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Thermodynamics Lecture Series
email: [email protected]://www5.uitm.edu.my/faculties/fsg/drjj1.html
Applied Sciences Education Research Group (ASERG)
Faculty of Applied SciencesUniversiti Teknologi MARA
Pure substances – Pure substances – Property tables and Property tables and Property DiagramsProperty Diagrams
QuotesQuotes
“You do not really understand something unless you can explain it to your grandmother.”
(Albert Einstein)
IntroductionIntroduction
Objectives:
1. State the meaning of pure substances
2. Provide examples of pure and non-pure substances.
3. Read the appropriate property table to determine phase and other properties.
4. Sketch property diagrams with respect to the saturation lines, representing phase and properties of pure substances.
FIGURE 1–5Some application areas of thermodynamics.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1-1
Application
Example: A steam power cycle.Example: A steam power cycle.
SteamTurbine
Mechanical Energyto Generator
Heat Exchanger
Cooling Water
Pump
Fuel
Air
CombustionProducts
System Boundaryfor ThermodynamicAnalysis
System Boundaryfor ThermodynamicAnalysis
Steam Power Plant
Steam Power PlantSteam Power Plant
FIGURE 1–17A control volume may involve fixed, moving, real, and imaginary boundaries.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1-5
Open system devices
Open system devicesOpen system devices
Heat Exchanger
Throttle
CHAPTER
2
Properties of Pure Substances
Title:
Pure Substances
• Pure substancesPure substances– Substance with fixed chemical compositionSubstance with fixed chemical composition
• Can be single element: Such as, NCan be single element: Such as, N22, H, H22, O, O22
• Compound: Such as Water, HCompound: Such as Water, H22O, CO, C44HH1010,,
• Mixture such as Air, Mixture such as Air,
• 2-phase system such as H2-phase system such as H22O.O.
– Responsible for the receiving and removing dynamic Responsible for the receiving and removing dynamic energy (working fluid)energy (working fluid)
• Pure substancesPure substances– Substance with fixed chemical compositionSubstance with fixed chemical composition
• Can be single element: Such as, NCan be single element: Such as, N22, H, H22, O, O22
• Compound: Such as Water, HCompound: Such as Water, H22O, CO, C44HH1010,,
• Mixture such as Air, Mixture such as Air,
• 2-phase system such as H2-phase system such as H22O.O.
– Responsible for the receiving and removing dynamic Responsible for the receiving and removing dynamic energy (working fluid)energy (working fluid)
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
H2OSat.
liquid
Qin
P = 100 kPa
T = 99.6 C
P = 100 kPa
T = 99.6 C
Water interacts with thermal energyWater interacts with thermal energy
99.6
2 =
f@
100 kPa
T, C
30, m3/kg
1
H2O:C. liquid
P = 100 kPa
T = 30 C
P = 100 kPa
T = 30 C
Qin
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
H2OSat.
liquid
Qin
P = 100 kPa
T = 99.6 C
P = 100 kPa
T = 99.6 C
Water interacts with thermal energyWater interacts with thermal energy
H2O:Sat. Liq.
Sat. VaporSat. Vapor
Qin
99.6
2 =
f@
100 kPa
T, C
30
, m3/kg1
3
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
Water interacts with thermal energyWater interacts with thermal energy
4 =
g@
100 kPa
99.6
2 =
f@
100 kPa
T, C
30, m3/kg
1
3
P = 100 kPa
T = 99.6 C
P = 100 kPa
T = 99.6 C
H2O:Sat. Vapor
H2O:Sat. Vapor
Qin
H2O:Sat. Liq.
Sat. VaporSat. Vapor
Qin
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
Water interacts with thermal energyWater interacts with thermal energy
150
5
99.6
2 =
f@
100 kPa
T, C
30, m3/kg
1
4 =
g@
100 kPa
3
5 = @100 kPa, 150°C
3 = [f + x f g]@100 kPa
1 = f@T1
H2O:SuperVapor
H2O:SuperVapor
P = 100 kPa
T = 150 C
P = 100 kPa
T = 150 C
Qin
P = 100 kPa
T = 99.6 C
P = 100 kPa
T = 99.6 C
H2O:Sat. Vapor
H2O:Sat. Vapor
Qin
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
Water interacts with thermal energyWater interacts with thermal energy
H2O:Sat. Liq.
Sat. VaporSat. Vapor
P = 100 kPa
T = 99.6 C
P = 100 kPa
T = 99.6 C
Qin
P = 100 kPa
T = 99.6 C
P = 100 kPa
T = 99.6 C
H2O:Sat. Vapor
H2O:Sat. Vapor
Qin
P = 100 kPa
T = 150 C
P = 100 kPa
T = 150 C
H2O:SuperVapor
H2O:SuperVapor
Qin
P = 100 kPa
T = 30 C
P = 100 kPa
T = 30 C
H2O:C. liquid
Qin
P = 100 kPa
T = 99.6 C
P = 100 kPa
T = 99.6 C
H2OSat.
liquid
Qin
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
99.6
2 =
f@
100 kPa
T, C
30, m3/kg
1
4 =
g@
100 kPa
3
5 = @100 kPa, 150°C
3 = [f + x f g]@100 kPa
1 = f@T1
150
100
kPa
5
Compressed liquidCompressed liquid: Good : Good estimation for properties estimation for properties by taking yby taking y = y = yf@T f@T where where
y can be either y can be either , u, h or , u, h or s.s.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2-1
FIGURE 2-11T-v diagram for the heating process of water at constant pressure.
Phase Change of WaterPhase Change of WaterPhase Change of WaterPhase Change of Water
T, C
, m3/kg
99.6
f@
100 kPa
g@
100 kPa
100
kPa
179.9
45.8
10 k
Pa
1000
kPa
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2-2
FIGURE 2-16T-v diagram of constant-pressurephase-change processes of a puresubstance at various pressures(numerical values are for water).
99.6
45.8
179.9
T –v diagram: Multiple P
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2-3
FIGURE 2-18T-v diagram of a pure substance.
T –v diagram: Multiple P
T, C
, m3/kg
T – v diagram - Example
70
=f@70 C = 0.001023
81.3
3.240
50 kPa
P, kPa T, C
50 70
Psat, kPa Tsat, C
81.33
Phase, Y?
Compressed Liquid,
T < Tsat
, m3/kg
f@70 C
T – v diagram - Example
T, C
, m3/kgf@200 kPa
= 0.001061
200
kPa
P, kPa , m3/kg
200 1.5493
T- diagram with respect to the saturation
lines
T- diagram with respect to the saturation
lines
Phase, Why?
Sup. V., Sup. V., > >gg
Psat, kPa Tsat, C
120.2120.2
374.1
400
= 1.5493
120.23
g@200 kPa
= 0.8857
T, C
400400
T – v diagram - Example
T, C
, m3/kg
1,00
0 kP
a
P, kPa u, kJ/kg
1,000 2,000
T- diagram with respect to the saturation
lines
T- diagram with respect to the saturation
lines
Phase, Why?
Wet Mix., uWet Mix., uf f < u < u< u < ugg
Psat, kPa Tsat, C
179.9179.9
374.1
f@1,000 kPa
= 0.001127
179.9
g@1,000 kPa
= 0.19444
T, C
179.9179.9
= [f + x f g]@1,000 kPa
Property TableSaturated water – Pressure table
Pressure
P, kPa
10
50
P, MPaP, MPa
0.100
1.00
10
22.09
Specific internal energy, kJ/kg
uf, kJ/kg ufg, kJ/kg ug, kJ/kg
191.82 2246.1 2437.9
340.44 2143.4 2483.9
417.36 2088.7 2506.1
761.68 1822.0 2583.6
1393.04 1151.4 2544.4
2029.6 0 2029.6
Specific volume, m3/kg
f, m3/kg g, m3/kg
0.001010 14.67
0.001030 3.240
0.001043 1.6940
0.001127 0.19444
0.001452 0.018026
0.003155 0.003155
Sat. temp.
Tsat, C
45.81
81.3381.33
99.6399.63
179.91179.91
311.06311.06
374.14374.14