2. B Power Cycles with Two-Phase Media (Vapor Power Cycles) [SB&vw-chapter 3, Chapter ll, Sections 11.1 to 11.7] In this section, we examine cycles that use two-phase media as the working fluid. These can be combined with gas turbine cycles to provide combined cycles which have higher efficiency than either alone. They can also be used by themselves to provide power sources for both terrestrial and space applications. The topics to be covered are i Behavior of two-phase systems: equilibrium, pressure temperature relations 11) Carnot cycles with two-phase media 1 Rankine cycles D Combined cycles 2. B. 1 Behavior of Two-Phase Systems The definition of a phase, as given by SB&vw, is"a quantity of matter that is liquid and its vapor and a glass of ice water. A system which has three phases is a container w 9 homogeneous throughout. Common examples of systems that contain more than one phase are ice, water, and water vapor We wish to find the relations between phases and the relations that describe the change of phase (from solid to liquid, or from liquid to vapor) of a pure substance, including the work done and the heat transfer. To start we consider a system consisting of a liquid and its vapor in equilibrium, which are enclosed in a container under a moveable piston, as shown in Figure 2B-1 The system is maintained at constant temperature through contact with a heat reservoir at temperature T, so there can be heat transfer to or from the system. Water vapor Liquid water Liquid water Figure 2B-1: Two-phase system in contact with constant temperature heat reservoir For a pure substance, as shown at the right, there is a one-to-one correspondence between the temperature at which vaporization occurs and the pressure These values are called the saturation pressure and saturation temperature (see Ch 3 in SB&vw) Temperature P-T relation for liquid-vapor system
2B-1 2.B Power Cycles with Two-Phase Media (Vapor Power Cycles) [SB&VW – Chapter 3, Chapter 11, Sections 11.1 to 11.7] In this section, we examine cycles that use two-phase media as the working fluid. These can be combined with gas turbine cycles to provide combined cycles which have higher efficiency than either alone. They can also be used by themselves to provide power sources for both terrestrial and space applications. The topics to be covered are: i) Behavior of two-phase systems: equilibrium, pressure temperature relations ii) Carnot cycles with two-phase media iii) Rankine cycles iv) Combined cycles 2.B.1 Behavior of Two-Phase Systems The definition of a phase, as given by SB&VW, is “a quantity of matter that is homogeneous throughout”. Common examples of systems that contain more than one phase are a liquid and its vapor and a glass of ice water. A system which has three phases is a container with ice, water, and water vapor. We wish to find the relations between phases and the relations that describe the change of phase (from solid to liquid, or from liquid to vapor) of a pure substance, including the work done and the heat transfer. To start we consider a system consisting of a liquid and its vapor in equilibrium, which are enclosed in a container under a moveable piston, as shown in Figure 2B-1. The system is maintained at constant temperature through contact with a heat reservoir at temperature T, so there can be heat transfer to or from the system. (a) Liquid water Liquid water Water vapor Water vapor (b) (c) Figure 2B-1: Two-phase system in contact with constant temperature heat reservoir For a pure substance, as shown at the right, there is a one-to-one correspondence between the temperature at which vaporization occurs and the pressure. These values are called the saturation pressure and saturation temperature (see Ch. 3 in SB&VW). P-T relation for liquid-vapor system
This means there is an additional constraint for a liquid-vapor mixture, in addition to the equation of state. The consequence is that we only need to specify one variable to determine the state of the system. For example, if we specify Then P is set. In summary, for two phases in equilibrium P=P(T). If both phases are present, any quasi-static process at constant T is also at constant P Let us examine the pressure-volume behavior of a liquid-vapor system at constant temperatu For a single-phase perfect gas we know that the curve would be Pv=constant. For the two-phase system the curve looks quite different, as indicated in Figure 2B-2 D Critical point Mixture of口 iquid and口 LIquid saturation curve curve Volume v Figure 2B-2-P-v diagram for two-phase system showing isotherms can coexist. This is roughly dome-shaped and is thus often referred to as the"vapor dome,"por Several features of the figure should be noted. First, there is a region in which liquid and vapor Outside of this regime, the equilibrium state will be a single phase. The regions of the diagram in which the system will be in the liquid and vapor phases respectively are indicated. Second is th steepness of the isotherms in the liquid phase, due to the small compressibility of most liquids Third, the behavior of isotherms at temperatures below the"critical point"(see below )in the region to the right of the vapor dome approach those of an ideal gas as the pressure decreases and the ideal gas relation is a good approximation in this region The behavior shown is found for all the isotherms that go through the vapor dome. At a high enough temperature, specifically at a temperature corresponding to the pressure at the peak of the vapor dome, there is no transition from liquid to vapor and the fluid goes continuously from a liquid-like behavior to a gas-type behavior. This behavior is unfamiliar, mainly because the temperatures and pressures are not ones that we typically experience; for water the critical temperature is 374C and the associated critical pressure is 220 atmospheres
2B-2 This means there is an additional constraint for a liquid-vapor mixture, in addition to the equation of state. The consequence is that we only need to specify one variable to determine the state of the system. For example, if we specify T then P is set. In summary, for two phases in equilibrium, P PT = ( ). If both phases are present, any quasi-static process at constant T is also at constant P. Let us examine the pressure-volume behavior of a liquid-vapor system at constant temperature. For a single-phase perfect gas we know that the curve would be Pv = constant. For the two-phase system the curve looks quite different, as indicated in Figure 2B-2. Volume, V Pressure, P Liquid phase Mixture of liquid and vapor Liquid saturation curve Vapor saturation curve Critical point Vapor phase Gas phase Critical isotherm D B A C Figure 2B-2 – P-v diagram for two-phase system showing isotherms Several features of the figure should be noted. First, there is a region in which liquid and vapor can coexist. This is roughly dome-shaped and is thus often referred to as the “vapor dome”. Outside of this regime, the equilibrium state will be a single phase. The regions of the diagram in which the system will be in the liquid and vapor phases respectively are indicated. Second is the steepness of the isotherms in the liquid phase, due to the small compressibility of most liquids. Third, the behavior of isotherms at temperatures below the “critical point” (see below) in the region to the right of the vapor dome approach those of an ideal gas as the pressure decreases and the ideal gas relation is a good approximation in this region. The behavior shown is found for all the isotherms that go through the vapor dome. At a high enough temperature, specifically at a temperature corresponding to the pressure at the peak of the vapor dome, there is no transition from liquid to vapor and the fluid goes continuously from a liquid-like behavior to a gas-type behavior. This behavior is unfamiliar, mainly because the temperatures and pressures are not ones that we typically experience; for water the critical temperature is 374o C and the associated critical pressure is 220 atmospheres
There is a distinct nomenclature used for systems with more than one phase. In this, the terms "vapor"and"gas "seem to be used interchangeably. In the zone where both liquid and vapor exist there are two bounding situations. When the last trace of vapor condenses, the state becomes turated liquid. When the last trace of liquid evaporates the state becomes saturated vapor(or di vapor). If we put heat into a saturated vapor it is referred to as superheated vapor. Nitrogen at room temperature and pressure(at one atmosphere the vaporization temperature of nitrogen is 77 K) is a superheated vapor Figure 2B-3: Constant pressure curves in T-V coordinates showing vapor dome Figure 2B-3 shows lines of constant pressure in temperature-volume coordinates. Inside the vapor dome the constant pressure lines are also lines of constant temperature It is useful to describe the situations encountered as we decrease the pressure or equivalently increase the specific volume, starting from a high pressure-low specific volume state(the upper left-hand side of the isotherm in Figure 2B-2). The behavior in this region is liquid-like with very little compressibility. As the pressure is decreased, the volume changes little until the boundary of the vapor dome is reached. Once this occurs, however, the pressure is fixed because the temperature is constant. As the piston is withdrawn, the specific volume increases through more liquid evaporating and more vapor being produced. During this process, since the expansion is sothermal we specified that it was), heat is transferred to the system. The specific volume will increase at constant pressure until the right hand boundary of the vapor dome is reached. At this point, all the liquid will have been transformed into vapor and the system again behaves as a single-phase fluid For water at temperatures near room temperature, the behavior would be essentially that of a perfect gas in this region. To the right of the vapor dome, as mentioned above, the behavior is qualitatively like that of a perfect gas Referring to Figure 2B-4, we define notation to be used in what follows. The states a and c denote he conditions at which all the fluid is in the liquid state and the gaseous state respectively 2B-3
2B-3 There is a distinct nomenclature used for systems with more than one phase. In this, the terms “vapor” and “gas” seem to be used interchangeably. In the zone where both liquid and vapor exist, there are two bounding situations. When the last trace of vapor condenses, the state becomes saturated liquid. When the last trace of liquid evaporates the state becomes saturated vapor (or dry vapor). If we put heat into a saturated vapor it is referred to as superheated vapor. Nitrogen at room temperature and pressure (at one atmosphere the vaporization temperature of nitrogen is 77 K) is a superheated vapor. Figure 2B-3: Constant pressure curves in T-v coordinates showing vapor dome Figure 2B-3 shows lines of constant pressure in temperature-volume coordinates. Inside the vapor dome the constant pressure lines are also lines of constant temperature. It is useful to describe the situations encountered as we decrease the pressure or equivalently increase the specific volume, starting from a high pressure-low specific volume state (the upper left-hand side of the isotherm in Figure 2B-2). The behavior in this region is liquid-like with very little compressibility. As the pressure is decreased, the volume changes little until the boundary of the vapor dome is reached. Once this occurs, however, the pressure is fixed because the temperature is constant. As the piston is withdrawn, the specific volume increases through more liquid evaporating and more vapor being produced. During this process, since the expansion is isothermal (we specified that it was), heat is transferred to the system. The specific volume will increase at constant pressure until the right hand boundary of the vapor dome is reached. At this point, all the liquid will have been transformed into vapor and the system again behaves as a single-phase fluid. For water at temperatures near room temperature, the behavior would be essentially that of a perfect gas in this region. To the right of the vapor dome, as mentioned above, the behavior is qualitatively like that of a perfect gas. Referring to Figure 2B-4, we define notation to be used in what follows. The states a and c denote the conditions at which all the fluid is in the liquid state and the gaseous state respectively
Critical point Mixture Liquid State of liquid (constant T line) In mixture t ga Figure 2B-4: Specific volumes at constant temperature and states within the vapor dome in a liquid-vapor system The specific volumes corresponding to these states are vr-specific volume of liquid phase v,- specific volume of gas phase For conditions corresponding to specific volumes between these two values, i.e., for state b, the system would exist with part of the mass in a liquid state and part of the mass in a gaseous(vapor) state. The average specific volume for this condition is v-average specific volume of two-phase system mass that exists in the two phases as follows The total mass of the system is givens, r and the We can relate the average specific volume to the specific volumes for liquid and vap total mass m= liquid mass vapor mass =m, + mg The volume of the system is Volume of liquid=V Volume of vapor=V8=mgg Total volume=V=m//+mgg The average specific volume, V, is the ratio of the total volume to the total mass of the system mgvf tmg average specific volume 2B-4
2B-4 State of liquid in mixture State of vapor in mixture Liquid + gas vf v vg Mixture Liquid "state" Standard-liquid line Critical point Saturated-vapor line Gas (vapor) (constant T line) T v a b c Figure 2B-4: Specific volumes at constant temperature and states within the vapor dome in a liquid-vapor system The specific volumes corresponding to these states are: v f - specific volume of liquid phase vg - specific volume of gas phase For conditions corresponding to specific volumes between these two values, i.e., for state b, the system would exist with part of the mass in a liquid state and part of the mass in a gaseous (vapor) state. The average specific volume for this condition is v - average specific volume of two-phase system. We can relate the average specific volume to the specific volumes for liquid and vapor and the mass that exists in the two phases as follows. The total mass of the system is given by total mass = liquid mass + vapor mass = m mm = +f g. The volume of the system is Volume of liquid =V mv f ff = Volume of vapor =V mv g gg = Total volume =V mv mv = + f f g g . The average specific volume, v , is the ratio of the total volume to the total mass of the system v mv mv m m f f g g f g = + + = average specific volume
The fraction of the total mass in the vapor phase is called quality, and denoted by X quality of a liquid-vapor system In terms of the quality and specific volumes, the average specific volume can be expressed as V=X (1-X) In reference to Figure 2B-5,ab=v-Vf,ac=vg-vI Ity T (b) Figure 2B-5: Liquid vapor equilibrium in a two-phase medium 2. B2-work and Heat Transfer with Two-Phase Media We examine the work and heat transfer in quasi-static processes with two-phase systems For definiteness, consider the system to be a liquid-vapor mixture in a container whose volume car be varied through movement of a piston, as shown in Figure 2B-5. The system is kept at constant temperature through contact with a heat reservoir at temperature T. The pressure is thus also constant, but the volume, v, can change. For a fixed mass, the volume is proportional to the specific volume v so that point b in Figure 2B-5 must move to the left or the right as V changes This implies that the amount of mass in each of the two phases, and hence the quality, also changes because mass is transferred from one phase to the other. We wish to find the heat and work transfer associated with the change in mass in each phase. The change in volume can be related to he changes in mass in the two phases as dV=vdm。+v,dm
2B-5 The fraction of the total mass in the vapor phase is called quality, and denoted by X. X m m m g f g = + = quality of a liquid-vapor system. In terms of the quality and specific volumes, the average specific volume can be expressed as: v Xv X v =⋅ +− ⋅ g f ( ) 1 In reference to Figure 2B-5, ab v v ac v v =− = − f g f , . ab ac v v v v X f g f = − − = = quality. (a) T L L-V T T V p a b c vf v v vg (b) Figure 2B-5: Liquid vapor equilibrium in a two-phase medium 2.B.2 - Work and Heat Transfer with Two-Phase Media We examine the work and heat transfer in quasi-static processes with two-phase systems. For definiteness, consider the system to be a liquid-vapor mixture in a container whose volume can be varied through movement of a piston, as shown in Figure 2B-5. The system is kept at constant temperature through contact with a heat reservoir at temperature T. The pressure is thus also constant, but the volume, V, can change. For a fixed mass, the volume is proportional to the specific volume v so that point b in Figure 2B-5 must move to the left or the right as V changes. This implies that the amount of mass in each of the two phases, and hence the quality, also changes because mass is transferred from one phase to the other. We wish to find the heat and work transfer associated with the change in mass in each phase. The change in volume can be related to the changes in mass in the two phases as, dV v dm v dm = + g g f f
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?
Neglect the changes in kinetic energy at inlet and outlet of the turbine and pump a)For the reversible cycl thermal nce 293 =0489 573 To find the work in the pump(compression process)or in the turbine, we need to find the enthalpy changes between states b and c, Ahbe, and the change between a and d, Ah had. To obtain these the approach is to use the fact that s=constant during the expansion to find the quality at state c and then, knowing the quality, calculate the enthalpy as h=Xh, +(1-X)h. We know the conditions at state b, where the fluid is all vapor, i.e., we know Tb, hb, Sb hb=vapor( 300C)=he(300C)=2749 kJ/kg s=5(900005-05k/kgK in the isentropic expansion pi We now need to find the quality at state c, X. Using the definition of quality given in Section 2.B. 1, and noting that se=Xse+(1-x)s, we obtain, s4-s/(T)s-s/(T) The quantity s, is the mass-weighted entropy at state c, which is at temperature Tc The quantity s, (T )is the entropy of the liquid at temperature T The quantity s(T)is the entropy of the gas(vapor)at temperature T The quantity△s(T) We know Sb=5.7045 kJ/kg S/g=83706k/kg-k se=0.2966 kJ/kg-K The quality at state c is thus 5.7045-0.2966 0.646 8.3706 The enthalpy at state c is
2B-10 Neglect the changes in kinetic energy at inlet and outlet of the turbine and pump. a) For the reversible cycle, η η thermal Carnot T T = =− =− = 1 1 293 573 0 489 1 2 . To find the work in the pump (compression process) or in the turbine, we need to find the enthalpy changes between states b and c, ∆hbc , and the change between a and d, ∆had . To obtain these the approach is to use the fact that s = constant during the expansion to find the quality at state c and then, knowing the quality, calculate the enthalpy as h Xh X h = +− g f ( ) 1 . We know the conditions at state b, where the fluid is all vapor, i.e., we know Ths b bb , , : hh Ch C b vapor o g o = ( ) 300 300 2749 kJ/kg = ( ) = s s Cs C b vapor o g o = ( ) 300 300 5 7045 = ( ) = . kJ/kg - K s s b c = in the isentropic expansion process. We now need to find the quality at state c, Xc . Using the definition of quality given in Section 2.B.1, and noting that s Xs X s c cg c = +− ( ) f 1 , we obtain, X s sT sT sT s sT s T c c f c g c f c c f c fg c = − ( ) ( ) − ( ) = − ( ) ( ) . The quantity sc is the mass-weighted entropy at state c, which is at temperature Tc. The quantity s T f ( ) c is the entropy of the liquid at temperature Tc . The quantity s T g c ( ) is the entropy of the gas (vapor) at temperature Tc . The quantity ∆ ∆ sT s T fg ( ) c = liquid gas → c at . We know: s s c b = = 5.7045 kJ/kg - K sfg = 8.3706 kJ/kg - K sf = 0.2966 kJ/kg - K. The quality at state c is thus, Xc = − = 5 7045 0 2966 8 3706 0 646 . . . . . The enthalpy at state c is