Modul Thermodynamic

FAKULTI PENDIDIKAN TEKNIKAL DAN
VOKASIONAL

THERMODYNAMIC

BBM 20203

UNIVERSITI TUN HUSSEIN ONN MALAYSIA

COPYRIGHT PAGE

FIRST EDITION (THERMODYNAMIC MODULE) 2020

MUHAMMAD FAIQ BIN MOHAMED RUDUAN
DR WAN HANIM NADRAH BINTI WAN MUDA

All rights reserved. All parts of this module may not reproduce
any part of the illustration, content and articles in any form or
manner whether electronically, recording, photocopying, etc.
before obtaining permission from the author and the party
concerned.
The author strives to ensure that this published book has
accurate complete information from the learning topics
discussed. There are sections that contain text in addition to
writing the Malay language to prevent the true meaning of the
text. If expert reference and assistance is needed, please seek
professional advice and services from a qualified specialist.

"Nothing in life is certain except death,
taxes and the second law of
thermodynamics."

PUBLISHED AND PRINTED BY

MUHAMMAD FAIQ BIN MOHAMED RUDUAN

BACHELOR OF VOCATIONAL EDUCATION (GENERAL MACHINING) WITH
HONORS

FACULTY OF TECHNICAL AND VOCATIONAL EDUCATION
UNIVERSITI TUN HUSSEIN ONN MALAYSIA, 86400,
JOHOR DARUL TA’ZIM

[email protected]
0145191922

Table Of Contents

SECTION A CONTENTS I
CHAPTER 1 INTRODUCTION ABOUT MODULE II
DEFINITION AND BASIC CONCEPT 1
CHAPTER 2 1.0 Introduction 1
1.1 Thermodynamic System 2
1.2 Properties of A System 3
1.3 State And Equilibirium 3
1.4 Processes and Cycles 4
1.5 Basic Unit in Thermodynamic 5
5
1.5.1 Multiples Unit 7
1.5.2 Pressure 10
1.5.3 Pressure Measurement Devices 11
1.5.4 Temperature 13
Exercisess 15
Answer 18
PURE SUBSTANCES 19
2.0 Introduction 19
2.0.1 Phase Of A Pure Substance 20
2.1 Relationship Between Pressure (P),
Specific Volume ( ) and Temperature (T) 20
2.1.1 Pressure (P), Specific Volume ( ) and
20
Temperature (T) Surface 21
2.1.2 One Phase Area 22
2.1.3 Two-Phase Area 25
2.1.4 Projection of P-v-T Surface
2.2 The Process of Exchanging Pure Substances

CHAPTER 3 Phases 26
2.2.1 Liquid Compressed 27
2.2.3 Saturated Liquid-Vapor Mixture 28
2.2.4 Superheated Steam 28
2.3 Thermodynamic Properties Table 31
2.3.1 Superheated Steam Table 32
2.3.2 Compressed Liquid 33
2.3.3 Additional Notes use of Property Tables 37
2.4 The Ideal-Gas Equation of State 39
2.4.1 Ideal Gas Condition Diagram 41
Exercises 42
Answer 47
ENERGY, HEAT AND WORK 48
3.0 Introduction 49
3.1 Energy 49
3.1.1 Potential Energy 50
3.1.2 Kinetic Energy 51
3.1.3 Internal Energy 51
3.1.4 Internal Energy, Enthalpy and Specific Heat
52
Ideal Gases 52
3.2 Heat 52
53
3.2.1 Heat Transfer 53
3.2.2 Modes of Heat Transfer 54
3.3 Work 55
3.3.1 Electric Power 55
3.3.2 Shaft Work
3.3.3 Spring Work
3.3.4 Boundary Work

3.3.5 Boundary Work for Close System 56

Exercises 62

CHAPTER 4 THE FIRST LAW OF THERMODYNAMIC 63
CHAPTER 5
CHAPTER 6 4.0 Introduction 64

4.1 Principles of Energy Eternity 65

4.2 First Law of Thermodynamic 66

4.3 Closed System Heat Transfer 71

4.3.1 Isometric Process 71

4.3.2 Isobaric Process 73

4.3.3 Isothermal Process 75

Exercises 77

Answer 78

THE FIRST LAW OF THERMODYNAMIC (OPEN SYSTEM) 81

5.0 Introduction 82

5.1 Principle of Energy Conservation for Open System 83

5.2 The Steady Flow Equation 84

5.3 Function The Steady Flow Equation 86

5.3.1 Nozzle 86

5.3.2 Diffuser 86

5.3.3 Turbine 89

5.3.4 Compressors and Pumps 92

5.3.5 Heat Exchanger 94

Exercises 96

Answer 97

THE SECOND LAW OF THERMODYNAMIC 102

6.1 Introduction 103

6.2 Heat Reservoir 104

6.3 The Second Legal Statement Thermodynamics 104

6.4 Heat Engine 107
6.5 Reverse Heat Engine 109
6.6 Cooling 111
6.7 Heat Pump 112
113
6.7.1 Reversible Processes 114
6.7.2 Irreversible Processes 115
6.7.3 Carnot Principle 116
6.7.4 Maximum performance of heat engine and
118
Reversible heat engine 119
Exercises
Answer

CHAPTER 1

DEFINITION & BASIC CONCEPT

OBJECTIVE
At the end of this chapter, student will be:

 Defines thermodynamic terms
 Relate the use of thermodynamic concepts

in daily life
 Explain the common terms used in the

thermodynamics field
 Uses basis units in the thermodynamic

CHAPTER 1
DEFINITION & BASIC CONCEPT

1.0 Introduction

Thermodynamics, science of the relationship between heat, work, temperature, and energy.
In broad terms, thermodynamics deals with the transfer of energy from one place to
another and from one form to another. The key concept is that heat is a form of energy
corresponding to a definite amount of mechanical work.
Thermodynamic

 Therme (heat) & dunamic (power) (Greek word)
 Conservation of energy principle
 During interaction, energy can change from one
 form to another but total amount of energy
 remain contant.
 Energy cannot be create or destroyed

1

1.1 Thermodynamic System
• System defined as a quantity of matter or a region in space chosen for study
• The mass or region outside the system is called the surroundings
• The real or imaginary surface that separate the system from its surroundings is called boundry.
Refer figure 1.1

Figure 1.1
Closed system
In thermodynamics, a closed system can exchange energy (as heat or work) but not matter, with its
surroundings. An isolated system cannot exchange any heat, work, or matter with the surroundings,
while an open system can exchange energy and matter.
Closed system

 Control mass
 consist of a fixed amount of mass
 no mass can across its boundry but energy (heat or work) can cross the boundry
 In special case, even energy is not allowed to cross the boundry, that call isolated system
Example : air in tire, gas in cylinder
Open system
An open system is one that freely allows both energy and matter to be transferred in an out of a
system.
For example, boiling water without a lid.Heat escaping into the air. Steam (which is matter) escaping
into the air.

2

1.2 Properties of A System

 Characteristic of system is called a property
 Presure (P), temperature (T), volume (V), mass (m)
 Density, = / (kg/m3)
 Specific volume, = / = / (m3/kg)

Properties of A System devide to 3 types

1. Intensive properties
 It does not depend on the size and material contained in the system
 Can not be mixed
 Example; density, pressure, temperature

1. Extensive properties
 Changed by system size and can be mixed
 Example; mass, volume and internal energy

2. Specific properties
 Extensive properties per unit mass
 Example; specific volume
 = / = / = m3/kg

1.3 State & Equilibirium

State
All the properties that can be measured throught the entire system and completely describe the
condition we call state of the system. If value of properties changes, the state will change

Thermal Equilibrium
 The temperature is the same throughout the entire system.
 Mechanical Equilibirium-related to pressure
 Phase Equilibirium-mass of each phase are Equilibirium level
 Chemical Equilibirium-chemical composition does not change with time

Figure 1.2 State Equilibrium
3

1.4 Processes & Cycles

Process

Any change that a system undergoes from one equilibrium state to another.

Path

The series of states through which a system passes during a process. To decribe a process
completely, one should specify the initial and final states, as well as the path it follows and the
interactions with the surroundings.

Quasistatic or quasi-equilibrium process

When a process proceeds in such a manner that the system remains infinitesimally close to an
equilibrium state at all times.

Properties Process – the process in which the particular properties of the material during the process
take place are ;-

Property Constant Process Name
Temperature Isotermal (sesuhu)
Pressure Isobar (setekanan)
Volume Isometric (seisipadu)
Entrophy Isentropik (seentropi)
Entalphy Isentalpik (seentalpi)

Cycle of Thermodynamic

 Some processes are combined in series and form a closed cycle. Refer figure 1.3
 The situation returns to its original state in the process.
 The sequential process in terms of process path and direction
 The net change in its properties is zero
 Total change of pressure = ( 2– 1) + ( 3 – 2) + ( 4– 3) + ( 1− 4) = 0

Figure 1.3
4

1.5 Basic Unit in thermodynamic

Parameter Mass Length Time
Second, s
SI Unit Kilogram, kg Meter, m Second, s

Imperial Unit Paun-jisim, Ibm feet, ft

 Relationship between the two systems above

 1 lbm= 0.45359 kg

 1 ft = 0.3048 m

There are several other published units often used in thermodynamics;

Quantiti Unit Symbol
Work Joule J
Heat Joule J
Power Watt W
Pressure Newton/meter square N/m²
Specific volume Meter square/Kilogram m³/kg

1.5.1 Multiples Unit

Methods are used to summarize or reduce the writing of numbers if the value is too large or too
small.

examples;
 10000000 = 1×10⁶
 0.00000123 = 1.23×10‾⁶

10⁶= mega dan 10‾⁶ = micro. You must know to change a value using multiple units of different.

Example : 1MW = 1000 kW

5

Si Unit

Prefixes Value Standard form Symbol
Tera T
Giga 1 000 000 000 000 1012 G
Mega M
Kilo 1 000 000 000 109 k
Deci d
Centi 1 000 000 106 c
Milli m
Micro 1 000 103 ᶙ
Nano n
Pico 0.1 10-1 p

0.01 10-2

0.001 10-3

0.000 001 10-6

0.000 000 001 10-9

0.000 000 000 001 10-12

Si Unit Table and Simbol

Example Calculation

Example 1 : Change 1 km/j To m/s

You Known

 1 km = 1000 m
 1 jam = 60 X 60 = 3600 s

Solution :

 1 × 1000 × 1 = 1000 = . /
1 3600 3600

Example 2 : Change 15km/j to cm/s

You Known

 1 km = 1000 m dan 1 m = 100cm
 1jam = 3600s

Solution :

 15 × 1000 × 100 × 1 = 15 ×105 = . /
1 1 3600 3600

6

Example 3 : Change 25 g/mm3 to kg/m3

You Known

 1 kg = 1000 g
 1 m³ = 1000 x 1000 x 1000 = 1x10⁹ mm³

Solution :

 25 × 1 × 1×109 3 = 25×109 = × /
3 1000 1 3 1000 3

Example 4 : Change 3N/cm² to kN/m²

You Known

 1 kN = 1000 N
 1 m² = 100 x 100 = 1 x 10⁴ cm²

Solution :

 3 × 1 × 1×104 2 = 3×104 = /
2 1000 1 2 1000 2

1.5.2 Pressure

Pressure is a measure of the force exerted per unit area on the boundaries of a substance (or
system). It is caused by the collisions of the molecules of the substance with the boundaries of
the system. The force exerted by a fluid/gas per unit area.

Pressure, =



 1 N/m² =1 Pascal (Pa)
 1 kPa =10³ Pa
 1 MPa =10⁶ Pa
 1 bar =10⁵

7

Example 1 :

The mass of 50kg acts on one piston with 100cm² area. Calculate the intensity of the pressure
on the water found under the piston on equilibrium conditions.

Solution :

Given ; Mass= 50kg

Area = 100cm² 0.01m²

Pressure acting on the piston.

 = 50 ×9.81 = .
2

0.01 2

Pressure measurement Figure 1.4

= ℎ

where

 P = pressure (N/m²)
 = density (kg/m²)
 g = gravity accelaration(m/s²)
 h = Height (m)

Example 2 :

A diver is 10m deep sea level. If local atmospheric pressure is 760mmHg and the density of
seawater is 1050 kg / m3, determine the pressure acting on diver body. Take 760mmHg = 1.014
bar.

Solution :

Given :-

 water = 1050
 g (gravity) 3
 ℎ ( height)
= 9.81
2

= 10m

 atmosfera = 1.014 bar

= ℎ

= [10501××91.8015×10 ] + 1.014 = .

8

Figure 1.5

Four commonly used types of pressure.

 Gauge pressure
 Atmospheric pressure
 Absolute pressure
 Vacuum pressure

Gauge pressure : Pgauge = Pabsolute – Patm Figure 1.6
Pabsolute = Pgauge + Patm p > Patm

Vacuum pressure : Pvacuum = Patm – Pabsolute p < Patm
Example 1 : Pabsolute = Patm - Pvacuum

A vacuum gauge connected to one the chamber shows 40 kPa where the pressure the
atmosphere is 99.95 kPa. Determine absolute pressure in the chamber.

Solution :
Pabsolute = Patm – Pvacuum
= 99.95 – 40 kPa

= 59.95 kPa

9

1.5.3 Pressure measurement devices

a) Manometer
One of the earliest pressure measuring instruments is still in wide use today because of

its inherent accuracy and simplicity of operation. It's the U-tube manometer, which is a U-shaped
glass tube partially filled with liquid. This manometer has no moving parts and requires no
calibration.. As shown in Figure 1, with each leg of a U-tube manometer exposed to the
atmosphere, the height of liquid in the columns is equal. Using this point as a reference and
connecting each leg to an unknown pressure, the difference in column heights indicates the
difference in pressures (see Figure 2).

` Figure 1 Figure 2

b) Bourdon gauge
Bourdon tube pressure gauges are used for the measurement of relative pressures from 0.6 to

7,000 bar. They are classified as mechanical pressure measuring instruments, and thus operate without
any electrical power. Bourdon tube pressure gauge are radially formed tubes with an oval cross-section.
The pressure of the measuring medium acts on the inside of the tube and produces a motion in the non-
clamped end of the tube. This motion is the measure of the pressure and is indicated via the movement.

10

c) Barometer
When atmospheric pressure is measured by a barometer, the pressure is also referred

to as the "barometric pressure". Assume a barometer with a cross-sectional area A, a height h,
filled with mercury from the bottom at Point B to the top at Point C. The pressure at the bottom
of the barometer, Point B, is equal to the atmospheric pressure. The pressure at the very top,
Point C, can be taken as zero because there is only mercury vapour above this point and its
pressure is very low relative to the atmospheric pressure. Therefore, one can find the
atmospheric pressure using the barometer and this equation is

Patm = ρgh.

Figure 1 Figure 2

1.5.4 Temperature

 Level of temperature qualitatively with word, cold and hot.
 Heat is transferred from higher to lower temperature.
 At one point, heat transfer stop, and reached thermal equilibirium.
 Zero law of thermodynamic
 Temparature scale

 T(°C) = T(K) -273
 T(R) = 1.8 T(K)
 T(°F) = T(R) – 460
 T(°F) = 1.8 T(°C) + 32

11

Example :
Convert the following temperature values to ° C;
a) 300K
b) 70 °F

Solution :
a) T(°C) = T(K) – 273
= 300 – 273 = 27°C
b) T(°F) = 1.8 T(°C) + 32
= [T(°F) – 32]/1.8
= [ 70 – 32]/1.8 = 21.11°C

12

Exercises (Thermodynamic System)
Question 1
The thermodynamic system can be divided into two. State both systems and explain each one
briefly. Give two examples for each.
Question 2
State whether the water that is boiling in the kettle is a closed system or an open system.
Explain
Question 3
A student studies the behavior of a gas in a pressure vessel. What system should be used,
closed or open system? Explain why.
Question 4
In a closed jar, the water temperature at the bottom and top of the container varies. Is the
system in thermal equilibrium?

Exercises (Basic Unit)
Question 5
Water weighing 7984 kg is filled in a container (V = 0.8m3) until full. At 25 ° C, 1 atmosphere,
determine the following quantity.
a) Density
b) Specific volume in units of cm3/kg
Question 6
An object has a mass of 50 kg. Determine its weight, in N in an area with a gravitational speed
of 9.75m/s2

13

Exercises (Pressure)
Question 7
A mercury barometer (Figure 1) is used to measure atmospheric pressure. The height of the
mercury inside the tube is 755mm. By taking the mercury density and gravitational speed as
13600 kg/m3 and 9.81 m/s2, calculate the atmospheric pressure in the bar

Figure 1
Question 8
A submarine is 200m deep above sea level. Seawater density, gravitational speed and
atmospheric pressure were 1040kg/m3, 9.87 m/s2 and 1.013 bar respectively. Determine the
absolute pressure acting on the vessel.

Figure 1

14

Exercises (Temperature)

Question 9

Determine the following temperature values from °F to°C

a) 60℉
b) -10℉
c) 0℉
d) 200℉

Question 10

Convert the following temperature values to K.

a) 50℃
b) 500R
c) 180℉
d) 200℃

Answer

Question 1

1. Closed System
 Mass control.
 Contains a fixed amount of mass and no mass flow through the boundary.
 Heat and work can cross borders and volumes can change.
 Eg; vehicle tires, gas in a cylinder.

2. Open System
 Volume control
 involves the flow of mass in and out across the boundaries of the system
 Heat and work can cross borders
 Eg; water pipes, water tanks, turbines, compressors, and water pumps.

Question 2

The water that is boiling in the kettle is an open system because there is a mass flow out of
the system where the water will evaporate into steam and be released during the boiling
process.

15

Question 3

The system is closed, because the gas is in a pressure vessel that has a constant volume and
the gas cannot be released.

Question 4

No, the system is said to be in thermal equilibrium when the overall temperature of the system
is uniform. In this case, the water temperature at the top and bottom is not uniform.

Question 5

a) Density

=
7984
= 0.8 3
= ⁄

b) Specific volume in units of cm3/kg

= = 0.8 3 = 1.002 × 10−4 3⁄
7984

= 1.002 × 10−4 3 × (100 × 100 × 100) 3 = . ⁄
1 3

Question 6

Weight = mass × gravity acceleration

= 50 × 9.75 ⁄ 2 = .

Question 7

= ℎ = 13600 ⁄ 3 × 9.81 ⁄ 2 × 755 × 10−3
1

= 100.73 × 100 = 1.0073
Question 8

= +

o = . + ( ⁄ × . ⁄ × )
o = . + .
o = . + ( . × )



16

Question 8
= +

o = 1.013 + (1040 ⁄ 3 × 9.81 ⁄ 2 × 200 )
o = 1.013 + 2040.48
o = 1.013 + (2040.48 × 1 )

100

o = 1.013 + 20.405 = .
Question 9

a) 60−32 = 15.6℃

1.8

b) −10−32 = −23.3℃

1.8

c) 0−32 = −17.8℃

1.8

d) 200−32 = 93.3℃

1.8

Question 10
a) 50 + 273 = 323
b) 500 = 277.8

1.8

c) 180−32 = 82.22 + 273 = 355.2

1.8

d) 200 + 273 = 473

17

CHAPTER 2
PURE SUBSTANCES

OBJECTIVE
At the end of this chapter, student will be:

 Defines pure substances
 Relationship Between Pressure (P), Specific

Volume ( ) and Temperature (T)
 Explain The Process of Exchanging Pure

Substances Phases
 Uses thermodynamic properties table
 What is ideal-gas equation of state

CHAPTER 2
PURE SUBSTANCES

2.0 Introduction
What is a Pure Substances
A substance that has a fixed chemical composition throughout is called pure substance.
Water, helium carbon dioxide, nitrogen are examples.
It does not have to be a single chemical element just as long as it is homogeneous
throughout, like air. A mixture of phases of two or more substance is can still a pure
substance if it is homogeneous, like ice and water (solid and liquid) or water and steam (liquid
and gas).

14

2.0.1 Phase Of A Pure Substance
There are three principle phases – solid, liquid and gas, but a substance can have several other phases
within the principle phase. Examples include solid carbon (diamond and graphite) and iron (three solid
phases). Nevertheless, thermodynamics deals with the primary phases only. Refer figure 2.1
In general:

 Solids have strongest molecular bonds.
 Solids are closely packed three dimensional crystals.
 Their molecules do not move relative to each other

 Intermediate molecular bond strength
 Liquid molecular spacing is comparable to solids but their molecules can float about in groups.
 There is molecular order within the groups

 Weakest molecular bond strength.
 Molecules in the gas phases are far apart, they have no ordered structure
 The molecules move randomly and collide with each other.
 Their molecules are at higher energy levels, they must release large amounts of energy to

condense or freeze.
3 main phase

Figure 2.1

19

2.1 Relationship Between Pressure (P), Specific Volume ( ) and Temperature (T)

 To investigate the properties of the pure substance, we need to note the most important
thermodynamic relationship which is Pressure (P), Specific Volume ( ) and Temperature (T)

 Temperature and Specific volume is independent variable.

2.1.1 Pressure (P), Specific Volume ( ) and Temperature (T) Surface
 P-υ-T surface is a 3-dimensional graph showing the position of the phase of a pure substance
is in a state of equilibrium.
 Phases about liquid, solid, vapor and mixed phase. (Refer figure 2.2)
 Using this diagram the position of the phase of a material can be seen clearly.

Surface of a substance that Surface of a substance that
contrast on freezing expand on freezing (like water)

Figure 2.2

2.1.2 One phase area

 Pressure, specific volume and temperature are independent of each other (one can be changed
without involving a change in the phase of the material.)

 The overall condition of the material in this area is determined by any 2 of the above properties.

20

2.1.3 Two-phase area
 The two-phase regions where two phases exist in equilibrium separate the single-phase
regions. The two-phase regions are: liquid-vapor, solid-liquid, and solid-vapor. Temperature and
pressure are dependent within the two-phase regions.
 Once the temperature is specified, the pressure is determined and vice versa. The states within
the two-phase regions can be fixed by specific volume and either temperature or pressure.
 If the pressure is changed then the temperature must be changed at the same rate. Otherwise
a phase change will occur.
 Conditions cannot be determined by simply knowing the pressure and temperature. Conditions
can be determined in full if the volume is certain and either pressure or temperature are known.

Figure 2.3
Saturated condition

The term saturation defines a condition in which mixture of vapor and liquid can exist together
at a given temperature and pressure. The temperature at which vaporization (boiling) starts to
occur for a given pressure is called the saturation temperature or boiling point. Phase change
begins and ends.

21

Critical point

In thermodynamics, a critical point (or critical state) is the end point of a
phase equilibrium curve. The most prominent example is the liquid–vapor critical point, the end
point of the pressure–temperature curve that designates conditions under which a liquid and
its vapor can coexist. At higher temperatures, the gas cannot be liquefied by pressure alone. At
the critical point, defined by a critical temperature Tc and a critical
pressure pc, phase boundaries vanish. Other examples include the liquid–liquid critical points in
mixtures. Refer figure 2.4

Saturated Liquid Saturated Steam
saturated liquid
ssaturated liquid

Figure 2.4
2.1.4 Projection of P-υ-T Surface

• In most P-υ-T surface analysis is difficult to use. It is projected to be a 2-dimensional diagram.
• From the 3-dimension of p-υ-T, we get a 2-dimensional diagram (plan view, front and side)

Unjuran permukaan p-υ-T

22

Phase Diagram
• The P-υ-T surface projected on the pressure-temperature plane is named as the phase
diagram.
• The material at any point on this line at a certain pressure and temperature consists of 2
phases.
• The 3-line phase in the P-υ-T surface is seen as a point & named after a 3-phase point
• Pressure and temperature of 3 phase of water is 0.6113kPa and 273.16K

Solid-Liquid Line

Liquid-Vapor Line

Solid-Vapor Line

Figure 2.5

The P-v Diagram

 During the vaporization process/ boiling process, both the temperature and the pressure remain
constant, but the specific volume is increases.

 Once the last drop of liquid is vaporized, further reduction in pressure results in a further
increase in specified volume.

 During the phase-change process, the weight is not remove, If the weights is removed, the
pressure and temperature will drop [since Tsat= f(Psat)], and the process would no longer be
isothermal.

 If the process is repeated for other temperatures, similar paths will be obtained for the phase-
change processes. Connecting the saturated liquid and the saturated vapor states by a curve,
we obtain the P-v diagram of a pure substance, as shown in Fig. 2.6.

23

Figure 2.6 P-v diagram of a pure substance
The T-v diagram

 As increases pressure, the shorter the saturation line ; pressure reaches 22.09 MPa(water) and
at this point, it is called critical point.

critical point; the point at which the saturated liquid and the saturated vapor states are identical
 The temperature, pressure and specific volume of a substance at the critical point are called

respectively, the critical temperature Tcr, critical pressure Pcr and critical specific volume vcr
 Important line is Isobar line. In a 1-phase area– (compressed liquid & superheated vapor) isobar

line in curved line.
 When the volume increases at constant pressure, the temperature increases. In the area of

saturated liquid mixtures– isobar line are horizontal (liquid phasevapor phase)
 During the conversion process the pressure & temperature phase is constant. (saturated

preassure & saturated temperature)

Figure 2.7 T-v diagram of a pure substance
24

2.2 The process of exchanging pure substances phases

Practical situations – two phases coexist in equilibrium
- water exists as a mixture of liquid and vapor in the boiler and condenser of a
steam power plant
- the refrigerant turns from liquid to vapor in evaporator of a refrigerator

Following terms/properties/state will be discuss base on Figure. 2.8
 compressed liquid
 saturated liquid
 saturated vapor
 superheated vapor

Figure 2.8
State 1

 Water at temperature 20°C and 1 atm pressure.
 Temperature & Specific volume start to increase–slowly
 State 1 2 water are in liquid compressed phase
State 2
 Temperature achive at 100°C
 The addition of heat will convert liquid to vapor
 Water are in saturated liquid phase.
 Any added heat will cause the water to turn into vapor

25

State 3

 Water turns into vapor without changing temperature with a big change in υ.
 Along in saturated liquid vapor region.

State 4

 All liquid changed to vapor
 saturated vapor

State 5

 The phase change process is complete– vapor phase
 If heated temperature will rise again
 The vapor outside of the saturated vapor line is called a superheated vapor
 Setemperature drops will not change the water phase as long as the temperature is above

100 ° C

Figure 2.9

2.2.1 Liquid Compressed

 A pure substance will be in a state of compressed liquid if its temperature is less than its
temperature at a certain pressure.

 Example; at a pressure of 7 bar and a temperature of 100 ° C, the water is in a state of
compressed liquid because its temperature is lower than its temperature at the same pressure
of 165 ° C

 The difference between the saturated temperature and the temperature of the liquid is
compressed at the same pressure called the sub-cold degree

26

 ∆ = −
 165-100 = 65℃
 In one-phase conditions, the pressures and temperatures of the materials are independent of

each other.
 The overall condition of the material can be determined by knowing one of its thermodynamic

properties.

Figure 2.10

2.2.3 Saturated liquid-vapor mixture

 Also known as wet steam. Liquids and vapors coexist in equilibrium
 For analysis - the amount of saturated liquid and saturated vapor in the system needs to be

known. (dryness or quality fraction) with the symbol x.
 The dryness fraction is intensive or equal to the mass of saturated vapor divided by the total

mass of the mixture.

 = = 0 < < 1

+ +

 If x is equal to 0.6 or 60% (60% saturated vapor, 40% saturated liquid)

 1 − =

+

 Dryness fractions can also be used to determine some other thermodynamic properties when

the material is in a state of saturated liquid-vapor mixture;

Specific Volume
 = + when = −

Specific Enthalpy
 ℎ = ℎ + ℎ when ℎ = ℎ − ℎ

Specific Entropy

 = + when = −

27

2.2.4 Superheated steam

 An explanation of the properties and uses of superheated steam (such as for electricity
generation). Including explanations of the Rankine and Carnot thermodynamic cycles,
superheated steam tables and the Mollier (H-S) chart.

 If the saturated steam produced in a boiler is exposed to a surface with a higher temperature,
its temperature will increase above the evaporating temperature.

 The steam is then described as superheated by the number of temperature degrees through
which it has been heated above saturation temperature.

 Superheat cannot be imparted to the steam whilst it is still in the presence of water, as any
additional heat simply evaporates more water. The saturated steam must be passed through an
additional heat exchanger. This may be a second heat exchange stage in the boiler, or a
separate superheater unit. The primary heating medium may be either the hot flue gas from the
boiler, or may be separately fired.

 Example; at a pressure of 2 bar and a temperature of 200 ° C, the water is in a state of
superheated steam because the temperature of the saturation at the same pressure is 120.2 °
C.

 = −
 200 – 120.2 = 79.8℃

2.3 Thermodynamic Properties Table

 To determine the condition of a material we need to know at least 2 properties of the material.
There are several ways (mathematical equations, tables, charts, computer programs)

 For learning purposes, attributes in tabular form are still widely used. The skill of using a table
of material properties is very important because thermodynamic analysis uses it extensively.

 The properties table can be divided into 3 parts, namely the table of saturated liquids,
compressed liquids and superheated vapors.

 Since the nature of a substance in the state of compressed liquid is small and close to the nature
of its saturated liquid, it is usually referred to the nature of the substance in the state of saturated
liquid.

 In the case of a saturated liquid-vapor mixture, the temperature & pressure of a material are
equal to its temperature and saturation pressure. In the diagrams of p-v & T-v, both the
temperature and pressure lines for the saturated liquids overlap with each other.

28

Table A–4: Saturation properties of water under temperature.
Table A–5: Saturation properties of water under pressure.

A partial list of Table
A–4.

At low pressures and volumes, a certain volume of saturated liquid
can be taken as 0.001m3 / kg
Example 1
 Determine the temperature of boiling water at pressures of 1 bar and 50 bar.
 Boiling water temperature = begins to turn into steam = saturated temperature
 1 bar = 99.63, 50 bar = 263.9
Example 2
 Determine the specific volume in the state of saturated vapor, at a temperature of 50 ℃ and
at a pressure of 5 bar
 at 50 ℃ = 12.03m3/kg
 at 5bar = 0.3749m3/kg
Example 3
A system contains a liquid-vapor-saturated mixture of water at a temperature of 100 ℃ and a
fraction of dryness, x = 0.8. Determine the specific volume of the mixture.

29

Solution
 At 100 ℃ ,
 = 1.0435 × 10−3m3/kg
 = 1.673m3/kg
 = + when = −
 =(1.0435 × 10−3)+ 0.8(1.673 − 1.0435 × 10−3)
 = . m3/kg

Example 4
Determine the fraction of dryness, definite volume and energy in water course at 7 bar and
enthalpy 2600 kJ/kg.

Solution
When 7 bar ,

 ℎ = 2763.5 / because h<hg, then the water is in a state of wet vapor
 ℎ = 697.22 /
 ℎ = ℎ + ℎ when ℎ = ℎ − ℎ
 2600 = 697.22+(x)(2763.5-697.22)
 X = 0.921
Ignoring the specific volume of water in a saturated liquid state, then
 = = 0.921(0.2729) = 0.2513 3/
 = +
 = 696.44 + 0.921(2572.5 − 696.44)
 = . /

30

2.3.1 Superheated steam table

Figure 2.11
In these tables, the properties are listed versus temperature for selected pressures starting with the
saturated vapor data. The saturation temperature is given in parentheses following the pressure
value (Fig. 2.6 and Fig. 2.7). Superheated vapourhas;

 Lower pressure ( P < given T)
 Higher temperatures ( T > a given P)
 Higher specific volumes ( v > a given P or T)
 Higher internal energy ( u > a given P or T)
 Higher enthalpies ( h > ℎ a given P or T)

Figure 2.12

31

Example 5

Determine the specific volume and energy in the steam when at 10 bar and 300 ℃.

Solution

at 10 bar is 179.9 ℃. Therefore, the water is in a state of extreme hot steam due to > .
From the superheated steam table at 10 bars and 300 ℃ are available.

 = 0.2579 3/
 = 2793.2 /

2.3.2 Compressed Liquid

 The difference in the nature of the pure material in the state of saturated liquids and compressed
liquids especially at low pressures is very small.

 The nearest method is used to determine the thermodynamic properties of compressed liquids.
 The most commonly used method is to take the value of properties in saturated liquids with

reference to a given temperature.
 = at a given temperature or = (T)
 = at a given temperature or = (T)
 ℎ = ℎ at a given temperature or ℎ = ℎ (T)

 Generally the specific volume of a substance in a state of compressed liquid is very small.
Therefore, the difference between the specific volume of compressed liquid and saturated liquid
( - ) at the same pressure is also too small. Therefore, the value of can be ignored.

 =

Example 6

Determine the specific volume and specific enthalpy of water at pressures of 10 bar and 90 ℃.

Solution

 at 10 bar is 179.9 ℃.
 Water is in a state of compressed liquid due to T <
 Therefore the specific volume and specific enthalpy also can be determined nearest using.
 = at 90℃ = 1.0360 × 10−3m3/kg
 ℎ = ℎ at 90℃ = 376.92 kJ/kg

32

2.3.3 Additional notes on the use of property tables

 If specific internal energy, is not given the use of the equation = −
 In certain circumstances the properties given in a problem are not in the table. Use the

interpolation method.

− 1 = − 1
2 − 1 2 − 1

= ൬ 2 − 11൰ ( 2 − 1) + 1
−

Example 7
Determine the temperature of saturated vapor, water at a pressure of 153 bar.

Solution
 The pressure of 153 bar is not in the table and lies between the pressure of 150 bar and 160
bar. Using the method of internalization.

pT − 342.1 = 153 − 150
150 342.1 347.3 − 342.1 160 − 150
153 Ts
160 347.3 153 − 150
= ൬160 − 150൰ × (347.3 − 342.1) + 342.1

= . ℃

33

Example 8

Determine the specific volume and specific enthalpy of water at a pressure of 14bar and a
temperature of 420 ℃.

Solution

 water at a pressure of 14 bar is 195.07 ℃
 water is in a state of superheated steam because T > .
 Referring to the table at a pressure of 14bar, it is found that the temperature at 420 ℃ is not in

the table. Therefore, the calibration was performed at a pressure of 14bar between 400 ℃ and
500 ℃.

T 420 = 420 − 400 × (0.2521 − 0.2178) + 0.2178
400 0.2178 3257.5 ൬500 − 400൰
420
500 420 ℎ420 = . /
0.2521 3474.1

ℎ420 = 420 − 400 × (3474.1 − 3257.5) + 3257.5
൬500 − 400൰

= . /

Example 9

3 kg of water vapor at 30 bar and 350 ℃ is contained in the cylinder and piston system. Water
is cooled at a constant volume until the temperature reaches 195 ℃. Then it is compressed at
a constant temperature to a pressure of 20 bar.

 Sketch the processes that take place in the diagrams T- and p- .
 Determine the fraction of dryness and pressure at condition 2 and the volume of the

cylinder at condition 3.

Solution

State 1

 Determine the phase of water in condition 1.
 Given P = 30 bar and T = 350 ℃
 Through table A1 at p = 30 bar, Ts = 233.9 ℃; 350> Ts (superheated steam)

34

State 2
 The air is cooled at a constant volume until the temperature reaches 195 ℃.
 The first process is the isometric process, 1 = 2
 Through table A3 at p = 30 bar, T= 350 ℃; = 1 = 0.0905 3⁄

State 3
 Determine the water condition in condition 2.
 By comparing the specific volume at condition 2, 2 with the specific volume of
saturated vapor, at a temperature of 195 ℃.
 From table A1 at 195 ℃ , = 0.1411 3⁄
 > 2 then condition 2 is in the wet vapor area.

 dryness fraction

= = 0.0905 = 0.642 ( ignored)
0.1411

 Since condition 2 is a mixture liquid and saturated vapor, then p2=ps at195 ℃ which is

13.98 bar

35

State 4
 Determine the phase of water in condition 3. Then it is compressed at a constant
temperature to a pressure of 20 bar.
 Compare the pressure at condition 3 (20 bar) with the pressure at condition 2 (13.98 bar).
 p3 > p2 , therefore condition 3 is in a state of compressed fluid. To obtain a certain volume
in condition 3, the nearest method is used.
≅ at 195 ℃
 Refer to table A1 found at a temperature of 195 ℃
≅ = . × − ⁄
 Therefore, the volume in condition 3
= = × . × − = .

36

Figure T-v

2.4 The Ideal-Gas Equation of State
 Any relation among the pressure, temperature, and specific volume of a substance is called an
equation of state.
 The property relations which involve other properties of a substance at equilibrium states are
also referred to as equation of state
 There are several equations of state, some simple and other are complex. The simplest and
best known equation of state is the ideal-gas equation of state, give as

 Pv=mRT

where R is the gas constant. This equation call also be called ideal-gas relation; and a gas which
obeys this relation is called an ideal gas. In this equation, P is the absolute pressure, T is the
absolute temperature and v is the specific volume, m is about mass.
 This equation predicts the P-v-T behavior of a gas quite accurately within some properly

selected region
 Gas and vapor are often used as a synonymous words. The vapor phase of a substance is

called a gas when it is above the critical temperature. Vapor usually implies a gas which is not
far from a state of condensation.

37

The P-v-T behavior of a substance can be represented more accurately by more complex equation
of state. Three of best known are;

 Van der Waals Equation of State
 Beattie-Bridgeman Equation of State
 Benedict-Webb-Rubin Equation of State
 Virial Equation of State

Example 10

Determine the mass of air in the room measuring 4m x 5m x 6m at 100kPa and 25 ℃. Take the
mass of air molecules as 29kg / kmol.

Solution
 = = 8.3143

29

= . /
 = , =



= 100×(4×5×6)

0.2867(25+273)

= .

Example 11

0.01 kg of a ideal gas contained in a pressure vessel containing a volume of 0.003 3. In this
case the absolute pressure and gas temperature are 5 bar and 120 ℃. The gas is expanded
until the pressure and volume are 1 bar and 0.02 3 respectively. determine the molecular mass
and temperature of the final state of the gas.

Solution

Initial condition (m =0.01kg, V= 0.003m3, p=5x102kPa, T = 393K)

 = = 5×102×(0.003) = 0.3817 /
0.01(393)

 = , = = 8.3143 = 21.78 /

0.3817

 = = 1×102 ×0.02
0.01×0.3817

 = .

38

2.4.1 Ideal Gas Condition Diagram
 Ideal gas exists in only one phase i.e. the gas phase. in this phase there is no saturated curve.
 Diagram of P-v and T-s for ideal gas

Example 12
1 kg of air and 1 thermodynamic cycle with different 3 processes as described below

 1-2: constant volume
 2-3: expansion at constant temperature
 3-1: compression at constant pressure
In condition 1, the temperature is 25 ℃ and the pressure is 1 bar and in condition 2 the pressure is
2 bar. Sketch the cycle above the p-v state diagram and using ideal gas, determine.

a) Temperature at 2 in K
b) Volume in condition 3, m3
Solution
Sketch the situation 1 = 1 bar, 25℃

39

Sketch the situation 2 = 2 bar, T=?
 Processes 1-2: constant volume

Sketch the situation 3 =
 Processes 2-3: expansion at constant temperature

 Processes 3-1: compression at constant pressure

a) Temperature at 2 in K

2 = 2 2, 2= 1 prosese 1-2 isometric



1 = 1
1

0.287 × 298
= 1 × 102

= 0.8556 3⁄

Therefore 2 = 2×102×0.8556 =
0.287

40

b) Volume in condition 3, m3

Proses 2-3 isothermal; T2 = T3 and P3 = P1=1 bar

3 = 3
3

1 × 0.287 × 596
= 1 × 102

= .

Exercises
Question 1
Sketch a diagram of the T-υ and P-υ conditions of water and show
a) Saturated liquid and vapor lines
b) Areas of compressed liquid, liquid-saturated vapors and superheated vapors.
c) Critical point and isobar @ isotermal.
Question 2
Determine the water phase for the following conditions. Show the points on the diagrams P-υ
and T-υ and label them completely.
a) p = 500 kPa, T = 200℃
b) p = 5 MPa, T = 260℃
c) T = 200℃, p = 0.9 Mpa
d) p = 20 MPa, T = 100℃

41

Question 3

Fill in the table below for water in the given conditions

T(℃) p(bar) (m3/kg) u(kJ/kg) h(kJ/kg) x phase
0.75
a) 70 2.0

b) 2.0 1500

c) 175.4

d) 10.0 2475

e) 350 40.0

f) 80 5.0

Question 4
Mark the dots in question 3 above in the T-υ and P-υ state diagrams.
Question 5
A 500 liter volume rigid tank contains 2 kg of water at 200 kPa. Determine the temperature,
internal energy and mass of saturated liquids and saturated vapors.

Answer
Question 1

42