
Unit operation- In chemical engineering and related fields, a unit operation is a basic step in a process.
Unit operations involve a physical change or chemical transformation such as separation,
crystallization, evaporation, filtration, polymerization, isomerization, and other reactions. For example,
in milk processing, homogenization, pasteurization, and packaging are each unit operations which are
connected to create the overall process. A process may require many unit operations to obtain the
desired product from the starting materials, or feed stocks.
Chemical engineering involves the conversion of raw materials into your desired product at least
possible cost. To convert raw materials into products there are two processes involved, unit operations
and unit processes.
Unit process: it involves the conversion of raw material into product chemically (by a chemical
reaction) means your raw material which had some chemical compound is changed to your product
which has different chemical compound.
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Unit operations: it involves the separation (by means of heat transfer, mass transfer, solubility, barrier,
transportation, mixing, blending, size reduction of components only physically (no chemical reaction).
Unit operations can be used before and after unit process according to the desired property of the raw
material and product.
Unit operations are processes related to physical changes of the materials involved:

Mixing (Solid-Solid, Solid-Liquid, Solid-Gas, Liquid-Liquid, Liquid-Gas, Gas-Gas)

Separating (Sieving, Filtration, Distillation, Fractional crystallization, Fractional freezing,
Centrifugation)

Heating / Cooling of materials

Transporting (Solids, Granulates, Liquids, Slurries, Pastes, Gases)

Measuring (Weighing, Volumetrizing, Counting)
Unit processes are those related to chemical changes of the materials involved:

Neutralization

Oxidation / Reduction

Esterification / Saponification

Nitration

Sulfonation

Polymerization

Diazotization

Chlorination

Alcylation
A. Mass and Density
Density of a substance is defined as the ratio of the mass of the substance to its volume.
Units are:
SI → kg/m3
CGS → g/cm3
American Eng. → lbm/ft3
Density is denoted mathematically with the Greek symbol ρ
For solid and liquids, density is independent of pressure and changes relatively small with
temperature.
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Densities of liquids and solids are assumed to be constant (incompressible). For Gases, density is a
strong function of pressure and temperature. Effects of temperature and pressure on gaseous
systems can’t be ignored.
Density of a substance can be used as conversion factor to relate the mass and volume of a quantity of
the substance.
The specific gravity is the ratio of the density of a substance ρ to the density of a reference substance
ρref.
SG =
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑎 𝑠𝑢𝑏𝑠𝑡𝑎𝑛𝑐𝑒
𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑜𝑓 𝑟𝑒𝑓𝑒𝑟𝑒𝑛𝑐𝑒
⍴
= ⍴𝑟𝑒𝑓
Reference substance is water in the liquid form at 4OC:
ρH2O(l)(4OC) = 1000 kg/m3
=1.000 g/cm3
= 62.43 lbm/ft3
Note:
The specific gravity of a substance at 20 OC relative to water at 4 OC is 0.6.
Example
1. The density of carbon tetrachloride is 1.595 g/cm3, what is the mass of 20.0 cm3 of CCl4?
2. A liquid has a specific gravity of 0.50. What is its density in g/cm3? What is its specific volume in
cm3/g? What is its density in Ibm/ft3? What is the mass of 30.0 cm3 of this liquid? What volume is
occupied by 18g?
B. Flow Rate
Once they are connected, water will flow at a certain rate (Quantity/ time)
Quantity can be expressed in terms of mass or volume to give:
1. Mass flow rate.
2. Volumetric flow rate.
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Physical meaning of flow rates
Mass flow rate is denoted as m
Volumetric flow rate is denoted as v
Flow rate is defined as the amount of fluid in mass or volume crossing a perpendicular cross-sectional
area in a specific time.
Mass and volumetric flow rates are related to fluid density through:
Fluid density is used to convert between mass flow rate and volumetric flow rates
Example:
The mass flow rate of n-hexane (ρ = 0.659 g/cm3) in a pipe is 6.59 g/s. what is the volumetric flow rate
of hexane?
C. Mass, Mole Fractions and Average Molecular Weight:
Composition of a mixture of substances can be expressed in terms of:
1. Mass fraction: It is defined for each component in the mixture as the ratio of the mass of a
component to the total mass of the mixture.
2. Mole fraction: It is defined for each component in the mixture as the ratio of the mole of a
component to the total mole of the mixture.
Example
Conversions Using Mass and Mole Fractions 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.
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2. Calculate the mass flow rate of A in a stream of solution flowing at a rate of 53 lbm/h.
Exercise
The specific gravity of gasoline is approximately 0.70.
A. Determine the mass (kg) of 50.0 liters of gasoline.
B. If the mass flow rate of gasoline delivered by a gasoline pump is 1150 kg/min, what is the
corresponding volumetric flow rate in liters/s.
Average molecular weight is the ratio of the mass of a sample to the number of moles of all species in
the sample
Example
Calculation the Average Molecular Weight air:
1. From its approximate molar composition of 79% N2, 21% O2.
Parts per Million and Parts per Billion
They are used to express concentration of traces, i.e. TOO SMALL CONCENTRATION
ppmi = yi × 106
ppbi = yi × 109
Mass fractions are used for liquids while mole fraction are used for gases
Example: concentration of SO2 in air is15.0 ppm
It means in every 106 moles of air, there are 15 moles of SO2
Example:
Concentrations
A solution of common salt in water is prepared by adding 20 kg of salt to 100 kg of water, to make a
liquid of density 1323 kg/m3. Calculate the concentration of salt in this solution as a
A. weight fraction,
B. weight/volume fraction,
C. mole fraction,
D. molal concentration.
Solution
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A. Weight fraction:
20 / (100 + 20) = 0.167: % weight / weight = 16.7%
B. Weight/volume:
A density of 1323kg/m3 means that lm3 of solution weighs 1323kg, but 1323kg of salt solution
contains (20 x 1323 kg of salt) / (100 + 20) = 220.5 kg salt /m3
1m3solution contains 220.5 kg salt.
Weight/volume fraction = 220.5 / 1000 = 0.2205
And so weight / volume = 22.1%
C. Moles of water = 100 / 18 = 5.56
Moles of salt = 20 / 58.5 = 0.34
Mole fraction of salt = 0.34 / (5.56 + 0.34) = 0.058
D. The molar concentration (M) is 220.5/58.5 = 3.77 moles in m3
Example:
Air Composition
If air consists of 77% by weight of nitrogen and 23% by weight of oxygen calculate:
A. the mean molecular weight of air,
B. the mole fraction of oxygen,
C. the concentration of oxygen in mole/m3 and kg/m3if the total pressure is 1.5 atmospheres and the
o
temperature is 25 C.
Solution
A. Taking the basis of 100 kg of air: it contains 77/28 moles of N2 and 23/32 moles of O2
Total number of moles = 2.75 + 0.72 = 3.47 moles.
So mean molecular weight of air = 100 / 3.47 = 28.8
Mean molecular weight of air = 28.8
B. The mole fraction of oxygen = 0.72 / (2.75 + 0.72) = 0.72 / 3.47 = 0.21
Mole fraction of oxygen = 0.21
C. In the gas equation, where n is the number of moles present:
The value of R is 0.08206 m3 atm/mole K and at a temperature of 25OC = 25 + 273 = 298 K, and
where V= 1 m3
pV = nRT
and so, 1.5 x 1 = n x 0.08206 x 298
n = 0.061 mole/m3
Weight of air = n x mean molecular weight
= 0.061 x 28.8 = 1.76 kg /m3
and of this 23% is oxygen, so weight of oxygen = 0.23 x 1.76 = 0.4 kg in 1 m3
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Concentration of oxygen = 0.4kg/m3
or 0.4 / 32 = 0.013 mole/m3
Mass balance and Energy Balance
A mass balance, also called a material balance, is an application of conservation of mass to the analysis
of physical systems. By accounting for material entering and leaving a system, mass flows can be
identified which might have been unknown, or difficult to measure without this technique. Material
quantities, as they pass through processing operations, can be described by material balances. Such
balances are statements on the conservation of mass. Similarly, energy quantities can be described by
energy balances, which are statements on the conservation of energy. If there is no accumulation, what
goes into a process must come out. This is true for batch operation. It is equally true for continuous
operation over any chosen time interval.
Material and energy balances are very important in an industry. Material balances are fundamental to
the control of processing, particularly in the control of yields of the products. The first material
balances are determined in the exploratory stages of a new process, improved during pilot plant
experiments when the process is being planned and tested, checked out when the plant is commissioned
and then refined and maintained as a control instrument as production continues.
The increasing cost of energy has caused the industries to examine means of reducing energy
consumption in processing. Energy balances are used in the examination of the various stages of a
process, over the whole process and even extending over the total production system from the raw
material to the finished product. Material and energy balances can be simple, at times they can be very
complicated, but the basic approach is general. Experience in working with the simpler systems such
as individual unit operations will develop the facility to extend the methods to the more complicated
situations, which do arise. The increasing availability of computers has meant that very complex mass
and energy balances can be set up and manipulated quite readily and therefore used in everyday
process management to maximize product yields and minimize costs.
The law of conservation of mass leads to what is called a mass or a material balance.
Mass In = Mass Out + Mass Stored
Raw Materials = Products + Wastes + Stored Materials.
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Σ mR = ΣmP + Σ mW + ΣmS
(Where Σ (sigma) denotes the sum of all terms).
ΣmR = ΣmR1 + Σ mR2 + ΣmR3 = Total Raw Materials
ΣmP = ΣmP1 + Σ mP2 + ΣmP3 = Total Products.
ΣmW= ΣmW1 + Σ mW2 + ΣmW3 = Total Waste Products
ΣmS = ΣmS1 + Σ mS2 + ΣmS3 = Total Stored Products.
If there are no chemical changes occurring in the plant, the law of conservation of mass will apply also
to each component, so that for component A:
mA in entering materials = mA in the exit materials + mA stored in plant.
For example, in a plant that is producing sugar, if the total quantity of sugar going into the plant is not
equaled by the total o f the purified sugar and the sugar in the waste liquors, then there is something
wrong. Sugar is either being burned (chemically changed) or accumulating in the plant or else it is
going unnoticed down the drain somewhere.
Raw Materials = Products + Waste Products + Stored Products + Losses
Where Losses are the unidentified materials.
Just as mass is conserved, so is energy conserved in food-processing operations. The energy coming
into a unit operation can be balanced with the energy coming out and the energy stored.
Energy In = Energy Out + Energy Stored
ΣER = ΣEP + ΣEW + ΣEL + ΣES
Where
ΣER = ER1 + ER2 + ER3 + ……. = Total Energy Entering
ΣEp = EP1 + EP2 + EP3 + ……. = Total Energy Leaving with Products
ΣEW = EW1 + EW2 + EW3 + … = Total Energy Leaving with Waste Materials
ΣEL = EL1 + EL2 + EL3 + ……. = Total Energy Lost to Surroundings
ΣES = ES1 + ES2 + ES3 + ……. = Total Energy Stored
Energy balances are often complicated because forms of energy can be inter converted, for example
mechanical energy to heat energy, but overall the quantities must balance.
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Law of conservation of mass and energy
Law of conservation of mass and energy states that, mass and energy can neither be created nor be
destroyed. It can be transformed from one form to another. So, the total mass or energy applied to a
process remains same. If one kilogram of mango pulp is put in a dryer, the sum of the amount of dried
pulp and the water evaporated will be one kilogram. The general formula for mass and energy balance
adopted in chemical or food processes is given below.
𝐼𝑛𝑓𝑙𝑜𝑤=𝑂𝑢𝑡 𝑓𝑙𝑜𝑤+𝐴𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑖𝑜𝑛
Accumulation means, the mass or energy absorbed or stored in a process. If the accumulation becomes
zero, the inflow will be equal to out flow. This kind of processes are said to be in steady state where as
the processes with non zero accumulation are called unsteady state processes. In the process of
evaporation of milk, milk passes to the calendria and thick consistent milk comes out leaving the
moisture as vapour form.
The mass balance around the evaporator will be;
𝑅𝑎𝑤 𝑚𝑖𝑙𝑘 = 𝑚𝑖𝑙𝑘 𝑎𝑐𝑐𝑢𝑚𝑢𝑙𝑎𝑡𝑒𝑑 𝑖𝑛𝑠𝑖𝑑𝑒 + 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑒𝑑 𝑚𝑖𝑙𝑘 + 𝑣𝑎𝑝𝑜𝑢𝑟
The energy equation can be written as;
𝑒𝑛𝑒𝑟𝑔𝑦 𝑖𝑛 = 𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑢𝑡 + 𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑜𝑠𝑠
The chemical processes can be classified as batch, continuous and semi-batch:
1. Batch process:
The feed materials are placed into the system (reactor, mixer, filter,….etc) at the beginning of the
process. These materials are held for a period of time known as "residence time" or " retention
period" during which the required physical and/or chemical changes are occurred. The products are
removed all at once after this time. No masses crossed the system boundary during this time. Batch
process fall into the category of closed systems.
The basis used in such processes is usually "one batch", and the material balance for physical batch
processes in which there is no chemical reaction can be written as:
Input (Initial quantity) = Output (Final quantity)
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Figure 1: The initial state of a batch mixing process
Figure 2: The final state of a batch mixing process
2. Semi-batch process:
In this type, all quantity of one reactant is initially put in the reactor, and then other reactants are
continuously fed. Only flows enter the systems, and no leave, hence the system is an unsteady state.
This arrangement is useful when the heat of reaction is large. The heat evolved can be controlled by
regulating the rate of addition of one of the reactants.
Figure 3: Semi-batch reactor (Stirred type reactor)
3. Continuous process (Flow process):
The input and output materials are continuously transferred across the system boundary; i.e. the feed
continuously enters the system and the product continuously leaves the system. The physical and/or
chemical changes take place during the flow of materials through the effective parts of equipments
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(packing, sieve plate, filter cloth,…etc). A convenient period of time such as minute, hour, or day must
chosen as a basis over which material balance calculations be made.
This type of processes can be classified as "steady state" and "unsteady state" processes.
a- Steady state process:
The steady state process can be defined as that process in which all the operating conditions
(temperature, pressures, compositions, flow rate,…..etc.) remains constant with time. In such process
there is no accumulation in the system, and the equation of material balance can be written as:
Input = Output
b- Unsteady state process:
For an unsteady state process, not all of the operating conditions in the process (e.g., temperature,
pressure, compositions, flow rate,… etc.) remain constant with time, and/or the flows in and out of the
system can vary with time, hence the accumulation of materials within can be written as follows:
Input - Output = Accumulation
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Example
It is required to prepare 1250 kg of a solution composed of 12 wt.% ethanol and 88 wt.% water.
Two solutions are available, the first contains 5 wt.% ethanol, and the second contains 25 wt.%
ethanol. How much of each solution are mixed to prepare the desired solution?
Solution
Ethanol balance
Input = output
5
25
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A 100 + B 100= M 100
0.05 A + 0.25 B = 0.12 M
A=
150−0.25 𝐵
0.05
=300-5 𝐵……………………………………………………………….….𝟏
Water balance
Input = output
0.95 A + 0.75 B = 0.88 M = 0.88 (1250) =1100
0.95 A+ 0.75 B=1100…………………………………………………………………….2
Sub. (1) in (2)
0.95(300-5 B) + 0.75 B = 1100
2850 - 4.75 + 0.75 B =1100
4 B =1750
B= 437.5 kg
Sub. B in (1)
A= 3000 – 5(437.5) = 812.5 kg
3. Checking: Total material balance (T.M.B.), Input = A + B = 437.5 + 812.5 =1250 kg
Output = M= 1250 kg
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