Slide 1
Pure Substance Pure Substances-Homogeneous and constant chemical
compositionMay exist in more than one phase but chemical
composition is same in all phases.Properties of pure substances
required for design and sizing of various devicesCompressible
Substances: Magnetic, electrical and surface effects
insignificant.Examples: Water (solid, liquid, and vapor phases)
Mixture of liquid water and water vapor Carbon Dioxide (CO2), N2
Mixtures of gases, such as air, as long as there is no change of
phase.Note: Mixture of liquid air and gaseous air cannot be pure
substance due to difference in their compositionsVapor Liquid Solid
Phase Equilibrium in a Pure SubstanceSequence of Phenomena on
heating 1 kg of water in the piston/cylinder arrangement at a
constant pressure of 0.1MPa and initial temperature of 20oC as a
system.
0.1MPa, T=20oC , specific volume v1 0.1MPa, T=99.6oC,
v2>>v1 0.1MPa, T>99.6oC, v>v2(a) 0.1MPa, T=20oC ,
specific volume v1 (b) 0.1MPa, T=99.6oC, v2>>v1 (c) 0.1MPa,
T>99.6oC, v>v2Saturation Temperature and Saturation
Pressure:The temperature at which vaporization takes place at a
given pressure is called saturation temperature. This pressure is
called the saturation pressure for the given temperature. Thus, for
water at 99.60C the saturation is 0.1 MPa, and for water at 0.1 MPa
the saturation temperature is 99.60C.
vapor-pressure curve-Plot between saturation temperature and
saturation pressure for a given substance. Important
DefinitionsSaturated Liquid-If a substance exists as liquid at the
saturation temperature and pressure.Compressed/Subcooled liquid:
Liquid at a temperature lower than the saturation temperature for a
given pressureTwo Phase substance: Substance exists as part liquid
and part vapor at the saturation temperature. Its quality (x) is
defined as the ratio of the mass of vapor to the total
mass.Saturated Vapor: If a substance exists as vapor at the
saturation temperature, it is called saturated vapor. Superheated
Vapor: Vapor at a temperature greater than the saturation
temperature for a given pressure. Temperature Volume diagram for
water showing liquid and vapor phases
Critical Point:
Its a point of inflection of zero slope in a T-V diagram
The conditions at that point are called critical temperature
(Tc), critical pressure (Pc) and critical volume (vc).
Two phases never exist during a constant pressure process for
P>Pc. Its a continuous substance called fluid
But, arbitrarily it is considered as compressed liquid at T <
Tc and super heated vapor at T>TcAt P>Pc, liquid and gas
phases never exist in equilibrium.
Critical Point:
Two phase mixtures-Quality conceptQuality of two phase mixture
is mass fraction of vaporSo for saturated vapor, x=1 Saturated
liquid, x=0For two phase mixture, x is between 0 and 1
Quality of two phase mixture, x = mass of vapor/(mass of liquid
+ mass of vapor), so for saturated vapor, x=100% and for saturated
liquid, x=0Estimation of two phase properties using quality
where Estimation of two phase properties using quality
Pressure Temperature diagram for waterExperiment to understand
P-T diagram
On heating 1 kg ice at -20oC, 100kPa (line EF) in a
piston-cylinder arrangement, temperature increases to 0oC, at which
it melts (during melting specific volume of most substances
increases except for water) and the temp remains constant. This is
saturated solid. After all ice is melted to liquid, on further
heating temp of liquid increases.
If the initial condition of ice is -20oC and 0.26kPa (line AB),
on heating, its temperature increases to -10oC when ice gets
transformed directly to vapor (sublimation)
If the initial condition is -20oC, 0.6113kPa (line CD), on
heating its temperature increases to 0.01oC when ice gets
transformed partly to liquid and partly to vapor i.e all three
phases coexist (triple point)
Though the above plot is for water, all other substances behave
similarily except that their critical and triple point data shall
differ.
Experiment to understand PT diagramTriple Point Data
Independent Properties of Pure Substance: The state of a
compressible pure substance is defined by two independent
properties.
For saturated liquid and vapor states, they have same
Temperature and Pressure. Hence pressure and temperature cannot be
independent properties. So pressure and specific volume or pressure
and quality are required to specify a saturation state of a pure
substance.
Independent Properties of Pure SubstanceThermodynamic
TablesOrganisationB.1.1T Psat vf vfg vg (similarily u, h, s)
B.1.2 P Tsat --do
B.1.3 Superheated Vapor series of T as f(P)
B.1.4Compressed liquid do
B.1.5Solid + vapor T P vi, vig , vg ( other properties like u, h
and s)
Thermodynamic TablesRecap
Pure substancesTsat, Psat, Vap Pressure CurveDef of Sat liquid,
Sat vap, SHV, SCL/CL, Two phase mixTwo phase propertiesCritical
pointThermodynamic Diagrams-T-V, P-V, P-TIndependent
propertiesOrganization of Tables
Recap:Equation of StateAny equation that relates the
temperature, pressure, and specific volume of a substance is called
the equation of state. The simplest and convenient equation of
state for substances in a gas phase is the ideal-gas equation of
state. This is valid at low densities when molecules are wide apart
(P.E neglected) and independent of one another. A gas tends to be
close to ideal gas at low pressures and high temperatures.As the
state of a substance is further removed from saturated region, it
is close to ideal gas i.e in superheated region.
Ideal Gas EquationTwo forms:
Where R is kJ/Kg.K (literature), is specific volume in
m3/kgWhere is is in kJ/Kmol.K (8.314), is molar specific volume in
m3/kmol. The two gas constants are related by P in kPa, T in
Kelvin
Where R is J/Kg.K, is specific volume in m3/kg
T - v Diagram for WaterCompressibility FactorSignifies the
deviation of the actual relation from the ideal gas equation of
state and it is defined as
or
It can also be expressed as
For an ideal gas Z = 1, and the deviation of Z from unity
measures the deviation of the actual P-V-T relation from the ideal
gas equation of state.
Compressibility Chart for NitrogenThree Observations from the
Chart:
At all temperatures, Z1 (ideal gas behaviour)as , P 0
At T 300K, Z is nearly unity upto P=10MPa
At low T and high P, Z deviates significantly from 1
Compressibility diagrams of other pure substances are similar in
trend.
Three Observations from the ChartGeneralized Compressibility
Charts:
Since the compressibility diagrams of all pure substances are
similar qualitatively and different quantitatively due to their
different critical values, the compressibility chart of Z as f
(reduced temp Tr = T/Tc, reduced pressure Pr =P/Pc) is same for all
substances, hence called Generalized Compressibility charts.
Theorem of Corresponding States:
Substances having same Tr and Pr have same Z value and hence
deviate from the ideal gas behaviour to the same extent.
Compressibility Chart based on Data for a Large Number of
Gases
Lee-Kesler Simple Fluid Compressibility FactorIdeal Gas Model
may be assumed if
Pvg Superheated vapor (B.4.1)From B.4.2, v of 0.05242 is lying
between pressures of 500 and 600kPa at 40oC. So interpolation.
(600-500)/(0.04628-0.05636) =
(P-500)/(0.05242-0.05636)P=535kPa
(u600-u500)/v600-v500 = u-u500/v-v500
(249.48-250.51)/(0.04628-0.05636)=u-250.51/0.05242-0.05636(or)600-500/u600-u500=
535-500/u-u500
u=250.2kJ/kg
Ex 2 for InterpolationQDetermine whether water at each of the
following states is a compressed liquid, superheated vapor, or a
mixture of saturated liquid and vapor.10 MPa, 0.003 m3/kg1 MPa,
1900C 2000C, 0.1 m3/kg 10 kPa, 100CQCalculate the specific volume
of saturated steam at 2000C, if the quality of steam is 70%.
QCalculate the specific volume of saturated steam at 2000C, if
the 1 m3 closed vessel contains 1:9 volumetric ratio of liquid and
vapor.
Determine the mass of vapor and liquidFind qualityCompute the %
vapor present by volume and also by mass, what is the
difference?QFor water at 100 kPa with a quality of 10% find the
volume fraction of vapor. This is a two-phase state at a given
pressure: Table B.1.2: vf = 0.001043 m3/kg, vg = 1.6940 m3/kg Vol
fn of vapor=Vvap/(Vvap+Vliq)=mgvg/(mgvg+mfvf)Divide with total mass
and using the definition of quality=xvg/(xvg+(1-x)vf)= 0.1 1.6940
(0.1 1.694 + 0.9 0.001043) = 0.16940/17034 = 0.9945
Notice that the liquid volume is only about 0.5% of the total.QA
glass jar is filled with saturated water at 500 kPa of quality 25%,
and a tight lid is put on. Now it is cooled to -10oC. What is the
mass fraction of solid at this temperature?
Comparison of Table, Ideal Gas and other equation of state
Specific volume calculationAmmonia at 500kPa, 40C
Table 5.2.2, v=0.2923m3/kgUsing ideal gas eqn
v=RT/P=0.48819*313/500=0.3056m3/kgerror=
0.2923-0.3056/0.2923=4.5%Compressibility factor -Given Tc=405.5K,
Pc=11.35MPaTr=0.772, Pr=0.044 From fig D1,
Z=0.97v=ZRT/P=0.97*0.48819*313/500=0.2964 Error: 1.4% w r t
Tables
Note: Table D4 gives Zf, Zg, Prsat as a f(Tr) to get vf, vg and
hence two phase volume
Comparison of Table, Ideal Gas and other equation of state