UNIT-I

BASIC CONCEPTS AND FIRST LAW

Thermodynamics
Thermodynamics is a branch of science that deals with the relationship between the heat and
mechanical energy.
Thermodynamics Laws and Applications
Thermodynamics laws are formulated based on the experience and the results of experiments.
These laws govern the principles of energy transfer. These laws are used in the field of energy transfer,
like, steam and nuclear power plants, gas turbines, internal combustion engines, air conditioning plants,
refrigeration, jet propulsion, compressors, chemical process plants, etc.
Macroscopic and Microscopic concepts
In macroscopic approach, the behaviour of the matter is studied considering a certain quantity of
matter (consisting of many molecules), without the events occurring in the molecular level. Action of
many molecules can be perceived by human senses.
In microscopic approach, the matter is composed of myriads of the molecules and the behaviour
of the matter is described by summing up behaviour of the individual molecules.
Concept of Continuum
In macroscopic analysis we consider a small amount (small volume) of matter. This small volume
is very large compared to molecular dimensions. Even a very small volume of a matter contains large
number of molecules.
Consider the mass δm in a volume δV. The average mass density is defined as δm/δV.

Domain of Domain of
molecular effect continuum
δm/δV
δm/δV

δm/δV ρ

δV, δm δV’
δV
If δV = 0, δm/δV = ∞
If δm = 0, δm/δV = 0
The smallest volume which may be regarded as continuous is δV’. The density ρ of the system at
a point is defined as
Thermodynamic System, Surroundings and Boundary
A thermodynamic system is a specified quantity of matter or a region in a space upon which the
attention is focused on to analyse the problem.

System

Surroundings
Boundary
Everything outside the system is called surroundings. The system and surroundings is separated
by a boundary. The boundary may be real or imaginary, fixed or moving.
There are three types of systems:
1. Open system
2. Closed system
3. Isolated system

Open system

System Mass out

Mass transfer

Boundary Energy transfer

The open system is one in which both matter (mass) and energy (work or heat) cross the
boundary of the system.
Examples: Steam turbine, Gas turbine, Rotary compressors, Heat exchangers, Centrifugal pump, Water
turbine, etc.

∫
Work Transfer = W = − V dp

Closed system
The closed system is a system of fixed mass. In closed system, there is no mass transfer, but
energy transfer takes place.
System

No mass transfer

Boundary Energy transfer

Examples: I.C.engine cylinder with the both valves closed, Reciprocating air compressor with both the
valves closed.

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 Work transfer = W = ∫ p dV
Isolated system
In isolated system, there is no mass or energy transfer across the boundary of the system.

System

No mass transfer

Boundary No energy transfer

Universe
The system and surroundings together is called “Universe”.
Thermodynamic Equilibrium
The system under Mechanical, Thermal and Chemical equilibrium is said to be under
Thermodynamic equilibrium.
A system is said to be in mechanical equilibrium when there is no unbalanced force within the
system. (Uniformity of force).
A system is said to be in chemical equilibrium when there is no chemical reaction between
different parts of the system. (Absence of chemical reaction).
A system is said to be in thermal equilibrium when there is no temperature change between
different parts of the system. (Uniformity of temperature).
The system under mechanical and thermal equilibrium but not under chemical equilibrium is said
to be under meta-stable equilibrium.
Thermodynamic Property
Any observable characteristic of the system under thermodynamic equilibrium is called a
property.
Examples: Pressure, Temperature, Volume, Density.
Intensive and Extensive property
The property which is not depending on the mass of the system is called intensive or intrinsic
property.
Examples: Pressure, Temperature, Density, Specific volume (m3/kg), Specific enthalpy (J/kg)
The property which is depending on the mass of the system is called extensive or extrinsic
property.
Examples: Volume (m3), Mass (kg), Enthalpy (J)
Consider a system composed of some fluid. Measure all the properties of the system, like,
pressure (p1), volume (V1), temperature (T1), density (ρ1), enthalpy (H1), mass (m1), specific volume (v1),
specific enthalpy (h1), etc. Now, disturb the mass by placing a partition and measure all the properties of a
part (p2, V2, T2, ρ2, H2, m2, v2, h2).

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 It can be seen that,
p1 = p2 ; T1 = T2 ; V1 ≠ V2 ; ρ1 = ρ2 ; H1 ≠ H2 ; m1 ≠ m2 ; v1 = v2 ; h1 = h2

System

Partition
Thermodynamic State

p State

.
V
A state is the condition of the system at an instant of time as described or measured by its
properties. Or each unique condition of the system is called state.
Example: p = 8 bar & V = 0.5 m3
Thermodynamic Path and Process
The succession of states passed through during a change of state is called the “Path”. When the
path is completely specified, the change of state is called a process.

Quasi-static process
Quasi means ‘almost’. A quasi-static process is a succession of equilibrium states and infinite
slowness is its characteristic feature.
Stops

Final position (2)

Piston Weight
Initial position (1)
Gas P
4
 Consider a system of gas contained in a cylinder. The force exerted by the gas is balanced by the
weight. The gas is in equilibrium in the initial state. If the weight is removed, the piston will move up
until it hits the stops. The system again comes to equilibrium state. The intermediate states between 1 and
2 are non-equilibrium states and cannot be represented by the thermodynamic coordinates.
p1 1

Locus of Non-equilibrium states

p

p2 2

V1 V V2
Now if single weight on the piston is made up of very small pieces of weights and these weights
are removed one by one very slowly from the top of the piston, at any instant of the upward travel of the
piston, the departure of the state of the system from equilibrium state will be infinitesimally small. So,
each intermediate state will be an equilibrium state. Such a process which is locus of all the equilibrium
points is called “Quasi-static process”.
1

x Locus of equilibrium states

x
p x
x
x2

V
Thermodynamic Cycle
A single process cannot take place continuously without end. To get continuous work, a set of
processes has to be repeated again and again. The set of processes which brings the system to the original
state is called a cycle. Continuous work transfer is possible only with a cycle.
Examples: Otto cycle, Diesel cycle, Dual cycle, etc.
Work

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 Work is one form of energy transfer. Work is said to be done by a system if the sole effect on
things external to the system can be reduced to the raising of weight. The weight may not actually be
raised, but the net effect external to the system would be the raising of a weight. Consider a Battery-motor
system driving a fan as shown in fig. The system is doing work upon the surroundings.
When the fan is replaced by a pulley and a weight as shown, the weight may be raised. The sole
effect on the things external to the system is then raising of a weight.

Work done by a system Æ +ve

Work done on the system Æ - ve
Unit of work Æ Nm (or) J
pdV work (or) Displacement work
Consider a cylinder-piston arrangement. Let p1 and V1 be pressure and volume of the gas in the
cylinder at initial position 1. Let the piston moves to a new final position 2 specified by pressure p2 and
volume V2. At any point in the travel of the piston, let the pressure be p and the volume V.

Work done = Force x Distance moved

dW = F x dl = p A dl = p dV
When the piston moves out from position 1 to position 2 with the volume changing from V1 to V2,
the amount of work W done by the system will be
V2
W1− 2 = ∫ p dV
V1

Work transfer = Area under the curve on a p-V diagram

Electrical work
When a current flows through a resistor, there is work transfer into the system. This is because the
current can drive a motor, the motor can drive a pulley and the pulley can raise a weight.

6
 Work transfer, W = E I
E Æ Potential difference (Voltage)
I Æ Current
Shaft work

When the shaft is rotated by a motor, there is work transfer into the system. This is because the
shaft can rotate a pulley which can raise a weight. If T is the torque applied to the shaft and ω is the
angular velocity, work transfer can be written as
W=Tω
Paddle-wheel work or stirring work

As the weight is lowered, and the paddle wheel turns, there is work transfer into the fluid system
which gets stirred. Since the volume of the system remains constant, pdV work is zero. If m is the mass of
the weight lowered through a distance dZ and T is the torque transmitted by the shaft in rotating through
an angle of dθ, the work transfer is given by,
2 2
W = ∫1
W ' dz = ∫
1
T dθ

W’ Æ Weight lowered
Flow work
The flow is significant in open system. This work represents the energy transferred across the
system boundary as result of the energy imparted to the fluid by a pump, blower or compressor to make
the fluid to flow across the control volume.
Flow work is given by, W = p V
Work done in stretching a wire
Let us consider a wire of length L, subjected to a tension force T. The infinitesimal amount of
work is done on the wire which makes the wire to stretch to a length L + dL.
2
W =− ∫
1
T dL

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 T T
L

Negative work Æ The work is done on the wire.

Work done in changing the area of a surface film
A film on the surface of a liquid has a surface tension, which is a property of the liquid and the
surroundings. The surface tension acts to make the surface are of the liquid a minimum. The work done
on a homogeneous liquid film in changing its surface area by an infinitesimal amount dA is
2
W =− ∫ 1
σ dA
σ Æ Surface tension (N/m)
Magnetization of a paramagnetic solid
The work done per unit volume on a magnetic material through which the magnetic and
magnetization fields are uniform is
I2
W = ∫I1
H dI

H Æ Field strength
I Æ Component of the magnetization field in the direction of the field
Heat
Heat is defined as the form of energy that is transferred across the boundary by virtue of a
temperature difference. The temperature difference is ‘potential’ and heat transfer per unit area is ‘flux’.
Zeroth Law of Thermodynamics
If the bodies A and B are in thermal equilibrium with the third body C, then these two bodies A
and B will be in thermal equilibrium with each other. This law is the basis for measurement of
temperature.

A B

C
If TA = TC & TB = TC then TA = TB
Application of Zeroth law
It is the basis for temperature measurement.
Ideal and Real gases
An ideal gas or perfect gas is a hypothetical gas consisting of identical particles of zero volume,
with no intermolecular forces. Additionally, the constituent atoms or molecules undergo perfectly elastic
collisions with the walls of the container. Real gases do not exhibit these exact properties. Gases are most
ideal at high temperatures and low pressure.
An ideal gas obeys the perfect gas law. The specific heats are constant.

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 Perfect gas law, p V = n Ru T = m R T
p Æ Pressure in Pa
V Æ Volume in m3
n Æ Amount of gas in kg-mole = m / M
M Æ Molecular weight
Ru Æ Universal gas constant = 8314 J/kg-mole K
m Æ Amount of gas in kg
R Æ Characteristic gas constant in J/kg K
T Æ Temperature of gas in K
In reality there is no ideal or perfect gas. At a very low pressure and at a very high temperature,
real gases like nitrogen, hydrogen, oxygen, helium, etc., behave as perfect gases. These gases are called
‘Semi perfect’ or ‘Permanent’ gases. For real gases specific heats vary appreciably with temperature
and little with pressure.
Internal energy and enthalpy
Internal Energy Æ Energy possessed by the system.
Joule’s law states that the specific internal energy of a gas depends only on the temperature of the gas and
is independent of both pressure and volume.
u = f(T)
Enthalpy Æ Total heat content of the flowing fluid
From definition of enthalpy, h = u + pv
But pv = RT
Therefore, h = u + RT and h also function of temperature only.
h = f(T)
Specific heats
Specific heat at constant volume is defined as the amount of heat required to rise the temperature of 1 kg
of fluid through 1oC when the volume is kept constant.
⎛ ∂u ⎞
The specific heat capacity at constant volume is defined as C v = ⎜ ⎟
⎝ ∂T ⎠ v
du
We know that u = f(T). Therefore C v =
dT
Specific heat at constant pressure is defined as the amount of heat required to rise the temperature of 1 kg
of fluid through 1oC when the pressure is kept constant.
⎛ ∂h ⎞
The specific heat capacity at constant pressure is defined as C p = ⎜ ⎟
⎝ ∂T ⎠ p
dh
We know that h = f(T). Therefore C p =
dT

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First Law of thermodynamics
I-Corollary:
For a cycle:
Whenever a system undergoes a cyclic change, the algebraic sum of heat transfer is proportional
to the algebraic sum of work transfer.
Net heat transfer = Net work transfer

∫ dQ = ∫ dW
For a process:
dQ = dW + dU

or Q = W + ΔU

ΔU = Change in internal energy = m Cv Δ T

m = Mass of the working fluid

Cv = Specific heat at constant volume

Δ T = Change in temperature

II-Corollary: (Law of Conservation of Energy)

In an isolated system, the energy of the system remains constant.

Q=W=0 & ΔU=0

III-Corollary: (PMM-I)
PMM-I is impossible. (Perpetual Motion Machine of First kind produces work continuously
without any input).
PMM-I

Q=0
Application of I-Law to Non-Flow or Closed system
Let,
m = Mass of the working fluid
p1 = Initial pressure of the working fluid
p2 = Final pressure of the working fluid
T1 = Initial temperature of the working fluid
T2 = Final temperature of the working fluid
U1 = Initial internal energy of the fluid
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 U2 = Final internal energy of the fluid
W = Work transfer
Q = Heat transfer
Cp = Specific heat at constant pressure
Cv = Specific heat at constant volume
Constant volume process (Isometric process)
In a constant volume process the working fluid is contained in a closed vessel. The boundary of
the system is immovable and hence no work transfer is possible through boundary of the system.
Consider a vessel containing m kg of certain gas. Q J of heat is supplied to the gas and there will
be pressure rise, but the volume remains constant.

m kg of gas

Q 1

V
During constant volume process, dV = 0
pV
General gas equation Æ =C
T
p
For constant volume process, =C
T
Work transfer = W = ∫ p dV = 0 ;
As per I-law for a process, Q=W+ΔU
Q = Δ U = m Cv (T2 – T1)
Constant pressure process (Isobaric process)

Piston

Cylinder
m kg of gas
p 1 2

Q
V

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 Consider a cylinder-piston arrangement as shown in fig. The piston is free to move up and down.
Let the cylinder contain m kg of certain gas. The heat is added to the gas. Since the piston is free to move,
the pressure remains constant and there is increase of volume. In this case the boundary of the system is
movable.
During constant pressure process, p = C
pV
General gas equation Æ =C
T
V
For constant pressure process, =C
T
2

Work transfer W =
∫ p dV = p ∫ dV = p (V
1
2 − V1 )

As per I-law for a process, Q=W+ΔU

Q = p (V2 – V1) + m Cv (T2 – T1)
= m R (T2 – T1) + m Cv (T2 – T1)
= m (R + Cv) (T2 – T1)
= m Cp (T2 – T1)
Constant temperature process (Isothermal process or Hyperbolic process)
A process at a constant temperature is called an isothermal process. When a working substance in
a cylinder behind a piston expands from a high pressure to a low pressure there is a tendency for the
temperature to fall. In an isothermal expansion heat must be added continuously in order to keep the
temperature at the initial value. Similarly in an isothermal compression heat must be removed from
working fluid continuously during the process.
During constant temperature process, T = C
pV
General gas equation Æ =C
T

For constant temperature process, pV = C

Piston p1 1

Cylinder
m kg of gas p

Q p2 2

V1 V V2

C dV ⎡V ⎤
Work transfer, W = ∫ pdV = ∫ V dV = C ∫ V = pV ln ⎢ 2 ⎥
⎣ V1 ⎦

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 ⎡V ⎤ ⎡V ⎤
= p1V1 ln ⎢ 2 ⎥ = p 2V2 ln ⎢ 2 ⎥
⎣ V1 ⎦ ⎣ V1 ⎦
As per I-law for a process, Q=W+ΔU

⎡V ⎤
= pV ln ⎢ 2 ⎥ + m C v (T2 − T1 )
⎣ V1 ⎦
⎡V ⎤ ⎡V ⎤
= pV ln ⎢ 2 ⎥ + 0 = p1V1 ln ⎢ 2 ⎥
⎣ V1 ⎦ ⎣ V1 ⎦
During isothermal process, Q = W
Note: The isothermal and hyperbolic processes are identical only in the case of a perfect gas and not for a
vapour. For example the isothermal expansion of wet steam is not hyperbolic.
Constant entropy process (Isentropic process)
In an isentropic process, the heat transfer between the working fluid and surroundings is zero.
During isentropic process, Q = 0
The governing equation for isentropic or reversible adiabatic process is,
p Vγ = C
γ Æ Specific heat ratio = Cp/Cv
W

Piston p1 1

Cylinder
m kg of gas p

p2 2
Q=0

V1 V V2
Work transfer
2 2

∫ ∫ Vγ ∫ Vγ
C dV
W = p dV = dV =C
1 1
V2
⎡ V −γ +1 ⎤
=C ⎢ ⎥
⎣ − γ + 1 ⎦ V1
⎡ V −γ + 1 − V −γ + 1 ⎤
=C ⎢ 2 1 ⎥
⎢⎣ − γ + 1 ⎥⎦

⎡V −γ +1 − V −γ +1 ⎤
= pV γ ⎢ 2 1 ⎥
⎢⎣ −γ +1 ⎥⎦
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 pVγ = p1V1γ = p2V2γ = C
⎡ p V − p 1V 1 ⎤
W =⎢ 2 2 ⎥
⎣ 1− γ ⎦
Heat transfer = Q = W + ΔU = 0
Therefore we can write, W = - ΔU = - m Cv (T2 – T1)
Polytropic process
In polytropic process, there is heat transfer between the working fluid and surroundings.
Pressure, volume and temperature are variables during a process.
The governing equation for isentropic or reversible adiabatic process is,
p Vn = C
n Æ Polytropic index

Piston p1 1

Cylinder
m kg of gas p

p2 2
Q≠0

V1 V V2
Work transfer
2 2

∫ ∫V ∫V
C dV
W = p dV = dV = C
n n
1 1
2
⎡ V − n +1 ⎤
=C ⎢ ⎥
⎣ − n + 1 ⎦1
⎡V −n +1 − V −n +1 ⎤
=C ⎢ 2 1 ⎥
⎢⎣ − n + 1 ⎥⎦

−n +1
⎡
n ⎢V2 − V1−n +1 ⎤
= pV ⎥
⎢⎣ − n +1 ⎥⎦
pVn = p1V1n = p2V2n
⎡ p V − p1V1 ⎤ ⎡ mR (T 2 − T1 ) ⎤
W =⎢ 2 2 ⎥=⎢ ⎥
⎣ 1− n ⎦ ⎣ 1− n ⎦
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 Heat transfer = Q = W + Δ U

⎡ mR (T 2 − T1 ⎤
Q =⎢ ⎥ + m C v (T 2 − T1 )
⎣ 1− n ⎦
⎡ mR (T 2 − T 1 ) ⎛ C (1 − n ) ⎞ ⎤
=⎢ ⎜⎜ 1 + v ⎟⎟ ⎥
⎣ 1− n ⎝ R ⎠⎦
⎡ mR (T 2 − T 1 ) ⎛ R + C v − C v n ) ⎞ ⎤
=⎢ ⎜⎜ ⎟⎟ ⎥
⎣ 1− n ⎝ R ⎠⎦
⎡ mR (T − T ) ⎛ C p − Cv + Cv − Cv n) ⎞⎤
=⎢ 2 1 ⎜ ⎟⎥
⎢ 1− n ⎜ C p − Cv ⎟⎥
⎣ ⎝ ⎠⎦
⎡ mR (T − T ) ⎛ C p − C v n ) ⎞ ⎤
=⎢ 2 1 ⎜ ⎟⎥
⎢ 1 − n ⎜ C − C v ⎟⎠ ⎥⎦
⎣ ⎝ p
Dividing both numerator and denominator by Cv
⎡ mR (T 2 − T 1 ) ⎛ γ − n ) ⎞ ⎤
Q =⎢ ⎜⎜ ⎟⎟ ⎥
⎢⎣ 1− n ⎝ γ − 1 ⎠ ⎥⎦

⎡⎛ γ − n ) ⎞ ⎤
Q = ⎢ ⎜⎜ ⎟⎟ W ⎥
⎢⎣ ⎝ γ − 1 ⎠ ⎥⎦

Free expansion process (Constant internal energy process)

Consider two vessels A and B interconnected by a short pipe with a valve ‘V’, and perfectly
thermally insulated. Initially let the vessel A be filled with a fluid at a certain pressure and let vessel B be
completely evacuated. When the valve V is opened, the fluid in vessel A will expand rapidly to fill the
vessel B. The pressure finally will be lower than the initial pressure in the vessel A. This is known as
“Free or Unrestricted expansion”. The process is highly irreversible, since the fluid is eddying
continuously during the process.

Perfect insulation
In free expansion process, W = 0; Q = 0;

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 As per I-Law, Q=W+ΔU
ΔU=0
∴ U2 = U1

Steady Flow Energy Equation (SFEE) (I-Law for open system)

Consider a system as shown in fig. Let m kg/s of working fluid is entering and leaving the
system.
Let, m = Mass flow rate of fluid through the system --- kg/s
U1 = Internal energy of the fluid at inlet --- W
C1 = Velocity of the fluid at inlet --- m/s
p1 = Pressure of the fluid at inlet -- Pa
V1 = Volume flow rate of fluid at inlet --- m3/s
Z1 = Height of the inlet section from datum --- m
Q = Heat transfer through the system --- W
W = Work transfer through the system --- W
U2, C2, p2, V2 and Z2 are corresponding values at outlet
(1)

Z1

(2)
Z2
Datum Line
In addition to internal energy, other forms of energy associated with the mass entering and
leaving the system is considered here.
Kinetic energy of fluid = m C2/2
Potential energy of the fluid = m g Z
Flow energy of fluid = pV
Between inlet and outlet, the energy equation can be written as,

C2 C 2
U1 + p1V1 + m 1 + mgZ1 + Q = U 2 + p2V2 + m 2 + mgZ2 + W
2 2

C2 C 2
mu1 + mp1v1 + m 1 + mgZ1 + Q = mu 2 + mp2v 2 + m 2 + mgZ 2 + W
2 2

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 C12 C 2
mh1 + m + mgZ 1 + Q = mh 2 + m 2 + mgZ 2 + W
2 2
Where, v Æ Specific volume = V/m
u Æ Specific internal energy = U/m
h Æ Specific enthalpy = u + pv
Application of I-Law to open system
Water turbine
In a water turbine water is supplied from a height. The potential energy of water is converted into
kinetic energy when it enters into the turbine and part of it is converted into useful work which is used to
generate electricity.
SFEE is given by,

C2 C 2
mu1 + mp1v1 + m 1 + mgZ 1 + Q = mu 2 + mp 2v 2 + m 2 + mgZ 2 + W
2 2
Generally in water turbine,
Heat transfer (Q) = 0; Z2 = 0; v1 = v2; T1 = T2 or U1 = U2
Energy equation becomes,

C2 C 2
mp1v1 + m 1 + mgZ1 = mp2v 2 + m 2 + W
2 2

(1)

Z1

Generator (2) Z2 = 0

Turbine

Note: W is positive since work is done by the system.

Steam or Gas turbine
In a steam or gas turbine, steam or gas is passed through the turbine and part of its energy is
converted into work in the turbine. The output of the turbine runs a generator to produce electricity.
Steam or gas in
Generator (1)

(2)
Exhaust steam or gas

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Generally in steam or gas turbine,
Z2 – Z1 = 0; Q = 0 (insulated turbine); Q = - ve if not perfectly insulated.
Energy equation becomes,
C 2 C 2
mh1 + m 1 − Q = mh 2 + m 2 + W
2 2
Note: W is positive since work is done by the system. Q is negative since the heat is transferred from the
hot casing to low temperature surroundings.
Water pump
(2)
Tank

Generator Pump

(1)
Sump
A water pump draws water from a lower level and pumps it to higher level. Work is required to
run the pump and this may be supplied from an external source such as an electric motor or a diesel
engine.
Generally in water pump,
Heat transfer (Q) = 0; v1 = v2; T1 = T2 or U1 = U2
Energy equation becomes,

C 2 C 2
mp1v 1 + m 1 + mgZ 1 = mp 2 v 2 + m 2 + mgZ 2 − W
2 2
Note: W is negative since work is done on the system.
Centrifugal air compressor
A centrifugal compressor compresses air and supplies the same at moderate pressure and in large
quantity.
Compressed air out
(2)

Generator

(1)
Air from atmosphere
Generally in centrifugal compressor,
Z2 – Z1 = 0; Q = - ve if not perfectly insulated.

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Energy equation becomes,

C12 C22
mh1 + m − Q = mh 2 + m −W
2 2
Note: W is negative since work is done on the system. Q also is negative since the heat is transferred from
the hot casing to low temperature surroundings.
Reciprocating air compressor
Air in Reservoir

Cylinder

Piston

The reciprocating air compressor draws in air from atmosphere and supply it at relatively higher
pressure and in small quantity. The velocity of air entering and leaving the compressor is generally very
small and is neglected.

Generally in reciprocating compressor,

Z2 – Z1 = 0; C1 = C2; Q = - ve if not perfectly insulated.
Energy equation becomes,

C 2 C 2
m 1 −Q = m 2 −W
2 2
Note: W is negative since work is done on the system. Q also is negative since the heat is transferred from
the hot casing to low temperature surroundings.
Heat Exchangers
A heat exchanger is a device to transfer heat from one fluid to another fluid through walls. There
is enthalpy change.
Examples: Boiler, Condenser, Evaporator.
Generally in heat exchangers,

Z2 – Z1 = 0; C1 = C2; Q = + ve in boilers and evaporators

Q = - ve in evaporators

Fluid in (1) (2) Fluid out

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Boilers:
A boiler produces high temperature vapour absorbing heat from the external source. The heat is
transferred to liquid.
Energy equation becomes, mh 1 + Q = mh 2
Condensers:
A condenser is a device to condense the vapour by rejecting its heat to the cooling medium. Here
the heat is rejected by the system.
Energy equation becomes, mh 1 − Q = mh 2

Evaporators:
An evaporator is device which produces low temperature vapour by absorbing heat from
relatively hot source. Here the heat is absorbed by the system.
Energy equation becomes, mh 1 + Q = mh 2

Steam nozzle
Steam nozzles are used in steam engines and steam power plants. It converts the pressure energy
into kinetic energy. Generally the nozzles are insulated. In steam nozzles the steam is expanded
isentropically.

Steam in Steam out

Generally in nozzles, Z2 – Z1 = 0; Q = 0; W = 0;

C 12 C 2
Energy equation becomes, mh 1 + m = mh 2 + m 2
2 2

2
The exit velocity of the steam can be written as, C2 = C1 + 2(h1 − h2 )

Throttling process (Constant enthalpy process)

When a fluid flows through a constricted passage, like a partially opened valve, an orifice, or a
porous plug, there is an appreciable drop in pressure and the flow is said to be throttled. The fig shows the
process of throttling by an orifice. The pipe is perfectly insulated.

(1) (2)

Orifice
Heat transfer (Q) = 0
Work transfer (W) = 0
C1 = C2 ; Z1 = Z2
Energy equation becomes, h1 = h 2
20
The throttling process is commonly used for the following process:
(i) For determining the condition of steam (dryness fraction)
(ii) For controlling the speed of the turbine
(iii) Used in refrigeration plant for reducing the pressure of the refrigerant before entry into the
evaporator
Path and Point function
Consider a process 1-2 as shown in fig (a). The process 1-2 may follow a path A, B or C. The
area under a curve in p-V diagram represents the work transfer and the area under a curve in a T-s
diagram represents the heat transfer as shown in fig (b). Work or heat depends on the path of the system,
but not on the end states. For this reason, work or heat is called a path function. W1-2 or Q1-2 is an inexact
or imperfect differential.
2
∫
1
dW = W2 − W1

Thermodynamic properties are point functions, since for a given state, there is a definite value for
each property. The change in a thermodynamic property of a system in a change of state is independent of
the path of the system and depends only on the end states of the system. The differentials of point
functions are exact or perfect differentials.
V2
∫V1
dV = V2 − V1

Note: In a cyclic process, the change in any property is zero.

∫ dV = 0, ∫ dp = 0, ∫ dT = 0

(a)

(b)
21
Internal energy – A property of the system
Consider a system which changes its state from state 1 to state 2 by following the path A and
returns to its original state 1 following (i) the path B and (ii) the path C.
Consider a cycle 1-A-2-B-1.

As per I-law for a cycle, ∫ dQ = ∫ dW

QA + QB = WA + WB
QA – WA = WB – QB ---------- (1)
For a process 1-A-2,
QA = WA + ΔUA
QA – WA = ΔUA --------- (2)

For a process 2-B-1,

QB = WB + ΔUB
WB – QB = - ΔUB --------- (3)
Therefore, from (1), (2) and (3), we can write,
ΔUA = - ΔUB --------- (4)
Consider a cycle 1-A-2-C-1.
QA + QC = WA + WC
QA – WA = WC – QC --------- (5)
For a process 2-C-1,
QC = WC + ΔUC
WC – QC = - ΔUC --------- (6)
Therefore, from (2), (5) and (6), we can write,
ΔUA = - ΔUC --------- (7)
From (4) and (7), ΔUC = ΔUB
From the above, it can be concluded that the internal energy depends on end states and
independent of path. Therefore the internal energy is a property.
Molar volume (Vm)
Molar volume Vn of a substance is the ratio of the volume V of a sample of that substance to the
number of moles in the sample.

22
 The molar volume for an ideal gas at 298.15 K and 1 bar is 0.024788 m³mol-1 or 24.788 l/mol.

Vm = V / Number of moles (n) ------ m3/kmol

V Æ Volume in m3
M Æ Molar mass in kg/kmol
ρ Æ Density in kg/m3
p V = m R T = n Ru T
p Vm = Ru T
R Æ Characteristic gas constant in J/kg-K
Ru Æ Universal gas constant in J/kmol-K
Characteristic and Universal Gas constant
Characteristic gas constant (R) = Cp – Cv J/kg-K
Universal gas constant (Ru) = Cp(mole) – Cv(mole) J/kmol-K
Cp(mole) Æ Molar specific heat at constant pressure = M Cp J/kmol-K
Cv(mole) Æ Molar specific heat at constant volume = M Cv J/kmol-K
Molar mass and Molecular mass (M)
Molar mass is the mass of one mole of the substance and is defined in kg/kmol. Molecular mass
(M) is the mass of one molecule of the substance and is defined in amu (automic mass unit). Molar mass
and molecular mass differs only in units.
1 amu ≈ 1.660538782(83) × 10−27 kg
Note:
It is advisable to use appropriate units while numerical problems are being solved.
Pressure (p) N/m2 (or) Pa
Temperature (T) K
Volume (V) m3
∗
Volume flow rate ( V ) m3/s
Mass (m) kg
∗
Mass flow rate ( m ) kg/s
Velocity (C) m/s
Angular velocity (ω) rad/s
Speed (N) rpm
Energy Nm (or) J
Force (F) N
Power (P) J/s (or) W
Specific heat (Cp or Cv) J/kg-K (or) J/kg oC
Characteristic Gas constant (R) J/kg-K
Universal gas constant (Ru) 8134 J/kmol-K

23
Molar volume or Molecular volume (Vm) m3/kmol
Molecular mass (M) amu
Enthalpy (H) J
Specific enthalpy (h) J/kg
Entropy (S) J/K
Specific entropy (s) J/kg-K
Molar specific heat (Cp(mole) or Cv(mole)) J/kmol-K
Gravitational acceleration (g) 9.81 m/s2
Limitations of I-law of thermodynamics
The I-law states that, in carrying out a process, heat & work are mutually convertible, a balance
of energy must hold as internal energy, the energy is neither gained nor lost in a process, it only
transforms.
But, this law does not place any distinction on the direction of the process, under consideration.
According to the I-law, it is assumed that any change of thermodynamic state can take place in
either direction. But it has been found that this is not the case particularly in the inter-conversion of heat
and work.
The processes naturally proceed in certain directions and not in the opposite directions, even
though the reversal of the processes does not violate the I-law.
Example: If two metal blocks at temperatures T1 & T2 (T1 > T2) are brought into contact with each other,
the heat flows from the high temperature block to the low temperature block till the temperature of both
the blocks are equal. The heat flows from low temperature block to the high temperature block is
impossible, i.e., the original temperatures T1 & T2 cannot be restored.
Therefore the I-law is a necessary but not sufficient due to the following restrictions:
(i) No restriction on the direction of energy flow
(ii) It does not deal with the portion of heat that may be converted into useful work

24
 PROBLEMS
1. Following amount of heat transfer occurs during a cycle comprising of four processes.
Calculate the workdone of the cycle and indicate about the type of work. +120 kJ, -20 kJ,
+16 kJ and +24 kJ.
Given: Cycle with four processes
Heat transfer during process 1 – 2 = Q1-2 = 120 kJ
Heat transfer during process 2 – 3 = Q2-3 = - 20 kJ
Heat transfer during process 3 – 4 = Q3-4 = 16 kJ
Heat transfer during process 4 – 1 = Q4-1 = 24 kJ
Required: Workdone and type of work
Solution:
First law for cycle Æ Net heat transfer = Net work transfer

∫ dQ = ∫ dW
∫ dW =120 + (−20) + 16 + 24 =140 kJ
The workdone is positive, therefore the work is done by the system.
2. The following data refer to a closed system which undergoes a thermodynamics cycle
consisting of four processes. Show that the data is consistent with the I-law of
thermodynamics and calculate, (a) Net rate of work output in kW and (b) Change in
internal energy.

Process Heat Transfer Work Transfer

(kJ/min) (kJ/min)

1–2 Nil - 1000

2–3 40,000 Nil

3–4 - 4,000 26,000

4–5 - 12,000 - 1000

Given: Q1-2 = 0, Q2-3 = 40000 kJ/min, Q3-4 = -4000 kJ/min, Q4-1 = -12000 kJ/min
W1-2 = -1000 kJ/min, W2-3 = 0, W3-4 = -4000 kJ/min, W4-1 = -12000 kJ/min
Required: (a) Net work in kW (b) Δ U
Solution:
(a) First law for cycle Æ Net heat transfer = Net work transfer

∫ dQ = ∫ dW
∫ dW = − 1000 + 0 + 26000 − 1000 = 24000 kJ / min
= 24000 / 60 = 400 kW

25
 ∫ dQ = 40000 − 4000 − 120000 = 24000 kJ / min
= 24000 / 60 = 400 kW

∫ dQ = ∫ dW
Therefore, the data is consistent with the I-law of thermodynamics.
(b) Change in internal energy during process 1 – 2 = U2 – U1
I-law for a process
Q=W+ΔU
U2 – U1 = Q1-2 – W1-2
= 0 – (-1000) = 1000 kJ/min
U3 – U2 = Q2-3 – W2-3
= 40000 – 0 = 40000 kJ/min
U4 – U3 = Q3-4 – W3-4
= -4000 - 26000 = -30000 kJ/min
U1 – U4 = Q4-1 – W4-1
= -12000 – (-1000) = -11000 kJ/min
3. Calculate the workdone when the volume changes from 4 m3 to 8 m3 through a non-flow
quasi-static process in which the pressure p is given by, p = (4 V – 5) bar.
Given: A process
Initial volume (V1) = 4 m3
Final volume (V2) = 8 m3
p = (4 V – 5) bar
Required: Workdone
Solution
2 2

Work transfer in non-flow process is W=

∫ p dV = ∫ (4V − 5) dV x 10
1 1
5

⎡ 4V 2 ⎤
= ⎢ − 5V ⎥
⎣ 2 ⎦ 1

= [ 4 (V2 – V12) / 2 – 5 (V2 – V1)] x 105

2

= [4 x (82 – 42)/2 – 5 x (8 – 4)] x 105

= 76 x 105 J – Ans
4. The internal energy of a certain substance is given by the equation u = 3.56 p v + 84, where
u is given in kJ/kg, p is in kPa and v is in m3/kg. A system composed of 3 kg of this
substance expands from an initial pressure of 500 kPa and a volume of 0.22 m3 to a final
pressure of 100 kPa in a process in which pressure and volume are related by pV1.2 = C. (a)
If the expansion is quasi-static, find heat transfer, workdone and change in internal energy
for the process (b) In another process the same system expands according to the same
pressure-volume relationship as in part (a), and from the same initial state to the same final
26
state as in part (a), but the heat transfer in this case is 30 kJ. Find the work transfer for this
process (c) Explain the difference in work transfer in part (a) and part (b).
Given:
Specific internal energy = u = 3.56 p v + 84 kJ/kg
Mass of the working fluid (m) = 3 kg
Initial pressure (p1) = 500 kPa
Initial volume (V1) = 0.22 m3
Final pressure (p2) = 100 kPa
Index of expansion (n) = 1.2
Required: (a) Q, W, Δ U (b) W (c) Difference W in (a) and (b)
Solution:
(a) Heat transfer during a process, Q = W + Δ U
u = 3.56 p v + 84
u1 = 3.56 p1 v1 + 84
u2 = 3.56 p2 v2 + 84
Δu = 3.56 (p2 v2 – p1 v1)
ΔU = m Δu and V=mv
ΔU = 3.56 (p2 V2 – p1 V1)
To find V2
1/ n
V2 ⎛p ⎞
= ⎜⎜ 1 ⎟
⎟
V1 ⎝ p2 ⎠

1 / 1. 2
V2 ⎛5⎞
=⎜ ⎟
0.22 ⎜⎝ 1 ⎟⎠

V2 = 0.841 m3
∴ ΔU = 3.56 (100 x 0.841 – 500 x 0.22)
= - 92.204 kJ --- Ans
Process follows the law pV1.2 = C.
W = (p2V2 – p1V1) / (1 – n)
= (100 x 103 x 0.841 – 500 x 103 x 0.22) (1 – 1.2)
= 129500 J = 129.5 kJ --- Ans
Q1-2 = W1-2 + Δ U
= 129.5 – 92.204 = 37.296 kJ --- Ans
(b) Heat transfer = 30 kJ
W1-2 = Q1-2 – Δ U
= 30 – (- 92.204) = 122.204 kJ --- Ans
(c) The work in part (b) is not equal to ∫p dV. The process in part (b) is not quasi-static.

27
5. A fluid is confined in a cylinder by a spring loaded, frictionless piston so that the pressure in
the fluid is a linear function of the volume (p = a + bV). The internal energy of the fluid is
given by the equation, U = 34 + 3.15 p V, where, U is in kJ, p is in kPa and V is in m3. If the
fluid changes from an initial state of 170 kPa, 0.03 m3 to a final state of 400 kPa, 0.06 m3,
with no work other than that done on the piston, find the direction and magnitude of the
work and heat transfer.
Given:
Pressure p = a + bV
Internal energy (U) = 34 + 3.15 pV kJ
Initial pressure (p1) = 170 kPa
Initial volume (V1) = 0.03 m3
Final pressure (p2) = 400 kPa
Final volume (V2) = 0.06 m3
Required: W and Q
Solution:
U1 = 34 + 3.15 p1V1
U2 = 34 + 3.15 p2V2
Δ U = 3.15 (p2V2 – p1V1)
= 3.15 x (400 x 103 x 0.06 – 170 x 103 x 0.03)
= 59535 J
p1 = a + b V1
170 = a + 0.03 b --- (1)
p2 = a + bV2
400 = a + 0.06 b --- (2)
From (1) and (2) 230 = 0.03 b, b = 7666.67 kPa
Substituting in (1), a = - 60 kPa
2

Work transfer is W=
∫ p dV
1

∫
= (a + bV ) dV
1

= [a V + b V2/2] = a (V2 – V1) + b (V22 – V12)/2

= - 60 x (0.06 – 0.03) + 7666.67 x (0.062 – 0.032) /2
= 8.55 kJ = 8550 J --- Ans
Work is positive, i.e., Work is done by the system.
Q1-2 = W1-2 + Δ U
= 8550 + 59535 = 68085 J --- Ans
Heat transfer is positive, i.e., Heat is supplied to the system.

28
6. A steam turbine operates under steady flow conditions. It receives 7200 kg/h of steam from
the boiler. The steam enters the turbine at enthalpy of 2800 kJ/kg, a velocity of 4000 m/min
and an elevation of 4 m. The steam leaves the turbine at enthalpy of 2000 kJ/kg, a velocity
of 8000 m/min and an elevation of 1 m. Due to radiation, heat losses from the turbine to the
surroundings amount to 1580 kJ/h. Calculate the output of the turbine.
Given:
Mass flow rate of steam (m) = 7200 kg/h = 2 kg/s
Initial specific enthalpy (h1) = 2800 kJ/kg
Initial velocity (C1) = 4000 m/min = 66.67 m/s
Elevation of inlet (Z1) =1m
Final specific enthalpy (h2) = 2000 kJ/kg
Final velocity (C2) = 8000 m/min = 133.3 m/s
Elevation of outlet (Z2) = 4 m
Heat losses from turbine (Q) = -1580 kJ/h = 0.4388 kJ/s
Required: Work output
Solution:
SFEE is given by, m (h1 – h2) + m (C12 – C22)/2 + m g (Z1 – Z2) + Q = W
W = 2 (2800 – 2000) x 103 + 2 x (66.672 – 133.32)/2
+ 2 x 9.81 x (1 – 4) – 0.4388 x 103
= 1586178.3 W --- Ans
7. Steam enters a nozzle at a pressure of 7 bar and 20oC (initial enthalpy = 2850 kJ/kg) and
leaves at a pressure of 1.5 bar. The initial velocity of steam at the entrance is 40 m/s and the
exit velocity of steam from nozzle is 700 m/s. The mass flow rate through the nozzle is 1400
kg/h. The heat loss from the nozzle is 11705 kJ/h. Determine the final enthalpy of steam and
nozzle area if the specific volume at outlet is 1.24 m3/kg.
Given:
Initial pressure (p1) = 7 bar
Initial velocity (C1) = 40 m/s
Initial enthalpy (h1) = 2850 kJ/kg
Final velocity (C2) = 700 m/s
Specific volume at outlet (v2) = 1.24 m3/kg
Mass flow rate (m) = 1400 kg/h = 0.389 kg/s
Heat loss from nozzle (Q) = -11105 kJ/h = -3251.4 J/s
Required: h2 and A2
Solution:
SFEE is given by, m (h1 – h2) + m (C12 – C22) / 2 + m g (Z1 – Z2) + Q = W

Assume Z1 – Z2 = 0 if not given. For nozzle W = 0

m (h1 – h2) + m (C12 – C22) + Q = 0
0.389 x (2850 - h2) x 103 + 0.389 x (402 – 7002) / 2 – 3251.4 = 0
29
 h2 = 2614.16 kJ/kg ---- Ans
Mass flow rate (m) = A2 C2 / v2 = A1 C1 / v1
0.389 = A2 x 700 / 1.24
A2 = 0.000689 m2 --- Ans
8. Air flows at the rate of 0.5 kg/s through an air compressor, entering at 7 m/s, 100 kPa and
0.95 m3/kg and leaving at 5 m/s, 700 kPa, and 0.19 m3/kg. The internal energy of air leaving
is 90 kJ/kg greater than that of the air entering. Cooling water in the compressor jackets
absorbs heat from the air at the rate of 58 kW. (a) Compute the rate of shaft work input to
the air in kW (b) Find the ratio of the inlet pipe diameter to outer pipe diameter.
Given:
Mass flow rate (m) = 0.5 kg/s
Initial velocity (C1) = 7 m/s
Initial specific volume (v1) = 0.95 m3/kg
Initial pressure (p1) = 100 kPa = 1 bar
Final velocity (C2) = 5 m/s
Final pressure (p2) = 700 kPa = 7 bar
Final specific volume (v2) = 0.19 m3/kg
u2 – u1 = 90 kJ/kg
Heat liberated by the air (Q) = 58 kJ/s ( - ve)
Required: (a) W (b) d1 / d2
Solution:
(a) SFEE is given by,
m (u1 – u2) + m (p1v1 – p2v2) + m (C12 – C22) / 2 + m g (Z1 – Z2) + Q = W
Assume Z2 – Z1 = 0 if not given.
0.5 x (-90) x 103 + 0.5 x (1 x 0.95 x 105 – 7 x 0.19 x 105) + 0.5 (72 – 52) / 2 – 58000 = W
W = - 121994 W
∴ Work input to the compressor = 121994 W --- Ans
(b) Mass flow rate is given by,
m = A1 C1 / v1 = A2 C2 / v2
A1 / A2 = C2 v1 / (C1v2)
= 5 x 0.95 /(7 x 0.19) = 3.57
A1 / A2 = d12 / d22
d1 / d2 = √3.57 = 1.8898 --- Ans
9. In a steam power station, steam flows steadily through a 0.2 diameter pipeline from the
boiler to the turbine. At the boiler end, the steam conditions are found to be 4 MPa, 400oC,
specific enthalpy 3214.6 kJ/kg and specific volume 0.073 m3/kg. At the turbine end, the
conditions are found to be 3.5 MPa, 392oC, specific enthalpy 3202.6 kJ/kg and specific
volume 0.084 m3/kg. There is a heat loss of 8.5 kJ/kg from the pipeline. Calculate the steam
flow rate.

30
 Given:
Diameter of pipeline (d) = 0.2 m
Pressure of the steam at inlet (p1) = 4 MPa = 40 bar
Initial temperature (T1) = 400o C = 673 K
Initial specific enthalpy (h1) = 3213.6 kJ/kg
Initial specific volume (v1) = 0.073 m3/kg
Pressure of the steam at outlet (p2) = 3.5 MPa = 35 bar
Final temperature (T2) = 392o C = 563 K
Final specific enthalpy (h2) = 3202.6 kJ/kg
Initial specific volume (v1) = 0.073 m3/kg
Heat losses from turbine (q) = -8.5 kJ/kg
Required: Steam flow rate
Solution:
This is the problem of heat exchanger.
Heat exchanger Turbine end

Boiler Turbine

Boiler end

SFEE is given by, m (h1 – h2) + m (C12 – C22) / 2 + m g (Z1 – Z2) + Q = W

Assume Z1 – Z2 = 0 if not given. For heat exchanger, W = 0 ;
Mass flow rate (m) = A1 C1 / v1 = A2 C2 / v2
A1 = A2
∴ C1 = C2v1 / v2 = C2 0.073 / 0.084 = 0.869 C2 --- (1)
SFEE for heat exchanger becomes,
m (h1 – h2) + m (C12 – C22) / 2 + m q = 0
(h1 – h2) + (C12 – C22) / 2 + q = 0
(3213.6 – 3202.6) x 103 + ((0.869C2)2 – C22)/2 -8.5 x 103 = 0
C2 = 142.9 m/s
m = A2 C2 / v2 = (π / 4) d2 C2 / v2
= π x 0.22 x 142.9 / (4 x 0.084) = 53.45 kg/s --- Ans
10. A piston and cylinder machine contains a fluid system which passes through a complete
cycle of four operations.

Process Q (kJ/min) W (kJ/min) ΔU (kJ/min)

a–b 0 2170 ---

31
 b–c 21000 0 ---

c–d -2100 --- -36600

d–a --- --- ---

During cycle the sum of all heat transfers is -170 kJ. The system operates 100 cycles per
min. Complete the following table showing the method for each item and compute the net
rate if output in kW.
Given:
Net heat transfer = -170 kJ
No of cycles = 100 / min
Required: To complete the table and to determine the net work,
Solution:
ΔU Æ Change in internal energy
I-law for a cycle, Net work transfer = Net heat transfer
∴ Net work output = - 170 kJ
= - 170 x no of cycles per second
= -170 x 100 / 60 = -283.33 kW --- Ans
I-law for a process is Q = W + ΔU
(ΔU)a-b = Qa-b – Wa-b
= 0 – 2170 = -2170 kJ/min --- Ans
(ΔU)b-c = Qb-c – Wb-c
= 21000 – 0 = -21000 kJ/min --- Ans
Wc-d = Qc-d – (ΔU)c-d
= -2100 +36600 = 34500 kJ/min --- Ans
Net Heat transfer, Qd-a = Qa-b + Qb-c + Qc-d + Qd-a
= -17000 – 0 – 21000 + 2100
= -35900 kJ/min --- Ans
Net work transfer, Wd-a = Wa-b + Wb-c + Wc-d + Wd-a
= -17000 – 2170 – 0 – 34500
= -53670 kJ/min --- Ans
For a cycle, sum of ΔU = 0
(ΔU)a-b + (ΔU)b-c + (ΔU)c-d + (ΔU)d-a = 0
-2170 + 21000 – 36600 + (ΔU)d-a = 0
(ΔU)d-a = 17770 kJ/min ---- Ans

Process Q (kJ/min) W (kJ/min) ΔU (kJ/min)

a–b 0 2170 -2170

32
 b–c 21000 0 21000

c–d -2100 34500 -36600

d–a -35900 -53670 17770

11. 12 kg of air per min is delivered by a centrifugal compressor. The inlet and outlet conditions
of air are C1 = 12 m/s, p1 = 1 bar, v1 = 0.5 m3/kg and C2 = 90 m/s, p2 = 8 bar, v2 = 0.14 m3/kg.
The increase in enthalpy of air passing through compressor is 150 kJ/kg and heat loss to the
surroundings is 700 kJ/min. Find motor power required to drive the compressor and ratio
of inlet and outlet pipe diameters. Assume that inlet and discharge lines are at the same
level.
Given:
Mass flow rate of air (m) = 12 kg/min = 0.2 kg/s
Increase in enthalpy (h2 – h1) = 150 kJ/kg
Heat loss to the surroundings (Q) = 700 kJ/min = 11666.7 J/s (-ve)
Z1 – Z2 =0
Required: Power and Diameter ratio
Solution:
SFEE is given by, m (h1 – h2) + m (C12 – C22) / 2 + m g (Z1 – Z2) + Q = W
0.2 x (-150 x 103) + 0.2 x (122 – 902) /2 – 11666.7 = W
W = -42462.3 W
Work in W is Power. ∴ P = -42462.3 W
Power required to drive the compressor = 42262.3 W --- Ans
Mass flow rate (m) = A1 C1 / v1 = A2 C2 / v2
A1 / A2 = v1 C2 / (v2 C1)
= 0.5 x 90 / (0.14 x 12) = 26.7857
d1 / d2 = √(A1 / A2) = √26.7857 = 5.175 --- Ans
12. At the inlet of a certain nozzle, the enthalpy of the fluid passing is 3000 kJ/kg and the
velocity is 60 m/s. At the discharge end, the enthalpy is 2762 kJ/kg. The nozzle is horizontal
and there is negligible heat loss from it. (a) Find the velocity of fluid at exit (b) If the inlet
area is 0.1 m2 and the specific volume at inlet is 0.187 m3/kg, find the mass flow rate (c) If
the specific volume at the nozzle exit is 0.498 m3/kg, find the exit area of the nozzle.
Given:
Nozzle
Initial enthalpy (h1) = 3000 kJ/kg
Initial velocity (C1) = 60 m/s
Heat transfer (Q) =0
Required: (a) C2 (b) m (c) A2
Solution:
(a) In nozzle flow, W = 0; Z1 – Z2 = 0 (if not given);
SFEE is given by, m (h1 – h2) + m (C12 – C22) / 2 + m g (Z1 – Z2) + Q = W
33
 For given nozzle, m (h1 – h2) + m (C12 – C22) / 2 = 0
(h1 – h2) + (C12 – C22) / 2 = 0
C22 = C12 + 2 (h1 – h2)
= 602 + 2 x (3000 – 2762) x 103
C2 = 692.5 m/s --- Ans
(b) A1 = 0.1 m ; v1 = 0.187 m3/kg
2

Mass flow rate (m) = A1 C1 / v1 = A2 C2 / v2

= 0.1 x 60 / 0.187 = 32.08 kg/s --- Ans
3
(c) v2 = 0.498 m /kg ; m = A2 C2 / v2
32.08 = A2 x 692.5 / 0.498
A2 = 0.02307 m2 --- Ans
13. 85 kJ of heat is supplied to a system at constant volume. The system rejects 90 kJ of heat at
constant pressure and 20 kJ of work is done on it. The system is brought to its original state
by adiabatic process. Determine the adiabatic work. Determine also the values of internal
energy at all end states if initial value is 100 kJ.
Given:
There are three processes.
1 – 2 Æ Constant volume process Æ HS = 85 kJ
2 – 3 Æ Constant pressure process Æ HR = 90 kJ & W = -20 kJ
3 – 1 Æ Adiabatic process
Initial internal energy (U1) = 100 kJ
Required: W3-1, U2 & U3
Solution: 3 2

p 1

V
I-law for a process, Q = W + Δ U
1 – 2 Æ Constant volume process, W1-2 = 0
Q1-2 = W1-2 + U2 – U1
85 = 0 + U2 – 100
U2 = 85 kJ --- Ans
2 – 3 Æ Constant pressure process
Q2-3 = W2-3 + U3 – U2
-90 = -20 + U3 – 185
U3 = 115 kJ --- Ans
3 – 1 Æ Adiabatic process, Q3-1 = 0
Q3-1 = W3-1 + U1 – U3

34
 0 = W3-1 + 100 – 115
W3-1 = 15 kJ --- Ans
14. 2 m of hydrogen at a pressure of 1 bar and 20oC is compressed isentropically to 4 bar. The
3

same gas is expanded to original volume by constant temperature process and reached
initial pressure and temperature by constant volume heat rejection process. Determine (a)
pressure, volume and temperature at each end of operation (b) the heat transferred during
the isothermal process (c) the heat rejected during constant volume process and (d) change
in internal energy during each process. Assume R = 4.206 kJ/kg-K and Cp = 14.25 kJ/kg-K.
Given:
There are three processes.
1 – 2 Æ Isentropic process
2 – 3 Æ Constant temperature process
3 – 1 Æ Constant volume process
Initial volume of hydrogen (V1) = 2 m3 = V3
Initial pressure (p1) = 1 bar
o
Initial temperature (T1) = 20 C = 293 K
Pressure after isentropic compression (p2) = 4 bar
3
Volume after isothermal expansion (V3) = 2 m
Required: (a) V2, T2, p3
Solution: 2

3
p 1

V
(a) R = Cp – Cv
Cv = 14.25 – 4.206 = 10.044 kJ/kg-K
γ = Cp / Cv = 14.25 / 10.044 = 1.419
For adiabatic process,
(γ −1) / γ
T2 ⎡ p 2 ⎤
=⎢ ⎥
T1 ⎣⎢ p1 ⎥⎦
( 1 . 419 − 1 ) / 1 . 419
T2 ⎡4⎤
= ⎢ ⎥ T2 = 441.2 K --- Ans = T3
293 ⎣1 ⎦
1/ γ
V2 ⎡ p1 ⎤
=⎢ ⎥
V1 ⎣ p2 ⎦

35
 1 / 1 .4
V2 ⎡ 1 ⎤
=⎢ ⎥ V2 = 0.753 m3 --- Ans
2 ⎣4⎦
2 – 3 Æ Constant temperature process
p2 V2 = p3 V3
4 x 0.753 = p3 x 2
p3 = 1.506 bar --- Ans
(b) Heat transferred during isothermal process (Q2-3)
Q2-3 = p2V2 ln[V3/V2]
= 4 x 105 x 0.753 x ln [2/0.753]
= 294223.4 J --- Ans
(c) Heat rejected during constant volume process (Q3-1)
Q3-1 = m Cv (T1 – T3)
To find m
p1 V1 = m R T1
1 x 105 x 2 = m x 4206 x 293
m = 0.1623 kg
∴ Q3-1 = 0.1623 x 1004.4 x (293 – 441.2)
= -241586.9 J --- Ans
(d) U2 – U1 = -W1-2 = (p2V2 – p1V1) / (1 – γ)
= - (4 x 105 x 0.753 – 1 x 105 x 2) / (1 – 1.419)
= 241527.4 J --- Ans
U3 – U2 = 0
U1 – U3 = Q3-1 = -241586.9 J ---- Ans
15. 0.5 m of air at 30 C and 1 bar is compressed polytropically to 0.08 m3. Find the final
3 o

pressure and temperature and workdone, change in internal energy and enthalpy, when the
index of compression has the value of 1.5. Take for air Cp = 1.005 kJ/kg-K and Cv = 0.718
kJ/kg-K.
Given:
Polytropic process
Initial volume (V1) = 0.5 m3
Initial temperature (T1) = 30oC = 303 K
Final volume (V2) = 0.08 m3
Index of compression (n) = 1.5
Required: p2, T2, W, Δ U, Δ H
Solution:
(1− n )
T2 ⎡ V1 ⎤
=⎢ ⎥
T1 ⎣V2 ⎦
36
 (1−1.5)
T2 ⎡ 0.5 ⎤
=
303 ⎢⎣ 0.08 ⎥⎦
T2 = 757.5 K --- Ans
n
p2 ⎡ V1 ⎤
=⎢ ⎥
p1 ⎣ V2 ⎦

1.5
p2 ⎡ 0.5 ⎤
=⎢ ⎥ p2 = 15.625 bar --- Ans
1 ⎣ 0.08 ⎦
W1-2 = (p2V2 – p1V1) / (1 – n)
= (15.625 x 105 x 0.08 – 1 x 105 x 0.5) / (1 – 1.5)
= -150000 J ---- Ans
U2 – U1 = m Cv (T2 – T1)
To find m
p1 V1 = m R T1
R = Cp – Cv = 1.005 – 0.718 = 0.387 kJ/kg-K
∴ 1 x 105 = m x 287 x 303
m = 0.5749 kg
∴ Δ U = 0.5749 x 0.718 x (757.5 – 303) = 187.6 KJ --- Ans
Δ H = H2 – H1 = m Cp (T2 – T1)
= 0.5749 x 1.005 x (757.5 – 303) = 262.6 kJ --- Ans
16. 3 kg of an ideal gas is expanded from a pressure of 8 bar and volume of 1.5 m3 to a pressure
of 1.6 bar and volume of 4.5 m3. The change in internal energy is 450 kJ. The specific heat
at constant volume for the gas is 0.7 kJ/kg-K. Determine (a) Gas constant (b) Index of
polytropic expansion (c) Workdone during polytropic expansion and (e) Initial and final
temperatures.
Given:
Mass of gas (m) = 3 kg
Initial pressure (p1) = 8 bar
Initial Volume (V1) = 1.5 m3
Final pressure (p2) = 1.6 bar
Final volume (V2) = 4.5 m3
Change in internal energy (U2 – U1) = 450 kJ
Specific heat at constant volume (Cv) = 0.7 kJ/kg-K
Required: (a) R (b) n (c) W (d) T1 & T2
Solution:
(a) p1 V1 = m R T1
8 x 105 x 1.5 = 3 x R T1

37
 R T1 = 400000 --- (1)
p2 V2 = m R T2
5
1.6 x 10 x 4.5 = 3 x R T2
R T2 = 240000 --- (2)
From (1) and (2) R (T2 – T1) = -160000
T2 – T1 = -160000 / R --- (3)
Also U2 – U1 = - 450 kJ (-ve on expansion)
U2 – U1 = m Cv (T2 – T1)
-450 = 3 x 0.7 x (T2 – T1)
Substituting (3), -450 = 3 x 0.7 x -160000 / R
R = 746.6 J/kg-K ---- Ans
n
(b) p1V1 = p2V2n
∴ n = ln [p1/p2] / ln [V2/V1]
= ln [8/1.6] / ln [4.5/1.5] = 1.465 ---- Ans
(c) W = (p2V2 – p1V1) / (1 – n)
= (1.6 x 4.5 x 105 – 8 x 1.5 x 105) / (1 – 1.465)
= 1032258 J ---- Ans
(d) p1 V1 = m R T1
8 x 105 x 1.5 = 3 x 746.6 x T1
T1 = 535.8 K ---- Ans
p2 V2 = m R T2
1.6 x 105 x 4.5 = 3 x 746.6 x T2
T2 = 321.5 K ---- Ans
17. A gas mixture obeying perfect gas law has molar mass of 26.7 kg/kmol. The gas mixture is
compressed to a compression ratio of 12 according to the law pV1.25 = C, from initial
conditions of 0.9 bar and 333 K. Assume a mean molar specific heat at a constant volume of
21.1 kJ/kmolK, find per kg of mass, the workdone and heat flow across the cylinder walls.
For the above gas, determine the value of characteristic gas constant, molar specific heat at
constant pressure and ratio of specific heats.
Given:
Polytropic process
Index of compression (n) = 1.25
Molar mass (M) = 26.7
Compression ratio (p2/p1) = 12
Initial pressure (p1) = 0.9 bar
Initial temperature (T1) = 333 K
Cv(mole) = 21.1 kJ/kmolK
Mass of gas (m) = 1 kg
Required: W, Q, R, Cp(mole) , Cp/Cv
38
 Solution:
Work done during polytropic process is given by,

p V − p1V1 m R (T2 − T1)

W= 2 2 =
1− n 1− n
To find R
Universal gas constant (Ru) = 8314 J/kg-K
Characteristic gas constant (R) = Ru/M = 8314/26.7 = 311.38 J/kg-K --- Ans
To find T2
(n −1) / n
T2 ⎡ p2 ⎤
=⎢ ⎥
T1 ⎣ p1 ⎦

= [12](1.25−1) / 1.25
T2
333
T2 = 547.4 K

1 x 313.38 x (547.4 − 333)

W=
1 − 1.25
= - 268754.7 J --- Ans

Q=
γ −n 1 .4 − 1 .25
Heat transfer (Q)
1− n
xW = x − 268754 .7
1 − 1 .25
= 100783 J --- Ans
Gas constant R
R = Cp – Cv
Cv = Cv(mole) / M
Cp = Cp(mole) / M
∴ Cv = 21100 / 26.7 = 790.26 J/kg-K
313.38 = Cp – 790.26
Cp = 1101.64 J/kg-K
∴ Cp(mole) = 1101.64 x 26.7 = 29413.8 J/kmolK ---- Ans
Specific heat ratio
Cp/Cv = 1101.64/790.26 = 1.394 --- Ans
18. A centrifugal pump delivers 2750 kg of water per min from initial pressure of 0.8 bar
absolute to a final pressure of 2.8 bar absolute. The suction is 2 m below and the delivery is
5 m above the centre of pump. If the suction and delivery pipes are of 15 cm and 10 cm
diameters respectively, make calculation for power required to run the pump.
Given:
Centrifugal pump
Flow process
39
Mass flow rate of water (m) = 2750 kg/min = 2750/60 kg/s
Initial pressure (p1) = 0.8 bar (abs)
Final pressure (p2) = 2.8 bar (abs)
Suction below the centre of the pump =2m
Delivery above the centre of the pump = 5 m
Suction pipe diameter (d1) = 15 cm = 0.15 m
Delivery pipe diameter (d2) = 10 cm = 0.1 m
Required: W
Solution:
SFEE is given by,
* ⎡ C12 ⎤ * ⎡ C 22 ⎤
m ⎢h1 + + g Z1 ⎥ + Q = m ⎢h2 + + g Z2 ⎥ +W
⎣ 2 ⎦ ⎣ 2 ⎦
* ⎡ C12 ⎤ * ⎡ C 22 ⎤
m ⎢u1 + p1v1 + + g Z1 ⎥ + Q = m ⎢u 2 + p 2 v 2 + + g Z2 ⎥ +W
⎣ 2 ⎦ ⎣ 2 ⎦
Consider datum from suction (1), Z1 = 0 & Z2 = 2 + 5 = 7 m
Generally for any liquid, u1 = u2, v1 = v2 = v and Q = 0

Therefore, SFEE is reduced to,

* ⎡ C2 ⎤ *⎡ C2 ⎤
m ⎢ p1v + 1 + g Z 1 ⎥ = m ⎢ p 2 v + 2 + g Z 2 ⎥ + W
⎣ 2 ⎦ ⎣ 2 ⎦
Water Out
(2)

Z2
Pump

(1)
Water in
To find C1 and C2

40
 * A1C1 A2 C 2
m= =
v1 v2
For water ρ = 1000 kg/m3 and v = 1/ρ = 1/1000 m3/kg
π π
A1 = d12 = x 0.15 2
4 4
π π
A2 = d 22 = x 0.12
4 4
π / 4 x 0.15 2 C1
2750 / 60 =
1 / 1000
C1 = 2.593 m/s
π / 4 x 0.12 C 2
2750 / 60 =
1 / 1000
C2 = 5.835 m/s
Therefore,
⎡ 2.593 2 ⎤
(2750 / 60) ⎢0.8 x 10 5 x (1 / 1000) + + 9.81 x 0⎥
⎣ 2 ⎦
⎡ 5.835 2 ⎤
= (2750 / 60) ⎢2.8 x 10 5 x (1 / 1000) + + (9.81 x 7)⎥ + W
⎣ 2 ⎦
W = 12833 W --- Ans
19. In an isentropic flow through nozzle, air flows at the rate of 600 kg/h. At the inlet to the
nozzle , pressure is 2 MPa and temperature is 127oC. The exit pressure is 0.5 MPa. Initial
velocity is 300 m/s. Determine, (i) Exit velocity of air and (ii) Inlet and exit area of nozzle.
Given:
Nozzle
Flow process
Fluid Æ Air
Mass flow rate of air (m) = 600 kg/h = 600/3600 kg/s
Initial pressure (p1) = 2 MPa = 2 x 106 Pa
Initial temperature (T1) = 1270C = 400 K
Final pressure (p2) = 0.5 MPa = 0.5 x 106 Pa
Initial velocity (C1) = 300 m/s
Required: (i) C2 (ii) A1 & A2
Solution:
(i)
SFEE is given by,
* ⎡ C2 ⎤ * ⎡ C2 ⎤
m ⎢h1 + 1 + g Z1 ⎥ + Q = m ⎢h2 + 2 + g Z 2 ⎥ + W
⎣ 2 ⎦ ⎣ 2 ⎦

41
Take Z1 = Z2
The flow is isentropic, Q = 0
For nozzle, W = 0
Therefore, SFEE is reduced to,

⎡ C12 ⎤ ⎡ C 22 ⎤
⎢h1 + ⎥ = ⎢h2 + ⎥
⎣ 2 ⎦ ⎣ 2 ⎦

C 2 = C12 + 2(h1 − h2 )

C 2 = C12 + 2 C p (T1 − T2 )

To find T2
( γ −1) / γ
T2 ⎡ p 2 ⎤
=⎢ ⎥
T1 ⎣ p1 ⎦
For air, γ = 1.4, R = 287 J/kg-K & Cp = 1005 J/kg-K
(1 . 4 − 1 ) / 1 . 4
T2 ⎡ 0 .5 ⎤
=⎢
400 ⎣ 2 ⎥⎦
T2 = 269.2 K
Therefore,

C 2 = 300 2 + 2 x 1005 ( 400 − 269.2)

C2 = 593.06 m/s ---- Ans
(ii)
* A1C1 A2 C 2
m= =
v1 v2
To fin v1 & v2
p1 v1 = R T1
6
2 x 10 x v1 = 287 x 400
v1 = 0.0574 m3/kg
p2 v2 = R T2
6
0.5 x 10 x v2 = 287 x 269.2
V2 = 0.1545 m3/kg
A1 x300
600 / 3600 =
0.0574
A1 = 3.1889 x 10-5 m2 --- Ans
A2 x593.06
600 / 3600 =
0.1545
A2 = 4.3419 x 10-5 m2 --- Ans
42
 UNIT-II
SECOND LAW AND AVAILABILITY ANALYSIS
Second law of Thermodynamics
Kelvin-Planck statement
It is impossible to construct an engine to work in a cyclic process whose sloe effect is to convert
all the heat supplied into an equivalent of work.
T1 T1 > T2 T1
Source Q2 = 0 Source

HS = Q1 Q1
W = HS = Q1 HE W = Q1 – Q2
HE
Q2
Impossible Kelvin Planck statement
Sink (T2)
Possible Kelvin statement
Clausius statement
It is impossible to construct a device to work in a cyclic process whose sole effect is the transfer
of heat from a body at a lower temperature to a body at a higher temperature.
T2 T2 < T1 T2
Source W=0 Source

HS = Q2 Q2
W
RHE

HR = Q1 = Q2 Q1 = Q2 + W
Sink Sink
T1 T1
Impossible Clausius statement Possible Clausius statement
Equivalence of Kelvin and Clausius statements
Consider the combination as shown in fig. The heat engine contradicted by Kelvin statement is
producing work. We assume that it is possible. Now it is combined with the RHE which is a regular heat
pump obeying Clausius statement. The combination results in a RHE which is contradicting the Clausius
statement. This means that if Kelvin statement is contradicted, Clausius statement is automatically
contradicted.
Net effect Æ + Q1 – Q1 + Q2 = W = Q1
Q1 = Q2 (impossible)

Source (T1) Sink (T1) T1

Q1 Q1 = W + Q2 Q2
W = Q1 W RHE RHE
HE + =
Q2
Q2
Source (T2) T2

Impossible Kelvin + Possible Clausius = Impossible Clausius

Similarly the impossible Clausius and possible Kelvin can be combined and the resulting engine
will contradict the Kelvin statement.
So, if one of the statements is contradicted, the other is automatically contradicted. So, the effect
of these two apparently different statements are basically the same.
Perpetual Motion Machine of Second kind (PMM-II)
- A machine which continuously absorb heat from a single reservoir and would convert this heat
completely into work.
PMM-II

W=Q

Efficiency of PMM-II is 100 %.

PMM-II obeys I-law but violates the II-law.
Reversibility and Irreversibility
The II-law of thermodynamics enables us to divide all processes into two classes:
1. Reversible or ideal process
2. Irreversible or natural process
A reversible process is a process that, after it has taken place, can be reversed and causes no
change in either the system or its surroundings. A reversible process is carried out infinitely slowly with
an infinitesimal gradient, so that every state passed through by the system is an equilibrium state. So, a
reversible process coincides with a quasi-static process.
Examples of reversible processes:
• Frictionless adiabatic process (Isentropic process)
2
 • Frictionless isothermal process
• Condensation and boiling of liquids
A process that is not reversible is termed irreversible. In an irreversible process, finite changes are
made; therefore the system is not at equilibrium throughout the process. At the same point in an
irreversible cycle, the system will be in the same state, but the surroundings are permanently changed
after each cycle.
Examples of irreversible processes:
• Combustion process
• Mixing of two fluids
• Heat flow from a higher temperature to lower temperature
• Flow of electric current through a resistor
Conditions for reversibility
1. The process should not involve friction.
2. Heat transfer should not take place with finite temperature difference.
3. The energy transfer as heat and work during the forward process should be identically equal to
energy transfer as heat and work during the reversal of the process.
4. There should be no free or unrestricted expansion.
5. There should be no mixing of the fluids.
6. The process must proceed in a series of equilibrium states.
Heat engine
Heat engine is a device used to convert heat energy into useful work when operating in a cyclic
process.
Examples: I.C. engines like, Petrol engines, Diesel engines, E.C. engines like, Steam power plants, Gas
turbine power plants.
WT
Boiler
HS Turbine Generator

HR
Condenser
Feed Pump
WP
• Boiler produces the steam and supplies to the turbine. Heat is supplied in the boiler.
• Turbine develops the work.
• Condenser condenses the exhaust steam from the turbine. Heat is rejected from the steam.

3
 • Pump takes the condensate to the boiler at high pressure. Work is given to the pump

As per I-law, Net heat transfer is equal to the net work transfer. T1
HS – HR = WT – WP Source
Thermal efficiency = Net work transfer / Heat supplied
= (WT – WP) / HS HS = Q1
= (HS – HR) / HS W
HS Æ Heat Supplied HE
HR Æ Heat Rejected HR (Q2)
HE Æ Heat Engine Sink
T2
Reversed heat engine (RHE)
A RHE is a device to absorb heat from low temperature body with the expenditure of work.
Heat pump:
Heat pump is used to take out the heat from low temperature source (atmosphere) at T2 and
supplies it to a sink (room) at high temperature T1 for heating of room (Winter air conditioning).
As per I-law, Net heat transfer is equal to the net work transfer. T2
HR – HS = W Source
COP Æ Coefficient of performance
COP = Output / Input Q2 (HS)
= Heat given to the room / Work W
= Q1 / (Q1 – Q2) RHE
Q1 (HR)
Sink
T1
Refrigerator:
Refrigerator is used to take out the heat from low temperature source (Room) at T2 and supplies it
to a sink (atmosphere) at high temperature T1 for cooling of room (Summer air conditioning).
As per I-law, Net heat transfer is equal to the net work transfer. T2
HR – HS = W Source
COP Æ Coefficient of performance
COP = Refrigeration effect / Input Q2 (HS)
= Heat taken from the room / Work W
= Heat supplied to working fluid / Work RHE
= Q2 / (Q1 – Q2) Q1 (HR)
4
 Sink
T1

Air Standard Cycles

Air Standard cycles: Cycles using a perfect gas, having the properties of air useful in the study of the
I. C. Engine because they represent a limit to which actual cycle may approach and they are subjected to
simple mathematical and explanatory treatment.

Assumptions made for analysis:

• The properties of the working medium can be calculated by the application of the perfect gas
equation.
i.e., pV = mRT
• The Specific heat of the substance remains same during all the processes in the cycle.
i.e.,Cp & Cv are unchanged.
• The cycles are composed of reversible processes.
• The gas does not undergo any chemical changes.
Various cycles
1. Carnot cycle 2. Otto cycle 3. Diesel cycle 4. Stirling cycle
5. Brayton cycle 6. Erricson cycle 7. Dual cycle
Air standard efficiency of a cycle
The thermal efficiency of an ideal air standard cycle is called the “Air standard efficiency”.
In an ideal air standard cycle, the working fluid is air. The petrol and diesel engines working on
Otto cycle and diesel cycle use petrol and diesel oil with air. This air fuel mixture behaves like air before
the combustion takes place. The properties of combustion products are also not different from those of
air. Therefore the efficiencies of petrol and diesel engines are calculated assuming them working on air
standard cycles.

The efficiency of a cycle is given by,

Output Net work output

η = ---------- = ----------------------
Input Heat Supplied

Heat supplied – Heat rejected

= --------------------------------------
Heat supplied
Carnot Cycle:
The Carnot cycle is a particular thermodynamic cycle, modeled on the Carnot heat engine,
studied by Nicolas Léonard Sadi Carnot in the 1820s and expanded upon by Benoit Paul Émile Clapeyron
in the 1830s and 40s.
This cycle has the highest possible efficiency and consists of four simple operations:
(i) Isothermal expansion
(ii) Isentropic expansion
5
 (iii) Isothermal compression
(iv) Isentropic compression
Operation: Consider a cylinder piston arrangement as shown in the fig. Let m kg of air is enclosed in a
cylinder. The cylinder head is made of perfect heat conductor or perfect heat insulator. The cylinder is
perfectly insulated.

Cylinder head

Cylinder

Piston

T1 3 4
4
p T

2 T2 2 1

V s

The cylinder is full of air when the piston is at BDC. The ‘perfect heat conductor cylinder head’
(Cold body) is brought in contact with the cylinder. The air is compressed at constant temperature from
(1) to (2) during its travel towards TDC. During this process the heat is rejected from the air. The ‘cold
body’ is removed and ‘perfect heat insulator head’ is brought in contact with the cylinder. Now the air is
compressed isentropically (2-3) till the piston reaches the TDC. Then the perfect insulator is removed.
The perfect conductor (Hot body) is brought in contact with the cylinder and the heat is supplied to the
air. Now expansion proceeds at constant temperature (3-4) during its travel towards BDC. The perfect
insulator is brought and further expansion proceeds isentropically upto BDC (4-1). Thus the cycle is
completed.

Efficiency of cycle

Process – 1 – 2 Æ Isothermal compression

∴ Heat transferred = Q1-2 = m T2 (s2 – s1) = HR

Note: The area under the curve in T-s diagram is heat transferred.

Process – 2 – 3 Æ Isentropic compression

6
 ∴ Heat transferred = Q2-3 = 0;

Process – 3 – 4 Æ Isothermal expansion

∴ Heat transferred = Q3-4 = m T1 (s2 – s1) = HS

Process – 4 – 1 Æ Isentropic expansion

∴ Heat transferred = Q4-1 = 0

∴ Cycle efficiency (η) = [Heat supplied – Heat rejected]/Heat supplied

m T1 (s2 – s1) – m T2 (s2 – s1)

= -------------------------------------
m T1 (s2 – s1)

T1 – T2 Tmax – Tmin
= ---------- = ---------------
T1 Tmin

The Carnot cycle efficiency is depending only on the temperatures T1 and T2.
Reversed Carnot cycle

Cylinder head

Cylinder

Piston

T1 3 2
2
p T

4 T2 4 1

V s
If a machine working on reversed Carnot cycle is driven from an external source, will work as a
refrigerator.
7
 p-V and T-s diagrams of reversed Carnot cycle are shown. The cylinder is full of air at
temperature T2 when the piston is at BDC. The ‘perfect heat insulator cylinder head’ is brought in contact
with the cylinder. The air is compressed isentropically from (1) to (2) during its travel towards TDC.
During this process the temperature of air is raised to T1. The ‘perfect heat insulator is removed and
‘perfect heat conductor head’ (Cold body) is brought in contact with the cylinder. Now the air is
compressed at constant temperature (2-3) till the piston reaches the TDC. Then the cold body is removed.
The perfect insulator is brought in contact with the cylinder. Now the air expanded isentropically from (3)
to (4) during its travel towards BDC. During this process the temperature of air is lowered to T2. The
perfect conductor (Hot body) is brought and further expansion proceeds at constant temperature upto
BDC (4-1). Thus the cycle is completed.
Efficiency of cycle
Process – 1 – 2 Æ Isentropic compression

∴ Heat transferred = Q1-2 = 0

Note: The area under the curve in T-s diagram is heat transferred.

Process – 2 – 3 Æ Isothermal compression

∴ Heat transferred = Q2-3 = m T1 (s2 – s3) = HR;

Process – 3 – 4 Æ Isentropic expansion

∴ Heat transferred = Q3-4 = 0;

Process – 4 – 1 Æ Isothermal expansion

∴ Heat transferred = Q4-1 = m T2 (s1 – s4) = m T2 (s2 – s3) = HS;

∴ COP = Refrigeration effect / Work input

= Heat supplied to the air / (Heat rejected – Heat supplied)

m T2 (s2 – s2)
= -----------------------------------
m T1 (s2 – s3) – m T2 (s2 – s3)
T2 Tmin
= ---------- = -------------
T1 – T2 Tmax – Tmin

The Carnot cycle COP is depending only on the temperatures T1 and T2.
Absolute Thermodynamic temperature scale (Kelvin scale)
The efficiency of any heat engine receiving heat Q1 and rejecting heat Q2 is given by
Wnet Q1 − Q2 Q
η= = = 1− 2
Q1 Q1 Q11

The efficiency of any reversible heat engine is given by

Wnet Q1 − Q2 Q T
η rev = = = 1− 2 = 1− 2
Q1 Q1 Q11 T1

8
 Q2
1− = f (T1 , T2 )
Q1
Q1
Or in terms of new function F, = F (T1 , T2 )
Q2
Consider reversible heat engines operating in series.

Q1
We can write, = F (T1 , T2 )
Q2
Q2
= F (T2 , T3 )
Q3
Q1 Q1 / Q3 F (T1 , T3 ) φ (T1 )
= = =
Q2 Q2 / Q3 F (T2 , T3 ) φ (T2 )
Q1 T1
= Æ Kelvin scale
Q2 T2
Clausius inequality

9
 Let us consider a cycle ABCD as shown. Let AB be the general process, either reversible or
irreversible, while the others are reversible. Let the cycle be divided into a number of small cycles as
shown.
For one of the elementary cycles,
dQ2
η = 1−
dQ
dQ Æ The heat supplied at T
dQ2 Æ Heat supplied at T2
The efficiency of the general cycle will be less than or equal to the efficiency of the reversible
cycle.
dQ2 ⎛ dQ2 ⎞
1− ≤ ⎜1 − ⎟⎟
dQ ⎜⎝ dQ ⎠ rev
dQ2 ⎛ dQ2 ⎞
≥⎜ ⎟
dQ ⎜⎝ dQ ⎟⎠ rev

dQ ⎛ dQ ⎞
≤⎜ ⎟
dQ2 ⎜⎝ dQ2 ⎟⎠ rev

⎛ dQ ⎞ T
But ⎜⎜ ⎟⎟ =
⎝ dQ2 ⎠ rev T2
dQ T
≤
dQ2 T2

dQ dQ2
≤ for any process
T T2
dQrev dQ2
For reversible process, dS = =
T T
dQ
Therefore, ≤ dS
T
10
 By definition, cyclic integral of any property is zero.

Therefore, ∫ dS = 0
dQ
∫ T
≤ 0 Æ Clausius inequality

dQ
∫ T
= 0 Æ The cycle is reversible

dQ
∫ T
< 0 Æ The cycle is irreversible and possible

dQ
∫ T
> 0 Æ The cycle is impossible

Entropy
All the heat is not equally valuable for converting into work. Heat that is supplied to a substance
at high temperature has a greater possibility of conversion into work than heat supplied to a substance at a
lower temperature.
Entropy is a function of a quantity of heat which shows the possibility of conversion of that heat
into work. The increase in entropy is small when heat is added at a higher temperature and is greater when
heat is added at lower temperature.
The change of entropy is defined mathematically as
dQ
dS =
T
Characteristics of entropy
• It increases when heat is supplied irrespective of the fact whether temperature changes or not.
• It decreases when heat is removed irrespective of the fact whether temperature changes or not.
• It remains unchanged in all adiabatic frictionless processes.
Change in entropy for a closed system and entropy chart
General expression
Consider heating of m kg of gas.
Let, p1 Æ Initial pressure of the gas
T1 Æ Initial temperature of the gas
V1 Æ Initial volume of the gas
Cp Æ Specific heat at constant pressure
Cv Æ Specific heat at constant volume
R Æ Gas constant
p2, T2, V2 are corresponding final conditions
I-law for a process Æ Q = W + ΔU
dQ = dW + dU

11
 dQ dW dU
Divide by T Æ = +
T T T
dQ
dS = ; W = ∫ p dV ; dW = p dV ; dU = m C v dT
T
pdV dT
dS = + mC v
T T
p mR
We know that pV = mRT Æ =
T V

dV dT
Therefore, dS = mR + mCv
V T
Integrating between (1) and (2),

2 2 2

∫ ∫ ∫
dV dT
dS = mR + mC v
V T
1 1 1

⎡V ⎤ ⎡T ⎤
S 2 − S1 = mR ln ⎢ 2 ⎥ + mC v ln ⎢ 2 ⎥
⎣ V1 ⎦ ⎣ T1 ⎦

Constant Volume process

2

S1 S S2
During constant volume process, V1 = V2

⎡T ⎤
S2 − S1 = mCv ln⎢ 2 ⎥
⎣ T1 ⎦
Constant pressure process
During constant pressure process, p1 = p2

V1 V2
General gas equation can be written as =
T1 T2

V2 T2
Or =
V1 T1
12
 2

S1 S S2

⎡T ⎤ ⎡T ⎤
S2 − S1 = mR ln⎢ 2 ⎥ + mCv ln⎢ 2 ⎥
⎣ T1 ⎦ ⎣ T1 ⎦
⎡T ⎤
= (mR + mCv ) ln⎢ 2 ⎥
⎣ T1 ⎦
⎡T ⎤
= mC p ln⎢ 2 ⎥
⎣ T1 ⎦
Isothermal process

During isothermal process, T1 = T2

⎡V ⎤
S 2 − S 1 = mR ln ⎢ 2 ⎥
⎣ V1 ⎦

1 2

S1 S S2

Isentropic process
During isentropic process, Q = 0 & ΔS = 0

S1 = S 2

Polytropic process

13
 2

S1 S S2
The governing equation is given by, pVn = C

⎡V ⎤ ⎡T ⎤
S2 − S1 = mR ln⎢ 2 ⎥ + mCv ln⎢ 2 ⎥
⎣ V1 ⎦ ⎣ T1 ⎦
1 /( n − 1) − 1 /( n − 1)
⎡ V 2 ⎤ ⎡ T1 ⎤ ⎡T ⎤
We know that, ⎢ ⎥=⎢ ⎥ =⎢ 2⎥
⎣ V1 ⎦ ⎣ T 2 ⎦ ⎣ T1 ⎦
−1 /(n −1)
⎡T ⎤ ⎡T ⎤
Therefore, S 2 − S1 = mR ln⎢ 2 ⎥ + mCv ln⎢ 2 ⎥
⎣ T1 ⎦ ⎣ T1 ⎦
−1 /( n − 1)
⎡T ⎤ ⎡T ⎤
= m(C p − Cv ) ln ⎢ 2 ⎥ + mC v ln ⎢ 2 ⎥
⎣ T1 ⎦ ⎣ T1 ⎦

⎛ − 1 ⎞ ⎡ T2 ⎤ ⎡T ⎤
= m(C p − Cv )⎜ ⎟ ln ⎢ ⎥ + mC v ln ⎢ 2 ⎥
⎝ n − 1 ⎠ ⎣ T1 ⎦ ⎣ T1 ⎦
Cp ⎛ − 1 ⎞ ⎡T2 ⎤ ⎡T ⎤
= mCv ( − 1)⎜ ⎟ ln⎢ ⎥ + mCv ln⎢ 2 ⎥
Cv ⎝ n − 1 ⎠ ⎣ T1 ⎦ ⎣ T1 ⎦

⎛ − 1 ⎞ ⎡ T2 ⎤ ⎡T ⎤
= mCv (γ − 1)⎜ ⎟ ln⎢ ⎥ + mCv ln⎢ 2 ⎥
⎝ n − 1 ⎠ ⎣ T1 ⎦ ⎣ T1 ⎦
⎡T ⎤⎡ γ − 1⎤
= mCv ln⎢ 2 ⎥⎢1− ⎥
⎣ T1 ⎦⎣ n − 1⎦
⎡T ⎤ ⎡ n − 1 − γ + 1⎤
= mCv ln⎢ 2 ⎥ ⎢
⎣ T1 ⎦ ⎣ n − 1 ⎥⎦
⎡T ⎤⎡ n − γ ⎤
= mCv ln⎢ 2 ⎥⎢ ⎥
⎣ T1 ⎦⎣ n −1⎦
Principle of increase of entropy
dQ dQ
For any process, ≤ dS (or) dS ≥
T T
For isolated system, dQ = 0, dS iso ≥ 0

For reversible process, dS iso = 0 ; S = constant

14
 For an irreversible process, dS iso > 0

It is thus proved that the entropy of an isolated system can never decrease. It always increases and
remains constant only when the process is reversible. This is known as the ‘principle of increase of
entropy’ or simply the ‘entropy principle’.
dS iso = dS uni = dS sys + dS surr ≥ 0

There is no reversible process in nature. Therefore the entropy of the universe is going on
increasing.
Absolute entropy
It is possible to find the amount by which the entropy of the system changes in a process, but not
the value of absolute entropy. When it is required to find the absolute entropy, a zero value of entropy of
the system at an arbitrary chosen standard state is assigned, and the entropy changes are calculated with
reference to this standard state.
Carnot theorem
It states that of all heat engines operating between a given constant temperature source and a
given constant temperature sink, none has a higher efficiency than a reversible engine.
Available energy (Exergy) and Unavailable energy (Anergy)
Available energy (AE) Æ Part of the low grade energy available for conversion into work.
Unavailable energy (UE) Æ Part of the low grade energy which must be rejected to environment and can
not be converted into shaft work.
Heat supplied, Q1 = AE + UE
Wmax = AE
Maximum work can be obtained from ideal or Carnot engine.
T1 − T2 T
η= = 1− 2
T1 T1
T1 Æ Higher temperature (Source temperature)
T2 Æ Lower temperature (Sink temperature)
For maximum efficiency T2 should be equal to atmospheric temperature (Ta).
Ta W AE
∴ η = 1− = =
T1 Q1 Q1

⎛ T ⎞
AE = Q1 ⎜⎜1 − a ⎟⎟
⎝ T1 ⎠
2

Q1
AE
T 1

15
 UE

To
S
Unavailable energy (UE) = To (S2 – S1)
Decrease in available energy when heat is transferred through a finite temperature difference
Q1
T1
Q1’
T1 `
T

To Increase in UE

ΔS
ΔS’
Let us consider a reversible heat engine operating between T1 and To.
Q1 = T1 ΔS; Q2 = To ΔS
AE = (T1 –To) ΔS = T1 ΔS – To ΔS
Now assume that the heat Q1 is transferred through a finite temperature difference from the
source at T1 to the engine absorbing heat at T1` lower than T1.
Q1 = T1’ ΔS’ ; Q2’ = To ΔS’
AE = (T1’ –To) ΔS’ = T1’ ΔS’ – To ΔS’
Decrease in AE = T1 ΔS – To ΔS – T1’ ΔS’ – To ΔS’
But T1 ΔS = T1’ ΔS’’
Availability
Availability of a given system is defined as the maximum useful work that is obtainable in a
process in which the system comes to equilibrium with its surroundings.
Open system: Wmax = (h1 − h2 ) − To ( s1 − s 2 )

Wu max = Wmax

Availability, ψ 1 − ψ 2 = (h1 − h2 ) − To ( s1 − s 2 )

Closed system: Wmax = (u1 − u 2 ) − To ( s1 − s 2 )

Availability, φ1 − φ 2 = (u1 − u 2 ) − To ( s1 − s 2 ) + p o (v1 − v 2 )

III-Law of thermodynamics (Nernst theorem)

16
 - States that the entropy of a pure substance at absolute zero temperature (0 K) is zero. Or it is
impossible to attain absolute zero temperature.
Consider a reversed heat engine to remove heat from low temperature body to attain 0 K. The
work required to run the RHE (refrigerator) will be minimum if it operates on Carnot cycle.
Carnot COP = T2 / (T1 – T2) = Heat extracted / Work input
Work input = Heat extracted x (T1 – T2) / T2
T1 Æ Sink temperature (high)
T2 Æ Source temperature (low)

If T2 is 0 K, the work input will become infinitive. Infinitive work input is impossible and
therefore attaining absolute zero temperature is impossible.

Problems
1. A heat engine receives heat at the rate of 1500 kJ/min and gives an output of 8.2 kW.
Determine the thermal efficiency and the rate of heat rejection.
Given:
Heat received by engine (HS) = 1500 kJ/min = 25 kW
Work output (W) = 8.2 kW
Required: ηth and HR
Solution:
ηth = Work output / Heat Supplied = 8.2 / 25 = 0.328 ---- Ans
Workdone = HS – HR
8.2 = 25 – HR
∴ Heat rejected (HR) = 16.8 kW ---- Ans
2. A house requires 2 x 105 kJ/h for heating the house in winter. Heat pump is used to absorb
heat from cold air outside in winter and send heat to the house. Work required to operate
the heat pump is 3 x 104 kJ/h. Determine heat abstracted from outside and COP.
17
 Given:
Heat pump
Heat to be supplied to the house (HR) = 2 x 105 kJ/h
Work input (W) = 3 x 104 kJ/h
Required: HS and COP
Solution:
HS = Heat absorbed from the atmosphere
HR = Heat rejected by the heat pump = Heat supplied to the room
Heat pump

HS HR (To house)
(From atmosphere)
W
For heat pump, - W = HS + HR
-3 x 104 = HS – 2 x 105
HS = 1.7 x 105 kJ/h ---- Ans
COP = Out put / Input
= Heat given to house / work input to heat pump
= 2 x 105 / 3 x 104 = 6.66 --- Ans
3. 0.04 kg of CO2 (M = 44) is compressed from 1 bar, 20oC, until the pressure is 9 bar and the
volume is then 0.003 m3. Calculate the change of entropy. Take Cp for CO2 as 0.88 kJ/K and
assume CO2 to be a perfect gas.
Given:
Mass of gas (m) = 0.04 kg
Initial pressure (p1) = 1.05 bar
Initial temperature (T1) = 20oC = 293 K
Final pressure (p2) = 9 bar
Final volume (V2) = 0.003 m3
Required: Δ S
Solution:
S2 – S1 = m R ln [V2/V1] + m Cv ln [T2/T1] Æ General equation
Universal gas constant (Ru) = 8314 J/kg-K
Characteristic gas constant (R) = Ru/M = 8314/44 = 188.95 J/kg-K
To find m
p1 V1 = m R T1
5
1 x 10 x V1 = 0.04 x 188.95 x 293
V1 = 0.02214494 m3
To find T2
18
 p2 V2 = m R T2
5
9 x 10 x V2 = 0.04 x 188.95 x T2
T2 = 357.24 K
To find Cv
Cp – Cv = R
880 – Cv = 188.95
Cv = 691.05 J/kg-K
∴ S2 – S1 = (0.04) (188.95) ln[0.003/0.02214494] + (0.04) (691.05) ln [357.24/293]
= -9.62847 J/kg-K --- Ans
4. A perfect gas is compressed according to the law pV1.25 = C from an initial pressure of 1 bar
and volume of 0.9 m3 to a final volume of 0.6 m3. Determine the final pressure and change of
entropy per kg of gas during the process. Take γ = 1.4 and R = 287 J/kg-K.
Given:
Polytropic process
Index of compression (n) = 1.25
Initial pressure (p1) = 1 bar
Initial volume (V1) = 0.9 m3
Final volume (V2) = 0.6 m3
Required: p2 & Δ S per kg
Solution:
S2 – S1 = m R ln [V2/V1] + m Cv ln [T2/T1] Æ General equation
To find p2
p2 / p1 = [V1/V2]n
p2 / 1 = [0.9/0.6]1.25
p2 = 1.66 bar --- Ans
To find Cv
Cp – Cv = R and Cp/Cv = γ
Cp = γ Cv
γ Cv – Cp = R
1.4 Cv – Cv = 287
Cv = 717.5 J/kg-K
∴ Cp = 1.4 x 717.5 = 1004.5 J/kg-K
To find T1 & T2
m = 1 kg
p1 V1 = m R T1
5
1 x 10 x 0.9 = 1 x 287 x T1
T1 = 313.6 K
p2 V2 = m R T2
19
 1.66 x 105 x 0.6 = 1 x 287 x T2
T2 = 347.04 K
∴ S2 – S1 = 1 x 287 x ln [0.6/0.9] + 1 x 717.5 x ln [347.04/313.6]
= -43.62 J/kg-K --- Ans
Note: Δ S can be found out using, (S2 – S1) = (n – γ)/(n – 1) x m Cv ln [T2/T1] during polytropic
process.
5. Calculate the change of entropy of 1 kg of air expanding polytropically in a cylinder behind
a piston from 7 bar and 600oC to 1.05 bar. The index of expansion is 1.25.
Given:
Polytropic process
Index of compression (n) = 1.25
Mass of air (m) = 1 kg
Initial pressure (p1) = 7 bar
Initial pressure (p1) = 1.05 bar
Initial temperature (T1) = 600oC = 873 K
Required: Δ S
Solution:
S2 – S1 = m R ln [V2/V1] + m Cv ln [T2/T1] Æ General equation
(n −1) / n
T2 ⎡ p2 ⎤
=⎢ ⎥
T1 ⎣ p1 ⎦

(1.25 −1) / 1.25

T2 ⎡ 1.05 ⎤
=
T1 ⎢⎣ 7 ⎥⎦
= 0.68425
1/ n
V2 ⎡ p1 ⎤
=⎢ ⎥
V1 ⎣ p2 ⎦

1/ 1.25
V2 ⎡ 7 ⎤
=
V1 ⎢⎣ 1.05 ⎥⎦
= 4.5617
Cv = 717 J/kg-K & R = 287 J/kg-K (Taken)
∴ S2 – S1 = 1 x 287 x ln [4.5617] + 1 x 717 x ln [0.68425]
= 163.52 J/kg-K --- Ans
6. A Carnot engine is operated between two reservoirs at temperatures of 450 K and 325 K. If
the engine receives 300 kJ of heat from the source in each cycle, calculate the amount of
heart rejected to the sink in each cycle. Calculate the efficiency of the engine and the
workdone by the engine in each cycle.
20
 Given:
Carnot engine
Higher temperature (T1) = 450 K
Lower temperature (T2) = 325 K
Heat supplied (HS) = 300 kJ
Required: HR, η, W
Solution:
ηc = (T1 – T2) / T1
= (450 – 325) / 450 = 0.278 ---- Ans
Also ηc = (HS – HR) / HS = W / HS
W = η x HS = 0.278 x 300 = 83.4 kJ --- Ans
W = HS – HR
83.4 = 300 – HR
HR = 216.66 kJ --- Ans
7. In a Carnot cycle, the maximum pressure and temperature are limited to 18 bar and 410oC.
The ratio of isentropic compression is 6 and isothermal expansion is 1.5. Assuming the
volume of the air at the beginning of isothermal expansion as 0.18 m3, determine (i) the
pressure and temperature at main points (ii) change in entropy during isothermal
expansion (iii) mean thermal efficiency of the cycle (iv) mean effective pressure of the cycle
and (v) the theoretical power if there are 210 working cycles per min.
Given:
Maximum pressure (p3) = 18 bar
Maximum temperature (T3) = 410oC = 683 K = T4
Ratio of isentropic compression (V2/V3) = 6
Ratio of isothermal expansion (V4/V3) = 1.5
Volume at the beginning of isothermal expansion (V3) = 0.18 m3
Working cycle per min (n) = 210
Required : (i) p1, T1, p2, T2, p4 (ii) Δ S (iii) η (iv) mep (v) P
Solution:
(i) To find p2
(p2 / p3) = [V3 / V2]γ
∴ p2 / 18 = [1/6]1.4
∴ p2 = 1.465 bar --- Ans
To find T2
(T2 / T3) = [V3 / V2]γ-1
∴ T2 / 683 = (1/6)1.4-1
∴ T2 = 333.55 K --- Ans = T1
To find p4
p3V3 = p4V4
V4 = 1.5 V3 = 1.5 x 0.18 = 0.27 m3
∴ 18 x 0.18 = p4 x 0.27
p4 = 12 bar --- Ans

21
 To find p1
(p1 / p4) = [V4 / V1]γ = [V3/V2]γ
∴ p1 / 12 = [1/6]1.4
∴ p1 = 0.977 bar --- Ans

T1 3 4
4
p T

2 T2 2 1

V s

(ii) Change in entropy during isothermal expansion

Δ S = m R ln [V4/V3]
To find m
p1 V1 = m R T1
To find V1
V1 = 6 x 0.27 = 1.62 m3
5
0.977 x 10 x 1.62 = m x 287 x 333.55
m = 1.653 kg
∴ Δ S = 1.653 x 287 x ln [1.5] = 192.36 J/K ---- Ans
(iii) η = (Tmax - Tmin) / Tmax = (683 – 333.55) / 683
= 0.5116 ---- Ans
(iv) pm = W / Vs
W = Area of T-s chart = Δ S (Tmax – Tmin)
= 192.36 x (683 – 333.55)
= 67220.2 J
Vs = V1 – V3
= 1.62 – 0.18 = 1.44 m3
pm = 67220.2 / 1.44 = 46680.7 bar --- Ans
(v) Power = W x n / 60 = 67220.2 x 210 / 60 = 2357270.7 W --- Ans
8. A domestic food freezer maintains a temperature of -15oC. The ambient air temperature is
30oC. If heat leaks into the freezer at the continuous rate of 1.75 kJ/s, what is the least
power necessary to pump this heat out continuously?
Given:
Freezer (Cold space)
Freezer temperature (T2) = -15oC = 258 K
Ambient temperature (T1) = 30oC = 303 K
Heat into the freezer = Heat absorbed by the refrigerant = Q2 = 1.75 kJ/s
Required: Least power (W)
Solution:
Refrigerator working on Carnot cycle requires least power input for its operation.

22
 Q2 T2
COP = =
W T1 − T2
258
COP = = 5.733
303 − 258
1.75
5.733 =
W
W = 0.305 kW --- Ans
9. A reversible heat engine operates between two reservoirs at temperatures of 600oC and
40oC. The engine drives a reversible refrigerator which operates between reservoirs at
temperatures of 40oC and -20oC. The heat transfer to the engine is 2000 kJ and the net work
output of the combined engine-refrigerator plant is 360 kJ. (a) Evaluate the heat transfer to
the refrigerant and the net heat transfer to the reservoir at 40oC. (b) Reconsider (a) given
that the efficiency of the heat engine and the COP of the refrigerator are each 40% of their
maximum possible values.
Given:
Combined engine-refrigerator plant
Engine source temperature (T1) = 600oC = 873 K
Engine sink temperature (T2) = 40oC = 313 K
Heat transfer to the engine (Q1) = 2000 kJ
Refrigerator source temperature (T4) = -20oC = 253 K
Refrigerator sink temperature (T3) = 40oC = 313 K
Net work transfer (W1 – W2) = 360 kJ
Required: (a) For reversible engine, Heat transfer to the refrigerant (Q4) and Net heat
transfer to the reservoir at 40oC (Q2 + Q3)
(b) For irreversible engine with η = 0.4 (ηrev) and COP = 0.4 (COPrev),
Q4 and Q2 + Q3

Solution:
T1 − T2 W1 Q1 − Q2
(a) η rev = = =
T1 Q1 Q1
Source Source
T1 T4

Q1 Q4
W1 W2
HE Ref
23
 Q2 Q3

T2 T3
Sink Sink
873 − 313
η rev = = 0.0414
873
W1
0.6414 = W1 = 1282.8 kJ
2000
W1 = Q1 – Q2
1282.8 = 2000 – Q2 Q2 = 717.2 kJ
W1 – W2 = 360
1282.8 – W2 = 360 W2 = 922.8 kJ
Q4 Q T4
COPrev = = 4 =
Q3 − Q4 W2 T3 − T4
253
COPrev = = 4.2166
313 − 253
Q4
4.2166 = Q4 = 3891 kJ ----- Ans
922.8
W2 = Q3 – Q4
922.8 = Q3 – 3891 Q3 = 4813.8 kJ
Therefore, Q3 + Q4 = 4813.8 + 717.2 = 5531 kJ ------ Ans
(b)
η = 0.4η rev = 0.4 (0.6414) = 0.256
COP = 0.4 COPrev = 0.4 (4.2166) = 1.686

W1
0.256 = W1 = 512 kJ
2000
W1 = Q1 – Q2
512 = 2000 – Q2 Q2 = 1488 kJ
W1 – W2 = 360
512 – W2 = 360 W2 = 152 kJ
Q4
1.686 = Q4 = 256.3 kJ ----- Ans
152
W2 = Q3 – Q4
152 = Q3 – 256.3 Q3 = 408.3 kJ
Therefore, Q3 + Q4 = 1488 + 408.3 = 1896.3 kJ ------ Ans
10. A heat engine operating between two reservoirs at 1000 K and 300 K is used to drive a heat
pump which extracts heat from the reservoir at 300 K at a rate twice that at which the
24
engine rejects heat to it. If the efficiency of the engine is 40% of the maximum possible and
COP of the heat pump is 50% of the maximum possible, what is the temperature of the
reservoir to which the heat pump rejects heat? What is the rate of heat rejection from the
heat pump if the rate of heat supply to the engine is 50 kW?
Given:
Combined engine-heat pump plant
Engine source temperature (T1) = 1000 K
Engine sink temperature (T2) = 300 K
Heat supply to the engine (Q1) = 50 kW
Heat pump source temperature (T4) = 300 K
Efficiency of the engine (η) = 0.4 ηrev
COP of the heat pump = 0.5 COPrev
Hear extraction by the heat pump (Q4) = 2 x Heat rejection by the engine (Q2)
Required: Temperature of heat pump sink (T3) and Heat rejection from the pump (Q3)
Solution:
Source Source
T1 T3

Q1 Q3
W1 W2
HE HP
Q2 Q4

T2 T4
Sink Sink
(a)
T1 − T2 W1 Q1 − Q2
η rev = = =
T1 Q1 Q1
1000 − 300
η rev = = 0 .7
1000
η = 0.4η rev = 0.4 (0.7) = 0.28
W1
η=
Q1
W1
0.28 = W1 = 14 kW
50
W1 = Q1 – Q2
14 = 50 – Q2 Q2 = 36 kW

25
 Q4 = 2 (36) = 72 kW
Q3 Q T3
COPrev = = 3 =
Q3 − Q4 W2 T3 − T4
W2 = Q3 – Q4
14 = Q3 – 72 Q3 = 86 kW ---- Ans
Note: W1 = W2
Q3 Q
COP = = 3
Q3 − Q4 W2
86
COP = = 6.1428
14
COP 6.1428
COPrev = = = 12.286
0 .5 0 .5
T3
12.286 = T3 = 326.6 K ----- Ans
T3 − 300
11. Two reversible heat engines A and B are arraged in series. Engine-A rejecting heat directly
to engine-B and receives 200 kJ at a temperature of 421oC from a hot source, while engine-
B is in communication with a cold sink at a temperature of 4.4oC. If the work output of A is
twice that of B, find (i) the intermediate temperature between A and B, (ii) the efficiency of
each engine and (iii) the heat rejected to the cold sink.
Given:
Two engines in series
Heat received by engine-A (Q1) = 200 kJ = 200000 J
Temperature of source for the engine-A (T1) = 421OC = 694 K
Temperature of cold sink (T3) = 4.4oC = 277.4 K
Work out put of engine-A (W1) = 2 W2
Required: (i) T2 (ii) ηA and ηB (iii) Q3
Solution:
T1 − T2 W1 Q1 − Q2
ηA = = =
T1 Q1 Q1
694 − T2 200000 − Q2
=
694 200000

T1

Q1

A W1

26
 Q2
T2
Q2

B W2
Q3

T3

T2 Q2
1− = 1−
694 200000
200000 T2 – 694 Q2 = 0
694 Q2
T2 = ----- (1)
200000
T2 − T3 W2 Q2 − Q3
ηB = = =
T2 Q2 Q2
T2 − 277.4 0.5 (Q1 − Q2 )
=
T2 Q2
T2 − 277.4 0.5 x 200000 − 0.5 Q2
=
T2 Q2
274 100000
1− = − 0 .5
T2 Q2
274 x 200000 100000
1− = − 0 .5
694 Q2 Q2
Solving the above, Q2 = 119961 J
Therefore, T2 = 694 x 119961 / 200000 = 416.26 K ----- Ans
694 − 416.26
ηA = = 0.4 ------ Ans
694
416.26 − 277.4
ηB = = 0.333 ----- Ans
416.26
119961 − Q3
Also, ηB = = 0.333
119961
Q3 = 80014 J --- Ans
12. 50 kg of water is at 313 K and enough ice is at -5oC is mixed with water in an adiabatic
vessel such that at the end of the process all the ice melts and water at 0oC is obtained. Find
the mass of ice required and the entropy change of water and ice. Given Cp of water is 4.2
kJ/kg-K, Cp of ice is 2.1 kJ/kg-K and latent heat of ice is 333.3 kJ/kg.

27
Given:
Mass of water (mw) = 50 kg
Initial temperature of water (Tw,1) = 313 K
Initial temperature of ice (Tice,1) = -5oC = 268 K
Final temperature of water (Tw,2) = 0oC = 273 K
Cp of water = 4200 J/kg-K
Cp of ice = 2100 J/kg-K
Latent heat of ice = 333.3 x 103 J/kg
Required: Mass of ice, Change of entropy of ice and water
Solution:
Ice (268 K)

Water (273 K)
Water (313 K)

mice = Mass of ice

Energy of the water before mixing = mw Cpw Tw1
= (50) (4200) (313) = 65730000 J
Energy of ice before mixing = mice Cpice Tice,1
= mice (2100) (268) = 562800 mice J
Energy of mixture (water at 0oC) = (mw + mice) Cpw Tw,2
= (50 + mice) (4200) (273)
Energy of water before mixing + Energy of ice before mixing
= Energy of mixture after mixing
65730000 + 562800 mice = (50 + mice) (4200) (273)
mice = 14.388 kg ---- Ans
⎛ Tw,1 ⎞
Entropy change of water = m w C pw ln⎜ ⎟ = (50) (4200) ln⎛⎜ 313 ⎞⎟
⎜T ⎟ ⎝ 273 ⎠
⎝ w, 2 ⎠
= 28713.6 J/K ----- Ans
Entropy change of ice (ΔS ) ice = (ΔS ) I + (ΔS ) II

SH LH
II
T I

28
 Heat (Q)
⎛ Tice ,1 ⎞
(ΔS ) I = mice C pice ln⎜⎜ ⎟ = (14.388) (2100) ln⎛⎜ 268 ⎞⎟ = −558.5 J / K
⎟ ⎝ 273 ⎠
⎝ Tice , 2 ⎠

dQ mice x Latent heat of fusion (14.388)(333.3 x 10 3 )

(ΔS ) II = = = = 17566 J / K
T T 273
(ΔS ) ice = (−558.5) + (17566) = 17007.5 J / K ------ Ans
13. 1 kg of air is contained in a piston cylinder assembly at 10 bar pressure and 500 K
temperature. The piston moves outwards and the air expands to 2 bar pressure and 350 K
temperature. Determine the maximum work obtainable. Assume the environmental
conditions to be 1 bar and 290 K. Also make calculations for the availability in the initial
and final states.
Given:
Mass of air (m) = 1 kg
Initial pressure of air (p1) = 10 bar = 10 x 105 Pa
Initial temperature of air (T1) = 500 K
Final pressure of air (p2) = 2 bar = 2 x 105 Pa
Final temperature of air (T2) = 350 K
Environment pressure (po) = 1 bar = 1 x 105 Pa
Environment temperature (To) = 290 K
Required: (a) Wmax (b) Ф1 – Фo and Ф2 – Фo
Solution:
(a) Wmax = (U 1 − U 2 ) − To (S1 − S 2 )

U 1 − U 2 = m C v (T1 − T2 ) = (1) (718) (500 − 350) = 107700 J

⎛V ⎞ ⎛T ⎞
S1 − S 2 = m R ln⎜⎜ 1 ⎟⎟ + m C v ln⎜⎜ 1 ⎟⎟
⎝ V2 ⎠ ⎝ T2 ⎠
V1 pT 2 x500
= 2 1 = = 0.2857
V2 p1T2 10 x 350

⎛ 500 ⎞
S1 − S 2 = (1) (287) ln (0.2857 ) + (1) (718) ln⎜ ⎟ = −103.46 J / K
⎝ 350 ⎠
Wmax = (107700) − 290 (− 103.46) = 137703.4 J − − − Ans

(b) φ1 − φ o = (U 1 − U o ) − To (S1 − S o ) + po (V1 − Vo )

U 1 − U o = m C v (T1 − To ) = (1) (718) (500 − 290) = 150780 J

29
 ⎛V ⎞ ⎛T ⎞
S1 − S o = m R ln⎜⎜ 1 ⎟⎟ + m C v ln⎜⎜ 1 ⎟⎟
⎝ Vo ⎠ ⎝ To ⎠
V1 pT 1 x500
= o 1 = = 0.1724
Vo p1To 10 x 290

⎛ 500 ⎞
S1 − S o = (1) (287) ln (0.1724 ) + (1) (718) ln⎜ ⎟ = −113.4 J / K
⎝ 290 ⎠
m R T1 m R To
V1 = Vo =
p1 po

⎛T T ⎞ ⎛ 500 290 ⎞
V1 − Vo = m R ⎜⎜ 1 − o ⎟⎟ = (1) (287) ⎜⎜ 5
− ⎟ = −0.6888 m 3
5 ⎟
⎝ p1 p o ⎠ ⎝ 10 x 10 1 x 10 ⎠
∴ φ1 − φ o = (150780) − (290)(− 113.4) + (1 x 10 5 )(− 0.6888)
= 114786 J − − − − Ans
φ 2 − φ o = (U 2 − U o 2 ) − To (S 2 − S o ) + po (V2 − Vo )
U 2 − U o = m C v (T2 − To ) = (1) (718) (350 − 290) = 43080 J

⎛V ⎞ ⎛T ⎞
S 2 − S o = m R ln⎜⎜ 2 ⎟⎟ + m C v ln⎜⎜ 2 ⎟⎟
⎝ Vo ⎠ ⎝ To ⎠
V2 p oT2 1 x350
= = = 0.6034
Vo p 2To 2 x 290

⎛ 350 ⎞
S 2 − S o = (1) (287) ln (0.6034 ) + (1) (718) ln⎜ ⎟ = −9.964 J / K
⎝ 290 ⎠
m R T2 m R To
V2 = Vo =
p2 po

⎛T T ⎞ ⎛ 350 290 ⎞
V2 − Vo = m R ⎜⎜ 2 − o ⎟⎟ = (1) (287) ⎜⎜ 5
− ⎟ = −0.33 m 3
5 ⎟
⎝ p 2 po ⎠ ⎝ 2 x 10 1 x 10 ⎠
∴ φ 2 − φ 2 = (43080) − (290)(− 9.964) + (1 x 10 5 )(− 0.33)
= 12969.56 J − − − − Ans
φ1 − φ 2 = (φ1 − φ o ) − (φ 2 − φ o )
= 114786 – 12969.56 = 101816.44 J --- Ans

14. Two kg of air at 500 kPa, 80oC expands adiabatically in a closed system until its volume is
doubled and its temperature becomes equal to that of the surroundings which is at 100 kPa,
5oC. For this process, determine (a) the maximum work, (b) the change in availability, and
(c) the irreversibility.

30
 Given:
Closed system
Mass of air (m) = 2 kg
Initial pressure of air (p1) = 500 kPa = 500 x 103 Pa
Initial temperature of air (T1) = 80oC = 353 K
Final pressure of air (p2) = 100 kPa = 100 x 103 Pa
Final temperature of air (T2) = 5oC = 278 K
V2 = 2 V1
Required: (a) Wmax (b) Ф1 – Ф2 (c) I
Solution:
(a) Wmax = (U 1 − U 2 ) − To (S1 − S 2 )

U 1 − U 2 = m C v (T1 − T2 ) = 2 (718) (353 − 278) = 107700 J

⎛V ⎞ ⎛T ⎞
S1 − S 2 = m R ln⎜⎜ 1 ⎟⎟ + m C v ln⎜⎜ 1 ⎟⎟
⎝ V2 ⎠ ⎝ T2 ⎠
⎛1⎞ ⎛ 278 ⎞
= 2 (287 ) ln⎜ ⎟ + 2 (718) ln⎜ ⎟ = −54.87 J / K
⎝2⎠ ⎝ 353 ⎠
∴ Wmax = (107700) − 278(− 54.87 ) = 122953.86 J ---- Ans

(b) φ1 − φ 2 = (U 1 − U 2 ) − To (S1 − S 2 ) + po (V1 − V2 )

p o (V1 − V2 ) = p o (V1 − 2V1 ) = − p oV1

m R T1 2 (287) (353)
V1 = = = 0.405244 m 3
p1 500 x 10 3

∴ φ1 − φ 2 = (107700) − 278(− 54.87 ) − 100 x 10 3 (0.405244)

= 82429.46 J − − − − Ans
(c) Irreversibility, I = Wmax - Wact
Q = Wact + (U2 – U1)
Q = 0 Æ Adiabatic process
Wact = (U1 – U2) = 107700 J
I = 122953.86 – 107700 = 15253.86 J ---- Ans
15. Air expands through a turbine from 500 kPa, 520oC to 100 kPa, 300oC. During expansion 10
kJ/kg of heat is lost to the surroundings which is at 98 kPa, 20oC. Neglecting the KE and PE
changes, determine per kg of air (a) the decrease in availability, (b) the maximum work,
and (c) the irreversibility.
Given:
Open system
Initial pressure of air (p1) = 500 kPa = 500 x 103 Pa
Initial temperature of air (T1) = 520oC = 793 K
31
Final pressure of air (p2) = 100 kPa = 100 x 103 Pa
Final temperature of air (T2) = 300oC = 573 K
Environment pressure (po) = 98 kPa = 98 x 103 Pa
Environment temperature (To) = 293 K
Heat loss to the surroundings (q) = 10 kJ/kg = 10 x 103 J/kg
Required: (a) Ф1 – Ф2 (b) Wmax (c) I
Solution:
(a) ψ 1 − ψ 2 = (H 1 − H 2 ) − To (S1 − S 2 )
= (h1 − h2 ) − To (s1 − s 2 ) per kg
h1 − h2 = C p (T1 − T2 ) = (1005) (793 − 573) = 221100 J / kg

⎛V ⎞ ⎛T ⎞
s1 − s 2 = R ln⎜⎜ 1 ⎟⎟ + C v ln⎜⎜ 1 ⎟⎟
⎝ V2 ⎠ ⎝ T2 ⎠
p1V1 p 2V2
=
T1 T2
V1 p T 100 x 793
= 2 1 = = 0.2768
V2 p1T2 500 x 573

⎛ 793 ⎞
∴ s1 − s 2 = (287) ln (0.2768) + (718) ln⎜ ⎟ = −135,34 J / kgK
⎝ 573 ⎠
ψ 1 − ψ 2 = (221100) − 293 (− 135.34) = 260754.62 J / kg --- Ans
(b) ∴ wmax = ψ 1 − ψ 2 = 260754.52 J / kg ---- Ans
(c) Irreversibility, I = wmax - wact
SFEE equation is given by,
m h1 + m q = m h2 + m wact
wact = q + (h1 – h2) = (-10 x 103) + (221100)
= 211100 J/kg
I = 260754.62 – 211100 = 49654.62 J /kg ---- Ans

32
 UNIT-III
PROPERTIES OF PURE SUBSTANCE AND STEAM POWER CYCLE
Pure Substance
A ‘Pure substance’ is defined as a homogeneous material, which retains its chemical composition
even though there may be a change in phase.
Water which is one of the pure substances, is an important engineering fluid. It exists in all three
phases, i.e., solid (ice), liquid (water) and gas (steam)
Other examples: Carbon dioxide, Oxygen, Hydrogen, Argon, Ammonia, etc..
Phases
In general, a pure substance exists in three forms. i) Solid phase, ii) Liquid phase & iii) Vapour or
gaseous phase.
When ice melts there is a transformation of phase from solid to liquid. This is called melting or
fusion of ice.
When water freezes there is a transformation of phase from liquid to solid. This is called freezing
or solidification of water.
When water vapourises there is transformation of phase from liquid to vapour. This is called
vapourisation.
When vapour condenses there is transformation of phase from vapour to liquid. This is called
condensation.
Steam generation
Consider 1 kg of ice contained in the cylinder under a pressure of ‘p’ exerted by the weight ‘W’
placed on the frictionless piston.

Let us assume ps = 5 bar & T1 = -10oC

Let heat be added continuously. First the ice temperature rises and there is no phase change, till
the temperature little lower than 0oC. The fusion temperature of water decreases slightly with increase in
pressure.
Fusion or freezing temperature = 0oC at atmospheric pressure. The transformation curve is shown
below.
I – Sensible heat addition
II – Latent heat of fusion
III – Sensible heat addition
IV – Latent heat of vapourisation
V – Heat of super heat

Super heated Vapour

T
(oC) D Vapourisation E
L+V

Liquid
B Melting C
Solid S+L
A
Heat added
I II III IV V

The warming up process is shown by ‘AB’. After this the ice begins to melt and two phase
mixture of ice & water remains at the same temperature. No rise in temperature is recorded while heat is
being added. This is shown by ‘BC’. At ‘C’, all the ice has melted and there is only one phase-liquid
water under the pressure ‘p’.
The quantity of heat required to transform ice into water while there is no change in temperature is
called ‘heat of fusion’.
If heat is further added to water, its temperature starts rising again and continues to do so till the
water reaches the boiling temperature (vapourisation temperature). This process is shown by ‘CD’.
The boiling point is function of pressure.
Tboiling = 100oC at 1 atm
= 151.84oC at 5 bar
This temperature is ‘Saturation temperature’ and the pressure corresponding to this temperature is
called ‘Saturation pressure’.

2
 After water has reached the saturation temperature, it starts vapourising while temperature remains
same and once again the two-phase mixture of water and water vapour is formed.
Two phase system can not be defined only by its temperature and pressure. If pressure is known,
the corresponding temperature of the two-phase system is known. Thus the state of vapour is defined by
either the pressure or temperature and the quality called ‘dryness fraction’ (x). Dryness fraction is defined
as the fraction of vapour present in the whole mass.
The process of vapourisation is shown by ’DE’. At ‘E’ all the water has vapourised and saturated
vapour at pressure ‘p’ is obtained as shown. And the heat required to vapourise the liquid to vapour at
constant temperature is known as the ‘Latent heat of vapourisation’.
Further heating, i.e., heating after the quality of the vapour is 100 %, a further rise in temperature
will rise the volume. The vapour is said to be ‘Superheated’. Therefore, Superheated vapour is vapour at
any temperature above the saturation temperature.
Critical point
As the pressure increased above atmospheric, the change of volume accompanying evaporation
decreases, till at sufficiently higher pressure, it is zero. This is referred as the ‘critical point’ and the
properties at this point for water are,
pc = 221.2 bar Tc = 374.15oC vc = 0.00317 m3/kg
Latent heat of vapourisation = 0

Critical temperature and pressure

The critical temperature of a liquid is that temperature beyond which the liquid cannot exist, no
matter how much pressure is placed on it. The pressure that is needed to cause the gas to condense at the
critical temperature is the critical pressure.
Triple point
At low pressures there is slight rise in melting points, but a marked drop in boiling points and also
very appreciable increase in the change of volume.
At pressure 0.006112 bar, the melting and boiling point temperatures become equal and the
change of phase ice-water-steam is shown by single horizontal line’BCF’. This line is called ‘Triple point
line’.
The temperature at this takes place is 0oC.

3
 Therefore only at 0oC & 0.006112 bar, ice, water & steam coexist in thermodynamic equilibrium
in a closed vessel.
p-v-T surface

(a)

4
(b)

(c)

(d)
5
 Fig (a) shows the p-v-T surface of a substance such as water that expands on freezing, in three-
dimensional view.
Fig (b) shows the p-v-T surface of a substance that contracts on freezing. Almost all the
substances behave in this manner.
There are regions on the p-v-T surfaces in both the cases labeled, solid, liquid and vapour. In
these single phase regions, the state is fixed by any two of the properties, pressure, specific volume and
temperature, since all are independent when they are in single phase. Located between the single phase
regions are two phase regions, where two phases exist in equilibrium, namely, liquid-vapour, solid-liquid
and solid-vapour. In the two phase region, the pressure and temperature are not independent properties.
One cannot be changed without changing the other. But, the state can be fixed by specific volume and
either pressure or temperature.
Gibb’s Phase rule
Gibbs' phase rule describes the possible number of degrees of freedom (F) in a closed system at
equilibrium, in terms of the number of separate phases (P) and the number of chemical components (C) in
the system.
Gibbs' rule then follows, as:
F=C−P+2
Where F is the number of degrees of freedom, C the number of chemical components, and P is the
number of phases that cannot be shared.
Consider water, the H2O molecule, C = 1.
When three phases are in equilibrium, P = 3, there can be no variation of the (intensive) variables
ie. F = 0. Temperature and pressure must be at exactly one point, the 'triple point' (temperature of 0.01oC
and pressure of 611.73 Pa). Only at the triple point can three phases of water exist at the same time. At
this one point, Gibbs rule states: F = 1 - 3 + 2 = 0.
When two phases are in equilibrium, P = 2, such as along the melting or boiling boundaries, the
(intensive) variable pressure is a determined function of (intensive) variable temperature, ie. one degree of
freedom. Along these boundaries, Gibbs rule states: F = 1 - 2 + 2 = 1.
Away from the boundaries of the phase diagram of water, only one phase exists (gas, liquid, or
solid), P = 1. So there are two degrees of freedom. At these points, Gibbs rule states: F = 1 - 1 + 2 = 2
Note that if you are considering three (intensive) variables: pressure, temperature, and volume of a gas
(ie. one phase, P = 1) then only two of the variables can be independent. This fact is illustrated by the
universal gas law:
pV=mRT
Another Example - For instance, a balloon filled with carbon dioxide has one component and one
phase, and therefore has two degrees of freedom - in this case temperature and pressure. If you have two
phases in the balloon, some solid and some gas, then you lose a degree of freedom - and indeed this is the
case, in order to keep this state there is only one possible pressure for any given temperature.
It is important to note that the situation gets more complicated when the (intensive) variables go
above critical lines or point in the phase diagram. At temperatures and pressure above the critical point,
the physical property differences that differentiate the liquid phase from the gas phase become less
defined. This reflects the fact that, at extremely high temperatures and pressures, the liquid and gaseous
phases become indistinguishable. In water, the critical point (thermodynamics) occurs at around 647K
(374°C or 705°F) and 22.064 MPa .
Type of Steam
Wet steam : Wet steam is defined as steam which is partly vapour and partly liquid suspended in it. It
means that evaporation of water is not complete.

6
Dry saturated steam : When the wet steam is further heated, and it does not contain any suspended
particles of water, it is known as dry saturated steam.
Superheated steam : When the dry steam is further heated at constant pressure, thus raising its
temperature, it is called superheated steam.
Steam properties

Saturated vapour
line
Saturated liquid
line
T hfg hg, sg, vg
hf, sf, vf

Heat added
The following are the steam properties :
• Dryness fraction
• Specific enthalpy
• Specific volume
• Specific internal energy
• Specific entropy
Dryness Fraction (Quality of steam)
It is the fraction of steam that is in the vapour form in the mixture.
Let, mg = mass of dry steam/kg of mixture
mf = mass suspended liquid/kg of mixture
x = dryness fraction of steam
mg
Therefore x=
mg + m f

For wet steam, x < 1

For dry steam, x = 1

Specific Enthalpy
Wet steam : If 1 kg of wet steam is heated mf kg receives the latent heat of vapourisation and mg kg is
unaffected.
∴ Latent heat absorbed by the wet steam = x hfg
hfg = Latent heat of vapourisation
∴ Specific enthalpy of wet steam (hwet) = hf + x hfg
hf = Specific enthalpy of saturated liquid

7
Dry steam
All the ‘hfg’ will be absorbed by the mg.
∴ Specific enthalpy of dry steam (hdry) = hf + hfg = hg since [hfg = hg – hf]
Superheated steam
Superheated steam is produced by adding sensible heat to dry steam.
∴ Specific enthalpy of superheated steam (hsup) = hf + hfg + Cp(Tsup – ts)
= hg + Cp(Tsup – ts)
Cp = Specific heat at constant pressure = 0.4 to 0.5 kJ/kg-K
Tsup = Temperature of superheated steam
ts = Saturation temperature
Tsup – ts = Degree of superheat
Specific volume
Wet steam
Let, vf = specific volume of saturated liquid
vg = specific volume of dry steam
∴ specific volume of wet steam (vwet) = vf + x vfg
= vf + x (vg – vf)
Dry saturated steam
Specific volume of dry steam (vdry) = vg
Superheated steam
Dry steam & Superheated steams behave like a perfect gas. Since the production of steam is
constant pressure process, we can write,
ps vg psup vsup
=
ts Tsup
vg vsup
=
ts Tsup
Tsup
vsup = v g
t sat
Specific internal energy
Wet steam
We know that specific enthalpy is given by,
h = u + pv
∴ u = h - pv
uwet = hwet - pvwet
Dry steam
u(dry) = h(dry) – pv(dry)

8
Super heated steam
u(sup) = h(sup) – pv(sup)
Specific entropy
Wet steam
s(wet) = sf + x sfg
Dry steam
s(dry) = sf + sfg = sg
Super heated steam
s(sup) = sg + Cp ln [Tsup/ts]
Tsup = Super heated steam temperature
ts = Saturation temperature
External Work of Evaporation
When he water is evaporated to form saturated steam, its volume increases from vf to vg at a
constant pressure, and thus external work is done by steam due to increases in volume. The energy for
doing the work is obtained during the absorption of latent heat. This work is called external work of
evaporation.
Wet Steam: Work of evaporation = p (x vg – vf)
x Æ Dryness fraction of steam
Dry Steam: Work of evaporation = p (vg – vf)
Processes of vapour
Constant pressure process

Non-Flow process:
2
w = ∫ p dv = p (v 2 − v1 )
1

q = w + (u 2 − u1 )
= p(v2 − v1 ) + (u 2 − u1 )
= p 2 v2 − p1v1 + u 2 − u1

9
 = h2 − h1
Flow process:
2
w = − ∫ v dp = 0
1

q = w + (h2 − h1 )
= 0 + (h2 − h1 ) = (h2 − h1 )
Therefore the heat transfer is the same in flow and non-flow processes.
Constant volume process

Non-Flow process:
2
w = ∫ p dv = 0
1

q = w + (u 2 − u1 )
= 0 + (u 2 − u1 )
= u 2 − u1
Flow process:
2
w = − ∫ v dp = v ( p 2 − p1 )
1

q = w + (h2 − h1 )
= v ( p 2 − p1 ) + (h2 − h1 )
Constant temperature process
Non-Flow process:
q = w + (u 2 − u1 )
w = q + (u1 − u 2 )
= T ( s 2 − s1 ) + (u1 − u 2 )
10
 q = T ( s 2 − s1 )

Flow process:
q = w + (h2 − h1 )
w = q + (h1 − h2 )
= T ( s 2 − s1 ) + (h1 − h2 )
q = T ( s 2 − s1 )
Adiabatic process
Non-Flow process:
q = 0 for adiabatic process
q = w + (u 2 − u1 )
w = (u1 − u 2 ) Æ Reversible
w = (u1 − u 2' ) Æ Irreversible
Flow process:
q = w + (h2 − h1 )
w = (h1 − h2 ) Æ Reversible
w = (h1 − h2' ) Æ Irreversible

11
Polytropic process

Non-Flow process:
2
p 2 v 2 − p1v1
w = ∫ p dv =
1
1− n
q = w + (u 2 − u1 )
p 2 v 2 − p1v1
= (u 2 − u1 ) +
1− n
Flow process:
2
n ( p 2 v 2 − p1v1 )
w = − ∫ v dp =
1
1− n
q = w + (h2 − h1 )
n ( p 2 v 2 − p1v1 )
= (h2 − h1 ) +
1− n
p 2 v 2 − p1v1
= (u 2 − u1 ) +
1− n
Therefore the heat transfer is the same in flow and non-flow processes.
Throttling process (Isenthalpic process)
Workdone = 0

12
 Heat transfer = 0
h1 = h2

Saturation curve

1 2
h

s
Rankine Cycle (Complete expansion cycle)
A power cycle continuously converts heat into work, in which a working fluid repeatedly
performs a succession of processes.
In vapour power cycle, the working fluid, which is water, undergoes a change of phase. Heat is
transferred to water in the boiler from an external source (furnace) to raise steam. The high pressure, high
temperature steam leaving the boiler expands in the turbine to produce shaft work. The steam leaving the
turbine condenses into water in the condenser rejecting the heat, and then the water is pumped back to the
boiler by feed pump absorbing some work.
A power cycle continuously converts heat into work, in which a working fluid repeatedly
performs a succession of processes.
In vapour power cycle, the working fluid, which is water, undergoes a change of phase. Heat is
transferred to water in the boiler from an external source (furnace) to raise steam. The high pressure, high
temperature steam leaving the boiler expands in the turbine to produce shaft work. The steam leaving the
turbine condenses into water in the condenser rejecting the heat, and then the water is pumped back to the
boiler by feed pump absorbing some work.
It consists of four ideal processes.
1. Reversible constant pressure process – this is heating process of water to form steam.
2. Reversible adiabatic process – this is expansion process of steam in the steam turbine.
3. Reversible constant pressure process – this is process of condensation of steam till it becomes
saturated liquid.
4. Reversible adiabatic process- this is compression process of saturated water in the feed pump.
Since all these four processes are ideal, the cycle is an Ideal cycle, called Rankine cycle. This is a
reversible cycle.
For the purpose of analysis the Rankine cycle is assumed to be carried out in a steady flow
operation.
For Steam turbine (Process 1-2 – Isentropic expansion)

SFEE is given by, m h1 + m C12/2 + Q1-2 = m h2 + m C22 + W1-2

But, C1 ≈ C2 and Q1-2 = 0

∴ W1-2 = m (h1 – h2)

13
 Workdone by the turbine = WT = m (h1 – h2)

Steam turbine
1
1
4 Boiler

Condenser
Feed pump 3

Simple steam power plant

For Condenser (Process 2-3 – Constant pressure condensation)

SFEE is given by, m h2 + m C22/2 + Q2-3 = m h3 + m C32 + W2-3

But, C2 ≈ C3 and W2-3 = 0

∴ Q2-3 = m (h3 – h2)

Heat rejected in the condenser = - m (h3 – h2) = m (h2 – h3)

14
For Feed pump (Process 3-4 – Isentropic compression)

SFEE is given by, m h3 + m C32/2 + Q3-4 = m h4 + m C42 + W3-4

But, C3 ≈ C4 and Q3-4 = 0

∴ W3-4 = m (h3 – h4)

Workdone on the Pump = WP = - m (h3 – h4) = m (h4 – h3)

For Boiler (Process 4-1 – Constant pressure heat addition)

SFEE is given by, m h4 + m C42/2 + Q4-1 = m h1 + m C12 + W4-1

But, C4 ≈ C1 and Q4-1 = 0

∴ Q4-1 = m (h1 – h4)

Heat supplied in the boiler = m ( h1 – h4)

Output Net work output

Rankine cycle efficiency = ηR = ---------- = ----------------------
Input Heat supplied

Net work output = WT – WP = m (h1 – h2) – m (h4 – h3)

Heat supplied = Q4-1 = m (h1 – h4)

m (h1 – h2) – m (h4 – h3)

∴ ηR = --------------------------------
m (h1 – h4)

(h1 – h2) – (h4 – h3)

∴ ηR = --------------------------- Æ considering pump work
(h1 – h4)

But pump work is very small & it can be neglected.

h1 – h2
= ---------- Æ neglecting pump work
h1 – h4

Note : Here h4 ≈ h3 = hf3 at condenser pressure.

Other parameters
Specific steaming rate (SSC)– Rate of steam flow required to produce unit shaft power output & usually
expressed in ‘kg/kWh’.

m
SSC = ----------- x 3600
WT –WP

3600
15
 = -----------
wT - wP

Note : WT – WP = m (wT – wP)

Specific heat rate (SHR) – Rate of heat input required to produce unit shaft power output. It is usually
expressed in ‘kJ/kWh’.

Heat supplied
SHR = -------------------- x 3600
WT - WP

m (h1 – h4)
= --------------- x 3600
WT - WP

h1 – h4
= ----------- x 3600
wT - wP

= 3600 / ηR
Work Ratio
wT - wP
WR = ------------
wT
Power developed
P = m (wT – wP)
Effect of initial temperature (T1)

1’

1
T

3 2 2’

`
s

As the temperature increases, h1’, wT & x2 increases.

• At the outlet of turbine it is required to maintain quality of steam in order to prevent the
damage of blades due to impingement of water particles.
• The maximum temperature of steam is fixed from metallurgical considerations.

Effect of increase of pressure

16
 Increase in pressure for particular Tmax,
• decreases the quality of exhaust steam
• decreases the heat supplied
If the moisture content of steam in the later stages of the turbine is higher, the entrained water
particles along with the vapour coming out from the nozzles with high velocity strike the blades and erode
their surfaces. From a consideration of the erosion of blades, in the later stages of the turbine, the
maximum moisture content at the turbine exhaust is not allowed to exceed 15 % or the quality to fall
below 85 %. Therefore like maximum temperature, maximum pressure at the turbine inlet also gets fixed.

1’’ 1’ 1 Tmax

T 4

3 2

x = 0.85

Reheat Cycle Reheater

LP Turbine

Boiler HP Turbine

Condenser

Feed Pump

If a steam pressure higher than pmax is used, in order to limit the quality to 0.85 at the turbine
exhaust, reheat has to be adopted.
The steam from the HP turbine is reheated in the reheater using flue gases at constant pressure to
initial temperature. Reheat pressure is an important factor. Low reheat pressure may bring down the T1
and hence efficiency. A high reheat pressure increases the moisture content at turbine exhaust. Thus the
reheat pressure is optimized.

17
 For most of the modern power plants, the reheat pressure is 0.2 to 0.25 p1.

Reheat -- increases the turbine work

-- reduces moisture content at turbine outlet
h2 – h2’ < h3 – h4
Total expansion = (h1 – h2) + (h3 – h4)
The use of reheat only gives a marginal increase in cycle efficiency, but it increases the net work
output by making possible the use of higher pressures, keeping the quality of steam at turbine exhaust
within a permissible limit.
Heat supplied = (h1 – h6) + (h3 – h2)

Pump work = h6 – h5

wT - wP (h1 – h2) + (h3 – h4) – (h6 – h5)

ηreheat = ------------ = ---------------------------------------
HS (h1 – h6) + (h3 – h2)

3600
Steam rate = ----------------------------------- ------ kg/kWh
(h1 – h2) + (h3 – h4) – (h6 – h5)

Advantages of Reheating
• There is an increased output of the turbine.
• Erosion and corrosion problems in steam turbine are eliminated since the quality of steam is
increased.
• There is an improvement in the thermal efficiency of the cycle.
18
 • Final dryness fraction of steam is improved.
• There is an increase in the nozzle and blade efficiencies.
Disadvantages
• Reheating requires more maintenance.
• The increase in thermal efficiency is depressed by the more cost of the heater.

Regeneration
It may be observed in the Rankine cycle that the condensate which is at a fairly low temperature
is pumped to the boiler. There is irreversible mixing of the cold condensate with hot boiler water. This
results in loss of cycle efficiency. Methods are adopted to heat the feed water from the condenser
reversibly by interchange of heat within the system and thus improving the cycle efficiency. This method
of heating is called regenerative feed heating and the cycle is called regenerative cycle.
The principle of regeneration can be practically utilized by extracting steam from the turbine at
several locations and supplying it to the regenerative heaters. The heating arrangement comprises of (i)
for medium capacity turbines – not more than 3 heaters (ii) for high pressure high capacity turbines – not
more than 7 heaters and (iii) for turbines of super critical parameters 8 to 9 heaters. The final condensate
heating temperature is kept 50 to 60oC below the boiler saturated steam temperature so as to prevent
evaporation of feed water.
Superheater

Boiler Turbine

m1 m2

m – m1 – m2

HPH LPH
Condenser

Pump(1) Pump (2) Pump (3)

Consider a cycle with 2 feed water heaters.

Let m = Mass of steam through the boiler.

m1 = Mass of steam through heater-I
m2 = Mass of steam through heater-II
∴ Mass of steam through the condenser = m – m1 – m2
5 – 6 Æ Pumping of (m – m1 – m2) kg of feed water by pump (3)

6 – 7 Æ Heating of (m – m1 – m2) kg of feed water by m2 kg of bled steam by LPH

7 – 8 Æ Pumping of (m – m1) kg of feed water by pump (2)

8 – 9 Æ Heating of (m – m1) kg of feed water by m1 kg of bled steam by HPH

19
 9 – 10 Æ Pumping of (m) kg of feed water by pump (1)

10 – 1 Æ Heating of (m) kg of feed water by supply of external heat in the boiler

m m
10
9 m1 2
8
T m2 m – m1
7 3
6
m – m1 – m2
5 4

Workdone
Efficiency = ------------------
Heat supplied

Workdone = m (h1 – h2) + (m – m1) (h2 – h3) + (m – m1 – m2) (h3 – h4)

Heat supplied = Heat supplied during process 10 – 1

= m (h1 – h10)
Advantages
• The heating process in the boiler tends to become reversible.
• The thermal stress problems due to temperature difference between feed water from the
condenser and boiler feed water are minimized.
• The thermal efficiency is improved.
• Heat supplied is reduced.
• Size of the condenser is reduced since it handles only (m – m1 – m2) kg of steam.
Disadvantages
• The plant becomes more complicated.
• Because of addition of heaters greater maintenance is required.
• Since the power output is slightly reduced due to extraction of steam, for a given power
output, large capacity boiler is required.
• The heaters are costly. The gain in thermal efficiency is depressed by heavier costs.
Steam Calorimeters
Steam calorimeters are used for measuring ‘dryness fraction’ of steam.
Types
• Tank or Bucket Calorimeter
20
 • Throttling Calorimeter
• Separating Calorimeter
• Separating & Throttling Calorimeter
• Electrical Calorimeter
Carnot cycle

4 1

T
3 2

s
1-2 – Steam turbine
2-3 – Condenser
3-4 – Feed pump
4-1 – boiler
Carnot cycle is independent of the working fluid. It consists of two isothermal & two isentropic
processes.

The vapour from wet region (3) is compressed isentropically from a lower temperature to the
upper temperature of the cycle by means of a pump. At this higher temperature, heat is added at constant
temperature in the boiler. The dry vapour is then expanded isentropically in a turbine. The wet vapour
from the turbine outlet is condensed in a condenser at constant temperature, thereby completing a cycle.

As we know that the Carnot cycle efficiency is depending on temperature limits, the efficiency is
given by,
T1 – T2
ηC = ---------
T1
Difficulties with Carnot cycle
1. Carnot cycle requires heat addition and heat rejection at constant temperatures and a constant
temperature heat transfer is not possible when the working medium is a gas or a superheated vapour.
2. The vapour compression process is difficult to accomplish.
3. Cycle efficiency is depending on temperature limits. The lower temperature is limited by the ambient
conditions, The upper temperature is limited by the metallurgical considerations. Modern metals can
withstand temperatures upto 650oC. The highest possible temperature at which a Carnot cycle can
operate with saturated steam is below critical temperature (374.15oC) and the critical pressure being
221.2 bar. Thus present day metals can not be used advantageously in a Carnot cycle with steam as
working fluid.

21
 PROBLEMS

1. Calculate the dryness fraction of steam which has 1.5 kg of water in suspension with 50 kg
of steam.
Given
Mass of water (mf) = 1.5 kg
Mass of steam (mg) = 50 kg
Required : Dryness fraction (x)
Solution
mg
Dryness fraction (x) = ------------
mg + mf

50
= -------------- = 0.971 --- Ans
50 + 1.5

2. Determine the amount of heat, which should be supplied to 2 kg of water at 25oC to convert
it into steam at 5 bar and 0.9 dry.
Given:
Mass of water (mf) = 2 kg
Initial temperature of water (T1) = 25oC
Pressure of steam (p) = 5 bar
Dryness fraction of steam (x) = 0.9
Required : Heat required to convert water into steam.
Solution:

151.5oC

25oC
Heat added

Heat required

Heat required = Sensible heat addition + Latent heat addition

22
 Sensible heat addition = m Cpw (ts – T1)

ts = saturation temperature = 151.5oC at 5 bar from steam table

Cpw = Specific heat at constant pressure = 4.186 kJ/kg

∴ Sensible Heat addition = 2 x 4.186 (151.5 – 25)

= 1059.058 kJ

Latent heat addition / kg = x hfg

Latent heat (hfg) = 2107.4 kJ/kg from steam table at 5 bar

Latent heat addition for ‘m’ kg = m x hfg

= 2 x 0.9 x (2107.4)

= 3793.32 kJ

∴ Total heat required = 1059.058 + 3793.32

= 4852.372 kJ --- Ans

3. Steam is generated at 8 bar from water at 32oC. Determine the heat required to produce 1
kg of steam (a) when the dryness fraction is 0.85 (b) when steam is dry saturated and (c)
when the steam is superheated to 305oC. The specific heat of superheated steam may be
taken as 2.093 kJ/kg-K.
Given:
Steam pressure (p) = 8 bar
Initial temperature of water (T1) = 32oC
Mass of steam (m) = 1 kg
Required: Heat required when (a) x = 0.85 (b) x = 1 (c) Tsup = 305oC
Solution:
(a)

SH LH

T
170.4oC

32oC

Heat added

Heat required

23
 Heat required = Sensible heat addition + Latent heat addition
Sensible heat addition = m Cpw (ts – T1)
ts = saturation temperature = 170.4oC at 8 bar from steam table
Cpw = Specific heat at constant pressure = 4.186 kJ/kg (Taken)
∴ Sensible Heat addition = 1 x 4.186 x (170.4 – 32)
= 579.34 kJ/kg
Latent heat addition / kg = x hfg
Latent heat (hfg) = 2046.5 kJ/kg from steam table at 8 bar
Latent heat addition for ‘m’ kg = m x hfg
= 1 x 0.85 x (2046.5)
= 1739.525 kJ/kg
∴ Total heat required = 579.34 + 1739.525
= 2318.865 kJ/kg --- Ans
(b)

SH LH

T
170.4oC

32oC
Heat added

Heat required

Heat required = Sensible heat addition + Latent heat addition

Latent heat addition / kg = x hfg
Latent heat (hfg) = 2046.5 kJ/kg from steam table at 8 bar
Latent heat addition for ‘m’ kg = m x hfg
= 1 x 1 x (2046.5)
= 2046.5 kJ/kg
∴ Total heat required = 579.34 + 2046.5
= 2625.84 kJ/kg --- Ans
(c)
LH
SH 305oC

170.4oC

24
 32oC
Heat added

Heat required

Heat required = Sensible heat addition + Latent heat addition + Sensible Heat addition
Sensible heat addition to superheated steam = m Cpv (Tsup – ts)
= 1 x 2.093 x (305 – 170.4)
= 281.72 kJ/kg
Latent heat addition / kg = hfg
Latent heat (hfg) = 2046.5 kJ/kg from steam table at 8 bar
∴ Total heat required = 579.34 + 2046.5 + 281.72
= 2907.56 kJ/kg --- Ans
4. Find specific volume, enthalpy and internal energy of wet steam at 18 bar, dryness fraction
0.9.
Given:
Pressure of steam (p) = 8 bar
Dryness fraction of steam (x) = 0.9
Required : v, h, u
Solution:
Specific volume (v) = vf + x vfg

From steam table, at 18 bar,

vf = 0.001168 m3/kg vg = 0.11033 m3/kg hf = 884.5 kJ/kg

hfg = 1910.3 kJ/kg hg = 2794.8 kJ/kg

∴ Specific volume (v) = 0.001168 + 0.9 (0.11033 – 0.001168)

= 0.0994138 m3/kg ---- Ans
Specific enthalpy (h) = hf + x hfg
= 884.5 + 0.9 (1910.3)
= 2603.77 kJ/kg ---- Ans
Specific internal energy (u) = h – pv
= 2603.77 – (18 x 105 x 0.0994138 / 1000)
= 2424.825 kJ/kg ---- Ans
Note : Divide the term ‘pv’ by 1000 to convert J/kg into kJ/kg.
5. Find the dryness fraction, specific volume and internal energy of the steam at 7 bar and
enthalpy 2600 kJ/kg.

25
 Given:
Pressure of steam (p) = 7 bar
Enthalpy of steam (h) = 2600 kJ/kg
Required : x, v, u
Solution:
We know , h(wet) = hf + x hfg
At 7 bar, from steam table, hf = 697.1 kJ/kg hfg = 2064.9 kJ/kg
3
vf = 0.001108 m /kg vg = 0.27268 m3/kg
2600 = 697.7 + x (2064.9)
x = 0.92 ---- Ans
Specific volume (v) = vf + x vfg
= 0.001108 + 0.92 (0.27268 – 0.001108)
= 0.25095 m3/kg ---- Ans
Specific internal energy (u) = h – pv
= 2600 – (7 x 105 x 0.25095 / 1000)
= 2424.335 kJ/kg ---- Ans
3
6. A vessel of volume 0.04 m contains a mixture of saturated water & saturated steam at a
temperature of 250oC. The mass of the liquid present is 9 kg. Find the pressure, mass, the
specific volume, the enthalpy, the entropy and the internal energy.
Given:
Total volume (V) = 0.04 m3
Temperature of saturated steam (ts) = 250oC
Mass of the liquid (mf) = 9 kg
Required : p, m, v, h, s, u
Solution:
From steam table at 250oC, ps = 39.73 bar --- Ans
From steam table at 250oC,
vf = 0.0012512 m3/kg vg = 0.05013 m3/kg hf = 1085.36 kJ/kg
hfg = 1716.2 kJ/kg sf = 2.7927 kJ/kg-K sfg = 3.2802 kJ/kg-K
Mass of steam (m) = mg + mf
Mass of dry steam (mg) = Vg / vg
Vg = Volume of dry steam = V – Vf
= V – mf (vf)
= 0.04 – 9 x 0.0012512
= 0.02874 m3
∴ Mass of dry steam (mg) = 0.02874 / 0.05013 = 0.5733 kg
∴ Mass of steam (m) = 0.5733 + 9 = 9.5733 ---- Ans
Specific volume (v) = vf + x vfg
26
 mg 0.5733
But, x = ---------- = -------------- = 0.06
mg + mf 0.5733 + 9
∴ v = 0.0012512 + 0.06 (0.5013 – 0.0012512)
= 0.0041839 m3/kg ---- Ans
Specific enthalpy (h) = hf + x hfg
= 1085.36 + 0.06 (1716.2)
= 1188.332 kJ/kg ---- Ans
Specific entropy (s) = sf + x sfg
= 2.7927 = 0.06 (3.2802)
= 2.9895 kJ/kg ---- Ans
Specific internal energy (u) = h – pv
= 1188.332 – (39.73 x 105 x 0.0041839 / 1000)
= 1171.709 kJ/kg ---- Ans
7. Steam has a pressure of 1 bar and temperature of 150oC after throttling. The pressure
before throttling is 14 bar. Find the dryness fraction of steam before throttling. Also find
the change in entropy.
Given:
Pressure after throttling (p2) = 1 bar
Temperature after throttling (T2) = 150oC
Pressure before throttling (p1) = 14 bar
Required : Dryness fraction (x) & (s2 – s1)
Solution:
The condition of steam will be superheated after throttling.
For throttling process, h1 = h2
We know, for wet steam, h1 = hf1 + x hfg1
At 14 bar, from steam table, hf = 830 kJ/kg
hfg = 1960 kJ/kg
For super heated steam, at 1 bar & 150oC, from steam table,
h2 = 2776.3 kJ/kg
∴ 830 + x (1960) = 2776.3
x = 0.993 --- Ans
Change in entropy = (s2 – s1)
At 14 bar, from steam table, sf = 2.284 kJ/kg-K
sfg = 4.185 kJ/kg-K
∴ s1 = 2.284 + 0.993 (4.185)
= 6.439705 kJ/kg-K
At 1 bar & 150oC , s2 = 7.614 kJ/kg-K

27
 ∴ (s2 – s1) = 7.614 – 6.439705 = 1.174295 --- Ans
8. Calculate the state of steam, i.e., whether it is wet, dry or superheated for the following
cases.
i) Steam has a pressure of 15 bar and specific volume of 0.12 m3/kg
ii) Steam has a pressure of 10 bar and temperature of 250oC
Given:
i) p =15 bar & v = 0.12 m3/kg
ii) p = 10 bar & T = 250oC
Required : To identify whether the steam is wet, dry or superheated.
Solution:
i) Assume that the steam is wet.
We know, for wet steam, v = vf + x (vg – vf)
At 15 bar, from steam table, vf = 0.00115 m3/kg
vg = 0.13177 m3/kg
∴ 0.12 = 0.00115 + x (0.13177 – 0.00115)
x = 0.91
x < 1, ∴ The steam is wet ---- Ans
ii) From steam table, at 10 bar, ts = 179.91oC.
But the given steam temperature is greater than saturation temperature.
∴ the steam is superheated. ---- Ans
9. Steam at 10 bar, 0.95 dry from a boiler is sent to a superheater, where it is heated to 400oC
at constant pressure. Using Mollier chart only, find heat supplied and change in internal
energy.
Given:
Pressure of steam (p) = 10 bar
Dryness fraction of steam (x) = 0.95
Final temperature (T2) = 400oC
Required : Heat supplied & Change in internal energy
Solution:
Heat supplied = h2 - h1
Change in internal energy = u2 – u1
= (h2 – p2v2) – (h1 – p1v1)
From Mollier chart, at 10 bar & 0.95 dry,
h1 = 2675 kJ/kg v1 = 0.195 m3/kg
At 400oC & 10 bar, h2 = 3250 kJ/kg v2 = 0.35 m3/kg
∴ (h2 – h1) = 3250 – 2675 = 575 kJ/kg ---- Ans
(u2 – u1) = [3250 – (10 x 105 x 0.35/1000)] –
[2900 – (10 x 105 x 0.195/1000)]
= 420 kJ/kg ---- Ans
28
10. A 0.25 m3 vessel contains steam at 10 bar, 200oC. Find i) the pressure, if the contents are
cooled to dry saturated state, ii) final state such that thermometer fixed on the vessel shows
130oC.
Given:
Total volume (V) = 0.25 m3
Initial pressure (p1) = 10 bar
Initial temperature (T1) = 200oC
Required : i) p2 if cooled to dry ii) ‘x’ if final temperature is 130oC
Solution:
i) Constant volume cooling, since the vessel contains vapour.
∴ v1 = v2
For solving this part, it is better to use Mollier chart. On Mollier chart locate point (1) at 10 bar &
200oC. Follow the constant volume line upto saturation line, since the steam is dry after cooling,
to locate point (2).
∴ p2 = 9.7 bar ---- Ans
ii) For solving this part we can use steam table.
We know, for wet steam, v1 = vf2 + x vfg2
v1 = 0.2 m /kg at 10 bar & 200oC
3

At 130oC, vf = 0.00107 m3/kg

vg = 0.66814 m3/kg
∴ 0.2 = 0.00107 + x (0.66814 – 0.00107)
x = 0.2982 ---- Ans
11. Find the volume, enthalpy and internal energy of steam when the condition of steam is i)
500 kPa and 0.75 dry ii) 1 MPa and 425oC
Given:
i) Pressure of steam (p) = 500 kPa = 5 bar
Dryness fraction (x) = 0.75
ii) Pressure of steam (p) = 10 bar
Temperature of steam (T) = 425oC
Required : v, h & u
Solution:
i)
We know that, for wet steam, v = vf + x(vg – vf)
h = hf + x(hfg)
u = h – pv
At 5 bar, from steam table,
vf = 0.001093 m3/kg vg = 0.37466 m3/kg hf = 640.1 kJ/kg hfg = 2107.4 kJ/kg
∴ v = 0.001093 + 0.75 (0.37466 – 0.001093)
= 0.281268 m3/kg ---- Ans
29
 h = 640.1 + 0.75 (2107.4)
= 2220.65 kJ/kg ---- Ans
u = 2220.65 – [5 x 105 x 0.281268 / 1000]
= 2080.016 kJ/kg ----- Ans
ii) the steam is superheated, since the saturation temperature is 179.91oC.
vsup = vg [Tsup/ts]
At 10 bar, vg = 0.19444 m3/kg
vsup = 0.19444 x [(425 + 273) / (179.91 + 273)]
= 0.29966 m3/kg ----- Ans
hsup = hg + Cp(sup) (Tsup – ts)
or From steam table, at 10 bar & 425oC,
hsup = 3317.55 kJ/kg ---- Ans
usup = hsup – p vsup
= 3317.55 – (10 x 105 x 0.29966)/1000
= 3017.89 kJ/kg ----- Ans
12. Steam at the rate of 400 kg/min enters the steam turbine of a simple Rankine cycle at 3 MPa
and 500oC. The condenser pressure is 8 kPa. Find (a) Turbine work and Pump work (b)
Power developed (c) Cycle efficiency (d) Specific steaming rate (e) Specific heat rate (f)
Work ratio.
Given:
Mass flow rate of steam (ms) = 400 kg/min = 6.667 kg/s
Boiler pressure (pb) = 3 MPa = 30 bar
Temperature of steam (T1) = 500oC
Condenser pressure (pc) = 8 kPa = 0.08 bar
Required : (a) wT & wP (b) P (c) ηR (d) SSC (e) SHR (f) WR
Solution:
Using steam table

(a)
Turbine work = wT = h1 – h2

h 4

30
 s
1-2 Æ Turbine 2-3 Æ Condenser 3-4 Æ Feed pump 4-1 Æ Boiler
o
h1 = specific enthalpy at turbine inlet (i.e., at 30 bar & 500 C)
From super heated steam table, h1 = 3456.2 kJ/kg
h2 = specific enthalpy at turbine outlet (i.e., at condenser pressure of 0.08 bar)
h2 = hf2 + x hfg2
From steam table at 0.08 bar
hf = 173.9 kg/kg hfg = 2403.2 kJ/kg sf = 0.598 kJ/kg-K
3
sfg = 7.637 kJ/kg-K vf = 0.001008 m /kg
To find ‘x’
In the turbine Isentropic expansion is taking place.
i.e., s1 = s2
s1 = 7.235 kJ/kg-K at 30 bar & 500oC
s2 = sf2 + x sfg2
7.235 = 0.598 + x (7.637)
∴ x = 0.869
∴ h2 = 173.9 + (0.869) (2403.2)
= 2262.28 kJ/kg
∴ wT = 3456.2 – 2262.28 = 1193.92 kJ/kg ---- Ans
Pump work = wP = vf3 (pb – pc) = h4 – h3
vf3 = specific volume of liquid at condenser pressure
= 0.001008 m3/kg
wP = 0.001008 (30 – 0.08) x 105
= 3015.936 J/kg = 3.016 kJ/kg ---- Ans
(b) Power developed (P) = ms (wT – wP)
= 6.667 x (1193.92 – 3.016)
= 7939.75 kW --- Ans
(h1 – h2) – (h4 – h3)
(c) Cycle efficiency (ηR) = ------------------------- Æ Considering pump work
h1 – h4

wT - wP
= ------------
h1 – h4
h4 ≈ h3 = hf3 at condenser pressure = 173.9 kJ/kg

1193.92 – 3.016
∴ ηR = --------------------
3456.2 – 173.9

= 0.363 --- Ans

31
 (d) Specific steaming rate (SSC)

3600 3600
= --------- = ------------------- = 3.0229 kg/kWh --- Ans
wT - wP 1193.92 – 3.016

3600
(e) Specific heat rate (SHR) = -------- = 3600/0.363 = 9917.35 kJ/kWh ---- Ans
ηR

wT - wP 1193.92 – 3.016
(f) Work ratio = ---------- = ---------------------- = 0.9974 --- Ans
wT 1193.92
Using Mollier chart
• Locate point (1) at 30 bar & 500oC.
At point (1) h1 = 3465 kJ/kg
• Draw a straight vertical line to meet at 0.08 bar to locate point (2).
At point (2) h2 = 2260 kJ/kg
• But point (3) can’t be located on the chart. For taking the values at point (3) refer steam table.
∴ h3 = hf3 = 173.9 kJ/kg
(a) Turbine work(wT) = h1 – h2 = 3465 – 2260 = 1205 kJ/kg ---- Ans
Pump work (wP) = h4 – h3 = vf3 (pb – pc)
= 0.001008 x (30 – 0.08) x 105
= 3.016 kJ/kg --- Ans
(b) Power developed (P) = ms (wT – wP)
= 6.667 (1205 – 3.016)
= 8013.63 kW ---- Ans
(h1 – h2) – (h4 – h3)
(c) Cycle efficiency (ηR) = -------------------------- Æ Considering pump work
h1 – h4

wT - wP
= -----------
h1 – h4

1205 – 3.016
= ------------------- = 0.365 --- Ans
3465 – 173.9

d) Specific steaming rate (SSC)

3600 3600
= ---------- = ----------------- = 2.995 kg/kWh --- Ans
wT - wP 1205 – 3.016

3600
(e) Specific heat rate (SHR) = -------- = 3600/0.365 = 9863.01 kJ/kWh ---- Ans
ηR
32
 wT - wP 1205 – 3.016
(f) Work ratio = ----------- = ------------------ = 0.9975 --- Ans
wT 1205

13. Determine the Rankine cycle efficiency working between 6 bar and 0.4 bar when supplied
with saturated steam. By what percentage is the efficiency increased by superheating to
300oC.
Given:
Boiler pressure (pb) = 6 bar
Condenser pressure (pc) = 0.4 bar
Inlet condition = Dry steam
Required : ηR & % increase in ηR if inlet temperature is 300oC
Solution:
h1 – h 2
Cycle efficiency (ηR) = ---------- Æ Neglecting pump work
h 1 – h4

h1 = hg at 6 bar = 2755.5 kJ/kg

h2 = specific enthalpy at turbine outlet (i.e., at condenser pressure of 0.4 bar)

h2 = hf2 + x hfg2

2
h 4

s
1-2 Æ Turbine 2-3 Æ Condenser 3-4 Æ Feed pump 4-1 Æ Boiler
From steam table at 0.4 bar
hf = 317.7 kg/kg hfg = 2319.2 kJ/kg sf = 1.026 kJ/kg-K sfg = 6.645 kJ/kg-K
To find ‘x’
In the turbine Isentropic expansion is taking place.
i.e., s1 = s2
s1 = sg = 6.758 kJ/kg-K at 6 bar
s2 = sf2 + x sfg2
6.758 = 1.026 + x (6.645)
∴ x = 0.862
33
 ∴ h2 = 317.7 + (0.862) (2319.2)
= 2316.85 kJ/kg
h4 = hf3 = 317.7 kJ/kg at 0.4 bar
2755.5 – 2316.85
∴ ηR = ------------------------ = 0.18 ---- Ans
2755.5 – 317.7
(ii) Inlet condition Æ 6 bar & 300oC
Outlet condition Æ 0.4 bar
From superheated steam table, h1 = 3062.3 kJ/kg
s1 = 7.374 kJ/kg-K
h2 = specific enthalpy at turbine outlet (i.e., at condenser pressure of 0.4 bar)
h2 = hf2 + x hfg2
To find ‘x’
In the turbine Isentropic expansion is taking place.
i.e., s1 = s2
s2 = sf2 + x sfg2
7.374 = 1.026 + x (6.645)
∴ x = 0.955
∴ h2 = 317.7 + (0.955) (2319.2)
= 2532.5 kJ/kg
h4 = hf3 = 317.7 kJ/kg at 0.4 bar
3062.3 – 2532.5
∴ ηR = ---------------------- = 0.193
3062.3 – 317.7

∴ % increase = (0.193 – 0.18)x 100/0.18

= 7.22 % --- Ans
14. In a single heater regenerator cycle, the steam enters the turbine at 30 bar, 400oC and the
exhaust pressure is 0.1 bar. The feed water heater is a direct contact type which operates at
5 bar. Find the efficiency and the steam rate of the cycle with and without regeneration.
Pump work may be neglected.

Given:
Pressure at turbine inlet (p1) = 30 bar
Temperature at turbine inlet (T1) = 400oC
Exhaust steam pressure (p3) = 0.1 bar
Feed heater pressure (p2) = 5 bar
Required: η & Steam rate with and without regeneration
Solution:
With regeneration

34
 Workdone
Efficiency (η) = ----------------------
Heat supplied
Workdone = m (h1 – h2) + (m – m1) (h2 – h3)
Take m = 1 kg

m m
7
6 m1 2
5
T m – m1 m – m1

4 3

h 2

s
From Mollier Chart
• Locate the point (1) at 30 bar & 400oC
• Draw the vertical straight line to get point (2) at 5 bar and point (3) at 0.1 bar
∴ h1 = 3240 kJ/kg h2 = 2800 kJ/kg h3 = 2200 kJ/kg
To find m1
Heat lost by m1 kg of bled steam = Heat gain by (m – m1) kg of feed water from pump (1)
i.e., m1 (h2 – h6) = (1 – m1) (h6 – h5)
h5 ≈ h4 = hf at 0.1 bar = 191.8 kJ/kg from steam table
Note: Always use steam table for finding out hf values.
h6 = hf at 5 bar = 640.1 kJ/kg from steam table
∴ m1 (2800 – 640.1) = (1 – m1) (640.1 – 191.8)
m1 = 0.1788 kg
35
 Workdone = 1 x (3240 – 2800) + (1 – 0.1788) (2800 – 2200)
= 932.72 kJ/kg
Heat supplied = m (h1 – h7)
h7 ≈ h6 = hf at 5 bar = 640.1 kJ/kg
∴ Heat supplied = 1 x (3200 – 640.1) = 2599.9 kJ/kg
Efficiency = 932.72 / 2599.9 = 0.359 --- Ans
Steam rate = 3600 / (wT – wP)
= 3600 / 932.72 = 3.86 kg/kWh ---- Ans
Without regeneration

1
T

4
3 2

s
h1 = 3240 kJ/kg
h2 = 2200 kJ/kg
Workdone = m (h1 – h2) = 1 x (3240 – 2200) = 1040 kJ/kg
Heat supplied = m (h1 – h4)
h4 ≈ h3 = hf at 0.1 bar = 191.8 kJ/kg
∴ Heat supplied = 1 x (3240 – 191.8) = 3008.2 kJ/kg
Efficiency = 1040 / 3008.2 = 0.341 ---- Ans
Steam rate = 3600 / wnet = 3600 / 1040 = 3.46 kg/kWh --- Ans
15. Steam is supplied to a turbine at a pressure of 30 bar and 400oC and is expanded
isentropically to a pressure of 0.1 bar. The steam is trapped from turbine at two different
location one at 10 bar and another at 3 bar, to supply feed water heaters. Find the thermal
efficiency and steam rate of the cycle. Neglect pump work.
Given:
Cycle with two feed water heaters
Pressure at turbine inlet (p1) = 30 bar
Temperature at turbine inlet (T1) = 400oC
Exhaust steam pressure (p4) = 0.1 bar
I-Feed heater pressure (p2) = 10 bar
II-Feed heater pressure (p3) = 3 bar
Required: η & Steam rate

36
Solution:

m 30 bar m
10
9 m1 10 bar 2
8
T m2 3 bar m – m1
7 3
6
m – m1 – m2 0.1 bar
5 4

Workdone
Efficiency = ------------------
Heat supplied
Workdone = m (h1 – h2) + (m – m1) (h2 – h3) + (m – m1 – m2) (h3 – h4)
Heat supplied = Heat supplied during process 10 – 1
= m (h1 – h10)
Take m = 1 kg
From Mollier chart,
h1 = 3240 kJ/kg h2 = 2950 kJ/kg h3 = 2700 kJ/kg h4 = 2200 kJ/kg
To find m1 & m2
Heat lost by m1 kg of bled steam = Heat gain by (m – m1) kg of feed water from pump (2)
i.e., m1 (h2 – h9) = (1 – m1) (h9 – h8)
Heat lost by m2 kg of bled steam = Heat gained by (m – m1 – m2) kg of feed water from pump (1)
m2 (h3 – h7) = (1 – m1 – m2) (h7 – h6)
h9 = hf at 10 bar = 762.6 kJ/kg from steam table
h8 ≈ h7 = hf at 3 bar = 561.5 kJ/kg
h6 ≈ h5 = hf at 0.1 bar = 191.8 kJ/kg
Note: Always use steam table for finding out hf values.
m1 (2950 – 762.6) = (1 – m1) (762.6 – 561.5)
m1 = 0.0842 kg
and m2 (2700 – 561.5) = (1 – 0.0842 – m2) (561.5 – 191.8)
m2 = 0.135 kg
Workdone = 1 x (3240 – 2950) + (1 – 0.0842) (2950 – 2700)
+ (1 – 0.0842 – 0.135) (2700 – 2200)
= 909.35 kJ/kg

37
 Heat supplied = m (h1 – h10)
H10 ≈ h9 = hf at 10 bar = 762.6 kJ/kg
∴ Heat supplied = 1 x (3240 – 762.6) = 2477.4 kJ/kg
Efficiency = 909.35 / 2477.4 = 0.367 --- Ans
Steam rate = 3600 / (wT – wP) = 3600 / 2477.4 = 1.4531 kg/kWh ---- Ans
16. In a steam turbine steam at 20 bar, 360oC is expanded to 0.08 bar. It is then enters a
condenser, where it is condensed to saturated liquid. The pump feeds back the water into
the boiler. Assume ideal processes, find per kg of steam, the net work and the cycle
efficiency.
Given:
Turbine inlet pressure (p1) = 20 bar
Turbine exhaust pressure (p2) = 0.08 bar
Turbine inlet temperature (T2) = 360oC
Required: wT – wP & η
Solution:
For solving this problem we can use either steam table or Mollier chart.
Using steam table
Net Work
Efficiency (η) = -------------------
Heat supplied
Net work = wT - wP
Turbine work = wT = h1 – h2
Pump work = wP = (p1 – p2) vf2

T
4

3 2

s
h1 = 3160.74 kJ/kg from superheated steam table at 360oC & 20 bar
h2 = hf2 + x2 hfg2
hf2 = 173.9 kJ/kg from steam table at 0.08 bar
hfg2 = 2403.2 kJ/kg
To find x2
s1 = s2

38
 s1 = 6.994 kJ/kg from steam table at 360oC & 20 bar
s2 = sf2 + x2 sfg2
sf2 = 0.593 kJ/kg-K from steam table at 0.08 bar
sfg2 = 7.637 kJ/kg-K from table
∴ 6.994 = 0.593 + x2 (7.637)
x2 = 0.838
h2 = 173.9 + 0.838 x 2403.2 = 2187.8 kJ/kg
∴ wT = 3160.74 – 2187.8 = 972.94 kJ/kg
3
vf2 = 0.001008 m /kg from table at 0.08 bar
∴ wP = 0.001008 x (20 – 0.08) x 105
= 2008 J/kg = 2.008 kJ/kg
∴ wnet = 972.94 – 2.008 = 970.932 kJ/kg --- Ans
Heat supplied = h1 – h4
To find h4
wP = h4 – h3 = 2.008
∴ h4 = h3 + 2.008
h3 = hf at 0.08 bar = 173.9 kJ/kg
∴ h4 = 2.008 + 173.9 = 175.908 kJ/kg
∴ HS = 3160.74 – 175.908 = 2984.832 kJ/kg
η = 972.94 / 2984.832 = 0.326 --- Ans

17. In a Rankine cycle, the steam at inlet to turbine is saturated at a pressure of 35 bar and the
exhaust pressure is 0.2 bar. Determine, the pump work, the turbine work, the Rankine
efficiency, the condenser heat flow and dryness at the end of expansion. Assume flow rate of
9.5 kg/s.
Given:
Turbine inlet pressure (p1) = 35 bar
Turbine exhaust pressure (p2) = 0.2 bar
Dry steam at inlet
Mass flow rate of steam (m) = 9.5 kg/s
Required: wT, wP, η, Q2-3 & x2
Solution:
Using steam table

1
T

39
 4

3 2

Net Work
Efficiency (η) = --------------------
Heat supplied
Net work = wT - wP
Turbine work = wT = h1 – h2
Pump work = wP = (p1 – p2) vf2
Heat flow through the condenser = Q2-3 = m (h2 – h3)
h1 = hg at 35 bar from steam table = 2802 kJ/kg
h2 = hf2 + x2 hfg2
hf2 = 251.5 kJ/kg from steam table at 0.2 bar
hfg2 = 2358.4 kJ/kg
To find x2
s1 = s2
s1 = sg at 35 bar from steam table = 6.123 kJ/kg-K
s2 = sf2 + x2 sfg2
sf2 = 0.832 kJ/kg-K from steam table at 0.2 bar
sfg2 = 7.077 kJ/kg-K from table
∴ 6.123 = 0.832 + x2 (7.077)
x2 = 0.7476 --- Ans
h2 = 251.5 + 0.7476 x 2358.4 = 2014.64 kJ/kg
∴ wT = 2802 – 2014.64 = 787.36 kJ/kg
Turbine work = 9.5 x 787.36 = 7479.92 kW ---- Ans
vf2 = 0.001017 m3/kg from table at 0.2 bar
∴ wP = 0.001017 x (35 – 0.2) x 105
= 3539.16 J/kg = 3.53916 kJ/kg
Pump work = 9.5 x 3.53916 = 33.622 kW ---- Ans
∴ wnet = 787.36 – 3.53916 = 783.82084 kJ/kg
Heat supplied = h1 – h4
To find h4
wP = h4 – h3 = 3.53916
∴ h4 = h3 + 3.53916
h3 = hf at 0.2 bar = 251.5 kJ/kg
∴ h4 = 3.53916 + 251.5 = 255.04 kJ/kg
40
 ∴ HS = 2802 – 255.04 = 2546.96 kJ/kg
η = 783.82084 / 2546.96 = 0.308 --- Ans
Condenser heat flow = 9.5 x (2014.64 – 251.5) = 16749.83 kW ---- Ans
18. In a steam power plant operating on an reheat Rankine cycle, the steam enters the high
pressure turbine at 3 MPa and 400oC. After expansion to 0.6 MPa, the steam is reheated to
400oC and then expanded in the low pressure turbine to the condenser pressure of 10 kPa.
Determine the thermal efficiency of the cycle and the quality of the steam at the outlet of the
low pressure turbine.
Given:
Reheat cycle
Pressure at inlet of HP turbine (p1) = 3 MPa = 30 bar
Temperature at inlet of HP turbine (T1) = 400oC = 673 K
Pressure at the inlet of reheater (p2) = 0.6 MPa = 6 bar
Temperature at the exit of reheater (T3) = 400oC = 673 K
Pressure at the exit of the LP turbine (p4) = 10 kPa = 0.1 bar
Required: Thermal efficiency and Dryness fraction at exit of LP turbine
Solution:
Using Mollier diagram:
• Locate the point (1) at 400oC and 30 bar in the Mollier chart
• 1-2 Æ Isentropic expansion in HP turbine Æ Draw the vertical line to meet 6 bar line and
locate point (2)
• 2-3 Æ Constant pressure heat addition Æ Follow the constant pressure line (6 bar) and reach
400oC and mark point (3)
• 3-4 Æ Isentropic expansion in LP turbine Æ Draw the vertical line to meet 0.1 bar line and
mark point (4)
1 3

2
h
6 4

5
s
h1 = 3230 kJ/kg h2 = 2830 kJ/kg h3 = 3275 kJ/kg h4 = 2445 kJ/kg
x4 = 0.94 --- Ans
(h1 − h2 ) + (h3 − h4 ) − (h6 − h5 )
η=
(h1 − h6 ) + (h3 − h2 )
(h6 – h5) = vf5 (p6 – p5)
From steam table,
41
 vf5 = 0.00101 m3/kg
hf5 = 191.8 kJ/kg = h5
(h6 – h5) = 0.00101 x (30 – 0.1) x 105/103 = 3.0199 kJ/kg
h6 = 3.0199 + 191.8 = 194.82 kJ/kg
(3230 − 2830) + (3275 − 2445) − (3.0199)
η= = 0.3525 − − − − Ans
(3230 − 194.82) + (3275 − 2830)
Using Steam table:
From superheated steam table at 30 bar and 400oC,
h1 = 3232.5 kJ/kg s1 = 6.925 kJ/kg-K
To check type of steam at (2)
s1 = s2 = sf2 + x2 sfg2
From steam table at 6 bar,
ts2 = 158.8oC sf2 = 1.931 kJ/kg-K sfg2 = 4.827 kJ/kg-K hg2 = 2755.5 kJ/kg
Therefore, 6.925 = 1.93 + x2 (4.827)
x2 = 1.0348
Therefore the steam is superheated at HP turbine outlet.
h2 = hg2 + Cpsup (T2 – ts2)
ts2 = 158.8oC
There are two unknowns, T2 and Cpsup
At s2 = 6.925 kJ/kg-K and 6 bar, from superheated steam table, the temperature, T2 is obtained.
s = sg2 = 6.758 kJ/kg-K, at 158.8oC,
= 6.966 kJ/kg-K, at 200oC
= 6.925 kJ/kg-K, at T2
(158.8 − 200)
Therefore, T2 = (6.925 − 6.966) + 200
(6.758 − 6.966)
= 191.9oC = 464.9 K
To find Cpsup
s1 = s2 = sg2 + Cpsup ln (T2/ts2)
6.925 = 6.758 + Cpsup ln (464.9/431.8)
Cpsup = 2.261 kJ/kg-K
Therefore, h2 = 2755.5 + 2.261x (191.9 – 158.8) = 2830.3 kJ/kg
From superheated steam table, at 6 bar and 400oC,
h3 = 3270.6 kJ/kg s3 = 7.709 kJ/kg-K
From saturated table, at 0.1 bar
hf4 = 191.8 kJ/kg
hfg4 = 2392.9 kJ/kg
sf4 = 0.649 kJ/kg-K
42
 sfg4 = 7.502 kJ/kg-K
To find x4
s4 = sf4 + x4 sfg4
Process 1-4 is isentropic, s4 = s3
7.709 = 0.649 + x4 (7.502)
x4 = 0.941 ----- Ans
Therefore, h4 = hf4 + x4 hfg4
= 191.8 + 0.941 (2392.9) = 2443.52 kJ/kg
(h6 – h5) = vf5 (p6 – p5)
From steam table,
vf5 = 0.00101 m3/kg
hf5 = 191.8 kJ/kg = h5
(h6 – h5) = 0.00101 x (30 – 0.1) x 105/103 = 3.0199 kJ/kg
h6 = 3.0199 + 191.8 = 194.82 kJ/kg
(3232.5 − 2830.3) + (3270.6 − 2443.52) − (3.0199)
η= = 0.3526 − − − − Ans
(3232.5 − 194.82) + (3270.6 − 2830.3)

43
 UNIT-IV
IDEAL AND REAL GASES, THERMO DYNAMIC RELATIONS
Ideal gas
A perfect gas or an ideal gas is defined as a gas having no intermolecular forces. A gas which
follows the gas laws at all ranges of pressures and temperatures can be considered as an ideal gas, but no
such gas exists in nature.
Boyle’s law
Boyle’s law states that if the temperature of a gas in a closed system is maintained constant
during a process, the volume of the gas will vary increasingly with absolute pressure during the change of
state.
V α 1/p or pV = Constant when T is constant
Charle’s law
If the pressure of the gas in a closed system is maintained constant during a process, the volume
of gas will vary directly with the absolute temperature.
VαT or V/T = C
Avogadro’s law
A mole of a substance has a mass numerically equal to the molecular weight of the substance.
Example: 1 kg mole of O2 has a mass of 32 kg
1 kg mole of N2 has a mass of 28 kg
Avogadro’s law states that the volume of a 1 kg mole of all gases at STP (0oC and 1.01325 bar) is
the same and is equal to 22.4 m3.
Example: 1 kg mole of O2 has a mass of 32 kg and volume of 22.4 m3
1 kg mole of N2 has a mass of 28 kg and volume of 22.4 m3
At p = 1.01325 bar and T = 273.15 K, for 1 kg mole of gas V = 22.4 m3
vm = 22.4 m3/kg mole
p vm = Ru T
5
1.01325 x 10 x 22.4 = Ru x 273.15
Ru = 8314.3 J/kg mol K
Ru Æ Universal gas constant; vm Æ Molar volume
Joule’s law
This law states that there is no change of temperature when a gas expands without doing external
work and without receiving or rejecting heat.
u = f(T) only for ideal gas
Equation of state
The functional relationship among the properties, pressure p, molar or specific volume v and
temperature T, is known as an equation of state, which may be expressed in the form,
f (p, v, T) = 0
If two properties of a gas are known, the third can be evaluated from the equation of state.
pv=RT
Real gas
 Most of the real gases obey Boyle’s and Charle’s law at low pressures and temperatures. But the
actual behaviour of real gases at elevated pressures and at low temperatures deviates considerably.
The ideal gas equation p v = R T can be derived analytically using the kinetic theory of gases by
making the following assumptions:
i. A finite volume of gas contains large number of molecules.
ii. The collision of molecules with one another and with the walls of the container is perfectly
elastic.
iii. The molecules are separated by large distances compared to their own dimensions.
iv. The molecules do not exert forces on one another except when they collide.
As long as the above assumptions are valid the behaviour of a real gas approaches closely that of
an ideal gas.
Van der Waals equation
The ideal gas equation p v = R T is being used with two important assumptions that there is little
or no attraction between the two molecules of the gas and that the volume occupied by the molecules
themselves is negligibly small compared to the volume of the gas. When the pressure is very small or
temperature is very large, the intermolecular attraction and the volume of the molecules compared to the
total volume of the gas are not of much importance, and the real gas obeys very closely the ideal gas
equation. But as pressure increases, the intermolecular forces of attraction and repulsion increase, and
also the volume of the molecules becomes appreciable compared to the total volume. Then the real gases
deviate considerably from the ideal gas equation. Van der Waals by applying the laws of mechanics to the
individual molecules, introduced two corrections terms in the ideal gas equation.
The Van der Waals equation for a real gas may be written as
⎛ a ⎞
⎜ p + 2 ⎟ (v − b ) = R T
⎝ v ⎠
aÆ a constant to account for the existence of mutual attraction between the molecules.
a/v2 Æ the force of attraction
bÆ the coefficient introduced to account for the volumes of molecules, and is known as
co-volume
27 R 2 Tc2 R Tc
a= b=
64 p c 8 pc
Tc Æ Critical temperature
pc Æ Critical pressure
Unique feature
• Van der Waals equation qualitatively accounts for the heating effect observed at ordinary
temperatures.
Limitations
• The values of a and b (which are assumed to be constant) are found to vary with temperature.
Thus the results obtained from the equation are incorrect when the variation of a and b is large
with respect to temperature.
• The equation is not accurate enough in the critical region.
Virial expansions

2
 The virial expansion, also called the virial equation of state, is the most interesting and versatile
of the equations of state for gases. The virial expansion is a power series in powers of the variable, B/V,
and has the form,
pv
= Ao + A1 p + A2 p 2 + A3 p 3 + − − − − − −
RT
pv B B B
= Bo + 1 + 22 + 33 + − − − − − −
RT v v v
The coefficient, Ao, A1, A2, ……. and Bo, B1, B2, …….. are called the virial coefficients and are
functions of temperature. Ao, Bo are first virial coefficient and A1, B1 are second virial coefficients and so
on.
Law of corresponding states
For each substance there is a compressibility factor. It would be very convenient if one chart
could be used for all substances. The general shapes of the vapour dome and of the constant temperature
lines on the p-v plane are similar for all substances, although the scales may be different. This similarity
can be exploited by using dimensionless properties called ‘reduced properties’.
p
Reduced pressure, p r =
pc
T
Reduced temperature, Tr =
Tc
v
Reduced volume, v r =
vc
Generally for all substances, vr = f(pr, Tr) and Z = f(pr, Tr)
The relation among the reduced properties, pr, Tr and vr is known as the law of corresponding
states.
Compressibility
In thermodynamics and fluid mechanics, compressibility is a measure of the relative volume
change of a fluid or solid as a response to a pressure (or mean stress) change.
1 ∂v
K= ----- (1)
v ∂p
where v is specific volume and p is pressure. The above statement is incomplete, because for any object
or system the magnitude of the compressibility depends strongly on whether the process is adiabatic or
isothermal. Accordingly we define the isothermal compressibility as:
1 ⎛ ∂v ⎞
K T = − ⎜⎜ ⎟⎟ ----- (2)
v ⎝ ∂p ⎠ T
where the subscript T indicates that the partial differential is to be taken at constant temperature. The
adiabatic compressibility as:
1 ⎛ ∂v ⎞
K S = − ⎜⎜ ⎟⎟ ----- (3)
v ⎝ ∂p ⎠ S
where S is entropy. For a solid, the distinction between the two is usually negligible.
Coefficient of volume expansion is defined as,
3
 1 ⎛ ∂v ⎞
β= ⎜ ⎟ ----- (4)
v ⎝ ∂T ⎠ p
The inverse of the compressibility is called the bulk modulus.

⎛ ∂p ⎞ ⎛ ∂v ⎞ ⎛ ∂T ⎞
We know that, ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ = −1 ----- (5)
⎝ ∂v ⎠ T ⎝ ∂T ⎠ p ⎝ ∂p ⎠V

⎛ ∂v ⎞
βv=⎜ ⎟ from (4)
⎝ ∂T ⎠ p

⎛ ∂v ⎞
− K T v = ⎜⎜ ⎟⎟ from (2)
⎝ ∂p ⎠ T
⎛ −1 ⎞ ⎛ ∂T ⎞
Substituting in (5), ⎜⎜ ⎟⎟ (β v )⎜⎜ ⎟⎟ = −1
⎝ K T v ⎠T ⎝ ∂p ⎠V
β ⎛ ∂p ⎞
=⎜ ⎟ ----- (6)
K T ⎝ ∂T ⎠V
Compressibility Chart

The compressibility factor (Z) is used to alter the ideal gas equation to account for the real gas
behaviour. The compressibility factor is usually obtained from the compressibility chart. Mathematically,
it is defined as,
p v v actual
Z= =
R T videal
p Æ the pressure,
v Æ the specific volume of the gas,
T Æ the temperature, and
4
R Æ the gas constant.
Z = 1 for ideal gas
p c vc
Critical compressibility factor, Z c =
R Tc
Z can, in general, be either greater or less than unity for a real gas. The deviation from ideal gas
behavior tends to become particularly significant (or, equivalently, the compressibility factor strays far
from unity) near the critical point, or in the case of high pressure or low temperature. In these cases, a
generalized Compressibility chart or an alternative equation of state better suited to the problem must be
utilized to produce accurate results.
Rules on partial derivatives
Theorem 1 (Exact differential)
If a relation exists among the variables x, y and z, then z may be expressed as function of x and y.
z = f (x, y)

⎛ ∂z ⎞ ⎛ ∂z ⎞
dz = ⎜ ⎟ dx + ⎜⎜ ⎟⎟ dy
⎝ ∂x ⎠ y ⎝ ∂y ⎠ x

⎛ ∂z ⎞ ⎛ ∂z ⎞
Let, M = ⎜ ⎟ and N = ⎜⎜ ⎟⎟
⎝ ∂x ⎠ y ⎝ ∂y ⎠ x
dz = M dx + N dy
Differentiating M partially with respect to y, and N with respect to x,
⎛ ∂M ⎞ ∂2z
⎜⎜ ⎟⎟ =
⎝ ∂y ⎠ x ∂x ∂y

⎛ ∂N ⎞ ∂2z
⎜ ⎟ =
⎝ ∂x ⎠ y ∂y ∂x

⎛ ∂M ⎞ ⎛ ∂N ⎞
⎜⎜ ⎟⎟ = ⎜ ⎟
⎝ ∂y ⎠ x ⎝ ∂x ⎠ y
This is the condition of exact or perfect differential.
Theorem 2
If a quantity f is a function of x, y and z and a relation exists among x, y and z, then f is a function
of any two of x, y and z.
If f = f(x, y, z)
⎛ ∂x ⎞ ⎛ ∂y ⎞ ⎛ ∂z ⎞
⎜⎜ ⎟⎟ ⎜ ⎟ ⎜ ⎟ = 1
⎝ ∂y ⎠ f ⎝ ∂z ⎠ f ⎝ ∂x ⎠ f
Theorem 3
Among the variables x, y and z, any one variable may be considered as a function of the other
two.
If x = x (y, z) y = y (x, z) z = z(x, y)

5
 ⎛ ∂x ⎞ ⎛ ∂z ⎞ ⎛ ∂y ⎞
⎜⎜ ⎟⎟ ⎜ ⎟ ⎜ ⎟ = −1
⎝ ∂y ⎠ z ⎝ ∂x ⎠ y ⎝ ∂z ⎠ z
Among the thermodynamics variables, p, v and T, the following relation holds good.

⎛ ∂p ⎞ ⎛ ∂v ⎞ ⎛ ∂T ⎞
⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ = −1
⎝ ∂v ⎠ T ⎝ ∂T ⎠ p ⎝ ∂p ⎠V

Maxwell’s equations
A pure substance existing in a single phase has only two independent variables. Of the eight
quantities, p, v, T, s, u, h, F (Helmholts function) and G (Gibbs function), any one may be expressed as a
function of any two others.
(i) Consider a non flow process.
q = w + Δu
dq = dw + du
du = dq – dw
= T ds – p dv ----- (1)
Therefore, u = f (s, v)

⎛ ∂u ⎞ ⎛ ∂u ⎞
du = ⎜ ⎟ ds + ⎜ ⎟ dv ----- (2)
⎝ ∂s ⎠V ⎝ ∂v ⎠ S
Comparing (1) and (2)
⎛ ∂u ⎞ ⎛ ∂u ⎞
T =⎜ ⎟ and − p=⎜ ⎟
⎝ ∂s ⎠V ⎝ ∂v ⎠ S

⎛ ∂T ⎞ ∂ 2u ⎛ ∂p ⎞ ∂ 2u
⎜ ⎟ = and ⎜− ⎟ =
⎝ ∂v ⎠ S ∂v ∂s ⎝ ∂S ⎠ S ∂s ∂v

⎛ ∂ 2u ⎞ ⎛ ∂ 2u ⎞
But, ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ ∂v ∂s ⎠ ⎝ ∂s ∂v ⎠
⎛ ∂T ⎞ ⎛ ∂p ⎞
Therefore, ⎜ ⎟ = −⎜ ⎟ Æ I – Maxwell equation
⎝ ∂v ⎠ S ⎝ ∂s ⎠V
The rate of increase in temperature with respect to volume during isentropic process is equal to
the rate of decrease in pressure with respect to entropy during constant volume process.
(ii) Consider a flow process.
q = w + Δh
dq = dw + dh
dh = dq – dw
= T ds + v dp ----- (1)
Therefore, h = f (s, p)

6
 ⎛ ∂h ⎞ ⎛ ∂h ⎞
dh = ⎜ ⎟ ds + ⎜⎜ ⎟⎟ dp ----- (2)
⎝ ∂s ⎠ p ⎝ ∂p ⎠ S
Comparing (1) and (2)

⎛ ∂h ⎞ ⎛ ∂h ⎞
T =⎜ ⎟ and v = ⎜⎜ ⎟⎟
⎝ ∂s ⎠ p ⎝ ∂p ⎠ S
⎛ ∂T ⎞ ∂ 2h ⎛ ∂v ⎞ ∂ 2h
⎜⎜ ⎟⎟ = and ⎜ ⎟ =
⎝ ∂p ⎠ S ∂p ∂s ⎝ ∂s ⎠ p ∂s ∂p

⎛ ∂ 2h ⎞ ⎛ ∂ 2h ⎞
But, ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ ∂p ∂s ⎠ ⎝ ∂s ∂p ⎠
⎛ ∂T ⎞ ⎛ ∂v ⎞
Therefore, ⎜⎜ ⎟⎟ = ⎜ ⎟ Æ II – Maxwell equation
⎝ ∂p ⎠ S ⎝ ∂s ⎠ p
The rate of increase in temperature with respect to pressure during isentropic process is equal to
the rate of increase in volume with respect to entropy during constant pressure process.
(iii) F = u – Ts
dF = du – (T ds + s dT)
= du – T ds – s dT
But, du = T ds – p dv
Therefore, dF = T ds – p dv – T ds – s dT
= - p dv – s dT ----- (1)
Therefore, F = f (v, T)
⎛ ∂F ⎞ ⎛ ∂F ⎞
dF = ⎜ ⎟ dT + ⎜ ⎟ dv ----- (2)
⎝ ∂T ⎠V ⎝ ∂v ⎠ T
Comparing (1) and (2)
⎛ ∂F ⎞ ⎛ ∂F ⎞
−s=⎜ ⎟ and − p=⎜ ⎟
⎝ ∂T ⎠V ⎝ ∂v ⎠ T

⎛ ∂s ⎞ ∂2F ⎛ ∂p ⎞ ∂2F
−⎜ ⎟ = and −⎜ ⎟ =
⎝ ∂v ⎠ T ∂v ∂T ⎝ ∂T ⎠V ∂T ∂v

⎛ ∂2F ⎞ ⎛ ∂2F ⎞
But, ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ ∂v ∂T ⎠ ⎝ ∂T ∂v ⎠
⎛ ∂s ⎞ ⎛ ∂p ⎞
Therefore, ⎜ ⎟ =⎜ ⎟ Æ III – Maxwell equation
⎝ ∂v ⎠ T ⎝ ∂T ⎠V
The rate of increase in entropy with respect to volume during isothermal process is equal to the
rate of increase in pressure with respect to temperature during constant volume process.
(iv) g = h – Ts
dg = dh – (T ds + s dT)
7
 = dh – T ds – s dT
But, dh = T ds + v dp
Therefore, dg = T ds + v dp – T ds – s dT
= v dp – s dT ----- (1)
Therefore, g = f (p, T)

⎛ ∂g ⎞ ⎛ ∂g ⎞
dg = ⎜ ⎟ dT + ⎜⎜ ⎟⎟ dp ----- (2)
⎝ ∂T ⎠ p ⎝ ∂p ⎠ T

Comparing (1) and (2)

⎛ ∂g ⎞ ⎛ ∂g ⎞
−s =⎜ ⎟ and v = ⎜⎜ ⎟⎟
⎝ ∂T ⎠ p ⎝ ∂p ⎠ T
⎛ ∂s ⎞ ∂2g ⎛ ∂v ⎞ ∂2g
− ⎜⎜ ⎟⎟ = and ⎜ ⎟ =
⎝ ∂p ⎠ T ∂p ∂T ⎝ ∂T ⎠ p ∂T ∂p

⎛ ∂2g ⎞ ⎛ ∂2g ⎞
But, ⎜⎜ ⎟⎟ = ⎜⎜ ⎟⎟
⎝ ∂p ∂T ⎠ ⎝ ∂T ∂p ⎠
⎛ ∂s ⎞ ⎛ ∂v ⎞
Therefore, ⎜⎜ ⎟⎟ = −⎜ ⎟ Æ IV – Maxwell equation
⎝ ∂p ⎠ T ⎝ ∂T ⎠ p
The rate of increase in entropy with respect to pressure during isothermal process is equal to the
rate of decrease in volume with respect to temperature during constant pressure process.
TdS equations
(i) Let s = f (T, v) Æ T and v independent variables
⎛ ∂s ⎞ ⎛ ∂s ⎞
ds = ⎜ ⎟ dT + ⎜ ⎟ dv ------ (1)
⎝ ∂T ⎠V ⎝ ∂v ⎠ T
Let u = f (T, v)
⎛ ∂u ⎞ ⎛ ∂u ⎞
du = ⎜ ⎟ dT + ⎜ ⎟ dv
⎝ ∂T ⎠V ⎝ ∂v ⎠ T
⎛ ∂u ⎞
= CV dT + ⎜ ⎟ dv ------ (2)
⎝ ∂v ⎠ T
Also, du = T ds – p dv ------ (3)
Equating (2) and (3),
⎛ ∂u ⎞
T ds − p dv = CV dT + ⎜ ⎟ dv
⎝ ∂v ⎠ T
CV 1 ⎡⎛ ∂u ⎞ ⎤
ds = dT + ⎢⎜ ⎟ + p ⎥ dv ------ (4)
T T ⎣⎝ ∂v ⎠ T ⎦
8
 ⎛ ∂s ⎞ C
Comparing (1) and (4), ⎜ ⎟ = V ------ (5)
⎝ ∂T ⎠V T

⎛ ∂s ⎞ ⎛ ∂p ⎞
From Maxwell III-equation, ⎜ ⎟ =⎜ ⎟ ------ (6)
⎝ ∂v ⎠ T ⎝ ∂T ⎠V
Substituting (5) and (6) in (1),
CV ⎛ ∂p ⎞
ds = dT + ⎜ ⎟ dv
T ⎝ ∂T ⎠V
⎛ ∂p ⎞
T ds = CV dT + T ⎜ ⎟ dv Æ First T ds equation
⎝ ∂T ⎠V
β
⎛ ∂p ⎞
We know that, =⎜ ⎟
K T ⎝ ∂T ⎠V

⎛ β ⎞
T ds = CV dT + T ⎜⎜ ⎟⎟ dv Æ First T ds equation
⎝ KT ⎠
(ii) Let s = f (T, p) Æ T and p are independent variables

⎛ ∂s ⎞ ⎛ ∂s ⎞
ds = ⎜ ⎟ dT + ⎜⎜ ⎟⎟ dp ------ (1)
⎝ ∂T ⎠ p ⎝ ∂p ⎠ T
Let h = f (T, p)

⎛ ∂h ⎞ ⎛ ∂h ⎞
dh = ⎜ ⎟ dT + ⎜⎜ ⎟⎟ dp
⎝ ∂T ⎠ p ⎝ ∂p ⎠ T
⎛ ∂h ⎞
= C p dT + ⎜⎜ ⎟⎟ dp ------ (2)
⎝ ∂p ⎠ T
Also, dh = T ds + v dp ------ (3)
Equating (2) and (3),
⎛ ∂h ⎞
T ds + v dp = C p dT + ⎜⎜ ⎟⎟ dp
⎝ ∂p ⎠ T
Cp 1 ⎡⎛ ∂h ⎞ ⎤
ds = dT + ⎢⎜⎜ ⎟⎟ − v ⎥ dp ------ (4)
T T ⎣⎢⎝ ∂p ⎠ T ⎥⎦

⎛ ∂s ⎞ Cp
Comparing (1) and (4), ⎜ ⎟ = ------ (5)
⎝ ∂T ⎠ p T

⎛ ∂s ⎞ ⎛ ∂v ⎞
From Maxwell IV-equation, ⎜⎜ ⎟⎟ = −⎜ ⎟ ------ (6)
⎝ ∂p ⎠ T ⎝ ∂T ⎠ p
Substituting (5) and (6) in (1),

9
 Cp ⎛ ∂v ⎞
ds = dT − ⎜ ⎟ dp
T ⎝ ∂T ⎠ p

⎛ ∂v ⎞
T ds = C p dT − T ⎜ ⎟ dp Æ Second-T ds equation
⎝ ∂T ⎠ p
1 ⎛ ∂v ⎞
We know that, β= ⎜ ⎟
v ⎝ ∂T ⎠ p

⎛ ∂v ⎞
βv=⎜ ⎟
⎝ ∂T ⎠ p
T ds = C p dT − T β v dp Æ Second T ds equation

(iii) Let u = f (p, v) Æ p and v are independent variables

⎛ ∂u ⎞ ⎛ ∂u ⎞
du = ⎜⎜ ⎟⎟ dp + ⎜ ⎟ dv ------ (1)
⎝ ∂p ⎠V ⎝ ∂v ⎠ p

⎛ ∂u ⎞ ⎛ ∂u ⎞ ⎛ ∂T ⎞
⎜⎜ ⎟⎟ = ⎜ ⎟ ⎜⎜ ⎟⎟
⎝ ∂p ⎠ v ⎝ ∂T ⎠V ⎝ ∂p ⎠V
KT
= Cv ------ (2)
β
Also, u = h – pv
⎛ ∂u ⎞ ⎛ ∂h ⎞
⎜ ⎟ =⎜ ⎟ − p
⎝ ∂v ⎠ p ⎝ ∂v ⎠ p

⎛ ∂h ⎞ ⎛ ∂T ⎞
=⎜ ⎟ ⎜ ⎟ −p
⎝ ∂T ⎠ p ⎝ ∂v ⎠ p
1
= Cp −p ------ (3)
vβ
Substituting (2) and (3) in (1),

KT ⎡Cp ⎤
du = C v dp + ⎢ − p ⎥ dv ------ (4)
β ⎣v β ⎦
We know that T ds = du + p dv

KT ⎡Cp ⎤
T ds = C v dp + ⎢ − p ⎥ dv + p dv
β ⎣v β ⎦
KT Cp
= Cv dp + dv ------ (5)
β vβ

10
 KT 1
T ds = CV dp + C p dv Æ Third T ds equation
β βv
Change in internal energy
du = T ds – p dv

⎛ ∂p ⎞
T ds = CV dT + T ⎜ ⎟ dv Æ First T ds equation
⎝ ∂T ⎠V
⎛ ∂p ⎞
du = CV dT + T ⎜ ⎟ dv − p dv
⎝ ∂T ⎠V
⎡ ⎛ ∂p ⎞ ⎤
du = CV dT + ⎢T ⎜ ⎟ − p ⎥ dv
⎣ ⎝ ∂T ⎠V ⎦

Internal energy of an ideal gas is function of temperature only

⎡ ⎛ ∂p ⎞ ⎤
du = CV dT + ⎢T ⎜ ⎟ − p ⎥ dv
⎣ ⎝ ∂T ⎠V ⎦
For an ideal gas, pv=RT
RT
p=
v
⎛ ∂p ⎞ R p
⎜ ⎟ = =
⎝ ∂T ⎠ v v T
⎡ ⎛ p⎞ ⎤
du = CV dT + ⎢T ⎜ ⎟ − p ⎥ dV = C v dT
⎣ ⎝ T ⎠V ⎦
Therefore the internal energy is depending only on temperature.
Change in internal energy when a gas obeys Van der Waals equation

⎡ ⎛ ∂p ⎞ ⎤
We know that, du = CV dT + ⎢T ⎜ ⎟ − p ⎥ dv
⎣ ⎝ ∂T ⎠V ⎦
⎛ a ⎞
Van der Waals equation is, ⎜ p + ⎟ (v − b ) = R T
⎝ v2 ⎠
RT a
p= − 2
v−b v
⎛ ∂p ⎞ R
⎜ ⎟ =
⎝ ∂T ⎠ v v − b

11
 ⎡ ⎛ R ⎞ RT a⎤
Therefore, du = CV dT + ⎢T ⎜ ⎟− + 2 ⎥ dv
⎣ ⎝v−b⎠ v−b v ⎦
⎡a⎤
du = CV dT + ⎢ 2 ⎥ dv
⎣v ⎦
2 2 2
dv
∫ du = CV ∫ dT + a ∫
1 1 1 v2

⎡1 1 ⎤
u 2 − u1 = CV (T2 − T1 ) + a ⎢ − ⎥
⎣ v1 v 2 ⎦
Change in entropy when a gas obeys Van der Waals equation
⎛ ∂p ⎞
T ds = CV dT + T ⎜ ⎟ dv Æ First T ds equation
⎝ ∂T ⎠V
dT ⎛ ∂p ⎞
ds = CV +⎜ ⎟ dv
T ⎝ ∂T ⎠V

⎛ a ⎞
Van der Waals equation is, ⎜ p + ⎟ (v − b ) = R T
⎝ v2 ⎠
RT a
p= − 2
v−b v
⎛ ∂p ⎞ R
⎜ ⎟ =
⎝ ∂T ⎠ v v − b
dT ⎛ R ⎞
Therefore, ds = CV +⎜ ⎟ dv
T ⎝v−b⎠
2 2 2
dT dv
∫1 ds = CV ∫1 T + R ∫1 v − b
⎛T ⎞ ⎛v −b⎞
s 2 − s1 = C v ln⎜⎜ 2 ⎟⎟ + R ln⎜⎜ 2 ⎟⎟
⎝ 1⎠
T ⎝ 1
v − b ⎠
Change in enthalpy
dh = T ds + v dp
⎛ ∂v ⎞
T ds = C p dT − T ⎜ ⎟ dp Æ Second T ds equation
⎝ ∂T ⎠ p

⎛ ∂v ⎞
dh = C p dT − T ⎜ ⎟ dp + v dp
⎝ ∂T ⎠ p

⎡ ⎛ ∂v ⎞ ⎤
dh = C p dT − ⎢T ⎜ ⎟ − v ⎥ dp
⎣⎢ ⎝ ∂T ⎠ p ⎥⎦

12
 ⎡ ⎛ ∂v ⎞ ⎤
dh = C p dT + ⎢v − T ⎜ ⎟ ⎥ dp
⎢⎣ ⎝ ∂T ⎠ p ⎥⎦
Enthalpy of an ideal gas is function of temperature only
⎡ ⎛ ∂v ⎞ ⎤
dh = C p dT + ⎢v − T ⎜ ⎟ ⎥ dp
⎢⎣ ⎝ ∂T ⎠ p ⎥⎦
For an ideal gas, pv=RT
RT
v=
p
⎛ ∂v ⎞ R v
⎜ ⎟ = =
⎝ ∂T ⎠ v p T
⎡ ⎛v ⎞⎤
dh = C p dT + ⎢v − T ⎜ ⎟ ⎥ dp = C p dT
⎣ ⎝T ⎠ ⎦
Therefore the enthalpy is depending only on temperature.
Difference in heat capacities (Cp – Cv)
2
⎛ ∂v ⎞ ⎛ ∂p ⎞ T vβ2
To derive C p − C v = −T ⎜ ⎟ ⎜ ⎟ or (Cp – Cv) = R or C p − C v =
⎝ ∂T ⎠ p ⎝ ∂v ⎠ T KT
Note: Derive First and Second T ds equations and proceed as follows:
⎛ ∂p ⎞
T ds = CV dT + T ⎜ ⎟ dv Æ First T ds equation
⎝ ∂T ⎠V
⎛ ∂v ⎞
T ds = C p dT − T ⎜ ⎟ dp Æ Second T ds equation
⎝ ∂T ⎠ p
Equating the above equations,
⎛ ∂p ⎞ ⎛ ∂v ⎞
CV dT + T ⎜ ⎟ dv = C p dT − T ⎜ ⎟ dp
⎝ ∂T ⎠V ⎝ ∂T ⎠ p

⎛ ∂p ⎞ ⎛ ∂v ⎞
(C − C v )dT = T ⎜ ⎟ dv + T ⎜ ⎟ dp
⎝ ∂T ⎠ v ⎝ ∂T ⎠ p
p

⎛ T ⎞⎛ ∂p ⎞ ⎛ T ⎞⎛ ∂v ⎞
dT = ⎜ ⎟⎜ ⎟ dv + ⎜ ⎟⎜
⎟⎝ ∂T ⎟⎠
dp ----- (1)
⎜C −C ⎟⎝ ∂T ⎠ ⎜C −C
⎝ p v ⎠ v ⎝ p v ⎠ p

We can write, T = f (p, v)

⎛ ∂T ⎞ ⎛ ∂T ⎞
dT = ⎜⎜ ⎟⎟ dp + ⎜ ⎟ dv ----- (2)
⎝ ∂p ⎠V ⎝ ∂v ⎠ p
Comparing (1) and (2),

13
 ⎛ ∂T ⎞ T ⎛ ∂p ⎞
⎜ ⎟ = ⎜ ⎟ ----- (3)
⎝ ∂v ⎠ p C p − C v ⎝ ∂T ⎠ v

⎛ ∂p ⎞ ⎛ ∂v ⎞
From (3), C p − C v = T ⎜ ⎟ ⎜ ⎟ ----- (4)
⎝ ∂T ⎠ v ⎝ ∂T ⎠ p

⎛ ∂p ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞
But, ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ = −1
⎝ ∂T ⎠ v ⎝ ∂v ⎠ p ⎝ ∂p ⎠ T

⎛ ∂p ⎞ ⎛ ∂v ⎞ ⎛ ∂p ⎞
⎜ ⎟ = −⎜ ⎟ ⎜ ⎟ ----- (5)
⎝ ∂T ⎠ v ⎝ ∂T ⎠ p ⎝ ∂v ⎠ T
Substituting (5) in (4),
2
⎛ ∂v ⎞ ⎛ ∂p ⎞
C p − C v = −T ⎜ ⎟ ⎜ ⎟ ----- (6)
⎝ ∂T ⎠ p ⎝ ∂v ⎠ T
pv=RT
RT
v=
p
⎛ ∂v ⎞ R
⎜ ⎟ = ----- (7)
⎝ ∂T ⎠ p p
RT
p=
v
⎛ ∂p ⎞ − RT
⎜ ⎟ = 2 ----- (8)
⎝ ∂v ⎠ T v
Substituting (7) and (8) in (6),
R2 ⎛ − RT ⎞ R3 T 2
C p − C v = −T ⎜ ⎟=
p2 ⎝ v2 ⎠ p2 v2
pv=RT
R3 T 2
C p − Cv = =R
R2 T 2
Cp – Cv = R ----- (9)
1 ⎛ ∂v ⎞ ⎛ ∂v ⎞
We know that, β = ⎜ ⎟ Æ⎜ ⎟ =βv
v ⎝ ∂T ⎠ p ⎝ ∂T ⎠ p

1 ⎛ ∂v ⎞ ⎛ ∂v ⎞
K T = − ⎜⎜ ⎟⎟ Æ ⎜⎜ ⎟⎟ = −v K T
v ⎝ ∂p ⎠ T ⎝ ∂p ⎠ T
Therefore equation (6) can be written as,
−1
C p − C v = −Tβ 2 v 2
v KT

14
 T vβ2
C p − Cv = ----- (10)
KT
Ratio of heat capacities
Note: Derive First and Second T ds equations and proceed as follows:
⎛ ∂p ⎞
T ds = CV dT + T ⎜ ⎟ dv Æ First T ds equation
⎝ ∂T ⎠V
⎛ ∂v ⎞
T ds = C p dT − T ⎜ ⎟ dp Æ Second T ds equation
⎝ ∂T ⎠ p
Consider isentropic process, ds = 0
⎛ ∂v ⎞
C p dT = T ⎜ ⎟ dp
⎝ ∂T ⎠ p

⎛ ∂v ⎞ dp
Cp = T⎜ ⎟
⎝ ∂T ⎠ p dT

⎛ ∂p ⎞
CV dT = −T ⎜ ⎟ dv
⎝ ∂T ⎠V
⎛ ∂p ⎞ dv
CV = −T ⎜ ⎟
⎝ ∂T ⎠V dT
Cp ⎛ ∂v ⎞ ⎛ ∂T ⎞ dp dT
= −⎜ ⎟ ⎜⎜ ⎟⎟
Cv ⎝ ∂T ⎠ p ⎝ ∂p ⎠ v dT dv
Cp ⎛ ∂v ⎞ ⎛ ∂T ⎞ dp
= −⎜ ⎟ ⎜⎜ ⎟⎟
Cv ⎝ ∂T ⎠ p ⎝ ∂p ⎠ v dv
dp ⎛ ∂p ⎞
But, = ⎜ ⎟ for isentropic process
dv ⎝ ∂v ⎠ s

Cp ⎛ ∂v ⎞ ⎛ ∂T ⎞ ⎛ ∂p ⎞
Therefore, = −⎜ ⎟ ⎜⎜ ⎟⎟ ⎜ ⎟ ----- (1)
Cv ⎝ ∂T ⎠ p ⎝ ∂p ⎠ v ⎝ ∂v ⎠ s

⎛ ∂p ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞
We can write, ⎜ ⎟ ⎜ ⎟ ⎜⎜ ⎟⎟ = −1
⎝ ∂T ⎠ v ⎝ ∂v ⎠ p ⎝ ∂p ⎠ T

⎛ ∂p ⎞ ⎛ ∂T ⎞ ⎛ ∂v ⎞
⎜ ⎟ ⎜ ⎟ = −⎜⎜ ⎟⎟
⎝ ∂T ⎠ v ⎝ ∂v ⎠ p ⎝ ∂p ⎠ T
Cp ⎛ ∂v ⎞ ⎛ ∂p ⎞
Substituting in (1), = ⎜⎜ ⎟⎟ ⎜ ⎟ ----- (2)
C v ⎝ ∂p ⎠ T ⎝ ∂v ⎠ s

1 ⎛ ∂v ⎞ ⎛ ∂v ⎞
K T = − ⎜⎜ ⎟⎟ Æ ⎜⎜ ⎟⎟ = −v K T
v ⎝ ∂p ⎠ T ⎝ ∂p ⎠ T

15
 1 ⎛ ∂v ⎞ ⎛ ∂v ⎞
K S = − ⎜⎜ ⎟⎟ Æ ⎜⎜ ⎟⎟ = −v K S
v ⎝ ∂p ⎠ S ⎝ ∂p ⎠ S
Cp KT
Substituting in (2), = ----- (3)
Cv KS
Joule-Thomson coefficient
A gas is made to undergo continuous throttling process by a valve, as shown. The pressures and
temperatures of the gas in the insulated pipe upstream and downstream of the valve are measured with
suitable manometers and thermometers.
⎛ ∂T ⎞
μ J = ⎜⎜ ⎟⎟
⎝ ∂p ⎠ h

Joule-Thomson expansion
Let p1 and T1 be the arbitrarily chosen pressure and temperature before throttling and let them be kept
constant. By operating the valve manually the gas is throttled successively to different pressures and
temperatures. These are then plotted on the T-p coordinates. The curve passing through all these points is
an isenthalpic curve.

Isenthalpic state of a gas

16
 Isenthalpic curves and the inversion curve

The initial pressure and temperature of the gas (before throttling) are then set to new values, and
by throttling to different states, a family of isenthalpes is obtained for the gas. The curve passing through
the maxima of these isenthalpes is called the inversion curve.
The numerical value of the slope of an isenthalpe on a T-p diagram at any point is called the
‘Joule-Thomson coefficient’ and is denoted by μJ.
At inversion curve, μJ = 0
At right side of the inversion curve, dT is positive, dp is negative.
μJ = - ve (Heating region)
At left side of the inversion curve, dT is negative, dp is negative.
μJ = + ve (Cooling region)
dh = T ds + v dp ----- (1)
Second T ds equation is given by,
⎛ ∂v ⎞
T ds = C p dT − T ⎜ ⎟ dp
⎝ ∂T ⎠ p

Substituting in (1),
⎛ ∂v ⎞
dh = C p dT − T ⎜ ⎟ dp + v dp
⎝ ∂T ⎠ p

⎡ ⎛ ∂v ⎞ ⎤
= C p dT − ⎢T ⎜ ⎟ − v ⎥ dp
⎣⎢ ⎝ ∂T ⎠ p ⎥⎦
The second term in the above equation stands only for a real gas, because for an ideal gas,
dh = Cp dT
dh = 0 for isenthalpic process

17
 ⎡ ⎛ ∂v ⎞ ⎤
Therefore, C p dT − ⎢T ⎜ ⎟ − v ⎥ dp = 0
⎢⎣ ⎝ ∂T ⎠ p ⎥⎦

⎡ ⎛ ∂v ⎞ ⎤
C p dT = ⎢T ⎜ ⎟ − v ⎥ dp
⎢⎣ ⎝ ∂T ⎠ p ⎥⎦

dT 1 ⎡ ⎛ ∂v ⎞ ⎤
= ⎢T ⎜ ⎟ −v⎥
dp C p ⎢⎣ ⎝ ∂T ⎠ p ⎥⎦

⎛ ∂T ⎞ 1 ⎡ ⎛ ∂v ⎞ ⎤
⎜⎜ ⎟⎟ = ⎢T ⎜ ⎟ −v⎥
⎝ ∂p ⎠ h C p ⎢⎣ ⎝ ∂T ⎠ p ⎥⎦

⎛ ∂T ⎞ 1 ⎡ ⎛ ∂v ⎞ ⎤
Therefore, μ J = ⎜⎜ ⎟⎟ = ⎢T ⎜ ⎟ −v⎥
⎝ ∂p ⎠ h C p ⎢⎣ ⎝ ∂T ⎠ p ⎥⎦
For an ideal gas
pv=RT
RT
v=
v
⎛ ∂v ⎞ R v
⎜ ⎟ = =
⎝ ∂T ⎠ p p T

⎛ ∂T ⎞ 1 ⎡ ⎛v⎞ ⎤
μ J = ⎜⎜ ⎟⎟ = ⎢T ⎜ ⎟ − v ⎥ = 0
⎝ ∂p ⎠ h C p ⎣ ⎝ T ⎠ ⎦
There is no change in temperature when an ideal gas is made to undergo a throttling.
Another form

⎛ ∂T ⎞ 1 ⎡ ⎛ ∂v ⎞ ⎤
μ J = ⎜⎜ ⎟⎟ = ⎢T ⎜ ⎟ −v⎥
⎝ ∂p ⎠ h C p ⎢⎣ ⎝ ∂T ⎠ p ⎥⎦
1 ⎛ ∂V ⎞
But, β = ⎜ ⎟
V ⎝ ∂T ⎠ p

1
μJ = [T β v − v ]= v [T β − 1]
Cp Cp
1
β= for an ideal gas since μJ = 0
T
Clausius-Clapeyron equation
Clausius-Clapeyron equation is a relationship between the saturation pressure, temperature, the
enthalpy of evaporation and the specific volume of the two phases involved. This equation provides a
basis for calculations of properties in a two phase region. It gives the slope of a curve separating the two
phases in the p-T diagram.

18
 Let us consider then change of state from saturated liquid to saturated vapour of a pure substance
which takes place at constant temperature. During the evaporation, the pressure and temperature are
independent of volume.
⎛ ∂p ⎞ ⎛ ∂s ⎞
III-Maxwell equation, ⎜ ⎟ =⎜ ⎟
⎝ ∂T ⎠ v ⎝ ∂v ⎠ T

⎛ dp ⎞ ⎛ ds ⎞ s g − s f s fg
⎜ ⎟=⎜ ⎟= =
⎝ dT ⎠ ⎝ dv ⎠ v g − v f v fg
sg = specific entropy of saturated vapour
sf = specific entropy of saturated liquid
vg = specific volume of saturated vapour
vf = specific volume of saturated liquid
hfg = increase in specific entropy
vfg = increase in specific volume
h fg
Also, sg − s f =
T
⎛ dp ⎞ s g − s f h fg h fg
⎜ ⎟= = =
⎝ dT ⎠ v g − v f v fg T v fg
This equation is known as Clausius-Clapeyron equation for evaporation of liquids. The derivative
dp/dT is the slope of vapour pressure versus temperature curve. Knowing this slope and the specific
volume vg and vf from experimental data, we can determine the enthalpy of evaporation, hfg.

19
20
 UNIT-V
GAS MIXTURES AND PSYCHROMETRY
Gas Mixtures
A pure substance is defined as a substance having a constant and uniform chemical composition.
A homogeneous mixture of gases which do not react with one another is considered to a pure substance.
Air is a homogeneous mixture of nitrogen, oxygen, and traces of other substances like, argon, helium,
carbon dioxide, etc, and as they do not react with one another, air is regarded a pure substance. The
properties of the mixture are determined from the properties of the constituent gases.
Dalton’s law of partial pressures
Let us consider a homogeneous mixture of ideal gases at a temperature T, a pressure p and a
volume V.
Suppose there are n1 moles of gas A1, n2 moles of gas A2, ……… nc moles of gas Ac.

p, T, V

n1, n2,…

p V = (n1 + n2 + ……… + nc) Ru T

n1 Ru T n2 Ru T nRT
p= + + ............. + c u
V V V
= p1 + p 2 + ............... + p c
p1, p2, ………. p3 are partial pressure of individual components
p = Total pressure
Ru = Universal gas constant = 8314 J/kmol-K
Dalton’s law of partial pressures states that the total pressure of a mixture of ideal gases is equal
to the sum of the partial pressures.
p = p1 + p2 + ……… + pc
Properties of gas mixture

∑n = n 1 + n2 + .......... + nc

n1
x1 = Æ Mole fraction or volume fraction of component-1
∑n
n2
x2 = Æ Mole fraction or volume fraction of component-2, ……
∑n
m1 R1T n1 Ru T
Partial pressure of component-1, p1 = =
V V
m1 R1T n1 Ru T
Partial volume of component-1, V1 = =
p p
Mass of component-1, m1 = M1 n1 = mf1 m
Gas constant of component-1, R = Ru/M1
R1 = Characteristic gas constant (J/kg-K)
M1 = Molar mass (Molecular weight) of component-1
n1 = Number of moles of component-1
mf1 = Mass fraction of component-1
Mass of gas mixture, m = m1 + m2 + …………+ mc
Volume of gas mixture, V = V1 + V2 + ………… + Vc
m1u1 + m2 u 2 + .......... + mc u c
Internal energy of mixture, u=
m1 + m2 + ......... + mc
m1 h1 + m2 h2 + .......... + mc hc
Enthalpy of mixture, h=
m1 + m2 + ......... + mc
m1C p1 + m2 C p 2 + .......... + mc C pc
Specific heat of mixture, Cp =
m1 + m2 + ......... + mc
m1C v1 + m2 C v 2 + .......... + mc C vc
Cv =
m1 + m2 + ......... + mc
n1 M 1 + n2 M 2 + .......... + nc M c
Molar mass of mixture, M =
n1 + n2 + ......... + nc
= x x M 1 + x 2 M 2 + ............ + xc M c

m1 R1 + m2 R2 + .......... + mc Rc Ru
Gas constant of mixture, R= =
m1 + m2 + ......... + mc M
p
Density of mixture, ρ=
RT
Specific heat ratio of mixture, γ = x1γ 1 + x 2 γ 2 + ............ + xc γ c

Note: Molecular weight of the mixture is also called as apparent molecular weight.

PROBLEMS
1. Air contains 21% O2 and 79% N2 by volume. Determine the molecular weight, gas constant
and density of air at STP.
Given
Mole fraction of O2 (xO2) = 0.21
Mole fraction of N2 (xN2) = 0.79
Temperature of mixture (T) = 273 K (Std temperature)
Pressure of mixture (p) = 1 atm = 1.01325 bar (Std pressure)
Required: M, R, ρ
Solution

2
 Molecular weight, M = xCO 2 M CO 2 + x N 2 M N 2
M = 0.21 (32) + 0.79 ( 28) = 28.84 ----- Ans
mO 2 RO 2 + m N 2 R N 2 Ru
Gas constant, R= =
mO 2 + m N 2 M
= 8314 / 28.84
= 288.3 J/kg-K ----- Ans
p 1.01325 x 10 5
Density of mixture, ρ= = = 1..287 kg / m 3 ----- Ans
RT 288.3 x 273
2. 0,45 kg of CO and 1 kg of air is contained in a vessel of volume 0.4 m3 at 15oC. Air has
23.3% of O2 and 76.7% of N2 by mass. Calculate the partial pressure of each constituent
and total pressure in the vessel. Molar masses of CO, O2 and N2 are 28, 32 and 28 kg/kmol.
Given:
Mass of CO (mCO) = 0.45 kg
Mass of air (ma) = 1 kg
Volume of mixture (V) = 0.4 m3
Temperature of mixture (T) = 15oC = 288 K
Mass fraction of O2 (mfO2) = 0.233
Mass fraction of N2 (mfN2) = 0.767
Molar mass of CO (MCO) = 28
Molar mass of O2 (MO2) = 32
Molar mass of N2 (MN2) = 28
Required: pCO, pO2, pN2 and p
Solution:
Mass of O2 (mO2) = mfO2 ma = 0.233 x 1 = 0.233 kg
Mass of N2 (mN2) = mfN2 ma = 0.767 x 1 = 0.767 kg
Mass of mixture (m) = mCO + mO2 + mN2
= 0.45 + 0.233 + 0.767 = 1.45 kg
Gas constant of CO (RCO) = Ru / MCO = 8324 / 28 = 296.93 J/kg-K
Gas constant of O2 (RO2) = Ru / MO2 = 8314 / 32 = 259.81 J/kg-K
Gas constant of N2 (RN2) = Ru / MN2 = 8314 / 28 = 296.93 J/kg-K
mCO RCOT 0.45 x 296.93 x 288
Partial pressure of CO, pCO = =
V 0.4
= 96205.3 Pa ---- Ans
mO 2 RO 2T 0.233 x 259.81 x 288
Partial pressure of O2, pO 2 = =
V 0..4
= 43585.7 Pa ---- Ans

3
 m N 2 R N 2T 0.767 x 296.93 x 288
Partial pressure of N2, p N 2 = =
V 0.4
= 163976.6 Pa ---- Ans
Total pressure, p = pCO + pO2 + pN2
= 96205.3 + 43585.7 + 163976.6
= 303767.6 Pa = 3.04 bar ----- Ans
3. A vessel of 0.35 m3 capacity contains 0.4 kg of carbon monoxide (Molecular weight = 28)
and 1 kg of air at 20oC. Calculate the partial pressure of each component and the total
pressure in the vessel. The gravimetric analysis of air is to be taken as 23.3% O2 and 76.7%
N2.
Given:
Mass of CO (mCO) = 0.4 kg
Mass of air (ma) = 1 kg
Volume of mixture (V) = 0.35 m3
Temperature of mixture (T) = 20oC = 293 K
Mass fraction of O2 (mfO2) = 0.233
Mass fraction of N2 (mfN2) = 0.767
Molar mass of CO (MCO) = 28
Required: pCO, pO2, pN2 and p
Solution:
Mass of O2 (mO2) = mfO2 ma = 0.233 x 1 = 0.233 kg
Mass of N2 (mN2) = mfN2 ma = 0.767 x 1 = 0.767 kg
Mass of mixture (m) = mCO + mO2 + mN2
= 0.4 + 0.233 + 0.767 = 1.4 kg
Gas constant of CO (RCO) = Ru / MCO = 8324 / 28 = 296.93 J/kg-K
Gas constant of O2 (RO2) = Ru / MO2 = 8314 / 32 = 259.81 J/kg-K
Gas constant of N2 (RN2) = Ru / MN2 = 8314 / 28 = 296.93 J/kg-K
mCO RCOT 0.4 x 296.93 x 293
Partial pressure of CO, pCO = =
V 0.35
= 99429.1 Pa ---- Ans
mO 2 RO 2T 0.233 x 259.81 x 293
Partial pressure of O2, pO 2 = =
V 0..35
= 50677 Pa ---- Ans
m N 2 R N 2T 0.767 x 296.93 x 293
Partial pressure of N2, p N 2 = =
V 0.35
= 190655.6 Pa ---- Ans
Total pressure, p = pCO + pO2 + pN2
= 99429.1 + 50677 + 190655.6
4
 = 340761.7 Pa = 3.407 bar ----- Ans
4. A mixture of ideal gases consists of 3 kg of nitrogen and 5 kg of carbon dioxide at a pressure
of 300 kPa and a temperature of 20oC. Find (i) mole fraction of each constituent, (ii) the
equivalent molecular weight of the mixture, (iii) the equivalent gas constant of the mixture,
(iv) the partial pressures and partial volumes, and (v) the volume and density of the
mixture.
Given:
Mass of nitrogen (mN2) = 3 kg
Mass of Carbon dioxide (mCO2) = 5 kg
Mixture pressure (p) = 300 kPa
Mixture temperature (T) = 20oC = 293 K
Required: (i) xN2, xCO2 (ii) M (iii) R (iv) pN2, pCO2, VN2, VCO2 (v) V, ρ
Solution:
mN 2 3
(i) Number of moles of N2, nN 2 = = = 0.1071
M N 2 28
mCO 2 5
Number of moles of CO2, nCO 2 = = = 0.1136
M CO 2 44
nN 2 0.1071
Mole fraction of N2 x N 2 = = = 0.4853 ----- Ans
n N 2 + nCO 2 0.1071 + 0.1136
nCO 2 0.1136
Mole fraction of N2 xCO 2 = = = 0.5147 ----- Ans
n N 2 + nCO 2 0.1071 + 0.1136
(ii) Equivalent molecular weight, M = xCO 2 M CO 2 + x N 2 M N 2

= 0.5147 ( 44) + 0.4853 ( 28) = 36.23 ----- Ans

mCO 2 RCO 2 + m N 2 R N 2 Ru
(iii) Equivalent gas constant, R = =
mCO 2 + m N 2 M
= 8314 / 36.23
= 229.48 J/kg-K ----- Ans
(iv) Partial pressure of N2, pN2 = xN2 p = 0.4853 (300) = 145.59 kPa ----- Ans
Partial pressure of CO2, pCO2 = xCO2 p = 0.5147 (300) = 154.41 kPa ----- Ans
m N 2 R N 2T
Partial volume of N2, VN 2 =
p
RN2 = Ru / MN2 = 8314 / 28 = 296.93 J/kg-K
3 (296.93) (293)
VN 2 = 3
= 0.87 m 3 ----- Ans
300 x 10
mCO 2 RCO 2T
Partial volume of CO2, VCO 2 =
p
RCO2 = Ru / MCO2 = 8314 / 44 = 188.95 J/kg-K
5
 5 (188.95) (293)
VCO 2 = = 0.923 m 3 ----- Ans
300 x 10 3
(v) Total volume, V = VN2 + VCO2 = 0.87 + 0.923 = 1.793 m3 ----- Ans

p 300 x 10 3
Density of the mixture, ρ = = = 4.4618 kg / m 3 ----- Ans
R T 229.48 x 293

PSYCHROMETRY
Atmospheric air always contains water vapour. The water vapour content in air plays an
important role in comfort air conditioning. The partial pressure of water vapour in atmospheric air is very
low and the vapour exists either in superheated or saturated.
The science which deals with the study of the behaviour of air and water vapour mixture is
known as Psychrometry. The properties of air and water vapour mixture are known as psychrometric
properties.
Dry air : Air with no water vapour is called ‘dry air’.
O2 = 21 % ; N2 = 78.1 % ; CO2 = 0.03 %
Since dry air is never found, it always may contain some water vapour.
Moisture : The water vapour present in the air is known as moisture.
Moist air : It is a mixture of dry air and water vapour.
Unsaturated and Saturated air : The moist air in which the water vapour exists in superheated state, is
known as Unsaturated air and such air will be invisible
If the water vapour is added to dry air or unsaturated air, a limit will be reached when the air will
be saturated and can hold no more water vapour. Such air will also be invisible and the air is called
‘saturated air’. But if more water is added, the drops of water may remain in suspension and make air
misty or foggy. Thus such drops are the condensed particles of water vapour. This will happen only
beyond saturation limit.
According to the Dalton’s law of partial pressures,
pb = pa + pv
pb = Barometric pressure
pa = Partial pressure of dry air
pv = Partial pressure of water vapour
Dry Bulb Temperature (DBT) : The temperature of the air measured by ordinary thermometer whose
bulb is dry, is known as DBT of the air.
Wet Bulb Temperature (WBT) : The temperature of the air measured by a thermometer when its bulb is
covered with wet cloth and is exposed to a current of air is known as WBT of air.
Wet bulb depression = DBT – WBT
Dew Point Temperature (DPT) : It is the temperature of air at which water vapour in the air starts
condensing when the is cooled. Thus this temperature will corresponds to the saturation temperature at
partial pressure of water vapour. DPT can be obtained from steam table at pv.
Dew point depression = DBT – DPT
Note:
For saturated air, DBT = WBT = DPT
6
Specific humidity (W)
It is the mass of water vapour present with one kg of dry air.
pv
W = 0.622 kg / kg dry air
pb − pv
C pa (Tw − Td ) + Ww h fgW ρ v mv
Also, W= = =
hgd − h fw ρ a ma
p sw
Ww = 0.622
pb − p sw
psw = Saturation pressure at WBT
hfgw = hfg at WBT in kJ/kg
hgd = hg at DBT in kJ/kg
hfw = hf at WBT in kJ/kg
Cpa = Specific heat of air = 1.005 kJ/kg-K
ρv = Density of water vapour in the mixture
ρa = Density of dry air in the mixture
mv = Mass of water vapour
ma = Mass of dry air
Density of dry air and water vapour
Density of dry air can be calculated as,
ρa = pa / (Ra Td)
Td = Dry bulb temperature in K
Ra = 287 J/kg-K
pa = Partial pressure of dry air in Pa
Density of water vapour can be calculated from,
ρv = W ρa
Mass of dry air and water vapour
Mass of dry air (ma) is calculated from,
pa V = ma Ra Td
Mass of water vapour (mv) is calculated from,
pv V = mv Rv Td
If gas constant (Rv) for water vapour is not available, mv can be calculated from,
mv = W ma
Specific humidity of saturated air (Ws)
It is the mass of water vapour present with 1 kg of dry air when the air is saturated.
ps
Ws = 0.622 kg / kg dry air
pb − p s

7
 ps Æ Saturation pressure at DBT
Degree of saturation (or) Saturation ratio (μ)
W p ( p − ps )
μ= = v b
Ws p s ( pb − p v )
If Φ = 0, pv = 0, W = 0, μ = 0
If Φ = 100%, pv = ps, W = Ws, μ = 1
Therefore μ varies between 0 and 1.
Relative humidity (φ)
It is the ratio of actual mass of water vapour in a given volume of moist air to the mass of water
vapour in the same volume of saturated air at the same temperature & pressure.
pv μ
φ= =
ps ps
1 − (1 − μ )
pb
pv can be calculated from, Carrier’s equation,
( pb − p sw ) (Td − Tw )
p v = p sw −
1527.4 − 1.3 Tw
Note: Tw Æ WBT in oC
psw = Saturation pressure at WBT pb = Barometric pressure
pv
pv also can be determined from, W = 0.622 knowing W.
pb − pv
Enthalpy of moist air
h = C pa Td + W [ h gd + 1.88(Td − Tdp )]
Td = DBT in oC Cpa = 1.005 kJ/kg-K W = Specific humidity in kg/kg dry air
Tdp = DPT in oC
Psychrometric Chart

8
 DBT - lines Î Uniformly spaced vertical lines.
W - lines Î Uniformly spaced horizontal lines.
DPT – lines Î Non-uniformly spaced horizontal lines.
WBT – lines Î Non-uniformly spaced inclined lines.
φ - lines Î Curved lines.

Psychrometric processes
Sensible cooling
Out Cooling coil (Td3)

9
 Air out
Air in (Td1) (Td2)

3 2 1 W 1 = W2

Refrigerant in
Td3 Td2 Td1

The cooling of air without change in its specific humidity is known as ‘Sensible cooling’.
Td 2 − Td 3
By pass factor ( BPF ) =
Td 1 − Td 3
Td3 = Coil temperature
Coil efficiency (ηc) = 1 – BPF

Sensible heating

The heating of air without change in its specific humidity is known as ‘Sensible heating’.
Td 3 − Td 2
By pass factor ( BPF ) =
Td 3 − Td 1

Steam out

Heating coil (Td3)

Air out
Air in (Td1) (Td2)

1 2 3 W1 = W2

Steam in
Td1 Td2 Td3

Td3 = Coil temperature

Coil efficiency (ηh) = 1 – BPF

Dehumidification
The removal of moisture from the air, without change in its dry bulb temperature is known as
‘dehumidification’. Perforated plate is used as ‘dehumidifier’ which removes water particles from the air.

10
 Perforated plate
h1

h2

1
W1 Air in Air out

2 W2

Td1 = Td2

Chemical dehumidification
Some substances like silica gel and activated alumina have great affinity with water vapour. They
are called absorbents. When air passes through a bed of silica gel, water vapour molecules get absorbed
on its surface. Latent heat of condensation is released.
For vapour to condense, it has to lose its heat to surrounding air. So, the DBT of air increases.
This process is called ‘Chemical dehumidification’.

1 W1

2 W2

Td1 Td2
Humidification
Addition of moisture to the air, without change in its dry bulb temperature is known as
‘Humidification’.

h2

h1

2
W2 Air in Air out

11
 1 W1
Nozzles

Td1 = Td2
Cooling water in

Adiabatic dehumidification and humidification

h1= h2

Td
Dehumidification or humidification of air at constant enthalpy with no heat transfer is called

‘adiabatic dehumidification’ or ‘adiabatic humidification’.

Cooling & Dehumidification

This process is generally used in Summer air conditioning to cool and dehumidify the air.

Sensible heat
Sensible heat factor (SHF) = -------------------
Total heat

h1 – h2
= ----------
h1 – h3

h1

h2

h3

2 1 W1 = W2

3 W3

Td2 = Td3
Heating & Humidification
This process is generally used in Winter air conditioning to warm & humidify the air.
12
 h3

h2

h1

3 W3

1 2 W1 = W2

Td1 Td2 = Td3

Sensible heat
Sensible heat factor (SHF) = --------------------
Total heat

h2 – h1
= ----------
h3 – h1
Adiabatic evaporative cooling
A large quantity of water is constantly circulated through a spray chamber. The moist air is
passed through this spray water. Some water evaporates into moist air. During this evaporation the water
absorbs latent heat from the air reducing air temperature and increasing air specific humidity. This
process is called ‘Evaporative cooling’ of air. If the air is cooled in an insulated chamber, then the cooling
is known as ‘adiabatic evaporative cooling’.
Cooling tower utilizes the phenomenon of evaporative cooling to cool warm water below the
DBT of the air.
Air out

Insulator
Water

Air in

Adiabatic mixing of two streams

m1, h1 W1

13
 m3, h3, W3

m2, h2 W2

m1 = Mass flow rate of dry air of First stream

h1 = Specific enthalpy of moist air/kg of dry air of first stream
W1 = Specific humidity of moist air of first stream
m2, h2, W2 Î Corresponding parameters of second stream
m3, h3, W3 Î Corresponding parameters of mixed stream
We can write,

m1 + m2 = m3 ------ (1)

m1 h1 +m2 h2 = m3 h3 ------ (2)

m1 W1 + m2 W2 = m3 W3 -------(3)

From (1) & (2), m1 W1 + m2 W2 = (m1 + m2) W3

m1(W1 – W3) = m2 (W3 – W2)

m1 W3 – W2
---- = -----------
m2 W1 – W3

And similarly we can write,

m1 h3 – h2
---- = ----------
m2 h1 – h3

h1

h3
1 W1
h2
3
W2

2 W3

Td2 Td3 Td1

Adiabatic saturation

14
 When unsaturated air flows over a long sheet of water in an insulated chamber, the water
evaporates, and the specific humidity of the air increases. Both the air and water are cooled as evaporation
takes places. The process continues until the energy transferred from the air to the water is equal to the
energy required to evaporate the water. When this point is reached, thermal equilibrium exists with
respect to the water, air and water vapour, and correspondingly the air is saturated. This process is called
adiabatic saturation. And the enthalpy is constant as there is no heat transfer during the process.

PROBLEMS
5. 5 gm of water vapour per kg of atmospheric air at 35oC, 60% RH and 1.013 bar is removed
and temperature of air after removing the water vapour is 25oC DBT. Determine, RH and
DPT.
Given:
Mass of water vapour (mv) = 5 gm
Initial DBT (Td1) = 35oC
Initial RH (Φ1) = 60%
Total pressure of air (pb) = 1.013 bar
Final DBT (Td2) = 25oC
Required: Φ2 and Tdp2
Solution:

15
 1

25oC 35oC
p v1
φ1 =
p s1
ps1 = 0.05628 bar at Td1 = 35oC from steam table
p v1
Therefore, 0.6 =
0.05628
pv1 = 0.03377 bar
p v1
W1 = 0.622
pb − p v1
0.03377
W1 = 0.622 = 0.02145 kg / kg dry air
1.013 − 0.03377
= 21.45 gm/kg dry air
Therefore, W2 = 21.45 – 5 = 16.45 gm/kg dry air
ps2 = 0.03169 bar at Td2 = 25oC from steam table
pv 2
W2 = 0.622
pb − p v 2
pv 2
0.01645 = 0.622 x
1.013 − p v 2
0.02679 – 0.026447 pv2 = pv2
pv2 = 0.0261 bar
pv 2 0.0261
φ1 = = = 0.823 = 82.3 % ---- Ans
p s 2 0.03169
From steam table, Tdp2 = 22oC at pv2 = 0.0261 bar ----- Ans
6. The sling psychrometer reads 40oC DBT and 28oC WBT. Calculate, (i) specific humidity, (ii)
relative humidity (iii) vapour density of air, (iv) DPT and (v) enthalpy of mixture. Assume
atmospheric pressure to be 1.013 bar.
Given:
Atmospheric pressure (pb) = 1.013 bar
Dry bulb temperature (Td) = 40oC
Wet bulb temperature (Tw) = 28oC
16
Required: (i) W (ii) Φ (iii) ρv (iv) Tdp (v) h
Solution:
pv
(i) W = 0.622
pb − pv

C pa (Tw − Td ) + Ww h fgW
Also, W=
hgd − h fw

p sw
Ww = 0.622
pb − p sw
psw = Saturation pressure at WBT = 0.03778 bar from steam table
0.03778
Ww = 0.622 = 0.024096 kg / kg dry air
1.013 − 0.03778
hfgw = hfg at WBT = 2435.4 kJ/kg from steam table at 28oC
hgd = hg at DBT = 2574.4 kJ/kg from steam table at 40oC
hfw = hf at WBT = 117.3 kJ/kg from steam table at 28oC
Cpa = Specific heat of air = 1.005 kJ/kg-K
1.005 (28 − 40) + 0.024096(2435.4)
W= = 0.018975 kg / kg dry air
2574.4 − 117.3
pv
(ii) φ=
ps
pv
W = 0.622
pb − pv
pv
0.018975 = 0.622 x
1.013 − pv
pv = 0.02998 bar
ps = 0.07375 bar at 40oC from steam table
Therefore, Φ = 0.02998/0.07375 = 0.4065 = 40 % ----- Ans
(iii) Density of water vapour can be found out from, W = ρv / ρa
pa
We know that, ρ a =
Ra Ta
Ra = 287 J/kg-K
Ta = DBT
pa = Partial pressure of dry air = pb – pv
pa = 1.013 – 0.02998 = 0.98302 bar

17
 0.98302 x 10 5
ρa = = 1.0943 kg / m 3
287 x (40 + 273)
Therefore, ρv = 0.018975 x 1.0943 = 0.020764 kg/m3 ----- Ans
(iv) At pv = 0.020764 bar from steam table, Tdp = 24oC ----- Ans
(v) h = C pa Td + W [ h gd + 1.88(Td − Tdp )]

= 1.005 ( 40) + 0.018975 [ 2574 .4 + 1.88( 40 − 24)]

= 89.62 kJ/kg dry air ---- Ans
Note: Substitute Td in oC
7. Atmospheric air at 1.0132 bar has a DBT of 32oC and WBT of 26oC. Compute (a) the
partial pressure of water vapour, (b) the specific humidity, (c) the DPT, (d) the RH, (e) the
degree of saturation, (f) the density of air in the mixture, (g) the density of vapour in the
mixture and (h) the enthalpy of the mixture.
Given:
Atmospheric pressure (pb) = 1.0132 bar
Dry bulb temperature (Td) = 32oC
Wet bulb temperature (Tw) = 26oC
Required: (a) pv (b) W (c) Tdp (d) Φ (e) μ (f) ρa (g) ρv (h) h
Solution:
(a)
I-Method (Using Carrier’s equation)
( pb − p sw ) (Td − Tw )
p v = p sw −
1527.4 − 1.3 Tw
Note: Tw Æ WBT in oC
psw = Saturation pressure at WBT = 0.03360 bar from steam table
(1.0132 − 0.03360) (32 − 26)
p v = 0.03360 −
1527.4 − 1.3 ( 26)
= 0.0296 bar ----- Ans
II-Method
C pa (Tw − Td ) + Ww h fgW
W=
hgd − h fw

p sw
Ww = 0.622
pb − p sw
psw = Saturation pressure at WBT = 0.03360 bar from steam table
0.03360
Ww = 0.622 = 0.021334 kg / kg dry air
1.0132 − 0.03360
hfgw = hfg at WBT = 2440.2 kJ/kg from steam table at 26oC
hgd = hg at DBT = 2560 kJ/kg from steam table at 32oC
18
hfw = hf at WBT = 108.9 kJ/kg from steam table at 26oC
Cpa = Specific heat of air = 1.005 kJ/kg-K
1.005 (26 − 32) + 0.021334(2440.2)
W= = 0.01878 kg / kg dry air
2560 − 108.9
pv
Also, W = 0.622
pb − pv
pv
0.01878 = 0.622 x
1.0132 − p v
pv = 0.0297 bar ---- Ans
pv
(b) W = 0.622
pb − pv
0.0296
= 0.622 = 0.01872 kg / kg dry air ---- Ans
1.0132 − 0.0296
(c) Tdp = 24oC at pv from steam table ----- Ans
pv
(d) φ=
ps
ps = 0.04753 bar at 32oC from steam table
Therefore, Φ = 0.0296/0.04753 = 0.623 = 62.3 % ----- Ans
W p ( p − ps )
(e) μ= = v b
Ws p s ( pb − p v )
0.0296 (1.0132 − 0.04753)
= = 0.6114 ----- Ans
0.04753 (1.0132 − 0.0296)
pa
(f) We know that, ρa =
Ra Ta
Ra = 287 J/kg-K
Ta = DBT
pa = Partial pressure of dry air = pb – pv
pa = 1.0132 – 0.0296 = 0.9836 bar
0.9836 x 10 5
ρa = = 1.1236 kg / m 3 ----- Ans
287 x (32 + 273)
(g) Density of water vapour can be found out from, W = ρv / ρa
Therefore, ρv = 0.01872 x 1.1236 = 0.021 kg/m3 ----- Ans
(h) h = C pa Td + W [ h gd + 1.88(Td − Tdp )]
At pv = 0.0296 bar from steam table, Tdp = 24oC ----- Ans
h = 1.005 (32) + 0.01872 [2560 + 1.88(32 − 24)]

19
 = 80.365 kJ/kg dry air ---- Ans
8. Air at 20 C, 40 % RH is mixed adiabatically with air at 40oC, 40 % RH in the ratio of 1 kg
o

of the former with 2 kg of latter (on dry basis). Find the final condition of air. Draw the
process in chart also as diagram.
Given:
DBT of I-stream (Td1) = 20oC
RH of I-stream (Φ1) = 40 %
DBT of II-stream (Td2) = 40oC
RH of II-stream (Φ2) = 40 %
ma1/ma2 = 1/2
Mass of dry air in the I-stream (ma1) = 1 kg
Mass of dry air in the II-stream (ma2) = 2 kg
Required: Final condition of air (h3, W3)
Solution:
1

We can write,
ma1 h1 + ma2 h2 = (ma1 + ma2) h3
ma1 W1 + ma2 W2 = (ma1 + ma2) W3

h2

h3
h1 2 W2
3 W3

1 W1

20
 Td1 Td3 Td2
Using Steam table and equations
p v1
φ1 =
p s1
ps1 = 0.02337 bar at Td1 = 20oC from steam table
pv1
0.4 =
0.02337
pv1 = 0.009348 bar
p v1
W1 = 0.622
pb − p v1
0.009348
= 0.622 x = 0.0058 kg / kg dry air
1.0132 − 0.009348
pv 2
φ2 =
ps2
ps2 = 0.07375 bar at Td2 = 40oC from steam table
pv 2
0.4 =
0.07375
pv2 = 0.0295 bar
pv 2
W2 = 0.622
pb − p v 2
0.0295
= 0.622 x = 0.01865 kg / kg dry air
1.0132 − 0.0295
h1 = C pa Td 1 + W1 [ h gd 1 + 1.88 (Td 1 − Tdp1 )]
At pv1 = 0.09348 bar from steam table, Tdp1 = 6oC
hgd1 = 2538.2 kJ/kg
h1 = 1.005 (20) + 0.0058[2538.2 + 1.88(20 − 6)]
= 34.97 kJ/kg dry air
h2 = C pa Td 2 + W2 [ h gd 2 + 1.88 (Td 2 − Tdp 2 )]
At pv2 = 0.0295 bar from steam table, Tdp2 = 24oC
hgd2 = 2574.4 kJ/kg
h2 = 1.005 (40) + 0.01865[2574.4 + 1.88(40 − 24)]
= 88.8 kJ/kg dry air
1 (34.97) + 2 (88.8) = (1 + 2) h3
h3 = 70.86 kJ/kg dry air ----- Ans
1 (0.0058) + 2 (0.01865) = (1 + 2) W3

21
 W3 = 0.01436 kg/kg dry air ----- Ans
Using chart
• Locate the point (1) on the chart at Φ1 = 40% and Td1 = 20oC
• Locate the point (2) at Φ2 = 40% and Td2 = 40oC
• Get h1, h2, W1 and W2 from chart
h1 = 35 kJ/kg dry air W1 = 0.0058 kg/kg dry air
h2 = 90 kJ/kg dry air W2 = 0.0187 kg/kg dry air
1 (35) + 2 (90) = (1 + 2) h3
h3 = 71.67 kJ/kg dry air ----- Ans
1 (0.0058) + 2 (0.0187) = (1 + 2) W3
W3 = 0.0144 kg/kg dry air ----- Ans
We can also get other properties from chart.
• Locate point (3) on the chart at W3 or h3
• Point (3) lies between (1) and (2) and on the line joining (1) and (2)
• Get WBT and DPT on the saturation curve
• Get DBT on the x-axis
• Get RH and specific volume at point (3)
Tw3 = 23.8oC, Td3 = 33.1oC, Tdp3 = 19.6oC, v3 = 0.887 m3/kg dry air, Φ3 = 46%

WBT
3
DPT
DBT
9. Air-water vapour mixture at 0.1 MPa, 30 C, 80% RH has a volume of 50 m3. Calculate the
o

specific humidity, DPT, WBT, mass of dry air and mass of water vapour.
Given:
Total pressure of mixture (pb) = 0.1 MPa = 1 bar
DBT of mixture (Td) = 30oC
RH of mixture (Φ) = 80%
Volume of mixture (V) = 50 m3
Required: W, DPT, WBT, ma and mw
Solution:
pv
W = 0.622
pb − pv
pv
φ=
ps
22
 ps = 0.04242 bar at Td = 30oC from steam table
pv = 0.8 (0.04242) = 0.03394 bar
0.03394
W = 0.622 x = 0.02185 kg / kg dry air ---- Ans
1 − 0.03394
Tdp = 26oC at pv = 0.03394 bar from steam table ---- Ans
• Using Carrier’s equation, only by trial and error we can find WBT. Therefore it is advisable
to refer chart for obtaining the WBT.
• Locate the point on the chart at 30oC and 80% RH
• Get WBT on saturation curve
Tw = 27oC ---- Ans
To find mass of dry air
pa = pb – pv = 1 – 0.03394 = 0.96606 bar
pa V = ma Ra Ta
5
0.96606 x 10 x 50 = ma x 287 x (30 + 273)
ma = 55.54 kg ----- Ans
To find mass of water vapour
ρ v mv
W= =
ρ a ma
mv = 0.02185 (55.54) = 1.2136 kg ---- Ans
Note: If we know the value of Rw, mw can be found out from, pw V = mw Rw Tv
Ta = Tv =Td
10. Saturated air at 20oC at a rate of 1.16 m3/s is mixed adiabatically with the outside air at
35oC and 50% RH at a rate of 0.5 m3/s. Assuming adiabatic mixing condition at 1 atm,
determine specific humidity, relative humidity, dry bulb temperature and volume flow rate
of the mixture.
Given:
Saturated air, i.e., Φ1 = 100%
DBT of I-stream (Td1) = 20oC
Volume flow rate of I-stream (V1) = 1.16 m3/s
DBT of II-stream (Td2) = 35oC
RH of II-stream (Φ2) = 50%
Volume flow rate of II-stream (V2) = 0.5 m3/s
Total pressure (pb) = 1 atm = 1.01325 bar
Required: W3, Φ3 Td3 and V3
Solution:
1

23
 2

h2
h3
h1
2 W2
3 W3
1 W1

Td1 Td3 Td2

We can write,
ma1 h1 + ma2 h2 = (ma1 + ma2) h3
ma1 W1 + ma2 W2 = (ma1 + ma2) W3
Using chart
• Locate the point (1) on the chart at Φ1 = 100% and Td1 = 20oC
• Locate the point (2) at Φ2 = 50% and Td2 = 35oC
• Get h1, h2, W1 and W2 from chart
h1 = 58 kJ/kg dry air W1 = 0.015 kg/kg dry air v1 = 0.851 m3/kg dry air
h2 = 81.5 kJ/kg dry air W2 = 0.018 kg/kg dry air v2 = 0.898 m3/kg dry air
V1 1.16
m1 = = = 1.3631 kg / s
v1 0.851
V2 0.5
m2 = = = 0.5568 kg / s
v 2 0.898
1.3631 (0.015) + 0.5568 (0.018) = (1.3631 + 0.5568) W3
W3 = 0.0159 kg/kg dry air ----- Ans
• Locate point (3) on the chart at W3
• Point (3) lies between (1) and (2) and on the line joining (1) and (2)
• Get DBT on the x-axis
• Get RH and specific volume at point (3)
From chart, Φ3 = 80% ----- Ans
Td3 = 25oC ------ Ans
v3 = 0.866 m3/kg dry air
V3 = m3 v3 = (1.3631 + 0.5568) (0.866)
= 1.6626 m3/s ----- Ans

24
11. Air at 16oC and 25% RH passes through a heater and then through a humidifier to reach
final DBT of 30oC and 50% RH. Calculate the heat and moisture added to the air. What is
the sensible heat factor?
Given:
Initial DBT of air (Td1) = 16oC
Initial RH of air (Φ1) = 25%
Final DBT of air (Td3) = 30oC
Final RH of air (Φ3) = 50%
Process Æ Heating and humidification
Required: Heat added, Moisture added and SHF
Solution:

h3
h2
3 W3
h1
W 1 = W2
1 2

Td1 Td2 = Td3

Using chart
• Locate the point (1) at Td1 and Φ1
• Locate the point (3) at Td3 and Φ3
• Draw the horizontal line (heating) from (1) and Vertical line (humidification) from (3) and
get intersection point (2)
• Get h1, h2, h3, W1, W2 and W3
h1 = 23 kJ/kg dry air h2 = 38 kJ/kg dry air h3 = 64 kJ/kg dry air
W1 = W2 = 0.0033 kg/kg dry air W3 = 0.0134 kg/kg dry air
Heat added = SH + LH = (h2 – h1) + (h3 – h2)
= h3 – h1
= 64 – 23 = 41 kJ/kg dry air ---- Ans
Moisture added = W3 – W2 = 0.0134 – 0.0033
= 0.0101 kg/kg dry air ----- Ans
SH h2 − h1 h −h
Sensible heat factor (SHF) = = = 2 1
SH + LH (h2 − h1 ) + (h3 = h2 ) h3 − h1

25
 38 − 23
= = 0.366 ----- Ans
64 − 23
12. For a hall to be air conditioned, the following conditions are given:
Outdoor condition Æ 40oC DBT, 20oC WBT
Required comfort condition Æ 20oC DBT, 60% RH
Seating capacity of hall = 1500
Amount of air supplied = 0.3 m3/min per person
If the required condition is achieved first by adiabatic humidification and then by cooling,
estimate (a) capacity of the cooling coil in tones, and (b) the capacity of the humidifier.
Given:
Process Æ Adiabatic humidification and Cooling
Before adiabatic humidification
DBT of air (Td1) = 40oC
WBT of air (Tw1) = 20oC
After adiabatic humidification
RH of air (Φ2) = 60%
After cooling (sensible)
DBT of air (Td3) = 20oC
Amount of air supplied (Va1) = 0.3 m3/min per person
Seating capacity = 1500
Required: (a) Capacity of cooling coil (b) Capacity of humidifier
Solution:

h2 = h1
h3
3 2 W2 = W3
1 W1

Td3 Td2 Td1

• Locate the point (1) at Td1 = 40 C and Tw1 = 20oC
o

• Locate the point (3) at Td3 = 20oC and RH = 60 %

• During adiabatic humidification, h = constant. During cooling, W = constant. Therefore
follow the constant WBT line from point (1) and constant W line from point (3) to get point
(2).
From chart, h1 = h2 = 57 kJ/kg dry air

26
 h3 = 42 kJ/kg dry air
W1 = 0.0065 kg/kg dry air
W2 = W3 = 0.0088 kg/kg dry air
v1 = 0.896 m3/kg dry air
(a) Capacity of cooling coil = ma (h2 – h3)
Volume flow rate of air (Va1) = 0.3 x 1500/60 = 7.5 m3/s
Mass flow rate of air (ma) = Va1/v1 = 7.5 / 0.896 = 8.37 kg/s
Capacity of cooling coil = 8.37 (57 – 42) = 125.55 kJ/s
= 125.55 / 3.89 = 32.27 TR ---- Ans
Capacity of humidifier = ma (W2 – W1) = 8.37 (0.0088 – 0.0065)
= 0.019251 kg/s
= 69.3 kg/h ----- Ans

27
 ME18301 ENGINEERING THERMODYNAMICS
SOLUTIONS TO THE PROBLEMS- UNIT-I

2. 1 Kg of air at a pressure of 1bar and 25 oC is heated at constant volume till the pressure is doubled.
It is then expanded isothermally to the original pressure and then cooled to the initial condition at
constant pressure. Show the process on P-V and T-S diagrams and calculate the work and heat
interactions during the cycle. (AU May 2006)
Data Given
m=1kg 2 3 2
P1=1bar=1x105 N/m2
T1=25 oC=298K T P
 P2 
   2 1 1 3
 P1 
S V
To Find: work and Heat transfer
Solution:
PROCESS 1-2 – CONSTANT VOLUME PROCESS

During this process work done is zero, since dV=0

Heat transfer=change in Internal Energy
U= m Cv (T2 –T1)
To Find T2:
T2  P2 
For constant Volume process,  
T1  P1 
T2 = 2  x298 =596K
U=1x712x(596-298)=212.176kJ
Q=212.176kJ
PROCESS 2-3 – CONSTANT TEMPERATURE PROCESS

In Iso- thermal process change in Internal Energy is zero. Hence Heat transfer=work transfer
Work transfer W2-3= Q2-3 = mR T2 ln (V3/ V2) = mR T2 ln (P2/ P3)
P 
P3= P1 , Since  2   2 & P1= 1bar , P2=2bar
 P1 
Therefore W2-3= Q2-3=1x287x596 ln(2)=118.564kJ
PROCESS 3-1 – CONSTANT PRESSURE PROCESS
Work Transfer= W3-1= P3(V1 –V3)=mR(T1 –T3)
T1 is given . Since 2-3 is isothermal process T2=T3
W3-1=1x287x(298-596)= -85.526kJ
Heat transfer Q=mx Cpx(T1 –T3)=1x1005x(298-596)= -299.490kJ

Net Heat Transfer During The Cycle=§Q=Q1-2 +Q2-3 +Q3-1

=212.17+118.56+(-299.49)
§Q =31.24kJ
According to First law of thermodynamics For a Cyclic process
Net Heat Transfer=Net work Transfer

1
4. Air at 1.02 bar, 22oC, initially occupying a cylinder volume of 0.015 m3, is compressed
reversibly and adiabatically by a piston to a pressure of 6.8 bar. Calculate: (a) The final
temperature (b) The final volume (c) The work done on the mass of air in the cylinder.
Data Given :
2
P1=1.02 bar=1.02x105 N/m2
V1=0.015m3 P
T1=22oC=22+273=295K
P2=6.8 bar=6.8x105 N/m2 1

V
To Find:
T2 , V2 &W
SOLUTION:
 1  .41.4 
T2  P2   6.8 
  = 
T1  P1   1.02 
.286 
 6.8 
T2 =   x295 =507.53K
 1.02 


 P1  V 2 
    
 P2   V1 
 1 
  1
 P  
x V1 =  1.02 
1.4
V2   1  -3
x.015 =3.871x10 m
3

 P2   6.8 
To find mass of Air:

P1 V1=mRT1
P1 V1 1.02x105 x 0.015
m= = = 0.0181kg
T1 287 x295

Workdone during reversible Adaiabatic process = m Cv (T2 –T1)

=0.0181x718x(295-507.5)
W = -2761.61kJ
Work done on the system is 2761.61kJ

2
6. 1 kg of gas at 1.1 bar,27 oC is compressed to 6.6bar as per the law PV1.3=constant. Calculate work
and heat transfer, if
(1) When the gas is ethane (C2H6) with molar mass 30kg/k mol and Cp of 2.1kJ/kgK
(2) When the gas is argon (Ar) with molar mass 40kg/k mol and Cp of 0.52kJ/kgK ( AU DEC2005)

Data Given:
m=1kg
5 2
P1=1.1bar=1.1x10 N/m
o
T1=27 C=300K
5 2
P2=6.6bar=6.6x10 N/m
1.3
PV =constant,n=1.3

To Find: Heat and Work Transfer For Ethane and argon

Solution:
For Polytropic Process, Work Transfer is
m R (T1 –T2)
W 1-2 =
n-1

Heat transfer during Polytropic process: Q1-2 = W1-2 x (-n/-1)

Where =adiabatic Index
n=Index of Compression
To find Final Temperature after Polytropic Compression:
n 1n  .31.3
T2  P2   6.6 
  = 
T1  P1   1.1 
.31.3
 6.6 
T2 =   x300 =453.81K
 1.1 

Gas Constant R=Universal Gas Constant/ Molecular weight

For the gas Methane:

For Methane M=30;Cp=2.1kJ/kgK (Given)
R=8314/30=277.13J/kg.moleK
Cv= Cp-R=2100-277.13=1822.87J/kgK
= Cp/ Cv=2100/1822.87=1.15
1x 277.13x (300  453.81)
W 1-2= = -142.084kJ
0.3
Q1-2= (-142.084) x (1.15-1.3)/0.15
Q1-2 = 702kJ
For the gas Argon:
For argon M=40;Cp=0.52kJ/kgK (Given)
R=8314/40=207.85J/kg.moleK
Cv= Cp-R=520-207.85=312.15J/kgK

3
= Cp/ Cv=520/312.15=1.67
1x 207.8513x (300  453.81)
W 1-2= = -106.565kJ
0.3
Q1-2= (-106.565) x(1.67-1.3)/0.67
Q1-2 = -58.84kJ
8. A certain quantity of air initially at state 1 has a pressure of 5bar,volume of 0.43 m3 and
temperature of 227 oC . Heat is added at constant pressure till the air reaches state 2 where its
volume is three times that at state 1.It is now expanded to state 3 according to PV1..3=constant.
An isothermal process brings the air from state 3 to state 1 completing a cycle. (i) Sketch the
cycle on p-v diagram (ii) Find heat rejected /received during each of the three processes (iii)
determine the network done during the cycle

1 P=C 2
Data Given PV1.3=C
P PV=C
P1=5 bar=5x105 N/m2 3

V1=0.43m3
T1=227oC=227+273=500K
V2=3V1
V
To Find
Q1-2 , Q2-3 , Q3-1 & Net Work done
Solution

To find Mass of Air

P1 V1=mRT1

P1 V1 5x105 x 0.43
m= = 1.50kg
RT1 287 x500
To find Q1-2 :
Heat transfer during Constant Pressure Process=mCp(T2 –T1)

For the process 1-2 Q1-2 =1.5 x1005 x (1500-500)

=1507.5 kJ
(T2/ T1) = (V2/ V1)
Heat received during the process 1-2 is 1507.5 kJ
T2 =(V2/ V1) x T1
= 3x 500=1500K
To find Q2-3 :

Heat transfer during Polytropic process PV1.3=C: Q2-3 = W2-3 x (-n/-1)

m R (T2 –T3)
Work done during the process W2-3 = =1.5 x 287 x (1500 –500)
n-1 =1435.005 kJ
0.3

4
 Q2-3 = 1435.005 x (1.4-1.3/0.4)
=358.775kJ
Heat received during the process 2-3 is 358.775 kJ
To find Q3-1 :
Heat transfer during Isothermal process = Work done during that process

Q3-1 = W3-1= mR T3 ln (V1/ V3) = mR T3 ln (P3/ P1)

[V1/ V3= P3/ P1]

To find P3: =1.5 x 287 x 500 x ln (0.043/5)

In the process 2-3 Q3-1 = - 1023.735 kJ

(T3/T2) = (P3/ P2)(n-1/n)

P3 =(T3/T2) (n/n-1) x P2

= (500/1500)(1.3/0.3) x 5
=0.043 bar

Heat rejected during the process 3-1 is 1023.735 kJ

THE NETWORK DONE DURING THE CYCLE :

According to First law of Thermodynamics
 W   Q
The Network done during the cycle = The Heat transfer during the cycle
=1507.5 kJ + 358.775kJ - 1023.735 kJ
=842.54 kJ
Net work done during the cycle 1-2-3-1 is 842.54 kJ

5
In a certain steady flow process, 12kg of fluid per minute enters at a pressure of
1.4bar,density25kg/m3 ,velocity 120m/sec and internal energy 920kJ/kg. The fluid properties
at exit are pressure 5.6bar,density5kg/m3 , velocity 180m/sec and internal energy 720kJ/kg.
During the process, the fluid rejects 60kJ/sec of heat and rises through 60m. Determine work
done during the process in kW.
Data Given:
m= 12kg/min=0.2kg/sec
5.6bar,
5kg/m3
At Entry: 180m/sec,
720kJ/kg
P1=1.4bar
1=25kg/m3:v1=1/1 STEADY FLOW Exit
V1=120m/sec DEVICE
U1=920kJ/kg=920 x103J/kgZ2 –Z1= 60m
At Exit: 1.4bar, 25kg/m3, 120m/sec
P2=5.6bar Entry 920kJ/kg
2=5kg/m3:v2 =1/2
V2=180m/sec
U2=720kJ/kg=920 x103J/kg
Q=60kJ/s
Q= -60kJ/sec= -60 x103J/sec
Z2 –Z1= 60m
To Find: W
Solution:
Applying steady flow energy equation

Q + m (U1+ P1 v1 +V12/2 +gZ1) = m (U2+ P2 v2 +V22/2 +gZ2) +W

 V1 2  V2 2 
W= m [ U 1  U 2   P1v1  P2 v 2      g Z 1  Z 2  ] + Q
 2 
 
 1.4 5.6  120 2  180 2
=0.2 [ 920  720  10 3      10 
5
 9.81 60 ] –60x103 (v1=1/1:
 25 5  2
v2 =1/2)
3 5 3
=0.2[200x10 + (-1.064x10 ) +(-9000) +(-588.6)]- 60x10

=0.2[200x103 –115988.6] - 60x103

W = -43.198 kW

WORK INPUT TO THE STEADY FLOW DEVICE IS 43.198kW

6
ADDITIONAL PROBLEMS
1. A system receives 200kJ of energy as heat at constant volume. Then it is cooled at constant
pressure when 50kJ of work was done on the system while it rejects 70kJ of heat. Supposing
the system is restored to the initial state by an adiabatic process, how much of work and heat
will be done by the system. (AU May 2006)
3
Data Given: 2

Q1-2= 200kJ
W2-3= -50kJ P
Q2-3= -70kJ
1
To Find:
Heat and work interactions in the process 3-1 V
Solution:
Since 3-1 is an adiabatic process heat transferred in the process is zero.
Net Heat Transfer During The Cycle=§Q=Q1-2 +Q2-3 +Q3-1
=200-70-0
=130kJ
According to First law of thermodynamics For a Cyclic process
Net Heat Transfer=Net work Transfer
130kJ=W1-2 +W2-3 +W3-1
Since 1-2 is a constant Volume process W1-2=0
130=0 +(-50) +W3-1
W3-1=180kJ
180kJ of work will be done by the system to restore the initial position
2. Air undergoes a cyclic process in a cylinder and piston arrangement. Atmospheric air at 1bar and
27 oC is compressed adiabatically to 10bar,expanded isothermally to initial pressure and bought to
initial condition at constant pressure. Find the change in internal energy; enthalpy change heat
transfer and work transfer for each process and for the cycle. (AU may 2003)

Data Given:
2
m=1kg
5 2 P
P1=1bar=1x10 N/m
o
T1=27 C=300K
5 2 3 1
P2=10bar=10x10 N/m

V
To Find:
Change in Internal energy& Enthalpy ,Heat transfer & work transfer for the process and cycle.
Solution:

7
Process 1-2: Reversible adiabatic Compression
 1  .4 
T2  P2    10  1.4
  = 
T1  P1  1
T2 = 10
.286 
x300 =579.6K

U= m Cv (T2 –T1)= 1x 718x ( 579.6-300) = 200.745kJ/kg

In Reversible adiabatic process, Q=0 & W= -U, Hence
W= -200.745kJ/kg
h= m Cp (T2 –T1)=1x1000x(579.36-300)=279.6kJ/kg
Process 2-3: Isothermal expansion

T3=579.6K ,P3=1bar

In isothermal Process U=0 & h=0

Work transfer W2-3= Q2-3 = mR T2 ln (V3/ V2) = mR T2 ln (P2/ P3)

=1x287x579.6 ln(10/1)= 383.024kJ

Process 3-1: constant pressure process

W3-1= P3 (V1 –V3)=mxR x(T1-T3)=1x287x(300-579.6)= -80.245kJ

Q3-1=m Cp (T1-T3)=1x 1000x (300-579.6)= -279.6kJ

In Constant pressure process h =Q
h= -279.6kJ
U= m Cv (T1 –T3)=1x718x(300-579.6)= -200.745kJ/kg

FOR THE CYCLE

§h=h1-2 +h2-3 +h3-1

=279.6 +0 + (-279.6)
=0
§U=U1-2 +U2-3 +U3-1
= 200.745 +0 +(-220.745)
=0
Since U and H are property of the system its cyclic integral will be always zero.
§W= -200.745 + 383.024 + (-80.245)
§W= 102.034kJ

8
3. A fluid undergoes a reversible adiabatic compression from 0.5Mpa, 0.2m3 to 0.05m3 according to
the law PV1.3 =constant. Determine the change in enthalpy, Internal Energy and entropy. (AU DEC2005)
2
Data Given
V1=0.2m3
P1=0.5MPa =0.5x106 N/m2 P
V2=0.05 m3 T
PV1.3 =constant
To Find: 1
h, U, S
Solution: V S

Work done during adiabatic process is given by

P V  P2V2
W= 1 1
 1
To find P2:
 1.3
 P2   V1   0.2 
     = 
 P1   V2   0.05 
1.3
 0.2  6 6 2
P2=   x0.5x10 =3.03x10 N/m
 0.05 

W=
0.5 x0.2  3.03x0.05x10 6 =  51500 = -171.67kJ
0.3 0.3
Change in Enthalpy h=h2-h1
=(U2+P2V2)-(U1+P1V1)
=(U2- U1)-(P1V1- P2V2)
In Reversible adiabatic Process U= -W
There fore U=(U2- U1)= 171.67kJ
Then h=171.67-(-51500)
h =223.17kJ
For Reversible adiabatic Process S=0