Engineering thermodynamics is a branch of thermodynamics that deals with the practical application of thermodynamic principles and concepts. One of the fundamental topics in engineering thermodynamics is the properties of pure substances. A pure substance is a material that has a fixed and constant chemical composition, regardless of its physical state. This means that a pure substance cannot be separated into two or more different substances by physical means. Examples of pure substances include water, oxygen, and carbon dioxide. The properties of a pure substance are critical in thermodynamics because they are used to calculate important thermodynamic parameters, such as enthalpy, entropy, and internal energy. These parameters are used to understand the behavior of systems and predict their response to changes in temperature, pressure, and other conditions. One of the key properties of a pure substance is its temperature-pressure phase diagram, which provides information about the physical state of the substance under different conditions. For example, water can exist as a solid (ice), a liquid (water), or a gas (steam) depending on the temperature and pressure conditions. This information is critical for understanding the behavior of a substance in different thermodynamic systems, such as power plants, refrigeration and air-conditioning systems, and chemical processes. Another important property of a pure substance is its enthalpy of vaporization, which is the amount of energy required to convert a unit mass of the substance from a liquid to a gas at a constant temperature. This property is critical in many applications, such as the design of steam power plants, which use the energy stored in steam to generate electricity. The specific heat capacity of a pure substance is another critical property. It represents the amount of energy required to raise the temperature of a unit mass of the substance by a unit temperature. This property is used to calculate the heat transfer in thermodynamic systems, such as refrigeration and air-conditioning systems. Another important property of a pure substance is its thermal conductivity, which represents its ability to transfer heat. This property is critical in the design of heat exchangers, where heat is transferred from one fluid to another. In conclusion, the properties of pure substances play a critical role in engineering thermodynamics. They provide valuable information about the behavior of a substance under different conditions and are used to calculate important thermodynamic parameters, such as enthalpy, entropy, and internal energy. This information is critical for the design and operation of a wide range of thermodynamic systems, such as power plants, refrigeration and air-conditioning systems, and chemical processes.

1. Engineering Thermodynamics:
Properties of Pure Substances
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2. Outline
2
• 2−1 Pure Substances
• 2−2 Phases of a Pure Substance
• 2−3 Phase Change processes of a Pure Substance
• 2−4 Property Diagrams for Phase Change Processes
• 2−5 Property Tables
• 2-6 The Ideal gas Equation of State
• 2−7 Compressibility Factor
• 2-9 Specific Heats
• 2-10&11 Internal Energy and Enthalpy Formulation For
Ideal Gases and Condensed Phases (Solid and Liquids)
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3. Pure Substances
3
A pure substance does not change chemical composition but can change phase.
Is the working fluid a two-phase substance?
H2O
R-134a
An “ideal” gas?
Air
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4. Phases of a Pure Substance
liquid
vapor
2
Pure H O
vapor
liquid
4
Water Air
Not pure: different condensation
temperatures for different
components
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5. Phase Change Processes of Pure Substances
5
Terminology
• Compressed liquid -- not about to evaporate
• Saturated liquid -- about to evaporate
• Saturated liquid-vapor mixture --two phase
• Saturated Vapor -- about to condense
• Superheated Vapor -- not about to condense
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6. Heating Water- Phases
6
Compressed
Liquid
Saturated
Liquid
Saturated Liquid-Vapor
Mixture
Liquid Liquid Liquid/Vapor

7. Heating Water- Phases
Vapor 7
Vapor
Saturated Vapor Superheated Vapor
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8. T-v Diagram for Heating Water
Isobaric process
8
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9. Saturation
9
• At a given pressure, a substance changes phase at a fixed
temperature, called the saturation temperature.
• At a given temperature, the pressure at which a substance
changes phase is called the saturation pressure.
• T and P are dependant during the phase change, i.e., Tsat
= f(Psat)
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10. Latent Heat
10
• Latent heat is the amount of energy absorbed or released during
phase change
• Latent heat of fusion: melting/freezing. Equals 333.7 kJ/kg for 1 atm
H2O
• Latent heat of vaporization: boiling/condensation. Equals 2257.1 kJ/kg
for 1 atm H2O
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11. Property Diagrams for Phase Change Processes
T-v diagram
11
A point beyond which T  Tcr
, a
liquid-vapor transition is no
longer possible at constant
pressure. If T  Tcr , the
substance cannot be liquefied, no
matter how great the pressure.
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12. 12
Scenarios for Phase Change Processes
• Compressed liquid or subcooled liquid:
- If heat is added, the temperature will rise but there will be no phase change.
- “Not about to vaporize”
- Example: liquid water at 20oC/1 atm.
• Saturated liquid:
- If heat is added, the temperature does not change but some of the liquid vaporizes.
- Example: liquid water at 100oC/1 atm.
• Saturated liquid/vapor mixture:
- As heat is added, liquid turns to vapor.
- Example: vapor water at 100oC/l atm.
• Saturated vapor:
- If heat is removed, the temperature does not change, but some of the
vapor condenses.
- Example: vapor water at 100oC/1 atm.
• Superheated vapor:
- If heat is added, the temperature will rise but there will be no phase change.
- Example: vapor vapor at 120oC/1 atm.
QH

13. Phase-Change: P-T Diagram
13
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14. P-T Diagram: Isothermal Process
14
GAS @ g
Weight
GAS
State d
LIQUID
GAS
Weight
LIQUID @ a
P
Compressed
Liquid
a
d
Superheated
Vapor
g
T
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15. P-T Diagram: Isobaric Process
15
Gas @ b
Q
GAS
STATE f
LIQUID
Q
GAS
LIQUID
Q
P
T
b
f
Superheated
Vapor
Subcooled
Liquid
a
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16. P-V Diagram: Substance that Contracts on Freezing
16
For water, the triple point is at 273.16 K (32.018 F) and 0.6113 kPa (0.0887 psia)

17. P-V Diagram: Substance that Expands on Freezing
17

18. Using T-v Diagrams
18
Sat. Vapor
line
Sat.
Liquid
line
T
v
1. Draw saturation lines
2. Note given information on state properties
T
P
v
U
H
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19. Using T-v Diagrams
19
v
P
const.
T
v
T const.
P
3. Draw constant pressure or constant temperature lines
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20. Using T-v Diagrams
20
T const.
P
v
P sat.
vf; uf; hf vg; ug; hg
critical point
v
P
const.
T
vf; uf; hf vg; ug; hg
T sat.
critical point
4. Note saturation temperature or saturation pressure, sat. liquid and sat. vap. states
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21. T
v
vf; uf; hf vg; ug; hg
P >
P
const.
T sat.
P
v
P sat. T>
T const.
vf; uf; hf vg; ug; hg
Using T-v Diagrams
5. Add lines showing higher or lower temperature or pressure
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22. Property Tables
22
Enthalpy – A Combination Property
H = U + PV
or
h = u + Pv
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23. Using Property Tables
23
Saturated Liquid and saturated vapor water Tables A-4 and A-5
Example:
A rigid tank contains 50 kg of
saturated liquid water at 90oC.
Determine the pressure in the tank
and the volume of the tank.
(Answers: 70.14 kPa, 0.0518 m)
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24. Dependence of Tsat and Psat
24
Elevation
(m)
Atmospheric
Pressure (kPa)
Boiling Temp.
( oC)
0 101.33 100.0
1000 89.50 96.3
5000 54.05 83.0
10,000 26.50 66.2
Substance Tsat (oC)
H20 100
R-134a -26
N2 -196
@ Patm
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25. Quality- practical application to finding properties
25
v – vf
x =
vg – vf
u – uf
=
ug – uf
h – hf
hg – hf
v = vf + x vfg , (vfg = vg – vf)
Same procedure applies in finding specific internal energy or enthalpy values.
=
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26. Quality- An Alternative Approach
26
P or T
vf vf<v<vg vg v
Sat. vapor
vg
Sat. liquid
vf
•
Since mtotal = mliquid + mvapor ,
f
or m = m + mg
Also, V = Vf + Vg
& V = mv
mv = mf vf + mg vg,
or
mv = (m-mg) vf + mgvg
v = (1-x) vf + xvg = vf + x vfg
Since Quality is a mass fraction of the vapor content in mixture, we define
x = mg / m
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27. Example:
Determine the specific volume of 2 kg of water at 100oC with a quality of 70%,
i.e., moisture content of 30%? Also, obtain mass fraction of the vapor?
Ans.
From Table (A-4): vf=0.0010435 m3/kg & vg=1.673 m3/kg
v = vf + x.vfg = 0.0010435 + 0.7(1.673 -0.0010435)
v = 1.7141 m3/kg
Since, x = mg/m then mg = 0.7x2 = 1.4 kg
Quality: Example Problem
27
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28. Fixing the State
28
Is P > Psat. or T < Tsat. ?
Yes: Sub-cooled liquid (vf @T; uf @T; hf @T)
No: Liquid-vapor mix. or Super-heated vapor
P = Psat. T = Tsat.
Liquid-vapor mixture
Quality calculation required
Is P < Psat. or T > Tsat. ?
Super-heated vapor
(super-heated vapor tables)
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29. Using Superheated Vapor Table
29
Table (A-6)
In the region to the right of the saturated
vapor line, a substance exists as
superheated vapor.
Example:
Determine the temperature of water at a
state of P = 0.5 MPa and h = 2890
kJ/kg.
(Answers: 216.4 oC)
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30. - In the absence of compressed liquid tables, a general rule of thumb for properties v &
u is to treat compressed liquid as saturated liquid at the given temperature:
y = yf @T
i.e., v = vf @T & u = uf @T
- For ‘h’ values in the compressed region:
h = hf @T + vf (P - Psat)
Compressed Liquid
30
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31. Determine the specific enthalpy of liquid water at 25oC and 300 kPa?
Compressed Liquid: Example Problem
31
Ans. The state is compressed liquid as Psat@25C = 3.169 kPa
However, the pressure of 300 kPa is below the minimum pressure (0.5 MPa) listed in Table (A-7):
So we will use the approximation!
v = vf @T = 0.001003 m3/kg
h = hf @T + vf (P - Psat)
= 104.89 + 0.001003(300 – 3.169) = 105.19 kJ/kg
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32. 32
Reference State and Reference Values
• Values of u and h are arbitrary since we are only interested in changes of these properties.
• For water, the reference state is u = 0 for saturated liquid @ 0.01°C.
• For R-134a and many other substances, the reference state is h = 0 for the saturated
liquid at -40 °C.
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33. 33
Basics of Interpolation
Assuming a linear variation in the examined property over a relatively short range.
Temp.
T (oF)
Sp. Volume
v (ft3/lbm)
Enthalpy
h (Btu/lbm)
80 0.5408 113.47
93 v=? h=?
100 0.5751 118.39
(TableA-13E) Superheated R-134a
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34. 34
Steam Table: Example Problem
Complete the following steam table. Locate each point on a PV diagram.
# T,oC P
, kPa v, m3/kg u, kJ/kg h, kJ/kg x Phase
1 160 1300
2 150 1.0
3 200 1000
4 200 4000
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35. 35
The Ideal Gas Equation of State
M = mass of one mol of the gas (TableA-1) = m / N
• Equation of State: Property relation that involves other properties of a substance at equilibrium
states. Ideal (imaginary) Gas Equation is the simplest among other equation of states
• History Background:
- Boyle (1662): P α 1/v =➔ for constant temperature (P1v1=P2v2)
- Charles (1802): for const pressure =➔ v/T=const.
or combined: Pv = RT at low pressures
- Ideal gas is an imaginary substance that could be resembled by gases at low densities (low P
& high T)
M Molar mass (molecular weight)
Ru = 8.314 kJ/(kmol.K) = 1.986 Btu/(lbmol.R)
(kJ/kg.K)
• Gas Constant R =
Ru Universal gas constant
=
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36. 36
Property Diagrams for Ideal Gases
P const.
T
v
P >
T const.
P
v
T >
Boyle
P α 1/v =➔ for constant temperature
Charles
for const pressure =➔ v/T=const.
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37. Special processes:
- Constant Pressure (Isobaric)
- Constant Volume (Isochoric)
- Constant Temperature (Isothermal)
- Polytropic (PV n = C)
- Ideal Gas
moves along lines of constant
pressure
Moves along lines of constant
specific volume (if mass constant)
Moves along lines of constant
temperature
As pressure increases, volume
decreases (non-linear relationship)
P2 v2
=
P1 v1
P v = RT , R =
T1 T2
Drawing Process Paths
37
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38. 38
Drawing Process Path (cont.)
Read the problem statement and relate to the property diagram !
Using other information:
“ Pressure increases to three times its present …..”
“ Volume decreases to ¼ its present value …..”
“ Heat losses cause temperature to decrease ….”
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39. 39
Water Vapor and Ideal Gas
Percentage of error involved in assuming
steam to be an ideal gas, and the region
where steam can be treated as an ideal gas
with less than 1 percent error.
Q. Is water vapor an Ideal gas?
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40. ▪ Z represents the volume ratio or compressibility.
▪ Z < 1 or Z > 1 for real gases.
RT
40
ideal
Compressibility Factor
The deviation from the ideal-gas behavior can
be corrected by compressibility factor Z.
Z 
P =
actual
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41. 41
Principle of corresponding states
• The compressibility factor Z is approximately the same for all gases at the same reduced temperature
and reduced pressure.
Z = Z(PR,TR) for all gases
PR = P/Pcr & TR = T/Tcr
where,
PR and TR are reduced values.
Pcr and Tcr are critical properties (Table A-1).
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42. 42
Reduced Properties
cr
not v !
• This works great if you are given a gas with known P &T and asked to find the v.
• However, if you were given P and v and asked to find T (or T and v and asked to find
P), you need to use:
pseudo-reduced specific volume, which is defined as:
• Ideal gas approximation (Z=1), if (see Fig. A-30):
✓ PR < 10 and TR > 2
✓ PR << 1
R
v =
vactual
RTcr / Pcr
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43. 43
Ideal Gas Example
Calculate the specific volume of nitrogen at
(a) 300 K and 8.0 MPa (b) 200 K and 20.0 MPa.
Tcr = 126.2 K and Pcr = 3.39 MPa for nitrogen (T
able A-1)
Part (a):
TR = T/Tcr = 300 K / 126.2 K =2.38
PR = P/Pcr = 8.0 MPa / 3.39 MPa = 2.36
Since, TR > 2 and PR < 10, we can use ideal gas equation of state.
ZRT
P
m3
kg
m3
kPa
kJ
kJ
)(300K)
kg.K
8,000 kPa
(0.2968
= 0.01113
=
v =
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44. ZRT
P
m3
kg
m3
kPa
kJ
kJ
)(200K)
kg.K
20,000 kPa
(0.91)(0.2968
= 0.002701
=
v =
Part (b):
TR = T/Tcr = 200 K / 126.2 K =1.58
PR = P/Pcr = 20.0 MPa / 3.39 MPa = 5.90
From Figure (A-30)b: Z=0.91
Ideal Gas Example
44
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45. Specific Heats
45
m = 1 kg
T = 1o
C
Specific heat = 5 kJ / kg.o
C
Definition:
The amount of energy required to raise the temperature of a unit mass of a substance by one
degree.
5 kJ
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46. Internal Energy & Enthalpy Formulation Using
Specific Heats
46
v
 v

T

C =  u 
p
C
 p

T

=  h 
Cv & Cp are properties
J/kg.K OR Btu/lbm.R
du = Cv dT
Constant volume process
dh = Cp dT
Constant pressure process
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47. Specific Heat of Ideal Gases
For real gases, specific heats vary with pressure but for most practical purposes a suitable average value may be
used.
h = u + Pv = u + RT dh = du + RdT
Cp = Cv + R (kJ/kg.K)
Cp = Cv + Ru (kJ/kmol.K)
Introducing another ideal gas property called specific heat ratio:
k = Cp / Cv > 1
In general, k is about 1.4, 1.6 and 1.3 for diatomic gases (CO, N2, O2), monatomic gases (Ar, He) and triatomic
gases (CO2, SO2), respectively.
47
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48. Special Cases for a Polytropic Process (PVn = C)
48
Exponent ‘n’ Value Process Type
0 Constant Pressure
1 Constant Temperature
k = Cp / Cv Adiabatic
 Constant Volume
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49. Specific Heat of Ideal Gases
49
du = Cv dT
2
Δ u = Cv(T) dT = Cv,average (T2 – T1)
1
dh = Cp dT
2
Δ h = Cp(T) dT = Cp,average (T2 – T1)
1
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50. 50
Specific Heats of Incompressible Solids & Liquids
Δ u = Cavg (T2 – T1)
Constant pressure process:
Constant temperature process (for liquids):
Incompressible means  or v is essentially constant.
Similar to ideal gases, specific heat depends only on temperature.
Δh = Δ u + v ΔP
Δ h = Cavg ΔT + v ΔP
Δ P = 0 Δh = Cavg ΔT
Δ T = 0 Δh = v ΔP
Cv = Cp = Cavg
h = u + Pv
dh = du + d(Pv)
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