Transcript No Slide Title

Thermodynamics Lecture
Series
Pure substances – Property
tables and Property
Diagrams & Ideal Gases
Applied Sciences Education Research Group
(ASERG)
Faculty of Applied Sciences
Universiti Teknologi MARA
email: [email protected]
http://www5.uitm.edu.my/faculties/fsg/drjj1.html
CHAPTER
2
Pure substance
Properties of Pure
Substances- Part 2
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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
Introduction
Objectives:
1. Choose the right property table to read and to
determine phase and other properties.
2. Derive and use the mathematical relation to
determine values of properties in the wet-mix
phase
3. Sketch property diagrams with respect to the
saturation lines, representing phases, processes
and properties of pure substances.
Introduction
Objectives:
4. Use an interpolation technique to determine
unknown values of properties in the
superheated vapor region
5. State conditions for ideal gas behaviour
6. Write the equation of state for an ideal gas in
many different ways depending on the units.
7. Use all mathematical relations and skills of
reading the property table in problem-solving.
Example: A steam power cycle.
Steam Power Plant
Combustion
Products
Steam
Turbine
Fuel
Air
Pump
Mechanical Energy
to Generator
Heat
Exchanger
Cooling Water
System Boundary
for Thermodynamic
Analysis
Phase Change of Water - Pressure Change
T = 30 C
T = 30 C
P = 100 kPa
P = 4.246 kPa
P, kPa
1 = f@ 30 °C
100
2 = f @ 30 °C
H2O:
C. liquid
H2O:
Sat. liquid
4.246
, m3/kg
2 = f@30 °C
1
Tsat@100 kPa = 99.63 C
Psat@30 C = 4.246 kPa
Water when pressure is reduced
Phase Change of Water - Pressure Change
T = 30 C
P, kPa
P = 4.246 kPa
1 = f@ 30 °C
100
2 = f@30 °C
3 = [f + x f g]@ 30 °C
Sat. Vapor
H2O:
Sat. liquid
4.246
H2O:
Sat. Liq.
, m3/kg
3
2 = f@ 30 °C
1
Tsat@100 kPa = 99.63 C
Psat@30 C = 4.246 kPa
Water when pressure is reduced
Phase Change of Water - Pressure Change
T = 30 C
1 = f@ 30 °C
P, kPa
2 = f@ 30 °C
P = 4.246 kPa
3 = [f + x f g]@ 30 °C
100
Sat. Vapor
H2O:
Sat. Liq.
4 = g@ 30 °C
4.246
H2O:
Sat. Vapor
, m3/kg
4 = g@ 30 °C
3
2 = f@ 30 °C
1
Tsat@100 kPa = 99.63 C
Psat@30 C = 4.246 kPa
Water when pressure is reduced
Phase Change of Water - Pressure Change
T = 30 C
T = 30 C
P = 4.246 kPa
P = 2 kPa
1 = f@ 30 °C
P, kPa
2 = f@ 30 °C
3 = [f + x f g]@ 30 °C
100
4 = g@ 30 °C
5= @2kPa, 30 °C
4.246
, m3/kg
5
4 = g@ 30 °C
3
2 = f@
30 °C
f@100
Tsat@100 kPa = 99.63 C
Psat@30 C = 4.246 kPa
2
kPa
1
H2O:
Sat. Vapor
H2O:
Super
Vapor
Water when pressure is reduced
Phase Change of Water - Pressure Change
T = 30 C
T = 30 C
T = 30 C
T = 30 C
P = 100 kPa P = 4.246 kPa P = 4.246 kPa P = 4.246 kPa
T = 30 C
P = 2 kPa
Tsat@100 kPa = 99.63 C
Psat@30 C = 4.246 kPa
Sat. Vapor
H2O:
C. liquid
H2O:
Sat. liquid
H2O:
Sat. Liq.
H2O:
Sat. Vapor
Water when pressure is reduced
H2O:
Super
Vapor
Phase Change of Water- Pressure Change
P, kPa
Tsat@100 kPa = 99.63 C
Psat@30 C = 4.246 kPa
100
1 = f@ 30 °C
2 = f@ 30 °C
3 = [f + x f g]@ 30 °C
4 = g@ 30 °C
5= @2kPa, 30 °C
4.246
2
, m3/kg
5
4 = g@ 30 °C
3
2 = f@ 30 °C
1
Compressed liquid: Good
estimation for properties
by taking y = yf@T where
y can be either , u, h or
s.
Phase Change of Water
P, C
1,553.8
101.35
1.2276
, m3/kg
g@100 C
f@100 C
Phase Change of Water
P, C
P-  diagram
with respect to
the saturation
lines
22,090
1,553.8
101.35
1.2276
, m3/kg
g@100 C
f@100 C
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 2-19
P-v diagram
of a pure
substance.
2-4
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
FIGURE 2-25
P-T diagram
of pure
substances.
2-7
Property Table
Saturated water – Temperature table
Temp
Tsat, C
Sat. P.
Specific volume,
m3/kg
Specific internal energy,
kJ/kg
P, kPa f, m3/kg g, m3/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg
10
1.2276 0.001000
106.38
42.00
2347.2
2389.2
50
12.349 0.001012
12.03
209.32
2234.2
2443.5
P, MPa
100
0.10235 0.001044
1.6729
418.94
2087.6
2506.5
200
1.5538 0.001157
0.13736
850.65
1744.7
2593.3
300
8.581 0.001404
0.02167
1332.0
1231.0
2563.0
22.09 0.003155 0.003155
2029.6
0
2029.6
374.14
Property Table
Saturated water – Temperature table
Specific internal energy,
Specific enthalpy, kJ/kg
Temp
kJ/kg
Tsat, C uf, kJ/kg ufg, kJ/kg ug, kJ/kg hf, kJ/kg hfg, kJ/kg hg, kJ/kg
10
42.00
2347.2
2389.2
42.01
2377.7
2519.8
50
209.32
2234.2
2443.5
209.33
2382.7
2592.1
100
418.94
2087.6
2506.5
419.04
2257.0
2676.1
300
1332.0
1231.0
2563.0
1332.0
1940.7
2793.2
374.14
2029.6
0
2029.6
2099.3
0
2099.3
Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal
energy and flow energy. Latent heat of vaporization.
T – v diagram - Example
P, kPa
T,  C
Tsat, C
Psat, kPa
, m3/kg
50
70
81.33
31.19
f@70 C
Phase, Y?
T, C
50 kPa
Compressed Liquid,
T < Tsat or P > Psat
81.3
70
, m3/kg
0.001030
=f@70 C = 0.001023
3.240
P – v diagram - Example
P, kPa
T,  C
Psat, kPa
Tsat, C
, m3/kg
50
70
31.19
81.33
f@70 C
Phase, Y?
P, kPa
Compressed Liquid,
P > Psat or T < Tsat
50
31.19
, m3/kg
0.001023
=f@70 C = 0.001023
5.042
P – v diagram - Example
P, kPa T,  C
200
400
Phase, Why?
Psat, kPa
Tsat, C
, m3/kg
Sup. Vap., T >Tsat
NA
120.2
1.5493
P, kPa
P-  diagram
with respect to
the saturation
lines
22,090.0
200
f@200 kPa
g@200 kPa
= 0.001061
= 0.8857
, m3/kg
 = 1.5493
T – v diagram - Example
P, kPa
u, kJ/kg
1,000
2,000
Psat, kPa
Tsat, C
T,  C
179.9
179.9
T, C
Phase, Why?
Wet Mix., uf < u < ug
T-  diagram
with respect to
the saturation
lines
374.1
179.9
 = [f + x f g]@1,000 kPa
f@1,000 kPa
= 0.001127

, m3/kg
g@1,000 kPa
= 0.19444
Saturated Liquid-Vapor Mixture
Given the pressure, P, then
T = Tsat, yf < y <yg
Vapor Phase:, Vg, mg, g, ug, hg
Liquid Phase:, Vf, mf, f, uf, hf
Mixture:, V, m, , u, h, x
V  V f  Vg
mt  m f  mg
Specific volume of
mixture?? Since V=m 
Sat. Vapor
H2O:
Sat. Liq.
Mixture’s quality
x
mg
mt
More vapor, higher quality
x = 0 for saturated liquid
x = 1 for saturated vapor
mt avg  m f  f  mg g
Saturated Liquid-Vapor Mixture
Given the pressure, P, then
T = Tsat, yf < y <yg
Vapor Phase:, Vg, mg, g, ug, hg
Sat. Vapor
H2O:
Sat. Liq.
Liquid Phase:, Vf, mf, f, uf, hf
Mixture:, V, m, , u, h, x
Mixture’s quality
mt avg  m f  f  mg g  mt  mg  f  mg g
 avg
 mg
 1 
 mt
x
mg
mt
Divide by
total mass, mt

 f  x g   f  x g  x f    x
f
fg

where  fg   g  f
Saturated Liquid-Vapor Mixture
Given the pressure, P, then
T = Tsat, yf < y <yg
Vapor Phase:, Vg, mg, g, ug, hg
Sat. Vapor
H2O:
Sat. Liq.
Liquid Phase:, Vf, mf, f, uf, hf
Mixture:, V, m, , u, h, x
   f  x fg
where
y  y f  xy fg
where
Mixture’s quality
 fg   g  f
y fg  yg  y f
x
mg
mt
y can be , u, h
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.
Interpolation: Example – Refrigerant-134a
P, kPa , m3/kg
200
0.10600
Phase, Why?
Psat, kPa
Tsat, C
T,  C
Sup. Vap.,  >  g
-
-10.09
??
Assume properties are linearly dependent.
T, C Perform interpolation in superheated vapor phase.
slope m1  slope m2
T  TL TH  TL

  L  H  L
TH
T = ??

TH  TL   L 
T  TL 
 H  L 
m2
m1
, m3/kg
TL
L

H
Interpolation: Example – Refrigerant-134a
P, kPa , m3/kg
200
0.10600
Phase, Why?
Psat, kPa
Tsat, C
T,  C
Sup. Vap.,  >  g
-
-10.09
??
T,  C
, m3/kg
u, kJ/kg
TL= 0
L = 0.10438
uL = 229.23
0 < TL < TH
 = 0.10600
0 < uL < uH
TH= 10
H = 0.10922
uH = 237.05

TH  TL   L 
T  TL 
 H  L 
Assume
properties
are linearly
dependent.
Perform
interpolation
in
superheated
vapor phase.

uH  uL   L 
u  uL 
 H  L 
Interpolation: Example – Refrigerant-134a
P, kPa , m3/kg
200
0.10600
Phase, Why?
Psat, kPa
Tsat, C
T,  C
Sup. Vap.,  >  g
-
-10.09
3.35
u, kJ/kg
231.85
P, kPa
P-  diagram
with respect to
the saturation
lines
22,090.0
Psat ??
200
f@200 kPa
g@200 kPa
= 0.0007532
= 0.0993
 = 0.10600
, m3/kg
Property Table
Saturated water – Temperature table
Temp
Tsat, C
Sat. P.
Specific volume,
m3/kg
Specific internal energy,
kJ/kg
P, kPa f, m3/kg g, m3/kg uf, kJ/kg ufg, kJ/kg ug, kJ/kg
10
1.2276 0.001000
106.38
42.00
2347.2
2389.2
50
12.349 0.001012
12.03
209.32
2234.2
2443.5
P, MPa
100
0.10235 0.001044
1.6729
418.94
2087.6
2506.5
200
1.5538 0.001157
0.13736
850.65
1744.7
2593.3
300
8.581 0.001404
0.02167
1332.0
1231.0
2563.0
22.09 0.003155 0.003155
2029.6
0
2029.6
374.14
Property Table
Saturated water – Temperature table
Specific internal energy,
Specific enthalpy, kJ/kg
Temp
kJ/kg
Tsat, C uf, kJ/kg ufg, kJ/kg ug, kJ/kg hf, kJ/kg hfg, kJ/kg hg, kJ/kg
10
42.00
2347.2
2389.2
42.01
2377.7
2519.8
50
209.32
2234.2
2443.5
209.33
2382.7
2592.1
100
418.94
2087.6
2506.5
419.04
2257.0
2676.1
300
1332.0
1231.0
2563.0
1332.0
1940.7
2793.2
374.14
2029.6
0
2029.6
2099.3
0
2099.3
Enthalpy H = U+PV, kJ. or h=u+P, kJ/kg: Sum of internal
energy and flow energy. Latent heat of vaporization.
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.
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
Water
30
Water
300
Tsat, C
f, m3/kg
g,m3/kg
(12 marks)
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
0.001022
5.229
Water
300
133.55
0.001073
0.6058
(12 marks)
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.]
Sample Questions
Substance
P
kPa
Psat@T
kPa
Water
200
-
T
C
Tsat@
P
C
Quali
ty
x
u
kJ/kg
Phase
&
Reaso
n
0.5
(5 marks)
Sample Questions
ANSWER:
Substance
Water
P
kPa
200
Psat@T
kPa
-
T
C
120.2
3
Tsat@
P
C
120.
23
Quali
ty
x
0.5
u
kJ/kg
u=uf+x
ufg
Phase
&
Reaso
n
Wet
mix
0<x
<1
(5 marks)
Sample Questions
iii) Refrigerant 134a
P
kPa
200
Psat@T
kPa
T
C
Tsat@P
C
Quality
x
u
kJ/kg
Phase
&
Reason
-16
(6 marks)
Sample Questions
ANSWER:
P
kPa
200
Psat@T
kPa
157.48
T
C
-16
Tsat@P
C
-10.09
Quality
x
NA
u
kJ/kg
Phase
&
Reason
uf@-16C
Comp.
Liquid
T<Tsat or
P>Psat
= 29.18
(6 marks)
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.
Sample Questions
Accepted pairs of saturated values and must be marked
correctly on the graph in m3/kg are:
P, kPa
f@200 kPa = 0.0007532
g @200 kPa = 0.0993
f@-16 C = 0.0007428
g @-16 C = 0.1247
200
-10.09 C
157.48
-16 C
, m3/kg
 = f =0.0007428
g = 0.1247
(8 marks)
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.
Sample Questions
Accepted pairs of saturated values and must be marked correctly on the
graph in m3/kg are:
f@100 C = 0.001044
f@400 kPa = 0.001084
g @100 C = 1.6729
g @400 kPa = 0.4625
f@50 kPa = 0.001030
g @50 kPa = 3.240
T, C
400 kPa
101.35 kPa
50 kPa
143.63
1
2
100
81.33
1 = f@100 C
= 0.001044
1 = 0.001084 0.4625
2 = 3.418
, m3/kg
(12 marks)
Ideal Gases
Properties of Pure
Substances- Ideal Gases
Equation of State
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
Ideal Gases
• Equation of State for ideal gas
– Boyle’s Law: Pressure of gas is inversely P  1 / V
proportional to its specific volume P
• Equation of State for ideal gas
– Charles’s Law: At low pressure, volume is
V T
proportional to temperature
Ideal Gases
• Equation of State for ideal gas
– Combining Boyles and Charles laws: P  RT , kJ / kg
Ru
where gas constant R is R 
M
and where M is molar mass
and where Ru is universal gas constant Ru = 8.314 kJ/kmol.K
EQOS: Since the total volume
is V = m, so :  = V/m
EQOS: since the mass m = MN
where N is number of moles:
So,
V
P  RT
m
So, PV  mRT, kJ
Ru
PV  MN
T
M
PV  NRuT , kJ
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
P  RT , kJ / kg
where,
PV  mRT, kJ PV  NRuT , kJ
Ru = 8.314 kJ/kmol.K, m = MN V = m and
Ru
R
M
Gas Mixtures – Ideal Gases
Real Gases
Low density (mass in 1 m3)
gases
Molecules are further apart
High density
Real gases satisfying condition
Pgas << Pcrit; Tgas >> Tcrit ,
have low density and can
be treated as ideal gases
Low density
Molecules far
Gas Mixtures – Ideal Gases
Real Gases
Equation of State - P--T
behaviour
High density
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.
Low density
Molecules far apart
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
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 = 8.314 kJ/kmolK
High density
Low density
Equations of States
Other EQOS
Van Der Waals Equation of State
Considers intermolecular interactions,
a/2, which then increases the pressure:
P replaced by P+a/2
Considers volume occupied by
molecules, b, at high pressures: Replace
 by  -b, then
High density
a 

EQOS is:  P  2   b   RT
 

2
where
2
27R Tcr
RTcr
a
; b
64Pcr
8Pcr
Low density
T – v diagram - Example
P, kPa
u, kJ/kg
1,000
2,000
Psat, kPa
Tsat, C
T,  C
179.9
179.9
T, C
Phase, Why?
Wet Mix., uf < u < ug
T-  diagram
with respect to
the saturation
lines
374.1
179.9
 = [f + x f g]@1,000 kPa
f@1,000 kPa
= 0.001127

, m3/kg
g@1,000 kPa
= 0.19444