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TRANSCRIPT
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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 Sciences
Universiti Teknologi MARA
Pure substances
Propertytables and Property
Diagrams
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Quotes
You do not really understand somethingunless you can explain it to yourgrandmother.
(Albert Einstein)
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Introduction
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 thesaturation lines, representing phase and
properties of pure substances.
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FIGURE 15
Some application areas of
thermodynamics.
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
1-1
Application
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Example: A steam power cycle.
Steam
TurbineMechanical Energy
to Generator
Heat
Exchanger
Cooling Water
Pump
Fuel
Air
Combustion
Products
System Boundary
for Thermodynamic
Analysis
Steam Power Plant
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Steam Power Plant
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FIGURE 117
A 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
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Open system devices
Heat ExchangerThrottle
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CHAPTER
2
Properties of Pure
Substances
Title:
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Pure Substances
Pure substances Substance with fixed chemical composition
Can be single element: Such as, N2, H2, O2
Compound: Such as Water, H2O, C4H10,
Mixture such as Air,
2-phase system such as H2O.
Responsible for the receiving and removing dynamic
energy (working fluid)
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Phase Change of Water
H2O
Sat.
liquid
Qin
P = 100 kPa
T = 99.6 C
Water 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
Qin
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Phase Change of Water
H2O
Sat.
liquid
Qin
P = 100 kPaT = 99.6 C
Water interacts with thermal energy
H2O:Sat. Liq.
Sat. Vapor
Qin
99.6
2=f@100
kPa
T, C
30
, m3/kg1
3
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Phase Change of Water
Water interacts with thermal energy
4=g@100kPa
99.6
2=f@100
kPa
T, C
30
, m3/kg
1
3
P = 100 kPa
T = 99.6 C
H2O:
Sat. Vapor
Qin
H2O:Sat. Liq.
Sat. Vapor
Qin
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Phase Change of Water
Water interacts with thermal energy
150
5
99.6
2=f@100kPa
T, C
30
, m3/kg
1
4=g@100kPa
3
5 = @100 kPa, 150C
3 = [f+ x fg]@100kPa1 = f@T1
H2O:
Super
Vapor
P = 100 kPa
T = 150 C
Qin
P = 100 kPa
T = 99.6 C
H2O:
Sat. Vapor
Qin
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Phase Change of Water
Water interacts with thermal energy
H2O:Sat. Liq.
Sat. Vapor
P = 100 kPa
T = 99.6 C
Qin
P = 100 kPa
T = 99.6 C
H2O:
Sat. Vapor
Qin
P = 100 kPa
T = 150 C
H2O:
Super
Vapor
Qin
P = 100 kPa
T = 30 C
H2O:C. liquid
Qin
P = 100 kPa
T = 99.6 C
H2O
Sat.
liquid
Qin
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Phase Change of Water
99.6
2=f@100kPa
T, C
30, m3/kg
1 4=g@100kPa
3
5 = @100 kPa, 150C
3
= [f
+ x f
g
]@100kPa
1 = f@T1
150
5 Compressed liquid: Good
estimation for properties
by taking y = yf@T where
y can be either , u, h or
s.
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Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2-1
FIGURE 2-11
T-v diagram for the
heating process of
water at constant
pressure.
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Phase Change of Water
T, C
, m3
/kg
99.6
f@100kPa
g@100kPa
179.9
45.8
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Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2-2
FIGURE 2-16
T-v diagram of
constant-
pressure
phase-changeprocesses of a
pure
substance at
various
pressures
(numerical valuesare for water).
99.6
45.8
179.9
Tv diagram: Multiple P
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Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
2-3
FIGURE 2-18
T-v diagram of a
pure substance.
Tv diagram: Multiple P
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T, C
, m3/kg
Tv 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
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Tv diagram - Example
T, C
, m3/kg
f@200 kPa
= 0.001061
P, kPa , m3/kg
200 1.5493
T- diagram
with respect to
the saturation
lines
Phase, Why?
Sup. V., >g
Psat
, kPa Tsat
, C
120.2
374.1
400
= 1.5493
120.23
g@200 kPa
= 0.8857
T, C
400
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Tv diagram - Example
T, C
, m3/kg
P, kPa u, kJ/kg
1,000 2,000
T- diagram
with respect to
the saturation
lines
Phase, Why?
Wet Mix., uf < u < ug
Psat
, kPa Tsat
, C
179.9
374.1
f@1,000 kPa
= 0.001127
179.9
g@1,000 kPa
= 0.19444
T, C
179.9
= [f+ x fg]@1,000 kPa
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Property Table
Saturated waterPressure table
Pressure
P, kPa
10
50
P, 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, m
3/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
81.33
99.63
179.91
311.06
374.14