Evaluating Properties
General consideration & Ideal gas model
Jongsup Hong
(email : jongsup.hong@yonsei.ac.kr)
Professor
Department of Mechanical Engineering
Yonsei University
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Evaluating Properties : Phase and Pure Substance
▪ Phase
•
A quantity of matter that is homogeneous throughout in both chemical
composition and physical structure
•
Homogeneity in physical structure is all solid, all liquid, or all gas
▪ Pure substance
•
One that is uniform and invariable in chemical composition
•
Can exist in more than one phase, but its chemical composition must be the
same in each phase
Solid-Liquid-Gas
Supercritical State
▪ A supercritical fluid is any substance at a temperature and pressure above its
critical point, where distinct liquid and gas phases do not exist
State Principle
▪ Two independent, intensive, thermodynamic properties are required to fix the
state of a simple compressible system
•
For example :
- P and v
- T and u
- x and h
▪ One independent property for heat transfer as one way of varying the energy,
another independent property for work transfer to change energy
▪ Intensive thermodynamic properties
P-v-T Relation
▪ Single phase regions
▪ Two phase regions : two phases exist in equilibrium (e.g., liquid-vapor),
possibly during changes in phase such as vaporization, melting, and sublimation
▪ Triple line (point) : three phases exist in equilibrium
▪ Saturation state : a state at which a phase change begins or ends
▪ Critical point :
•
Critical temperature : maximum temperature at which liquid and vapor phases
coexist in equilibrium
•
Critical pressure, critical specific volume
vapor dome
P-v-T Relation
▪ Saturation temperature Tsat : temperature at which a pure substance changes
phase at a given pressure
▪ Saturation pressure Psat : pressure at which a pure substance changes phase at a
given temperature
P-v-T Relation
Contracts on freezing
Expands on freezing (water)
P-T diagram (phase diagram)
P-v Diagram
v
T-v Diagram
Isobar
T-v Diagram
Phase Change
(a)
(b)
(c)
Constant pressure phase change of water :
a) Subcooled or compressed liquid
: (1
b) Two-phase liquid-vapor mixture (steam) : (f
c) Superheated vapor
: (g
f)
g)
s)
Phase Change
Two-phase : Quality
▪ For saturated mixture (liquid-vapor) region
•
Quality (x): an intensive property
•
x gives a fraction of vapor (gas)
•
(1-x) gives a fraction of moisture content
Quality Relations
Steam Tables – Single-phase Regions
Properties tabulated in Tables A-4 and A-5 include
► Temperature (T)
► Pressure (p)
Table A-5/A-5E
► Specific volume (v)
► Specific internal energy (u)
► Specific enthalpy (h), which is a
sum of terms that often appears in
thermodynamic analysis:
Table A-4/A-4E
h = u + pv
Enthalpy is a property because it is defined in terms of
properties; physical significance is associated with it in Chapter 4.
► Specific entropy (s), an intensive property developed in
Chapter 6
Steam Tables – Single-phase Regions
▪ Example: Properties associated with superheated water vapor at 10 MPa and
400°C are found in Table A-4.
►v = 0.02641 m3/kg
►h = 3096.5 kJ/kg
►u = 2832.4 kJ/kg
►s = 6.2120 kJ/kg∙K
Table A-4
T
C
o
v
m /kg
u
kJ/kg
3
h
kJ/kg
s
kJ/kg∙K
p = 80 bar = 8.0 MPa
(Tsat = 295.06oC)
v
m /kg
3
u
kJ/kg
h
kJ/kg
s
kJ/kg∙K
p = 100 bar = 10.0 MPa
(Tsat = 311.06oC)
Sat.
320
360
0.02352
0.02682
0.03089
2569.8
2662.7
2772.7
2758.0
2877.2
3019.8
5.7432
5.9489
6.1819
0.01803
0.01925
0.02331
2544.4
2588.8
2729.1
2724.7
2781.3
2962.1
5.6141
5.7103
6.0060
400
440
480
0.03432
0.03742
0.04034
2863.8
2946.7
3025.7
3138.3
3246.1
3348.4
6.3634
6.5190
6.6586
0.02641
0.02911
0.03160
2832.4
2922.1
3005.4
3096.5
3213.2
3321.4
6.2120
6.3805
6.5282
Steam Tables – Linear Interpolation
▪ When a state does not fall exactly on the grid of values provided by property
tables, linear interpolation between adjacent entries is used
▪ Example: Specific volume associated with superheated water vapor at 10 bar
and 215°C is found by linear interpolation between adjacent entries in
Table A-4.
(0.2275 – 0.2060) m3/kg (v – 0.2060) m3/kg
→
slope =
=
(240 – 200)oC
(215 – 200)oC
v = 0.2141 m3/kg
Table A-4
T
C
o
v
m /kg
3
u
kJ/kg
h
kJ/kg
s
kJ/kg∙K
p = 10 bar = 1.0 MPa
(Tsat = 179.91oC)
Sat.
200
240
0.1944
0.2060
0.2275
2583.6
2621.9
2692.9
2778.1
2827.9
2920.4
6.5865
6.6940
6.8817
Example
▪ Heating Ammonia at constant pressure
Assumptions:
- Ammonia is a closed system
- States 1 and 2 are equilibrium states
- The process occurs at constant pressure
Steam Tables – Two-phase Region
► Tables A-2/A-2E
(Temperature Table) and
A-3/A-3E (Pressure Table)
provide
►saturated liquid (f) data
►saturated vapor (g) data
Table note: For saturated liquid specific volume, the table heading is vf×103.
At 8oC, vf × 103 = 1.002 → vf = 1.002/103 = 1.002 × 10–3.
Table A-2
Temp
o
C
Press.
bar
.01
4
5
6
8
0.00611
0.00813
0.00872
0.00935
0.01072
Specific Volume
Sat.
Liquid
Sat.
Vapor
Internal Energy
kJ/kg
Sat.
Sat.
Liquid
Vapor
vf× 103
1.0002
1.0001
1.0001
1.0001
1.0002
vg
206.136
157.232
147.120
137.734
120.917
uf
0.00
16.77
20.97
25.19
33.59
m /kg
3
ug
2375.3
2380.9
2382.3
2383.6
2386.4
Enthalpy
kJ/kg
Sat.
Liquid
hf
0.01
16.78
20.98
25.20
33.60
Evap.
Sat.
Vapor
Entropy
kJ/kg∙K
Sat.
Sat.
Liquid
Vapor
hfg
2501.3
2491.9
2489.6
2487.2
2482.5
hg
2501.4
2508.7
2510.6
2512.4
2516.1
sf
0.0000
0.0610
0.0761
0.0912
0.1212
sg
9.1562
9.0514
9.0257
9.0003
8.9501
Temp
o
C
.01
4
5
6
8
Steam Tables – Two-phase Region
▪ Since pressure and temperature are NOT independent properties in the twophase liquid-vapor region, they cannot be used to fix the state in this region
▪ The property, quality (x), defined only in the two-phase liquid-vapor region, and
either temperature or pressure can be used to fix the state in this region
▪ For example, the total volume of the mixture is V = Vliq + Vvap
▪ Then, the specific volume of the mixture is
𝒎𝒍𝒊𝒒
𝒎𝒗𝒂𝒑
𝑽 𝑽𝒍𝒊𝒒 𝑽𝒗𝒂𝒑
𝒗= =
+
=
𝒗𝒇 +
𝒗𝒈 = 𝟏 − 𝒙 𝒗𝒇 + 𝒙𝒗𝒈
𝒎
𝒎
𝒎
𝒎
𝒎
▪ For others,
v = (1 – x)vf + xvg = vf + x(vg – vf)
u = (1 – x)uf + xug = uf + x(ug – uf)
h = (1 – x)hf + xhg = hf + x(hg – hf)
Steam Tables – Two-phase Region
▪ Example: A system consists of a two-phase liquid-vapor mixture of water at 6°C
and a quality of 0.4. Determine the specific volume, in m3/kg, of the
mixture.
► Apply
v = vf + x(vg – vf)
Substituting values from Table 2:
vf = 1.001×10–3 m3/kg and vg = 137.734 m3/kg:
v = 1.001×10–3 m3/kg + 0.4(137.734 – 1.001×10–3) m3/kg
v = 55.094 m3/kg
Table A-2
Temp
o
C
Press.
bar
.01
4
5
6
8
0.00611
0.00813
0.00872
0.00935
0.01072
Specific Volume
Sat.
Liquid
Sat.
Vapor
Internal Energy
kJ/kg
Sat.
Sat.
Liquid
Vapor
vf× 103
1.0002
1.0001
1.0001
1.0001
1.0002
vg
206.136
157.232
147.120
137.734
120.917
uf
0.00
16.77
20.97
25.19
33.59
m /kg
3
ug
2375.3
2380.9
2382.3
2383.6
2386.4
Enthalpy
kJ/kg
Sat.
Liquid
hf
0.01
16.78
20.98
25.20
33.60
Evap.
Sat.
Vapor
Entropy
kJ/kg∙K
Sat.
Sat.
Liquid
Vapor
hfg
2501.3
2491.9
2489.6
2487.2
2482.5
hg
2501.4
2508.7
2510.6
2512.4
2516.1
sf
0.0000
0.0610
0.0761
0.0912
0.1212
sg
9.1562
9.0514
9.0257
9.0003
8.9501
Temp
o
C
.01
4
5
6
8
Example
▪ Heating water at constant volume
Assumptions:
- Water is a closed system
- States 1, 2 and 3 are equilibrium states
- The volume of the container remains constant
Enthalpy and Specific Heat (Heat Capacity)
▪ Enthalpy is a property constructed due to the frequent occurrence of the
combination of properties shown below
▪ Specific heat
Example
▪ Analyzing two processes in series
Assumptions:
- Water is a closed system
- The piston is the only work mode
- Negligible change in kinetic or potential energy
Approximation for Liquids
Approximation for Liquids
1) Specific volume (v) and energy (u) vary only gradually as pressure changes at
fixed temperature
𝑣(𝑇, 𝑝) ≈ 𝑣𝑓 (𝑇)
𝑢(𝑇, 𝑝) ≈ 𝑢𝑓 (𝑇)
2) Then, the approximate value of specific enthalpy (h) can be obtained as follows
ℎ 𝑇, 𝑝 = 𝑢 𝑇, 𝑝 + 𝑝𝑣(𝑇, 𝑝)
≈ 𝑢𝑓 𝑇 + 𝑝𝑣𝑓 (𝑇)
≈ 𝑢𝑓 𝑇 + 𝑝𝑠𝑎𝑡 𝑣𝑓 𝑇 − 𝑝𝑠𝑎𝑡 𝑣𝑓 𝑇 + 𝑝𝑣𝑓 (𝑇)
≈ ℎ𝑓 𝑇
≈ ℎ𝑓 𝑇
+ 𝑣𝑓 𝑇 (𝑝 − 𝑝𝑠𝑎𝑡 )
If this term is very small..
Approximation for Liquids
▪ Using ‘Incompressible Substance Model’
•
A substance idealized by the assumption that the specific volume is constant
and the specific internal energy vary only with temperature
𝛻 ∙ 𝑣റ = 0
•
𝑢(𝑇, 𝑝) ≈ 𝑢(𝑇)
Given the constant volume,
𝜕𝑢
𝑑𝑢
𝑐𝑣 =
=
𝜕𝑇 𝑣 𝑑𝑇
•
Then, the enthalpy and specific heat, cp, can be expressed as
ℎ 𝑇, 𝑝 = 𝑢 𝑇, 𝑝 + 𝑝𝑣(𝑇, 𝑝)
= 𝑢 𝑇 + 𝑝𝑣
𝜕ℎ
𝑑𝑢 𝑣𝑑𝑝 𝑝𝑑𝑣 𝑑𝑢
𝑐𝑝 =
=
+
+
=
= 𝑐𝑣 = 𝑐
𝜕𝑇 𝑝 𝑑𝑇 𝑑𝑇
𝑑𝑇
𝑑𝑇
Approximation for Liquids
•
Given the single specific heat,
𝑇2
𝑑𝑢
𝑐𝑣 =
𝑑𝑇
•
𝑑𝑢 = 𝑐𝑣 𝑑𝑇
𝑢2 − 𝑢1 = න 𝑐𝑑𝑇
𝑇1
Then, the enthalpy change can be evaluated as
ℎ(𝑇, 𝑝) = 𝑢 𝑇 + 𝑝𝑣
ℎ2 𝑇, 𝑝 − ℎ1 𝑇, 𝑝 = 𝑢2 𝑇 − 𝑢1 𝑇 + 𝑣(𝑝2 − 𝑝1 )
•
If the specific heat is constant
𝑇2
𝑢2 − 𝑢1 = න 𝑐𝑑𝑇 = 𝑐(𝑇2 − 𝑇1 )
𝑇1
ℎ2 − ℎ1 = 𝑢2 𝑇 − 𝑢1 𝑇 + 𝑣 𝑝2 − 𝑝1
= 𝑐(𝑇2 − 𝑇1 ) + 𝑣 𝑝2 − 𝑝1
Ideal Gas
Ideal Gas
▪ Thermodynamic properties
Reference States
Reference States
Example
▪ Polytropic process of air as an ideal gas
Assumptions:
- The air is a closed system
- The air behaves as an ideal gas
- The compression is polytropic with n = 1.3
- Negligible change in kinetic or potential energy