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Chapter 3
PROPERTIES OF PURE SUBSTANCES
Pure Substances, Phase Change Processes, Property Diagrams
3-1C Yes. Because it has the same chemical composition throughout.
3-2C A liquid that is about to vaporize is saturated liquid; otherwise it is compressed liquid.
3-3C A vapor that is about to condense is saturated vapor; otherwise it is superheated vapor.
3-4C No.
3-5C No.
3-6C Yes. The saturation temperature of a pure substance depends on pressure. The higher the pressure, the higher the saturation or boiling temperature.
3-7C The temperature will also increase since the boiling or saturation temperature of a pure substance depends on pressure.
3-8C Because one cannot be varied while holding the other constant. In other words, when one changes, so does the other one.
3-9C At critical point the saturated liquid and the saturated vapor states are identical. At triple point the three phases of a pure substance coexist in equilibrium.
3-10C Yes.
3-11C Case (c) when the pan is covered with a heavy lid. Because the heavier the lid, the greater the pressure in the pan, and thus the greater the cooking temperature.
3-12C At supercritical pressures, there is no distinct phase change process. The liquid uniformly and gradually expands into a vapor. At subcritical pressures, there is always a distinct surface between the phases.
Property Tables
3-13C A given volume of water will boil at a higher temperature in a tall and narrow pot since the pressure at the bottom (and thus the corresponding saturation pressure) will be higher in that case.
3-14C A perfectly fitting pot and its lid often stick after cooking as a result of the vacuum created inside as the temperature and thus the corresponding saturation pressure inside the pan drops. An easy way of removing the lid is to reheat the food. When the temperature rises to boiling level, the pressure rises to atmospheric value and thus the lid will come right off.
3-15C The molar mass of gasoline (C 8 H 18 ) is 114 kg/kmol, which is much larger than the molar mass of air that is 29 kg/kmol. Therefore, the gasoline vapor will settle down instead of rising even if it is at a much higher temperature than the surrounding air. As a result, the warm mixture of air and gasoline on top of an open gasoline will most likely settle down instead of rising in a cooler environment
3-16C Ice can be made by evacuating the air in a water tank. During evacuation, vapor is also thrown out, and thus the vapor pressure in the tank drops, causing a difference between the vapor pressures at the water surface and in the tank. This pressure difference is the driving force of vaporization, and forces the liquid to evaporate. But the liquid must absorb the heat of vaporization before it can vaporize, and it absorbs it from the liquid and the air in the neighborhood, causing the temperature in the tank to drop. The process continues until water starts freezing. The process can be made more efficient by insulating the tank well so that the entire heat of vaporization comes essentially from the water.
3-17C Yes. Otherwise we can create energy by alternately vaporizing and condensing a substance.
3-18C No. Because in the thermodynamic analysis we deal with the changes in properties; and the changes are independent of the selected reference state.
3-19C The term hfg represents the amount of energy needed to vaporize a unit mass of saturated liquid at a specified temperature or pressure. It can be determined from hfg = hg - hf.
3-20C Yes; the higher the temperature the lower the hfg value.
3-21C Quality is the fraction of vapor in a saturated liquid-vapor mixture. It has no meaning in the superheated vapor region.
3-22C Completely vaporizing 1 kg of saturated liquid at 1 atm pressure since the higher the pressure, the lower the hfg.
3-23C Yes. It decreases with increasing pressure and becomes zero at the critical pressure.
3-24C No. Quality is a mass ratio, and it is not identical to the volume ratio.
3-25C The compressed liquid can be approximated as a saturated liquid at the given temperature. Thus
v T , P ≅v f @ T.
3-26 [ Also solved by EES on enclosed CD ] Complete the following table for H 2 O :
T , ° C P , kPa v, m^3 / kg Phase description
50 12.352 4.16 Saturated mixture 120.21 200 0.8858 Saturated vapor 250 400 0.5952 Superheated vapor 110 600 0.001051 Compressed liquid
p=Pressure(Fluid$,T=T,x=x) h=enthalpy(Fluid$,T=T,x=x) s=entropy(Fluid$,T=T,x=x) v=volume(Fluid$,T=T,x=x) u=IntEnergy(Fluid$,T=T,x=x) endif Endif If Prop1$='Pressure, kPa' Then p=Value If Prop2$='Pressure, kPa' then Call Error('Both properties cannot be Pressure, p=xxxF2',p) if Prop2$='Temperature, C' then T=value h=enthalpy(Fluid$,T=T,P=p) s=entropy(Fluid$,T=T,P=p) v=volume(Fluid$,T=T,P=p) u=intenergy(Fluid$,T=T,P=p) x=quality(Fluid$,T=T,P=p) endif if Prop2$='Enthalpy, kJ/kg' then h=value T=Temperature(Fluid$,p=p,h=h) s=entropy(Fluid$,p=p,h=h) v=volume(Fluid$,p=p,h=h) u=intenergy(Fluid$,p=p,h=h) x=quality(Fluid$,p=p,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value T=Temperature(Fluid$,p=p,s=s) h=enthalpy(Fluid$,p=p,s=s) v=volume(Fluid$,p=p,s=s) u=intenergy(Fluid$,p=p,s=s) x=quality(Fluid$,p=p,s=s) endif if Prop2$='Volume, m^3/kg' then v=value T=Temperature(Fluid$,p=p,v=v) h=enthalpy(Fluid$,p=p,v=v) s=entropy(Fluid$,p=p,v=v) u=intenergy(Fluid$,p=p,v=v) x=quality(Fluid$,p=p,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value T=Temperature(Fluid$,p=p,u=u) h=enthalpy(Fluid$,p=p,u=u) s=entropy(Fluid$,p=p,u=u) v=volume(Fluid$,p=p,s=s) x=quality(Fluid$,p=p,u=u) endif if Prop2$='Quality' then x=value T=Temperature(Fluid$,p=p,x=x) h=enthalpy(Fluid$,p=p,x=x) s=entropy(Fluid$,p=p,x=x) v=volume(Fluid$,p=p,x=x)
u=IntEnergy(Fluid$,p=p,x=x) endif Endif If Prop1$='Enthalpy, kJ/kg' Then h=Value If Prop2$='Enthalpy, kJ/kg' then Call Error('Both properties cannot be Enthalpy, h=xxxF2',h) if Prop2$='Pressure, kPa' then p=value T=Temperature(Fluid$,h=h,P=p) s=entropy(Fluid$,h=h,P=p) v=volume(Fluid$,h=h,P=p) u=intenergy(Fluid$,h=h,P=p) x=quality(Fluid$,h=h,P=p) endif if Prop2$='Temperature, C' then T=value p=Pressure(Fluid$,T=T,h=h) s=entropy(Fluid$,T=T,h=h) v=volume(Fluid$,T=T,h=h) u=intenergy(Fluid$,T=T,h=h) x=quality(Fluid$,T=T,h=h) endif if Prop2$='Entropy, kJ/kg-K' then s=value p=Pressure(Fluid$,h=h,s=s) T=Temperature(Fluid$,h=h,s=s) v=volume(Fluid$,h=h,s=s) u=intenergy(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,s=s) endif if Prop2$='Volume, m^3/kg' then v=value p=Pressure(Fluid$,h=h,v=v) T=Temperature(Fluid$,h=h,v=v) s=entropy(Fluid$,h=h,v=v) u=intenergy(Fluid$,h=h,v=v) x=quality(Fluid$,h=h,v=v) endif if Prop2$='Internal Energy, kJ/kg' then u=value p=Pressure(Fluid$,h=h,u=u) T=Temperature(Fluid$,h=h,u=u) s=entropy(Fluid$,h=h,u=u) v=volume(Fluid$,h=h,s=s) x=quality(Fluid$,h=h,u=u) endif if Prop2$='Quality' then x=value p=Pressure(Fluid$,h=h,x=x) T=Temperature(Fluid$,h=h,x=x) s=entropy(Fluid$,h=h,x=x) v=volume(Fluid$,h=h,x=x) u=IntEnergy(Fluid$,h=h,x=x) endif endif If Prop1$='Entropy, kJ/kg-K' Then
end "Input from the diagram window" {Fluid$='Steam' Prop1$='Temperature' Prop2$='Pressure' Value1= value2=101.3}
Call Find(Fluid$,Prop1$,Prop2$,Value1,Value2:T,p,h,s,v,u,x,State$)
T[1]=T ; p[1]=p ; h[1]=h ; s[1]=s ; v[1]=v ; u[1]=u ; x[1]=x "Array variables were used so the states can be plotted on property plots."
ARRAYS TABLE
h KJ/kg
P
kPa
s kJ/kgK
T
C
u KJ/kg
v m^3 /kg
x
2964.5 400 7.3804 250 2726.4 0.5952 100
0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,
0
100
200
300
400
500
600
700
s [kJ/kg-K]
T
C [
] 8600 kPa 2600 kPa 500 kPa 45 kPa
Steam
10 -4^10 -3^10 -2^10 -1^100 101 102
0
100
200
300
400
500
600
700
v [m^3 /kg]
T
C [
] 8600 kPa 2600 kPa 500 kPa 45 kPa
Steam
10 -3^10 -2^10 -1^10 0 101
100
101
102
103
104
105
v [m^3 /kg]
P
k [ P a
]
250 C
170 C
110 C 75 C
Steam
0 500 1000 1500 2000 2500 3000
100
101
102
103
104
105
h [kJ/kg]
P
k [ P a
]
250 C
170 C
110 C 75 C
Steam
0,0 1,0 2,0 3,0 4,0 5,0 6,0 7,0 8,0 9,0 10,
0
500
1000
1500
2000
2500
3000
3500
4000
s [kJ/kg-K]
h
k [ J
k /
g ]
8600 kPa 2600 kPa 500 kPa 45 kPa
Steam
3-31 Complete the following table for Refrigerant-134a :
T, ° C P, kPa v , m^3 / kg Phase description
-8 320 0.0007569 Compressed liquid 30 770.64 0.015 Saturated mixture -12.73 180 0.11041 Saturated vapor 80 600 0.044710 Superheated vapor
3-32 Complete the following table for Refrigerant-134a :
T, ° C P, kPa u , kJ / kg Phase description 20 572.07 95 Saturated mixture -12 185.37 35.78 Saturated liquid 86.24 400 300 Superheated vapor 8 600 62.26 Compressed liquid
3-33E Complete the following table for Refrigerant-134a :
T, ° F P, psia^ h , Btu / lbm^ x^ Phase description 65.89 80 78 0.566 Saturated mixture 15 29.759 69.92 0.6 Saturated mixture 10 70 15.35 - - - Compressed liquid 160 180 129.46 - - - Superheated vapor 110 161.16 117.23 1.0 Saturated vapor
3-34 Complete the following table for H 2 O :
T, ° C P, kPa v , m^3 / kg Phase description
140 361.53 0.05 Saturated mixture 155.46 550 0.001097 Saturated liquid (^125 750) 0.001065 Compressed liquid 500 2500 0.140 Superheated vapor
3-35 Complete the following table for H 2 O :
T, ° C P, kPa u , kJ / kg Phase description 143.61 400 1450 Saturated mixture 220 2319.6 2601.3 Saturated vapor 190 2500 805.15 Compressed liquid 466.21 4000 3040 Superheated vapor
3-36 A rigid tank contains steam at a specified state. The pressure, quality, and density of steam are to be determined.
Properties At 220°C v f = 0.001190 m^3 /kg and v g = 0.08609 m^3 /kg (Table A-4).
Analysis ( a ) Two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. The pressure of the steam is the saturation pressure at the given temperature. Then the pressure in the tank must be the saturation pressure at the specified temperature,
P = T sat@220°C= 2320 kPa
Steam 1.8 m^3 220 °C
( b ) The total mass and the quality are determined as
×
×
- 2 13. 94 518. 1 kg
1 3.94kg 0.08609m/kg
2/3 (1.8m)
504 .2kg 0.001190m/kg
1/3 (1.8m)
3
3
3
3
t
g
t f g
g
g g
f
f f
m
m x
m m m
m
m
v
V
v
V
( c ) The density is determined from
= = = 287.8kg/m^3
( ) 0. 001190 ( 0. 0269 )( 0. 08609 ) 0. 003474 m^3 /kg
v
v v v v
f x g f
3-37 A piston-cylinder device contains R-134a at a specified state. Heat is transferred to R-134a. The final pressure, the volume change of the cylinder, and the enthalpy change are to be determined.
Analysis (a) The final pressure is equal to the initial pressure, which is determined from
= 90.4^ kPa
2 2 1 atm (^21000) kg.m/s
1 kN (0.25m)/
(12kg)(9.81m/s) 88 kPa
π D /4 π
mg P P P p
(b) The specific volume and enthalpy of R-134a at the initial state of 90.4 kPa and -10°C and at the final state of 90.4 kPa and 15°C are (from EES)
v 1 = 0.2302 m^3 /kg h 1 = 247.76 kJ/kg
R-134a Q
0.85 kg -10°C
v 2 = 0.2544 m^3 /kg h 2 = 268.16 kJ/kg
The initial and the final volumes and the volume change are
∆ = − = 0.0205m^3
( 0. 85 kg)(0.2544m/kg) 0. 2162 m
( 0. 85 kg)(0.2302m/kg) 0. 1957 m 3 3 2 2
3 3 1 1
V
V v
V v
m
m
(c) The total enthalpy change is determined from
∆ H = m ( h 2 − h 1 )=( 0. 85 kg)(268.1 6 −247.76)kJ/kg= 17.4 kJ/kg
3-40 A person cooks a meal in a pot that is covered with a well-fitting lid, and leaves the food to cool to the room temperature. It is to be determined if the lid will open or the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up.
Assumptions 1 The local atmospheric pressure is 1 atm = 101.325 kPa. 2 The weight of the lid is small and thus its effect on the boiling pressure and temperature is negligible. 3 No air has leaked into the pan during cooling.
Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A-4).
Analysis Noting that the weight of the lid is negligible, the reaction force F on the lid after cooling at the pan-lid interface can be determined from a force balance on the lid in the vertical direction to be
PA +F = PatmA
or,
= 6997 mPa= (since 1 Pa= 1 N/m )
( 101 , 325 2339. 2 )Pa 4
( 0. 3 m)
2 2
2
2
6997 N
F APatm P π D Patm P
P (^) 2.3392 kPa
Patm = 1 atm The weight of the pan and its contents is
W = mg =( 8 kg)(9.81m/s^2 )= 78.5 N
which is much less than the reaction force of 6997 N at the pan-lid interface. Therefore, the pan will move up together with the lid when the person attempts to open the pan by lifting the lid up. In fact, it looks like the lid will not open even if the mass of the pan and its contents is several hundred kg.
3-41 Water is boiled at sea level (1 atm pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of the heat generated to the water. The rate of evaporation of water is to be determined.
Properties The properties of water at 1 atm and thus at the saturation temperature of 100°C are h fg = 2256.4 kJ/kg (Table A-4).
Analysis The net rate of heat transfer to the water is
Q &= 0. 60 × 3 kW= 1. 8 kW H 2 O Noting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid 100 °C water, the rate of evaporation of water is determined to be
= =0.80× 10 −^ kg/s= 2.872 kg/h
4 kJ/kg
8 kJ/s = 3 fg
evaporation h
Q
m
3-42 Water is boiled at 1500 m (84.5 kPa pressure) in a pan placed on top of a 3-kW electric burner that transfers 60% of the heat generated to the water. The rate of evaporation of water is to be determined.
Properties The properties of water at 84.5 kPa and thus at the saturation temperature of 95°C are hfg = 2269.6 kJ/kg (Table A-4).
Analysis The net rate of heat transfer to the water is (^) H 2 O
Q &^ = 0 60. × 3 kW =1 8. kW^95 °C
Noting that it takes 2269.6 kJ of energy to vaporize 1 kg of saturated liquid water, the rate of evaporation of water is determined to be
= = 0.79 3 × 10 −^^3 kg/s = 2.855 kg/h
6 kJ/kg
8 kJ/s evaporation = h fg
Q
m
3-43 Water is boiled at 1 atm pressure in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined.
Properties The properties of water at 1 atm and thus at a saturation temperature of T sat = 100°C are h fg =
2256.5 kJ/kg and v f = 0.001043 m^3 /kg (Table A-4).
Analysis The rate of evaporation of water is
H 2 O 1 atm
001742 kg/s 45 60 s
704 kg
704 kg
( / 4 ) (0.25m) / 4
evap evap
2 2 evap evap
×
t
m m
D L
m f f
&
v v
V
Then the rate of heat transfer to water becomes
Q & = m &evap hfg =( 0. 001742 kg/s)(2256.5kJ/kg)= 3.93 kW
3-44 Water is boiled at a location where the atmospheric pressure is 79.5 kPa in a pan placed on an electric burner. The water level drops by 10 cm in 45 min during boiling. The rate of heat transfer to the water is to be determined.
Properties The properties of water at 79.5 kPa are T sat = 93.3°C, h fg = 2273.9 kJ/kg and vf = 0.
m^3 /kg (Table A-5).
Analysis The rate of evaporation of water is H 2 O 79.5 kPa
001751 kg/s 45 60 s
727 kg
727 kg
( / 4 ) (0.25m) / 4
evap evap
evap^22 evap
×
t
m m
D L
m f f
&
v v
V
Then the rate of heat transfer to water becomes
Q & = m &evap hfg =( 0. 001751 kg/s)(2273.9kJ/kg)= 3.98 kW
3-49 EES Problem 3-48 is reconsidered. Using EES (or other) software, the effect of the mass of the lid on the boiling temperature of water in the pan is to be investigated. The mass is to vary from 1 kg to 10 kg, and the boiling temperature is to be plotted against the mass of the lid. Analysis The problem is solved using EES, and the solution is given below.
"Given data" {P_atm=101[kPa]} D_lid=20 [cm] {m_lid=4 [kg]}
"Solution"
"The atmospheric pressure in kPa varies with altitude in km by the approximate function:" P_atm=101.325(1-0.02256z)^5. "The local acceleration of gravity at 45 degrees latitude as a function of altitude in m is given by:" g=9.807+3.3210^(-6)z*convert(km,m)
"At sea level:" z=0 "[km]" A_lid=piD_lid^2/4convert(cm^2,m^2) W_lid=m_lidgconvert(kgm/s^2,N) P_lid=W_lid/A_lidconvert(N/m^2,kPa) P_water=P_lid+P_atm T_water=temperature(steam_iapws,P=P_water,x=0)
mlid [kg] Twater [C] 1 100. 2 100. 3 100. 4 100. 5 100. 6 100. 7 100. 8 100. 9 100. 10 100.8 1 2 3 4 5 6 7 8 9 10
100
m (^) lid [kg]
T^ w ater
[C]
r
0 1 2 3 4 5 6 7 8 9
30
40
50
60
70
80
90
100
110
z [km]
P^ w ate
[kPa]
Effect of altitude on boiling pressure of w ater in pan w ith lid
mass of lid = 4 kg
P w^ at er
[kPa]
0 1 2 3 4 5 6 7 8 9
70
75
80
85
90
95
100
105
z [km]
T^ w ater
[C]
mass of lid = 4 kg
Effect of altitude on boiling temperature of water in pan with lid
T^ w ater
[C]
3-50 A vertical piston-cylinder device is filled with water and covered with a 20-kg piston that serves as the lid. The boiling temperature of water is to be determined.
Analysis The pressure in the cylinder is determined from a force balance on the piston,
PA = P atm A + W
P
P atm
W = mg
or,
119.61 kPa
1000 kg/ms
1 kPa 0.01m
(20kg)(9.81m/s) (100 kPa) 2 2
2
atm
A
mg P P
The boiling temperature is the saturation temperature corresponding to this pressure,
T = T sat@119.61kPa= 104.7 ° C (Table A-5)
3-51 A rigid tank that is filled with saturated liquid-vapor mixture is heated. The temperature at which the
liquid in the tank is completely vaporized is to be determined, and the T- v diagram is to be drawn.
Analysis This is a constant volume process ( v = V / m = constant),
H 2 O 75 °C
and the specific volume is determined to be
0.1667m /kg 15 kg
2.5 m^33 = = = m
V
v
When the liquid is completely vaporized the tank will contain saturated vapor only. Thus,
T
1
v 2 = v g =0.1667 m^3 /kg^2
The temperature at this point is the temperature that
corresponds to this v g value,
= (^) sat@ = 0.1667m (^3) /kg= 187.0 ° C (Table A-4) vg
T T v
3-52 A rigid vessel is filled with refrigerant-134a. The total volume and the total internal energy are to be determined.
Properties The properties of R-134a at the given state are (Table A-13).
R-134a 2 kg 800 kPa 120 °C
0.037625m /kg
327 .87kJ/kg 120 C
800 kPa =^3
v
u T
P
o
Analysis The total volume and internal energy are determined from
655.7 kJ
0.0753 m^3 = = =
(2kg)(327.87 kJ/kg)
(2 kg)(0.0376 25 m^3 /kg) U mu
V m v
3-55 [ Also solved by EES on enclosed CD ] A piston-cylinder device contains a saturated liquid-vapor mixture of water at 800 kPa pressure. The mixture is heated at constant pressure until the temperature rises to 350°C. The initial temperature, the total mass of water, the final volume are to be determined, and the P-
v diagram is to be drawn.
Analysis (a) Initially two phases coexist in equilibrium, thus we have a saturated liquid-vapor mixture. Then the temperature in the tank must be the saturation temperature at the specified pressure,
T = T sat@ 800 kPa= 170.41 ° C
(b) The total mass in this case can easily be determined by adding the mass of each phase,
= + = + = 93.45 kg
3.745kg 0.24035m/kg
0.9m
89.704kg 0.001115m/kg
0.1m
3
3
3
3
t f g
g
g g
f
f f
m m m
m
m
v
V
v
V
P
(^1 ) (c) At the final state water is superheated vapor, and its specific volume is
0.35442m/kg (Table A-6) 350 C
800 kPa 3 2 2
o^ v T
P v
Then,
= =(93.45 kg)(0.3544 2 m^3 /kg)= 33.12m^3
V 2 mt v 2
3-56 EES Problem 3-55 is reconsidered. The effect of pressure on the total mass of water in the tank as the pressure varies from 0.1 MPa to 1 MPa is to be investigated. The total mass of water is to be plotted against pressure, and results are to be discussed.
Analysis The problem is solved using EES, and the solution is given below.
P[1]=800 [kPa] P[2]=P[1] T[2]=350 [C] V_f1 = 0.1 [m^3] V_g1=0.9 [m^3] spvsat_f1=volume(Steam_iapws, P=P[1],x=0) "sat. liq. specific volume, m^3/kg" spvsat_g1=volume(Steam_iapws,P=P[1],x=1) "sat. vap. specific volume, m^3/kg" m_f1=V_f1/spvsat_f1 "sat. liq. mass, kg" m_g1=V_g1/spvsat_g1 "sat. vap. mass, kg" m_tot=m_f1+m_g V[1]=V_f1+V_g spvol[1]=V[1]/m_tot "specific volume1, m^3" T[1]=temperature(Steam_iapws, P=P[1],v=spvol[1])"C" "The final volume is calculated from the specific volume at the final T and P" spvol[2]=volume(Steam_iapws, P=P[2], T=T[2]) "specific volume2, m^3/kg" V[2]=m_tot*spvol[2]
mtot [kg] P 1 [kPa] 96.39 100 95.31 200 94.67 300 94.24 400 93.93 500 93.71 600 93.56 700 93.45 800 93.38 900 93.34 1000 10 -3^10 -2^10 -1^10 0 10 1 10
10 0
10 1
10 2
10 3
10 4
10 5
v [m 3 /kg]
P [kPa]
350 C
Steam
1 2 P=800 kPa
100 200 300 400 500 600 700 800 900 1000
93
93.
94
94.
95
95.
96
96.
P[1] [kPa]
m
tot
[kg]
