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Revision 2 June 2016 Thermodynamic Cycles Student Guide GENERAL DISTRIBUTION GENERAL DISTRIBUTION: Copyright © 2016 by the National Academy for Nuclear Training. Not for sale or for commercial use. This document may be used or reproduced by Academy members and participants. Not for public distribution, delivery to, or reproduction by any third party without the prior agreement of the Academy. All other rights reserved. NOTICE: This information was prepared in connection with work sponsored by the Institute of Nuclear Power Operations (INPO). Neither INPO, INPO members, INPO participants, nor any person acting on behalf of them (a) makes any warranty or representation, expressed or implied, with respect to the accuracy, completeness, or usefulness of the information contained in this document, or that the use of any information, apparatus, method, or process disclosed in this document may not infringe on privately owned rights, or (b) assumes any liabilities with respect to the use of, or for damages resulting from the use of any information, apparatus, method, or process disclosed in this document. iii Table of Contents INTRODUCTION ....................................................................................................................... 2 TLO 1 COMPRESSION PROCESSES .......................................................................................... 2 Overview ........................................................................................................................... 2 ELO 1.1 Gas Laws ............................................................................................................ 4 ELO 1.2 Compression Process ........................................................................................ 11 TLO 1 Summary .............................................................................................................. 14 TLO 2 THE SECOND LAW OF THERMODYNAMICS ................................................................ 17 Overview ......................................................................................................................... 17 ELO 2.1 Thermodynamic Entropy .................................................................................. 18 ELO 2.2 Carnot’s Principle of Thermodynamics ............................................................ 22 ELO 2.3 Thermodynamics of Ideal and Real Processes ................................................. 29 ELO 2.4 Thermodynamic Power Plant Efficiency .......................................................... 31 TLO 2 Summary .............................................................................................................. 53 THERMODYNAMIC PROCESSES SUMMARY ............................................................................ 57 THERMODYNAMIC CYCLES KNOWLEDGE CHECK ANSWERS .................................................. 1 ELO 1.1 Gas Laws ............................................................................................................ 1 ELO 1.2 Compression Process .......................................................................................... 2 ELO 2.1 Thermodynamic Entropy .................................................................................... 2 ELO 2.2 Carnot’s Principle of Thermodynamics .............................................................. 3 ELO 2.3 Thermodynamics of Ideal and Real Processes ................................................... 4 ELO 2.4 Thermodynamic Power Plant Efficiency ............................................................ 4 iv This page is intentionally blank. iv Thermodynamic Cycles Revision History Revision Date Version Number Purpose for Revision Performed By 11/7/2014 0 New Module OGF Team 12/11/2014 1 Added signature of OGF Working Group Chair OGF Team 1/16/2016 2 Incorporated OGF Team Changes OGF Team Rev 2 1 Introduction We have not yet discussed processes performed by gases as we have focused on the steam cycle, yet many applications of the use of gases are occurring all the time during plant operation. The compression of a gas results in different final states than the compression of a saturated vapor such as steam. Gases follow laws that relate their volume, pressure, and temperature unlike steam, which undergoes phase changes if temperature or pressure varies sufficiently. These laws must be understood to ensure plant equipment is maintained within design limits. The second law of thermodynamics does not take into account the feasibility of the process or the efficiency of the energy transformation studied. The second law of thermodynamics allows us to determine the maximum efficiency of the operating system so that design comparisons maximize the system’s efficiency. Objectives At the completion of this training session, the trainee will demonstrate mastery of this topic by passing a written exam with a grade of 80 percent or higher on the following Terminal Learning Objectives (TLOs): 1. Explain compression processes and the laws associated with them. 2. Apply the Second Law of Thermodynamics to analyze real and ideal systems and components. TLO 1 Compression Processes Overview Gas is another working fluid used throughout the plant. Gas responds differently to temperature, pressure, and volumetric changes than steam, so it requires additional explanation. A gas is a state of matter distinguished from the solid and liquid states by the following: Relatively low density and viscosity Relatively great expansion Contraction with changes in pressure and temperature Ability to diffuse readily Spontaneous tendency to distribute uniformly throughout any container. A vapor is often confused with a gas. Vapor has evaporated from a liquid or solid such as water. Rev 2 2 Most familiar gases are colorless and odorless and include the following: The oxygen and nitrogen of the atmosphere The bubbles of carbon dioxide that rise in a glass of soda The helium gas that is used to fill balloons. A few gases have characteristic color, such as: Nitrogen dioxide is red-brown Iodine vapor has a beautiful violet color Anything that we can smell can exist in the gaseous state because our sense of smell reacts only to gases. Gases have many observable physical properties. They fill whatever space is available, but applying pressure can compress them into a smaller volume. Temperature affects them; they can expand and contract, or exert different pressures, depending on the temperature. It is obvious from the force of the wind on a stormy day that gases can flow readily from place to place and that they have mass. However, gases are not very dense; a vessel filled with air floats on the surface of a pond because the air is less dense than the water. Temperature, pressure, and volume must always be specified when gases are discussed because of their interrelated effect. The quantitative relationships among the temperature, pressure, and volume of a gas are expressed in the gas laws, which were first explored in the eighteenth and nineteenth centuries. Compression and pressurization processes are very common in many types of industrial plants. These processes vary from being the primary function of a piece of equipment, such as an air compressor to an incidental result of another process, such as filling a tank with water without first opening the tank's vent valve. Operators maintain important plant parameters such as the safety injection accumulators within required legal and design limits using gas compression and expansion. Understanding the relationships between temperature, pressure, and volume of gases are important to ensure operating within required parameters. Objectives Upon completion of this lesson, you will be able to do the following: 1. Describe the ideal gas laws and how to solve for an unknown pressure, temperature, or volume. 2. Describe the effects of pressure and temperature changes on confined fluids. Rev 2 3 ELO 1.1 Gas Laws Introduction Because of their interrelated effect, temperature, pressure, and volume must always be specified when gases are discussed. The quantitative relationships among the temperature, pressure, and volume of a gas are expressed in the gas laws, which were first explored in the eighteenth and nineteenth centuries. The gas laws are useful because at low pressures all real gases behave like a perfect gas. Monatomic gas behavior is very similar to perfect gas behavior; the Ideal Gas Law is therefore accurate for predicting the gas behavior. Accuracy will decrease with diatomic and polyatomic gases. Still, the Ideal Gas Law is useful to develop the behavior of even these gases with experimentally derived corrections made to produce the desired accuracy. Charles’s Laws Charles’s law or the law of volumes is an experimental gas law that describes how gases tend to expand when heated. The figure below shows a piston and cylinder assembly filled with a gas at absolute temperature (T1 = 300°K) and volume (V1 as shown). The piston is free to move against a constant external pressure. Figure: Charles’s Law for Constant Pressure Adding heat causes the temperature of the gas to increase. The volume increases and applies pressure against the piston causing the piston to move outward as the gas temperature increases. The piston will continue to rise until the cylinder pressure on the internal piston face equalizes to the external pressure on the piston, restoring system equilibrium. Rev 2 4 The initial and final pressures are the same but the absolute temperature (T2 = 600°K) is higher and volume (V2) is twice as large as (V1) since the absolute temperature was doubled. Repeating the process of adding heat to the gas, causing the piston to move farther upward, and re-measuring the process variables of gas volume and temperature, will lead to the following conclusion: "The volume of a gas at constant pressure is directly proportional to the temperature of the gas at low pressures. Charles, as the result of experimentation also concluded that the "pressure of a gas varies directly with temperature when the volume is held constant". The mathematical expressions of Charles’s Law are: 𝑉1 𝑇1 𝑃1 𝑇1 = 𝑜𝑟 = 𝑉2 𝑇2 𝑃2 𝑇2 Boyle's Law Now imagine that the piston and cylinder assembly without the heater. Gas fills the cylinder to a gas at volume (V1), temperature (T1), and at an absolute pressure (P1). The cylinder adds no heat to the gas, so the gas temperature remains constant. We move the piston physically to a new position by adding or removing weights, as shown in the figure below. The volume (V2) and absolute pressure (P2) are measured, and the procedure repeated. Examining the measured variables, we develop the following conclusion about the gas: Figure: Boyle's Law – V and P Rev 2 5 Boyle's Law "At low-pressures, the volume of a gas at constant temperature is inversely proportional to the absolute pressure of the gas." This statement is Boyle's Law, written mathematically as: 𝑃1 𝑉1 = 𝑃2 𝑉2 = 𝑃𝑉 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 Combined Gas Law Charles’s Law and Boyle's Law are valid for ideal gases and real gases in the pressure range that the real gas behaves like an ideal gas. Therefore, any real gas at low pressure will obey both these laws, stated as follows: "For a given mass of any gas, the product of the absolute pressure and volume occupied by the gas, divided by its absolute temperature, is a constant." The figure below shows how Boyle’s and Charles’s Laws relate to compression and temperature increases of gases. Figure: Combined Gas Law This statement is the Combined Gas Law, written mathematically as: 𝑃1 𝑉1 𝑃2 𝑉2 𝑃𝑉 = = 𝑇1 𝑇2 𝑇 NOTE: Pressure and temperature MUST be expressed in absolute values. The P-T-V diagram on the next page shows all three relationships from the above equation. Rev 2 6 Figure: PTV Diagram for Combined Gas Law Example A compressor discharges into an air receiver, and cycles off when the pressure in the receiver reaches 160 psia. During the compression, the compressor added heat to the air and its temperature in the receiver is 140°F. Assuming no air loads are in service, at what temperature (°F) should the compressor restart to maintain the receiver above 150 psia? Solution: Assuming an ideal gas: 𝑃1 𝑉1 𝑃2 𝑉2 = 𝑇1 𝑇2 The receiver volume is constant, therefore: 𝑃1 𝑃2 = 𝑇1 𝑇2 𝑇1 (°𝑅) = 460°𝐹 + 140℉ = 600°𝑅 𝑃1 = 160 𝑝𝑠𝑖 𝑃2 = 150 𝑝𝑠𝑖 160 𝑝𝑠𝑖 150 𝑝𝑠𝑖 = 600°𝑅 𝑇2 Rev 2 7 𝑇2 = 600 × 150 160 𝑇2 (°𝑅) = 562.5°𝑅 𝑇2 (°𝐹) = 562.5°𝑅 − 460° = 102.5℉ Ideal Gas Law By combining the results of Charles’s and Boyle's experiments the following is obtained using the specific volume (v = V/M): 𝑃𝑣 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑇 This constant is the ideal gas constant, designated by R0. Pressure, volume, and temperature determines the state of an amount of gas, according to the equation: 𝑃𝑣 = 𝑅0 𝑇 Where: P = the absolute pressure (Pa) v = the specific volume of the vessel containing n moles of gas R0 = the ideal gas constant T = the temperature in degrees Kelvin (°K) Mole It is common practice to discuss quantities of substances in terms of a measurement called a mole. In order to define a mole, we must first define another term, Avogadro's number. Avogadro's number is the number of carbon atoms in 12 grams of carbon-12. The experimentally determined value of Avogadro's number is 6.023 x 1023 atoms. One mole of any substance is equal to the amount of that substance having Avogadro's number of atoms. Normally, atomic mass units (amu) quantify a substance's atomic mass (atomic weight). One-twelfth (1/12) of the mass of a carbon-12 atom defines one amu, which is equivalent to 1.6604 x 10-24 grams. One mole of an element is equal to the atomic mass number of that element in grams. 1 𝑚𝑜𝑙𝑒 = 6.023 × 1023 𝑎𝑡𝑜𝑚𝑠 6.023 × 1023 𝑎𝑡𝑜𝑚𝑠 (12 Rev 2 𝑎𝑚𝑢 ) = 72.27 × 1023 𝑎𝑚𝑢 𝑎𝑡𝑜𝑚𝐶12 8 𝑔 = 72.27 × 1023 𝑎𝑚𝑢 (1.6604 × 10−24 𝑎𝑚𝑢) when 𝑃𝑣 𝑇 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 = 12 𝑔𝑟𝑎𝑚𝑠 Using the element's atomic weight improves the accuracy of the calculation, but the added accuracy is insignificant and is not usually required. Simplifying the relationship yields: 𝑁𝑢𝑚𝑏𝑒𝑟 𝑜𝑓 𝑚𝑜𝑙𝑒𝑠 = 𝑀𝑎𝑠𝑠(𝑔𝑟𝑎𝑚𝑠) (𝐴𝑡𝑜𝑚𝑖𝑐 𝑀𝑎𝑠𝑠)( 𝑔𝑟𝑎𝑚𝑠 ) 𝑚𝑜𝑙𝑒 , or, or mass = (# moles)(GMW) An ideal gas was defined as one in which Pv/T = a constant under all circumstances. Though no such gas exists, the fact that a real gas behaves approximately like an ideal gas provides a specific target for theories for the gaseous state. Experimenters found the constant, in terms of the number of moles (n) of gas in a sample, by making use of the fact that the molar volume of a gas at standard temperature and pressure (STP) is 22.4 liters. At STP: 𝑇𝑒𝑚𝑝𝑒𝑟𝑎𝑡𝑢𝑟𝑒(𝑇) = (0°𝐶 = 460°𝑅) 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒(𝑃) = 1 𝑎𝑡𝑚 𝑉𝑜𝑙𝑢𝑚𝑒(𝑉) = (𝑛) ( Thus, 𝑃𝑣 𝑇 22.4 𝑙𝑖𝑡𝑒𝑟𝑠 ) 𝑚𝑜𝑙𝑒 = 𝐾, 𝐾 = 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑃𝑉 22.4 𝑙𝑖𝑡𝑒𝑟𝑠 1 = (𝓃)(1 𝑎𝑡𝑚) ( )( ) = 𝑛𝑅 𝑇 𝑚𝑜𝑙𝑒 273°𝐾 The figure below shows the classic Ideal Gas Law expression: Figure: Ideal Gas Law Rev 2 9 The figure below expresses the mole-volume relationship of an ideal gas at standard temperature and pressure: Figure: Mole - Volume Relationship We cannot refer to gases that do not obey this law as ideal. Water vapor DOES NOT obey the ideal gas laws. An ideal gas has properties that are constant throughout its mass and whose molecular movements are not influenced by chemical reactions or external forces. There is no known ideal gas. The ideal gas equation is a good approximation to real gases at sufficiently high temperatures and low pressures; that is, PV = RT, where P is the pressure, V is the volume per mole of gas, T is the temperature, and R is the gas constant. At low pressures, all real gases behave like a perfect gas. The ideal gas law is the most accurate for monatomic gases at high temperatures and low pressures. This follows because the law neglects the size of the gas molecules and the intermolecular attractions. Neglecting molecular size becomes less important for larger volumes, i.e., for lower pressures. The relative importance of intermolecular attractions diminishes with increasing thermal kinetic energy. Engineers use the ideal gas law because it is simple to use and approximates real gas behavior. Most physical conditions of gases used by man fit this description. Knowledge Check (Answer Key) According to Charles’s Law, at low pressure, the _____ of a gas at constant _____ is directly proportional to the temperature of the gas. Rev 2 A. density; pressure B. volume; pressure C. pressure; volume D. weight; volume 10 Knowledge Check (Answer Key) Calculate the value of the missing property. P1= 100 psia; P2 = ? V1 = 50 ft3; V2 = 25 ft3 T1 = 60°F; T2 = 70°F A. 233 psi B. 210 psi C. 204 psi D. 200 psi ELO 1.2 Compression Process Introduction The most common use of gas behavior is during the compression process using ideal gas approximations. A compression process may occur at constant temperature (ΔT = constant), constant volume (ΔV = constant), or adiabatic (no heat transfer). The amount of work that results from it depends upon the process, as brought out in the study of the first law of thermodynamics. When using ideal gas assumptions, the compression process results in work performed on the system, and is essentially the area under a P-V curve. Maintaining constant temperature or maintaining constant pressure results in different amounts of work as shown in the figure below. Figure: Pressure-Volume Diagram Rev 2 11 Compressibility A fluid is any substance that conforms to the shape of its container; a fluid may be either a liquid or a gas. A fluid is considered incompressible when the velocity of the fluid is greater than one-third of the speed of sound for the fluid, or if the fluid is a liquid. We assume that such a fluid has a constant density. The variation of density of the fluid with changes in pressure is the primary factor considered in deciding whether a fluid is incompressible. Because compressible fluids experience density changes, their property relationships vary more than incompressible fluids. In addition, it is easy to determine the state of a liquid if you know its temperature and pressure. The process becomes more difficult once the substance becomes a gas. Figure: P-V Diagram for Gas Constant Pressure Process As shown on the above P-V diagram, the work done in a constant pressure process is the product of the pressure and the change in volume. 𝑊1−2𝑎 = 𝑃(𝛥𝑉) Constant Temperature Process The work done in a constant temperature process is the product of the temperature and the change in volume. 𝑊1−2𝑏 = 𝑇(𝛥𝑉) Rev 2 12 Constant Volume Process The work done in a constant volume process is the product of the volume and the change in pressure. 𝑊1−2𝑐 = 𝑉(𝛥𝑃) The above equation also applies to liquids. The power requirement for pumps that move incompressible liquids (such as water) can be determined by replacing the volume (V) with the product of the specific volume and the mass. Power Requirements 𝑊1−2𝑐 = 𝑚𝑣(𝛥𝑃) Taking the time rate of change of both sides determines the power requirements of the pump. 𝑊̇1−2𝑐 = 𝑚̇𝑣(∆𝑃) Effects of Pressure Changes on Fluid Properties The predominant effect of a pressure increase in a compressible fluid, such as a gas, is an increase in the fluid density. A pressure in an incompressible fluid will not result in a significant effect on the density. For example, increasing the pressure of 100°F water from 15 psia to 15,000 psia will only increase its density by approximately 6 percent. Therefore, in engineering calculations, we assume that the density of incompressible fluids remains constant. Effects of Temperature Changes on Fluid Properties An increase in temperature will tend to decrease the density of any fluid. The effect of a temperature change will depend on whether the fluid is compressible if the fluid is confined in a container of fixed volume. A gas will respond to a temperature change in a manner predicted by the ideal gas laws. A 5 percent increase in absolute temperature will result in a 5 percent increase in the absolute pressure. If the fluid is an incompressible liquid in a closed container, a fluid temperature increase will have a tremendously greater and potentially catastrophic effect. The fluid tries to expand, but the walls of the container prevent its expansion as the fluid temperature increases. Because the fluid is incompressible, this results in a tremendous pressure increase for a relatively minor temperature change. The change in specific volume for a given change in temperature is not the same at various beginning temperatures. Resultant pressure changes will vary. Rev 2 13 A useful thumb rule for water is that pressure in a water-solid system will increase about 100 psi for every 1°F increase in temperature. Knowledge Check (Answer Key) When can a fluid be considered incompressible? A. if it is liquid B. if it is steam but not flowing C. if it is a saturated vapor D. if it is a superheated steam Knowledge Check (Answer Key) A contained fluid is heated. The resulting change in pressure will be… A. greater for an incompressible fluid. B. greater for a compressible fluid. C. the same for both fluids. D. the same for both fluids only if the volume is held constant. TLO 1 Summary Review each ELO with the class by using good questioning techniques. Example of an effective method for asking directed questions: Rev 2 I have a question...I will select someone to respond" Ask the question Pause Select an individual to answer the question Ensure that everyone heard and understood the response, amplify and re-state as necessary 14 ELO 2.1 What is Charles’s Law, and what is the mathematical relationship? The volume of a gas at constant pressure is directly proportional to the temperature of the gas at low pressures. 𝑉1 𝑇1 𝑃1 𝑇1 = 𝑜𝑟 = 𝑉2 𝑇2 𝑃2 𝑇2 What is Boyle's Law, and what is the mathematical relationship? At low-pressures, the volume of a gas at constant temperature is inversely proportional to the absolute pressure of the gas (𝑃1 )(𝑉1 ) = (𝑃2 )(𝑉2 ) = (𝑃3 )(𝑉3 ) = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 What is the mathematical expression of the combined law? 𝑃1 𝑉1 𝑃2 𝑉2 𝑃𝑉 = = 𝑇1 𝑇2 𝑇 What pressures and temperatures do we use in solving the gas laws? Temperature and Pressure MUST BE IN ABSOLUTE. Explain the ideal gas law. Ideal gases follow the above laws. A specific gas constant is used to account for difference in gases atomic structure (R). 𝑃𝑣 = 𝑅𝑇 What is a mole of gas? One mole of any substance is that amount having Avogadro's Number of 6.023 x 1023 atoms. We use moles to account for large volumes of gas in the ideal gas equation. Rev 2 15 ELO 2.2 When can we assume that a fluid is incompressible? A fluid may be considered incompressible when the velocity of the fluid is greater than one-third of the speed of sound for the fluid, or if the fluid is a liquid. What is the major parameter used to determine incompressibility? Variation of fluid density with pressure changes is a primary factor considered in deciding whether a fluid is incompressible. An increase in the pressure of an incompressible fluid will not have a significant effect on the fluid density. Increasing the pressure of 100°F water from 15 psia to 15,000 psia will only increase the water density by approximately 6 percent. Explain how a compressible fluid responds to temperature changes. For a fluid in a closed container of fixed volume, the effect of a temperature change will depend on whether the fluid is compressible. If the fluid is a compressible gas, it will respond to a temperature change in a manner predicted by the ideal gas laws. A 5 percent increase in absolute temperature will result in a 5 percent increase in the absolute pressure. If the fluid is an incompressible liquid in a closed container, an increase in the temperature will have a tremendously greater and potentially catastrophic effect. Rule of thumb for water: pressure in a water-solid system will increase about 100 psi for every 1°F increase in temperature. Objectives Now that you have completed this lesson, you should be able to: 1. Describe the ideal gas laws and explain how to solve for an unknown pressure, temperature, or volume. 2. Describe the effects of pressure and temperature changes on confined fluids. Rev 2 16 TLO 2 The Second Law of Thermodynamics Overview The first law of thermodynamics requires a balance of the various forms of energy as they pertain to the specified thermodynamic system and/or control volume studied. However, this first law of thermodynamics does not take into account the feasibility of the process or the efficiency of the energy transformation studied. The second law of thermodynamics allows us to determine the maximum efficiency of the operating system so that design comparisons maximize the system’s efficiency. The second law of thermodynamics states that it is impossible to construct a device that operates within a cycle that can convert all the heat supplied it into mechanical work. Recognizing that even the most thermally and mechanically perfect cycles must reject some heat defines thermodynamic power cycle efficiency. Maximizing the power cycle efficiency is a major part of the operator's job. There are many important and interdependent plant parameters affecting plant efficiency. To produce the maximum electrical power output for the allowed core thermal power input, the operator must continually monitor these parameters and adjust plant conditions as necessary. Advancements in sensing these key plant parameters give the operator real time data of actual plant efficiency and core power levels, so the operator is able to make immediate adjustments to maintain the plant within licensed limitations, set by the Nuclear Regulatory Commission (NRC). The ability to relate current plant conditions to the plant’s thermal efficiency is a fundamental operator attribute. Precise control of plant parameters requires continual oversight by the operator and adjustments as needed. Objectives Upon completion of this lesson, you will be able to do the following: 1. Explain the second law of thermodynamics using the term entropy. 2. Given a thermodynamic system, determine the: a. Maximum efficiency of the system b. Efficiency of the components within the system 3. Differentiate between the path for an ideal process and that for a real process that for a real process on a T-s or h-s diagram. 4. Describe how individual factors affect system or component efficiency. Rev 2 17 ELO 2.1 Thermodynamic Entropy Introduction We must first understand that the first law of thermodynamics governs all cycles when we examine thermodynamic cycles. The first law states energy can be neither created nor destroyed, but only altered in form. This means that all of the energy added to a cycle must be accounted for in its entirety. The first law of thermodynamics places no restrictions on how conversions from heat to work or vice versa take place or to what extent these conversions may proceed, which is addressed by the second law of thermodynamics. The second law is based on experimental evidence and observations of actual processes. It suggests that processes proceed in a certain direction but not in the opposite direction. The second law, which implies that all real processes are irreversible, governs all real processes. With the second law of thermodynamics, the limitations imposed on any process can be studied to determine the maximum possible efficiencies of such a process then a comparison can be made between the maximum possible efficiency and the actual efficiency achieved. Energy-Conversion Systems One of the areas of application of the second law is the study of energyconversion systems. For example, it is not possible to convert all the energy obtained from a nuclear reactor into electrical energy. There must be losses in the conversion process. As shown the figure below on the left as heat flows from the heat source to the heat sink, it is capable of doing work. As shown below in the figure on the right, not all of the heat transfers into work. This is the second law: there must be rejected heat. Figure: Second Law of Thermodynamics for a Heat Engine Rev 2 18 Considering those losses, the second law of thermodynamics can be used to derive an expression for the maximum possible energy conversion efficiency. The second law denies the possibility of completely converting into work all of the heat supplied to a system operating in a cycle, no matter how perfectly designed the system may be. The restriction placed by the second law requires that some of the heat supplied (QS) to the engine must be rejected as heat (QR). The difference between the heat supplied and the heat rejected is the net amount of work produced in the cycle (WNET). The cycle efficiency is the percentage of energy input to a cycle that is converted to net work output. The concept of the second law is best stated using Max Planck's description: Figure: Kelvin-Planck's Second Law of Thermodynamics The first law of thermodynamics does not define the energy conversion process completely. The first law relates to and evaluates the various energies involved in a process. However, no information about the direction of the process can be obtained by the application of the first law. Early in the development of the science of thermodynamics, investigators noted that while work could be converted completely into heat, the converse was never true for a cyclic process. Certain natural processes were observed always to proceed in a certain direction; for example, heat transfer occurs from a hot to a cold body. The second law was developed as an explanation of these natural phenomena. Figure: Heat Flow Direction Rev 2 19 Entropy The physical property of matter called entropy (S) explains the second law of thermodynamics. The change in entropy determines the direction in which a given process proceeds. Entropy also measures the unavailability of heat to perform work in a cycle. The second law predicts that not all heat provided to a cycle can be transformed into an equal amount of work; some heat rejection must take place. The change in entropy is the ratio of heat transferred during a reversible process to the absolute (abs) temperature of the system. ∆𝑆 = ∆𝑄 (𝑓𝑜𝑟 𝑎 𝑟𝑒𝑣𝑒𝑟𝑠𝑖𝑏𝑙𝑒 𝑝𝑟𝑜𝑐𝑒𝑠𝑠) 𝑇𝑎𝑏𝑠 Where: • ∆𝑆 = the change in entropy of a system during some process (BTU/°R [degrees Rankine]) • ∆𝑄 = the amount of heat added to the system during the process (BTU) (∆Q is change in heat) • 𝑇𝑎𝑏𝑠 = the absolute temperature at which the heat was transferred (°R) Figure: Entropy Entropy (S) is a natural process that starts in one equilibrium state, ends in another state, and will go in the direction that causes the entropy of the system plus the environment to increase for an irreversible process and to remain constant for a reversible process. Therefore, Sf = Si (reversible) and Sf > Si (irreversible). The second law of thermodynamics is also expressed as ΔS ≥ 0 for a closed cycle. In other words, entropy must increase or stay the same for a cyclic system; it can never decrease. Rev 2 20 Entropy is an extensive property of a system and like the total internal energy or total enthalpy, may be calculated from specific entropies based on a unit mass quantity of the system, so that S = ms. Values of the specific entropy are tabulated along with specific enthalpy, specific volume, and other thermodynamic properties of interest in the steam tables described in a previous module. The property of specific entropy is used advantageously as one of the coordinates when representing a reversible process graphically. The area under a reversible process curve on the T-s diagram represents the quantity of heat transferred during the process. Figure: T-s Diagram With Rankine Cycle Reversible processes are often used in thermodynamic problems by comparison to the real irreversible process to aid in a second law analysis. Reversible processes can be depicted on diagrams such as h-s and T-s, shown below in the figure. The dashes represent the intermediate state of the fluid is not determined, only the beginning and end states are known. In both real processes shown below, entropy increases, the dashed lines slant to the right. The compression of the fluid by the pump and the expansion of the fluid through the turbine both increase entropy compared to the ideal case. Figure: T-s and h-s Diagrams for Expansion and Compression Processes Rev 2 21 Knowledge Check (Answer Key) The second law of thermodynamics can also be expressed as ________ for a closed cycle. A. Sf = Si B. ΔS ≥ 0 C. ΔT < 0 D. ΔS < 0 ELO 2.2 Carnot’s Principle of Thermodynamics Introduction In 1824 Nicolas Léonard Sadi Carnot, a French military engineer and physicist known as the father of thermodynamics, advanced the study of the second law by using reversible processes that disclosed a principle consisting of the following: No engine can be more efficient (η) than a reversible (ideal) engine operating between the same high temperature heat source and low temperature heat sink. Efficiencies of all reversible engines operating between the same constant temperature reservoirs are the same. Efficiency of a reversible engine depends only upon the temperatures of the heat source and heat receiver. Figure: Carnot's Efficiency Principle Rev 2 22 Carnot Cycle Guidelines The Carnot cycle can best be described using an ideal frictionless thermally isolated piston operating between a constant heat source and heat sink. A P-v and T-s diagram shown below illustrates the cycle as the heat source is applied to the piston, causing a reversible isothermal expansion between point 1 and 2. The piston then moves doing an amount of work (w1-2) due to the isothermal (constant temperature) expansion of the gas, shown on the figure as the line between points 1-2. The gas is allowed to finish expanding adiabatically between point 2 and 3 and an amount of work (w2-3) is done, shown as the line between points 2-3. Next, the heat sink is applied to the piston, and a reversible isothermal compression of the gas occurs between points 3 and 4. The piston is used to compress the gas and an amount of heat (q3-4) is transferred to the heat sink through the cylinder head. This isothermal compression requires some amount of work (w3-4) to be done on the piston, shown as the line between points 3-4. In process 4, the cylinder is removed from the heat sink. The piston returns to its initial state by undergoing adiabatic compression requiring some amount of work (w4-1). The cycle is completed when the cylinder is again placed in contact with the heat source, shown in the figure below as the line between points 4-1: Figure: Single Piston Carnot Engine Cycle Rev 2 23 During the Carnot cycle just described, a certain amount of heat and work were added or removed from the system. Work is done on the system when the piston travels into the cylinder and compresses the gas. Work is done by the system as the gas expands to force the piston out of the cylinder. Heat is added to the system to cause the piston to move outward (𝑄𝐴 = 𝑞1−2 ). Heat is added, entropy increases, and the process line goes from left to right. Heat is removed from the system as the piston is compressed in the final portion of the cycle (𝑄𝑅 = 𝑞3−4 ). Entropy decreases, and the process line goes from right to left as heat is removed from the system. Two ideal assumptions are made that result in the Carnot cycle having the highest possible efficiency: 1. Both work processes occur with no friction and thus there is no change in entropy. 2. The heat addition and heat rejection occur with no change in the temperature of the working fluid. Therefore, the temperature difference (T) between the working fluid and the heat source and the heat sink remains constant. We define thermodynamic cycle efficiency by analyzing the energy output or work (W) produced compared to the energy input (QA). The greater the percentage of energy input converted to work, the greater the cycle efficiency. The figure below shows a Carnot cycle representation. The heat input (QH) is the area under line 2-3. The heat rejected (QC) is the area under line 1-4. The difference between the heat added and the heat rejected is the net work (sum of all work processes), represented as the area of rectangle 1-2-3-4. Figure: Carnot Cycle Representation In a perfectly efficient cycle, all of the energy put into the cycle converts to a useful work output. However, as stated previously, heat must be rejected for the cycle to be continuous. Rev 2 24 The efficiency (η) of the cycle is the ratio of the net work of the cycle to the heat input to the cycle. This ratio can be expressed by the following equation: 𝜂= 𝑊𝑛𝑒𝑡 ℎ𝑒𝑎𝑡 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 𝜂= (𝑄𝐴 − 𝑄𝑅 ) 𝑄𝐴 𝜂= (𝑄𝐻 − 𝑄𝐶 ) (𝑇𝐻 − 𝑇𝐶 ) = 𝑄𝐻 𝑇𝐻 𝑇 = 1 − (𝑇 𝐶 )Where: 𝐻 • η = cycle efficiency • TC = designates the low-temperature reservoir (°R) • TH = designates the high-temperature reservoir (°R) This equation shows that the maximum possible efficiency exists when TH is at its highest possible value or when TC is at its lowest value. The above represents an upper limit of efficiency for any given system operating between the same two temperatures since all practical systems and processes are irreversible. The system's maximum possible efficiency would be that of a Carnot cycle, but because Carnot cycles represent reversible processes, the real system cannot reach the Carnot efficiency value. Thus, the Carnot efficiency serves as an unattainable upper limit for any real system's efficiency. The following example demonstrates the above principles. Example 1: Carnot Efficiency An inventor claims to have an engine that receives 100 BTUs of heat and produces 25 BTUs of useful work when operating between a source at 140 °F and a receiver at 0 °F. Is the claim a valid claim? Solution 1: 𝑇ℎ = 140 ℉ + 460 = 600 °𝑅 𝑇𝑐 = 0 ℉ + 460 = 460 °𝑅 𝜂= 600 − 460 × 100 = 23.3% 600 Claimed efficiency = 25/100 = 25 percent; this exceeds the Carnot efficiency value. Therefore, the claim is invalid. Rev 2 25 The second law determines the maximum possible efficiencies obtained from a power system. Actual efficiencies will always be less than this maximum. Real systems have losses, such as friction, that are not reversible and that preclude real systems from obtaining the maximum possible efficiency. An illustration of the difference that may exist between the ideal and actual efficiency is presented in the figure below and in the following example: Example 2: Actual Versus Ideal Efficiency The actual efficiency of a steam cycle is 18.0 percent. The facility operates from a steam source at 340 °F and rejects heat to atmosphere at 60 °F. Compare the Carnot efficiency to the actual efficiency. Figure: Real Process Cycle Compared to Carnot Cycle Solution: Solve for the Carnot maximum efficiency: 𝑇𝑐 𝜂 = 1−( ) 𝑇ℎ 𝜂 = 1− 60 + 460 340 + 460 𝜂 = 1− 520 800 𝜂 = 1 − 0.65 𝜂 = 35% Therefore, ideal Carnot efficiency of 35 percent compared to 18.0 percent actual efficiency. Rev 2 26 The second law equations are treated in much the same manner as the first law equations. An isolated, closed, or open system used in the analysis depends on the types of energy that cross the boundary. The open system analysis is still the more general case, with the closed and isolated systems being special cases of the open system. The approach used to solve second law problems is similar to that used in the first law analysis. A control volume using the second law is shown below in the figure. In this diagram, the fluid moves through the control volume from the inlet section to the outlet section while work is delivered externally to the control volume. We assume that the boundary of the control volume is at some environmental temperature and that all of the heat transfer (Q) occurs at this boundary. Entropy is a property that may be transported with the flow of the fluid into and out of the control volume, just like enthalpy or internal energy. The entropy flow into the control volume resulting from mass transport, 𝑚̇𝑖𝑛 𝑠𝑖𝑛 , and the entropy flow out of the control volume is 𝑚̇𝑜𝑢𝑡 𝑠𝑜𝑢𝑡 , assuming that the properties are uniform at sections in and out. Entropy may also be added to the control volume because of heat transfer at the boundary of the control volume. Figure: Control Volume for Second Law Analysis A simple demonstration of the use of this form of system in second law analysis gives the student a better understanding of its use. Rev 2 27 Example 3: Open System Second Law Steam enters the nozzle of a steam turbine with a velocity of 10 ft/sec at a pressure of 100 psia and temperature of 500 °F. At the nozzle discharge, the pressure and temperature are one (1) atmospheric pressure (atm) at 300 °F. What is the increase in entropy for the system if the mass flow rate is 10,000 lbm/hr? Solution: 𝑚̇𝑠𝑖𝑛 + ∆𝑠 = 𝑚̇𝑠𝑜𝑢𝑡 Where: • ∆𝑠 = entropy added to the system • ∆𝑠 = 𝑚̇(𝑠𝑜𝑢𝑡 − 𝑠𝑖𝑛 ) • sin = 1.7088 BTUs/lbm-°R (from steam tables) • sout = 1.8158 BTUs/lbm-°R (from steam tables) • ∆𝑠 • ∆𝑠 • ∆𝑠 = 10,000(0.107) • ∆𝑠 = 1,070 𝑚̇ 𝑚̇ 𝐵𝑇𝑈 = 𝑠𝑜𝑢𝑡 − 𝑠𝑖𝑛 = 1.8158 − 1.7088 𝑙𝑏𝑚-°𝑅 𝐵𝑇𝑈 = 0.107 𝑙𝑏𝑚-°𝑅 𝐵𝑇𝑈 𝑙𝑏𝑚-°𝑅 = 𝑒𝑛𝑡𝑟𝑜𝑝𝑦 𝑎𝑑𝑑𝑒𝑑 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑦𝑠𝑡𝑒𝑚 The second law of thermodynamics gives a maximum efficiency limit (which is never reached in physical systems) that an ideal thermodynamic system can perform. The efficiency is determined by knowing the inlet and exit absolute temperatures of the overall system (one that works in a cycle) and applying Carnot's efficiency equation. Rev 2 28 Knowledge Check (Answer Key) The steam generators produce dry saturated steam at 1,000 psig. The main condenser is operating with zero subcooling at 1 psia. What is the maximum efficiency obtainable? A. 44 percent B. 34 percent C. 28 percent D. 23 percent Knowledge Check (Knowledge Check) Determine the Carnot Efficiency of a steam engine that is supplied with saturated steam at 300 psia and exhausts to atmosphere… A. 44 percent B. 56 percent C. 42 percent D. 35 percent ELO 2.3 Thermodynamics of Ideal and Real Processes Introduction It is convenient to arrange the various thermodynamic processes on a property diagram in evaluating the various cycles present in a nuclear power plant. The most common set of coordinates used is a plot of temperature versus specific entropy, a T-s diagram. Using this type of diagram, we can analyze the various processes that take place and how these processes effect the entire cycle as well as the amount of heat and work, both of which occur during the processes. Any ideal thermodynamic process can be drawn as a path on a property diagram. A real process that approximates the ideal process can also be represented on the same diagrams, usually by dashed lines. Rev 2 29 Carnot Cycle Guidelines Entropy is constant in an ideal expansion or compression process. Isentropic processes are represented by vertical lines on T-s and h-s diagrams, shown below in the figures. A real expansion and real compression process operating between the same pressures as the ideal process are shown by dashed lines and will slant slightly toward the right, since the entropy increases from the start to the end of the real process. All real processes are irreversible. It is helpful to compare real processes to ideal processes in system design. The reversible process indicates a maximum work output for a given input, which compares to real work output for efficiency purposes. The h-s diagram clearly shows that the real expansion process (turbine) results in a smaller change in enthalpy, meaning less energy is extracted in the real turbine than the ideal turbine. The figure below also shows that more enthalpy must be added during the compression (pump) process, meaning more energy must be added to the real system than the ideal system. These differences in enthalpy across various components tell us how efficient the real process compares to the ideal process of the Carnot cycle. Figure: T-s and h-s Diagrams for Expansion and Compression Processes Rev 2 30 Knowledge Check (Answer Key) Why are real processes shown with dotted lines on property diagrams? A. They occur faster than real processes. B. The value of entropy during the process is not determined. C. The entropy values during the process are the same as the real process until the outlet from the process. D. You would not be able to distinguish between real and ideal processes if the real process was a solid line. ELO 2.4 Thermodynamic Power Plant Efficiency Introduction The actual construction of the steam cycle in nuclear power plants is considerably more complex than the basic cycles that we have covered so far, including numerous required auxiliary systems and instrumentation and control equipment. However, adding this detail to the steam cycle does not contribute significantly to an understanding of the energy transfer characteristics of the overall plant. A simplified steam cycle for a typical steam-electric plant is shown below in the figure. Notice that the cycle shown includes two components we have not yet discussed, the Moisture Separator Reheater (MSR) and the Feedwater Preheater. We discuss both of these components and their effect on cycle efficiency later in this module. To analyze a complete power plant steam cycle, it is first necessary to analyze the elements that make up the cycle. Although specific designs differ, there are three basic types of elements in power cycles: (1) turbines, (2) pumps, and (3) heat exchangers. Each of these three elements imparts a characteristic change in the properties of the working fluid. Typical Steam Cycle Previously we calculated system efficiency by knowing the temperature of the heat source and the heat sink. It is also possible to calculate the efficiencies of each individual component by comparing the actual work produced by the component to the work that produced by an ideal component operating isentropically between the same inlet and outlet conditions. Rev 2 31 Figure: Typical Steam Cycle Steam Turbine Efficiency Guidelines A steam turbine extracts energy from the working fluid (steam) to do work in the form of rotating the turbine shaft. The steam works as it expands through the turbine. The shaft work converts to electrical energy by the generator. In the application of the first law general energy equation to a simple turbine under steady flow conditions, demonstrates that the decrease in the enthalpy of the working fluid Hin - Hout equals the work done by the working fluid in the turbine (Wt). Figure: Turbine Work Rev 2 32 𝐻𝑖𝑛 − 𝐻𝑜𝑢𝑡 = 𝑊𝑡 𝑚̇(ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 ) = 𝑤̇𝑡 Where: • Hin = enthalpy of the working fluid entering the turbine (BTU) • Hout = enthalpy of the working fluid leaving the turbine (BTU) • Wt = work done by the turbine (ft-lbf) • 𝑚̇ = mass flow rate of the working fluid (lbm/hr) • hin = specific enthalpy of the working fluid entering the turbine (BTU/lbm) • hout = specific enthalpy of the working fluid leaving the turbine (BTU/lbm) • 𝑤̇𝑡 = power of the turbine (BTU/hr) Ideal Versus Real Turbine The calculation of turbine work using only the enthalpy change is valid because the change of kinetic and potential energy and the amount of heat lost by the working fluid while in the turbine are negligible. These assumptions are valid for most practical applications. However, to apply these relationships, one additional definition is necessary. In any ideal case, the working fluid does work reversibly by expanding at constant entropy. In an ideal turbine, the entropy of the working fluid entering the turbine Sin equals the entropy of the working fluid leaving the turbine. 𝑆𝑖𝑛 = 𝑆𝑜𝑢𝑡 or 𝑠𝑖𝑛 = 𝑠𝑜𝑢𝑡 Where: • Sin = entropy of the working fluid entering the turbine (BTU/°R) • Sout = entropy of the working fluid leaving the turbine (BTU/°R) • sin = specific entropy of the working fluid entering the turbine (BTU/lbm-°R) • sout = specific entropy of the working fluid leaving the turbine (BTU/lbm-°R) An ideal turbine performs the maximum amount of work theoretically possible, and therefore provides a basis for analyzing the performance of real turbines. Rev 2 33 Because of friction losses in the blades, steam leakage past the blades and to a lesser extent mechanical friction, a real turbine does less work than an ideal turbine. Turbine efficiency (ηt), is defined as the ratio of the actual work done by the turbine (Wt.actual) to the work that would be done by the turbine if it were an ideal turbine (Wt.ideal). It is shown that Wi is larger than Wa, meaning more enthalpy is extracted from the steam in an ideal turbine operating between the same temperatures (pressures) than in a real turbine. Figure: h-s Diagram for Ideal and Real Turbines An actual turbine does less work than an ideal turbine because of factors such as: Friction losses in the turbine blades Steam leakage past the turbine blades Mechanical friction Turbine efficiency (ηt) is the ratio of the actual work done by the turbine Wt.actual to the work that would be done by the turbine if it were an ideal turbine Wt.ideal Generally, turbine efficiency is 60% - 80% for small turbines and 90% for large turbines such that: 𝜂𝑡 = 𝑊𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 𝑊𝑡.𝑖𝑑𝑒𝑎𝑙 𝜂𝑡 = (ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 )𝑎𝑐𝑡𝑢𝑎𝑙 (ℎ𝑖𝑛 − ℎ𝑜𝑢𝑡 )𝑖𝑑𝑒𝑎𝑙 Where: • 𝜂𝑡 = turbine efficiency (no units) Rev 2 34 • Wt.actual = actual work done by the turbine (ft-lbf) • Wt.ideal = work done by an ideal turbine (ft-lbf) • (hin – hout)actual = actual enthalpy change of the working fluid (BTU/lbm) • (hin – hout)ideal = actual enthalpy change of the working fluid in an ideal turbine (BTU/lbm) Figure: Entropy Diagram Measures System Efficiency A vertical line on the T-s diagram is a constant entropy ideal process. Entropy increases in the actual turbine process. The smaller the increase in entropy, the closer the turbine efficiency (ηt) is to 1.0 or 100 percent. If an ideal turbine became a real turbine, its output would decrease due to losses such as friction, windage, moisture, and tip leakage. To raise the turbine back to its original output, the turbine steam supply valves would open, increasing the mass flow rate of the steam going into the turbine. The work of the turbine increases to overcome the losses. Opening the steam supply valves decreases steam generator pressure and adding more heat raises the steam generator pressure back to its original value. However, since the heat added is greater than the increase in work from the turbine, the cycle efficiency decreases. Rev 2 35 Pump Efficiency Guidelines A pump performs work on a system's working fluid to overcome the head loss and keep the fluid moving. Like the turbine, the application of the first law general energy equation to a simple pump under steady flow conditions results in the work of the pump (Wp), equals the change in working fluid enthalpy across the pump ( Hout - Hin). Figure: Work Done by the Pump Equals Change in Enthalpy Across the Ideal Pump 𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 = 𝑊𝑝 𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 ) = 𝑊̇𝑝 Where: • Hout = enthalpy of the working fluid leaving the pump (BTU) • Hin = enthalpy of the working fluid entering the pump (BTU) • Wp = work done by the pump on the working fluid (ft-lbf) • 𝑚̇ = mass flow rate of the working fluid (lbm/hr) • hout = specific enthalpy of the working fluid leaving the pump (BTU/lbm) • hin = specific enthalpy of the working fluid entering the pump (BTU/lbm) • 𝑊̇𝑝 = power of pump (BTU/hr) Real Versus Ideal Pump As in the turbine, the kinetic and potential energy changes and the heat lost by the working fluid while in the pump are negligible. These are valid assumptions along with the assumption that the working fluid is incompressible. For the ideal case, it can be shown that the work done by the pump (Wp) equals the change in enthalpy across the ideal pump. Rev 2 36 𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 = (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙 𝑊̇𝑝.𝑖𝑑𝑒𝑎𝑙 = 𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙 Where: • Wp = work done by the pump on the working fluid (ft-lbf) • Hout = enthalpy of the working fluid leaving the pump (BTU) • Hin = enthalpy of the working fluid entering the pump (BTU) • 𝑊̇𝑝 = power of pump (BTU/hr) • 𝑚̇ = mass flow rate of the working fluid (lbm/hr) • hout = specific enthalpy of the working fluid leaving the pump (BTU/lbm) • hin = specific enthalpy of the working fluid entering the pump (BTU/lbm) The ideal pump provides a basis for analyzing the performance of actual pumps, which requires more work because of unavoidable losses due to friction and fluid turbulence. The work done by a pump Wp equals the change in enthalpy across the actual pump. 𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = (𝐻𝑜𝑢𝑡 − 𝐻𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙 𝑊̇𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑚̇(ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙 Pump Efficiency Now, consider a real pumping process opposed to the ideal pumping process. In the real pumping process with friction taken into account, the entropy increases across the pump. Recall that entropy increases in all real processes. Because the work of the pump (WPUMP) increases to make up for the frictional losses, the net work (WNET) decreases. Since the net work decreases, the overall cycle efficiency also decreases. Pump efficiency (ηp) is the ratio of the work required by the pump if it were an ideal pump (Wp.ideal) to the actual work required by the pump (Wp.actual). 𝜂𝑝 = 𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 Also shown as: 𝜂𝑝 = 𝐹𝑙𝑜𝑤 Rev 2 𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 𝐵𝑟𝑎𝑘𝑒 𝐻𝑜𝑟𝑠𝑒𝑝𝑜𝑤𝑒𝑟 37 Example: A pump operating at 75 percent efficiency has an inlet specific enthalpy of 200 BTU/lbm. The exit specific enthalpy of the ideal pump is 600 BTU/lbm. What is the exit specific enthalpy of the actual pump? Solution: Using the equation above: 𝜂𝑝 = 𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 = 𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 𝜂𝑝 (ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑎𝑐𝑡𝑢𝑎𝑙 = ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = (ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙 𝜂𝑝 (ℎ𝑜𝑢𝑡 − ℎ𝑖𝑛 )𝑖𝑑𝑒𝑎𝑙 + ℎ𝑖𝑛.𝑎𝑐𝑡𝑢𝑎𝑙 𝜂𝑝 (600 𝐵𝑇𝑈 𝐵𝑇𝑈 − 200 ) 𝑙𝑏𝑚 𝑙𝑏𝑚 + 200 𝐵𝑇𝑈 0.75 𝑙𝑏𝑚 ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = 533.3 𝐵𝑇𝑈 𝐵𝑇𝑈 + 200 𝑙𝑏𝑚 𝑙𝑏𝑚 ℎ𝑜𝑢𝑡.𝑎𝑐𝑡𝑢𝑎𝑙 = 733.3 𝐵𝑇𝑈/𝑙𝑏𝑚 Pump efficiency (ηp) relates the minimum amount of work theoretically possible to the actual work required by the real pump. However, the work required by a pump is normally only an intermediate form of energy. Usually, a motor or turbine runs the pump. Pump efficiency does not account for losses in this motor or turbine. An additional efficiency factor, motor efficiency (ηm) is the ratio of the actual work required by the pump to the electrical energy input to the pump motor, when expressing both in the same units. Rev 2 38 𝜂𝑚 = 𝑊𝑝.𝑎𝑐𝑡𝑢𝑎𝑙 𝑊𝑚.𝑖𝑛 𝐶 Where: • 𝜂𝑚 = motor efficiency (no units) • Wp.actual = actual work required by the pump (ft-lbf) • Wm.in = electrical energy input to the pump motor per kilowatt hour (kWh) • C = conversion factor = 2.655 x 106 ft-lbf/kWh Motor Efficiency Guidelines Like pump efficiency, motor efficiency is always less than 1.0 or 100 percent for an actual pump motor. The combination of pump efficiency and motor efficiency relates the ideal pump to the electrical energy input to the pump motor. 𝜂𝑚 𝜂𝑝 = 𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 𝑊𝑚.𝑖𝑛 𝐶 Where: • 𝜂𝑚 = motor efficiency (no units) • 𝜂𝑝 = pump efficiency (no units) • 𝑊𝑝.𝑖𝑑𝑒𝑎𝑙 = ideal work required by the pump (ft-lbf) • 𝑊𝑚.𝑖𝑛 = electrical energy input to the pump motor (kWh) • C = conversion factor = 2.655 x 106 ft-lbf/kWh Real Versus Ideal Cycle Efficiency Guidelines In the preceding sections, we discussed the Carnot cycle, cycle efficiencies, and component efficiencies. In this section, we apply this information to compare and evaluate various ideal and real cycles. This determines how modifying a cycle affects the cycle's available energy that can be extracted for work. A Carnot cycle's efficiency depends solely on the temperature of the heat source and the heat sink. To improve a cycle's efficiency, all we have to do is increase the temperature of the heat source and decrease the temperature of the heat sink. In the real world, the ability to do this is limited by the following constraints. Rev 2 39 For a real cycle, the heat sink is limited by the fact that the earth is our final heat sink, and therefore is fixed at about 60 °F (520 °R). The heat source is limited to the combustion temperatures of the fuel burned or the maximum limits placed on nuclear fuels by their structural components (pellets, cladding, etc.). In the case of fossil fuel cycles, the upper limit is ~ 3,040 °F (3,500 °R). However, even this temperature is not attainable due to the metallurgical restraints of the boilers; and therefore, they are limited to about ~ 1,500 °F (~ 1,960 °R) for a maximum heat source temperature. Using these limits to calculate the maximum efficiency attainable by an ideal Carnot cycle gives the following: 𝜂= 𝑇𝑆𝑂𝑈𝑅𝐶𝐸 − 𝑇𝑆𝐼𝑁𝐾 1,960 °𝑅 − 520 °𝑅 = = 73.5% 𝑇𝑆𝑂𝑈𝑅𝐶𝐸 1,960 °𝑅 This calculation indicates that the Carnot cycle, operating with ideal components under real world constraints, should convert almost threequarters of the input heat into work. This ideal efficiency is beyond the present capabilities of any real systems. Heat Rejection By analyzing a T-s diagram, we will understand why an efficiency of 73 percent is not possible for real components. The energy added to a working fluid during the Carnot isothermal expansion is given by qs. Not all of this energy is available for use by the heat engine since a portion of it (qr) must be rejected to the environment. This is given by: 𝑞𝑟 = 𝑇𝑜 𝛥𝑠 in units of BTU/lbm Where: • To = the average heat sink temperature of 520 °R The available energy (A.E.) for the Carnot cycle may be given as: • 𝐴. 𝐸. = 𝑞𝑠 − 𝑞𝑟 Substituting the above equation for qr gives: 𝐴. 𝐸. = 𝑞𝑠 − 𝑇𝑜 𝛥𝑠 in units of BTU/lbm, (Δs is change in absolute entropy) and equals the area of the shaded region labeled available energy in the figure on the next page between the temperatures 1,962 °R and 520 °R. Rev 2 40 Figure: Entropy Measures Temperature as Pressure and Volume Increase Typical Power Cycle A typical power cycle employed by a fossil fuel plant as shown on the above below. The working fluid is water, which places certain restrictions on the cycle. If we wish to limit ourselves to operation at or below 2,000 psia, it is readily apparent that constant heat addition at our maximum temperature of 1,962 °R is not possible (2 to 4). The nature of water and certain elements of the process controls require us to add heat in a constant pressure process instead (1-2-3-4). Because of this, the average temperature where we add heat is far below the maximum allowable material temperature Rev 2 41 Figure: Carnot Cycle Versus Typical Power Cycle Available Energy The actual available energy (area under the 1-2-3-4 on the above curve) is less than half of what is available from the ideal Carnot cycle (area under 12'-4) operating between the same two temperatures. Typical thermal efficiencies for fossil plants are 40 percent while nuclear plants have efficiencies of 31 percent. These numbers are less than half the maximum thermal efficiency of the ideal Carnot cycle calculated earlier. The figure on the following page shows a proposed Carnot steam cycle superimposed on a T-s diagram. Several problems make it undesirable as a practical power cycle. Significant pump work is required to compress a two-phase mixture of water and steam from point 1 to the saturated liquid state at point 2. Cavitation in the pump would occur if the inlet fluid was at saturation. Third, a condenser designed to produce a two-phase mixture at the outlet (point 1) would pose technical problems. Rev 2 42 Figure: Ideal Carnot Cycle Rankine Cycle Steam engines drove the industrial revolution, so early thermodynamic developments centered on improving the performance of contemporary steam engines. It was desirable to construct a steam cycle as close to reversible as possible and would take better advantage of the characteristics of steam than does the Carnot cycle. The Rankine cycle was developed as a more practical version of the Carnot cycle. The Rankine cycle, shown below in the figure, confines the isentropic compression process to the liquid phase only (points 1 to 2). This minimizes the amount of work required to attain operating pressures and avoids the mechanical problems associated with pumping a two-phase mixture. The compression process shown between points 1 and 2 is exaggerated*. In reality, a temperature rise of only 1 °F occurs in compressing water from 14.7 psig at a saturation temperature of 212 °F to 1,000 psig. Figure: Rankine Cycle Rev 2 43 *The constant pressure lines converge rapidly in the subcooled or compressed liquid region and it is difficult to distinguish them from the saturated liquid line without artificially expanding the constant pressure lines away from it. Like the Carnot cycle, in a Rankine cycle available and unavailable energy on a T-s diagram is represented by the areas under the curves; therefore, the larger the unavailable energy, the less efficient the cycle. The same loss of cycle efficiency is seen when two Rankine cycles are compared as shown below in the figure. Figure: Rankine Cycle Efficiency Comparisons On a T-s Diagram If the ideal turbine was replaced with a real turbine, the efficiency of the cycle will be reduced. This is because the non-ideal turbine incurs an increase in entropy increases the area under the T-s curve for the cycle. However, the increase in the area of available energy (3-2-3') is less than the increase in area for unavailable energy (a-3-3'-b). Figure: Rankine Cycle With Real Versus Ideal Turbine Rev 2 44 Rankine Cycle Efficiencies Cycle efficiency compares by contrasting the amount of rejected energy to available energy of both cycles. The comparison shows that Cycle b, above, has less heat available for work and more rejected heat, making it less efficient. An h-s diagram also compares systems and helps determine their efficiencies. Typical Steam Cycle A simplified version of the major components of a typical steam plant cycle is shown below in the figure. This figure does not contain the exact detail found at most power plants; however, it adequately demonstrates understanding the basic operation of a power cycle. Figure: Typical Steam Cycle The typical steam cycle is comprised of the following processes: Rev 2 1-2: Heat is added to the working fluid in the steam generator under a constant pressure condition 2-3: Saturated steam from the steam generator is expanded in the high-pressure (HP) turbine to provide shaft work output at constant entropy 3-4: Moist steam from the HP turbine exit is dried and superheated in the moisture separator reheater (MSR) 4-5: Superheated steam from the MSR is expanded in the lowpressure (LP) turbine to provide shaft work at constant entropy 5-6: Steam exhaust from the turbine is condensed in the condenser by cooling water under a constant vacuum condition 6-7: Condensate is compressed as a liquid by the condensate pump 7-8: Condensate is preheated by the low-pressure feedwater heaters 45 8-9: Condensate is compressed as a liquid by the feedwater pump 9-1: Feedwater is preheated by the high-pressure heaters 1-2: Cycle starts again and heat is added to the working fluid in the steam generator under a constant pressure condition The typical steam cycle is shown below on the T-s diagram. The numbered points on the cycle correspond to the numbered points on the above figure. The Rankine cycle is an ideal cycle and does not exactly represent the real processes in the plant but is a closer approximation than the Carnot cycle. Real pumps and turbines would exhibit an entropy increase across them when shown on a T-s diagram. Figure: Rankine Steam Cycle (Ideal) A T-s diagram of a cycle that closely approximates actual plant processes is shown below in the figure. The pumps and turbines in this cycle are real pumps and turbines and show an entropy increase across them. A small amount of subcooling is evident in the condenser as demonstrated by the small dip down to point 5. Subcooling is the process of cooling condensed vapor beyond what is required for the condensation process. This small amount of subcooling decreases cycle efficiency because additional heat was removed from the cycle to the cooling water as heat rejected. This additional heat rejected must then be added back in the steam generator. Condenser subcooling decreases the cycle’s efficiency and is sometimes required to have an adequate suction head to prevent the condensate pumps from cavitating. By controlling the temperature or flow rate of the cooling water to the condenser, the operator directly affects the overall cycle efficiency. Rev 2 46 Figure: Steam Cycle (Real) The Mollier diagram plots and illustrates the energy transferred to or from the steam during the cycle. The numbered points on the Mollier diagram shown below correspond to the numbered points on the previous Rankine cycle diagrams. The Mollier diagram is limited to plotting only the saturated or superheated form of the working fluid; the liquid portion of the steam cycle is not indicated on this type of diagram. The following conditions are visible on the Mollier diagram: Point 1: Saturated steam at 540 °F Point 2: 82.5 percent quality at exit of HP turbine Point 3: Temperature of superheated steam is 440 °F Point 4: Condenser vacuum is 1 psia The solid lines represent the conditions for ideal turbines as verified by the fact that no entropy change shows across the turbines. The dotted lines represent the path taken if real turbines were considered, in which case an increase in entropy is evident. Rev 2 47 Figure: Mollier Diagram (Use student copy during lesson) Causes of Inefficiency In this section, we compare some of the types and causes for the inefficiencies of real components and cycles to that of their ideal counterparts. Components In real systems, a percentage of the overall cycle inefficiency is due to the losses by the individual components. Turbines, pumps, and compressors all behave non-ideally due to heat losses, friction and windage losses. All of these losses contribute to the non-isentropic behavior of real equipment. As previously explained, these losses are disclosed as an increase in the system's entropy or amount of energy that is unavailable for use by the cycle. Rev 2 48 Cycles Some compromises are made in real systems due to cost and other factors in the design and operation of the cycle. In a large power generating station, the condensers are designed to subcool the liquid by 8 °F to 10 °F. This compromise allows the condensate pumps to pump without cavitation but each degree of subcooling is energy that must be put back by reheating the water. This energy used in reheating does no useful work, which decreases cycle efficiency. Imperfect thermal insulation results in a heat loss to the environment; again this is energy lost to the system and therefore unavailable to do work. Both resistance to fluid flow and mechanical friction in machines are other real world losses that result in decreased cycle efficiency. Secondary System Parameters Operators should have a thorough understanding of how changing plant parameters affect plant operation and overall efficiency. The loss of efficiency causes a change in power output from the reactor. The parameters that could affect thermodynamic efficiency are discussed below. Increasing Steam Temperature at the Turbine Entrance A higher turbine inlet steam temperature raises the available work that can be extracted from the turbine. More work increases plant efficiency. Increasing Feedwater Heating Raising feedwater temperature to near Tsat for the existing steam generator pressure increases the plant’s efficiency. Less energy from fission process is needed to raise feedwater to operating temperatures. We must account for the energy added to the feedwater. Overall plant efficiency increases. Increasing Condenser Vacuum More work is extracted from the turbine due to the lower pressure in the main condenser. Condensing steam increases condenser vacuum. Overall plant efficiency increases. Increasing Circulating Water System Flow Rate Increasing the circulating water system flow rate raises the differential temperature between condensate and cooling medium. This reduces condenser temperature, lowering the pressure at the exhaust of the turbine. Lower pressure at the exhaust of the turbine increases plant efficiency. Rev 2 49 However, the additional heat removed by more condensate depression must be replenished. Lowering Circulating Water System Inlet Temperature By decreasing the circulating water system inlet temperature, the differential temperature is increased between condensate and cooling medium. Larger differential temperature results in larger heat transfer rate, which lowers condenser temperature and pressure. Lower pressure at the exhaust of the turbine increases plant efficiency. However, the additional heat removed by more condensate depression must be replenished. Reducing Condensate Depression Any amount of condensate depression causes more required heat addition to raise the condensate temperature to the operating temperature. If less energy is rejected to the circulating water, the feedwater requires less energy from the fission process to raise the temperature to operating temperatures. Controlling condensate depression to the minimum required raises the plant’s efficiency. Remove Air and Non-Condensable Gases Excessive air and non-condensable gases within the main condenser minimize the heat transfer area. More energy is required to achieve the same cooling or achieving insufficient condensate depression. Removal of non-condensable gases increases the plant’s efficiency. Knowledge Check (Answer Key) The rate of heat transfer between two liquids in a heat exchanger will increase if the … (Assume specific heats do not change.) Rev 2 A. inlet temperature of the hotter liquid decreases by 20 °F. B. inlet temperature of the colder liquid increases by 20 °F. C. flow rates of both liquids decrease by 10 percent. D. flow rates of both liquids increase by 10 percent. 50 Knowledge Check (Answer Key) Which one of the following pairs of fluids undergoing heat transfer in typical cross-flow design heat exchangers yields the greatest heat exchanger overall heat transfer coefficient? Assume comparable heat exchanger sizes and fluid flow rates. A. Oil to water in a lube oil cooler B. Steam to water in a feedwater heater C. Water to air in a ventilation heating unit D. Water to water in a cooling water heat exchanger Knowledge Check (Answer Key) A nuclear power plant is operating at 90 percent of rated power. Main condenser pressure is 1.7 psia and hotwell condensate temperature is 120 °F. Which one of the following describes the effect of a 5 percent decrease in cooling water flow rate through the main condenser? Rev 2 A. Overall steam cycle efficiency increases because the work output of the turbine increases. B. Overall steam cycle efficiency increases because condensate depression decreases. C. Overall steam cycle efficiency decreases because the work output of the turbine decreases. D. Overall steam cycle efficiency decreases because condensate depression increases. 51 Knowledge Check (Answer Key) Which one of the following actions decreases overall nuclear power plant thermal efficiency? A. Reducing turbine inlet steam moisture content B. Reducing condensate depression C. Increasing turbine exhaust pressure D. Increasing temperature of feedwater entering the steam generators Knowledge Check (Answer Key) Which one of the following changes causes an increase in overall nuclear power plant thermal efficiency? Rev 2 A. decreasing the temperature of the water entering the steam generators B. decreasing the superheat of the steam entering the lowpressure turbines C. decreasing the circulating water flow rate through the main condenser D. decreasing the concentration of non-condensable gases in the main condenser 52 TLO 2 Summary During this lesson, you learned about the Second Law of Thermodynamics, which states that it is impossible to construct a device that operates within a cycle that can convert all the heat supplied it into mechanical work. Recognizing that even the most thermally and mechanically perfect cycles must reject some heat defines thermodynamic power cycle efficiency. The listing below provides a summary of sections in this TLO. 1. Review ELO 2.1 by asking students the following: • Planck's statement of the second law of thermodynamics, which is that it is impossible to construct an engine that will work in a complete cycle and produce no other effect except the raising of a weight and the cooling of a heat reservoir. • Entropy is a measure of the unavailability of heat to perform work in a cycle. This relates to the second law since the second law predicts that not all heat provided to a cycle can be transformed into an equal amount of work, some heat rejection must take place. • Second law of thermodynamics demonstrates that the maximum possible efficiency of a system is the Carnot efficiency written as: 𝜂= 𝑇𝐻 − 𝑇𝐶 𝑇𝐻 2. Review ELO 2.2 by having student explain the following statements: • Maximum efficiency of a closed cycle can be determined by calculating the efficiency of a Carnot cycle operating between the same values of high and low temperatures. • Efficiency of a component can be calculated by comparing the work produced by the component to the work that would have been produced by an ideal component operating isentropically between the same inlet and outlet conditions. • An isentropic expansion or compression process is represented as a vertical line on a T-s or h-s diagram. A real expansion or compression process looks similar, but is slanted slightly to the right. • Maximizing the ΔT and ΔP between the source and the heat sink ensures the highest possible cycle efficiency. Rev 2 53 • The second law of thermodynamics gives a maximum efficiency limit (which is never reached in physical systems) that an ideal thermodynamic system can perform. The efficiency is determined by knowing the inlet and exit absolute temperatures of the overall system and applying Carnot's efficiency equation. • Cycle efficiency = 1 − (𝑇 𝐶 ) (temperature in degrees R) 𝑇 𝐻 3. Review ELO 2.3 by having students plot their plant's steam cycle on a T-s diagram. • A T-s diagram is frequently used to analyze energy transfer system cycles. Work done by or on the system and heat added to or removed from the system can be visualized on the T-s diagram. • Use the following as a guideline: 𝜂= 𝑇𝐻 − 𝑇𝐶 𝑇𝐻 4. Review ELO 2.4 by having the students list the major components of the steam cycle and discuss operating conditions that improve the plant’s efficiency. Figure: Typical Steam Cycle and T-s Diagram • Rev 2 Discuss conditions yielding improved cycle efficiency, shown below in the table: 54 Improved Cycle Efficiency Condition Effect Discussion Superheating More Efficient With More Superheating Increased heat added results in more net work from the system, even though more heat is rejected. Moisture Separator Reheater (MSR) Use of MSR Has Minor Effect On Efficiency More work is done by the low-pressure (LP) turbine since inlet enthalpy is higher but more heat is rejected. The principle benefit of MSR use is protection of the final blading stages in LP turbine from water droplet impingement. Feedwater Preheating More Efficient With Less heat must be added Feedwater Preheating from the heat source (reactor) since the feedwater enters the steam generator closer to saturation temperature. Condenser Vacuum More Efficient With Higher Vacuum (Lower Backpressure) Net work output is higher and heat rejection is lower as condenser pressure is lowered. Condensate Depression More Efficient With Minimal Condensate Depression Minimal condensate depression reduces both the amount of heat rejected and the amount of heat that must be supplied to the cycle. Rev 2 55 Condition Effect Discussion Steam More Efficient At At higher steam Temperature/Pressure Higher Steam temperature, the inlet and Temperature/Pressure exit entropy from the turbine are lower so less heat is rejected. Steam density increases as pressure increases, so more turbine work is done. Steam Quality More Efficient At Higher Steam Quality Enthalpy content increases as moisture content decreases and more net work is done. • Hotwell is the area at the bottom of the condenser where the condensed steam collects to pump back into the system feedwater. • Condensate depression is the amount the condensate in a condenser that is cooled below saturation (degrees subcooled). • Condensers operate at a vacuum to ensure the temperature (and thus the pressure) of the steam is as low as possible. • Causes of decreased efficiency include the following: — — — — — — — Rev 2 Presence of friction Heat losses Cycle inefficiencies – Subcooling – Tsat of the steam generator Turbine service lifetime is affected by moisture impingement on the blades and other internal parts Removing as much moisture from the steam limits moisture content at every stage of the turbine Feedwater heater is a power plant component used to preheat water delivered to a steam generating boiler Moisture separator reheaters improve the plant’s efficiency and are used to avoid the erosion corrosion and droplet impingement erosion in the LP turbine, to remove moisture, and to superheat the steam 56 Objectives Now that you have completed this lesson, you should be able to do the following: 1. Explain the second law of thermodynamics using the term entropy. 2. Given a thermodynamic system, determine the: a. Maximum efficiency of the system b. Efficiency of the components within the system 3. Differentiate between the path for an ideal process and that for a real process on a T-s or h-s diagram. 4. Describe how individual factors affect system or component efficiency. Thermodynamic Processes Summary In this module, you learned about applying the First and Second Laws of Thermodynamics to processes, systems, diagram principles, and energy balances on major components within a nuclear power generation plant or facility. During this lesson, you learned about the First Law of Thermodynamics, which states that energy can be neither created nor destroyed, but only altered in form. The energy forms may not always be the same but the total energy in the system remains constant. You learned about open, closed, isolated, and steady flow systems. You studied processes including thermodynamic, cyclic, reversible, irreversible, adiabatic, isentropic, and isenthalpic. All of the energies entering and leaving the control volume boundary, any work done on or by the control volume, and any heat transferred into and out of the control volume boundaries are in the energy balance equation. The Second Law of Thermodynamics states that it is impossible to construct a device that operates within a cycle that can convert all the heat supplied it into mechanical work. Recognizing that even the most thermally and mechanically perfect cycles must reject some heat defines thermodynamic power cycle efficiency. You studied entropy—a measure of the unavailability of heat to perform work in a cycle—to explain the second law, and that the change in entropy determines the direction a given process will proceed. The Carnot cycle represents an upper limit of efficiency for any given system operating between the same two temperatures since all practical systems and processes are irreversible. The system's maximum possible efficiency would be that of a Carnot cycle, but because Carnot cycles represent reversible processes, the real system cannot reach the Carnot efficiency value. Thus, the Carnot efficiency serves as an unattainable upper limit for any real system's efficiency. Rev 2 57 You studied the ideal Rankine cycle, which is an ideal cycle where no increase in entropy occurs as work is done on and by the system. You learned about the thermodynamics of ideal and real systems by arranging the various thermodynamic processes on a property diagram to evaluate the various cycles present in a nuclear power plant. The most common set of coordinates used is a plot of temperature versus specific entropy is a T-s diagram. You also studied an h-s diagram, which compares systems and determines their efficiencies, and a Mollier diagram, which plots and illustrates the energy transferred to or from the steam during the cycle. Now that you have completed this module, you should be able to demonstrate mastery of this topic by passing a written exam with a grade of 80 percent or higher on the following TLOs: 1. Apply the First Law of Thermodynamics to analyze thermodynamic systems and processes. 2. Apply the Second Law of Thermodynamics to analyze real and ideal systems and components. Rev 2 58 Thermodynamic Cycles Knowledge Check Answer Key Thermodynamic Cycles Knowledge Check Answers ELO 1.1 Gas Laws Knowledge Check According to Charles’s Law, at low pressure, the _____ of a gas at constant _____ is directly proportional to the temperature of the gas. A. density; pressure B. volume; pressure C. pressure; volume D. weight; volume Knowledge Check Calculate the value of the missing property. P1= 100 psia; P2 = ? V1 = 50 ft3; V2 = 25 ft3 T1 = 60°F; T2 = 70°F Rev 2 A. 233 psi B. 210 psi C. 204 psi D. 200 psi 1 Thermodynamic Cycles Knowledge Check Answer Key ELO 1.2 Compression Process Knowledge Check When can a fluid be considered incompressible? A. if it is liquid B. if it is steam but not flowing C. if it is a saturated vapor D. if it is a superheated steam Knowledge Check A contained fluid is heated. The resulting change in pressure will be… A. greater for an incompressible fluid. B. greater for a compressible fluid. C. the same for both fluids. D. the same for both fluids only if the volume is held constant. ELO 2.1 Thermodynamic Entropy Knowledge Check The second law of thermodynamics can also be expressed as ________ for a closed cycle. Rev 2 A. Sf = Si B. ΔS ≥ 0 C. ΔT < 0 D. ΔS < 0 2 Thermodynamic Cycles Knowledge Check Answer Key ELO 2.2 Carnot’s Principle of Thermodynamics Knowledge Check The steam generators produce dry saturated steam at 1,000 psig. The main condenser is operating with zero subcooling at 1 psia. What is the maximum efficiency obtainable? A. 44 percent B. 34 percent C. 28 percent D. 23 percent Knowledge Check Determine the Carnot Efficiency of a steam engine that is supplied with saturated steam at 300 psia and exhausts to atmosphere… Rev 2 A. 44 percent B. 56 percent C. 42 percent D. 35 percent 3 Thermodynamic Cycles Knowledge Check Answer Key ELO 2.3 Thermodynamics of Ideal and Real Processes Knowledge Check Why are real processes shown with dotted lines on property diagrams? A. They occur faster than real processes. B. The value of entropy during the process is not determined. C. The entropy values during the process are the same as the real process until the outlet from the process. D. You would not be able to distinguish between real and ideal processes if the real process was a solid line. ELO 2.4 Thermodynamic Power Plant Efficiency Knowledge Check The rate of heat transfer between two liquids in a heat exchanger will increase if the … (Assume specific heats do not change.) Rev 2 A. inlet temperature of the hotter liquid decreases by 20 °F. B. inlet temperature of the colder liquid increases by 20 °F. C. flow rates of both liquids decrease by 10 percent. D. flow rates of both liquids increase by 10 percent. 4 Thermodynamic Cycles Knowledge Check Answer Key Knowledge Check Which one of the following pairs of fluids undergoing heat transfer in typical cross-flow design heat exchangers yields the greatest heat exchanger overall heat transfer coefficient? Assume comparable heat exchanger sizes and fluid flow rates. A. Oil to water in a lube oil cooler B. Steam to water in a feedwater heater C. Water to air in a ventilation heating unit D. Water to water in a cooling water heat exchanger Knowledge Check A nuclear power plant is operating at 90 percent of rated power. Main condenser pressure is 1.7 psia and hotwell condensate temperature is 120 °F. Which one of the following describes the effect of a 5 percent decrease in cooling water flow rate through the main condenser? Rev 2 A. Overall steam cycle efficiency increases because the work output of the turbine increases. B. Overall steam cycle efficiency increases because condensate depression decreases. C. Overall steam cycle efficiency decreases because the work output of the turbine decreases. D. Overall steam cycle efficiency decreases because condensate depression increases. 5 Thermodynamic Cycles Knowledge Check Answer Key Knowledge Check Which one of the following actions decreases overall nuclear power plant thermal efficiency? A. Reducing turbine inlet steam moisture content B. Reducing condensate depression C. Increasing turbine exhaust pressure D. Increasing temperature of feedwater entering the steam generators Knowledge Check Which one of the following changes causes an increase in overall nuclear power plant thermal efficiency? Rev 2 A. decreasing the temperature of the water entering the steam generators B. decreasing the superheat of the steam entering the lowpressure turbines C. decreasing the circulating water flow rate through the main condenser D. decreasing the concentration of non-condensable gases in the main condenser 6