Fundamentals of Thermodynamics PDF

Fundamentals of THERMODYNAMICS Partha Haldar  Fundamentals of THERMODYNAMICS PARTHA HALDAR Assistant Professor of Mechanical Engineering Government College of Engineering & Ceramic Technology Kolkata Dedicated to my son Rivan Ther...

Fundamentals of THERMODYNAMICS Partha Haldar  Fundamentals of THERMODYNAMICS PARTHA HALDAR Assistant Professor of Mechanical Engineering Government College of Engineering & Ceramic Technology Kolkata Dedicated to my son Rivan Thermodynamics: Thermodynamics can be defined as the science of energy. Although everybody has a feeling of what energy is, it is difficult to give a precise definition for it. Energy can be viewed as the ability to cause changes. The name thermodynamics stems from the Greek words ‘therme’ (heat) and ‘dynamis’ (power). One of the most fundamental laws of nature is the conservation of energy principle. It states that during an interaction, energy can change from one form to another but the total amount of energy remains constant. That is, energy cannot be created or destroyed. The first law of thermodynamics is simply an expression of the conservation of energy principle, and it asserts that energy is a thermodynamic property. The second law of thermodynamics asserts that energy has quality as well as quantity, and actual processes occur in the direction of decreasing quality of energy. For example, a cup of hot coffee left on a table eventually cools, but a cup of cool coffee in the same room never gets hot by itself. Thermodynamic System: A system is defined as a quantity of matter or a region in space chosen forstudy. The mass or region outside the system is called the surroundings.The real or imaginary surface that separates the system from its surroundingsis called the boundary. Theboundary of a system can be fixed or movable. Note that the boundary is thecontact surface shared by both the system and the surroundings. Mathematically speaking, the boundary has zero thickness, and thus it can neither contain any mass nor occupy any volume in space. Fig. System, surrounding and boundary Classification of Systems: Systems may be considered to be closed or open, depending on whether afixed mass or a fixed volume in space is chosen for study. A closed system(also known as a control mass) consists of a fixed amount of mass, and nomass can cross its boundary. That is, no mass can enter or leave a closedsystem. But energy, in the form of heat or work, cancross the boundary. Fig.: Mass cannotcross the boundary of a closedsystem but energy can.  If, as a special case, even energy is not allowed to cross the boundary,that system is called an isolated system.No work is done in an isolated system. An open system, or a control volume, as it is often called, is a properlyselected region in space. It usually encloses a device that involvesmass flow such as a compressor, turbine, or nozzle. Flow through thesedevices is best studied by selecting the region within the device as thecontrol volume. Both mass and energy can cross the boundary of a control volume. The boundaries of a control volume are called a control surface, and they can be real or imaginary. In the case of a nozzle, the inner surface of the nozzle forms the real part of the boundary, and the entrance and exit areas form the imaginary part, since there are no physical surfaces there. A control volume can be fixed in size and shape, as in the case of a nozzle, or it may involve a moving boundary. A control volume can also involve heat and workinteractions just as a closed system, in addition to mass interaction. Fig.: A control volume can involve fixed, moving, real, and imaginary boundaries. Properties of a System: Any characteristic of a system is called a property. Some familiar propertiesare pressure P, temperature T, volume V, and mass m. Properties are considered to be either intensive or extensive. Intensiveproperties are those that are independent of the mass of a system, such astemperature, pressure, and density. Extensive properties are those whosevalues depend on the size or extent of the system. Total mass, total volumeand total momentum are some examples of extensive properties. Extensive properties per unit mass are called specific properties. Someexamples of specific properties are specific volume (v =V/m) and specific energy (e =E/m). Fig.: Criterion to differentiate intensive andextensive properties. State and Equilibrium: Consider a system not undergoing any change. At this point, all the propertiescan be measured or calculated throughout the entire system, whichgives us a set of properties that completely describes the condition, or thestate, of the system. At a given state, all the properties of a system havefixed values. If the value of even one property changes, the state will changeto a different one. Fig.: A system at two different states. Thermodynamics deals with equilibrium states. The word equilibrium implies a state of balance. In an equilibrium state there are no unbalancedpotentials (or driving forces) within the system. There are many types of equilibrium, and a system is not in thermodynamicequilibrium unless the conditions of all the relevant types of equilibriumare satisfied. A system is in thermal equilibrium if thetemperature is the same throughout the entire system. Mechanical equilibrium is related to pressure,and a system is in mechanical equilibrium if there is no change in pressureat any point of the system with time. If a system involves two phases, it is in phaseequilibrium when the mass of each phase reaches an equilibrium level andstays there. Finally, a system is in chemical equilibrium if its chemicalcomposition does not change with time, that is, no chemical reactions occur.A system will not be in equilibrium unless all the relevant equilibrium criteria are satisfied. The State Postulate: As noted earlier, the state of a system is described by its properties. But we know from experience that we do not need to specify all the properties in order to fix a state. Once a sufficient number of properties are specified, the rest of the properties assume certain values automatically. That is, specifying a certain number of properties is sufficient to fix a state. The number of properties required to fix the state of a system is given by the state postulate: The state of a simple compressible system is completely specified by two independent, intensive properties. The state postulate requires that the two properties specified be independent to fix the state. Two properties are independent if one property can be varied while the other one is held constant. Temperature and specific volume, for example, are always independent properties, and together they can fix the state of a simple compressible system.  Fig.: The state of nitrogen is fixed by two independent, intensive properties. Thermodynamic Processes: Any change that a system undergoes from one equilibrium state to another is called a process, and the series of states through which a system passes during a process is called the path of the process. Fig.: A process between states 1 and 2 and the process path. To describe a process completely, one should specify the initial and final states of the process, as well as the path it follows, and the interactions with the surroundings. When a process proceeds in such a manner that the system remains infinitesimally close to an equilibrium state at all times, it is called a quasistatic, or quasi-equilibrium, process. A quasi-equilibrium process can be viewed as a sufficiently slow process that allows the system to adjust itself internally so that properties in one part of the system do not change any faster than those at other parts. When a gas in a piston-cylinder device is compressed suddenly, the molecules near the face of the piston will not have enough time to escape and they will have to pile up in a small region in front of the piston, thus creating a high-pressure region there. Because of this pressure difference, the system can no longer be said to be in equilibrium, and this makes the entire process non-quasi-equilibrium. However, if the piston is moved slowly, the molecules will have sufficient time to redistribute and there will not be a molecule pileup in front of the piston. As a result, the pressure inside the cylinder will always be nearly uniform and will rise at the same rate at all locations. Since equilibrium is maintained at all times, this is a quasi-equilibrium process. It should be pointed out that a quasi-equilibrium process is an idealized process and is not a true representation of an actual process.  Fig.: Quasi-equilibrium and non-quasi-equilibrium compression processes. But many actual processes closely approximate it, and they can be modelled as quasi- equilibrium with negligible error. Engineers are interested in quasi-equilibrium processes for two reasons. First, they are easy to analyze; second, work-producing devices deliver the most work when they operate on quasi-equilibrium processes. Therefore, quasi-equilibrium processes serve as standards to which actual processes can be compared. An isothermal process is a process during which the temperature T remains constant. An isobaricprocess is a process during which the pressure P remains constant. An isochoric (or isometric)process is a process during which the specific volume vremains constant. A system is said to have undergone a cycle if it returns to its initial state at the end of the process. That is, for a cycle the initial and final states are identical. A system is said to have undergone a cycle if it returns to its initial state at the end of the process. That is, for a cycle the initial and final states are identical. The Steady-Flow Process: The term steady implies no change with time. The opposite of steady is unsteady, or transient. The term uniform, however, implies no change with location over a specified region. Steady-flow process can be defined as a process during which a fluid flows through a control volumesteadily. Fig.: During a steady flow process, fluid properties within the control volume may change with position but not with time. Steady-flow conditions can be closely approximated by devices that are intended for continuous operation such as turbines, pumps, boilers, condensers, and heat exchangers or power plants or refrigeration systems. Some cyclic devices, such as reciprocating engines or compressors, do not satisfy any of the conditions stated above since the flow at the inlets and the exits will be pulsating and not steady. Reversible and Irreversible Processes: A reversible process is defined as a process that can be reversed without leaving any trace on the surroundings. That is, both the system and the surroundings are returned to their initial states at the end of the reverse process. This is possible only if the net heat and net work exchange between the system and the surroundings is zero for the combined (original and reverse) process. Processes that are not reversible are called irreversible processes. Irreversibilities: Irreversibilities include friction, unrestrained expansion, mixing of two fluids, heat transfer across a finite temperature difference, electric resistance, inelasticdeformation of solids, and chemical reactions. The presence of any of these effects renders a process irreversible. Internally and Externally Reversible Processes: A process is called internally reversible if no irreversibilities occur within the boundaries of the system during the process. The quasi-equilibrium process is an example of an internally reversible process. Consider a can of cold soda left in a warm room. Heat is transferred from the warmer room air to the cooler soda. The only way this process can be reversed and the soda restored to its original temperature is to provide refrigeration, which requires some work input. At the end of the reverse process, the soda will be restored to its initial state, but the surroundings will not be. So this is an internally reversible process if the can is considered as the system. A process is called externally reversible if no irreversibilities occur outside the system boundaries during the process. Heat transfer between a reservoir and a system is an externally reversible process if the outer surface of the system is at the temperature of the reservoir. A process is called totally reversible, or simply reversible, if it involves no irreversibilities within the system or its surroundings. Atotally reversible process involves no heat transfer through a finite temperature difference, no non-quasi-equilibrium changes, and no friction or other dissipative effects. The Zeroth Law of Thermodynamics: The zeroth law of thermodynamics states that if two bodies are in thermal equilibrium with a third body, they are also in thermal equilibrium with each other. It may seem silly that such an obvious fact is called one of the basic laws of thermodynamics. However, it cannot be concluded from the other laws of thermodynamics, and it serves as a basis for the validity of temperature measurement. By replacing the third body with a thermometer, the zeroth law can be restated as two bodies are in thermal equilibrium if both have the same temperature reading even if they are not in contact. The zeroth law was first formulated and labelled by R. H. Fowler in 1931. As the name suggests, its value as a fundamental physical principle was recognized more than half a century after the formulation of the first and the second laws of thermodynamics. It was named the zeroth law since it should have preceded the first and the second laws of thermodynamics. Forms of Energy of a System: Energy can exist in numerous forms such as thermal, mechanical, kinetic, potential, electric, magnetic, chemical, and nuclear, and their sum constitutes the total energy (E) of a system. The total energy of a system can be divided in two groups: macroscopic and microscopic. The macroscopic forms of energy are those a system possesses as a whole with respect to some outside reference frame, such as kinetic and potential energies. The microscopic forms of energy are those related to the molecular structure of a system and the degree of the molecular activity, and they are independent of outside reference frames. The sum of all the microscopic forms of energy is called the internal energy of a system and is denoted by U. = = + + = + + 2 For stationary systems, the changes in kinetic and potential energies are zero. Therefore the total energy change is equal to change in internal energy. Most closed systems remain stationary during a process and thus experience no change in their kinetic and potential energies. Closed systems whose velocity and elevation of the center of gravity remain constant during a process are frequently referred to as stationary systems. Problem 1: Wind speed is 8.5 m/sec. wind energy per unit mass =? 1 Answer: Wind energy per unit mass = & = 36.125 +/- 2 Some Physical Insight to Internal Energy: Internal energy is defined as the sum of all the microscopic forms of energy of a system. It is related to the molecular structure and the degree of molecular activity and can be viewed as the sum of the kinetic and potential energies of the molecules. To have a better understanding of internal energy, let us examine a system at the molecular level. The molecules of a gas move through space with some velocity, and thus possess some kinetic energy. This is known as the translational energy. The atoms of polyatomic molecules rotate about an axis, and the energy associated with this rotation is the rotational kinetic energy. The atoms of a polyatomic molecule may also vibrate about their common center of mass, and the energy associated with this back-and-forth motion is the vibrational kinetic energy. For gases, the kinetic energy is mostly due to translational and rotational motions, with vibrational motion becoming significant at higher temperatures. The electrons in an atom rotate about the nucleus, and thus possess rotational kinetic energy. Electrons at outer orbits have larger kinetic energies. Electrons also spin about their axes, and the energy associated with this motion is the spin energy. Other particles in the nucleus of an atom also possess spin energy. The portion of the internal energy of a system associated with the kinetic energies of the molecules is called the sensible energy. The average velocity and the degree of activity of the molecules are proportional to the temperature of the gas. Therefore, at higher temperatures, the molecules possess higher kinetic energies, and as a result the system has a higher internal energy. If sufficient energy is added to the molecules of a solid or liquid, the molecules overcome these molecular forces and break away, turning the substance into a gas. This is a phase- change process. Because of this added energy, a system in the gas phase is at a higher internal energy level than it is in the solid or the liquid phase. The internal energy associated with the phase of a system is called the latent energy. The internal energy associated with the atomic bonds in a molecule is called chemical energy. The tremendous amount of energy associated with the strong bonds within the nucleus of the atom itself is called nuclear energy. We need not be concerned with nuclear energy in thermodynamics unless, of course, we deal with fusion or fission reactions. Mechanical Energy: The mechanical energy can be defined as the form of energy that can be converted to mechanical work completely and directly by an ideal mechanical device such as an ideal turbine. Kinetic and potential energies are the familiar forms of mechanical energy. Thermal energy is not mechanical energy, however, since it cannot be converted to work directly and completely. Therefore, the mechanical energy of a flowing fluid can be expressed on a unit mass basis as: + = + 2./01 2 =flow energy + kinetic energy + potential energy Then the mechanical energy change of a fluid during incompressible (2 = 4 5 ) flow becomes − 9 − 9 ∆ = + + (: − :9 )./01 2 2 In the absence of any losses, the mechanical energy change represents the mechanical work supplied to the fluid (∆./01 > 0) or extracted from the fluid (∆./01 < 0). Mechanisms of Energy Transfer to or from a system, Ein and Eout: Energy can be transferred to or from a system in three forms: heat, work and mass flow. Heat Transfer: Heat is defined as the form of energy that is transferred between two systems (or a system and its surroundings) by virtue of a temperature difference. Heat transfer to a system (heat gain) increases the energy of the molecules and thus the internal energy of the system and heat transfer from a system (heat loss) decreases it. Heat is energy in transition. It is recognized only as it crosses the boundary of a system. Consider the hot baked potato,the potato contains energy, but this energy is heat transfer only as it passes through the skin of the potato (the system boundary) to reach the air, as shown in Fig. Once in the surroundings, the transferred heat becomes part of the internal energy of the surroundings. Thus, in thermodynamics, the term heat simply means heat transfer.  Fig.: Energy is recognized as heat transfer only as it crosses the system boundary A process during which there is no heat transfer is called an adiabatic process. There are two ways a process can be adiabatic: Either the system is well insulated so that only a negligible amount of heat can pass through the boundary or both the system and the surroundings are at the same temperature and therefore there is no driving force (temperature difference) for heat transfer. An adiabatic process should not be confused with an isothermal process. Even though there is no heat transfer during an adiabatic process, the energy content and thus the temperature of a system can still be changed by other means such as work. Fig.: During an adiabatic process, a system exchanges no heat with its surroundings Work Transfer: If the energy crossing the boundary of a closed system is not heat, it must be work. The generally accepted formal sign convention for heat and work interactions is as follows: heat transfer to a system and work done by a system are positive; heat transfer from a system and work done on a system are negative. Mass Flow: When mass enters a system, the energy of the system increases because mass carries energy with it. Likewise, when some mass leaves the system, the energy contained within the system decreases because the leaving mass takes out some energy with it. For example, when some hot water is taken out of a water heater and is replaced by the same amount of cold water, the energy content of the hot-water tank (the control volume) decreases as a result of this mass interaction.  Fig.: The energy content of a control volume can be changed by mass flow as well as heat and work interactions More on Heat and work Transfer: Heat is easy to recognize: Its driving force is a temperature difference between the system and its surroundings. Then we can simply say that an energy interaction that is not caused by a temperature difference between a system and its surroundings is work. Heat and work are energy transfer mechanisms between a system and its surroundings, and there are many similarities between them: a) Both are recognized at the boundaries of a system as they cross the boundaries. That is, both heat and work are boundary phenomena. b) Systems possess energy, but not heat or work. c) Both are associated with a process, not a state. Unlike properties, heat or work has no meaning at a state. d) Both are path functions (i.e., their magnitudes depend on the path followed during a process as well as the end states). Path functions have inexact differentials designated by the symbol >. Therefore, a differential amount of heat or work is represented by >?or >@, respectively, instead of dQ or dW. Properties, however, are point functions (i.e., they depend on the state only, and not on how a system reaches that state), and they have exact differentials designated by the symbol d. A small change in volume, for example, is represented by dV, and the total volume change during a process between states 1 and 2 is AB = − 9 =∆ 9 Fig.: Properties are point functions; but heat and work are path functions (their magnitudes depend on the path followed) The total work done during process 1–2, however, is A >@ = @9 (C ∆@) 9 That is, the total work is obtained by following the process path and adding the differential amounts of work (>@) done along the way. The integral of >@ is not @ − @9 (i.e., the work at state 2 minus work at state 1), which is meaningless since work is not a property and systems do not possess work at a state. Problem 1: A candle is burning in a well-insulated room. Taking the room (the air plus the candle) as the system, determine (a) if there is any heat transfer during this burning process and (b) if there is any change in the internal energy of the system. Answer: (a) Heat is recognized as it crosses the boundaries. Since the room is well insulated, we have an adiabatic system and no heat will pass through the boundaries. Therefore, Q=0 for this process. (b) The internal energy involves energies that exist in various forms (sensible, latent, chemical and nuclear). During the process just described, part of the chemical energy is converted to sensible energy. Since there is no increase or decrease in the total internal energy change of the system =0. Problem 2: A potato initially at room temperature (25°C) is being baked in an oven that is maintained at 200°C, as shown in Fig. Is there any heat transfer during this baking process? Answer: This is not a well-defined problem since the system is not specified. Let us assume that we are observing the potato, which will be our system. Then the skin of the potato can be viewed as the system boundary. Part of the energy in the oven will pass through the skin to the potato. Since the driving force for this energy transfer is a temperature difference, this is a heat transfer process. Problem 3: A well-insulated electric oven is being heated through its heating element. If the entire oven, including the heating element, is taken to be the system, determine whether this is a heat or work interaction. Answer: For this problem, the interior surfaces of the oven form the system boundary, as shown in Fig. The energy content of the oven obviously increases during this process, as evidenced by a rise in temperature. This energy transfer to the oven is not caused by a temperature difference between the oven and the surrounding air. Instead, it is caused by electrons crossing the system boundary and thus doing work. Therefore, this is a work interaction. Problem 4: A well-insulated electric oven is being heated through its heating element. If the system is taken as only the air in the oven without the heating element, determine whether this is a heat or work interaction. Answer: The system boundary will include the outer surface of the heating element and will not cut through it, as shown in Fig. Therefore, no electrons will be crossing the system boundary at any point. Instead, the energy generated in the interior of the heating element will be transferred to the air around it as a result of the temperature difference between the heating element and the air in the oven. Therefore, this is a heat transfer process. Discussion: For both cases, the amount of energy transfer to the air is the same. These two examples show that an energy transfer can be heat or work, depending on how the system is selected. THE FIRST LAW OF THERMODYNAMICS: The first law of thermodynamics is also known as the conservation of energy principle. The first law of thermodynamics states that energy can be neither created nor destroyed during a process; it can only change forms. So the first law of thermodynamics can be stated as follows for a process: If Q is the amount of heat transferred to the system and W is the amount of work transferred from the system during the process, the net energy transfer (Q-W) will be stored into the system. The energy stored into the system is neither heat nor work it is known as the internal energy of the system. ∴ Q − W = ∆E Or, Q = W + ∆E For an isolated system, there is no interaction of the system with the surrounding. For an isolated system dQ=0 dW=0 hence, dE=0 or, E=constant The energy of an isolated system is always constant. Example 1: First, we consider some processes that involve heat transfer but no work interactions. The potato baked in the oven is a good example. As a result of heat transfer to the potato, the energy of the potato will increase. If we disregard any mass transfer (moisture loss from the potato), the increase in the total energy of the potato becomes equal to the amount of heat transfer. That is, if 5 kJ of heat is transferred to the potato, the energy increase of the potato will also be 5 kJ.  Fig.: The increase in the energy of a potato in an oven is equal to the amount of heat transferred to it Example 2: Consider the heating of water in a pan on top of a range. If 15 kJ of heat is transferred to the water from the heating element and 3 kJ of it is lost from the water to the surrounding air, the increase in energy of the water will be equal to the net heat transfer to water, which is 12 kJ. Fig.: In the absence of any work interactions, the energy change of a system is equal to the net heat transfer Example 3: Now consider a well-insulated (i.e., adiabatic) room heated by an electric heater as our system (Fig.). As a result of electrical work done, the energy of the system will increase. Since the system is adiabatic and cannot have any heat transfer to or from the surroundings (Q=0), the conservation of energy principle dictates that the electrical work done on the system must equal the increase in energy of the system. Fig.: The work (electrical) done on an adiabatic system is equal to the increase in the energy of the system Example 4: Next, let us replace the electric heater with a paddle wheel (Fig.). As a result of the stirring process, the energy of the system will increase. Again, since there is no heat interaction between the system and its surroundings (Q=0), the shaft work done on the system must show up as an increase in the energy of the system.  Fig.: The work (shaft) done on an adiabatic system is equal to the increase in the energy of the system Example 5: Many of you have probably noticed that the temperature of air rises when it is compressed (Fig.). This is because energy is transferred to the air in the form of boundary work. In the absence of any heat transfer (Q=0), the entire boundary work will be stored in the air as part of its total energy. The conservation of energy principle again requires that the increase in the energy of the system be equal to the boundary work done on the system. Fig.: The work (boundary) done on an adiabatic system is equal to the increase in the energy of the system. Example 6: We can extend these discussions to systems that involve various heat and work interactions simultaneously. For example, if a system gains 12 kJ of heat during a process while 6 kJ of work is done on it, the increase in the energy of the system during that process is 18 kJ (Fig.). That is, the change in the energy of a system during a process is simply equal to the net energy transfer to (or from) the system.  Fig.: The energy change of a system during a process is equal to the net work and heat transfer between the system and its surroundings Energy Balance: The net change (increase or decrease) in the total energy of the system during a process is equal to the difference between the total energy entering and the total energy leaving the system during that process. ( I ℎ 5 5 )−( &I ℎ 5 5 ) = (Kℎ I L ℎ 5 5 ) , MN − OPQ = ∆ RSRQ/. The change in the total energy of a system during a process is the sum of the changes in its internal, kinetic, and potential energies and can be expressed as: ∆ =∆ +∆ +∆ Noting that energy can be transferred in the forms of heat, work and mass, and that the net transfer of a quantity is equal to the difference between the amounts transferred in and out, the energy balance can be written more explicitly as: MN − OPQ = (?MN − ?OPQ ) + (@MN − @OPQ ) + (.TRR MN −.TRR OPQ ) = ∆ RSRQ/. Case 1: For stationary systems, the changes in kinetic and potential energies are zero. Therefore the total energy change is equal to change in internal energy. Case 2: For a closed system undergoing a cycle, the initial and final states are identical, and thus ∆ RSRQ/. = − 9 = 0. Noting that a closed system does not involve any mass flow across its boundaries, the energy balance for a cycle can be expressed in terms of heat and work interactions as: @N/Q OPQ = ?N/Q MN That is, the net work output during a cycle is equal to net heat input. Fig.: Energy balance for a cyclic process So the first law of thermodynamics for a closed system undergoing a cycle can be stated as: When a closed system (control mass) undergoes a cycle, the cyclic integral of heat is always proportional to the cyclic integral of work. Prove that internal energy is a property: Let us assume that we have one system which is undergoing a change of state from initial state 1 to another state 2 via following the path A as shown in following figure. System is returning to initial state i.e. state 1 from state 2 via following the path B. Here, we can say that system is undergoing in a cycle 1-2-1 as shown in figure. Let us recall the “first law of thermodynamics for a system undergoing a change of state” and apply for path A, where system is changing its state from state 1 to state 2. We will have following equation QA- WA= ∆EA Similarly, we will have following equation when system is changing its state from state 2 to state 1 via following the path B. QB- WB= ∆EB We have already seen that system is undergoing in a cycle 1-2-1 as displayed in above figure. Hence, we will use the concept of “first law of thermodynamics for a system undergoing a cycle”. Let us see the equation for system which constitutes a cycle 1-2-1 and we will have following equation. WA+WB= QA+QB WB- QB= QA- WA - (QB- WB) = QA- WA - (∆EB) = ∆EA Let us assume that system is returning to initial state 1 from state 2 via following the path C, in that case we will go ahead similarly as we have gone above and finally we will have following equation - (∆EC) = ∆EA Now if we will look the end result for first case where system is returning to initial state by following the path B and of second case where system is returning to initial state by following the path C, what we will secure here that change in system energy is same in both cases and it will not depend over the path followed by the system to return to its initial state. Therefore we can conclude that system energy will have some definite magnitude for each state of the system and it will not depend over the path followed by the system and hence energy will be considered as a point function and also a property of the system. Problem 1: A rigid tank contains a hot fluid that is cooled while being stirred by a paddle wheel. Initially, the internal energy of the fluid is 800 kJ. During the cooling process, the fluid loses 500 kJ of heat, and the paddle wheel does 100 kJ of work on the fluid. Determine the final internal energy of the fluid. Neglect the energy stored in the paddle wheel.  UV − VX MNVWV OPQ = ∆ UVVWV VX RSRQ/. Y/Q /N/Z[S QZTNR\/Z ]S 1/TQ,^OZ_ TN`.TRR a1TN[/ MN MNQ/ZNTb,_MN/QM0,cOQ/NQMTb /N/Z[S @R1T\Q,MN − ?OPQ = ∆ = − 9 100 − 500 = − 800 , = 400 -+ Therefore, the final internal energy of the system is 400 kJ. Problem: A paddle wheel used for mixing and stirring of fluids turns 600 rpm when 2·5 Nm torque is applied to it. What is power transmitted to the liquid by the wheel? 600 1 Answer: @fg1T\Q = 2h f = 2h × × 2.5 × -@ = 0.157 -@ 60 1000 During one cycle the working fluid in an engine engages in the work interaction 15 kJ to the fluid and 44 kJ from the fluid and there are three heat interactions, two of which are known as 75 kJ to the fluid and 40 kJ from the fluid. Find out the magnitude and direction of the third heat transfer. Answer:@9 = −15 -+, @ = 44 -+, ?9 = 75 -+, ? = −40 -+, ?k =? For the cyclic process we know that, m B? = m B@ , 75 − 40 + ?k = −15 + 44 , ?k = −6 -+ So the heat will flow out from the system. A system contained in a piston and cylinder machine passes through a complete cycle of four processes. The sum of all heat transferred during a cycle is -350kJ. The system completes 200 cycles per minute. Complete the following table showing the method for each item, and compute the net rate of work output in kW. Process Q (kJ/min) W (kJ/min) ∆n (op/qrs) 1-2 0 4340 - 2-3 42000 0 -  3-4 -4200 - -73200 4-1 - - - Answer: For the process 1 − 2, ?9t = ∆ + @9t , 0 = ∆ + 4340 , ∆ = −4340 -+/ I For the process 2 − 3, ? tk = ∆ + @ tk , 42000 = ∆ + 0 , ∆ = 42000 -+/ I For the process 3 − 4, ?ktu = ∆ + @ktu , −4200 = −73200 + @ktu , @ktu = 69000 -+/ I The sum of all heat transferred during a cycle is -350kJ. So, ∮ ? = −350 × 200 = −70000 -+/ I ∴ ?9t + ? tk + ?ktu + ?ut9 = −70000 , ?ut9 = −107800 -+/ I Now, internal energy being a property, ∮ x = 0 , −4340 + 42000 − 73200 + ∆ ut9 =0 ,∆ ut9 = 35540 -+/ I Now, ?ut9 = ∆ + @ut9 , @ut9 = −1433340 -+/ I Total Energy of a Flowing Fluid As we have seen the total energy of a simple compressible system consists of three parts: internal, kinetic, and potential energies. On a unit-mass basis, it is expressed as & =x+- +y =x+ + (-+/- ) 2 where V is the velocity and is the elevation of the system relative to some external reference point. The fluid entering or leaving a control volume possesses an additional form of energy-the flow energy, Pv. P is the fluid pressure. Then the total energy of a flowing fluid on a unit- mass basis becomes & z= &+ = &+x+ + 2 But the combination & + x has been defined as the enthalpy h. So the relation reduces to: & z=ℎ+ + (-+/- ) 2 MOVING BOUNDARY WORK: - During the expansion or compression of a gas in a piston-cylinder device, part of the boundary (the inner face of the piston) moves back and forth. Therefore, the expansion and compression work is often called moving boundary work, or simply boundary work. Consider the gas enclosed in the piston–cylinder device shown in Fig. The initial pressure of the gas is P, the total volume is V, and the cross sectional area of the piston is A. If the piston is allowed to move distance ds in a quasi-equilibrium manner, the differential work done during this process is, >@] = {B5 = |B5 = B The total boundary work done during the entire process as the piston moves is obtained by adding all the differential works from the initial state to the final state: @] = A B 9 The area under the process curve on a P-V diagram is equal, in magnitude, to the work done during a quasi-equilibrium expansion or compression process of a closed system. The boundary work done during a process depends on the path followed as well as the end states. If work were not a path function, no cyclic devices (car engines, power plants) could operate as work-producing devices. The work produced by these devices during one part of the cycle would have to be consumed during another part, and there would be no net work output. The cycle shown in Fig. produces a net work output because the work done by the system during the expansion process (area under path A) is greater than the work done on the system during the compression part of the cycle (area under path B), and the difference between these two is the net work done during the cycle (the colored area). Fig. The net work done during a cycle is the difference between the work done by the system and the work done on the system Problem 1: A rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively. Determine the boundary work done during this process. Answer: A sketch of the system and the P-V diagram of the process are shown in Fig. The boundary work can be determined from equation to be @] = A B& = 0 5I 4 B& I5. 9 Problem 2: A non flow quasi static reversible process occurs for which p=(-2V+16) bar, where V is the volume in m3. What is the work done when volume changes from 2 m3 to 6 m3. Answer: } ~ € = } ‚ ƒ/„… † † @9t = A yB = A (−2 + 16) × 10‡ = 32 × 10‡ + x Boundary Work for a Constant-Pressure Process: @] = A B& = ˆA B& = ˆ (& − &9 ) 9 9 Isothermal Compression of an Ideal Gas: A piston-cylinder device initially contains 0.4 m3 of air at 100 kPa and 80°C. The air is now compressed to 0.1 m3 in such a way that the temperature inside the cylinder remains constant. Determine the work done during this process. For an ideal gas at constant temperature T0,  &= ‰ ˆ =K K , = & K B& & & & & @] = A B& = A B& = K A =K = 9 &9 = & = ‰ 9 9 & 9 & &9 &9 &9 ˆ &9 Š Similarly, the ratio Š‹ can be replaced by Œ Œ ‹ @] = −55.5 -+ The negative sign indicates that this work is done on the system (a work input), which is always the case for compression processes. Work done by a substance in a reversible non-flow manner is in accordance with }‚ Ž= „ , where P is in bar. Determine the work done on or by the system as pressure increases from 10 to 100 bar. Indicate whether it is a compression process or expansion process. Answer: Given, 150 = 9‡ˆ 9‡ˆ 9 = = 15 k = = 1.5 k 9ˆ 9ˆˆ So and 9‡ˆ 9‡ˆ = ’ = × 10‡ C/ ‘ ‘ Again 9.‡ 9.‡ 9‡ˆ So work done @9t = “9‡ B = “9‡ ” × 10‡ B = −3.454 × 10– + x ‘ This work is done on the system. It is therefore compression work. R=0.2968 kJ/Kg.K Polytropic Process: - During actual expansion and compression processes of gases, pressure and volume are often related by & N = K , where n and C are constants. A process of this kind is called a polytropic process. Below we develop a general expression for the work done during a polytropic process. The pressure for a polytropic process can be expressed as, = K& tN & tN—9 − &9tN—9 & − 9 &9 ‰( − 9) @] = A B& = A K& tN B& = K = = 9 9 − +1 1− 1− For the special case of n=1 the boundary work becomes equivalent to the isothermal process. A mass of gas is compressed in a quasi static process from 80 kPa, 0.1 m3 to 0.4 MPa, 0.03 m3. Assuming that the pressure and volume are related by Ž˜ = ™š˜›œ ˜œ, find the work done by the gas system. Answer: 9 = 80 - , 9 = 0.1 k , = 0.4 = 400 - , = 0.03 k We know that, 9 9 N = N Or, = 1.33 & tN—9 − &9tN—9 & − 9 &9 @] = A B& = A K& tN B& = K = = −12.12 + 9 9 − +1 1− Consider a gas contained in a piston cylinder assembly as the system. The gas is initially at a pressure of 500 kPa occupies a volume of 0.2 m3. The gas is taken to the final state where pressure is equal to 100 kPa by the following two different processes. a) The volume of the gas is inversely proportional to the process. b) The process follow the path y& N = 4 5 , where n=1.4. Calculate the work done by the gas in each case.  Ans: (a) 9 = 500 - = 100 - 9 = 0.2 k &=K K , = & K B& & & @] = A B& = A B& = K A =K = 9 &9 = 9 &9 = 160.94 -+ 9 9 9 & 9 & &9 &9 Ans: (b) &N = K ∴ 9 9 N = N Œ , =ž Ÿ 9 9 , = 0.6314 k & − 9 &9 @] = = 92.15 -+ 1− In a cylinder piston arrangement, 3 kg of an ideal gas are expanded adiabatically from a temperature of 135℃ to 35℃ and it is found to perform 160 kJ of work during the process while its enthalpy change is 315 kJ. Determine the specific heats at constant volume and constant pressure and characteristic gas constant. Answer: The given mass of ideal gas m=3kg 9 = 135 + 273 = 408 = 35 + 273 = 308 @ -B = @ = 160 -+ ℎ y 4ℎ = ¢ = 315 -+ We have to find out: Kc =? , KŠ =? B ‰ =? Work done during adiabatic process= & − 9 &9 ‰( − 9) @] = = 1−£ 1−£ 3 × ‰ (308 − 408) , 160000 = 1 − 1.4  , ‰ = 0.213 -+/-. Again, ¢= Kc ( 9 − ) , 315 = 3 × Kc × 100 , Kc = 1.05 -+/-. Further we know that, Kc − KŠ = ‰ , KŠ = 0.837 -+/-. A fluid at a pressure of 3.5 bar and with specific volume of 0.2 m3/kg is contained in a piston-cylinder arrangement. The fluid expands reversibly to a pressure of 0.8 bar according to the law ¤¥… = ¦š˜›œ ˜œ. Determine the work done by the fluid on the piston. B 1 1 @ = A yB = K A =K§ − ¨ 9 9 9 Now, K = y9 9 = 0.14 K =© = 0.418 k y ∴ @ = 36.507 -+/- INTERNAL ENERGY, ENTHALPY AND SPECIFIC HEATS OF IDEAL GASES Internal energy: For an ideal gas the internal energy is a function of the temperature only. x = x( )  Fig.: Schematic of the experimental apparatus used by Joule In his classical experiment, Joule submerged two tanks connected with a pipe and a valve in a water bath, as shown in Fig. Initially, one tank contained air at a high pressure and the other tank was evacuated. When thermal equilibrium was attained, he opened the valve to let air pass from one tank to the other until the pressures equalized. Joule observed no change in the temperature of the water bath and assumed that no heat was transferred to or from the air. Since there was also no work done, he concluded that the internal energy of the air did not change even though the volume and the pressure changed. Therefore, he reasoned, the internal energy is a function of temperature only and not a function of pressure or specific volume. (Joule later showed that for gases that deviate significantly from ideal-gas behavior, the internal energy is not a function of temperature alone.) SPECIFIC HEATS: The specific heat is defined as the energy required raising the temperature of a unit mass of a substance by one degree. In thermodynamics, there are two kinds of specific heats: specific heat at constant volume KŠ and specific heat at constant pressure Kc. The specific heat at constant volume can be viewed as the energy required raising the temperature of the unit mass of a substance by one degree as the volume is maintained constant. >x KŠ = ž Ÿ > Š This implies the change in internal energy with temperature at constant volume. Bx = KŠ ( )B ∆x = x − x9 = A KŠ ( )B 9 The specific heat at constant pressure can be viewed as the energy required raising the temperature of the unit mass of a substance by one degree as the pressure is maintained constant. >ℎ Kc = ž Ÿ > c This implies the change in enthalpy with temperature at constant pressure. Bℎ = Kc ( )B ∆ℎ = ℎ − ℎ9 = A Kc ( )B 9 For ideal gas: CP and CV are function of T. For perfect gas CP and CV are constant. ENTHALPY: Enthalpy of a substance, ℎ = x + & = x + ‰ It is an intensive property of a system. Since R is constant and x = x( ), it follows that the enthalpy of an ideal gas, is also a function of temperature only: ℎ = ℎ( ). Since x and ℎ depend only on temperature for an ideal gas, the specific heats CV and CP also depend, at most, on temperature only. Special Case: We know that for a closed stationary system of unit mass, Bª = Bx + B& At constant pressure, PB& = B( &) Hence, (Bª)c = Bx + B( &) Or, (Bª)c = B(x + &) = Bℎ Specific Heat Relations of Ideal Gases: A special relationship between Kc and KŠ for ideal gases can be obtained by differentiating the relation ℎ = x + ‰ , which yields Bℎ = Bx + ‰ B Or, Kc B = KŠ B + ‰B Or, Kc = KŠ + ‰ Ideal-gas property called the specific heat ratio £, defined as: Kc £= KŠ The specific ratio also varies with temperature, but this variation is very mild. For monatomic gases, its value is essentially constant at 1.667. Many diatomic gases, including air, have a specific heat ratio of about 1.4 at room temperature. Problem 1: During certain process, the heat capacity of a system is given by ™˜ =. « +. «¬ -®/-¯℃. If the mass of the gas is 8 kg and its temperature changes from 20℃ to 120℃ find: (a) heat transferred, (b) mean specific heat of the gas. Answer: 9 ˆ (a) ? = “ KN B = 8 “ ˆ (0.4 + 0.004 )B = 544 -+ (b) We know that, ?= KN B , 544 = 8 × KN × (120 − 20) , KN = 0.68 -+/- ℃ Problem 2: An insulated rigid tank initially contains 0.7 kg of helium at 27°C and 350 kPa. A paddle wheel with a power rating of 0.015 kW is operated within the tank for 30 min. Determine (a) the final temperature and (b) the final pressure of the helium gas. -+ { ¢ Ix 5 4Š,TŠ[ = 3.15 -. Constant specific heats can be used for helium. The system is stationary and thus the kinetic and potential energy changes are zero, ∆KE=∆PE=0 and ∆E=∆U. The volume of the tank is constant, and thus there is no boundary work. The system is adiabatic and thus there is no heat transfer. We take the contents of the tank as the system. This is a closed system since no mass crosses the system boundary during the process. The amount of paddle-wheel work done on the system is: @R1 = (0.015 × 30 × 60)-+ = 27 -+ Under the stated assumptions and observations, the energy balance on the system can be expressed as: UV − VX MNVWVOPQ = ∆ UVVWV VX RSRQ/. Y/Q /N/Z[S QZTNR\/Z ]S 1/TQ,^OZ_ TN`.TRR a1TN[/ MN MNQ/ZNTb,_MN/QM0,cOQ/NQMTb /N/Z[S ∴ @R1 = ∆ = (x − x9 ) = 4Š,TŠ[ ( − 9) , 27 = 0.7 × 3.15 × ( − 300) , = 39.24℃ Now, The final pressure is determined from the ideal-gas relation: 9 &9 & = 9 350 , = ( x I I 4 5 ) 300 312.24 , = 364.29 - Flow work and Total Energy of a Flowing Fluid: Control volumes involve mass flow across their boundaries, and some work is required to push the mass into or out of the control volume. This work is known as the flow work, or flow energy, and is necessary for maintaining a continuous flow through a control volume. To obtain a relation for flow work, consider a fluid element of volume as shown in Fig. The fluid immediately upstream forces this fluid element to enter the control volume; thus, it can be regarded as an imaginary piston. The fluid element can be chosen to be sufficiently small so that it has uniform properties throughout. Fig.: Schematic for flow work Fig.: Flow work is the energy needed to push a fluid into or out of a control volume, and it is equal to & If the fluid pressure is P and the cross-sectional area of the fluid element is A, the force applied on the fluid element by the imaginary piston is {= | To push the entire fluid element into the control volume, this force must act through a distance L. Thus, the work done in pushing the fluid element across the boundary (i.e., the flow work) is @°bO^ = {± = |± = The flow work relation is the same whether the fluid is pushed into or out of the control volume. The total energy of a simple compressible system consists of three parts: internal, kinetic, and potential energies. On a unit-mass basis for a non-flowing fluid, it is expressed as =x+- +y =x+ + 2 The fluid entering or leaving a control volume possesses an additional form of energy—the flow energy &. Then the total energy of a flowing fluid on a unit-mass basis (denoted by z) becomes z= &+ = & + (x + - + y ) = & + x + + =ℎ+ + 2 2 Energy Analysis of Steady-Flow Systems: Let us consider a certain set of assumptions that lead to a reasonable model for this type of process, which we refer to as the steady-state steady flow process. 1. The control volume does not move relative to the coordinate frame. 2. The state of the mass at each point in the control volume does not vary with time. 3. As for the mass that flows across the control surface, the mass flux and the state of this mass at each discrete area of flow on the control surface do not vary with time. During a steady-flow process, the total energy content of a control volume remains constant (ECV = constant), and thus the change in the total energy of the control volume is zero (∆ECV=0). Therefore, the amount of energy entering a control volume in all forms (by heat, work, and mass) must be equal to the amount of energy leaving it. Then the rate form of the general energy balance reduces for a steady-flow process to [?f + @f + ³ f (ℎ + + )]MNb/Q = [?f + @f + ³ f (ℎ + + )]OPQb/Q 2 2 Some Steady-Flow Engineering Devices Many engineering devices operate essentially under the same conditions for long periods of time. The components of a steam power plant (turbines, compressors, heat exchangers, and pumps), for example, operate nonstop for months before the system is shut down for maintenance. Therefore, these devices can be conveniently analyzed as steady-flow devices. In this section, some common steady-flow devices are described, and the thermodynamic aspects of the flow through them are analyzed. The conservation of mass and the conservation of energy principles for these devices are illustrated with examples. Nozzles and Diffusers Nozzles and diffusers are commonly utilized in jet engines, rockets, spacecraft, and even garden hoses. A nozzle is a device that increases the velocity of a fluid at the expense of pressure. A diffuser is a device that increases the pressure of a fluid by slowing it down. That is, nozzles and diffusers perform opposite tasks. The cross-sectional area of a nozzle decreases in the flow direction for subsonic flows and increases for supersonic flows. The reverse is true for diffusers. surroundings is usually very small (?f = 0) since the fluid has high velocities, and thus it The rate of heat transfer between the fluid flowing through a nozzle or a diffuser and the Nozzles and diffusers typically involve no work (@f = 0) and any change in potential energy does not spend enough time in the device for any significant heat transfer to take place. is negligible (∆y = 0). But nozzles and diffusers usually involve very high velocities, and as a fluid passes through a nozzle or diffuser, it experiences large changes in its velocity. Therefore, the kinetic energy changes must be accounted for in analyzing the flow through these devices (∆- ≠ 0). Fig.: Nozzles and diffusers are shaped so that they cause large changes in fluid velocities and thus kinetic energies Problem: At the inlet to a certain nozzle the specific enthalpy of the fluid is 3025 kJ/kg and the velocity is 60 m/s. At the exit from the nozzle the enthalpy is 2790 kJ/kg. The nozzle is horizontal and there is negligible heat loss from it. i) Find the velocity at the nozzle exit. ii) If the inlet area is 0.1 m2 and specific volume at inlet is 0.19 m3/kg, find the rate of flow of fluid. iii) If the specific volume at the nozzle exit is 0.5 m3/kg, find the exit area of the nozzle. Answer: (i) Under stated assumptions and observations, the energy balance for this steady-flow system can be expressed in the rate form as: [?f + @f + ³ f (ℎ + + )]MNb/Q = [?f + @f + ³ f (ℎ + + )]OPQb/Q 2 2 { ℎI5 4 5 , ℎ9 + =ℎ + 9 2 2  60 3025000 + = 2790000 + 2 2. = 688.2 /5 ¶· ¶ (ii) |9 = 0.1 , &9 = 0.19 , V9 = 60 ¸¹ » 1 - f 9 = |9 9 29 = 0.1 × 60 × = 31.6 0.19 5 ¶· (iii) & = 0.5 ¸¹ | f =| 2 = & | × 688.2 , 31.6 = 0.5 , | = 0.0229 R=0.287 kPa.m3/kg.K Turbines and Compressors In steam, gas, or hydroelectric power plants, the device that drives the electric generator is the turbine. As the fluid passes through the turbine, work is done against the blades, which are attached to the shaft. As a result, the shaft rotates, and the turbine produces work. Compressors, as well as pumps and fans, are devices used to increase the pressure of a fluid. Work is supplied to these devices from an external source through a rotating shaft. Therefore, compressors involve work inputs. Even though these three devices function similarly, they do differ in the tasks they perform. A fan increases the pressure of a gas slightly and is mainly used to mobilize a gas. A compressor is capable of compressing the gas to very high pressures. Pumps work very much like compressors except that they handle liquids instead of gases. Note that turbines produce power output whereas compressors, pumps, and fans require power input. Heat transfer from turbines is usually negligible (?f ≈ 0) since they are typically well insulated. Heat transfer is also negligible for compressors unless there is intentional cooling. Potential energychanges are negligible for all of these devices (∆y ≅ 0). The velocitiesinvolved in these devices, with the exception of turbines and fans, are usually too low to cause any significant change in the kinetic energy (∆- ≅ 0). The fluid velocities encountered in most turbines are very high, and the fluid experiences a significant change in its kinetic energy. However, this change is usually very small relative to the change in enthalpy, and thus it is often disregarded. A turbine operates under steady flow conditions, receiving steam at the following state: pressure 1.2 MPa, temperature 188℃, enthalpy 2785 kJ/kg, velocity 33.3 m/s and elevation 3 m. The steam leaves the turbine at the following state: pressure 20 kPa, enthalpy 2512 kJ/kg, velocity 100 m/s and elevation 0 m. Heat lost to the surroundings at the rate of 0.29 kJ/kg. If the rate of steam flow through the turbine is 0.42 kg/s, what is the power output of the turbine in kW? (10) Answer: [?f + @f + ³ f (ℎ + + )]MNb/Q = [?f + @f + ³ f (ℎ + + )]OPQb/Q 2 2 33.3 − 100 , @fOPQb/Q = 0.42 ¾(2785 − 25512) × 1000 + + (9.81 × 3) − 0.29¿ 2 = 112.67 -@ Throttling Valves: Throttling valves are any kind of flow-restricting devices that cause a significant pressure drop in the fluid is often accompanied by a large drop in temperature. Some familiar examples are ordinary adjustable valves, capillary tubes, and porous plugs. For single-stream steady-flow device while throttling the equation is ℎ ≅ ℎ9. ℎ9 = x9 + 9 &9 = x + & =ℎ Thus the final outcome of a throttling process depends on which of the two quantities increases during the process. If the flow energy increases during the process ( & > 9 &9 ), it can do so at the expense of the internal energy. As a result, internal energy decreases, which is usually accompanied by a drop in temperature and vice versa. The temperature of an ideal gas does not change during a throttling (ℎ = 4 5 ) process since ℎ = ℎ( ). Introduction to the Second Law of Thermodynamics: The first law places no restriction on the direction of a process, but satisfying the first law does not ensure that the process can actually occur. Processes occur in a certain direction, and not in the reverse direction. A process cannot occur unless it satisfies both the first and the second laws of thermodynamics. Fig.: A process must satisfy both the first and second laws of thermodynamics to proceed It is common experience that a cup of hot coffee left in a cooler room eventually cools off. This process satisfies the first law of thermodynamics since the amount of energy lost by the coffee is equal to the amount gained by the surrounding air. Now let us consider the reverse process-the hot coffee getting even hotter in a cooler room as a result of heat transfer from the room air. We all know that this process never takes place. Yet, doing so would not violate the first law as long as the amount of energy lost by the air is equal to the amount gained by the coffee. The use of the second law of thermodynamics is not limited to identifying the direction of processes, however. The second law also asserts that energy has quality as well as quantity. The first law is concerned with the quantity of energy and the transformations of energy from one form to another with no regard to its quality. Thermal Energy Reservoirs It is very convenient to have a hypothetical body with a relatively large thermal energy capacity (mass × specific heat) that can supply or absorb finite amounts of heat without undergoing any change in temperature. Such a body is called a thermal energy reservoir. A two-phase system can be modeled as a reservoir also since it can absorb and release large quantities of heat while remaining at constant temperature. Another familiar example of a thermal energy reservoir is the industrial furnace. A reservoir that supplies energy in the form of heat is called a source, and one that absorbs energy in the form of heat is called a sink. The Second Law of Thermodynamics: There are two classical statements of the second law, the Kelvin-Planckstatement, which is related to heat engines and the Clausius statement, which is related to refrigerators or heat pumps. Kelvin-Planck Statement: It is impossible for any device that operates on a cycle to receive heat from a single reservoir and produce a net amount of work. Clausius Statement It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body to a higher-temperature body. Equivalence of the Two Statements The Kelvin–Planck and the Clausius statements are equivalent in their consequences, and either statement can be used as the expression of the second law of thermodynamics. Any device that violates the Kelvin–Planck statement also violates the Clausius statement, and vice versa. This can be demonstrated as follows.  Fig. a: A refrigerator that is powered by a 100 percent efficient heat engine Fig. b: The equivalent refrigerator Consider the heat-engine-refrigerator combination shown in Fig. 6–27a,operating between the same two reservoirs. The heat engine is assumed to have, in violation of the Kelvin– Planck statement, a thermal efficiency of100 percent, and therefore it converts all the heat it receives to work W.This work is now supplied to a refrigerator that removes heat in the amountof from the low-temperature reservoir and rejects heat in the amount of + to the high-temperature reservoir. During this process, the high temperature reservoir receives a net amount of heat (the differencebetween + and ). Thus, the combination of these two devices canbe viewed as a refrigerator, as shown in Fig. b, that transfers heat in an amount of from a cooler body to a warmer one without requiring anyinput from outside. This is clearly a violation of the Clausius statement. Therefore, a violation of the Kelvin–Planck statement results in the violation of the Clausius statement. It can also be shown in a similar manner that a violation of the Clausius statement leads to the violation of the Kelvin–Planck statement. Therefore, the Clausius and the Kelvin–Planck statements are two equivalent expressions of the second law of thermodynamics. Alternative Solution for Equivalence of the Two Statements Statements: To prove that violation of the Kelvin-Planck Kelvin Planck Statement leads to a violation of the Clausius Statement, let us assume that Kelvin-Planck Kelvin statement is incorrect. Consider a cyclically working device 1, which absorbs energy Q1 as heat from a thermal reservoir at TH. Equivalent amount of work W(W=Q1) is performed. Consider another device 2 operating as a cycle, which absorbs energy QL as heat from a low temperature thermal reservoir at TL and rejects energy QH (QH=QL+W). Such a device does not violate Clausius statement. If the two devices are now combined, the combined device (enclosed by the dotted boundary) transfers heat QL from the low temperature reservoir at TL to a high temperature re reservoir at TH with out receiving any aid from an external agent, which is the violation of the Clausius statement. Likewise let us assume that the Clausius statement is incorrect. So we have a device 1, cyclically working transferring heat Q from a low temperature reservoir at TL to a high temperature thermal reservoir at TH. Consider another device 2, which absorbs heat Q1 from a high temperature reservoir at TH does work W and rejects energy Q as heat to the low temperature reservoir at TL as shown in figure. If the two devices are combined (shown in figure by a dotted enclosure), then the combined device receives energy (Q1-Q) as heat from a thermal reservoir and delivers equivalent work (W=Q1-Q) in violation of the Kelvin-Planck statement. Therefore violation of Clausius statement leads to the violation of the Kelvin-Planck statement. Hence, these two statements are equivalent. Heat Engines Work can easily be converted to other forms of energy, but converting other forms of energy to work is not that easy. The mechanical work done by the shaft shown in Fig, for example, is first converted to the internal energy of the water. This energy may then leave the water as heat. We know from experience that any attempt to reverse this process will fail. That is, transferring heat to the water does not cause the shaft to rotate. Fig.: Work can always be converted to heat directly and completely, but the reverse is not true We conclude that work can be converted to heat directly and completely, but converting heat to work requires the use of some special devices. These devices are called heat engines. Heat engines differ considerably from one another, but all can be characterized by the following: 1. They receive heat from a high-temperature source (solar energy, oil furnace, nuclear reactor). 2. They convert part of this heat to work (usually in the form of a rotating shaft). 3. They reject the remaining waste heat to a low-temperature sink (the atmosphere, rivers). 4. They operate on a cycle. So cyclic heat engine is a thermodynamic device in which there is net heat transfer to the system and net work transfer from the system. Fig.: Part of the heat received by a heatengine is converted to work, while the rest is rejected to a sink =1− Refrigerator: We all know from experience that heat is transferred in the direction of decreasing temperature, that is, from high-temperature mediums to low temperature ones. This heat transfer process occurs in nature without requiring any devices. The reverse process, however, cannot occur by itself. The transfer of heat from a low-temperature medium to a high-temperature one requires special devices called refrigerators. 1 = = = = $% !", − −1 $& Notice that the value of COPR can be greater than unity. This is in contrast to the thermal efficiency, which can never be greater than 1. In fact, one reason for expressing the efficiency of a refrigerator by another term-the coefficient of performance-is the desire to avoid the oddity of having efficiencies greater than unity. Problem 1: COPR=4 then find QH/QL=? 1 = = $% − −1 $& Or, QH/QL=1.25 Heat Pumps: Another device that transfers heat from a low-temperature medium to a high-temperature one is the heat pump, shown schematically in Fig. Refrigerators and heat pumps operate on the same cycle but differ in their objectives. The objective of a refrigerator is to maintain the refrigerated space at a low temperature by removing heat from it. Discharging this heat to a higher-temperature medium is merely a necessary part of the operation, not the purpose. The objective of a heat pump, however, is to maintain a heated space at a high temperature. This is accomplished by absorbing heat from a low-temperature source, such as well water or cold outside air in winter, and supplying this heat to the high-temperature medium such as a house. An ordinary refrigerator that is placed in the window of a house with its door open to the cold outside air in winter will function as a heat pump since it will try to cool the outside by absorbing heat from it and rejecting this heat into the house through the coils behind it. 1 ' = = = = $& ! " , − 1− $% ∴ +1= ' Two reversible heat engines A and B are arranged in series, Engine A is rejecting heat directly to engine B. Engine A receives 200kJ at a temperature of 421°C from a hot source, while engine B is in communication with a cold sink at a temperature of 4.4°C. If the work output of A is twice that of B find (a) the intermediate temperature between A and B, (b) the efficiency of each engine, and (c) the heat rejected to the cold sink. Perpetual-Motion Machines: A process cannot take place unless it satisfies both the first and second laws of thermodynamics. Any device that violates either law is called a perpetual-motion machine. A device that violates the first law of thermodynamics is called a perpetual-motion machine of the first kind (PMM1), and a device that violates the second law of thermodynamics is called a perpetual-motion machine of the second kind (PMM2). PMM1: Consider the steam power plant as shown in Fig. It is proposed to heat the steam by resistance heaters placed inside the boiler, instead of by the energy supplied from fossil or nuclear fuels. Part of the electricity generated by the plant is to be used to power the resistors as well as the pump. The rest of the electric energy is to be supplied to the electric network as the net work output. The inventor claims that once the system is started, this power plant will produce electricity indefinitely without requiring any energy input from the outside. Well, here is an invention that could solve the world’s energy problem-if it works, of course. A careful examination of this invention reveals that the system enclosed by the shaded area is continuously supplying energy to the outside at a rate of ) + !) " , without receiving any energy. That is, thissystem is creating energy at a rate of ) + !) " , which is clearly a violation of the first law. Therefore, this wonderful device is nothing more than a PMM1 and does not warrant any further consideration. PMM2: - Well, the possibility of doubling the efficiency would certainly be very tempting to plant managers and, if not properly trained, they would probably give this idea a chance, since intuitively they see nothing wrong with it. A student of thermodynamics, however, will immediately label this device as a PMM2, since it works on a cycle and does a net amount of work while exchanging heat with a single reservoir (the furnace) only. It satisfies the first law but violates the second law, and therefore it will not work. THE CARNOT CYCLE The efficiency of a heat-engine cycle greatly depends on how the individual processes that make up the cycle are executed. The net work, thus the cycle efficiency, can be maximized by using processes that require the least amount of work and deliver the most, that is, by using reversible processes. It is no surprise that the most efficient cycles are reversible cycles, that is, cycles that consist entirely of reversible processes. Probably the best known reversible cycle is the Carnot cycle, first proposed in 1824 by French engineer Sadi Carnot. The Carnot cycle is composed of four reversible processes-two isothermal and two adiabatic and it can be executed either in a closed or a steady-flow system. Consider a closed system that consists of a gas contained in an adiabatic piston–cylinder device, as shown in Fig. The insulation of the cylinder head is such that it may be removed to bring the cylinder into contact with reservoirs to provide heat transfer. The four reversible processes that make up the Carnot cycle are as follows: Reversible Isothermal Expansion (process 1-2, * =constant). Initially(state 1), the temperature of the gas is * and the cylinder head is in close contact with a source at temperature *. The gas is allowed to expandslowly, doing work on the surroundings. As the gas expands, thetemperature of the gas tends to decrease. But as soon as the temperaturedrops by an infinitesimal amount dT, some heat is transferred from the reservoir into the gas, raising the gas temperature to*. Thus, the gastemperature is kept constant at *. Since the temperature differencebetween the gas and the reservoir never exceeds a differential amount dT,this is a reversible heat transfer process. It continues until the piston reaches position 2. The amount of total heat transferred to the gas during this process is. Reversible Adiabatic Expansion (process 2-3, temperature drops from * to * ). At state 2, the reservoir that was in contact with the cylinder head is removed and replaced by insulation so that the system becomes adiabatic. The gas continues to expand slowly, doing work on the surroundings until its temperature drops from * to * (state 3). Thepiston is assumed to be frictionless and the process to be quasi-equilibrium, so the process is reversible as well as adiabatic. Reversible Isothermal Compression (process 3-4, * =constant). At state3, the insulation at the cylinder head is removed, and the cylinder is brought into contact with a sink at temperature *. Now the piston ispushed inward by an external force, doing work on the gas. As the gas iscompressed, its temperature tends to rise. But as soon as it rises by aninfinitesimal amount dT, heat is transferred from the gas to the sink, causing the gas temperature to drop to *. Thus, the gas temperatureremains constant at *. Since the temperature difference between the gasand the sink never exceeds a differential amount dT, this is a reversible heat transfer process. It continues until the piston reaches state 4. The amount of heat rejected from the gas during this process is. Reversible Adiabatic Compression (process 4-1, temperature rises from * to* ). State 4 is such that when the low-temperature reservoir isremoved, the insulation is put back on the cylinder head, and the gas iscompressed in a reversible manner, the gas returns to its initial state (state1). The temperature rises from * to * during this reversible adiabatic compression process, which completes the cycle. The P-V diagram of this cycle is shown in Fig. Remembering that on a P-V diagram the area under the process curve represents the boundary work for quasi-equilibrium processes, we see that the area under curve 1-2-3 is the work done by the gas during the expansion part of the cycle, and the area under curve 3-4-1 is the work done on the gas during the compression part of the cycle. The area enclosed by the path of the cycle (area 1-2-3-4-1) is the difference between these two and represents the net work done during the cycle. Carnot principles: 1. The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs. 2. The efficiencies of all reversible heat engines operating between the same two reservoirs are the same. THE CARNOT HEAT ENGINE The hypothetical heat engine that operates on the reversible Carnot cycle is called the Carnot heat engine. The thermal efficiency of any heat engine, reversible or irreversible, is given by =1− =1− For reversible heat engines, the heat transfer ratio in the above relation can be replaced by the ratio of the absolute temperatures of the two reservoirs, as given by * =1− * So for reversible heat engine * ( )-". = * The temperatures are in Kelvin scale, and the temperatures on this scale are called absolute temperatures. On the Kelvin scale, the temperature ratios depend on the ratios of heat transfer between a reversible heat engine and the reservoirs and are independent of the physical properties of any substance. The Carnot Refrigerator and Heat Pump A refrigerator or a heat pump that operates on the reversed Carnot cycle is called a Carnot refrigerator, or a Carnot heat pump. The coefficient of performance of any refrigerator or heat pump, reversible or irreversible, is given by: 1 = = $% − −1 $& 1 ' = = $ − 1 − $& % The COPs of all reversible refrigerators or heat pumps can be determined by replacing the heat transfer ratios in the above relations by the ratios of the absolute temperatures of the high- and low-temperature reservoirs. * 1 = = /% * −* −1/& * 1 ' = = / * −* 1 − /& % Which is the more effective way to increase the efficiency of a Carnot engine: to increase 01 , keeping 02 constant; or to decrease 02 , keeping 01 constant? / The efficiency of a Carnot engine is given by = 1 − /3 4 Let *5 be decreased by ∆* with *7 remaining the same. /3 8∆/ Thus, 7 =1− /4 If *7 is increased by the same ∆* with *5 remaining the same 3 / Then 5 = 1 − / 9∆/ 4 3/ /3 8∆/ (/4 8/3 )∆/9∆/ 3 Thus 7 − 5 = / 9∆/ − = 4 /4 /4 (/4 9∆/) Since *7 > *5 , 7 − 5 > 0 The more effective way to increase the cycle efficiency is to decease *5. Pure Substance: A substance that has a fixed chemical composition throughout is called a pure substance. Water, nitrogen, helium, and carbon dioxide, for example, are all pure substances. Even homogeneous mixtures can be called pure substances. However, a mixture of oil and water is not a pure substance. Since oil is not soluble in water, it will collect on top of the water, forming two chemically dissimilar regions. A mixture of two or more phases of a pure substance is still a pure substance as long as the chemical composition of all phases is the same. A mixture of ice and liquid water, for example, is a pure substance because both phases have the same chemical composition. A mixture of liquid air and gaseous air, however, is not a pure substance since the composition of liquid air is different from the composition of gaseous air, and thus the mixture is no longer chemically homogeneous. This is due to different components in air condensing at different temperatures at a specified pressure. Phases of a Pure Substance: Even though there are three principal phases: solid, liquid, and gas, a substance may have several phases within a principal phase, each with a different molecular structure. Carbon, for example, may exist as graphite or diamond in the solid phase. Helium has two liquid phases; iron has three solid phases. Ice may exist at seven different phases at high pressures. Compressed Liquid and Saturated Liquid: Consider a piston–cylinder device containing liquid water at 20°C and 1 atm pressure. Under these conditions, water exists in the liquid phase, and it is called a compressed liquid, or a subcooled liquid, meaning that it is not about to vaporize. Heat is now transferred to the wateruntil its temperature rises to, say, 40°C. As the temperature rises, the liquidwater expands slightly, and so its specific volume increases. To accommodate this expansion, the piston moves up slightly. The pressure in the cylinder remains constant at 1 atm during this process since it depends on theoutside barometric pressure and the weight of the piston, both of which areconstant. Water is still a compressed liquid at this state since it has notstarted to vaporize. As more heat is transferred, the temperature keeps rising until it reaches 100°C. At this point water is still a liquid, but any heat addition will cause some of the liquid to vaporize. That is, a phase-change process from liquid to vapour is about to take place. A liquid that is about tovaporize is called a saturated liquid. Saturated Vapour and Superheated Vapour: Once boiling starts, the temperature stops rising until the liquid is completely vaporized. Any heat loss from this vapour will cause some of the vapour to condense (phase change from vapour to liquid). A vapour that is about to condense is called a saturated vapour. A vapour that is not about to condense (i.e., not a saturated vapour) is called a superheated vapour. Saturation Temperature and Saturation Pressure: It probably came as no surprise to you that water started to boil at 100°C.Strictly speaking, the statement “water boils at 100°C” is incorrect. The correct statement is “water boils at 100°C at 1 atm pressure.” The only reasonwater started boiling at 100°C was because we held the pressure constant at1 atm (101.325 kPa). If the pressure inside the cylinder were raised to 500kPa by adding weights on top of the piston, water would start boiling at151.8°C. That is, the temperature at which water starts boiling depends onthe pressure.At a given pressure, the temperature at which a pure substance changesphase is called the saturation temperature,. Likewise, at a given temperature, the pressure at which a pure substance changes phase is called thesaturation pressure,. At a pressure of 101.325 kPa, is 99.97°C. It takes a large amount of energy to melt a solid or vaporize a liquid. The amount of energy absorbed or released during a phase-change process is called the latent heat. More specifically, the amount of energy absorbed during melting is called the latent heat of fusion and is equivalent to the amount of energy released during freezing. Similarly, the amount of energy absorbed during vaporization is called the latent heat of vaporization and is equivalent to the energy released during condensation. The magnitudes of the latent heats depend on the temperature or pressure at which the phase change occurs. At 1 atm pressure, the latent heat of fusion of water is 333.7 kJ/kg and the latent heat of vaporization is 2256.5 kJ/kg. The T-v Diagram: Let us add weights on top of the piston until the pressure inside the cylinder reaches 1 MPa. At this pressure, water has a somewhat smaller specificvolume than it does at 1 atm (101kPa=0.1MPa) pressure. As heat is transferred to the water atthis new pressure, the process follows a path that looks very much like theprocess path at 1 atm pressure, as shown in Fig, but there are somenoticeable differences. First, water starts boiling at a much higher temperature (179.9°C) at this pressure. Second, the specific volume of the saturated liquid is larger and the specific volume of the saturated vapour is smaller than the corresponding values at 1 atm pressure. That is, the horizontal line that connects the saturated liquid and saturated vapour states is much shorter. As the pressure is increased further, this saturation line continues to shrink, as shown in Fig, and it becomes a point when the pressure reaches 22.06 MPa for the case of water. This point is called the critical point and it is defined as the point at which the saturated liquid and saturated vapour states are identical. The temperature, pressure, and specific volume of a substance at the critical point are called, respectively, the critical temperature critical pressure and critical specific volume. The critical-point properties of water are =22.06 MPa, =373.95°C and =0.003106 m3/kg. At pressures above the critical pressure, there is not a distinct phase change process. Instead, the specific volume of the substance continually increases, and at all times there is only one phase present. Eventually, it resembles a vapour, but we can never tell when the change has occurred. Above the critical state, there is no line that separates the compressed liquid region and the superheated vapour region. However, it is customary to refer to the substance as superheated vapour at temperatures above the critical temperature and as compressed liquid at temperatures below the critical temperature. The P-v Diagram: - The general shape of the P-v diagram of a pure substance is very much like the T-v diagram. Consider again a piston–cylinder device that contains liquid water at 1 MPa and 150°C. Water at this state exists as a compressed liquid. Now the weights on top of the piston are removed one by one so that the pressure inside the cylinder decreases gradually. The water is allowed to exchange heat with the surroundings so its temperature remains constant. Asthe pressure decreases, the volume of the water increases slightly. When the pressure reaches the saturation-pressure value at the specified temperature (0.4762 MPa), the water starts to boil. During this vaporization process, both the temperature and the pressure remain constant, but the specific volume increases. Once the last drop of liquid is vaporized, further reduction in pressure results in a further increase in specific volume. The P-T Diagram: - Figure shows the P-T diagram of a pure substance. This diagram is often called the phase diagram since all three phases are separated from each other by three lines. The sublimation line separates the solid and vapour regions, the vaporization line separates the liquid and vapour regions, and the melting (or fusion) line separates the solid and liquid regions. These three lines meet at the triple point, where all three