Lecturer : Maizan bin Sulaiman Telephone No : 07-2612488 (ext 1715),

019-7512123E-mail : maizans4@gmail.com,

maizan@pis.edu.my

COURSE OUTLINE

COURSE OUTLINE

Penggunaan “Kad Comel”

• Saya, Maizan bin Sulaiman merekabentuk dan mengunapakai “Kad Comel” ini khas untuk tujuan pemantauan sahaja iaitu bagi kelas JJ207 Thermodynamic.

•Selain itu, diharap pelajar mendapat munafaat dari Kad Comel ini dengan memberi komen-komen, keluhan atau aduan tentang apa sahaja mengenai kursus/kelas ini.

• Insya-Allah saya akan mengambil jalan yang sewajarnya daripada maklumbalas Kad Comel ini untuk kebaikan bersama.

1. Penggunaan: Tidak wajib, tetapi sangat perlu untuk rujukan pensyarah.2. Jenis Dokumen: Bukan dokumen rasmi PIS.3. Kekerapan: Digunakan setiap kali pertemuan.4. Perhatian pelajar: diminta memberi komen(JIIKA ADA),menjawab soalan-soalan pendek (jawab jika tahu, tiada markah

akan diberi, tak perlu jawab jika masih tidak jelas/tak mampu kerana ruangan komen disediakan untuk melontar keluhan anda).

Terima Kasih.

CHAPTER 1BASIC THERMODYNAMICS

INTRODUCTIONWhen engineering calculations are performed,

it is necessary to be concerned with the units of the physical quantities involved.

A unit is any specified amount of a quantity by comparison with which any other quantity of

the same kind is measured. For example, meters, centimeters and

millimeters are all units of length. Seconds, minutes and hours are alternative time units.

1.1 Fundamental and Derived Quantities

1. In the present discussion, we consider the system of units called SI (International System of Units) and it is a legally accepted system in many countries.

2. SI units will be used throughout this module.

1.Length. 2.mass. 3.time. 4.electric current. 5.thermodynamic temperature. 6.luminous intensity.

These six quantities are absolutely independent of one another. They are also called the ‘Indefinables’ of mechanics.

Fundamental Physical Quantities.Six fundamental physical quantities.

NOTE; SUKAR UNTUK DITAKRIFKAN

Quantity Unit Symbol

Mass kilogram kgTime second s

Length meter mThermodynamic

temperaturedegree Kelvin K

Electric current ampere ALuminous intensity candela cd

Six Fundamental units and its symbols

Derived unitsPhysical quantities like area, volume, density, velocity, acceleration, force, energy, power, torque etc. are called derived quantities since they depend

on one or more of these fundamental quantities

Note: 1. Fundamental units: Unit asas.2. Derived units: Unit terbitan.

Example: Area = Wide x LengthThe unit: m2 = m x m

Derived unitsQuantity Unit Symbo

lNotes

Area meter square m2

Volume meter cube m3 1 m3 = 1 x 103 litreVelocity meter per second m/sAcceleration

Meter per second squared

m/s2

Density kilogram / meter cube

kg/m3

Force N 1 N = 1 kgm/s2

Pressure Newton/meter square

N/m2 1 N/m2 = 1 Pascal1 bar = 105 N/m2 = 102 kN/m2

1.ForceNewton’s second law may be written as force α (mass x acceleration), for a body of a constant mass, ie: F = kmaWhere: k is constant. In a coherent(logical) system of units such as SI, k = 1,

Hence: F = ma The unit : kgm/s2 = Newton, N

The SI unit of force is therefore kgm/s2. This composite unit is called the Newton, N. i.e. 1 N = 1 kg.m/s2

2.Energy1. Heat and work are both forms of energy.2. The work done by a force is the product of the force

and the distance moved in the same direction. 3. The SI unit of work = force x distance in the Newton

meter, Nm.4. A general unit for energy is introduced by giving the

Newton meter the name Joule, J.i.e. 1 Joule = 1 Newton x 1 meter

or

1 J = 1 Nm

3. Power

1. The use of an additional name for composite units is extended further by introducing the

Watt,W as the unit of power.

2. Power is the rate of energy transfer (or work done) by or to a system.

i.e.1 Watt, W = 1 J/s = 1 Nm/s

4. Pressure1. Pressure is the force exerted (apply) by a fluid per

unit area, only when we deal with gas or liquid.2. The pressure on a surface due to forces from

another surface or from a fluid is the force acting at 90o to the unit area of the surface.

3. i.e. pressure = force/ area4. P = F/A5. The unit of pressure, is N/m2 and this unit is

sometimes called the Pascal, Pa.

1 bar = 105 N/m2 = 105 Pa

5. Density

1. Density is the mass of a substance per unit volume.

2. The unit of density is kg/m3 .

volumemassDensity

Vm

Force, F = maPressure, P = F/A

Work, W = F x L Density, = m/V

Summary

Example 1.1

2N/m 05.49 0.01

9.81 x 50

areaforce (P) Pressure

Force = mass x accelerationPressure = force/area

Solution to Example 1.1

Example 1.2

Solution to Example 1.2We should end up with the unit of kilograms. Putting the given information into perspective, we have

= 850 kg/m3 and V = 2 m3

It is obvious that we can eliminate m3 and end up with kg by multiplying these two quantities. Therefore, the formula we are looking for is

Vm

Thus, m = V

= (850 kg/m3)(2 m3)

= 1700 kg

TEST YOUR UNDERSTANDING

1.1 What is the work done by an expanding gas if the force resisting the motion of the piston is 700 N and the length of the stroke is 0.5 m ? 1.2 What is the force required to accelerate a mass of 30 kg at a rate of 15 m/s2 ? 1.3 The fuel tank of a large truck measures 1.2m x 0.9m x 0.6m. How many litres of fuel are contained in the tank when it is full? (you need to know the conversion form m3 into litres vice-versa) 1.4 A weather research instrument is suspended below a helium filled balloon which is a 3.8m diameter sphere. If the specific volume of helium is 5.6m3/kg, what is the weight of helium in the balloon? Explain briefly why the balloon rises in the atmosphere.

Answer

1.1 Work = Force x Distance= (700 N)(0.5 m)= 350 Nm or J 1.2 Force = mass x acceleration

F = ma = (30 kg)(15 m/s2) = 450 kg.m/s2 or N

1.3 Volume = 1.2 x 0.9 x 0.6 = 0.648 m3

Since 1m3 = 1000 litres

Then, contents of full tank = 0.648 x 1000= 648 litres

1.4 Radius of volume, r = = 1.9 m

Volume of balloon, V =

= 28.73 m3

Mass of helium in balloon, m = = 28.73/5.6

= 5.13 kg.

w = mg = 5.13 x 9.81

= 50.3 N

Density of helium, = =

= 0.1786 kg/m3

The balloon rises in the atmosphere because the density of helium is less than the density of atmosphere.

2d

3 34 r

vV

v1

6.51

1.2 Unit Conversions 1. Measurements that describe physical

quantities may be expressed in a variety of different units.(Quantity=length, Unit=m or ft.)

2. As a result, one often has to convert a quantity from one unit to another.

3. For example, we would like to convert, say, 49 days into weeks.

4. One approach is to multiply the value by ratios of the equivalent units.

5. The ratios are formed such that the old units are cancelled, leaving the new units.

1. Despite (even though, biarpun) their causing us errors, units/dimensions can be our friends.

2. All terms in an equation must be dimensionally homogeneous (uniform).

3. That is, we can’t add apples to oranges…4. Neither can we add J/mol to J/kg s.5. By keeping track of our units/dimensions, we can

automatically do a reality check on our equations.6. But the fun doesn’t stop there…7. A dimensional analysis can help to determine the

form of an equation that we may have forgotten.(example in solution of 1.4)

The Dimensional Homogeneity

The example of unit conversions are:

•1 kg = 1000 g.

•1 m = 100 cm = 1000 mm.

•1 km = 1000 m = (100 000 cm @ 105 cm) = (1 000 000 mm @ 106 mm).

•1 hour = 60 minutes = 3600 seconds.

•1 m3 = 1000 litre, or 1 litre = 1 x 10-3 m3.

•1 bar = 1 x 105 N/m2 = 1 x 102 kN/m2.

Multiple and sub-multiple of the basic units are formed by means of prefixes, and the ones most commonly used are shown in the following

table:Multiplying Factor Prefix Symbol

1 000 000 000 000 1012 tera T 1 000 000 000 109 giga G 1 000 000 106 mega M 1 000 103 kilo k 100 102 hector h 10 101 deca da

0.1 10-1 desi d0.01 10-2 centi c0.001 10-3 milli m0.000 001 10-6 micro 0.000 000 001 10-9 nano n0.000 000 000 001 10-12 pico p

Multiplying factors

Multiplying Factor Prefix Symbol

1 000 000 000 000 1012 Tera(meter) T 1 000 000 000 109 Giga(meter) G 1 000 000 106 Mega(meter) M 1 000 103 Kilo(meter) k 100 102 Hector(meter) h 10 101 Deca(meter) da

0.1 10-1 desi(meter) d0.01 10-2 centi(meter) c0.001 10-3 milli(meter) m0.000 001 10-6 micro(meter) 0.000 000 001 10-9 nano(meter) n0.000 000 000 001 10-12 pico(meter) p

Multiplying Factor Prefix Symbol

1 000 000 000 000 1012 Tera(meter) T 1 000 000 000 109 Giga(meter) G 1 000 000 106 Mega(meter) M 1 000 103 Kilo(meter) k 100 102 Hector(meter) h 10 101 Deca(meter) da

1 100 Meter -0.1 10-1 desi(meter) d0.01 10-2 centi(meter) c0.001 10-3 milli(meter) m0.000 001 10-6 micro(meter) 0.000 000 001 10-9 nano(meter) n0.000 000 000 001 10-12 pico(meter) p

The application of prefix into basic units of “Meter”

EXAMPLE: PREFIX for basic units of “Meter”

Example 1.3

Convert 1 km/h to m/s.

Solution to Example 1.3

m/s 278.0 s 3600m 1000

s 3600j 1 x

km 1m 1000 x

jkm 1

jkm 1

Example 1.4

Convert 25 g/mm3 to kg/m3

Solution to Example 1.4

1 kg = 1000 g1 m = 1000 mm

1 m3 = 1000 x 1000 x 1000 mm3 = 109 m3

36

3

9

3

39

33

kg/m 10 x 25 m 1000

kg 1x 10 x 25

g 1000kg 1 x

m 1mm 10 x

mmg 25

mmg 25

TEST YOUR UNDERSTANDING

1.5 Convert the following data:a) 3 N/cm2 to kN/m2

b) 15 MN/m2 to N/m2

1.6 Convert 15 milligram per litre to kg/m3.

1.5 a) 1 kN = 1000 N

1 m2 = 100 x 100 = 104 cm2

2

2

4

2

24

22

kN/m 30 m 1000kN10 x 3

N 1000kN 1x

m 1cm10x

cmN 3

cmN 3

b) 1 MN = 106 N/m2

26

6

22

N/m 10 x 15 MN 1

N10x mMN 15

mMN 15

1.6 1 kg = 1 000 000 mg 1 m3 = 1000 litre

33-

3

kg/m 10 x 15

m 1litre 1000 x

mg 000 000 1kg 1 x

litremg 15

litremg 15

Past years Final exam CollectionsConvert the following unit

1. 15km/h cm/s [416.667 cm/s]2. 15MN/m2 N/m2 [ 15X 106 N/m2 ]3. 25g/mm3 kg/m3 [25X 106 kg/m3]4. 15 N/cm2 kN/m2 [kN/m2]5. 20 mg/litre kg/m3 [0.02kg/m3 = 20 x 10-3kg/m3]6. 35 g/mm3 kg/m3 [25X 106 kg/m3]

7. 18 mg/litre kg/m3 [18X 10-3 kg/m3]

Past years Final exam CollectionsConvert the following unit

1. 15km/h cm/s [416.667 cm/s]2. 15MN/m2 N/m2 [ 15X 106 N/m2 ]3. 25g/mm3 kg/m3 [25X 106 kg/m3]4. 15 N/cm2 kN/m2 [kN/m2]5. 20 mg/litre kg/m3 [0.02kg/m3 = 20 x 10-3kg/m3]6. 35 g/mm3 kg/m3 [25X 106 kg/m3]

7. 18 mg/litre kg/m3 [18X 10-3 kg/m3]

Precise definition of the basic concepts forms a sound foundation for the development of science and

prevents possible misunderstandings.

In this Chapter, the systems that will be used are reviewed, and the basic concepts of thermodynamics such as system, energy, property, state, process, cycle,

pressure and temperature are explained.

Careful study of these concepts is essential for a good understanding of the topics in the following Chapter.

Basic concepts of thermodynamics

Definitions of system, boundary, surrounding, open system and close system

A thermodynamic system, or simply a system, is defined as a quantity of matter or a region in space chosen for study.

The fluid contained by the cylinder head, cylinder walls and the piston may be said to be the system.

SYSTEM: Is a gases in

cylinder

The mass or region outside the system is called the surroundings. The surroundings may be

affected by changes within the system.

The boundary is the surface of separation between the system and its surroundings. It may be the cylinder and the piston or an imaginary surface drawn as in Fig. 2.1-1, so

as to enable an analysis of the problem under consideration to be made.

A close system1. Also known as a control mass.2. Consists of a fixed amount of mass, and

no mass can cross its boundary.3. No mass can enter or leave a close

system, But energy, in the form of heat or work can cross the boundary.

4. The volume of a close system does not have to be fixed.(The example is refer to expansion and compression stroke in IC engine)

An open system1. It is a Control volume.2. It is often called, is a properly selected

region in space. (refer to air flowing in and out of the turbine continuously, only which in the system or in the turbine will be considered.)

3. It usually encloses a device, which involves mass flow such as a boiler, compressor, turbine or nozzle.

4. Both mass and energy can cross the boundary of a control volume

Fluid Inlet

Fluid Outlet

SYSTEM

SURROUNDINGS

BOUNDARY

Gas turbine engine, the example of an open system Simplified diagram

An open system

Properties1. Properties are macroscopic characteristics of a

system such as mass, volume, energy, pressure, and temperature to which numerical values can be assigned at a given time without

knowledge of the history of the system. 2. Many other properties are considered during

the course of our study of engineering thermodynamics. Thermodynamics also deals

with quantities that are not properties, such as mass flow rates and energy transfers by work

and heat

Properties are considered to be either intensive or extensive

1. Intensive properties(bersifat mendalam) are those which are independent of the size of the

system such as temperature, pressure and density.

2. Extensive properties (bersifat meluas)are those whose values depend on the size or extent of

the system. Mass, volume and total energy are some examples of extensive properties.

Criterion to differentiate intensive and extensive properties.

Refer to 3D sketch for better understanding

State (Peringkat)The word state refers to the condition of system as

described by its properties. Since there are normally relations among the properties of a system, the state

often can be specified by providing the values of a subset of the properties.

Example:State 1, with its properties are normally being written as

P1=10 bar , T1= 32 o C , V1= 1 m3 State 2, with its properties are normally being written as

P1=20 bar , T1= 60 o C , V1= 2 m3

Process1. When there is a change in any of the properties

of a system, the state changes and the system are said to have undergone a process.

2. A process is a transformation from one state to another.

3. Processes may be reversible or actual (irreversible). A reversible process is one that is wholly theoretical

4. Processes may be constrained to occur at constant temperature (isothermal), constant pressure, constant volume, polytropic and adiabatic (with no heat transfer to the surroundings).

Zeroth law of thermodynamicsThe "zeroth law" states that if two systems are at the same time in

thermal equilibrium with a third system, they are in thermal equilibrium with each other.

If A and C are each in thermal equilibrium with B, A is also in equilibrium with C. Practically this means that all three are at the same

temperature, and it forms the basis for comparison (evaluation) of temperatures. It is so named because it logically precedes (come first)

the First and Second Laws of Thermodynamics.

1

2

3

What are the ordinary and absolute temperature scales in the SI and the English system?

Absolute : Kelvin in SI system & Rankine in English (Imperial) system ,

Ordinary : Celsius in SI system & Fahrenheit in English (Imperial) system.

TEST YOUR UNDERSTANDING

2.1 Fill in the blanks with suitable names for the close system in the diagram below.

ii. _________

i. _____________

iii. _____________

2.2 Study the statements in the table below and decide if the statements are TRUE (T) or FALSE (F).

STATEMENT TRUE or FALSEi. The mass or region inside the system is called

the surroundings.ii. In a close system, no mass can enter or leave a

system.iii. Intensive properties are those which are

independent of the size of the systemiv. Mass, volume and total energy are some

examples of intensive properties.

2.1 i.Surroundings

SystemBoundary

2.2 i.FalseTrueTrueFalse

THE END

Thank You