BOE_Tutorials.pdf

INDIA BOILER DOT COM
TUTORIAL
FOR
BOILER OPERATION ENGINEER’S
PROFICIENCY EXAMINATION

WELCOME NOTE
Dear Reader, India Boiler Dot Com welcomes you to the Tutorial
Programme and we wish you purposeful learning with all satisfaction.
We have structured the tutorial with a view not only to enable you to
successfully appear at the ensuing examination leading to the Boiler
Engineer’s Proficiency Certificate Class-I and but also to enable you to
become a really confident, competent and capable Boiler Engineer for
achieving proper, safe, efficient, economical and reliable operation in which
ever plant you work.
You already possess basic knowledge of English language, Science and
Mathematics to the extent necessary for fully understanding the principles
and practices of Boiler Operation and Maintenance. As there is no definite
syllabus, which covers the said examination, we have tried to ensure that the
study materials cover for the typical questions that have been asked during
the past 12 years. We would respect your discretion to formulate the answers
by either elaborating or making it more precise depending upon the type of
question asked. In addition to that we have also included that information
which according to our judgment every Boiler Engineer should posses in
order to successfully handle the operation and maintenance of a Boiler.
These lessons will also include numerical and model questions fashioned in
the trend of those asked during previous examinations along with the model
answers for some of them. Some of these questions will have to be answered
by you during our direct training schedule for our assessment of your
progress. If you have any question on a particular lesson, you will have to
ask the same during these training days.

TOTAL STUDY MATERIAL CONTENTS
CHAPTER – 1 FUNDAMENTAL OF PHYSICS
CHAPTER – 2 PROPERTIES OF STEAM
CHAPTER – 3 FUELS AND COMBUSTION
PART - I
CHAPTER – 4 THERMODYNAMICS & HEAT ENGINE CYCLES
CHAPTER – 5 BOILERS
CHAPTER – 6 BOILER MOUNTINGS, ACCESSORIES AND
AUXILIARIES
PART - II
CHAPTER – 7 FIRING IN BOILERS AND TYPES OF FURNACES
CHAPTER – 8 STRENGTH OF MATERIALS
CHAPTER – 9 DRAUGHT SYSTEM OF BOILERS
CHAPTER – 10 BOILER PERFORMANCE
CHAPTER – 11 STEAM ENGINES AND CONDENSERS
CHAPTER – 12 BOILER OPERATION & MAINTENANCE
CHAPTER – 13 BOILER SAFETY
PART - III
CHAPTER – 14 DESIGN FUNDAMENTALS OF BOILER
CHAPTER – 15 BOILER METALLURGY
CHAPTER – 16 FUNDAMENTALS OF INSTRUMENTATION
CHAPTER – 17 BOILER PROTECTION AND INTERLOCKS
CHAPTER – 18 BOILER CONTROLS
CHAPTER – 19 FURNACE SAFEGUARD SYSTEM
CHAPTER – 20 BOILER WATER CHEMISTRY AND CHEMICAL
TREATMENT
CHAPTER – 21 BOILER CHEMICAL CLEANING PROCESS & EROSION
AND CORROSION CONTROL
CHAPTER – 22 BOILER POLLUTION CONTROL
CHAPTER – 23 LIQUID FUEL HANDLING
CHAPTER – 24 SOLID FUEL HANDLING
CHAPTER – 25 ASH HANDLING
CHAPTER – 26 ELECTROSTATIC PRECIPITATORS
PART - IV
CHAPTER – 27 INDIAN BOILER REGULATION

INDIABOILER DOT COM
TUTORIAL FOR SECOND CLASS BOILER ENGINEER’S PROFICIENCY EXAMINATION
DLP/BOE-II/ I/1692001
A-1
CHAPTER - 1
Fundamentals of Physics
1.1 Units for measurement
Physical measurement always requires specification of both a value (i.e., a
number representing “how much”) and a unit (i.e., “of what”). Systems of
measurement are formal strategies for indexing amounts of specified physical
quantities.
1.1.1 Unit:
To express the magnitude of a physical quantity a standard is chosen which is
of the same kind as physical quantity. This standard is taken as reference to
measure a physical quantity which is known as unit.
Therefore the process of measurement of a physical quantity involves.
i) The selection of the unit and
ii) Number of times the unit is contained in that physical quantity
In general,
measure of a physical quantity = numerical value of the quantity X size of its
unit
1.1.2 Fundamental and derived units:
Fundamental units are those units, which can neither be derived from one
another, nor can they be further resolved into any other units.
The three fundamental units are
(i) Mass (ii) Length and (iii) Time
Derived units are units of all such physical quantities which can be expressed
in terms of the fundamental units of mass, length and time.
Ex. unit of area = (metre)2
unit of volume = (metre)3
hence all derived units can be obtained by writing it in terms of fundamental
units.
1.1.3 System of Units:
The common system of units are:
(i) CGS system: It was set up in France and is based on centimetre, gram and
second as the fundamental units of length, mass and time respectively. It is a
metric system of unit
(ii) FPS system or British system of units: - It is based on foot, pound and
second as the fundamental units of length mass and time.
(iii) MKS system: It was also set up in France and is based on metre, kilogram
and second as the fundamental units of mass, length and time.

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1.1.4 SI system of unit:
In 1960 The General Conference of Weights and Measures introduced a new
system of, units known as SI units. It is based on seven basic and two
supplementary units given as:
1.1.4.1 Base Units:
Name of the
Property
Unit of Measurement Symbol
Length “metre” “m”
Mass “kilogram” “kg”
Time “second” “s”
Electric Current “ampere” “A”
Thermodynamic
Temperature
“kelvin” “K”
Amount of Substance “mole” “mol”
Luminous Intensity “candela” “cd”
1.1.4.2 Supplementary Units:
Name of the
Property
Plane angle “radian” “rad”
Solid angle “steradian” “sr”
All properties of interest to a Boiler Engineer can be derived from the above
Base and Supplementary Units as can be seen from the following:
1.1.4.3 Derived Units:
Name of the
Property
Acceleration “metre per sec2
” “m/s2
”
Area “square metre” “m2
”
Density “kilogram per cubic metre” “kg/m2
”
Energy “joules” “J” = “N.m”
Entropy “joules per kelvin” “J/K”
Specific entropy joule per kilogram kelvin J/(kg·K)
Force “newton” “N” = “kg.m/s2
Power (Rate of Work) “watt” “W” “J/s”
Pressure “pascal” “Pa” = “N/m2
”
Quantity of Heat “joules” “J” = “N.m”
Stress (Same as
Pressure)
“pascal” “Pa” = “N/m2
”

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Velocity “metre per second” “m/s”
Work “joules” “J” = “N.m”
Moment of force newton meter N·m
specific energy joule per kilogram J/kg
1.1.4.4 Advantage of SI
(i) It is a rational system of units - Its makes use of only one unit for one
physical quantity. Ex. all types of energies are expressed in Joules. Whereas in
MKS system different units are used for different types of energies. For ex.
mechanical energy is measured in Joule, heat energy in calorie and electrical
energy in watt hour.
(ii) SI is a coherent system of units i.e all derived units can be obtained by
dividing and multiplying the basic and supplementary units and no numerical
factors are introduced as used to be the case with certain units of the CGS and
MKS systems.
(iii) SI is a metric system. The multiples and sub-multiples can be expressed
as the powers of 10.
1.1.4.5 The Prefixes of the S I
The S I allows the sizes of units to be made bigger or smaller by the use of
appropriate prefixes. For example, the electrical unit of a watt is not a big unit
even in terms of ordinary household use, so it is generally used in terms of
1000 watts at a time. The prefix for 1000 is kilo so we use kilowatts[kW] as
our unit of measurement. For makers of electricity, or bigger users such as
industry, it is common to use megawatts [MW] or even gigawatts [GW]. The
full range of prefixes with their [symbols or abbreviations] and their
multiplying factors are given below.
yotta [Y] = 1024
zetta [Z] = 1021
exa [E] = 1018
peta [P] = 1015
tera [T] = 1012
giga [G] = 109
(a thousand millions = a billion)
mega [M] = 106
(a million)
kilo [k] = 1000 (a thousand)
hecto [h] = 100
deca [da] = 10
1 deci [d] = 0.1
1 centi [c] = 0.01
1 milli [m] = 0.001 (a thousandth)
1 micro [µ] = 10-6
(a millionth)
1 nano [n] = 10-9
(a thousand millionth)
1 pico [p] = 10-12
1 femto [f] = 10-15
1 atto [a] = 10-18
1 zepto [z] = 10-21
1 yocto [y] = 10-24

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1.1.5 Rules of working with Units of Measure:
Rules
a. Derived units, when written in words, are not written with a capital letter
unless they are at the beginning of a sentence. The exception is the Celsius
degree which is always written with a capital letter.
Symbols are written with the following:
• no full stop except at the end of a sentence;
• as singular even though the written term is plural;
• a space between the numerical value and the symbol, e.g. 12 m; 1.2
kg/m 3
and 0.12 rad;
• in lower case unless the symbol has been taken from a proper name, e.g.
ampere is written as "A"; kelvin as "K"; and volt as "V";
• in lower case unless the prefix is mega, giga, tera, peta or exa, e.g.
megawatt is written as MW; gigamole as Gmol and petahertz as PHz;
• the product in a compound unit should be indicated by "." eg "N.m"; and
"cd.sr";
• "." and "/" are only used with symbols and not with unit names written
in full eg m/s and not "metre/second"; and kW.s and not
"kilowatt.second"; and
• when a unit involves the division of one symbol by another it can be
written in one of three ways eg "m/s"; "m.s-1"
; and " m "
b. The combination of a prefix and a unit is written as one word, eg,
millimetre, microgram or nanosecond.
c. Prefixes are generally used in combination with the units so that, for
convenience, the number portion of the measurement is greater than 0.1
and usually below 1,000. For example, 0.005 grams is written as 5
milligrams or 5 mg.
d. When writing a numerical measurement, the symbol for the prefix is placed
in front of the symbol for the unit eg, ninety nine millimetres = 99 mm,
nine point nine micrograms = 9.9 mg and nine nanoseconds = 9 ns.
e. Where a compound unit has both a numerator and a denominator, any
prefix is preferably attached to the symbol in the numerator eg kJ/mol and
not J/mmol; and mA/mol and not A/kmol.
1.1.6 Some of the useful conversions for SI Units:
Force:
1 kgf = 9.81 N
Pressure:
1 bar = 105
Pa (N/m2
)
1.02
kgf/cm2
750
mmHg
1.02 x 104
mmwc
0.987
atm
14.504
psi

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Heat, work & energy:
1 MJ = 106
Nm
238.85
k cal
948.0
Btu
101.97 x 103
kg m
737.6 x 103
ft lb
0.278
kW h
Power:
1 MJ/s = 106
Watt
238.85
k cal/s
101.97 x 103
kg m/s
73.76 x 103
ft lb/s
1360
metric hp
56880
Btu/min
Calorific value and Specific Enthalpy:
1 MJ/kg = 238.85
k cal/kg
101.97 x 103
kg m/kg
334.55 x 103
ft lb/lb
429.92 x 103
Btu/lb
Specific heat and Specific Entropy:
1 kJ/kg K = 0.2388 Kcal/kgo
C 0.2388 Btu/lbo
F
HP:
1 metric HP = 735.75 watt* 1 British HP = 746
* In Boiler Engineering calculations, HP is to be considered as metric HP
Heat:
4.187 kJ = 1 kcal = 2.205 CHU = 3.969 BTU
Mechanical Equivalent of heat(J):
= 427 kgm = 4.187 kJ = 1 kcal
1.2 Matter:
The Matter exists in three states:
Solid, Liquid and Vapour.
All matter comprises of atoms and molecules. The extent to which these
atoms are bonded to each other decides the state of the matter. As long as the
bonding force between the atoms is large, the matter remains Solid and retains
its shape and tends to oppose any cause, which tends to change its shape. This
is the reason that solids have the property of elasticity and are stressed if
strained. The Bonding Forces are of several types such as gravitational pull
between atoms and Covalent Bonds.
The atoms and molecules of any matter is in a constant state of random
vibrations and the vibrations’ amplitude increases with temperature of the
matter. Depending on temperature the Inter-atomic Distance varies and when
the Inter-atomic Distance is such that force of attraction between tow atoms
equals force of repulsion, a threshold state is reached when state of matter
changes from Solid to Liquid and the matter no longer exhibits the property
of elasticity and does not resist a change in its shape.

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At a further higher temperature the atoms and molecules reach a state of
vibration when there exist no cohesive force between its atoms or molecules.
At such a temperature the matter changes its state from Liquid to Vapour.
The temperature at which a solid matter changes its state to liquid is called its
Melting Point and the temperature at which the matter in Liquid state changes
to Vapour is called Boiling Point. These temperatures depend upon pressure.
To the extent that the three phases of matter relate to the subject of Boiler
Technology and Engineering, the reader is advised to recall that Water is
Solid when it is Ice. When heated, Ice (which is water in solid state) changes
its state to Liquid (Water) and when further heated, water changes to the third
state, which is vapour and called Steam.
Under atmospheric pressure Melting Point of Ice is 00
C or 320
F and Boiling
Point of Water is 1000
C or 2120
F. However, these points of temperature
depend on pressure. The change in melting point of ice is not relevant to
boilers and steam engines and turbines hence we will not deal with it further.
However, the boiling point of water varies greatly with its pressure.
This matter is covered in greater details under Properties of Steam in this
tutorial. The characteristics of change of state or phase of water to steam are
derived experimentally and are published by various renowned international
bodies (such as ASME) as “Steam Tables”.
It is considered relevant here to mention that in a boiler just above the surface
of water in the boiler drum, contains tiny particles of water which though
suspended above the surface of water in the drum, is not actually in vapour
state and needs to absorb heat in order to get converted into vapour or steam.
That is why the steam just above the water surface in boiler drums is called
“wet steam”. When each and every tiny water particle has got fully converted
into vapour or steam it is called “dry saturated steam”.
1.2.1 Mass:
Mass is a property of physical objects that, roughly speaking, measures the
amount of matter they contain.
Strictly speaking, there are three different quantities called mass:
• Inertial mass is a measure of an object's inertia: its resistance to changing
its state of motion when a force is applied. An object with small inertial
mass changes its motion more readily, and an object with large inertial
mass does so less readily.
• Passive gravitational mass is a measure of the strength of an object's
interaction with the gravitational field. Within the same gravitational field,
an object with a smaller passive gravitational mass experiences a smaller
force than an object with a larger passive gravitational mass. (This force is
called the weight of the object. In informal usage, the word "weight" is
often used synonymously with "mass", because the strength of the
gravitational field is roughly constant everywhere on the surface of the

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Earth. In physics, the two terms are distinct: an object will have a larger
weight if it is placed in a stronger gravitational field, but its passive
gravitational mass remains unchanged.)
• Active gravitational mass is a measure of the strength of the gravitational
field due to a particular object. For example, the gravitational field that
one experiences on the Moon is weaker than that of the Earth because the
Moon has less active gravitational mass.
The SI unit for mass is the kilogram (kg), which is equal to the mass of the
international prototype kilogram.
1.2.2 Mole:
The mole is the amount of substance. The unit is call the mole (mol), and it’s
defined as the number of molecules present in 0.012 kilograms of carbon-12.
In other words, 1 mole of carbon-12 has a mass of 12 grams. 1 kmol = 103
mol
1.2.3 Volume:
Volume of gas or any substance is defined as the space, which it occupies.
Unit of volume of any substance is cubic centimeter or cubic meter.
The volume is also expressed in litre.
1 liter = 1000 cm3
=106
mm3
= 10-3
m3
1.2.4 Specific volume:
The specific volume of a substance is its volume per unit mass i.e m3
/kg.
The unit of specific volume is m3
/kg.
One kilogram of air at 00
C and under an absolute pressure of 1.0332 kg/cm2
(760 mm of Hg) has volume of 0.7734 m3
.
Therefore the specific volume of air under these conditions is 0.7734 m3
/kg. It
is denoted by v
1.2.5 Density & Specific Gravity:
Density of a substance signifies how densely it is packed with mass.
Mathematically it is expressed as “mass per unit volume” i.e kg/m3
it also
termed as “Mass density” and denoted by ρ.
Density or Mass density, ρ = m/v
Depending upon the Units of Measurements, the Density is expressed in
various units such as “gm/cc” r “kg/ m3
or “lb/ ft3
”. For example water has a
Density of “1 gm/cc” or “1000 kg/m3
” or “62.4 lb/cft”. Thus the numeric
value of Density of a substance is different for the same substance in different
Units of Measurements.
The Specific Gravity of a substance is its “density compared with that of
water”. Since Specific Gravity is only a comparison of Density of a substance

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with the Density of water, it has no unit and its value remains same
irrespective of the Units of Measurement. Specific Gravity is generally used
for liquids.
1.2.6 Concentration:
Concentration is the amount of a substance contained in a given volume.
"Amount of a substance" and "given volume" can take many forms.
1.2.6.1 Mole/Volume
when a known number of moles of a substance is dissolved or dispersed in a
liquid to give a known volume of solution or suspension.
Moles per litre (mol/L) and moles per cubic centimetre (mol/cc) express
concentration in the terms of mole per unit volume (mol/v).
Five mol/L is equal to five mole of substance in one litre of solution.
The number of moles of a substance in one litre of solution is called the
molarity of that solution.
1.2.6.2 Mass/Volume
when a mass of a substance is dissolved or suspended in a liquid to give a
known volume of solution or suspension.
Kilograms per litre (kg/L), grams per litre (g/L), milligrams per litre (mg/L)
and grams per cubic centimetre ( g/cc) express concentration in the terms of
mass per unit volume which is usually referred to as weight/volume (w/v).
1.2.6.3 Mass/Mass
when a mass of substance is dispersed in another mass to give a known
resultant mass.
Grams per kilogram (g/kg) and milligrams per kilogram (mg/kg) express
concentration in the terms of mass per unit mass, which is usually referred to
as weight/weight (w/w).
1.2.6.4 Volume/Volume
when a volume of substance is dispersed in another volume of substance to
give a known resultant volume.
Millilitres per litre expresses concentration in the terms of volume per unit
volume, which is usually referred to as volume/volume (v/v).
Thus, if 80 millilitres of alcohol is diluted to 2000 millilitres with water the
result is a solution of 2000 millilitres or 2 litres.
1.2.6.5 Parts Per Million (ppm)
A part per million is one of a quantity in one million of another quantity. Parts
per million is abbreviated to "ppm".
To convert a concentration from v/v to ppm the concentration must be in
ml/ml, L/L, etc... and then multiply by 106
to get ppm v/v.

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In the same way w/w concentration can only be converted to ppm using the
factor "106
" if the concentration is expressed in grams per gram (g/g),
kilograms per kilograms (kg/kg), etc.
The volume of a substance in the gaseous form can be equated to the mass of
the substance if the exact volume is known. Thus, the volume /volume
concentration of a substance can be equated to the mass/volume concentration
of that substance.
1.2.6.6 Percent
Percent (%) is also used to express concentration much the same as parts per
million (ppm). The difference is that percent relates to one in a hundred
compared to ppm, which relates to one in a million.
1.2.7 Pressure:
Pressure is defined as ‘Force per Unit Area’. Take a jar of glass with a flat
bottom, filled with water and keep it over a table. The weight of water in the
jar exerts a force on the surface of the table. If this force is measured over a
unit area of the surface, then it is called the pressure. Therefore pressure can
be defined as the force exerted by an object over the surface of unit Area.
i.e. pressure = force / area
In practice it is expressed or measured in following units:
N/m2
Kg/cm2
Lb/in2
bar (1 bar=105
N/m2
)
pascal (1Pascal= 1 N/m2
)
height of liquid column (Normally water & Mercury)
1.2.8 Atmospheric Pressure:
The atmosphere, surrounding the earth, exerts a pressure on its surface
equivalent to the weight of air acting over unit area of the earth's surface and it
is known as atmospheric pressure. At sea level, the weight of air over a weight
of unit area of earth’s surface is equivalent to weight of a column of 76 cm
(760 mm) of mercury column (Hg) at 00
C. It is taken as the standard
barometric pressure. This is also known as a physical atmosphere or
barometric atmosphere.
The density of mercury is 13. 595 grams per cubic centimeter
Therefore standard barometric pressure = 76 x 13. 595 = 1033. 32 grams / sq.
cm. i.e. 1.03322 kg/sq.cm
1 ata = 1 metric or technical atmosphere (1 kg/cm2
ab.) = 760 /1.0332 =735.6
mm of Hg
Pressure is also measured in the unit of mm of water column. One
Atmospheric pressure is equivalent to 760x13.595mm of water column. That
is 10332.2mm of water. This is 10.332 meter of water.
1kg/cm2
= 735.6mm of Hg = 735.6x13.595 = 10meter of water
We know that the atmosphere exerts pressure as mentioned. However, if we
take a pressure gauge in our hand, it reads Zero, even though the atmospheric

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pressure is present. Thus pressure Gauge measures the pressure with reference
to Atmospheric pressure (Ref. Fig.-1). The pressure indicated by the gauge
above atmosphere is known as gauge pressure. The pressure indicated by the
gauge when the system pressure is less than atmospheric is termed as
Vacuum, which is generally measured in mm of water or mercury column
and the gauges are known as vacuum gauges.
Fig. 1
Thus the Absolute pressure = Atmospheric pressure + Gauge pressure
Pab = Pat + Pg
or
Absolute pressure = Atmospheric pressure – Vacuum
Pab = Pat - Vacuum
1.2.9 Barometer:
The barometer is the simplest instrument for measuring atmospheric pressure.
The earth's atmosphere at sea level has a weight of 14.7 pounds over a square
inch of surface. This is the weight of a column of air that extends from sea
level at the earth's surface to the edge of the atmosphere. This weight changes
as the temperature and composition of the air mass changes. A barometer uses
a substitute column of mercury fluid in place of the air. One atmosphere in a
mercury barometer is equaled by a column of only 760 mm Hg.
Fig.2
Absolute Pressure
= Atm. Pressure + Gauge Pressure
Vacuum =
Atm. Pressure - Absolute Pressure
Absolute Pressure
= Atm. Pressure - Vacuum
Absolute zero Pressure
Atmospheric Pressure
1.0332 kg/cm2
Vacuum Gauge Pressure
Pressure Gauge Reading
Gauge Pressure
ρ
h = 760 mm of
A simple Mercury

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1.2.10 Manometer:
The manometer is one of the simplest tools for measuring gas pressure
differences. A manometer is a u-tube. One side of the "U" is open to
atmosphere and the other side is connected to a closed container. The "U" is
filled with a fluid. If both sides of the "U" have the same liquid levels then the
pressure inside and the pressure outside are the same. The difference between
the liquid levels equals the pressure difference between inside and outside.
The mercury level will be lower on the side with greater pressure. The higher
pressure "pushes" the mercury down. The Manometer measures the Gauge
pressure. Mercury is particularly convenient for use in manometers (and
barometers) because at room temperature it has low vapor pressure, does not
wet glass, and has a high density.
Fig-3
1.2.11 Properties of Gases
1.2.11.1 The Kinetic Molecular Theory
The Kinetic Molecular Theory is the basis of the many properties of gases.
The five postulates to the Kinetic Theory are as follows:
• Gases are composed of molecules whose size is negligible compared to the
average distance between them.
• Molecules move randomly in straight lines in all directions and at various
speeds.
• The forces of attraction or repulsion between two molecules in a gas are very
weak or negligible, except when they collide.
• When molecules collide with one another, the collisions are elastic; no kinetic
energy is lost.
• The average kinetic energy of a molecule is proportional to the absolute
temperature.
h
h is how much lower the gas
pressure is than the pressure in
the atmosphere
h is how much higher the
gas pressure is than the
pressure in the atmosphere h
GAS GAS
Manometer

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1.2.11.2 Boyle's Law
Boyle's Law states the volume of a definite quantity of dry gas is inversely
proportional to the pressure, provided the temperature remains constant.
Mathematically Boyle's law can be expressed as P1V1 = P2V2
• V1 is the original volume
• V2 is the new volume
• P1 is original pressure
• P2 is the new pressure
1.2.11.3 Charles's Law
Charles's Law can be stated as the volume occupied by any sample of gas at a
constant pressure is directly proportional to the absolute temperature.
V / T =constant
• V is the volume
• T is the absolute temperature (measured in Kelvin)
Charles's Law can be rearranged into two other useful equations.
V1 / T1 = V2 / T2 & V2 = V1 (T2 / T1)
• V1 is the initial volume
• T1 is the initial temperature
• V2 is the final volume
• T2 is the final temperature
Charles's Law only works when the pressure is constant.
Note: Charles's Law is fairly accurate but gases tend to deviate from it at very
high and low pressures.
1.2.11.4 NTP
NTP stands for Normal Temperature and Pressure. NTP is 0o
Celcius and 1
atmospheric pressure. Gases properties can be compared using NTP as a
reference.
1.2.11.5 Combined Law
The combined gas law is a combination of Boyle's Law and Charles's Law;
hence its name the combined gas law. In the combined gas law, the volume of
gas is directly proportional to the absolute temperature and inversely
proportional to the pressure.
This can be written as PV / T = constant.
Therefore we can write P1V1 / T1 = P2V2 / T2.
• P1 is the initial pressure
• V1 is the initial volume
• T1 is the initial temperature (in Kelvin)
• P2 is the final pressure
• V2 is the final volume
• T2 is the final temperature (in Kelvin)
Also Vt = V0 { 1 + t / 273 }.
Where ‘V0’ is the volume at 00
C and‘t’ is the temperature in 0
C

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1.2.11.6 Ideal Gas Law
The ideal gas law is a combination of all the gas laws. The ideal gas law can
be expressed as PV = mRsT.
• P is the pressure in atm
• V is the volume in liters
• m is the mass of the gas considered
• Rs is a constant
• T is the temperature in Kelvin
This equation is known as Characteristic equation of a gas.
Sometimes R is called the Characteristic or specific gas constant.
1.2.11.7 Universal gas constant:
If the molecular mass of any gas is multiplied by its specific gas constant Rs it
will be found that the product is the same for all gases. This constant is termed
as Universal gas constant.
For SI system the value of universal gas constant is 8.3143 kJ/ kmol K.
Thus Ru = MRs = 8.3143, kJ/kmol K.
Where M is the molecular mass of the gas in kg/ kmol.
[It should be noted that in any calculations involving the gas laws, absolute
pressures and absolute temperatures must be used.]
Avogadro's Law:
At a given temperature and pressure, equal volumes of gas contain equal numbers
of moles.
V = constantAL n
For example, It has been found that at NTP, 22.4 m3
of Hydrogen gas has a mass of
2 kg, therefore 1 kmol of Hydrogen.
As per Avogadro’s Law, at NTP 22.4 m3
of all other gas will have the
corresponding mass of 1 kmol of that gas.
1.2.11.8 Perfect Gas: A perfect gas or ideal gas may be considered as one that
obeys the laws of Boyle and Charles and the Characteristic equation of a gas
which is obtained by combining the above laws.
No gas is perfect, but many gases can approach this standard within the
temperature limits of applied thermodynamics.
1.3 Temperature:
Temperature is the measure of the relative warmth or coolness of an object.
The temperature of a substance does not measure its heat content but rather
the average kinetic energy of its molecules resulting from their motions. A
one-pound block of iron and a two-pound block of iron at the same
temperature do not have the same heat content. Because they are at the same
temperature the average kinetic energy of the molecules is the same; however,

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the two-pound block has more molecules than the one-pound block and thus
has greater heat energy.
For measurement of Temperature there are two scales of measurements, one is
“Fahrenheit” and the other is “Centigrade” or “Celsius”. The arbitrary
reference taken is the freezing point of water under atmospheric conditions.
This point at which water freezes to a solid state is considered as ZERO in
Celsius or Centigrade Scale. Again the point of reference of water boiling at
atmospheric condition and transforming to vapor stage is taken as 100 in
Celsius or Centigrade Scale. In the Fahrenheit Scale the point corresponding
to temperature at freezing of water is taken as 32, for water boiling point as
212
In MKS systems, the unit of temperature is degree Centigrade (C)
In FPS system the unit of temperature is degree. Fahrenheit (F)
In SI system the unit of temperature is degree Celsius (C)
10
Centigrade = 10
Celsius.
A temperature reading on one scale can be converted into a reading on the
other scale by the following formula:
C/100 = (F-32)/180 or C=5/9 (F-32) or F= 1.8C+32
where, C is temperature in Celsius or Centigrade and F is temperature in
Fahrenheit.
1.3.1 Absolute temperature scale:
We know that temperature is the effect causes by internal energy of a
substance due to random motion of molecules of a substance. A body that is
hotter has its molecules moving more vigorously than that of a body which is
colder. Thus, there can be a state when there is absolutely no random motion
of the molecules of a substance. There is one particular temperature at which
the molecular random motion of each substance totally stops. This
temperature is called ‘Absolute Zero’ because there can not be a temperature
lower than this (since the molecules can not be more stationery than being in
no motion at all).
Absolute zero is the temperature at which all vibratory, translatory and
rotational motions of the molecule of a substance is supposed to cease i.e.
when internal energy becomes zero. A gas on cooling will contract in volume
as the temperature falls. Charles found with perfect gases, the decrease in
volume per degree Centigrade decrease in temperature is 1/273rd
of its initial
volume at 00
C, pressure remaining constant. Thus, the volume of gas will be
zero at temperature –2730
C. This temperature 2730
C below 00
C (or -2720
C)
is called the Absolute Zero of temperature. The absolute temperature is the
temperature measured above the point of Absolute Zero. Absolute temperature
is expressed by the capital latter ‘K’ and the scale using the Absolute Zero is
called ‘Kelvin’ Scale. By adding 273 to the temperature in degree Centigrade

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we get the temperature in degrees of the Kelvin scale or 0
K.
Temperature K = Temperature 0
C+ 273
i.e. K = C + 273
Absolute temperature in degree Fahrenheit is known as degree Rankine or 0
R
and the Absolute Zero in degree Fahrenheit occurs at –4600
F.
Thus, Temperature 0
R = Temperature 0
F + 460
i.e. R = F + 460
1.3.2 Thermometer:
An instrument used for measuring temperature is called a thermometer and is
constructed by using one of the following principles:
• the change of length, such as length of a mercury column,
• the change of volume, such as volume of a fixed mass of gas at constant
pressure,
• the change of pressure, such as pressure of a fixed mass of gas at constant
volume,
• the change in electric resistance, as in a thermistor,
• the flow of electricity due to Seebeck effect, as in a thermocouple,
• the radiation, as in radiation pyrometers.
1.3.2.1 Glass Bulb (Mercury thermometer): Most common for measuring air
temperature is the liquid-in-glass thermometer, which consists of a glass tube
enlarged at the bottom into a bulb that is partially filled with mercury(or
organic liquid). The tube's bore is extremely small—less than 0.02 inch (0.5
millimeter) in diameter. Thus a small amount of expansion or contraction of
the mercury in the bulb, caused by heating or cooling, produces a noticeable
rise or fall in its level in the tube.
1.3.2.2 Bimetal thermometer: Two different metals are bonded together with one
end attached to an indicating needle which aligns with a circular scale on the
face of the instrument. Since the metals expand at different rates, movement
occurs depending on the temperature fluctuation and the needle moves.
1.3.2.3 Indicating Material: A variety of “crayons” and pellets are available that
melt at specific temperatures. These do not really measure temperature
directly, but do indicate the maximum temperature that a material was
exposed to.
1.3.2.4 Vapor/Gas Filled: Such thermometer operates on a similar principle to the
glass bulb type thermometer.
1.3.2.5 Galileo thermometer: These tend to be used in decorative settings around
the home or office. These interesting models operate based on principles of
specific gravity.
1.3.2.6 RTD and Thermistor: These are based on the change in resistance of a
conductor when the temperature of the wire changes. In both the instruments
temperatures are digitally displayed and have better accuracies.

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1.3.2.7 Thermocouple: These operate based on the temperature change that occurs
at the junction of two dissimilar wires. When the temperature changes a small
current is generated by the junction. This current is then compared to a
reference junction (calibrated standard or ice water bath) and converted to a
temperature by electric or electronic means. So the system includes the
thermocouple itself, connecting wiring and some method (Generally a digital
meter) to display the temperature reading.
Another significant advantage of the thermocouple is that the indicating
instrument can be a very long distance from the thermocouple environment.
Once the system is calibrated and the current from the thermocouple is
captured, a variety of electrical options are available for getting the
information to a display unit.
1.3.3 Pyrometer: It is a non-contacting device intercepting and measuring thermal
radiation emitted from an object to determine surface temperature. Pyrometer
is derived from the Greek word pyro, meaning fire.
The temperature of a material, affects the color. The infrared light spectrum
works very well for this and is the basis for the infrared thermometer or
pyrometer. These units do not require a contact with the material and are
available as hand held units. They can sense a very high range of
temperatures.
Some applications of pyrometers
Item Instrument used to measure temperature
Boiler combustion space Optical pyrometer
Economiser, feed water heaters and
chimney gases
Base metal thermo-couple
Incandescent filaments Optical pyrometer
Incandescent gas mantels Radiation pyrometer
1.4 Work:
If a heavy mass is to be moved from one place to other, one has to apply force
or spend energy. The Force applied to a body multiplied by the distance
moved is the amount of work done or amount of energy spent.
Work = Force x distance (traveled in the direction of force)
Work only involves the useful part of a force, namely the part that is effective
in causing the motion.
[Suppose a pail of water weighing 7 N is carried over a distance of 10 m. In
order to hold the pail up against gravity a vertical force of 7 N is exerted on
the pail. The motion, however, is horizontal, and the force exerted does no
work, even though one might get tired of holding the pail after a while.]
In SI system, the unit for work done is Newton-metre (Nm), which is the
product of a unit force (one Newton) acting through a 1-metre distance. This
unit of work done is also called joules (J).
1 J = 1 Nm
1 kg.m = 9.81 Nm = 9.81 Joules (J)
Work can also be measured in foot pounds or Kg metres

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1.5 Power:
Suppose a weight is lifted off the floor at a fixed a distance. The work done in
this case would be the product of the force exerted times the distance covered,
independent of how fast the weight was lifted. Now if the same weight is
lifted faster that is in lesser period, then one might be tempted to say that more
``work'' is done. Actually the work done in both the cases is same and it is the
Power that is different.
The power exerted by a force is defined as change in work done over a period.
Power exerted = work / time = force x distance / time = force x velocity.
The SI Unit for Power is “Watt”. In other words Jules/sec = Watt.
In British Units the Unit of Power is “Horse Power” or “HP”, which
corresponds to a rate of work of 550 ft-lb/sec or 1 British HP = 746 Watts.
But unless and other wise mentioned, HP should be considered as Metric HP
1 Metric HP = 75 kg m/ s = 735.75 Watts.
1.6 Energy:
In mechanics is defined as “capacity of doing work”. Units of Energy and
Work are same. Energy exists in two forms, namely, Potential Energy and
Kinetic Energy.
1.6.1 Potential Energy is possessed by a body due to its position relative to
other body or of parts of the same body under the action of a force or forces
tending to alter their relative position. For example, a body which is allowed
to fall towards earth may be made to do work; hence before it begins to fall it
possesses potential energy, or energy due to its position relative to earth.
Fig 4
A compressed spiral spring has potential energy because if it is allowed to
resume its unstrained form it can be made to do work. Likewise compressed
air possesses potential energy. The energy stored in a piece of coal is potential
energy, and under favourable conditions the atoms of the constituents of the
coal and atoms of oxygen of the air will rush together and produce heat which
may be converted into work. . If a body of W kg weight is allowed to fall from
an elevation L2 to an elevation L1, the change in potential energy.
∆PE = PE 2 - PE 1 = W (L2 – L1)
The unit of potential energy is Kilograms meter (MKS) and Newton metre
(SI)
h
PE=mg
PE=0
Gravitational
Potential

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1.6.2 Kinetic Energy of a body is due to its being in motion with respect to another
body. A kilogram of water at rest at a height of 100 metres above level of the
sea possesses 100 kg.m of potential energy and if this water is allowed to fall
freely to the level of the sea, without doing work on the way it will in every
position of its fall possess 100 kg.m of energy, but as it descends its potential
energy will diminish, and the remainder of 100 kg.m will be stored in water as
kinetic energy. When the 1kg of water would have fallen 25 metres its
potential energy would be reduced by 25 kg.m to be only 75 kg.m and its
kinetic energy would then be 25 kg.m so that total is 75 kg.m (Potential
Energy) + 25 kg.m (Kinetic Energy) = 100 kg.m.
A body of weight ‘w’ kg, moving with a velocity ‘v’ possesses a certain
amount of kinetic energy (KE) with reference to earth gravitational force,
K.E = w. v2
/2g.
The unit of K.E. is also kg.m in MKS and Newton metre in SIS.
Other forms of energy are also different manifestations of these two forms.
For example Electricity stored in a Capacitor having a Capacitance of C
Farads and charged to a Voltage of V Volts is a Potential Energy and its value
is ½CV2
Joules. Similarly, the Energy in an Inductor having Inductance of L
Henry (and passing an electrical current of I Amps)has a Kinetic Energy equal
to ½LI2
Joules.
All other forms of energy such as Magnetic, Light etc. are similarly
explainable in terms of Potential or Kinetic Energy.
1.7 Internal Energy: The molecules of all substances are continuously in motion.
The movement of molecules is more in gases than in liquids. Even when a gas
is stored in a closed vessel and is stagnant, that is not moving, it possesses a
considerable amount of internal Kinetic Energy due to motion of its molecules
within the limits of its containing vessel. In addition of the Internal Kinetic
Energy substances also have Internal Potential Energy due to the relative
position of their molecules. Thus, the Internal Energy, E of a substance may
be defined as the algebraic sum of Internal Kinetic Energy and Internal
Potential Energy of its molecules.
The internal energy of substance increases with increases of temperature of
substance due to increases of molecular activity. Thus Internal Energy is a
function of Temperature and its value increases or decreases by adding heat to
or subtracting heat from the substance.
1.8 Torque:
Torque is a measure of the 'strength' being used in turning (or attempting to
turn) something.
A common example is that of a spanner being used to move a nut. A force is
being applied at one end of the spanner. That force is multiplied by the
distance between it and the turning-point (which, in this case, is the centre of
the nut) to give a measure of the torque which is being applied. This seems to

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be the same as for work which is also a force being multiplied by a distance
but look closely, in the definition for torque there is no mention of the force
moving as there is in the full definition for work.
So, they are different things even though the units are the same, and no work
is done until, in this case, the spanner moves - and even then it is a matter of
how far the force moves, and not its distance from the centre.
The SI preferred unit for torque is newton metres [Nm] and for work is
joules [J].
1.9 Specific Energy :
This is a measure of the amount of energy contained in a unit quantity of some
substance. The unit quantity may be either of mass or of volume. For unit
mass, usually referred to as Specific Energy, Its units are [J/kg] or [kJ/kg]
For unit volume, usually referred to as Calorific Value, units such as [kJ/m³]
or [MJ/m³]. should be used .
1.10 Mechanical Equivalent Of Heat: Heat and Work are mutually convertible
from one form into another. In a heat engine the heat produced by combustion
of the fuel used is converted into the work done by the engine. When the
brakes are applied to the wheels of a moving train, in order to bring it to rest,
the kinetic energy of the train is converted into heat at the rubbing surfaces of
the brake blocks and wheels, or if the wheels skid the heat is produced at the
rubbing surfaces of wheels and rails. Careful experiments have shown that a
certain definite number J or foot pound of work is equivalent to one unit of
heat.
In British Units J is 778 ft.lb. for 1 Btu.
and in metric units, 4.187 Kilojoules = 1 Kilocalories &
1 Kilocalories = 427 kg-m
1.11 HEAT AND HEAT TRANSFER
1.11.1 Heat
Heat is believed to be “a mode of motion”. It is supposed that a body
possessing heat has its particles or molecules in a state of motion, the rate of
motion increasing as the body gets warmer and diminishing as the body cools.
As to the character of motion of the molecules it may be imagined to be an
oscillatory motion in the case of solids and liquids, but in the case of gases it
is supposed to be a motion of translation.
It is found that all the phenomenon of Heat may be explained by this theory.
For example, it is well known that in general the effect of heat on matter is to
enlarge it. A piece of iron when heated gets longer, wider and thicker (due to
thermal expansion). Now it is natural to expect that if the molecule of iron
have more motion as the iron gets hotter they will require more room and will

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therefore push one another further apart and consequently cause the whole
body to get larger, just as a crowd of people take up more space when they
jostle one another than they do when standing still or when jostling to a less
extent.
For the purpose of explaining how heat is transmitted through space the latter
is supposed to be filled or permeated with an invisible imponderable (and
imaginary) fluid called the ether which takes up the motion of the molecules
of a body and transmits them to the molecules of other bodies in space, just as
when a body is made to move up and down in a trough of water a similar
motion is given to a cork floating in the water at the same distance.
Different units of measurement of heat are as given bellow.
In British System:
British Thermal Unit (BTU):
The quantity of heat required to raise the temperature of one pound of water
through 1°F is defined as a BTU.
In MKS Units:
Centigrade Heat Unit (CHU):
The quantity of heat required to raise the temperature of one pound of water
through 1°C is defined as a CHU.
The quantity of heat required to raise the temperature of one kilograms of
water through 1°C is defined as one Kilo-Calorie
Since the amount of heat required per degree centigrade varies at different
points on the temperature scale, a more precise definition is the amount of
heat required to raise the temperature of one kg of water initially at 14.5c to
15.5c while maintained at constant pressure of 760mm of hg.
1 Kcal = 2.205 CHU = 3.969 BTU
The unit for heat in SI system is measured in Joules (J)
1 Kcal = 4187 Joules = 4.187 Kilo Joules
1.11.2 Specific Heat of a substance may be defined as the amount of heat that must
be supplied to the substance to raise the temperature of unit mass of the
substance through one degree. When a body is heated, the heat energy is used
to speed up the internal motion of its molecules and also to provide the work
necessary to expend the body. In a solid or a liquid, the amount of expansion
is very small and the work of expansion is similarly small. When a gas is
heated, expansion is considerably more and values of specific heat will
depend on nature of heating process i.e., whether the heating is at Constant
Volume or at Constant Pressure. Thus gas has a two important types of
Specific Heat, namely:
(1) Specific Heat at constant volume ( Cv)
(2) Specific Heat at constant pressure (Cp)

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1.11.2.1 Specific Heat at Constant Volume: Consider 1 kg of gas being heated in a
closed vessel so that no expansion of gas is allowed. The number of kcal
required to raise the temperature of 1 kg of gas through 10
C under these
condition is called the Specific Heat at Constant Volume and is denoted by
‘Cv’.
In this case there is no work due to expansion of gas, because the gas is
contained in closed vessel and all the heat supplied is used only to increase the
Internal Energy i.e. Kinetic Energy and Potential Energy of molecules of the
gas.
1.11.2.2 Specific Heat at Constant Pressure: Consider 1 kg of gas being heated in a
cylinder fitted with a movable piston which exerts a constant pressure on the
gas. When the gas is heated it will expand and move the piston through some
distance in this case. Therefore, in this case, in addition to the heat required
for increasing the kinetic energy of the molecules, further heat must be added
to perform the work of moving the piston through the distance. The Specific
Heat at Constant Pressure is denoted by ‘Cp’.
The value of the specific heat of gas at constant pressure will therefore always
be greater than that at constant volume by the amount of expansive work
done.
The unit of specific heat in MKS System of units is kcal/kg 0
C and in the SI
system of units it is kJ/ kg K.
1.11.2.3 Ratio of specific heats: The ratio of two specific heats, ‘Cp’ and ‘Cv’ of any
given gas is assumed to be constant. It is expressed by the symbol ‘ϒ’
(gamma). It is called ‘Gas Constant’ and is an universal constant for ideal
gases. It has no units for measurement.
ϒ = Cp/Cv
For air, Cp = 0.24 kcal/kg 0
C or 1.0035 kJ/ kg 0
K
and Cv = 0.172 kcal/kg 0
C or 0.7165 kJ/kg 0
K
hence, ϒ = 0.24/0.172 = 1.4
1.11.3 Enthalpy (H) or Total Heat or Heat Content: Enthalpy is nothing but total
heat energy content in a substance. It is denoted by ‘H’ and is defined as
follows:
H = E + (PV/J) kcal
Where, E is the Internal Energy, P is the Absolute Pressure, V is the Volume
in m3
and J = 427 kg-m.
1.11.4 HEAT TRANSFER:
In Boiler heat energy is released from the combustion of fossil fuels and the
heat is transferred to different fluids in the system and a part of it is lost or left
out as unutilized. It is therefore essential to study the general principle of heat

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transfer for understanding the behaviour of boiler in relation to heat transfer
during different conditions of operation.
Let us take an example of a kettle of water being heated under fire. When fire
is applied the water in the kettle gets heated. Heat to water is passed through
the metal wall of the kettle. Now remove fire. The water in the kettle cools
down. The heat is now given to the air which surrounds the kettle.
In this process the heat is transferred from fire to kettle then to the water. On
cooling the transfer has taken place from water to kettle and from kettle to
surrounding air. The transfer of heat first has taken by way of Conduction
within the Kettle walls and then heat by the process of Convection transferred
to water in centre of the Kettle from the water immediately in contact with
walls of the kettle. When the fire is put out, the water started cooling down as
transfer of heat in water occurs by Convection from centre of the Kettle to the
water layer immediately in contact with Kettle walls and by Conduction
within walls of the Kettle. Outer surface of the Kettle transfers heat to the
surroundings by way of Convection & Radiation.
.
In Boiler generally the heat transfer takes place in all the three modes of heat
transfer process namely Conduction, Convection and the Radiation.
1.11.4.1 Conduction:
Conduction is the process of transfer of heat through solids from one part of
the body to the other, by physical contact, without the molecules moving, but
imparting vibration from one molecule to the neighbouring one.
In a Boiler the water tubes are exposed to fire. The heat travels by Conduction
from outer surface to inner surface of water tubes and then transfers to water
at centre of the tubes and in the drum from the water immediately in contact
with inner surface of the tubes by convection. In a metal the heat transfer
takes place by passing on heat from particle to particle by contact without any
physical movement of the particles themselves.
The quantity of heat conducted depends on:
a. the differential temperature between combustion chamber and the water
inside the tube,
b. the thickness of the tube,
c. surface area of the tube,
d. the characteristics of the metal and
e. the cleanliness of the surface.
1.11.4.2 Convection:
This process can occur only in fluids or gases. This process of heat transfer
takes place when the molecules are displaced physically. The fluid or gas
when heated expands, becomes less dense and raises up causing movement
and allowing the colder and more dense gas or liquid to replace it. In the

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Boiler the heat from tube metal goes to water flowing inside. Similarly when
gas or liquid is heated, it expands, becomes less dense and rises up causing
movement and allowing the colder and denser gas or liquid to replace it.
Mainly in Superheater, Reheater and Economizer the heat from hot gas is
getting transferred to metal outer surface by way of convection process.
Heat transfer by convection depends on the specific characteristics of the
medium i.e. gas or liquid.
1.11.4.3 Radiation:
Heat when it travels from source to another substance through an empty space
(often imagined as ether) or through vacuum or gas or air in straight lines, the
process of heat transfer is called radiation. The tube metal surface at the top of
the furnace of a Boiler gets heat by way of radiation. We get heat from Sun by
radiation.
All substances emit heat energy by radiation depending on their temperature.

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Examples
1. Question: Defined absolute temperature scale
(Feb-88, Aug-89)
Answer:
• Absolute temperature scale or Kelvin temperature scale is based on absolute zero
of temperature.
• Absolute zero, or 00
K, is the temperature at which molecular energy is a
minimum and it corresponds to a temperature of −273° on the Celsius temperature
scale.
• At absolute temperature a perfect gas is considered to have a zero volume.
• Absolute temperature is expressed by the capital latter ‘K’ and at this scale the
freezing point of water (0°C) is 273 K, and the boiling point of water (100°C), is
373K, respectively.
2. Question: The temperature of water while entering an economiser is 20ºC, while
leaving the economiser is 80ºC. If the rate of flow of water is 500 kg per minute,
how much quantity of heat is supplied in the economiser? Specific heat of water
may be taken as 4.182 kJ/ kg.
Answer:
Quantity of water flowing m = 500 kg/minute
Rise in temperature of circulating water ∆t = 80 – 20 = 60ºC.
Specific heat of water CP = 4.182 kJ/kg
Therefore, quantity of heat supplied to water in the economiser per minute
mCP ∆t = 500 x 4.182 x 60 = 125460 kJ/minute i.e. 125.46 MJ/minute.
3. Question: A certain gas occupies 3 cubic metres at a temperature of 150ºC.
The pressure of the gas is 7 bar. The gas expands in such a manner that the volume
becomes 9 cu metre and the temperature is 10ºC. What is the pressure of the gas?
Answer:
Considering the pressure given in absolute,
P1 = 7 bar Ab.; V1 = 3 cu meter; V2 = 9 cu meter
T1 = 150 + 273 = 423 K and T2 = 273 + 10 = 283 K
Using the relation P1V1/ T1 = P2V2/ T2
or, (7 x 3) / 423 = P2 x 9 / 283
Or P2 = 1.561 bar Ab.
4. Question: The compression ratio of an engine is 12 to 1, the pressure at the
commencement of the compression stroke is 100 kN/m2
and the temperature 115ºC.
Calculate the absolute pressure at the end of compression stroke if the temperature
has then risen to 180ºC.

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Answer:
From the relation
P1V1/ T1 = P2V2/ T2
We have,
P1 = 100 kN/m2
(considering pressure given in Absolute);
T1 = 115 + 273 = 388 K; T2 = 180 + 273 = 453 K
V2 = V1/12
On substitution of values, we get
100 x V1/ 388 = P2 x V1/ (12 x 453)
or P2 = 1401 kN/m2
(14.01 bar) absolute
5. Question: Calculate the molecular volume of all gases at 200 kN/ m2
and 30ºC.
According to the characteristic equation of a gas we have pV = mRT where p is the
pressure of the gas in N/m2
, V is the volume of mass m kg of gas in cu metre, R is
the characteristic gas constant and T is the absolute temperature of the gas in
Kelvin.
Answer:
If ‘m’ is the molecular mass, then V will be molecular volume of the gas and
mR = 8.3143 kJ/ kg mole K
We have, P = 200 kN/ m2
; T = 273 + 30 = 303 K
Therefore, 200 x V = 8.3143 x 303
V = (8.3143 x 303) / 200 =12.596 m3
6. Question: A steel cylinder of 77 litres of capacity contains CO2 at 27 Deg. C and
pressure of 110 ata. Calculate the weight of gas contained in the steel cylinder.
Solution: Here,
Volume = 77 liters = 77 x 10-3
m3
Pressure = 110 ata = 110 kg/cm2
ab = 110 x 0.981 = 107.91 bar ab.
= 107.91 x 102
kN/ m2
Temperature = 27o
C = 27 + 273 = 300 K
We know MR = 8.3143 kJ/kg mol K and M of CO2 = 44 kg
∴ Sp. Gas constant R of CO2 = 8.3143/44 = 0.189 kJ/ kg K
From the relation PV = mRT,
107.91 x 102
x 77 x 10-3
= m x 0.189 x 300
or, m = (107.91 x 102
x 77 x 10-3
)/ (0.189 x 300) = 14.65 kg

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7. Question: Define a new temperature scale, say o
D in which the boiling and
freezing points of water are 300o
D and 100o
D respectively. Correlate this scale with
the Centigrade scale. The o
D reading on this scale is a certain number of degrees on
a corresponding absolute temperature scale. What is this absolute temperature at
o
D?
Solution: Here, the freezing point and the boiling point are 100o
D and 300o
D
respectively.
Comparing it with centigrade scale, we see a rise of 100o
C (100o
C – 0o
C) will be
equal to a rise of 200o
D (300o
D – 100o
D) in the new scale.
So for every 1o
rise in the centigrade scale, there will be 2o
rise in the new scale.
Again 0o
C = 100o
D
Therefore the relation between these two scales will be
D = 100 + 2C
For example 25o
C will read in the new scale as 100 + 2 × 25 = 150o
D
We can write the above relation as C = (D-100)/ 2
In absolute scale, any reading in the new scale will read as (D-100)/ 2 + 273

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Examples for practice
1. Question: i) Differentiate between the manometer & barometer.
ii) Differentiate between pressure gauge and vacuum gauge.
2. Question: Fill the gap (Oct-95)
a. 1m3
= ____Cubic Feet
b. 1kW = _______ HP
c. 1lb = ________ Grain
d. 1000 cm3
= ______ dm3
e. 1m3
/hr = _____ cfm
f.
1 bar = ______ lb/sq. in
g. 1dm3
/sec = ______ cubic feet/s
h. 1kW h = _____ Kcal.
3. Question: Fill in the gap (Aug-96)
i) 1 kg/cm2
= --------- lb/m2
, 1 bar = --------- lb/sq. ft
ii) Temperature °K = Temp °----------- + ----------
Temperature °R = Temp °---------------- + --------------
iii) 1 metric horse power = --------- watts
1 British horse power = ------------- watts
4. Question: (a) Convert the following reading of pressure to ‘bar’.
(i) 800 kPa, (ii) 10 atm (iii) 2 Mpa (iv) 112000 Nm2
(v) 200 KN/m2
(b) (i) A vacuum guage on condensers reads 620 mm of mercury and at the same
time barometer reads 740 mm of mercury. What is the absolute pressure in the
condenser in Mpa.
(ii) Find the heat equivalent of work done in KJ, when a weight of 500 kg is
raised through height of 6000 cms.
(iii) A pressure gauge reads 2.3 Mpa and barometer reads 90 Kpa, calculate
absolute pressure in pascal. (Feb-2002)
5. Question: A locomotive engine having weight of 3500 kg running at 72 kmph
posses certain stored energy due to the motion. Determine the stored energy in kJ and
kW.
(b) A train weighing 1450 MT is pulled up a 2% degraded 4475 KW engine. Train
resistance is 8750 Kg. At what speed the train is running. (Feb-2002)
6. Question: The pressure of a gas supplied to an engine is measured as 100mm of
water gauge when barometer reads 756 mm of mercury. Determine the volume of
1.5kg of this gas if it’s temp. is 850
C. The gas constant of the gas is 0.686 kJ kg K.
(Aug-2002)

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7. Question: A body weights 50kg on earth. Find its weight on the (a) Moon where
gravitational acceleration is 1.7M/sec2
(b) Sun where gravitational acceleration is
270 M/sec2
8. Question: A thermometer immersed in fluid the value of temperature in F and C
shows same what will be value of temp.? Express this value of temp. in deg.R &
deg.K.
9. Question: A vacuum gauges reads the vacuum in a chamber as 300 MM of Hg,
what is the absolute pressure in the chamber if the atmospheric pressure is 760 mm
of Hg. The specific weight of mercury at this temp. is 13550 kg/M3
(Feb.91).
10. Question: A manometer joined to a gas cylinder indicates 20 kPa, while the
barometer reads 760 mm of mercury. What will be the reading of the manometer if
the barometric pressure drops to 730 mm mercury?
11. Question: A steam power plant develops 4460 kW. What is the equivalent of
this power in thermal unit?
12. Question: The gas used in gas engine trial was tested in a Boy’s calorimeter.
The pressure of gas supply was 70 mm of water column. What is the absolute
pressure of the gas if the barometric pressure is 760 mm of mercury?

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A-29
APPENDIX
HEAT TRANSFER CALCULATIONS
The general equation for heat flow rate by any of the above three modes of
heat transfer from one media to other may be written as
q = US ∆t
Where q = heat flow rate in K.Cal/hr
S = Surface area involved in the heat transfer in m2
∆t = Temperature difference causing heat flow in °C
U = Overall heat transfer coefficient in K.cal/m2
/hr/°C
= 1/R where R is overall resistance
Conduction:
If a flat plate is heated on one side and cooled on other side, heat will flow
from hot side to the cold. The heat flow rate q can be expressed as below:
q = KS (t1 – t2) / l
Where q = rate of heat flow – K.Cal/hr
K = Thermal conductivity for 1 cm thickness – K.cal/m2
/hr/°C
S = Heating surface in m2
t = temperature difference causing heat flow (t1 – t2) in °C
l = length or thickness of the plate in cm.
K/l is expressed as conductance and hence l/K is the resistivity.
Convection:
Heat transfer by convection between a fluid and a solid such as in a boiler
tube is expressed as below:
qc = Uc S ∆t .. (3)
Where qc = rate of heat flow by convection in K.Cal/hr
U = Convection film conductance in K.cal/m2
/hr/°C
S = heat transfer surface in m2
∆t = temperature difference between fluid bulk temperature
and solid surface temperature in °C.
Radiation:
Radiation emitted by a body depends upon its surface area and temperature.
The relationship between them is given by Stefan-Boltzman law
q = σ S T4
q – rate of heat flow
σ – Stefan-Boltzman constant
S – surface area of body
T – absolute temperature of the emitter

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For bodies other than black bodies whose emissivity will be less than 1, the
formula will be changed as
q = σ ES T4
where E is emissivity of the body
If we consider two parallel planes of infinite size and they are black bodies,
then heat transfer from the hot plane (at T1 °K) to the other plane (at T2 °K) is
given by the formula
q = σ S (T1
4
– T2
4
)
If all the radiation emitted by one does not fall on the other it is essential to
introduce an angle factor in the formula
In boiler the radiation becomes luminous by entrained particles such as
pulverised coal, soot etc. and calculation of luminous radiation is complex.
The gases such as oxygen and nitrogen absorbs or emit only slight amount of
radiation. But water vapours, carbon dioxide, sulphur dioxide and carbon
monoxide which are part of flue gases in the boiler also absorb and emit. They
emit and radiate only in certain wave length bands that lie outside of the
visible range and are called as non-luminous gas radiation.

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Miscellaneous Numerical Problems on Elementary Concepts
Example:1 A vertical tank contains water. The height of cylindrical tank is 120 cm
and diameter is 30 cm. The tank is 3/4TH
full of water. If a cubic piece of iron of
10cm sides is dropped in the tank, calculate the rise in height of water level and
final height of water in the tank. (Feb 1988)
Solution: As the tank is ¾ th full, the height of water in tank = 120 × ¾ = 90 cm.
Radius of the cylindrical tank r = d/ 2 = 30/ 2 = 15 cm
Volume of cubic piece of iron = 10 × 10 × 10 = 1000 cm3
The cubic piece of iron, when dropped in the tank will displace its equal volume of
water.
If the height of the water rises by say h cm after the piece of iron is dropped,
The increased volume of water in the cylinder = π × r2
× h = π ×152
× h
= 706.86 × h cm3
Now 706.86 × h cm3
= 1000 cm3
Or, h = 1.41 cm
And final height of water in a tank = 90 + 1.41 = 91.41 cm
Example:2 A fuel storage tank is in the form of a cylinder 2.6 m dia. with one end
hemisphere and the other end plain. Calculate the capacity of tank in litre and in kg,
if it contains oil having Sp. gravity 0.90. Length of the cylindrical portion is 6.0 m.
Also determine the area of the sheet metal used in construction.
Solution: Given ,
Dia. of cylinder = 2.6 m, r = 1.3 m
Length of cylindrical portion = 6 m
One end hemisphere & one end plain.
Specific gravity of oil = 0.9
Volume of hemispherical end = ½ × (4/3) πr3
= (2/3) πr3
Volume of cylindrical portion = π r2
L
∴Total volume = π r2
L + (2/3) π r3
= π (1.3)2
× 6 + (2/3) π (1.3)3
= 31.86 + 4.60
= 36.46 m3
∴The capacity of the tank = 36.46 × 1000 = 36460 liters.
Now specific gravity = density of oil/ density of water
∴ Density of oil = specific gravity × density of water
= 0.9 × 1000 kg/ m3
= 900 kg/ m3
∴The capacity of the tank = 36.46 × 900 = 32814 kg

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Area of shell metal used = Area of hemisphere + Area of cylindrical portion +
Area of plain end.
= 2πr2
+ 2πrL + πr2
= 2π(1.3)2
+ 2π × 1.3 × 6 + π(1.3)2
= 10.62 + 49.00 + 5.31 = 64.93 m2
Example:3 What will be the volume of earth removed by digging the well of 2.5
metre. The dia. of the well is 10 metre. If the earth is uniformly spread over an area
of 150 m2
. What will be rise in level ? Assume the area to be square.
( Feb. 1995)
Solution: Volume of earth removed = (π / 4) × d2
× h
= (π / 4) × 102
× 2.5 = 196.35 m3
Let us assume the rise in level is h m, when the earth is uniformly spread.
The volume of the earth = Area × height = 150 × h
∴150 h =196.35
or, h = 196.35/150 =1.309 m.
Example:4 Calculate the area of an elliptical manhole having major axis of 60 cm
and minor axis of 50 cm. (Feb.1995)
Solution: Area of ellipse = π × a × b,
Where a = semi major axis and b = semi minor axis.
Given, major axis = 60 cm and minor axis = 50 cm
∴Semi major axis a = 30 cm and semi minor axis b = 25 cm
∴Area = π × 30 × 25 = 2356.20 cm2
Example:5 A heap of coal is in the form of cone base diameter of which is 50
meter & height is 12 meters. Find out the Tonnage of coal in the heap as coal is
having bulk density of 1.1 tons / 1.5 m3
. (15-6-1998)
Solution: Here,
Base diameter of conical heap of coal = 50 m
Height of conical heap of coal = 12 m
Bulk density = 1.1 Tons / 1.5 m3
Volume of cone = Area of base × Perpendicular height/ 3
= π/4 ×502
× 12/ 3 = 7853.98 m3
∴ Weight of the coal heap = (1.1/ 1.5) × 7853.98 = 5759.58 Tons

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Example:6 A steel pipe 3.4 meters long with an outside diameter of 31.75 mm and
thickness of 5.25 mm. If density of materials is 7.6 gm/ cc. Find out the mass of
pipe.
Solution: Here
Length of pipe = 3.4 m = 340 cm
Outside diameter of pipe = D = 31.75 mm = 3.175 cm
Thickness of pipe = 5.25 mm
∴Inside diameter of pipe d = 31.75 – 2 × 5.25 = 21.25 mm = 2.125 cm.
Volume of pipe material = (π/4) × L × (D2
- d2
) = (π/4) × 340 × (3.1752
– 2.1252
)
= 1486.05 cm3
Density of the material = 7.60 g/ cc
∴ Mass of the pipe = 1486.05 × 7.60 = 11293.98 g = 11.294 kg
Example:7 A horizontal return tabular boiler is 180 cm in diameter and 600 cm
long contains 75 tubes of 75 mm OD × 70 mm ID. Find the boiler heating surface
area, where heat transfer may be taken as inner surface area of all tubes, half the
area of boiler shell and 2/3rd
of the tube plates less the area of the tube holes.
(Aug 1999)
Solution:
Given,
Length of the shell ( L )= 600cms = 6.0 m
Dia of shell (D) = 180cms = 1.8 m
I D of tube (DI) = 70 mm = 0.07 m
O D of tubes = 75 mm = 0.075 m
No.of tubes = 75
Now Inner heating surface area of 75 no. of tubes S1 = 75 × π × 0.07 × 6
= 98.96 m2
Half surface area of shell ( heating surface area of shell cylindrical portion) S2
= ½ × π × 1.8 × 6 = 16.96 m 2
Area of tube plate S3 (2 nos in case of horizontal return tubular boiler)
= 2 × 2/3 × [(π /4) × 1.802
– 75 × ( π/4)(0.075)2
]
=2.951 m2
Total heating surface area = S1 + S2 + S3 = 98.96 + 16.96 + 2.951 = 118.871 m2
Example:8 An immersion heater of 1500 watt is immersed in 200 kg of water. If no
heat loss is assumed, find the time taken to heat the water from 30 deg. C to 80 deg.
C. Find also the units of electricity consumed.

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A-34
Solution: Given, Heater capacity = 1500 Watt = 1.5 kW = 1.5 kJ/ sec
Mass of water m = 200 kg
Initial temp. of water = tI = 30o
C
Final temp. of water = t0 = 80o
C
Increase in temperature ∆ t = 80 – 30 = 50o
C
We know that specific heat of water CP = 4.187 kJ/ kg o
C
∴ Heat received by the water = mCP ∆ t = 200 × 4.187 × 50 = 41870 kJ
∴ Time taken to heat the water @ 1.5 kJ/ sec = 41870/ 1.5 = 27913.33 sec
= 27913.33/ 3600 = 7.753 hour
Units of electricity consumed = 1.5 × 7.753 = 11.63 kWH
Power of a pump:
The power of a pump driven by steam, water or air is to be found out as under.
We know that Power = rate of work done, where work = force × distance, where
force = Pressure × area
In a pump, the piston is driven by the working substance (i.e. steam or water or
air) under a pressure through the length of the cylinder.
Therefore the force acting on the piston with an area A, driven by the working
substance under a pressure P = P × A.
Now the work done by the piston while it moves through the length of the cylinder L
in a single stroke = force × distance = P × A × L
If the pump makes N working strokes per minute, then the power of the pump = the
rate of work done per minute = P × A × L × N
In SI units, if P is in N/ m2
, L is in m, A is in m2
and N is in number of the strokes per
minute, then rate of work done P × A × L × N will be in N × m2
× m = N-m/ min
m2
× min
= J/ min = J/ 60 sec = W/60 (J/ sec = Watt)
So we can see that the above relation when expressed as P × A × L × N/ 60 will give
us the power of the pump in Watt. If P is taken in kN/ m2
, then the result will be in
kW.
To find out the horse power of a pump, we can find out the power in Watt and then
find out the HP from the relation 1 HP = 735.75 Watt.
Alternatively we know that 1 HP = 75 kg-m/ sec.
Now if the pressure P is taken in kg/ cm2
, Length L in m, area A in cm2
and N is in
number of strokes per minute, then rate of work done P × A × L × N/ 60 will be in
kg × cm2
× m = kg-m/ sec
cm2
× sec
So we can see that the above relation when expressed as P × A × L × N/ (60 × 75) or
P × A × L × N/ 4500 will give us the power of the pump in HP. (as 1 HP = 75 kg-m/
sec)
[Unless HP is asked, while finding out the power of a pump or an engine, always
work out in SI unit and find out the power in Watt or kilo Watt]

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A-35
Example:9 What is the horse power developed in a simple steam pump cylinder
200 mm diameter 300 mm stroke, with pump making 50 working strokes per
minute, with steam pressure 7 kg/cm2
g.
( Feb.1988)
Solution: Given,
Dia. of piston = 200 mm = 20 cm.
Length of stroke =300 mm = 0.30 m
N =50 strokes /min.
Pressure P = 7 kg/ cm2
We know HP = P × A × L × N / 4500
Here area A = π × d2
/ 4 = π × 202
/ 4 = 314.16 cm2
∴HP = 7 × 314.16 × 0.30 × 50/ 4500 = 7.33 HP
Example:10 Find the HP at the shaft of the pump required to lift 50000 lit. of water
per hour through a total head of 100 meter. The efficiency of the pump is 60%.Find
the HP of the motor which driven a pump, if the efficiency of the motor is 80%.
(Feb 1988)
Solution:
Here pump discharge Q = 50000 l/ hr = 50 m3
/ hr
= 50/ 3600 = 0.0139 m3
/ s,
H = hρ = 100 × 1000 = 105
kg/ m2
Pump eff. ηPump = 60%
∴ HP at the shaft of the pump = HQ/ 75 ηPump
= (0.0139 × 105
)/(75 × 0.60) = 30.89 HP
since motor eff. ηmotor = 80%
∴ HP of motor = 30.89/ 0.80 = 38.61 HP
Example:11 1360 Litres of water is pumped into tank per minute under pressure of
1.41 kg/cm2 by pump. Find the H.P. required for pumping the water. Assume no
loss and pump efficiency is 85%.
Solution: Here,
Q = 1360 litres/minute = 1.36 m3
/ min = 1.36/ 60 = 0.023 m3
/sec.
H = 1.41 kg/cm2
= 1.41 × 104
kg/ m2
η = 85 %
∴ H.P. = (Q × H) / 75 × η = (0.023 × 1.41 × 104
) / 75 × 0.85 = 5.09

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Example:12 Determine the work in forcing 5 litres of water into a vessel against a
pressure of 1.2 MN / m2
.
Solution: As we know,
Work = Quantity of water in m3
× pressure in N/ m2
Given, 5 litres of water = 0.005 m3
& Pressure = 1.2 MN/m2
= 1.2 × 106
N / m2
∴ Work = 0.005 × 1.2 × 106
= 6000 N-m
Example:13 A duplex feed pump supplies water to a boiler working at pressure14
bar. The loss of head of the water in feed lines economies etc. of 20% of the boiler
working pressure. The diameter and length of the stroke for the pump are 5.0 cm
and 8.0 cm respectively. The pump runs at 60 strokes per minute. Find the brake
horse power to drive the pump if the pump efficiency is 65%. Neglect the loss due
to leakage and spillage. (Feb.2000)
Solution:
Given , Boiler working pressure = 14 bar = 14 × 102
kN/ m2
Dia. of plunger = 5 cm = 0.05 m.
Length of stroke = 8 cm = 0.08 m
N = 60 strokes per minute
ηpump = 65 %
Total loss of head is 20 % of boiler working pressure.
∴ Required pressure for the pump to deliver water in boiler = 14 × 1.2
= 16.80 × 102
kPa
Since work out put of a duplex pump is double the single cylinder pump,
Break Power required to drive the pump = 2 × PLAN/ 60 × 0.65
= 2 × 1680 × 0.08 × (π/4) ×0.052
× 60/ 60 × 0.65 = 0.812 kW
BHP = 0.812 × 103
/ 735.75 = 1.10
Alternatively,
Pressure P = 14 × 1.0197 = 14.27 kg/ cm2
Required pressure to deliver water in boiler = 14.27 × 1.2 = 17.124 kg/ cm2
Dia. of plunger = 5 cm, area A = (π/4) ×52
= 19.63 cm2
Length of stroke L = 8 cm = 0.08 m
And ηpump = 65 %
∴BHP = 2 × PLAN/ 4500 × ηpump = 2 × 17.124 × 0.08 × 19.63 × 60/ 4500 × 0.65
= 1.10 HP

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Example:14 A steam pipe of internal dia. 12 cm carries steams at 12 kgf/cm2
ab. It
is lagged to a radius of 12 cm with asbestos of thermal conductivity of 1 Kcal / m-hr
deg.C. The temp. of the surrounding is 25 deg. C and the loss from the lagging
surface is 12 Kcal / m2
.hr deg.C. Calculate the loss in Kcal per meter length of the
pipe. The heat loss through an insulated pipe per meter length is given by
(T1-T2)
Q = 2 π K
Log e (r2/r1)
= 2 π K (T1-T2)/ 2.3 log10(r2/r1)
where T1 & T2 are inner and outer surface temp. of insulated pipe and r1 & r2 are
the inner and outer radius of insulation and K is the conductivity of insulating
material.
Solution: At 12 kg / cm2
ab. pressure.
T1 = temp. of dry saturated steam = 187.08 deg.C (From steam table)
T2 = 25 deg.C
r2 = 12 cm = 0.12 m
r1 = 6 cm = 0.06 m
Q = 2 π K × (T1-T2) / { 2.3 log10(r2/r1) }
= 2 π ×1 × (187.08-25) / { 2.3 log10(0.12/.06) }
= 2 π × 162.08/ 0.692
= 1471.64 Kcal / m hr
Example:15 Train “X” 125 metre Long is running at a speed of 80 Km per hr. In
what time will it pass another train “Y” 80 metre long running at 45 Km per hr.
When
(i)Moving in opposite direction, (ii) Moving in same direction and (iii)Y Standing
on a station.
Solution: Given, Train X length = 125 m, Speed of train = 80 km/hr.
Train Y length = 80 m, Speed of train = 45 km/hr.
Case – I
When trains moving in the opposite direction
V= relative velocity = 80 + 45 = 125 km/hr. = 34.722 m/sec.
S = Distance to be traveled = 125 + 80 = 205 m.
Using the relation, S = Vt

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A-38
34.722 × t = 205,
∴ t = 205 /34.722 = 5.90 sec.
Case – II
When trains moving in the same direction.
V= relative velocity = 80 - 45 = 35 km/hr. = 9.722 mt/sec.
S = Distance to be traveled = 125 + 80 = 205 mt.
Using the relation, S = Vt
9.722 × t = 205
∴ t = 205 /9.722 = 21.09 sec.
Case - III
When train Y standing on station.
V = relative velocity = 80 + 0 = 80 km/hr. = 22.22 m/sec.
S = Distance to be traveled = 125 + 80 = 205 mt.
Using the relation, S = Vt , 22.22 × t = 205
∴ t = 205 / 22.22 = 9.23 seconds
Example:16 Two trains whose lengths are 100 m and 50 m respectively are
moving on parallel tracks and take 5 seconds to cross each other completely. If the
shorter train is moving with double the speed of the longer one find the speed of the
train.
(i) When bath the trains are moving in the same direction.
(ii) When the trains are moving in opposite direction.
Solution: Here,
Length of Train A = 100 m.
Length of Train B = 50 m
Let speed of train `A’ is x m/sec
Speed of train `B’ = 2 x m/sec
Case – I
When both Trains are moving in same direction, then relative velocity of train `B’
= 2x – x = x m/sec
To cross the train `A’, distance is required to be traveled by train B = 100 + 50 =
150 m
From the relation S = Vt
150 = x × 5, or x = 150/ 5 = 30 m/ sec.
∴ Speed of train A = 30 m/ sec.
And speed of train B = 60 m/ sec
Case – II
When both Trains are moving in the opoosite direction, then relative velocity of
train `B’ = 2x + x = 3x m/sec

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A-39
To cross the train `A’, distance is required to be traveled by train B = 100 + 50 =
150 m
From the relation S = Vt
150 = 3x × 5, or x = 150/ 15 = 10 m/ sec.
∴ Speed of train A = 10 m/ sec.
And speed of train B = 20 m/ sec
Example:17 Two train A & B leaves the same station on parallel lines. A starts
with uniform acceleration of 1/ 6 meter/ sec2
and attain a speed of 25 km/ hr. When
the speed is reached to maintain the speed of A train constant at 25 KM/ hr. B train
leaves 40 sec after A train left with uniform acceleration of 1/3 meter/ sec2
to attain
a maximum speed of 50 km /hr.. When will B train overtake A train.? (Feb 1999)
Solution:
Given, Train A
fA = uniform acceleration = 1/6 m /s2
vA = Final speed = 25km/hr = 6.944 m/ sec
u = Initial speed = 0 m/sec.
Train B
f B = uniform acceleration = 1/3 m /s2
vB = Final speed = 50 km/hr = 13.888 m/ sec
u = Initial speed = 0 m/sec.
We know that, v = u + ft
Therefore the time taken by train A to attain its final speed is
tA = vA/ fA = 6.944×6 = 41.66 sec and
at a distance SA = 6.944×41.66/2 = 144.64 m
Similarly, the time taken by train B to attain its final speed
tB = 13.888×3 = 41.66 sec and SB = 13.888×41.66/2 = 289.28 m
Let us consider the time T = 0, when train A starts from the station. It reaches its
final speed when T = 41.66 s and at time T=81.66 sec the train B reaches its final
speed.
By the time train B reaches its final speed, the train A moves [144.64(during 41.66
sec) + 6.944×40 (during 40 sec at cont. speed)] = 422.40 m away from the station.
Position of both the train at T= 81.66 sec is
422.4 m
289.28 m 133.12 m
A
B

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A-40
Train B is 133.12 m behind the train A and both the trains are traveling at their
corresponding uniform speed.
Let us assume they meet at a distance after time t,
The relative speed of train B with respect to train A will be 13.888 – 6.944
= 6.944 m/ s.
Therefore 6.944 x t = 133.12, or t = 133.12/6.944 = 19.17 s
So
The train B will overtake the train A after 81.66 + 19.17 = 100.83 sec after
train A leaves the station
Example:18 A Train is uniformly accelerated and passes successive kilometer
stones with velocity of 18 KMPH and 36 KMPH respectively. Calculate the
velocity when it passes the third kilometer stone. Also find out the time taken for
each of these two intervals of one kilometer stone.
Solution: Using the relation, between first and second kilometer stone
S = {(U + V)/2} × t
Here,
S = 1 km = 1000 m
U = 18 KMPH = 18/ 36 = 5 m/ s
V = 36 KMPH = 36/ 36 = 10 m/ s
1000 = (5 + 10) × t, or t = 1000/15 = 66.67 seconds
We can find uniform acceleration using the relation
V = u + ft, where f = acceleration in m/ s2
10 = 5 + f × 66.67, f = 5/66.67 = 0.075 m/ s2
Now between second and third stone
U = 10 m/s
S = 1000 m
And f = 0.075 m/s2
We get can find out velocity at 3rd
stone by using the relation
V2
= U2
+ 2 f S
V2
= 102
+ 2 × 0.075 × 1000
V2
= 250, or V = √ 250 = 15.81 m/s = 15.81 × 3600/ 1000 = 56.92 kmph
Again, using the relation, S = {(U + V)/2} × t
Where t is the time taken to travel from 2nd
stone to 3rd
stone.
U = 10 m/ s
V = 15.81 m/s
1000 = (10 + 15.81) × t, or t = 38.74 seconds

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A-41
Example for practice
Example:1 A fuel storage tank is in the form of cylinder with hemispherical ends.
The diameter of cylinder is 2.7 Mtrs. The length of cylinder is 4.7 Mtrs. The tank is
required to contain oil of specific gravity of 0.91. Determine the capacity of the
tank in Ltrs. and kg. of oil. Also determine the length of welding as the tank is
fabricated by welding process. Hemispherical end is made of deep forged hat
process.
Example 2: A cylindrical vertical tank contains water. The height of the tank is 135
cms and inside diameter is 33 cms. The tank is 2/3 full of water. If can iron cube of
13 cms is dropped in the water. Calculate the raise in the height of water level and
final height of water in the tank if the tank is fabricated from 8 mm thick M.S.plate
find the total weight of the tank with water and iron cube. Assuring tank is open at
top and density of M.S. as 7.85.
( Feb.1999)
Example:3 Find out the horse power required to lift 90000 liters of water per hour.
To a pressure of 160 bar. Assume the efficiency of feed pump is 87 %. Find out also
the horse power of electric motor is 82.5%.
(Aug 1999)
Example:4 Train “X” 125 mtrs long is running at a speed of 85 km/hr. In what time
will it pass another train “Y” 60 mtrs long running at 50 km/hr. When
(1) Moving in opposite direction
(2) Moving in same direction
(3) Standing on a station.
Example:5 Two trains A & B leave the same station on parallel lines. The trains A
starts with a uniform acceleration of 0.2 m / s2
and attains a speed of 45 KMPH.
When the steam is reduced to keep the speed constant. The Train B leave one
minute after with a uniform acceleration of 1.4 m / s2
to attain a speed of 72 KMPH.
When the train B will overtake the train A ?
(Feb.2000)

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B- 1
CHAPTER – 2
Properties of Steam
1.0 Introduction:
Water vapour is called “Steam”. It is the most commonly used working substance
in the operation of steam engine and steam turbines. Steam which has just been
formed at the saturation temperature is a vapour which is a partially evaporated
liquid carrying in it particles of liquid and under these circumstances, it can be
liquefied back into water by minor changes in temperature or pressure. Steam as a
vapour would not obey the laws of perfect gases unless it is in a highly dried
condition. Steam in such a dried state is known as Superheated Steam.
Superheated Steam is assumed to behave like a perfect gas when highly
superheated. It possesses properties like those of gases – relationship between
pressure, volume, temperature, internal energy, enthalpy and entropy. But the
pressure, volume and temperature of steam as a vapour is not connected by any
simple relationship such as are expressed by the characteristic equation for a
perfect gas.
Pressure, temperature and volume can be given their actual absolute value; where
as enthalpy and entropy are purely relative quantities. They are measured
relatively from convenient datum condition and calculated per kg of steam. For
steam, datum point is arbitrarily fixed as the condition of the water at 0 0
C. Thus,
the enthalpy, the internal energy and the entropy of water at 00
C are measured to
be zero. All their values measured above this temperature are considered positive
and those measured below are taken as negative. The properties of steam and the
changes in the properties have been determined by laboratory experiments by
competent bodies such as ASME and others and have been published in form of
Steam Tables or Graphically in the form of Steam Charts.
1.1 Formation of steam from water at constant pressure:
We know that there are three phases of substance namely solid, liquid and
gaseous. Water, is in a liquid phase, Ice is its solid phase and Steam is the gaseous
phase. The changes of phases take place with change in its temperature and
pressure on it.
Fig. 1 shows different phases of substance “water”. By adding heat AB, the
temperature of solid substance (ice) increased from –ve 0
C to 00
C at constant
atmospheric pressure (1.0332 kg/cm2
or 1.01325 bars). At point b solid phase
starts changing from solid to liquid on further addition of heat. At point c it
completely changes into liquid phase.
The addition of heat BC at constant temperature (00
C) is called Latent Heat of
Liquefaction of Ice. Its value is 80 kcal (334.96 kJ) per kg of water. From point c,
further heat addition causes rise in temperature of water. At point ‘d’, on further
adding of heat phase change starts from liquid to vapour at constant temperature
1000
C at constant pressure 1.01325 bar or 1.0332 kg/cm2
. At point ‘e’, liquid

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B- 2
Fig. 1
phase completely changes to vapour or gaseous phase, steam. Heat CD is called
Sensible Heat. Heat DE is called latent heat of evaporation or latent heat. The
condition of water at point ‘d’ is called Saturated Water. At this condition
vapour formation starts on minor addition of heat and on removal of heat,
formation of water particles starts. At point E there are no water particles in the
steam and the steam is said to be dry and is known as Dry Saturated Steam. As
heating continuous further, the temperature of steam begins to rise and steam is
now known as Superheated Steam and behaves more or less as a perfect gas.
Heat EF is known as Superheat.
Fig. 2 Shows graphically that what happens when heat is added to one kilogram
of water initially at 0 0
C
The temperature corresponding to point B is called saturated temperature.
Saturation temperature increases with increase of pressure and decreases with
decrease of pressure. The addition of heat “ab” is known as sensible heat and is
denoted by the letter “h”. (As specific heat of water is 1 kcal/ kg 0
C or 4.187 kJ /
kg K, heat required for raising the temperature of 1 kg of water through 10
C is
4.187 kJ) On further addition of heat, no temperature rise takes place up to point
‘C’. At point B steam begins to be formed and at point C all water is converted
into steam. All added heat BC is used to increase the kinetic and potential energy
of water molecules. This heat is known as latent heat of evaporation or latent heat
and is denoted by the letter “L”.
Tsup
T
A B E
C D F
Temperature
e
d
ts
b c
00
c
f
H

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B- 3
Fig. 2
Latent heat decreases with increase of pressure. The value of latent heat becomes
zero at pressure 225.4 kg/cm2
abs. or 220.9 bar abs. This State is known as
Critical State. This pressure is called Critical Pressure and corresponding
saturated temperature, 374.140
C, is called Critical Temperature of steam.
Between point B and C content of water will decrease as we go toward point C.
The steam which contains water particles in suspension is known as Wet Steam.
At point C, all the water, including those particles of water, held in suspension
will be evaporated, the steam is said to be dry and is known as Dry Saturated
Steam.
On heating further from point C, the temperature of steam begins to rise again and
steam is now known as superheated steam. The temperature corresponding to
point D is known as Superheat Temperature (tsup). The temperature difference
between temperature of Superheated Steam and Saturation Temperature (tsat) i.e.,
(tsup – tsat) is known as a degree of superheat. Greater the degree of superheat, the
more will the steam acquire the properties of a perfect gas. Superheating is
assumed to take place at constant pressure. The value of specific heat of
superheated steam (Cp) varies with the pressure and degree of superheat. It
increases with the increase of pressure and decreases with the increase of degree
of superheat. For calculation purpose it can be considered as 2.09 kJ/kg K unless
specified.
Tsup
Liquid Vapour Gas
Evaporation
Sensible
Heat
Latent Heat Super
Heat
Ts
B
C
D
a b c α
Heat Added
Temperature

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B- 4
1.2 Enthalpy:
The total heat content of the substance comprises three components namely; heat
of water or Sensible Heat, heat of evaporation or Latent Heat and the additional
heat imparted to the dry saturated steam to make it superheated called Super
Heat. It is also known as enthalpy of water, enthalpy of evaporation and enthalpy
of super heated steam.
1.2.1 Enthalpy of Water:
The amount of heat absorbed by one kilogram of water being heated from the
freezing point (00
C) to the boiling point is known as the Enthalpy of the Saturated
Water (sensible heat of water). It is denoted by the symbol ‘h’.
Sensible heat of 1 kg of water at 00
C is zero and sensible heat of 1 kg of water at
1000
C is 4.187 x 100 = 418.7 kJ/kg.
1.2.2 Enthalpy of Evaporation:
The enthalpy of evaporation (or Latent heat) is defined as the amount of heat
required to convert one kilogram of water at the saturation temperature tsat
corresponding to its pressure into steam at the same temperature and pressure.
Latent heat varies with the pressure. Its value decrease with increase of pressure.
Its value at 1 bar absolute pressure is 2258 kJ/kg. Value of Latent heat
corresponding to any pressure can be directly obtained from the steam tables.
1.2.3 Enthalpy of Dry Saturated Steam:
It is the sum of Enthalpy of Water and Enthalpy of Evaporation (i.e., Latent Heat).
It is defined as the quantity of heat required to raise the temperature of one
kilogram of water from freezing point to the saturation temperature and then
convert it into dry saturated steam at that temperature and pressure. It is denoted
by the symbol ‘Hs’.
Thus, Hs = h + L
This value ‘Hs’ can be directly obtained from the Steam Tables corresponding to
given value of pressure or temperature of Steam.
1.2.4 Enthalpy of wet steam:
It is the amount of heat required to raise the temperature of one kilogram of water
from freezing point (00
C) to the boiling point ‘tsat’ and then to convert the boiling
water into wet steam. Obviously, such wet steam shall not have absorbed full
quantity of Latent Heat, the content of which will be proportional to the fraction
of Dry Saturated Steam in the Wet Steam. If ‘x’ is the per unit content of Dry
Saturated Steam in Wet Steam then, Hwet = h + x.L kJ/kg. The term ‘x’ is called
Dryness Fraction and its value lies between 0 and 1. This is further explained
below:
Wet steam is a mixture of water particle in suspension and dry steam. Dryness
fraction of wet steam is a ratio of the mass of actual dry steam to mass of wet
steam containing it.
If, ms= Mass of dry steam contained in the steam considered and
m = mass of water in suspension in the steam considered;

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B- 5
then, dryness fraction , x = ms/(ms + m)
1.2.5 Enthalpy of Superheated Steam:
It is defined as the amount of heat required to convert one kg of water at freezing
temperature (00
C) into superheated steam at a given pressure and temperature
above saturation temperature. It is denoted by symbol Hsup.
Hsup = h + L + Cp (tsup – tsat) kJ/kg
or
Hsup = Hs + Cp (tsup – tsat) KJ/kg
1.3 Specific Volume of Steam:
The volume of dry saturated steam in cubic metre per kg (m3
/kg) is known as
Specific Volume of dry Saturated Steam. Its symbol is Vs.
1.3.1 Specific Volume of Wet Steam:
One kg of wet steam will consist of ‘x’ kg of dry steam and (1-x) kg of water in
suspension where ‘x’ is a dryness fraction of wet steam.
Specific volume of wet stream, having a dryness fraction of ‘x’
= Volume of dry steam + volume of water particles
= x.Vs + (1-x).Vw
Where Vs and Vw denote the specific volume of steam and water respectively. As
the volume of water at low pressure is very small compared with the steam, the term
“(1-x).Vw” will become still smaller and can be neglected.
Hence, the specific volume of wet stream, neglecting volume of water particles
= x.Vs m3
/kg
and
density of wet steam = 1/(x.Vs) kg/m3
1.3.2 Specific Volume of Super Heated Steam (Vsup):
Vsup = Vs . (Tsup/ Ts) m3
/kg
Tsup = absolute temp of superheated steam
= absolute temp of saturated steam at same pressure.
Vs = Specific volume of dry saturated steam.
1.4 Steam Tables:
Steam Tables contain the properties of Steam in a tabular form. The properties of
1 kg of water, dry saturated steam and Specific Heat of Steam at Constant
Pressure (Cp) corresponding to various temperatures as also specific volume are
given in Steam Tables. The information given in these tables has been obtained
experimentally by various competent bodies of international repute such as
ASME. In the steam tables, the initial temperature of water is taken as 00
C. The
information contained in steam tables is as under:
• Pressure of steam ( absolute) in bar p
• Saturation temperature of steam ts
• Specific volume of dry saturated steam in m3
/kg vg
• Enthalpy of saturated water in kJ / kg hf

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B- 6
• Enthalpy of dry saturated steam in kJ / kg hg
• Enthalpy of evaporation of dry steam in kJ / kg hfg
• Entropy of water in kJ / kg K φw
• Entropy of dry saturated steam in KJ / kg K φs
1.5 Entropy:
1.5.1 Entropy of Water:
It is an important thermodynamic property of a working substance, which
increases with the addition of heat and decreases with its removal.
The increase or decrease of entropy, when multiplied by the absolute temperature,
gives the heat absorbed or rejected by the working substance.
δQ = δφ x T or, δφ = δQ/T
In general, addition of heat δQ to one kg of water will cause its temperature to rise by
δT.
Then δQ = Cp. δT where Cp= specific heat of water.
∴ δφ= Cp (δT / T)
The total increase entropy of water from 0o
C to a saturation temperature of T K
can be obtained by integrating between temperature limits of 273 and T
φ2 - φ1 = Cp 273∫
T
dT/T = Cp {loge(T)}T
273 = Cp loge(T /273)
Entropy of water at 00
C (freezing temperature T0 = 273 K) is taken as zero.
Entropy of water at boiling temperature Ts is written as φw
φw = Cp loge(Ts /T0) = Cp loge(Ts /273)
This value can be obtained directly from the steam tables.
1.5.2 Entropy of Evaporation (φe):
Heat addition during evaporation takes place at constant temperature if pressure is
maintained constant. If the steam formed is wet, having dryness fraction ‘x’,
Entropy of evaporation, φe = xL / Ts
If the steam formed is dry saturated, x=1,
then the entropy of evaporation is φe = L / Ts
The value of φe = φs - φw may also be obtained from steam table.
1.5.3 Entropy of Dry Saturated Steam:
The entropy φs of dry saturated steam, reckoned from the freezing point of water
(00
C) is equal to the sum of water entropy (φw) and evaporation entropy (φe)
φs = φw + φe = Cp loge(Ts /273) + L / Ts
The value of φs may also be obtained directly from the Steam Tables.
1.5.4 Entropy of Superheated Steam:
Entropy of Superheated Seam is calculated as follows:

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