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Chapter 5
Heat & ThermodynamicsDr. R. D. Senthilkumar,
Asst. Head, Dept. of Math & Appl. Sciences
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
Learning Outcomes
After completing this chapter, you will be able to
Define heat, temperature, thermodynamicsand thermal conductivity.
Recall the laws of thermodynamics.
Demonstrate knowledge of flow of heat andheat engine
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
What is Heat?
Heatis a thermal energy which gives us the
sensation of warmth or hotness.
Heat is measured in Joules.
A body becomes hotter when it gains heat
energy and becomes colder when it emits
heat energy. Thus measure of the degree of
hotness or coldness of a body is known as its
temperature.
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
Thermodynamics
A branch of heat in which we study heat in
motion is called thermodynamics.
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
Thermal conductivity
A measure of the materials ability to conduct
heat is called thermal conductivity.
It is measured in the unit of, in SI, W /m K
(Watts per meter Kelvin)
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
Thermal conductivity (K)
The amount of heat ()flowingfrom one face to another dependsupon the following factors:
Directly proportional to the face
area, i.e., Directly proportional to the time of
conduction, i.e.,
Directly proportional to the
difference in temperature, i.e., ( )
Inversely proportional to thethickness of the slab, i.e., 1/
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Thermal conductivity
Therefore,
=
where, is the coefficient of thermalconductivity and it depends upon the nature of
the material. Thus,
=
W/m K
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Sample Exercise1
A copper rod 20 cm long and of 0.785
area of cross-section thermally insulated is
heated at one end through 100 while the
other end is kept at 30 . Calculate theamount of heat which will flow in 20 minutes
along the way. Thermal conductivity of
copper is 380 W/m K.
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Solution to Sample Ex1
Given values are,
= 0.785 10
= 0.2
= 100 30 = 70
= 20 60 = 1200 K = 380 W/mK
Formula,
=
= 380 0.785 10
70 12000.2
= 1.3 10J ()
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Determination of Q by
Lees Disc method
Heat energy conducted, = 12
J
In case of poor conductors whose K is very small thequantity of heat conducted will also be very small.
Thus to increase the value of , either area of cross-sectionshould be increasedor thickness should bedecreasedor the temperature difference (1 2)should be decreased.
As the difference in temperature cannot be increasedbeyond a certain safe limit, hence is increased bytaking thin discs of large area of cross-section.
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Working
A constant current ()is passed through the heater coil and the
potential difference across the coil is measured.
The heat generated by the heating element is conducted through
1 3and then passed through 1 2.
The discs 1 3ensure uniform distribution of heat so that flow
of heat through 1 2is normal and steady.
After some time when steady state is reached the temperatures are
measured.
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Calculations
Let the thickness of 1 = 1and the thickness of 2 = 2
Heat produced by the heater coil in one second =
4.2/ - (1)
Heat passing through 1in one second = 12
1
Heat passing through 2in one second =
3
42
Heat passing through 1 2in one second = 12
1+
34
2 (2)
Equating (1) and (2), we have
4.2=
12
1+
34
2 .(3)
From the above expression, the value of K can be determined.
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Radial Flow of Heat
(Spherical shell method) In this case the specimen under test is
enclosed in between two concentric spheres
of radii 1 2, at the centre of these two
shells a heating element is placed.
The heat is conducted through specimen
from the inner to outer shell. Let 1and 2
be the temperature of inner and outer shell
when the steady state is reached.
Thermal conductivity can be determined by
using =
2
1412(12) W/mK
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Thermal Insulations in buildings
Thermal insulation is the reduction of heat transfer
between objects in thermal contact or in range of
radiative influence.
Thermal Insulation of buildings helps to keep heat induring the winter and out in summer to improve comfort
and save energy.
Insulation could add additional benefits such as acoustics
and waterproofing. Effective draught proofing, moisture control and
ventilation are important at design stage.
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Methods of Thermal Insulation
The principle of thermal insulation is the
resistance to heat flowis achieved by:
bulk insulation method,
reflective insulation method or
a combination of both.
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Bulk Insulation
It resists the heat transfer by
conduction and convection, depends on
pockets of trapped air or low
conductive gasses within its structure.
The following materials are used for
bulk insulation: glass fiber, slag wool,
rock fiber, cellulose fiber, polyester
fiber, polystyrene, and polyurethane.
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Reflective Insulation
Reflective Foil Insulation resists mostly the
radiant heat flow
It is effective only when installed/applied in
combination with air spaces.
Reflective Foils have air spaces, together with
high reflective/low emissive surfaces facing the
air spaces.
The reflective surfaces should be positioned to
face the brighter side downwards.
The thermal resistance of reflective insulation
varies with the direction of heat flow through it,i.e. vertical, horizontal or sloped, the number of
air spaces and defined thicknesses of the air
spaces.
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Laws of Thermodynamics
Thermodynamics is a study of heat conduction
in the materials.
There are two well-known laws of
thermodynamics in physics and they are
discussed below.
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First Law of Thermodynamics
The first law of thermodynamics is based onthe principle of conservation of energy i.e.,energy cannot be created or destroyed but
can be converted from one form to another.
According to first law of thermodynamics, a
definite amount of mechanical work isneeded to produce definite amount of heatand vice versa,
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The ratio of the work done and the heat
produced is always constant.
Let Wis work done in producing Hamount of
heat, thus according to the above law,
or =
=
where, is the proportionality constant and isknown as a Joules constant.
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Sample Ex
A car is brought to halt by applying brakes in
50 m. If the average frictional force which is
stopping the car is 7.5 , how much heat
will be produced?
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solution
Given values are,
Frictional force = 7.5 = 7.5 10
The distance moved = 50 m
Work done (W) =
= 7.5 10
50W = 375 10
We know that, / =
=
= 375 10
4.2 /
= 89.286
= 89.286 4.2 kJ =
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Second Law of Thermodynamics
Claussis Statement:It is impossible for a self-acting machine (i.e., a
machine without the support by any external agency), working in a
cyclical process, to transfer heat from a body at lower temperature
to a body at higher temperature.
i.e., Heat cannot spontaneously flow from cold body to hot body
without the performance of work by an external agency. This is
evident from ordinary experience of refrigeration. For example, in a
refrigerator, heat flows from cold to hot region, but only when
forced by an external agent, which is called the refrigeration
system.
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
Kelvins Statement:It is impossible to derive a
continuous supply of work by cooling a body to a
temperature lower than that of the coldest of its
surroundings. The above statement can be explained as it is
impossible to extract energy by heat from a high-
temperature energy source and then convert all of
the energy into work. i.e., no heat engine can covertwhole of the heat energy supplied to it into useful
work.
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
Heat Engine
A heat engine is a device that converts heatenergy into mechanical energy.
Examples include a) steam engines, b) steam
and gas turbines, c) spark-ignition and dieselengines, and external combustion engine.
These engines are used to provide the
mechanical power for transportation,operating machinery, and producingelectricity.
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
All heat engines works in a cyclicprocess of absorbing heat from thesource of heat , pressurizingthe working fluid or gas,
performing a mechanical work (W)and releasing unused heat to thecooler .
Ex: In a car engine, the source of
heat is the combustion of petrol ordiesel, and the cooler is the air inthe atmosphere.
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W
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
Efficiency of Heat Engine
The efficiency () of a heat engine is defined as the ratio of
the work performed by the engine (work done) to the heat
supplied to the engine by the source (input heat)
=
100%
=
100%
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W
,
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
Carnot Engine
A Carnot heat engine is a hypothetical engine
that operates on the reversible Carnot cycle.
The basic model of this engine was developed
by Nicolas Leonard Sadi Carnot, a Frenchengineer, in 1824.
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
The four reversible processes are:
1. Isothermal expansion(by placing the system in contactwith a heat reservoir with temperature TH).
2. Adiabatic expansionto TC< TH.
3. Isothermal compression(by placing the system in
contact with a heat reservoir with temperature TC).
4. Adiabatic compressionfrom TCto TH.
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Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC
The product of
pressure and
volume represents
a quantity of work.
This is represented
by the area below
a P-V curve.
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Thank You