Higher Technological Institute

Mechanical Engineering Dep.

Thermodynamics (2) (Lectures)

Dr. Ahmed Abd EL badie Eng. Mahmoud Essam Eng. Abd El Rahman Essam

Jan.2019 بسم هللا الرحمن الرحيم

Thermodynamics (2) Lectures Chapter (1)

The Second Law of Thermodynamics

1.1 Introduction:

In the preceding course we applied the first law of thermodynamics, or the conservation of energy principle, to processes involving closed and open systems. As pointed out repeatedly in this course, energy is a conserved property, and no process is known to have taken place in violation of the first law of thermodynamics. Therefore, it is reasonable to conclude that a process must satisfy the first law to occur. However, as explained here, satisfying the first law alone does not ensure that the process will actually take place. It is common experience that a cup of hot coffee left in a cooler room eventually cools off (Fig. 1.1). 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. As another familiar example, consider the heating of a room by the passage of electric current through a resistor (Fig. 1.2). Again, the first law dictates that the amount of electric energy supplied to the resistance wires be equal to the amount of energy transferred to the room air as heat. Now let us attempt to reverse Fig.1.1 this process. It will come as no surprise that transferring some heat to the wires does not cause an equivalent amount of electric energy to be generated in the wires. Finally, consider a paddle-wheel mechanism that is operated by the fall of a mass (Fig. 1.3). The paddle wheel rotates as the Fig.1.2 mass falls and stirs a fluid within an insulated container. As a result,

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 1 the potential energy of the mass decreases, and the internal energy of the fluid increases in accordance with the conservation of energy principle. However, the reverse process, raising the mass by transferring heat from the fluid to the paddle wheel, does not occur in nature, although doing so would not violate the first law Fig.1.3 of thermodynamics. It is clear from these arguments that processes proceed in a certain direction and not in the reverse direction. 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. This inadequacy of the first law to identify whether a process can take place is remedied by introducing another general principle, the second law of thermodynamics. We show later in this chapter that the reverse processes discussed above violate the second law of thermodynamics. This violation is easily detected with the help of a property, called entropy, defined in the next chapter. A process cannot occur unless it satisfies both the first and the second laws of thermodynamics (Fig. 1.4). There are numerous valid statements of the second law of thermodynamics. Two such statements are presented and discussed later in this chapter in relation to some engineering devices that operate on cycles. The use of the second law of thermodynamics is not limited to identifying the direction of processes. 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 Fig.1.4 transformations of energy from one form to another with no regard to its quality. Preserving the quality of energy is a major concern to engineers, and the second law provides the necessary means to determine the quality as well as the degree of degradation of energy during a process. As discussed later in this chapter, more of high-temperature energy can be converted to work, and thus it has a higher quality than the same amount of energy at a lower temperature. The second law of thermodynamics is also used in determining the theoretical limits for the performance of commonly used engineering systems, such as heat engines and refrigerators, as well as predicting the degree of completion of chemical reactions. The second law is also closely associated with the concept of perfection. In fact, the second law defines perfection for thermodynamic processes. It can be used to quantify the level of perfection of a process, and point the direction to eliminate imperfections effectively.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 2

1.2 Thermal Energy Reservoirs: In the development of the second law of thermodynamics, it is very convenient to have a hypothetical body with a relatively large thermal energy capacity (mass x specific heat) that can supply or absorb finite amounts of heat without undergoing any change in temperature. Such a body is called a thermal energy reservoir, or just a reservoir. In practice, large bodies of water such as oceans, lakes, and rivers as well as the atmospheric air can be modeled accurately as thermal energy reservoirs because of their large thermal energy storage capabilities or thermal masses (Fig. 1.5). The atmosphere, for example, does not warm up as a result of heat losses from residential buildings in winter. Likewise, mega joules of waste energy dumped in large rivers by power plants do not cause any significant change in water temperature. A two-phase system can also be modeled as a reservoir 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. The temperatures of most furnaces are carefully controlled, and they are capable of supplying large quantities of thermal energy as heat in an essentially isothermal manner. Therefore, they can be modeled as reservoirs. A reservoir that supplies energy in the form of heat is called a source, and one that absorbs energy in the form of heat Fig 1.5 is called a sink (Fig. 1.6). Thermal energy reservoirs are often referred to as heat reservoirs since they supply or absorb energy in the form of heat.

Fig. 1.6

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 3

1.3 Heat Engines: As pointed out earlier, 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. 1.7, 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. From this and other Fig. 1.7 observations, 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 (Fig. 1.8): 1. They receive heat from a high-temperature source (solar energy, oil furnace, nuclear reactor, etc.). 2. They convert part of this heat to work (usually in the form of a rotating shaft). 3. They reject the remaining waste heat to a low-temperature sink (the atmosphere, rivers, etc.).

4. They operate on a cycle. Fig. 1.8

Heat engines and other cyclic devices usually involve a fluid to and from which heat is transferred while undergoing a cycle. This fluid is called the working fluid. The term heat engine is often used in a broader sense to include work producing devices that do not operate in a thermodynamic cycle. Engines that involve internal combustion such as gas turbines and car engines fall into this category. These devices operate in a mechanical cycle but not in a thermodynamic cycle since the working fluid (the combustion gases) does not undergo a complete cycle. Instead of being cooled to the initial temperature, the exhaust gases are purged and replaced by fresh air-and-fuel mixture at the end of the cycle. The work-producing device that best fits into the definition of a heat engine is the steam power plant, which is an external-combustion engine. That is, combustion takes place outside the

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 4 engine, and the thermal energy released during this process is transferred to the steam as heat. The schematic of a basic steam power plant is shown in Fig. 1.9. This is a rather simplified diagram, and the discussion of actual steam power plants is given in later chapters. The various quantities shown on this figure are as follows:

Qin = amount of heat supplied to steam in boiler from a high-temperature source (furnace).

Qout = amount of heat rejected from steam in condenser to a low temperature sink (the atmosphere, a river, etc.).

Wout = amount of work delivered by steam as it expands in turbine.

Win = amount of work required to compress water to boiler pressure.

Fig. 1.9

The network output of this power plant is simply the difference between the total work output of the plant and the total work input:

Wnet,out = Wout - Win (kJ) (1–1) The network can also be determined from the heat transfer data alone. The four components of the steam power plant involve mass flow in and out, and therefore should be treated as open

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 5 systems. These components, together with the connecting pipes, however, always contain the same fluid (not counting the steam that may leak out, of course). No mass enters or leaves this combination system, which is indicated by the shaded area on Fig. 1.9; thus, it can be analyzed as a closed system. Recall that for a closed system undergoing a cycle, the change in internal energy ∆U is zero, and therefore the network output of the system is also equal to the net heat transfer to the system:

Wnet,out = Qin - Qout (kJ) (1–2)

1.3.1 Thermal Efficiency

In Eq. 1–2, Qout represents the magnitude of the energy wasted in order to complete the cycle. But Qout is never zero; thus, the network output of a heat engine is always less than the amount of heat input. That is, only part of the heat transferred to the heat engine is converted to work. The fraction of the heat input that is converted to network output is a measure of the performance of a heat engine and is called the thermal efficiency Ƞth (Fig. 1.10). For heat engines, the desired output is the network output, and the required input is the amount of heat supplied to the working fluid. Then the thermal efficiency of a heat engine can be expressed as

(1-3)

Or (1-4)

It can also be expressed as (1-5)

Since Wnet,out = Q in - Q out Cyclic devices of practical interest such as heat engines, refrigerators, and heat pumps operate between a high-temperature medium (or reservoir) at temperature TH and a low-temperature medium

(or reservoir) at temperature TL. To bring uniformity to the treatment of heat engines, refrigerators, and heat pumps, we define these two quantities:

QH = magnitude of heat transfer between the cyclic Fig. 1.10

device and the high temperature medium at temperature TH.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 6

QL = magnitude of heat transfer between the cyclic device and the low temperature medium at

temperature TL.

Notice that both QL and QH are defined as magnitudes and therefore are positive quantities. The direction of QH and QL is easily determined by inspection. Then, the network output and thermal efficiency relations for any heat engine (shown in Fig. 1.11) can also be expressed as

Wnet,out = QH - QL and

or (1-6)

The thermal efficiency of a heat engine is always less than unity since both QL and QH are defined as positive quantities. Thermal efficiency is a measure of how efficiently a heat engine converts the heat that it receives to work. Heat engines are built for the purpose of converting heat to work, and engineers are constantly trying to improve the efficiencies of these devices since increased efficiency means less fuel consumption and thus lower fuel bills and less Fig. 1.11 pollution. The thermal efficiencies of work-producing devices are relatively low. Ordinary spark- ignition automobile engines have a thermal efficiency of about 25 percent. That is, an automobile engine converts about 25 percent of the chemical energy of the gasoline to mechanical work. This number is as high as 40 percent for diesel engines and large gas-turbine plants and as high as 60 percent for large combined gas-steam power plants. Thus, even with the most efficient heat engines available today, almost one-half of the energy supplied ends up in the rivers, lakes, or the atmosphere as waste or useless energy.

Example 1.1 Heat is transferred to a heat engine from a furnace at a rate of 80 MW. If the rate of waste heat rejection to a nearby river is 50 MW, determine the net power output and the thermal efficiency for this heat engine. Solution

̇ ̇ Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 7

The net power output of this heat engine is

Then the thermal efficiency is easily determined to be ̇ ̇ ̇

̇

( ) ̇ 1.3.2 The Second Law of Thermodynamics: Kelvin–Planck Statement Under the ideal conditions, a heat engine must reject some heat to a low-temperature reservoir in order to complete the cycle. That is, no heat engine can convert all the heat it receives to useful work. This limitation on the thermal efficiency of heat engines forms the basis for the Kelvin–Planck statement of the second law of thermodynamics, which is expressed as follows: 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. That is, a heat engine must exchange heat with a low- temperature sink as well as a high-temperature source to keep operating. The Kelvin–Planck statement can also be expressed as no heat engine can have a thermal efficiency of 100 percent (Fig. 1.12), or as for a power plant to operate, the working fluid must exchange heat with the environment as well as the furnace. Note that the impossibility of having a 100 percent efficient heat engine is not due to friction or other dissipative Fig. 1.12 effects. It is a limitation that applies to both the idealized and the actual heat engines. Later in this chapter, we develop a relation for the maximum thermal efficiency of a heat engine. We also demonstrate that this maximum value depends on the reservoir temperatures only.

1.4 Refrigerators and Heat Pump: 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

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 8 occur by itself. The transfer of heat from a low-temperature medium to a high-temperature one requires special devices called refrigerators. Refrigerators, like heat engines, are cyclic devices. The working fluid used in the refrigeration cycle is called a refrigerant. The most frequently used refrigeration cycle is the vapor-compression refrigeration cycle, which involves four main components: a compressor, a condenser, an expansion valve, and an evaporator, as shown in Fig. 1.13.

Fig. 1.13 Fig. 1.14

The refrigerant enters the compressor as a vapor and is compressed to the condenser pressure. It leaves the compressor at a relatively high temperature and cools down and condenses as it flows through the coils of the condenser by rejecting heat to the surrounding medium. It then enters a capillary tube where its pressure and temperature drop drastically due to the throttling effect. The low-temperature refrigerant then enters the evaporator, where it evaporates by absorbing heat from the refrigerated space. The cycle is completed as the refrigerant leaves the evaporator and reenters the compressor. In a household refrigerator, the freezer compartment where heat is absorbed by the refrigerant serves as the evaporator, and the coils, usually behind the refrigerator where heat is dissipated to the kitchen air, serve as the condenser. A refrigerator is shown schematically in Fig.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 9

1.14. Here QL is the magnitude of the heat removed from the refrigerated space at temperature

TL, QH is the magnitude of the heat rejected to the warm environment at temperature TH, and

Wnet,in is the network input to the refrigerator. As discussed before, QL and QH represent magnitudes and thus are positive quantities.

1.4.1 Coefficient of Performance The efficiency of a refrigerator is expressed in terms of the coefficient of performance

(COP), denoted by COPR. The objective of a refrigerator is to remove heat (QL) from the refrigerated space. To accomplish this objective, it requires a work input of Wnet,in. Then the COP of a refrigerator can be expressed as

(1-7)

This relation can also be expressed in rate form by replacing QL by and Wnet,in by . The conservation of energy principle for a cyclic device requires that. ̇ ̇ (KJ) (1-8)

Then the COP relation becomes (1-9)

Notice that the value of COP R can be greater ⁄ than unity. That is, the amount of heat removed from the refrigerated space can be greater than the amount of work input. 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.

1.4.2 Heat Pump Another device that transfers heat from a low-temperature medium to a high temperature one is the heat pump, shown schematically in Fig. 1.15. 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

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 10 water or cold outside air in winter, and supplying this heat to the high-temperature medium such as a house (Fig. 1.16).

Fig. 1.15 Fig. 1.16 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. The measure of performance of a heat pump is also expressed in terms of the coefficient of performance COPHP, defined as (1-10) which can also be expressed as

(1-11)

A comparison of Eqs. 1–7 and 1– 10 reveals that ⁄ (1-12)

Q Q for fixed values of L and H. This relation implies that the coefficient of performance of a heat pump is always greater than unity since COPR is a positive quantity. That is, a heat pump will function, at worst, as a resistance heater, supplying as much energy to the house as it consumes. In reality, however, part of QH is lost to the outside air through piping and other devices, and COPHP may drop below unity when the outside air temperature is too low. When

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 11 this happens, the system usually switches to a resistance heating mode. Most heat pumps in operation today have a seasonally averaged COP of 2 to 3. Most existing heat pumps use the cold outside air as the heat source in winter, and they are referred to as air-source heat pumps. The COP of such heat pumps is about 3.0 at design conditions. Air-source heat pumps are not appropriate for cold climates since their efficiency drops considerably when temperatures are below the freezing point. In such cases, geothermal (also called ground-source) heat pumps that use the ground as the heat source can be used. Geothermal heat pumps require the burial of pipes in the ground 1 to 2 m deep. Such heat pumps are more expensive to install, but they are also more efficient (up to 45 percent more efficient than air-source heat pumps). The COP of ground-source heat pumps can be as high as 6 in the cooling mode. Air conditioners are basically refrigerators whose refrigerated space is a room or a building instead of the food compartment. A window air-conditioning unit cools a room by absorbing heat from the room air and discharging it to the outside. The same air-conditioning unit can be used as a heat pump in winter by installing it backwards. In this mode, the unit absorbs heat from the cold outside and delivers it to the room. Air-conditioning systems that are equipped with proper controls and a reversing valve operate as air conditioners in summer and as heat pumps in winter.

Example 1.2 The food compartment of a refrigerator, shown in Fig. 1.17, is maintained at 4ºC by removing heat from it at a rate of 360 kJ/min. If the required power input to the refrigerator is 2 kW, determine (a) the coefficient of performance of the refrigerator and (b) the rate of heat rejection to the room that houses the refrigerator.

Solution

Fig. 1.17

̇ ̇ ̇ ⁄

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 12

Example 1.3 A heat pump is used to meet the heating requirements of a house and maintained at 20ºC. On a day when the outdoor air temperature drops to -2ºC, the house is estimated to lose heat at a rate of 80,000 kJ/h. If the heat pump under these conditions has a COP of 2.5, determine (a) the power consumed by the heat pump and (b) the rate at which heat is absorbed from the cold outdoor air. Solution

̇ ̇ ⁄

̇ ̇ ̇ ⁄ 1.4.3 The Second Law of Thermodynamics: Clausius Statement There are two classical statements of the second law the Kelvin–Planck statement, which is related to heat engines and discussed in the preceding section, and the Clausius statement, which is related to refrigerators or heat pumps. The Clausius statement is expressed as follows: 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. It is common knowledge that heat does not, of its own volition, transfer from a cold medium to a warmer one. The Clausius statement does not imply that a cyclic device that transfers heat from a cold medium to a warmer one is impossible to construct. In fact, this is precisely what a common household refrigerator does. It simply states that a refrigerator cannot operate unless its compressor is driven by an external power source, such as an electric motor (Fig. 1.18). This way, the net effect on the surroundings involves the consumption of some energy in the form of work, in addition to the transfer of heat from a colder body to a warmer one. That is, it leaves a trace in the surroundings. Therefore, a household refrigerator is in complete Fig. 1.18 compliance with the Clausius statement of the second law.

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Both the Kelvin–Planck and the Clausius statements of the second law are negative statements, and a negative statement cannot be proved. Like any other physical law, the second law of thermodynamics is based on experimental observations. To date, no experiment has been conducted that contradicts the second law, and this should be taken as sufficient proof of its validity.

1.4.4 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. Consider the heat-engine-refrigerator combination shown in Fig. 1.19a, operating between the same two reservoirs. The heat engine is assumed to have, in violation of the Kelvin–Planck statement, a thermal efficiency of 100 percent, and therefore it converts all the heat QH it receives to work W. This work is now supplied to a refrigerator that removes heat in the amount of QL from the low-temperature reservoir and rejects heat in the amountof QL + QH to the high- temperature reservoir. During this process, the high-temperature reservoir receives a net amount of heat QL (the difference between QL+ QH and QH). Thus, the combination of these two devices can be viewed as a refrigerator, as shown in Fig. 1.19b, that transfers heat in an amount of QL from a cooler body to a warmer one without requiring any input 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.

Fig. 1.19a Fig. 1.19b

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 14

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 (Fig. 1.20). Therefore, the Clausius and the Kelvin– Planck statements are two equivalent expressions of the second law of thermodynamics.

Fig. 1.20

1.5 Perpetual Motion Machines We have repeatedly stated that 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, and despite numerous attempts, no perpetual-motion machine is known to have worked. But this has not stopped inventors from trying to create new ones. A device that violates the first law of thermodynamics (by creating energy) 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). Consider the steam power plant shown in Fig. 1.21. 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 network 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

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 15 continuously supplying energy to the outside at a rate of without receiving any energy. That is, this system 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.

Fig. 1.21 Fig. 1.22

Now let us consider another novel idea by the same inventor. Convinced that energy cannot be created, the inventor suggests the following modification that will greatly improve the thermal efficiency of that power plant without violating the first law. Aware that more than one- half of the heat transferred to the steam in the furnace is discarded in the condenser to the environment, the inventor suggests getting rid of this wasteful component and sending the steam to the pump as soon as it leaves the turbine, as shown in Fig. 1.22. This way, all the heat transferred to the steam in the boiler will be converted to work, and thus the power plant will have a theoretical efficiency of 100 percent. The inventor realizes that some heat losses and friction between the moving components are unavoidable and that these effects will hurt the efficiency somewhat, but still expects the efficiency to be no less than 80 percent (as opposed to 40 percent in most actual power plants) for a carefully designed system.

1.6 Reversible and Irreversible Processes The second law of thermodynamics states that no heat engine can have an efficiency of 100 percent. Then one may ask, what is the highest efficiency that a heat engine can possibly have? Before we can answer this question, we need to define an idealized process first, which is called the reversible process.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 16

A reversible process is defined as a process that can be reversed without leaving any trace on the surroundings (Fig. 1.23). That is, both the system and the surroundings are returned to their initial states at the end of the reverse process. Irreversible process is defined as a process that cannot be reversed without leaving any trace on the surroundings. It should be pointed out that a system can be restored to its initial state following a process, regardless of whether the process is reversible or irreversible. But for reversible processes, this restoration is made without leaving any net change on the surroundings, whereas for irreversible processes, the surroundings usually do some work on the system and therefore does not return to their original state. Reversible processes actually do not occur in nature. They are merely idealizations of actual processes. Reversible processes can Fig. 1.23 be approximated by actual devices, but they can never be achieved. That is, all the processes occurring in nature are irreversible. You may be wondering, then, why we are bothering with such fictitious processes. There are two reasons. First, they are easy to analyze, since a system passes through a series of equilibrium states during a reversible process. Second, they serve as idealized models to which actual processes can be compared. Engineers are interested in reversible processes because work-producing devices such as car engines and gas or steam turbines deliver the most work, and work-consuming devices such as compressors, fans, and pumps consume the least work when reversible processes are used instead of irreversible ones. Reversible processes can be viewed as theoretical limits for the corresponding irreversible ones. Some processes are more irreversible than others. We may never be able to have a reversible process, but we can certainly approach it. The more closely we approximate a reversible process, the more work delivered by a work-producing device or the less work required by a work-consuming device. The concept of reversible processes leads to the definition of the second law efficiency for actual processes, which is the degree of approximation to the corresponding reversible processes. This enables us to compare the performance of different devices that are designed to do the same

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 17 task on the basis of their efficiencies. The better the design, the lower the irreversibilities and the higher the second-law efficiency.

1.6.1 Irreversibilities The factors that cause a process to be irreversible are called irreversibilities. They include friction, unrestrained expansion, mixing of two fluids, heat transfer across a finite temperature difference, electric resistance, inelastic deformation of solids, and chemical reactions. The presence of any of these effects renders a process irreversible. A reversible process involves none of these. Some of the frequently encountered irreversibilities are discussed briefly below. Friction: is a familiar form of irreversibility associated with bodies in motion. When two bodies in contact are forced to move relative to each other (a piston in a cylinder, for example, as shown in Fig. 1.24), a friction force that opposes the motion develops at the interface of these two bodies, and some work is needed to overcome this friction force. The energy supplied as work is eventually converted to heat during the process and is transferred to the bodies in contact, as evidenced by a temperature rise at the interface. When the direction of the motion is reversed, the bodies are restored to their original position, but the interface does not cool, and heat is not converted back to work. Instead, more of the work is converted to heat while overcoming the friction forces that Fig. 1.24 also oppose the reverse motion. Since the system (the moving bodies) and the surroundings cannot be returned to their original states, this process is irreversible. Therefore, any process that involves friction is irreversible. The larger the friction forces involved, the more irreversible the process is.

W12= Wideal - WFr (1-13)

W21= -(Wideal + WFr) (1-14)

W12 + W21 = - 2WFr ≠ 0 (1-15)

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Unrestrained expansion: of a gas separated from a vacuum by a membrane, as shown in Fig. 1.25. When the membrane is ruptured, the gas fills the entire tank. The only way to restore the system to its original state is to compress it to its initial volume, while transferring heat from the gas until it reaches its initial temperature. From the conservation of energy considerations, it can easily be shown that the amount of heat transferred from the gas equals the amount of work done on the gas by the surroundings. The restoration of the surroundings involves conversion of this heat completely to work, which would violate Fig. 1.25 the second law. Therefore, unrestrained expansion of a gas is an irreversible process.

Heat transfer: can occur only when there is a temperature difference between a system and its surroundings. Therefore, it is physically impossible to have a reversible heat transfer process. But a heat transfer process becomes less and less irreversible as the temperature difference between the two bodies approaches zero. Then, heat transfer through a differential temperature difference dT can be considered to be reversible. As dT approaches zero, the process can be reversed in direction (at least theoretically) without requiring any refrigeration. Notice that reversible heat transfer is a conceptual process and cannot be duplicated in the real world. The smaller the temperature difference between two bodies, the smaller the heat transfer rate will be. Any significant heat transfer through a small temperature difference requires a very large surface area and a very long time. Therefore, even though approaching reversible heat transfer is desirable from a thermodynamic point of view, it is impractical and not economically feasible.

1.6.2 Internally and Externally Reversible Processes Internally Reversible process: is defined as there is no irreversibilities occur within the boundaries of the system during the process.

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Externally Reversible process: is defined as there is no irreversibilities occur outside the system boundaries during the process.

Totally Reversible process or (Simply Reversible): is defined as there are no irreversibilities within the system or its surroundings.

Fig. 1.26 1.7 The Carnot Cycle We mentioned earlier that heat engines are cyclic devices and that the working fluid of a heat engine returns to its initial state at the end of each cycle. Work is done by the working fluid during one part of the cycle and on the working fluid during another part. The difference between these two is the network delivered by the heat engine. The efficiency of a heat-engine cycle greatly depends on how the individual processes that make up the cycle are executed. The network, 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. Therefore, it is no surprise that the most efficient cycles are reversible cycles, that is, cycles that consist entirely of reversible processes. Reversible cycles cannot be achieved in practice because the irreversibilities associated with each process cannot be eliminated. However, reversible cycles provide upper limits on the performance of real cycles. Heat engines and refrigerators that work on reversible cycles serve as models to which actual heat engines and refrigerators can be compared. Reversible cycles also serve as starting points in the development of actual cycles and are modified as needed to meet certain requirements. Probably the best known reversible cycle is the Carnot cycle, first proposed in 1824 by French engineer Said Carnot. The theoretical heat engine that operates on the Carnot cycle is called the Carnot heat engine. The Carnot cycle is composed of four reversible processes two

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 20 isothermal and two adiabatic and it can be executed either in a closed or a steady flow system. The four reversible processes that make up the Carnot cycle are as follows:

1. Reversible Isothermal Expansion (process 1-2, TH = constant). 2. Reversible Adiabatic Expansion (process 2-3).

3. Reversible Isothermal Compression (process 3-4, TL = constant). 4. Reversible Adiabatic Compression (process 4-1).

Fig. 1.27 Fig. 1.28

The P-v diagram of this cycle is shown in Fig. 1.27. Remembering that on a P-v diagram the area under the process curve represents the boundary work for quasi-equilibrium (internally reversible) 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.

18.1 The Reversed Carnot Cycle The Carnot heat engine cycle just described is a totally reversible cycle. Therefore, all the processes that comprise it can be reversed, in which case it becomes the Carnot refrigeration cycle. This time, the cycle remains exactly the same, except that the directions of any heat and work interactions are reversed: Heat in the amount of QL is absorbed from the low-temperature medium, heat in the amount of QH is rejected to a high-temperature medium, and a work input of

Wnet,in is required to accomplish all this.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 21

The P-v diagram of the reversed Carnot cycle is the same as the one given for the Carnot cycle, except that the directions of the processes are reversed, as shown in Fig. 1.28.

1.8 The Carnot Principles The second law of thermodynamics puts limits on the operation of cyclic devices as expressed by the Kelvin–Planck and Clausius statements. A heat engine cannot operate by exchanging heat with a single reservoir, and a refrigerator cannot operate without a net energy input from an external source. We can draw valuable conclusions from these statements. Two conclusions pertain to the thermal efficiency of reversible and irreversible (i.e., actual) heat engines and they are known as the Carnot principles (Fig. 1.29), expressed as follows:

The First Carnot Principle: The efficiency of an irreversible heat engine is always less than the efficiency of a reversible one operating between the same two reservoirs.

Ƞth,irrev ˂ Ƞth,rev , Wirrev ˂ Wrev , QL,irrev ˃ QL.rev (1-16)

The Second Carnot Principle: The efficiencies of all reversible heat engines operating between the same two reservoirs are the same.

Ƞth.1 = Ƞth,2 , W1 = W2 , QL1 = QL2 (1-17)

Fig. 1.29

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 22

1.9 The Thermodynamic Temperature Scale A temperature scale that is independent of the properties of the substances that are used to measure temperature is called a thermodynamic temperature scale. Such a temperature scale offers great conveniences in thermodynamic calculations, and its derivation is given below using some reversible heat engines. The second Carnot principle discussed in Section 1.8 states that all reversible heat engines have the same thermal efficiency when operating between the same two reservoirs. That is, the efficiency of a reversible engine is independent of the working fluid employed and its properties, the way the cycle is executed, or the type of reversible engine used. Since energy reservoirs are characterized by their temperatures, the thermal efficiency of reversible heat engines is a function of the reservoir temperatures only. That is,

(Ƞth)rev = g(TH,TL) or = g(TH,TL) (1-18)

So ( ) (1-19)

Kelvin purposed( that:) ( )

Q(TH) = TH , Q(TL) = TL or (1-20)

This temperature scale is called the 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. On this scale, temperatures vary between zero and infinity.

1.10 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 Eq. 1–6 as

where Q is heat transferred to the heat engine from a high-temperature reservoir at T , H H and QL is heat rejected to a low-temperature reservoir at TL. For reversible heat engines, the heat transfer ratio in the above relation can be replaced by the ratio of the absolute temperatures of the

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 23 two reservoirs, as given by Eq. 1–20. Then the efficiency of a Carnot engine, or any reversible heat engine, becomes

(1-21)

This relation is often referred to as the Carnot efficiency, since the Carnot heat engine is the best known reversible engine. This is the highest efficiency a heat engine operating between the two thermal energy reservoirs at temperatures TL and TH can have. All irreversible (i.e., actual) heat engines operating between these temperature limits (TL and TH) have lower efficiencies. An actual heat engine cannot reach this maximum theoretical efficiency value because it is impossible to completely eliminate all the irreversibilities associated with the actual cycle.

Note that TL and TH in Eq. 1–21 are absolute temperatures. Using ºC or ºF for temperatures in this relation gives results grossly in error. The thermal efficiencies of actual and reversible heat engines operating between the same temperature limits compare as follows:

(1-22)

{

Example 1.4 A Carnot heat engine, receives 500 kJ of heat per cycle from a high-temperature source at 652ºC and rejects heat to a low-temperature sink at 30ºC. Determine (a) the thermal efficiency of this Carnot engine and (b) the amount of heat rejected to the sink per cycle.

Solution

( )

( ) ( ) ( ) ( ) 1.11 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 Eqs. 1–9 and 1–11 as

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 24

and

⁄ ⁄

Where QL is the amount of heat absorbed from the low-temperature medium and QH is the amount of heat rejected to the high-temperature medium. The COP’s 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 medium, as expressed by Eq. 1–21. Then the COP relations for reversible refrigerators and heat pumps become These are the highest coefficients of performance that a refrigerator or a heat pump operating between the temperature limits of TL and TH can have. All actual refrigerators or heat pumps operating between these temperature limits (TL and TH) have lower coefficients of performance.

(1-23)

And ⁄ (1-24)

The coefficients of performance of actual and reversible ⁄ refrigerators operating between the same temperature limits can be compared as follows:

(1-25)

{

A similar relation can be obtained for heat pumps by replacing all COPR’s in Eq. 1–25 by

COPHP. The COP of a reversible refrigerator or heat pump is the maximum theoretical value for the specified temperature limits. Actual refrigerators or heat pumps may approach these values as their designs are improved, but they can never reach them.

As a final note, the COPs of both the refrigerators and the heat pumps decrease as TL decreases. That is, it requires more work to absorb heat from lower-temperature medium. As the temperature of the refrigerated space approaches zero, the amount of work required to produce a finite amount of refrigeration approaches infinity and COPR approaches zero.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 25

Example 1.5 A Carnot refrigeration cycle is executed in a closed system in the saturated liquid–vapor mixture region using 0.8 kg of refrigerant-134a as the working fluid. The maximum and the minimum temperatures in the cycle are 20 and -8ºC, respectively. It is known that the refrigerant is saturated liquid at the end of the heat rejection process, and the network input to the cycle is 15 kJ. Determine the fraction of the mass of the refrigerant that vaporizes during the heat addition process, and the pressure at the end of the heat rejection process. Solution

The pressure at the end of heat rejection process is simply the saturation pressure at heat rejection temperature,

Example 1.6 A heat pump is to be used to heat a house during the winter. The house is to be maintained at 21ºC at all times. The house is estimated to be losing heat at a rate of 135,000 kJ/h when the outside temperature drops to -5ºC. Determine the minimum power required to drive this heat pump. Solution

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 26

Sheet (1)

(Second Law of Thermodynamics)

1- a) What is a thermal energy reservoir? Give some examples.

b) Is it possible for a heat engine to operate without rejecting any waste heat to a low- temperature reservoir? Explain.

c) What are the characteristics of all heat engines?

2- Heat is transferred to a heat engine from a furnace at a rate of 90 MW. If the rate of waste heat rejection to a nearby river is 60 MW, determine the net power output and the thermal efficiency for this heat engine.

3- A 600-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of heat transfer to the river water. Will the actual heat transfer rate be higher or lower than this value? Why?

4- A steam power plant receives heat from a furnace at a rate of 280 GJ/h. Heat losses to the surrounding air from the steam as it passes through the pipes and other components are estimated to be about 8 GJ/h. If the waste heat is transferred to the cooling water at a rate of 145 GJ/h, determine (a) net power output and (b) the thermal efficiency of this power plant.

5- An automobile engine consumes fuel at a rate of 28 L/h and delivers 60 kW of power to the wheels. If the fuel has a heating value of 44,000 kJ/kg and a density of 0.8 g/cm3, determine the efficiency of this engine.

6- a) What is the difference between a refrigerator and a heat pump?

b) What is the difference between a refrigerator and an air conditioner?

c) In a refrigerator, heat is transferred from a lower temperature medium (the refrigerated space) to a higher temperature one (the kitchen air). Is this a violation of the second law of thermodynamics? Explain.

d) Show that the Kelvin–Planck and the Clausius expressions of the second law are equivalent.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 27

7- A household refrigerator with a COP of 1.2 removes heat from the refrigerated space at a rate of 60 kJ/min. Determine (a) the electric power consumed by the refrigerator and (b) the rate of heat transfer to the kitchen air.

8- Determine the COP of a heat pump that supplies energy to a house at a rate of 8000 kJ/h for each kW of electric power it draws. Also, determine the rate of energy absorption from the outdoor air.

9- Determine the COP of a refrigerator that removes heat from the food compartment at a rate of 5040 kJ/h for each kW of power it consumes. Also, determine the rate of heat rejection to the outside air.

10- A heat pump with a COP of 2.5 supplies energy to a house at a rate of 60,000 kJ/h. Determine (a) the electric power drawn by the heat pump and (b) the rate of heat absorption from the outside air.

11- A Carnot heat engine operates between a source at 1000 K and a sink at 300 K. If the heat engine is supplied with heat at a rate of 800 kJ/min, determine (a) the thermal efficiency and (b) the power output of this heat engine.

12- A heat engine is operating on a Carnot cycle and has a thermal efficiency of 55 percent. The waste heat from this engine is rejected to a nearby lake at 40°C at a rate of 800 kJ/min. Determine (a) the power output of the engine and (b) the temperature of the source.

13- An inventor claims to have developed a heat engine that receives 700 kJ of heat from a source at 500 K and produces 300 kJ of network while rejecting the waste heat to a sink at 290 K. Is this a reasonable claim? Why?

14- An experimentalist claims that, based on his measurements, a heat engine receives 300 kJ of heat from a source of 900 k, converts 160 kJ of it to work, and rejects the rest as waste heat to a sink at 540 k. Are these measurements reasonable? Why?

15- A Carnot refrigerator operates in a room in which the temperature is 22°C and consumes 2 kW of power when operating. If the food compartment of the refrigerator is to be maintained at 3°C, determine the rate of heat removal from the food compartment.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 28

16- An inventor claims to have developed a refrigeration system that removes heat from the closed region at -12°C and transfers it to the surrounding air at 25°C while maintaining a COP of 6.5. Is this claim reasonable? Why?

17- During an experiment conducted in a room at 25°C, a laboratory assistant measures that a refrigerator that draws 2 kW of power has removed 30,000 kJ of heat from the refrigerated space, which is maintained at -30°C. The running time of the refrigerator during the experiment was 20 min. Determine if these measurements are reasonable.

18- A Carnot heat engine receives heat from a reservoir at 900°C at a rate of 800 kJ/min and rejects the waste heat to the ambient air at 27°C. The entire work output of the heat engine is used to drive a refrigerator that removes heat from the refrigerated space at -5°C and transfers it to the same ambient air at 27°C. Determine (a) the maximum rate of heat removal from the refrigerated space and (b) the total rate of heat rejection to the ambient air.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 29

Chapter (2)

Entropy

2.1 Entropy The second law of thermodynamics often leads to expressions that involve inequalities. An irreversible (i.e., actual) heat engine, for example, is less efficient than a reversible one operating between the same two thermal energy reservoirs. Likewise, an irreversible refrigerator or a heat pump has a lower coefficient of performance (COP) than a reversible one operating between the same temperature limits. Another important inequality that has major consequences in thermodynamics is the Clausius inequality. It was first stated by the German physicist R. J. E. Clausius (1822–1888), one of the founders of thermodynamics, and is expressed in 1865 as

That is, the cyclic integral of δQ/T is always less than or equal to zero. This inequality is valid for all cycles, reversible or irreversible. The symbol (integral symbol with a circle in the middle) is used to indicate that the integration is to be performed over the entire cycle. Any heat transfer to or from a system can be considered to consist of differential amounts of heat transfer. Then the cyclic integral of δQ/T can be viewed as the sum of all these differential amounts of heat transfer divided by the temperature at the boundary. To demonstrate the validity of the Clausius inequality, consider a system connected to a thermal energy reservoir at a constant thermodynamic (i.e., absolute) temperature of TR through a reversible cyclic device (Fig. 2.1). The cyclic device receives heat δQR from the reservoir and supplies heat δQ to the system whose temperature at that part of the boundary is T (a variable) while producing work δWrev. The system produces work δWsys as a result of this heat transfer.

Applying the energy balance to the combined system Fig. 2.1 identified by dashed lines yields

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 30

Where δWC is the total work of the combined system (δWrev + δWsys) and dEC is the change in the total energy of the combined system. Considering that the cyclic device is a reversible one, we have

Where the sign of δQ is determined with respect to the system (positive if to the system and negative if from the system) and the sign of δQR is determined with respect to the reversible cyclic device. Eliminating δQR from the two relations above yields

We now let the system undergo a cycle while the cyclic device undergoes an integral number of cycles. Then the preceding relation becomes

Since the cyclic integral of energy (the net change in the energy, which is a property, during a cycle) is zero. Here WC is the cyclic integral of δWC, and it represents the network for the combined cycle.

It appears that the combined system is exchanging heat with a single thermal energy reservoir while involving (producing or consuming) work WC during a cycle. On the basis of the Kelvin–Planck statement of the second law, which states that no system can produce a net amount of work while operating in a cycle and exchanging heat with a single thermal energy reservoir, we reason that WC cannot be a work output, and thus it cannot be a positive quantity.

Considering that TR is the thermodynamic temperature and thus a positive quantity, we must have

(2-1)

Which is the Clausius inequality. This inequality is valid for all thermodynamic cycles, reversible or irreversible, including the refrigeration cycles. If no irreversibilities occur

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 31 within the system as well as the reversible cyclic device, then the cycle undergone by the combined system is internally reversible. As such, it can be reversed. In the reversed cycle case, all the quantities have the same magnitude but the opposite sign.

Therefore, the work WC, which could not be a positive quantity in the regular case, cannot be a negative quantity in the reversed case. Then it follows that WC,int,rev = 0 since it cannot be a positive or negative quantity, and therefore

(2-2)

( ) for internally reversible cycles. Thus, we conclude that the equality in the Clausius inequality holds for totally or just internally reversible cycles and the inequality for the irreversible ones. To develop a relation for the definition of entropy, let us examine Eq. 2–2 more closely. Here we have a quantity whose cyclic integral is zero. Let us think for a moment what kind of quantities can have this characteristic. We know that the cyclic integral of work is not zero. (It is a good thing that it is not. Otherwise, heat engines that work on a cycle such as steam power plants would produce zero network.) Neither is the cyclic integral of heat. Now consider the volume occupied by a gas in a piston–cylinder device undergoing a cycle, as Fig. 2.2 shown in Fig. 2.2. When the piston returns to its initial position at the end of a cycle, the volume of the gas also returns to its initial value. Thus, the net change in volume during a cycle is zero. This is also expressed as (2-3) That is, the cyclic integral of volume (or any other property) is zero. Conversely, a quantity whose cyclic integral is zero depends on the state only and not the process path, and thus it is a property. Therefore, the quantity (δQ/T)int rev must represent a property in the differential form. Clausius realized in 1865 that he had discovered a new thermodynamic property, and he chose to name this property entropy. It is designated S and is defined as

(2-4)

( ) ( ⁄ ) Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 32

Entropy is an extensive property of a system and sometimes is referred to as total entropy. Entropy per unit mass, designated s, is an intensive property and has the unit kJ/kg·K. The term entropy is generally used to refer to both total entropy and entropy per unit mass since the context usually clarifies which one is meant. The entropy change of a system during a process can be determined by integrating Eq. 2–4 between the initial and the final states: (2-5)

Notice that we have actually defined the∫ change( ) in entropy instead ( ⁄of )entropy itself, just as we defined the change in energy instead of the energy itself when we developed the first-law relation. Absolute values of entropy are determined on the basis of the third law of thermodynamics, which is discussed later in this chapter. Engineers are usually concerned with the changes in entropy. Therefore, the entropy of a substance can be assigned a zero value at some arbitrarily selected reference state, and the entropy values at other states can be determined from Eq. 2–5 by choosing state 1 to be the reference state (S = 0) and state 2 to be the state at which entropy is to be determined. To perform the integration in Eq. 2–5, one needs to know the relation between Q and T during a process. This relation is often not available, and the integral in Eq. 2–5 can be performed for a few cases only. For the majority of cases we have to rely on tabulated data for entropy. Note that entropy is a property, and like all other properties, it has fixed values at fixed states. Therefore, the entropy change ∆S between two specified states is the same no matter what path, reversible or irreversible, is followed during a process (Fig. 2.3). Also note that the integral of δQ/T gives us the value of entropy change only if the integration is carried out along an internally reversible path between the two states. The integral of δQ/T along an irreversible path is not a property, and in general, different values will be obtained when the integration is carried out along different Fig. 2.3 irreversible paths. Therefore, even for irreversible processes, the entropy change should be determined by carrying out this integration along some convenient imaginary internally reversible path between the specified states.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 33

A Special Case: Internally Reversible Isothermal Heat Transfer Processes Recall that isothermal heat transfer processes are internally reversible. Therefore, the entropy change of a system during an internally reversible isothermal heat transfer process can be determined by performing the integration in Eq. 2–5:

Which reduces to (2-6)

( ⁄ )

Where T0 is the constant temperature of the system and Q is the heat transfer for the internally reversible process. Equation 2–6 is particularly useful for determining the entropy changes of thermal energy reservoirs that can absorb or supply heat indefinitely at a constant temperature. Notice that the entropy change of a system during an internally reversible isothermal process can be positive or negative, depending on the direction of heat transfer. Heat transfer to a system increases the entropy of a system, whereas heat transfer from a system decreases it. In fact, losing heat is the only way the entropy of a system can be decreased.

Example 2.1 A piston–cylinder device contains a liquid–vapor mixture of water at 300 K. During a constant- pressure process, 750 kJ of heat is transferred to the water. As a result, part of the liquid in the cylinder vaporizes. Determine the entropy change of the water during this process.

Solution

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 34

2.2 The Increase of Entropy Principle Consider a cycle that is made up of two processes: process 1-2, which is arbitrary (reversible or irreversible), and process 2-1, which is internally reversible, as shown in Figure 2.4. From the Clausius inequality,

Or

Fig. 2.4

The second integral in the previous relation is recognized as the entropy change S1 - S2. Therefore,

Which can be rearranged as

(2-7)

∫ It can also be expressed in differential form as

(2-8)

Where the equality holds for an internally reversible process and the inequality for an irreversible process. We may conclude from these equations that the entropy change of a closed system during an irreversible process is greater than the integral of δQ/T evaluated for that process. In the limiting case of a reversible process, these two quantities become equal. We again emphasize that T in these relations is the thermodynamic temperature at the boundary where the differential heat δQ is transferred between the system and the surroundings.

The quantity ∆S = S2 - S1 represents the entropy change of the system. For a reversible process, it becomes equal to ∫ δQ/T, which represents the entropy transfer with heat.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 35

The inequality sign in the preceding relations is a constant reminder that the entropy change of a closed system during an irreversible process is always greater than the entropy transfer. That is, some entropy is generated or created during an irreversible process, and this generation is due entirely to the presence of irreversibilities. The entropy generated during a process iscalled entropy generation and is denoted by Sgen. Noting that the difference between the entropy change of a closed system and the entropy transfer is equal to entropy generation, Eq. 2–7 can be rewritten as an equality as

(2-9)

∫

Note that the entropy generation Sgen is always a positive quantity or zero. Its value depends on the process, and thus it is not a property of the system. Also, in the absence of any entropy transfer, the entropy change of a system is equal to the entropy generation. Equation 2–7 has far-reaching implications in thermodynamics. For an isolated system (or simply an adiabatic closed system), the heat transfer is zero, and Eq. 2–7 reduces to (2-10)

This equation can be expressed as the entropy of an isolated system during a process always increases or, in the limiting case of a reversible process, remains constant. In other words, it never decreases. This is known as the increase of entropy principle. Note that in the absence of any heat transfer, entropy change is due to irreversibilities only, and their effect Fig. 2.5 is always to increase entropy. Entropy is an extensive property, thus the total entropy of a system is equal to the sum of the entropies of the parts of the system. An isolated system may consist of any number of subsystems (Fig. 2.5). A system and its surroundings, for example, constitute an isolated system since both can be enclosed by a sufficiently large arbitrary boundary across which there is no heat, work, or mass transfer (Fig. 2.6). Fig. 2.6 Therefore, a system and its surroundings can be viewed as the two subsystems of an isolated

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 36 system, and the entropy change of this isolated system during a process is the sum of the entropy changes of the system and its surroundings, which is equal to the entropy generation since an isolated system involves no entropy transfer. That is,

(2-11)

Where the equality holds for reversible processes and the inequality for irreversible ones. Note that ∆Ssurr refers to the change in the entropy of the surroundings as a result of the occurrence of the process under consideration. Since no actual process is truly reversible, we can conclude that some entropy is generated during a process, and therefore the entropy of the universe, which can be considered to be an isolated system, is continuously increasing. The more irreversible a process, the larger the entropy generated during that process. No entropy is generated during reversible processes (Sgen = 0). Entropy increase of the universe is a major concern not only to engineers but also to philosophers, theologians, economists, and environmentalists since entropy is viewed as a measure of the disorder (or “mixed-up-ness”) in the universe. The increase of entropy principle does not imply that the entropy of a system cannot decrease. The entropy change of a system can be negative during a process, but entropy generation cannot. The increase of entropy principle can be summarized as follows:

This relation serves as a criterion in determining whether a process is reversible, irreversible, or impossible. Things in nature have a tendency to change until they attain a state of equilibrium. The increase of entropy principle dictates that the entropy of an isolated system increases until the entropy of the system reaches a maximum value. At that point, the system is said to have reached an equilibrium state since the increase of entropy principle prohibits the system from undergoing any change of state that results in a decrease in entropy.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 37

Example 2.2 A heat source at 800 K loses 2000 kJ of heat to a sink at (a) 500 K and (b) 750 K. Determine which heat transfer process is more irreversible. Solution (a) For the heat transfer process to a sink at 500 K:

And

(b) Repeating the calculations in part (a) for a sink temperature of 750 K, we obtain

and

The total entropy change for the process in part (b) is smaller, and therefore it is less irreversible. This is expected since the process in (b) involves a smaller temperature difference and thus a smaller irreversibility.

2.3 Entropy Change of Pure Substance Entropy is a property, and thus the value of entropy of a system is fixed once the state of the system is fixed. Specifying two intensive independent properties fixes the state of a simple compressible system, and thus the value of entropy and of other properties at that state. Starting with its defining relation, the entropy change of a substance can be expressed in terms of other properties (see Sec. 2.7). But in general, these relations are too complicated and are not practical to use for hand calculations. Therefore, using a Fig. 2.7

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 38 suitable reference state, the entropies of substances are evaluated from measurable property data following rather involved computations, and the results are tabulated in the same manner as the other properties such as v, u, and h (Fig. 2.7). The entropy values in the property tables are given relative to an arbitrary reference state. In steam tables the entropy of saturated liquid sf at 0.01ºC is assigned the value of zero. For refrigerant-134a, the zero value is assigned to saturated liquid at -40ºC. The entropy values become negative at temperatures below the reference value. The value of entropy at a specified state is determined just like any other property. In the compressed liquid and superheated vapor regions, it can be obtained directly from the tables at the specified state. In the saturated mixture region, it is determined from

Where x is the quality and sf and sfg values are listed in the saturation tables. In the absence of compressed liquid data, the entropy of the compressed liquid can be approximated by the entropy of the saturated liquid at the given temperature:

The entropy change of a specified mass m (a closed system) during a process is simply (2-12)

( ) ( ⁄ ) This is the difference between the entropy values at the final and initial states. When studying the second-law aspects of processes, entropy is commonly used as a coordinate on diagrams such as the T-s and h-s diagrams. The general characteristics of the T-s diagram of pure substances are shown in Fig. 2.8 using data for water. Notice from this diagram that the constant-volume lines are steeper than the constant-pressure lines and the constant pressure lines are parallel to the constant- Fig. 2.8 temperature lines in the saturated liquid–vapor mixture region. Also, the constant-pressure lines almost coincide with the saturated liquid line in the compressed liquid region.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 39

Example 2.3 A rigid tank contains 5 kg of refrigerant-134a initially at 20ºC and 140 kPa. The refrigerant is now cooled while being stirred until its pressure drops to 100 kPa. Determine the entropy change of the refrigerant during this process.

Solution

The refrigerant is a saturated liquid–vapor mixture at the final state since vf ˂ v2 ˂ vg at 100 kPa pressure. Therefore, we need to determine the quality first:

Thus,

Then, the entropy change of the refrigerant during this process is

The negative sign indicates that the entropy of the system is decreasing during this process. This is not a violation of the second law, however, since it is the entropy generation Sgen that cannot be negative.

2.4 Isentropic Processes We mentioned earlier that the entropy of a fixed mass can be changed by (1) heat transfer and (2) irreversibilities. Then it follows that the entropy of a fixed mass does not change during a process that is internally reversible and adiabatic (Fig. 2.9). A process during which the entropy remains constant is called an isentropic process. It is characterized by

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 40

Isentropic process: (2-13)

( ⁄ ) That is, a substance will have the same entropy value at the end of the process as it does at the beginning if the process is carried out in an isentropic manner. Many engineering systems or devices such as pumps, turbines, nozzles, and diffusers are essentially adiabatic in their operation, and they perform best when the irreversibilities, such as the friction associated with the process, are minimized. Therefore, an isentropic process can serve as an appropriate model for actual processes. Also, isentropic processes enable us to define efficiencies for processes to compare the actual performance of these devices Fig. 2.9 to the performance under idealized conditions.

It should be recognized that a reversible adiabatic process is necessarily isentropic (S2 =

S1), but an isentropic process is not necessarily a reversible adiabatic process. (The entropy increase of a substance during a process as a result of irreversibilities may be offset by a decrease in entropy as a result of heat losses, for example.) However, the term isentropic process is customarily used in thermodynamics to imply an internally reversible, adiabatic process.

Example 2.4 Steam enters an adiabatic turbine at 5 MPa and 450ºC and leaves at a pressure of 1.4 MPa. Determine the work output of the turbine per unit mass of steam if the process is reversible.

Solution

The inlet state is completely specified since two properties are given. But only one property (pressure) is given at the final state, and we need one more property to fix it. The second

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 41 property comes from the observation that the process is reversible and adiabatic, and thus isentropic. Therefore, S2 = S1, and

Then, the work output of the turbine per unit mass of the steam becomes

2.5 Property Diagram Involving Entropy Property diagrams serve as great visual aids in the thermodynamic analysis of processes. We have used P-v and T-v diagrams extensively in previous course in conjunction with the first law of thermodynamics. In the second law analysis, it is very helpful to plot the processes on diagrams for which one of the coordinates is entropy. The two diagrams commonly used in the second-law analysis are the temperature-entropy and the enthalpy- entropy diagrams. Fig. 2.10 Consider the defining equation of entropy (Eq. 2–4). It can be rearranged as

(2-14)

T-S As shown in Fig. 2.10, δQint rev corresponds to a differential ( ) area on a diagram. The total heat transfer during an internally reversible process is determined by integration to be

(2-15)

This corresponds to the area under ∫ the process (curve ) on a T-S diagram. Therefore, we conclude that the area under the process curve on a T-S diagram represents heat transfer during an internally reversible process. This is somewhat analogous to reversible boundary work being represented by the area under the process curve on a P-v diagram. Note that the area under the

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 42 process curve represents heat transfer for processes that are internally (or totally) reversible. The area has no meaning for irreversible processes. Equations 2–14 and 2–15 can also be expressed on a unit-mass basis as

(2-16) and ( ⁄ ) (2-17)

∫ ( ⁄ ) To perform the integrations in Eqs. 2–15 and 2–17, one needs to know the relationship between T and s during a process. One special case for which these integrations can be performed easily is the internally reversible isothermal process. It yields

(2-18)

Or ( ) (2-19)

( ⁄ ) Where T0 is the constant temperature and ∆S is the entropy change of the system during the process. An isentropic process on a T-s diagram is easily recognized as a vertical-line segment. This is expected since an isentropic process involves no heat transfer, and therefore the area under the process path must be zero (Fig. 2.11). The T-s diagrams serve as valuable tools for visualizing the second-law aspects of processes and cycles, and thus they are frequently used in thermodynamics. Another diagram commonly used in engineering is the enthalpy-entropy diagram, which is quite valuable in the analysis of steady-flow devices such as turbines, compressors, and nozzles. The coordinates of an h-s diagram represent two properties of major interest: enthalpy, which is a primary property in the first-law analysis of the steady-flow devices, and entropy, which is the property that accounts for irreversibilities during adiabatic processes. In analyzing the steady flow of steam through an adiabatic turbine, for example, the vertical distance between the inlet and the exit states ∆h is a measure of the work output of the turbine, and the horizontal distance ∆s is a measure of the irreversibilities associated with the process (Fig. 2.12).

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 43

Fig. 2.11 Fig. 2.12

The h-s diagram is also called a Mollier diagram after the German scientist R. Mollier (1863– 1935).

The T-S Diagram of Carnot Cycle

The Carnot cycle is to be shown on a T-S diagram, and the areas that represent QH, QL, and

Wnet,out are to be indicated. Recall that the Carnot cycle is made up of two reversible isothermal (T = constant) processes and two isentropic (S = constant) processes. These four processes form a rectangle on a T-S diagram, as shown in Fig. 2.13. On a T-S diagram, the area under the process curve represents the heat transfer for that process. Thus the area A12B represents

QH, the area A43B represents QL, and the difference between these two (the area in color) represents the network since Fig. 2.13

Therefore, the area enclosed by the path of a cycle (area 1234) on a T-S diagram represents the network. Recall that the area enclosed by the path of a cycle also represents the net work on a P-v diagram.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 44

2.6 What is Entropy? It is clear from the previous discussion that entropy is a useful property and serves as a valuable tool in the second-law analysis of engineering devices. But this does not mean that we know and understand entropy well. Because we do not. In fact, we cannot even give an adequate answer to the question, what is entropy? Not being able to describe entropy fully, however, does not take anything away from its usefulness. We could not define energy either, but it did not interfere with our understanding of energy transformations and the conservation of energy principle. Granted, entropy is not a household word like energy. But with continued use, our understanding of entropy will deepen, and our appreciation of it will grow. The next discussion should shed some light on the physical meaning of entropy by considering the microscopic nature of matter. Entropy can be viewed as a measure of molecular disorder, or molecular randomness. Third law of thermodynamics: the entropy of a pure crystalline substance at absolute zero temperature is zero. The third law of thermodynamics provides an absolute reference point for the determination of entropy. The entropy determined relative to this point is called absolute entropy, and it is extremely useful in the thermodynamic analysis of chemical reactions.

2.7 The T ds Relations

Recall that the quantity (δQ/T)int rev corresponds to a differential change in the property entropy. The entropy change for a process, then, can be evaluated by integrating δQ/T along some imaginary internally reversible path between the actual end states. For isothermal internally reversible processes, this integration is straightforward. But when the temperature varies during the process, we have to have a relation between δQ and T to perform this integration. Finding such relations is what we intend to do in this section. The differential form of the conservation of energy equation for a closed stationary system (a fixed mass) containing a simple compressible substance can be expressed for an internally reversible process as

(2-20)

But

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 45

Thus,

(2-21)

Or ( ) (2-22)

( ⁄ ) This equation is known as the first T ds, or Gibbs, equation. Notice that the only type of work interaction a simple compressible system may involve as it undergoes an internally reversible process is the boundary work. The second T ds equation is obtained by eliminating du from Eq. 2–22 by using the definition of enthalpy (h = u + Pv):

(2-23)

}

( ) Equations 2–22 and 2–23 are extremely valuable since they relate entropy changes of a system to the changes in other properties. Unlike Eq. 2–4, they are property relations and therefore are independent of the type of the processes. These Tds relations are developed with an internally reversible process in mind since the entropy change between two states must be evaluated along a reversible path. However, the results obtained are valid for both reversible and irreversible processes since entropy is a property and the change in a property between two states is independent of the type of process the system undergoes. Equations 2–22 and 2–23 are relations between the properties of a unit mass of a simple compressible system as it undergoes a change of state, and they are applicable whether the change occurs in a closed or an open system. Explicit relations for differential changes in entropy are obtained by solving for ds in Eqs. 2– 22 and 2–23:

(2-24) and

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 46

(2-25)

The entropy change during a process can be determined by integrating either of these equation between the initial and the final states. To perform these integrations, however, we must know the relationship between du or dh and the temperature (such as du = cv dT and dh = cp dT for ideal gases) as well as the equation of state for the substance (such as the ideal-gas equation of state Pv = RT). For substances for which such relations exist, the integration of Eq. 2–24 or 2–25 is straightforward. For other substances, we have to rely on tabulated data. The T ds relations for non-simple systems, that is, systems that involve more than one mode of quasi-equilibrium work, can be obtained in a similar manner by including all the relevant quasi-equilibrium work modes.

2.8 Entropy Change of Liquids and Solids Recall that liquids and solids can be approximated as incompressible substances since their specific volumes remain nearly constant during a process. Thus, dv ≈ 0 for liquids and solids, and Eq. 2–24 for this case reduces to (2-26)

Since cp = cv = c and du = c dT for incompressible substances. Then the entropy change during a process is determined by integration to be

Liquids, Solids (2-27)

∫ ( ) ( ⁄ )

Where cavg is the average specific heat of the substance over the given temperature interval. Note that the entropy change of a truly incompressible substance depends on temperature only and is independent of pressure. Equation 2–27 can be used to determine the entropy changes of solids and liquids with reasonable accuracy. However, for liquids that expand considerably with temperature, it may be necessary to consider the effects of volume change in calculations. This is especially the case when the temperature change is large.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 47

A relation for isentropic processes of liquids and solids is obtained by setting the entropy change relation above equal to zero. It gives

Isentropic: (2-28)

Example 2.5 Liquid methane is commonly used in various cryogenic applications. The critical temperature of methane is 191 K (or -82ºC), and thus methane must be maintained below 191 K to keep it in liquid phase. Determine the entropy change of liquid methane as it undergoes a process from 110 K and 1 MPa to 120 K and 5 MPa (a) using tabulated properties and (b) approximating liquid methane as an incompressible substance. What is the error involved in the latter case?

Solution

Therefore,

(b) Approximating liquid methane as an incompressible substance, its entropy change is determined to be

Since

Therefore, the error involved in approximating liquid methane as an incompressible substance is

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 48

2.9 The Entropy Change of Ideal Gases An expression for the entropy change of an ideal gas can be obtained from Eq. 2–24 or 2–25 by employing the property relations for ideal gases. By substituting du = cv dT and P = RT/v into Eq. 2–24, the differential entropy change of an ideal gas becomes

(2-29)

The entropy change for a process is obtained by integrating this relation between the end states:

(2-30)

∫ ( ) A second relation for the entropy change of an ideal gas is obtained in a similar manner by substituting dh = cp dT and v = RT/P into Eq. 2–25 and integrating. The result is

(2-31)

∫ ( ) The specific heats of ideal gases, with the exception of monatomic gases, depend on temperature, and the integrals in Eqs. 2–30 and 2–31 cannot be performed unless the dependence of cv and cp on temperature is known.

Even when the cv(T) and cp(T) functions are available, performing long integrations every time entropy change is calculated is not practical. Then two reasonable choices are left: either perform these integrations by simply assuming constant specific heats or evaluate those integrals once and tabulate the results. Both approaches are presented next. a) Constant Specific Heats (Approximate Analysis) Assuming constant specific heats for ideal gases is a common approximation, and we used this assumption before on several occasions. It usually simplifies the analysis greatly, and the price we pay for this convenience is some loss in accuracy. The magnitude of the error introduced by this assumption depends on the situation at hand. For example, for monatomic ideal gases such as helium, the specific heats are independent of temperature, and therefore the

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 49 constant-specific-heat assumption introduces no error. For ideal gases whose specific heats vary almost linearly in the temperature range of interest, the possible error is minimized by using specific heat values evaluated at the average temperature (Fig. 2.14). The results obtained in this way usually are sufficiently accurate if the temperature range is not greater than a few hundred degrees.

The entropy change relations for ideal gases under the constant specific heat assumption are easily obtained by Fig. 2.14 replacing cv(T) and cp(T) in Eqs. 2–30 and 2–31 by cv,avg and cp,avg, respectively, and performing the integrations. We obtain

(2-32) and ( ⁄ ) (2-33)

( ⁄ ) Entropy changes can also be expressed on a unit-mole basis by multiplying these relations by molar mass:

(2-34) and ̅ ̅ ̅ ( ⁄ ) (2-35)

̅ ̅ ̅ ( ⁄ ) b) Variable Specific Heats (Exact Analysis When the temperature change during a process is large and the specific heats of the ideal gas vary nonlinearly within the temperature range, the assumption of constant specific heats may lead to considerable errors in entropy change calculations. For those cases, the variation of specific heat with temperature should be properly accounted for by utilizing accurate relations for the specific heats as a function of temperature. The entropy change during a process is then

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 50 determined by substituting these cv(T) or cp(T) relations into Eq. 2–30 or 2–31 and performing the integrations. Instead of performing these laborious integrals each time we have a new process, it is convenient to perform these integrals once and tabulate the results. For this purpose, we choose absolute zero as the reference temperature and define a function sº as

(2-36)

∫ ( ) Obviously, sº is a function of temperature alone, and its value is zero at absolute zero temperature. The values of sº are calculated at various temperatures, and the results are tabulated in the appendix as a function of temperature for air. Given this definition, the integral in Eq. 2– 31 becomes

(2-37)

∫ ( )

Where is the value of sº at T2 and is the value at T1. Thus,

(2-38)

( ⁄ ) It can also be expressed on a unit-mole basis as

(2-39)

̅ ̅ ̅ ̅ ( ⁄ ) Note that unlike internal energy and enthalpy, the entropy of an ideal gas varies with specific volume or pressure as well as the temperature. Therefore, entropy cannot be tabulated as a function of temperature alone. The sº values in the tables account for the temperature dependence of entropy. The variation of entropy with pressure is accounted for by the last term in Eq. 2–38. Another relation for entropy change can be developed based on Eq. 2–30, but this would require the definition of another function and tabulation of its values, which is not practical.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 51

Example 2.6 Air is compressed from an initial state of 100 kPa and 17ºC to a final state of 600 kPa and 57ºC. Determine the entropy change of air during this compression process by using (a) property values from the air table and (b) average specific heats.

Solution (a) The properties of air are given in the air table (Table A–17). Reading sº values at given temperatures and substituting, we find

(b) The entropy change of air during this process can also be

determined approximately from Eq. 2–33 by using a cp value at the average temperature of 37ºC (Table A–2b) and treating it as a constant:

2.9.1 Isentropic Processes of Ideal Gases Several relations for the isentropic processes of ideal gases can be obtained by setting the entropy change relations developed previously equal to zero. Again, this is done first for the case of constant specific heats and then for the case of variable specific heats. a) Constant Specific Heats (Approximate Analysis) When the constant-specific-heat assumption is valid, the isentropic relations for ideal gases are obtained by setting Eqs. 2–32 and 2–33 equal to zero. From Eq. 2–32,

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 52

Which can be rearranged as

(2-40)

⁄ or ( )

(ideal gas) (2-41)

( ) ( )

Since R = cp - cv , k = cp /cv , and thus R/cv = k - 1. Equation 2–41 is the first isentropic relation for ideal gases under the constant specific heat assumption. The second isentropic relation is obtained in a similar manner from Eq. 2–33 with the following result:

(ideal gas) (2-42) ⁄

The third isentropic relation is (obtained ) by substituting ( ) Eq. 2–42 into Eq. 2–41 and simplifying:

(ideal gas) (2-43)

( ) ( ) Equations 2–41 through 2–43 can also be expressed in a compact form as (2-44)

(ideal gas) (2-45) ( )⁄ (2-46)

The specific heat ratio k, in general, varies with temperature, and thus an average k value for the given temperature range should be used. Note that the ideal-gas isentropic relations above, as the name implies, are strictly valid for isentropic processes only when the constant specific heat assumption is appropriate. b) Variable Specific Heats (Exact Analysis) When the constant-specific-heat assumption is not appropriate, the isentropic relations developed previously yields results that are not quite accurate. For such cases, we should use an isentropic relation obtained from Eq. 2–38 that accounts for the variation of specific heats with temperature. Setting this equation equal to zero gives

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 53

or

(2-47) where is the value at the end of the isentropic process.

Relative Pressure and Relative Specific Volume Equation 2–47 provides an accurate way of evaluating property changes of ideal gases during isentropic processes since it accounts for the variation of specific heats with temperature. However, it involves tedious iterations when the volume ratio is given instead of the pressure ratio. This is quite an inconvenience in optimization studies, which usually require numerous repetitive calculations. To remedy this deficiency, we define two new dimensionless quantities associated with isentropic processes. The definition of the first is based on Eq. 2–47, which can be re arranged as

or

The quantity exp(sº/R) is defined as the relative pressure Pr. With this definition, the last relation becomes

(2-48)

( )

Note that the relative pressure Pr is a dimensionless quantity that is a function of temperature only since sº depends on temperature alone. Therefore, values of Pr can be tabulated against temperature. This is done for air in Table A–17. Sometimes specific volume ratios are given instead of pressure ratios. This is particularly the case when automotive engines are analyzed. In such cases, one needs to work with volume ratios. Therefore, we define another quantity related to specific volume ratios for isentropic processes. This is done by utilizing the ideal-gas relation and Eq. 2–48:

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 54

The quantity T/Pr is a function of temperature only and is defined as relative specific volume vr. Thus,

(2-49)

Equations 2–48 and 2–49 are( strictly) valid for isentropic processes of ideal gases only. They account for the variation of specific heats with temperature and therefore give more accurate results than Eqs. 2–41 through 2–46. The values of Pr and vr are listed for air in Table A–17.

Example 2.7 Air is compressed in a car engine from 22ºC and 95 kPa in a reversible and adiabatic manner. If the compression ratio V1/V2 of this engine is 8, determine the final temperature of the air.

Solution

Alternative Solution The final temperature could also be determined from Eq. 2–41 by assuming constant specific heats for air:

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Sheet (2)

(Entropy)

1- An insulated piston–cylinder device contains 5 L of saturated liquid water at a constant pressure of 150 kPa. An electric resistance heater inside the cylinder is now turned on, and 2200 kJ of energy is transferred to the steam. Determine the entropy change of the water during this process.

2- A rigid tank is divided into two equal parts by a partition. One part of the tank contains 1.5 kg of compressed liquid water at 300 kPa and 60°C while the other part is evacuated. The partition is now removed, and the water expands to fill the entire tank. Determine the entropy change of water during this process, if the final pressure in the tank is 15 kPa.

3- A 0.5-m3 rigid tank contains refrigerant-134a initially at 200 kPa and 40 percent quality. Heat is transferred now to the refrigerant from a source at 35°C until the pressure rises to 400 kPa. Determine (a) the entropy change of the refrigerant, (b) the entropy change of the heat source, and (c) the total entropy change for this process.

4- A well-insulated rigid tank contains 2 kg of a saturated liquid–vapor mixture of water at 100 kPa. Initially, three-quarters of the mass is in the liquid phase. An electric resistance heater placed in the tank is now turned on and kept on until all the liquid in the tank is vaporized. Determine the entropy change of the steam during this process.

5- A heavily insulated piston–cylinder device contains 0.05 m3 of steam at 300 kPa and 150°C. Steam is now compressed in a reversible manner to a pressure of 1 MPa. Determine the work done on the steam during this process.

6- An insulated piston–cylinder device contains 0.05 m3 of saturated refrigerant-134a vapor at 0.8-MPa pressure. The refrigerant is now allowed to expand in a reversible manner until the

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 56

pressure drops to 0.4 MPa. Determine (a) the final temperature in the cylinder and (b) the work done by the refrigerant.

7- A rigid tank contains 5 kg of saturated vapor steam at 100°C. The steam is cooled to the ambient temperature of 25°C. (a) Sketch the process with respect to the saturation lines on a T-v diagram. (b) Determine the entropy change of the steam, in kJ/K.

(c) For the steam and its surroundings, determine the total entropy change or Sgen associated with this process, in kJ/K.

8- An insulated piston–cylinder device initially contains 300 L of air at 120 kPa and 17°C. Air is now heated for 15 min by a 200-W resistance heater placed inside the cylinder. The pressure of air is maintained constant during this process. Determine the entropy change of air, assuming (a) constant specific heats and (b) variable specific heats.

9- Air is compressed in a piston–cylinder device from 100 kPa and 17°C to 800 kPa in a reversible, adiabatic process. Determine the final temperature and the work done during this process, assuming (a) constant specific heats and (b) variable specific heats for air.

10- Oxygen gas is compressed in a piston–cylinder device from an initial state of 0.8 m3/kg and 25°C to a final state of 0.1 m3/kg and 287°C. Determine the entropy change of the oxygen during this process. Assume constant specific heats.

11- A piston–cylinder device contains 1.2 kg of nitrogen gas at 120 kPa and 27°C. The gas is now compressed slowly in a polytropic process during which PV1.3 = constant. The process ends when the volume is reduced by one-half. Determine the entropy change of nitrogen during this process.

12- An insulated rigid tank is divided into two equal parts by a partition. Initially, one part contains 5 kmol of an ideal gas at 250 kPa and 40°C, and the other side is evacuated. The

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 57

partition is now removed, and the gas fills the entire tank. Determine the total entropy change during this process.

13- Air enters a nozzle steadily at 280 kPa and 77°C with a velocity of 50 m/s and exits at 85 kPa and 320 m/s. The heat losses from the nozzle to the surrounding medium at 20°C are estimated to be 3.2 kJ/kg. Determine (a) the exit temperature and (b) the total entropy change for this process.

14- Air at 800 kPa and 400°C enters a steady-flow nozzle with a low velocity and leaves at 100 kPa. If the air undergoes an adiabatic expansion process through the nozzle, what is the maximum velocity of the air at the nozzle exit, in m/s?

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 58

2.10 Reversible Steady flow Work The work done during a process depends on the path followed as well as on the properties at the end states. Recall that reversible (quasi-equilibrium) moving boundary work associated with closed systems is expressed in terms of the fluid properties as

We mentioned that the quasi-equilibrium work interactions lead to the maximum work output for work producing devices and the minimum work input for work consuming devices. It would also be very insightful to express the work associated with steady flow devices in terms of fluid properties. Taking the positive direction of work to be from the system (work output), the energy balance for a steady-flow device undergoing an internally reversible process can be expressed in differential form as

But

( ) } Substituting this into the relation above (and canceling ) dh yield

Integrating, we find

(KJ/Kg) (2-50)

When the changes in kinetic ∫and potential energies are negligible, this equation reduces to

(KJ/Kg) (2-51)

∫ Equations 2–50 and 2–51 are relations for the reversible work output associated with an internally reversible process in a steady flow device. They will give a negative result when work is done on the system. To avoid the negative sign, Eq. 2–50 can be written for work input to steady-flow devices such as compressors and pumps as

(2-52)

∫ Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 59

The resemblance between the vdP in these relations and Pdv is striking. They should not be confused with each other, however, since Pdv is associated with reversible boundary work in closed systems (Fig. 2.15). Obviously, one needs to know v as a function of P for the given process to perform the integration. When the working fluid is incompressible, the specific volume v remains constant during the process and can be taken out of the integration. Then Eq. 2–50 simplifies to

(KJ/kg) (2-53)

( ) For the steady flow of a liquid through a device that involves no work interactions (such as a nozzle or a pipe section), the work term is zero, Fig. 2.15 and the equation above can be expressed as

(2-54)

( ) ( ) Which is known as the Bernoulli equation in fluid mechanics. It is developed for an internally reversible process and thus is applicable to incompressible fluids that involve no irreversibilities such as friction or shock waves. This equation can be modified, however, to incorporate these effects. Equation 2–51 has far-reaching implications in engineering regarding devices that produce or consume work steadily such as turbines, compressors, and pumps. It is obvious from this equation that the reversible steady flow work is closely associated with the specific volume of the fluid flowing through the device. The larger the specific volume, the larger the reversible work produced or consumed by the steady-flow device. This conclusion is equally valid for actual steady-flow devices. Therefore, every effort should be made to keep the specific volume of a fluid as small as possible during a compression process to minimize the work input and as large as possible during an expansion process to maximize the work output.

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Example 2.8 Determine the compressor work input required to compress steam isentropically from 100 kPa to 1 MPa, assuming that the steam exists as (a) saturated liquid and (b) saturated vapor at the inlet state. Solution (a) In this case, steam is a saturated liquid initially, and its specific volume is

Which remains essentially constant during the process. Thus,

(b) This time, steam is a saturated vapor initially and remains a vapor during the entire compression process.

This result could also be obtained from the energy balance relation for an isentropic steady flow process. Next we determine the enthalpies:

Thus,

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2.11 Minimizing The Compressor Work We have just shown that the work input to a compressor is minimized when the compression process is executed in an internally reversible manner. When the changes in kinetic and potential energies are negligible, the compressor work is given by (Eq. 2–52) (2-55)

Obviously one way of minimizing the ∫ compressor work is to approximate an internally reversible process as much as possible by minimizing the irreversibilities such as friction, turbulence, and nonquasi-equilibrium compression. The extent to which this can be accomplished is limited by economic considerations. A second (and more practical) way of reducing the compressor work is to keep the specific volume of the gas as small as possible during the compression process. This is done by maintaining the temperature of the gas as low as possible during compression since the specific volume of a gas is proportional to temperature. Therefore, reducing the work input to a compressor requires that the gas be cooled as it is compressed. To have a better understanding of the effect of cooling during the compression process, we compare the work input requirements for three kinds of processes: an isentropic process (involves no cooling), a polytropic process (involves some cooling), and an isothermal process (involves maximum cooling). Assuming all three processes are executed between the same pressure levels (P1 and P2) in an internally reversible manner and the gas behaves as an ideal gas (Pv = RT) with constant specific heats, we see that the compression work is determined by performing the integration in Eq. 2–55 for each case, with the following results: Isentropic (Pv k = constant):

(2-56a) ( ) ( )⁄ Polytropic (Pv n = constant): [( ) ]

(2-56b) ( ) ( )⁄ Isothermal (Pv = constant): [( ) ] (2-56c)

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 62

The three processes are plotted on a P-v diagram in Fig. 2.16 for the same inlet state and exit pressure. On a P-v diagram, the area to the left of the process curve is the integral of vdP. Thus it is a measure of the steady flow compression work. It is interesting to observe from this diagram that of the three internally reversible cases considered, the adiabatic compression (Pv k = constant) requires the maximum work and the isothermal compression (T = constant or Pv = constant) requires the minimum. The work input requirement for the Fig. 2.16 polytropic case (Pvn = constant) is between these two and decreases as the polytropic exponent n is decreased, by increasing the heat rejection during the compression process. If sufficient heat is removed, the value of n approaches unity and the process becomes isothermal. One common way of cooling the gas during compression is to use cooling jackets around the casing of the compressors.

Multistage Compression with Intercooling It is clear from these arguments that cooling a gas as it is compressed is desirable since this reduces the required work input to the compressor. However, often it is not possible to have adequate cooling through the casing of the compressor, and it becomes necessary to use other techniques to achieve effective cooling. One such technique is multistage compression with intercooling, where the gas is compressed in stages and cooled between each stage by passing it through a heat exchanger called an intercooler. Ideally, the cooling process takes place at constant pressure, and the gas is cooled to the initial temperature T1 at each intercooler. Multistage compression with intercooling is especially attractive when a gas is to be compressed to very high pressures. Fig. 2.17

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 63

The effect of intercooling on compressor work is graphically illustrated on P-v and T-s diagrams in Fig. 2.17 for a two-stage compressor. The gas is compressed in the first stage from P1 to an intermediate pressure Px, cooled at constant pressure to the initial temperature T1, and compressed in the second stage to the final pressure P2. The compression processes, in general, can be modeled as polytropic (Pv n = constant) where the value of n varies between k and 1. The colored area on the P-v diagram represents the work saved as a result of two-stage compression with intercooling. The process paths for single-stage isothermal and polytropic processes are also shown for comparison. The size of the colored area (the saved work input) varies with the value of the intermediate pressure Px, and it is of practical interest to determine the conditions under which this area is maximized. The total work input for a two-stage compressor is the sum of the work inputs for each stage of compression, as determined from Eq. 2–56b:

(2-57) ( ) ( )⁄ ( )⁄ [( ) ] [( ) ]

The only variable in this equation is Px. The Px value that minimizes the total work is determined by differentiating this expression with respect to Px and setting the resulting expression equal to zero. It yields

(2-58) ⁄ ( ) That is, to minimize compression work during two-stage compression, the pressure ratio across each stage of the compressor must be the same. When this condition is satisfied, the compression work at each stage becomes identical, that is, wcomp I,in = wcomp II,in.

Example 2.9 Air is compressed steadily by a reversible compressor from an inlet state of 100 kPa and 300 K to an exit pressure of 900 kPa. Determine the compressor work per unit mass for (a) isentropic

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 64 compression with k = 1.4, (b) polytropic compression with n = 1.3, (c) isothermal compression, and (d) ideal two-stage compression with intercooling with a polytropic exponent of 1.3. Solution (a) Isentropic compression with k = 1.4:

(b) Polytropic compression with n = 1.3:

(c) Isothermal compression:

(d) Ideal two-stage compression with intercooling (n = 1.3): In this case, the pressure ratio across each stage is the same, and its value is

The compressor work across each stage is also the same. Thus the total compressor work is twice the compression work for a single stage:

Discussion Of all four cases considered, the isothermal compression requires the minimum work and the isentropic compression the maximum. The compressor work is decreased when two

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 65 stages of polytropic compression are utilized instead of just one. As the number of compressor stages is increased, the compressor work approaches the value obtained for the isothermal case.

2.12 Isentropic Efficiencies of Steady Flow Devices We mentioned repeatedly that irreversibilities inherently accompany all actual processes and that their effect is always to downgrade the performance of devices. In engineering analysis, it would be very desirable to have some parameters that would enable us to quantify the degree of degradation of energy in these devices. In the last chapter we did this for cyclic devices, such as heat engines and refrigerators, by comparing the actual cycles to the idealized ones, such as the Carnot cycle. A cycle that was composed entirely of reversible processes served as the model cycle to which the actual cycles could be compared. This idealized model cycle enabled us to determine the theoretical limits of performance for cyclic devices under specified conditions and to examine how the performance of actual devices suffered as a result of irreversibilities. Now we extend the analysis to discrete engineering devices working under steady-flow conditions, such as turbines, compressors, and nozzles, and we examine the degree of degradation of energy in these devices as a result of irreversibilities. However, first we need to define an ideal process that serves as a model for the actual processes. Although some heat transfer between these devices and the surrounding medium is unavoidable, many steady-flow devices are intended to operate under adiabatic conditions. Therefore, the model process for these devices should be an adiabatic one. Furthermore, an ideal process should involve no irreversibilities since the effect of irreversibilities is always to downgrade the performance of engineering devices. Thus, the ideal process that can serve as a suitable model for adiabatic Fig. 2.18 steady-flow devices is the isentropic process (Fig. 2.18). The more closely the actual process approximates the idealized isentropic process, the better the device performs. Thus, it would be desirable to have a parameter that expresses quantitatively how efficiently an actual device approximates an idealized one. This parameter is the isentropic or adiabatic efficiency, which is a measure of the deviation of actual processes from the corresponding idealized ones.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 66

Isentropic efficiencies are defined differently for different devices since each device is set up to perform different tasks. Next, we define the isentropic efficiencies of turbines, compressors, and nozzles by comparing the actual performance of these devices to their performance under isentropic conditions for the same inlet state and exit pressure.

2.12.1 Isentropic Efficiency of Turbines For a turbine under steady operation, the inlet state of the working fluid and the exhaust pressure are fixed. Therefore, the ideal process for an adiabatic turbine is an isentropic process between the inlet state and the exhaust pressure. The desired output of a turbine is the work produced, and the isentropic efficiency of a turbine is defined as the ratio of the actual work output of the turbine to the work output that would be achieved if the process between the inlet state and the exit pressure were isentropic:

(2-59)

Usually the changes in kinetic and potential energies associated with a fluid stream flowing through a turbine are small relative to the change in enthalpy and can be neglected. Then, the work output of an adiabatic turbine simply becomes the change in enthalpy, and Eq. 2–59 becomes

(2-60)

Where h2a and h2s are the enthalpy values at the exit state for actual and isentropic processes, respectively (Fig. 2.19).

The value of ηT greatly depends on the design of the individual components that make up the turbine. Well- designed, large turbines have isentropic efficiencies above 90 percent. For small turbines, however, it may drop even below 70 percent. The value of the isentropic efficiency of a turbine is determined by measuring the actual work output of the turbine and by calculating the isentropic work output for the measured inlet conditions and the exit pressure. This value can then be Fig. 2.19 used conveniently in the design of power plants.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 67

Example 2.10 Steam enters an adiabatic turbine steadily at 3 MPa and 400ºC and leaves at 50 kPa and 100ºC. If the power output of the turbine is 2 MW, determine (a) the isentropic efficiency of the turbine and (b) the mass flow rate of the steam flowing through the turbine.

Solution (a) The enthalpies at various states are

The exit enthalpy of the steam for the isentropic process h2s is determined from the requirement that the entropy of the steam remain constant (s2s = s1):

Obviously, at the end of the isentropic process steam exists as a saturated mixture since sf ˂ s2s˂ sg. Thus, we need to find the quality at state 2s first:

and

By substituting these enthalpy values into Eq. 2–60, the isentropic efficiency of this turbine is determined to be

(b) The mass flow rate of steam through this turbine is determined from the energy balance for steady-flow systems:

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 68

2.12.2 Isentropic Efficiencies of Compressors and Pumps The isentropic efficiency of a compressor is defined as the ratio of the work input required to raise the pressure of a gas to a specified value in an isentropic manner to the actual work input:

(2-61)

Notice that the isentropic compressor efficiency is defined with the isentropic work input in the numerator instead of in the denominator. This is because ws is a smaller quantity than wa, and this definition prevents ηC from becoming greater than 100 percent, which would falsely imply that the actual compressors performed better than the isentropic ones. Also notice that the inlet conditions and the exit pressure of the gas are the same for both the actual and the isentropic compressor. When the changes in kinetic and potential energies of the gas being compressed are negligible, the work input to an adiabatic compressor becomes equal to the change in enthalpy, and Eq. 2–61 for this case becomes (2-62)

Where h2a and h 2s are the enthalpy values at the exit state for actual and isentropic compression processes, respectively, as illustrated in Fig. 2.20. Again, the value of ηC greatly depends on the design of the compressor. Well- designed compressors have isentropic efficiencies that range from 80 to 90 percent. Fig. 2.20

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 69

When the changes in potential and kinetic energies of a liquid are negligible, the isentropic efficiency of a pump is defined similarly as

(2-63) ( )

When no attempt is made to cool the gas as it is compressed, the actual compression process is nearly adiabatic and the reversible adiabatic (i.e., isentropic) process serves well as the ideal process. However, sometimes compressors are cooled intentionally by utilizing fins or a water jacket placed around the casing to reduce the work input requirements. In this case, the isentropic process is not suitable as the model process since the device is no longer adiabatic and the isentropic compressor efficiency defined above is meaningless. A realistic model process for compressors that are intentionally cooled during the compression process is the reversible isothermal process. Then we can conveniently define an isothermal efficiency for such cases by comparing the actual process to a reversible isothermal one:

(2-64)

Where wt and wa are the required work inputs to the compressor for the reversible isothermal and actual cases, respectively.

Example 2.11 Air is compressed by an adiabatic compressor from 100 kPa and 12ºC to a pressure of 800 kPa at a steady rate of 0.2 kg/s. If the isentropic efficiency of the compressor is 80 percent, determine (a) the exit temperature of air and (b) the required power input to the compressor.

Solution (a) We know only one property (pressure) at the exit state, and we need to know one more to fix the state and thus determine the exit temperature. The property that can be determined with

minimal effort in this case is h2a since the isentropic efficiency of the compressor is given. At the compressor inlet,

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 70

The enthalpy of the air at the end of the isentropic compression process is determined by using one of the isentropic relations of ideal gases,

And

Substituting the known quantities into the isentropic efficiency relation, we have

Thus,

(b) The required power input to the compressor is determined from the energy balance for steady-flow devices,

Discussion Notice that in determining the power input to the compressor, we used h2a instead of h2s since h2a is the actual enthalpy of the air as it exits the compressor. The quantity h2s is a hypothetical enthalpy value that the air would have if the process were isentropic.

2.12.3 Isentropic Efficiency of Nozzles Nozzles are essentially adiabatic devices and are used to accelerate a fluid. Therefore, the isentropic process serves as a suitable model for nozzles. The isentropic efficiency of a nozzle is defined as the ratio of the actual kinetic energy of the fluid at the nozzle exit to the kinetic energy value at the exit of an isentropic nozzle for the same inlet state and exit pressure. That is,

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 71

(2-65)

Note that the exit pressure is the same for both the actual and isentropic processes, but the exit state is different. Nozzles involve no work interactions, and the fluid experiences little or no change in its potential energy as it flows through the device. If, in addition, the inlet velocity of the fluid is small relative to the exit velocity, the energy balance for this steady flow device reduces to

Then the isentropic efficiency of the nozzle can be expressed in terms of enthalpies as

(2-66)

Where h2a and h2 s are the enthalpy values at the nozzle exit for the actual and isentropic processes, respectively (Fig. 2.21). Isentropic efficiencies of nozzles are typically above 90 percent, and nozzle efficiencies above 95 percent are not Fig. 2.21 uncommon.

Example 2.12 Air at 200 kPa and 950 K enters an adiabatic nozzle at low velocity and is discharged at a pressure of 110 kPa. If the isentropic efficiency of the nozzle is 92 percent, determine (a) the maximum possible exit velocity, (b) the exit temperature, and (c) the actual exit velocity of the air. Assume constant specific heats for air.

Solution The temperature of air will drop during this acceleration process because some of its internal energy is converted to kinetic energy. This problem can be solved accurately by using property data from the air table. But we will assume constant specific heats (thus sacrifice some accuracy) to demonstrate their use. Let us guess the average temperature of the air to be about 850 K. Then, the average values of cp and k at this anticipated average temperature are determined from Table

A–2b to be cp = 1.11 kJ/kg·K and k = 1.349.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 72

(a) The exit velocity of the air will be a maximum when the process in the nozzle involves no irreversibilities. The exit velocity in this case is determined from the steady-flow energy equation. However, first we need to determine the exit temperature. For the isentropic process of an ideal gas we have:

Or

This gives an average temperature of 882 K, which is somewhat higher than the assumed average temperature (850 K). This result could be refined by reevaluating the k value at 882 K and repeating the calculations, however, it is not warranted since the two average temperatures are sufficiently close (doing so would change the temperature by only 0.6 K, which is not significant). Now we can determine the isentropic exit velocity of the air from the energy balance for this isentropic steady-flow process:

Or

(b) The actual exit temperature of the air is higher than the isentropic exit temperature evaluated above and is determined from

Or

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 73

That is, the temperature is 11 K higher at the exit of the actual nozzle as a result of irreversibilities such as friction. It represents a loss since this rise in temperature comes at the expense of kinetic energy. (c) The actual exit velocity of air can be determined from the definition of isentropic efficiency of a nozzle,

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 74

Sheet (3) (Reversible Steady-Flow Work and Isentropic Efficiencies of Steady-Flow Devices)

1- Water enters the pump of a steam power plant as saturated liquid at 20 kPa at a rate of 45 kg/s and exits at 6 MPa. Neglecting the changes in kinetic and potential energies and assuming the process to be reversible, determine the power input to the pump.

2- Saturated refrigerant-134a vapor at 120 kPa is compressed reversibly in an adiabatic compressor to 800 kPa. Determine the work input to the compressor. What would your answer be if the refrigerant were first condensed at constant pressure before it was compressed?

3- Liquid water enters a 25-kW pump at 100-kPa pressure at a rate of 5 kg/s. Determine the highest pressure the liquid water can have at the exit of the pump. Neglect the kinetic and potential energy changes of water, and take the specific volume of water to be 0.001 m3/kg.

4- Consider a steam power plant that operates between the pressure limits of 10 MPa and 20 kPa. Steam enters the pump as saturated liquid and leaves the turbine as saturated vapor. Determine the ratio of the work delivered by the turbine to the work consumed by the pump. Assume the entire cycle to be reversible and the heat losses from the pump and the turbine to be negligible.

5- Nitrogen gas is compressed from 80 kPa and 27°C to 480 kPa by a 10-kW compressor. Determine the mass flow rate of nitrogen through the compressor, assuming the compression process to be (a) isentropic, (b) polytropic with n =1.3, (c) isothermal, and (d) ideal two- stage polytropic with n=1.3.

6- Helium gas is compressed from 80 kPa and 20°C to 600 kPa at a rate of 0.3 m3/s. Determine the power input to the compressor, assuming the compression process to be (a) isentropic, (b) polytropic with n = 1.2, (c) isothermal, and (d) ideal two-stage polytropic with n = 1.2.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 75

7- Steam enters an adiabatic turbine at 8 MPa and 500°C with a mass flow rate of 3 kg/s and leaves at 30 kPa. The isentropic efficiency of the turbine is 0.90. Neglecting kinetic energy change of the steam, determine (a) the temperature at the turbine exit and (b) the power output of the turbine.

8- Steam enters an adiabatic turbine at 7 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 150°C, and 140 m/s. If the power output of the turbine is 6 MW, determine (a) the mass flow rate of the steam flowing through the turbine and (b) the isentropic efficiency of the turbine.

9- Combustion gases enter an adiabatic gas turbine at 1400K and 800 kPa and leave at 400 kPa with a low velocity. Treating the combustion gases as air and assuming an isentropic efficiency of 82 percent, determine the work output of the turbine.

10- Refrigerant-134a enters an adiabatic compressor as saturated vapor at 120 kPa at a rate of 0.3 m3/min and exits at 1-MPa pressure. If the isentropic efficiency of the compressor is 80 percent, determine (a) the temperature of the refrigerant at the exit of the compressor and (b) the power input, in kW. Also, show the process on a T-s diagram with respect to saturation lines.

11- Air enters an adiabatic compressor at 100 kPa and 17°C at a rate of 2.4 m3/s, and it exits at 257°C. The compressor has an isentropic efficiency of 84 percent. Neglecting the changes in kinetic and potential energies, determine (a) the exit pressure of air and (b) the power required to drive the compressor.

12- Air is compressed by an adiabatic compressor from 95 kPa and 27°C to 600 kPa and 277°C. Assuming variable specific heats and neglecting the changes in kinetic and potential energies, determine (a) the isentropic efficiency of the compressor and (b) the exit temperature of air if the process were reversible.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 76

13- Carbon dioxide enters an adiabatic compressor at 100 kPa and 300 K at a rate of 1.8 kg/s and exits at 600 kPa and 450 K. Neglecting the kinetic energy changes, determine the isentropic efficiency of the compressor.

14- Air enters an adiabatic nozzle at 400 kPa and 547°C with low velocity and exits at 290 m/s. If the adiabatic efficiency of the nozzle is 90 percent, determine the exit temperature and pressure of the air.

15- Hot combustion gases enter the nozzle of a turbojet engine at 260 kPa, 747°C, and 80 m/s, and they exit at a pressure of 85 kPa. Assuming an isentropic efficiency of 92 percent and treating the combustion gases as air, determine (a) the exit velocity and (b) the exit temperature.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 77

Chapter (3)

Gas Power Cycles

3.1 Basic Considerations In The Analysis Of Power Cycles Most power-producing devices operate on cycles, and the study of power cycles is an exciting and important part of thermodynamics. The cycles encountered in actual devices are difficult to analyze because of the presence of complicating effects, such as friction, and the absence of sufficient time for establishment of the equilibrium conditions during the cycle. To make an analytical study of a cycle feasible, we have to keep the complexities at a manageable level and utilize some idealizations. When the actual cycle is stripped of all the internal irreversibilities and complexities, we end up with a cycle that resembles the actual cycle closely but is made up totally of internally reversible processes. Such a cycle is called an ideal cycle (Fig. 3.1). A simple idealized model enables engineers to study the effects of the major parameters that dominate the cycle without getting bogged down in the details. The cycles discussed in this chapter are somewhat idealized, but they still retain the general characteristics of the actual cycles they represent. The conclusions reached from the analysis of ideal cycles are also applicable to actual cycles. The thermal efficiency of the Otto cycle, the ideal cycle for spark-ignition automobile engines, for example, increases with the compression ratio. This is also the case for actual automobile engines. The numerical values Fig. 3.1 obtained from the analysis of an ideal cycle, however, are not necessarily representative of the actual cycles, and care should be exercised in their interpretation. The simplified analysis presented in this chapter for various power cycles of practical interest may also serve as the starting point for a more in-depth study. Heat engines are designed for the purpose of converting thermal energy to work, and their performance is expressed in terms of the thermal efficiency ηth, which is the ratio of the network produced by the engine to the total heat input:

(3-1)

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 78

Recall that heat engines that operate on a totally reversible cycle, such as the Carnot cycle, have the highest thermal efficiency of all heat engines operating between the same temperature levels. That is, nobody can develop a cycle more efficient than the Carnot cycle. Then the following question arises naturally: If the Carnot cycle is the best possible cycle, why do we not use it as the model cycle for all the heat engines instead of bothering with several so-called ideal cycles? The answer to this question is hardware related. Most cycles encountered in practice differ significantly from the Carnot cycle, which makes it unsuitable as a realistic model. Each ideal cycle discussed in this chapter is related to a specific work-producing device and is an idealized version of the actual cycle. The ideal cycles are internally reversible, but, unlike the Carnot cycle, they are not necessarily externally reversible. That is, they may involve irreversibilities external to the system such as heat transfer through a finite temperature difference. Therefore, the thermal efficiency of an ideal cycle, in general, is less than that of a totally reversible cycle operating between the same temperature limits. However, it is still considerably higher than the thermal efficiency of an actual cycle because of the idealizations utilized. The idealizations and simplifications commonly employed in the analysis of power cycles can be summarized as follows: 1. The cycle does not involve any friction. Therefore, the working fluid does not experience any pressure drop as it flows in pipes or devices such as heat exchangers. 2. All expansion and compression processes take place in a quasi-equilibrium manner. 3. The pipes connecting the various components of a system are well insulated, and heat transfer through them is negligible. Neglecting the changes in kinetic and potential energies of the working fluid is another commonly utilized simplification in the analysis of power cycles. This is a reasonable assumption since in devices that involve shaft work, such as turbines, compressors, and pumps, the kinetic and potential energy terms are usually very small relative to the other terms in the energy equation. Fluid velocities encountered in devices such as condensers, boilers, and mixing chambers are typically low, and the fluid streams experience little change in their velocities, again making kinetic energy changes negligible. The only devices where the changes in kinetic energy are significant are the nozzles and diffusers, which are specifically designed to create large changes in velocity.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 79

In the preceding chapters, property diagrams such as the P-v and T-s diagrams have served as valuable aids in the analysis of thermodynamic processes. On both the P-v and T-s diagrams, the area enclosed by the process curves of a cycle represents the network produced during the cycle (Fig. 3.2), which is also equivalent to the net heat transfer for that cycle. The T-s diagram is particularly useful as a visual aid in the analysis of ideal power cycles. An ideal power cycle does not involve any internal irreversibilities, and so the only effect that can change the entropy of the working fluid during a process is heat transfer. In the preceding chapters, property diagrams such as the P- v and T-s diagrams have served as valuable aids in the analysis of thermodynamic processes. On both the P-v and T-s diagrams, Fig. 3.2 the area enclosed by the process curves of a cycle represents the network produced during the cycle (Fig. 3.2), which is also equivalent to the net heat transfer for that cycle. The T-s diagram is particularly useful as a visual aid in the analysis of ideal power cycles. An ideal power cycle does not involve any internal irreversibilities, and so the only effect that can change the entropy of the working fluid during a process is heat transfer. On a T-s diagram, a heat-addition process proceeds in the direction of increasing entropy, a heat-rejection process proceeds in the direction of decreasing entropy, and an isentropic (internally reversible, adiabatic) process proceeds at constant entropy. The area under the process curve on a T-s diagram represents the heat transfer for that process. The area under the heat addition process on a T-s diagram is a geometric measure of the total heat supplied during the cycle qin, and the area under the heat rejection process is a measure of the total heat rejected qout. The difference between these two (the area enclosed by the cyclic curve) is the net heat transfer, which is also the network produced during the cycle. Therefore, on a T-s diagram, the ratio of the area enclosed by the cyclic curve to the area under the heat addition process curve represents the thermal efficiency of the cycle. Any modification that increases the ratio of these two areas will also increase the thermal efficiency of the cycle. Although the working fluid in an ideal power cycle operates on a closed loop, the type of individual processes that comprises the cycle depends on the individual devices used to execute the cycle. In the Rankine cycle, which is the ideal cycle for steam power plants, the working

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 80 fluid flows through a series of steady-flow devices such as the turbine and condenser, whereas in the Otto cycle, which is the ideal cycle for the spark-ignition automobile engine, the working fluid is alternately expanded and compressed in a piston– cylinder device. Therefore, equations pertaining to steady-flow systems should be used in the analysis of the Rankine cycle, and equations pertaining to closed systems should be used in the analysis of the Otto cycle.

3.2 Air Standard Assumptions In gas power cycles, the working fluid remains a gas throughout the entire cycle. Spark- ignition engines, diesel engines, and conventional gas turbines are familiar examples of devices that operate on gas cycles. In all these engines, energy is provided by burning a fuel within the system boundaries. That is, they are internal combustion engines. Because of this combustion process, the composition of the working fluid changes from air and fuel to combustion products during the course of the cycle. However, considering that air is predominantly nitrogen that undergoes hardly any chemical reactions in the combustion chamber, the working fluid closely resembles air at all times. Even though internal combustion engines operate on a mechanical cycle (the piston returns to its starting position at the end of each revolution), the working fluid does not undergo a complete thermodynamic cycle. It is thrown out of the engine at some point in the cycle (as exhaust gases) instead of being returned to the initial state. Working on an open cycle is the characteristic of all internal combustion engines. The actual gas power cycles are rather complex. To reduce the analysis to a manageable level, we utilize the following approximations, commonly known as the air-standard assumptions: 1. The working fluid is air, which continuously circulates in a closed loop and always behaves as an ideal gas. 2. All the processes that make up the cycle are internally reversible. 3. The combustion process is replaced by a heat-addition process from an external source (Fig. 3.3). 4. The exhaust process is replaced by a heat-rejection process that restores the working fluid to its initial state.

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Another assumption that is often utilized to simplify the analysis even more is that air has constant specific heats whose values are determined a room temperature (25ºC, or 77ºF). When this assumption is utilized, the air-standard assumptions are called the cold-air-standard assumptions. A cycle for which the air-standard assumptions are applicable is frequently referred to as an air-standard cycle. The air-standard assumptions previously stated provide considerable simplification the analysis without significantly Fig. 3.3 deviating from the actual cycles. This simplified model enables us to study qualitatively the influence of major parameters on the performance of the actual engines.

3.3 An Overview Of Reciprocating Engines Despite its simplicity, the reciprocating engine (basically a piston–cylinder device) is one of the rare inventions that have proved to be very versatile and to have a wide range of applications. It is the powerhouse of the vast majority of automobiles, trucks, light aircraft, ships, and electric power generators, as well as many other devices. The basic components of a reciprocating engine are shown in Fig. 3.4 The piston reciprocates in the cylinder between two fixed positions called the top dead center (TDC) the position of the piston when it forms the smallest volume in the cylinder and the bottom dead center (BDC) the position of the piston when it forms the Fig. 3.4 largest volume in the cylinder. The distance between the TDC and the BDC is the largest distance that the piston can travel in one direction, and it is called the stroke of the engine. The diameter of the piston is called the bore. The air or air fuel mixture is drawn into the cylinder through the intake valve, and the combustion products are expelled from the cylinder through the exhaust valve. Fig. 3.5 The minimum volume formed in the cylinder when the piston is at TDC is called the clearance volume (Fig. 3.5). The volume displaced by the piston as it moves between TDC and BDC is called the displacement volume. The ratio of the maximum

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 82 volume formed in the cylinder to the minimum (clearance) volume is called the compression ratio r of the engine:

(3-2)

Notice that the compression ratio is a volume ratio and should not be confused with the pressure ratio. Another term frequently used in conjunction with reciprocating engines is the mean effective pressure (MEP). It is a fictitious pressure that, if it acted on the piston during the entire power stroke, would produce the same amount of network as that produced during the actual cycle (Fig. 3.6). That is,

Wnet = MEP x Piston area x Stroke = MEP x Displacement volume Or

(3-3)

( ) The mean effective pressure can be used as a parameter to compare the performances of reciprocating engines of equal size. The engine with a larger value of MEP delivers more network per cycle and thus performs better. Reciprocating engines are classified as spark-ignition (SI) engines or compression-ignition (CI) engines, depending on how the combustion process in the cylinder is initiated. In SI engines, the combustion of the air–fuel mixture is initiated by a Fig.3.6 spark plug. In CI engines, the air–fuel mixture is self-ignited as a result of compressing the mixture above its self-ignition temperature. In the next two sections, we discuss the Otto and Diesel cycles, which are the ideal cycles for the SI and CI reciprocating engines, respectively.

3.4 Otto Cycle: The Ideal Cycle For Spark-Ignition Engines The Otto cycle is the ideal cycle for spark-ignition reciprocating engines. It is named after Nikolaus A. Otto, who built a successful four-stroke engine in 1876 in Germany using the cycle

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 83 proposed by Frenchman Beau de Rochas in 1862. In most spark-ignition engines, the piston executes four complete strokes (two mechanical cycles) within the cylinder, and the crankshaft completes two revolutions for each thermodynamic cycle. These engines are called four-stroke internal combustion engines. A schematic of each stroke as well as a P-v diagram for an actual four-stroke spark-ignition engine is given in Fig. 3.7a. Initially, both the intake and the exhaust valves are closed, and the piston is at its lowest position (BDC). During the compression stroke, the piston moves upward, compressing the air– fuel mixture. Shortly before the piston reaches its highest position (TDC), the spark plug fires and the mixture ignites, increasing the pressure and temperature of the system. The high-pressure gases force the piston down, which in turn forces the crankshaft to rotate, producing a useful work output during the expansion or power stroke. Towards the end of expansion stroke, the exhaust valve opens and the combustion gases that are above the atmospheric pressure rush out of the cylinder through the open exhaust valve. This process is called exhaust blowdown, and most combustion gases leave the cylinder by the time the piston reaches BDC. The cylinder is still filled by the exhaust gases at a lower pressure at BDC. Now the piston moves upward one more time, purging the exhaust gases through the exhaust valve (the exhaust stroke), and down a second time, drawing in fresh air–fuel mixture through the intake valve (the intake stroke). Notice that the pressure in the cylinder is slightly above the atmospheric value during the exhaust stroke and slightly below during the intake stroke.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., ThermodynamicsFig. 3.7 (2), MEC 151, Jan 2019 Page 84

In two-stroke engines, all four functions described above are executed in just two strokes: the power stroke and the compression stroke. In these engines, the crankcase is sealed, and the outward motion of the piston is used to slightly pressurize the air–fuel mixture in the crankcase, as shown in Fig. 3.8. Also, the intake and exhaust valves are replaced by openings in the lower portion of the cylinder wall. During the latter part of the power stroke, the piston uncovers first the exhaust port, allowing the exhaust gases to be partially expelled, and then the intake port, allowing the fresh air–fuel mixture to rush in and drive most of the remaining exhaust gases out of the cylinder. This Fig. 3.8 mixture is then compressed as the piston moves upward during the compression stroke and is subsequently ignited by a spark plug. The two-stroke engines are generally less efficient than their four-stroke counterparts because of the incomplete expulsion of the exhaust gases and the partial expulsion of the fresh air–fuel mixture with the exhaust gases. However, they are relatively simple and inexpensive, and they have high power-to-weight and power-to-volume ratios, which make them suitable for applications requiring small size and weight such as for motorcycles, chain saws, and lawn mowers. Advances in several technologies such as direct fuel injection, stratified charge combustion, and electronic controls brought about a renewed interest in two stroke engines that can offer high performance and fuel economy while satisfying the stringent emission requirements. For a given weight and displacement, a well-designed two-stroke engine can provide significantly more power than its four stroke counterpart because two-stroke engines produce power on every engine revolution instead of every other one. In the new two-stroke engines, the highly atomized fuel spray that is injected into the combustion chamber toward the end of the compression stroke burns much more completely. The fuel is sprayed after the exhaust valve is closed, which prevents unburned fuel from being ejected into the atmosphere. With stratified combustion, the flame that is initiated by igniting a small amount of the rich fuel air mixture near the spark plug propagates through the combustion chamber filled with a much leaner mixture, and this results in much cleaner combustion. Also, the advances in electronics have made it possible to ensure the optimum operation under varying engine load and speed conditions. Major car companies have

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 85 research programs underway on two-stroke engines which are expected to make a comeback in the future. The thermodynamic analysis of the actual four-stroke or two-stroke cycles described is not a simple task. However, the analysis can be simplified significantly if the air-standard assumptions are utilized. The resulting cycle, which closely resembles the actual operating conditions, is the ideal Otto cycle. It consists of four internally reversible processes: 1-2 Isentropic compression 2-3 Constant-volume heat addition 3-4 Isentropic expansion 4-1 Constant-volume heat rejection The execution of the Otto cycle in a piston–cylinder device together with a P-v diagram is illustrated in Fig. 3-7b. The T-s diagram of the Otto cycle is given in Fig. 3.9. The ideal Otto cycle shown in Fig. 3.7b has one shortcoming. This ideal cycle consists of two strokes equivalent to one mechanical cycle or one crankshaft rotation. The actual engine operation shown in Fig. 3.7a, on the other hand, involves four strokes equivalent to two mechanical cycles or two crankshaft rotations. This can be corrected by including intake and exhaust strokes in the ideal Otto cycle, as shown in Fig. 3.10. In this Fig. 3.9 modified cycle, air-fuel mixture (approximated as air due to air- standard assumptions) enters the cylinder through the open intake valve at atmospheric pressure P0 during process 0-1 as the piston moves from TDC to BDC. The intake valve is closed at state 1 and air is compressed isentropically to state 2. Heat is transferred at constant volume (process 2-3); it is expanded isentropically to state 4; and heat is rejected at constant volume (process 4-1). Fig. 3.10. Exhaust gases (again approximated as air) are expelled through the open exhaust valve (process 1-0) as the pressure remains constant at P0. The modified Otto cycle shown in Fig. 3.10 is executed in an open system during the intake and exhaust processes and in a closed system during the remaining four processes. We should point out that the constant-volume heat addition process (2-3) in the ideal Otto cycle replaces the

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 86 combustion process of the actual engine operation while the constant-volume heat rejection process (4-1) replaces the exhaust blowdown. The work interactions during the constant-pressure intake (0-1) and constant- pressure exhaust (1-0) processes can be expressed as

These two processes cancel each other as the work output during the intake is equal to work input during the exhaust. Then, the cycle reduces to the one in Fig. 3.7b. Therefore, inclusion of the intake and exhaust processes has no effect on the network output from the cycle. However, when calculating power output from the cycle during an ideal Otto cycle analysis, we must consider the fact that the ideal Otto cycle has four strokes just like actual four-stroke spark- ignition engine. This is illustrated in the last part of Example 3.1. The Otto cycle is executed in a closed system, and disregarding the changes in kinetic and potential energies, the energy balance for any of the processes is expressed, on a unit-mass basis, as (3-4)

( ) ( ) ( ⁄ ) No work is involved during the two heat transfer processes since both take place at constant volume. Therefore, heat transfer to and from the working fluid can be expressed as

(3-5a)

And ( ) (3-5b)

( ) Then the thermal efficiency of the ideal Otto cycle under the cold air standard assumptions becomes

Processes 1-2 and 3-4 are isentropic, and v2 = v3 and v4 = v1. Thus,

(3-6)

( ) ( ) Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 87

Substituting these equations into the thermal efficiency relation and simplifying give

(3-7)

Where (3-8)

Is the compression ratio and k is the specific heat ratio cp /cv. Equation 3-7 shows that under the cold-air-standard assumptions, the thermal efficiency of an ideal Otto cycle depends on the compression ratio of the engine and the specific heat ratio of the working fluid. The thermal efficiency of the ideal Otto cycle increases with both the compression ratio and the specific heat ratio. This is also true for actual spark-ignition internal combustion engines. A plot of thermal efficiency versus the compression ratio is given in Fig. 3.11 for k = 1.4, which is the specific heat ratio value of air at room temperature. For a given compression ratio, the thermal efficiency of an actual spark-ignition engine is less than Fig. 3.11 that of an ideal Otto cycle because of the irreversibilities, such as friction, and other factors such as incomplete combustion. We can observe from Fig. 3.11 that the thermal efficiency curve is rather steep at low compression ratios but flattens out starting with a compression ratio value of about 8. Therefore, the increase in thermal efficiency with the compression ratio is not as pronounced at high compression ratios. Also, when high compression ratios are used, the temperature of the air–fuel mixture rises above the auto ignition temperature of the fuel (the temperature at which the fuel ignites without the help of a spark) during the combustion process, causing an early and rapid burn of the fuel at some point or points ahead of the flame front, followed by almost instantaneous inflammation of the end gas. This premature ignition of the fuel, called auto ignition, produces an audible noise, which is called engine knock. Auto ignition in spark- ignition engines cannot be tolerated because it hurts performance and can cause engine damage.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 88

The requirement that auto ignition not be allowed places an upper limit on the compression ratios that can be used in spark ignition internal combustion engines. Improvement of the thermal efficiency of gasoline engines by utilizing higher compression ratios (up to about 12) without facing the auto ignition problem has been made possible by using gasoline blends that have good antiknock characteristics, such as gasoline mixed with tetraethyl lead. Tetraethyl lead had been added to gasoline since the 1920s because it is an inexpensive method of raising the octane rating, which is a measure of the engine knock resistance of a fuel. Leaded gasoline, however, has a very undesirable side effect: it forms compounds during the combustion process that are hazardous to health and pollute the environment. In an effort to combat air pollution, the government adopted a policy in the mid-1970s that resulted in the eventual phase-out of leaded gasoline. Unable to use lead, the refiners developed other techniques to improve the antiknock characteristics of gasoline. and the compression ratios had to be lowered to avoid engine knock. The ready availability of high octane fuels made it possible to raise the compression ratios again in recent years. Also, owing to the improvements in other areas (reduction in overall automobile weight, improved aerodynamic design, etc.), today’s cars have better fuel economy and consequently get more miles per gallon of fuel. This is an example of how engineering decisions involve compromises, and efficiency is only one of the considerations in final design. The second parameter affecting the thermal efficiency of an ideal Otto cycle is the specific heat ratio k. For a given compression ratio, an ideal Otto cycle using a monatomic gas (such as argon or helium, k = 1.667) as the working fluid will have the highest thermal efficiency. The specific heat ratio k, and thus the thermal efficiency of the ideal Otto cycle, decreases as the molecules of the working fluid get larger (Fig. 3.12). At room temperature it is 1.4 for air, 1.3 for carbon dioxide, and 1.2 for ethane. The working fluid in actual engines Fig. 3.12 contains larger molecules such as carbon dioxide, and the specific heat ratio decreases with temperature, which is one of the reasons that the actual cycles have lower thermal efficiencies than the ideal Otto cycle. The thermal efficiencies of actual spark-ignition engines range from about 25 to 30 percent.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 89

Example 3.1 An ideal Otto cycle has a compression ratio of 8. At the beginning of the compression process, air is at 100 kPa and 17ºC, and 800 kJ/kg of heat is transferred to air during the constant-volume heat-addition process. Accounting for the variation of specific heats of air with temperature, determine (a) the maximum temperature and pressure that occur during the cycle, (b) the network output, (c) the thermal efficiency, and (d) the mean effective pressure for the cycle. (e) Also, determine the power output from the cycle, in kW, for an engine speed of 4000 rpm (rev/min). Assume that this cycle is operated on an engine that has four cylinders with a total displacement volume of 1.6 L.

Solution (a) The maximum temperature and pressure in an Otto cycle occur at the end of the constant- volume heat-addition process (state 3). But first we need to determine the temperature and pressure of air at the end of the isentropic compression process (state 2), using data from Table A–17:

Process 1-2 (isentropic compression of an ideal gas):

Process 2-3 (constant-volume heat addition):

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 90

(b) The network output for the cycle is determined either by finding the boundary (Pdv) work involved in each process by integration and adding them or by finding the net heat transfer that is equivalent to the net work done during the cycle. We take the latter approach. However, first we need to find the internal energy of the air at state 4: Process 3-4 (isentropic expansion of an ideal gas):

Process 4-1 (constant-volume heat rejection):

Thus,

(c) The thermal efficiency of the cycle is determined from its definition:

Under the cold-air-standard assumptions (constant specific heat values at room temperature), the thermal efficiency would be (Eq. 3-7)

(d ) The mean effective pressure is determined from its definition, Eq. 3-3:

Where

Thus,

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 91

(e) The total air mass taken by all four cylinders when they are charged is

The network produced by the cycle is

That is, the network produced per thermodynamic cycle is 0.8037 kJ/cycle. Noting that there are two revolutions per thermodynamic cycle (nrev = 2 rev/cycle) in a four-stroke engine (or in the ideal Otto cycle including intake and exhaust strokes), the power produced by the engine is determined from

3.5 Diesel Cycle: The Ideal Cycle For Compression-Ignition Engines The Diesel cycle is the ideal cycle for CI reciprocating engines. The CI engine, first proposed by Rudolph Diesel in the 1890s, is very similar to the SI engine discussed in the last section, differing mainly in the method of initiating combustion. In spark-ignition engines (also known as gasoline engines), the air–fuel mixture is compressed to a temperature that is below the auto ignition temperature of the fuel, and the combustion process is initiated by firing a spark plug. In CI engines (also known as diesel engines), the air is compressed to a temperature that is Fig. 3.13 above the auto ignition temperature of (Fig. 3.13). In gasoline engines, a mixture of air and fuel is compressed during the compression stroke, and the compression ratios are limited by the onset of auto ignition or engine knock. In diesel engines, only air is compressed during the compression stroke, eliminating the possibility of auto ignition. Therefore, diesel engines can be designed to operate at much higher compression ratios, typically between 12 and 24. Not having to deal with the problem of auto ignition has another

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 92 benefit: many of the stringent requirements placed on the gasoline can now be removed, and fuels that are less refined (thus less expensive) can be used in diesel engines. The fuel injection process in diesel engines starts when the piston approaches TDC and continues during the first part of the power stroke. Therefore, the combustion process in these engines takes place over a longer interval. Because of this longer duration, the combustion process in the ideal Diesel cycle is approximated as a constant-pressure heat-addition process. In fact, this is the only process where the Otto and the Diesel cycles differ. The remaining three processes are the same for both ideal cycles. That is, process 1-2 is isentropic compression, 2-3 is constant-pressure heat addition, 3-4 is isentropic expansion, and 4-1 is constant-volume heat rejection. The similarity between the two cycles is also Fig. 3.14 apparent from the P-v and T-s diagrams of the Diesel cycle, shown in Fig. 3.14. Noting that the Diesel cycle is executed in a piston cylinder device, which forms a closed system, the amount of heat transferred to the working fluid at constant pressure and rejected from it at constant volume can be expressed as

( ) ( ) (3-9a) And ( ) ( ) (3-9b)

( ) Then the thermal efficiency of the ideal Diesel cycle under the cold-airstandard assumptions becomes

( ⁄ )

( ) ( ⁄ )

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 93

We now define a new quantity, the cutoff ratio rc, as the ratio of the cylinder volumes after and before the combustion process:

(3-10)

Utilizing this definition and the isentropic ideal-gas relations for processes 1-2 and 3-4, we see that the thermal efficiency relation reduces to

(3-11)

* ( )+ Where r is the compression ratio defined by Eq. 3-8. Looking at Eq. 3-11 carefully, one would notice that under the cold-air-standard assumptions, the efficiency of a Diesel cycle differs from the efficiency of an Otto cycle by the quantity in the brackets. This quantity is always greater than 1. Therefore,

(3-12)

When both cycles operate on the same compression ratio. Also, as the cutoff ratio decreases, the efficiency of the Diesel cycle increases (Fig. 3.15). For the limiting case of rc = 1, the quantity in the brackets becomes unity (can you prove it?), and the efficiencies of the Otto and Diesel cycles become identical. Remember, though, that diesel engines operate at much higher compression ratios and thus are usually more efficient than the spark ignition (gasoline) engines. The diesel engines also burn the fuel more completely since they usually operate at lower revolutions per minute and the air–fuel mass ratio is much higher than spark-ignition engines. Thermal efficiencies of large diesel engines range from about 35 to 40 Fig. 3.15 percent.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 94

The higher efficiency and lower fuel costs of diesel engines make them attractive in applications requiring relatively large amounts of power, such as in locomotive engines, emergency power generation units, large ships, and heavy trucks. As an example of how large a diesel engine can be, a 12-cylinder diesel engine built in 1964 by the Fiat Corporation of Italy had a normal power output of 25,200 hp (18.8 MW) at 122 rpm, a cylinder bore of 90 cm, and a stroke of 91 cm. In modern high-speed compression ignition engines, fuel is injected into the combustion chamber much sooner compared to the early diesel engines. Fuel starts to ignite late in the compression stroke, and consequently part of the combustion occurs almost at constant volume. Fuel injection continues until the piston reaches the top dead center, and combustion of the fuel keeps the pressure high well into the expansion stroke. Thus, the entire combustion process can better be modeled as the combination of constant-volume and constant-pressure processes. The ideal cycle based on this concept is called the Fig. 3.16 dual cycle and P-v diagram for it is given in Fig. 3.16. The relative amounts of heat transferred during each process can be adjusted to approximate the actual cycle more closely. Note that both the Otto and the Diesel cycles can be obtained as special cases of the dual cycle. Dual cycle is a more realistic model than diesel cycle for representing modern, high-speed compression ignition engines.

Example 3.2 An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process, air is at 95 kPa and 27ºC. Accounting for the variation of specific heats with temperature, determine (a) the temperature after the heat-addition process, (b) the thermal efficiency, and (c) the mean effective pressure.

Solution (a) Process 1-2: isentropic compression.

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 95

Process 2-3: P = constant heat addition.

(b)

Process 3-4: isentropic expansion.

Process 4-1: v = constant heat rejection.

(c)

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 96

References

Yunus A. Cengle, Michael A. Boles, (2015) "Thermodynamics an engineering approach", eighth edition .

Dr. Ahmed Abd EL badie, HTI, Mech. Eng. Dep., Thermodynamics (2), MEC 151, Jan 2019 Page 97