Engineering Thermodynamics

Jegathishkumar R
Properties of Pure Substance
• 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.
• Introduce the compressibility factor, which accounts for the deviation of real gases
from ideal-gas behavior.
• Rankine cycle – Thermodynamic analysis of process

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 What is a 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 Examples:
1. Water (solid, liquid, and vapor phases) 2. Mixture of liquid water and water vapor
3. Carbon dioxide, CO2 4. Nitrogen, N 2 5. Mixtures of gases, such as air, as long as there is no change of phase

❖ The substances exist in different phases, e.g. at room temperature and pressure, copper is solid and
mercury is a liquid.
❖ There are 3 Principal phases
• solid
• Liquid
• gas
Each with different molecular structures.
Phase-change Processes of Pure Substances
❖ There are many practical situations where two phases of a
pure substances coexist in equilibrium.

❖ E.g. water exists as a mixture of liquid and vapor in the

boiler and etc.
❖ Solid: strong intermolecular bond
❖ Liquid: intermediate intermolecular bonds
❖ Gas: weak intermolecular bond

Solid Liquid Gas

The molecules in a solid are kept at their positions by the
large spring like inter-molecular forces. In a solid, the
attractive and repulsive forces between the molecules
tend to maintain them at relatively constant distances
from each other.
PHASE-CHANGE PROCESSES OF PURE SUBSTANCES

• Compressed liquid (sub-cooled liquid): A substance that

it is not about to vaporize.
• Saturated liquid: A liquid that is about to vaporize.

At 1 atm 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).

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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).

At 1 atm pressure, the As more heat is

As more heat is transferred, temperature remains constant transferred, the
part of the saturated liquid at 100°C until the last drop of temperature of the
vaporizes (saturated liquid is vaporized (saturated vapor starts to rise
liquid–vapor mixture). vapor). (superheated vapor).
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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 100°C 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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 The variation of
the temperature
of fruits and
vegetables with
pressure during
vacuum cooling
from 25°C to 0°C.

In 1775, ice was made by

evacuating the air space
in a water tank.
If the entire process between state 1 and 5 described in the figure is
reversed by cooling the water while maintaining the pressure at the same
value, the water will go back to state 1, retracing the same path, and in so
doing, the amount of heat released will exactly match the amount of heat
added during the heating process.

T-v diagram for the

heating process of
water at constant
pressure.
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Latent Heat

❖ 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.

❑ 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.
 s

At supercritical pressures ( P > Pcr), Critical point : The point at

there is no distinct phase-change which the saturated liquid
(boiling) process. and saturated vapor states
are identical.
T-V curve of pure substance
• If all of the saturated liquid states are connected, the saturated liquid
line is established. If all of the saturated vapor states are connected,
the saturated vapor line is established.
• These two lines intersect at the critical point and form what is often
called the “steam dome.” The critical point of water is 373.95oC,
22.064 Mpa.
• The region between the saturated liquid line and the saturated
vapor line is called by these terms: saturated liquid-vapor mixture
region, wet region (i.e., a mixture of saturated liquid and saturated
vapor), two-phase region, and just the saturation region.
• Notice that the trend of the temperature following a constant
pressure line is to increase with increasing volume
the trend of the pressure following a
constant temperature line is to decrease
with increasing volume.
Extending the Diagrams For water,
Ttp = 0.01°C
to Include Ptp = 0.6117 kPa
the Solid Phase
At triple-point pressure
and temperature, a
substance exists in three
phases in equilibrium.

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1 – 2 : Volume increases
from V1 to V2 as ice
temperature increases to 0 C.
2 – 3 : The volume of water
decreases as ice melts at 0 C.
3 – 3’ : The volume of water
slightly decreases from 0 C
to 4 C.
3’ – 4 : The volume of water
increases due to thermal
expansion.
4 – 5 : There is a large
increase in volume of water
due to formation of steam.
5 – 6 : The volume of vapour
increases.
The triple point of
water is 0.01oC, 0.6117 kPa

At low pressures (below the

triple-point value), solids
evaporate without melting
first (sublimation).
The P-v-T surfaces present a great deal of information at once, but in a
thermodynamic analysis it is more convenient to work with two-dimensional
diagrams, such as the P-v and T-v diagrams.
Is Water Vapor an Ideal Gas?
• At pressures below 10 kPa, water vapor can be
treated as an ideal gas, regardless of its
temperature, with negligible error (less than 0.1
percent).
• At higher pressures, however, the ideal gas
assumption yields unacceptable errors,
particularly in the vicinity of the critical point
and the saturated vapor line.
• In air-conditioning applications, the water vapor
in the air can be treated as an ideal gas. Why?
• In steam power plant applications, however, the
pressures involved are usually very high;
therefore, ideal-gas relations should not be used.
Percentage of error ([|v table - v ideal|/v table] ×100) 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.
P-V-T surface for real substance
• Real substances that readily change phase from solid to liquid to gas
such as water, refrigerant-134a, and ammonia cannot be treated as
ideal gases in general.
• The pressure, volume, temperature relation, or equation of state for
these substances is generally very complicated, and the
thermodynamic properties are given in table form.
• The properties of these substances may be illustrated by the
functional relation F(P,v,T)=0, called an equation of state.
• The above two figures illustrate the function for a substance that
contracts on freezing and a substance that expands on freezing.
Constant pressure curves on a temperature-volume diagram
 Saturated water-Temperature table
Temp., T °C Sat. Specific volume, Internal energy, Enthalpy, Entropy,
Press., m 3 /kg kJ/kg kJ/kg kJ/kg⋅K
Psat kPa Sat. liquid, Sat. Sat. Evap., Sat. Sat. Evap., hfg Sat. Sat. Evap., s fg Sat.
vf vapor, liquid, ufg vapor, ug liquid, vapor, hg liquid, s f vapor,
vg uf hf sg
0.01 0.6117 0.001000 206.00 0.00 2374.9 2374.9 0.00 2500.9 2500.9 0.0000 9.1556 9.155
6
5 0.8725 0.001000 147.03 21.02 2360.8 2381.8 21.02 2489.1 2510.1 0.0763 8.9487 9.024
9
10 1.228 0.001000 106.32 42.02 2346.6 2388.7 42.02 2477.2 2519.2 0.1511 8.7488 8.899
9
15 1.706 0.001001 77.885 62.98 2332.5 2395.5 62.98 2465.4 2528.3 0.2245 8.5559 8.780
3
20 2.339 0.001002 57.762 83.91 2318.4 2402.3 83.91 2453.5 2537.4 0.2965 8.3696 8.666
1
25 3.170 0.001003 43.340 104.83 2304.3 2409.1 104.83 2441.7 2546.5 0.3672 8.1895 8.556
7
30 4.247 0.001004 32.879 125.73 2290.2 2415.9 125.74 2429.8 2555.6 0.4368 8.0152 8.452
0
 Saturated water-Pressure table
Press. Sat. Specific volume, Internal energy, Enthalpy, Entropy,
Temp.,
P kPa m 3 /kg kJ/kg kJ/kg kJ/kg⋅K
T sat °C
Sat. Sat. Sat. Evap., Sat. Sat. Evap., Sat. Sat. Evap., Sat.
liquid,
liquid, vapor, liquid, ufg vapor, liquid, hfg vapor, sfg vapo
sf r,
vf vg uf ug hf hg
sg

0.6117 0.01 0.001000 206.00 0.00 2374.9 2374.9 0.00 2500.9 2500.9 0.0000 9.1556 9.15
56

1.0 6.97 0.001000 129.19 29.30 2355.2 2384.5 29.30 2484.4 2513.7 0.1059 8.8690 8.97
49

1.5 13.02 0.001001 87.964 54.69 2338.1 2392.8 54.69 2470.1 2524.7 0.1956 8.6314 8.82
70

2.0 17.50 0.001001 66.990 73.43 2325.5 2398.9 73.43 2459.5 2532.9 0.2606 8.4621 8.72
27

2.5 21.08 0.001002 54.242 88.42 2315.4 2403.8 88.42 2451.0 2539.4 0.3118 8.3302 8.64
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Quality

❖ When a substance exists as part liquid and

part vapor at saturation conditions, its quality
(x) is defined as the ratio of the mass of the
vapor to the total mass of both vapor and
liquid.
❖ The quality is zero for the saturated liquid
and one for the saturated vapor (0 ≤ x ≤ 1)

❖ For example, if the mass of vapor is 0.2 g and

the mass of the liquid is 0.8 g, then the quality
is 0.2 or 20%.
 Quality

Mixture of liquid and vapor

We note

Recall the definition of quality x

Then

Note, quantity 1- x is often given the name moisture. The specific volume of the saturated
mixture becomes

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The form that we use most often is

It is noted that the value of any extensive property per unit mass in the saturation region is
calculated from an equation having a form similar to that of the above equation. Let Y be
any extensive property and let y be the corresponding intensive property, Y/m, then

The term yfg is the difference between the saturated vapor and the saturated liquid values of
the property y; y may be replaced by any of the variables v, u, h, or s.

We often use the above equation to determine the quality x of a saturated liquid-vapor state.

The following application is called the Lever Rule:

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Property Table
Example 2.1
Determine the saturated pressure, specific volume,
internal energy and enthalpy for saturated water vapor
at 45oC and 50oC.
Example 2.2
Determine the saturated pressure, specific volume,
internal energy and enthalpy for saturated water vapor
at 47⁰ C .
Solution:

❑ Extract data from steam

table
T Psat v u h
45 9.595 15.25 2436. 2582.
3 1 1 4
47 Psat v u h
50 12.35 12.02 2442. 2591.
2 6 7 3
❑ Interpolation for Psat

Interpolation Scheme
for Psat

❑ Do the same principal to

others!!!!
Example 1
Determine the enthalpy of
1.5 kg of water contained
in a volume of 1.2 m3 at
❑ Find the quality
200 kPa.
Solution:
❑ Specific volume for water

❑ From table A-5:

❑ The enthalpy
Example 2
Determine the internal energy of refrigerant-134a at a
temperature of 0°C and a quality of 60%.

Solution:

❖ From table A-5: ❖ The internal energy of R

134a at given condition:
Example 3

Consider the closed, rigid

container of water as shown. m g , Vg
The pressure is 700 kPa, the Sat. Vapor
mass of the saturated liquid is
1.78 kg, and the mass of the
saturated vapor is 0.22 kg. Heat m f , Vf
is added to the water until the Sat. Liquid

pressure increases to 8 MPa.

Find the final temperature,
enthalpy, and internal energy of
the water
Solution:
State 2:
❖ Theoretically:
❖ Information :

❖ The quality before

pressure increased ❖ From table A-5:
(state 1).

❖ Since that it is in
❖ Specific volume at superheated region, use
state 1 table A-6:
COMPRESSIBILITY FACTOR—A MEASURE OF DEVIATION FROM IDEAL-GAS
BEHAVIOR
Compressibility factor (Z) A The farther away Z is from unity, the more the gas
factor that accounts for the deviates from ideal-gas behavior.
deviation of real gases from Gases behave as an ideal gas at low densities (i.e.,
ideal-gas behavior at a given low pressure, high temperature).
temperature and pressure. 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 temperature or
pressure.
Reduced Reduced Pseudo-reduced
pressure temperature specific volume
Z can also be determined from a
knowledge of PR and vR .

Comparison of Z factors for various gases. 37

Validation of ideal gas assumptions
• The compressibility factor is expressed as a function of the reduced
pressure and the reduced temperature. The Z factor is approximately
the same for all gases at the same reduced temperature and reduced
pressure.
• The ideal gas equation of state is used when (1) the pressure is small
compared to the critical pressure or pressure is less than 10 times
the critical pressure (2) when the temperature is twice the critical
temperature. The critical point is that state where there is an
instantaneous change from the liquid phase to the vapor phase for a
substance
• For instance the critical pressure and temperature for oxygen are
5.08 MPa and 154.8 K, respectively. For temperatures greater than
300 K and pressures less than 50 MPa (1 atmosphere pressure is
0.10135 MPa) oxygen is considered to be an ideal gas.
Calculate the specific volume of nitrogen at 300 K and 8.0 MPa and compare the result
with the value given in a nitrogen table as v = 0.011133 m 3/kg.

Tcr = 126.2 K, P cr = 3.39 MPa R = 0.2968 kJ/kg-K

Since T > 2Tcr and P < 10Pcr, we use the ideal gas equation of state

Nitrogen is clearly an ideal gas at this state.

OTHER EQUATIONS OF STATE
Several equations have been proposed to represent the P-v-T behavior of substances accurately over a larger
region with no limitations.

Van der Waals Equation

of State
Critical isotherm
of 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.
Beattie-Bridgeman Equation of State

Benedict-Webb-Rubin Equation of State

The constants are given in Table 3–4. This equation can handle substances at
densities up to about 2.5 ρ cr.

Virial Equation of State

The coefficients a(T), b(T), c(T), and so on, that are functions of
temperature alone are called virial coefficients.
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