Chapter Two Properties of Pure Substances

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Chapter TwoProperties of Pure Substances

• Ted Huddleston, Ph.D., P.E.
• Dept of Chemical Engineering
• University of South Alabama
• Mobile, Alabama

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Properties

• Tables
• Charts
• Equations

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Enthalpy
H U PV Enthalpy is an extensive
property Units? PV has units of
Divide by total mass to obtain specific values.
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Enthalpy
Enthalpy is a thermodynamic state property.
H U PV h
u Pv Do not concern yourself with its real
meaning or physical significance. It is
simply a property that is defined in the above
equations and makes many calculations
simpler. Page 77
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One Saturated Systems
Mixture of Liquid and Vapor both phases can
coexist. Temperature and Pressure are not
independent. Each is a monotonically increasing
function of the other. At a given temperature,
the corresponding saturation pressure is also
known as the vapor pressure. Page 78
T P
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Saturated System
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Saturated System
V mf vf mg vg where mf is mass in liquid
phase vf is specific volume (m3/kg) of
liquid mg is mass in vapor phase vg is
specific volume (m3/kg) of vapor Also V m
v Where m is total mass mf mg
v total volume / total mass or the
average specific volume of tank
contents.
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Saturated System
V mf vf mg vg Now divide both sides by
total mass m mf mg.
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Saturated System
v (1-x) vf x vg 0 lt x lt1 x
is the quality of the system and is the mass
fraction of the system that is in the vapor
phase. (1-x) is sometimes called the wetness
and is the mass fraction of the system that is in
the liquid phase. v V / mass, the average
specific volume of the container contents.
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Saturated System
v (1-x) vf x vg Define vfg vg
vf Thus v (1-x) vf x vg vf x vf x vg
vf x(vg vf) vf x vfg v (1-x) vf x
vg vf x vfg Specific enthalpy, specific
internal energy, and specific entropy for
saturated mixtures can be written in the same
form. h (1-x) hf x hg hf x hfg u
(1-x) uf x ug uf x ufg s (1-x) sf x
sg sf x sfg
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A partial list of Table A4.
Page 830
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Saturated System
Table A-4 - page 830
Saturated Water Temperature table
Temperature is listed in the left column as the
independent variable. Table A-5 - page
832 Saturated Water Pressure table
Pressure is listed in the left column as
the independent variable. Use
whichever table is convenient.
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Saturated System Example
A tank contains 10 kg of water. The pressure is
0.200 MPa and the quality is 0.20 Find the
temperature and the total internal energy in the
tank. Solution From Table A-5 Tsat
120.23 oC uf 504.49 kJ/kg
ug 2529.5 kJ/kg u (1-x)uf x ug
(1-0.2) 504.49 (0.2)2529.5 909.49
kJ/kg U 10 kg 909.49 kJ/kg
9094.9 kJ
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Saturated System Example (cont)
Find the volume of the tank. vf
0.001061 vg 0.8857 m3/kg
Vf 8 kg (vf) 0.008488 m3 Vg 2 kg
(vg) 1.771 m3 V Vf Vg 1.779 m3
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Saturated System Example (cont)
Find the internal thermal energy of each phase.
uf 504.49 kJ/kg ug
2529.5 kJ/kg Uf 8 kg (504.49 kJ/kg)
4035.9 kJ Ug 2 kg (2529.5 kJ/kg) 5059.0
kJ U Uf Ug 4035.9 5059.0 9094.9
kJ
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Saturated Systems - Interpolation
What is the saturated pressure for water
corresponding to a temperature of 143 oC ? From
Table A4 Temp Sat Press
140 0.3615 145
0.4154 We need to interpolate.
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Linear Interpolation
x1 y1 x y ? x2 y2 Given
table entries x1, x2, y1, y2, and given the
value x such that x1 lt x lt x2, we calculate
the value of y by ratio and proportion.
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Saturated Systems - Interpolation
Temp Sat Press 140
0.3615 143 145
0.4154
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Two Superheated Systems
The system is all vapor. There is no liquid.
P and T are independent. Page 83.
P T
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A partial listing of Table A6.
Page 834
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Superheated System - Example
What is the specific enthalpy of water at 0.80
MPa and 400 oC? From Table A-6 page 834
h 3267.1 kJ/kg
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Three Compressed Liquid
Tank contains only liquid there is no vapor
phase. Temperature and pressure are independent.
Properties vary with temperature, but
liquid properties vary hardly at all with
pressure. Page 84
P T
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Compressed Liquid
Because properties are so insensitive to pressure
changes the following approximation is often used
for properties of compressed liquids. Given a
compressed liquid at T and P, use the property
of the saturated liquid at the given
temperature. u uf _at_T v vf _at_T h hf _at_T
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Compressed Liquid
For very high pressures, there is a table of
properties see Table A-7 on page 838. The
value of enthalpy at T and P at high pressures
can be reasonably approximated by h
hf_at_T vf_at_T(P Psat_at_T) However, for this
class (EG 270 Huddleston) the approximation on
the previous slide is sufficient.vf_at_T(P

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Compressed Liquid-Example
What is the specific internal energy of water at
100 oC and 240 kPa? At 100 oC the saturation
pressure of water is 101.35 kPa from Table A-4.
The pressure is higher than the saturation
pressure. This system is a compressed
liquid. Find uf of water at 100 oC from Table
A-4. u uf at 100 418.94 kJ/kg
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Example 2-9 page 85
Note that in each part of the problem, values of
two properties are given. We need two pieces
of information to use the tables. The State
Postulate page 12 The state of a
simple compressible system is completely
specified by two independent, intensive
properties.
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Example 2-9 page 85
(a) P 200 kPa or 0.200 MPa
x 0.6
Quality is 0.6, which means the system is a
saturated mixture of liquid and vapor. Use Table
A-5. Tsat 120.23 oC
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Example 2-9 (a) (cont)P 0.2 MPa x 0.6
Find specific internal energy. u
(1-x) uf x ug From Table A-5 uf
504.49 ug 2529.5 kJ/kg u
(1-0.6) 504.49 (0.6)2529.5 1719.5 kJ/kg
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Example 2-9 (B)T 125 oC u 1600 kJ/kg

• Procedure Go to Saturated Water Temperature
  Table (Table A-4) and read values of uf and ug.
• If u lt uf system is compressed liquid (page
  84)
• If uf lt u lt ug system is saturated a mixture
  of liquid and vapor (page 78)
• If ug lt u system is superheated (page 83)

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Example 2-9 (b) (cont)T 125 oC u 1600
kJ/kg
From Table A-4 uf 524.74
ug 2534.6 kJ/kg The value u 1600 is
between the value of a saturated liquid and the
value of a saturated vapor. We have a saturated
mixture of liquid and vapor phases. Calculate
the quality. u (1-x)uf x ug can be
solved for x (u-uf)/(ug-uf) x
(1600-524.74)/(2534.6-524.74) 0.535 x
0.535 Also Psat 0.2321 MPa
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Example 2-9 ( c)P 1 MPa u 2950
kJ/kg
Again we do not know what phases are present. We
read uf and ug from Table A-5 and apply the same
tests as we did in part (b). uf 761.68
ug 2583.6 kJ/kg u gt ug which
means the system is superheated. Only a dry
vapor phase is present. All the mass is in the
vapor phase but the quality x does not apply
here. The quality x is used only for saturated
mixtures. Use Table A-7. On page 835 in the
block for P 1.00 MPa the u value falls
between 400 and 500 oC. We must interpolate.
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Example 2-9 ( c)P 1 MPa u 2950 kJ/kg
T u 350
2875.2 ? 2950 400 2957.3
Linear Interpolation
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Example 2-9 (d)T 75 oC P 500 kPa

• From Table A-4 at 75 oC
  Psat 38.58 kPa. Psat lt 500
  Therefore system is a compressed liquid. (see
  page 84).
• From Table A-5 at 500 kPa (0.5 MPa) Tsat
  151.86 oC 75 oC lt Tsat Therefore
  system is a compressed liquid. (see page 84).

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Example 2-9 (e)P 850 kPa x 0.0
A quality of zero is given. This means that
conditions exist such that a mixture of liquid
and vapor could exist but here the mixture has
zero vapor. The system is a saturated liquid.
Use Table A-5. At P 850 kPa or 0.85 MPa
u uf 731.27 kJ/kg
T Tsat 172.96 oC.
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Example TH 101Water

• T 390 oC P 2.8 MPa h ?
• Table A-5 P 3.0 MPa Tsat 233.90 oC System
  is superheated.
• Table A-6 has entries at pressures of 2.5
  MPa and 3.0 MPa and at temperatures of 350 oC and
  400 oC.

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Example TH 101 (cont)T 390 oC P 2.8 MPa
h ?
P 2.5
P 3.0 T 350 3126.3
3115.3 T 400
3239.3 3230.9
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Example TH 101 (cont)T 390 oC P 2.8 MPa
h ?
P 2.5 P 2.8
P 3.0 T 350 3126.3
3119.7 3115.3 T 400
3239.3
3230.9 At 350 oC interpolate to find h at 2.8
MPa.
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Example TH 101 (cont)T 390 oC P 2.8 MPa
h ?
P 2.5 P 2.8
P 3.0 T 350 3126.3
3119.7 3115.3 T 400
3239.3 3234.3 3230.9 At 400
oC interpolate to find h at 2.8 MPa.
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Example TH 101 (cont)T 390 oC P 2.8 MPa
h ?
P 2.5 P 2.8
P 3.0 T 350 3126.3
3119.7 3115.3 T 390
3211.4 T 400
3239.3 3234.3 3230.9 At 2.8
MPa interpolate to find h at 390 oC.
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T-v diagram of a pure substance
Page 72
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P-v diagramof a pure substance
Page 72