1. Thermodynamics Lecture Series
Assoc. Prof. Dr. J.J.
Pure substances – Property tables and Property Diagrams & Ideal Gases
Applied Sciences Education Research Group (ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA

2. Properties of Pure Substances- Part 2
CHAPTER 2
Pure substance
Properties of Pure Substances- Part 2
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3. "What we have to learn to do, we learn by doing." -Aristotle
Quotes
"Education is an admirable thing, but it is well to remember from time to time that nothing that is worth knowing can be taught." - Oscar Wilde
"What we have to learn to do, we learn by doing." -Aristotle

4. Introduction
Objectives:
Choose the right property table to read and to determine phase and other properties.
Derive and use the mathematical relation to determine values of properties in the wet-mix phase
Sketch property diagrams with respect to the saturation lines, representing phases, processes and properties of pure substances.

5. State conditions for ideal gas behaviour
Introduction
Objectives:
Use an interpolation technique to determine unknown values of properties in the superheated vapor region
State conditions for ideal gas behaviour
Write the equation of state for an ideal gas in many different ways depending on the units.
Use all mathematical relations and skills of reading the property table in problem-solving.

6. Example: A steam power cycle.
Turbine
Mechanical Energy
to Generator
Heat
Exchanger
Cooling Water
Pump
Fuel
Air
Combustion
Products
System Boundary
for Thermodynamic
Analysis
Steam Power Plant

7. Phase Change of Water - Pressure Change
100
P, kPa
, m3/kg 1
30 C
T = 30 C
P = 100 kPa
H2O:
C. liquid
T = 30 C
P = kPa
H2O:
Sat. liquid
2 = 30 °C
1 = 30 °C
2 = °C
4.246
kPa = C
C = kPa
Water when pressure is reduced

8. Phase Change of Water - Pressure Change
T = 30 C
P = kPa
H2O:
Sat. liquid
2 = 30 °C
, m3/kg 1
100
P, kPa
4.246
30 C
2 = °C
1 = 30 °C
H2O:
Sat. Liq.
Sat. Vapor
3 = [f + x f 30 °C 3
kPa = C
C = kPa
Water when pressure is reduced

9. Phase Change of Water - Pressure Change
H2O:
Sat. Liq.
Sat. Vapor
T = 30 C
P = kPa 1 3
P, kPa
, m3/kg
2 = 30 °C
100
4.246
30 C
3 = [f + x f 30 °C
1 = 30 °C
2 = 30 °C
H2O:
Sat. Vapor
4 = 30 °C
kPa = C
C = kPa
4 = 30 °C
Water when pressure is reduced

10. Phase Change of Water - Pressure Change
4 = 30 °C
2 = kPa
, m3/kg 1 3
2 = 30 °C
100
P, kPa
4.246
30 C
3 = [f + x f 30 °C
4 = 30 °C
1 = 30 °C
2 = 30 °C
H2O:
Sat. Vapor
T = 30 C
P = kPa
H2O:
Super
Vapor
T = 30 C
P = 2 kPa
5= 30 °C 2 5
kPa = C
C = kPa
Water when pressure is reduced

11. Phase Change of Water - Pressure Change
T = 30 C
P = 100 kPa
H2O:
C. liquid
T = 30 C
P = kPa
H2O:
Sat. liquid
H2O:
Sat. Liq.
Sat. Vapor
T = 30 C
P = kPa
H2O:
Sat. Vapor
T = 30 C
P = kPa
H2O:
Super
Vapor
T = 30 C
P = 2 kPa
kPa = C
C = kPa
Water when pressure is reduced

12. Phase Change of Water- Pressure Change
, m3/kg 1
100
P, kPa
30 C
kPa = C
C = kPa
1 = 30 °C
2 = 30 °C
2 = 30 °C
4.246
3 = [f + x f 30 °C
4 = 30 °C
5= 30 °C 2 5 3
4 = 30 °C
Compressed liquid: Good estimation for properties by taking y = where y can be either , u, h or s.

13. Phase Change of Water C
P, C
, m3/kg
101.35
100 C C
1,553.8
200 C
1.2276
10 C

14. P-  diagram with respect to the saturation lines
Phase Change of Water
P, C
, m3/kg
101.35 C
1,553.8
1.2276
200 C
10 C
100 C C
22,090
P-  diagram with respect to the saturation lines

15. FIGURE 2-19 P-v diagram of a pure substance.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2-4

16. FIGURE 2-25 P-T diagram of pure substances.
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 2-25 P-T diagram of pure substances.
2-7

17. Property Table Saturated water – Temperature table
Tsat, C 10 50
100
200
300
374.14
Sat. P.
P, kPa
1.2276
12.349
P, MPa
1.5538
8.581
22.09
Specific volume, m3/kg
f, m3/kg
g, m3/kg
106.38
12.03
1.6729
Specific internal energy, kJ/kg
uf, kJ/kg
ufg, kJ/kg
ug, kJ/kg
42.00
2347.2
2389.2
209.32
2234.2
2443.5
418.94
2087.6
2506.5
850.65
1744.7
2593.3
1332.0
1231.0
2563.0
2029.6

18. Property Table Saturated water – Temperature table
Tsat, C 10 50
100
300
374.14
Specific internal energy, kJ/kg
uf, kJ/kg
ufg, kJ/kg
ug, kJ/kg
42.00
2347.2
2389.2
209.32
2234.2
2443.5
418.94
2087.6
2506.5
1332.0
1231.0
2563.0
2029.6
Specific enthalpy, kJ/kg
hf, kJ/kg
hfg, kJ/kg
hg, kJ/kg
42.01
2377.7
2519.8
209.33
2382.7
2592.1
419.04
2257.0
2676.1
1332.0
1940.7
2793.2
2099.3
Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal energy and flow energy. Latent heat of vaporization.

19. T < Tsat or P > Psat
T – v diagram - Example
P, kPa
T,  C 50 70
Tsat, C
81.33
Psat, kPa
31.19
, m3/kg C
Phase, Y?
Compressed Liquid,
T < Tsat or P > Psat
T, C
, m3/kg
50 kPa
81.3
3.240 70
C =

20. P > Psat or T < Tsat
P – v diagram - Example
P, kPa
T,  C 50 70
Psat, kPa
31.19
Tsat, C
81.33
, m3/kg C
Phase, Y?
Compressed Liquid,
P > Psat or T < Tsat
P, kPa
, m3/kg
70 C
C = 50
31.19
5.042

21. P-  diagram with respect to the saturation lines
P – v diagram - Example
P, kPa
T,  C
200
400
Phase, Why?
Sup. Vap., T >Tsat
Psat, kPa
Tsat, C NA
120.2
, m3/kg
1.5493
P, kPa
, m3/kg
22,090.0
400 C
P-  diagram with respect to the saturation lines
200
kPa =
kPa = C
 =

22. T-  diagram with respect to the saturation lines
T – v diagram - Example
P, kPa
u, kJ/kg
1,000
2,000
Psat, kPa
Tsat, C
179.9
T,  C
179.9
Phase, Why?
Wet Mix., uf < u < ug
T, C
, m3/kg
1,000 kPa
T-  diagram with respect to the saturation lines
374.1
kPa =
179.9
kPa = 
 = [f + x f kPa

23. Saturated Liquid-Vapor Mixture
Given the pressure, P, then
T = Tsat, yf < y <yg
H2O:
Sat. Liq.
Sat. Vapor
Vapor Phase:, Vg, mg, g, ug, hg
Liquid Phase:, Vf, mf, f, uf, hf
Mixture:, V, m, , u, h, x
Mixture’s quality
More vapor, higher quality
x = 0 for saturated liquid
x = 1 for saturated vapor
Specific volume of mixture?? Since V=m 

24. Saturated Liquid-Vapor Mixture
Given the pressure, P, then
T = Tsat, yf < y <yg
H2O:
Sat. Liq.
Sat. Vapor
Vapor Phase:, Vg, mg, g, ug, hg
Liquid Phase:, Vf, mf, f, uf, hf
Mixture:, V, m, , u, h, x
Mixture’s quality
Divide by total mass, mt
where

25. Saturated Liquid-Vapor Mixture
Given the pressure, P, then
T = Tsat, yf < y <yg
H2O:
Sat. Liq.
Sat. Vapor
Vapor Phase:, Vg, mg, g, ug, hg
Liquid Phase:, Vf, mf, f, uf, hf
Mixture:, V, m, , u, h, x
Mixture’s quality
where
y can be , u, h
where
If x is known or has been determined, use above relations to find other properties. If either , u, h are known, use it to find quality, x.

26. Interpolation: Example – Refrigerant-134a
P, kPa
, m3/kg
200
Phase, Why?
Sup. Vap.,  >  g
Psat, kPa
Tsat, C -
-10.09
T,  C ??
Assume properties are linearly dependent.
Perform interpolation in superheated vapor phase.
T, C
, m3/kg TH L TL
 H
T = ?? m2  m1

27. Interpolation: Example – Refrigerant-134a
P, kPa
, m3/kg
200
Phase, Why?
Sup. Vap.,  >  g
Psat, kPa
Tsat, C -
-10.09
T,  C ??
T,  C
, m3/kg
u, kJ/kg
TL= 0
L =
uL =
0 < TL < TH
 =
0 < uL < uH
TH= 10
H =
uH =
Assume properties are linearly dependent. Perform interpolation in superheated vapor phase.

28. Interpolation: Example – Refrigerant-134a
P, kPa
, m3/kg
200
Phase, Why?
Sup. Vap.,  >  g
Psat, kPa
Tsat, C -
-10.09
T,  C
3.35
u, kJ/kg
231.85
P, kPa
, m3/kg
22,090.0
 =
T = 3.35 C
Psat ??
kPa =
kPa =
200 C
P-  diagram with respect to the saturation lines

29. Property Table Saturated water – Temperature table
Tsat, C 10 50
100
200
300
374.14
Sat. P.
P, kPa
1.2276
12.349
P, MPa
1.5538
8.581
22.09
Specific volume, m3/kg
f, m3/kg
g, m3/kg
106.38
12.03
1.6729
Specific internal energy, kJ/kg
uf, kJ/kg
ufg, kJ/kg
ug, kJ/kg
42.00
2347.2
2389.2
209.32
2234.2
2443.5
418.94
2087.6
2506.5
850.65
1744.7
2593.3
1332.0
1231.0
2563.0
2029.6

30. Property Table Saturated water – Temperature table
Tsat, C 10 50
100
300
374.14
Specific internal energy, kJ/kg
uf, kJ/kg
ufg, kJ/kg
ug, kJ/kg
42.00
2347.2
2389.2
209.32
2234.2
2443.5
418.94
2087.6
2506.5
1332.0
1231.0
2563.0
2029.6
Specific enthalpy, kJ/kg
hf, kJ/kg
hfg, kJ/kg
hg, kJ/kg
42.01
2377.7
2519.8
209.33
2382.7
2592.1
419.04
2257.0
2676.1
1332.0
1940.7
2793.2
2099.3
Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal energy and flow energy. Latent heat of vaporization.

31. Sample Questions
(a)Briefly, give a description that best fit each of the following:
i)process ii)cyclic process
iii) saturated vapor iv) kinetic energy
ANSWER: (2 marks each)
(i) A property change or a change of state of a system
(ii) A process where the initial and the final state is the same.
(iii) Vapor ready to condense.
(iv) Energy associated with the movement(speed) of a system.

32. Sample Questions
Using the property table, read the saturation temperatures and the saturated volumes for water at pressures of 30 kPa and 300 kPa, respectively.
ANSWER:
System
P, kPa
Tsat, C
f, m3/kg
g,m3/kg
Water 30
300
(12 marks)

33. Sample Questions
Using the property table, read the saturation temperatures and the saturated volumes for water at pressures of 30 kPa and 300 kPa, respectively.
ANSWER:
System
P, kPa
Tsat, C
f, m3/kg
g,m3/kg
Water 30
69.1
5.229
300
133.55
0.6058
(12 marks)

34. Sample Questions
Using the property table, determine the missing properties in the table below. [Note: The symbols used are the following: P for pressure, T for temperature, and u for the specific internal energy. When indicating the quality, use NA for the compressed liquid and the superheated vapor phase and use 0 < x < 1 for the wet mix phase. When indicating properties such as , u, h or s in the wet mix phase, just write down an expression on how to get it. NO CALCULATIONS ARE REQUIRED.]

35. Sample Questions (5 marks) Substance P kPa Psat@T T C Tsat@P Quality
Substance P
kPa T C
Quality x u
kJ/kg
Phase
& Reason
Water
200 -
0.5
(5 marks)

36. Sample Questions ANSWER: (5 marks) Substance P kPa Psat@T T C Tsat@P
Substance P
kPa T C
Quality x u
kJ/kg
Phase
& Reason
Water
200 -
120.23
0.5
u=uf+xufg
Wet mix
0 < x <1
(5 marks)

37. Sample Questions iii) Refrigerant 134a (6 marks) P kPa Psat@T T CP
kPa T C
Quality x u
kJ/kg
Phase
& Reason
200
-16
(6 marks)

38. T<Tsat or P>Psat
Sample Questions
ANSWER: P
kPa T C
Quality x u
kJ/kg
Phase
& Reason
200
157.48
-16
-10.09 NA
= 29.18
Comp.
Liquid
T<Tsat or P>Psat
(6 marks)

39. Sample Questions
In the question that follows, you need to use the data in the table above along with the property table to sketch property diagrams with respect to the saturation lines to indicate the state of the system. In your diagram, draw and label the temperature or the pressure lines. Place an X mark on the graph to indicate the state. Insert values for T, Tsat, P, Psat, , f and g.
Use the space below to draw a T -  diagram for the system in part (iii) of the table.

40. Sample Questions (8 marks) P, kPa 200 -10.09 C 157.48 -16 C , m3/kg
Accepted pairs of saturated values and must be marked correctly on the graph in m3/kg are:
kPa =
kPa =
C =
C =
200
157.48
 = f =
g =
-16 C
, m3/kg
P, kPa C
(8 marks)

41. Sample Questions
Heat is isothermally transferred into a piston-cylinder device containing water at 100 oC, 400 kPa until the pressure drops to 50 kPa. Use the space below to draw a T -  diagram with respect to the saturation lines for this process. Indicate the initial and final states and the direction of the process. Draw and label the temperature lines and insert values for T1, T2, Tsat, P1, P2, 1, 2, f, and g.

42. Sample Questions (12 marks) T, C 400 kPa 101.35 kPa 50 kPa 143.63 1
Accepted pairs of saturated values and must be marked correctly on the graph in m3/kg are:
C =
C =
kPa =
kPa =
kPa =
kPa = 3.240
T, C
143.63
81.33
1 =
50 kPa
400 kPa
, m3/kg
0.4625
2 = 3.418
100
kPa 2 1
1 = C =
(12 marks)

43. Properties of Pure Substances- Ideal Gases Equation of State

44. Ideal Gases
Equation of State
An equation relating pressure, temperature and specific volume of a substance.
Predicts P- -T behaviour quite accurately
Any properties relating to other properties
Simplest EQOS of substance in gas phase is ideal-gas (imaginary gas) equation of state

45. Equation of State for ideal gas
Ideal Gases
Equation of State for ideal gas
Boyle’s Law: Pressure of gas is inversely proportional to its specific volume P
Equation of State for ideal gas
Charles’s Law: At low pressure, volume is proportional to temperature

46. and where M is molar mass
Ideal Gases
Equation of State for ideal gas
Combining Boyles and Charles laws:
where gas constant R is
and where M is molar mass
and where Ru is universal gas constant Ru = kJ/kmol.K
EQOS: Since the total volume is V = m, so :  = V/m
So,
EQOS: since the mass m = MN where N is number of moles:
So,

47. Hence real gases satisfying conditions P << Pcr, T >> Tcr
Ideal Gases
Equation of State for ideal gas
Real gases with low densities behaves like an ideal gas
Hence real gases satisfying conditions
P << Pcr, T >> Tcr
Obeys EQOS
where,
Ru = kJ/kmol.K,
m = MN
V = m and

48. Gas Mixtures – Ideal Gases
High density
Low density
Molecules far apart
Real Gases
Low density (mass in 1 m3) gases
Molecules are further apart
Real gases satisfying condition
Pgas << Pcrit; Tgas >> Tcrit , have low density and can be treated as ideal gases

49. Gas Mixtures – Ideal Gases
Real Gases
Equation of State - P--T behaviour
P = RT (energy contained by 1 kg mass) where  is the specific volume in m3/kg, R is gas constant, kJ/kgK, T is absolute temp in Kelvin.
High density
Low density
Molecules far apart

50. Gas Mixtures – Ideal Gases
Real Gases
Equation of State - P--T behaviour
P=RT , since = V/m then,
P(V/m) =RT. So,
PV = mRT, in kPam3=kJ.
Total energy of a system.
High density
Low density

51. Gas Mixtures – Ideal Gases
Real Gases
Equation of State - P--T behaviour
PV =mRT = NMRT = N(MR)T
Hence, can also write PV = NRuT where
N is no of kilomoles, kmol,
M is molar mass in kg/kmole and
Ru is universal gas constant; Ru=MR.
Ru = kJ/kmolK
High density
Low density

52. Other EQOS EQOS is: Equations of States
Van Der Waals Equation of State
Considers intermolecular interactions, a/2, which then increases the pressure: P replaced by P+a/2
High density
Considers volume occupied by molecules, b, at high pressures: Replace  by  -b, then
EQOS is:
where
Low density

53. T-  diagram with respect to the saturation lines
T – v diagram - Example
P, kPa
u, kJ/kg
1,000
2,000
Psat, kPa
Tsat, C
179.9
T,  C
179.9
Phase, Why?
Wet Mix., uf < u < ug
T, C
, m3/kg
1,000 kPa
T-  diagram with respect to the saturation lines
374.1
kPa =
179.9
kPa = 
 = [f + x f kPa