Thermodynamics

2. Thermodynamics-I
Lecture No.05
Akbar Ali Qureshi
Lecturer
Email: akbaraliqureshi@bzu.edu.pk
Contact no: 0335-6387138
Mechanical Engineering Department
UCE&T, BZU Multan.

3. Sequence
• The working fluid
• Liquid, vapor and gas
• The use of vapor tables
• Properties of wet vapor
• Problems

4. The working fluid
• The matter contained within the boundaries of the
system is defined as the working fluid.
• In thermodynamic systems the working fluid can be
in the liquid, vapor or gaseous phase.
• All substances can exist in any of these phases, but
we tend to identify all substances with the phase in
which they are in equilibrium at atmospheric
temperature and pressure.
• For example, substances such as oxygen and nitrogen
are gases; H2O is liquid or vapor (water or steam);
H2O can become a gas at very high temperature.

5. Liquid, vapor and gas .
• The points P, Q and R represents the boiling
points of the liquid.
Boiling points plotted on PV diagram

6. Liquid, vapor and gas...Contd.
 At points P’, Q’ and R’ the liquid is
completely changed into vapor.
Points of complete vaporization plotted on PV diagram

7. Liquid, vapor and gas … Contd.
 Saturation State: A state at which a change of phase may
occur without the change of temperature or pressure.
 Saturated Liquid line: The boiling points P, Q and R are
saturation states, and series of such boiling points joined
up is called the saturated liquid line.
 Saturated vapor line: The points P’,Q’ and R’ are saturation
states, at which liquid is completely changed into vapor,
and the series of such points joined up is called the
saturated vapor line.
 Wet vapor: The substance existing at a state point inside
loop consists of a mixture of liquid and dry vapor and is
known as wet vapor.

8. Liquid, vapor and gas ...Contd.
 Specific enthalpy of vaporization: The additional heat
supplied which changes the phase of the substance from
liquid to vapor at constant temperature and pressure is
called specific enthalpy of vaporization.

9. Liquid, vapor and gas … Contd.
 Isotherms: The lines of constant temperatures.
 Superheated: When a dry saturated vapor is heated at constant
pressure, its temperature rises and it becomes superheated.
Isothermals for a vapor plotted on PV diagram

10. Liquid, vapor and gas ...Contd.
• Dryness fraction: The condition or quality of a wet vapor
is most frequently defined by its dryness fraction.
Dryness fraction, x= mass of dry vapor in 1kg of mixture.
Note: For dry saturated vapor, x=1
For dry saturated liquid, x=0

11. The use of vapor tables.
• The tables which will be used in this course are arranged
by Rogers and Mayhew.
• The tables are mainly concerned with steam, but some
properties of refrigerants are also given.
• For example, the table for wet steam is shown below.

12. Properties of wet vapor.
• The properties of wet vapor are.

13. Problems.
Problem1: Calculate the specific volume, specific enthalpy
and specific internal energy of wet steam at 18 bar,
dryness fraction 0.9.(v=0.0994m3/kg, h=2605.8kJ/kg and
u=2426.5kJ/kg)
Problem 2: Calculate the dryness fraction, specific volume
and specific internal energy of steam at 7 bar and
specific enthalpy 2600kJ/kg.(x=0.921, v=0.2515m3/kg
and u=2420kJ/kg)

14. Sequence
• The properties of superheated vapor
• Problems
• Interpolation
• The perfect gas
• Problems

15. The properties of superheated
vapor
• For steam in superheated region, temperature and
pressure are independent properties.
• When the temperature and pressure are given for
superheated steam then the state is defined and all the
other properties can be found.
• For example, steam at 2 bar and 200C is superheated
since the saturation temperature at 2 bar is 120.2C,
which is less than the actual temperature. The steam
in this state has a degree of superheat of 220-
120.2=79.8C

16. Problems.
Problem 1: Steam at 110 bar has a specific volume of
0.0196m3/kg. Calculate the temperature, the specific
enthalpy and specific internal energy.
Problem 2: Steam at 150 bar has a specific enthalpy of
3309kJ/kg. Calculate the temperature, the specific
volume and specific internal energy

17. Interpolation.
 For properties which are not tabulated
exactly in the tables it is necessary to
interpolate between the values tabulated.
 For example to find the temperature at
9.8 bar, it is necessary to interpolate
between the values given in the tables.

18. The perfect gas.
 The characteristic equation of state:
At temperatures that are considerably in excess of the
critical temperature of a fluid, and also at very low
pressures, the vapor of the fluid tends to obey the
equation.
• The constant R is called the specific gas constant and the
units of R are Nm/kgK or kJ/kgK
• Perfect gas: An imaginary ideal gas which obeys the law is
called a perfect gas.
The characteristic equation of state of a perfect gas

19. The perfect gas ...Contd.
 For a mass m and occupying a volume V,
……………………(a)
 Another form of the characteristic equation can be derived by
using the amount of substance (mol).
 The amount of substance of a system is that quantity which
contains as many elementary entities as there are atoms in 0.012kg
of C-12.
 The normal unit for the amount of substance is ‘mol’. In SI it is
convenient to use ‘kmol’
 Molar mass: The mass of any substance per amount of substance
is known as the molar mass.
where m is the mass and n is the amount of substance and the units of
molar mass is kg/kmol .

20. The perfect gas ...Contd.
 Substituting for m in eq. (a) gives,
 Now Avogadro’s hypothesis states that the volume of 1
mol of any gas is same as volume of 1 mol of any other
gas, when the gases are at the same temperature and
pressure .
 Therefore V/n is same for all gases at same value of p
and T. i.e. the quantity pV/nT is constant for all gases .
this constant is called the molar gas constant. And is
given the symbol R̃ and the value of R is 8.3145kJ/kmolK

21. Problems.
Problem1: A vessel of volume 0.2m3 contains nitrogen at
1.013bar and 15C. If 0.2kg of nitrogen is now pumped
into the vessel, calculate the new pressure when the
vessel has returned to its initial temperature. The molar
mass of nitrogen is 28kg/kmol, and it may be assumed to
be a perfect gas. (Answer 1.87 bar)
Problem 2: A certain perfect gas of mass 0.01 kg occupies a
volume of 0.003m3 at a pressure of 7 bar and
temperature of 131C. The gas is allowed to expand until
the pressure is 1 bar and the final volume is 0.02m.
Calculate the molar mass of the gas and the final
temperature. (Answers 16 kg/kmol; 111.5C)

22. Specific Heat
Specific Heat at Constant Pressure (CP) :
The Energy required to raise the temperature of a unit mass of a substance by 1
degree, as the Pressure is maintained CONSTANT.
Specific Heat at Constant Volume (CV) :
The Energy required to raise the temperature of a unit mass of a substance by 1
degree, as the Volume is maintained CONSTANT.
m = 1 kg
∆T = 1 ºC
Sp. Heat = 5 kJ/kg ºC
5 kJ
DEFINITION :
The Energy required to raise the temperature of a
unit mass of a substance by 1 degree.

23. Specific heat

24. Specific heat at constant volume Cv
Hence, CV is change in Internal
Energy of a substance per unit
change in temperature at constant
Volume.

25. Specific Heat at constant Pressure
Hence, CP is change in Enthalpy of a
substance per unit change in temperature
at constant Pressure.

26. Relation between Specific Heat

27. The End