Heat n Thermodynamics

8/13/2019 Heat n Thermodynamics

1/34

Chapter 5

Heat & ThermodynamicsDr. R. D. Senthilkumar,

Asst. Head, Dept. of Math & Appl. Sciences


2/34

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

12/24/2012 2


3/34


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.

12/24/2012 3


4/34


Thermodynamics

A branch of heat in which we study heat in

motion is called thermodynamics.

12/24/2012 4


5/34


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)

12/24/2012 5


6/34


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/

12/24/2012 6


7/34Dr. R.D. Senthilkumar, Asst. Head, Dept. of Math & Appl. Sciences, MEC

Thermal conductivity

Therefore,

=

where, is the coefficient of thermalconductivity and it depends upon the nature of

the material. Thus,

=

W/m K

12/24/2012 7



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.

12/24/2012 8



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 ()

12/24/2012 9



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.

12/24/2012 10


11/34



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.

12/24/2012 12



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.

12/24/2012 13



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

12/24/2012 14



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.

12/24/2012 15



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.

12/24/2012 16



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.

12/24/2012 17



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.

12/24/2012 18



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.

12/24/2012 19



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,

12/24/2012 20



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.

12/24/2012 21



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?

12/24/2012 22



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 =

12/24/2012 23



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.

12/24/2012 24


25/34


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.

12/24/2012 25


26/34


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.

12/24/2012 26


27/34


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.

12/24/2012 27

W


28/34


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%

12/24/2012 28

W

,


29/34


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.

12/24/2012 29


30/34


31/34


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.

12/24/2012 31


32/34


The product of

pressure and

volume represents

a quantity of work.

This is represented

by the area below

a P-V curve.

12/24/2012 32


33/34


34/34

Thank You

Heat n Thermodynamics

Documents

Transcript of Heat n Thermodynamics