Modul Thermodynamic

Example 5

A pump is used to pump water at a rate of 10kg/s through a pipe. At the entrance, the pressure
is 1.0bar, temperature 25° & velocity 3m/s. Conditions in the output section are at 1.5bar, 30°
& velocity 15m/s. The outlet is located 15m above the inlet. Determine the power required by
the pump in kW. Take g=9.81m/s2.

Solution

( ̇ − ̇ ) = ̇ (( − ) + ( ̅ − ̅ ) + ( − ))

 In this case the heat transfer can be ignored and Determine the phase for the input and output

section.

̇ = ̇ ((ℎ1 − ℎ2) + ( ̅2 12 − ̅ 22 ) + ( 1 − 2))
2

Input section Output section
̅ = / ̅ = /

= . , Ts = 99.63℃ > T1 = . , Ts = 111.4℃ > T⅔
phase of compressed fluid phase of compressed fluid

= ℃ = . / = 30℃ = 125.79 /

̇ = 10 ((104.89 − 125.79) + 32 − 152 + 9.81(0 − 15))
(2 2)

= − .

93

5.3.5 Heat exchanger
A heat exchanger is a system used to transfer heat between two or more fluids. Heat
exchangers are used in both cooling and heating processes. The fluids may be separated by a
solid wall to prevent mixing or they may be in direct contact. They are widely used in space
heating, refrigeration, air conditioning, power stations, chemical plants, petrochemical
plants, petroleum refineries, natural-gas processing, and sewage treatment.
The main process for heat transfer at constant pressure. Another example is the
heat sink, which is a passive heat exchanger that transfers the heat generated by an electronic
or a mechanical device to a fluid medium, often air or a liquid coolant.
 During the heat transfer process, the work is zero i.e. no work is supplied or produced.
Therefore, the steady flow equation of this system is
 The change in Kinetic Energy and Potential Energy of heat exchangers is small when
compared to heat transfer and neglected.

̇ = ̇ ( − ) (kW)

To know more about heat
exchanger, Please scan!!

94

Example 6

In a car cooling system, air is used to cool the cooling liquid inside the radiator. Air flows in
through the radiator at 1,013 bar, 30 ℃ and out at 50 ℃. The flow rate of air volume at the input
section is 0.5 3/ . Assuming air is ideal gases and ignores changes in Kinetic Energy and
Potential Energy, calculate the heat transfer that occurs between air and cooling liquid in kJ/kg
and air mass flow rate.

Solution

 Because ideal gas, then equation is :

(h2-h1) = cp(T2-T1)

= ℎ2 − ℎ1 = ( 2 − 1)
= 1.005(50 − 30)

= . /

 The air mass flow rate can be determined using a formula.

̇ = 1̇ = 1̇ 1
1 1

0.5 × 1.013 × 102
= 0.287 × 303

= . /

The End

95

Exercises
Question 1
Air undergoes two processes in series as below;
Process 1-2: polytropic compression with n= 1.3, from p1=100 kPa, υ1= 0.04m3/kg to υ2=
0.02m3/ kg
Process 2-3: constant pressure up to υ3= υ1
Sketch the above processes on the p- υ diagram and determine the amount of work per unit of
air mass.
Question 2
A closed system contains 0.8 kg of air at 1 bar and 25℃. This air is compressed to a pressure
of 5bar. During the process a certain amount of heat is removed to set the air temperature in
the system. Determine the work involved during this process. (R= 0.287 kJ/kgK)
Question 3
A closed system containing 1 kg of water undergoes two processes in series as follows;
Process 1-2: compression at a constant volume from 5 bar, 200 ℃ to a pressure of 10 bar
Process 2-3: cooling at constant pressure until the specific volume of water is 0.150 m3/kg.
Sketch both processes on diagrams p-υ and T-υ and determine the temperature at condition 2,
the dryness fraction of the final condition and the clean work cycle in kJ.
Question 4
A closed system contains 2 kg of water undergoing two series thermodynamic processes as
follows;
Process 1-2: expansion with pυ = constant from saturated vapor state at 100 bar to 10 bar
Process 2-3: constant pressure up to υ3=υ1
Sketch the process on the diagrams p-υ and T-υ and calculate the amount of work in kJ.

96

Exercises
Question 5
An object of mass 2 kg moves with a velocity of 50 m/s. What is the kinetic energy of the
object in kJ.

Answer
Question 1

Determine P2

 1 = ( 2)1.3

2 1

 2 = 1 = 100 = 246.23
( 12)1.3 (00..0024)1.3

Work for isothermal processes, W12;

12 = 1 1 − 2 2 = (100 × 0.04) − (246.23 × 0.02) = −3.082 /
− 1 1.3 −1

Work for isobaric processes, W23;

23 = ( 3 − 2) = 246.23(0.04 − 0.02) = 4.9246 /

Total work done Wtotal

= 12 + 23 = −3.082 + 4.9246 = . /

97

Answer

Question 2

State 1

 P1 = 1bar
 T1 =25°C = 298K

State 2 = constant temperature

P2 = 5bar

Give air m = 0.8kg

Work for isothermal prosses, W12

12 = ( 12) = 0.8 × 0.287 × 298 × 1 = − .
(5)

Question 3

State 1

1 = 5
1 = 200℃; = 151.9℃; 1 > ; ℎ
From steam table; 1 = 0.4249 3⁄
State 2

2 = 10
2 = 1 = 0.4249 (Isometric Prosses)
at 10 bar = 0.1944; 2> ; superheated steam phase

Determination of temperature at condition 2 through internal determination;

Temperature, T Specific volume, ν 2 − 600 = 0.4249 − 0.4011
600 0.4011 700 − 600 0.4478 − 0.4011
T2 0.4249
700 0.4478 = ℃

98

State 3

3 = 2 = 10 (Isobaric prosses)
3 = 0.150 3⁄ ; = 0.1944 3⁄ ; > 3 Phase: a mixture of saturated liquid and vapor

Dryness coefficient;

= + ;

3 = ; = 3 = 0.150 = 0.772
0.1944

Total work done = 12 + 23
12 = 0 (Isometric Prosses)
23 = ( 3 − 2) = 1000 (0.150 − 0.4249) = −274.9 /
= 12 + 23 = 0 + (−274.9) = − . /
Diagrams of p-υ and T-υ

99

Question 4
The given parameter = 2 kg of water

State 1

 1 = 100
 1 = = 311.1℃

From steam table = = . ⁄

State 2

 2 = 10
 = 0.1944 3⁄

 2 =?

1 1 = 2 2; 2 = 1 1 = 100 × 0.01803 = 0.1803 3⁄
2 10

2 = 0.1803 3⁄ ; > 2 ∴Phase of a mixture of liquid and saturated vapors

Dryness coefficient;

2 = ; = 2 = 0.1803 = 0.93
0.1944

State 3

Phase = ?

 3 = 2 = 10
 = 0.1944 3⁄
 3 = 1 = 0.01803 3⁄ ; > 3 ∴ ℎ

100

Determination of work;

Work fo Isotermal Prosses, W12;

12 = 2 2 ( 12) = 10 × 102 × 0.1803 × 2 × 0.1803 = 830.3
(0.01803)

Work for Isobaric Prosses, W23;

23 = ( 3 − 2) = 10 × 102 × 2 × (0.01803 − 0.1803) = −162.27

Total work done, Wtotal;

ℎ = 12 + 23 = 830.3 + (−162.27) = .
Diagrams of p-υ and T-υ

Question 5

 m = 2 kg
 = 50 ⁄

= 1 2 = 1 × 2 × 502 × 1 / = .
2 2 (1000 2⁄ 2)

101

CHAPTER 6

The Second Law of Thermodynamics

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

 Defines the second law of thermodynamic
 Defines of the second legal statement

thermodynamics
 What is heat engine, reverse heat engine,

cooling and heat pump
 How to calculated heat engine, reverse

heat engine, cooling and heat pump
energy.

CHAPTER 6
The Second Law of Thermodynamics

6.0 Introduction

What is the first law of thermodynamics?
The First Law of Thermodynamics states that heat is a form of energy, and
thermodynamic processes are therefore subject to the principle of
conservation of energy. This means that heat energy cannot be created or
destroyed.
∑ =∑

What is the second law of thermodynamics?
Heat transfer occurs spontaneously from higher- to lower-temperature
bodies but never spontaneously in the reverse direction. Another way of
stating this: It is impossible for any process to have as its sole result heat
transfer from a cooler to a hotter object.
_ > _ ℎ

103

6.2 Heat reservoir

 Defined as a system that has the ability to receive & dissipate heat in an infinite amount
without undergoing any temperature change.

 Sea, lake, river & atmospheric air.
 Example; the hot water discharged by the steam power plant into the lake or sea will not raise

the temperature of the lake or sea water as a whole. Due to the enormous volume of lake or
sea water.
 There are 2 types of reservoirs. Hot heat reservoir (heat supply, QP) or so-called source.
 Cold heat reservoir (receiving heat dissipated, QS by the system) or called sink.
 The direction of heat conduction can determine the type of heat reservoir.

Figure 6.1

6.3 The second legal statement thermodynamics.
The most commonly referred statement is.

 Kelvin-Plank statement
Form of heat exchange to work (Principle of heat engine) = a tool for producing work.

 Clausius statement
The concept of heat transfer between 2 heat reservoirs (Principle of reverse heat engine).

104

Kelvin-Plank statement
"It is impossible for any system operating in a thermodynamic cycle to produce work to the
environment by only receiving heat from a heat reservoir only"

Figure 6.2
The work-producing system will not work by simply receiving heat supply from a heat reservoir
only.
 Summary; not all heat supplied from a hot heat reservoir is converted to work form because
some parts need to be removed to a cold heat reservoir. The heat dissipated is heat that
cannot be converted into work.
 Therefore, all power cycles have less than 100% efficiency. To achieve 100%, it is necessary
to convert all the heat supplied to work. (contradicts Kelvin-Planck's statement).
 Examples of combustion in the engine.

Figure 6.3
105

Clausius statement
"It is impossible for any system to operate in a thermodynamic cycle by only producing heat
transfer from a cold source to a hotter source"

Figure 6.4
In the above figure heat pump and refrigerator,both engines break Clausius statement
theory.Because this type of supply heat from a cold body to hot body without taking any external
help.This type of device is called perpetual motion machine of the second kind. Refrigerator and
heat engine both are transferring heat from cold body to hot body in cyclic process but in different
surroundings.Refrigerator operating in a cyclic process,maintain a cold body at a temperature
lower than the temperature of the outside temperature.Heat pump operating in a cyclic
process,maintain a hot body at temperature higher than the temperature of the outside
temperature.

To more info about second law thermodynamic :
Please scan me!!

106

6.4 Heat Engine

What is heat engine in thermodynamics?
In thermodynamics and engineering, a heat engine is a system that converts heat or thermal
energy to mechanical energy, which can then be used to do mechanical work. It does this by
bringing a working substance from a higher state temperature to a lower state temperature.
 Not all heat supplied is converted to work. The remaining heat is discharged to the second
heat reservoir.
 Examples of heat engines: steam power plants and vehicle engines.
 In a steam power plant, the heat reservoir is the combustion chamber in the boiler and the
cold heat reservoir is the sea, river or lake water to cool the condenser.

Figure 6.5
 Heat supplied from the heat reservoir = Qp
 Work produced = Wclean
 Heat dissipated to cold heat reservoir = Qs
 The First Law of Thermodynamics = ∑ = ∑

− = ℎ
 The Second Law stipulates that coarse heat must be greater than clean work,

> ℎ
 To produce ℎ then the value of must be greater than

107

Heat engine criteria

 Receive heat from high temperature sources (combustion, solar energy, furnaces & nuclear
reactors). Among the systems that meet these criteria are vehicle engines, gas turbine power
plants & steam power plants.

 Convert part of heat to work (shaft rotation)
 Eliminate waste heat to low temperature sinks (sea, river & air)
 Operates in a thermodynamic cycle.

Refer Figure 6.5 about criteria heat engine

Boilers And Condenser

The main function of a boiler is to heat water to generate steam. Boilers are pressure vessels
designed to heat water or produce steam, which can then be used to provide space heating and/or
service water heating to a building. In most commercial building heating applications, the heating
source in the boiler is a natural gas fired burner.

The function of the condenser in a refrigeration system is to transfer heat from the refrigerant
to another medium, such as air and/or water. By rejecting heat, the gaseous refrigerant
condenses to liquid inside the condenser.

Figure 6.6

 Heat engine performance is expressed using the term thermal efficiency, ℎ

 ℎ = . = ℎ
ℎ ℎ

 ℎ = − = −


 Since p is always greater than s, then the thermal efficiency of a system is always less than

1 or 100%.

108

Example 1

A car engine produces 120kW of power from the heat supplied by 300kW as a result of fuel
combustion. Determine the heat efficiency of the engine and the amount of heat dissipated to the
environment.

Solution

 = = = . @ %


 The amount of heat eliminated = − = 300 − 120

=180

6.5 Reverse Heat Engine

The reverse heat engine is a device that transfers energy from an object at a lower
temperature to an object at a higher temperature by doing work on the system. This is essentially
reverse of the process that takes place inside a heat engine, in which energy flows from a higher
to a lower temperature and work is generated as a result. There are three common examples of
everyday devices that operate on the principle of the reverse heat engine. Refrigerators are
reverse heat engines that pump energy out of the inside of the fridge and into the surrounding
area by the way of an electric motor. In other words, the motor takes some quantity of energy
from the inside of the fridge and brings it to the outside.

Figure 6.7 More about Reverse Heat Engine
Please Scan Me!!

109

Air conditioners are another example of reverse heat engines. The air conditioner takes
energy from some system (a room perhaps) and pumps that energy into the surrounding area
(outside), which is at a higher temperature. Refrigerators and air conditioners are rated by a
quantity known as the coefficient of performance, which is the ratio of the quantity of energy
pumped by the motor to the work done by the motor. A third example of a reverse heat engine is
a heat pump. Heat pumps are usually used in the winter months to bring energy into the house
and thereby heat the house. A heat pump uses a motor to do work and bring energy from the
outside (at a lower temperature) into the house (at a higher temperature). The coefficient of
performance for a heat pump is the ratio of total energy that flows into the house to the work done
by the heat pump.

Figure 6.8 Figure 6.9

 Can be divided into 2 types namely refrigerant and heat pump.
 The difference is in terms of the effect of heat used.
 The coolant emphasizes the cooling effect, while the heat pump emphasizes the heating

effect.

110

6.6 Cooling

 An important component, evaporator = absorbs heat from the cooled space causing room

temperature to drop.

 Example; refrigerator & air conditioning system.

 Coolant performance is expressed using the term coefficient of performance, PPP

 Since the refrigerant serves to cool the space at Ts temperature, Qs is the preferred quantity.

Qs is the amount of heat that needs to be removed from space.

= = =
−

 More heat is removed from the cold heat reservoir, s so the coefficient of cooling performance

is higher.

 The value of usually exceeds 1.

Figure 6.10
Example :

The storage space of a refrigerator is set at a temperature of 4 ℃ by removing heat at a rate of
360kJ / min. If the power required by the refrigerator is 2kW, determine.

a) Refrigeration performance coefficient.
b) The rate of heat dissipated by the environment.

111

Solution :

 ̇ = 360 × 1 =
60

 = ̇ = 6 = 3
̇ 2

 ̇ = ̇ − ̇

 ̇ = ̇ + ̇ = 2 + 6 = 8

= 480 /

6.7 Heat Pump

 Transfer heat from the cold reservoir to the hot reservoir. The main function is to heat a
space and set the temperature of the space at a specified level.

 An example of a heat pump is a heater used to heat a room on a cold winter.
 Heat pumps can also be used as a drying system for certain materials that require a drying

process at a temperature not too high.
 In some cases, there are also systems in which both heating and cooling effects are

applied simultaneously.

 For heat pumps, the heat supplied to the hot reservoir, Qp is the preferred quantity.

 = =


 = =
−

 For areas with winter & summer, dual season air conditioning system is used. In winter it

will heat the room (acts as a heat pump) & in summer it cools the space (acts as a cooler).

 For dual-function systems − =



 Thus − = atau = +
 112

6.7.1 Reversible Process

In thermodynamics, a reversible process is a process whose direction can be reversed
to return the system to its original state by inducing infinitesimal changes to some
property of the system's surroundings. Throughout the entire reversible process, the system
is in thermodynamic equilibrium with its surroundings. Having been reversed, it leaves no
change in either the system or the surroundings. Since it would take an infinite amount of
time for the reversible process to finish, perfectly reversible processes are impossible. However,
if the system undergoing the changes responds much faster than the applied change, the
deviation from reversibility may be negligible. In a reversible cycle, a cyclical reversible
process, the system and its surroundings will be returned to their original states if one half cycle
is followed by the other half cycle.

Thermodynamic processes can be carried out in one of two ways: reversibly or irreversibly.
Reversibility means the reaction operates continuously at quasiequilibrium. In an ideal
thermodynamically reversible process, the energy from work performed by or on the system
would be maximized, and that from heat would be zero. However, heat cannot fully be
converted to work and will always be lost to some degree (to the surroundings). (This is
true only in case of a cycle. In case of an ideal process, heat can be completely converted into
work, e.g., isothermal expansion of an ideal gas in a piston–cylinder arrangement.) The
phenomenon of maximized work and minimized heat can be visualized on a pressure–
volume graph as the area beneath the equilibrium curve, representing work done. In order
to maximize work, one must follow the equilibrium curve precisely

Figure 6.11 Figure 6.12

113

6.7.2 Irreversible Processes
Irreversible processes, on the other hand, are a result of straying away from the curve,

therefore decreasing the amount of overall work done, an irreversible process can be
described as a thermodynamic process that departs from equilibrium. Irreversibility is defined
as the difference between the reversible work and the actual work for a process. When
described in terms of pressure and volume, it occurs when the pressure (or the volume) of a
system changes so dramatically and instantaneously that the volume (or the pressure) does not
have time to reach equilibrium. A classic example of irreversibility is allowing a certain
volume of gas to be released into a vacuum. By releasing pressure on a sample and thus
allowing it to occupy a large space, the system and surroundings are not in equilibrium during the
expansion process and there is little work done. However, significant work will be required, with
a corresponding amount of energy dissipated as heat flow to the environment, in order to
reverse the process (compressing the gas back to its original volume and temperature)

Figure 6.13
(a) Heat transfer through a temperature difference is irreversible, and (b) the reverse process is

impossible.

114

6.7.3 Carnot Principle

 Carnot Principle: describes the limits of thermal efficiency for both reversible and irreversible
processes.
a) The heat efficiency of an irreversible heat engine is usually less than the efficiency of a
reversible heat engine.
b) All heat engines operating between 2 identical heat reservoirs have the same heat
efficiency. (for reversible process)

 Under certain circumstances, it is possible that one of them has a higher heat efficiency despite
receiving the same amount of heat from the heat reservoir. (working material type)

 Carnot's principle rejects this possibility (theory only) regardless of the internal factors of the heat
engine.

All reversible heat
engines operating
between the same two
reservoirs have the
same efficiency.

Figure 6.14 For reversible cycles, the heat
, transfer ratio QH /QL can be
replaced by the absolute
temperature ratio TH /TL.

 This temperature scale is called the Kelvin scale, and the temperatures on this scale are called
absolute temperatures.

115

6.7.4 Maximum performance of heat engine and reversible heat engine.

a) The cycle that operates in a reversible manner will have the highest performance.

b) From the definition of Kelvin temperature scale.

( ) =



c) Maximum heat efficiency of heat engine.

= −


d) Maximum performance coefficient.

=
−

=
−

Example :
A power cycle operating between 2 heat reservoirs receives heat, Qp from a hot heat

reservoir at a temperature of 2000K and removes heat to a cold heat reservoir at 400K. For each
of the following cases, determine whether the cycle operates in a reversible, irreversible or
impossible manner.

a. Qp = 1000kJ, W = 850kJ
b. Qp = 2000kJ, Qs = 400kJ
c. W = 1600kJ, Qs = 500kJ

116

Solution :

 In all the above cases, the temperature of the hot water reservoir and the temperature of
the cold heat reservoir are the same. Thus, the maximum efficiency of all power cycles is
the same, that is :-

= 1 − = 1 − 400 = 0.8
2000

 Actual efficiency is determined

a. = = 850 = 0.85 > (impossible)
1000

b. = 1 − = 1 − 400 = 0.8 = (reversible)
2000

c. Qp = W+Qs = 1600+500 = 2100kJ

= = 1600 = 0.76 < (Irreversible)
2100

117

Exercises
Question 1
State the principle of energy immortality of a closed system in the form of energy equations
Question 2
A closed system undergoes a thermodynamic cycle consisting of 2 processes. During the first
process, 30kJ of heat is supplied to the system while the system performs work of 50kJ. During
the second process, 35kJ of work is done on the system. Determine;
a) Heat transfer during the second process
b) Clean work
c) Total heat transfer
Question 3
A rigid tank contains 5kg of air at 100kPa and 27 ℃. The air is heated until the pressure increases
by 2 times. Determine the volume of the tank and the amount of heat transfer.
Question 4
Air undergoes 2 processes in series as below;
Process 1-2: polytropic compression with n = 1.3, than pressure p1 = 100kpa, v1 = 0.04 m3 / kg
to v2 = 0.02 m3 / kg
Process 2-3: constant pressure up to v3 = v1
Sketch the above processes on the p-v diagram and determine the amount of work per unit of air
mass.
Question 5
A rigid tank contains 5 kg of air at 100kPa and 27 ℃. The air is heated until the pressure increases
2 times. Determine the volume of the tank and the amount of heat transfer.

118

Answer

Question 1

System energy change = incoming energy - outgoing energy

Question 2

Process 1 Process 2

Q12 = 30 kJ Q21 = ?

W12 = 50 kJ W21 = -35 kJ

 Heat supplied (+ Q)

 Heat released (-Q)

 System does work (+ W)

 Work is done on the system (-W)

a) Heat transfer during the second process

12 + 21 = 12 + 21

21 = 12 + 21 − 21
21 = 50 + (−35) − 30 = −

b) Clean work
= 12 + 21 = 50 − 35 =

c) Total heat transfer
= 12 + 21 = 30 − 15 =

119

MODULES DEVELOPED TO COMPLETE THE BACHELOR PROJECT SPECIFICALLY FOR
STUDENTS MECHANICAL FACULTY OF TECHNICAL AND VOCATIONAL EDUCATION (FPTV)

WHO ARE TAKING THE THERMODYNAMIC THEORY. THE DEVELOPMENT OF THIS
MODULE IS THE RESULT OF PRELIMINARY SURVEY OF BACHELOR STUDENTS AT FACULTY

OF TECHNICAL AND VOCATIONAL EDUCATION WHO HAVE DIFFICULTY
UNDERSTANDING SOME OF THE SUBTOPICS OF THERMODYNAMICS THEREFORE, THE
CONTENT CONTAINED IN THIS MODULE DESCRIBES THE CALCULATION STEP BASED ON
THEIR RESPECTIVE SUBTOPICS AS WELL AS THE PREPARATION OF ACTIVITIES AT THE

END OF THE SUBTOPICS.

FAIQ RUDUAN