Thermodynamics
By : Ihsan H. Dakhil
"Properties of Pure Substances"
- Pure Substances
A substance that has a fixed chemical composition throughout
is called a pure substance. Water, nitrogen, helium, and carbon
dioxide are all pure substances.
A mixture of ice and liquid water, for example, is a pure
substance
because
both
phases
have
the
same
chemical
composition but a mixture of oil and water is not a pure substance,
since oil is not soluble in water.
- Phase Change Process of Pure Substance:
A piston cylinder containing liquid water at 20°C and 1 atm
(state 1). Under these conditions water exists in the liquid phase, and
it is called a compressed liquid, or a subcooled liquid. Heat is now
transferred to the water until its temperature rises to 40°C the liquid
water expands slightly and so its specific volume increases. The
pressure in the cylinder remains constant at 1 atm during this
process. Therefore, the piston moves up slightly. As more heat is
transferred, the temperature keeps rising until it reaches 100°C (state
2). At this point water is still a liquid, but any heat addition will cause
some of the liquid to vaporize. A liquid that is about to vaporize is
called a saturated liquid.
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
Once boiling starts, the temperature stops rising until the liquid is
completely vaporized. The thermometer will always read 100°C. At
Midway about the vaporization line (state 3) the cylinder contains
equal amounts of liquid and vapor ( saturated mixture ). The
vaporization process continues until the last drop of liquid is
vaporized (state 4). At this point, the entire cylinder is filled with
vapor. A vapor that is about to condense is called a saturated vapor.
When more heat transfer results in an increase in both the
temperature and the specific volume. At state 5, the temperature of
the vapor is, let us say 300°C. A vapor is called a superheated vapor.
the phase change due to heating is shown in the figure below:
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
During a vaporization process, we will have a mixture of
saturated liquid and saturated vapor. To analyze this mixture
properly, we need a new property called the quality (x) or
dryness ratio.
(Quality has significance for saturated mixtures only)
x: is the ratio of the mass of vapor to the total mass of the
mixture.
x
m vapor
mtotal

mg
m
where:
mtotal = mliquid + mvapor = mf + mg
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
Consider a tank that contains a saturated liquid–vapor mixture. The
volume occupied by saturated liquid is Vf and the volume occupied
by saturated vapor is Vg . The total volume V is the sum of the two:
V = Vf + Vg
since: V = m ʋ
mf = m – mg
m v = mf vf + mg vg
m v = (m - mg) vf + mg vg
Dividing by m yields :
v = ( 1 – x ) vf + x v g
since: x = mg /m , vfg = vg - vf
v = vf + x vfg
solving for x :
x
v vf
v fg
Critical point: A point has liquid and vapor phases can’t be
distinguish because they have the same property. Its has ( Tc ) and
( Pc ) at which a pure substance can exit in vapor / liquid equilibrium.
The critical point properties of water are Pc 22.09 MPa, Tc 374.14 K,
and vc 0.003155 m3/kg.
- The T- v Diagram:
The phase change process of water at different pressures will
describe to develop the T-v diagram in following Figure below. At
pressures above the critical pressure [super critical pressures (P >
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
Pcr)] there is no distinct phase change (boiling) process. The specific
volume of the substance continually increases.
Above the critical state, there is no line that separates the
compressed liquid region and the superheated vapor region. The
substance refers as superheated vapor at temperatures above the
critical temperature and as compressed liquid at temperatures below
the critical temperature as Fig. below:
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
- The P-v Diagram:
The general shape of the P-v diagram of a pure substance is
very much like the T-v diagram, but the T is constant lines on this
diagram as shown in following Fig.:
Steam tables:
Steam is the vapor form of water and is invisible when pure and
dry. 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 called
steam tables.
It consists of two tables of the energy transfer properties of water
and steam, saturated steam tables and superheated steam tables. It
can be used to determine various properties of water such as specific
(volume, internal energy, enthalpy, entropy).
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
Note: When a value is required from a table at conditions which lie
between listed values, linear interpolation is necessary.
Example1: A tank contains 50 kg of saturated liquid water at 90°C.
Determine the pressure in the tank and the volume of the tank?
Solution: Since saturation conditions in the tank, the pressure must
be the saturation pressure at 90°C:
P = Psat @ 90 C = 70.11 kPa
The specific volume of the saturated liquid at 90°C is,
ʋ = ʋ f @90 C = 1.036*10 -3 m3/kg
Then the total volume of the tank becomes:
V = m ʋ = 50 kg * 0.001036 m3/kg = 0.0518 m3
Ex.2 : A vessel 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 vessel and (b) the volume of the vessel?
Sol. : a) we have a saturated mixture, and the pressure must be the
saturation pressure at the given temperature
P = Psat @ 90°C = 70.11 kPa
(b) At 90 °C, we have vf = 1.036*10-3 m3/kg and vg = 2.3613 m3/kg.
One way of finding the volume of the vessel is to determine the
volume occupied by each phase and then add them:
V = Vf + Vg = mf vf + mg vg
= ( 8 * 0.001036 ) + ( 2 * 2.3613 ) = 4.73 m3
Another way is to first determine the quality x, then the average
specific volume v, and finally the total volume:
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
x= mg / m = 2 / 10 = 0.2
v = vf + x vfg
= 0.001036 + 0.2 * (2.3613 - 0.001036) = 0.473 m3/ kg
V = m v = 10 * 0.473 = 4.73 m3
Ex. 3 : Determine the temperature of water at a state of P = 0.5 MPa
and h = 2890 kJ/kg?
Sol. : At 0.5 MPa, the enthalpy of saturated water vapor is
hg = 2747.5 kJ/kg
since h > hg
We have superheated vapor.
From super heated table, at 0.5 MPa search on (h) value was given
corresponded with Temp.
The temperature is between 200 and 220°C. By linear interpolation it
is determined to be:
T = 216.3 oC
Example 4 :Four kilograms of water are placed in an enclosed
volume of 1 m3. Heat is added until the temperature is 150°C. Find the
(a) pressure (b) mass of the vapor, and (c) volume of the vapor.
Solution :
At 150 oC , vg = 0.3924 m3/kg
The volume of 4 kg of saturated vapor at 150°C is :
0.3924 × 4 = 1.57 m3
Since the given volume is less than this, therefore, the
state is saturated mixture.
a ) From steam table @ 150 oC , P sat. = 476 kPa
b ) To find the mass of the vapor we must determine the quality.
v = x vg + ( 1 – x ) vf
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
0.25 = 0.3924 x + 0.001091 ( 1 – x )
x = 0.636
mass of the vapor : mg = m x = 4 * 0.636 = 2.544 kg
c ) volume of the vapor : Vg = mg vg = 2.544 * 0.3924 = 0.998 m 3
Throttling Valves :
When a fluid flows through a valve or other restriction such as
an orifice, a partially open valve, or a porous plug that results a
pressure drop in the fluid without involving any work, change in
kinetic or potential energy. Throttling valves are usually small
devices and the flow through them may be assumed to be adiabatic
(Q = 0) since there is neither sufficient time nor large enough area for
any effective heat transfer to take place. Therefore; the conservation
of energy equation for this flow device reduces to :
H2 = H1 , ∆H = 0
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‫ أحسان حبيب داخل‬.‫م‬
Thermodynamics
By : Ihsan H. Dakhil
Example: Steam enters a throttling valve at 8000 kPa and 300°C and
leaves at a pressure of 2000 kPa. Determine the final temperature and
specific volume of the steam ?
Solution :
The enthalpy of the steam at enters ( 8000 kPa , 300 oC ) is found from
the superheat steam table to be h1 = 2786.8 kJ/kg.
This must equal the exiting enthalpy.
[ h2 = h1 At throttling process ]
since at 2000 kPa hg = 2797.2 kJ/kg ,The exiting steam is in the quality
region
Thus the final temperature T sat2 = 212.4 oC
h2 = x hg + ( 1- x ) hf
2786.8 = 2797.2 x + ( 1 - x ) 908.6
x = 0.994
v2 = x vg + ( 1- x ) vf
= (0.994) 0.099536 + ( 1- 0.994 ) 0.001177
= 0.0989 m3/kg
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