The document discusses properties of substances such as water. It defines key concepts like state principle, phase diagrams, and phase changes. It provides examples of using property tables to determine specific volumes and enthalpies at given temperatures and pressures. The document also covers the ideal gas model and its applications, such as using the gas equation of state and determining work in polytropic processes.

1. Lesson 3:
Evaluating Properties
1By Meng Chamnan GIM

2. State Principle
• State principle has been developed as a guide in
determining the number of independent properties
required to fix the state of a system.
• the state principle indicates that the number of
independent intensive properties of a simple
compressible substance is two.
2By Meng Chamnan GIM

3. Diagram p-v-T
3By Meng Chamnan GIM
Fig. 1: p-v-T surface and
projections for water. (a)
Three-dimensional view. (b)
Phase diagram. (c) p-v
diagram.

4. T–v Diagram
By Meng Chamnan GIM 4

5. Phase Change
5By Meng Chamnan GIM

6. Phase Change
6By Meng Chamnan GIM

7. - Liquid State
-Two-phase liquid–vapor mixture, the ratio of the
mass of vapor present to the total mass of the
mixture is its quality, x. In symbols,
The value of the quality ranges from zero to unity:
at saturated liquid states, x = 0, and at saturated
vapor states, x = 1
- Vapor state
By Meng Chamnan GIM 7
vapor
liquid vapor
m
x
m m



8. Vapor and Liquid Tables
By Meng Chamnan GIM 8
a critical point, also called a critical state, specifies the conditions (temperature, pressure
and sometimes composition) at which a phase boundary ceases to exist

9. Example1:
To gain more experience with Tables T-4 and T-5
verify the following: Table T-4 gives the specific
volume of water vapor at 10.0 MPa and 600oC as
0.03837 m3/kg. At 10.0 MPa and 100oC, Table T-5
gives the specific volume of liquid water as
1.0385×10-3 m3/kg. Table T-4E gives the specific
volume of water vapor at 500 lbf/in.2 and 600oF as
1.158 ft3/lb. At 500 lbf/in.2 and 100oF, Table T-5E
gives the specific volume of liquid water as
0.016106 ft3/lb.
By Meng Chamnan GIM 9

10. Example2:
Let us determine the specific volume of water vapor at a state
where p = 10 bar and T = 215oC. Shown in this Figure is a
sampling of data from Table T-4. At a pressure of 10 bar, the
specified temperature of 215oC falls between the table values
of 200 and 240oC, which are shown in bold face. The
corresponding specific volume values are also shown in bold
face. Determine the specific volume v corresponding to
215oC.
By Meng Chamnan GIM 10

11. Evaluating Specific Volume
By Meng Chamnan GIM 11
   1 f g f g fv x v x v v x v v      
liq vapV V V 
liq vapV VV
v
m m m
  
liq vap
f g
m m
v v v
m m
   
    
   
liq liq fV m v 
vap vap gV m v 
/ 1liqm m x /vapx m m

12. Example3:
Consider a system consisting of a two-phase
liquid–vapor mixture of water at 100oC and a quality
of 0.9. What is the specific volume of the mixture?
By Meng Chamnan GIM 12

13. Example4:
Let us determine the pressure of water at each of three states
defined by a temperature of 100oC and specific volumes,
respectively, of v1 = 2.434 m3/kg, v2 = 1.0 m3/kg, and v3 =
1.0423 × 10-3 m3/kg. And try to find the vapor quality, x, at
state 2?
By Meng Chamnan GIM 13

14. Example5:
A closed, rigid container of volume 0.5 m3 is placed on a hot plate.
Initially, the container holds a two-phase mixture of saturated liquid
water and saturated water vapor at p1 = 1 bar with a quality of 0.5.
After heating, the pressure in the container is p2 = 1.5 bar. Indicate
the initial and final states on a T–v diagram, and determine:
(a) the temperature, in oC, at each state.
(b) the mass of vapor present at each state, in kg.
(c) If heating continued, determine the pressure, in bar, when the
container holds only saturated vapor.
By Meng Chamnan GIM 14

15. Evaluating Pressure, Specific Volume,
and Temperature inside the dome
By Meng Chamnan GIM 15
   1 f g f g fu x u x u u x u u      
   1 f g f g fh x h x h h x h h      
h u pv  h u pv 
H U pV 

16. Example6:
We determine the specific enthalpy of Refrigerant
134a when its temperature is 12oC and its specific
internal energy is 132.95 kJ/kg. Referring to Table
T-6, the given internal energy value falls between uf
and ug at 12oC, so the state is a two-phase liquid–
vapor mixture.
By Meng Chamnan GIM 16

17. Example7:
A well-insulated rigid tank having a volume of 10 ft3
contains saturated water vapor at 212oF. The water
is rapidly stirred until the pressure is 20 lbf/in.2.
Determine the temperature at the final state, in oF,
and the work during the process, in Btu.
By Meng Chamnan GIM 17

18. Example8:
Water contained in a piston–cylinder assembly undergoes two processes
in series from an initial state where the pressure is 10 bar and the
temperature is 400oC.
Process 1–2: The water is cooled as it is compressed at a constant
pressure of 10 bar to the saturated vapor state.
Process 2–3: The water is cooled at constant volume to 150oC.
(a) Sketch both processes on T–v and p–v diagrams.
(b) For the overall process determine the work, in kJ/kg.
(c) For the overall process determine the heat transfer, in kJ/kg.
By Meng Chamnan GIM 18

19. Evaluating Specific Heats cv and cp
By Meng Chamnan GIM 19
v
v
u
c
T
 
 
 
p
p
h
c
T
 
 
 
p
v
c
k
c
 property k, called the specific heat ratio
kJ/kg ∙ K or kJ/kmol ∙ K or Btu/lb ∙ oR or Btu/lbmol ∙ oR

20. Universal Gas Constant
20By Meng Chamnan GIM
8.314 kJ/kmol K
1.986 Btu/lbmol R
1545 ft lbf/lbmol R
R


 
  

21. Evaluating Properties Using the
Ideal Gas Model
By Meng Chamnan GIM 21
pV mRT pV nRT
Ideal Gas Model
pv RT
( ) ( )h h T u T RT  

22. Example9:
One pound of air undergoes a thermodynamic cycle
consisting of three processes.
Process 1–2: constant specific volume
Process 2–3: constant-temperature expansion
Process 3–1: constant-pressure compression
At state 1, the temperature is 540oR, and the
pressure is 1 atm. At state 2, the pressure is 2 atm.
Employing the ideal gas equation of state,
(a) sketch the cycle on p–v coordinates.
(b) determine the temperature at state 2, in oR.
(c) determine the specific volume at state 3, in ft3/lb.
By Meng Chamnan GIM 22

23. Polytropic Process for Ideal Gas
23By Meng Chamnan GIM
constantn
pV  1 1 2 2
2 1
1 2
n n
n
pV p V
p V
p V

 
  
 
When n= 0, the process is an isobaric (constant-pressure) process, and when n=
±∞ , the process is an isometric (constant-volume) process
n is a constant

24. Ideal gas equation of state
By Meng Chamnan GIM 24

25. Example10:
Air undergoes a polytropic compression in a piston–
cylinder assembly from p1 = 1 atm, T1 = 70oF to p2 = 5
atm. Employing the ideal gas model, determine the work
and heat transfer per unit mass, in Btu/lb, if n = 1.3.
By Meng Chamnan GIM 25