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- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/01%3A_Basic_Concepts_and_Definitions/1.01%3A_Chapter_introduction_and_learning_objectivesIt lays the foundation for a comprehensive analysis of different thermodynamic processes and cycles to be presented in this book. Explain the basic scope of engineering thermodynamics and its common a...It lays the foundation for a comprehensive analysis of different thermodynamic processes and cycles to be presented in this book. Explain the basic scope of engineering thermodynamics and its common areas of application Demonstrate an understanding of fundamental concepts, such as system and its surroundings, closed and open systems, extensive and intensive properties, equilibrium state, quasi-equilibrium process, and cycle
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/01%3A_Basic_Concepts_and_Definitions/1.06%3A_Chapter_reviewThermodynamics has a wide range of applications in many engineering fields, in particular, the fields related to energy conversion and conservation. An important skill that students need to develop, w...Thermodynamics has a wide range of applications in many engineering fields, in particular, the fields related to energy conversion and conservation. An important skill that students need to develop, when performing thermodynamic analysis on devices, is to identify the system, its surroundings, and their interactions. There are three types of systems: closed system, open system, and isolated system, defined in terms of their ability to transfer mass and energy with the surroundings.
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/01%3A_Basic_Concepts_and_Definitions/1.04%3A_Extensive_and_intensive_propertiesThe interactions between the molecules are so frequent that the physical or bulk properties of the system do NOT depend on the behaviour of individual molecules. The pressure and temperature of the ai...The interactions between the molecules are so frequent that the physical or bulk properties of the system do NOT depend on the behaviour of individual molecules. The pressure and temperature of the air are not affected by the changing mass in each compartment; therefore, pressure and temperature are intensive properties. On the other hand, the mass and volume of the air in each of the compartments are different from the original values in the container.
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/05%3A_The_First_Law_of_Thermodynamics_for_a_Control_Volume/5.01%3A_Chapter_introduction_and_learning_objectivesA common feature of these devices is that they all have inlets and outlets, through which a working fluid transfers both mass and energy into and out of the devices. Examples are given to illustrate t...A common feature of these devices is that they all have inlets and outlets, through which a working fluid transfers both mass and energy into and out of the devices. Examples are given to illustrate the applications of the first law of thermodynamics in typical SSSF devices such as turbines, compressors, heat exchangers, expansion valves, and mixing chambers. Apply the conservation equations of mass and energy to steady-state, steady flow devices
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/07%3A_Appendices/7.06%3A_Appendix_F_-_Triple_Point_of_Selected_SubstancesK (°C) kPa (atm) 192.4 K (−80.7 °C) 120 kPa (1.2 atm) 195.40 K (−77.75 °C) 6.060 kPa (0.05981 atm) 83.8058 K (−189.3442 °C) 68.9 kPa (0.680 atm) 134.6 K (−138.6 °C) 7×10 −4 kPa (6.9×10 −6 atm) 216.55 ...K (°C) kPa (atm) 192.4 K (−80.7 °C) 120 kPa (1.2 atm) 195.40 K (−77.75 °C) 6.060 kPa (0.05981 atm) 83.8058 K (−189.3442 °C) 68.9 kPa (0.680 atm) 134.6 K (−138.6 °C) 7×10 −4 kPa (6.9×10 −6 atm) 216.55 K (−56.60 °C) 517 kPa (5.10 atm) 68.10 K (−205.05 °C) 15.37 kPa (0.1517 atm) 89.89 K (−183.26 °C) 1.1×10 −3 kPa (1.1×10 −5 atm) 150 K (−123 °C) 4.3×10 −7 kPa (4.2×10 −9 atm) 104.0 K (−169.2 °C) 0.12 kPa (0.0012 atm) 173.08 K (−100.07 °C) 26.60 kPa (0.2625 atm) 13.8033 K (−259.3467 °C)
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/02%3A_Thermodynamic_Properties_of_a_Pure_Substance/2.06%3A_Chapter_reviewAn important step in thermodynamic analysis is to predict the thermodynamic properties, in particular, the intensive properties at different states of a process or a cycle. In this chapter, we have in...An important step in thermodynamic analysis is to predict the thermodynamic properties, in particular, the intensive properties at different states of a process or a cycle. In this chapter, we have introduced common thermodynamic properties of pure substances, and how to use phase diagrams and thermodynamic tables to determine the phase of a fluid and its corresponding properties at a given state.
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/zz%3A_Back_Matter/20%3A_GlossaryIt is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body (heat sink) to a higher-temperature body (heat sour...It is impossible to construct a device that operates in a cycle and produces no effect other than the transfer of heat from a lower-temperature body (heat sink) to a higher-temperature body (heat source). A state refers to a specific condition of a system that is described by a unique set of thermodynamic properties, such as pressure, temperature, specific volume, specific enthalpy, and so on.
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/04%3A_The_First_Law_of_Thermodynamics_for_Closed_SystemsThis chapter explains the concepts of heat, work, and the first law of thermodynamics for closed systems. Through examples, students will learn how to apply the first law of thermodynamics to engineer...This chapter explains the concepts of heat, work, and the first law of thermodynamics for closed systems. Through examples, students will learn how to apply the first law of thermodynamics to engineering problems involving closed systems.
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/04%3A_The_First_Law_of_Thermodynamics_for_Closed_Systems/4.04%3A_WorkWe will assume the expansion process is quasi-equilibrium, and the piston moves up an infinitesimal distance . The boundary work done by the gas to the surroundings in this infinitesimal process is ; ...We will assume the expansion process is quasi-equilibrium, and the piston moves up an infinitesimal distance . The boundary work done by the gas to the surroundings in this infinitesimal process is ; therefore, the total boundary work between two states in a process can be written as
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/07%3A_Appendices
- https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Introduction_to_Engineering_Thermodynamics_(Yan)/07%3A_Appendices/7.04%3A_Appendix_D_-_Thermodynamic_Properties_of_Carbon_DioxideP = 800 kPa (Sat. T = -46.00 o C) P = 1000 kPa (Sat. T = -40.12 o C) P = 1400 kPa (Sat. T = -30.58 o C) P = 2000 kPa (Sat. T = -19.50 o C) P = 3000 kPa (Sat. T = -5.55 o C) T ( o C) P = 6000 kPa (Sat....P = 800 kPa (Sat. T = -46.00 o C) P = 1000 kPa (Sat. T = -40.12 o C) P = 1400 kPa (Sat. T = -30.58 o C) P = 2000 kPa (Sat. T = -19.50 o C) P = 3000 kPa (Sat. T = -5.55 o C) T ( o C) P = 6000 kPa (Sat. T = 21.97 o C) Enthalpy H = 200 kJ/kg at 0°C for saturated liquid. Entropy S = 1 kJ/kg-K at 0°C for saturated liquid. Friend, “Thermophysical Properties of Fluid Systems” in NIST Chemistry WebBook, NIST Standard Reference Database Number 69, Eds.
