Concepts in Material & Energy Balance

Computations
3130508

Dr. Vyomesh M. Parsana

Assistant Professor
Chemical Engineering Department
VVP Engineering College, Rajkot

Concepts in MEBC Dr. Vyomesh M. Parsana 1/202

 Teaching & Examination Scheme

Concepts in MEBC Dr. Vyomesh M. Parsana 2/202

Reference Books

Basic Principles & Calculations in Chemical Engineering,

D.M.Himmelblau. 6th Ed., 2004
Stoichiometry, B.I.Bhatt & Thakore, Tata McGraw Hill Book
Company, 5th Ed, 2010
Chemical Process Principles, Vol.1, O.A.Hougen,
K.M.Watson, R.A.Ragatz., Indian print, CBS Publishers, 2nd
Ed., 1995
Stoichiometry & Process Calculations, Narayanan K.V., &
Lakshmikutti B., Prentice Hall, 2006
Process Calculations, V Venkataramani and N Anantharaman,
PHI Learning, 2004
Chemical Process Calculations Manual, David Carr
Igbinoghene, McGraw Hill Professional, 2004
Optimization of Chemical Processes, T F Edgar, D M
Himmelblau and L S Lasden, Tata McGraw Hill, 2001
 Course Outcomes

Concepts in MEBC Dr. Vyomesh M. Parsana 4/202

 Introduction

What is MEBC?
It is basics of Mathematics and Chemistry. The main objective of
course is to make a clear conceptualized knowledge regarding
various unit operations carried out in Chemical Engineering. This
will provide a background for applying these principles to industrial
problems.

Why do we need to Study MEBC?

Concepts in MEBC Dr. Vyomesh M. Parsana 5/202

 Unit
Units are the means of expressing the Dimensions, such as feet(ft)
& centimeter(cm) for length or hours(h) & seconds(s) for time.

Dimension
Dimensions are our basic concepts of physical measurement such
as length(L), mass(M), time(T), temperature(θ) and so on.

Concepts in MEBC Dr. Vyomesh M. Parsana 7/202

 Dimensional Consistency

The equations written in terms of dimensions are called

Dimensional Equations.
A basic principle states that the equations must be
dimensionally constant.
Consistent use of dimensional equations is utmost important.
In Engineering and Science, Dimensional analysis is the
analysis of the relationships between different physical
quantities by identifying their base quantities and units of
measure and tracking these dimensions as calculations or
comparisons are performed.

Concepts in MEBC Dr. Vyomesh M. Parsana 8/202

 Different System of Units
FPS
Foot Pound Second
This system was developed in England.
The modern name of this unit system is US Customery unit
system.

MKS
Meter Kilogram Second
In 1791, in France, a system of units entirely based on the
unit of length, the metre was created.
Because of its foundation being entirely based on the metre,
this system got the name Metric System.
It was adopted in the year 1889.
Concepts in MEBC Dr. Vyomesh M. Parsana 9/202
 Different System of Units

CGS
Centimeter Gram Second
It is a subsidiary system which was derived from the MKS
system.

Concepts in MEBC Dr. Vyomesh M. Parsana 10/202

 Different System of Units

SI
The International System of Units (SI, abbreviated from the
French Système International (d’unités)) is the modern form
of the Metric System.
The above 3 systems were not been accepted by all the
countries as a part of common Unit System.
For better understanding, in particular in Science and
Technology and in international and trade relations, a need for
an International System of units was felt.
So, in the 11th general conference on Weights and Measures
held in the year 1960 at Paris in France, a common system of
units was accepted by all the countries and was called
International System Of Units i.e. SI units.

Concepts in MEBC Dr. Vyomesh M. Parsana 11/202

 Physical Quantity

It is a property of a material or system that can be quantified

by measurement.
A physical quantity can be expressed as the combination of a
numerical value and a unit.
Basically the physcial quantities are categorized into two parts
i.e. Fundamental Quantities and Derived Quantities.

Concepts in MEBC Dr. Vyomesh M. Parsana 12/202

 Fundamental Quantity Derived Quantity
It is independent These are the quantities
Physical Quantity which which are calculated
we can’t express in other from two or more
Physical Quantity. measurements.
These quantites can be These quantities can be
measured directly. computed only.
As per the SI unit There are huge number
system, basically there of Derived Quantites
are 7 Fundamental available.
Quantities. Ex.:- Area, Density....
Ex.:- Length, Mass....

Concepts in MEBC Dr. Vyomesh M. Parsana 13/202

 Fundamental Quantity Units Dimensions
Length Metre L
Mass Kilogram M
Time Second T
Temperature Kelvin θ
Current Ampere I
Amount of Substance Mole η
Luminous Intesity Candela -

Derived Quantity Units Dimensions

Area (Metre)2 L2
Density Kilogram/(Metre)3 M 1 L−3
Velocity Metre/Second L1 T −1
Work N ∗ Metre M 1 L2 T −2
Freqency 1/Second T −1
Length Volume
1 ft = 0.3048 m 1 US gal = 3.7854 L
1 mile = 1.609 km 1 UK gal (Imperial gal) = 4.5461 L
1 in = 2.54 cm 1 m3 = 1000 L
1 ft = 12 in

Energy /Power Force

1 Cal (IT ) = 4.186 J 1 kgf = 9.8066 N
1 hp = 745.7 W 1 lbf = 4.4482 N
1 BTU = 1055.04 J
1 W = 1 J/s

Mass
1 lb = 0.4536 kg
Pressure
1 atm = 1.01325 bar = 1.01325 × 105 N/m2 (Pa) = 101.325 kPa =
760 mmHg (torr ) = 14.7 psi = 1.033 kgf /cm2 = 407.189 inch. WC

Temperature
T (K ) = T (◦ C ) + 273.15
T (◦ F ) = 1.8T (◦ C ) + 32
T (R) = 1.8T (K )
T (◦ F ) = 1.8T (◦ K ) − 459.67
∆(◦ F ) = ∆1.8T (◦ C )
∆(◦ F ) = ∆1.8T (K )
Numericals based on Units & Dimensions

Example
1100 ft/s to miles/hour

Example
400 in3 /day to cm3 /min
Numericals based on Units & Dimensions

Example
42 ft 2 /hr to cm2 /sec

Example
25 psig to psia
Numericals based on Units & Dimensions

Example
1 mm/ ◦ F
Numericals based on Units & Dimensions

Example
In a double effect evaporator plant, the second effect is maintained
under vaccum of 475 torr (mm Hg). Find the absolute pressure in
kPa, bar and psi.
Numericals based on Units & Dimensions

Example
Iron metal weighing 500 lb occupies a volume of 29.25 lit.
Calculate the density of Fe in g /cm3 .
Numericals based on Units & Dimensions

Example
The diameter and height of a vertical cylindrical tank are 5 ft and
6 ft 6 inch respectively. It is full up to 75% height with carbon
tetrachloride (CCl4 ), the density of which is 1.6 kg/lit. Find the
mass in kg.
Numericals based on Units & Dimensions
Example
In case of liquids, the local heat-transfer coefficient for long tubes
and using bulktemperature properties is expressed by the empirical
equation
G 0.8 k 0.67 CP0.33
h = 0.023 D 0.2 µ0.47

where,
Btu
h = heat-transfer coefficient, h.ft 2 .◦ F
lb
G = mass velocity of liquid, ft 2 .s
Btu
CP = heat capacity, lb.◦ F
Btu
k = thermal conductivity, h.ft.◦ F
D = diameter of tube, ft
lb
µ = viscosity of liquid, ft.s
Is the given equation dimensionally consistent? If yes, convert the
equation into SI units.
Numericals based on Units & Dimensions

Example
The empirical equation for laminar flow heat transfer to flat plate
is given by
1/3 1/2
k 2/3 CP u0 ζ 1/2
hx = 0.332 q
x 3/4
x 1/2 µ1/6 1−( x0 )

where,
Btu
hx = heat transfer coefficient, s.ft 2 .◦ F
Btu
CP = heat capacity, lb.◦ F
lb
ζ = density, ft 3
Btu
k = thermal conductivity, s.ft.◦ F
lb
µ = viscosity of liquid, ft.s
Example
x = distance from leading edge of plate or from the tube
entrance, ft
x0 = distance at heated section, ft
Convert the empirical equation into metric units.
Numericals based on Units & Dimensions

Example
The conductance of a fluid –flow sytem is defined as teh
volumetric flow rate, reffered to a pressure of one torr (133.322
Pa.). For an orifice, the conductance C can be computed from
q
T 3
C = 89.2A M ft /s

where,
A = area of opening, ft 2
T = Temperature, ◦ R
M = Molecular Weight
Convert the empirical equation into SI units.
Numericals based on Units & Dimensions

Example
Vapour pressure of benzene in the temperature range of 280.65 K
to 377.15 K can be calculated using the following Antoine equation
1211
log10 P = 6.9057 − T +220.8

where,
P = Vapour pressure in torr
T = Temperature in ◦ C
Convert the above equation in SI units.
 Mole

The amount of substance which contains as many elementary

entities as there are atoms in 0.012 kg of Carbon-12 (C12 ) is
defined as a mole.

What do we mean by mole?

A mole of a substance or a mole of particles is defined as
exactly 6.02214076 × 1023 particles, which may be atoms,
molecules, ions, or electrons.
∴ 1mol = 6.02214076 × 1023

weight of A
η of A= molecular weight of A

Concepts in MEBC Dr. Vyomesh M. Parsana 29/202

 Atomic Weight

What is Atomic Weight?

The atomic mass is a weighted average mass of all of the
isotopes of that element.
The atomic mass of a single atom is simply its total mass and
is typically expressed in atomic mass units or amu.

Example
H=1 amu
C=12 amu
O=16 amu
Cl=35.5 amu

Concepts in MEBC Dr. Vyomesh M. Parsana 30/202

 Molar Mass

What is Molar Mass?

The molar mass is the mass of a given chemical element or
chemical compound (g) divided by the amount of substance
(mol).

How to Calculate Molar Mass?

The molar mass of a compound can be calculated by adding
the standard atomic masses (in g /mol) of the constituent
atoms.

Concepts in MEBC Dr. Vyomesh M. Parsana 31/202

 Molecular Weight

What is Molecular Weight?

Molecular Weight is a measure of the sum of the atomic
weight values of the atoms in a molecule.

Applications & Units

It is used in chemistry to determine stoichiometry in chemical
reactions and equations.
It is either unitless or expressed in terms of atomic mass units
(amu) or Daltons (Da).
It is calculated in practice by summing the atomic weights of
the atoms making up the substance’s molecular formula.

Concepts in MEBC Dr. Vyomesh M. Parsana 32/202

 Example
Let us Calculate M.W. of C6 H14 i.e. Hexane.
The subscripts indicate the number of each type of atom, so
there are 6 carbon atoms and 14 hydrogen atoms in each
hexane molecule.
M.W. = (number of C atoms)(C atomic weight) +
(number of H atoms)(H atomic weight)
=(6Ö12)+(14Ö1)
=72+14
=86 amu

Concepts in MEBC Dr. Vyomesh M. Parsana 33/202

Numericals based on Mole, Atomic Weight & Molecular
Weight

Example
How many grams of NH4 Cl are there in 5 mol?

Example
How many moles of K2 CO3 will contain 117 kg K?
Numericals based on Mole, Atomic Weight & Molecular
Weight

Example
Convert CuSO4 .5H2 O into mole. Find equivalent mole of CuSO4
in the crystals.
 Equivalent Weight

What is Equivalent Weight?

The equivalent weight of an element or a compound is equal
to the atomic weight or molecular weight divided by the
valence.
It is the mass of a given substance which will combine with or
displace a fixed quantity of another substance.
The valency of an element or a compound depends on the
number of hydrogen ions accepted or the hydroxyl ions
donated for each atomic weight or molecular weight.
One equivalentweight of an element or a compound has
precisely the same power for chemical combination as one
equivalent weight of any other element or compound.

Concepts in MEBC Dr. Vyomesh M. Parsana 36/202

 M.W .
Eq.Wt. = Valency

Example Example

For − HCl For − Na2 CO3

M.W . M.W .
Eq.Wt. = Eq.Wt. =
Valency Valency
36.5 106
= =
1 2
=36.5 =53
For − CaCl2 For − H3 PO4
M.W . M.W .
Eq.Wt. = Eq.Wt. =
Valency Valency
111 98
= =
2 3
=55.5 =32.66
 Percentage Composition of Solids

In a mixture of two compounds A and B.

Weight % of A
weight of A
= weight of A + weight of B × 100

Mole % of A
moles of A
= moles of A + moles of B × 100

Concepts in MEBC Dr. Vyomesh M. Parsana 38/202

Numericals based on Percentage Composition

Example
Sodium Chloride weighing 600 kg is mixed with 200 kg Potassium
Chloride. Find the composition of the mixture in weght % and
mole %.
Solution:
Molecular weight of NaCl=40
Molecular weight of KCl=74.5
weight of NaCl
Weight % of NaCl = × 100
weight of NaCl + weight of KCl
600
= × 100
600 + 200
600
= × 100
800
=75%
Numericals based on Percentage Composition

Example

weight of KCl
Weight % of KCl = × 100
weight of NaCl + weight of KCl
200
= × 100
600 + 200
200
= × 100
800
=25%

ηNaCl = 10.26
ηKCl = 2.68
ηNaCl
Mole % of NaCl = × 100
ηNaCl + ηKCl
10.26
= × 100 =79.29%
10.26 + 2.68
Numericals based on Percentage Composition

Example

ηKCl
Mole % of NaCl = × 100
ηNaCl + ηKCl
2.68
= × 100
10.26 + 2.68
=20.71%
Numericals based on Percentage Composition

Example
A saturated solution of salicylic acid in methanol contains 64 kg of
salicylic acid per 100 kg methanol at 298K. Find the weight% and
mole% composition of the solution.
Numericals based on Percentage Composition

Example
Find nitrogen (N) content of 100 kg urea sample containing
96.43% urea.
Numericals based on Percentage Composition

Example
What will be the % Na2 O content of lye containing 73% caustic
soda?
 Solution

What is Solution?
It is a special type of homogeneous mixture composed of two
or more substances.
In such a mixture, a solute is a substance dissolved in another
substance, known as a solvent.
The solute can be a solid, a liquid, or a gas.

Solubility
The maximum amount of solid which can be dissolved in the
solvent will be equal to its solubility at that particular temperature.

Concepts in MEBC Dr. Vyomesh M. Parsana 45/202

 Molarity

It is defined as number of moles of solute dissolved in 1 litre

of solution.
moles of solute
Molarity (M) = volume of solution (in lit)

Concepts in MEBC Dr. Vyomesh M. Parsana 46/202

 Normality

It is defined as the number of gram equivalents dissolved in 1

litre of solution.
equivalent weight of solute
Normality (N) = volume of solution (in lit)

Concepts in MEBC Dr. Vyomesh M. Parsana 47/202

 Molality

It is defined as the number of moles of solute dissolved in 1 kg

of solvent.
moles of solute
Molality (m) = kg of solvent

Concepts in MEBC Dr. Vyomesh M. Parsana 48/202

Numericals based on Molarity, Normality & Molality

Example
A solution of Caustic Soda in water contains 20% NaOH (by wt.)
at 333 K. The density of the solution is 1.196 kg/l. Find Molarity,
Normality and Molality of the solution.
Solution:
Basis - 1ltr of Solution
Mass of Solution = 1.196 kg
Mass of NaOH = 1.196×0.2
=239.2 g
As Valency of NaOH is 1
∴ Molecular Weight = Equivalent Weight
239.2
η of NaOH = 40
=5.98
Numericals based on Molarity, Normality & Molality
Example
Molarity

moles of NaCl
Molarity (M) =
volume of solution (in lit)
5.98
=
1
=5.98 M

Normality

equivalent weight of NaCl

Normality (N) =
volume of solution (in lit)
5.98
=
1
=5.98 N
Numericals based on Molarity, Normality & Molality

Example
Molality
Mass of Solvent = Mass Solution - Mass Solute
=1.196-0.2392
=0.9568 kg
moles of NaCl
Molality (m) =
kg of H2 O
5.98
=
0.9568
=6.25 N
Numericals based on Molarity, Normality & Molality

Example
A H2 SO4 solution has a Molarity of 11.24 and Molality of 94.
Calculate the Density of the solution.
Solution:
Basis - 1ltr of Solution
moles of H2 SO4
Molarity (M) = volume of solution (in lit)
∴ η of H2 SO4 = 11.24
Molecular Weight of H2 SO4 = 98
Mass of H2 SO4 (Solute) = η of H2 SO4 × Molecular Weight of
H2 SO4
=98×11.24
=1101.52 g
moles of H2 SO4
Molality (m) = kg of H2 O
∴ Mass of H2 O(Solvent) = 119.6 g
Numericals based on Molarity, Normality & Molality

Example

Mass of Solution = Mass of Solute + Mass of Solvent

= 1.10152 + 0.1196
= 1.2211 kg
Mass
Density =
Volume
1.2211
=
1
= 1.2211 kg/l
Numericals based on Molarity, Normality & Molality

Example
Aqueous solution of Triethanolamine (TEA), i.e.,
N(CH2 CH2 OH)3 , contains 50% TEA by weight. Find the molarity
of the solution if the density of the solution is 1.05 kg/l.
Numericals based on Molarity, Normality & Molality

Example
A chemist is interested in preparing 500 ml of 1 N, 1 M and 1 m
solution of H2 SO4 . Assuming the density of H2 SO4 solution to be
1.075 g /cm3 , calculate the quantities of H2 SO4 to be taken to
prepare these solutions.
Numericals based on TOC & ThOD

Example
Glycerine (CH2 OH.CHOH.CH2 OH), weighing 600 mg is dissolved
in pure water to make a final solution of 1 litre. Find the TOC,
ThOD of the solution.
 Laws for Gas

Ideal Gas
It states that, for a given mass at constant Volume of an ideal gas,
the Pressure exerted on the sides of it’s container is directly
proportional to it’s absolute Temperature.

PV = nRT

P→Pressure in atm
V→Volume in litre(l)
n→Number of moles
R→Universal Gas Constant 8.314 J/mol
T→Temperature in Kelvin(K)

Concepts in MEBC Dr. Vyomesh M. Parsana 57/202

 Laws for Gas

Boyle’s Law
It states that, the Pressure of a given quantity of gas varies
inversely with it’s Volume at constant Temperature.

PV = Constant

Charle’s Law
It states that, the Volume of a given quantity of gas is directly
proportional to it’s absolute Temperature at constant Pressure.

V
T = Constant

Concepts in MEBC Dr. Vyomesh M. Parsana 58/202

 Laws for Gas
Gay-lussac’s Law
It states that, the Pressure of a given quantity of gas is directly
proportional to it’s absolute Temperature at constant Volume.

P
T = Constant

Combined Gas Law

It states that the ratio of the product of Pressure, Volume and the
absolute Temperature of a gas is equal to a constant. The
constant is a true constant if the number of moles of the gas
doesn’t change.

PV
T = Constant

Concepts in MEBC Dr. Vyomesh M. Parsana 59/202

 Laws of Gas

VanderWalls Equation of State

(P + Va2 )(V − b) = RT

P→Pressure
T→Absolute Temperature
V→Molar Volume
R→Universal Gas Constant
n→Number of moles
27R 2 TC2
a= 64PC
RTC
b= 8PC

Concepts in MEBC Dr. Vyomesh M. Parsana 60/202

 Gas Mixture

Amagat’s Law
P
Volume Total Volume of gas, V = Vi .
Where Vi is Volume of pure component ‘i 0 .
Pressure Partial Pressure of component, Pi = Pxi .
Where P is total Pressure and xi is mole fraction of
i th component.
P P
Pi P
= P xi
Now P xi = 1
∴ P = Pi

Dalton’s Law
It states that the total Pressure is the sum of the partial Pressure
exerted by each component.

Concepts in MEBC Dr. Vyomesh M. Parsana 61/202

 Gas Mixture

For an ideal gas mixture

Volume% = Mole% = Pressure%

Average Molecular Weight

P
M = Mi xi

Mi →Molecular Weight of ith component

xi →Moles of ith component

Concepts in MEBC Dr. Vyomesh M. Parsana 62/202

Numericals based on Gas Mixture

Example
Calculate the specific volume of superheated steam at 10 Mpa a
and 623 K using
the ideal gas law
the Vander-Waals equation
If the actual specific volume of steam at the above conditions is
0.02242 m3 /kg , find the % error in the above cases.
Numericals based on Gas Mixture

Example
Cracked gas from a petroleum refinery has the following
composition by volume: Methane 45%, Ethane 10%, Ethylene
25%, Propane 7%, Propylene 8%, n-Butane 5%.
Find:
Average molecular weight of the mixture
the composition by wt%
Numericals based on Gas Mixture

Example
A mixture of H2 and O2 contains 11.1% H2 by weight.
Calculate:
average molecular weight of gas mixture
partial pressure of O2 and H2 at 100 kPa and 303 K
Numericals based on Gas Mixture

Example
A mixture of CH4 and C2 H6 has the average molecular weight of
22.4. Find mole% CH4 and C2 H6 in the mixture.
Numericals based on Gas Mixture

Example
The average molar mass of a flue gas sample is calculated by two
different engineers. One engineer uses the correct molar mass of 28
for N2 and determines the average molar mass to be 30.08, the
other engineer, using an incorrect value of 14, calculates the
average molar mass to be 18.74.
Calculate the volume % of N2 in the flue gases
If the remaining components of the flue gases are CO2 and
O2 , calculate the volume % of each of them
 Gas-Liquid Mixture

An ideal liquid-liquid mixture obeys Raoult’s law. The same law

also applies to a gas-liquid mixture, i.e., the vapour pressure of a
gas-liquid mixture at given temperature equals the mole fraction of
a liquid multiplied by the vapour pressure of pure liquid at the
same temperature.
Pi = yi P = xi Pisat

At low concentration of gas in the liquid, Raoult’s law does not

hold. For such non-ideal behaviour, Henry’s law is found to be
useful. If Pi is the partial pressure of the solute gas i
Pi = Hi xi
Hi →Henry’s law constant
xi →Mole fraction of the ith component in the solution
Concepts in MEBC Dr. Vyomesh M. Parsana 68/202
Numericals based on Gas-Liquid Mixture

Example
A solution containing 55% benzene, 28% toluene and 17% xylene
by weight is in contact with its vapour at 373 K. Calculate the
total pressure and molar composition of the liquid and vapour.
The vapour pressure data at 373 K:
Benzene: 178.6 kPa, Toluene: 74.6 kPa, Xylene: 28 kPa
Numericals based on Gas-Liquid Mixture

Example
The liquid mixture contains n-butane, 1-butane and furfural. It is
boiled at 338 K and 671.325 kPa pressure. The mole fraction of
n-butane in the ternary vapour mixture in equilibrium with the
liquid is found to be 49.1 % by volume. Assuming ideal behavior of
the liquid and vapour mixtures, find the composition of the vapour
mixture.
Data:
Vapour pressure of furfural at 333 K = 3.293 kPa
Mole fraction of furfural in liquid mixture = 0.7734
 Humidity & Saturation

One of the most important psychrometric operations is the

air-water contact operation. These operations are also referred to
as Humidification operations.

Basic Terminology
Dry-Bulb Temperature (DB)
The temperature of the vapour-gas mixture recorded
by the immersion of a thermometer in the mixture is
termed as Dry-Bulb Temperature (DB).
Absolute Humidity (H)
The weight of vapour present in a unit weight of dry
air is termed as absolute humidity.
vapour in kg
H = dry gas(non−condensable) in kg

Concepts in MEBC Dr. Vyomesh M. Parsana 71/202

 Humidity & Saturation

Molal Humidity (Hm )

It is defined as the ratio of moles of vapour
(condensable) to the moles of dry (non-condensable)
gas.
vapour in kmol
Hm = dry gas in kmol
% Humidity or % Absolute Humidity or % Saturation
It is defined as the ratio of the actual absolute
humidity to the saturation humidity.
%Humidity = HHs × 100
PA
= P−P A
× P−P
PS × 100
S

Concepts in MEBC Dr. Vyomesh M. Parsana 72/202

 Humidity & Saturation

Relative Humidity or Relative Saturation (RH)

It is defined as the ratio of partial pressure of water
vapour in air to the vapour pressure of water at the
Dry-Bulb Temperature (DB).
RH = PPAS × 100
Humid Heat (CS )
It is defined the heat capacity 1 kg dry air and the
moisture contained in it.
CS = 1.006 + 1.84H, kg drykJ air .K

Concepts in MEBC Dr. Vyomesh M. Parsana 73/202

 Humidity & Saturation

Humid Volume (VH )

It is the volume of a mixture of air and water vapour
per kg of dry air. This is also known as
psychrometric
 volume.

VH = MA + M1B × 22.4136 × 273.15
H DB
× 101.325
P ,
m3
kg dry air
Wet-Bulb Temperature (WB)
It is the steady state temperature attained by a
wet-bulb thermometer exposed to a rapidly moving
stream of the vapour gas mixture. The bulb of the
wet-bulb thermometer is covered with wet wick i.e.
with the same liquid that forms the vapour-gas
mixture. WB is always equal to or lower than DB.
Concepts in MEBC Dr. Vyomesh M. Parsana 74/202
 Humidity & Saturation

Dew Point (DP)

It is the temperature at which the air-water vapour
mixtures become saturated when the mixture is
cooled at constant total pressure in the absence of
liquid water. This means that partial pressure of
water vapour in the mixture equals the vapour
pressure of water at DP. Dew point is always less
than or equal to the dry bulb temperature.

Concepts in MEBC Dr. Vyomesh M. Parsana 75/202

Numericals based on Humidity & Saturation

Example
The dry bulb temp & dew point of ambient air were found to be
302 K & 291 K respectively. The barometer reads 100 kPa.
Calculate:
the absolute molal humidity
the absolute humidity
% RH
% saturation
humid heat
humid volume
Data:
Vapour pressure of water at 291 K = 2.0624 kPa
Vapour pressure of water at 302 K = 4.004 kPa
 Heat

Sensible Heat
It is heat exchanged by a body or thermodynamic system in which
the exchange of heat changes the temperature of the body or
system, leaving certain other macroscopic variables of the body or
system unchanged, such as volume or pressure.

Latent Heat
It is defined as the heat or energy that is absorbed or released
during a phase change of a substance. It could either be from a
gas to a liquid or liquid to solid and vice versa.

Concepts in MEBC Dr. Vyomesh M. Parsana 78/202

 Graph

Concepts in MEBC Dr. Vyomesh M. Parsana 79/202

 Explanation
A → B is a Solid Phase & the heat supplied is known as
Sensible Heat
B → C is a phase conversion having Solid + Liquid Mixture
& the heat supplied is known as Latent Heat
C → D is a Liquid Phase & the heat supplied is known as
Sensible Heat
D → E ia a phase conversion having Liquid + Vapour
Mixture the heat supplied is known as Latent Heat
E → F is a Vapour(Gaseous) Phase

Concepts in MEBC Dr. Vyomesh M. Parsana 80/202

 Specific Heat

Specific heat is defined by the amount of heat needed to raise

the temperature of 1 gram of a substance 1 degree Celsius
(◦ C).
RT ◦
Q = T12 Cp dT
◦
Cp = a + bT + cT 2 + dT 3
Here a, b, c & d are constants.

Concepts in MEBC Dr. Vyomesh M. Parsana 81/202

Numerical based on Specific Heat
Example
Heat capacity data for gaseous SO2 are reported in standard data
book and also by following equation Cp◦ = 43.458 + 10.634 Ö
5
10−3 T – 5.945×10
T2
. Calculate the heat required to raise the
temperature of 1 kmol pure SO2 from 300K to 1000K, using the
above equation.
Solution:
Basis - n=1 kmol pure SO2
T1 = 300K T2 = 1000K
5.945×105
Cp◦ = 43.458 + 10.63410−3 T – T2
Z T2
◦
Q=n Cp dT
T1
1000
5.945 × 105
Z
= 43.458 + 10.63410−3 T –
300 T2
=33871.91 KJ
Numerical based on Specific Heat

Example
Calculate the enthalpy change in 24gm of N2 if heated from 300K
to 1500K at constant pressure.
Cp◦ = 29 + 0.2199 × 10−2 T + 0.5723 × 10−5 T 2 − 2.871 × 10−9 T 3
Solution:
n= 24
28
=0.857 mol of N2
T1 = 300K T2 = 1500K
Cp = 29 + 0.2199 × 10−2 T + 0.5723 × 10−5 T 2 − 2.871 × 10−9 T 3
◦
RT ◦
Q = n T12 Cp dT
R 1500
= 0.857 300 29 + 0.2199 × 10−2 T + 0.5723 × 10−5 T 2
−2.871 × 10−9 T 3
=34224.44 J
Numerical based on Specific Heat

Example
If cooling tower water available at 298K is used at a rate of
1500Kg/h for heat duty of 119647.78KJ/h. Calculate the outlet
temperature of water assuming specific heat of water to be
4.187KJ/(Kg*K).
Solution:
Basis
T1 = 298K
Water flow rate = 1500Kg/h
Q = 119647.78KJ/h
Cp = 4.187KJ/(Kg*K)

Q = mCp ∆T
119647.78 = 1500 × 4.187(T2 − 298)
T2 =317.05 K
 Standard Heat of Formation

It is the change in Enthalpy during the formation of 1 mole of

substance from it’s constituent element with standard state.

Example
1
2H2 + C + 2 O2 → CH3 OH

Concepts in MEBC Dr. Vyomesh M. Parsana 85/202

 Standard Heat of Combustion

When 1 mole of compound is completely burnt in Oxygen with all

product & resultant in their standard condition, the change in
enthalpy is called standard Heat of Combustion.

Example
C2 H6 O + 3O2 → 2CO2 + 3H2 O

Concepts in MEBC Dr. Vyomesh M. Parsana 86/202

 Standard Heat of Reaction

It is the change in Enthalpy in a system when substance is used to

transform by a given Chemical Reaction, when all products &
reactants are at standard state.

Example
1
SO2 + 2 O2 → SO3

Concepts in MEBC Dr. Vyomesh M. Parsana 87/202

Numericals based on Heat of Formation, Combustion &
Reaction

Example
Calculate the Heat of Reaction of following reaction.
4NH3 + 5O2 → 4NO + 6H2 O
Data:
∆HNH3 = −11020 cal/mol ∆HNO = 21570 cal/mol
∆HH2 O = −57796 cal/mol
Solution:

∆HR at 298 K = ∆Hf of Product − ∆Hf of Reactant

= 4(21570) − 6(57796) + 4(11020) − 5(0)
=-216416 cal/mol
Numericals based on Heat of Formation, Combustion &
Reaction

Example
Calculate the Heat of Formation of Glycerol Liquid at 298K from
it’s elements using Hess Law.
Data:
∆Hf CO2 =-393.51 cal/mol ∆Hf H2 O=-285.83 cal/mol
∆HR at 298 K=-1659.10 cal/mol
Solution:
7
Balanced Equation: C3 H8 O3 + 2 H2 O → 3CO2 + 4H2 O

∆HR at 298 K = ∆Hf of Product − ∆Hf of Reactant

−1659.10 = −∆Hf of (C3 H8 O3 ) − 3(393.51) − 4(285.83)
∆Hf of (C3 H8 O3 ) =-664.75 cal/mol
 Antoine Equation

It is a empirical correlations describing the relation between

vapor pressure and temperature for pure substances.
It is derived from the Clausius–Clapeyron relation.
It is worth noting that this equation is still widely used today
because of its accuracy.

Equation
B
log Pv = A − T +C
Pv → Pressure in bar
T → Temperature in Kelvin
A, B & C are constants which are different for different
Substances.

Concepts in MEBC Dr. Vyomesh M. Parsana 90/202

 Numericals based on Antoine Equation

Example
Calculate the Vapor Pressure of Acetaldehyde at 250K
Data:
A = 7.134 B = 1600 C = 18.65
Solution:
B
log10 Pv = A −
T +C
1600
= 7.134 −
250 + 18.65
= 1.17
Pv = 101.17
=14.79 bar

Concepts in MEBC Dr. Vyomesh M. Parsana 91/202

 Numericals based on Antoine Equation

Example
Calculate the Vapor Pressure of n-Hexane at 32 ◦ C
Data:
A = 4.00266 B = 1171.53 C = −48.782
Solution:
B
log10 Pv = A −
T +C
1171.53
= 4.00266 −
305 − 48.782
= −0.56
Pv = 10−0.56
=0.27 bar

Concepts in MEBC Dr. Vyomesh M. Parsana 92/202

 Equation of Latent Heat

We have three empirical equations to determine latent heat of

vaporization.

Watson Equation
 0.38
λV TC −T
λV 1 = TC −T1
λV → Latent Heat of Vaporization in KJ/Kmol at
Temperature T.
TC → Critical Temperature in Kelvin
T1 → Temperature at which Latent Heat is known
λV 1 →Latent Heat of Vaporization in KJ/Kmol at
Temperature T1 .

Concepts in MEBC Dr. Vyomesh M. Parsana 93/202

 Riedel Equation
λV 1.092(ln PC −5.6182)
R TB = 0.93−TBr
λV → Latent Heat of Vaporization in KJ/Kmol at
Temperature T.
R → Universal Gas Constant i.e. 8.314 KJ/Kmol
PC → Critical Pressure in KPa
TBr → Reduced Temperature at Temperature T
TBr = T /TC

Concepts in MEBC Dr. Vyomesh M. Parsana 94/202

 NIST Equation
λV = Ae −αTBr (1 − TBr )β
A, α & β → are constants which are different for different
Substances.
TBr → Reduced Temperature at Temperature T
TBr = T /TC

Concepts in MEBC Dr. Vyomesh M. Parsana 95/202

Numericals based on Latent Heat
Example
For O-xylene Calculate:
Latent Heat of vaporization at TB using Riedel equation.
Latent Heat of vaporization at 25 ◦ C using Watson equation.
PC = 3732 KPa TC = 630.3 K TB = 417.6 K
Solution:
Using Riedel Equation
TB
TBr = T C
417.6
= 630.3
= 0.66
λV 1.092(lnPC − 5.6182)
=
R × TB 0.93 − TBr
λV 1.092(ln3732 − 5.6182)
=
8.314 × 417.6 0.93 − 0.66
λV =36600.5 KJ/Kmol
Numericals based on Latent Heat

Example
Using Watson Equation

 0.38
λV TC − T
=
λV 1 TC − T1
 0.38
λV 630.3 − 298
=
36600.5 630.3 − 417.6
λV =43362.75 KJ/Kmol
Numericals based on Latent Heat

Example
For Ethanol Calculate:
Latent Heat of Vaporization at TB using Riedel and NIST
equation.
◦
Latent Heat of Vaporization at 25 using Watson and NIST
equation.
Data:
For Ethanol
PC = 61.37 bar TC = 514 K TB = 351.4 K
NIST equation constants are
A=50430 α=-0.4475 β=0.4969
Numericals based on Latent Heat

Example
Solution:
Using Riedel and NIST equation
TB
TBr = T C

= 351.4
514
= 0.68
λV 1.092(lnPC − 5.6182)
=
R × TB 0.93 − TBr
λV 1.092(ln6137 − 5.6182)
=
8.314 × 351.4 0.93 − 0.68
λV =39609.64 KJ/Kmol
Numericals based on Latent Heat

Example

λV = Ae −αTBr (1 − TBr )β
= 50430e −(−0.4475)(0.68) (1 − 0.68)0.4969
=38810.9 KJ/Kmol
Numericals based on Latent Heat

Example
Using Watson and NIST equation
TB
TBr = T C

= 298
514
= 0.0.58
 0.38
λV TC − T
=
λV 1 TC − T1
 0.38
λV 514 − 298
=
38810.9 514 − 351.4
λV =43233.4 KJ/Kmol
Numericals based on Latent Heat

Example

λV = Ae −αTBr (1 − TBr )β
= 50430e −(−0.4475)(0.58) (1 − 0.58)0.4969
=42478.6 KJ/Kmol
Numericals based on Latent Heat
Example
100 kg of Cadmium at 27◦ C is to be melted. The heat is supplied
by steam. Calculate mass of steam to be supplied.
Data:
Melting Point of Cadmium is 320.9◦ C
Atomic Weight of Cadmium is 112
CP = 6 + 0.005T Kcal/Kmol◦ C where T is in ◦ C
Latent Heat of Fusion = 2050 Kcal/Kmol
Latent Heat of Steam = 210 Kcal/Kg
Solution:
Z 320.9 
Q=n CP dT + 2050
27
Z 320.9 
100
= 6 + 0.005T dT + 2050
112 27
= 3633.04 Kcal
Numericals based on Latent Heat

Example

Q = Mass of Steam × Latent Heat of Steam

3633.04
M=
210
= 19 Kg
 Hess Law

It states that the net enthalpy change in a given chemical

reaction is the same whether the reaction takes place in one
step or in several steps.
This law permits us to treat thermochemical equations as
algebraic equations.
Enthalpy of reaction remains same whether reaction occurs in
one step or multi step at same temperature and pressure.

Concepts in MEBC Dr. Vyomesh M. Parsana 105/202

 Block Diagram

Concepts in MEBC Dr. Vyomesh M. Parsana 106/202

 A → B ∆HR
A → C ∆H1
C → D ∆H2
D → E ∆H3
E → B ∆H4

∴ by Hess Law ∆HR = ∆H1 + ∆H2 + ∆H3 + ∆H4

Concepts in MEBC Dr. Vyomesh M. Parsana 107/202

Numerical based on Hess Law

Example
Obtain the expression relating the heat of reaction and the
temperature of reaction.
SO2 + 12 O2 → SO3

Gas ∆HR KJ/molK a b×103 c×106

SO2 -296.81 24.75 62.95 -44.26
O2 0 26.026 11.755 -2.34
SO3 -395.72 22.04 121.6 -91.87
Numerical based on Hess Law
Example
Solution:
∆HR = −395.72 + 296.81
= 98910J
∆a = 22.04 − 26.026
2 − 24.75
= −15.72
∆b = 121.6 − 11.755 − 62.95 × 10−3


2
= 52.77 × 10−3
∆c = −91.87 + 2.34 −6


2 + 44.26 × 10
= −48.78 × 10−6
RT
∆HT + ∆HR = 298 CP dT
RT
= 298 ∆a + ∆bT + ∆cT 2 dT

∆HT = 98190 − 15.72(T − 298) + 26.385 × 10−3 (T 2 − 2982 )

−16.26 × 10−6 (T 3 − 2983 )
 Introduction
A process design starts with the development of a Process
Flow Sheet or Diagram.
For the development of such a diagram, Material & Energy
Balance Calculations are necessary.
These balances follow the laws of conservation of Mass and
Energy.
Under this portion, there is a balancing of Mass without the
occurence of any Chemical Reaction.
It means we can have Material Balance without Chemical
Reaction.
Here the concept of ”Block Diagram” is very useful.

Concepts in MEBC Dr. Vyomesh M. Parsana 111/202

 One should have proper understanding and knowledge of
these Block Diagrams.
The study of block diagram becomes essential due to the
understanding of different unit operations used in Chemical
Engineering.

Block Diagram
It is a simplified form of a Process Flow Sheet.
They are useful in estimating feed and product streams along
with required utilities.
It is useful at an early stage of the process development.

Concepts in MEBC Dr. Vyomesh M. Parsana 112/202

 Process Flowsheet
A process It is one in which all incoming and outgoing
materials and utilities are shown.
Each flowsheet is presented in different form.
One process engineer may wish to give the component
balance on the flow diagram while another may prefer to give
it in a tabular form at the bottom.
Some flowsheets may include safety aspects, such as
explosibility, flammability, toxicity, corrosivity, radioactivity
hazards and so on, while others may have the sequence of
flow of a batch process in a tubular form.
It should be clearly understood that such a diagram is
different form piping and instrumentation (P & I) diagram.

Concepts in MEBC Dr. Vyomesh M. Parsana 113/202

 Process Flowsheet Contains....
Flow rate or quantity of each stream; compositions of the
streams are also recommended.
Operating conditions of each stream, such as pressure and
temperature.
Heat added/removed in a particular equipment.
Flow rates of utilities such as steam, cooling water, brine, etc.
Any specific information which is useful in understanding the
process.

Concepts in MEBC Dr. Vyomesh M. Parsana 114/202

Classification of Material Balances
 There are 3 general methods of solving Material
Balance problems Without Chemical Reactions
Make the balance of a tie material, the quantity of which does
not change during the particular operation. eg. solids in
evaporation.
By making balance of inert materials. Ex- ash in coal,
nitrogen in air combustion.
When two or more compounds are present in a system and all
are affected simultaneously it is required that the material
balance equations be solved by satisfying simultaneous
equations.

Concepts in MEBC Dr. Vyomesh M. Parsana 116/202

 From the law of conservation of mass, for any process
Input – accumulation = output or disappearance
When the accumulation of the material is constant or nil
Input = Output
In any given problem, first one has to determine the particular
class under which the problem falls. Then a definite basis is
assumed.
Often, the basis is defined in the statement of the problem
itself.
If the basis is convenient, it may be adopted, otherwise, a
new, more convenient basis can be selected.

Concepts in MEBC Dr. Vyomesh M. Parsana 117/202

 Degree of Freedon
The concept of degree of freedom is well known to Chemical
Engineers.
This is an index which fixes the number of independent
equations that are required to be solved for finding the
specified number of unknowns.
If the number of equations is less than the number of
unknowns, the system is considered under-defined. An
optimum solution can be found by fixing some unknowns
based on a judgement.
If the degree of freedom is negative, the system is
over-defined. For such calculations, redundant information
should be discarded to obtain a unique solution.

Concepts in MEBC Dr. Vyomesh M. Parsana 118/202

 Different types of Unit Operations
Evaporation
Mixing/Blending
Distillation
Filtration
Dissolution
Crystallization
Absorption & Stripping
Extraction (liquid-liquid) & Leaching
Extraction of Oil from Seeds
Drying
Recycling & Bypassing

Concepts in MEBC Dr. Vyomesh M. Parsana 119/202

 Evaporation

Overall Mass Balance

Weak Liquor = Water Evaporated + Thick Liquor +
Crystallized Product(if any )
Mass Balance of ’Solids’
Solids in Weak Liquor = Solids in Thick Liquor
Mass Balance of ’Water’
Water in Weak Liquor =
Water Evaporated + Water in Thick Liquor
Concepts in MEBC Dr. Vyomesh M. Parsana 120/202
Numericals based on Evaporation

Example
A single effect evaporator is fed with 1000 kg/h of weak liquor
containing 15% caustic by weight and is concentrated to get thick
liquor containing 40% by weight caustic (NaOH).
Calculate:
kg/h of water evaporated
kg/h of thick liquor obtained
Numericals based on Evaporation

Example
Single effect evaporator concentrating weak liquor containing 4%
solids to 55% solids (by weight) is fed with 5000 kg/h of weak
liquor.
Calculate:
water evaporated per hour
flow rate of thick liquor
Numericals based on Evaporation

Example
An evaporator system concentrating weak liquor from 5% to 50%
solids handles 100 kg of solids per hour. If the same system is to
concentrate a weak liquor from 4% to 35%, find the capacity of
the system in terms of solids that can be handled per hour
assuming water evaporation capacity to be same in both cases.
Numericals based on Evaporation

Example
An evaporator is fed with 15000 kg/h of a solution containing 10%
NaCl, 15% NaOH and rest water. In operation, water is evaporated
and NaCl is precipitated as crystals. The thick liquor leaving the
evaporator contains 45% NaOH, 20% NaCl and rest water.
Numericals based on Evaporation

Example
A spent lye sample contains 9.6% glycerol and 10.3% NaCl salt. It
is concentrated at the rate of 5000 kg/h in a double effect
evaporator until the final solution contains 80% glycerol and 6%
salt. 4.5% glycerol is lost by entrainment. All the percentage are
by mass.
Determine:
the evaporation taken place in the system.
the amount of salt crystallized out from the evaporator
 Mixing/Blending

Overall Mass Balance

Feed 1 + Feed 2 + Feed 3 = Desired Product
Mass Balance of Component ’A’
‘A0 in the Feed 1 + ‘A0 in the Feed 2 + ‘A0 in the Feed 3 =
‘A0 in the Desired Product

Concepts in MEBC Dr. Vyomesh M. Parsana 126/202

Numericals based on Mixing/Blending

Example
The dilute acid containing 25% H2 SO4 is concentrated by
commercial grade sulfuric acid containing 98% H2 SO4 to obtain
desired acid containing 65% H2 SO4 . Find the quantities of the
acids required to make 1000 kg of desired acid.
Numericals based on Mixing/Blending

Example
It is desired to make up 1000 kg of a solution containing 35% of a
substance ‘A’. Two solutions are available, one containing 10% ‘A’
and other containing 50% ‘A’. How many kg of each solution will
be required? All % are by weight.
Numericals based on Mixing/Blending

Example
The waste acid from a nitrating process containing 20% HNO3 ,
55% H2 SO4 and 25% H2 O by weight is to be concentrated by
addition of concentrated Sulphuric acid containing 95% H2 SO4
and concentrated nitric acid containing 90% HNO3 to get desired
mixed acid containing 26% HNO3 and 60% H2 SO4 . Calculate the
quantities of waste acid and concentrated acids required for 1000
kg of desired acid.
Numericals based on Mixing/Blending

Example
It is required to make 1000 kg mixed acid containing 60% H2 SO4 ,
32% HNO3 , and 8% water by blending
the spent acid containing 11.3% HNO3 , 44.4% H2 SO4 , 44.3%
H2 O
aqueous 90% HNO3
aqueous 98% H2 SO4
All % are by weight. Calculate the quantities of each of the three
acids required for mixing.
Numericals based on Mixing/Blending

Example
If in the above question 10% oleum is used instead of concentrated
sulfuric acid; calculate the quantities of the three acids required for
obtaining 1000 kg of mixed desired acid.
Numericals based on Mixing/Blending

Example
It is desired to have a mixed acid containing 40% HNO3 , 43%
H2 SO4 and 17% H2 O by weight. Sulfuric acid of 98% by wt. is
readily available.
Calculate:
the strength of nitric acid
the weight ratio of sulfuric acid to nitric acid
 Distillation

It is an operation in which one or more compounds of the

liquid mixtures of two or more components are separated
using thermal energy.
Concepts in MEBC Dr. Vyomesh M. Parsana 133/202
 Basically, the difference in vapour pressures of different
components at the same temperature is responsible for such a
separation.
Overall Mass Balance
Feed = Distillate + Bottoms
Mass Balance of Component ‘A’
‘A0 in the Feed = ‘A0 in the Distillate + ‘A0 in the Bottoms

Concepts in MEBC Dr. Vyomesh M. Parsana 134/202

Numericals based on Distillation

Example
The feed to a continuous fractionating column analyses by weight
28% benzene and 72% toluene. The analysis of the distillate shows
52% benzene and 5% benzene was found in the bottom product.
Calculate the amount of distillate and bottom product per 1000
kg/h of feed. Also calculate the % recovery of benzene.
Numericals based on Distillation

Example
A solution of ethyl alcohol containing 8.6% alcohol is fed at rate of
1000 kg/h to a continuous distillation column. The product is a
solution containing 95.5% alcohol. The waste solution from the
column carries 0.1% alcohol. All % are by wt.
Calculate:
the mass flow rate of top & bottom product in kg/h
the % loss of alcohol
Numericals based on Distillation

Example
5000 kg of benzene and toluene containing 50 mol% benzene is
distilled to get an overhead product containing 95% by mole
benzene and a residue containing 90% by mole toluene. Calculate
benzene and toluene in feed, distillate, and residue.
 Filtration

Overall Mass Balance

Feed Slurry = WetSolids + Filtrate
Mass Balance of ‘Solids’
Solids in Slurry =
Solids in Wet Product + Solids in Filtrate(ifany )

Concepts in MEBC Dr. Vyomesh M. Parsana 138/202

Numericals based in Filtration

Example
Centrifuge is fed with a slurry containing 25% solids by weight and
wet solids obtained after filtration are analyzed to contain 8%
moisture by weight and filtrate is found to contain 200 ppm solids.
If centrifuge machine produces 100 kg per hour of desired wet
product and quantity of slurry to be handled is 5000 kg per batch.
Calculate:
the time required for filtration of slurry
loss of solids in filtrate per batch
 Dissolution

It is same like mixing.

They are carried out in agitated vessels.

Concepts in MEBC Dr. Vyomesh M. Parsana 140/202

Numericals based on Reverse Osmosis

Example
The feed water to reverse osmosis plant has dissolved solids to the
extent of 5000 mg/L. The feed to product ratio (by mass) is 4:3.
The treated water from the plant contains 600 mg/L of solids.
Find the dissolved solids in the reject stream.
 Crystallization

It is a unit operation in which the dissolved solids of the

solution are separated out by solubility difference at different
temperatures and/or concentration of the solution (by
evaporation).
In this operation, the final mother liquor is always saturated.

Concepts in MEBC Dr. Vyomesh M. Parsana 142/202

 Overall Mass Balance
Feed Solution = Saturated Solution + Crystals
Mass Balance of ‘Crystals’
‘Crystals 0 in Feed Solution =
‘Crystals 0 Obtained + ‘Crystals 0 in Saturated Solution

Concepts in MEBC Dr. Vyomesh M. Parsana 143/202

Numericals based on Crystallization

Example
What will be the yield of Glauber salt (Na2 SO4 .10H2 O) if a pure
32% solution is cooled to 293 K without any loss due to
evaporation?
Data
Solubility of Na2 SO4 in water at 293 K is 19.4 kg per 100 kg water.
Numericals based on Crystallization

Example
1000 kg of sodium carbonate solution containing 25% sodium
carbonate is subjected to evaporative cooling, during which process
15% of the water present in the solution is evaporated. From the
concentrated solution Na2 CO3 .10H2 O crystallizes out. Calculate
how much crystals would be produced if the solubility of
Na2 CO3 .10H2 O is 21.5 gm per 100 gm of water.
Numericals based on Crystallization

Example
Crystals of MgCl2 .6H2 O have a solubility of 190 g per 100 g
ethanol at 298.15 K. It is desired to make 1000 kg of saturated
solution. Calculate the quantities of the crystals and ethanol
required to make above solution. Also, find the composition of the
saturated solution by mass.
 Absorption & Stripping

Absorption is a unit operation in which a mixture of gases is

brought in contact with liquid.
A definite component of the gas mixture is dissolved in the
liquid. The operation is sometimes also termed as Scrubbing.

Concepts in MEBC Dr. Vyomesh M. Parsana 147/202

 It may be physical absorption only or it may be accompanied
by a chemical reaction.
Stripping or desorption is an operation in which a dissolved
gas of a solution is stripped off from the liquid using a
stripping medium.
Inert Gas Balance
Inert gas in inlet gas mixture =
Inert gas in lean gas(gas leaving the tower )
Solute Gas Balance
Solute gas in inlet gas mixture +
Solute gas in solvent at inlet =
Solute gas in lean gas + Solute gas in solution
Solute Gas Removed by Absorption
Solute gas in inlet gas mixture − Solute gas in lean gas =
Solute in solution − Solute in solvent at inlet
Concepts in MEBC Dr. Vyomesh M. Parsana 148/202
Numericals based on Absorption & Stripping

Example
The NH3 air mixture containing 0.2 kg NH3 per kg air enters
absorption system where NH3 is absorbed in water. The gas
leaving the system is found to obtain 0.004 kg NH3 per kg of air.
Find the % recovery of ammonia.
Numericals based on Absorption & Stripping

Example
A gas mixture containing 15 mol% ‘A’ and 85 mol% inerts is fed
to an absorption tower where it is contacted with liquid solvent ‘B’
which absorbs ‘A’. The mole ratio of solvent to gas entering tower
is 2:1. The gas leaving the absorber contains 2.5% ‘A’, 1.5% ‘B’
and rest inerts on mole basis.
Calculate:
The % recovery of solute ‘A’
The fraction of solvent ‘B’ fed to column lost in gas leaving
the tower
During process same solvent evaporates and gets added in gas
leaving the tower.
Numericals based on Absorption & Stripping

Example
Isothermal & isobaric absorption of SO2 is carried out in a packed
tower containing raching rings. The gases enter the bottom of the
tower containing 14.8% SO2 by vol. water is distributed at the top
of the col at the rate of 16.5 lps. The total vol of the gas handled
at 101.3 kPa and 303 K is 1425 m3 /h, the gases leaving the tower
are found to contain 1% SO2 by vol. Calculate the % SO2 by wt in
outlet tower.
 Extraction (liquid-liquid) & Leaching

Overall Mass Balance

Feed solution + Solvent = Extract phase + Raffinate phase
Mass Balance of component ‘A’
‘A0 in the feed = ‘A0 in Extract phase + ‘A0 in Raffinate phase
Concepts in MEBC Dr. Vyomesh M. Parsana 152/202
 When a mixture of liquids is not easily separable by
distillation, extraction is employed.
In this operation, a “solvent” is added to the liquid-liquid
mixture.
As a result, two immiscible layers are formed, both of which
may contain varying amounts of different components.
Those isolated layers are removed as Extract phase and
Raffinate phase using density difference.
Invariably, distillation has to follow extraction for the recovery
of the solvent for reuse. For example, Furfural is common
solvent for the extraction operations in a petroleum refinery.
Leaching is also basically a solid-liquid extraction operation in
which a particular component of the solid is leached out with
the help of a solvent. A common example is the leaching of
oil from the oil cake using hexane as a solvent.
Concepts in MEBC Dr. Vyomesh M. Parsana 153/202
Numericals based on Extraction (liquid-liquid) & Leaching

Example
A mixture of phenol and water forms two separate liquid phase,
one rich in phenol and other rich in water, composition of layers is
70% and 9% (by weight) phenol respectively. If 500 kg of phenol
and 700 kg of water are mixed and layers are allowed to separate,
what will be the weights of two layers?
Numericals based on Extraction (liquid-liquid) & Leaching

Example
An aqueous solution of pyridine containing 27% (by wt.) pyridine
and 73% water is to be extracted with chloro-benzene. The feed
and solvent are mixed well in batch extractor and the mixture is
then allowed to stand for phase separation. The extract phase
contains 11% pyridine, 88.1% chlorobenzene and 0.9% water (by
wt.) The raffinate phase contains 5% pyridine and 95% water (by
wt.).
Calculate:
the quantities of two phases(layers)
the wt. ratio of solvent to feed based on 100 kg of feed
Numericals based on Extraction (liquid-liquid) & Leaching

Example
A mixture containing 47.5% acetic acid & 52.5% water (by wt) is
being separated by the extraction in a counter current multistage
unit. The operating temp. is 297 K & the solvent used is pure
iso-propyl ether. Using the solvent in the ratio of 1.3 kg/kg feed.
The final extraction composition on a solvent free basis is found to
be 82% by wt of acetic acid. The Raffinate is found to contain
14% by wt of acetic acid on a solvent-free basis. Calculate the %
of acid of the original feed which remains unextracted.
Numericals based on Extraction (liquid-liquid) & Leaching
Example
A 100 kg mixture of 27.8% of acetone (A) and 72.2% of
chloroform (B) by mass is to be batch extracted with a mixed
solvent at 25 ◦ C. The mixed solvent of an unknown composition is
known to contain water (S1) and acetic acid (S2). The mixture of
the original mixture and the mixed solvent is shaken well, allowed
to attain equilibrium, and separated into two layers.
The compositions of the two layers are given below:

Layer A B S1 S2
Upper Layer 7.5 3.5 57.4 31.6
Lower Layer 20.3 67.3 2.8 9.6

Calculate:
the quantities of the two layers
the mass-ratio of the mixed solvent to the original mixture
the composition of the mixed solvent (mass basis)
Numericals based on Extraction (liquid-liquid) & Leaching

Example
The analysis of a sample of Babul bark (of northern India) yields
5.8% moisture, 12.6% tannin, and 8.3% soluble non-tannin organic
matter and rest lignin. In order to extract tannin out of bark, a
counter-current extraction process is employed. The residue from
the extraction process is analyzed and found to contain 0.92%
tannin and 0.65% solute non-tannin organic matter on a dry basis.
Find the % of tannin recovered on the basis of the original tannin
present in the bark. All analyses are given on mass basis.
 Extraction of Oil from Seeds

Mass Balance of Solids

Solids in seeds =
solids in meal cake(assuming no solids in solvent)
Mass Balance of Oil
Oil in seeds = Oil in solvent(extracted oil) + Oil in cake
Concepts in MEBC Dr. Vyomesh M. Parsana 159/202
Numericals based on Extraction of Oil from Seeds

Example
The groundnut seeds containing 45% oil and 55% solids are fed to
expeller, the cake coming out of expeller is found to contain 95%
solids and 5% oil. Find the % recovery of oil.
Numericals based on Extraction of Oil from Seeds

Example
Soyabeen seeds are extracted with n-hexane in batch extractors.
The flaked seeds contain 18.6% oil, 69.0% solids and 12.4%
moisture. At the end of the extraction process, de-oiled cake
(DOC) analysis yields 0.8% oil, 87.7% solids and 11.5% moisture.
Find the percentage recovery of oil. All percentages are by mass.
 Drying

Mass balance of moisture

Moisture removed from solids = moisture added in air
∴ Moisture in wet solids–Moisture in dried solids =
Moisture in outlet air –Moisture in inlet air

Concepts in MEBC Dr. Vyomesh M. Parsana 162/202

Numericals based on Drying

Example
2000 kg of wet solids containing 70% solids by wt. are fed to a
tray dryer and dried by air. The product finally obtained is found
to contain 1% moisture by weight.
Calculate:
kg of water removed from wet solids
kg of product obtained
Numericals based on Drying

Example
Tray dryer is fed with 1000 kg of wet ortho-nitro aniline containing
10% water. The dried product contains 99.5% ortho-nitro aniline
and rest water. Find the percentage of original water that is
removed in the dryer.
Numericals based on Drying

Example
Slabs of building boards contain 16% moisture. They are dried to
a water content of 0.5% by circulating hot air over them. The
outgoing air contain 0.09 kg water vapor per kg dry air. Calculate
the quantity of fresh air required per 1000 kg/h net dry board if
the fresh air is supplied at 301 K and 101.325 kPa containing 0.2
kg/kg dry air humidity.
Numericals based on Drying

Example
Wood containing 40% moisture is dried to 5% moisture. What
mass of water in kilograms is evaporated per kg of dry wood.
 Recycling & Bypassing

These operations are performed due to some of the following

reasons.
To utilize the valuable reactants to their maximum and avoid
wastage. e. g. Chemical reactions.
To utilize the heat being lost in the outgoing stream, e.g. hot
air dryers, calcining the lime in the kiln, etc.
To improve the performance of the equipment, e. g. SO3
cannot be easily dissolved in the Oleum.
To control the operating variables in a reaction, e. g.
pressure, temperature, etc.
To improve selectivity of a product, e.g. the manufacture of
chloromethanes.

Concepts in MEBC Dr. Vyomesh M. Parsana 167/202

Numericals based on Recycling & Bypassing

Example
Fresh juice contains 15% solids and 85% water (by wt.) is to be
concentrated to contain 40% solids. In a single evaporation
system, it is found that volatile constituents of juice escape with
water leaving the concentrated juice with a flat taste. To overcome
this problem, part of the fresh juice bypasses the evaporator
operation is shown schematically in figure.
Example
Calculate:
The fraction of juice that bypasses the evaporator
The concentrated juice produced (containing 40% solids) per
100 kg of fresh juice fed to the process
Numericals based on Recycling & Bypassing
Example
In a textile industry, it is desired to make 24% solution (by weight)
of caustic soda for a mercerization process. Due to very high heat
of dissolution of caustic soda in water, the above solution is
prepared by two step process.
First, in a dissolution tank, soda is dissolved in the correct quantity
of water to produce 50% (by wt.) solution. After complete
dissolution and cooling, the solution is taken to dilution tank where
some more water is added to produce 24% solution. Assuring no
evaporation loss of water in dissolution tank, calculate the weight
ratio of water fed to dissolution tank (W1 ) to bypassed water to
dilution tank (W2 ).
Numericals based on Recycling & Bypassing

Example
In a particular drying operation, it is necessary to hold the moisture
content of feed to a calciner to 15% (by weight) to prevent lumping
and sticking. This is accomplished by mixing the feed having 30%
moisture (by weight) with a recycle stream of dried material having
3% moisture (by weight). The drying operation is shown in the
figure. What fraction of the dried product must be recycled?
 Material Balance of Unsteady-state Operations

In unsteady-state conditions, input and output parameters

change with respect to time.
In certain batch operations, at the end of each batch, the
material balance changes with respect to a particular
component, thereby putting the system in an unsteady-state.
In most such cases, material balance calculations are made by
considering small time intervals (∆θ) and finding the change
the parameter taking place during the period (∆θ).
Thus, integral calculus is very useful in solving such problems.

Concepts in MEBC Dr. Vyomesh M. Parsana 172/202

Numericals based on Material Balance of Unsteady-state
Operations

Example
A storage tank of DM water has a holding capacity of 2000 m3
upto an overflow point. The inflow of DM water to the tank is 25
litre/second having silica (as SiO2 ) content of 0.005 mg/l. The
supply of DM water to the high-pressure boilers from the tank
amounts to 25 litre/second. With time the DM water quality
deteriorates and the silica content in the feed DM water increased
to 0.02 mg/l. Assume that inflow into and outflow from the tank
remains constant at 25 litre/second. Calculate the time for the
silica content in the storage tank to increase to 0.012 mg/l.
Numericals based on Material Balance of Unsteady-state
Operations

Example
A storage tank of a demineralized (DM) water has a holding
capacity of 1500 m3 upto an overflow point. The inflow of DM
water to the tank is 25 lit/sec having silica (as SiO2 ) content of
0.005 mg/lit. The supply of DM water to the high-pressure boilers
from the tank amount to 25 dm3 /sec. With time the DM water
quality deteriorates and the silica content in the feed DM water
increases to 0.02 mg/lit. Assume that the inflow into and the
outflow from the tank remains constant at 25 lit/sec. Calculate
the time for the silica content in storage tank to increase to 0.01
mg/lit. Change in the concentration during the time interval ∆θ.
Consider θ=0 when the silica level in the incoming DM water
increases to 0.02 mg/lit.
 Introduction
Under this portion, there is a balancing of Mass without the
occurence of any chemical reaction.
It means we can have material balance without chemical
reaction.
Here the concept of ”Block Diagram” is very useful.
One should have proper understanding and knowledge of
these block diagrams.
The study of block diagram becomes essential due to the
understanding of different unit operations used in Chemical
Engineering.

Concepts in MEBC Dr. Vyomesh M. Parsana 176/202

 A+B →C
Consider the above given reaction.
0 A0
and 0 B 0 are the reactants which react to give us 0 C 0 as a
product.
We can take the example of Formation of Ammonia.
Based on this reaction, we can understand different terms
associated with it.

Example
N2 + 3H2 → 2NH3

Concepts in MEBC Dr. Vyomesh M. Parsana 177/202

 Different Terms

A+B →C
Stoichiometric Equation
It indicates relative moles of reactants and products
taking part in a reaction.
Stoichiometric Co-efficient
It is the number that precede the formula of each
component involved in chemical reaction.
Stoichiometric Ratio
It is the ratio of stoichiometric coefficients of two
components in the balance reaction equation.

Concepts in MEBC Dr. Vyomesh M. Parsana 178/202

 Different Terms
A+B →C
In actual practice the reactants are not taken/fed in stoichiometric
proportion because of economical/technical or safety aspects.
Based on that we have two definations:
Limiting Reactant
The quantity which is present in a lesser proportion
in a given reaction then it’s theoratical requirenment
is called limiting reactant. When the given reaction
tends towards the completion, then the limiting
reactant will be the first one to be disappear.
Excess Reactant
The quantity present in a larger portion in a reaction
is known as excess reactant. It is present in more
amount than the theoratical requirement.
Concepts in MEBC Dr. Vyomesh M. Parsana 179/202
 % excess of ’B’
h 0 0
i
or fed−moles of 0 B 0 theoretically
= moles of B supplied
moles of 0 B 0 theoretically required
required
× 100

% conversion of ’A’
moles(or mass) of 0 A0 reacted
= moles(or mass) of 0 A0 fed × 100

Concepts in MEBC Dr. Vyomesh M. Parsana 180/202

 Different Terms
A+B →C
Fed/Feed
It simply means the proportion of reactants send to
the reactor for the reaction. It is the total of reacted
amount and unreacted amount.
Selectivity
The selectivity of a reaction is the ratio of the desired
product formed (in moles) to the undesired product
formed (in moles).
Yield
According to the Elements of Chemical Reaction
Engineering manual, yield refers to the amount of a
specific product formed per mole of reactant
consumed.
Concepts in MEBC Dr. Vyomesh M. Parsana 181/202
 Different Terms

A+B →C
Theoretical Requirenment
It is the value of a reactant corresponding to its
stoichiometric proportion.

Concepts in MEBC Dr. Vyomesh M. Parsana 182/202

 Yield of ’C’
moles of 0 A0 reacted to produce 0 C 0
= moles of 0 A0 totally reacted × 100

A+B →C +D
C →Desired
D →Undesired

Selectivity of ’C’ relative to ’D’

moles of 0 C 0 formed
= moles of 0 D 0 formed

Concepts in MEBC Dr. Vyomesh M. Parsana 183/202

Numericals based on Material Balance Involving Chemical
Reaction

Example
The carbon monoxide is reacted with hydrogen to produce
methanol. Calculate from the reaction:
the stoichiometric ratio of H2 to CO
kmol of CH3 OH produced per kmol CO reacted
the weight ratio of CO to H2 if both are fed to reactor in
stoichiometric proportions
the quantity of CO required to produce 1000 kg CH3 OH
Numericals based on Material Balance Involving Chemical
Reaction

Example
In production of sulphur trioxide, 100 kmol of SO2 and 200 kmol
of O2 are fed to reactor. The product stream is found to contain
80 kmol SO3 . Find the percentage conversion of SO2 .
Numericals based on Material Balance Involving Chemical
Reaction

Example
In manufacture of acetic acid by oxidation of acetaldehyde, 100
kmol of acetaldehyde is fed to the reactor per hour. The product
leaving the reactor contains 14.81% acetaldehyde, 59.26% acetic
acid, and rest oxygen (mole basis). Find the percentage conversion
of acetaldehyde.
Numericals based on Material Balance Involving Chemical
Reaction

Example
The feed containing A, B and inerts enter a reactor. The reaction
taking place is:
2A + B → C
The product stream leaving the reactor is having following
composition by mole:
A : 23.07%
B : 11.54%
C : 46.15%
Inerts : 19.24%
Find the analysis of feed on mole basis.
Numericals based on Material Balance Involving Chemical
Reaction

Example
Formaldehyde is produced from methanol catalytic reactor. The
production rate of formaldehyde is 1000 kg/h. If conversion of
methanol is 65%, calculate the required feed rate of methanol.
Numericals based on Material Balance Involving Chemical
Reaction

Example
Monochloroacetic acid (MCA) is manufactured in a semi batch
reactor by the action of glacial acetic acid with Cl2 gas at 373 K in
presence of PCl3 catalyst. MCA thus formed will further react with
Cl2 form dichloroacetic acid (DCA). To prevent the formation of
DCA, excess acetic acid is used. A small scale unit which produces
5000 kg/day MCA, requires 536 kg/day of Cl2 gas. Also, 263
kg/day of DCA is separated in the crystallizer to get almost pure
MCA product. Find % conversion of Cl2 , % yield of MCA and
selectivity.
Numericals based on Material Balance Involving Chemical
Reaction

Example
Tallow is essentially glyceryltristearate
(CH2 OOCH35 C17 − CHOOCH35 C17 − CH2 OOCH35 C17 ). It is
desired to saponify the tallow with caustic soda. For 100 kg tallow,
Calculate:
the theoretical requirement of caustic soda
the amount of glycerine liberated
Numericals based on Material Balance Involving Chemical
Reaction

Example
In a silver electroplating plant, silver nitrate is used. When 1130
amperes were passed through AgNO3 solution for 32,400 sec, it
was found that 2.0 m3 O2 (at NTP) was liberated at the anode.
Calculate:
the amount of silver liberated in kg
the current efficiency of the cell
Numericals based on Material Balance Involving Chemical
Reaction

Example
Formaldehyde is produced by dehydrogenation of Methanol.
CH3 OH → HCHO + H2
The conversion per pass is 67%. The product leaving the reactor is
fed to separation unit battery where formaldehyde is separated
from methanol and hydrogen. The separated methanol is recycled
to reactor. If the production rate of formaldehyde is 1000 kg/h.
Calculate:
combined feed ratio
flow rate of methanol required to the process as fresh feed
Numericals based on Material Balance Involving Chemical
Reaction

Example
Calculate the following for the reaction:
C2 H4 + 2Cl2 → C2 HCl3 + H2 + HCl
the stoichiometric ratio of Cl2 to C2 H4
if 4 kmol Cl2 is used per kmol C2 H4 , find the % excess Cl2
the amount of HCl produced from 50 kg of C2 H4 assuming
reaction goes to completion
Numericals based on Material Balance Involving Chemical
Reaction

Example
Methane oxidation reactions are:
CH4 + O2 → HCHO + H2 O
CH4 + 2O2 → CO2 + 2H2 O
100 kmol of methane is charged. If product stream is found to
contain 10 kmol CO2 and 40 kmol formaldehyde.
Calculate:
% conversion of methane
yield of formaldehyde
Numericals based on Material Balance Involving Chemical
Reaction

Example
SO2 is oxidised to SO3 . If conversion is 70% and air is used in 80%
excess over theoretical requirement.
Calculate:
the kmol air fed per kmol SO2
composition of gases leaving reactor (on vol. basis)
Numericals based on Material Balance Involving Chemical
Reaction

Example
CO and steam are fed to a reactor for production of H2 and CO2 .
The product gas is found to contain 38.46 mol% H2 , 38.46 mol%
CO2 and 23.08 mol% H2 O. Find the mole ratio of steam to CO.
Numericals based on Material Balance Involving Chemical
Reaction

Example
The analysis of a gas entering the secondary converter in a contact
H2 SO4 plant is 4% SO2 , 13% O2 and 83% N2 (on vol. basis). The
gas leaving the converter contains 0.45% SO2 on SO3 free basis
(by vol.).
Calculate:
the % SO2 entering the converter getting converted to SO3
the actual analysis of gases leaving converter on vol. basis
Numericals based on Material Balance Involving Chemical
Reaction

Example
In the Deacon process for manufacturing chlorine, HCl gas is
oxidized with air. The reaction taking place is:
4HCl + O2 → 2H2 O + 2Cl2
If the air is used in excess of 30% of that theoretically required,
and if the oxidation is 80% complete, calculate the composition by
volume of dry gases leaving the reaction chamber.
Numericals based on Material Balance Involving Chemical
Reaction

Example
The gaseous reaction A → 2B + C takes place isothermally in a
constant pressure reactor. Starting with a mixture of 75% A and
25% inerts (by volume), in a specified time the volume doubles.
Calculate the conversion achieved.
Numericals based on Material Balance Involving Chemical
Reaction

Example
What will be the composition of gases obtained by burning pure
FeS2 with 60% excess air? Assume that the reaction proceeds in
the following manner.
4FeS2 + 11O2 → 2Fe2 O3 + 8SO2
Numericals based on Material Balance Involving Chemical
Reaction

Example
One gm of an alloy of aluminum and magnesium reacts with excess
HCl acid to form AlCl3 , MgCl2 and H2 gas. The H2 gas collected
over mercury at 273 K, occupied 1200 ml at 93.26 kPa. Find the
composition of the alloy by wt.
Numericals based on Material Balance Involving Chemical
Reaction

Example
Ethyl alcohol is industrially produced by fermentation of molasses.
The molasses sample contains 45% by wt. fermentable sugar in the
form of sucrose. The reactions taking place in the fermenter are:
invertase
C12 H22 O11 + H2 O −−−−−→ C6 H12 O6 + C6 H12 O6
zymase
C6 H12 O6 −−−−→ 2C2 H5 OH + 2CO2
Calculate the theoretical production of rectified spirit having
density of 0.785 kg/l in liters per tonne of molasses.