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7/27/2019 Chapter 3 Lecture Thermodynamics
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Chapter 3
PROPERTIES OF PURESUBSTANCES
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Objectives
Introduce the concept of a pure substance.
Discuss the physics of phase-change processes.
Illustrate the P-v, T-v, and P-Tproperty diagrams and P-v-T
surfaces of pure substances.
Demonstrate the procedures for determining thermodynamic
properties of pure substances from tables of property data.
Describe the hypothetical substance ideal gas and the
ideal-gas equation of state.
Apply the ideal-gas equation of state in the solution of typical
problems.
Introduce the compressibility factor, which accounts for the
deviation of real gases from ideal-gas behavior.
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PURE SUBSTANCE
Pure substance: A substance that has a fixed (not
necessarily single) chemical composition throughout.Any homogeneous mixture is considered a pure substance.
Non-homogeneous mixture (oil and water) is not a pure
substance.
Nitrogen and gaseous air arepure substances. A mixture of liquid water and
gaseous water is a pure
substance, but a mixture of liquid
air and gaseous air is not.
AIR
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PHASES OF A PURE SUBSTANCE
A phase has a distinct molecular arrangement.
Three principal phases
Solid
Liquid
Gas
Phase change processes are Melting: solid liquid
Solidification: Liquid solid
Vaporization: Liquid gas
Condensation: Gas
liquid Internal energy of the substance changes during phase
change.
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PHASES OF A PURE SUBSTANCE
In many applications, we have a mixture of the two
phases (phase change).
Boiler, condenser, refrigerator, etc.
The properties we are interested in:
P Pressure
specific volume T Temperature
U internal energy
Cv, Cp specific heats
X quality (% of mass converted to vapor) We will learn to determine these properties during
thermodynamic processes.
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PHASE-CHANGE PROCESSES OF PURE
SUBSTANCES
Compressed liquid (subcooled liquid): A substance that it
is not about to vaporize. Saturated liquid: A liquid that is about to vaporize.
At 1 atm and 20C,
water exists in the
liquid phase(compressed l iqu id).
At 1 atm pressure
and 100C, water
exists as a liquid
that is ready to
vaporize
(saturated l iqu id).
Due to increased T, v also increases slightly as P remains constant
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Saturated vapor: A vapor that is about to condense.
Saturated liquidvapor mixture:The state at which the liquid andvapor phases coexistin equilibrium.
Superheated vapor: A vapor that is not about to condense (i.e., not a
saturated vapor).
As more heat is transferred,
part of the saturated liquid
vaporizes (saturated liqu idvapor mix ture).
At 1 atm pressure, the
temperature remainsconstant at 100C until the
last drop of liquid is vaporized
(saturated vapo r).
As more heat is
transferred, thetemperature of the
vapor starts to rise
(superheated vapor).
v5 > v4 > v3 > v2 > v1
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Saturation Temperature and Saturation Pressure The temperature at which water starts boiling depends on the pressure;
therefore, if the pressure is fixed, so is the boiling temperature.
Saturation temperature Tsat: The temperature at which a pure substance
changes phase at a given pressure.
Saturation pressure Psat: The pressure at which a pure substance changes
phase at a given temperature.
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Exercise
What is the saturation pressure for water at 270oC?
From Table A-4 (page 915)
Psat = 5503.0 kPa
Consider water at 270oC and 4000 kPa, is it liquid,
gas, or solid
Tsat = 250.35oC at P = 4000 kPa (A-5) so it is
vapor or gas state.
10
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PROPERTY DIAGRAMS FOR PHASE When water is kept under different pressures in a piston-cylinder
system, upon heating, water follows different process paths
As the pressure is increased, the saturation line continues to shrinkand it becomes a point, called critical point.
T-vdiagram of constant-
pressure phase-change
processes of a pure substance
at various pressures
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T-vdiagram of a pure substance.
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At supercritical pressures (P >Pcr),
there is no distinct phase-change
process.
Critical point: The point at
which the saturated liquid
and saturated vapor states
are identical.
For water: Table A-1 forother pure substances
Pcr= 22.06 MPa
Tcr= 373.95oC
Vcr= 0.003106 m3/kg
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P-vdiagram of a pure substance.
T-vdiagram of a pure substance.
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Exercise Plot T-v diagrams for water going through the following process
Isothermal process. T = 200oC, P1 = 1 MPa, P2 = 5 MPa
For T = 200oC, Psat = 1.5549 MPa (Table A-4) P1 < Psat, so initially (1) we have vapor, but P2>Psat, so at (2), we
have compressed liquid
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Exercise contd Plot P-v diagrams for water going through the following process
Isothermal process. T = 200oC, P1 = 1 MPa, P2 = 5 MPa
For T = 200oC, Psat = 1.5549 MPa (Table A-4) P1 < Psat, so initially (1) we have vapor, but P2>Psat, so at (2), we
have compressed liquid
16
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Exercise contd What are the initial and final specific volumes, v1 and v2
At state 1, we have superheated vapor then use Table A-6, page
918 Use T1 = 200
oC and P1 = 1 MPa to read v1 = 0.20602 m3/kg
At state 2, we have compressed liquid, then use Table A-7, page
922 Use T2 = 200
oC and P2 = 5 MPa to read v2 = 0.0011531 m3/kg
17
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Extending the
Diagrams to Include
the Solid Phase
P-vdiagram of a substance thatcontracts on freezing.
P-vdiagram of a substance that
expands on freezing (such as water).
At triple-point pressureand temperature, a
substance exists in three
phases in equilibrium.
For water,
Ttp = 0.01C
Ptp = 0.6117 kPa
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Sublimation: Passing from
the solid phase directly into
the vapor phase.
At low pressures (belowthe triple-point value),
solids evaporate without
melting first (sublimation).
P-Tdiagram of pure substances.
Table 3.3 for Triple Point temperatures and
pressures
Phase Diagram
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P-v-Tsurface of a substancethat contracts on freezing.
P-v-Tsurface of a substance thatexpands on freezing (like water).
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PROPERTY TABLES For most substances, the relationships among thermodynamic properties are too
complex , therefore, properties are frequently presented in the form of tables.
Some thermodynamic properties can be measured easily, but others cannot and
are calculated by using the relations between them and measurable properties.
EnthalpyA Combination Property
The combination
u +Pvis
frequently
encountered in
the analysis of
control volumes.
The productpressure
volume has energy units.
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Saturated Liquid and Saturated Vapor States
Table A4: Saturation properties of water under temperature.
Table A5: Saturation properties of water under pressure.
A partial list of Table A4.
Enthalpy of vaporization, hfg(Latent
heat of vaporization): The amount of
energy needed to vaporize a unit mass
of saturated liquid at a given
temperature or pressure.
E l A i id t k t i 50 k f
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Examples: A rigid tank contains 50 kg ofsaturated liquid at 90oC. Determine the
pressure in the tank and the volume of the
tank.
Solution:
m= 50 kg, saturated liquid water
T = 90oC
Find pressure from Table A-4
At T = 90oC Psat = 70.183 kPa
Find vfat 90oC: vf= 0.001036 m
3/kg
V = vf.m = (0.001036)(50) = 0.0518 m3
E amples A f 200 f t t d li id
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Examples: A mass of 200 g of saturated liquidwater is completely vaporized at a constant
pressure of 100 kPa. Determine (a) the volume
change and (b) the amount of energy transferred
to the water.
Solution: Table A-5 at 100 kPa
vfg = vg vf= 1.6941 0.001043 = 1.6931 m3/kg
Change in volume = m.vfg = 0.3386 m3
The amount of energy needed to vaporize a unitmass is the enthalpy of vaporization at that
pressure, which is hfg = 2257.5 kJ/kg for water at
100 kPa.
Thus, total energy: H = m.hfg
= (0.2 kg)(2257.5
kJ/kg) = 451.5 kJ
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Saturated LiquidVapor MixtureQuality, x: The ratio of the mass of vapor to the total mass of the mixture.
Quality is between 0 and 1 0: sat. liquid; 1: sat. vapor.
A two-phase system can be treated as a
homogeneous mixture for convenience.
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Quality is related
to the horizontal
distances on P-v
and T-v
diagrams.
The vvalue of a saturated
liquidvapor mixture lies
between the vfand vgvalues at
the specified TorP.
Vf= Specific volume of saturated liquid
Vg = Specific volume of saturated vapor
Vavg = Specific volume of saturated liquid-
vapor mixture
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y v, u, orh.
We can use quality x to determine the
enthalpy, h, and internal energy, u, of the
saturated liquid-vapor mixture
Review examples 3-4 page 128
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A rigid tank contains 10 kg of water at 90oC. If 8 kg of the water is in
the liquid form and the rest is in the vapor form, determine (a) the
pressure in the tank and (b) the volume of the tank.
Review examples 3-4, page 128
Review examples 3-5 page 129
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An 80L vessel contains 4 kg of refrigerant-134a at a pressure of 160
kPa. Determine (a) the temperature, (b) the quality, the enthalpy of
the refrigerant, and (d) the volume occupied by the vapor phase
Review examples 3 5, page 129
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Superheated VaporIn the region to the right of thesaturated vapor line and at
temperatures above the critical
point temperature, a substance
exists as superheated vapor.In this region, temperature and
pressure are independent
properties.
A partial
listing of
Table A6.
At a specified
P, superheated
vapor exists at
a higherh than
the saturatedvapor.
Compared to saturated vapor,
superheated vapor is characterized by
The compressed liquid properties
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Compressed LiquidCompressed liquid is characterized by
yv, u, orh
A more accurate relation forh
A compressed liquid
may be approximated
as a saturated liquid at
the given temperature.
At a given P
and T, a puresubstance will
exist as a
compressed
liquid if
The compressed liquid properties
depend on temperature much more
strongly than they do on pressure.
Table A-7 is the only table exists in
this text.
Interpolation
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Interpolation
In most applications, we cannot read the properties of water directly off the tables.
Interpolation Example
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Interpolation Example
Determine the temperature of water at P = 0.5 MPa and h = 2890 kJ/kg.
From Table A-5, we determine that we have superheated water vapor. Why?
Use Table A-6, superheated vapor at P = 0.5 MPa
T1 = 200oC h1 = 2855.8 kJ/kg
T2 = 250oC h2 = 2961.0 kJ/kg
(h-h1)/(h2-h1) = (T-T1)/(T2-T1)
h2 h1 = 105.2 kJ/kg
h h1 = 2890 2855.8 = 34.2 kJ/kg
(h-h1)/(h2 h1) = 34.2/105.2 = 0.3250
T2-T1 = 250oC-200oC = 50oC
(h-h1)/(h2-h1).(T2-T1) = 0.3250 (50) = 16.44oC
T = T1 + (h-h1)/(h2-h1).(T2-T1) = 200oC = 16.44oC = 216.44oC
Problem
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Problem
A piston-cylinder device contains 0.1m3 liquid water an 0.9m3 of watervapor in equilibrium at 800 kPa. Heat is transferred at constant pressureuntil T= 350oC.
(a) What is the initial temperature?
(b) Determine the total mass of water.
(c) Calculate the final volume.
(d) Show the process on a p-v diagram with respect to saturation lines.
P1 = 800 kPa P2 = 800 kPa
T1 = ? T2 = 350oC
Vapor
Liquid
(a) Initially two phases coexist, thus we have a saturated liquid-vapormixture. Then the temperature in the tank must be the saturationtemperature
T = Tsat@800kPa = 170.41oC Table A-5 Page 917
Problem
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Problem
(b) To find total mass we can add the mass of each phase
mf= Vf/vf= 0.1 m3/0.001115 m3/kg = 89.69 kg
mg= Vg/vg = 0.9 m3/0.2404 m3/kg = 3.74 kg
mTotal = mf+ mg = 93.43 kg
(c) At final state, water is superheated vapor. P2 = 800 kPa and T2 = 350oC
Using Table A-6, v2 = 0.3544 m3/kg, thus, V2 = mv2 = (93.43) (0.3544) =
0.33.1 m3
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Problem 3-26
Complete this table for H2O
36
T, C P, kPa h, kJ / kg x Phase description200 0.7
140 1800
950 0.0
80 500
800 3162.2
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Problem 3-26
Complete this table for H2O
37
T, C P, kPa h, kJ / kg x Phase description120.21 200 2045.8 0.7 Saturated mixture
140 361.53 1800 0.565 Saturated mixture
177.66 950 752.74 0.0 Saturated li quid
80 500 335.37 - - - Compressed liquid
350.0 800 3162.2 - - - Superheated vapor
THE IDEAL GAS EQUATION OF STATE
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THE IDEAL-GAS EQUATION OF STATE
Instead of using tables to obtain properties, using simple equations
(if we have them available) would be easier.
Equation of state: Any equation that relates the pressure, temperature,and specific volume of a substance.
The simplest and best-known equation of state for substances in the gas
phase is the ideal-gas equation of state. This equation predicts the P-v-T
behavior of a gas quite accurately within some properly selected region.
R: gas constant
M: molar mass (kg/kmol)
Ru: universal gas constant
Ideal gas equation of state
Different substances have differentgas constants.
P: absolute pressure
T: absolute temperature in Kelvin (K)V: Specific volume
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The ideal-gas relation often is
not applicable to real gases;
thus, care should be exercised
when using it.
Mass = Molar mass Mole number
Various expressions of idealgas equation
Ideal gas equation
at two states for afixed mass
Real gases behave as an ideal gas at low
densities (i.e., low pressure, high
temperature).
Is Water Vapor an Ideal Gas?
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Is Water Vapor an Ideal Gas?
At pressures below 10 kPa, watervapor can be treated as an ideal
gas, regardless of its temperature,with negligible error (less than 0.1percent).
At higher pressures, however, theideal gas assumption yieldsunacceptable errors, particularly in
the vicinity of the critical point andthe saturated vapor line.
In air-conditioning applications, thewater vapor in the air can betreated as an ideal gas. Why?
In steam power plant applications,
however, the pressures involvedare usually very high; therefore,ideal-gas relations should not beused.
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OF DEVIATION FROM IDEAL-GAS BEHAVIOR
The compressibility factor is
unity for ideal gases.
Compressibility factorZ
A factor that accounts for
the deviation of real gases
from ideal-gas behavior at
a given temperature and
pressure.
The farther away Zis from unity, the more the
gas deviates from ideal-gas behavior.
Gases behave as an ideal gas at low densities
(i.e., low pressure, high temperature).
Question: What is the criteria for low pressure
and high temperature?
Answer: The pressure or temperature of a gas
is high or low relative to its critical temperatureor pressure.
At very low pressures, all gases approach
ideal-gas behavior (regardless of their
temperature).
How to determine Z
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42Figure A-15 (page 932) Comparison ofZfactors for various gases.
Reduced
temperature
Reduced
pressurePseudo-reduced
specific volume
Zcan also be determined from
a knowledge ofPRand vR.
o to dete e
Deviation from Ideal Behavior
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At very low pressure, PR>1)
The deviation from ideal gas
behavior is greatest in the vicinity
of the critical point
Deviation from Ideal Behavior
Example 3-11
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p
Determine the specific volume v, of refrigerant -134a at P = 1 MPa and at
50oC.
(a) Use ideal gas equation
(b) Use compressibility chart
Solution
(a) We need to find gas constant, R, for refrigerant.
From Table A-1, R = 0.08149 kJ/kgK
Ideal gas equation, Pv = RT
v = RT/P = 0.08149 (273+50)/1000 = 0.02632 m3/kg
(b) To use compressibility chart, we need to calculate PR and TR
PR
= 1 MPa/4.059 MPa = 0.246
TR = 323K/374.2K = 0.863
Using the chart (Fig. 3.49), we find Z = 0.84
Then v = Z videal = 0.0221 m3/kg
Note that more accurate value of v is obtained from table A-13 asv = 0.02179 m3/k
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OTHER EQUATIONS
OF STATE
Several equations have been proposed torepresent the P-v-Tbehavior of substances
accurately over a larger region with no
limitations.
Van der Waals
Equation of State Critical isothermof a pure
substance has
an inflection
point at the
critical state.
This model includes two effects not considered
in the ideal-gas model: the intermolecular
attraction forces and the volume occupied by the
molecules themselves. The accuracy of the van
der Waals equation of state is often inadequate.
S
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Summary Pure substance
Phases of a pure substance
Phase-change processes of pure substances
Compressed liquid, Saturated liquid, Saturated vapor, Superheated vapor
Saturation temperature and Saturation pressure
Property diagrams for phase change processes
The T-vdiagram, The P-vdiagram, The P-Tdiagram, The P-v-Tsurface
Property tables
Enthalpy
Saturated liquid, saturated vapor, Saturated liquid vapor mixture,Superheated vapor, compressed liquid
Reference state and reference values
The ideal gas equation of state
Is water vapor an ideal gas?
Compressibility factor
Other equations of state
Problem
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Problem A piston-cylinder contains 50 L liquid water at 25oC and 300 kPa. Heat is
added at constant pressure until entire liquid is vaporized.
(a) What is the mass of water?
(b) What is the final temperature?
(c) Determine the total enthalpy change
T1 = 25oC T1 < Tsat at 300 kPa
P1 = 300 kPa Initially we have compressed
water water
For this case, P1 = 300 kPa this is too low to use Table A-7
Then, v1 = vf@25C = 0.001003 m3/kg
h1 = hf@25C = 104.83 kJ/kg
(a) m = V1/v1 = 0.05 m3 / 0.001003 m3/kg = 49.85 kg
(b) At the final state, the cylinder contains saturated vapor
Final temperature must be saturation temperature. For 300 kPa, T = 133.52oC
(c) Final enthalpy is h2 = hg@300kPa = 2724.9 kJ/kg