Volumetric and Thermodynamic Properties of Pure Fluids PDF

VOLUMETRIC AND THERMODYNAMIC PROPERTIES OF PURE FLUIDS PVT Behavior of Pure Substances Topic Learning Outcome: At the end of this topic, you should be able to: Derive thermodynamic relationships among thermodyn...

VOLUMETRIC AND THERMODYNAMIC PROPERTIES OF PURE FLUIDS PVT Behavior of Pure Substances Topic Learning Outcome: At the end of this topic, you should be able to: Derive thermodynamic relationships among thermodynamic properties using steam tables, pressure-enthalpy, temperature-entropy, and enthalpy-entropy charts. A substance that has a fixed chemical composition throughout is called a pure substance. Water, nitrogen, helium, and carbon dioxide, for example, are all pure substances. A pure substance does not have to be of a single chemical element or compound, however. A mixture of various chemical elements or compounds also qualifies as a pure substance as long as the mixture is homogeneous. Air, for example, is a mixture of several gases, but it is often considered to be a pure substance because it has a uniform chemical composition. However, a mixture of oil and water is not a pure substance. Since oil is not soluble in water, it will collect on top of the water, forming two chemically dissimilar regions. A mixture of two or more phases of a pure substance is still a pure substance as long as the chemical composition of all phases is the same. A mixture of ice and liquid water, for example, is a pure substance because both phases have the same chemical composition. A mixture of liquid air and gaseous air, however, is not a pure substance since the composition of liquid air is different from the composition of gaseous air, and thus the mixture is no longer chemically homogeneous. This is due to different components in air condensing at different temperatures at a specified pressure. Digital Image. Thoughtco. htps://www.thoughtco.com/examples-of-pure-substances-608350 A Pure Substance consists of only one type of atom or molecule or compound. TIP: If you can write a chemical formula for that substance, it is a pure substance. 1 CHE 2231/L: Chemical Engineering Thermodynamics Phases of a Pure Substance When studying phases or phase changes in thermodynamics, one does not need to be concerned with the molecular structure and behavior of different phases. However, it is very helpful to have some understanding of the molecular phenomena involved in each phase, and a brief discussion of phase transformations follows. Intermolecular bonds are strongest in solids and weakest in gases. One reason is that molecules in solids are closely packed together, whereas in gases they are separated by relatively large distances. The molecules in a solid are arranged in a three-dimensional pattern (lattice) that is repeated throughout. Because of the small distances between molecules in a solid, the attractive forces of molecules on each other are large and keep the molecules at fixed positions. The molecular spacing in the liquid phase is not much different from that of the solid phase, except the molecules are no longer at fixed positions relative to each other and they can rotate and translate freely. In the gas phase, the molecules are far apart from each other and a molecular order is nonexistent. Gas molecules move at random, continually colliding with each other and the walls of the container they are in. The arrangement of atoms in different phases: (a) molecules are at relatively fixed positions in a solid, (b) groups of molecules move about each other in the liquid phase, and (c) molecules move about at random in the gas phase. Phase Change Processes of Pure Substances There are many practical situations where two phases of a pure substance coexist in equilibrium. Water exists as a mixture of liquid and vapor in the boiler and the condenser of a steam power plant. The refrigerant turns from liquid to vapor in the freezer of a refrigerator. Even though many home owners consider the freezing of water in underground pipes as the most important phase-change process, attention in this section is focused on the liquid and vapor phases and their mixture. As a familiar substance, water is used to demonstrate the basic principles involved. Remember, however, that all pure substances exhibit the same general behavior. Consider a piston-cylinder device containing liquid water at 200C and 1atm pressure. 2 CHE 2231/L: Chemical Engineering Thermodynamics  (1) (2) (3) (4) (5) Initially, heat is added at constant pressure to the liquid water at 20 °C. At this state (1), this liquid is what we call compressed or subcooled liquid. As heat is added, the water temperature rises until it reaches 100 °C (2) the liquid at this point is saturated liquid. When you continue to add heat in the system, saturated liquid will turn to saturated vapor, the temperature would be constant at 100°C. At point (3) there exists both saturated liquid and saturated vapor, this is what we call saturated mixture. As heat is continuously added, eventually all of the saturated liquid will turn to saturated vapor (4). When this point is reached and heat is still added, temperature now will rise (>100°C). The vapor will become superheated (5). This constant pressure heat addition is illustrated in a TV diagram as shown below: Property Diagrams for Phase-Change Processes The variations of properties during phase-change processes are best studied and understood with the help of property diagrams. Next, we develop and discuss the T-v, P-v, and P- T diagrams for pure substances. 3 CHE 2231/L: Chemical Engineering Thermodynamics  T-V Diagram Water was used as a reference substance for the diagram below. The diagram is at constant pressure at various pressures. If you take a look at the diagram, the horizontal line that connects the saturated liquid and saturated vapor is what we call the saturation line. As the pressure is increase, this saturation line continuous to shrink and it becomes a point when the pressure reaches 22.05 MPa for the case of water. This point is called the critical point, and it is defined as the point at which the saturated liquid and saturated vapor states are identical. At pressures above the critical pressure, there is not a distinct phase change process. Instead, the specific volume of the substance continually increases, and at all times there is only one phase present. Eventually, it resembles a vapor, but we can never tell when the change has occurred. 4 CHE 2231/L: Chemical Engineering Thermodynamics  P-V Diagram The general shape of the P-v diagram of a pure substance is very much like the T-v diagram, but the T = constant lines on this diagram have a downward trend, as shown in the figure below P-T Diagram This diagram is often called the phase diagram since all three phases are separated from each other by three lines. The sublimation line separates the solid and vapor regions, the vaporization line separates the liquid and vapor regions, and the melting (or fusion) line separates the solid and liquid regions. These three lines meet at the triple point, where all three phases coexist in equilibrium. The vaporization line ends at the critical point because no distinction can be made between liquid and vapor phases above the critical point. Substances that expand and contract on freezing differ only in the melting line on the P-T diagram. The amount of heat added to effect the various phase changes is equal to the change in enthalpy. These various enthalpy differences across the phase boundaries have certain names. The change of enthalpy between a solid and liquid phase is the latent heat of fusion. The change of enthalpy between a liquid and vapor phase is the latent heat of vaporization, and finally the change of enthalpy in going from solid to vapor phase is the latent heat of sublimation. 5 CHE 2231/L: Chemical Engineering Thermodynamics Note:  At a given pressure, the temperature at which a pure substance changes phase is called the saturation temperature.  At a given temperature, the pressure at which a pure substance changes phase is called the saturation pressure. Gas Laws These laws relate the pressure, volume, and temperature of a gas. Boyle’s Law This states at constant temperature, the product of the pressure and volume of a given mass of an ideal gas in a closed system is always constant. 1 𝑉𝑉 ∝ 𝑃𝑃 𝑃𝑃1 𝑉𝑉1 = 𝑃𝑃2 𝑉𝑉2 Charles’ Law This states that, for a given mass of an ideal gas at constant pressure, the volume is directly proportional to its absolute temperature, assuming in a closed system. 𝑉𝑉 ∝ 𝑇𝑇 𝑉𝑉1 𝑉𝑉2 = 𝑇𝑇1 𝑇𝑇2 Gay-Lussac’s Law This law also sometimes called as Amonton’s Law, states that for a given mass and constant volume of an ideal gas, the pressure exerted on the sides of its container is directly proportional to its absolute temperature. 𝑃𝑃 ∝ 𝑇𝑇 6 CHE 2231/L: Chemical Engineering Thermodynamics  𝑃𝑃1 𝑃𝑃2 = 𝑇𝑇1 𝑇𝑇2 Avogadro’s Law This states that the volume occupied by an ideal gas is directly proportional to the number of molecules of the gas present in the container. This gives rise to the molar volume of a gas, which at STP (273.15 K, 1 atm) is about 22.4 L. 𝑉𝑉1 𝑉𝑉2 = 𝑛𝑛1 𝑛𝑛2 Combined Gas Law It shows the relationship between the pressure, volume, and temperature for a fixed mass (quantity) of gas. 𝑃𝑃1 𝑉𝑉1 𝑃𝑃2 𝑉𝑉2 = 𝑇𝑇1 𝑇𝑇2 THE IDEAL GAS An ideal gas is a hypothetical gas whose molecules occupy negligible space and have no interactions, and which consequently obeys the gas laws exactly. Equations for Process Calculations: Ideal Gases Isochoric Process (Constant volume) Equations applicable to a mechanically reversible constant-volume process are as follows. For a mole or unit mass of an ideal gas, 𝑄𝑄 = ∆𝑈𝑈 = 𝑛𝑛𝑛𝑛𝑛𝑛∆𝑇𝑇 𝑊𝑊 = 0 ∆𝐻𝐻 = 𝑛𝑛𝑛𝑛𝑛𝑛∆𝑇𝑇 Where: 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶 + 𝑅𝑅 Isobaric Process (Constant pressure) 𝑄𝑄 = ∆𝐻𝐻 = 𝑛𝑛𝑛𝑛𝑛𝑛∆𝑇𝑇 ∆𝑈𝑈 = 𝑛𝑛𝑛𝑛𝑛𝑛∆𝑇𝑇 𝑊𝑊 = −𝑃𝑃∆𝑉𝑉 = −𝑅𝑅∆𝑇𝑇 Isothermal Process (Constant temperature) ∆𝑈𝑈 = 0 ∆𝐻𝐻 = 0 𝑄𝑄 = −𝑊𝑊 𝑉𝑉2 𝑃𝑃2 𝑄𝑄 = 𝑅𝑅𝑅𝑅 ln = −𝑅𝑅𝑅𝑅 ln 𝑉𝑉1 𝑃𝑃1 𝑉𝑉2 𝑃𝑃2 𝑊𝑊 = −𝑅𝑅𝑅𝑅 ln = 𝑅𝑅𝑅𝑅 ln 𝑉𝑉1 𝑃𝑃1 7 CHE 2231/L: Chemical Engineering Thermodynamics  Adiabatic Process (Q = 0) Note: 𝐶𝐶𝐶𝐶 = 𝐶𝐶𝐶𝐶 + 𝑅𝑅 𝐶𝐶𝐶𝐶 Let: = 𝛾𝛾 𝐶𝐶𝐶𝐶 The relationship for pressure, temperature and volume for an adiabatic process would be 𝛾𝛾−1 𝛾𝛾−1 𝑇𝑇1 𝑉𝑉1 = 𝑇𝑇2 𝑉𝑉2 1−𝛾𝛾 1−𝛾𝛾 𝑇𝑇1 𝑃𝑃1 𝛾𝛾 = 𝑇𝑇2 𝑃𝑃2 𝛾𝛾 𝛾𝛾 𝛾𝛾 𝑃𝑃1 𝑉𝑉1 = 𝑃𝑃2 𝑉𝑉2 Therefore, 𝑄𝑄 = 0 𝑊𝑊 = ∆𝑈𝑈 = 𝑛𝑛𝑛𝑛𝑛𝑛∆𝑇𝑇 ∆𝐻𝐻 = 𝑛𝑛𝑛𝑛𝑛𝑛∆𝑇𝑇 Work can also be computed as 𝛾𝛾−1 𝛾𝛾−1 𝑃𝑃1 𝑉𝑉1 𝑃𝑃2 𝛾𝛾 𝑅𝑅𝑇𝑇1 𝑃𝑃2 𝛾𝛾 𝑊𝑊 = − 1 = − 1 𝛾𝛾 − 1 𝑃𝑃1 𝛾𝛾 − 1 𝑃𝑃1 Polytropic Process This is any reversible process on any open or closed system of gas or vapor which involves both heat and work transfer, such that a specified combination of properties were maintained constant throughout the process. For such a process, the following equations generally apply 𝑃𝑃𝑉𝑉 𝛿𝛿 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑇𝑇𝑉𝑉 𝛿𝛿−1 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 1−𝛿𝛿 𝑇𝑇𝑃𝑃 𝛿𝛿 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝛿𝛿−1 𝑅𝑅𝑇𝑇1 𝑃𝑃2 𝛿𝛿 𝑊𝑊 = − 1 𝛿𝛿 − 1 𝑃𝑃1 𝛿𝛿−1 (𝛿𝛿 − 𝛾𝛾)𝑅𝑅𝑇𝑇1 𝑃𝑃2 𝛿𝛿 𝑄𝑄 = − 1 (𝛿𝛿 − 1)(𝛾𝛾 − 1) 𝑃𝑃1 Sample Problems: 1. Air is compressed from an initial condition of 1 bar and 25 °C to a final state of 5 bar and 25 °C by three different mechanically reversible processes in a closed system a. Heating at constant volume followed by cooling at constant pressure. b. Isothermal compression. c. Adiabatic compression followed by cooling at constant volume. At these conditions, air may be considered an ideal gas with the constant heat capacities, CV = (5/2)R and CP = (7/2)R. Calculate the work required, heat transferred and the changes in internal energy and enthalpy of the air for each process. 2. An ideal gas undergoes the following sequence of mechanically reversible processes in a closed system: a. From an initial state of 700C and 1 bar, it is compressed adiabatically to 1500C. b. It is then cooled from 1500C to 700C at constant pressure. c. Finally, it is expanded isothermally to its original state. 8 CHE 2231/L: Chemical Engineering Thermodynamics  Calculate ΔU, ΔH, Q and W for each of the three processes and for the entire cycle. Take CV = (3/2)R and CP = (5/2)R. 3. An ideal gas initially at 30 °C and 100 kPa, undergoes the following cyclic processes in a closed system. a. In mechanically reversible processes, it is first compressed adiabatically to 500 kPa, then cooled at a constant pressure of 500 kPa to 30 °C and finally expanded isothermally to its original state. b. The cycle traverses exactly the same changes of state, but each step is irreversible with an efficiency of 80% compared with the corresponding mechanically reversible process. Calculate Q, W, ∆U & ∆H for each step of the process and for the cycle. Take CP = 7/2 R. Note:  The work of an irreversible process is calculated by a two step procedure. First, W is determined for a mechanically reversible process that accomplishes the same change of state as the actual irreversible process. Second, this result is multiplied or divided by an efficiency to give the actual work. If the process produces work, the absolute value for the reversible process is too large and must be multiplied by an efficiency. If the process requires work, the value of the reversible process is too small and must be divided by an efficiency. 4. One mole of an ideal gas with CP = 7/2 R and CV = 5/2 R expands from P1=8 bar and T1 = 600 K to P2=1bar by each of the following paths. (a) Constant volume. (b) Constant temperature. (c) Adiabatically. Assuming mechanical reversibility, calculate W, Q, ∆H for each process. Sketch each path on a single PV diagram. 5. One cubic meter of an ideal gas at 600 K and 1000 kPa expands to five times its initial volume as follows: (a) By a mechanically reversible, isothermal process. (b) By a mechanically reversible, adiabatic process. (c) By an adiabatic, irreversible process in w/c expansion is against a restraining pressure of 100 kPa. For each case, calculate the final temperature, pressure and the work done by the gas. CP = 21 J/mol-K. 6. A three-process cycle uses 3 kg of air and undergoes the following processes: polytropic compression from state 1 to state 2, where P1 = 150 kPa, T1 = 360 K, P2 = 750 kPa, and δ = 1.2; constant-pressure cooling from state 2 to state 3; and constant temperature heating from state 3 to state 1, completing the cycle. Find the temperatures, pressure, and volumes at each state and determine the ∆H, ∆U, Q & W for each process and for the entire cycle. Assume CP = 5/2 R. References Smith, J. M., H. C. Van Ness and M. M. Abbott (2001). Chemical Engineering Thermodynamics. McGraw Hill. Cengel, Y. and M. Boles. Thermodynamics: An Engineering Approach. 9 CHE 2231/L: Chemical Engineering Thermodynamics