ENGINEERING
THERMODYNAMICS

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ENGINEERING
THERMODYNAMICS

An Introduction

M. S. Kassim, Ph.D.
D. S. Khudhur, Ph.D.
H. M. Hussain, Ph.D.
L. J. Habeeb, Ph.D.

MERCURY LEARNING AND INFORMATION
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Boston, Massachusetts
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M. Kassim, D. Khudhur, H. Hussain, L. Habeeb. Engineering Thermodynamics.
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CONTENTS

Chapter 1:Introduction

1.1Definition and Scope of Thermodynamics

1.2Heat and Work

1.3Thermodynamic Systems

1.4Macro and Micro Approach to Study the System Matter

1.5System Stability

1.6Property Relations

Review Questions

Numerical Exercises

Chapter 2:Laws of Thermodynamics

2.1Introduction

2.2Zeroth Law

2.3First Law of Thermodynamics

2.4First Law of Thermodynamics for a Closed System

2.5First Law of Thermodynamics for an Open System

2.6Second Law of Thermodynamics

Review Questions

Numerical Exercises

Chapter 3:Second Law of Thermodynamics and Entropy

3.1Introduction

3.2Reversible and Irreversible Processes

3.3Entropy

3.4Entropy Balance for Closed Systems

3.5Entropy Balance for Open Systems

3.6Entropy of the Ideal Gas

3.7Maxwell Equations

3.8Euler’s Cyclic Chain Rule

3.9Entropy Differentials

3.10Temperature–Entropy (T-s) Diagram

3.11Enthalpy–Entropy (h-s) Diagram

3.12Third Law of Thermodynamics (Nernst Law)

Review Questions

Numerical Exercises

Chapter 4:Energy Analysis of Systems

4.1Introduction

4.2Definition of Exergy

4.3Exergy Analysis of a System

4.4Exergy Loss of a System

4.5Exergy Equation

Review Questions

Numerical Exercises

Chapter 5:Thermodynamic Relations

5.1Types of Thermodynamic Properties

5.2Equations of State

5.3Maxwell’s Relations

5.4Flow Exergy of Matter

5.5Exergy Efficiencies

5.6Exergy Transfer by Heat

Review Questions

Numerical Exercises

Chapter 6:Thermodynamic Cycles

6.1Introduction

6.2Cycle Definition

6.3Performance Parameters

6.4Carnot Vapor Power Cycle

6.5Principles of the Carnot Cycle

6.6Ideal Rankine Cycle

6.7Stirling Cycle

6.8Ericsson Cycle

6.9Refrigeration Cycle

Review Questions

Numerical Exercises

Chapter 7:Thermodynamic Applications

7.1Engineering Applications of Steady Flow Energy Equation

7.2Diffusers and Nozzles

7.3Water Turbine

7.4Throttling Devices

7.5Mixing Chambers

7.6Gas Turbine or Steam Turbine

7.7Compressors and Pumps

7.8Fans and Blowers

7.9Centrifugal Water Pump

7.10Centrifugal Compressor

7.11Reciprocating Compressor

7.12Boiler

7.13The Heat Exchangers

7.14Condenser

7.15Evaporator

7.16Flow in Ducts and Pipes

7.17Unsteady Flow Processes

7.18Exergy Efficiency of Heat Exchanger

7.19Exergy Efficiency of Expansion (Turbine) and Compression (Compressor)

Review Questions

Numerical Exercises

Chapter 8:Measurement Theory and Instruments

8.1Introduction

8.2Performance and Purpose of Measurement Systems

8.3Measurement Variables

8.4Measurement of Temperature

8.5Pressure Measurement

8.6Measurement of Velocity

8.7Flow Measurement

8.8Measurement System Error

Review Questions

Numerical Exercises

Appendix: Thermodynamic Tables SI Units

References

Index

CHAPTER 1

INTRODUCTION

1.1 DEFINITION AND SCOPE OF THERMODYNAMICS

Thermodynamics is a basic science that deals with the conversion of heat into mechanical work. Nowadays, the common concept of thermodynamics is that it is concerned with the understanding of changes in behavior and properties of matter affected by temperature change.

This concept is also known as energy science, establishing relations among work, heat, and features of the systems, which are in equilibrium. It depicts the changes in the physical systems through an interaction of energy. Energy is neither formed nor annihilated and total energy remains constant.

This concept deals with real physical systems that might be solid, fluid, mixtures, or even empty space with electromagnetic waves. The system should be some type of vessel where no chemical reaction takes place.

This basic science studies changes in the properties or behavior of the fluid cooled down or heated. The fluid may be a gas (such as air), vapor (such as water vapor), liquid, or a mixture of the substances provided they do not react chemically with each other.

It also studies the relationship between changes of fluid properties and quantities of work and heat causing this change.

Thermodynamics is said to be based on four basic principles or laws found through experience and not through mathematical derivation.

These laws are as follows:

1.Zeroth law deals with thermal equilibrium under which temperature is defined. It is called zeroth law because it was drafted after the first law.

2.First law is the law of the conservation of energy, that is, energy can neither be created nor destroyed.

3.Second law determines the direction of the process or the direction of the energy flow and the percentage of energy conversion.

4.Third law determines the entropy and shows the impossibility to reach absolute zero entropy. Entropy has zero value at absolute zero temperature.

Engineers use this science to design thermal engines such as power plants, reciprocating engines, jets and missiles, gas and steam turbines, steam boilers, compressors, air conditioners, and other machines and plants. It is very necessary for engineers to understand the concepts and laws of thermodynamics.

The conversion of heat into mechanical work has been known since the 18th century. In the middle of the 19th century, a scientist named James Prescott Joule created the relationship between mechanical work and thermal energy. Many scientists have contributed to the development of thermodynamics such as Carnot, Kelvin, Clausius, and others. The present-day thermodynamics deals with all thermal machines and air conditioners, which are consumers of energy.

1.2 HEAT AND WORK

Heat is the energy transferred between the system and its surroundings due to the difference of temperature only. There are three methods for the transfer of heat: convection, radiation, and conduction.

Heat is also considered as a method of altering the energy content of the system by the effective variance of temperature. Another method for altering the energy of a system is called work. We can define work as electric work (in an electric motor), mechanical work (in a piston-cylinder and engine), and chemical work.

1.2.1  Types of Mechanical Work

In mechanics, when a force F travels through a distance dx, work is being done. The examples of mechanical work are as follows:

1.the elasticity of a spring,

2.moving system boundary or displacement or flow of energy,

3.surface tension, and

4.shaft rotation on shaft energy.

The concept of thermodynamics was developed to understand the limitation of the efficiency of an engine. Modern power plants work on complicated cycles when compared with the operation of original steam engines, which had only a small percentage of effectiveness in the conversion of heat into work, whereas new utility steam power plants or gas turbine plants have efficiency higher than 60%.

1.2.2  Forms of Energy

There are many forms of energy such as potential, kinetic, thermal, magnetic, chemical, mechanical, nuclear, electrical, etc.

If E is the total energy of a system, the specific energy indicated by e is stated as energy per unit mass:

Energy cannot be destroyed or created. It can only be transferred or changed from one form to another, that is, energy is conserved. Electrical energy, thermal energy, or mechanical energy can be transformed from one form to another. Heat engines convert thermal energy into mechanical energy.

1.2.3  Displacement Work

Work is done when the point of application of a force moves in the direction of force:

W = F × x

where W represents work, F represents force, and x represents displacement.

If the force changes with displacement, then

(1.1)

or

(1.2)

Figure 1.1 shows a piston-cylinder arrangement. In position 1, a gas at pressure p is confined inside a cylinder. “A” represents the cross-sectional area of the cylinder. Distance X represents the top of the piston in the cylinder. Some weights may be placed on the piston top to balance gas pressure in a stable state.

FIGURE 1.1  Piston cylinder-arrangement.

In state 2, the piston moves to a new height X + dX with dp + p as a new pressure in the cylinder to balance a new force of weight.

Obviously work is done as a force acts through distance dX. The amount of work done is given as follows:

(1.3)

But, F varies during the process.

For state 1, force F = p × A

For state 2, force F = (p + dp) × A

The average value of force is given as follows:

Neglect dp dX as very small

(1.4)

Now, , where dV is the difference in volume, and the significant equation will be given as:

(1.5)

By integration,

(1.6)

1W_2 represents the area under a curve in a p − V graph. Figure 1.2 shows two processes A and B joining the same points 1 and 2.

FIGURE 1.2  p − V graph of work for two diverse procedures joining the similar state.

The work 1W_2 depends on the selected path, and not the endpoints. The work is the area on p − V diagram beneath the process line.

1.2.4  Polytropic Processes

The polytropic process is explained by the following equation:

n is identified as the polytropic index as presented in Figure 1.3.

FIGURE 1.3  p − V diagram for various polytropic processes.

Work is done by a system if the sole effect on the surrounding (i.e., the things exterior to the system) could be reduced to the rising of a weight.

Sign convention

●Work done by the system on the surrounding is positive.

●Work done by the surrounding system is negative.

1.2.5  Other Forms of Work

In addition to pressure forces, there are other types of forces. These forces also can do the work:

●A wire is elongated by a force of tension τ through length dL. The work done is given as follows:

(1.7)

●Surface tension δ. The differential work is given as follows:

(1.8)

●Electrical work in a system having ε as potential and electrical charge, q of an atom, and the distance as x, will be

(1.9)

In general,

(1.10)

The work done by gas expanding in a vacuum is given as:

Because it is a non-steady process.

1.2.6  Work Summary

TABLE 1.1  Work summary for an ideal gas (unit mass)

EXAMPLE 1.1

In a piston-cylinder arrangement, the gas is initially at 150 kPa and occupies a volume of 0.03 m^3. The gas is heated so that the volume of gas increases to 0.1 m^3. Calculate the work done by the gas if the volume is inversely proportional to the pressure.

Solution:

EXAMPLE 1.2

Compute the work which a pump shall have to do upon water in an hour to just force the water into a tank (closed) having a pressure of 1.0 MP at the rate of 1.5 m^3/min horizontally from an open well.

Solution:

EXAMPLE 1.3

The flow energy of 0.124 m^3/min of a fluid crossing a boundary of a system is 18 kW. Find the pressure at this point.

Solution:

EXAMPLE 1.4

Determine the size of a spherical balloon filled with hydrogen at 30°C and atmospheric pressure for lifting 400 kg payload. Atmospheric air is at a temperature of 27°C and the barometer reading is 75 cm of Hg.

Solution:

Gas constant for hydrogen:

Apply equation of state:

Mass of hydrogen filling the balloon:

EXAMPLE 1.5

Three kg of air kept at 100 kPa and 300 K is compressed polytropically to 1500 kPa and 500 K.

Calculate:

(i) index,

(ii) final volume,

(iii) work done, and

(iv) heat exchanged.

Solution: (i) Index

(ii) Final volume

(iii) Work done

Work is done in the air during compression.

(iv) Heat exchanged

Heat is rejected by the system.

EXAMPLE 1.6

Calculate the work done in a piston-cylinder arrangement during an expansion process, where the pressure is given by the equation:

The volume changes from 1 m^3 to 4 m^3 during expansion.

Solution:

EXAMPLE 1.7

The cylinder of a compressor contains 1 kg of NH_3. The initial pressure of NH_3 is 2 MPa, and the temperature is 180°C. Cooling the gas to 40°C, then to 20°C, at this state the NH_3 quality is 50%. Determine the overall work for the process, which is a direct function of the difference of V against p.

Solution:

The area under the curve of the graph p − V gives total work:

FIGURE 1.4

1.2.7  Heat

Heat is a form of energy, which is transferred across the boundary of a system at a definite temperature to another system at a different temperature due to the difference in temperature between the two.

There are three modes of transfer of heat:

1.Conduction is also called heat spread due to some factors. Bacon is cooked by conduction. This is explained by Fourier’s law.

For one dimensional heat conduction:

2.Convection: The conduction is enhanced by the flow of fluid as seen in Figure 1.5 Convective effects are explained by Newton’s law of cooling:

FIGURE 1.5  Convective heat transfer from the hot body.

where h is a constant.

3.Thermal radiation: The sun heats the earth by radiation. This is due to the remote effects of energy diffusion. The rate of heat transfer by radiation is given by:

σ is called Stefan–Boltzmann constant.

The sign convection of heat transfer is as follows:

●Heat entering into the system is +ve.

●Heat leaving from the system is −ve.

EXAMPLE 1.8

A piston/cylinder engine contains water vapor of quality 0.7. The pressure is 200.00 kPa at an initial volume of 0.1 m^3. The temperature of the system is raised to 200.00°C. Estimate the heat and the work transfer by the system.

Solution: Control volume contains the water vapor of 0.7 dryness fraction.

Equation of continuity:

Equation of energy 5.11:

When p is constant

Stage 1: Refer to steam Table B.1.2

Saturation temperature at

From the first conditions, the mass, m, is given by:

From the steam table, we get

From the equation of energy, the work is:

The value of heat is given by:

1.3 THERMODYNAMIC SYSTEMS

A system in thermodynamics is a quantity of matter for the study and analysis of a problem. The system needs more surveys. Systems can be a simple cylinder containing gas or as complicated as a nuclear power plant. The whole region outside the system is called the surrounding. Systems are separated from the surroundings by real or imaginary limits known as boundaries as shown in Figure 1.6.

FIGURE 1.6  Thermodynamic system.

The boundary may be fixed or movable and real or imaginary.

The boundary may be in motion or at rest.

●Heat could be exchanged between the system and the surrounding.

●Work could be exchanged between the system and the surrounding.

Thermodynamic systems could be classified according to constant mass or a constant volume of the system.

1.3.1  Closed System

It is a control mass system that contains a fixed amount of matter. No mass can leave or enter a closed system border as seen in Figure 1.7.

FIGURE 1.7  Closed system (control mass).

Mass cannot cross the borders of a closed system. The energy can be exchanged at the boundary.

Energy can be either heat or work. The volume of a closed system can change and may not be constant. As a special case, an isolated system is defined as a system that is not affected by its surroundings and there is no mass or energy exchange.

1.3.2  Open System

In an open system, both energy (heat and work) and mass can cross the boundary, for example, nozzle, turbine, or compressor. These equipments are studied as the control volume system, where energy and mass can pass its border (Figure 1.8) but volume remains constant or fixed.

FIGURE 1.8  Open system.

Figure 1.9 shows energy interaction and one or many of the system properties may change. The system might reach a new equilibrium state.

FIGURE 1.9  System energy.

A large number of engineering devices are open systems where the flow of matter takes place into the system and out of the system. These are designed as control volume systems. For example, a car radiator, a turbine, a compressor, and a water heater.

(i) Control Volume system: It has constant volume with mass passing through its border.

(ii) Control Surface: The border of the control volume system is called Control Surface.

1.3.3  Isolated System

Figure 1.10 shows an isolated system. In this system, there is no exchange or transfer of mass and energy between the system and the surrounding.

FIGURE 1.10  Isolated system.

1.3.4  Adiabatic System

It is a system that is thermally insulated at its boundary. It has a fixed thermal reservoir. If there is no work exchange, it can be an insulated system.

1.3.5  Similar System (Homogeneous System)

A phase can be defined as a matter with a similar physical structure and chemical composition. The system with single-phase matter is defined as a similar or homogeneous system. For example, a mixture of nitric acid with water, heptane with octane, and vapor of water with air.

1.3.6  Dissimilar System (Heterogeneous System)

A system that contains matter in two or more phases is called a heterogeneous system. Example: mixture of steam with water or water with ice.

1.4 MACRO AND MICRO APPROACH TO STUDY THE SYSTEM MATTER

A thermodynamic system may be a rod of steel or a vessel containing gas at some pressure. Most of the systems in engineering are gas containers. As per Avogadro’s Hypothesis, an equal volume of all gases under identical conditions of temperature and pressure contain same number of molecules. The number of molecules may be 10^23 moving randomly in all directions causing a collision of the order of 10^9 per unit time. They are separated from each other by a distance of the order of main-free path. The matter is neither continuous nor homogeneous from a microscopic point of view.

The thermodynamic properties are found out by statistically averaging the behavior of individual molecules. The results obtained are very accurate. For research and study of the law of thermodynamics deeply, statistical thermodynamics is used.

The microscopic approach is rather complex, cumbersome and time-consuming. However, engineering systems working at high pressure contain the large number of molecules in comparison to small volumes. The linear dimensions of the system are quite large as compared with inter-molecular distance, which can be neglected. It is assumed that the system contains the continuous distribution of matter with no voids or cavities. The system is regarded as a continuum. The macroscopic approach is based on the continuum hypothesis. All intensive properties such as pressure, temperature and density are regarded as definite values. The results shown are sufficiently accurate. The study is simple and quick.

The variance between microscopic and macroscopic approaches is given in Table 1.2.

TABLE 1.2

Macroscopic approach

Microscopic approach

A certain amount of material is measured without taking into consideration the behavior of the individual molecule. This approach is concerned with gross or overall behavior of system material.

This is recognized as conventional thermodynamic or classical thermodynamics.

The system is made of a large number of separate particles called molecules. These molecules have different velocities and energies. The values of these energies are constantly changing with time.

The approach is related to statistical thermodynamics.

1.5 SYSTEM STABILITY

In mechanics, a body is stable if it is in equilibrium. The principle of equilibrium states, “A stationary body subjected to a system of forces will be in equilibrium if the resultant of the forces and resultant moment are zero.”

The properties of the system are measured and processes are studied when the systems are stable or in thermodynamic equilibrium. A system having the same values of properties (pressure, temperature, velocity, elevation, composition, etc.) throughout is in thermodynamic equilibrium. The system should be simultaneously in the state of mechanical, thermal, electrical, and chemical equilibrium.

1.Thermal equilibrium denotes uniformity of temperature or absence of temperature gradient or heat flow.

2.Mechanical equilibrium denotes uniformity of pressure or absence of unbalanced forces.

3.Electrical equilibrium denotes the absence of electrical potential and the absence of current flow.

4.Chemical equilibrium denotes the absence of phase change or chemical reaction and there is no mass diffusion.

1.6 PROPERTY RELATIONS

The functional relationship among the thermodynamic properties such as pressure, volume and temperature is known as the equation of state or characteristic gas equation.

The equation can be developed by combining Boyle’s Law and Charles’s Law for a perfect gas:

(1.17)

The above equations are called characteristic gas equations:

1.6.1  Universal Gas Constant

The universal gas constant or molar constant of a gas is the product of the characteristic gas constant (R) and the molecular mass (M) of the gas.

The value of is the same for all gases and can be derived from Avogadro’s law where:

EXAMPLE 1.9

A tank of 0.35 m^3 capacity contains H_2S gas at 300 K. When 2.5 kg of gas is withdrawn, the temperature and pressure in the tank become 288 K and 10.5 bar, respectively. Calculate the mass of the gas initially present in the tank. Also, determine the initial pressure of the gas.

Solution:

The molecular mass of H_2S gas is given as:

Applying equation of state for final condition

Applying equation of state for the initial condition

REVIEW QUESTIONS

1.Define thermodynamics. Discuss its scope. List the laws of thermodynamics.

2.Differentiate between work and heat transfer.

3.Explain in detail about displacement work. What are other forms of work?

4.What is a thermodynamic system? List various types of systems.

5.Differentiate between:

a.Open and closed systems.

b.Insulated and adiabatic systems.

c.Similar and dissimilar systems.

6.Explain the following:

a.Continuum

b.Phase

c.Elastic work

d.Polytropic index

7.Compare macro and micro approaches to the study of system matter.

8.Write notes on:

a.Thermodynamic equilibrium

b.Universal gas constant

9.Establish a relationship for an ideal gas. What is the significance of the equation of state?

NUMERICAL EXERCISES

1.Estimate the heat required to raise the temperature of fluid from 300 K to 400 K if specific heat is given by

The mass of fluid is 2 kg. What will be the mean specific heat?(180 kJ, 0.9 kJ/kg K)

2.One mole of an ideal gas at 0.1 MPa and 300 K is heated at constant pressure till the volume is doubled and then it is allowed to expand at constant temperature until the volume is doubled again. Calculate the work done by the gas.(6.15 MJ)

3.Calculate the change of enthalpy as 1 kg of oxygen is heated from 500 K to 2,000 K. The value of specific heat constant pressure is given as

(232,559.55 kJ/kg)

4.An engine cylinder has a piston area of 0.12 m^2 and contains gas at a pressure of 1.5 MPa. The gas expands according to a process, which is represented by a straight line on a pressure-volume diagram. The final pressure is 0.15 MPa. Calculate the work done by the gas on the piston if the stroke is 0.3 m.(23.4 kJ)

5.10 kg mol of a gas occupies a volume of 603.1 m^3 at a temperature of 140°C while its density is 0.464 kg/m^3. Find its molecular weight and gas constant and its pressure.(28,296.9 J/kg.K)

CHAPTER 2

LAWS OF THERMODYNAMICS

2.1 INTRODUCTION

The laws of thermodynamics define the conversion of energy from one form to another. The first law of thermodynamics deals with the conservation of energy. It discusses the irreversible and reversible, steady and unsteady processes. It expresses the equations of energy for various types of engineering devices. The applications of second law are also discussed.

2.2 ZEROTH LAW

If there are three systems in thermal equilibrium and two of them have the same temperature, these two systems will have same temperature as that of the third one.

The idea of thermal equilibrium or temperature has been given by the zeroth law of thermodynamics.

The basis for temperature measurement is also given by the same law.

If the temperatures of two bodies 1 and 2 are compared with the help of a third body 3, the temperatures of bodies 1 and 2 are same even in the absence of thermal contact between them. Thermometer refers to body 3.

2.3 FIRST LAW OF THERMODYNAMICS

2.3.1  Definition

Energy can only be transformed, or changed from one form to another. It cannot be generated or crushed. Therefore, energy can neither be created nor destroyed. Total energy is conserved.

First law of thermodynamics gives the foundation for relations among the different types of energies and interactions of energies.

2.3.2  Energy Conservation

The total energy of the system is denoted by E. The unit of energy is J. It contains the following types of energy forms:

●thermal,

●kinetic,

●electrical,

●chemical,

●potential, and

●magnetic energies.

2.3.3  Energy Balance

The difference between the total energy entering the system and the total energy leaving the system is equal to the change in total energy of the system.

Or,

This relationship is called the equation of energy balance and is applicable to all types of systems undergoing any process.

Change of energy of a system due to a process can also be calculated by differences between energy at the initial and at the final states.

Change of energy = E final state = E initial state.

Or,

If the state of system does not change by the process, change of energy of the system will be zero.

Different kinds of energies are prevailing some of which are internal (chemical, latent, nuclear, and sensible), magnetic, electric, potential, and kinetic. If surface tension, magnetic, and electric effects are absent, the total energy of a system will be the sum of its potential, kinetic, and internal energies.

Or

Figure 2.1 represents the first law of thermodynamics.

FIGURE 2.1  First law of thermodynamics

For unit mass,

● The internal energy per unit mass, or specific internal energy =

●The transfer of heat per unit mass,

●The work per unit mass,

Most system has zero elevation or velocity. Therefore, changes in potential and kinetic energies are 0 (i.e., )

EXAMPLE 2.1

A system of mass m has and . This is a system with fixed internal energy, held at stationary height, and there are no heat interchanges with its surroundings. The first law gives equilibrium between variations in work and kinetic energy. The system is being speeded by a straight force F through a distance x. There is no force of friction. The system can be shown by Figure 2.2.

FIGURE 2.2  System speed by a force

Solution: The first law of thermodynamics is given by:

Newtonian mechanics is expressed as follows:

Work is done by the system on the surrounding, hence the sign of work is negative.

Newton’s second law of thermodynamics shows that

Divide both sides by dt,

Assume that v_1 = 0 and the kinetic energy is zero at this position,

EXAMPLE 2.2

Cold water is heated from 15°C to 75°C by steam at 300°C and 500 kPa for home hot water supply. If there is a condition that the steam must not condense, what is the quantity of steam required per kg of liquid water?

Solution:

There is transfer of heat from steam into liquid water. No work is done.

Energy equation for water is given by:

Properties of the liquid water can be taken from steam tables.

Energy of steam line has identical transfer of heat but it comes out of the steam

Energy equation for steam:

For properties of the steam, see steam table at a pressure of 500 kPa

2.3.4  Mechanisms of Energy Transfer, Ein and Eout

Transfer of energy from or to a system can take place in three forms: mass flow, heat, and work. Work and heat transfer represents only two forms of energy transfer in a closed system of constant mass.

1.Transfer of Heat: Heat transfer to a system (heat earn) raises the molecular energy and hence the internal energy of the system, and heat transfer from a system (heat loss) decreases the energy of the molecules of the system and hence the internal energy of the system.

2.Transfer of Work: When work is supplied to a system, it raises the system’s energy (i.e., work transfer from a system). The internal energy of the system changes accordingly.

3.Flow of Mass: Mass flow represents an extra mechanism of transfer of energy. Because mass carries energy with it, thus when mass goes into a system, the energy confined within the system drops because the exit mass carries out some of the energy with it.

2.4 FIRST LAW OF THERMODYNAMICS FOR A CLOSED SYSTEM

For adiabatic closed systems, transfer of heat is zero. The transfer of work is zero for systems with no interaction of work. The energy transfer with mass is zero for systems that have no flow of mass through their boundary (i.e., close systems). A simple closed system is shown in Figure 2.3, where all the three possibilities are specified.

FIGURE 2.3  Closed system with energy E replacing heat Q and work W with its surrounding

Let U_1 be the internal energy of a closed system in equilibrium at state 1. If a quantity of heat 1Q_2 is transported through its boundary and an amount of work 1W_2 is applied by the system and the system again is in equilibrium state 2, then,

If u is the specific internal energy of the system and m is the mass of the system then,

where 1w_2 and 1q_2 are work and heat transfer per unit mass of the system.

2.5 FIRST LAW OF THERMODYNAMICS FOR AN OPEN SYSTEM

(i) Flow Work

In an open system, fluid enters and leaves the system. The flow work of the fluid is needed to push to the system against the pressure of the system. To eject the fluid from the system, flow work is the product of specific volume, v and pressure, p.

(ii) Enthalpy

In the study of open systems, it is convenient to add the flow work “pv” with internal energy “u” to raise the system energy. The specific enthalpy h of the system can be defined as the total of flow work and internal energy. Therefore, specific enthalpy h is given as:

The mass crosses the boundary of an open system in addition to interaction of work and heat with the surroundings. The flow of fluid carries kinetic and potential energies.

In Figure. 2.4, V_2 and V_1 are the outlet and inlet velocities, m_2 and m_1 are the rates of mass flow at outlet and inlet, h_2 and h_1 are the enthalpies at outlet and inlet and z_2 and z_1 are the outlet and inlet elevations with reference to a datum; w and q are the rates of work and heat transfer to the system. Total energy of the system is E.

FIGURE 2.4  First law of thermodynamics of the open system

Energy can be transported in different mode such as mass, heat, and work. The net transfer of energy is the difference between the quantities transported at inlet and outlet. The energy equation becomes as follows:

where the subscripts “out” and “in” indicate amounts at exit and inlet of the system, respectively.

The first law of thermodynamics for a control volume can be defined as follows:

2.1

Total enthalpy is explained as:

The total enthalpy is the sum of kinetic, potential, internal, and flow work.

From Eq. (2.1),

or,

where i is at inlet and e at exit of the system, cv is control volume.

Rearranging the above equation, first law of thermodynamics becomes as follows:

For a steady-state open system

For steady-state steady-flow process, mass is constant, and the time rate of change of all quantities becomes equal to 0. The total energy of the system and the mass do not change with time, term dE/dt is 0 and for constant mass, m=m_2=m_1, the energy equation will be as follows:

2.2

where w and q are work and heat rate.
Eq. (2.2) is the equation for first law of thermodynamics.

2.6 SECOND LAW OF THERMODYNAMICS

First law of thermodynamics does not indicate the degree of transformation of cyclic heat into cyclic work. It indicates the possibilities of operations that can happen in nature, which preserve total energy. Clearly it is not the entire story concerning energy. As per first law, there will be no deficiency of energy as energy can be recycled.

First law of thermodynamics creates equality between the amount of the mechanical work and heat applied but does not identify the state conditions, under which transformation of heat into work is probable and not the direction in which transfer of heat can happen. The deficiency has been overcome by the second law of thermodynamics. The second law limits the machines from transforming whole of heat into work.

Entropy S can be defined as a thermodynamic property, which determines the amount of energy destruction.

FIGURE 2.5  Work

2.6.1  Heat Engine

In a cyclic process, the initial state of a system is equal to the final state. The total change of the internal energy U will be zero.

A process is reversible if the heat supplied is equal to the work produced by the system through a cycle. The work is negative if it is supplied to the system.

A cyclic process can be regarded as a heat engine. Consider a heat engine operating between T_1>T_2.

If Q is the heat supplied to the engine from a heat reservoir at temperature T_1, Q_1, is converted into work W, Q_2 represents the part of heat that is delivered to a second reservoir at T_2<T_1 (condenser).

According to the first law of thermodynamics:

A heat engine is shown in Figures. 2.6 and 2.7. There are different types of heat engines. They undergo the following processes.

FIGURE 2.6  Heat engine

FIGURE 2.7  Heat engine of steam power plant

1.Heat engines take heat from a high temperature heat source (nuclear reactor, oil furnace, boiler, etc.)

2.Part of this heat is converted into work

3.The remaining heat is thrown into a low-temperature reservoir (condenser, atmosphere, lake, river)

4.Engine works in a cycle.

The ratio of work produced to heat supplied is called efficiency.

Efficiency = advantage/outlay (cost)

The engines have low thermal efficiencies. For example, engines of steam power plants have a thermal efficiency nearly 40%, sparkle-ignition motor car engine has about 20%, and diesel engines have roughly 30%.

The heat Qout in power cycle cannot be saved. Without cooling in the condenser, the power cycle cannot be completed. All heat engines (even an ideal cycle) must reject some energy to basin at lower temperature so as to complete the cycle.

2.6.2  Heat Pumps and Refrigerators

Heat passes from areas of high temperature to ones of low temperature. The reverse cannot happen passively. Because such transfers need special machines called refrigerators.

Refrigerators are cyclical engines, and the fluids used in the cycles are called refrigerants. Heat pumps are engines that work to transfer heat from a medium at low temperature to a region at higher temperature.

Heat pumps and refrigerators are shown in Figure 2.8. These are basically similar machines but with different purpose only. Refrigerator maintains a region at a low temperature, whereas a heat pump takes heat from a region at low temperature and supplies it to a warmer region.

FIGURE 2.8  Heat pump and the refrigerator

2.6.3  Coefficient of Performance (COP)

The coefficient of performance (COP) can be defined as the performance of heat pumps and refrigerators that is expressed by the following formula:

It can be realized that air conditioners are refrigerators, which cool the regions such as a building or a room.

The quantity of heat removed from the cooled region in BTU’s units for one Watt-hour is called the Energy Efficacy Rating (EER).

In general, air conditioners have an EER in the range from 8.00 to 12.00 (Coefficient of Performance 2.300 to 3.500).

2.6.4  Second Law of Thermodynamics: Clarification of Clausius Statement

It is not possible for a machine that works in a cycle to transfer heat from a body at lower temperature to a body at higher temperature without the aid of external work. In other words, a refrigerator can work only when its compressor is supplied with external power. Clausius and Kelvin-Planck statements of the second law of thermodynamics are contrary to each other. The two explanations of the second law are similar as presented in Figure 2.9. In other words any machine that contradicts the statement must conduct as that of Kelvin-Planck and Clausius and vice-versa.

FIGURE 2.9  Clausius statement and Kelvin-Planck statement

Any machine that contradicts the second law of thermodynamics is called a perpetual motion machine of the second type (PMM2) and the machine that contradicts the first law of thermodynamics (by generating energy) is called a perpetual motion of the first type (PMM1).

EXAMPLE 2.3

A heat engine is supplied with 2512 kJ/min of heat at 650°C. Heat rejection takes place at 100°C. Specify the following results of heat rejection.

(i) 867 kJ/min

(ii) 1015 kJ/min

(iii) 1494 kJ/min

Solution: The schematic diagram of the engine is shown in Figure 2.10

FIGURE 2.10  Heat engine

(i) Q2 = 867 kJ/min

Applying Clausius inequality

The cycle is not possible.

(ii) Q_2 = 1015 kJ/min

The cycle is reversible

(iii) Q_2 = 1494 kJ/min

The cycle is irreversible and possible.

REVIEW QUESTIONS

1. What is zeroth law of thermodynamics? How does the concept of thermal equilibrium lead to invention of thermometer?

2.State the First Law of Thermodynamics. Derive energy equation in different forms.

3.Define:

(i) Total energy

(ii) Flow energy

(iii) Enthalpy

4.Define the conditions for application of First Law of Thermodynamics to a closed system.

5.Derive energy equation for steady-state steady-flow open system.

6.Why is the Second Law of thermodynamics needed?

7.Define engine efficiency and coefficient of performance (COP) of a heat pump and a refrigerator. Prove that COP of a heat pump is more than that of a refrigerator.

8.Explain Perpetual Motion Machines of First Type and Second Type.

NUMERICAL EXERCISES

1.A reversible heat engine operates between reservoirs at 420 and 280 K. If the engine output is 2.5 kJ, determine the efficiency of the engine and its heat interactions with the two reservoirs. Subsequently, the engine is reversed and made to operate as heat pump between the same reservoirs. Calculate the coefficient of the heat pump and power input required when the heat transfer rate from the 280-K reservoir is 4 kW. (5 kJ, 2.5 kW)

2.Obtain the coefficient of performance of the composite refrigerator system in which two reversible refrigerator B only.

3.A Carnot engine E_1 operates between temperatures T_1 and T_2 and engine E_2 operates between temperatures T_2 and T_3 receiving heat rejected from engine E_1. What is the relationship of η_2 with η_3 of engine E_3 working between T_1 and T_3 (η_3 = η_1 + η_2 − η_1η_2)

4.A cold storage of 100 tonnes of refrigeration capacity runs at 1/4th of its Carnot coefficient of performance. Inside temperature is –15°C and atmospheric temperature is 35°C. Determine the power required to run the plant. Take one metric ton of refrigeration as 3.52 kW. (281.6 kW)

5.The steam supply to an engine comprises two streams, which mix before entering the engine. One stream is supplied at the rate of 0.01 kg/s with the enthalpy and a velocity of 20 m/s. The other stream is supplied at the rate of 0.1 kg/s with an enthalpy of 2665 kJ/kg and a velocity of 120 m/s. At the exit from the engine, the fluid leaves as two streams, one of water at the rate of 0.001 kg/s with an enthalpy 421 kJ/kg and the other of stream. The fluid velocities at the exit are negligible. The engine develops a shaft power of 25 kW. The heat transfer is negligible. Estimate the enthalpy of second exit steam. (2489 kJ/kg)

CHAPTER 3

SECOND LAW OF THERMODYNAMICS AND ENTROPY

3.1 INTRODUCTION

Entropy might give an indication about the second law of thermodynamics. The concept of entropy is as fundamental to physics as energy or temperature. However, it is a distinct concept related to the amount of disorder of a system. As the disorder of the system increases, the entropy increases. The system’s molecular motion increases with the increase in thesystem’s internal energy. Heat addition increases the system disorder by increasing the molecular motion randomness. In general, heat addition increases the system disorder. However, heat addition effect on the cold (low temperature) and hot (high temperature) systems is not the same. In other words, for the same amount of the added heat, disorder level in the cold systems is much higher than that of the hot systems. The entropy change for a reversible system, which can be defined as the system in which the variation in the entropy takes place due to the variation in the system’s internal energy, can be calculated as shown in Equation 3.1.

where ∆S, ∆Q, and T represents the entropy change (joule/Kelvin or calories/Kelvin), heat change of the system (Joules or calories), and temperature (Kelvin), respectively.

Second law of thermodynamics may be simplified as follows:

1.The entropy of the system will naturally increase or there will be no change in the system’s entropy if the system is already maintained at the maximum level of entropy.

2.If the system is left to itself, it naturally tends to be in equilibrium within the surroundings.

3.When the system and the surroundings are maintained in equilibrium, there is no changes in the entropy of a system.

For a real process, energy quantity is preserved; however, it changes from one form to another and its quality is certain to decrease. Disorganization is the main result of the process due to the changes in the energy levels and in the forms. Therefore, disorganized energy measurement and the entropy terms are not disconnected. In other words, the terms entropy, energy, and exergy are inter-dependent, and one affects the other in some way as shown in Figure 3.1.

FIGURE 3.1  Interaction between the energy, the exergy, and the entropy.

Figure 3.1 describes the relationship between these three terms so that the unknown can be easily obtained if the other two are known. This can be expressed as follows:

(Entropy balance) = (Energy balance) − (Exergy balance)(3.2)

3.2 REVERSIBLE AND IRREVERSIBLE PROCESSES

According to the second law of thermodynamics, a heat engine cannot achieve 100% efficiency. Then, what is the maximum efficiency that can be achieved by a heat engine? It depends on the following two types of processes: reversible process (an ideal process) and irreversible process. The reversible process can be defined as the process in which the system and the surrounding (both of them) can return to their original state (return of process is possible). Whereas, the irreversible process is defined as the process in which the system and the surrounding (both of them) cannot return to their original state (return process is impossible). Restoring both the system and the surrounding together is the necessary condition for the reversible process. In other words, if any one of them failed to restore back to their original state while the other did, it is said to be an irreversible process. The reversible process is only an assumption for ideal processes, which is difficult to achieve. With certain assumptions, the systems can be approximated to get reasonable results. Quasi-static process is used for two reasons: (a) it is easier for the analysis as long as the system involves several equilibrium states, and (b) it works as the boundaries (limits) for comparison with the actual process. No reversible process is possible to happen in the nature. All the processes happening in the nature are irreversible processes.

The reversible process cannot be accomplished, but the processes are designed to get closer. The closer the process, the more output work delivered or the minimum heat input consumed by the system.

The concept of reversible process and the definition of efficiency of the second law of thermodynamics are not independent for the actual process. System efficiency allows to compare between different machines (designed to do the same task) based on the comparison between their efficiencies. The machines should be designed for the lowest level of irreversibility.

3.2.1  Irreversibility

Irreversibility can be defined as the factors that shift any process from reversible to irreversible process. Some of the good examples of these are heat transfer, frictional force, chemical reactions, electric resistance, and an inelastic deformation of solids. Any one of these factors in a process can lead to irreversibility.

3.2.2  Friction

Friction can be defined as the resistance to the motion for two or more bodies in motion. It is a very familiar form of irreversibility. The frictional force can be developed between any two surfaces in contact, such as the force between the surfaces of a piston moving inside a cylinder. This force consumes some amount of system’s energy and therefore, specific amount of the work is required to substitute the reduction in the system’s energy. In the piston–cylinder example, the frictional energy can be recognized by the heat developed between the two surfaces in contact. Temperature increasing at the same surfaces is another indication about friction. The developed heat does not depend on the direction of motion. Heat develops whatever the direction of motion is. This process can be considered as an irreversible process because the system and its surrounding are not able to return to their original state (irreversible process). The degree of irreversibility varies with the value of force of friction and is linearly dependent.

The force of friction results at the interface between two solid bodies. It can also develop at the interface between a fluid and a solid and even between the layers of the fluid through a fluid flow due to velocity difference. The drag force of rigid surface motion in air is a good example of force of friction that cannot be overcome without applying an external source of work and energy.

3.2.3  Factors Causing an Irreversible Process

There are several factors which make a process irreversible. They are as follows:

●the chemical action and reaction,

●process involves mixing,

●heat transfer,

●unrecovered deformation,

●compression and unrestrained expansion and compression.

3.2.4  Internal and External Reversibilities

The internal irreversibilities occur within the system boundaries. The system may be at several states of equilibrium. It will pass through all these states when returned to its original state.

FIGURE 3.2  Reversible and internally reversible processes.

The external irreversibilities occur outside the system boundaries. For example, heat transfer occur between a system and a reservoir both at the same temperature. In a totally reversible process, both internal and external irreversibilities should be absent.

3.3 ENTROPY

Calculation of entropy requires information about the number of states of molecules energy. However, in classical thermodynamics, entropy can be estimated in terms of properties at the macroscopic level.

3.3.1  Mathematical Definition

For an internally reversible process, the condition = 0 must be valid.

Therefore, for a reversible process,

For convenience, the subscript “int” is ignored.

3.3.2  Characteristics of Entropy

An alternative definition of entropy is an energy distribution measurement at the molecular levels of the matter involved in a system. Higher value of the entropy indicates larger number of levels of energy distribution. T-s diagram can be used to depict the processes as shown in Figure 3.3. The amount of heat transferred (reversible heat) can be represented by the area under the curve’s connecting points 1 and 2 on the T-s diagram. For the adiabatic and reversible process, the term indicates that there is no change in the entropy during the process.

FIGURE 3.3  A process represented on the temperature-entropy (T-s) diagram.

Figure 3.4 shows the Carnot cycle on T-s diagram. The rectangular representation of the cycle means that changes in the entropy of both heat absorption (∆SH) and heat rejection (∆SL) are the same. However, change of entropy for the total system ∆S can be calculated by adding the changes of the entropies of the subsystems together.

FIGURE 3.4  A Carnot cycle represented on a T-s diagram.

3.3.3  Entropy as a Thermodynamic Property

In a closed system, the first law of thermodynamics of a reversible process can be written as follows:

This equation shows that the entropy S and the volume V are the two state variables for which the differential of the internal energy U = U(S, T) becomes exact. One basic pair of state variables will transform to another, namely from (S, V) to (T, V) when one of the variables (entropy S) takes the role of a thermodynamic property. The differential of the entropy can be derived by inserting dS and dU/T of the forms.

By substituting in Equation (3.3), it leads to

where the terms is defined by CV. Partial derivatives of the entropy, as a state function, with respect to T and V can be written as follow:

3.4 ENTROPY BALANCE FOR CLOSED SYSTEMS

3.4.1  Closed Systems

Heat transfer increases the entropy of the system (system disorder) according to the statement presented by Equation (3.5) so that,

(3.5)

The degree of irreversibility is the reason for the change of the entropy of an irreversible process. The entropy generation for an adiabatic system due to irreversible process can be given by:

(3.6)

Equation (3.6) states that entropy generation depends on the degree of the process irreversibility and the factor 1/T_0. However, entropy generation will be zero (Sgen = 0) when the process is reversible and adiabatic. In this case, the process is called “isentropic process.” Equation (3.6) shows that for an isolated system, the system entropy does not change if the process is a reversible process. For , is always valid for an isolated system.

EXAMPLE 3.1

Find the relationship between the pressure and the volume of your process for an ideal gas in piston–cylinder device knowing that the process is isentropic.

Solution: Since the process is isentropic, the change of the entropy must be zero, that is S_2 − S_1 = 0. Therefore, . cp / cv is equal to k so that k = cp / cv. Hence, the process can be described by the expression pvk= constant.

EXAMPLE 3.2

For a substance flow inside your pipe (incompressible flow), derive a mathematical expression for the isentropic shaft work.

Solution: Flow inside your pipe may be affected by several parameters, such as a pump work to overcome the losses due to friction, pipe diameter and cross section, changes in pipe elevation, etc. Taking most of these parameters into consideration, the flow equation (energy equation) can be written as follows:

For incompressible flow,

For frictionless and incompressible flow, the fluid flows due to elevation, velocity, and pressure changes. Steady flow energy in the absence of work done by the pump can be given by,

Or

The last equation, called the Bernoulli’s equation, describes the isentropic flow inside a pipe for incompressible fluids.

3.4.2  Thermodynamic Cycles

The cyclic integral of entropy with respect to time may be expressed as follows:

(3.7)

As entropy is a system property, the integral (cyclic) of entropy of the occupied flared must be zero, that is . For all cyclic processes, the irreversibility cannot be negative; then, from Equation (3.7), Equation (3.8) must be valid for all possible cycles, that is

(3.8)

The inequality shown in Equation (3.8) is called Clausius inequality. It was introduced by R.J.E. Clausius, German physicist in 1870. This inequality is used to test the validity of the second law of thermodynamics for a cycle. This inequality states that is negative for the real cycles, while equals to zero for a reversible process.

FIGURE 3.5  Validity of a heat engine

EXAMPLE 3.3

A heat engine withdraws 325 kJ of heat energy drawn from a high temperature reservoir at 1000K and passes 125 kJ of the heat energy to a low temperature reservoir at 400K.

The required work output (net) of the engine is 200 kJ. Is that possible?

Solution: The first law of thermodynamics states that:

It is possible for the engine to provide the required energy. However, this engine might not be valid in the light of the second law of thermodynamics. Application of Equation (3.7) yields,

Therefore, this engine is not possible.

3.5 ENTROPY BALANCE FOR OPEN SYSTEMS

For open systems, the net entropy transfer because of mass flow can be given as follows:

…(3.9)

In general, the rate of a system’s entropy change can be calculated as follows:

…(3.10)

It is important to notice that system energy and system energy of the steady state steady flow (SSSF) system are constant. For this process, substituting into Equation (3.10) and with some mathematical manipulation results in,

…(3.11)

For an adiabatic SSSF of single input and single output system, Equation (3.11) can be modified into Equation (3.12) as follows:

(3.12)

The quantity Sgen is a positive value due to fluid and surface friction. For the adiabatic SSSF, the inequality Se>Si is always valid and true. This means that the losses in system energy correspond to gains in entropy through the process. In other words, the process at which both energy and entropy are conserved and showed no increase in value is called the principle of entropy increase.

EXAMPLE 3.4

Figure 3.5 shows duct of a ventilation system working in a well-insulated steady state system. The added pressure is 1 atm. Air with cp = 1.005 kJ/kg·K as assumed a perfect gas flows inside the duct. By neglecting potential and kinetic energy, find:

A. air temperature at the exit of the duct,

B. pipe diameter at the exit of the duct,

C. the entropy production rate in the duct measured as kW/K.

FIGURE 3.6  A ventilation system.

Solution:

A.The energy equation for the airflow inside the duct shown in Figure 3.6 can be written as,

The mass conservation yields,

After substituting the numerical values, and .

For an ideal gas, h = cpT. Using this expression in the equation of the energy above, air temperature at the exit can be calculated as,

B.Since and the diameter of the circular cross section becomes D_3 = 1.73 m.

C.For the insulated steady state system,

Using the ideal gas entropy relation, the entropy generated is equal to 2.412103 kW/K.

3.6 ENTROPY OF THE IDEAL GAS

The specific heat Cv and the free energy U of the ideal gas are given as follows:

Using the partial derivatives and the equation of the state PV = nRT of the ideal gas leads to the expression

The entropy difference is given by Equation (3.13) as follows:

(3.13)

where n is the number of moles (constant in this equation).

3.7 MAXWELL EQUATIONS

The commutativity of the differentiation operations, Schwarz’s theorem, can be used to derive relations between thermodynamic quantities.

For the case of the differential, this implies that

where an inversion for the last step was used. The relation is called Maxwell’s equation.

3.7.1 Energy Equation

The entropy is not an experimentally controllable variable, such as T, V, and P, which allowed to the thermal equation measurement of the state P = P(T, V).

However, Equation (3.1) is used to deduce the energy equation, which allows to determine the equation of state U= U(T, V).

Energy equation: Derivatives of the entropy was used as:

Cancelling identical terms leads to:

(3.15)

This is called energy equation.

3.8 EULER’S CYCLIC CHAIN RULE

The partial derivatives in the energy equation (Equation (3.15)) may be replaced by thermodynamic coefficients. The involved type of variable transformation can be applied to a large set of thermodynamic quantities.

Implicate variable dependencies: In general, equation of state is described by a set of variables, example P, V, and T, so that:

From the relative partial derivatives of the state variables,

This relation can be rewritten as:

(3.16)

The importance of Equation (3.16), Euler’s chain rule, lies in the fact that one does not know the equation-of-state function f (P, V, T) explicitly.

Expansion and compression coefficients: Using Equation (3.16), we get:

(3.17)

The thermodynamic coefficients used are as follows:

Using Equation (3.17), the energy equation (Equation (3.15)) leads to Meyer’s relation between Cp and CV.

(3.18)

The energy equation (Equation (3.18)) can be used to rewrite Mayer’s relation, which can be derived by considering the chain rule for (∂U / ∂T)P, as:

This leads to:

(3.19)

3.9 ENTROPY DIFFERENTIALS

The rewritten energy equation (Equation (3.18)) can be used to write the differential (Equation (3.20)) of the entropy as follows:

3.9.1  The Two Independent Variables, T and V

Equation (3.20) of the differential of the entropy implies that the absorbed heat δQ can be expressed likewise in terms of directly measurable coefficients,

(3.21)

where T and V are the independent variables.

3.9.2  The Two Independent Variables, T and P

Using Maxwell’s Equation (3.14),

Entropy can be given in terms of T and P as,

(3.22),

where T and P are the independent variables.

3.9.3  The Two Independent Variables, V and P

The term dT can be rewritten as

Using Equation (3.18), via , and inserting dT into Equation (3.19) leads to:

(3.23)

For T dS, V and P are the two independent pairs of state variables.

3.10 TEMPERATURE–ENTROPY (T-s) DIAGRAM

The work due to system boundary movement can be expressed as follows:

where P is the applied pressure at the boundary. The temperature, pressure, and the other properties of a system involved in an ideal process must be uniform. If not, energy loss occurs because of the internal irreversibility. In this case, volume change of the system requires performing of an external work on the system equivalent to . The pressure in this equation is not the pressure applied at the boundary; but it is the pressure of the system.

Mathematically, the work of an ideal process can be calculated by the area under the p-V curve of the process shown in Figure 3.7. However, the work done for an irreversible process is smaller than the area under the curve for an ideal process. This is because of the non-uniform thermodynamic properties.

FIGURE 3.7  The mechanical work for an ideal process of a pistonÐcylinder system shown by the shaded area

Heat transfer for a reversible process ina closed system can be calculated by the shaded area under the T-s curve as shown in Figure 3.8 (a). Figure 3.8 (b) shows isentropic, isochoric, isobaric, and isothermal processes, on a T-s diagram.

FIGURE 3.8  The ideal heat transfer to a system and T-s presentation of various ideal processes

3.11 ENTHALPY–ENTROPY (h-s) DIAGRAM

Enthalpy–entropy diagram for the SSSF process represents an important diagram.

The term “enthalpy” is an important factor in energy analysis of the systems with fluid flow because it is used to define the exergy of the system or the available energy in the system. Therefore, the processes involving exergy and energy interactions can be illustrated by h-s diagram. This diagram is called Mollier diagram developed by R dart Mollier, a German scientist. Figure 3.9 shows the general characteristics of a Mollier diagram.

FIGURE 3.9  Mollier diagram

On T-s diagram the constant pressure line is a horizontal line for a saturated state,while the constant pressure curves keep declining even for saturated states. In this graph, at a specific pressure, the temperature of the saturation can be found by the crossing point of constant temperature and constant pressure lines at the line of saturated vapor.

The enthalpy change, ∆h, can be obtained by measuring the vertical distance between the required states on this diagram. This represents the consumed work of an adiabatic compressor or the work of the shaft of an adiabatic turbine. It can be the change in the kinetic energy of a fluid flows through a nozzle or a diffuser. On the other hand, the entropy generated ∆s for a fluid flows adiabatically can be obtained by measuring the horizontal distance between two particular states on this diagram.

3.12 THIRD LAW OF THERMODYNAMICS (NERNST LAW)

Statistical study for more experiments done by Nernst showed that the entropy in the limit T → 0 becomes a constant and independent of the other parameters, such a pressure and volume. This principle can be expressed as,

(3.24)

The entropy is defined as follows:

(3.25)

This equation is equivalent in statistical mechanics, where most of the states of matter are referred to a unique ground state. Macroscopically, at ground state leads tolimit T = 0. Entropies are observed only for exotic phases of matter. Heat capacities vanish as T → 0. The heat capacities disappear at T = 0 as a consequence of Equation (3.25) so that

(i) The ideal gas does not fulfil the third law: The heat capacities of the ideal gas are constants and can be given by,

This contradicts the third law which states that gas at high temperature is the limit of the state of matter, which undergoes as gas → liquid → solid upon cooling.

(ii) No thermal expansion for T → 0: Maxwell’s equation, Equation (3.24),

The last step shows that any derivative of a constant is zero.

FIGURE 3.10  Lined method of cooling.

(iii) The absolute T = 0 (zero point) is unattainable: It is required to reach lower temperatures by performing adiabatic and isothermal transformations, subsequently.

Using a gas as a working substance, cooling can be achieved by the lined method through a sequence of isothermal and adiabatic transformations.

(iv) A → B isothermal compression: Work is performed on the gas and an amount of heat Q_1 < 0 is transferred from the matter to be cooled (characterized by a low temperature T_1) to the reservoir (having a higher temperature) in a reversible process. The entropy of the substance being cooled diminishes consequently into

(v) B → C adiabatic expansion: The gas cools by performing work. The entropy remains however (with δQ = 0) constant.

FIGURE 3.11  Isothermal compression and adiabatic expansion.

Note that all entropy curves converge to S (T → 0) → 0. Hence, the process becomes progressively ineffective and then infinite number of lined iterations would be needed to reach the limit T → 0.

REVIEW QUESTIONS

1.What is entropy? What are its characteristics? How is it related to second law of thermodynamics?

2.Differentiate between a reversible and an irreversible process.

3.What is irreversibility? What are its sources? Differentiate between internal and external irreversibilities with examples.

4.Prove that entropy is a thermodynamic property.

5.Derive an equation of entropy balance for a closed system.

6.Derive an expression for an ideal gas undergoing isentropic process in a closed system.

7.Derive Bernoulli’s equation for an incompressible flow of fluid inside a pipe.

8.Derive an equation of entropy balance for an open system.

9.Define the following:

a.Entropy of an ideal gas.

b.Maxwell’s equation.

c.Euler’s cyclic chain rule.

10.What is the difference between T-s diagram and h-s diagram?

11.Explain the third law of thermodynamics

12.Write notes on:

a.Isothermal compression

b.Adiabatic expansion

NUMERICAL EXERCISES

1.A heat engine working on Carnot cycle absorbs heat from three reservoirs at 1000 K, 800K, and 600K. The engine works at 10 kW of network and rejects 400 kJ/min of heat to a heat sink at 300K. If heat supplied by the reservoir at 1000 K is 60% of the heat supplied by the reservoir at 600K, calculate the quantity of heat absorbed from each reservoir(312.5 kJ/min, 500 kJ/min, 187.5 kJ/min)

2.0.25 kg/s of water is heated from 30°C to 60°C by hot gases that enter at 180°C and leave at 80°C. Calculate the mass flow rate of gases where Cp = 1.08kJ/kg·K. Find the entropy change of water and hot gases. Take specific heat of water as 4.186 kJ/kg·K. Also find the increase of unavailable energy if the ambient temperature is 27°C.(0.29 kg/s, 0.0988 kJ, 0.078 kJ, 23.436 kJ)

3.10 m^3 of air at 175°C and 5 bar is expanded to a pressure of 1 bar while temperature is 30°C. Calculate the enthalpy change for the process.(3.445 kJ)

4.5 kg of ice at −10°C is kept in atmosphere which is at 30°C. Calculate the change in entropy of the universe where it mails and comes into thermal equilibrium with atmosphere. Take latent heat of fusion as 335 kJ/kg and specific heat of ice as half that of water.(0.652 kJ)

5.Using an engine of 30% efficiency to drive a refrigerator having a COP of 5, which is the heat input into the heat engine for each MJ removed from the cold body by refrigerator. If the system is used as a heat pump, how many MJ of heat would be available for heating for each MJ of heat input to the engine.(0.67 MJ, 1.87 MJ)

CHAPTER 4

ENERGY ANALYSIS OF SYSTEMS

4.1 INTRODUCTION

The second law of thermodynamics is a very powerful tool for the analysis and the optimization of complex systems. In this chapter, a general introduction to exergy (availability) is given and explained as the maximum useful work that can be drawn from a system enclosed in a specific environment. Reversible work is defined as the maximum work that can be drawn from a process of a system between two specific states. The principle of irreversibility (destruction of energy or the lost work) is presented as a function of irreversibility. Second-law efficiency is defined. Energy balance is developed for control volume and closed system. The datum term condition of the dead state used in the calculations of the exergy or the available energy of a system is taken as the ambient conditions of the pressure P_0 and the temperature T_0, when a system is in mechanical and thermal equilibrium.

4.2 DEFINITION OF EXERGY

The word “exergy” was first used by Rant in 1956, and related functions had been presented by Keenan and Gibbs. Exergy of a system can be defined as the maximum theoretical work that can be obtained from a specific amount of energy. The term “exergy” (at a given time) can be defined as the maximum useful work that can be obtained from a system at equilibrium with the surrounding. Both energy and exergy are system properties. The more complete definition for the exergy is the work that can be obtained from the system which works ideally (under ideal conditions, reversible process) and the environment is the heat reservoir. Availability is another term used to define the exergy of a system especially involved with the heat exchange. Energy is the maximum work obtained from a system undergoing heat exchange process maintained at equilibrium with the environment.

Exergy of a system is useful to check the possibility of energy harvesting from the system, for example, from exhaust temperature of the power plants. The combination of the first and the second laws of thermodynamics is used for exergy computation. Mathematically, it can be expressed as follows:

(4.1)

4.3 EXERGY ANALYSIS OF A SYSTEM

Energy contents of a system cannot be totally converted into useful work. Some portion of the system’s energy will have to be rejected as waste energy. Hence, it is necessary to have a system property that enables to control the useful work potential of a specific energy of the system.

The characteristics of exergy can be stated as follows:

1.Exergy defines the maximum work that can be transferred by a system at a given time, t. For a given state of a system, there is only a single value of the maximum work, irrespective of the path taken by the system to reach its present state at time t. Exergy is a system property and it is independent on the path of the process. It is exactly like p, T, v, h, and u of the system. However, exergy depends on the state of the system and its environment.

2.Exergy of a system is zero when the system is at equilibrium with the surrounding. The exergy of a system is always positive. Negative exergy has no meaning.

3.As shown in Figure 4.1, as the system changes its state from an initial to an intermediate state, its exergy also changes. However, only some percentage of this change can be transformed into work. Regardless of the process and method, the useful work will always be less than the change in the exergy of the system.

4.System’s exergy increases with its difference from its environment. Exergy is destroyed whenever energy drops its quality.

5.The surroundings are assumed to be unaffected by the system changes. Surroundings are maintained at a constant state of pressure P_0 and temperature T_0. In order to estimate a system’s exergy, the state of surroundings is taken as the reference state.

FIGURE 4.1  The change in exergy between two states.

4.3.1 Exergy Analysis of a Closed System

For the analysis of a system’s exergy, the first step is to specify the initial state that makes it non-variable. Maximum work can be obtained from a process occurring between two specific states reversibly. The irreversibilities are ignored for the determination of work potential. A dead state is the state of a system that must be achieved at the end of the process to obtain the maximum work.

Exergy cannot be created, but can be destroyed.

Exergy balance of a system

(Total energy entering) − (Total exergy leaving) − (Total exergy destroyed) = (Change in the total exergy of the system)

Figure 4.2 shows exergy balance graphically. For more convenience, the exergy balance can be expressed as follows:

Xin − Xout − Xdestroyed = Xsystem(4.2)

FIGURE 4.2  Exergy balance.

Equation (4.2) shows the exergy balance. The change in a system’s exergy is equal to the difference between the system’s exergy that is transferred through its boundary (input and output) and the destroyed exergy within the boundaries due to the system’s irreversibility. The exergy is transferred either by mass, work, or heat. Therefore, exergy balance can be expressed as follows:

(4.3)

The exergy transfer rates are in terms of mass, work, and heat and can be expressed as follows:

(4.4)

The term “Xdestroyed” is cancelled out from the reversible process. To obtain the destroyed exergy, the term “Sgen,” that is entropy generated, must be obtained first. The destroyed exergy can be calculated as follows:

(4.5)

For the closed system, there is no mass transfer. The exergy balance for a closed system will be,

And,

(4.6)

4.3.2 Exergy Analysis of an Open System

The balance of exergy for an open and a steady state system with constant volume is presented in Figure 4.3.

FIGURE 4.3  Open steady state system.

And for a flow of mass:

(4.10)

4.4 EXERGY LOSS OF A SYSTEM

In all engineering processes, the following factors may cause a loss of exergy of the system:

a.Electrical resistance

b.Inelastic deformations

c.Viscous fluid flow

d.Friction: solid-to-solid, solid-to-fluid, fluid-to-fluid

e.Shockwaves

f.Damping of vibrating systems

g.Fluid behavior at sudden expansion

h.Fluid flow through a valve or throttling process

i.Heat transfer by temperature difference, “thermal friction”

j.Sudden occurrence of chemical reactions

k.Mixing of liquids or gases having different chemical compositions

l.Process of osmosis

m.Phase existing in another face

n.Mixing of fluids at different pressures and temperatures

The exergy loss of a system is zero if it undergoes a reversible process.

Since there is no loss in a system’s exergy for reversible process, the system’s exergy at time t shows the maximum possible work that can be obtained till the system comes into common equilibrium with the environment. The reverse is also true. For a reversible process without the need of any additional energy, the amount of work extracted from the system can be supplied back, and hence the initial state conditions of the system and the environment can be restored. Consequently, there will be no noticeable change in the system or in its environment. System irreversibility can be measured by the amount of exergy loss due to irreversible process. However, exergy loss is zero for a reversible process. The irreversibility can be classified into two categories: external and internal.

4.4.1 External Irreversibility

The system’s irreversibility that takes place due to the system reaction or system interaction with the environment is called external irreversibility. Examples are friction, pressure changes, heat transfer due to temperature difference, etc.

4.4.2 Internal Irreversibility

This irreversibility takes place when the properties of the system are distributed heterogeneously. Local temperature, pressure, and velocity gradients within the system lead to sudden processes, which lead to exergy loss. For example, the presence of viscosity of fluid flow inside a channel results in different particle velocities. The velocity differences develop a rubbing action between the particles producing heat. The presence of heat transfer results in exergy loss which leads to irreversibility. This is called internal irreversibility because it occurs inside the system.

4.4.3 Exergy Loss of an Open, Steady State, Constant Volume System

Exergy balance of an irreversible open steady state system is shown in Figure 4.4.

FIGURE 4.4  Irreversible open steady state system.

(4.11)

For the system under consideration:

From the first law, the energy balance of an open steady state system is:

The exergy balance becomes:

(4.14)

The entropy balances is:

By combining these two equations,

(4.15)

4.5 EXERGY EQUATION

Transferring system’s exergy can be defined as the system that allows the energy to pass through the boundaries provided that some of this energy can be transformed into a useful work. This definition indicates that exergy and energy are interrelated. In other words, exergy transfer cannot take place without energy transfer. The rate of exergy change for a system can be presented by Equation (4.16).

(4.16)

The energy transfer between two systems can be accomplished in three different modes. Similarly, the exergy transfer within a system can take place in three different modes, that is, the exergy transfer by convection, heat, and work.

4.5.1 Transfer of Exergy by Work

The exergy transfer by a property difference other than the temperature is called transfer of exergy by work. For example, in the case of work transfer accomplished by the boundaries motion, the useful work can be obtained by removing the portion of the work done by the environment. The exergy rate due to work can be calculated as follows:

(4.17)

Following example gives the procedure for calculation of the exergy loss of a system.

EXAMPLE 4.1

An insulated rigid tank, as shown in Figure 4.5, holds 1 kg of pressurized air at 1 bar and maintained at 27°C. A motor runs a fan to perform an external work so that the final temperature of the air is 527°C. This results in exergy increasing to 100 kJ. Find the losses in exergy and explain the reason for the same.

FIGURE 4.5  Adiabatic tank.

Solution: The energy equation for the adiabatic tank is as follows:

For ideal gas,

Rigid tank prevents volume change and thus no work transfer occurs through the boundaries. Exergy increment of the system will be 358 kJ for the reversible process. However, the change of exergy is given as Df = 100 kJ and the exergy transferred by the work is Dxw = −358 kJ.

The main reason of additional exergy losses is due to the friction developed between the following:

a.The fluid and the fan blades

b.The fluid and the surface of the tank

c.The fluid layers

d.The shaft housing and its surface

e.The bearing of the housing

Exergy cannot be conserved and that contradicts the principle of energy conservation. The friction is a critical factor for exergy loss during work transfer between two systems.

4.5.2 Transfer of Exergy by Heat

Temperature difference causes exergy transfer due to heat flow across the boundaries of two systems. Figure 4.6 shows a reversible power cycle, where heat is rejected to the environment at T_0. The maximum theoretical work from a reversible power cycle can be calculated as follows:

FIGURE 4.6  Reversible power cycle.

Heat is transferred to the system at varying temperature:

where is thermodynamic equivalent temperature of heat transfer to the cycle.

EXAMPLE 4.2

Figure 4.7 shows a reversible heat engine working between two thermal reservoirs at 227°C. The heat engine supplies work to a refrigeration system (reversible). The refrigeration system is used to cool the space at −50°C by rejecting the heat at 27°C to an environment. Calculate the ratio Q_1/Q_2.

FIGURE 4.7  A reversible heat engine.

Solution: For the reversible heat engine, the output work is calculated as follows:

The work consumed by the refrigeration system is calculated as follows:

Since,

The heat ratio becomes:

By substituting the temperature values as T_1 = 500 K, T_0 = 300 K and T_2 = 223 K,

EXAMPLE 4.3

Figure 4.8 shows a house heating system. It consists of a reversible heat pump. In order to maintain the temperature inside the house at 22°C, the required heat (2°C outside temperature) is 12 × 10^4 kJ/h. The work done by the compressor is 10 kW.

FIGURE 4.8  Heating of a house.

a.Calculate the number of hours for one day of pump work.

b.The house heating could be accomplished using electric heaters. Find the useful energy that can be conserved using the heat pump. The determined useful energy is required on a daily basis.

Solution: The inside and the outside temperatures are T = 295 K and T_0 = 275 K, respectively.

The heat loss of the house is:

a.Number of hours required:

b.The electrical energy (using the heat pump) can be calculated as follows:

The electrical energy (using the heaters) is

The difference between the electrical energy consumed by the heaters and the heat pumps (799.2 − 54.18 = 745.02 kWh) is the useful energy that can be gained.

4.5.3 Transfer of Exergy by Convection

Fluid flow that occurs in and out of a system causes exergy transfer by convection. The net exergy accumulation rate due to transfer of energies by heat, work, and conviction are as follows:

{The net rate of exergy transfer by work at an instant of time, t} =

{The net rate of exergy transfer by heat at an instant of time, t} =

{The net rate of exergy accumulation by convection at an instant of time, t} =

The net rate of exergy supplied into the system will be

(4.20)

As a convention, one can assume that the symbol represents the irreversibility rate at time t. It is known that the reversible process shows no irreversibility rate (). However, the inequality is true for the irreversible process. Based on that, Equation (4.20) can be rearranged as follows:

(4.21)

where f is the exergy accumulation rate in the system at time t and rate of the destroyed exergy is the irreversibility rate . This means,

(4.22)

For better understanding of Equation (4.20), all the terms must be found out.

REVIEW QUESTIONS

1.Define energy and explain its relevance.

2.With the help of switchable diagrams, explain the exergy analysis of the following:

a.Closed system

b.Open system

3.List out the various factors that cause exergy loss in a system.

4.Develop a suitable exergy equation for a system and explain the following:

a.Transfer of exergy by work

b.Transfer of exergy by heat

c.Transfer of exergy by convection

5.With a suitable example, prove that heating of a house with the help of a heat pump is more efficient than heating by electric heaters.

NUMERICAL EXERCISES

1.Calculate the physical exergy of a perfect gas (air, mass (m) = 3 kg, isobaric specific heat Cp = 1.0005 kJ/kg·K, and gas constant R = 0.287 kJ/kg·K) for a state defined by p_1 = 3 bar and T_1 = 398 K. The environmental parameters are p_0 = 1 bar and T_0 = 298 K.(323.4 kJ)

2.A flat plate collector heats water from 305 K to 335 K. Calculate the exergy efficiency of the collector if its efficiency is 48% and ambient temperature is 293 K.(8%)

3.Air is heated in a central receiver solar power plant from 693 K to 1073 K. Calculate the exergy efficiency if ambient temperature is 303 K and collector efficiency is 70%.(42.5%)

CHAPTER 5

THERMODYNAMIC RELATIONS

5.1 TYPES OF THERMODYNAMIC PROPERTIES

The condition of thermodynamic system can be defined by its characteristics called properties. The properties can be essential, derived, or measured. Relationships can be developed to correlate the changes in the derived properties with reference to the measured properties. Some of the measured properties are V, P, T, cp, cv, and composition. Small letters are used to indicate the specific quantities, for example, v denotes the specific volume. Using first and second laws of thermodynamics, the internal energy (u) and entropy (s) are considered as the fundamental derived properties. As per first law, the energy is conserved and as per second law, entropy of the universe continues to increase. The derived properties are used for promoting the energy balance of systems where internal energy and other properties are frequently combined.

The mass flow in the open systems that crosses the boundary between the system and the surroundings provides two types of energy: flow work (pv) and internal energy (u).

An enthalpy (h) can be defined as follows:

(5.1)

Or total enthalpy can be defined as follows:

(5.2)

For an open system, flow work is included in the enthalpy along with internal energy (U).

In an isothermal system, heat can be extracted from the surroundings for free. For any environment having constant temperature T, Helmholtz free energy (A) is very suitable.

(5.3)

A denotes work if the system is created from nothing. The heat extracted from the surroundings is T∆S = T (Sf − Si) = TSf, where Sf is the final entropy of the system, while Si represents its zero-initial entropy. If a system is terminated with an initial entropy Si, A denotes the recovered work, as some heat needs to be dumped (TSi) into the environment for eradicating the entropy of the system. Equation (5.3) covers all work including the work performed by the surroundings. It is more convenient to use Gibbs free energy in an isobaric and isothermal environment.

(5.4)

Gibbs free energy can be defined as the work that is needed for generating a system out of nothing within an environment having constant pressure (P) and temperature (T).

The changes in the state of a system are more practical for which the changes in A and G should be considered. Changing A and keeping the temperature constant leads to:

(5.5)

Q denotes the heat added and W represents the system’s work. For reversible process, Q is equal to T∆S and the change in A is equivalent to the work done by the system. However, for irreversible process, Q is lower than T∆S and ∆A is lower than W. The change of A is less than the work done by the system.

At constant temperature (T) and pressure (P), the change in G for the environment is calculated as follows:

(5.6)

For any process,

(similar signs for processes with reversibility) (5.7)

The work (W) includes the work done to the environment (–P∆V) and any “other” work achieved on the system.

(5.8)

Substituting Equations (5.7) and (5.8) into Equation (5.6), the following was obtained.

(5.9)

5.2 EQUATIONS OF STATE

For the calculation of entropy, enthalpy, and energy of a matter, a correlation among volume, pressure, and temperature is required. In addition to the graphical and tabular configuration of the p-v-T diagram, there are analytical expressions termed as equations of state, which represents another approach of the p-v-T correlation. The state equations are used for calculating h, u, s, and other thermodynamic properties.

5.2.1 Virial Equation of State

The Virial equation of state can be derived from the static mechanics by relating the p-v-T behavior of any gas to the forces between its molecules. The Virial equation of state defines the term as a power-series in the inversion of molar volume ().

(5.10)

where B, C, and D are Virial coefficients and acts as the functions of temperature.

Simplifying Equation (5.10), the following was obtained:

(5.11)

B(T) can be calculated as follows:

(5.12)

ɷ denotes the Pitzer acentric factor. This parameter describes the polarity and the geometry of a molecule. A program written by T. K. Nguyen named “comp4.exe” could be used to find the Pitzer acentric factor for over 1000 compounds.

Comp4.exe can be found in the CHE302 course Distribution Folder. In some cases, where the molecular interactions do not exist, all Virial coefficients are equal to zero. Since the coefficients are equal to zero, Equation (5.10) becomes,

(5.13)

Equation (5.13) is the ideal gas equation of state.

5.2.2 Van der Waals Equation of State

Van der Waals equation of state could be used to show the assessment of thermodynamic properties. It looks easier than Virial equation of state.

Van der Waals equal of state is as follows:

(5.14)

where b is a constant related with the occupied finite volume by the molecules, and is related with the attractive forces between molecules.

a and b can be calculated using critical properties because the critical isotherm has an inflection point as shown in Figure 5.1. In the case of critical point, it will be as follows:

(5.15)

The isotherm that passes through the critical point is as follows:

(5.16)

The first and second derivatives of P with give the following:

(5.17)

FIGURE 5.1  Van der Waals equation for isotherms.

Multiplying Equation (5.17) by 2 and Equation (5.18) by and adding together to get the following:

(5.19)

(5.20)

If b = is substituted into Equation (5.18), the following was obtained:

In the case of critical point, it will be as follows:

(5.21)

Equation (5.21) can be used for solving a and b in terms of the critical pressure and critical temperature. When and b = are substituted into Equation (5.21), the following was obtained:

5.3 MAXWELL’S RELATIONS

5.3.1 Energy and Entropy

Entropy S(V, U, N) and energy U(V, S, N) can give complete information about thermodynamics.

The important equation of thermodynamics for energy is as follows:

This can be used to find the chemical potential, temperature, and pressure at equilibrium conditions. The partial derivatives are as follows:

Important equation of thermodynamics for entropy is as follows:

(5.26)

The partial derivatives will be as follows:

Maxwell’s relations can be found by taking the second partial derivatives of these qualities. Such relationships can be used for connecting the partial derivatives, of properties which can be measured with those which cannot be measured.

Starting from:

Determine:

Under appropriate conditions,

Equation (5.31) is known as the Maxwell’s relation. By performing the other second partial derivatives, another two Maxwell’s relations can be found.

In terms of energy,

In terms of entropy,

And the results are as follows:

The Maxwell’s relations are as follows:

5.3.2 Enthalpy H (S, p, N)

The enthalpy has been defined as H = U + PV. Its differential can be combined with the basic thermodynamic equation to get dH in terms of S, p_0, and N.

or           (5.35)

H = H (S, p, N), where S, p, and N are the natural properties of H.

Differentiating partially,

The Maxwell’s relations are as follows:

5.3.3 Helmholtz Free Energy F (T, V, N)

To transform from U to F, replace independent variable S by its conjugate T. In a Legendre transformation, a new function is described by adding or subtracting conjugate product to replace an independent variable with its conjugate. Therefore, by subtracting TS from U, the new function F was defined.

Therefore, F = F (T, V, N) was preferred. By following the same procedures, the following can be defined:

The Maxwell’s relations are as follows:

5.3.4 Gibbs Free Energy G (T, N, p)

The independent variables V and S are replaced by their conjugates p and T, while transforming from U to G. This represents a twice Legendre transformation of U or one Legendre transformation of either F or H.

Following the same procedures, the following can be defined:

The Maxwell’s relations are as follows:

5.3.5 The Grand Potential Ω (T, V, µ)

Here, the independent variables N and V are replaced by their conjugates µ and p, while transforming from U to Ω. It can be done by a double transformation of U Legendre or a single transformation of F. For engineering work, the ground potential is less frequently used than the other potentials.

In an open system, where particles are exchanged with the surrounding, ground potential is used.

Therefore, Ω = Ω (T, V, µ).

Differentiating again

The Maxwell’s relations are as follows:

EXAMPLE 5.1

Blender blades are churning the fluid at 97°C. The work is transferred to the fluid at a rate of 360 kJ/h. The blender is shown in Figure 5.2. The bowl of the blender is insulated circumferentially. The bottom surface of the bowl passes heat at a rate of 360 kJ/h to the surrounding at 27°C. Calculate:

FIGURE 5.2  Blender.

A.The net energy rate supplied to the system,

B.Energy accumulation rate of the system

C.Irreversibility rate of the process.

Solution:

A.The heat transfer and work rates, respectively, are

Energy balance of the system shows that

No change in the system’s energy and it is a steady state system.

B.The constant energy system shows no exergy.

⸫

The exergy transfer by work is calculated as follows:

Ans.

The transfer of exergy by heat is calculated as follows:

Ans.

C.From the equation of exergy balance:

5.4 FLOW EXERGY OF MATTER

The characteristics of a system used to find the flow energy of matter are as follows:

●System with only reversible processes.

●Steady state flow in an open system with constant volume.

●Heat is transferred only to and from the environment at T_0.

●System maintains the equilibrium between the matter and the environment, as shown in Figure 5.3.

FIGURE 5.3  Matter in equilibrium.

5.5 EXERGY EFFICIENCIES

Energy efficiency can be given by

where and have to be specified for each type of system.

5.6 EXERGY TRANSFER BY HEAT

Figure 5.4 shows the schematic diagram of exergy transfer by heat.

FIGURE 5.4  Exergy transfer by heat.

REVIEW QUESTIONS

1.Discuss various types of thermodynamic properties and their correlations.

2.Explain the following:

a.Virial equation of state.

b.Van der Waals equation of state.

3.Derive Maxwell’s relation in terms of:

a.Energy and entropy

b.Enthalpy

c.Helmholtz free energy

d.Gibbs free energy

e.Ground potential

4.Explain the following with the schematic diagrams:

a.Exergy transfer by heat.

b.Flow exergy of matter.

5.Define exergy efficiency.

NUMERICAL EXERCISES

1.Determine the reversible work and the irreversibility associated with a process in which a metallic ball of 2 kg is cooled down from 300°C to 150°C by keeping it in the open space at 20°C. Specific heat of the metal may be taken as 0.45 kJ/kg·K.(143 kJ, 55 kJ)

2.5 kg of air at 550 K and 4 bar is enclosed in a closed vessel.

(i) Determine the availability of the system if the surrounding’s pressure and temperature are 1 bar and 290 K, respectively.

(ii) If the air is cooled at a constant pressure to the atmospheric temperature, determine the availability and effectiveness. (576.7 kJ, 64.8%)

3.5 Kg of ice at −10°C is kept in atmosphere which is at 30°C. Calculate the change in entropy of universe when it melts and comes into thermal equilibrium with the atmosphere. Take latent heat of fusion as 335 kJ/kg and specific heat of ice as half that of water.(0.652 kJ)

4.0.25 kg/s of water is heated from 30°C to 60°C by hot gases that enter at 180°C and leave at 80°C. Calculate the mass flow rate of gases when its Cp = 1.08 kJ/kg·K. Find the enthalpy change of water and of hot gases. Take the specific heat of water as 4.186 kJ/kg·K. Also find the increase of unavailable energy if the ambient temperature is 27°C. (0.0988 kJ, 23.436 kJ)

CHAPTER 6

THERMODYNAMIC CYCLES

6.1 INTRODUCTION

The majority of power-producing machines work on thermodynamic cycles and therefore research and development work in the area of thermodynamics has assumed to gain very high importance.

Every power engine operates as a steady-state device and does not store power. The working fluid passes through a series of processes forming a closed-loop cycle. The cycle of processes is repeated again and again as the engine operates.

Thermodynamic cycles are classified as gas cycles and vapor cycles. In gas cycles, the fluid remains in gaseous phase, whereas the face may change in vapor cycles. If steam is the working fluid, it will be gas in some processes and liquid in other processes.

Thermodynamic cycles are also classified as an open cycle and a closed cycle. In a closed cycle, the same working fluid is moved from initial to fluid state and then recirculates. In the case of open cycles, the working fluid is changed in each cycle and not being recirculated. In a car engine, the burnt gases can be exhausted and a fresh mixture of air and fuel is changed at the end of every cycle.

There are engines where the working fluid does not undergo an entire thermodynamic cycle.

Thermodynamic cycles are also categorized as per their utility.

1.Power cycles

2.Heat pump and refrigeration cycles.

1.  Power cycles: thermodynamic cycles that are used in engine for power production are called power cycles. Power cycles may be either gas power cycles or vapor power cycles.

The gas power cycles are further classified as follows:

a.Stirling cycle

b.Dual cycle

c.Carnot gas power cycle

d.Diesel cycle

e.Ericsson cycle

f.Otto cycle

g.Brayton cycle

The vapor power cycles are further classified as follows:

a.Carnot vapor power cycle

b.Rankine cycle

c.Reheat cycle

d.Regenerative cycle

Carnot, reheat, and Rankine cycles will be discussed in detail.

2.  Heat pump and refrigeration cycles: these thermodynamic cycles are used for heat pump and refrigeration. They may be either vapor cycles or air cycles depending upon the working fluid used.

6.2 CYCLE DEFINITION

A thermodynamic cycle consists of a series of processes and normally the working fluid returns to its original state. It is also called as a closed cycle.

Figure 6.1 shows a thermodynamic cycle that is considered to be a closed system. A thermodynamic cycle may not be a closed cycle. For instance, a jet engine repeatedly receives fresh ambient air and releases hot burnt gases to the atmosphere. The burnt gas is cooled down in the atmosphere to ambient temperature.

FIGURE 6.1  Thermodynamic cycle

The concept of thermodynamic cycle, as shown in Figure 6.2, can be explained as follows:

FIGURE 6.2  The system’s cycle with the addition of heat.

Consider a fluid contained in a piston−cylinder apparatus that undergoes heat interaction with two heat reservoirs at different temperatures. The fluid exercises the following four processes and regains its original state.

1−2: adiabatic compression to raise the temperature to high-temperature reservoir.

2−3: the fluid at the temperature of high-temperature reservoir is isothermally heated.

3−4: adiabatic expansion to the temperature of lower-temperature reservoir.

4−1: isothermal cooling of the gas to the initial pressure and the temperature.

The cycle produced by these four processes is called the Carnot cycle.

6.3 PERFORMANCE PARAMETERS

A few performance parameters of power cycles are explained below.

Thermal effectiveness: it is the limit of the quantity of heat entering the engine which is converted to the work of the engine.

(6.1)

Ratio of back work: the ratio of the pump work (−W) required to the turbine work (+W) produced.

(6.2)

Usually, ratio of back work is small indicating the pump work (−W) required to the turbine work (+W) produced.

Work ratio:

(6.3)

Specific steam consumption (SSC): it is the amount of steam required to produce unit power. It is usually expressed as kg/kWh and has a numerical value from 3 to 5 kg/kWh.

(6.4)

6.4 CARNOT VAPOR POWER CYCLE

The Carnot cycle consists of four reversible processes: isothermal addition of heat, isentropic expansion, isothermal discharge of heat, and isentropic compressor as shown in Figure 6.3.

FIGURE 6.3  Steady-flow Carnot engine.

Carnot cycle is a perfect cycle having maximum thermodynamic effectiveness. Carnot vapor cycle contains vapor as the working fluid. The T-S and P-V diagrams of a Carnot cycle are plotted in Figure 6.4. The Carnot cycle works as a closed system.

FIGURE 6.4  Arrangement of Carnot cycle.

A cylinder−piston arrangement or a turbine with a gas or a vapor as the working fluid is used for Carnot cycle.

It consists of (i) two constant pressure processes (2−3) and (4−1) and (ii) two adiabatic processes without friction (1−2) and (3−4). The four steps can be described in detail as follows:

1.Process (2−3): At temperature T_2 and from dryness practices X_2, 1 kg of hot water is heated to saturated steam pressure P_3 and temperature T_3. The pressure (P_2 − P_3) and the temperature (T_2 − T_3) remain constant during this process.

2.Process (3−4): The vapor is expanded by isentropic process to temperature T_4 and pressure P_4. Point “4” indicates the state of vapor after expansion.

3.Process (4−1): Heat is rejected at a constant temperature (T_4 = T_1) and pressure (P_4 = P_1). The vapor is condensed into liquid.

4.Process (1−2): The wet vapor at “1” is compressed isentropically untill the vapor reaches its initial condition of P_2 and T_2. Then the cycle gets completed.

Heat added:

Heat rejected:

6.4.1 Plant Components

1. Pump

Water enters the pump at state 1 and is pumped isentropically to the working pressure of the boiler. The water temperature rises through this isentropic compression process due to a small reduction of specific volume of water.

2. Boiler

Water enters the boiler as a condensed liquid at state 2 and leaves as a superheated/saturated vapor at state 3. Heat is transferred to the water at a constant pressure. This heat may be generated in a nuclear reactor, burning gases, or any other sources of heat. When the vapor reaches the superheated state, the boiler is called a steam generator.

3. Turbine

Superheated steam enters the turbine, and expands isotropically producing work by rotating a shaft coupled to an electric generator.

4. Condenser

Condenser is a heat exchanger where vapor is condensed by abstracting the heat at a constant pressure P_4.

Vapor leaves the condenser as a saturated liquid and enters the pump to complete the cycle.

6.4.2 Thermal Analysis of Carnot Cycle

(6.5)

Net work per unit mass flow:

(6.6)

Heat added in boiler:

(6.7)

Heat rejected in condenser:

(6.8)

Thermal efficiency:

The heat addition and rejection can be expressed in terms of temperature and entropy as follows:

Since:

6.4.3 Limitations of Carnot’s Cycle

Thermodynamically, the Carnot cycle is simple, and when the values of T_2 and T_1 are known, the cycle reaches maximum thermal effectiveness. But it is very difficult to use this cycle in practice due to the following limitations:

1.The process of condensation of vapor to saturation condition is difficult.

2.To find the state 4 in the beginning of condensation is difficult.

3.The critical temperature of water vapor is fixed at 374°C, therefore, maximum possible temperature is bounded.

4.It is difficult to work with the superheated steam because providing the superheat at a constant temperature rather than constant pressure is difficult.

6.5 PRINCIPLES OF THE CARNOT CYCLE

The Carnot cycle is the best reversible cycle.

But there is irreversibility in the real process. Therefore, actual cycles are less efficient than the Carnot cycle.

The Carnot cycle has four processes as shown in Figure 6.5 for a closed system of piston−cylinder arrangement.

FIGURE 6.5  v-p graph of Carnot cycle.

1−2: Reversible isothermal expansion process

The gas expands at a constant temperature TH and the heat QH is transferred to gas.

2−3: Reverse adiabatic expansion process

The gas expands by the reversible process, and the temperature of gas reduces to TL from TH.

3−4: Reverse isothermal compression

Heat is rejected to the surroundings from the system (reversible) at a constant gas temperature T_2.

4−1: Reversible adiabatic compression process

The temperature of gas increases to TH from TL.

Carnot cycle works between two separate temperatures TH and T_2.

FIGURE 6.6  Carnot principle.

For all reversible and irreversible heat cycles,

(6.15)

For the Carnot cycle,

(6.16)

The efficiency of the Carnot cycle is always more than the efficiency of an irreversible (real) working between the same two reservoirs as shown in Figure 6.6.

(6.17)

6.5.1 The Reverse Carnot Cycle

The reverse Carnot cycle is used for refrigeration. QH represents the amount of heat rejected to a reservoir at high temperature, whereas QL represents the amount of heat received from a reservoir at low temperature. Work entering the system is important to complete the cycle.

The V-P diagram of the reversed Carnot cycle is similar to the usual Carnot cycle, but the directions of the process are reversed as shown in Figure 6.7.

FIGURE 6.7  P-V diagram of the reversed Carnot cycle.

6.5.2 Carnot Cycle for Heat Pump and Refrigerator

A heat pump or refrigerator working on a reversed Carnot cycle is called as refrigerator or a Carnot heat pump.

The coefficient of performance (COP) of any refrigerator or heat pump is calculated as follows:

The COP of the reversible refrigerators or heat pumps is calculated as follows:

Similar to the heat engine, one can conclude:

(6.18)

6.5.3 Carnot Cycle for Heat Engine

All heat engines working on the principles of Carnot cycle are called the Carnot heat engines. The thermal effectiveness of any heat engine (irreversible or reversible) is calculated as follows:

(6.19)

where QH is the heat transferred to the engine from the reservoir at a high temperature TH, and QL is the heat rejected to the reservoir at low temperature TL.

For the reversible heat engines, the ratio of heat transfer with reservoir is equal to the ratio of the temperatures of the two reservoirs.

Therefore, efficiency of a Carnot engine, or some reversible heat engine, will be as follows:

(6.20)

This is the maximum efficiency of heat engine working between thermal reservoirs TH and TL as shown in Figure 6.8.

All irreversible (real) heat engines working in the range of two temperature limits (TH and TL) have lesser efficiency.

The actual thermal efficiencies of the heat engines working between the same ranges of temperatures (Figure 6.9) are as follows:

(6.21)

FIGURE 6.8  Carnot heat engine.

FIGURE 6.9  Carnot principles.

EXAMPLE 6.1

A Carnot cycle operates in vapor in the range of pressure from 7 kPa to 7 MPa. Calculate the work of turbine, thermal efficiency, and work of compressing of vapor.

Solution:

FIGURE 6.10

6.6 IDEAL RANKINE CYCLE

Some of the limitations of the Carnot cycle can be removed by superheating the vapor in the boiler and a complete condensing in the condenser. As shown in Figure 6.11, the cycle is called the Rankine cycle. This is the ideal cycle for steam power plants.

FIGURE 6.11  Schematic layout of ideal Rankine cycle.

The T-S diagram of ideal Rankine cycle is shown in Figure 6.12.

FIGURE 6.12  T-S diagram of an ideal Rankine cycle.

The ideal Rankine cycle consists of the following four processes:

Process 1−2: isentropic compressor in a pump.

Process 2−3: addition of heat in a boiler at a constant pressure.

Process 3−4: development of isentropic work in an engine.

Process 4−1: rejection of heat in a condenser at a constant pressure.

6.6.1 Analysis of the Ideal Rankine Cycle

The four machines of Rankine cycle (condenser, turbine, boiler, and pump) have constant flow. Rankine cycle is considered as a constant-flow process.

There is no work for the condenser and the boiler. The pump and the turbine are taken as isentropic.

1−2: Pump work:

(6.22)

where p_2 and p_1 are in kPa.

The values of vf and p_2 can be obtained from this steam tables.

(6.23)

2−3: Heat supplied in boiler:

3−4: Turbine work:

where p_1 = pressure of condenser or low pressure of cycle, h_1 = enthalpy from steam table at p_1, p_2 = pressure of boiler or pressure at turbine inlet or high pressure of cycle, h_2 = enthalpy from steam table at P_2,= the mass flow rate of steam in kg/sec.

It is possible to use Mollier diagram to get h_2 and h_1.

4−1: Condensation process at constant pressure:

(i) Net work:

(6.30)

(ii) Thermal efficiency:

Ratio of the network to the heat supplied.

(iii) Specific steam consumption:

EXAMPLE 6.2

An ideal Rankine cycle has steam as the working fluid. Saturated steam enters the turbine at a pressure of 8000 kPa, whereas saturated liquid leaves the condenser at a pressure of 8 kPa. The cycle power output is 100,000 KW.

Calculate:

a.The thermal efficiency of the cycle.

b.The back work ratio of the cycle.

c.The flow rate of steam, in kg/h.

d.The flow rate of condenser cooling water, in kg/h, if cooling water enters the condenser at 15°C and leaves at 35°C.

FIGURE 6.13  Steam power plant.

Solution:

a.Thermal efficiency:

b.Back work ratio:

c.Mass flow rate of steam:

d.Flow rate of condenser cooling water:

6.6.2 Comparison between Carnot and Rankine Cycles

i.For the same temperature limits, Rankine cycle gives higher output than the Carnot cycle. Rankine cycle needs a lesser flow rate of vapor to obtain a known output. Rankine cycle needs higher rates of heat transfer in boiler and condenser.

ii.Efficiency is less than that of Carnot cycle. The efficiency of Rankine cycle can be more than that of Carnot cycle, if superheated steam is used.

iii.The compressor power is very high in Carnot cycle as compared to pumping power in Rankine cycle.

Figure 6.14 gives comparison of SSC and efficiency of two cycles.

FIGURE 6.14  Specific steam consumption and thermal efficiency of Carnot and ideal Rankine cycles.

6.6.3 Effect of Operating Conditions on Rankine Cycle Efficiency

The Rankine cycle efficiency can be increased by the following factors:

1. Lowering the condenser pressure (lowers Tlow, avg)

The effect of lower pressure of condenser on the Rankine cycle efficiency is shown in T-S diagram (Figure 6.15). The shaded area in Figure 6.15 shows the increase of power output by lowering the condenser pressure. The heat input shown by the area under the curve 2′-2 increases but very little. The thermal efficiency of the cycle increases with the decrease in condenser pressure.

FIGURE 6.15  Result of lowering the condenser pressure on the ideal Rankine cycle.

The vapor power cycles work in a closed loop and the condenser pressure is lower than the atmospheric pressure to increase the thermal efficiency of the cycle.

2. Superheating the steam to high temperature (increases Thigh, avg)

The steam temperature can be increased by superheating without the increase in boiler pressure as indicated by the shaded zone in Figure 6.16. The increase in the heat input and the higher work output are shown by the area under the curve 3-3′. This leads to higher thermal efficiency.

FIGURE 6.16  Ideal Rankine cycle showing the influence of superheating the vapor to higher temperatures.

There is an additional advantage of superheating the vapor to higher temperatures as the dryness fraction of steam at the turbine’s exit improves.

3. Raising the boiler pressure (increases Thigh, avg)

The temperature of heat-addition process in the boiler can be increased by raising the working pressure of the boiler. But the dryness fraction of steam at the turbine’s exit decreases as point 4 moves to the left.

These methods can be used to increase the thermal efficiency of the Rankine cycle.

i.Use of dual vapor.

ii.Extraction of water from steam.

iii.Feed water reheating or regeneration.

iv.Reheating of steam.

6.6.4 Regeneration Cycle

In Rankine cycle, the condensate at low temperature takes an irreversible addition of heat in a hot boiler thereby cycle efficiency decreases. The condensate can be heated with the steam extracted from the turbine before sending to the boiler.

FIGURE 6.17  Effect of increasing the boiler pressure on the ideal Rankine cycle.

This heating is called regeneration heating and the cycle is called regeneration cycle.

The regenerative heating is always utilized and a number of the heaters are provided. The process of steam extraction is called bleeding of steam. The number of feed water heaters (FWHs) depends on the size of plant.

Eight to nine heaters may be used for large thermal power plants.

A FWH is a heat exchanger to heat condensate by steam. There are two types of FWHs:

1.Open or direct contact type.

2.Closed type with:

a.Drains cascaded backward

b.Drains pumped forward.

FIGURE 6.18  Types of feed water heaters.

Advantages of regeneration cycle:

1.The heat supply in the boiler is decreased.

2.The temperature difference in the boiler is reduced.

3.The boiler heating approaches to reversible process.

4.The temperature of heat addition in boiler increases.

5.The thermal efficiency depends on the temperature of heat supply and therefore increases.

6.The size of condenser is reduced.

7.The steam flow in turbine becomes wet and water has to be separated by steam bleeding, and the need for water separation decreases along with the decrease in the corrosion of turbine blades.

Disadvantages of regeneration cycle:

1.The plant becomes complicated increasing capital cost and decreasing reliability.

2.Heaters need additional maintenance.

3.Boiler size increases.

4.The heaters are expensive and the gain in thermal efficiency is not much as compared to additional cost.

6.6.4.1  Open FWHs

The advantages of two types of FWHs are given in the following table:

Open FWHs

Closed FWHs

1. Simple and inexpensive.

2. More efficient heat transfer due to direct contact.

3. One pump can be used for many FWHs

1. Complex and expensive because of the internal tubing network.

2. Less efficient since the two streams are not allowed to come in direct contact.

3. One pump is used for each FWH

In an open-type FWH, the feed water from condenser and bled steam from turbine are directly mixed. The schematic diagram of a power plant along with T-S graph is shown in Figure 6.19.

FIGURE 6.19  Ideal regeneration Rankine cycle with open feed water heater.

In a cycle of Rankine ideal regeneration, vapor enters the turbine at the pressure of the boiler and decreases isentropically to a moderate pressure. Some steam is removed at this condition and same to the heater. The rest of vapor stays to expand isentropically to the pressure of condenser. This vapor goes to the condenser as a saturated liquid at the pressure of the condenser. The feed water enters the pump at an isentropic condition, wherever it is condensed to the pressure of heater of feed water (stage 2) and is in flying to the provide for water warmer, wherever feed water combines with the vapor extracted from the turbine.

Thermal analysis of power plant with open FWH:

Energy balance of FWH:

For 1 kg of working fluid,

Inlet of energy = outlet of energy

(6.35)

FIGURE 6.20

Energy entering the heater = Energy leaving the heater

(6.36)

EXAMPLE 6.3

In an ideal regeneration Rankine cycle of steam power plant working with one open feed water heaters, steam enters the turbine at 600.00°C and 15000 kPa and condensed at 10.000 kPa. Steam is heated at a pressure of 1.2 MPa and enters the feed water heater. Calculate the fraction of steam bled and the thermal efficiency of the cycle.

FIGURE 6.21  Schematic diagram of Example 6.3.

Solution:

FIGURE 6.22  T-S diagram.

Condenser outlet:

Pump outlet:

FWH outlet:

Boiler inlet:

Turbine inlet:

Bled steam:

Condenser inlet:

6.6.4.2  Closed FWHs

Closed FWHs are shell-and-tube heat exchangers in which feed water temperature increases as the bled steam condenses on the outer of the tubes with feed water flows inside the tubes.

The two steams are at different pressures and do not mix.

The bled steam condenses in the closed feed water while heating the feed water from the pump.

The heated feed water is sent to the boiler and the condensates from the FWH are also pumped to the boiler.

There are two types of closed feed water heating schemes as shown in Figure 6.23.

FIGURE 6.23  Feed water heating schemes.

i.Closed FWH with drains cascaded backward

•This scheme of FWHs is simple and mostly used.

•It is shell and tube heat exchanger.

•The feed water flows in the tubes and the bled steam flows in the shell.

•Only one pump is needed as the steam does not mix with the feed water.

•If decreator is used, another pump may be used before the boiler feed pump.

•The bled steam is drained back to previous low-pressure FWH by reducing the pressure or led back to the condenser.

•Terminal Temperature Difference (TTD) is defined as the difference between temperature of bled steam entering the feed water and the sub-cooled water temperature. TTD is shown in Figure 6.24, as:

TTD = Saturation temperature of bled steam − exit water temperature. The TTD is positive and is often of the order of 0°C–5°C.

FIGURE 6.24  Terminal temperature difference.

FIGURE 6.24(A)  Power plant with closed FWH and drains cascaded backward.

FIGURE 6.25  Closed FWH with drains cascaded backward.

Energy balance of the high-pressure FWH:

The energy balance of the high-pressure FWH is given as follows:

Energy entering the heater = Energy leaving the heater

For throttling process, enthalpy is constant.

Energy balance of the low-pressure FWH:

Energy entering the heater = Energy leaving the heater

For throttling process, enthalpy is constant.

h_11 = h_12

Thermal analysis of cycle:

For boiler:       (6.37)

For condenser:

(6.38)

For pump:       (6.39)

For turbine:

(6.40)

For cycle:       (6.41)

Thermal efficiency:     (6.42)

EXAMPLE 6.4

In a steam power plant working on the ideal regenerative Rankine cycle with two closed feed water heaters, steam enters at 600.00°C and pressure of 15000 kPa and it is condensed at 10.00 kPa. At 4000 kPa, a small amount of steam is bled from the turbine. At 500 kPa, steam enters closed feed water heater. Calculate the cycle’s thermal efficiency and rate of mass flow of steam entering the two closed FWHs.

Solution:

Pump inlet:

Saturated liquid:

Pump outlet:

⸫

Turbine inlet:

Bled steam:

FWH inlet:

Condenser inlet:

At

At

The energy balance of the high-pressure FWH is given as follows:

Energy entering the regenerator = Energy leaving the regenerator

The energy balance of the low-pressure FWH is given as follows:

Energy entering the regenerator = Energy leaving the regenerator

For throttling process, enthalpy is constant.

i.Closed FWH with drains pumped forward

The field water is heated by bled steam without mixing. The pressures of two streams are different. The condensate of bled steam leaves the heater as saturated water. Figure 6.26 shows an ideal regenerative Rankine cycle with a closed FWH with drains pumped forward.

FIGURE 6.26  The ideal regenerative Rankine cycle with closed FWH with drains pumped forward.

Energy and mass balance of FWH

m_1 = mass flow rate of bled steam for the closed FWH

ms = m_7 + m_8

The energy balance of the closed FWH is given as follows:

E inlet FWH = E outlet FWH

m_1 h_7 − (m_1 − 1) h_2 = m_1h_3 − (m_1 − 1) h_9 (6.43)

Thermal analysis of cycle:

For boiler:

(6.44)

For condenser:

(6.45)

For pump:

(6.46)

Net work:

(6.47)

Thermal efficiency:

(6.48)

FIGURE 6.27  Closed FWH.

EXAMPLE 6.5

Repeat the solution of Example 6.4 using the same data with FWHs installed in forward pumped drain arrangement.

Solution: From previous example:

h_1 = hf, p_1 = 191.81 kJ/kg

h_2 = 206.95 kJ/kg

h_7 = 3582.3 kJ/kg

h_8 = 2113.2 kJ/kg

h_9 = 3152 kJ/kg

h_10 = 2687.5 kJ/kg

h_11 = 1087.29 kJ/kg

h_13 = 640.21 kJ/kg

Assume

At

At

For low-pressure FWH, the pump work:

The energy equilibrium of the high-pressure FWH is given as follows:

E incoming to the regenerator = E exiting from the regenerator

The energy balance of the low-pressure FWH is given as follows:

E incoming to the regenerator = E exiting from the regenerator

The energy balance of the low-pressure FWH is given as follows:

Energy entering the regenerator = Energy parting the regenerator

6.6.5 The Placement of FWHs

For maximum thermal efficiency, it is very important to decide the pressure of bled steam from turbine.

6.6.6 Reheat Cycle

The thermal effectiveness of the cycle can be increased by the following methods:

i.Regenerative feed water heating.

ii.Binary vapor cycle.

iii.Reheating of steam.

iv.Removal of water.

Reheating system

Figure 6.28(b) shows schematic diagram and T-S diagram of an ideal reheat Rankine cycle.

Steam is expanded in the high-pressure turbine isentropically to an intermoderate pressure and sent back to the boiler for reheating to initial temperature. Steam is expanded isentropically in the low-pressure turbine to the condenser pressure.

(6.49)

The improvement in thermal efficiency by reheating depends on the reheat pressure with reference to the inlet stem pressure.

FIGURE 6.28(A)  Ideal reheat Rankine cycle.

FIGURE 6.28(B)  T-S diagram of ideal reheat Rankine cycle.

Figure 6.29 shows the reheat pressure versus cycle efficiency.

FIGURE 6.29  Reheat pressure versus cycle efficiency.

Advantages of reheating:

1.Output power increases.

2.Heat supply increases.

3.Thermal efficiency increases.

4.The wetness drops and corrosion of blade becomes less. This leads to the increase in turbine life.

Disadvantages of reheating:

1.Added equipment and piping increase plant cost and operational reliability decreases.

2.The increase in thermal efficiency may not offset the disadvantages of plant complexity.

EXAMPLE 6.6

In an ideal reheat Rankine cycle steam power plant, steam vapor enters the turbine at a temperature of 600.00°C and pressure of 15000 kPa, and is condensed at 10.000 kPa. The vapor moisture at the exit of turbine should not exceed 10.400%, calculate the following:

a.Reheat pressure.

b.Thermal efficiency of cycle.

Assume the steam is reheated to the initial temperature.

FIGURE 6.30  Reheat Rankine cycle.

Solution:

a.The reheat pressure is determined from the requirement that the entropies at states 5 and 6 are the same:

Turbine outlet:

But

⸫

Thus,

Turbine inlet:

Therefore, steam should be reheated at a pressure of 4 MPa or lower to prevent a moisture content above 10.4%.

b.To determine the thermal efficiency, the enthalpies at all other states must be known:

Condenser outlet:

Boiler inlet:

High-pressure turbine inlet:

High-pressure turbine outlet:

Thus

and

6.6.7 Cogeneration Cycles

Cogeneration is the simultaneous generation of electricity and steam (or heat) in a single power plant. Chemical industries, paper mills, and municipalities that use district heating need to process heat or steam as well as electricity. These industries and municipalities can produce electricity more conveniently and cheaply by cogeneration.

The cogeneration plan efficiency is calculated as follows:

A cogeneration plant with adjustable load is shown in Figure 6.31.

FIGURE 6.31  Cogeneration plant with adjustable loads.

The boiler steam is used for both electric power generators and heat. Factor of utilization ∈u of cogeneration plant is calculated as follows:

Let m_4 = total mass flow rate of steam in kg/s

Or

FIGURE 6.32

EXAMPLE 6.7

In a cogeneration plant, boiler produces steam at temperature of 450.00°C and pressure of 10000 kPa at a rate of 5.00 kg·s−1. Steam expands to 500 kPa in the turbine, then it is supplied to the process heater. Before entering the boiler, it is heated in a feed water heater. The condenser works at 20.00 kPa.

a.Calculate the rate of power generation.

b.If only 60% of steam is used as process heat and the condenser receives the remaining steam, calculate the rate of power generation.

FIGURE 6.33  Schematic diagram of Example 6.7.

Solution: (a)

For pump P-I inlet:

Pump outlet:

For pump P-H inlet:

Pump outlet:

Turbine inlet:

Bled steam:

Turbine exit:

When the entire steam is routed to the process heater:

(b)

FIGURE 6.34

When only 60% of the steam is routed to the process heater:

Energy balance of FWH

Energy entering the heater = Energy leaving the heater

6.6.8 Binary Vapor Cycle

Binary vapor cycle uses two fluids. Figure 6.35 shows parts of a binary vapor power plant.

1.Mercury cycle: Mercury boiler, mercury turbine, mercury condenser or steam generator, and mercury feed pump.

2.Steam cycle: Steam generator or mercury condenser, steam super heater, steam turbine, steam condenser, and water feed pump.

A. Thermodynamic properties of mercury:

Mercury has its following thermodynamic properties:

1.The boiling pressure of 12.5 bar only is necessary to match the steam temperature of 540°C. High pressure is not necessary to obtain high main temperature of heat supply.

2.Mercury does not have any erosive or corrosive effects on the metal parts.

3.The boiling point is 354.4°C and the freezing point is 3.3°C at atmospheric pressure.

4.Mercury can be safely used with high-temperature metals.

5.The saturation state of liquid can be obtained by vertical expansion almost isentropic as for Carnot cycle.

FIGURE 6.35  Elements of binary vapor power plant.

B. Binary vapor cycle analysis:

Work done per kilogram of steam

Heat supplied per kilogram of steam produced in the mercury condenser or steam boiler

Thermal efficiency of steam cycle.

Mass of mercury in the Hg cycle per kilogram of steam produced in the steam cycle.

Thermal efficiency of Hg cycle.

Heat required for Hg per kilogram of steam in mercury boiler.

Work produced per kilogram of Hg cycle

Heat removed from Hg per kilogram of steam in the mercury condenser.

Work done in mercury cycle per kilogram of water flow:

(6.62)

Work done in steam cycle per kilogram of water flow:

(6.63)

Heat supplied in Hg boiler per kilogram of water flow:

(6.64)

Overall work done in binary vapor cycle:

(6.65)

⸫ Total efficiency of binary vapor cycle:

(6.66)

Thermal efficiency of Hg cycle:

(6.67)

(6.68)

FIGURE 6.36  Binary vapor cycle on T-S graph.

Heat supplied by the Hg vapor = Heat received by the steam

(6.69)

But

(6.70)

Thermal efficiency of steam cycle:

(6.71)

Combining overall thermal effectiveness, the following was obtained:

(6.72)

There is a problem in the design of binary cycle plants. Special plant design is required to avoid leakage of dangerous Hg vapor. Cost of Hg inventory is very high. Pumping of Hg and heat transfer are difficult.

6.7 STIRLING CYCLE

The Stirling cycle is similar to Carnot cycle which was invented in 1816 by Dr. Robert Stirling. Figure 6.37 shows an ideal Stirling cycle on T-S and P-V diagrams and consists of the following four thermodynamic processes:

1.1−2: reversible isothermal compression.

2.2−3: heat addition by reversible isochoric process.

3.3−4: reversible isothermal expansion.

4.4−1: reversible isochoric heat rejection.

The working material for Stirling cycle is a gas such as hydrogen, air, etc. and not water vapor. Similar to Carnot cycle, all processes in a perfect Stirling cycle are reversible. Heat pump or refrigerator works on reversed Stirling cycle. This cycle has the application for cryogenic engine.

FIGURE 6.37  Stirling cycle.

6.8 ERICSSON CYCLE

The Ericsson cycle is another perfect cycle of thermodynamics discovered by John Ericsson. T-S and P-V diagrams for ideal Ericsson cycle are shown in Figure 6.38. It consists of the following processes:

1.1−2: reversible isothermal compression process.

2.2−3: heat addition by reversible isobaric process.

3.3−4: reversible isothermal expansion process.

4.4−1: reversible isobaric heat rejection process.

FIGURE 6.38  Ericsson cycle.

This cycle is not used in engines but are used in gas turbine plants.

Stirling, Ericsson, and Carnot cycles are identical in thermodynamics. In theory, each cycle can obtain same maximum efficiency. None of these cycles have found applications in practical engineering.

6.9 REFRIGERATION CYCLE

In vapor refrigeration cycles, the heat is moved from lower temperature to higher temperature at the cost of external work. Figure 6.39 shows the mechanical components of a refrigeration cycle when used as air conditioner.

The COP is defined as follows:

(6.73)

where QL is the useful heat energy extracted from refrigerated space, and WC is the work energy consumed by the compressor. The reverse process is used as a basic for heat pump cycle that extracts heat from the ambient air in winter and heat is transferred to the indoor space. COP of a heat pump is expressed as follows:

(6.74)

FIGURE 6.39  Refrigeration cycle.

EXAMPLE 6.8

One kilogram of air contained in a piston−cylinder apparatus occupies a 0.4700 m^3·kg−1 of specific volume at 250°C and executes a cycle between two heat reservoirs which are, respectively, at 250°C and 150°C temperatures. As shown in Figure 6.40, the cycle proceeds as follows: (1−2): Expanding isothermally to specific volume of 1.19 m^3·kg. (2−3): Cooling to 150°C at a constant volume. (3−4): Isothermal compression to specific volume of 0.47 m^3·kg. Heating to 250°C at constant volume. Show that equation holds for this cycle. Calculate the thermal efficiency of the cycle.

FIGURE 6.40  Schematic diagram of Example 6.8.

Solution: The net heat transfer to the system may be evaluated as follows:

From Figure 6.40(b),

Net heat transported through this cycle is calculated as follows:

Net work transported by the cycle can be expressed as follows:

The processes (4−1) and (2−3) are similar. Work terms W_41 and W_23 are zero.

Also,

where QH is the heat received from the heat reservoir.

REVIEW QUESTIONS

1. Write a brief note on the types of thermodynamic cycles.

2.With the help of a diagram, explain the main elements of a heat cycle. Also define the main performance parameters.

3.With the help of suitable diagram, describe the Carnot Vapor Power Cycle. Also explain hCarnot and limitations of the cycle.

4.Explain in detail about the principle of Carnot cycle as applicable to heat engine and refrigerator.

5.What is Reverse Carnot cycle?

6.Describe in detail about the ideal Rankine cycle. Give its comparison with the Carnot cycle.

7.With the help of T-S diagram, explain the effects of the following operating conditions on the performance of Rankine cycle.

a.Lowering the condenser pressure.

b.Raising the boiler pressure.

c.Superheating the steam.

8.What is regeneration? Discuss the various types of regeneration system.

9.Describe a regeneration system with the following:

a.Open FWHs.

b.Closed FWHs.

10.Differentiate between the following:

a.Closed FWH system versus drains cascaded backward.

b.Closed FWH system versus drains pumped forward.

11.What is reheating? With the help of suitable diagrams, discuss a reheat cycle.

12.What is cogeneration? With the help of a diagram discuss a cogeneration system.

13.Describe an Hg-steam binary vapor power cycle. What are the advantages and limitations?

14.Write notes on the following:

a.Stirling cycle.

b.Ericsson cycle.

15.Explain the performance of a refrigeration cycle suitable for air conditioning.

NUMERICAL EXERCISES

1.Dry saturated steam is supplied to a steam turbine at 5 MPa. Condenser pressure is 5 kPa. Show the Rankine cycle on T-S diagram and determine the simple Rankine efficiency.(36.67%)

2.A steam power plant working on Rankine cycle has steam supply pressure of 20 bar and condenser pressure of 0.5 bar. If the initial condition of steam is dry and saturated, calculate the Carnot and Rankine efficiencies of the cycle neglecting the pump work.(27%, 33%)

3.A steam power plant working on Rankine cycle has a steam pressure of 100 bar and 550°C and condenser pressure of 0.05 bar at the turbine’s inlet. Determine the cycle’s efficiency, SSC, and work ratio, where all processes are reversible.(43%, 2.55 kg/kWh, 0.99)

4.A regenerative Ericsson hot air engine works between temperature limits of 318 K and 503 K. The expansion ratio is 2. Calculate the following:

i.Work done per kilogram of air.

ii.Cycle efficiency.(36.76 kJ/kg, 36.8%)

3.A 5-tonne refrigeration plant is working on R-12 with condenser and evaporator temperatures as 40°C and −10°C, respectively. Determine the following:

i.The refrigerant flow rate in kg/s.(0.18 kg/s)

ii.The volumetric flow rate of compressor in m^3/s.(0.0139 m^3/s)

iii.The compressor’s outlet temperature. (48°C)

iv.The heat rejection in the condenser in kW.(24.27 kW)

v.The pressure ratio.(4.39)

vi.The quality of refrigerant after throttling.(30.5%)

vii.The COP of plant.(4.14)

viii.The power required to drive the compressor.(4.72 kW)

ix.Compare the COP with that of a Carnot refrigerator working between 40°C and −10°C. (0.787)

CHAPTER7

THERMODYNAMIC APPLICATIONS

7.1 ENGINEERING APPLICATIONS OF STEADY FLOW ENERGY EQUATION

For a system at steady-state conditions, the energy remains constant with time. In general, the energy conservation principle for such system gives the following:

,(7.1)

where and are mass flow rates of incoming and outgoing fluids. Therefore, for solving Equation (7.1) for a system, the application of the principle of mass conservation is important.

The total mass flow at the inlet of the system is equal to the total mass flow at the outlet of the system.

Energy analysis of the important engineering equipment will be discussed. The conservation of mass and energy for the two-port systems is as follows:

(7.2)

,(7.3)

where q and w are the specific heat and work transfer of the system.

7.2 DIFFUSERS AND NOZZLES

Diffusers and nozzles are devices used for changing the velocity of a flowing stream. In steam or gas turbine applications, a nozzle increases the velocity of a fluid at the expense of a pressure drop in the direction of flow. A diffuser, however, increases the fluid pressure in the direction of flow at the expense of a decrease in velocity. In centrifugal compressors, the increase in pressure of an accelerated fluid is provided by such devices.

FIGURE 7.1  (a) Converging nozzle, (b) diverging nozzle, and (c) converging–diverging nozzle.

Under subsonic or supersonic flow conditions, the general shapes of a nozzle or a diffuser are shown in Figure 7.1. A diffuser for the supersonic flow or a nozzle for the supersonic flow must have a decreasing cross-sectional area in the direction of flow as shown in Figure 7.1(a). For subsonic flows, however, the opposite occurs. In Figure 7.1(b), the spray nozzle, which is used for the humidification processes, is a supersonic nozzle with an increasing cross-sectional area along the direction of flow. If a fluid has to be accelerated from a subsonic to a supersonic velocity, a converging–diverging nozzle, as shown in Figure 7.1(c), must be used. In such applications, the fluid assumes the sonic flow conditions are at the throat.

Since both of these devices are essentially ducts, no shaft work is involved, and for most conditions, due to the high velocity of the fluid, the heat transfer through the walls is negligible. The potential energy change may also be neglected. Owing to single inlet and outlet, Equations (7.2) and (7.3) are applicable to such devices. With these assumptions, Equation (7.3) may be simplified as follows:

FIGURE 7.2  Schematic of Example 7.1.

(7.4)

By rearranging, the velocity of a fluid at any cross section of a nozzle or a diffuser may be calculated as follows:

(7.5)

In the case of a nozzle, when the pressure and the enthalpy of the fluid are reduced together, acceleration of the fluid flow is accrued. The pressure of steam is changed to its kinetic energy supplied to the turbine.

EXAMPLE 7.1

Steam enters a nozzle with a stagnation enthalpy of 2780 kJ/kg and mass flow rate of 9 kg/min at 1. At the nozzle’s exit, the steam has a velocity of 1070 m/s and a specific volume of 18.75 m^3/kg. Determine, at the nozzle’s exit,

a.the enthalpy of the steam,

b.the cross-sectional area of the nozzle.

Solution:

a.A stagnation enthalpy of a fluid is the summation of the kinetic energy and the enthalpy by h.

b.For one-dimensional flow, considering the continuity of equation, the cross-sectional area of the nozzle at the exit can be determined as follows:

7.3 WATER TURBINE

Figure 7.1 shows a water turbine, receiving water from the height. The potential energy of water is converted into kinetic energy when it enters the turbine, and part of it is transformed into useful work that is used to produce electricity. Taking turbine shaft center as datum, the equation of energy can be written as follows:

FIGURE 7.3  Water turbine.

Work is done by the system, therefore W is positive.

7.4 THROTTLING DEVICES

A throttling device is used to reduce the pressure of the flow by an obstruction. An orifice, a valve, a long capillary tube, and a porous plug are some of the throttling devices.

A reduction in pressure by an orifice in Figure 7.4(a) is used to measure the velocity of a fluid flowing through a pipe. In Figure 7.4(b), by reducing the cross-sectional area of the flow by a valve, a greater pressure drop across the valve occurs and the flow rate decreases. The throttling devices are widely used in refrigeration units, and are also utilized in reducing the power of an engine.

In a throttling process, no work interaction is involved, and usually the transport of heat of the process is neglected. Furthermore, the changes of kinetic and potential energies can be neglected. Thus, for a throttling process, Equation (7.8) is simplified as follows:

(7.8)

By which it is stated that a throttling process can be considered as a constant enthalpy process.

FIGURE 7.4  Throttling processes.

EXAMPLE 7.2

High pressure water at the temperature of 200°C and 20 bar in pressure is adiabatically throttled to 15 bar by a valve (Figure 7.5). Determine:

a.The exit temperature and the quality of water.

b.The velocity if A_2 = 2A_1, and V_1 = 10 m/s.

FIGURE 7.5  Throttling devices.

Solution:

a.The enthalpy of compressed water and the enthalpy of saturated water may be assumed to be same at 200°C. From steam tables, take h_1 = 852.45 kJ/kg.

For saturated steam at p_2 = 15 bar, h_2f < h_2 < h_2g. Therefore, water at the exit of the valve is a mixture of liquid and vapor. Thus, the exit temperature becomes T_2 = 198.3°C and the quality of water is calculated as follows:

b.The specific volume at the exit of the valve is calculated as follows:

Considering the continuity of equation of one-dimensional flow, the exit velocity becomes

7.5 MIXING CHAMBERS

In chemical industry, certain fluids at different temperatures are mixed in appropriate ratios of mass so that a desired temperature is obtained. In a central air-conditioning system, outdoor air is mixed with cooled air for providing desired conditions in the indoor space. The mixing of fluids in such devices must be at the same pressure. There is no work transfer through such devices, and the chambers are usually insulated. The alteration in potential energy and kinetic energy of fluids entering and leaving the chamber may be neglected. Accordingly, the continuity and the energy equations become,

(7.9)

(7.10)

Figure 7.6 shows a steam–water mixer which provides low-pressure hot water by using cold water and steam.

FIGURE 7.6  Steam–water mixing device.

Water and steam valves are fixed at the entrance, and the water temperature is measured by a thermometer at the exit.

EXAMPLE 7.3

Hot water is supplied for a process by mixing steam at 10 bar and 0.2 in quality to the mixing chamber (Figure 7.6) at a rate of 1 kg/s. After throttling to 5 bar, the steam is mixed with 2 kg/s of water at 30°C. Measure the mass flow rate and the temperature of the mixture at the chamber’s outlet.

Solution: With respect to the boundary around the chamber, Equations (7.9) and (7.10) become,

By using the steam tables,

The enthalpy at the chamber’s exit will be h_3 = 365 kJ/kg. Since, h_3 < h_3f at p_3 = 5 bar, the water at the chamber’s outlet is a compressed liquid, and the temperature is

7.6 GAS TURBINE OR STEAM TURBINE

In a gas or steam turbine, gas or steam is passed into the turbine and portion of its energy is converted into work. The turbine runs a generator for production of electricity as shown in Figure 7.7. The gas or steam leaves the turbine at a low temperature and pressure.

FIGURE 7.7  Steam or gas turbine.

These machines are used in a large number of processes frequently encountered in industry varying from power protection to refrigeration, and from liquid transportation in pipe lines to air handling duct systems.

The energy content of the working fluid is converted into mechanical shaft work. As the fluid does not work on the rotating blades of this machine, the fluid’s temperature and pressure decrease in the direction of flow. A single-stage, axial flow steam turbine is presented in Figure 7.8. As shown in Figure 7.8, the steam is first accelerated by the nozzles, and rotates the blades that are attached to the turbine wheel. Rotating shaft in turn turns the generator. The change in the potential energy between the inlet and the outlet of a turbine is usually neglected. Unless stated, the heat losses through the turbine casing are small in comparison to the enthalpy changes. Therefore, for a turbine, Equation (7.3) is simplified as follows:

(7.11)

FIGURE 7.8  Single-stage steam turbine.

7.7 COMPRESSORS AND PUMPS

Compressors and pumps are utilized in compressing or raising the pressure of a fluid. A compressor that uses a gas or a vapor as the working fluid can be either rotating or reciprocating type. In a rotary compressor, the fluid is accelerated and the kinetic energy is increased by the shaft work done on the impeller. Then, in the diffuser section, deceleration of fluid causes an increase in fluid’s pressure. Figure 7.9(a) displays a rotary compressor which might be used for raising the pressure of the inlet air steam of a jet engine. In a household refrigerator, however, a reciprocating compressor is used to increase the refrigerant’s vapor pressure. A sectional view of a pump is shown in Figure 7.9(a). Pumps are used to raise the pressure of liquids.

FIGURE 7.9  Single-stage rotary compressor and pump.

Both machines consume work energy. Again, the change in the potential energy between the inlet and the outlet of a compressor is negligible. For compressors with a large power capacity either fins are attached to the compressor body or water is circulated through the jackets of the compressor’s cylinders for cooling. Hence, in general, the energy equation for a compressor might be written as follows:

(7.12)

7.8 FANS AND BLOWERS

Fans are almost universally used for circulation of air or other gases through low-pressure systems. As shown in Figure 7.10, a centrifugal fan is widely used for moving large or small quantities of air over an extended range of pressures at the expense of work consumption. The heat transfer through the housing of a fan and the change in the potential energy of the fluid are usually neglected. Therefore, Equation (7.3) is simplified as follows:

(7.13)

FIGURE 7.10  Centrifugal fan.

7.9 CENTRIFUGAL WATER PUMP

Pulling of water from a lower level and pumping it to higher level is done by centrifugal water pump as shown in Figure 7.11. Running the pump requires work, which may be provided by a diesel engine or electric motor.

FIGURE 7.11  Centrifugal water pump.

Here ∆u = 0 and Q = 0 because there is no change in the water temperature.

The equation of energy becomes:

(7.14)

Work is applied on the system so the signal of W is negative.

7.10 CENTRIFUGAL COMPRESSOR

Figure 7.12 shows a centrifugal compressor that is used to supply air at moderate pressure and large quantity.

∆Z = 0

FIGURE 7.12  Centrifugal compressor.

Work W is supplied to the system, therefore it is taken as negative and also heat Q is taken as negative as it is lost from the system.

The energy equation becomes:

(7.15)

EXAMPLE 7.4

A small steam turbine working at the partial load produces 100 kW of power at the flow rate of 0.3 kg/s. The steam (Figure 7.13) is throttled before entering the turbine from 1.5 MPa at 300°C to 1 MPa pressure. For an exhaust pressure of 10 kPa, calculate:

FIGURE 7.13  Schematic of Example 7.4.

a.The steam state at the outlet of the turbine.

b.The exhaust duct diameter for a steam velocity of 20 m/s.

Solution:

a.In Equation (7.13), neglecting the change in the kinetic energy, the enthalpy at the outlet becomes h_3 = h_1 − w, where h_1 = 3037.6 kJ/kg, and the turbine specific power is:

Thus, the enthalpy relation yields h_3 = 2704.3 kJ/kg at p_3 = 10 kPa.

Using the superheated steam tables, the temperature and the specific volume at the turbine’s outlet are as follows:

b.Assuming one-dimensional flow in the exhaust duct,

And the duct d_3 = 0.58 m Ans.

EXAMPLE 7.5

An axial flow compressor in Figure 7.14 intakes air at 100 kPa and 27°C, and compresses to 500 kPa and 227°C with an outlet velocity of 100 m/s. For a shaft power of 50 kW, determine the amount of air flowing though the compressor in one hour.

FIGURE 7.14  Schematic of Example 7.5.

Solution: Assuming the compression process to be adiabatic, Equation (7.11) is simplified as follows:

Since the atmospheric air is stagnant, V_1 = 0.

Specific shaft work w as

The mass flow rate becomes

The amount of air flowing through the compressor in one hour will be 871.2 kg. Ans.

7.11 RECIPROCATING COMPRESSOR

Figure 7.15 shows a reciprocating compressor which supplies the air at a high pressure in small quantities (comparing with centrifugal compressor). The reciprocating compressor can be considered as a steady flow system by using a receiver which decreases the flow instabilities completely.

FIGURE 7.15  Reciprocating compressor.

Applying the equation of energy to the process,

EXAMPLE 7.6

A reciprocating compressor (Figure 7.16) is cooled by water circulated through the water jacket of the compressor. The compressor intakes the air at 1 bar and 25°C and compresses to 7 bar and 80°C. The cooling water circulating at a rate of 0.3 kg/s enters at 15°C and leaves at 35°C. If the volumetric rate of flow for the air at the inlet is 0.05 m^3/s, determine the shaft power.

FIGURE 7.16  Reciprocating compressor cooling.

Solution:

The mass flow rate of supplied air:

,

Neglecting the change in the kinetic energy of air,

And the amount of heat removed by water is calculated as follows:

The heat loss of air during the compression process equals the heat removed by water.

Then,

And the transfer of heat per unit mass of air is calculated as follows:

From equation of energy, substituting these values

The shaft power of the compressor is calculated as follows:

7.12 BOILER

A boiler can be used to transform heat to the entering water and produce the steam. The boiler system is shown in Figure 7.17.

For this system,

FIGURE 7.17  Boiler.

Because there is no work is created and not absorbed so W is equal to 0.0

The system’s equation of energy becomes

(7.17)

7.13 THE HEAT EXCHANGERS

A heat exchanger is an apparatus which supplies heat energy to two or more fluids at various temperatures. Transfer of heat between the fluids happens by the separation wall.

The fluids do not mix because they are parted by a surface of heat transfer. Common examples of heat exchangers are the tube and shell exchangers, dry cooling towers, evaporators, automobile radiators, condensers, and air preheaters.

The heat exchanger in Figures 7.18 and 7.19 is composed of active heat exchanging sections as a matrix including the passive fluid distribution sections as baffles, headers, seals, entering and exiting nozzles, and the heat transfer surface. There are no moving parts in a heat exchanger.

The heat loss to the environment from a heat exchanger is usually small in comparison to the heat transfer between the two fluids. There is no work transfer and the changes of potential and kinetic energies of both fluids can be neglected.

FIGURE 7.18  Elements of heat exchanger.

FIGURE 7.19  Cut front of heat exchanger.

Equation (7.1) is simplified as follows:

(7.18)

7.14 CONDENSER

A condenser is used to condense the steam in steam power plants and the vapor of refrigerant in the refrigeration system by the use of air or water as a cooling medium. A condenser is shown in Figure 7.20.

FIGURE 7.20  Condenser.

For this system, ∆KE = 0.0 and ∆PE = 0.0.

W = 0.0 (no work is created or absorbed).

Using the equation of energy for steam flow,

(7.19)

One kilogram of steam flowing through the condenser has heat Q.

The transfer of heat between water and steam is the only transfer in the system.

Heat absorbed by the cooling water is calculated as follows:

From Equation (7.8),

,(7.20)

where mw = mass flow rate of cooling water and cw = specific heat of water.

7.15 EVAPORATOR

FIGURE 7.21  Evaporator.

For this system, ∆KE = 0.0 and ∆PE = 0.0

Work is neither absorbed nor provided, W = 0

By using the equation of energy to the system, the following is obtained.

(7.21)

The system’s temperature is less than the temperature of environment. Heat flows into the system, and therefore Q is taken as positive.

EXAMPLE 7.7

Figure 7.22(a) shows a layout of a room’s air conditioner in which refrigerant R22 is used as the working fluid. The warm room air at 100 kPa, 27°C enters the evaporator at a flow rate 6m^3/min. As shown in Figure 7.22(b), the refrigerant at 245 kPa pressure and 0.3 quality flows into the tubes of the evaporator at a rate of 1 kg/min, and at the same pressure leaves as saturated vapor. Calculate the temperature of air at the evaporator’s outlet.

Solution:

The hot and the cold fluids are, respectively, the room air and the refrigerant. The mass flow rate of air is calculated as follows:

The enthalpy change of the refrigerant is calculated as follows:

From Equation (7.18),

And solving for Te, the temperature of air at the evaporator’s exit becomes

Te = 5.03°C Ans.

FIGURE 7.22  Layout of a room’s air conditioner.

7.16 FLOW IN DUCTS AND PIPES

There is no need to signify the engineering importance of transporting a liquid or a gas between two stations. The pressure drop happens when fluids flow through a duct or a restraining channel. This drop of pressure is called fluid friction. The magnitude of fluid friction is influenced by different factors such as fluid density, velocity, viscosity, condition of its surface, shape or diameter of duct part, types of flow turbulent or viscous, and fluid temperature and pressure.

A piping or a duct system has only one inlet and outlet. Hence, the energy equation in the form of Equation (7.3) is applicable to such systems. In certain applications, in addition to flow friction, heat transfer takes place. There are numerous examples for such a case like flow of water through the tubes of a boiler, flow in pipes of a heat exchanger, or in flow of a refrigerant through the tubes of an evaporator or a condenser. In these applications, transfer of heat cannot be neglected in the energy equation. For a flow of air in distribution ducts of an air-conditioning system, however, the walls of the channel are insulated and the system is taken to the adiabatic condition. In cases of installing a fan, a pump, or an electric heater to a pipe system, then the work transfer in Equation (7.3) has to be retained.

Let us consider a flow of an incompressible fluid in an insulated duct. The enthalpy change for an incompressible fluid is calculated as follows:

(7.22)

Due to fluid friction, certain amount of heat will be produced at the conduit walls. Because of insulation, this heat is regained as an increase in the fluid’s internal energy. However, suppose the fluid is to be non-viscous, then there is no friction, and the process of flow becomes isothermal. Thus, for an incompressible, non-viscous flow of a fluid through an insulated duct together with Equation (7.22), Equation (7.23) is simplified as follows:

(7.23)

This is known as the Bernoulli’s equation.

EXAMPLE 7.8

A hair dryer is basically a duct in which a 100-W electric resistor and a 50-W fan are installed. The fan steadily sucks in the ambient air at 100 kPa, 20°C and forces it over a resistor such that the exit temperature of air is 50°C. Calculate:

a.Velocity of the air at the exit.

b.Volumetric flow rate of the air at the entrance.

The exit of dryer has a cross-sectional area of 20 cm^2.

FIGURE 7.23  Schematic of Example 7.8.

Solution:

a.The mass flow rate is calculated as follows:

The enthalpy change of the refrigerant is calculated as follows:

The total work transferred to the duct is calculated as follows:

There is no change in the potential energy of this flow, and the surface of the duct is assumed to be insulated. Therefore, Equation (7.3) becomes

Where the velocity of ambient air is taken to be zero, V_1 = 0. Rearranging this expression resulted as follows:

Which yields

b.The mass flow rate is calculated as follows:

With respect to the inlet conditions,

And the volumetric flow rate at the inlet is calculated as follows:

7.17 UNSTEADY FLOW PROCESSES

In engineering practice, the unsteady flow processes are as common as the steady flow processes. In a transient flow process, the average mass flow and energy inside and outside of the control volume are not same.

The process of charging of compressed gas tanks is not a steady flow process because during this process, fluid mass flow rate as well as the temperature and the pressure of the fluid in the tank will vary with time. The system is called transient system. One of the applications of the transient systems is to run a turbine by the pressurized air of a storage tank, for fulfilling the energy needs of a facility during a cutoff period.

In transient systems, the shape and the boundary might vary with time, and the boundary can be in motion with respect to a reference point. Besides, an accurate calculation of the terms is required for mass and energy equation. However, analytic method can still be devised to get results reasonable for a system under study. Thus, to get a solution, the following simplifications can be used for transient system:

1.The thermodynamic state of every point within the boundary of the system is the same at a given time. The properties of the system are single valued. However, the state of the system is time variant.

2.The state of the fluid flowing into the system is time invariant. This assumption can be fulfilled by extending the boundary of the system to a region where the fluid properties do not vary with time. The flow rate of incoming fluid, however, might be a function of time.

3.The fluid flowing out of the system is assumed to be at the same stage as the system. This assumption requires that the fluid is essentially in equilibrium at all times. Hence, a discharge of a fluid must be slow enough that the process is quasistatic.

In regard to these assumptions, the kinetic energy change is neglected. Unless indicated, the change in the potential energy is usually negligible.

Two subsequent cases will be studied:

1.Filling of a tank.

2.Draining of a tank.

7.17.1  Filling of a Tank

Assume,

T_1 = initial temperature of fluid

T_2 = final temperature of fluid

v_1 = initial specific volume of fluid

v_2 = final specific volume of fluid

p_1 = initial pressure of fluid

p_2 = final pressure of fluid

m_1 = initial mass of fluid

m_2 = final mass of fluid

u_1 = initial specific internal energy of fluid

u_2 = final specific internal energy of fluid

C′ = inlet velocity of fluid

v′ = inlet specific volume of fluid

T ′ = inlet temperature of fluid

P ′ = inlet pressure of fluid

h′ = inlet specific enthalpy of fluid

u′ = inlet specific internal energy of fluid

FIGURE 7.24  Filling of a tank.

The amount of fluid entering is calculated as follows:

The energy of the fluid entering is calculated as follows:

Heat transported to the control volume is calculated as follows:

(7.26)

If the tank is completely insulated, there is no transfer of heat.

Initially, if the tank is empty and completely insulated,

Therefore,

(7.28)

Similarly, by neglecting the kinetic energy in the pipe line, the following is obtained.

(7.29)

7.17.2  Draining of a Tank

Similar to the tank’s filing process, the energy equation can be taken as follows:

(7.30)

where C ′= outlet velocity of fluid and h′ = outlet specific enthalpy of fluid.

For a completely drained tank and no heat transfer and negligible exit velocity,

(7.31)

7.18 EXERGY EFFICIENCY OF HEAT EXCHANGER

The fluids do not mix because they are separated by a heat transfer surface. For adiabatic condition of heat exchanger and steady state, exergy equation will be as follows:

Decrease in the exergy of the hot fluid = increase in the exergy of the cold fluid

Exergy destruction = 0

The exergy efficiency of a heat exchanger is calculated as follows:

(7.32)

FIGURE 7.25  Exergy efficiency of heat exchanger.

EXAMPLE 7.9

An air heater has tubes of 10 cm diameter and water enters the tubes at a mass flow rate of 24 kg/min at 140°C and 0.5 MPa and exits at 60°C and 0.5 MPa. Air is the cold fluid that enters heat exchanger at volumetric flow rate of 100 m^3/min. The inlet conditions are 25°C and 110 kPa, and the used air velocity is 25 m/s. At the outlet, the pressure of air is 110 kPa. If the surrounding is at P_0 = 100 kPa, T_0 = 25°C, determine:

a.Outlet temperature of the air.

b.Efficiency of the heat exchanger.

FIGURE 7.26  Heat exchanger.

Solution:

a.At steady-state condition, the equation of energy is as follows:

Assuming air as an ideal gas,

Using thermodynamic properties of water,

b.By neglecting the change in the velocity of water, the change of exergy of water stream is calculated as follows:

The cross-sectional area of the air tubes at the inlet is assumed to be same as at the outlet.

The exergy change of air stream is calculated as follows:

The heat exchanger’s efficiency, from Equation (7.32), is calculated as follows:

It shows that 52.4% of the useful energy has been wasted at the inlet of heat exchanger, but from exact efficiency of energy, there is no waste, and energy efficiency is 100%. This is not correct as total transfer of useful energy from hot fluid to cold fluid cannot happen.

7.19 EXERGY EFFICIENCY OF EXPANSION (TURBINE) AND COMPRESSION (COMPRESSOR)

Turbines and compressors transform fluid energy. Assuming that there is no heat loss to the environment and the system operates at steady condition, energy recovery by a turbine can be determined by using the following equation:

(7.33)

Exergy supplied to a turbine is

The exergy efficiency is calculated as follows:

(7.34)

is the maximum useful work, that is, Wrev. This work can be extracted out from the stream of fluid at similar inlet and outlet conditions. Equation (7.34) gives two dissimilar definitions of exergetic efficiency of a turbine. A pump, fan, or compressor with no transfer of heat to the surrounding and at steady condition, taking, in Equation (7.35), the exergy equation will be as follows:

(7.35)

The percentage increase in the exergy of flow to the provided exergy represents the exergy efficiency. For a fan or a compressor at adiabatic condition,

Exergy efficiency of steam turbines is shown in Figure 7.27.

FIGURE 7.27  Steam turbine efficiency.

Exergy efficiency of compressors is shown in Figure 7.28.

FIGURE 7.28  Compressor.

EXAMPLE 7.10

Steam entering an adiabatic turbine with a velocity of 80 m/s at 600°C and 6 MPa and leaving at 100°C and 50 kPa is shown in Figure 7.28. The speed at the outlet is 140 m/s. The conditions of surrounding are 25°C and 100 kPa, and the shaft power of turbine is determined to be 5 MW. Calculate the exergy loss in the turbine.

FIGURE 7.29  Steam turbine.

Solution:

The exergy equation can be simplified as follows:

In the case of Example 7.10,

For a steady-state flow, the shaft work will be

The exergy relation is given as follows:

The mass flow rate is calculated as follows:

For

The wastage of exergy is calculated as follows:

7.19.1  Exergy Efficiency of Combustion

Exergy efficiency of adiabatic combustor is shown in Figure 7.30.

FIGURE 7.30  Adiabatic combustor.

If fuel and oxidant are supplied at environmental temperature,

And the equation is simplified as follows:

EXAMPLE 7.11

The burning gas flows at a speed of 80 m/s through the turbo-jet engine nozzle at 747°C and 260 kPa, and leaves at 5000°C and 70 kPa. The ambient temperature is 17°C. Calculate:

FIGURE 7.31  Flow through nozzle.

a.The speed of gas at the nozzle’s exit.

b.The exergy rate of reduction of the gas at a mass flow rate of 0.1 kg/s.

Suppose k = 1.300 and cp = 1.1500 kJ/kg·K of the burning gas.

Solution:

a.For this specific condition, there is neither work nor heat exchange across the nozzle’s boundary. The equation of energy at a steady-state condition is given as follows:

Using the given data, the velocity of gas leaving the nozzle is calculated as follows:

b.The equation of exergy is given as follows:

Suppose if the gas behaves as an ideal gas, the change of entropy can be determined as follows:

The exergy rate reduction due to mixing and friction of fluid atoms is given as follows:

REVIEW QUESTIONS

1.What are diffusions and nozzles? What are their types and uses?

2.Derive an energy equation for a water turbine.

3.Write notes on:

a.Throttling devices

b.Mixing chambers

c.Gas and steam turbines

d.Compressors and pumps

e.Fans and Blowers

4.Derive energy equation for the following:

a.Centrifugal water pump

b.Centrifugal air compressor

c.Reciprocating compressor

5.Write notes on heat exchangers such as boilers, condenser, and evaporators.

6.Write Bernoulli’s equation for flow through ducts and pipes.

7.Explain the energy equations for the following unsteady flow processes:

a.Filling a tank

b.Emptying a tank.

8.Describe the exergy efficiency of heat exchange, expansion, compression, and combustion processes.

NUMERICAL EXERCISES

1.In an isentropic flow through a nozzle, air flows at the rate of 600 kg/hr. At the nozzle’s inlet, pressure is 2 MPa and temperature is 127°C. The exit pressure is 0.5 MPa. If the initial air velocity is 300 m/s, determine:

a.The exit velocity of air.

b.The inlet and exit area of nozzle. (594 m/s, 31.88 mm^2, 43.35 mm^2)

2.Water vapor at 90 kPa and 150°C enters a subsonic diffuser with a velocity of 150 m/s and leaves the diffuser at 190 kPa with a velocity of 55m/s. During this process, 1.5 kJ/kg of heat is lost to the surrounding. Determine:

a.The final temperature.

b.The mass flow rate.

c.The exit diameter assuming the inlet diameter as 10 cm and steady flow.(154°C, 0.543 kg/s, 11.42 cm)

3.Following are the details of a steam turbine:

Steam flow rate

= 1 kg/sec

Inlet velocity of steam

= 100 m/sec

Exit velocity of steam

= 150 m/sec

Enthalpy at inlet

= 2900 kJ/kg

Enthalpy at outlet

= 1600 kJ/kg

Write the steam flow energy equation. Assuming that the change in the potential energy is negligible, determine the power available from the turbine.(1293.75 kW)

4.An air compressor compresses atmospheric air at 0.1 MPa and 27°C by ten times of inlet pressure. During compression, the heat loss to the surrounding is estimated to be 5% of compression work. Air enters the compressor with a velocity of 40 m/s and leaves with a velocity of 100 m/s. Inlet and exit from cross-sectional areas are 100 cm^2 and 20 cm^2, respectively. Estimate the temperature of air at exit from compressor and power input to the compressor.(1498 K, 5467.86 kW)

CHAPTER 8

MEASUREMENT THEORY AND INSTRUMENTS

8.1 INTRODUCTION

A measurement system is the one which connects the observer with the process. A typical measurement system is depicted in Figure 8.1. The measured variable is considered to be the information variable. The observer is offered with the values of the information variable. The true value of the variable is the input to measurement system and the measured value of the variable is the output of measurement system.

FIGURE 8.1  Measurement system

8.2 PERFORMANCE AND PURPOSE OF MEASUREMENT SYSTEMS

A process is known as a system, which is used to generate information. Examples are a jet fighter, a chemical reactor, a weather system, the human heart, a car, a submarine, a gas platform, etc.

Acceleration, velocity, and displacement variables are generated by a car and the composition, pressure, and temperature variables are generated by a chemical reactor.

In a perfect measurement system, the measured value should be the same as the true value. The precision of the system means the nearness between the true value and the measured value. A complete precise system is a theoretical system. A precision real system is quantified.

The error of measurement system (E) can be defined by the following equation:

E = (measured value) − (true value)
E = (system output) − (system input)

If the measured value of the gas flow rate inside your pipe is 11.0 m^3/h while the true value is 11.2 m^3/h, the error will be E = −0.2 m^3/h. If the measured value of the engine rotational velocity is 3140 rpm and the true value is 3133 rpm, E = +7 rpm. This is the main performance index for a measurement system.

8.3 MEASUREMENT VARIABLES

8.3.1  Calibration

Calibration is a comparison between the sensor or instrument output under test versus the instrument output of known precision when the same value of measurement input is used for both instruments.

This process is carried out for a range of inputs, which cover the full range of measurement of the sensor or instrument. Calibration assures that the measuring precision of all sensors and instruments used in a measurement system for the complete range of measurement, on one condition that the calibrated sensors and instruments are used in the same environment under which they were calibrated. If sensors and instruments are used under different environments, suitable rectifications have to be done. Calibration steps used for a single sensor will be applicable to all instruments.

8.3.2  Units of Measurement

The units of measurement were used in exchange trading to determine the sums to be paid and to make rules of the relative values of various merchandises. Early measuring systems were dependent on whatsoever was available as a measuring unit. The human torso was a suitable example for the objective of measuring the length and it granted the units of the cubit, the hand, and the foot. Although mostly enough for swap commerce systems, these units of measurement are inaccurate, changing from one body to another. There has been a progressive action towards units of measurement which must be more precise.

Many units were named for measuring the same physical variable. For example, length can be measured in meters, yards, etc. In addition to main length units, sub-division in units is available such as inches, feet, millimeters, and centimeters, with a relation between these units. For example, 1 feet is equal to 12 inches, 1 mile is equal to 1760 yards, and 1 yard is 3 feet. A metric system is now used which contains, for example, the meter unit and its sub-divisions (millimeter and centimeter) for measuring the value of length.

8.3.3  Measurement Uncertainty

The instrument accuracy means closeness of the instrument reading output to the true value. It is more customary to quote the inaccuracy value of measurement instead of the instrument precision value.

Inaccuracy of measurement is the range in which reading maybe erroneous and is quoted as a proportion of the full-scale reading of a device.

The maximum error of measurement is attached to the full-scale reading of the instrument. Measured values are less than the full-scale reading. This means that the potential error of measurement is magnified. It is important that devices are selected for a range, which will cover the values to be measured. If the pressure has to be measured between 0 and 1 bar, do not use the instrument for range of measurement between 0 and 10 bar.

EXAMPLE 8.1

The range of measurement of a pressure gauge is 0−10 bar and a quoted inaccuracy is ±1.0% on full-scale reading.

1.Find the maximum error obtained in measurement for this instrument.

2.Find the probable measurement error obtained as a proportion of the output reading if the pressure gauge is used to measure air pressure value of 1 bar.

Solution:

1.The maximum error obtained in all measurement reading is 1.0% of the full-scale reading, which is 10 bar for this device. Thus, the maximum probable error is 1.0% × 10 bar = 0.1 bar

2.The maximum error of measurement is a fixed value attached to full-scale reading of instrument, regardless of the amount that the device is really measuring. The maximum error value is 0.1 bar. Therefore, when measuring the pressure of 1 bar, the maximum error of this pressure is 10% of the measured value, i.e., 0.01 bar.

8.3.4  Measurement Sensitivity and Repeatability

The measurement sensitivity is defined as a measure of the variation in output of instrument when the amount being measured varies by a given quantity.

The measurement sensitivity is the slope of the straight line.

Repeatability shows the proximity of output readings when the same value of the input is applied frequently through a short time period, with the same device and observer, same conditions of measurement, same conditions of utilizing maintained all the time, and the same location.

8.4 MEASUREMENT OF TEMPERATURE

Temperature is the property of a system that indicates the potential for heat transfer with other systems. Therefore, two systems are said to have equal temperature when there is no heat transfer between them.

Temperature can be measured by a thermometer based on the molecular or internal energy of the body.

The first measurement of temperature was executed by Galileo at the end of 16th century. His thermometer was based on air expansion. A scale was connected to his device to point out “heat degrees.” It is necessary to have unanimous measurement units. For temperature, the degree of Celsius (C°) and the Kelvin for the international units.

8.4.1  Concept of Temperature

Temperature measures the change of kinetic energy with temperature in a material. Kinetic energy is the movement of molecules: the molecules of solids vibrate in place, but molecules of liquids and gases move freely. All the molecules do not move with the same speed. With higher temperature, the molecules move faster. The temperature is defined as a measure of potential of stored energy in a mass of matter.

Temperature is the potential used to produce heat transmit from a higher temperature point to a lower temperature point. The temperature does not measure the amount of heat in a matter because molecules have both potential and kinetic energy. Temperature can only measure kinetic energy and not potential energy. The heat transfer rate is a function of temperature change.

8.4.2  Temperature Scales

To compare and measure temperatures, a temperature scale should be available. These temperature scales are defined in an expression of physical properties that take place at fixed temperatures.

i.The Celsius temperature (°C)

The Celsius temperature scale is known by international convention in the expression of two fixed points, the steam under the ice points. The ice point temperature is defined as 0°C and the steam point temperature as 100°C. The ice point is a temperature of water and ice at a pressure of 1.013 × 10^5 N.m−2. Filtered water is used to prepare the ice flakes and blended with ice-cold filtered water.

The steam point is the boiling temperature of filtered water at a pressure of 1.0132 × 10^5 N.m−2. Temperature interval between the steam point and the ice point is 100°C.

Kelvin (K), thermodynamic or absolute temperature scale on the name of Lord Kelvin is a scale, which does not depend on properties of any specific material. Lord Kelvin divided the temperature interval between the ice point and steam point into one-hundred divisions thus 1 K is equal to same interval of temperature as 1°C. The Kelvin unit is defined as the fraction (1/273.16) of the temperature of the water triple point. This definition was adopted by 13th CGPM in 1967.

ii.Rankine and Fahrenheit scales

The Rankine and Fahrenheit scales are presently used in the United States and Britain. Steam tables, engineering data, etc. have been published utilizing the Rankine and Fahrenheit units. The steam point (212°F) and the ice point (32°F) are used on the Fahrenheit scale.

The Rankine cycle is a thermodynamic temperature corresponding to the Fahrenheit scale. The ice point is at 491.67°R while zero Fahrenheit is 459.67°R on the Rankine scale.

8.4.3  Temperature Measuring Instruments

These instruments are classified into two wide categories:

1.Non-electrical methods

a.By using the change in a liquid volume with change of temperature

b.By using the change of gas pressure with temperature change

c.By using the pressure change of vapor with temperature change

2.Electrical methods

a.Thermocouple

b.Resistance thermometers

c.By comparing the filament colors with that of the object whose temperature is to be found out, that is, optical pyrometers.

d.By achieving the power received by radiation, that is, radiation pyrometers.

The following instruments are used for temperature measurements in industries:

1. Thermistors, 2. thermocouples, 3. RTD (Resistance Temperature Detectors), 4. IC sensors, 5. bimetallic indicators, 6. optical sensors such as: (a) pyrometers, (b) infrared detectors, (c) liquid crystals, 7. liquid bulb thermometers, 8. gas bulb thermometers.

The most popular instruments are the thermistors, the thermocouples, the RTD’s and the IC sensors.

8.4.3.1  Classification of Thermometers

Thermometers are also classified as the following:

a.Expansion thermometers

i.Liquid-in-glass thermometers

ii.Bimetallic thermometers

b.Pressure thermometers

i.Vapor pressure thermometers

ii.Liquid-filled thermometers

iii.Gas-filled thermometers

a.Thermocouple thermometers

b.Resistance thermometers

c.Radiation pyrometers

d.Optical pyrometers

8.4.4  High-temperature Thermometers

Mercury boils at 357°C at atmospheric pressure. In order to increase the range of mercury-in-glass thermometer for high temperature use, thermometer bore is expanded into a bulb having a space of about 20 times as that of the stem bore. The bore above the mercury with the bulb is filled with carbon dioxide or nitrogen at high pressure to prevent the boiling of mercury at high temperatures.

8.4.4.1  Use of Liquids other than mercury

In industries where the mercury leakage from the cracked bulb might cause large damage to the products, other liquids are utilized to fill in the thermometer. Other liquids are also used where the temperature range of the mercury-in-glass thermometer is not suitable. Table 8.1 shows liquid with their suitable temperature range.

Toluene, pentane, and ethyl alcohol can be used for liquid-in-glass thermometers. These liquids are colorless, tincture is added for easy reading. Due to low freezing point, these are suitable for low-temperature thermometers.

TABLE 8.1 Thermometer liquids with temperature range

Liquid

Temperature range (°C)

Mercury

−35 to +510

Alcohol

−80 to +70

Toluene

−80 to +100

Pentane

−200 to +30

Creosote

−5 to +200

8.4.5  Thermocouples Material

Thermocouple materials are divided into two groups depending on material cost, low-cost metal thermocouples and base metal thermocouples.

Table 8.2 shows thermocouples along with their specifications as per British standard BS4937. The schematic diagram of thermocouple is shown in Figure 8.2.

TABLE 8.2 Thermocouples as per British Standards

FIGURE 8.2  Thermocouple diagram

The recommended colors of thermocouple wire are shown in Table 8.3.

TABLE 8.3 Recommended colors of thermocouple wire

8.4.6  Temperature Calibration

There are two types of baths for calibration depending upon the temperature limit:

i.water bath calibration

ii.sand bath calibration.

8.4.7  Comparison of Advantages and Disadvantages of Measurement Systems

Table 8.4 gives the advantages and disadvantages of three types of temperature measurement systems. The final selection of the technology depends upon design time, cost, circuit size, etc.

TABLE 8.4

a.Advantages

Thermocouple

Thermistor

RTD

Self-powered

High output

Most stable

Simple

Fast

Most accurate

Inexpensive

Tow-wire ohm measurement

Most linear than thermocouple

Variety of physical forms

b.Disadvantages

Thermocouple

Thermistor

RTD

Non-linear for a wide range

Non-linear

Expensive

Low voltage

Limited temperature range

Slow

Reference required

Fragile

Current source required

Least stable

Current source required

Small resistance change

Least sensitive

Self-heating

FIGURE 8.3  Advantages and disadvantages of three systems of temperature measurement with their differences

EXAMPLE 8.2

The changes of density of mercury are linear with the temperature as, ρm = 14277.5 − 2.5 T (K). Due to the influence of temperature, the same pressure difference will be measured by different manometer heights. Suppose in New York City, on a hot summer day the temperature is 40°C and the pressure is the same as the pressure measured on a cold winter day of −10°C. What will be the present deviation in the manometer reading?

Solution: Pressure is same for both cases,

The present deviation in height may be expressed as follows:

As T_1 = 313 K and T_2 = 263 K from given relation, ρ_1/ρ_2 = 0.9908. The percentage error becomes, Err% = 0.91%, which is less than 1 percentage.

8.5 PRESSURE MEASUREMENT

8.5.1  Definition of Pressure

The fluid exerts a force on the boundary of the system. The pressure is the force per unit area. There are three types of pressure measurements.

1.Absolute pressure is defined as the difference between the absolute zero pressure and the pressure at a specific point in a fluid. This can be measured by a barometer.

2.Gauge pressure: Pressure gauge is used to measure the difference between the local atmospheric pressure and unknown pressure.

3.Differential pressure: A pressure gauge is used to measure the difference between two unknown pressures.

8.5.2  Pressure Units

Pascal

Pa

Bar

bar

Standard atmosphere

atm

Kilogram force per square cm

Kgf/cm^2

Pound force per square cm

Inch Ib/in^2

Millimeter of water

mm H_2O

Millimeter of mercury

mm Hg

Inch of water

In H_2O

Inch of mercury

In Hg

8.5.3  Pressure Thermometers

Vapor, liquid, and gas pressure thermometers can be used. This fluid is heated and fluid pressure increases. The temperature can be measured by Bourdon pressure gauge. The thermometer bulb full of fluid is put in the area where the temperature needs to be measured.

Pressure thermometers are explained below:

i.Vapor pressure thermometer

A vapor pressure thermometer is shown in Figure 8.4. The fluid bulb is fixed in the area where the temperature needs to be measured. Some of the fluid vaporizes and raises the vapor pressure, which is measured on the Bourdon tube. The relationship between vapor pressure and temperature of a volatile liquid is of exponential shape.

FIGURE 8.4  Vapor pressure thermometer

ii.Liquid-filled thermometer

A liquid-filled thermometer is shown in Figure 8.5. The liquid expansion causes the indicator to move on the dial. Alcohol, toluene, glycerin, and mercury have been successfully used for pressure changes between 3 and 100 bar. These thermometers are used for temperatures up to 650°C.

FIGURE 8.5  Liquid-filled thermometer

iii.Gas-filled thermometer

The gas-filled thermometer has almost the same range as the liquid-filled thermometer. Helium (He) and nitrogen (N_2) are insert gases, with low specific heat and high expansion coefficient. Therefore, He and N_2 gases are commonly used. The construction is same as the liquid-filled thermometer but with bigger bulbs.

These thermometers are commonly utilized for pressures less than 35 bar.

8.6 MEASUREMENT OF VELOCITY

The flow rate and velocity of flow are commonly needed for the control of the industrial processes and thermodynamic analysis of systems. The flow speed is generally measured indirectly by the following instruments:

i.pitot tube

ii.vane anemometer

iii.hot wire anemometer

iv.laser Doppler anemometer.

8.6.1  Pitot Tube

The velocity measurement of fluid is very commonly carried out by a pitot tube.

Principle

A pitot tube is used to measure the change between static pressure and stagnation pressure at the point where the speed is to be calculated (Figure 8.6).

FIGURE 8.6  Pitot tube measures change of stagnation pressure

Figure 8.7 shows two types of tubes: a Brabbee tube and a Prandtl tube depending upon the tip shape. Fluid velocity near the tape is given as follows:

(8.1)

where C is the coefficient of speed and has to be calculated by calibrating the pitot tube. h = height of fluid in monometer. The velocity (V_1) of a perfect gas at the tip point,

(8.2)

where P_0 is the stagnation pressure, and (T_1, p_1) are the temperature and the pressure at a point where the speed is to be calculated.

The simplest pitot tube is a tube with an effective opening offer 3.125–6.35 mm diameter facing the approaching fluid. A normal upstream tab can be utilized for calculating the line pressure.

FIGURE 8.7  Pitot tube for velocity measurement

An industrial type of pitot tube cylindrical probe installed inside the air stream. At the upstream face, the fluid flow speed of the probe is taken to be zero. Velocity head transforms into short pressure, which is sensed by a small hole in upstream face. On the other side, a small hole of the probe measure static pressure. The instrument measures the differential pressure.

Figure 8.8 shows a pitot tube with the tabs of measuring static pressure.

The pitot tube produces no loss of pressure in the flow stream. It is installed over a nipple in the pipe. If installed over an isolation value, it can be moved backward and forward over the stream to create the flow velocity profile. For high-speed flow streams, it is necessary to use tubes of high strength and stiffness.

A tube installed in a high-speed stream has the possibility to fluctuate and get broken. Pitot tubes are commonly used for low to medium velocity flows.

FIGURE 8.8  Pitot tube installation

EXAMPLE 8.3

The air velocity in a duct is measured as 40.1 m/s by a pitot tube and the recorded pressure is 0.12 m of water column. Take air density as 1.2 kg/m^3, and determine the coefficient of velocity, C.

Solution: By applying Equation (8.1),

8.6.2  Vane Anemometer

It is shown in Figure 8.9, the moving stream forces the vane to rotate. The strength of force increases with the velocity of air. The rotational velocity of the vane anemometer also increases. The vane anemometers are used to measure the speed of air in large flow areas. The anemometer calibration is carried out by measuring the rotational speed of vane at known wind speeds.

FIGURE 8.9  Vane anemometer

8.6.3  Hot Wire Anemometer

A thin platinum wire of 0.005 mm diameter and 1 mm length is electrically heated and put perpendicular into a flow area as shown in Figure 8.10. Heat is transferred from the hot wire anemometry to the fluid stream by conviction.

FIGURE 8.10  Probe of a hot wire anemometer

If the wire current and its temperature Tw are kept constant, the equation between the power input (to the wire) and the difference of temperature between the fluid and the wires is as follows:

(8.3)

where R is the wire resistance, I is the current in the wire, ρ is the density of the free stream and V is the velocity of the free stream. B and A are constants and calculated by probe calibration curve. The hot wire anemometer is very suitable for measuring speeds in turbulent flow. The hot wire anemometer may record changes of velocity with a frequency (up-to-10^5 Hz), and speeds as small as 0.02 m/s. The hot wire anemometer is very suitable for small speed flows. It is used in the analysis of the boundary layer.

The hot wire anemometer has been used on a large scale for many years as a tool of research in fluid mechanics.

The hot wire anemometer is shown in Figures 8.11 and 8.12 as a small, electrically heated element. These elements are sensitive to heat transfer between the element and its surrounding and also structures and temperature changes can be sensed.

FIGURE 8.11  One type of hot wire anemometer

FIGURE 8.12  Another type of hot wire anemometer

EXAMPLE 8.4

The velocity data of air (ρ = 1.2 kg/m^3) flowing over a 24 cm, diameter pipe is recorded by the hot wire anemometer and given in the table below. Calculate the mean velocity, the mass flow rate of air, and the ratio of maximum to mean spread.

Solution: The mean velocity,

At each strip, the velocity is the rate of the top and the bottom speeds. Thus the mean speed is Vm = 4.71 m/s, Vmax = 9.7 m/s.

8.6.3.1  Principle of Operation

Consider a thin wire fixed to probes and subjected to a velocity U as shown in Figure 8.13. As current is passed through the wire, heat is produced (I^2RW). This is balanced with loss of heat to the surrounding under equilibrium condition. With change of velocity, the coefficient of convective heat transfer will change, the temperature of the wire will alter and finally arrive at a new equilibrium state.

FIGURE 8.13  Principle of operation of hot wire anemometer

8.6.3.2  Classification of Hot Wire Anemometer

Based on the number of sensors, the classification of anemometer is shown in Figure 8.14.

FIGURE 8.14  Classification of hot wire probes

8.6.4  Laser Doppler Anemometer (LDA)

In the case of a hot wire anemometer, dirt inside the flow may be deposited on the wire acting as insulation, or because of high wire temperature, fluid may be spoiled in the measurement area. Furthermore, because of the probe size, the floor may be distributed under such objectives. LDA can be used for perfect measurement of speed and turbulence in liquid and gas flows.

As shown in Figure 8.15, a fixed at wavelength laser acts as a light source and the visual components divided the beam of the layer into a second beam and a reference beam and then these beams intersect at measurement particle.

The lighter frequency is dispersed by moving particle by a quantity proportionate to the practical velocity. Laser Doppler anemometer uses this principle in measuring the particle speed. Each of the frequency following filter closes onto the modification frequency to obtain the frequency of Doppler. The frequency of Doppler is linear regarding the speed over the geometry of the visual system.

The particles should be of small size sufficient to pursue the flow. The inflow of liquid-like particles is normally attended, but the inflow of gas, particle with size 10−6 m, are grown. To transmit the light beam, the fluid surroundings should be diaphanous. The applications of a laser Doppler anemometer include flow inside engine cylinders, pipe flows, combustion process, and flow between blades of the pump impeller.

FIGURE 8.15  Operation of laser Doppler anemometer

8.7 FLOW MEASUREMENT

For a gaseous or liquid matter, flow can be measured as either the volume flow rate or the mass flow rate, per unit time. A mass flow rate gives more precise measurement.

8.7.1  Mass Flow Rate

The mass flow rate is measured by different techniques depending upon the fluid state, that is, gaseous, solid, or liquid.

8.7.1.1  Variable Area Flow Meter (Rotameter)

The area of orifice is used in rotameters to measure the flow rate. It is inexpensive, reliable, and used widely in industry, accounting for around 20% of all flow meters sold. For its use in automatic control charts, rotameter is combined with a row of fibers, which gives the float location by sensing the light emitted from it.

Figure 8.16 shows this instrument, which comprises a glass tube with tapered shape and a float is made of stainless steel that moves freely in side of the tube. The fluid flows over the float, the forces establish a balance float location in the inside of the tube.

FIGURE 8.16  Schematic of rotameter

The device should be fixed vertically, and basically, the motion of float is linear with the flow rate. Variable area flow meters are not suitable for high pressures and liquids with large particles. The inaccuracy of the low-cost device is usually ±5%; however, more costly rotameter offers precision as high as ±0.5%.

A simple rotameter (Figure 8.17) consists of a tapered glass tube and a float that takes a steady position when flooded. The float location is a measure of the flow rate.

FIGURE 8.17  Flow meters (Rotameters)

EXAMPLE 8.5

At standard conditions (p = 1 atm, T = 20°C), a Variable Area Flow meter is calibrated for water, (ρ = 998 kg/m^3). However, the rotameter is used to measure the oil flow rate (ρ = 880 kg/m^3) and the scale on the rotameter reads (12 L/s). Calculate the oil flow rate.

Solution: The equation of flow rate in the rotameter is , where A is the flow area between the float and the tube with a tapered shape. Thus, the ratio flow rate becomes The float of aerodynamic suspension is got by the same value of pressure drop through the float, Assuming that the pressure drop is commensurate with V_2,

8.7.1.2  Orifice Plate

The orifice plate is a metal disc having a hole in its center, which is installed inside the pipe, carrying the flowing fluid. Orifice plates are cheap and available in a board range of sizes. As a result, they represent around 50% of the devices used in the industry for measuring the volumetric flow rate. One restriction of the orifice plate is that its inaccuracy is usually ±2% and may approach ±5%. Furthermore, pressure loss caused in the measured fluid flow is between 50 and 90% of the pressure change, (P_1 − P_2). The other orifice plate problems are a progressive alteration in the coefficient of discharge through the hole edges and a tendency for particles inside the flowing fluid to block the hole. There may be decrease in diameter progressively as the particles stick. The last problem can be decreased by using an orifice plate with a hole at the center. If this hole is near to the pipe bottom, solids inside the flowing fluid incline to be swept across, and particles build up is decreased.

The same problem appears when there are gas or vapor bubbles in the flowing liquid. This difficulty can be removed by installing the orifice plate in the vertical portion of pipeline.

The hole shapes in orifice plates may be eccentric, concentric or segmental as illustrated in Figure 8.18. Orifice plates are damaged by erosion.

FIGURE 8.18  Eccentric, concentric, and segmental orifice plates

The discharge coefficient to Cd for a particular orifice meter is a function of the portion of the pressure tapes, the diameter ratio of the orifice to the inner pipe diameter, the Reynolds number (Re) in the pipeline, and the orifice plate thickness.

As shown in Figure 8.19, an orifice meter consists of a thin plate with an aperture in its center. The orifice aperture is commonly circular. The abrupt constriction of area in these devices produces a high loss of pressure compared with the other two devices. The downstream pressure is measured at a place of 0.5 D_1 length and upstream pressure is measured at a place D_1 length.

FIGURE 8.19  Orifice meter

As a function of pressure drop, the measured flow rate over the orifice plate is expressed as follows:

(8.6)

where K is the coefficient of the orifice plate and, depends upon the plate diameter ratio and the Reynolds number (Re) of the flow that is expressed as,

EXAMPLE 8.6

A pipe of diameter (30 cm) loads oil, (ρ = 880 kg/m^3, µ = 0.799 kg/m^2), and to determine the flow rate an orifice (diameter = 15 cm) is fitted to the pipe. A mercury manometer is used to read the pressure drop through the office as 0.95 m. Calculate the flow rate of oil.

Solution:

and assume high Reynolds number flow.

From Figure 8.20,

These values are substituted into Equation 8.6

The condition of Re must be checked.

FIGURE 8.20  Coefficients of discharge of the orifice [J. H. Perry, 1984]

With respect to Figure 8.20,

K = 0.72, and the corrected flow rate becomes,

8.7.1.3  Venturi Meter

The Venturi is an accurately machined tube. This shows measurement inaccuracy of just (±1%). However, it requires machining, which means cost. Pressure loss in the measuring system is (10–15%) of the pressure difference (P_1 − P_2).

The first use of Venturi meter was by J.B Venturi in 1797. As shown in Figure 8.21, reducing the flow cross-sectional area, some pressure head is transformed into speed.

The head differential has to be measured between the throat part and the upstream to evaluate the flow rate.

FIGURE 8.21  Cross-section of a Venturi meter

(8.7)

where P is the density of the fluid, and for Venturi meter, 0.25 < β < 0.50.

C is the coefficient of discharge and changes in the range of 0.935−0.988.

The upstream part closeness angle is 21° with the axis of pipe and the diverging part has an angle 5°−7°. In linking the Venturi meter to a pipeline, the upstream part of the Venturi should have a distance 10D_1 straight pipe. The total losses in a Venturi are approximately between 10 and 20 percentage of the total pressure drop.

EXAMPLE 8.7

A venturi meter was designed for measuring the flow rate of water in horizontal pipes of diameter 300 cm. The evaluated discharge over the pipe is 15 m^3/s, and the pressure drop is restricted to 250 kPa at the venturi. Determine the diameter of the venturi throat. Assume that the coefficient of discharge is 0.95.

Solution: From Equation 8.7, the relationship between A_2 and β is,

Substituting,

This relation is determined for A_2 by method of trial and error. Suppose β = 0.35, calculate (D_2), D_2 = 0.944 m, and β = 0.314, after that finally diameter of venturi throat is D_2 = 0.945 m.

8.7.2  Calibration of Flow Meters

The first consideration in selecting calibration method for flow measuring devices is to establish the precision level necessary. The calibration system should not be costly and should satisfy the statutory requirements needed.

The flow measurement precision is influenced greatly by the conditions of flow and the flowing fluid characteristics. Flow measuring devices are calibrated on-sight in the field. The calibration is achieved under the conditions of practical flow, which are difficult to re-produce in a laboratory. For the validity of calibration, it is necessary to replicate flow calibration tests so that the same value of reading is obtained in two sequential tests.

If onsite calibration is not possible or is not of sufficient precision, the device may be sent for calibration utilizing equipment supplier or device manufacturers or other calibration organization. However, this can be costly. Moreover, the calibration facility does not repeat the normal working conditions of the meter examined, and suitable compensation for variance between conditions of calibration and conditions of natural utilization should be applied.

8.8 MEASUREMENT SYSTEM ERROR

In a perfect measurement system, the measured value should be the same as the correct value. The system precision can be defined as nearness of the measured value to the correct value. A completely precision system is a theoretical system. Processor of a real system is quantified utilizing the error of measurement system (E),

where    E = [measured value] − [true value].

E = [system output] − [system input].

REVIEW QUESTIONS

1.What is a measurement system? How will you define its performance? Explain the following terms:

i.Calibration

ii.Measurement uncertainty

iii.Sensitivity and repeatability

iv.System error

2.Explain the following:

i.Temperature scales

ii.Temperature measuring instruments

iii.Thermocouples

3.How will you select your temperature measurement instrument?

4.Write a detailed note on pressure thermometer.

5.Explain the following devices for velocity measurement:

a.Pitot tube

b.Vane anemometer

c.Hot wire anemometer

d.Laser Doppler anemometer

6.Compare the following flow meters:

a.Rotameter

b.Orifice plate

c.Venturi meter

NUMERICAL EXERCISES

1.During temperature measurement, it is found that a thermometer gives the same temperature in °C and °F. Express the temperature value in R. (419.67 R)

2.A platinum resistance thermometer has resistance of 2.8 ohms at 0°C and 3.8 ohms at 100°C. Calculate the temperature when the resistance is indicated as 5.8 ohm. (300°C)

3.A certain thermometer using pressure as the thermometric property gives the value of pressure as 1.86 and 6.81 at ice point and steam point, respectively. The temperature of ice point and steam point are assigned the number 32 and 212, respectively.

Determine the temperature corresponding to p −2.5 if the temperature t is defined in terms of pressure p as t = a ln p + b, where a and b are constants. (73°C)

4.A turbine is supplied with steam at a pressure of 20 bar gauge. After expansion in the turbine, the steam passes to condenser, which is maintained at a vacuum of 250 mm of mercury by means of pumps. Express the inlet and exhaust steam pressure in N/m^2 and kPa. (2101396.2 N/m^2 or 2101.4 kPa 68024 N/m^2 or 68 kPa)

5.A vessel of cylindrical shape of 50 cm diameter and 75 cm height contains 4 kg of gas. The pressure gauge mounted on the vessel indicates 620 mm of Hg above atmosphere. If the borometer reading is 760 mm of Hg, calculate the absolute pressure of gas in bar. Also determine the density and specific volume of the gas. (1.84 bar, 27.21 kg/m^3, 0.036 m^3/kg)

APPENDIX

THERMODYNAMIC TABLES SI UNITTS

Table

Table A-1

Atomic or Molecular Weights and Critical Properties of Selected Elements and Compounds

Table A-2

Properties of Saturated Water (Liquid-Vapor): Temperature Table

Table A-3

Properties of Saturated Water (Liquid-Vapor): Pressure Table

Table A-4

Properties of Superheated Water Vapor

Table A-5

Properties of Compressed Liquid Water

Table A-6

Properties of Saturated Water (Solid-Vapor): Temperature Table

Table A-7

Properties of Saturated Refrigerant 22 (Liquid-Vapor): Temperature Table

Table A-8

Properties of Saturated Refrigerant 22 (Liquid-Vapor): Pressure Table

Table A-9

Properties of Superheated Refrigerant 22 Vapor

Table A-10

Properties of Saturated Refrigerant 134a (Liquid-Vapor): Temperature Table

Table A-11

Properties of Saturated Refrigerant 134a (Liquid-Vapor): Pressure Table

Table A-12

Properties of Superheated Refrigerant 134a Vapor

Table A-13

Properties of Saturated Ammonia (Liquid-Vapor): Temperature Table

Table A-14

Properties of Saturated Ammonia (Liquid-Vapor): Pressure Table

Table A-15

Properties of Superheated Ammonia Vapor

Table A-16

Properties of Saturated Propane (Liquid-Vapor): Temperature Table

Table A-17

Properties of Saturated Propane (Liquid-Vapor): Pressure Table

Table A-18

Properties of Superheated Propane Vapor

Table A-19

Properties of Selected Solids and Liquids: cp, ? and K

Table A-20

Ideal Gas Specific Heats of Some Common Gases

Table A-21

Variation of cp with Temperature for Selected Ideal Gases

Table A-22

Ideal Gas Properties of Air

Table A-23

Ideal Gas Properties of Selected Gases

Table A-24

Thermochemical Properties of Selected Substances at 298 K and 1 atm

Table A-25

Standard Molar Chemical Exergy of Selected Substances at 298 K and p0

TABLE A-1  Atomic or Molecular Weights and Critical Properties of Selected Elements and Compounds

TABLE A-2  Properties of Saturated Water (Liquid-Vapor): Temperature Table

TABLE A-3  Properties of Saturated Water (Liquid-Vapor): Pressure Table

TABLE A-4  Properties of Superheated Water Vapor

TABLE A-5  Properties of Compressed Liquid Water

TABLE A-6  Properties of Saturated Water (Solid-Vapor): Temperature Table

TABLE A-7  Properties of Saturated Refrigerant 22 (Liquid-Vapor): Temperature Table

TABLE A-8  Properties of Saturated Refrigerant 22 (Liquid-Vapor): Pressure Table

TABLE A-9  Properties of Superheated Refrigerant 22 Vapor

TABLE A-10  Properties of Saturated Refrigerant 134a (Liquid-Vapor): Temperature Table

TABLE A-11  Properties of Saturated Refrigerant 134a (Liquid-Vapor): Pressure Table

TABLE A-12  Properties of Superheated Refrigerant 134a Vapor

TABLE A-13  Properties of Saturated Ammonia (Liquid-Vapor): Temperature Table

TABLE A−14  Properties of Saturated Ammonia (Liquid-Vapor): Pressure Table

TABLE A-15  Properties of Superheated Ammonia Vapor

TABLE A-16  Properties of Saturated Propane (Liquid-Vapor): Temperature Table

TABLE A-17  Properties of Saturated Propane (Liquid-Vapor): Pressure Table

TABLE A-18  Properties of Superheated Propane Vapor

TABLE A-19  Properties of Selected Solids and Liquids: cp and K

TABLE A-20  Ideal Gas Specific Heats of Some Common Gases

TABLE A-21  Variation of cp with Temperature for Selected Ideal Gases

TABLE A-22  Ideal Gas Properties of Air

TABLE A−23  Ideal Gas Properties of Selected Gases

TABLE A-24  Thermochemical Properties of Selected Substances at 298 K and 1 atm

TABLE A-25  Standard Molar Chemical Exergy of Selected Substances at 298 K and p0

REFERENCES

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[22] J.P. O’Connell, J.M. Haile, 2005, “Thermodynamics Fundamentals for Applications”, Published in the United States of America by Cambridge University Press, New York.

[23] M. Tabatabaian, M. Tabatabaian, 2018, “Advanced Thermodynamics Fundamentals- Mathematics – Applications”, Mercury Learning and Information.

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INDEX

A

Adiabatic System, 16

Analysis of the Ideal Rankine Cycle, 99

B

Binary Vapor Cycle, 127

Boiler, 151

C

Calibration, 170

Calibration of Flow Meters, 195

Carnot Cycle for

Heat Engine, 95

Heat Pump and Refrigerator, 95

Carnot Vapor Power cycle, 90

Centrifugal

Compressor, 147

Water Pump, 146

Characteristics of Entropy, 43

Clarification, of Clausius Statement, 34

Closed Systems, 15, 44

Coefficient of Performance (COP), 33

Cogeneration Cycles, 122

Comparison between Carnot and Rankine Cycles, 103

Comparison of Advantages and Disadvantages of Measurement Systems, 178

Compressors and Pumps, 144

Concept of Temperature, 172

Condenser, 152

Cycle Definition, 88

D

Definition

of Exergy, 59

of Pressure, 179

and Scope of Thermodynamics, 1

Diffusers and Nozzles, 138

Displacement Work, 3

Dissimilar System, 16

Draining of a Tank, 160

E

Effect of Operating Conditions on Rankine Cycle Efficiency, 103

Energy

Balance, 24

Conservation, 24

Equation, 50

Energy and Entropy, 77

Engineering Applications of Steady Flow Energy Equation, 137

Enthalpy-Entropy (h-s) Diagram, 55

Enthalpy H(S, p, N), 79

Entropy, 42

Entropy as a Thermodynamic Property, 44

Entropy Balance for

Closed Systems, 44

Open Systems, 47

Entropy Differentials, 52

Entropy of the Ideal Gas, 49

Equations of State, 75

Ericsson Cycle, 130

Euler’s Cyclic Chain Rule, 51

Evaporator, 153

Exergy Analysis for

Open System, 62

Closed System, 61

a System, 60

Exergy Efficiencies, 84

Exergy Efficiency of

Combustion, 165

Expansion and Compression, 163

Heat Exchanger, 160

Exergy Equation, 65

Exergy Loss of a Open, Steady State, Constant Volume System, 64

Exergy Loss of a System, 63

Exergy Transfer by Heat, 84

External Irreversibility, 64

F

Factors Causing an Irreversible Process, 42

Fans and Blowers, 145

Filling of a Tank, 158

First Law of Thermodynamics, 23

Closed System, 28

Open System, 29

Flow Exergy of Matter, 83

Flow in Ducts and Pipes, 155

Flow Measurement, 188

Forms of Energy, 3

Friction, 41

G

Gas Turbine or Steam Turbine, 144

Gibbs Free Energy, 81

Grand Potential, 81

H

Heat, 12

Heat and Work, 2

Heat Engine, 31

Heat Exchangers, 151

Heat Pumps and Refrigerators, 33

Helmholtz Free Energy, 80

Heterogeneous System, 16

High-temperature Thermometers, 175

Homogenous System, 16

Hot Wire Anemometer (HWA), 185

I

Ideal Rankine Cycle, 98

Internal and External Reversibilities, 42

Internal Irreversibility, 64

Irreversibility, 41

Isolated System, 16

L

Laser Doppler Anemometer (LDA), 187

Limitations of Carnot’s Cycle, 93

M

Macro and Micro Approach to Study the System Matter, 16

Mass Flow Rate, 189

Mathematical Definition, 43

Maxwell

Equations, 50

Relations, 77

Measurement of

Sensitivity and Repeatability, 172

System Error, 195

Temperature, 172

Uncertainty, 171

Variables, 170

Velocity, 181

Mechanisms of Energy Transfer, 27

Mixing Chambers, 142

N

Nernst Law, 55

O

Open system, 15

Other Forms of Work, 5

P

Performance and Purpose of Measurement Systems, 169

Performance Parameters, 89

Pitot Tube, 182

Placement of Feed Water Heaters, 119

Plant Components, 91

Polytropic Processes, 4

Pressure

Measurement, 179

Thermometers, 180

Units, 180

Principles of the Carnot Cycle, 93

Property Relations, 18

R

Reciprocating Compressor, 149

Refrigeration Cycle, 88, 131

Reheat Cycle, 119

Reverse Carnot Cycle, 94

Reversible and Irreversible Processes, 40

S

Second Law of Thermodynamics, 31, 34

Similar System, 16

Stirling Cycle, 129

System Stability, 17

T

Temperature

Calibration, 177

Entropy (T-s) Diagram, 53

Measuring Instruments, 174

Scales, 173

Thermal Analysis of Carnot Cycle, 92

Thermocouples Material, 175

Thermodynamic

Cycles, 46

Systems, 14

Third Law of Thermodynamics, 55

Throttling Devices, 141

Transfer of Exergy by

Convection, 70

Heat, 67

Work, 66

Two Independent Variables, T and P, 53

Two Independent Variables, T and V, 52

Two Independent Variables, V and P, 53

Types of Thermodynamic Properties, 73

U

Units of Measurement, 170

Universal Gas Constant, 19

Unsteady Flow processes, 157

V

Van der Waals Equation of State, 76

Vane Anemometer, 184

Virial Equation of State, 75

W

Water Turbine, 140

Work Summary, 6

Z

Zeroth Law, 23