STUDY GUIDE
University of South Africa
Module Title:THERMODYNAMICS I
MODULE CODE: TDY2601
Department of Mechanical, Bioresources, and Biomedical
Engineering
STUDY GUIDE
TDY2601
OVERVIEW OF THE MODULE
Thermodynamics is an exciting and fascinating subject dealing with energy, which is
essential for substance of life. Thermodynamics has been an essential part of engineering
curricula worldwide for a long time. We assume that all students have an adequate
background in calculus and physics, since they need to understand the basic principles of the
subject to design, operate and maintain systems, such as motor vehicles and sea vessel
engines; steam and gas power plants, which generate electricity; aircraft engines; rocket
propulsion and military missile systems; refrigeration, air-conditioning and cryogenic
systems and so forth. It also explains natural phenomena such as weather systems, pollution,
life processes and so forth. In fact, thermodynamics is a ubiquitous subfield of physics
outside physics departments; and the thermodynamic laws, specifically the first and second
laws, are some of the most fundamental in nature. No process known to mankind has ever
violated these laws, whether it be science or humanities. Thermodynamics is an integral and
very important part of engineering courses; and students need to master these principles for
their future engineering career.
Prescribed and recommended text books:
Title:
Thermodynamics: An engineering approach (9th Edition in SI Units)
Author:
Yunus A. Cengel
Edition:
Any edition (Preferably the latest edition)
Publisher:
McGraw-Hill Publications
i
STUDY GUIDE
TDY2601
Contents
LEARNING UNIT 1: INTRODUCTION AND BASIC CONCEPTS .......................................... 1
1.1
LEARNING OUTCOMES ............................................................................................................ 1
1.2
THERMODYNAMICS AND ENERGY .......................................................................................... 1
1.3
IMPORTANCE OF DIMENSIONS AND UNITS ............................................................................ 2
1.4
DENSITY AND SPECIFIC GRAVITY ............................................................................................. 3
1.5
PROCESSES AND CYCLES ......................................................................................................... 3
1.6
TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS ............................................ 4
1.7
PRESSURE ................................................................................................................................ 6
1.8
THE MANOMETER ................................................................................................................... 8
1.9
THE BAROMETER AND ATMOSPHERIC PRESSURE .................................................................. 8
1.10
WORK ACTIVITIES .................................................................................................................... 9
1.11
SUMMARY ............................................................................................................................. 11
LEARNING UNIT 2: ENERGY TRANSPORT BY HEAT, WORK AND MASS ................... 13
2.1
LEARNING OUTCOMES .......................................................................................................... 13
2.2
FORMS OF ENERGY ............................................................................................................... 13
2.3
ENERGY TRANSFER BY HEAT ................................................................................................. 15
2.4
ENERGY TRANSFER BY WORK ............................................................................................... 17
2.5
ELECTRICAL WORK ................................................................................................................ 17
2.6
MECHANICAL FORMS OF WORK .......................................................................................... 18
2.7
THE FIRST LAW OF THERMODYNAMICS................................................................................ 18
2.8
MECHANISMS OF ENERGY TRANSFER (HEAT, WORK AND MASS FLOW)................................ 21
2.9
ENERGY CONVERSION EFFICIENCIES ..................................................................................... 22
2.10
ENERGY AND ENVIRONMENT ............................................................................................... 24
2.11
WORK ACTIVITIES .................................................................................................................. 25
2.12
SUMMARY ............................................................................................................................. 30
LEARNING UNIT 3: PROPERTIES OF PURE SUBSTANCES ............................................... 31
3.1
LEARNING OUTCOMES .......................................................................................................... 31
3.2
PURE SUBSTANCE.................................................................................................................. 31
3.3
PHASES-CHANGE PROCESSES OF PURE SUBSTANCE ............................................................ 32
3.4
PROPERTY DIAGRAMS FOR PHASE-CHANGE PROCESSES ..................................................... 34
3.5
PROPERTY TABLES ................................................................................................................. 36
3.6
THERMODYNAMICS PROCESSES ........................................................................................... 44
3.7
THE IDEAL GAS EQUATION OF STATE.................................................................................... 45
3.8
COMPRESSIBILITY FACTOR-A MEASURE OF DEVIATION FROM IDEAL GAS BEHAVIOUR ...... 47
ii
STUDY GUIDE
TDY2601
3.9
WORK ACTIVITIES .................................................................................................................. 48
3.10
SUMMARY ............................................................................................................................. 51
LEARNING UNIT 4: ENERGY ANALYSIS OF CLOSED SYSTEMS ...................................... 52
4.1
LEARNING OUTCOMES ......................................................................................................... 52
4.2
MOVING BOUNDARY WORK ................................................................................................. 52
4.3
ENERGY BALANCE FOR CLOSED SYSTEMS ............................................................................. 55
4.4
SPECIFIC HEATS ..................................................................................................................... 57
4.5
INTERNAL ENERGY, ENTHALPY AND SPECIFIC HEATS OF IDEAL GASES. ............................... 57
4.6
INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS AND LIQUIDS .................. 60
4.7 WORK ACTIVITIES ........................................................................................................................ 61
4.8
SUMMARY ............................................................................................................................. 66
LEARNING UNIT 5: MASS AND ENERGY ANALYSIS OF CONTROL VOLUMES ........ 68
5.1
LEARNING OUTCOMES ......................................................................................................... 68
5.2
CONSERVATION OF MASS ..................................................................................................... 68
5.3
FLOW WORK AND THE ENERGY OF A FLOWING FLUID ........................................................ 71
5.4
ENERGY ANALYSIS OF SOME STEADY-FLOW SYSTEMS ......................................................... 72
5.5
SOME STEADY FLOW ENGINEERING DEVICES ....................................................................... 74
5.6
ENERGY ANALYSIS OF UNSTEADY-FLOW PROCESSES ........................................................... 78
5.7
WORK ACTIVITIES ................................................................................................................. 79
5.8 SUMMARY ................................................................................................................................... 86
LEARNING UNIT 6: THE SECOND LAW OF THERMODYNAMICS ................................. 88
6.1 LEARNING OUTCOMES ................................................................................................................ 88
6.2
INTRODUCTION TO THE SECOND LAW ................................................................................ 88
6.3
THERMAL ENERGY RESERVOIRS ............................................................................................ 89
6.4
HEAT ENGINES ...................................................................................................................... 89
6.5
REFRIGERATORS AND HEAT PUMPS ..................................................................................... 92
6.6
THE CARNOT CYCLE .............................................................................................................. 93
6.7
THE CARNOT HEAT ENGINE .................................................................................................. 94
6.8
THE CARNOT REFRIGERATOR AND HEAT PUMP ................................................................... 95
6.9
WORK ACTIVITIES ................................................................................................................. 95
6.10
SUMMARY ............................................................................................................................. 98
LEARNING UNIT 7: ENTROPY .................................................................................................... 99
7.1
LEARNING OUTCOMES ......................................................................................................... 99
7.2
ENTROPY CHANGE OF PURE SUBSTANCES ........................................................................... 99
7.3
ISENTROPIC PROCESSES ........................................................................................................ 99
iii
STUDY GUIDE
TDY2601
7.4
ENTROPY CHANGE OF LIQUIDS AND SOLIDS ..................................................................... 100
7.5
THE ENTROPY CHANGE OF IDEAL GASES ............................................................................ 101
7.6
REVERSIBLE STEADY-FLOW WORK ..................................................................................... 101
7.7
ISENTROPIC EFFICIENCIES OF STEADY-FLOW DEVICES ....................................................... 102
7.8
ENTROPY BALANCE ............................................................................................................. 105
7.9
WORK ACTIVITIES ................................................................................................................ 106
7.10
SUMMARY .......................................................................................................................... 113
iv
STUDY GUIDE
TDY2601
LEARNING UNIT 1: INTRODUCTION AND BASIC CONCEPTS
1.1
LEARNING OUTCOMES
At the end of this unit, you should be able to:
•
Identify the unique terms associated with thermodynamics through the specific
definition of basic concepts to form a sound foundation for developing the principles
of thermodynamics.
1.2
•
Explain the metric SI and the English unit systems.
•
Discuss properties of a system, define density, specific gravity and specific weight.
•
Explain the concepts of temperature, temperature scales, pressure and absolute and
gauge pressure.
THERMODYNAMICS AND ENERGY
The study of thermodynamics is concerned with the ways in which energy is stored in a body;
and how energy transformations involving heat and work may take place. One of the most
fundamental laws of nature is the conservation of energy principle. It simply states that
during an energy interaction, energy can change from one form to another, but the total
amount of energy remains constant– energy cannot be created or destroyed.
Energy: The ability to cause changes.
The term "thermodynamics" stems from the Greek words therme (heat) and dynamis (power).
Conservation of energy principle: Energy can change from one form to another during an
interaction, but the total amount of energy remains constant.
Energy cannot be created or destroyed.
The first law of thermodynamics: An expression of the conservation of energy principle.
The first law asserts that energy is a thermodynamic property.
The second law of thermodynamics: It asserts that energy has quality as well as quantity;
and actual processes occur in the direction of decreasing quality of energy. For example, a
cup of hot coffee left on the table eventually cools down, but a cup of cool coffee in the same
room never gets hot by itself (figure 1.1).
On a larger scale, thermodynamics plays a major part in designing and analysing automotive
engines, rockets, jet engines, conventional or nuclear power plants, solar collectors; and
1
STUDY GUIDE
TDY2601
designing vehicles –and airplanes (figure 1.2). The energy-efficient home that one may be
living in, for example, is designed based on minimising heat loss in winter and heat gain in
summer. The size, location and the power input of the fan of your computer is also selected
after an analysis that involves thermodynamics.
Figure 1.1: Heat flows in the direction of decreasing temperature.
Figure 1.2: Application areas of thermodynamics.
1.3
IMPORTANCE OF DIMENSIONS AND UNITS
Dimensions can characterise any physical quantity.
Some basic dimensions such as mass m, length L, time t and temperature T are selected as
primary or fundamental dimensions, while others such as velocity V, energy E and volume
V are expressed in terms of the primary dimensions and are called secondary dimensions or
derived dimensions.
2
STUDY GUIDE
1.3.1
TDY2601
Unity conversion ratios
Like all non-primary dimensions that can be formed by suitable combinations of primary
dimensions, all non-primary units (secondary units) can be formed by combinations of
primary units. For example, force units can be expressed as:
𝑚
𝑁 = 𝑘𝑔 2
𝑠
1.4
DENSITY AND SPECIFIC GRAVITY
Density is defined as mass per unit volume (figure 1.3).
Density: 𝜌 =
𝑚
𝑉
(kg/m3)
Figure 1.3: Density is mass per unit volume; specific volume is volume per unit mass.
The reciprocal of density is the specific volume v, which is defined as volume per unit mass.
That is,
𝑣=
𝑉
𝑚
=
1
𝜌
Specific gravity or relative density is the ratio of the density of a substance to the density of
some standard substances at a specified temperature (usually water at 40C, the density of
which is 1000 kg/m3. That is, specific gravity: 𝑆𝐺 =
𝜌
𝜌𝑤𝑎𝑡𝑒𝑟
Specific weight is the weight of a unit volume of a substance.
Specific weight: 𝛾𝑠 = 𝜌𝑔
(N/m3),
Where g is the gravitational acceleration.
1.5
PROCESSES AND CYCLES
Process: Any change that a system undergoes from one equilibrium to another.
Path: The series of states through which a system passes during a process.
To fully describe a process, one should specify the initial and final states, the path it follows
and the interactions with the surroundings.
3
STUDY GUIDE
TDY2601
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.
•
Process diagrams plotted by employing thermodynamic properties as coordinates are
very useful in visualising the processes.
•
Some common properties that are used as coordinates are temperature T, pressure P and
volume V (or specific volume v).
•
The prefix iso- is often used to designate a process for which a particular property
remains constant.
•
Isothermal process: A process during which the temperature T remains constant.
•
Isobaric process: A process during which the pressure P remains constant.
•
Isochoric (or isometric) process: A process during which the specific volume v remains
constant.
•
Cycle: A process during which the initial and final states are identical.
Figure 1.4: The P-V diagram of a compression process.
1.6
•
TEMPERATURE AND THE ZEROTH LAW OF THERMODYNAMICS
The zeroth law of thermodynamics: If two bodies are in thermal equilibrium with a
third body, they are also in thermal equilibrium with each other.
•
By replacing the third body with a thermometer, the zeroth law can be restated because
two bodies are in thermal equilibrium if both have the same temperature reading
even if they are not in contact.
4
STUDY GUIDE
TDY2601
Figure 1.5: Two bodies reaching thermal equilibrium after being brought into contact in an
isolated enclosure.
1.6.1
Temperature scales
All temperature scales are based on some easily reproducible states, such as the freezing and
boiling points of water: ice point and steam point.
Ice point: A mixture of ice and water that is in equilibrium with air saturated with vapour at 1
atm pressure (0°C or 32°F).
Steam point: A mixture of liquid water and water vapour (with no air) in equilibrium at 1
atm pressure (100°C or 212°F).
Celsius scale: in SI unit system
Fahrenheit scale: an English unit system
Thermodynamic temperature scale: A temperature scale that is independent of the
properties of any substance.
Kelvin scale (SI) Rankine scale (E)
A temperature scale nearly identical to the Kelvin scale is the ideal-gas temperature scale.
The temperatures on this scale are measured, using a constant volume gas thermometer.
Figure 1.6: A constant volume gas thermometer would read 273.15°C at absolute zero
pressure.
The Kelvin scale is related to the Celsius scale by
T(K)=T(0C)+273.15
5
STUDY GUIDE
1.7
TDY2601
PRESSURE
Pressure is defined as a normal force exerted by a fluid per unit area.
1 Pa=1 N/m2
The pressure unit Pascal is too small for pressures encountered in practice. Therefore, its
multiples kilopascal (1 kPa=103 Pa) and megapascal (1MPa=106 Pa) are commonly used.
1 bar=105 Pa=0.1 MPa=100 kPa
1 atm=101,325 kPa=1.01325 bars
1 kgf/cm2=9.807 N/cm2=9.807 x 104 N/m2=9.807 x 104 Pa
1 kgf/cm2=0.9807 bar
1 kgf/cm2=0.9679 atm
1 atm=14.696 psi
1 MPa=106 Pa
The actual pressure at a given position is called the absolute pressure and it is measured
relative to absolute vacuum (absolute zero pressure); however, most devices measuring
pressure are calibrated to read zero in the atmosphere (figure 1.7); therefore, they indicate the
difference between the absolute pressure and the local atmospheric pressure. This difference
is called the gauge pressure. Pressures below the atmospheric pressure are called vacuum
pressures and are measured by vacuum gauges that indicate the difference between
atmospheric and absolute pressure.
Figure 1.7: Some basic pressure gauges.
Absolute, gauge and vacuum pressures are all positive quantities and are related to one
another as showed in figure 1.8.
𝑃𝑔𝑎𝑔𝑒 = 𝑃𝑎𝑏𝑠 − 𝑃𝑎𝑡𝑚
𝑃𝑣𝑎𝑐 = 𝑃𝑎𝑡𝑚 − 𝑃𝑎𝑏𝑠
6
STUDY GUIDE
TDY2601
Figure 1.8: Absolute, gauge and vacuum pressures.
1.7.1
Variation of pressure with depth
To obtain a relation for the variation of pressure with depth, consider a rectangular fluid
element of height ∆𝓏, length ∆𝑥 and unit depth (into the page) in equilibrium as shown in
figure 1.9. Assuming the density of the fluid 𝜌 to be constant, a force balance in the vertical
𝓏-direction gives:
∆𝑃 = 𝑃2 − 𝑃1 = 𝜌𝑔∆𝓏
Figure 1.9: Free body diagram of a rectangular fluid element in equilibrium.
If one takes point 1 to be a free surface of a liquid open to the atmosphere (figure 1.10) where
the pressure is the atmospheric pressure 𝑃𝑎𝑡𝑚 , then the pressure at a depth ℎ from the free
surface becomes:
𝑃 = 𝑃𝑎𝑡𝑚 + 𝜌𝑔ℎ or 𝑃 = 𝜌𝑔ℎ
Figure 1.10: Pressure in a liquid at rest increases linearly with distance from the free surface.
7
STUDY GUIDE
1.8
TDY2601
THE MANOMETER
A manometer is commonly used to measure small and moderate pressure differences. It
contains one or more fluids such as mercury, water, alcohol or oil.
Consider the manometer shown in figure 1.11 that is used to measure the pressure in the tank.
Since the gravitational effects of gases are negligible, the pressure anywhere in the tank and
at position 1 has the same value. Furthermore, since pressure in a fluid does not vary in the
horizontal direction within a fluid, the pressure at point 2 is the same as the pressure at
point1, P2=P1.
Figure 1.11: The basic manometer.
The pressure at any point can be determined by starting with a point of known pressure and
adding or subtracting 𝜌𝑔ℎ terms as one advance towards the point of interest.
1.9
THE BAROMETER AND ATMOSPHERIC PRESSURE
A device, called a barometer, measures atmospheric pressure; thus, the atmospheric pressure
is often referred to as barometric pressure.
A frequently used pressure unit is the standard atmosphere, which is defined as the pressure
produced by a column of mercury 760 mm in height at 0°C (Hg = 13,595 kg/m3) under
standard gravitational acceleration (g = 9.807 m/s2).
The Italian, Evangelista, Torricelli (1608-1647), was the first to conclusively prove that the
atmospheric pressure can be measured by inverting a mercury-filled tube into a mercury
container that is open to the atmosphere as shown in figure 1.12.
Figure 1.12: The basic barometer.
8
STUDY GUIDE
TDY2601
𝑃𝑎𝑡𝑚 = 𝜌𝑔ℎ
1.10
WORK ACTIVITIES
Activity 1.1:
Mass, force, and units
Question
What is the force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road
and (b) on an uphill road?
Answer
There is no acceleration; therefore, the net force is zero in both cases.
Question
A 3 kg rock is thrown upward with a force of 200 N at a location where the local
gravitational acceleration is 9.79 m/s2. Determine the acceleration of the rock, in m/s2.
Answer
Activity 1.2:
Systems, properties, state, and processes
Question
What is the difference between intensive and extensive properties?
Answer
Intensive properties do not depend on the size (extent) of the system, but extensive properties
do.
Question
Define the isothermal, isobaric and isochoric processes.
Answer
9
STUDY GUIDE
TDY2601
A process during which the temperature remains constant is called isothermal; a process
during which the pressure remains constant is called isobaric; and a process during which the
volume remains constant is called isochoric.
Activity 1.3:
Temperature
Question
What is the zeroth law of thermodynamics?
Answer
The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both
have the same temperature reading, even if they are not in contact.
Question
The temperature of a system rises by 450C during a heating process. Express this rise in
temperature in kelvins.
Answer
This problem deals with temperature changes which are identical in Kelvin and Celsius
scales. Thus
∆𝑇(𝐾) = ∆𝑇(0C)=45K
Activity 1.4 Pressure, manometer and barometer
Question
The pressure in a compressed air storage tank is 1500 kPa. What is the tank’s pressure in (a)
kN and m units; (b) kg, m, and s units; and (c) kg, km, and s units?
Answer
10
STUDY GUIDE
TDY2601
Question
The gauge pressure in a liquid at a depth of 3 m is read to be 42 kPa. Determine the gauge
pressure in the same liquid at a depth of 9m.
Answer
Question
A manometer containing oil (𝜌=850 kg/m3) is attached to a tank filled with air. If the oil level
difference between the two columns is 36 cm and the atmospheric pressure is 98 kPa,
determine the absolute pressure of the air in the tank.
Answer
1.11
•
SUMMARY
Thermodynamics and energy
✓ Application areas of thermodynamics
•
Importance of dimensions and units
✓ Some SI and English units, dimensional homogeneity and unity conversion
ratios
•
Density and specific gravity
•
Temperature and the zeroth law of thermodynamics
✓ Temperature scales
11
STUDY GUIDE
•
TDY2601
Pressure
✓ Variation of pressure with depth
•
The manometer and the atmospheric pressure
•
Work activities
12
STUDY GUIDE
TDY2601
LEARNING UNIT 2: ENERGY TRANSPORT BY HEAT,
WORK AND MASS
2.1
LEARNING OUTCOMES
At the end of this unit, you should be able to:
•
Identify the various forms of energy.
•
Explain the concept of heat and the terminology associated with energy transfer by
heat.
•
Explain the concept of work, including electrical work and several forms of
mechanical work.
•
Apply the first law of thermodynamics, energy balances and mechanisms of energy
transfer.
•
Identify energy conversion efficiencies.
•
Discuss the implications of energy conversion on the environment.
2.2
FORMS OF ENERGY
•
Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential,
electric, magnetic, chemical and nuclear; and their sum constitutes the total energy, E
of a system.
•
Thermodynamics deals only with the change of the total energy.
•
Macroscopic forms of energy: Those forms of energy a system possesses as a whole
regarding some outside frame of reference, such as kinetic and potential energies.
•
Microscopic forms of energy: Those forms of energy related to the molecular
structure of a system and the degree of the molecular activity.
•
Internal energy, U: The sum of all the microscopic forms of energy.
•
Kinetic energy, KE: The energy that a system possesses because of its motion
relative to some reference frame.
•
Potential energy, PE: The energy that a system possesses because of its elevation in
a gravitational field.
13
STUDY GUIDE
Kinetic energy:
TDY2601
KE = m
V2
2
ke =
Kinetic energy per unit mass:
(kJ)
V2
2
(kJ/kg)
Potential energy:
PE = mgz
(kJ)
Potential energy per unit mass:
pe = gz
(kJ/kg)
Total energy of a system:
V2
E = U + KE + PE = U + m
+ mgz
2
(kJ)
V2
+ gz
Energy of a system per unit mass: e = u + ke + pe = u +
2
Total energy per unit mass:
e=
E
m
(kJ/kg)
(kJ/kg)
Figure 2.1: Mass and energy flow rates associated with the flow of steam in a pipe of inner
diameter D with an average velocity of Vavg.
Mass flow rate:
𝑚̇ = 𝜌𝑣̇ = 𝜌𝐴𝑐 𝑉𝑎𝑣𝑔
(kg/s)
Energy flow rate:
𝐸̇ = 𝑚𝑒
̇
(kJ/s or kW)
14
STUDY GUIDE
2.2.1
TDY2601
Mechanical energy
Mechanical energy: The form of energy that can be converted to mechanical work
completely and directly by an ideal mechanical device, such as an ideal turbine.
Kinetic and potential energies: These are familiar forms of mechanical energy.
Mechanical energy of a flowing fluid per unit mass:
emech =
P
+

V2
+ gz
2
(kJ/kg)
Rate of mechanical energy of a flowing fluid:
P V2

E mech = m emech = m +
+ gz 
 2

.
.
.
(J)
Mechanical energy change of a fluid during incompressible flow per unit mass:
emech =
P2 − P1

V − V1
+ 2
+ g (z 2 − z1 )
2
2
2
(kJ/kg)
Rate of mechanical energy change of a fluid during incompressible flow:
 P2 − P1 V2 2 − V1 2

 E mech = m emech = m
+
+ g (z 2 − z1 )
2
 

.
2.3
.
.
(kW)
ENERGY TRANSFER BY HEAT
Energy can cross the boundary of a closed system in two distinct forms: heat and work as
shown in figure 2.2.
Heat: The form of energy that is transferred between two systems (or a system and its
surroundings) by a temperature difference.
15
STUDY GUIDE
TDY2601
Figure 2.2: Energy can cross the boundaries of a closed system in the form of heat and work.
Figure 2.3: Temperature difference is the driving force for heat transfer. The larger the
temperature difference, the higher the rate of heat transfer.
Heat transfer per unit mass:
q=
Q
m
(kJ/kg)
The amount of heat transfer when the heat transfer rate is constant:
.
Q = Q t
(kJ)
The amount of heat transfer when the rate of heat transfer changes with time:
.
Q =  t1 2 Q dt
t
(kJ)
A process during which there is no heat transfer is called an adiabatic process (figure 2.4).
The word adiabatic comes from the Greek word adiabatos, which means not to be passed.
There are two ways in which a process can be adiabatic: The system is well insulated so that
only a negligible amount of heat can pass through the boundary; or the system and its
surroundings are at the same temperature; therefore, there is no driving force (temperature
difference) for heat transfer.
16
STUDY GUIDE
TDY2601
Figure 2.4: Adiabatic process, a system exchanges no heat with its surroundings.
2.4
•
ENERGY TRANSFER BY WORK
Work: The energy transfer associated with a force acting through a distance.
✓ A rising piston, a rotating shaft and an electric wire crossing the system boundaries are all
associated with work interactions
•
Formal sign convention: Heat transfer to a system and work done by a system are
positive; heat transfer from a system and work done on a system are negative.
•
Alternative to sign convention is to use the subscripts in and out to indicate direction. This is
the primary approach in this text.
Work done per unit mass:
w=
W
m
(kJ/kg)
Figure 2.5: Specifying the directions of heat and work.
2.5
ELECTRICAL WORK
In an electric field, electrons in a wire move under the effect of electromotive forces, doing
work. When N coulombs of electrical charge move through a potential difference V, the
electrical work done is
𝑊𝑒 = 𝑽𝑁
Which can also be expressed in the rate form as
𝑊̇𝑒 = 𝑉𝐼
(W)
17
STUDY GUIDE
TDY2601
In figure 2.6, 𝑊̇𝑒 is the electrical power and I the number of electrical charges flowing per
unit time.
Figure 2.6: Electrical power in terms of resistance R, current I, and potential difference V.
When V and I remain constant during the time interval ∆𝑡 , the currents reduce to
𝑊̇𝑒 = 𝑉𝐼∆𝑡
2.6
(kJ)
MECHANICAL FORMS OF WORK
There are various ways of doing work, each in some way related to a force acting through a
distance as shown in figure 2.7.
•
There are two requirements for a work interaction to exist between a system and its
surroundings:
✓ There must be a force acting on the boundary.
✓ The boundary must move.
Work Done = Force x Distance
W = Fs
(kJ)
Figure 2.7: The work done is proportional to the force applied (F) and the distance travelled
(s).
2.7
•
THE FIRST LAW OF THERMODYNAMICS
The first law of thermodynamics (the conservation of energy principle) provides a
sound basis for studying the relationships among the various forms of energy and energy
interactions.
18
STUDY GUIDE
•
TDY2601
The first law states that 'energy can be neither created nor destroyed during a process; it
can only change forms'.
We all know that a rock at some elevation possesses some potential energy; and part of this
potential energy is converted to kinetic energy as the rock falls (figure 2.8). Experimental
data show that the decrease in potential energy (mg∆𝑧) exactly equals the increase in kinetic
energy [m(V22-V12)/2] when the air resistance is negligible; thus, confirming the conservation
of energy principle for mechanical energy.
We consider some processes that involve heat transfer but no work interactions. A potato that
is baked in the oven is a good example of this (figure 2.9). Because of heat transfer to the
potato, the energy of the potato will increase. If we disregard any mass transfer (moisture loss
from the potato), the increase in the total energy of the potato becomes equal to the amount of
heat transfer. If 5 kJ of heat is transferred to the potato, the energy increase of the potato will
also be 5 kJ.
The first Law: For all adiabatic processes between two specified states of a closed system,
the net work done is the same irrespective of the nature of the closed system and the details of
the process.
Figure 2.8: Energy cannot be created or destroyed; it can only change forms.
Figure 2.9: The increase in the energy of a potato in an oven is equal to the amount of heat
transferred to it.
Figure 2.10: In the absence of any work interactions, the energy change of a system is equal
to the net heat transfer.
19
STUDY GUIDE
TDY2601
Figure 2.11: The work (electrical) done on an adiabatic system is equal to the increase in the
energy of the system.
Figure 2.12: The work (shaft) done on an adiabatic system is equal to the increase in the
energy of the system.
2.7.1
Energy balance
The net change (increase or decrease) in the total energy of the system during a process is
equal to the difference between the total energy entering and the total energy leaving the
system
during that process.
𝐶ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑡ℎ𝑒 𝑡𝑜𝑡𝑎𝑙
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔
𝑇𝑜𝑡𝑎𝑙 𝑒𝑛𝑒𝑟𝑔𝑦 𝑙𝑒𝑎𝑣𝑖𝑛𝑔
(
)−(
)=(
)
𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
𝑒𝑛𝑒𝑟𝑔𝑦 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚
𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = ∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚
Figure 2.13: The work (boundary) done on an adiabatic system is equal to the increase in the
energy of the system
20
STUDY GUIDE
TDY2601
Figure 2.14: The energy change of a system during a process is equal to the net work and
heat transfer between the system and its surroundings
Energy change of a system, Esystem
2.7.2
Energy change=Energy at final state-Energy at initial state
Esystem = E final − Einitial = E2 − E1
E = U + KE + PE
U = m(u 2 − u1 )
KE =
(
1
m V 2 2 − V 21
2
)
PE = mg ( z 2 − z1 )
Internal, kinetic, and potential energy changes.
2.8
MECHANISMS OF ENERGY TRANSFER (HEAT, WORK AND MASS FLOW)
𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 = (𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡 ) + (𝑊𝑖𝑛 − 𝑊𝑜𝑢𝑡 ) + (𝐸𝑚𝑎𝑠𝑠,𝑖𝑛 − 𝐸𝑚𝑎𝑠𝑠,𝑜𝑢𝑡 ) = ∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚
Heat transfer:
.
Q = Q t
Work transfer:
.
W = W t
21
STUDY GUIDE
TDY2601
 dE 
E =  t
 dt 
Mass flow:
Figure 2.15: The energy content of a control volume can be changed by mass flow and by
heat and work interactions.
2.9
ENERGY CONVERSION EFFICIENCIES
Efficiency is one of the most frequently used terms in thermodynamics; and it indicates how
well an energy conversion or transfer process is accomplished. "Efficiency" is also one the
most frequently misused terms in thermodynamics and a source of misunderstandings.
Efficiency, in general, can be expressed in terms of the desired output and the required input
as:
Efficiency =
Desired output
Required input
Combustion efficiency =
Q
HV
=
Amount of heat released during combustion
Heating value of the fuel burned
A combustion efficiency of 100% indicates that the fuel is burned completely; and that the
stack gases leave the combustion chamber at room temperature; therefore, the amount of heat
released during the combustion process is equal to the heating value of the fuel.
Overall efficiency of a power plant:
ηoverall =
Ẇnet,electric
̇
HV X mnet
•
Generator: A device that converts mechanical energy to electrical energy.
•
Generator efficiency: The ratio of the electrical power output to the mechanical
power input.
•
Thermal efficiency of a power plant: The ratio of the net electrical power output to
the rate of fuel energy input.
22
STUDY GUIDE
2.9.1
TDY2601
Efficiencies of mechanical and electrical devices
The transfer of mechanical energy is usually accomplished by a rotating shaft; therefore,
mechanical work is often referred to as shaft work. A pump or a fan receives shaft work
(usually from an electric motor) and transfers it to the fluid as mechanical energy (less
frictional losses), while a turbine converts the mechanical energy of a fluid to shaft work. In
the absence of any irreversibility, such as friction, mechanical energy can be converted
entirely from one mechanical form to another; and the mechanical efficiency of a device or
process can be defined as follows:
ηmech =
Mechanical energy output
Mechanical energy input
=
Emech,out
Emech,in
E
= 1 − mech,loss
Emech,in
The degree of perfection of the conversion process between the mechanical work supplied or
extracted and the mechanical energy of the fluid is expressed by the pump efficiency and
turbine efficiency, defined as:
ηpump =
Mechanical energy increase of the fluid
Mechanical energy input
Ẇpump,u
∆Ėmech,fluid
=
Ẇshaft,in
Ẇpump
=
∆𝐸̇𝑚𝑒𝑐ℎ,𝑓𝑙𝑢𝑖𝑑 = 𝐸̇𝑚𝑒𝑐ℎ,𝑜𝑢𝑡 − 𝐸̇𝑚𝑒𝑐ℎ,𝑖𝑛
ηturbine =
Mechanical energy output
Mechanical energy decrease of the fluid
Ẇ
= |∆Ė shaft,out | =
mech,fluid
Ẇturbine
Ẇturbine,ℯ
The mechanical efficiency should not be confused with the motor efficiency and the
generator efficiency:
Motor efficiency:
ηmotor =
Mechanical power output
Electric power input
=
Ẇshaft,out
Ẇelect,in
Generator efficiency:
η𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟 =
Electric power output
Mechanical power input
=
Ẇelect,out
Ẇshaft,in
A pump is usually packaged with its motor and a turbine with its generator; therefore, we are
usually interested in the combined or overall efficiency of pump-motor and turbinegenerator combinations as shown in figure 2.16 defined as:
Pump motor’s overall efficiency
ηpump−motor = ηpump ηmotor =
Ẇpump,u
∆Ė
= mech,fluid
Ẇelect,in
Ẇelect,in
23
STUDY GUIDE
TDY2601
Turbine-generator overall efficiency:
ηturbine−gen = η𝑡𝑢𝑟𝑏𝑖𝑛𝑒 ηgenerator =
Ẇelect,out
Ẇ
= |∆𝐸̇ elect,out |
Ẇturbine,ℯ
𝑚𝑒𝑐ℎ,𝑓𝑙𝑢𝑖𝑑
Figure 2.16: The overall efficiency of a turbine generator is the product of the efficiency of
the turbine and the efficiency of the generator and represents the fraction of the mechanical
energy of the fluid converted to electric energy.
All the efficiencies defined above range between 0% and 100%. The lower limit of 0%
corresponds to the conversion of the entire mechanical or electric energy input to thermal
energy and the device, in this case, functions like a resistance heater. The upper limit of
100% corresponds to the case of perfect conversion with no friction or other irreversibility;
therefore, no conversion of mechanical or electric energy to thermal energy.
2.10
•
ENERGY AND ENVIRONMENT
The conversion of energy from one form to another often affects the environment and the
air we breathe in many ways; therefore, the study of energy is not complete without
considering its impact on the environment.
•
Pollutants emitted during the combustion of fossil fuels are responsible for ozone and
smog, acid rain and global warming.
The environmental pollution has reached such high levels that it has become a serious threat
to vegetation, wild life and human health.
Ozone: It irritates the eyes and damages the air sacs in the lungs where oxygen- carbon
dioxide exchange takes place, causing eventual hardening of this soft spongy tissue.
Smog: Smog is primarily made up of ground level ozone (O3), but it also contains numerous
other chemicals among which is carbon monoxide (CO); particulate matter, such as soot and
dust; volatile organic compounds (VOCs), such as benzene and butane and other
hydrocarbons.
24
STUDY GUIDE
TDY2601
Acid rain: The acid-laden droplets that can be as acidic as lemon juice, that are washed from
the air to the soil by rain or snow are known as acid rain.
Global warming or global climate change: The greenhouse effect makes life on earth
possible by keeping the earth warm. However, excessive amounts of these gases disturb the
delicate balance by trapping too much energy, which causes the average temperature of the
earth to rise and the climate at some localities to change. These undesirable consequences of
the greenhouse effect are referred to as global warming or global climate change.
2.11
WORK ACTIVITIES
Activity 2.1:
Forms of energy
Question
What is mechanical energy? How does it differ from thermal energy? What are the forms of
mechanical energy of a fluid stream?
Answer
Mechanical energy is the form of energy that can be converted to mechanical work
completely and directly by a mechanical device, such as a propeller. It differs from thermal
energy in that thermal energy cannot be converted to work directly and completely. The
forms of mechanical energy of a fluid stream are kinetic, potential, and flow energies.
Question
A water jet that leaves a nozzle at 60 m/s at a flow rate of 120 kg/s is to be used to generate
power by striking the buckets located on the perimeter of the wheel. Determine the power
generation potential of this water jet.
Answer
25
STUDY GUIDE
Activity 2.2:
TDY2601
Energy transfer by heat and work
Question
Consider an electric refrigerator located in a room. Determine the direction of the work and
heat interactions (in or out) when the following are taken as the system: (a) the contents of
the refrigerator, (b) all parts of the refrigerator including the contents; and (c) everything in
the room during a winter day.
Answer
(a) From the perspective of the contents, heat must be removed to reduce and maintain the
temperature of the contents. Heat is also added to the contents by the room temperature,
since the room temperature is hotter than the contents.
(b) Considering the system formed by the refrigerator box when the doors are closed, there
are three interactions, namely electrical work and two heat transfers. There is a transfer of
heat from the room temperature to the refrigerator through its walls. There is also a
transfer of heat from the hot portions of the refrigerator (back of the compressor where the
condenser is placed) system to the room temperature. Finally, electrical work is added to
the refrigerator through the refrigeration system.
(c) Heat is transferred through the walls of the room from the warm room air to the cold
winter air. Electrical work is done in the room through the electrical wiring leading into
the room.
Question
What is an adiabatic process? What is an adiabatic system?
26
STUDY GUIDE
TDY2601
Answer
An adiabatic process is a process during which there is no heat transfer. A system that does
not exchange any heat with its surroundings is an adiabatic system.
Activity 2.3 Mechanical forms of work
Question
A motor vehicle is accelerated from rest to 85 km/h in ten seconds. Would the energy
transferred to the motor vehicle be different if it were accelerated to the same speed in five
seconds?
Answer
The work done is the same, but the power is different
Question
Determine the energy required to accelerate an 800 kg motor vehicle from rest to 100 km/h
on a level road.
Answer
Analysis The work needed to accelerate a body and the change in kinetic energy of the body:
Activity 2.4:
The first law of thermodynamics
Question
On a hot summer day, a student turns on his or her fan when leaving his or her room in the
morning. Will the room be warmer or cooler than the neighbouring rooms when he or she
returns in the evening? Why? Assume all the doors and windows are kept closed.
Answer: Warmer, because energy is added to the room temperature in the form of electrical
work.
Question
A fan is used to accelerate quiescent air to a velocity of 8 m/s at a rate of 9 m3/s. Determine
the minimum power that must be supplied to the fan. Take the density of air to be 1.18 kg/m3.
Answer
Assumptions: The fan operates steadily.
Properties: The density of air is given to be ρ = 1.18 kg/m3.
Analysis: A fan transmits the mechanical energy of the shaft (shaft power) to mechanical
energy of air (kinetic energy). For a control volume that encloses the fan, the energy balance
can be written as:
27
STUDY GUIDE
TDY2601
Discussion: The conservation of energy principle requires the energy to be conserved as it is
converted from one form to another; and it does not allow any energy to be created or
destroyed during a process. The power required will be considerably higher, because of the
losses associated with the conversion of mechanical shaft energy to kinetic energy of air.
Activity 2.5:
Energy conversion efficiencies
Question: Define turbine efficiency, generator efficiency and combined turbine-generator
efficiency.
Answer:
Question
A 90-hp (shaft output) electric motor vehicle is powered by an electric motor mounted in the
engine compartment. If the motor has an average efficiency of 91%, determine the rate of
heat supply by the motor to the engine compartment at full load.
Answer
Assumptions: The motor operates at full load so that the load factor is 1.
Analysis: The heat gene rated by a motor is due to its inefficiency and is equal to the
difference between the electrical energy it consumes and the shaft power it delivers:
28
STUDY GUIDE
TDY2601
Discussion: Note that the electrical energy that is not converted to mechanical power is
converted to heat.
Activity 2.6:
Energy and environment
Question
How does energy conversion affect the environment? What are the primary chemicals that
pollute the air? What is the primary source of these pollutants?
Answer
Energy conversion pollutes the soil, the water and the air; and environmental pollution is a
serious threat to vegetation, wild life and human health. The emissions emitted during the
combustion of fossil fuels are responsible for smog, acid rain, global warming and climate
change. The primary chemicals that pollute the air are hydrocarbons (HC, also referred to as
volatile organic compounds, VOC), nitrogen oxides (NOx) and carbon monoxide (CO). The
primary source of these pollutants is the motor vehicles.
Question
What is the greenhouse effect? How does the excess CO2 gas in the atmosphere cause the
greenhouse effect? What are the potential long-term consequences of the greenhouse effect?
How can one combat this problem?
Answer
Carbon dioxide (CO2), water vapour and trace amounts of some other gases, such as methane
and nitrogen oxides, act like a blanket and keep the earth warm at night by blocking the heat
radiated from the earth. This is known as the greenhouse effect. The greenhouse effect
makes life on earth possible by keeping the earth warm. But excessive amounts of these gases
disturb the delicate balance by trapping too much energy, which causes the average
temperature of the earth to rise and the climate at some localities to change. These
undesirable consequences of the greenhouse effect are referred to as global warming or
global climate change. The greenhouse effect can be reduced by reducing the net production
of CO2; and consuming less energy (for example, by buying energy efficient cars and
appliances and planting trees).
29
STUDY GUIDE
2.12
•
TDY2601
SUMMARY
Forms of energy
✓ Macroscopic = kinetic + potential
✓ Microscopic = Internal energy (sensible + latent + chemical + nuclear)
•
Energy transfer by heat
•
Energy transfer by work
•
Mechanical forms of work
•
The first law of thermodynamics
✓ Energy balance
✓ Energy change of a system
✓ Mechanisms of energy transfer (heat, work, mass flow)
•
Energy conversion efficiencies
✓ Efficiencies of mechanical and electrical devices (turbines, pumps)
•
Energy and environment
✓ Ozone and smog
✓ Acid rain
✓ The Greenhouse effect: global warming
•
Work activities
30
STUDY GUIDE
TDY2601
LEARNING UNIT 3: PROPERTIES OF PURE SUBSTANCES
3.1
LEARNING OUTCOMES
At the end of this unit you should be able to:
3.2
•
Define pure substance.
•
Discuss the physics of phase change processes.
•
Demonstrate the procedures for determining thermodynamics properties of pure
substances from tables of property data.
•
Describe the hypothetical substance ideal gas and ideal gas equation of state.
•
Apply the ideal gas equation of state in the solution of typical problems.
•
Introduce the compressibility factor, which accounts for the deviation of real gases
from ideal gas behaviour.
PURE SUBSTANCE
Pure substance: A substance that has a fixed chemical composition throughout, for example
water, nitrogen, helium and carbon dioxide that are all pure substances.
Air is a mixture of several gases, but it is also considered to be a pure substance.
Figure 3.1: Nitrogen and gaseous air are pure substances.
Figure 3.2: A mixture of liquid and gaseous water is a pure substance, but a mixture of liquid
and gaseous air is not.
31
STUDY GUIDE
3.3
TDY2601
PHASES - CHANGE PROCESSES OF PURE SUBSTANCE
Compressed liquid (subcooled liquid): A liquid substance that it is not about to vaporise.
Saturated liquid: A liquid that is about to vaporise.
Saturated vapour: A vapour that is about to condense
Superheated vapour: A vapour that is not about to condense (not a saturated vapour).
Property diagrams for phase-change processes:
Pure substance states that the equilibrium state can be determined by specifying any two
independent intensive properties.
Consider a piston-cylinder device containing liquid water at 200C and 1atm pressure (state 1).
Under these conditions, water exists in the liquid phase and it is called a compressed or a
subcooled liquid, which means that it is not about to vaporise.
Process 1-2:
The temperature and specific volume will increase from a compressed liquid or subcooled
liquid, state 1 to the saturated liquid, state 2. In the compressed liquid region, the properties
of the liquid are approximately equal to the properties of the saturated liquid state at that
temperature.
Process 2-3:
At state 2, the liquid has reached a temperature at which it begins to boil, called the saturation
temperature and is said to exist as a saturated liquid.
Properties at a saturated liquid state are noted by the subscript f and v2 = vf. During the phase
change both the temperature and pressure remain constant (water boils at 100°C when the
pressure is 1 atmosphere or 101.325 kPa).
At state 3 the liquid and vapour phase are in equilibrium; and any point on the line between
states 2 and 3 has the same temperature and pressure.
32
STUDY GUIDE
TDY2601
Process 3-4:
A saturated vapour exists at state 4 and vaporisation is complete. The subscript g will always
denote a saturated vapour state. Note v4 = vg.
Thermodynamic properties at the saturated liquid state and saturated vapour state are
reflected as the saturated temperature table in table A-4 and as the saturated pressure table in
table A-5. These tables contain the same information. In table A-4 the saturation temperature
is the independent property, and in table A-5 the saturation pressure is the independent
property. The saturation pressure is the pressure at which phase change will occur for a given
temperature. In the saturation region the temperature and pressure are dependent properties; if
one is known, then the other is automatically known.
Process 4-5:
If the constant pressure heating is continued, the temperature will begin to increase above the
saturation temperature, 100°C in this example; and the volume also increases. State 5 is
called a superheated state, because T5 is greater than the saturation temperature for the
pressure and the vapour is not about to condense. Thermodynamic properties for water in the
superheated region are found in the superheated steam table, table A-6.
This constant pressure heating process is illustrated in the following figure.
33
STUDY GUIDE
TDY2601
Figure 3.3: T-v diagram for the heating process of water at constant pressure
3.4
PROPERTY DIAGRAMS FOR PHASE-CHANGE PROCESSES
If all the saturated liquid states are connected, the saturated liquid line is established. If all the
saturated vapour states are connected, the saturated vapour line is established. These two
lines intersect at the critical point and form what is often called the “steam dome.” The
following terms are used when referring to the region between the saturated liquid line and
the saturated vapour line: "saturated liquid-vapour mixture region", "wet region (a mixture of
saturated liquid and saturated vapour)" "two-phase region and just" the saturation region".
Notice that the trend of the temperature following a constant pressure line is to increase with
increasing volume; and the trend of the pressure following a constant temperature line is to
decrease with increasing volume.
Figure 3.4: T-v diagram of constant pressure phase-change processes of a pure substance at
various pressures (numerical values are for water)
34
STUDY GUIDE
TDY2601
Figure 3.5: T-v diagram of a pure substance
Figure 3.6: P-v diagram of a pure substance
Figure 3.7: P-T diagram of pure substances
35
STUDY GUIDE
TDY2601
Figure 3.8: P-v-T surface of a substance that expands on freezing (like water)
3.5
PROPERTY TABLES
In addition to the temperature, pressure and volume data, thermodynamic property tables
contain the data for the specific internal energy u, enthalpy h and entropy s. Enthalpy is a
combination of the properties of energy; it combines thermal internal energy (explained
earlier) and flow energy. The latter is energy of a substance by its internal pressure. As the
molecules of a substance move internally, they exert a force on a normal surface within or at
the boundaries of the system due to molecular collision with the surface or boundary. If this
boundary is free to move, energy in the form of mechanical work is normally transferred out
of the system. If there is an opening at the boundary, the substance will flow out if it is a
fluid, because of the pressure difference with the surroundings.
It can be proved that the specific flow energy is the product of pressure and specific volume
so that the fluid enthalpy is given by using:
H = U + PV
The enthalpy per unit mass is as follows:
h = u + Pv
We will find that the enthalpy h is quite useful when calculating the energy of mass streams
flowing into and out of control volumes. The enthalpy is also useful in the energy balance
during a constant pressure process for a substance contained in a closed piston-cylinder
36
STUDY GUIDE
TDY2601
device. The enthalpy has units of energy per unit mass, kJ/kg. The entropy s is a property
defined by the second law of thermodynamics and is related to the heat transfer to a system
divided by the system temperature; thus, the entropy has units of energy divided by
temperature.
3.5.1
Saturated liquid and saturated vapour states
Since temperature and pressure are dependent properties using the phase change, two tables
are given for the saturation region. Table A-4 indicates temperature as the independent
property; and table A-5 indicates pressure as the independent property. These two tables
contain the same information and often only one table is given.
Table A-4: Saturated water-temperature table
37
STUDY GUIDE
TDY2601
Table A-5: Saturated water-pressure table
Saturation pressure is the pressure at which the liquid and vapour phases are in equilibrium
at a given temperature.
Saturation temperature is the temperature at which the liquid and vapour phases are in
equilibrium at a given pressure.
The subscript fg used in tables A-4 and A-5 refers to the difference between the saturated
vapour value and the saturated liquid value region.
The quantity hfg is called the enthalpy of vaporisation (or latent heat of vaporisation). It
represents the amount of energy needed to vaporise a unit or mass of saturated liquid at a
given temperature or pressure. It decreases as the temperature or pressure increases and
becomes zero at the critical point.
38
STUDY GUIDE
3.5.2
TDY2601
Quality and saturated liquid-vapour mixture
Now, let us review the constant pressure heat addition process for water shown in figure 3.9.
Since state 3 is a mixture of saturated liquid and saturated vapour, how do we locate it on the
T-v diagram? To establish the location of state 3 a new parameter called, the quality x, is
defined as:
The quality is zero for the saturated liquid and one for the saturated vapour (0 ≤ x ≤ 1). The
average specific volume at any state 3 is given in terms of the quality as follows. Consider a
mixture of saturated liquid and saturated vapour. The liquid has a mass mf and occupies a
volume Vf. The vapour has a mass mg and occupies a volume Vg.
Figure 3.9: The relative amounts of liquid and vapour phases in a saturated mixture specified
by the quality x and a two-phase system of homogeneous mixture.
We note
V=Vf + Vg
mtotal =mf + mg
V=mv, Vf=mfvf, Vg=mgvg
mv=mfvf + mgvg
recalls the definition of quality x
then
39
STUDY GUIDE
TDY2601
Note, quantity 1- x is often given the name moisture. The specific volume of the saturated
mixture becomes
𝑣 = (1 − 𝑥)𝑣𝑓 + 𝑥𝑣𝑔
The form that we use most often is
𝑣 = 𝑣 𝑓 + 𝑥(𝑣𝑔 − 𝑣𝑓 )
It is noted that the value of any extensive property per unit mass in the saturation region is
calculated from an equation which has a form similar to that of the above equation. Let Y be
any extensive property and let y be the corresponding intensive property, Y/m, then
The term yfg is the difference between the saturated vapour and the saturated liquid values of
the property y which can be replaced by any of the variables v, u, h, or s.
We often use the above equation to determine the quality x of a saturated liquid-vapour state.
The following application is called the Lever Rule:
3.5.3
Superheated vapour
A substance is said to be superheated if the given temperature is greater than the saturation
temperature for the given pressure.
In superheated water table A-6, temperature and pressure are the independent properties. The
value of temperature to the right of the pressure is the saturation temperature for the pressure.
The first entry in the table is the saturated vapour state at the pressure.
40
STUDY GUIDE
TDY2601
Figure 3.10: Format of the superheated vapour table
Table A-6: Superheated water vapour
3.5.4
Compressed liquid water table
Compressed liquid tables are not as commonly available; and table A-7 is the only
compressed liquid table in this text. The format of table A-7 is very much like the format of
the superheated water tables. One reason for the lack of compressed liquid data is the relative
independence of compressed liquid property pressure.
Data for water compressed liquid states are found in the compressed liquid table, table A-7.
Table A-7 is arranged like table A-6, except for the saturation states that are the saturated
liquid states. Note that the data in table A-7 begin at 5 MPa or 50 times atmospheric pressure.
41
STUDY GUIDE
TDY2601
Table A-7: Compressed liquid water
At pressures below 5 MPa for water, the data are approximately equal to the saturated liquid
data at the given temperature. We approximate intensive parameter y, that is v, u, h and s
data as
𝑦 ≅ 𝑦𝑓@ 𝑇
The enthalpy is more sensitive to variations in pressure; therefore, at high pressures the
enthalpy can be approximated by
ℎ ≅ ℎ𝑓 @𝑇 + 𝑣𝑓 (𝑃 − 𝑃𝑠𝑎𝑡 )
For our work, the compressed liquid enthalpy may be approximated by
ℎ ≅ ℎ𝑓 @𝑇
3.5.5
Saturated ice water vapour table
When the temperature of a substance is below the triple point temperature, the saturated solid
and liquid phases exist in equilibrium. Here we define the quality as the ratio of the mass that
is vapour to the total mass of solid and vapour in the saturated solid vapour mixture. The
process of changing directly from the solid phase to the vapour phase is called sublimation.
Data for saturated ice and water vapour are given in table A-8. In table A-8, the term
"sublimation" refers to the difference between the saturated vapour value and the saturated
solid value.
42
STUDY GUIDE
TDY2601
Table A-8: Saturated ice water vapour
The specific volume, internal energy, enthalpy and entropy for a mixture of saturated ice and
saturated vapour are calculated similarly to that of saturated liquid vapour mixtures:
where the quality x of a saturated ice vapour state is
3.5.6
Quality of the working Substance
A pure substance is one, which is homogeneous and chemically stable. Thus, it can be a
single substance which is present in more than one phase, for example, liquid water and water
vapour contained in a boiler in the absence of any air or dissolved gases.
Phase – Is the state of the substance, such as solid liquid or gas.
Mixed phase–It is possible that phases may be mixed: ice + water, water + vapour and so
forth.
Quality of a mixed phase or dryness fraction (x):
43
STUDY GUIDE
TDY2601
The dryness fraction is defined as the ratio of the mass of pure vapour present to the total
mass of the mixture (liquid and vapour; say 0.9 dry for example). The quality of the mixture
may be defined as the percentage dryness of the mixture (90% dry).
Saturated state – A saturated liquid is a vapour, the dryness fraction of which is equal to
zero. A saturated vapour has a quality of 100% or a dryness fraction of one.
Superheated vapour – A gas is described as superheated when its temperature at a given
pressure is greater than the saturated temperature at the specific pressure; that is, the gas has
been heated beyond its saturation temperature.
Degree of superheat – The difference between the actual temperature of a given vapour and
the saturation temperature of the vapour at a given pressure.
Sub cooled liquid – A liquid is described as undercooled when its temperature at a given
pressure is lower than the saturated temperature at the specific pressure; that is, the liquid has
been cooled below its saturation temperature.
Degree of undercooling – The difference between the saturation temperature and the actual
temperature of the liquid is a given pressure.
Triple point – A state point in which all solid, liquid and vapour phases coexist in
equilibrium.
Critical Point – A state point at which transitions between liquid and vapour phases are not
clear.
3.6
THERMODYNAMICS PROCESSES
A process is a path in which the state of the system changes and some properties vary from
their original values. There are six types of processes associated with thermodynamics:
Adiabatic: no heat transfer from or to the fluid
Isothermal: no change in the temperature of the fluid
Isobaric: no change in the pressure of the fluid
Isochoric: no change in the volume of the fluid
Isentropic: no change in the entropy of the fluid
Isenthalpic: no change in the enthalpy of the fluid
3.6.1
Thermodynamics of working fluids
The behaviour of the working substance is a very essential factor in understanding
thermodynamics. In this chapter the emphasis is on pure substances, such as gases and steam
44
STUDY GUIDE
TDY2601
properties, and how they are interrelated. These relations are important in the design and
operation of thermal systems.
The ideal gas equation is a very well-known approximation when relating thermal properties
for a state point, or during a process. However, not all gases are perfect; and even the same
gas may behave as an ideal gas under certain circumstances and then changes into non-ideal,
or real under different conditions. There are other equations or procedures to deal with such
conditions. Steam or water vapour is not governed by simple equations but properties of
water and steam.
3.7
THE IDEAL GAS EQUATION OF STATE
An ideal gas is defined as a gas that obeys the following PvT relationship at all temperatures
and pressures:
The relationship among the state variables, temperature, pressure and specific volume is
called the equation of state. We now consider the equation of state for the vapour or gaseous
phase of simple compressible substances:
𝐹(𝑃, 𝑇, 𝑣) ≡ 0
Based on our experience in chemistry and physics, we recall that the combination of Boyle’s
and Charles’ laws for gases at low pressure result in the equation of state for the ideal gas as
𝑇
𝑃=𝑅 ( )
𝑣
Where R is the constant of proportionality and it is called the gas constant, it takes on a
different value for each gas. If a gas obeys this relation, it is called an ideal gas. We often
write this equation as:
Pv = RT
R is the specific gas constant; it has a fixed value for any particular gas and it is expressed in
kJ/kgK in the SI unit:
Rair = 0.287 kJ/kgK; Roxygen = 0.262 kJ/kgK
𝑅=
𝑅𝑢
𝑀
The mass, m, is related to the moles, N, of substance through the molecular weight or molar
mass, M, see table A-1. The molar mass is the ratio of mass to moles and has the same value,
regardless of the system of units:
𝑀𝑎𝑖𝑟 = 28.97
𝑘𝑔
𝑘𝑚𝑜𝑙
= 28.97
𝑙𝑏𝑚
𝑙𝑏𝑚𝑜𝑙
45
STUDY GUIDE
TDY2601
Since 1 kmol = 1000 gmol or 1000 gram-mole and 1 kg = 1000 g, 1 kmol of air has a mass of
28.97 kg or 28,970 gm = N M
The ideal gas equation of state may be written in several ways:
𝑃𝑣 = 𝑅𝑇
𝑃
𝑉
𝑚
= 𝑅𝑇
𝑃𝑉 = 𝑚𝑅𝑇
𝑃𝑉 =
𝑚
𝑀
(𝑀𝑅) 𝑇
𝑃𝑉 = 𝑁𝑅𝑢 𝑇
𝑃
𝑉
𝑁
= 𝑅𝑢 𝑇
𝑃𝑣̅ = 𝑅𝑢 𝑇
Here
P = absolute pressure in MPa, or kPa
𝑣̅ = molar specific volume in m3/kmol
T = absolute temperature in K
Ru = 8.314 kJ/ (kmol⋅K)
Some values of the universal gas constant are:
Universal Gas Constant, Ru = 8.314 kJ/ (kmol⋅K) = 8.314 kPa⋅m3/ (kmol⋅K)
Ideally, the behaviour of air is characterised by its mass; the volume it occupies; its
temperature and the pressure condition in which it is kept. An ideal gas is governed by the
perfect gas equation of state which relates to the state pressure, volume and temperature of a
fixed mass (m is constant) of a given gas (R is constant) as:
𝑃𝑉 = 𝑚𝑅𝑇
(1)
Where P – Pressure (Pa)
V – Volume (m3)
T – Absolute temperature (K)
T (K) = 273 + t (ºC )
m – Mass (kg)
R – Gas constant (J/kgK)
The equation of state can be written in the following forms, depending on what is needed to
be calculated as follows:
1. In terms of the pressure
𝑃 =
𝑚𝑅𝑇
𝑉
46
(2)
STUDY GUIDE
TDY2601
2. In terms of the volume
𝑉 =
3. In terms of the mass
𝑚 =
4. In terms of the temperature
𝑇 =
5. In terms of the gas constant
𝑅 =
6. In terms of the density
𝜌 =
𝑚𝑅𝑇
(3)
𝑃
𝑃𝑉
(4)
𝑅𝑇
𝑃𝑉
(5)
𝑚𝑅
𝑃𝑉
(6)
𝑚𝑇
𝑚
𝑉
=
𝑃
𝑅𝑇
(7)
The specific gas constant, R, is a property related to the molar mass (M) in kg/kmol of the gas
and the universal gas constant, Ro, as
R = Ro / M
(8)
Where Ro = 8314.3 J/kgK
The ideal gas equation can also be written on time basis, relating the mass flow rate (kg/s)
and the volumetric flow rate (m3/s) as follows:
𝑃𝑉𝑡 = 𝑚𝑡 𝑅 𝑇
3.8
(9)
COMPRESSIBILITY FACTOR-A MEASURE OF DEVIATION FROM
IDEAL GAS BEHAVIOUR
To understand the above criteria and to determine how much the ideal gas equation of state
deviates from the actual gas behaviour, we introduce the compressibility factor Z as follows:
𝑃𝑣̅ = 𝑍 𝑅 𝑇
Or
𝑍 =
𝑃𝑣
𝑅𝑇
For an ideal gas Z = 1 and the deviation of Z from unity measures the deviation of the actual
P-V-T relation from the ideal gas equation of state.
The compressibility factor is expressed as a function of the reduced pressure and the reduced
temperature. The Z factor is approximately the same for all gases at the same reduced
temperature and reduced pressure, which are defined as follows:
𝑃𝑅 =
𝑃
𝑃𝑐𝑟
And 𝑇𝑅 =
𝑇
𝑇𝑐𝑟
Where Pcr and Tcr are the critical pressure and temperature, respectively. The critical constant
data for various substances are given in table A-1 from thermodynamics textbooks. This is
47
STUDY GUIDE
TDY2601
known as the principle of corresponding states. Figure 3-11 gives a comparison of Z factors
for various gases and supports the principle of corresponding states.
Figure 3.11: Comparison of Z factor for various gases
3.9
WORK ACTIVITIES
Activity 3.1:
Pure substances, phase-change processes and property diagrams
Question
What is the difference between saturated vapour and superheated vapour?
Answer
A liquid that is about to vaporise is saturated liquid; or compressed liquid.
Question
A househusband is cooking beef stew for his family in a pan that is (a) uncovered, (b)
covered with a light lid, and (c) covered with heavy lid. In which case will the cooking time
be the shortest? Why?
Answer
Case (c) where the pan is covered with a heavy lid, because the heavier the lid, the greater the
pressure in the pan; therefore, the greater the cooking temperature.
Activity 3.2: Property tables
Question
Which process requires more energy: completely vaporising 1 kg of saturated liquid water at
1 atm pressure; or completely vaporising I kg of superheated liquid water at 8 atm pressure?
48
STUDY GUIDE
TDY2601
Answer
Completely vaporising 1 kg of saturated liquid at 1 atm pressure since the higher the pressure,
the lower the hfg.
Question
A 9 m3 container is filled with 300 kg of R-134a at 100C. What is the specific enthalpy of the
R-134a in the container?
Answer
Analysis: The specific volume is:
𝑣=
𝑣
𝑚
=
9 𝑚3
300 𝑘𝑔
= 0.03 𝑚3 /𝑘𝑔
Table A-11 in the textbook indicates this as a mixture of liquid and vapour.
Using the properties at 10°C line, the quality and the enthalpy are determined to be as
follows:
𝑥=
𝑣−𝑣𝑓
𝑣𝑓𝑔
(0.03−0.0007930)𝑚3 /𝑘𝑔
= (0.049403−0.0007930)𝑚3
/𝑘𝑔
= 0.6008
ℎ = ℎ𝑓 − 𝑥ℎ𝑓𝑔 = 65.43 + (0.6008)(190.73) = 180.02 𝑘𝐽/𝑘𝑔
Activity 3.3:
Ideal gas
Question
Propane and methane are commonly used for heating in winter; and the leakage of these fuels
poses fire danger for homes, even if it is for short periods. Which gas leakage do you think
poses a greater risk of fire? Explain.
Answer
Propane (molar mass = 44.1 kg/kmol) poses a greater fire danger than methane (molar
mass=16 kg/kmol), since propane is heavier than air (molar mass = 29 kg/kmol) and it will
settle near the floor. While, methane is lighter than air and will therefore rise and leak out.
Question
A 2 kg mass of helium is maintained at 300 kPa and 270C in a rigid container. How large
is the container in m3?
Answer
49
STUDY GUIDE
TDY2601
Question
A spherical balloon which is 9 m in diameter is filled with helium at 270C and 200 kPa.
Determine the mole number and the mass of the helium in the balloon.
Answer
A balloon is filled with helium gas. The mole number and the mass of helium in the balloon
are to be determined. Assumptions: In specified conditions helium behaves as an ideal gas.
Properties: The universal gas constant is Ru = 8.314 kPa.m3/kmol.K. The molar mass of
helium is 4.0 kg/kmol (table A-1).
Analysis: The volume of the sphere is:
Activity 3.4
Compressibility factor
Question
How are the reduced pressure and reduced temperature defined?
Answer
Reduced pressure is the pressure normalised with respect to the critical pressure; and reduced
temperature is the temperature normalised with respect to the critical temperature.
Question
Determine the specific volume of superheated water vapour at 3.5 MPa and 4500C based on
(a) the ideal gas equation, (b) the generalised compressibility chart and (c) the steam tables.
Determine the error involved in the first two cases.
Answer
50
STUDY GUIDE
TDY2601
3.10
SUMMARY
•
Pure substance
•
Phase change processes of pure substances
✓ Compressed liquid, saturated liquid, saturated vapour and superheated vapour
✓ Saturation temperature and saturation pressure
•
Property diagrams for phase change processes
✓ The T-v diagram, The P-v diagram, The P-T diagram and, The P-v-T surface
•
Property tables
✓ Enthalpy
✓ Saturated liquid, saturated vapour, saturated liquid vapour mixture,
superheated vapour and compressed liquid
✓ Quality of the working substance
•
Thermodynamics processes
•
✓ Thermodynamics of working fluid
The ideal gas equation of state for the hypothetical gas
•
The ideal gas equation of state
•
Compressibility factor
•
Work activities
51
STUDY GUIDE
TDY2601
LEARNING UNIT 4: ENERGY ANALYSIS OF CLOSED
SYSTEMS
4.1
LEARNING OUTCOMES
At the end of this unit, you should be able to
•
explain moving boundary work
•
use a formula to explain the polytropic process
•
calculate pressure, temperature, volume, heat transfer and work done in closed
systems
•
differentiate between specific heats at constant pressure and constant volume
•
calculate internal energy and enthalpy in a constant volume, pressure and specific
heats of gases, solids and liquids
4.2
MOVING BOUNDARY WORK
One form of mechanical work frequently encountered in practice is associated with the
expansion or compression of a gas in a piston-cylinder device. Part of the boundary (the inner
face of the piston) moves back and forth during this process. Therefore, the expansion and
compression work are often called moving boundary work, or simply boundary work as
shown in figure 4-1.
Figure 4-1: The work associated with a moving boundary is called boundary work
Consider the gas enclosed in the piston-cylinder device shown in figure 4-2. The initial
pressure of the gas is P, the total volume is V and the cross-sectional area is A. If the piston
can move a distance ds in a quasi-equilibrium manner, the differential work done during this
process is as follows:
52
STUDY GUIDE
TDY2601
𝛿𝑊𝑏 = 𝐹 𝑑𝑠 = 𝑃𝐴 𝑑𝑠 = 𝑃 𝑑𝑉
2
𝑊𝑏 = ∫1 𝑃 𝑑𝑉
(kJ)
Figure 4-2: A gas does a differential amount of work Wb as it forces the piston to move by a
differential amount ds.
The quasi-equilibrium expansion process described is indicated on a P-V diagram in figure 43.
In this diagram, the differential area d is equal to P dV, which is the differential. The total A
under the process curve 1-2 is obtained by adding these differential areas.
2
2
Area=𝐴 = ∫1 𝑑𝐴 = ∫1 𝑃 𝑑𝑉
Figure 4-3: The area under the process curve on a P-V diagram represents the boundary
work.
53
STUDY GUIDE
TDY2601
Figure 4-4: The boundary work done during a process depends on the path followed and the
end states.
Figure 4-5: The net work done during a cycle is the difference between the work done by the
system and the work done on the system.
4.2.1
Polytropic processes
During the actual expansion and compression process of gases, pressure and volume are often
related by PVn=C, where n and C are constants. A process of this kind is called a polytropic
process as shown in figure 4.6.
The pressure for a polytropic process can be expressed as follows:
Polytropic process: C, n (polytropic exponent) constants
P=CV-n
𝟐
𝟐
𝑽 −𝒏+𝟏 −𝑽𝟏 −𝒏+𝟏
𝑾𝒃 = ∫𝟏 𝑷𝒅𝑽 = ∫𝟏 𝑪𝑽−𝒏 𝒅𝑽 = 𝑪 𝟐
−𝒏+𝟏
=
𝑷𝟐 𝑽𝟐 −𝑷𝟏 𝑽𝟏
𝟏−𝒏
Since C=P1V1n = P2V2n. For an ideal gas (PV=mRT), this equation can also be written as:
54
STUDY GUIDE
TDY2601
For a special case of n=1 the boundary work becomes
2
2
𝑉
𝑊𝑏 = ∫1 𝑃 𝑑𝑉 = ∫1 𝐶𝑉 −1 𝑑𝑉 = 𝑃𝑉𝐼𝑛 ( 2)
𝑉1
Constant pressure process
2
2
𝑊𝑏 = ∫1 𝑃 𝑑𝑉 = 𝑃0 ∫1 𝑑𝑉 = 𝑃0 (𝑉2 − 𝑉1 )
Figure 4-6: Schematic and P-V diagram for a polytropic process
4.3
ENERGY BALANCE FOR CLOSED SYSTEMS
Energy balance for any system undergoing any process
Energy balance in the rate form
For constant rates, the total quantities during a time interval are related to the qualities per
unit time as:
̇ , 𝑊 = 𝑊∆𝑡
̇ , and ∆𝐸 = (𝑑𝐸⁄ )∆𝑡
𝑄 = 𝑄∆𝑡
𝑑𝑡
(kJ)
The energy balance can be expressed on a per unit mass basis as:
For a closed system undergoing a cycle, the initial and final states are identical; therefore,
∆𝐸𝑠𝑦𝑠𝑡𝑒𝑚 = 𝐸2 − 𝐸1 = 0. 𝐓𝐡𝐞𝐧 𝐭𝐡𝐞 𝐞𝐧𝐞𝐫𝐠𝐲 𝐛𝐚𝐥𝐚𝐧𝐜𝐞 𝐟𝐨𝐫 𝐚 𝐜𝐲𝐜𝐥𝐞 𝐬𝐢𝐦𝐩𝐥𝐢𝐟𝐢𝐞𝐬 𝐭𝐨 𝐸𝑖𝑛 −
𝐸𝑜𝑢𝑡 = 0 or Ein=Eout.. Take note that a closed system does not involve any mass flow across
55
STUDY GUIDE
TDY2601
its boundaries, since the energy balance for a cycle can be expressed in terms of heat and
work interactions, such as:
The energy balance relation for a closed system in this case becomes:
Energy balance when sign convention is used (heat input and work output are positive; and
heat output and work input are negative).
Figure 4-7: For a cycle E = 0, thus Q = W
Figure 4-8: Various forms of the first-law relation for closed systems when sign convention
is used
56
STUDY GUIDE
4.4
TDY2601
SPECIFIC HEATS
Specific heat is defined as the energy required to raise the temperature of a unit mass of a
substance by one degree (figure 4-9). In general, the energy depends on how the process is
executed. In thermodynamics, one is interested in two specific kinds of heats: specific heat at
constant volume (Cv) and specific heat at constant pressure (Cp).
Specific heat at constant volume, cv: The energy required to raise the temperature of the unit
mass of a substance by one degree as the volume is maintained to remain constant.
Specific heat at constant pressure, cp: The energy required to raise the temperature of the
unit mass of a substance by one degree as the pressure is maintained to remain constant.
Figure 4-9: Specific heat is the energy required to raise the temperature of a unit mass of a
substance by one degree in a specified way.
4.5
INTERNAL ENERGY, ENTHALPY AND SPECIFIC HEATS OF IDEAL
GASES.
We defined an ideal gas as a gas, the temperature, pressure and specific volume of which are
related by
PV=RT
Joule submerged two tanks connected with a pipe and a valve in a water bath, as shown in
figure 4-10. Initially one tank contained air at a high pressure and the other tank was
evacuated.
57
STUDY GUIDE
TDY2601
Figure 4-10: Schematic presentation of the experimental apparatus used by Joule
4.4.1
For ideal gases, u, h, cv, and cp vary with temperature only.
Internal energy and enthalpy change of an ideal gas:
At low pressures, all real gases approach ideal gas behaviour; therefore, their specific heats
depend on temperature only.
The specific heats of real gases at low pressures are called ideal-gas-specific heats or zeropressure-specific heats and are often denoted cp0 and cv0.
58
STUDY GUIDE
TDY2601
Accurate analytical expressions for ideal gas-specific heats, based on direct measurements or
calculations from statistical behaviour of molecules, are available and are given as thirddegree polynomials in the appendix (table A-2c from the textbook) for several gases. A plot
of cpo (T) data for some common gases given in figure 4-11.
Figure 4-11: Ideal gas constant pressure-specific heats for some gases (see table A–2c for cp
equations)
u and h data for a number of gases that have been tabulated).
These tables are obtained by choosing an arbitrary reference point and performing the
integrations by treating state 1 as the reference state.
Figure 4-12: In the preparation of ideal gas tables, 0 K is chosen as the reference temperature
Internal energy and enthalpy change when specific heat is taken to remain constant at an
average value
𝑢2 − 𝑢1 = 𝑐𝑣,𝑎𝑣𝑔 (𝑇2 − 𝑇1 )
ℎ2 − ℎ1 = 𝑐𝑝,𝑎𝑣𝑔 (𝑇2 − 𝑇1 )
(kJ/kg)
(kJ/kg)
59
STUDY GUIDE
TDY2601
Figure 4-13: The relation  u = cv T is valid for any kind of process, whether the volume is
constant.
Figure 4-14: Small temperature intervals, the specific heats assumed to vary linearly with
temperature
4.6
INTERNAL ENERGY, ENTHALPY, AND SPECIFIC HEATS OF SOLIDS
AND LIQUIDS
Incompressible substance: A substance, the specific volume (or density) of which is
constant. Solids and liquids are incompressible substances.
The specific volumes of solids and liquids essentially remain constant during a process
(figure 4-15):
Figure 4-15: The specific volumes of incompressible substances remain constant during a
process.
60
STUDY GUIDE
TDY2601
It can be shown that the constant volume and constant pressure-specific heats are identical for
incompressible substances (figure 4.16). Therefore, the subscripts on cp and cv for solids and
liquids can be dropped; and both specific heats can be represented by a single symbolic c.
That is,
Cp=Cv=C
Figure 4.16: The cv and cp values of incompressible substances are identical and are denoted
by c.
4.6.1 Internal energy changes
Like those of ideal gases, the specific heats of incompressible substances depend on
temperature only. Thus, the partial differentials in the defining equation of cv can be replaced
by ordinary differentials, which yield
Enthalpy changes
4.7 WORK ACTIVITIES
Activity 4.1: Moving boundary work
61
STUDY GUIDE
TDY2601
Question
The volume of 1 kg of helium in a piston-cylinder device is initially 7 m3. Now helium is
compressed to 3 m3 while its pressure is maintained to stay constant at 150 kPa. Determine
the initial and final temperatures of helium as well as the work required to compress it in kJ.
Answer
Question
A mass of 5 kg of saturated water vapour at 300 kPa is heated at constant pressure until the
temperature reaches 2000C. Calculate the work done by the steam during this process.
Answer
62
STUDY GUIDE
TDY2601
Activity 4.2: Closed-system energy analysis
Question
Saturated water vapour at 2000C is isothermally condensed to a saturated liquid in a pistoncylinder device. Calculate the heat transfer and the work done during this process in kJ/kg.
Answer
63
STUDY GUIDE
TDY2601
Question
Saturated water vapour in a closed system is condensed by cooling it at constant pressure to a
saturated liquid at 40 kPa. Determine the heat transferred and the work done during this
process, in kJ/kg.
Answer
Activity 4.3 Specific heats, ∆𝒖, and ∆𝒉 of ideal gases
Question
In relation ∆𝑢=mcv,avg∆𝑇, what is the correct unit of cv-kJ/kg.0C or kJ/kg. K?
Answer
It can be either. The difference in temperature in both the K and °C scales is the same.
Question
The temperature of 2 kg neon is increased from 20 to 1800C. Calculate the change in the total
internal energy of the neon in kJ. Would the internal energy change be any different if neon
were replaced with argon?
Answer
64
STUDY GUIDE
TDY2601
Activity 4.4 Closed-system energy analysis: ideal gases
Question
Is it possible to compress an ideal gas isothermally in an adiabatic piston-cylinder device?
Explain.
Answer
No, it is not, because the first law relation Q - W = ΔU reduces to W = 0 in this case, since the
system is adiabatic (Q = 0) and ΔU = 0 for the isothermal processes of ideal gases. Therefore,
this adiabatic system cannot receive any net-work at constant temperature.
Question
Argon is compressed in a polytropic process with n=1.2 from 120 kPa and 100C to 800 kPa in
a piston-cylinder device. Determine the work produced and heat transferred during this
compression process, in kJ/kg.
Answer
65
STUDY GUIDE
TDY2601
Activity 4.5 Closed-system energy analysis: solids and liquids
Question
A 1 kg block of iron is heated from 25 to 750C. What is the change in the iron’s total internal
energy and enthalpy?
Answer
Question
Carbon steel balls (𝜌=7833 kg/m3 and cp=0.465 kJ/kg. 0C) 8 mm in diameter are annealed by
heating them first to 9000C in a furnace; and then allowing them to cool down slowly to
1000C in ambient air at 350C. If 2500 balls are to be annealed per hour, determine the total
rate of heat transfer from the balls to the ambient air.
Answer
4.8
SUMMARY
•
Moving boundary work
✓ Wb for a polytropic process
•
Energy balance for closed systems
•
Specific heats
✓ Constant pressure-specific heat, cp
66
STUDY GUIDE
TDY2601
✓ Constant volume-specific heat, cv
•
Internal energy, enthalpy and specific heats of ideal gases
✓ Specific heat relations of ideal gases
•
•
Internal energy, enthalpy and specific heats of incompressible substances (solids and
liquids)
✓ Internal energy changes
Work activities
67
STUDY GUIDE
TDY2601
LEARNING UNIT 5: MASS AND ENERGY ANALYSIS OF
CONTROL VOLUMES
5.1
LEARNING OUTCOMES
At the end of this unit, you should be able to:
5.2
•
Apply the conservation of mass principle to various systems including steady and
unsteady flow-control volumes.
•
Identify the energy carried by fluid stream crossing a control surface as the sum of
internal energy, flow work, kinetic energy and potential energy of the fluid.
•
Solve energy balance problems for common steady-flow devices such as nozzles,
compressors, turbines, throttling valves, mixers, heaters and heat exchangers.
•
Apply the energy balance to general unsteady-flow processes with emphasis on the
uniform flow process as the model for commonly encountered charging and
discharging processes.
CONSERVATION OF MASS
The conservation of mass principle is one of the most fundamental principles in nature. We
are familiar with this principle and it is not difficult to understand. A person does not need to
be a rocket scientist to figure out how much vinegar and oil dressing will be obtained by
mixing 100 g of oil with 25 g of vinegar. Even chemical equations are balanced based on the
conservation of mass principles. When 16 kg of oxygen reacts with 2 kg of hydrogen, 18 kg
of water is formed (Figure 5.1).
Conservation of mass: Mass, like energy, is a conserved property and it cannot be created or
destroyed during a process.
Closed systems: The mass of the system remains constant during a process.
Control volumes: Mass can cross the boundaries and so we must keep track of the amount of
mass entering and leaving the control volume.
Figure 5.1: Mass is conserved even during chemical reactions.
68
STUDY GUIDE
TDY2601
Mass m and energy E can be converted to each other according to the well-known formula
proposed by Albert Einstein (1879-1955).
E=mc2
Where c is the speed of light in a vacuum, which is c = 2.9979  108 m/s.
This equation suggests that there is equivalence between mass and energy.
5.2.1
Mass and volume flow rates
The amount of mass flowing through a cross section per unit time is called the mass flow rate
and is denoted by 𝑚̇. The dot over a symbol is used to indicate time rate of change.
Average velocity
However, velocity is never uniform over a cross section of a pipe, because of the no-slip
condition at the walls. Rather, velocity varies from zero at the walls to some maximum value
at or near the centreline of the pipe. The average velocity Vavg is defined as the average
value of Vn across the entire cross section of the pipe (figure 5.2).
𝑉𝑎𝑣𝑔 =
1
∫ 𝑉 𝑑𝐴
𝐴𝑐 𝐴𝑐 𝑛 𝑐
Figure 5.2: The average velocity Vavg is defined as the average speed through a cross section
Ac–Area of the cross section normal to the flow direction.
Vn–Normal flow velocity.
Mass flow rate through the entire cross-sectional area of a pipe or duct is obtained by:
(𝑘𝑔/𝑠)
𝑚̇ = 𝜌𝑉𝑎𝑣𝑔 𝐴𝑐
̇
𝑉
𝑚̇ = 𝜌𝑉̇ =
𝑣
69
STUDY GUIDE
5.2.2
TDY2601
Conservation of mass principle
The conservation of mass principle for control volume can be expressed as the net mass
transfer to or from a control volume during a time interval; t is equal to the net change (increase or
decrease) in the total mass within the control volume during t.
(
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑒𝑛𝑡𝑒𝑟𝑖𝑛𝑔
𝑇𝑜𝑡𝑎𝑙 𝑚𝑎𝑠𝑠 𝑙𝑒𝑎𝑣𝑖𝑛𝑔
𝑁𝑒𝑡 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑚𝑎𝑠𝑠
)−(
)=(
)
𝑡ℎ𝑒 𝐶𝑉 𝑑𝑢𝑟𝑖𝑛𝑔 ∆𝑡
𝑡ℎ𝑒 𝐶𝑉 𝑑𝑢𝑟𝑖𝑛𝑔 ∆𝑡
𝑤𝑖𝑡ℎ𝑖𝑛 𝑡ℎ𝑒 𝐶𝑉 𝑑𝑢𝑟𝑖𝑛𝑔 ∆𝑡
or
(𝑘𝑔)
𝑚𝑖𝑛 − 𝑚𝑜𝑢𝑡 = ∆𝑚𝑐𝑣
Where ∆𝑚𝐶𝑉 = 𝑚𝑓𝑖𝑛𝑎𝑙 − 𝑚𝑖𝑛𝑖𝑡𝑖𝑎𝑙 is the change in the mass of the control volume during the
process (figure 5.3).
Figure 5.3: conservation of mass principle for an ordinary bathtub
5.2.3
Special Case: incompressible flow
The conservation of mass relations can be simplified even further when the fluid is
incompressible, which is usually the case for liquids. Cancelling the density from both sides
of the general steady-flow relation gives the following:
3
(𝑚 ⁄𝑠)
Steady, incompressible flow: ∑𝑖𝑛 𝑉̇ = ∑𝑜𝑢𝑡 𝑉̇
Steady, incompressible flow (single stream): 𝑉1̇ = 𝑉̇2 → 𝑉1 𝐴1 = 𝑉1 𝐴2
70
STUDY GUIDE
5.3
TDY2601
FLOW WORK AND THE ENERGY OF A FLOWING FLUID
Unlike closed systems, control volumes involve mass flow across their boundaries; and some
work is required to push the mass into or out of the control volume. This work is known as
the flow work, or flow energy and is necessary for maintaining a continuous flow through a
control volume.
Consider a fluid element of volume V indicated in figure 5.4 to obtain a relation for flow
work. The fluid immediately upstream forces this fluid element to enter the control volume;
thus, it can be regarded as an imaginary piston. The fluid element can be chosen to be
sufficiently small so that it has uniform properties throughout.
Figure 5.4: schematic for flow work
Workflow or flow energy is:
(𝑘𝐽/𝑘𝑔)
𝑊𝑓𝑙𝑜𝑤 = 𝑃𝑣
5.3.1 Energy transport by mass
Take note that 𝜃 is total energy per unit mass, the total energy of a mass m of fluid flowing is
simply𝑚𝜃, provided that the properties of the mass m are uniform. Also, when a fluid stream
with uniform properties is flowing at a mass flow rate of𝑚̇, the rate of energy flow with this
̇ (figure 5.5); that is,
specific stream is 𝑚𝜃
Amount of energy transport: 𝐸𝑚𝑎𝑠𝑠 = 𝑚𝜃 = 𝑚 (ℎ +
Rate of energy transport: 𝐸𝑚𝑎𝑠𝑠 = 𝑚̇𝜃 = 𝑚̇ (ℎ +
71
𝑉2
2
𝑉2
2
(𝑘𝐽)
+ 𝑔𝑧)
+ 𝑔𝑧)
(𝑘𝑊)
STUDY GUIDE
TDY2601
Figure 5.5: The product 𝑚̇ 𝑖 𝜃𝑖 is the energy transported into control volume by mass per unit
time.
When the properties of the mass at each inlet or exit change with time and over the cross
section:
5.4
ENERGY ANALYSIS OF SOME STEADY-FLOW SYSTEMS
A large number of engineering devices such as turbines, compressors and nozzles operate for
long periods of time under the same conditions once the transient start-up period is
completed; steady operation is established and classified as steady-flow devices (figure
5.6).
Figure 5.6: Power plants operate under steady conditions
During a steady-flow process, no intensive or extensive properties within the control volume
change with time. Thus, the volume V, the mass m, and the total energy content E of the
control volume remain constant (figure 5.7).
72
STUDY GUIDE
TDY2601
Figure 5.7: The mass and energy contents of a control volume remain constant under steadyflow conditions.
The fluid properties at an inlet or exit remain constant during a steady flow process. The
properties may, however, be different at various inlets and exits. They may even vary over
the cross section of an inlet or an exit. However, all properties, including the velocity and
elevation, must remain constant with time at a fixed point at an inlet or exit. It follows that
the mass flow rate of the fluid at an opening must remain constant during a steady-flow
process (figure 5.8).
Figure 5.8: The fluid properties at an inlet or exit remain constant (do not change with time),
under steady flow conditions.
The mass balance for a general steady flow system was given as:
The mass balance for a single-stream (one inlet and one outlet) steady-flow system was given
as:
𝑚̇ 1 = 𝑚̇ 2 → 𝜌1 𝑉1 𝐴1 = 𝜌2 𝑉2 𝐴2
For a steady-flow process, the rate form of the general energy balance reduces to:
73
STUDY GUIDE
TDY2601
Take note that energy can be transferred by heat, work and mass only; the energy balance for
a general steady flow system can also be written more explicitly as:
∆𝐤𝐞 = (𝐕𝟐𝟐 − 𝐕𝟏𝟐 )/𝟐 . The unit of kinetic energy is m2/s2, which is equivalent to J/kg. The
enthalpy is usually given in kJ/kg. To add these two quantities, the kinetic energy should be
expressed in kJ/kg.
∆𝐩𝐞 = (𝔃𝟐 − 𝔃𝟏 ). A similar argument can be given for the potential energy term. A potential
energy change of 1 kJ/kg corresponds to an elevation difference of 102 m. The only time the
potential energy term is significant is when a process involves pumping a fluid to high
elevations; and one is interested in the required pumping power.
5.5
SOME STEADY FLOW ENGINEERING DEVICES
Many engineering devices operate essentially under the same conditions for long periods.
The components of a steam power plant (turbines, compressors, heat exchangers and pumps),
for example, operate nonstop for months before the system is shut down for maintenance
(figure 5.9). Therefore, these devices can be conveniently analysed as steady-flow devices.
Figure 5.9: A modern land-based gas turbine used for electric power production. This is a
General Electric LM5000 turbine. It has a length of 6.2 m, it weighs 12.5 tons and produces
55.2 MW at 3600 rpm with steam injection.
74
STUDY GUIDE
5.5.1
TDY2601
Nozzles and diffusers
Nozzles and diffusers are commonly utilised in jet engines, rockets, spacecraft and even
garden hoses.
A nozzle is a device that increases the velocity of a fluid at the expense of pressure.
A diffuser is a device that increases the pressure of a fluid by slowing it down.
The cross-sectional area of a nozzle decreases in the flow direction for subsonic flows and
increases for supersonic flows. The reverse is true for diffusers.
Figure 5.10: Nozzles and diffusers are shaped so that they cause large changes in fluid
velocities and thus kinetic energies.
5.5.2
Turbines and compressors
Turbine drives the electric generator in steam, gas or hydroelectric power plants.
As the fluid passes through the turbine, work is done against the blades that are attached to
the shaft. As a result, the shaft rotates and the turbine produces work.
Compressors, pumps and fans are devices used to increase the pressure of a fluid. Work is
supplied to these devices from an external source through a rotating shaft.
A fan increases the pressure of a gas slightly and is mainly used to mobilise a gas.
A compressor can compress the gas to very high pressures.
Pumps work very much like compressors except that they handle liquids instead of gases.
75
STUDY GUIDE
TDY2601
Figure 5.11: Schematic example of compressors
Energy balance for the compressor of the compressor can be expressed in the rate form as:
5.5.3
Throttling valves
Throttling valves are any kinds of flow-restricting devices that cause a significant pressure
drop in the fluid.
What is the difference between a turbine and a throttling valve?
The pressure drop in the fluid is often accompanied by a large drop in temperature; and for
this reason throttling devices are commonly used in refrigeration and air-conditioning
applications.
5.5.4
Mixing chambers
In engineering applications, the section where the mixing process takes place is commonly
referred to as a mixing chamber.
Figure 5.12: The T-elbow of an ordinary shower serves as the mixing chamber for the hot
and the cold water streams.
76
STUDY GUIDE
TDY2601
Figure 5.13: Mixing chamber as the system
Energy balance for the adiabatic mixing chamber in the figure is:
5.5.5
Heat exchangers
Heat exchangers are devices where two moving fluid streams exchange heat without mixing.
Heat exchangers are widely used in various industries and they come in various designs.
When the entire heat exchanger is selected as the control volume 𝑄̇ becomes zero, since the
boundary for this case lies just beneath the insulation and little or no heat crosses the
boundary (figure 5.14a). However, if only one of the fluids is selected as control volume, the
heat will cross this boundary as it flows from one fluid to the other and 𝑄̇ will not be zero
(figure 5.14b).
Figure 5.14: The heat transfer associated with a heat exchanger may be zero or nonzero
depending on how the control volume is selected.
77
STUDY GUIDE
TDY2601
Mass and energy balances for the adiabatic heat exchanger in the figure are as follows:
5.6
ENERGY ANALYSIS OF UNSTEADY-FLOW PROCESSES
However, many processes of interest involve changes in the control volume with time. Such
processes are called unsteady-flow or transient-flow processes.
Most unsteady-flow processes can be represented reasonably well by the uniform-flow
process.
Uniform-flow process: The fluid flow at any inlet or exit is uniform and steady; therefore,
the fluid properties do not change with time or position over the cross section of an inlet or
exit. If they do, they are averaged and treated as constants for the entire process.
Mass balance
Energy balance
78
STUDY GUIDE
TDY2601
Figure 5.15: The energy equation of a uniform-flow system reduces to that of a closed
system when all the inlets and exits are closed.
Figure 5.16: A uniform-flow system may involve electrical, shaft and boundary work all at
once.
5.7
WORK ACTIVITIES
Activity 5.1:
Conservation of mass
Question
Consider a device with one inlet and one outlet. If the volume flow rates at the inlet and at the
outlet are the same, is the flow through this device necessarily steady? Why?
Answer
No, a flow with the same volume flow rate at the inlet and the exit is not necessarily steady
(unless the density is constant). To be steady, the mass flow rate through the device must
remain constant.
Question
Air enters the 1-m2 inlet of an aircraft engine at 100 kPa and 200C with a velocity of 180 m/s.
Determine the volume flow rate at the engine’s inlet in m3/s and the mass flow rate at the
engine’s exit in kg/s.
Answer
79
STUDY GUIDE
Activity 5.2:
TDY2601
Steady-flow energy balance: nozzles and diffusers
Question
A diffuser is an adiabatic device that decreases the kinetic energy of the fluid by slowing it
down. What happens to the lost kinetic energy?
Answer
It is mostly converted to internal energy as shown by a rise in the fluid temperature.
Question
Steam at 3 MPa and 4000C enters an adiabatic nozzle steadily with a velocity of 40 m/s and
leaves at 2.5 MPa and 300 m/s. Determine (a) the exit temperature and (b) the ratio of the
inlet to outlet area A1/A2.
Answer
80
STUDY GUIDE
Activity 5.3:
TDY2601
Turbines and compressors
Question
Consider an air compressor operating steadily. How would you compare the volume flow
rates of the air at the compressor inlet and exit?
Answer
The volume flow rate at the compressor inlet will be greater than that at the compressor exit.
Question
Steam enters an adiabatic turbine at 10 MPa and 5000C and leaves at 10 kPa with a quality of
90%. Determine the mass flow rate required for a power output of 5 MW by Neglecting the
changes in kinetic and potential energies.
Answers
81
STUDY GUIDE
Activity 5.4:
TDY2601
Throttling valves
Question
It is claimed, based on temperature measurements, that the temperature of a fluid rises in a
well-insulated valve with negligible friction during a throttling process. How do you evaluate
this claim? Does this process violate any thermodynamic laws?
Answer
The temperature of a fluid can increase, decrease or remain the same during a throttling
process. Therefore, the claim is valid since no thermodynamic laws are violated.
Question
A well-insulated valve is used to throttle steam from 8 MPa and 3500C to 2 MPa. Determine
the final temperature of the steam.
Answer
82
STUDY GUIDE
Activity 5.5:
TDY2601
Mixing chambers and heat exchangers
Question
Consider a steady-flow heat exchanger involving two different kinds of fluid streams. Under
what conditions will the amount of heat lost by one fluid be equal to the amount of heat
gained by the other?
Answer
Under the conditions of no heat and work interactions between the mixing chamber and the
surrounding medium.
Question
A stream of refrigerant-134a at 1 MPa and 200C is mixed with another stream at 1 MPa and
800C. If the mass flow rate of the cold stream is twice that of the hot one, determine the
temperature and the quality of the exit stream.
Answer
83
STUDY GUIDE
Activity 5.6:
TDY2601
Pipe and duct flow
Question
Saturated liquid water is heated in a steady-flow steam boiler at a constant pressure of 2 MPa
at a rate of 4 kg/s to an outlet temperature of 2500C. Determine the rate of heat transfer in the
boiler.
Answer
84
STUDY GUIDE
TDY2601
Question
Refrigerant-134a enters the condenser of a refrigerator at 900 kPa and 600C and leaves as a
saturated liquid at the same pressure. Determine the heat transfer from the refrigerant per unit
mass.
Answer
85
STUDY GUIDE
TDY2601
Activity 5.7 Charging and discharging processes
Question
An insulated rigid tank is initially evacuated. A valve is opened and atmospheric air at 95 kPa
and 170C enters the tank until the pressure in the tank reaches 95 kPa, at which point the
valve is closed. Determine the final temperature of the air in the tank. Assume constant
specific heats.
Answer
5.8 SUMMARY
• Conservation of mass
✓ Mass and volume flow rates
✓ Mass balance for a steady-flow process
✓ Mass balance for incompressible flow
•
Flow work and the energy of a flowing fluid
✓ Energy transport by mass
•
Energy analysis of steady-flow systems
•
Some steady-flow engineering devices
✓ Nozzles and diffusers
✓ Turbines and compressors
✓ Throttling valves
86
STUDY GUIDE
TDY2601
✓ Mixing chambers and heat exchangers
•
Energy analysis of unsteady-flow processes
•
Work activities
87
STUDY GUIDE
TDY2601
LEARNING UNIT 6: THE SECOND LAW OF THERMODYNAMICS
6.1 LEARNING OUTCOMES
At the end of this unit, you should be able to:
6.2
•
Identify valid processes as those that satisfy the first and second laws of
thermodynamics.
•
Discuss the thermal energy reservoirs, reversible and irreversible processes, heat
engines, refrigerators and heat pumps.
•
Describe the Carnot cycle.
•
Examine the idealised Carnot heat engines, refrigerators and heat pumps.
INTRODUCTION TO THE SECOND LAW
In learning units 4 and 5, we applied the first law of thermodynamics, or the conservation of
energy principle to processes involving closed and open systems. As pointed out repeatedly
in those chapters, energy is a conserved property; and no process is known to have taken
place in violation of the first law of thermodynamics. Therefore, it is reasonable to conclude
that a process must satisfy the first law to occur.
Figure 6.1: A cup of hot coffee does not get hotter in a cooler room.
Figure 6.2: Transferring heat to a paddle wheel will not cause it to rotate
Figure 6.3: Transferring heat to a wire will not generate electricity
These processes cannot occur even though they are not in violation of the first law.
88
STUDY GUIDE
6.2.1
TDY2601
Major uses of the second law
1. The second law may be used to identify the direction of processes.
2. The second law also asserts that energy has quality as well as quantity. The first law is
concerned with the quantity of energy and the transformations of energy from one form to
another with no regard to its quality. The second law provides the necessary means to
determine the quality and the degree of degradation of energy during a process.
3. The second law of thermodynamics is also used to determine the theoretical limits for
the performance of commonly used engineering systems such as heat engines,
refrigerators and predicting the degree of completion of chemical reactions.
6.3
•
THERMAL ENERGY RESERVOIRS
A hypothetical body with a relatively large thermal energy capacity (mass x specific
heat) that can supply or absorb finite amounts of heat without undergoing any change in
temperature is called a thermal energy reservoir or just a reservoir.
•
In practice, large bodies of water such as oceans, lakes and rivers and the atmospheric air
can be modelled accurately as thermal energy reservoirs, because of their large thermal
energy storage capabilities or thermal masses.
Figure 6.4: Bodies with relatively large thermal masses can be modelled as thermal energy
reservoirs.
Figure 6.5: A source supplies energy in the form of heat and a sink absorbs it.
6.4
HEAT ENGINES
The devices that convert heat to work.
1.
They receive heat from a high-temperature source (solar energy, oil furnace, nuclear
reactor etc.).
2.
They convert part of this heat to work (usually in the form of a rotating shaft).
89
STUDY GUIDE
3.
TDY2601
They reject the remaining waste heat to a low-temperature sink (the atmosphere, rivers,
etc).
4.
They operate in a cycle.
Heat engines and other cyclic devices usually involve a fluid to and from which heat is
transferred while undergoing a cycle. This fluid is called the working fluid.
Figure 6.6: Part of the heat received by a heat engine is converted to work, while the rest is
rejected to a sink.
A steam power plant
Figure 6.7: Schematic presentation of a steam power plant
𝑄in =Amount of heat supplied to steam in a boiler from a high-temperature source (furnace).
𝑄out =Amount of heat rejected from steam in a condenser to a low-temperature sink (the
atmosphere, a river etc).
𝑊out =Amount of work delivered by steam as it expands in a turbine.
𝑊in =Amount of work required to compress water to boiler pressure.
𝑊net,out = 𝑊𝑜𝑢𝑡 − 𝑊𝑖𝑛
(kJ)
𝑊net,out = 𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡
(kJ)
90
STUDY GUIDE
6.4.1
TDY2601
Thermal efficiency
The fraction of the heat input that is converted to net work output is a measure of the
performance of a heat engine and is called the thermal efficiency𝜂𝑡ℎ .
For heat engines, the desired output is the net work output; and the required input is the
amount of heat supplied to the working fluid. Then the thermal efficiency of a heat engine
can be expressed as:
Thermal efficiency =
ηth =
ηth =
Net work output
Total heat input
Wnet,out
Qin
Qout
Qin
𝑊net,out = 𝑄𝑖𝑛 − 𝑄𝑜𝑢𝑡
(kJ)
Figure 6.8: Schematic presentation of a heat engine
QH=Magnitude of heat transfer between the cyclic device and the high-temperature medium
at temperature TH.
QL=Magnitude of heat transfer between the cyclic device and the low-temperature medium at
temperature TL.
The net-work output and thermal efficiency relations for any heat engine (shown in figure
6.8) can expressed as follows:
𝑊net,out = 𝑄𝐻 − 𝑄𝐿
ηth =
Wnet,out
QH
ηth = 1 −
QL
QH
91
STUDY GUIDE
6.5
•
TDY2601
REFRIGERATORS AND HEAT PUMPS
The transfer of heat from a low-temperature to a high-temperature medium requires
special devices called refrigerators.
•
Like heat engines, refrigerators are cyclic devices.
•
The working fluid used in the refrigeration cycle is called a refrigerant.
•
The most frequently used refrigeration cycle is the vapour-compression refrigeration
cycle.
In a household refrigerator the freezer compartment, where heat is absorbed by the
refrigerant, serves as the evaporator; and the coils usually behind the refrigerator where heat
is dissipated to the kitchen air serve as the condenser.
Figure 6.9: Basic components of a refrigeration system and typical operating conditions
6.5.1 Coefficient of performance
The efficiency of a refrigerator is expressed in terms of the coefficient of performance
(COP).
The objective of a refrigerator is to remove heat (QL) from the refrigerated space.
COPR =
Desired output
Required input
=
𝑄𝐿
𝑊𝑛𝑒𝑡,𝑖𝑛
𝑊net,in = 𝑄𝐻 − 𝑄𝐿
COPR =
𝑄𝐿
𝑄𝐻 −𝑄𝐿
(kJ)
= 𝑄𝐻
1
⁄𝑄 −1
𝐿
6.5.2 Heat pumps
The objective of a heat pump is to supply heat 𝑄𝐻 to the warmer space as shown in figure
6.10.
92
STUDY GUIDE
TDY2601
Figure 6.10: Heat pump
The measure of performance of a heat pump is also expressed in terms of the coefficient of
performance COPHP, defined as follows:
COPHP =
Desired output
Required input
𝑄𝐻
=
𝑊𝑛𝑒𝑡,𝑖𝑛
Which is also be expressed as
COPHP =
𝑄𝐻
𝑄𝐻 −𝑄𝐿
=
1
𝑄
1− 𝐿⁄𝑄
𝐻
COPHP=COPR+1
6.6
THE CARNOT CYCLE
The Carnot cycle is composed of four reversible processes: two isothermal and two adiabatic;
and it can be executed in either a closed or a steady-flow system. Consider a closed system
that consists of a gas contained in an adiabatic piston-cylinder device. The insulation of the
cylinder head is as such that it may be removed to bring the cylinder into contact with
reservoirs to provide heat transfer. The four reversible processes that make up the Carnot
cycle are as follows:
•
reversible isothermal expansion (process 1-2, TH = constant)
•
reversible adiabatic expansion (process 2-3, temperature drops from TH to TL)
•
reversible isothermal compression (process 3-4, TL = constant)
•
reversible adiabatic compression (process 4-1, temperature rises from TL to TH)
93
STUDY GUIDE
TDY2601
Figure 6.11: P-V diagram of the Carnot cycle
Figure 6.12: P-V diagram of the reversed Carnot cycle
6.7
THE CARNOT HEAT ENGINE
The hypothetical heat engine that operates on the reversible Carnot cycle is called the Carnot
heat engine.
Figure 6.13: The Carnot heat engine is the most efficient of all heat engines operating
between the same high and low-temperature reservoirs.
Any heat engine
ηth = 1 −
QL
QH
Carnot heat engine
ηth = 1 −
TL
TH
94
STUDY GUIDE
6.8
TDY2601
THE CARNOT REFRIGERATOR AND HEAT PUMP
The refrigerator or a heat pump that operates on the reversed Carnot cycle is called a Carnot
refrigerator, or a Carnot heat pump. The coefficient of performance of any refrigerator or heat
pump, reversible or irreversible, is given by
COPR = 𝑄𝐻
1
⁄𝑄 −1
𝐿
COPHP =
1
𝑄𝐿
1− ⁄𝑄
𝐻
Carnot refrigerator or heat pump
COPHP,rev =
1
𝑇
1− 𝐿⁄𝑇
𝐻
COPR,rev = 𝑇𝐻
1
⁄𝑇 −1
𝐿
6.9
WORK ACTIVITIES
Activity 6.1: Second law of thermodynamics and thermal energy reservoirs.
Question
Describe an imaginary process that satisfies the first law of thermodynamics but violates the
second law.
Answer
Transferring 5 kWh of heat to an electric resistance wire and producing 5 kWh of electricity.
Question
Describe an imaginary process that satisfies the second law of thermodynamics but violates
the first law.
Answer
An electric resistance heater, which consumes 5 kWh of electricity and supplies 6 kWh of
heat to a room.
Activity 6.2:
Heat engines and thermal efficiency
Question
Does a heat engine with a thermal efficiency of 100% necessarily violate (a) the first law and
(b) the second law of thermodynamics? Explain.
Answer
(a)
No, (b) Yes. According to the second law of thermodynamics, no heat engine can
have and efficiency of 100%.
95
STUDY GUIDE
TDY2601
Question
A 600-MW steam power plant, which is cooled by a nearby river has a thermal efficiency of
40%. Determine the rate of heat transfer to the river water. Will the actual heat transfer rate
be higher or lower than this value? Why?
Answer
Activity 6.3:
Refrigerators and heat pumps
Question
What is the difference between a refrigerator and a heat pump?
Answer
The difference between the two devices is one of purpose. The purpose of a refrigerator is to
remove heat from a cold medium whereas the purpose of a heat pump is to supply heat to a
warm medium.
Question
Bananas
are
to
be
cooled
from
240C
to
130C
at
a
rate of 215 kg/h by a refrigeration system. The power input to the refrigerator is 1.4 kW.
Determine the rate of cooling in kJ/min, and the COP of the refrigerator. The specific heat of
banana above freezing is 3.35 kJ/kg.0C.
Answer
96
STUDY GUIDE
Activity 6.4:
TDY2601
Carnot heat engines
From a work production perspective, which is more valuable: (a) thermal energy reservoirs at
675 K and 325 K or (b) thermal energy reservoirs at 625 K and 275 K?
Answer
Question
A heat engine is operating on a Carnot cycle and has a thermal efficiency of 75%. The waste
heat from this engine is rejected to a nearby lake at 150C at a rate of 14 kW. Determine the
power output of the engine and the temperature of the source, in 0C.
Answer
Activity 6.5:
Carnot refrigerators and heat pumps
Question
A homeowner buys a new refrigerator and a new air conditioner. Which one of these devices
would you expect to have higher COP? Why?
Answer
The difference between the temperature limits is typically much higher for a refrigerator than
it is for an air conditioner. The smaller the difference between the temperature limits a
refrigerator operates on, the higher is the COP. Therefore, an air conditioner should have a
higher COP.
97
STUDY GUIDE
TDY2601
Question
A refrigerator operating on the reversed Carnot cycle has a measured work input of 200 kW
and heat rejection of 2000 kW to a heat reservoir at 270C. Determine the cooling load
supplied to the refrigerator in kW; and the temperature of the heat source in 0C.
Answer
6.10
SUMMARY
•
Introduction to the second law
•
✓ Major uses of the second law
Thermal energy reservoirs
•
Heat engines
✓ Thermal efficiency
•
Refrigerators and heat pumps
✓ Coefficient of performance (COP)
✓ Heat pumps
•
The Carnot cycle
•
The Carnot heat engine
•
The Carnot refrigerator and heat pump
•
Work activities
98
STUDY GUIDE
TDY2601
LEARNING UNIT 7: ENTROPY
7.1
LEARNING OUTCOMES
At the end of this unit, you should be able to:
7.2
•
Calculate the entropy changes that take place during processes for pure substances,
incompressible substances (solids and liquids) and ideal gases.
•
Develop the reversible steady-flow work relations.
•
Develop the isentropic efficiencies for various steady-flow devices.
•
Apply the entropy balance to various systems.
ENTROPY CHANGE OF PURE SUBSTANCES
Entropy is a property; therefore, the value of entropy of a system is fixed once the state of the
system is fixed.
Figure 7.1: the entropy of a pure substance is determined from the tables (like other
properties)
The value of entropy at a specified state is determined just like any other property. In the
compressed liquid and superheated vapour regions, it can be obtained directly from tables at
the specified state. In the saturated mixture region, it is determined from:
𝑠 = 𝑠𝑓 + 𝑥𝑠𝑓𝑔
(kJ/kg.K)
The entropy change of a specified mass m (a closed system) during a process is simply
∆𝑆 = 𝑚∆𝑠 = 𝑚(𝑠2 − 𝑠1 )
(kJ/K)
What is the difference between the entropy values at the final and initial states?
7.3
ISENTROPIC PROCESSES
A process during which the entropy remains constant is called an isentropic process.
It is characterised by
99
STUDY GUIDE
TDY2601
isentropic processes: ∆𝑠 = 0 𝑜𝑟 𝑠2 = 𝑠1
(kJ/kg.K)
Figure 7.2: During an internally reversible, adiabatic (isentropic) process, the entropy
remains constant.
Figure 7.3: The isentropic process appears as a vertical line segment on a T-s diagram.
7.4
ENTROPY CHANGE OF LIQUIDS AND SOLIDS
Recall that liquids and solids can be approximated as incompressible substances since their
specific volumes remain nearly constant during a process.
The entropy change during a process is determined by integration to be as follows:
2
Liquids, solids: 𝑠2 − 𝑠1 = ∫1 𝑐(𝑇)
𝑑𝑇
𝑇
𝑇
≅ 𝑐𝑎𝑣𝑔 𝑙𝑛 2
𝑇1
(kJ/kg.K)
Where 𝑐𝑎𝑣𝑔 is the average specific heat of the substance over the given temperature interval?
The above equation can be used to determine the entropy changes of solids and liquids with
reasonable accuracy.
A relation for isentropic processes of liquids and solids is obtained by setting the entropy
change relation above equal to zero. It gives
𝑇
Isentropic: 𝑠2 − 𝑠1 = 𝑐𝑎𝑣𝑔 𝑙𝑛 2 = 0 → 𝑇2 = 𝑇1
𝑇1
That is, the temperature of a truly incompressible substance remains constant during an
isentropic process.
100
STUDY GUIDE
7.5
TDY2601
THE ENTROPY CHANGE OF IDEAL GASES
The entropy change relations for ideal gases under the constant specific heat assumption are
easily obtained by
T
v
T1
v1
T
P
T1
P1
s2 − s1 = cv,avg ln 2 + Rln 2
(KJ/kg.K)
s2 − s1 = cp,avg ln 2 + Rln 2
7.6
(KJ/kg.K)
REVERSIBLE STEADY-FLOW WORK
The work done during a process depends on the path followed and on the properties at the
end states. Recall that reversible (quasi-equilibrium) moving boundary work associated with
closed systems is expressed in terms of the fluid properties as:
2
Wb = ∫1 Pd𝑣
The work input to steady-flow devices, such as compressors and pumps, as:
2
Wrev,in = ∫1 𝑣d𝑃 + ∆𝐾𝑒 + ∆𝑝𝑒
For the steady flow of a liquid through a device that involves no work interactions, such as a
nozzle or pipe section, the work term is zero and the equation can be expressed as:
𝑉 1 −𝑉12
𝑣(𝑃2 − 𝑃1 ) + 2
2
+ 𝑔(𝑧2 − 𝑧1 ) = 0
Which is Bernoulli equation in fluid mechanics.
It is developed for an internally reversible process; and, therefore, applies to incompressible
fluids that involve no irreversibility, such as friction or shock waves.
Figure 7.4: Reversible work relations for steady-flow and closed systems
101
STUDY GUIDE
7.7
TDY2601
ISENTROPIC EFFICIENCIES OF STEADY-FLOW DEVICES
Now we extend the analysis to discrete engineering devices working under steady-flow
conditions such as turbines, compressors and nozzles; and we examine the degree of
degradation of energy in these devices as a result of irreversibility.
Although some heat transfer between these devices and the surrounding medium is
unavoidable, many steady-flow devices are intended to operate under adiabatic conditions.
The ideal process that can serve as a suitable model for adiabatic steady-flow devices is
isentropic process (figure 7.4).
Figure 7.5: The isentropic process involves no irreversibility and serves as the ideal process
for adiabatic devices.
7.7.1
Isentropic efficiency of turbines
The isentropic efficiency of a turbine is defined as the ratio of the actual work output of the
turbine to the work output that would be achieved if the process between the inlet state and
the exit pressure were isentropic.
𝜂𝑇 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑤𝑜𝑟𝑘
𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑡𝑢𝑟𝑏𝑖𝑛𝑒 𝑤𝑜𝑟𝑘
=
𝑤𝑎
𝑤𝑠
Then the work output of an adiabatic turbine simply becomes the change in enthalpy:
𝜂𝑇 =
ℎ1 −ℎ2𝑎
ℎ1 −ℎ2𝑠
Where ℎ2𝑎 and ℎ2𝑠 are the enthalpy values at the exit state for actual and isentropic processes
respectively (figure 7.6).
102
STUDY GUIDE
TDY2601
Figure 7.6: The h-s diagram for the actual and isentropic processes of an adiabatic turbine.
7.7.2
Isentropic efficiencies of compressors and pumps
The isentropic efficiency of a compressor is defined as the ratio of the work input required to
raise the pressure of a gas to a specified value in an isentropic manner to the actual work
input.
𝜂𝐶 =
𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑤𝑜𝑟𝑘
𝐴𝑐𝑡𝑢𝑎𝑙 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑠𝑜𝑟 𝑤𝑜𝑟𝑘
=
𝑤𝑠
𝑤𝑎
When the changes in kinetic and potential energies of the gas being compressed are
negligible, the work input to an adiabatic compressor becomes equal to the change in
enthalpy,
𝜂𝑐 =
ℎ2𝑠 −ℎ1
ℎ2𝑎 −ℎ1
Where ℎ2𝑎 and ℎ2𝑠 are the enthalpy values at the exit state for actual and isentropic
compression processes respectively (figure 7.7).
Figure 7.7: The h-s diagram of the actual and isentropic processes of an adiabatic
compressor.
When the changes in kinetic and potential energies of a liquid are negligible, the isentropic
efficiency of a pump is defined similarly as:
𝜂𝑃 =
𝑤𝑠
𝑤𝑎
=
𝑣(𝑃2 −𝑃1 )
ℎ2𝑎 −ℎ1
103
STUDY GUIDE
TDY2601
Figure 7.8: Compressors are sometimes intentionally cooled to minimise the work input.
7.7.3
Isentropic efficiency of nozzles
The isentropic efficiency of a nozzle is defined as the ratio of the actual kinetic energy of the
fluid at the nozzle exit to the kinetic energy value at the exit of an isentropic nozzle for the
same inlet state and exit pressure.
𝜂𝑁 =
𝐴𝑐𝑡𝑢𝑎𝑙 𝐾𝐸 𝑎𝑡 𝑛𝑜𝑧𝑧𝑙𝑒 𝑒𝑥𝑖𝑡
𝐼𝑠𝑒𝑛𝑡𝑟𝑜𝑝𝑖𝑐 𝐾𝐸 𝑎𝑡 𝑛𝑜𝑧𝑧𝑙𝑒 𝑒𝑥𝑖𝑡
𝑉2
= 2𝑎
2
𝑉2𝑠
Then the isentropic efficiency of the nozzle can be expressed in terms of enthalpies as:
𝜂𝑁 ≅
ℎ1 −ℎ2𝑎
ℎ1 −ℎ2𝑠
where ℎ2𝑎 and ℎ2𝑠 are the enthalpy values at the nozzle exit for the actual and isentropic
processes, respectively (figure 7.9).
Figure 7.9: The h-s diagram of the actual and isentropic processes of an adiabatic nozzle
Figure 7.10: A substance leaves actual nozzles at a higher temperature (thus a lower
velocity) because of friction.
104
STUDY GUIDE
7.8
TDY2601
ENTROPY BALANCE
The property entropy is a measure of molecular disorder or randomness of a system; and the
second law of thermodynamics states that the entropy can be created but it cannot be
destroyed. Therefore, the entropy change of a system during a process is greater that the
entropy transfer by an amount equal to the entropy generated during the process within the
system; and the increase of entropy principle for any system is expressed as (figure 7.11):
Change in the
Total
Total
Total
( entropy ) − (entropy) + ( entropy ) = ( total entropy )
entering
leaving
generated
of the system
Figure 7.11: Energy and entropy balances for a system
Entropy change = entropy at final state–entropy at initial state
or
Entropy change of a system: ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚 = 𝑆𝑓𝑖𝑛𝑎𝑙 − 𝑆𝑖𝑛𝑖𝑡𝑖𝑎𝑙 = 𝑆2 − 𝑆1
7.8.1
Mechanisms of entropy transfer, Sin and Sout
Entropy can be transferred to or from a system by two mechanisms: heat transfer and mass
flow (in contrast, energy is transferred by work also).
Entropy transfer by heat transfer: Sheat =
Q
T
(T = Constant)
Entropy transfer by mass flow: 𝑆𝑚𝑎𝑠𝑠 = 𝑚𝑠
7.8.2
Entropy generation, 𝑺𝒈𝒆𝒏
The term 𝑆𝑔𝑒𝑛 represents the entropy generation within the system boundary only (figure
7.12) and not the entropy generation that may occur outside the system boundary during the
process because of external irreversibility.
The entropy balance can also be expressed on a unit-mass basis as:
(𝑆𝑖𝑛 − 𝑆𝑜𝑢𝑡 ) + 𝑆𝑔𝑒𝑛 = ∆𝑆𝑠𝑦𝑠𝑡𝑒𝑚
(kJ/kg.K)
105
STUDY GUIDE
TDY2601
Figure 7.12: Mechanisms of entropy transfer for a general system
7.9
WORK ACTIVITIES
Activity 7.1:
Entropy and the increase of entropy principle
Question
Is an isothermal process necessarily internally reversible? Explain your answer with an
example.
Answer
No. An isothermal process can be irreversible. Example: A system that involves paddlewheel work while losing an equivalent amount of heat.
Question
Heat in the amount of 100 kJ is transferred directly from a hot reservoir at 1200 K to a cold
reservoir at 600 K. Calculate the entropy change of the two reservoirs and determine if the
increase of entropy principle is satisfied.
Answer
Analysis: the entropy change of the source and sink is given by
∆S =
QH
TH
+
QL
TL
=
−100 kJ
1200 K
+
100 kJ
600 K
= 0.0833 kJ/K
Since the entropy of everything involved in this process has increased, the transfer of heat is
possible.
106
STUDY GUIDE
TDY2601
Activity 7.2 Entropy changes of pure substances
Question
Is a process that is internally reversible and adiabatic necessarily isentropic? Explain.
Answer
Yes, because an internally reversible adiabatic process involves no irreversibilities or heat
transfers.
Question
Refrigerant-134a is expanded isentropically from 800 kPa and 600C at the inlet of a steadyflow turbine to 100 kPa at the outlet. The outlet area is 1 m2 and the inlet area 0.5 m2.
Calculate the inlet and outlet velocities when the mass flow rate is 0.5 kg/s.
Answer
Activity 7.3:
Entropy change of incompressible substances
Question
A 25 kg iron block initially at 3500C is quenched in an insulated tank that contains 100 kg of
water at 180C. Assuming the water that vaporises during the process condenses back in the
tank, determine the total entropy change during this process.
Answer
107
STUDY GUIDE
Activity 7.4:
TDY2601
Entropy change of ideal gases
Question
An ideal gas undergoes a process between two specified temperatures, first at constant
pressure and then at constant volume. For which case will the ideal gas experience a larger
entropy change? Explain.
Answer
𝑇
∆𝑠 = 𝑐𝑝 𝑙𝑛 ( 2⁄𝑇 ) for a constant pressure process
1
𝑇
And ∆𝑠 = 𝑐𝑣 𝑙𝑛 ( 2⁄𝑇 ) for a constant volume
1
Note that 𝑐𝑝 > 𝑐𝑣 the entropy change will be larger for a constant pressure process.
Question
108
STUDY GUIDE
TDY2601
Air is compressed in a piston-cylinder device from 90 kPa and 220C to 900 kPa in a
reversible, adiabatic process. Determine the final temperature and the work done during this
process, assuming (a) constant specific heats and (b) variable specific heats for air.
Answer
Activity 7.5: Reversible steady-flow work
Question
Saturated water vapour at 1500C is compressed in a reversible steady-flow device to 1000
kPa while its specific volume remains constant. Determine the work required, in kJ/kg.
Answer
109
STUDY GUIDE
Activity 7.6:
TDY2601
Isentropic efficiencies of steady- flow devices
Question
Steam at 3 MPa and 4000C is expanded to 30 kPa in an adiabatic turbine with an isentropic
efficiency of 92%. Determine the power produced by this turbine in kW when the mass flow
rate is 2 kg/s.
Answer
110
STUDY GUIDE
TDY2601
Question
Steam at 4 MPa and 3500C is expanded in an adiabatic turbine to 120 kPa. What is the
isentropic efficiency of this turbine if the steam is exhausted as a saturated vapour?
Answer
Activity 7.7:
Entropy balance
Question
Oxygen enters an insulated 12-cm-diameter pipe with a velocity of 70 m/s. At the pipe
entrance, the oxygen is at 240 kPa and 200C; and at the exit it is at 200 kPa and 180C.
Calculate the rate at which entropy is generated in the pipe.
Answer
111
STUDY GUIDE
TDY2601
Question
An ordinary egg can be approximated as a 5.5 cm diameter sphere. The egg is initially at a
uniform temperature of 80C and is dropped into boiling water at 970C. Determine (a) how
much heat is transferred to the egg by the time the average temperature of the egg rises to
700C and (b) the amount of entropy generation associated with this heat transfer process by
taking the properties of the egg to be 𝜌 = 1020
𝑘𝑔
𝑚3
Answer
112
and 𝑐𝑝 =3.32 kJ/kg.0C.
STUDY GUIDE
7.10
TDY2601
SUMMARY
•
Entropy changes of pure substances
•
Isentropic processes
•
Entropy change of liquids and solids
•
The entropy changes of ideal gases
•
Reversible steady-flow work
•
Isentropic efficiencies of steady-flow devices
•
✓ Isentropic efficiency of turbines
✓ Isentropic efficiencies of compressors and pumps
✓ Isentropic efficiency of nozzle
Entropy balance
✓ Mechanisms of entropy transfer, Sin and Sout
✓ Entropy generation, Sgen
Work activities
•
113