chapter 2 properties of pure substances€¦ · · 2017-10-25water is a pure substance, but a...
TRANSCRIPT
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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-T property 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.
• Present some of the best-known equations of state.
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PURE SUBSTANCE
• Pure substance: A substance that has a fixed chemical
composition throughout.
• Air is a mixture of several gases, but it is considered to
be a pure substance (as along as it in the same phase).
Nitrogen and gaseous air are
pure substances. A mixture of liquid and gaseous
water is a pure substance, but a
mixture of liquid and gaseous air
is not.
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PURE SUBSTANCE PROPERTIES
DIAGRAM
• To study the phase change process for a
pure substance, we take water as an
example.
• Water, in liquid phase is filled in a piston
cylinder without any weight on the piston.
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PHASE-CHANGE PROCESSES OF PURE SUBSTANCES
At 1 atm
(≈1bar) and
20°C, water
exists in the
liquid phase
(compressed
liquid).
At 1 atm
pressure and
100°C, water
exists as a
liquid that is
ready to
vaporize
(saturated
liquid).
As more heat is
transferred, part of
the saturated liquid
vaporizes
(saturated liquid–
vapor mixture).
At 1 atm
pressure, the
temperature
remains
constant at
100°C until
the last drop
of liquid is
vaporized
(saturated
vapor).
As more
heat is
transferred,
the
temperatur
e of the
vapor starts
to rise
(superheat
ed vapor).
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• Compressed liquid (subcooled liquid): A substance that it
is not about to vaporize.
• Saturated liquid: A liquid that is about to vaporize.
T-v diagram for the heating process of water
at constant pressure.
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• Saturated vapor: A vapor that is about to condense.
• Saturated liquid–vapor mixture: The state at which the liquid and vapor phases coexist in equilibrium.
• Superheated vapor: A vapor that is not about to condense (i.e., not a saturated vapor).
T-v diagram for the heating process of water
at constant pressure.
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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.
• Water boils at 100C at 1 atm pressure.
• 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.
The liquid–
vapor
saturation
curve of a
pure
substance
(numerical
values are for
water).
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• Latent heat: The amount of energy
absorbed or released during a phase-
change process.
• Latent heat of fusion: The amount of
energy absorbed during melting. It is
equivalent to the amount of energy
released during freezing.
• Latent heat of vaporization: The amount
of energy absorbed during vaporization
and it is equivalent to the energy released
during condensation.
• The magnitudes of the latent heats
depend on the temperature or pressure at
which the phase change occurs.
• At 1 atm pressure, the latent heat of
fusion of water is 333.7 kJ/kg and the
latent heat of vaporization is 2256.5 kJ/kg.
• The atmospheric pressure, and thus the
boiling temperature of water, decreases
with elevation.
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PROPERTY DIAGRAMS FOR PHASE-
CHANGE PROCESSES • The variations of properties during phase-change processes are best
studied and understood with the help of property diagrams such as the T-v, P-v, and P-T diagrams for pure substances.
T-v diagram of
constant-pressure
phase-change
processes of a pure
substance at various
pressures
(numerical values
are for water).
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Why properties diagrams are important?
• From the diagram we can
show any process that a
system go through.
• For example: 1kg water is
heated in constant pressure
from state 1 to state 2. Then
the system is heated in
constant volume until state 3.
• The above process can be
shown in Pv diagram or Tv
diagram
• The value of a system
characteristics are also can
be defined at each state
involved in the process
3
1
2
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PROPERTY TABLES • For most substances, the relationships among thermodynamic properties are too
complex to be expressed by simple equations.
• 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.
• The results of these measurements and calculations are presented in tables in a
convenient format.
Enthalpy—A Combination Property
The
combination
u + Pv is
frequently
encountered
in the analysis
of control
volumes.
The product pressure
volume has energy units.
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Saturated Liquid and Saturated Vapor States
• Table A–4: Saturation properties of water under temperature.
• Table A–5: Saturation properties of water under pressure.
A partial list of Table A–4.
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.
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Examples Pressure of saturated
liquid
A rigid tank contains 50kg of
saturated liquid water at 90oc.
Determine the pressure in the tank
and the volume of the tank.
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P saturation at 90 ºC= 70.183 Kpa
vf= 0.001036 m3/kg,
Total volume= mass x specific volume= 50 kg x 0.001036 m3/kg
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Example: Temperature of saturated
vapor
A piston cylinder device contains 0.06m3
of saturated water vapor at 350kPa
pressure. Determine the temperature
and the mass of the vapor inside the
cylinder
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T saturation at 350 kPa = 138.86 C,
Specific volume at saturated vapor vg= 0.52422 m3/kg
Mass of the water vapor= Total volume (m3)/specific volume (m3/kg)
= 0.06/0.52442 kg =0.11 kg
• A mass of 200 g of saturated liquid water 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.
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From Table A5, we get vf= 0.001043 m3/kg, vg= 1.6941 m3/kg, hf=417.51 kJ/kg, hg=2675.0 kJ/kg,
so, Vchange= (vg-vf)xmass= (1.6941-0.001043)x0.2 m3=
Hchange= (hg-hf)xmass= (2675.0-417.51)x0.2 kJ
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Saturated Liquid–Vapor Mixture Quality, 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.
The properties of the saturated liquid are the same whether it exists alone or in
a mixture with saturated vapor.
The relative
amounts of
liquid and
vapor phases
in a saturated
mixture are
specified by
the quality x.
A two-phase system can be
treated as a homogeneous
mixture for convenience.
Temperature and
pressure are dependent
properties for a mixture.
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The v value of a
saturated liquid–
vapor mixture lies
between the vf and vg
values at the
specified T or P.
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Superheated Vapor In the region to the right of the
saturated 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.
Compared to saturated vapor,
superheated vapor is characterized by
An excellent tricks to memorize the above
relation:
• As you go right, the values of the
properties will increase except pressure.
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Compressed Liquid Compressed liquid is characterized by
Hence, a compressed liquid
may be approximated as a
saturated liquid at the given
temperature as below.
The compressed liquid properties
depend on temperature much more
strongly than they do on pressure.
An excellent tricks to memorize the above relation:
• As you go left, the values of the properties will
decrease except pressure.
23 P-v diagram of a pure substance.
T-v diagram of a pure substance.
Very important note:
• Compressed liquid region (use table
A7)
• Saturated liquid-vapor region (use
table A4 or A5)
• Super heated vapor region (use table
A6
Very important!
Steps to determine unknown properties:
1) Find regions from given properties or
phase description.
2) Then go into the relevant table.
3) Then determine the unknown
properties (either directly or by
interpolation or by equations).
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PRESSURE AND VOLUME OF
SATURATED MIXTURE
• A rigid tank contains 10kg of water at
90 ºC. If 8kg 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
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T, °C P, kPa v, m3/kg u, kJ/kg h, kJ/kg Quality,
x
Phase description
i 140 0.05 Saturated liquid vapor
mixture
ii 550 Saturated liquid
iii 125 750
iv 300 0.2579
v 400 1450
vi 220 Saturated vapor
vii 190 2050
By using data from Table A4, Table A5, Table A6 or Table A7
Complete this table for H2O
Solutions are in next slide
Solutions
i) From Table A4 (because the substance is in saturated stage), we find; P= 361.53 kPa,
vf= 0.001080 m3/kg, vg= 0.50850 m3/kg, uf=588.77 kJ/kg, ufg= 1960.9 kJ/kg, hf=589.16 kJ/kg,
hfg=2144.3 kJ/kg
vfg= vg-vfg= 0.50850-0.001080 m3/kg = 0.50782 m3/kg,
Here, given v= 0.05, we know v= vf+ x.vfg; so, x= (v-vf)/vfg = (0.050-0.001080)/0.50782 = 0.096,
h= hf+x.hfg= 589.16+0.096x2144.3 kJ/kg= 795.02 kJ/kg,
ii) By using table A5 (because water is saturated and pressure is given), Tsaurated= 155.48 ºC,
vf= 0.001097 m3/kg, uf= 655.32 kJ/kg, hf= 655.77 kJ/kg
For saturated liquid, we know that v=vf, u= uf, and h= hf and x=0,
Therefore, v= 0.001097 m3/kg, u=655.32 kJ/kg, h= 655.77 kJ/kg
iii) By checking table A4( because we are not sure yet which table we should use) we see that the
saturation pressure corresponding to the given temperature (i.e. 125 C) is 232.23 kPa, but our given
pressure is 750 kPa, that mean P> Psaturation , therefore the substance is compressed liquid.
So, we should use table A4, we find
vf= 0.001065 m3/kg, uf= 524.83 kJ/kg, hf= 525.07 kJ/kg,
For compressed liquid (it can be assumed that the properties are same as the saturated liquid at the given
temperature), i.e. v=vf, u=uf and h=hf
Therefore, v= 0.001065 m3/kg, u= 534.83 kJ/kg, h= 525.07 kJ/kg,
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iv) We can check the table A4 (we yet not know which table we should use), we see that at 300 ºC,
vg= 0.021659,
But the given v= 0.2579 m3/kg is greater that vg, therefore the substance is superheated vapour
(check lecture slides for understanding),
So, We will use table A6, we should now find the values which match with T=300 C and v=0.2579
m3/kg,
We get, P= 1 MPa, u= 2793.7 kJ/kg, h= 3051.6 kJ/kg
v) We will check table A5, we see that uf= 604.22 kJ/kg, ug= 2553.1 kJ/kg, but given u= 1450 kJ/kg,
therefore the given u is greater that uf and less that ug, and so, the substance is saturated liquid-
vapour mixture,
So we should continue to use table A5, we find T= 143.61 C, ufg= 1948.9 kJ/kg, vf= 0.001084 m3/kg ,
vfg=0.46133 m3/kg, hf= 604.66 kJ/kg, hfg=2133.4 kJ/kg
We know, x=(u-uf)/ufg, so, x=(1450-604.22)/1948.9 = 0.433,
v=vf+x.vfg= 0.001084+0.433x0.46133 m3/kg= 0.1997 m3/kg,
h=hf+x.hfg= 604.66+0.433x2133.4= 1528.42 kJ/kg,
vii) Given T= 190 C and h= 2050 kJ/kg
We can check table A4, we see that the hf=807.43 kJ/kg, hg=2785 kJ/kg, So, hf<h<hg, therefore the
substance is saturated liquid vapour mixture,
So we will continue to use table A4, the values are, P=1255.2 kPa, vf= 0.001141, vfg= 0.1552,
uf=806.00, ufg= 1783, hfg=1977.9, we can find from h=hf+x.hfg, so x= 0.6282
v=vf+x.vfg=0.09863 m3/kg, u=uf+x.ufg= 1926 kJ/kg, Done !
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THE IDEAL-GAS EQUATION OF STATE
• Equation of state: Any equation that relates the pressure, temperature,
and specific volume of a substance.
• The simplest : 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 different
gas constants.
Note , P is absolute pressure
T is absolute temperature
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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
Ideal gas equation at two
states for a fixed mass
Real gases
behave as an ideal
gas at low
densities (i.e., low
pressure, high
temperature). Pv= RT
PV=mRT
PV= NRuT
Different forms of equation of state
P1V1=mRT1
P2V2=mRT2
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When the Ideal gas equation is used
• The pressure is small compared to the
critical pressure
P< Pcr
• The temperature is twice the critical
temperature and pressure is less than 10
times of the critical pressure.
T=2Tcr and P<10Pcr
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Is Water Vapor an Ideal Gas?
• YES: at pressures below 10 kPa.
• NO: in actual application (higher pressure environment) such as Steam Power Plant
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4-7 COMPRESSIBILITY FACTOR—A MEASURE
OF DEVIATION FROM IDEAL-GAS BEHAVIOR Compressibility factor Z
A factor that accounts for
the deviation of real gases
from ideal-gas behavior at
a given temperature and
pressure.
Z bigger than 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).
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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-v diagram, The P-v diagram, The P-T diagram, The P-v-T surface
• 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?