Thermodynamics - Chapter 5: Pure substance - Ngo Thanh An

Nội dung text: Thermodynamics - Chapter 5: Pure substance - Ngo Thanh An

1. Chapter 5 – Pure substance • Pure substance: A pure substance has a homogenous and fixed chemical composition throughout, and may exist in more than one phase • Air is a mixture of several gases, but it is considered to be a pure substance. 2
2. Chapter 5 – Pure substance Solid: • The molecules in solid are arranged in a three-dimensional pattern (lattice) • The molecular spacing is close each other • The molecules cannot move, but they continually oscillate about their equilibrium position • Their velocity depends on the temperature => Increasing temperature leads to group molecules breaking away => melting process 4
3. Chapter 5 – Pure substance • Compressed liquid (subcooled liquid): A substance that it is not about to vaporize. • Saturated liquid: A liquid that is about to vaporize. At 1 atm and 20°C, water exists in the liquid phase (compressed liquid). At 1 atm pressure and 100°C, water exists as a liquid that is ready to vaporize (saturated liquid). 6
4. Chapter 5 – Pure substance If the entire process between state 1 and 5 described in the figure is reversed by cooling the water while maintaining the pressure at the same value, the water will go back to state 1, retracing the same path, and in so doing, the amount of heat released will exactly match the amount of heat added during the heating process. T-v diagram for the heating process of water at constant pressure. 8
5. Chapter 5 – Pure substance • Latent heat: The amount of energy absorbed or released during a phase-change process. • Latent heat of fusion: The amount of energy absorbed during melting. It is equivalent to the amount of energy released during freezing. • Latent heat of vaporization: The amount of energy absorbed during vaporization and it is equivalent to the energy released during condensation. • The magnitudes of the latent heats depend on the temperature or pressure at which the phase change occurs. • At 1 atm pressure, the latent heat of fusion of water is 333.7 kJ/kg and the latent heat of vaporization is 2256.5 kJ/kg. • The atmospheric pressure, and thus the boiling temperature of water, decreases with elevation. 10
6. Chapter 5 – Pure substance • saturated liquid line • saturated vapor line • compressed liquid region • superheated vapor region • saturated liquid–vapor mixture region (wet region) At supercritical pressures Critical point: The point at (P > Pcr), there is no which the saturated liquid distinct phase-change and saturated vapor states (boiling) process. are identical. 12
7. Chapter 5 – Pure substance The P-v-T surfaces present a great deal of information at once, but in a thermodynamic analysis it is more convenient to work with two-dimensional diagrams, such as the P-v and T-v diagrams. 14
8. Chapter 5 – Pure substance • For most substances, the relationships among thermodynamic properties are too complex to be expressed by simple equations. • Therefore, properties are frequently presented in the form of tables. • Some thermodynamic properties can be measured easily, but others cannot and are calculated by using the relations between them and measurable properties. • The results of these measurements and calculations are presented in tables in a convenient format. Enthalpy—A Combination Property The combination u + Pv is frequently encountered in the analysis of control volumes. The product pressure volume has energy units. 16
9. Chapter 5 – Pure substance Examples: Saturated liquid and saturated vapor states of water on T-v and P- v diagrams. 18
10. Chapter 5 – Pure substance y v, u, or h. 20
11. Chapter 5 – Pure substance Compared to saturated vapor, superheated In the region to the right of the vapor is characterized by saturated vapor line and at temperatures above the critical point temperature, a substance exists as superheated vapor. In this region, temperature and pressure are independent properties. At a specified P, superheated vapor exists at a higher h than the saturated vapor. 22
12. Chapter 5 – Pure substance Example 1. A rigid tank contains 50 kg of saturated liquid water at 90ºC. Determine the pressure in the tank and the volume of the tank. (P = 70.14 kPa, V = 0.0518 m3) Example 2. A mass of 200 g of saturated liquid water is completely vaporized at a constant pressure of 100 kPa. Determine the volume change and amount of energy added to the water Table A-5, at P=100 kPa, 3 3 vg=1.6950 m /kg & vf=0.001043 m /kg, hfg = 2258.0 kJ/kg 3 (ΔV =0.3387 m , mhfg = 451.6 kJ) 24
13. Chapter 5 – Pure substance Example 4 . Determine the temperature of water at a state of P = 0.5 MPa and h = 2890 kJ/kg. (at P = 0.5 MPa, hg = 2748.7 kJ/kg (T=216.3ºC)