The system mass is constant (m=m, +m,= constant so that for any changes We can define the quantity dme dmo dm. =-dm,= mass transferred from liquid to vapor In terms of dm f the volume change of the system is The work done is given by dw= pdy The change in internal energy, AU, can be found as follows. The internal energy of the system can be expressed in terms of the mass in each phase and the specific internal energy(internal energy per unit mass, u) of the phase as U=ume.m d0=urdm +u, dm Note that the specific internal energy can be expressed in a similar way as the specific volume in terms of the quality and the specific enthalpy of each phase X:a2+(1-X) Writing the first law for this process do=du+ dm ur)dme+p(vg-v,a Py Pve ldr mfg h -he ldm The heat needed for the transfer of mass is proportional to the difference in specific enthalpy between vapor and liquid. The pressure and temperature are constant, so that the specific internal energy and the specific enthalpy for the liquid phase and the gas phase are also constant. For a finite change in mass from liquid to vapor, m fe, therefore, the quantity of heat needed is 0=(n-h ) mg-AH (enthalpy change) 2B-6
2B-6 The system mass is constant ( mm m =+= f g constant) so that for any changes dm dm dm == +f g 0 . We can define the quantity dmfg dm dm dm fg = =g f - = mass transferred from liquid to vapor. In terms of dmfg the volume change of the system is dV v v dm = − ( ) g f fg . The work done is given by dW = PdV = − P v v dm ( ) g f fg . The change in internal energy, ∆U , can be found as follows. The internal energy of the system can be expressed in terms of the mass in each phase and the specific internal energy (internal energy per unit mass, u) of the phase as, U um um = + f f g g dU u dm u dm u u dm = + =− f f gg g ( )f fg . Note that the specific internal energy can be expressed in a similar way as the specific volume in terms of the quality and the specific enthalpy of each phase: u Xu X u =⋅ +− ⋅ g f ( ) 1 Writing the first law for this process: dQ = dU + dW = − ( ) u u dm P v g f fg + − ( ) g v dm f fg . = + [ ] ( ) u Pv u g g − + ( ) f f fg Pv dm = − ( ) h h dm g f fg. The heat needed for the transfer of mass is proportional to the difference in specific enthalpy between vapor and liquid. The pressure and temperature are constant, so that the specific internal energy and the specific enthalpy for the liquid phase and the gas phase are also constant. For a finite change in mass from liquid to vapor, mfg, therefore, the quantity of heat needed is Q h hm H = − ( ) g f fg = ∆ (enthalpy change)
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-7
2B-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
For the vapor dome, is there vapor and liquid inside the dome and outside is it just liquid or just gas? Is it interchangeable? Is it true for the plasma phase? (MP 2B.1 What is h How do we find it?(MP 2B.2 Reasoning behind the slopes for T=cst lines in the P-V diagram (MP 2B. 3) For a constant pressure heat addition why is g=Ah?(MP 2B. 4) What is latent heat?(MP 2B.5) Why is U a function of x?(MP 2B. 6) 2. B.3 The Carnot Cycle as a Two-Phase Power Cycle A Carnot cycle that uses a two-phase fluid as the working medium is shown below Figure 2B-6. Figure 2B-6a gives the cycle in P-v coordinates, 2B-6b in T-s coordinates, and 2B-6c in h-s coordinates. The boundary of the region in which there is liquid and vapor both present(the vapor dome)is also indicated. Note that the form of the cycle is different in the T-s and h-s representation; it is only for a perfect gas with constant specific heats that cycles in the two coordinate representations have the same shapes (a) p-v diagram T2 5? (b)T-s diagram (c)h-s diagram Figure 2B-6: Carnot cycle with two-phase medium. (a)cycle in P-v coordinates, (b)cycle in T- coordinates,(c)cycle in h-S coordinates The processes in the cycle are as follow i Start at state a with saturated liquid(all of mass in liquid condition). Carry out a reversible isothermal expansion to b(a b)until all the liquid is vaporized. During this process a quantity of heat qH per unit mass is received from the heat source at temperature T2 i) Reversible adiabatic(i. e, isentropic)expansion(b>c) lowers the temperature to T Generally state c will be in the region where there is both liquid and vapor 2B-8
2B-8 For the vapor dome, is there vapor and liquid inside the dome and outside is it just liquid or just gas? Is it interchangeable? Is it true for the plasma phase? (MP 2B.1) What is hfg ? How do we find it? (MP 2B.2) Reasoning behind the slopes for T=cst lines in the P-V diagram. (MP 2B.3) For a constant pressure heat addition, why is q=∆h? (MP 2B.4) What is latent heat? (MP 2B.5) Why is U a function of x? (MP 2B.6) 2.B.3 The Carnot Cycle as a Two-Phase Power Cycle A Carnot cycle that uses a two-phase fluid as the working medium is shown below in Figure 2B-6. Figure 2B-6a gives the cycle in P-v coordinates, 2B-6b in T-s coordinates, and 2B-6c in h-s coordinates. The boundary of the region in which there is liquid and vapor both present (the vapor dome) is also indicated. Note that the form of the cycle is different in the T-s and h-s representation; it is only for a perfect gas with constant specific heats that cycles in the two coordinate representations have the same shapes. a a b b c c d d e T T2 T2 T2 T1 T1 T1 s1 s2 s s h f g h a b d c p p2 p1 v (a) p-v diagram (b) T-s diagram (c) h-s diagram Figure 2B-6: Carnot cycle with two-phase medium. (a) cycle in P-v coordinates, (b) cycle in T-s coordinates, (c) cycle in h-s coordinates The processes in the cycle are as follows: i) Start at state a with saturated liquid (all of mass in liquid condition). Carry out a reversible isothermal expansion to b (a b) until all the liquid is vaporized. During this process a quantity of heat qH per unit mass is received from the heat source at temperature T2 . ii) Reversible adiabatic (i.e., isentropic) expansion (b c) lowers the temperature to T1. Generally state c will be in the region where there is both liquid and vapor
ii) Isothermal compression(c>d) at T to state d. During this compression, heat q, per unit mass IS Reversible adiabatic (i. e, isentropiccompression(d a) in which the vapor condenses to liquid and the state returns to In the T-s diagram the heat received, qH, is abef and the heat rejected, qL, is dcef. The net work sented by abcd. The thermal effi b In the h-s diagram, the isentropic processes are vertical lines as in the T-s diagram. The isotherms in the former, however, are not horizontal as they are in the latter. To see their shape we note that for these two-phase processes the isotherms are also lines of constant pressure(isobars), since P P(n. The combined first and second law is ra=dh-中 For a constant pressure reversible process, darey =Tds= dh. The slope of a constant pressure line in h-s coordinates is thus T=constant; slope of constant pressure line for two-phase medium The heat received and rejected per unit mass is given in terms of the enthalpy at the different states qL =hd-he.(In accord with our convention this is less than zero. The thermal efficiency is n="m=+=(-h)+(ba or, in terms of the work done during the isentropic compression and expansion processes, which correspond to the shaft work done on the fluid and received by the fluid (h-h)-(h2-h h Example: Carnot steam cycle: Heat source temperature = 300C Heat sink temperature= 20C What is the(i)thermal efficiency and (ii) ratio of turbine work to compression(pump) work if reversible? b) the turbine and the pump have adiabatic efficiencies of 0
2B-9 iii) Isothermal compression (c d) at T1 to state d. During this compression, heat qL per unit mass is rejected to the source at T1. iv) Reversible adiabatic (i.e., isentropic) compression (d a) in which the vapor condenses to liquid and the state returns to a. In the T-s diagram the heat received, qH , is abef and the heat rejected, qL , is dcef. The net work is represented by abcd. The thermal efficiency is given by η = = =− w q abcd abef T T net H Area Area 1 1 2 . In the h-s diagram, the isentropic processes are vertical lines as in the T-s diagram. The isotherms in the former, however, are not horizontal as they are in the latter. To see their shape we note that for these two-phase processes the isotherms are also lines of constant pressure (isobars), since P = P(T). The combined first and second law is Tds dh dp = − ρ . For a constant pressure reversible process, dq Tds dh rev = = . The slope of a constant pressure line in h-s coordinates is thus, ∂ ∂ h s T P = = constant; slope of constant pressure line for two-phase medium. The heat received and rejected per unit mass is given in terms of the enthalpy at the different states as, q hh H ba = − q hh L = −d c . (In accord with our convention this is less than zero.) The thermal efficiency is η = = + = ( ) − + − ( ) ( ) − w q q q q hh h h h h net H H L H ba dc b a , or, in terms of the work done during the isentropic compression and expansion processes, which correspond to the shaft work done on the fluid and received by the fluid, η = ( ) − − − ( ) ( ) − hh hh h h bc a d b a . Example: Carnot steam cycle: Heat source temperature = 300o C Heat sink temperature = 20o C What is the (i) thermal efficiency and (ii) ratio of turbine work to compression (pump) work if a) all processes are reversible? b) the turbine and the pump have adiabatic efficiencies of 0.8?