neighboring states. But we can not use the properties to do interpolation calculations to get the state properties of wet vapor based on the properties of the states on different sides of the bold line in the table for thermodynamic properties of unsaturated water and superheated steam Variation of properties of compressed liquid with pressure is very mild. 100 times increase in the pressure often causes properties to change only less than 1 percent. This is because the compressed liquid properties depend on temperature much more strongly than they do on pressure. In the absence of compressed liquid data, a general approximation is to treat compre ssed liquid as saturated liquid at 6.3.3 Saturated Liquid-Vapor Mixture A saturated mixture can be treated as a combination of the saturated liquid and the saturated vapor To analyze this mixture properly, we need to know the quality x, that is the ratio of the mass of vapor to the total mass of the mixture For saturated mixtures, quality can serve as one of the two independent intensive properties eeded to describe a state. Note x=m, / motal that the properties of the saturated liquid are the same whether it exists alone or in a mixture with saturated vapor:. Then the properties of this"mixtur will simply be the average properties of the saturated liquid-vapor mixture under consideration Thus, at a given pressure, or a given temperature, the properties of a saturated mixture can be ulated by the following formula Specific volume of wet vapor Vr=xv+(1-x)v (6-2) Enthalpy of wet vapor h, =xh"+(1-x)h (6-3) Entropy of wet vapor Sx=xs+(1-x)s 4 where h, vand s represent enthalpy, specific volume, entropy of saturated liquid, respectively. And h, v"and s"represent enthalpy, specific volume, entropy of saturated vapor, respect The internal energy of saturated liquid-vapor mixture u is determined by u,=h 6. 4 The Enthalpy -Entropy Diagram(h-s Diagram)for Steam On T-s and p-v diagrams, the area enclosed by process curves and the axes of coordinates indicates the amount of heat addition/rejection and work output/input, but it is inconvenient for use of calculation. If using enthalpy-entropy diagram (h-s diagram), the length of lines can represent the amount of heat and work directly It is very intuitive, and thus it has been widely used in engineering calculations. The h-s diagram for steam is shown in Fig 6-4. The h-s diagram is constructed follows the enthalpy is chosen to be the axis of dinate and the origin of coordinates is assumed to be the state Fig 6-4 h-s Diagram for Water Steam of water at the triple point, where s. =0 and 1=0. Making use of the data taken from the steam tables, the saturated liquid line
103 / v total x m m = Fig.6-4 h − s Diagram for Water Steam neighboring states. But we can not use the properties to do interpolation calculations to get the state properties of wet vapor based on the properties of the states on different sides of the bold line in the table for thermodynamic properties of unsaturated water and superheated steam,. Variation of properties of compressed liquid with pressure is very mild. 100 times increase in the pressure often causes properties to change only less than 1 percent. This is because the compressed liquid properties depend on temperature much more strongly than they do on pressure. In the absence of compressed liquid data, a general approximation is to treat compressed liquid as saturated liquid at the given temperature. 6.3.3 Saturated Liquid–Vapor Mixture A saturated mixture can be treated as a combination of the saturated liquid and the saturated vapor. To analyze this mixture properly, we need to know the quality x, that is the ratio of the mass of vapor to the total mass of the mixture. For saturated mixtures, quality can serve as one of the two independent intensive properties needed to describe a state. Note that the properties of the saturated liquid are the same whether it exists alone or in a mixture with saturated vapor. Then the properties of this “mixture” will simply be the average properties of the saturated liquid–vapor mixture under consideration. Thus, at a given pressure, or a given temperature, the properties of a saturated mixture can be calculated by the following formula. Specific volume of wet vapor (6-2) Enthalpy of wet vapor (6-3) Entropy of wet vapor (6-4) where h , v and s represent enthalpy, specific volume, entropy of saturated liquid, respectively. And h , v and s represent enthalpy, specific volume, entropy of saturated vapor, respectively. The internal energy of saturated liquid-vapor mixture u is determined by x x x u h pv = − 6.4 The Enthalpy - Entropy Diagram ( h s − Diagram) for Steam On T − s and p − v diagrams, the area enclosed by process curves and the axes of coordinates indicates the amount of heat addition/rejection and work output/input, but it is inconvenient for use of calculation. If using enthalpy-entropy diagram ( h − s diagram), the length of lines can represent the amount of heat and work directly. It is very intuitive, and thus it has been widely used in engineering calculations. The h − s diagram for steam is shown in Fig.6-4. The h − s diagram is constructed as follows: the enthalpy is chosen to be the axis of ordinate and the entropy is the abscissa. The origin of coordinates is assumed to be the state of water at the triple point, where 0 0 s h = = 0 and 0 . Making use of the data taken from the steam tables, the saturated liquid line v xv x v x = + (1− ) hx = xh+ (1− x)h s xs x s x = + (1− )
x=0 and saturated vapor linex=1, meeting at the critical point C, are firstly plotted on the diagram. The saturated liquid line initiates from the origin of coordinates. In the wet region the isobars and isotherms are identical and are inclined straight lines. In the superheated region the isobars and isotherms diverge: the isobars run upwards and the isotherms turn to approach horizontal lines. In the wet region, lines of constant quality (x=const)are also plotted, which merge at the critical point C. Isochors(constant-volume lines) are also plotted on the h-s diagram, which run more steeply ds, compared with the isobars, as shown by the dotted line in Fig.6- It is convenient to use h-s diagram to determine the properties of water vapor. However, the accuracy of the readings depends on the person who uses it. Therefore, in practice, it is often to analyze thermodynamic processes of water vapor with the aid of h-s diagram and tables for water vapor simultaneously to simplify the calculations and to ensure the accuracy Usually, the entire h-s diagram is just partly given out. It is the part of the wet region with quality greater than 0.6 because the lines in the region with quality less than 0.6 is too dense and the data seldom used in projects. The h-s diagram of steam for application in engineering practice is hown in Appendix Figure 2 6.5 Thermodynamic Processes of Water Vapor The main purpose of analyzing thermodynamic processes of water vapor is to determine the final state and state properties of the process and to determine the thermodynamic energy and enthalpy changes of steams to obtain the energy conversion relations, including those between work output and heat ddition during the process. In general, steam tables and h-s diagram are used. The analysis is based on the first and the second laws of thermodynamics. Here we just discuss reversible processes. The (1)Based on the given conditions, determine the initial state and related properties (2)Based on the characteristics of the process and one of the properties given for the final state determine the final state and its properties (3) Based on the initial and final states, calculate the q, Au and w during the process Energy conversion relationships of water vapor are discussed respectively for the four basic thermodynamic processes as following 6.5.1 Isochoric process Durin isochoric (constant-volume) process, when heat is transferred to water, its pressure and temperature increase. The work output is zero as the volume remains constant. The heat addition leads to the increase in the internal energy of the water. And the increment of internal energy is equal to the heat addition. On h-s diagram, the process is depicted by the curve 1-2 in Fig. 6-5 The work output can be calculated by w=pdv=0 le amount of heat addition is The change in internal energy of the steam is △a=h2-h1-(P2-P1) The technical work available is w,=-]vdp=v(P1-P2)
104 x = 0 and saturated vapor line x =1, meeting at the critical point C, are firstly plotted on the diagram. The saturated liquid line initiates from the origin of coordinates. In the wet region the isobars and isotherms are identical and are inclined straight lines. In the superheated region the isobars and isotherms diverge: the isobars run upwards and the isotherms turn to approach horizontal lines. In the wet region, lines of constant quality ( x =const) are also plotted, which merge at the critical point C. Isochors (constant-volume lines) are also plotted on the h − s diagram, which run more steeply upwards, compared with the isobars, as shown by the dotted line in Fig.6-4. It is convenient to use h − s diagram to determine the properties of water vapor. However, the accuracy of the readings depends on the person who uses it. Therefore, in practice, it is often to analyze thermodynamic processes of water vapor with the aid of h − s diagram and tables for water vapor simultaneously to simplify the calculations and to ensure the accuracy. Usually, the entire h − s diagram is just partly given out. It is the part of the wet region with quality greater than 0.6 because the lines in the region with quality less than 0.6 is too dense and the data seldom used in projects. The h − s diagram of steam for application in engineering practice is shown in Appendix Figure 2. 6.5 Thermodynamic Processes of Water Vapor The main purpose of analyzing thermodynamic processes of water vapor is to determine the final state and state properties of the process and to determine the thermodynamic energy and enthalpy changes of steams to obtain the energy conversion relations, including those between work output and heat addition during the process. In general, steam tables and h − s diagram are used. The analysis is based on the first and the second laws of thermodynamics. Here we just discuss reversible processes. The analyzing procedure is as following: ⑴ Based on the given conditions, determine the initial state and related properties. ⑵ Based on the characteristics of the process and one of the properties given for the final state, determine the final state and its properties. ⑶ Based on the initial and final states, calculate the q u w , and during the process. Energy conversion relationships of water vapor are discussed respectively for the four basic thermodynamic processes as following. 6.5.1 Isochoric Process During an isochoric (constant-volume) process, when heat is transferred to water, its pressure and temperature increase. The work output is zero as the volume remains constant. The heat addition leads to the increase in the internal energy of the water. And the increment of internal energy is equal to the heat addition. On h s − diagram, the process is depicted by the curve 1-2 in Fig.6-5. The work output can be calculated by 0 2 1 w = pdv = The amount of heat addition is q = u The change in internal energy of the steam is ( ) 2 1 p2 p1 u = h − h −v − The technical work available is = − = − 2 1 1 2 w vdp v( p p ) t
h Fig 6-5 Isochoric Process of Water Vapor Fig 6-6 Isobaric Process of Water Vapor 6.5.2 Isobaric Process On h-s diagram, the isobaric(constant-pressure) process is depicted by the curve 1-2 in Fig. 6-6 The change in internal energy of the steam is v2=v 1) The work output is w=p(V2-n)=q-△ The technical work available is PI-p The amount of heat addition is 6.5.3 Isothermal Process On h-s diagram, the isothermal (constant-temperature) process is depicted by the curve 1-2 in Fig. 6-7 The change in internal energy of steam is Fig 6-7 Isothermal Process M=l2-l1=(h2-P22)-(-P1n1) The amount of heat addition is 7(S2-S1) The expansion work output is The technical work available is 6.5. 4 Adiabatic Process The adiabatic process is realized without heat addition or rejection and the entropy of the working medium during a reversible adiabatic(isentropic) process remains constant. That is, s=const(as shown by a straight line 1-2 in Fig. 6-8). Superheated steam turns into saturated vapor, and further into wet vapor. An irreversible process is shown by the dotted line 1-2 in Fig. 6-9. During an adiabatic expansion process, the pressure and the temperature of the steam decreases, however, the entropy Icreases Under isentropic conditions, it is easy to determine the final state properties on h-s diagram The amount of heat addition is equal to zero, that is
105 Fig .6-5 Isochoric Process of Water Vapor Fig. 6-6 Isobaric Process of Water Vapor Fig. 6-7 Isothermal Process 6.5.2 Isobaric Process On h s − diagram, the isobaric (constant-pressure) process is depicted by the curve 1-2 in Fig.6-6. The change in internal energy of the steam is ( ) 2 1 2 1 2 1 u = u −u =h −h − p v −v The work output is w= p(v2 − v1 ) = q − u The technical work available is = − = − 2 1 1 2 w vdp v( p p ) t The amount of heat addition is q = h2 − h1 6.5.3 Isothermal Process On h s − diagram, the isothermal (constant-temperature) process is depicted by the curve 1-2 in Fig.6-7. The change in internal energy of steam is ( ) ( ) 2 1 2 2 2 1 1 1 u = u −u = h −p v − h − p v The amount of heat addition is ( ) 2 1 q =T s − s The expansion work output is w= q −u The technical work available is wt = q − h 6.5.4 Adiabatic Process The adiabatic process is realized without heat addition or rejection and the entropy of the working medium during a reversible adiabatic (isentropic) process remains constant. That is, s = const (as shown by a straight line 1-2 in Fig. 6-8). Superheated steam turns into saturated vapor, and further into wet vapor. An irreversible process is shown by the dotted line 1-2 in Fig. 6-9. During an adiabatic expansion process, the pressure and the temperature of the steam decreases, however, the entropy increases. Under isentropic conditions, it is easy to determine the final state properties on h s − diagram. The amount of heat addition is equal to zero, that is
q The work output or input during an adiabatic process can be determined from the following The change in internal energy is M=(h2-P2V2)-(h-Pv1) The technical work available h Fig. 6-8 Re (Isentropic Process K Example 6-11 Steam expands isothermally at t=300'Cfrom P,=1 MPa to p2=0.1MPa Determine the amount of heat added to the steam, the change in internal energy and the expansion (Solution] From the steam table, we can get to know the water is superheated vapor at state"I 1=300 C and P,=1 MPa. The other properties of th state V1=0.25793m3/kg,h=30504kJkg,S1=71216kJkg Draw an isothermal line through the initial point"Iand it meets the isobar of p,=0. 1 MPa at point We can find h,=3 073. 8 kJ/kg,s,=8.214 8 kJ/kg. K)when P2=0. 1 MPa and t=300 C The amount of the heat addition can be determined by q=7(s2-s1)=573×(82148-7.1216)=6264kJkg e change in internal energy of the steam is equal to △a=(h2-P2V2)-(h-P11) (3073.8 0.1×10°×26388 )-(305041×10×0.25793 1000 1000 The expansion work output is =q-A=6264-17.5=6089kJ/ks K Example 6-21 Superheated vapor at 8 MPa and 500 Experiences an isentropic expansion process to P2=0.1 MPa. Determine the final state of the steam, the change in internal energy and the work Solution: We can find on the h-s diagram that at the end of expansion process, the steam becomes saturated liquid-vapor mixture with the quality of x=0.84 The initial properties of the superheated vapor at 8 MPa and 500C are h=3400kJ/kg,n1=0045m3/kg
106 Fig. 6-8 Reversible Adiabatic Process Fig. 6-9 Irreversible Adiabatic Process (Isentropic Process) The work output or input during an adiabatic process can be determined from the following equation ( ) ( ) 1 2 1 1 1 2 2 2 w = u −u = h − p v − h − p v The change in internal energy is ( ) ( ) 2 2 2 1 1 1 u = h − p v − h − p v The technical work available is wt = −h 【Example 6-1】 Steam expands isothermally at t = 300 ℃from p1 =1 MPa to p2 = 0.1 MPa. Determine the amount of heat added to the steam, the change in internal energy and the expansion work output. 【Solution】From the steam table, we can get to know the water is superheated vapor at state “1”, t = 300 ℃ and p1 =1 MPa. The other properties of this state are 3 1 v = 0.257 93 m /kg , 1 h = 3 050.4 kJ/kg , 1 s = 7.121 6 kJ/kg . Draw an isothermal line through the initial point “1”and it meets the isobar of 2 p = 0.1 MPa at point “2”.We can find 2 h = 3 073.8 kJ/kg ， 2 s = 8.214 8 kJ/(kg K) when 2 p = 0.1 MPa and t = 300 ℃， The amount of the heat addition can be determined by 2 1 q T s s = − = − = ( ) 573 (8.214 8 7.121 6) 626.4 kJ/kg The change in internal energy of the steam is equal to 2 2 2 1 1 1 6 6 ( ) ( ) 0.1 10 2.638 8 1 10 0.257 93 (3 073.8 ) (3 050.4 ) 1 000 1 000 17.5 kJ/kg = − − − u h p v h p v = − − − = The expansion work output is w= q −u = 626.4−17.5= 608.9 kJ/kg 【Example 6-2】 Superheated vapor at 8 MPa and 500℃experiences an isentropic expansion process to p2 = 0.1 MPa. Determine the final state of the steam, the change in internal energy and the work output by using the h s − diagram Solution: We can find on the h s − diagram that at the end of expansion process, the steam becomes saturated liquid-vapor mixture with the quality of x = 0.84 The initial properties of the superheated vapor at 8 MPa and 500℃ are 1 h = 3 400 kJ/kg; 1 v = 0.045 m3 /kg q = 0
The final properties at p,=0. 1 MPa are h2=2135kJ/kg,v2=1.5m3/k The change in internal energy amounts to △=h2-h-(P22-P)=2135-340(0.1×10×1.58×10°×045 1000 000 The expansion work is equal to =-△a=1055kJ/kg Chapter 7 Atmospheric Air 7.1 Atmospheric Air and state Properties 7. 1. 1 Definition of Atmospheric air(Moist air) Air is a mixture of nitrogen, oxygen, and small amounts of some other gases. Air in the atmosphere
107 The final properties at 2 p = 0.1 MPa are 2 h = 2 135 kJ/kg ; 3 2 v = 1.5 m /kg The change in internal energy amounts to 6 6 2 1 2 2 1 1 0.1 10 1.5 8 10 0.045 ( ) 2 135 3 400 ( ) 1000 1000 -1 055 kJ/kg u h h p v p v = − − − = − − − = The expansion work is equal to w u = − =1 055 kJ/kg Chapter 7 Atmospheric Air 7.1 Atmospheric Air and State Properties 7.1.1 Definition of Atmospheric air (Moist air) Air is a mixture of nitrogen, oxygen, and small amounts of some other gases. Air in the atmosphere