The heat needed per unit mass, g, for transformation between the two phases is h The notation hfg refers to the specific enthalpy change between the liquid state and the vapor state The expression for the amount of heat needed, is a particular case of the general result that in any reversible process at constant pressure, the heat flowing into, or out of, the system is equal to the enthalpy change. Heat is absorbed if the change is from solid to liquid (heat of fusion), liquid to vapor(heat of vaporization), or solid to vapor(heat of sublimation) A numerical example is furnished by the vaporization of water at 100oC. 1 How much heat is needed per unit mass of fluid vaporized? 11) How much work is done per unit mass of fluid vaporized? i What is the change in internal energy per unit mass of fluid vaporized In addressing these questions, we make use of the fact that problems involving heat and work exchanges in two-phase media are important enough that the values of the specific thermodynamic properties that characterize these transformations have been computed for many different working fluids. The values are given in SB&vw in tables B 1.1 and B. 1. 2 for water at saturated conditions and in Tables B 1.3, B. 1. 4, and B 1.5 for other conditions, as well as for other working fluids. From these At 100C, the vapor pressure is 0.1013 MPa, The specific enthalpy of the vapor, he, is 2676 kJ/kg and the specific enthalpy of the liquid, h, is 419 kJ/kg The difference in enthalpy between liquid and vapor, hfg, occurs often enough so that it is tabulated also. This is 2257 kJ/kg The specific volume of the vapor is 1. 6729 m/kg and the specific volume of the liquid 0.001044 The heat input to the system is the change in enthalpy between liquid and vapor, hf, and is equal to2257x10°Jkg The work done is P(vg -,)which has a value of P{v-v)=0.1013x10x1629-0041=0169x10Jkg The change in internal energy per unit mass(uf)can be found from Au=q-w or from the tabulated values as 2.088 x 10 J/kg. This is much larger than the work done. Most of the heat input is used to change the internal energy rather than appearing as work. Muddy b。in 2B-72B-7 The heat needed per unit mass, q, for transformation between the two phases is q Q m hh h fg = =− ( ) g f fg = . The notation hfg refers to the specific enthalpy change between the liquid state and the vapor state. The expression for the amount of heat needed, q, is a particular case of the general result that in any reversible process at constant pressure, the heat flowing into, or out of, the system is equal to the enthalpy change. Heat is absorbed if the change is from solid to liquid (heat of fusion), liquid to vapor (heat of vaporization), or solid to vapor (heat of sublimation). A numerical example is furnished by the vaporization of water at 100o C: i) How much heat is needed per unit mass of fluid vaporized? ii) How much work is done per unit mass of fluid vaporized? iii) What is the change in internal energy per unit mass of fluid vaporized?. In addressing these questions, we make use of the fact that problems involving heat and work exchanges in two-phase media are important enough that the values of the specific thermodynamic properties that characterize these transformations have been computed for many different working fluids. The values are given in SB&VW in Tables B.1.1 and B.1.2 for water at saturated conditions and in Tables B.1.3, B.1.4, and B.1.5 for other conditions, as well as for other working fluids. From these: - At 100o C, the vapor pressure is 0.1013 MPa, - The specific enthalpy of the vapor, hg , is 2676 kJ/kg and the specific enthalpy of the liquid, hf , is 419 kJ/kg - The difference in enthalpy between liquid and vapor, hfg, occurs often enough so that it is tabulated also. This is 2257 kJ/kg, - The specific volume of the vapor is 1.6729 m3 /kg and the specific volume of the liquid is 0.001044. The heat input to the system is the change in enthalpy between liquid and vapor, hfg, and is equal to 2.257 x 106 J/kg. The work done is Pv v ( ) g − f which has a value of Pv v ( ) g − f =0.1013 x 106 x [1.629 – 0.001044] =0.169 x 106 J/kg. The change in internal energy per unit mass (ufg) can be found from ∆uqw = − or from the tabulated values as 2.088 x 106 J/kg. This is much larger than the work done. Most of the heat input is used to change the internal energy rather than appearing as work. Muddy points