PROPERTY TABLES +
EQUATION OF STATE
AGUS HARYANTO
20-02-2010
http://www.energimasadepan.wordpress.com
WHY TABLES?
 The relationships among thermodynamic properties are
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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.
Property tables of other substances are used in the same
manner.
For each substance, the thermodynamic properties are
listed in more than one table.
Enthalpy—A Combination Property
 Two new properties: enthalpy h and entropy s.
 Entropy is a property associated with the second law
of thermodynamics.
In the analysis of certain
processes, such in power
generation & refrigeration,
we encounter the combination
of properties u + Pv. This
combination is defined as
enthalpy, h:
h = u + Pv (kJ/kg)
Use of Enthalpy
 The widespread use of the property enthalpy is due
to Professor Richard Mollier (Germany), who
recognized the importance of the group u + Pv in the
analysis of steam turbines.
 The properties of the of steam are represented in
tabular and graphical form (as in the famous Mollier
chart).
 The term enthalpy (from Greek enthalpien, which
means to heat) was used in the 1930s .
Saturated Liquid and Saturated Vapor
 Table A–4  based on T
 Table A–5  based on P
Subscript used:
 f = for saturated liquid
 g = for saturated vapor.
 fg = difference of saturated
vapor and saturated liquid
 vf = v of saturated liquid
 vg = v of saturated vapor
 vfg = difference of vg and vf
vfg = vg - vf
Table A–4
EXAMPLE 3–1
A rigid tank contains
50 kg of saturated
liquid water at 90°C.
Determine the
pressure in the tank
and the volume of the
tank.
Schematic and T-v diagram
EXAMPLE 3–1 : Solution
Analysis Since saturation conditions exist in the
tank, the pressure must be the saturation pressure at
90°C:
 P = Psat @ 90°C = 70.183 kPa
(Table A–4)
The specific volume of the saturated liquid at 90°C is
v = vf @ 90°C = 0.001036 m3/kg
(Table A–4)
 Then the total volume of the tank becomes
V = mv = 50 kg (0.001036 m3/kg) = 0.0518 m3
EXAMPLE 3–3
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.
Schematic and P-v diagram
EXAMPLE 3–3 : Solution
 Analysis (a) The volume change per unit mass during a
vaporization process is vfg, which is the difference
between vg and vf. Using Table A–5 at 100 kPa and
substituting yield:
vfg = vg - vf = 1.6941 - 0.001043 = 1.6931 m3/kg
V = m.vfg = (0.2 kg)(11.6931 m3/kg) = 0.3386 m3
 (b) The amount of energy needed to vaporize a unit mass
of water at a given pressure is the enthalpy of
vaporization at that pressure, which is hfg = 2257.5 kJ/kg
for water at 100 kPa. Thus, the amount of energy
transferred is:
Hfg = m  hfg = (0.2 kg)(2257.5 kJ/kg) = 451.5 kJ
Saturated Liquid–Vapor Mixture
 To analyze this mixture properly, we need to know
the proportions of the liquid and vapor phases in the
mixture. A new property called the quality x as the
ratio of the mass of vapor to the total mass of the
mixture:
where:
Quality, x
 Quality has significance for saturated mixtures only.
 Its value is between 0 and 1.
 The quality of a system that consists of saturated liquid is 0
(or 0 percent), and the quality of a system consisting of
saturated vapor is 1 (or 100 percent).
 In saturated mixtures, quality can serve as one of the two
independent intensive properties needed to describe a state.
 Note that the properties of the saturated liquid are the same
whether it exists alone or in a mixture with saturated vapor.
 During the vaporization process, only the amount of saturated
liquid changes, not its properties. The same can be said about
a saturated vapor during condensation.
The Use of Quality, x
Consider a tank that contains a saturated liquid–vapor
mixture. The volume of saturated liquid is Vf, and the
volume of saturated vapor is Vg. The total volume V is :
V = Vf + V g
V = mv  mt.vavg = mf.vf + mg.vg
mf = mt – mg  mt.vavg = (mt – mg) vf + mg.vg
Dividing by mt gives:
vavg = (1 – x) vf + x.vg
since x = mg/mt. This relation can also be expressed as
vavg = vf + x.vfg
(m3/kg)
where vfg = vg – vf.
Solving for quality, we obtain
The Use of Quality, x
Analog:
 uavg = uf + x.ufg
(kJ/kg)
 havg = hf + x.hfg
(kJ/kg)
 yavg = yf + x.yfg
where y is v, u, or h.
 yf  yavg  yg
 The subscript “avg”
(average) is usually
dropped for simplicity.
 yf  y  yg
EXAMPLE 3–4 Pressure and Volume
A rigid tank contains 10
kg of water at 90°C. 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.
Schematic and T-v diagram
EXAMPLE 3–4 : Solution
 Analysis (a) Since the two phases coexist in
equilibrium, we have a saturated mixture, and the
pressure must be the saturation pressure at the given
temperature:
P = Psat @ 90°C = 70.183 kPa
(Table A–4)
 (b) At 90°C, we have vf = 0.001036 m3/kg and vg =
2.3593 m3/kg (Table A–4). To find the volume of the
tank is to determine the volumes of each phase and then
add them:
V = Vf + Vg = mf.vf + mg.vg
= (8 kg)(0.001036 m3/kg) + (2 kg)(2.3593 m3/kg)
= 4.73 m3
EXAMPLE 3–4 : Solution (Alt)
 (b) Another way is to first determine the quality x,
then the average specific volume v, and finally the
total volume:
and
EXAMPLE 3–5 Saturated Mixture
An 80-L vessel contains
4 kg of refrigerant-134a
at a pressure of 160 kPa.
Determine (a) the
temperature, (b) the
quality, (c) the enthalpy
of the refrigerant, and
(d) the volume occupied
by the vapor phase.
Schematic and P-v,h diagram
EXAMPLE 3–5 : Solution
 Analysis (a) At this point we do not know whether the
refrigerant is in the compressed liquid, superheated
vapor, or saturated mixture region. From the
information in the problem:
v = V/m = (0.080 m3)/4 kg = 0.02 m3/kg
From Table A–12 @ 160 kPa
 vf = 0.0007437 m3/kg
 vg = 0.12348 m3/kg
Obviously, vf < v < vg, and, the refrigerant is in the
saturated mixture region. Thus, the temperature
must be the saturation temperature:
T = Tsat @ 160 kPa = 15.60°C
EXAMPLE 3–5 : Solution (Cont’d)
 (b) Quality can be determined from
 (c) At 160 kPa, we also read from Table A–12 that hf
= 31.21 kJ/kg and hfg = 209.90 kJ/kg. Then,
EXAMPLE 3–5 : Solution (Cont’d)
 (d) The mass of the vapor is (remember definition of
x):
mg = x.mt = (0.157)(4 kg) = 0.628 kg
and the volume occupied by the vapor phase is
Vg = mg.vg = (0.628 kg)(0.12348 m3/kg)
Vg = 0.0775 m3 = 77.5 L
The rest of the volume (2.5 L) is occupied by the
liquid.
Superheated Vapor (Table A-6)
Superheated vapor is
characterized by:
 Lower pressure (P < Psat at a
given T)
 Higher temperature (T > Tsat
at a given P)
 Higher specific volume (v >
vg at a given P or T)
 Higher internal energy (u >
ug at a given P or T)
 Higher enthalpy (h > hg at a
given P or T)
Table A-6
EXAMPLE 3–6 Internal Energy
 Determine the internal energy of water at 20 psia
and 400°F.
 Analysis At 20 psia, the saturation temperature is
227.92°F. Since T > Tsat, the water is in the
superheated vapor region. Then the internal energy
at the given temperature and pressure is determined
from the superheated vapor table (Table A–6E) to be
u = 1145.1 Btu/lbm
EXAMPLE 3–7 T of Superheated Vapor
 Determine the temperature of
water at a state of P = 0.5 MPa
and h = 2890 kJ/kg.
 Analysis At 0.5 MPa, the
enthalpy of saturated water vapor
is hg = 2748.1 kJ/kg. Since h > hg,
then we have superheated
vapor. Under 0.5 MPa in Table
A–6 we read:
T is between 200 and 250°C.
By linear interpolation :
T = 216.3°C
Compressed Liquid (Table A-7)
 Compressed liquid tables are not commonly
available.
 Variation of properties due to P is insignificant.
Increasing the P 100 times, often causes properties
to change < 1 %.
 If no data for compressed liquid, compressed liquid
is treated as saturated liquid at the given T. Thus,
 To reduce error in h at low to moderate P and T :
Compressed Liquid
In general, a compressed liquid is characterized by
 Higher pressures (P > Psat at a given T)
 Lower temperatures (T < Tsat at a given P)
 Lower specific volumes (v < vf at a given P or T)
 Lower internal energies (u < uf at a given P or T)
 Lower enthalpies (h < hf at a given P or T)
EXAMPLE 3–8 Compressed Liquid
 Determine the internal
energy of compressed
liquid water at 80°C and 5
MPa, using (a) data from
the compressed liquid
table and (b) saturated
liquid data. What is the
error involved in the
second case?
Schematic and T-u diagram
EXAMPLE 3–8 : Solution
 Analysis At 80°C, the saturation pressure of water is
47.416 kPa, and since 5 MPa > Psat, we obviously have
compressed liquid.
 (a) From the compressed liquid table (Table A–7)
 (b) From the saturation table (Table A–4), we read
u = uf @ 80°C = 334.97 kJ/kg
 The error is:
< 1 percent.
Example 9 : Missing Properties
Example 9 : Solution
 (a) The quality is given to be x = 0.6, which implies
that 60 percent of the mass is in the vapor phase and
the rest is liquid. Therefore, we have saturated
liquid–vapor mixture at a P of 200 kPa. The T must
be the Tsat at the given pressure:
T = Tsat @ 200 kPa = 120.21°C
(Table A–5)
At 200 kPa, from Table A–5 : uf = 504.50 kJ/kg and
ufg = 2024.6 kJ/kg. The average internal energy :
Example 9 : Solution (Cont’d)
 (b) The temperature and the internal energy are given, but we do
not know which table to be used. First, we go to Table A–4 and find
uf and ug at 125°C, where uf = 524.83 kJ/kg and ug = 2534.3 kJ/kg.
Next we compare the given u to uf and ug. Mind that:
If
u < uf
 we have compressed liquid
If
uf < u < ug
 we have saturated mixture
If
u > ug
 we have superheated vapor
In our case u = 1600 is in between the uf and ug at 125°C. Therefore,
we have saturated liquid–vapor mixture. The pressure must be Psat
at the given temperature:
P = Psat @ 125°C = 232.23 kPa
(Table A–4)
The quality is determined from:
Example 9 : Solution (Cont’d)
 (c) This is similar to case (b), except pressure is given
instead of temperature. Following the argument given
above, we read the uf and ug values at the specified
pressure. At 1 MPa, we have uf = 761.39 kJ/kg and ug =
2582.8 kJ/kg. The specified u value is 2950 kJ/kg, which
is greater than the ug value at 1 MPa. Therefore, we have
superheated vapor, and the temperature at this state
is determined from the superheated vapor table by
interpolation to be:
T = 395.2°C
(Table A–6)
We would leave the quality column blank in this case
since quality has no meaning for a superheated vapor.
Example 9 : Solution (Cont’d)
 (d) In this case the temperature and pressure are given,
but again we cannot tell which table to use. We go to
Table A–5 to find Tsat at 500 kPa, where we have Tsat =
151.83°C. Compare T to Tsat, mind that:
T < Tsat @ given P  compressed liquid
T = Tsat @ given P  saturated mixture
T > Tsat @ given P  superheated vapor
In our case, T = 75°C < Tsat so we have
compressed liquid. But no data available,
that
we treat as saturated liquid at 75°C, and
Again, we would leave the quality column blank
Example 9 : Solution (Cont’d)
 (e) The quality is given to be x = 0, and thus we have
saturated liquid at the specified pressure of 850 kPa.
Then the temperature must be the saturation
temperature at the given pressure, and the internal
energy must have the saturated liquid value:
IDEAL-GAS EQUATION OF STATE
 Property tables provide very accurate information about the
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properties, but they are bulky and prone to typo errors.
A practical approach is to have some simple relations among
the properties that are sufficiently general and accurate.
Any equation that relates the pressure, temperature, and
specific volume of a substance is called an equation of state
(EOS).
Property relations that involve other properties of a substance
at equilibrium states are also referred to as EOS.
The simplest and best-known equation of state for substances
in the gas phase is the ideal-gas EOS.
This equation predicts the P-v-T behavior of a gas quite
accurately within some properly selected region.
Ideal-gas EOS
 Also known as Boyle-Gay Lussac Law
 Pv = RT
or
 A gas that obeys this relation is called an ideal gas
 R is called the gas constant, which is different for
each gas and is determined from:
 Ru = universal gas constant.
 M = molar mass (molecular weight). Eg. N2 = 28.
Ru = universal gas constant
 The constant Ru is the same for all substances, and
its value is:
Forms of Ideal-gas EOS
 The ideal-gas equation of state can be written in
several different forms:
Pv = RT
is the molar specific volume, that is, the volume
per unit mole (m3/kmol or ft3/lbmol).
Forms of Ideal-gas EOS
 By writing ideal gas EOS twice for a fixed mass and
simplifying, the properties of an ideal gas at two
different states are related to each other by
 An ideal gas is an imaginary substance that obeys
the relation Pv = RT
EXAMPLE 3–10 Mass of Air in a Room
 Determine the mass of the air in a
room whose dimensions are 4 m x 5 m
x 6 m at 100 kPa and 25°C.
 Analysis Assume air at that condition
can be treated as an ideal gas. From
Table A–1, gas constant of air is Rair =
0.287 kPa·m3/kg·K, and absolute
temperature is T = 25°C + 273 = 298
K. The volume of the room, V = 4 x 5 x
6 = 120 m3.
From the ideal-gas relation:
Is Water Vapor an Ideal Gas?
Percentage of error
involved in assuming
steam to be an ideal
gas, and the region
where steam can be
treated as an ideal gas
with less than 1
percent error.
COMPRESSIBILITY FACTOR, Z
 The ideal-gas equation is very simple and thus very
convenient to use.
 Real gases deviate from ideal-gas behavior
significantly at states near the saturation region and
the critical point.
 This deviation from ideal-gas behavior at a given
temperature and pressure can accurately be
accounted for by the introduction of a correction
factor called the compressibility factor Z:
or
or
Reduced Pressure and Temperature
 Gases behave differently at a given temperature and
pressure, but they behave very much the same at
temperatures and pressures normalized with respect to
their critical temperatures and pressures. The
normalization is done as
 PR = reduced pressure ; TR = reduced temp.
 The Z factor for all gases is same at the same PR and TR.
generalized compressibility chart
Ggeneralized compressibility chart
The following observations can be
made from the generalized
compressibility chart:
1. At very low pressures (PR << 1),
gases behave as an ideal gas (Z  1)
regardless of temperature (Fig. 3–
52).
2. At high temperatures (TR > 2),
ideal-gas behavior can be assumed
with good accuracy regardless of
pressure (except when PR >> 1).
3. The deviation of a gas from idealgas behavior is greatest in the
vicinity of the critical point (Fig.
3–53).
EXAMPLE 3–11 Use of Generalized Charts
 Determine the specific volume of refrigerant-134a at
1 MPa and 50°C, using (a) the ideal-gas equation of
state and (b) the generalized compressibility chart.
Compare the values obtained to the actual value of
0.021796 m3/kg and determine the error involved in
each case.
EXAMPLE 3–11 : Solution
Analysis From Table A–1 we have:
R = 0.0815 kPa.m3/kg.K
Pcr = 4.059 Mpa
Tcr = 374.2 K
 (a) Using ideal-gas assumption :
Error = (0.026325 - 0.021796)/0.021796 = 0.208,
or 20.8 percent.
EXAMPLE 3–11 : Solution (Cont’d)
 (b) To determine the correction factor Z from the
compressibility chart, we first need to calculate the
reduced pressure and temperature:
Thus:
Pseudo-reduced specific volume, vR
 When P and v, or T and v,
are given instead of P and T,
the generalized
compressibility chart can
still be used to determine
the third property by
defining pseudo-reduced
specific volume vR :
Go to Fig. A–15
EXAMPLE 3–12 Generalized Charts
 Determine the pressure of
water vapor at 600°F and
0.51431 ft3/lbm, using (a)
the steam tables, (b) the
ideal-gas equation, and (c)
the generalized
compressibility chart.
EXAMPLE 3–12 : Solution
From Table A–1E we have:
R = 0.5956 psia.ft3/lbm.R
Pcr = 3200 psia and Tcr = 1164.8 R
 (a) The pressure at the specified state is determined
from Table A–6E:
 (b) Using ideal-gas assumption:
EXAMPLE 3–12 : Solution (Cont’d)
error = (1228 - 1000)/1000 = 0.228, or 22.8 %
 (c) To determine the correction factor Z from the
compressibility chart :
Using Fig. A–15, at vR = 2.372 and TR = 0.91, we have
PR = 0.33. Thus,
P = PR.Pcr = (0.33)(3200 psia) = 1056 psia
Error = 5.6 percent
OTHER EQUATIONS OF STATE
The ideal-gas equation of state is very simple, but its
range of applicability is limited.
 Van der Waals EOS
 The constants a and b are provided in Table A–1.
OTHER EQUATIONS OF STATE
 Beattie-Bridgeman
 The constants for Beattie-Bridgeman EOS are given in
Table 3–4 for various substances.
 The Beattie-Bridgeman EOS is reasonably accurate for
densities up to about 0.8 cr ( at critical point).
OTHER EQUATIONS OF STATE
 Benedict-Webb-Rubin:
OTHER EQUATIONS OF STATE
 Virial Equation of State
EXAMPLE 3–13 Different EOS
 Predict the pressure of nitrogen gas at T = 175 K and
v = 0.00375 m3/kg on the basis of (a) the ideal-gas
equation of state, (b) the van der Waals equation of
state, (c) the Beattie-Bridgeman equation of state,
and (d) the Benedict-Webb-Rubin equation of state.
Compare the values obtained to the experimentally
determined value of 10,000 kPa.
EXAMPLE 3–13 : Solution
Analysis Gas constant for N2 (Table A–1) is R = 0.2968
kPa.m3/kg.K.
 (a) Using the ideal-gas EOS:
Error = 38.5 percent
 (b) Using van der Waals EOS. For nitrogen (Table A-1),
Tcr = 126.2 K and Pcr = 3.39 Mpa = 3390 kPa.
kPa.m3 2
27(0.2968
) (126.2 K)2
kg.K
=
64(3390 kPa) 2
a = 0.175 m6.kPa/kg2
EXAMPLE 3–13 : Solution (Cont’d)

27(0.2968 kPa.m3 / kg.K)(126. 2 K)
=
64(3390 kPa)
b = 0.00138 m3/kg
Error = 5.3 percent
 (c) The constants in the Beattie-Bridgeman EOS
(Table 3-4) : A = 102.29;B = 0.05378; C = 4.2  104.
EXAMPLE 3–13 : Solution (Cont’d)
Substituting the values into Equation, we obtain:
error = 1.1 percent.
 (d) The constants in the Benedict-Webb-Rubin
equation are (Table 3-4):
a = 2.54
A0 = 106.73
b = 0.002328
B0 = 0.04074
c = 7.379  104
C0 = 8.164  105
 = 1.272  104
 = 0.0053
EXAMPLE 3–13 : Solution (Cont’d)
Substituting these values into Equation gives:
Error = 0.09 percent
HOMEWORK