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BMM2523_Chapter 2 thermodynamics9-13 An air-standard cycle executed in a piston-cylinder system is...

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thermodynamic asdasddddddddddddddddddddddddddddddddddddddddddddddddddasdsa sadsaaaasdasd sadsa ds9-13An air-standard cycle executed in a piston-cylinder system is composed of three specified processes. The cycle is to besketcehed on theP-v andT -sdiagrams and the back work ratio are to be determined. Assumptions 1The air-standard assumptions are applicable.2Kinetic and potential energy changes are negligible.3Air isan ideal gas with constant specific heats. PropertiesThe properties of air are given as R= 0.287 kPa·m3 /kg·K,c p= 1.005 kJ/kg·K,cv = 0.718 kJ/kg·K, andk =1.4. Analysis(a) TheP-v andT -sdiagrams of the cycle are shown in the figures.(b) Process 1-2: Isentropic compression)(12,21T T mcwin−=−v s T 321v P32111212=⎟⎟ ⎠⎜⎜⎝ =r T T T v 11−− ⎞⎛ k k v rocess 2-3: Constant pressure heat additionT T mRP−=−=V V v he back wrk ratio isP 2,32Pd wout =∫−)()(232323To)()(2312,32,21T T mRT T mcwwr out inbw−−==−−v Noting that1 thus,and and −==−=k Rccck cc Rp pv v v From ideal gas relation,r T T ===212323v v v v Substituting these into back work relation,( )( )0.256=−−−=−−−=−⎟ ⎠ ⎞⎜⎝ ⎛ −−=−−−=−−−166114.11111111111)1 / () / 1(114.011232122r r k r r k T T T T T T Rk Rr k k bw PROPRIETARY MATERIALpreparation. If you are a student using this Manual, you are using it without permission. . © 2011 The McGraw-Hill Companies, Inc. Limited distribution permitted only to teachers and educators for course
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1 Chapter 9 and 10 GAS & VAPOUR POWER CYCLES (4 weeks) CO2: Analyse and apply thermodynamics concepts to solve problems in cycles involving various gas and vapour power schemes Hj. Amirruddin Bin Abdul Kadir Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Thermodynamics: An Engineering Approach, 7th Edition Yunus A. Cengel, Michael A. Boles McGraw-Hill, 2011 Quiz 1 1. How does useful work differ from actual work? For what kind of systems are these two identical? 2. What is the second low efficiency? How does it differ from the first-law efficiency? 3. Can a process for which the reversible work is zero be reversible? Can it be irreversible? Explain. 4. Consider a process during which no entropy is generated (S gen = 0). Does the exergy destruction for this process have to be zero? Quiz 1-Ans 1. Useful work differs from the actual work by the surroundings work. They are identical for systems that involve no surroundings work such as steady-flow system (3 Marks). 2. The second-law efficiency is a measure of the performance of a device relative to its performance under reversible conditions. It differs from the first law efficiency in that it is not a conversion efficiency (3 Marks). 3. A processes with Wrev = 0 is reversible if it involves no actual useful work. Otherwise it is irreversible (3 Marks). 4. yes (1 Marks). 4 Objectives Evaluate the performance of gas power cycles for which the working fluid remains a gas throughout the entire cycle. Develop simplifying assumptions applicable to gas power cycles. Review the operation of reciprocating engines. Analyze both closed and open gas power cycles. Solve problems based on the Otto, Diesel, Stirling, and Ericsson cycles. Solve problems based on the Brayton cycle; the Brayton cycle with regeneration; and the Brayton cycle with intercooling, reheating, and regeneration. Analyze jet-propulsion cycles. Identify simplifying assumptions for second-law analysis of gas power cycles. Perform second-law analysis of gas power cycles.
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  • 1Chapter 9 and 10GAS & VAPOUR

    POWER CYCLES(4 weeks)

    CO2: Analyse and apply thermodynamics concepts to solve problems in cycles involving various gas and vapour power schemes

    Hj. Amirruddin Bin Abdul Kadir

    Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

    Thermodynamics: An Engineering Approach, 7th EditionYunus A. Cengel, Michael A. Boles

    McGraw-Hill, 2011 Quiz 1 1. How does useful work differ from actual work? For

    what kind of systems are these two identical?

    2. What is the second low efficiency? How does it differ from the first-law efficiency?

    3. Can a process for which the reversible work is zero be reversible? Can it be irreversible? Explain.

    4. Consider a process during which no entropy is generated (Sgen = 0). Does the exergy destruction for this process have to be zero?

    Quiz 1-Ans 1. Useful work differs from the actual work by the surroundings work.

    They are identical for systems that involve no surroundings work such as steady-flow system (3 Marks).

    2. The second-law efficiency is a measure of the performance of a device relative to its performance under reversible conditions. It differs from the first law efficiency in that it is not a conversion efficiency (3 Marks).

    3. A processes with Wrev = 0 is reversible if it involves no actual useful work. Otherwise it is irreversible (3 Marks).

    4. yes (1 Marks).4

    Objectives Evaluate the performance of gas power cycles for which the

    working fluid remains a gas throughout the entire cycle. Develop simplifying assumptions applicable to gas power

    cycles. Review the operation of reciprocating engines. Analyze both closed and open gas power cycles. Solve problems based on the Otto, Diesel, Stirling, and

    Ericsson cycles. Solve problems based on the Brayton cycle; the Brayton cycle

    with regeneration; and the Brayton cycle with intercooling, reheating, and regeneration.

    Analyze jet-propulsion cycles. Identify simplifying assumptions for second-law analysis of

    gas power cycles. Perform second-law analysis of gas power cycles.

  • 25

    Objectives - continued Evaluate the performance of gas power cycles for

    which the working fluid remains a gas throughout the entire cycle.

    Analyze vapor power cycles in which the working fluid is alternately vaporized and condensed.

    Analyze power generation coupled with process heating called cogeneration.

    Investigate ways to modify the basic Rankine vapor power cycle to increase the cycle thermal efficiency.

    Analyze the reheat and regenerative vapor power cycles.

    Analyze power cycles that consist of two separate cycles known as combined cycles and binary cycles.

    6

    BASIC CONSIDERATIONS IN THE ANALYSISOF POWER CYCLES

    Modeling is a powerful engineering tool that provides great insight and simplicity at the expense of some loss in accuracy.

    Most power-producing devices operate on cycles.Ideal cycle: A cycle that resembles the actual cycle closely but is made up totally of internally reversible processes.Reversible cycles such as Carnot cycle have the highest thermal efficiency of all heat engines operating between the same temperature levels. Unlike ideal cycles, they are totally reversible, and unsuitable as a realistic model.

    Thermal efficiency of heat engines:

    The analysis of many complex processes can be reduced to a manageable level by utilizing some idealizations.

    7

    The ideal cycles are internally reversible, but, unlike the Carnot cycle, they are not necessarily externally reversible. 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.

    8

    The idealizations and simplifications in the analysis of power cycles:

    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.

    Care should be exercised in the interpretation of the results from ideal cycles.

    On both P-v and T-s diagrams, the area enclosed by the process curve represents the net work of the cycle.

    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.

  • 39

    THE CARNOT CYCLE AND ITS VALUE IN ENGINEERING

    P-v and T-s diagrams of a Carnot cycle.

    The Carnot cycle is composed of four totally reversible processes: isothermal heat addition, isentropic expansion, isothermal heat rejection, and isentropic compression.For both ideal and actual cycles: Thermal efficiency increases with an increase in the average temperature at which heat is supplied to the system or with a decrease in the average temperature at which heat is rejected from the system.

    A steady-flow Carnot engine.

    Example 1: 9.11The maximum possible thermal efficiency of a gas power cycle with specified reservoirs is to be determined. Analysis The maximum efficiency this cycle can have is

    11

    AIR-STANDARD ASSUMPTIONS

    The combustion process is replaced by a heat-addition process in ideal cycles.

    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.

    4. The exhaust process is replaced by a heat-rejection process that restores the working fluid to its initial state.

    Cold-air-standard assumptions: When the working fluid is considered to be air with constant specific heats at room temperature (25C).Air-standard cycle: A cycle for which the air-standard assumptions are applicable.

    Example 2: 9.14 The three processes of an ideal gas power cycle are described. The cycle is to be shown on the P-v and T-s diagrams, and the maximum temperature, expansion and compression works, and thermal efficiency are to be determined.

  • 4Example 2: 9.14 - continued Example 3: 9.15 The four processes of an air-standard cycle are described. The cycle is to be shown on P-v and T-s diagrams, and the net work output and the thermal efficiency are to be determined.

    Example 3: 9.15 - continued Example 4: 9.18 A Carnot cycle with the specified temperature limits is considered. The net work output per cycle is to be determined.

  • 5Example 5: 9.20 AAn ideal gas Carnot cycle with air as the working fluid is considered. The maximum temperature of the low-temperature energy reservoir, the cycle's thermal efficiency, and the amount of heat that must be supplied per cycle are to be determined.

    18

    AN OVERVIEW OF RECIPROCATING ENGINES

    Nomenclature for reciprocating engines.

    Spark-ignition (SI) engines Compression-ignition (CI) engines

    Compression ratio

    Mean effective pressure

    19

    OTTO CYCLE: THE IDEAL CYCLE FOR SPARK-IGNITION ENGINES

    Actual and ideal cycles in spark-ignition engines and their P-v diagrams. 20

    Schematic of a two-stroke reciprocating engine.

    The two-stroke engines are generally less efficient than their four-stroke counterparts but they are relatively simple and inexpensive, and they have high power-to-weight and power-to-volume ratios.

    T-s diagram of the ideal Otto cycle.

    Four-stroke cycle1 cycle = 4 stroke = 2 revolutionTwo-stroke cycle1 cycle = 2 stroke = 1 revolution

  • 621 22

    The thermal efficiency of the Otto cycle increases with the specific heat ratio k of the working fluid.

    Thermal efficiency of the ideal Otto cycle as a function of compression ratio (k = 1.4).

    In SI engines, the compression ratio is limited by autoignition or engine knock.

    Example 6: 9.30 An ideal Otto cycle is considered. The thermal efficiency and the rate of heat input are to be determined.

    Example 7: 9.34 A six-cylinder, four-stroke, spark-ignition engine operating on the ideal Otto cycle is considered. The power produced by the engine is to be determined.

  • 7Example 7: 9.34 - continued

    26

    DIESEL CYCLE: THE IDEAL CYCLEFOR COMPRESSION-IGNITION ENGINES

    In diesel engines, the spark plug is replaced by a fuel injector, and only air is compressed during the compression process.

    In diesel engines, only air is compressed during the compression stroke, eliminating the possibility of autoignition (engine knock). Therefore, diesel engines can be designed to operate at much higher compression ratios than SI engines, typically between 12 and 24.

    1-2 isentropic compression

    2-3 constant-volume heat addition

    3-4 isentropic expansion

    4-1 constant-volume heat rejection.

    27

    Thermal efficiency of the ideal Diesel cycle as a function of compression and cutoff ratios (k=1.4).

    Cutoff ratio

    for the same compression ratio

    Example 8: 9-3

  • 8Example 8: 9-3 - continued Example 9: 9-50 An air-standard Diesel cycle with a compression ratio of 18.2 is considered. The cutoff ratio, the heat rejection per unit mass, and the thermal efficiency are to be determined.

    Example 9: 9-50 - continued

    32

    QUESTIONS ???Diesel engines operate at higher air-fuel ratios than gasoline engines. Why?Despite higher power to weight ratios, two-stroke engines are not used in automobiles. Why?The stationary diesel engines are among the most efficient power producing devices (about 50%). Why?What is a turbocharger? Why are they mostly used in diesel engines compared to gasoline engines.

    P-v diagram of an ideal dual cycle.

    Dual cycle: A more realistic ideal cycle model for modern, high-speed compression ignition engine.

  • 9Example 10: 9-59 An ideal dual cycle has a compression ratio of 15 and cutoff ratio of 1.4. The net work, heat addition, and the thermal efficiency are to be determined.

    Example 10: 9-59 - continued

    Example 11: 9-60 A six-cylinder compression ignition engine operates on the ideal Diesel cycle. The maximum temperature in the cycle, the cutoff ratio, the net work output per cycle, the thermal efficiency, the mean effective pressure, the net power output, and the specific fuel consumption are to be determined.

    Example 11: 9-60 - continued

  • 10

    Example 11: 9-60 - continued

    38

    STIRLING AND ERICSSON CYCLES

    A regenerator is a device that borrows energy from the working fluid during one part of the cycle and pays it back (without interest) during another part.

    Stirling cycle1-2 T = constant expansion (heat addition from the external source)2-3 v = constant regeneration (internal heat transfer from the working fluid to the regenerator)3-4 T = constant compression (heat rejection to the external sink)4-1 v = constant regeneration (internal heat transfer from the regenerator back to the working

    fluid)

    T-s and P- vdiagrams of Carnot, Stirling, and Ericsson cycles.

    39The execution of the Stirling cycle. A steady-flow Ericsson engine.

    The Ericsson cycle is very much like the Stirling cycle, except that the two constant-volume processes are replaced by two constant-pressure processes.

    Both the Stirling and Ericsson cycles are totally reversible, as is the Carnot cycle, and thus:

    The Stirling and Ericsson cycles give a message: Regeneration can increase efficiency.

    Example 12: 9.69 An ideal steady-flow Ericsson engine with air as the working fluid is considered. The maximum pressure in the cycle, the net work output, and the thermal efficiency of the cycle are to be determined.

  • 11

    Example 13: 9.72 An ideal Ericsson cycle operates between the specified temperature limits. The rate of heat addition is to be determined.


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