1

EGERTON UNIVERSITY

DEPARTMENT OF INDUSTRIAL AND ENERGY ENGINEERING

COURSE: ITEC 112: THERMO-FLUIDS

LECTURE NOTES

Lecturer: Dr. L. K. Langat

Technologist: Mr. D. Chirchir

SEPTEMBER, 2011
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aircrafts, ships and land vehicles), energy conversion devices

CHAPTER ONE (e.g. fuel cells), energy sources, heat exchangers, production of
low temperatures, etc.
Introduction
Energy exists in many forms from the energy locked up in matter
Like all sciences, the basis of thermodynamics is experimental
itself to the intense radiant heat emitted by the sun, and between
observation and these findings have been formalized into certain
these limits energy sources are available such as chemical energy
basic laws which are known as the Zeroth, First, Second (and
fuels and the potential energy of large water masses evaporated
Third) laws of thermodynamics.
by the sun. in order for this energy to serve man‘s needs
(purpose) means must be found to transform the energy into a
A Heat Engine is the name given to a system, which by operating
convenient form.
in a cyclic manner produces net work from a supply of heat.
Thermodynamics is made up of two words ―Thermo‖ and
―Dynamics‖. These words come from Greek and have the
The laws of thermodynamics are natural hypotheses based on
meanings: ―Thermo‖ = hot or heat; ―Dynamics‖ = the study of
observations of the world in which we live. For example, heat
matter in motion. Thus ―Thermodynamics‖ can be defined as the
and work are mutually convertible forms of energy and this is the
study of heat related to matter in motion, i.e, it is the science of
basis of the first law of thermodynamics. Like a river never flows
the of the relationship between heat, work and the properties of
uphill unaided, heat never flows from an object at a low
systems. This means that it a science of the transformation of
temperature to one at a higher temperature unaided. This is the
energy, and the accompanying change in the state of matter. It is a
basis of the second law of thermodynamics, which can be used to
physical theory of great generality affecting practically every
show that a heat engine cannot convert all the heat supplied to it
phase of human experience. It is based on two master concepts,
into mechanical work but must always reject some at a lower
energy and entropy, and two great principles, the first and second
temperature.
law of thermodynamics.
The two branches of thermodynamics of interest are: (1) classical
Thermodynamic System, Heat and Work.
thermodynamics – which concerns itself with the phenomena that
A Thermodynamic System is a collection of matter within
occur in real macroscopic systems, e.g. changes in volume when
prescribed and identifiable boundaries. The boundaries may be
a gas is heated at constant pressure; and (2) applied (or
flexible (e.g the fluid in the cylinder of a reciprocating engine
engineering) thermodynamics – which is concerned with work
during the expansion stroke) or fixed and either real or imaginary.
producing or utilizing machines, e.g. engines, turbines or
The region outside the boundary is known as the surrounding
compressors together with the working substances used in such
machines. [Other branches of thermodynamics are kinetic theory Boundary Surrounding
and statistical thermodynamics, physical & chemical
thermodynamics.] Thus, Applied thermodynamics is the science
of the relationship between heat work and the properties of System
systems. It is concerned with the means necessary to convert heat
energy from available sources such as chemical fuels or nuclear
piles into mechanical work. Piston
There are three types of thermodynamic systems namely; a closed
The principles of thermodynamics apply in whole or in part to all system, an open system and an isolated system
engineering activities because every engineering operation 1. Closed system: A closed system allows energy
involves an interaction between energy and matter. They are used transfer across the boundaries, but the mass is
in the design, development and analysis of all power producing fixed ; for example the fluid expanding behind a
systems, e.g. internal combustion engines, gas turbines, steam piston of an engine.
power plants, nuclear power plants, refrigeration systems, air
conditioning systems, propulsion systems (for rockets, missiles,
 3

2. Open System: An open system is one in which temperature, internal energy, enthalpy and entropy. Thus, the
there is mass transfer across the boundaries. For state of a system may be represented by a point on a diagram of
example the fluid in a turbine at any instant. system properties as shown in fig 1.2.
3. An Isolated System: This is a system which can
neither exchange mass nor energy with its
P
surrounding e.g. P
Heat – Heat is a form of energy, which is transferred from a body 1 1
at a higher temperature to another body at a lower temperature by
2
virtue of the temperature difference between the bodies. Heat 2

may be said to be a transitory characteristic of a body. For

v T
example, if two systems A and B show no change in their
observable characteristics when they are brought into contact with T T
1
one another then they are in thermal equilibrium. However if a 1

third body C at a higher temperature is brought into contact with

2
A there will be heat transfer from C to A and hence an decrease in 2
the intrinsic energy of C and an increase in the intrinsic energy of
A until they are at equilibrium. v S

This principle of thermal equilibrium is called the Zeroth law of

Pressure: Pressure of a system is the force exerted by the system
thermodynamics. The possibility of devising a means of
on a unit area of its boundaries. Pressure is measured in N/m3 or
measuring temperatures rests upon this principle.
bar (1bar = 105 N/m2). Gauge pressure is the pressure read on a
measuring gauge. Absolute pressure is the sum of the gauge
Work: Work is the product of force and the distance moved in the
pressure and the atmospheric pressure. Vacuum pressure is the
direction of the force. Work can be done on the system by the
pressure below the atmospheric pressure. A barometer is an
surrounding or vice versa. Work is measured in Nm or Nm/kg for
instrument used to measure atmospheric pressure using, for
a unit mass of fluid. Note that heat and work are transitory
example, the height of column of liquid. Standard atmosphere
energies and should not be confused with the intrinsic energy
(101325 N/m2) gives a height of 760 mm of mercury or 10.326 m
possessed by a system.
of water.

Properties of the working fluid

To measure differences in pressure a U-tube manometer can be
In practice, the matter contained within the boundaries of a
used. When the pressure difference is small a liquid of a lower
thermodynamic system can either be liquid, vapour or gas, and is
density can be used and when the difference is even smaller, an
known as the working fluid. At any instant the state of the
inclined manometer can be used. When one end of the
working fluid may be defined by certain characteristics. These
manometer is open to the atmosphere, the pressure measured is
characteristics are referred to as thermodynamic properties of the
known as gauge pressure, which is the pressure above or below
working fluid.
the atmospheric pressure.

To define the state of a thermodynamic system, only two

The actual pressure (also known as absolute pressure) is given by:
thermodynamic properties are required. Any such property must
Absolute pressure = atmosphere pressure ± gauge pressure. When
be measurable and have a unique numerical value when the fluid
gauge pressures are too big such that the mercury manometer
is in any particular state. The value of a property must be
becomes too long, the Bourdon pressure gauge is used.
independent of the process through which the fluid has passed in
Specific Volume: This is the volume occupied by a unit mass of
reaching that state. This implies that a change in the value of a
the system and is denoted by, v, while V denotes the absolute
property depends only on the initial and final states of the system.
volume. Specific volume is measured in m3/kg.
Some of the common properties are pressure, volume,
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Temperature: Though a familiar term, its exact definition is

difficult. It is the degree of hotness of a body or system and is The second law of thermodynamics shows that there is the
measured using a thermometer. A thermometer employs the possibility of an absolute zero of temperature which suggests an
thermometric properties of a substance such as mercury and absolute temperature scale. An absolute zero of temperature
alcohol or an electrical resistance as in the case of a would be the lowest temperature possible and this would
thermocouple. The temperature scales are calibrated in degrees therefore be a more reasonable temperature to adopt as the zero
o o
Fahrenheit ( F) or degrees centigrade ( C). A thermodynamic for a temperature scale.
scale provides a basis for absolute temperature measurement in
Kelvin, K. Normally capital T is used for absolute temperature The absolute thermodynamic temperature scale is called the
while small t is used for other temperatures. Two bodies are said Kelvin scale and on this scale the unit of temperature is Kelvin
to have equal temperatures if there is no net heat transfer between (K) (not oK). The zero on the Celsius scale is defined as 0 0C =
them when they are brought into contact with one another; they 273.15K. The temperatures on the absolute scale are related to the
are then said to be in thermal equilibrium. temperatures on the Celsius scale by K = oC + 273.15, e.g. 373.15
K = 100oC.
The possibility of devising a means of measuring temperature On the absolute scale, there is only one fixed point which is
rests upon the principle of thermal equilibrium which is stated in known as the standard fixed point and is usually chosen as the
the Zeroth law of thermodynamics: If two bodies are separately in triple point of water, the temperature at which ice, liquid water
thermal equilibrium with a third body then they must be in and water vapour coexist in equilibrium.
thermal equilibrium with each other. The triple point of water has been chosen to be 273.16 K exactly
The temperatures of any group of bodies may be compared by and so if X =value of any thermometric property, then on the
bringing a particular system known as a thermometer into contact absolute scale;
with each in turn. The thermometer must posses an easily X
observable characteristic known as a thermometric property, e.g. X tp
Temperature θ K = 273.16
pressure of a gas in a closed vessel; the length of a column of
Where
mercury in a capillary tube; the resistance of a platinum wire, etc.
Xθ = value of thermometric property at temperature θ K
To report temperature, a scale must be devised so that any device
Xtp = value of thermometric property at temperature 273.16 K
used to measure temperature will record the same value when
used in the same conditions.
Reversibility
When a fluid undergoes a reversible process both the fluid and
Scale of temperature
it‘s surrounding can be restored back to their original state.
The two widely used temperature scales are Fahrenheit and
Celsius; the Celsius scale in more common. The lower fixed point
A reversible process is one in which a system changes in such a
on Celsius scale is the temperature of the melting of pure ice (ice
way that at any instant during the process, the state point can be
point) at standard atmosphere = 0oC. The upper fixed point is the
located on a diagram of system properties. A reversible process
temperature at which pure water boils (steam point) at standard
between two states can be drawn as a line on a diagram of
atmosphere = 100oC.
properties as shown below. The fluid undergoing the process
If X = value of any thermometric property, then on the Celsius
passes through a continuous series of equilibrium states
scale,
XT  X0
X 100  X 0
Temperature T oC =
Where
XT = value of property at temperature T
Xo = value of property at 0oC
X100 = value of property at 100oC
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P
(N/m2) 1
Fluid
F
Pressure

V (m/s)
Assume ideal conditions for a reversible process and that the
pressure and temperature of the fluid are uniform. Let the area of
Criteria for reversibility
cross-section of the piston be A and the pressure of the fluid at
A process is said to be reversible when;
any instance be P.
1. The process is frictionless This implies that the fluid
Let the piston move under the action of the force exerted by the
itself must have no internal friction and there must be
fluid pressure, a distance dl to the right.
no mechanical friction (eg. between cylinder and
The work done by the fluid on the piston is given by;
piston)
Work done = Force × Distance Moved
2. The difference in pressure between the fluid and its
= (PA) × dl
surroundings during the process is infinitesimally
small. This means that the process must take place But Adl = dV

infinitely slowly since the force to accelerate the Work done = PdV
boundaries of the system is infinitely small. Where dV is a small increase in volume
3. The difference in temperature between the fluid and its For a unit mass;
surroundings during the process is infinitesimally Work done – Pdv
small. This means that the heat supplied or rejected to Where v is the specific volume.
or from the fluid must be transferred infinitely slowly Hence when a reversible process takes place between states 1 and
2, the work done by a unit mass of fluid is given by;
In practice, all process are irreversible and are usually represented 2

by a dotted line joining the end state to indicate that the

Work done W= 
1
Pdv
intermediate states are indeterminate. The work done by the fluid during any reversible process is given
by the area under the line on a p-v diagram
P
(N/m2)
P
(N/m2) 1
1

P
2
2

V (m/s)
dV V(m3)

Consider an ideal frictionless fluid contained in a cylinder behind When a fluid undergoes a series of processes and finally returns
a piston shown below
to its initial state, then it is said to have undergone a
thermodynamic cycle.
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Example 1. Example 3
3
A fluid of volume 0.05m is contained behind a piston at a 1 Kg of a certain fluid is contained in a cylinder at an initial
pressure of 10 bar. After a reversible expansion at constant pressure of 20 bar. The fluid is allowed to expand reversibly
3
pressure, the final volume is 0.2 m . Calculate the work done by behind a piston according to a law pv2 = constant until the
the fluid. [150 KJ]. volume is doubled. The fluid is then cooled reversibly at constant
pressure until the piston regains the original position; heat is then
Example 2 supplied reversibly with the piston firmly locked in position until
A fluid at a pressure of 3 bar, and with a specific volume of 0.18 the pressure rises to the original value. Calculate the net work
m3/Kg, contained in a cylinder behind a piston expands reversibly done by the fluid for an initial volume of 0.05m3. [25 KJ]
according to a law, pv = C/v2, where C is a constant. Calculate
the work done by the fluid on the piston. [29.85KJ/kg] Example 4
A certain fluid at 10 bar is contained in a cylinder behind a
When a compression process takes place reversibly, the work piston, the initial volume being 0.05m3. Calculate the work
done on the fluid is given by the shaded area. done by the fluid when it expands reversibly;
P a. At a constant pressure to a final volume of 0.2m3
(N/m2) b. According to a linear law to a final volume of 0.2m3 and a
final pressure of 2 bar.
2
c. According to a law pv = constant to a final volume of
0.1m3
P d. According to a law pv3 = constant to a final volume of 0.
1 06m3
 a  b
e. According to a law p   2   to a final
v  v
dV
3
V( m ) volume of 0.1m3 and a final pressure of 1 bar if a and b
are constants.
1 Sketch all the process on a p-v diagram.
Work done on the fluid =  Pdv
2 [150 KJ; 90 KJ; 34.7KJ; 7.64K; 19.2KJ ]
The rule is that a process from left to right on the p-v diagram is
one in which the fluid does work on the surrounding (i.e W is The First Law of Thermodynamics
positive). Conversely a process from right to left is one in which
The hypothesis that states that energy can neither be created or
the fluid has work done on the fluid by the surrounding (i.e. W is
destroyed is basically principle of conservation of energy The
negative).
First Law of Thermodynamics is merely one statement of this
A reversible cycle consisting of four reversible processes 1 to 2, 2
general principle with particular reference to heat energy- and
to 3, 3 to 4, and 4 to1 is shown below.
mechanical energy (i.e. work).
P
1 The first law of thermodynamics which is a statement of the
2 Principle of Conservation of Energy states that:

When a system undergoes a thermodynamic cycle, then the net

heat supplied from its surroundings is equal to the net work done
4 by the system on the surroundings.
3
 dQ   dW
0 0


V
Where is the sum for a complete cycle.
The net work done is equal to the shaded area 0

In the statement of the first law of thermodynamics, it was

assumed that there is no change in the intrinsic energy at the end
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of the cycle. This intrinsic energy is known as internal energy and Then at inlet: At outlet:
is denoted by the symbol u for unit mass or U for mass m of the
Internal energy" = u. Internal energy = u2
fluid. It is a property which depends on the pressure and
Flow work = P1V1 Flow work = P2V2
temperature of the fluid.

If the final internal energy of the system is greater than the initial Kinetic energy (K.E.) = ½ C12 Kinetic energy (K.E.) = ½ C22

internal energy, then the difference between the net heat supplied Potential energy (P.E) = Z1g Potential energy (P.E) = Z2g
and the net work output is the increase in the internal energy. Heat supplied = Q Work Done = W

For steady flow, energy entering must be exactly equal to energy

leaving the system.
Thus Q=(U2 –U1)+ w

For 1 kg,

Q=(u2 –u1)+ W The sum of internal energy and the pv term is known as enthalpy
This equation is known as the Non-Flow Energy Equation and is denoted by the symbol h.
(NFEE). Its differential form can be written as, Thus enthalpy, h = u + pv
dQ = du + dW However, enthalpy for any mass, m, other than the unit mass is
2 denoted H and is given by H = mh
For a reversible non – flow process, W 
 pdv
1
Therefore, neglecting changes in elevation (i.e. Z1g = Z2g) the
equation becomes,
For small quantities, this expression becomes,
C12 C2
dW = pdv u1   p1v1  Q  u 2  2  p2 v2  W
2 2
Substituting this in 3 gives
The rate of mass flow in the steady flow system can be
dQ=du+ pdv
determined from the continuity of mass equation.
Therefore,
CA
2 m
Q  (u 2  u1 )   pdv v
1
where C is the velocity of fluid, A the area of cross section and v
The Steady-Flow Energy Equation, SFEE the specific volume.
In most practical problems, the rate at which a fluid flows through
a machine Or piece of apparatus is assumed constant. This type of Steam As A Working Fluid
flow is known as Steady Flow.
The matter contained within the boundaries of a thermodynamic
Consider a steady flow of one kg of fluid through a piece of system is defined as the working fluid. The working fluid may be
apparatus in figure below. liquid, vapour, or a gas. Some of the common working fluids are
water, refrigerants, and air

Consider a p-v diagram for any substance. When a liquid is

heated at anyone constant pressure there is one fixed temperature
at which bubbles of vapour form in the liquid; this phenomenon
is known as boiling. The higher the pressure of the liquid then the
higher the temperature at which boiling occurs. It is also found
that the volume occupied by 1 kg of a boiling liquid at a higher
pressure is slightly larger than the volume occupied by 1 kg of the
Let Q be the heat supplied to the system per kg of fluid and W be
same liquid when it is boiling at a low pressure. A series of
the work done by the fluid as it passes through the apparatus.
boiling points plotted on a p-v diagram will appear as a sloping
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line, as shown in below. The points P, Q, and R represent the

boiling points of a liquid at pressures PP,PQ and PR a respectively.

When a liquid at boiling point is heated further at constant

pressurhe te additional heat supplied changes the phase of the
substance from liquid to vapour; during this change of phase the
pressure and temperature remain constant. The heat supplied is
called the latent heat of vaporization. It is found that the higher
the pressure then the smaller is the amount of latent heat required.
There is a definite value of specific volume of the vapour at
anyone pressure, at the point at which vaporization is complete,
hence a series of points such as p', Q', and R' can be plotted and Lines of constant temperature, called isothermals, can be plotted
joined to form a line. on a p-v diagram as shown in fig. 3.4. The temperature lines
become horizontal between the saturated liquid line and the
saturated vapour line (e.g. between P and p', Q and Q', Rand R').
Thus there is a corresponding saturation temperature for each
saturation pressure. At pressure PP the saturation temperature is
T1, at pressure pQ the saturation temperature is T2, and at pressure
PR the saturation temperature is T3• The critical temperature line
Tc just touches the top of the loop at the critical point C.

When a dry saturated vapour is heated at constant pressure its

temperature rises and it becomes superheated. The difference
When the two curves already drawn are extended to higher pres- between the actual temperature of the superheated vapour and the
sures they form a continuous curve, thus forming a loop. The saturation temperature at the pressure of the vapour is called the
pressure at which the turning point occurs is called the critical degree of superheat. For example, the vapour at point S is
pressure and the turning point itself is called the critical point superheated at pQ and T3, and the degree of superheat is T3 - T2•
(point C on the diagram). It can be seen that at the critical point The condition or quality of a wet vapour is most frequently
the latent heat is zero. The substance existing at a state point defined by its dryness traction, and when this is known as well as
inside the loop consists of a mixture of liquid and dry vapour and the pressure or temperature then the state of the wet vapour is
is known as a wet vapour. fully defined.
A saturation state is defined as a state at which a change of phase Dryness fraction, x = The mass of dry vapour in 1 kg of the
may occur without change of pressure or temperature. Hence the mixture
boiling points P, Q, and R are saturation states, and a series of
(Sometimes a wetness fraction is defined as the mass of liquid in
such boiling points joined up is called the saturated liquid line.
1 kg of the mixture, i.e. Wetness fraction = (1-x.)
Similarly the points P', Q', and R', at which the liquid is
Note that for a dry saturated vapour, x = 1; and that for a
completely changed into vapour, are saturation states, and a series
saturated liquid, x = 0.
of such points joined up is called the saturated vapour line. The
word saturation as used here refers to energy saturation. For
example, a slight addition of heat to a boiling liquid changes The use of vapour tables
some of it into a vapour, and it is no longer a liquid but is now a Tables of properties of different working fluids are available (e.g.
wet vapour. Similarly when a substance just on the saturated those by Mayhew and Rogers) which can be used to determine
vapour line is cooled slightly, droplets of liquid will begin to the state properties during any thermodynamic process. In such
form, and the saturated vapour becomes a wet vapour. A saturated tables, the properties are designated with subscripts f for saturated
vapour is usually called dry saturated to emphasize the fact that no liquid; g for dry saturated; and hfg for the intermediate stages.
liquid is present in the vapour in this state. Superheat tables contain properties for superheated vapour.
 9

h at 0·01 °e and 0.006112 bar = 0  0.006112 10  0.0010002

Saturation state properties 5

The saturation pressures and corresponding saturation 103

temperatures of steam are tabulated in parallel columns in the first (where Vf at O'01°e is 0·0010002 m3/kg)
table, for pressures ranging from 0·006112 bar to the critical i.e h = 6.112 × 10-04 kJ/kg
pressure of 221·2 bar. The. specific volume, internal energy,
This is negligibly small and hence the zero for enthalpy may be
enthalpy, and entropy are also tabulated for the dry saturated
taken as 0.01oC.
vapour at each pressure and corresponding saturation
Note that at the other end of the pressure range tabulated in the
temperature. The suffix g is used to denote the dry saturated state.
first table the pressure of 221·2 bar is the critical pressure,
The following are some of the formulae used to calculate the
374·15°e is the critical temperature, and the latent heat, hfg, is
properties of a wet vapour:
zero.
(i) Volume, v, is given by
Properties of wet vapour
v = vf (1-x) + x vg
For a wet vapour the total volume of the mixture is given by the
Where x is the dryness fraction, vf volume of the liquid and vg the volume of liquid present plus the volume of dry vapour present.
volume of the dry saturated vapour.
Properties of superheated vapour
For most practical problems, volume of the liquid is usually
For steam in the superheat region temperature and pressure are
negligibly small compared with the volume of dry saturated
independent properties. When the temperature and pressure are
vapour.
given for superheated steam then the state is defined and all the
Therefore, v = xvg other properties can be found. For example, steam at 2 bar and
u = (l-x)uf + xug = uf + x(ug - uf) 200°C is superheated since the saturation temperature at 2 bar is

(iii.) Similarly enthalpy, h, is given by 120'2°C, which is less than the actual temperature. The steam in
this state has a degree of superheat of 200-120'2=79'8 K. The
h = (l-x)hf + xhg
tables of properties of superheated steam range in pressure from
= hf + x(hg-hf) = hf +xhfg
0·006112 bar to the critical pressure of 221·2 bar, and there is an
The change in specific enthalpy from hf to hg is given the symbol additional table of supercritical pressures up to 1000 bar. At each
hfg• pressure there is a range of temperatures up to high degrees of
When saturated water is changed to dry saturated vapour, from superheat, and the values of specific volume, internal energy,
equation enthalpy, and entropy are tabulated at each pressure and

Q=(u2-u1) + W = (ug – uf) + W temperature for pressures up to and including 70 bar; above this
pressure the internal energy is not tabulated. For reference the
Also W is. represented by the area under the horizontal line on the
saturation temperature is inserted in brackets under each pressure
P-V diagram,
in the superheat tables and values of Vg, Ug, hg and Sg are also given.
W= (vg – vf)p
Interpolation
So that Q = (ug-uf) + p(vg-vf)
For properties which are not tabulated exactly in the tables it is
= (ug + pvg) – (uf + pvf)
necessary to interpolate between the values tabulated.
But h = u + pv

Therefore Q = hg – hf = hfg
The Perfect gas and the Characteristic gas equation
Thus hfg is the heat required to change a saturated liquid to a dry
In practice, there is no perfect gas, however many gases tend
saturated vapour and is called the latent heat.
towards a perfect condition. A perfect gas is an imaginary ideal
In the case of steam tables, the internal energy of saturated liquid gas which obeys the law,
is taken to be zero at the Triple point (Le. at 0·01 °e and
PV
0·006112 bar). Then since, from equation 2.7, h=u+pv, we have,  Cons tan t  R
T

Where p and V are the pressure and volume ofthe gas

 10

respectively and R is the specific gas constant In a constant pressure process, the work done by the fluid is

PV=RT given by,

For gas occupying mass m kg, occupying volume V m3 the W = p(V2-V1) = mR(T2-T1)

equation becomes, Substituting this in the N.F.E.E we have

PV=mRT Q= (U2-U1)+W = mCv(T2-T1) + mR(T2-T1) =m(Cv+R)(T2-T1)

Another form of the characteristic equation is derived from the But Q = mCp(T2-T1), hence
kilogrammole. The kilogramme-mole is defined as a quantity of a Q = mCp(T2-T1) = m(Cv+R)(T2-T1)
gas equivalent to M kg of the gas (M is the molecular weight of
So that Cp = Cv + R
the gas.) Thus for oxygen gas whose molecular weight is 32, 1
And R = Cp - Cv
kg-mole is equivalent to 32 kg of oxygen.
Also the ratio of the specific heats is
From the of the kilogram-mole, for m kg of a gas
Cp
m=nM  
Cv
Where n is the number of moles
Also from Cp – Cv = R
Cp R
Hence, pV= nMRT or MR 
pV 1 
nT Cv Cv

However, V/n is the same for all gases at the same pressure, p,
R so that C  R
  1 
and temperature, T (Avogadro's hypothesis,). Hence the quantity Cv
v
 1
pV/nT is constant for all gases. This constant denoted by the
symbol Ro' is called the universal gas constant.
R
And Cp = ɤCv =
 1
pV=nRoT

From MR=Ro' R= Ro/M

Applications of the First Law of Thermodynamics
From empirical information, the volume of 1 mole of any perfect
gas at pressure of 1× 105 N/m2 and temperature 0oC is To appreciate the applications of the concepts relating to
approximately 22.71 m3• Therefore Ro is given by;
the first law of thermodynamics we now consider
pV 110  22.71
5
processes which are approximated to in practice.
Ro    8314.14Nm / moleK
nT 1 273.15

Reversible Non-flow processes

Relationship between specific heats

Specific heat is generally defined as Constant volume process

dQ=mCpdT
In a constant volume process the working substance is
For a perfect gas, the specific heat at constant pressure, cp' and the
contained in a rigid vessel, hence the boundaries of the
specific heat at constant volume, cy' can be assumed to be
constant at all pressures and temperatures. system are immovable and no work can be done on or by

Hence for a reversible non-flow process at constant pressure,

the system, other than paddlewheel work input. It will be
assumed that' constant volume' implies zero work-unless
Q = mCp(T2-T1)
stated otherwise.
And for a reversible non-flow process at constant volume,

Q = mCv(T2-T1)
From the non-flow energy equation,
It can be shown that for any process for a perfect gas, between
states 1 and 2 Q = (u2-u1)+ W
U2-U1 = mCv(T2-T1)
 11

Since no work is done, we therefore have P

P

Q = u2-u1 P1=P2 P1=P2

or for mass, m, of the working substance, Q = U2-U1

v1 V2 v1 V2
V(m3)
All the heat supplied in a constant volume process goes to
Fig ( c) Fig ( d)
increasing the internal energy.

From the non-flow energy equation, 2.2,

Fig a Fig b Hence for a reversible constant pressure process

A constant volume process for a vapour is shown on a p-v

diagram is shown in fig. (a). The initial and final states
Now enthalpy, h = u +pv, hence,
have been chosen to be in the wet region and superheat
region respectively. In fig.(b) a constant volume process is
or for mass, m, of a fluid,
shown on a p-v diagram for a perfect gas. For a perfect gas
we have,

Q = U2-U1 = mcV(T2 - T1) A constant pressure process for a vapour is shown on a p-v
diagram below. The initial and final states have been
Constant pressure process chosen to be in the wet region and the superheat region
respectively. In fig. (d) a constant pressure process for a
In a constant volume process the boundary of the system is
perfect gas is shown on a p-v diagram.
inflexible and thus, the pressure rises when heat is
supplied. Hence for a constant pressure process the For a perfect gas we have
boundary must move against an external resistance as heat
is supplied; for instance a fluid in a cylinder behind a Q = mCp(T2 - T1)
piston can be made to undergo a constant pressure process.
Since the piston is pushed through a certain distance by the Constant temperature or isothermal process

force exerted by the fluid, then work is done by the fluid

A process at constant temperature is called an isothermal
on its surroundings.
process. When a fluid in a cylinder behind a piston
expands from a high pressure to a low pressure there is a
 12

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 But C= PV = P1V1 = P2V2
isothermal compression heat must be removed from the
fluid continuously during the process. An isothermal
process for a vapour is shown on a P-V diagram in fig. (e). v 2 P1
Also since P1V1 = P2V2 then 
v1 P2

For a mss m of the gas

But for a unit mass of gas, P1 V1 = RT

P1
Fig (e) Fig ( f)
Hence, we have, W = RT ln . or for mass, m, of the gas
P2

From state 1 to state A the pressure remains at PI, since in P1

W = mRT ln
the wet region the pressure and temperature are the P2
corresponding saturation values. It can be seen therefore For a perfect gas from Joule's law, we have, U=CvT
that an isothermal process for wet steam is also at constant
pressure and heat supplied from state 1 to state A per kg of Hence for an isothermal process for a perfect gas, since T2

steam =hA –h1 In the superheat region the pressure falls to = T1 then U2-Ul = 0

P2 as shown
i.e. the internal energy remains constant in an isothermal
process for a perfect gas.
Fig. (f) shows an isothermal process for a perfect gas on a
From the non-flow energy equation for an isothermal
p-v diagram. The equation of the process is pv=constant,
process for a perfect gas.
which is the equation of a hyperbola. It must be stressed
that an isothermal process is only of the form pv = constant
for a perfect gas, because it is only for a perfect gas that an
equation of state, pv = RT, can be applied.

The work done by a perfect gas in expanding from state 1 Hence in an isothermal process for a perfect gas the heat
to state 2 isothermally and reversibly is given by the flow is equivalent to the work done.
shaded area on fig.(f) shown. Reversible adiabatic non-flow process
An adiabatic process is one in which no heat is transferred
to. or from the fluid during the process. Such a process can
be reversible or irreversible. For a reversible adiabatic non-

flow process
and for an adiabatic process Q = 0
 13

Therefore we have but

In an adiabatic expansion, the work done by the fluid is at

the expense of a reduction in the internal energy of the
fluid. Similarly in an adiabatic compression process all the
work done on the fluid goes to increasing the internal
energy of the fluid. For an adiabatic process to take place,
perfect thermal insulation for the system must be available.
For a vapour undergoing a reversible adiabatic process the
work done can be found by evaluating U1 and U2 from
tables and applying the foregoing equation. In order to fix
state 2, use must be made of the fact that the process is
reversible and adiabatic.
From the non-flow energy equation
We therefore have a simple relationship between P and v
for any perfect gas undergoing a reversible adiabatic
Also for a reversible process dW = p dv, hence for an process, each perfect gas having its own value of γ
adiabatic process From, pv=RT, we have,

u—cvT or du_—c,dT

Dividing through by T to give a form that can be Therefore for a reversible adiabatic process for a perfect

integrated, gas between states 1 and 2 we have:

we have T= (pv)/R, therefore substituting,

Dividing through by cv The work done in an adiabatic process is; W = (u2 – u1).
And for a perfect gas the gain in internal energy is;
 14

For a process in which pvn = constant, we have p = c/vn,

where c is a constant.

n
P1  v 2 
n n
 
P2  v1 
But, c = PV = P1V1 = P2V2 and

For a perfect gas;

Hence

A reversible adiabatic process for a perfect gas is shown on

a p-v diagram in fig. above. The work done is given by the
shaded area, and this area can be evaluated by integration, It can be seen that these equations are exactly similar to the
equations for a reversible adiabatic process for a perfect
gas. Hence the reversible adiabatic process for a perfect
gas is a particular case of a polytropic process with the
index, n, equal to γ.
Therefore, since pvγ = constant, c, then
Hence;

For a perfect gas expanding polytropically it is sometimes

more
convenient to express the work done in terms of the
temperatures at the end states. From
Polytropic processes
Many processes in practice approximate to a reversible law
and
of the form pv = constant, where n is a constant. Both
vapours and perfect gases obey this type of law closely in
many non-flow processes. Such processes are internally
reversible.
For any reversible process,

From the non-flow energy equation, the heat flow during

 15

the process is given by,

Hence substituting,

Thus,

Similarly, I to A‘ is constant pressure heating; 1 to B‘ is

In an expansion, work is done by the gas, and hence the isothermal expansion; 1 to C‘ is reversible adiabatic

term W is positive. Thus when the polytropic index n is expansion; 1 to D‘ is constant volume cooling. Note that,

less than γ, in an expansion, then the right-hand side of the since y is always greater than unity, then process 1 to C

equation is positive (i.e. heat is supplied during the must lie between processes 1 to B and 1 to D, similarly,

process). Conversely, when n is greater than γ in an process 1 to C‘ must lie between processes 1 to B‘ and 1 to

expansion, then heat is rejected by the gas. Similarly, the D‘.

work done in a compression process is negative, therefore

For a vapour a generalization such as the above is not
when n is less than γ, in compression, heat is rejected; and
possible.
when n is greater than γ, in compression, heat must be
supplied to the gas during the process.
One important process for a vapour should be mentioned
N.B γ for all perfect gases has a value greater than unity.
here. A vapour may undergo a process according to a law
In a polytropic process the index n depends only on the
pv = constant. In this case, since the characteristic equation
heat and work quantities during the process. The various
of state, pv = RT, does not apply to a vapour, then the
processes considered earlier are special cases of the
process is not isothermal. Tables must be used to find the
polytropic process for a perfect gas. For example,
properties at the end states, making use of the fact that P1v1
= P2v2.

Irreversible processes

In processes in which a fluid is enclosed in a cylinder

behind a piston, friction effects can be assumed to be

This is illustrated on a p-v diagram in fig.below. negligible. Certain processes cannot be assumed to be
 16

internally reversible, and the important cases will now be initial temperature is equal to the final temperature.
briefly discussed.
Throttling
Unresisted, or free, expansion
A flow of fluid is said to be throttled when there is some
This process was mentioned in Section 1.5 in order to restriction to the flow, when the velocities before and after
show that in an irreversible process the work done is not the restriction are either equal or negligibly small, and
given by J P dv. Consider two vessels A and B, when there is a negligible heat loss to the surroundings.
interconnected by a short pipe with a valve X, and The restriction to flow can be a partly open valve, an
perfectly thermally insulated. Initially let the vessel A be orifice, or any other sudden reduction in the cross-section
filled with a fluid at a certain pressure, and let B be of the flow.
completely evacuated. When the valve X is opened the
fluid in A will expand rapidly to fill both vessels A and B. An example of throttling is shown in fig below. The fluid,

the pressure finally will be lower than the initial pressure in flowing steadily along a well-lagged pipe, passes through

vessel A. This is known as an unresisted expansion or free an orifice at section X. Since the pipe is well lagged it can

expansion be assumed that no heat flows to or from the fluid.

The process is not reversible since external work would

have to be done to restore the fluid to its initial condition.

Q = (u2-u1) + W

Applying the steady flow equation between any two

sections of the flow,

Now in this process no work is done on or by the fluid, Now since Q=O, and W=O, then,
since the boundary of the system does not move. No heat
flows into or from the fluid since the system is lagged. C12 C2
h1   h2  2
The process is therefore adiabatic, but irreversible. 2 2
When the velocities Cl and C2 are small, or when Cl is very
u2 – u1= 0 or u2 = ul nearly equal to C2, then the kinetic energy terms may be
neglected. (Note that sections 1 and 2 can be chosen well
In a free expansion therefore the internal energy initially upstream and well downstream of the disturbance to the
equals the internal energy finally flow, so that this latter assumption is justified.)

For a perfect gas, we have, u = CvT Therefore for a throttling process, the enthalpy initially is
equal to the enthalpy finally.
Therefore for a free expansion of a perfect gas T1 = T2
The process is adiabatic, but is highly irreversible because
That is, for a perfect gas undergoing a free expansion the of the eddying of the fluid round the orifice at X. Between
 17

sections 1 and X the enthalpy drops and the kinetic energy Reversible flow processes
increases as the fluid accelerates through the orifice.
Between sections X and 2 the enthalpy increases as the Although flow processes in practice are usually highly

kinetic energy is destroyed by fluid eddies. irreversible it is sometimes convenient to assume that a
flow process is reversible in order to provide an ideal
For a perfect gas, from, h=CpT, therefore, comparison. Some work is done on or by the gas by virtue
of the forces acting between the moving gas and its
CpTl = CpT2 or Tl = T2 surroundings. For example, for a reversible adiabatic flow
process for a perfect gas,
For throttling of a perfect gas, therefore, the temperature
C12 C2
initially equals the temperature finally and no work is done h1   Q  h2  2  W
2 2
during the process. For a vapour, throttling can be used as
Then since Q = 0
a means of finding the dryness fraction of wet steam.
 C 2  C12 
W  h1  h2    2 
 2 
Adiabatic mixing
The mixing of two streams of fluid is quite common in Also since the process is assumed to be reversible, then for

engineering practice, and can usually be assumed to a perfect gas PVY = constant. This equation can be used to

occur adiabatically. Consider two streams of a fluid fix the end states.

mixing as shown in fig. below. Let the streams have NB: Even if the kinetic energies terms are negligibly

mass flow rates m1 and m2, and temperatures T1 and small, the work done in a reversible adiabatic flow process

T2• Let the resulting mixed stream have a temperature between two states is not equal to the work done in a

Ta. There is no heat flow to or from the. fluid, and no reversible adiabatic non-flow process between the same

work is done, hence neglecting changes in kinetic states. Note that the kinetic energy change is small

energy from the flow equation, we have, compared with the enthalpy change. This is often the case
in problems on flow processes and the change in kinetic
energy can sometimes be taken to be negligible.
For a perfect gas, from h=CpT, hence,
Non-steady-flow processes

There are cases in which the total energy of the system

within the boundary is not as constant as in the case of a

The mixing process is highly irreversible due to steady flow process, but varies with time. This happens
when;
the large amount of eddying and churning of the
 The rate of mass flow crossing the boundary of a
fluid that takes place.
system at inlet is not the same as the rate of mass
flow crossing the boundary of the system at
outlet.

 The rate at which work is done on or by the fluid,

and the rate at which heat is transferred to or from
the system is not necessarily constant with time.

Such a system is said undergo an unsteady process.

 18

Examples of Steady-flow processes

 
Thus 21 C22  C12   h1  h2 
- The steady flow energy equation is:
- When the fluid is a perfect gas
Q  W   h2  h1   1
2 C22  C12   g z 2  z1  1
2 C 2
2 
 C12  c p  T1  T2 
where Q and W are the heat and work
- The state of the fluid is usually known at the inlet of
transfers per unit mass flowing through
these devices but only one property e.g. pressure, is
the system.
known at the outlet.
- As in previous section, analysis will be done while
- To fix the final thermodynamic state of the fluid, and
considering the fluid to be either steam or air.
hence determine h2, it is necessary to assume the process
- Irreversibility due to viscous friction is not neglected
to be reversible; the flow is then isentropic and the fact
unless the velocity of flow is very small.
that s1 = s2 can be used.
- In many cases it is necessary to calculate the unknown
- Since the velocity of flow is very high in these devices,
quantity assuming the process to be reversible, and
the effect of friction cannot be neglected. One way of
multiply the result by a process efficiency.
accounting for this is to assume the process to be
- In other cases irreversibility is dealt with by treating the
reversible and then multiply the result by a process
process as a polytropic process.
efficiency to obtain a more realistic estimate.
- The value of the process efficiency must be determined
Boiler and condenser from tests carried out on a similar device.
For any fluid - For nozzles the process efficiency is given by
W=0 C 22
N 

Thus Q  h 2  h1   21 C22  C12   C 2  2
- The velocity in these devices is usually so small that the Where C2 is the actual outlet velocity and C‘2 is the outlet

effect of friction can be neglected and the process velocity which would have been achieved had the final

regarded as internally reversible. pressure been reached isentropically.

- This implies that there is no pressure drop due to friction, - For a diffuser the process efficiency is given by:

and the pressure can be assumed constant throughout the p 2  p1

D 
system. p 2  p1

Where p2 is the actual outlet pressure and p‘2 is the

Example 1: 1500 kg of steam are to be produced per hour pressure which would result from an isentropic process
at a pressure of 30 bar with 100 K of superheat. The feed leading to the same outlet velocity.
o
water is supplied to the boiler at a temperature of 40 C.
Find the rate at which heat must be supplied, assuming Example 2: A fluid expands from 3 bar to 1 bar in a
typical values for the velocity at inlet and outlet: 2 m/s in nozzle. The initial velocity is 90 m/s, the initial
the feed pipe and 45 m/s in the steam main. temperature is 150oC, and from experiments on similar
nozzles it has been found that the isentropic efficiency is

Nozzle and diffuser likely to be 0.95. Find the final velocity when the fluid is:
(a) steam; (b) air.
- For any fluid
Q = 0 (adiabatic process)
W=0
 19

Turbine and rotary compressor Reciprocating compressor (or expander)

For any fluid - The process is approximately isothermal and low

Q = 0 (adiabatic process) velocities can be assumed.

W = (h2 - h1) (Since the velocity at inlet is - Thus for any fluid

approximately the same as that at outlet.) Q – W = (h2 – h1)

When the fluid is a perfect gas - When the fluid is a perfect gas enthalpy is a function of

W = cp(T1 - T2) temperature and so h2 = h1 and Q = W. Also since T is

The process efficiencies of turbines and rotary compressors p2

constant, s 2  s1   R ln and hence
are given by: p1

T1  T2 T  T1 W  Q   RT ln
p2
T  and C  2 respectively
T1  T2 T2  T1 p1

Example 5: A fluid, initially at 155.5oC and 1 bar, is

Example 3: A fluid enters a turbine at the rate of 14 kg/s compressed reversibly and isothermally in a steady-flow
with an initial pressure and temperature of 3 bar and process to a state where the specific volume is 0.28 m3/kg.
150oC. If the final pressure is 1 bar and the isentropic Find the heat transferred and work done, per kg of fluid
efficiency of the turbine is 0.85, find the power developed when the fluid is: (a) steam (b) air.
and the change of entropy between inlet and outlet when Multi-stream steady-flow process
the fluid is: (a) steam; (b) air. Provided the streams do not react chemically with one
another, the steady flow energy equation can be applied to
Throttling give:

- For any fluid Q W   mh 

out
1
2
  mh 
C 2  gz 
in
1
2
C 2  gz 
h1 = h2
- When the fluid is a perfect gas and conservation of mass gives: m  m
out in
T1 = T2
Example 6: Steam is to be condensed by direct injection of
- A throttling calorimeter is used to determine the dryness
cold water. The steam enters the condenser at a rate of 450
fraction of a wet vapour. In this device, a sample of the
kg/h with a dryness fraction of 0.9 and a pressure of 1 atm.
vapour (which is not too wet) is throttled in a well lagged
The estimated heat loss from the condenser to the
chamber such that the final state is in the superheat
surroundings is 8500 kJ/h. If the cold water enters with a
region in which pressure and temperature are
temperature of 15oC, and the mixture of condensate and
independent. (Pressure and temperature are not
cooling water is to leave at 95oC, determine the rate of
independent properties in the wet region.)
flow of cooling water required.

Example 4: Air flows at the rate of 2.3 kg/s in a 15 cm

diameter pipe. It has a pressure of 7 bar and a temperature
of 95oC before it is throttled by a valve to 3.5 bar. Find the
velocity of the air downstream of the restriction, and show
that the enthalpy is essentially the same before and after
the throttling process. Also find the change of entropy.
 20

CHAPTER TWO  It is impossible to construct a device that operating in a

cycle will produce no effect other than the transfer of
Second Law of Thermodynamics heat from a cooler to a hotter body.
Consequences of the Second Law
Cycle efficiency
1. If a system is taken through a cycle and produces
For any closed system taken through a cycle, the First
work, it must be exchanging heat with at least two
Law of thermodynamics can be expressed symbolically as:
reservoirs at different temperatures.
Q1 - Q2 = W (2.1)
2. If a system is taken through a cycle while exchanging
Where
heat with only one reservoir, the work done must be
Q1 = heat supplied from a heat reservoir
either zero or negative.
Q2 = heat rejected to a heat sink
3. Since heat can never be converted continuously and
W = net work done by the system during the cycle
completely into work, whereas work can always be
A system operating in a cycle and producing a net
converted continuously and completely into heat,
quantity of work from a supply of heat is called a heat
work is a more valuable form of energy transfer than
engine. The greater the proportion of heat supply converted
heat.
into work, the better is the engine. Consequently, the cycle
efficiency of a heat engine is defined as: Reversibility and irreversibility

work done W Q1  Q 2 Q When a fluid undergoes a reversible process, both the

    1  2 (2.2)
heat supplied Q1 Q1 Q1 fluid and its surroundings can always be restored to their
There is nothing implicit in the First Law to say that original states. If either following phenomena is present
some proportion of the heat supplied to an engine must be during a process it cannot be reversible:
rejected, and therefore that the cycle efficiency cannot be - friction, and
unity; all that the First Law states is that net work cannot - heat transfer across a finite temperature difference.
be produced during a cycle without some supply of heat. Since at least one of these is always present in some
The Second Law is an expression of the fact that some heat degree, no real process can be reversible.
must always be rejected during the cycle, and therefore that A cycle is reversible if it consists only of reversible
the cycle efficiency is always less than unity. processes; if any of the processes in a cycle are
irreversible, the whole cycle is irreversible. The efficiency
Statements of the Second Law
of an engine in which irreversible processes occur must
Kelvin-Planck: It is impossible to construct a system
always be less than that of the hypothetical reversible
(device) which will operate in a cycle, extract heat from
engine, i.e. it is impossible to construct an engine operating
a reservoir, and do an equivalent amount of work on the
between only two reservoirs which will have a higher
surroundings.
efficiency than a reversible engine operating between the
Clausius: It is impossible to construct a system (device)
same reservoirs.
which will operate in a cycle and transfer heat from a
cooler to a hotter body without work being done on the Carnot cycle

system by the surroundings. The original concept of a reversible cycle is due to

Others: Carnot who thought of a particular cycle, called the Carnot
 It is impossible for a heat engine to produce a net work cycle, which is composed of four processes:
output in a complete cycle if it exchanges heat only with 1 A reversible isothermal heat addition from a source
a single energy reservoir. at temperature TH.
 21

2 A reversible adiabatic process in which work is the fraction 1/T0 of the interval from T = 0 to T 0. The
done by the system. temperature scale defined in this way is called the
3 A reversible isothermal heat rejection to a sink at thermodynamic scale, because it is dependent solely on the
temperature TL. laws of thermodynamics and not upon the properties of any
4 A reversible adiabatic process in which work is particular substance. It is an absolute scale because it
done on the system. presents the idea of an absolute zero, i.e. T = 0 when Q =
0. The Second Law implies that Q can in fact never be
Thermodynamic temperature scale
zero, and therefore it can be concluded that absolute zero is
All reversible engines operating between the same two
a conceptual limit and not a temperature that can ever be
reservoirs have the same efficiency. This efficiency must
reached in practice.
depend upon the only feature that is common to them all,
Consider a series of reversible engines shown below, each
viz. the temperatures of the reservoirs. This efficiency is
operating between only two reservoirs and each producing
called the Carnot efficiency.
the same quantity of work. Each sink is a source for the
Recall that
following engine, the heat entering a reservoir being equal
Q
  1 2 (2.3) to the heat leaving it.
Q1

and hence it follows that Q2/Q1 is a function only of the

temperatures of the reservoirs.
Let a positive number T0 be allotted to some reservoir
in a convenient reference state which is easily reproducible
(e.g. a pure substance melting at a definite pressure);
further, define the temperature T of any other reservoir by
the equation
Q
T  T0 (2.4)
Q0
If the temperatures of the reservoirs are defined in the way
The value T is then fully and uniquely determined by: suggested by equation (1.1), then
(a) the arbitrary choice of T0, and
(b) the ratio Q/Q0 which, as a consequence of the Second
Law, is a fixed and definite quantity for the two given
and the efficiencies of the engines therefore become
reservoirs.

But since

It follows that

If Q is measured for several sinks at different

Simplifying we have
temperatures and plotted against T, a definition of a scale
of temperature which is linear in Q is obtained. The slope
of the line is Q0/T0. The unit of temperature now follows as
 22

temperature of the atmosphere or the sea, it may be said

that a given quantity of heat is more useful for
Thus the differences between the temperatures of
producing work the higher the temperature of the
successive reservoirs are the same and can provide
source from which it is received.
intervals or units of temperature. T 0 may be as high as
we wish to make it, and the intervals may be made as Entropy

small as desired by increasing the number of engines in Definition

the series. The series of reservoirs could theoretically This is a property of a closed system such that a change
be used as a standard set of temperatures with which to in its value is equal to
calibrate any practical thermometer. 2


dQ
We define the unit of temperature by choosing two (2.5)
T
1
fixed points such as the ice point of water T i and the
for any reversible process undergone by the system
steam point Ts and define the number of degrees, i.e,
between state 1 and state 2.
Ts = Ti = 100
It is denoted by S and so
This is equivalent to placing 100 reversible engines in
2
 dQ 
  T 
series between the fixed points.
 S2  S1 (2.6)
When used in conjunction with the equation Ts  Qs , 1 rev

Ti Qi
or in differential form
this arbitrary choice defines a linear thermodynamic
 dQ 
dS    (2.7)
scale.  T  rev
Ti = 273.16 K, Where K – Kelvin
It is an extensive property (like internal energy or
Conversion of t to T is given by T = t + 273
enthalpy) which may be calculated from specific entropies
based on a unit mass of the system so that
Engines operating between more than two reservoirs
S = ms (2.8)

In many practical cycles the heat is received and rejected during Characteristics of entropy
processes which involve a continuous change in the temperature - If the process undergone by a system is a reversible
of the fluid. At any instant during a heating and cooling process, adiabatic one (dQ = 0), the entropy change will be zero,
heat must be exchanged between the system and a source or sink
and this is called an isentropic process; if the process is
which differs only infinitesimally in temperature from the fluid in
irreversible and adiabatic, then the entropy must
the system.
increase.
- The entropy of any closed system which is thermally
Characteristics of engines operating between only two
isolated from the surroundings either increases or, if the
reservoirs are:
process undergone by the system is reversible, remains
(a) All reversible engines, operating between a source at
constant.
temperature T1 and a sink at temperature T2, have an
- For a reversible isothermal process,
efficiency equal to (T1 - T2)/T1. [This is the efficiency
Q = TS (2.9)
of a Carnot cycle.]
(b) For a given value of T2, the efficiency increases with Determination of values of entropy
T1. Since the lowest possible temperature of a practical The property entropy arises as a consequence of the
infinite sink is fixed within close limits, i.e. the Second Law, in much the same way as the property
 23

internal energy arises from the First Law. There is, 2

This equation is analogous to W   pdv for any
however, an important difference. The change in internal 1

energy can be found directly from a knowledge of the heat reversible process

and work crossing the boundary during any non-flow

Thus, as there is a diagram on which areas represent work
process undergone by a closed system. The change in
done in a reversible process, there is also a diagram on
entropy, on the other hand, can be found from a knowledge
which areas represent heat flow in a reversible process.
of the quantity of heat transferred only during a reversible
These diagrams are the pv and the T-s diagrams
non-flow process.
respectively, as shown in figs. 2.2a and 2.2b. For a
reversible process 1-2 in fig. 2.2a, the shaded area p dv,
General equations represents work done; for a reversible process 1-2 in fig.
- As with internal energy, only changes of entropy are 2.2b, the shaded area
normally of interest. The entropy at any arbitrary T ds, represents heat flow.
reference state can be made zero, and the entropy at any - .

other state can be found by evaluating (dQ/T) for any

reversible process by which the system can change
from the reference state to this other state.
- Since no real process is reversible, values of entropy
cannot be found from measurements of Q and T in a
direct experiment. The entropy is a thermodynamic
property, however, and it can be expressed as a
Fig 2.2 a P-v diagram Fig 2.2b T-s diagram
function of other thermodynamic properties which can
be measured in experiments involving real processes.
- Two important relations of this kind can be obtained by (a) Entropy Change for a vapour
combining the equations expressing the First and
Second Laws. Thus the First Law yields the equation Thermodynamic processes can be represented on T-s

(dQ)rev = du + p dv (2.10) diagrams in a way that is dictated by the working fluid. In

and combining this with Eq. (2.7) gives its simplest form the T-s diagram consists of a series of

T ds = du + p dv (2.11) constant pressure lines and saturation curve.

- Alternatively, substituting du = dh - d(pv) gives

T ds = dh - v dp (2.12)
- Equations (2.11) and (2.12) are the general relations
between properties which apply to any fluid. Moreover,
when integrated they give the difference in entropy
between any two equilibrium states, regardless of
whether any particular process joining them is carried
out reversibly or not.
Where
For any reversible process AB – represents the liquid state
B – Saturated liquid state
- BC – Represents wet vapour
 24

C – Represents dry saturated steam

CD- superheated steam Reversible isothermal process
- For all vapour substances values of specific entropy A reversible isothermal process will appear as a straight
may be tabulated along with enthalpy, specific volume, line on a T-s diagram, and the area under the line must
and other thermodynamic properties of interest. represent the heat flow during the process.
- In the liquid-vapour saturation region the specific
entropy is obtained from saturation properties and
quality in the same manner as other properties so that
S = Sf + xSfg, Where Sfg = Sg – Sf
- The temperature-entropy (T-s) chart and Mollier chart,
which is a plot of enthalpy versus entropy (h-s), can be
used to ease calculations.
- Thus, in a reversible process, areas on the p-v diagram
represent work output while areas on a T-s diagram The figure above represents a reversible isothermal
represent heat supplied. expansion of wet steam into the superheat region. The
shaded area represents the heat supplied during the

Entropy change for perfect gas process,

i.e. Heat supplied = T(s2 —s1)
- For a perfect gas, enthalpy and internal energy are
Note that the absolute temperature must be used. The
functions of temperature alone (i.e. du = cvdT and dh =
temperature tabulated in steam tables is t°C, and care must
cpdT respectively) and pv = RT and so Eq. (2.11) can be
be taken to convert this into T K.
written as
dT dv
ds  c v R (2.13) Isothermal process for a perfect gas
T v
For constant specific heats, Eq. (2.13) may be A reversible isothermal process for a perfect gas is shown

integrated between two end states to give: on a T-s diagram in fig below. The shaded area represents
the heat supplied during the process,
T2 v
s2  s1  c v ln  R ln 2 (2.14)
T1 v1

- Alternatively, Eq. (2.12) may be written as

dT dp
ds  c p R (2.15)
T p
and integrated to give
T2 p
s2  s1  c p ln  R ln 2 (2.16)
T1 p1

i.e. Q = T(s2—s1)
Reversible processes on the T-s diagram
For a perfect gas undergoing an isothermal process it is
possible to evaluate s2—s1. From the non-flow for a
We now consider various reversible processes in relation to
reversible process we have,
the T-s diagram. The constant volume and constant
pressure processes have already been represented on the T-
s diagram.
 25

Polytropic process
To find the change of entropy in a polytropic process for a
vapour when the end states have been fixed using p1v —
p2v, the entropy values at the end states can be read
straight from tables.

A polytropic process is the general case for a perfect gas.

To find the entropy change for a perfect gas in the general
case, consider the non-flow energy equation for a
From previous derivation
reversible process,

dQ = du + pdv
Also for unit mass of a perfect gas from Joule‘s law du = c v
dT, and pv = RT.

Also

Note that this result is the same as that derived earlier

Hence between any two States 1 and 2

Reversible adiabatic process (or isentropic process)

This can be illustrated on a T-s diagram as in fig. below
For a reversible adiabatic process the entropy remains
constant, and hence the process is called an isentropic
process. Note that for a process to be isentropic it need not
be either adiabatic or reversible, but the process will
always appear as a vertical line on a T-s diagram. Cases in
which an isentropic process is not both adiabatic and
reversible occur infrequently and will be ignored for this
course
An isentropic process for superheated steam expanding Since in the process T2 < T1, then it is more convenient to
into the wet region is shown in fig. below. write

The first part of this expression for s2 – s1 is the change of

entropy in an isothermal process from v1 to v2 while the
second
part of the expression is the change of entropy in a constant
volume process from T1 to T2,
Using the fact that the entropy remains constant, the end
states can be found easily from tables. It can be seen therefore that in calculating the entropy
change in
 26

a polytropic process from state 1 to state 2 we have in change of entropy and the heat supplied. Sketch the
effect replaced the process by two simpler processes; from process on a T-s diagram. Take the isentropic index, ,
1 to A and then from A to 2. It is clear from figure that for nitrogen as 1.4 and assume that nitrogen is a
s2 - s1 = (SA—S1) - (SA-S2) perfect gas. [0.00125 kJ/K; 0.404 kJ]
Any two processes can be chosen to replace a polytropic 4. 1 kg of steam undergoes a reversible isothermal
process in order to find the entropy change. For example, process from 20 bar and 250oC to a pressure of 30 bar.
going from 1 to B and then from B to 2 as in figure above, Calculate the heat flow, stating whether it is supplied
we have or rejected, and sketch the process on a T-s diagram.
= (SB—s1)—(SB—s1) [- 135 kJ/kg]
At constant temperature between P1 and P2‘ 5. 1 kg of air is allowed to expand reversibly in a cylinder
behind a piston in such a way that the temperature
remains constant at 260oC while the volume is doubled.
and at constant pressure between T 1 and T2 we have The piston is then moved in, and heat is rejected by the
air reversibly at constant pressure until the volume is
the same as it was initially. Calculate the net heat flow
Hence
and the overall change of entropy. Sketch the process
on a T-s diagram. [-161.9 kJ/kg; -0.497 kJ/kg K]
6. 1 kg of air at 1.02 bar, 20oC, undergoes a process in
which the pressure is raised to 6.12 bar, and the
There are obviously a large number of possible equations volume becomes 0.25 m3. Calculate the change of
for the change of entropy in a polytropic process. Each entropy and mark the initial and final states on a T-s
case can be dealt with by sketching the T-s diagram and diagram. [0.083 kJ/kg K]
replacing the process by two other simpler reversible 7. Steam at 15 bar is throttled to 1 bar and a temperature
processes, as in foregoing figure. of 150oC. Calculate the initial dryness fraction and the
change of specific entropy. Sketch the process on a T-s

Problems diagram. [0.992; 1.202 kJ/kg K]

1. 1 kg of steam at 20 bar, dryness fraction 0.9 is heated 8. A turbine is supplied with steam at 40 bar, 400 oC,
reversibly at constant pressure to a temperature of which expands through the turbine in steady flow to an
300oC. Calculate the heat supplied, the change of exit pressure of 0.2 bar, and a dryness fraction of 0.93.
entropy, and show the process on a T-s diagram, The inlet velocity is negligible, but the steam leaves at
indicating the area which represents the heat flow. high velocity through a duct of 0.14 m2 cross-sectional
[415 kJ/kg; 0.8173 kJ/kg K] area. If the mass flow is 3 kg/s, and the mechanical
2. 0.05 kg of steam at 10 bar, dryness fraction 0.84, is efficiency is 90%, calculate the power output of the
heated reversibly in a rigid vessel until the pressure is turbine. Show that the process is irreversible and
20 bar. Calculate the change of entropy and the heat calculate the change of specific entropy. Heat losses
supplied. Show the area which represents heat flow on from the turbine are negligible.
a T-s diagram. [0.0704 kJ/kg K; 36.85 kJ] [2048 W; 0.643 kJ/kg K]
3. A rigid cylinder containing 0.006 m3 of nitrogen (molar
mass 28 kg/kmol) at 1.04 bar, 15oC, is heated
reversibly until the temperature is 90oC. Calculate the
 27

CHAPTER THREE 6. The source of heat supply and the sink for heat
Gas Power Cycles rejection are assumed to be external to the air.
Introduction

Many work producing devices (engines) use or utilize a

The air standard cycle
working fluid that is always air e.g the petrol engine, diesel
- Cycles in which the fuel is burned directly in the
engine, gas turbine etc. In these engines there is a change
working fluid are not heat engines in the true meaning
of the composition of the working fluid, because during
of the term since the system is not reduced to its initial
combustion it changes from air and fuel to combustion
state.
products. For this reason, such engines are called internal
- The working fluid undergoes a chemical change by
combustion engines.
combustion and the resulting products are exhausted to
The working fluid does not go through a complete the atmosphere.
thermodynamic cycle (even though the engine operates in a - In practice such cycles are used frequently and are
mechanical cycle and thus the internal combustion engine called internal-combustion cycles; the fuel is burned
operates on the so-called open cycle. In order to analyze directly in the working fluid, which is normally air.
internal combustion engines, closed cycles that closely - By supplying fuel inside the cylinder, higher
approximate to the open cycles are derived. One such temperatures for the working fluid can be attained; the
approach is the air standard cycle. maximum temperature of all cycles is limited by the
metallurgical limit of the material used and the
Inlet Exhaust
efficiency of the cooling system.
Cylinder
- Examples of internal combustion cycles are the open
Piston
cycle gas turbine unit, the petrol engine, the diesel or

Connecting oil engine, and the gas engine.

Rod Crank

o In the open cycle gas turbine the working fluid

flows at a steady rate from one component to

At intake – air + fuel another round the cycle.

Compression – Air and Fuel are compressed o In the petrol engine a mixture of air and petrol is

Power stroke – combustion products drawn into the cylinder, compressed by the
piston, then ignited by an electric spark. The hot
Exhaust Stroke
gases expand, pushing the piston back, and are
The air standard cycle is based on several assumptions
then swept out to exhaust, and the cycle
1. A fixed amount (Mass) of air is the working fluid
recommences with the introduction of a fresh
throughout the entire cycle. Thus there is no inlet
charge of petrol and air.
or outlet (exhaust) process
o In the diesel or oil engine, the oil is sprayed
2. Air is always an ideal gas
under pressure into the compressed air at the end
3. the cycle is completed by heat transfer from the
of the compression stroke, and the combustion is
surrounding
spontaneous due to the high temperature of the
4. All processes are internally reversible
air after compression.
5. Air has a constant specific heat
 28

o In a gas engine a mixture of gas and air is Indicated work = area of cycle = ɠW
induced into the cylinder, compressed and then Stroke volume of the diagram = V1-V2

W
ignited as in the petrol engine by an electric
spark. Mean effective Pressure (Pm) =
V1  V2
A cycle with a higher mean effective pressure will indicate
- The cycle can be represented on any diagram of
that it has better work characteristics than a cycle with a
properties, and is usually drawn on the p-v diagram,
lower mean effective pressure.
since this allows a more direct comparison to be made
with the actual engine machine cycle. We will now consider the following cycles starting with
- Note that an air standard cycle on a p-v diagram is a the Carnot cycle.
true thermodynamic cycle, whereas a record of the
pressure variations in an engine cylinder against piston
displacement is a machine cycle. Carnot cycle

The air standard cycle enables us to examine the influence An ideal theoretical cycle which is the most efficient

of a number of variables in performance conceivable is the Carnot cycle. This cycle forms a
reference point for the determination of the efficiencies of
Thermal efficiency = ( Net ).Work.Done the other power cycles. By calculating the thermal
Net.Heat. Re cieved
efficiency, it is possible to establish the maximum possible
Net.Work.Done
Work.Ratio 
Gross.Work.Output efficiency between the temperature limits taken (corollary
A cycle which has a good thermal efficiency and a good II of the 2nd law)
work ratio suggests good overall efficiency potential in a This cycle consists of two isothermal processes joined by
practical power producing plant two adiabatic/isentropic processes. It is most conveniently

Specific Fuel Consumption = Mass.of . fuel.used(kg / h) represented on a T-s and p-v diagrams as follows:
power.output(kw)
Process 1-2 = isothermal heat supply.
Low specific fuel consumption indicates a better energy Process 2-3 = isentropic expansion from T2 to T3
conversion. For reciprocating engines, a means of Process 3-4 = isothermal heat rejection
comparison between cycles can be made based on ―mean Process 4-1 = isentropic compression from T4 to T1
effective pressure‖ This is the theoretical pressure which if The cycle is completely independent of the working
it was maintained throughout the volume changes of the substance used.
cycle would give the same output of the work as that
P
obtained from the cycle. T
P1 1

P T1=T2 1 2 2
P2
Area of the cycle =ɠW
P4 4
T3=T4 3
4 3 P3

Pm A B S A V4 V2 B V

Shaded Area =ɠW The cycle efficiency is given by

V2 V1 V
 29

net work output occur reversibly at constant pressure. The expansion and

heat supplied compression processes are isentropic.
heat supplied  heat rejected

heat supplied
T s  s A   T2 s B  s A 
 1 B
T1 s B  s A 
T p2 p
T3 3 p2 2 3
T  T2 s B  s A 
 1
T1 s B  s A  T2 2
p1
T  T4
 1   2  T1 4 p1
 T1  1 1 4
There is no attempt to use the Carnot cycle with gas as
working substance in practice because of two reasons A B s v

1. The pressure of the gas changes continuously from p 4 Heat

supplied
to p1 during the isothermal heat supply, and from p 2
3
to p3 during the isothermal heat rejection. But in 2 Heater

practice it is much more convenient to heat a gas at Net work

output
approximately constant pressure or at constant Compressor Turbine
4
volume. 1 Cooler
2. The Carnot cycle, despite its high thermal efficiency,
Heat
has a small work ratio. [Work ratio is the ratio of the rejected
net work output (area 12341) to the gross work output
Neglecting velocity changes and applying the steady-flow
of the system (area 123BA1); the work done on the
energy equation to each part of the cycle gives:
gas is given by 341AB3.]
Work input to compressor = (h2 – h1) = cp(T2 – T1)
Work output from turbine = (h3 – h4) = cp(T3 – T4)
Examples
Heat supplied in heater = (h3 – h2) = cp(T3 – T2)
3.1. What is the highest possible theoretical efficiency of a
Heat rejected in cooler = (h4 – h1) = cp(T4 – T1)
heat engine operating with a hot reservoir of furnace
Thus
gases at 2000oC when the cooling water is available
c p T3  T2   c p T4  T1 
at 10oC? 
c p T3  T2 
3.2. A hot reservoir at 800oC and a cold reservoir at 15oC
are available. Calculate the thermal efficiency and the T4  T1
 1 (3.2)
work ratio of a Carnot cycle using air as the working T3  T2

fluid, if the maximum and minimum pressures in the Since process 1 to 2 and 3 to 4 are isentropic between the
cycle are 210 bar and 1 bar. same pressures P2 and P1, then
 1 / 
T2 T3  p 2 
    r p 1 / 
T1 T4  p1 
Joule Cycle
where rp is the pressure ratio, p2/p1.
This is also known as Brayton or Constant Pressure cycle
i.e. T3  T4 rp 1 /  and T2  T1rp 1 / 
and forms the basis for the closed cycle gas turbine unit. In
this cycle the heat supply and heat rejection processes so T3  T2  rp 1 /  T4  T1 
 30

Hence substituting in the expression for the efficiency Example

gives 3.3. In a gas turbine unit, air is drawn at 1.02 bar and 15 oC,
T4  T1 1 and is compressed to 6.12 bar. Calculate the thermal
  1  1   1 /  (3.3)
T4  T1 rp 1/  rp efficiency and the work ratio of the ideal cycle, when

Thus for the Joule cycle the cycle efficiency depends only the maximum cycle temperature is limited to 800 oC.

on the pressure ratio.

The Otto cycle
The work ratio (rw) is: This is the ideal air standard cycle for the petrol engine, the
net work output
rw  gas engine and the high speed oil engine. It consists of the
gross work output
following processes:
c p T3  T4   c p T2  T1  Process 1-2 = isentropic compression

c p T3  T4 
Process 2-3 = reversible constant volume heating

 1
T1  T2  (3.4)
Process 3-4 = isentropic expansion
T3  T4  Process 4-1 = reversible constant volume heating
Now, as previously P
T 3

 r p 1 / 
T2 T3 3

T1 T4 4
2
2 4
therefore
1

T2  T1rp 1 /  and T4   31 / 

1
T
v
rp V2 V1 S1 S3

To give a direct comparison with an actual engine the ratio

Hence substituting


T1 r p 1 /   1  of the specific volumes, v1/v2, is taken to be the same as the
rw  1 
 
T3 1  1 / r p 1 /   compression ratio of the actual engine, i.e.
Compression ratio,
T1  1 / 
 1 rp (3.5) rv 
v1
T3 v2
Thus the work ratio depends not only on the pressure ratio
swept volume  clearance volume

but also on the ratio of the minimum and maximum clearance volume
temperatures. For a given inlet temperature, T 1, the The heat supplied at constant volume between T 2 and T3 is
maximum temperature, T3, must be made as high as given by:
possible for a high work ratio. Q1 = cv(T3 – T2)
Similarly the heat rejected per unit mass at constant
For an open-cycle gas turbine unit the actual cycle is not volume between T4 and T1 is given by
such a good approximation to the ideal Joule cycle, since Q2 = cv(T4 – T1)
fuel is burned with the air, and a fresh charge is Processes 1 to 2 and 3 to 4 are isentropic and therefore
continuously induced into the compressor. The ideal cycle there is no heat flow.
nevertheless provides a good basis for comparison, and in Thus
many calculations for the ideal open-cycle gas turbine the c v T3  T2   cv T4  T1 

effects of the mass of fuel and the charge in the working c v T3  T2 
fluid are neglected.
 31

T4  T1 By substituting in the equation of thermal efficiency, i.e.,

 1 (3.6)
T3  T2 Q1  Q2

Now for processes 1 to 2 and 3 to 4, which are isentropic, Q1
 1  1
T2 T3  v1  v  At constant pressure process 2-3
     4   rv 1
T1 T4  v 2   v3

 Q1 =CpdT = Cp(T3 – T2)

Then T3  T4 rv 1 and T2  T1rv 1 At constant Volume process 4-1

Q2 = CvdT = Cv(T4 –T1)
Hence substituting gives
T4  T1 Cv (T4  T1 ) Cp
  1  diesel  1  but 
T4  T1 rv 1 C p (T3  T2 ) Cv

1
 1 (3.7) (T4  T1 )
rv 1  diesel  1 
 (T3  T2 )
Thus the thermal efficiency of the Otto cycle depends only
Process 3-4 is isentropic expansion
on the compression ratio, rv.
From the characteristic gas equation
P3V3 P4V4 and P V   P V 
 3 3 4 4
Example T3 T4
3.4. Calculate the ideal air standard cycle efficiency based   1  1
T4  V3  V4  V3  V V 
        3  2 
on the Otto cycle for a petrol engine with a cylinder T3  V4  V3  V4   V2 V4 
bore of 50 mm, a stroke of 75 mm and a clearance
volume of 21.3 cm3. If we denote r  V3 = Cut off ratio, and
c
V2
The Diesel Cycle V1 = Compression ratio
rv 
This is ideal air standard cycle for the original diesel V2
and given that V1 = V4
engine and consists of the following processes:
Process 1-2 = isentropic compression  1  1
Then T4   rc  So that T  T  rc 

Process 2-3 = reversible constant pressure heating 3
T3  rv 
4
 rv 
Process 3-4 = isentropic expansion
Process 4-1 = reversible constant volume cooling  1  1
Also T2   V1  so that T  T  V1   T1 rv 1
1 
T1  V2 
2
 V2 
P
PVɤ = C T 3
2 3
For constant pressure process 2-3
4 2 P2V2 P3V3 and P = P
 2 3
4 T2 T3
1 1
V2 V3 So that T3 V3
v    rc
V2 V1 S1 S3 T2 T3 T2 V2
Heat supplied
Q1 = cp(T3 – T2) Therefore T3 = T2 rc = T1 rv  1  rc
Heat rejected
r 
T4  T1 rv 
 1
Q2 = cv(T4 – T1) And  rc   c   1  T1 rc
 rv 
There is no heat flow in processes 1-2 and 3-4 since they
are isentropic.
 32

Substituting for T2 T3 and T4 in the equation for diesel name ‗dual-combustion‘. In order to get the thermal
efficiency
efficiency, three factors are necessary. These are:
(T  T ) (T1 rc  T1 )
 diesel  1 4 1 = 1 - The compression ratio, rv = v1/v2,
 (T3  T2 ) 
 T1 rc rv  1  T1 rv  1  - The ratio of pressure, rp = p3/p2, and
- The ratio of volumes,  = v4/v3.
=1  T1 (rc  1)

T1 . rv 
 1
rc  1 Then it can be shown that

rp    1
  1
rp  1  rp   1rv 1
(3.9)

= 1  (rc  1) (3.8)
 .rv rc  1
 1
Thus the thermal efficiency of a dual-combustion cycle
Hence the efficiency of a diesel engine depends on the cut-
depends not only on the compression ratio but also on the
off ratio as well as the compression ratio.
relative amounts of heat supplied at constant volume and at
constant pressure.
Example
Equation (3.9) is much too cumbersome to use, and the
3.5. A diesel engine has an inlet temperature and pressure
best method of calculating thermal efficiency is to evaluate
of 15oC and 1 bar respectively. The compression ratio
each temperature round the cycle and then get the total
is 12/1 and the maximum cycle temperature is 1100oC.
heat supplied (Q1) and the total heat rejected (Q2) as:
Calculate the air standard thermal efficiency based on
Q1 = cv(T3 – T2) + cp(T4 – T3)
the diesel cycle.
Q2 = cv(T5 – T1)

The dual-combustion cycle Note that when rp = 1 (i.e. p3 = p2), then Eq (3.9) reduces
This is also known as the limited-pressure or mixed cycle to the thermal efficiency of the diesel cycle.
and is the ideal air standard cycle of modern diesel and oil
engines. It consists of the following processes: Example
Process 1-2 = isentropic compression 3.6. An oil engine takes in air at 1.01 bar, 20 oC and the
Process 2-3 is reversible constant volume heating maximum cycle pressure is 69 bar. The compressor
Process 3-4 = reversible constant pressure heating ratio is 18/1. Calculate the air standard thermal
Process 4-5 = isentropic expansion efficiency and the mean effective pressure based on
Process 5-1 = reversible constant volume cooling the dual-combustion cycle. Assume that the heat
added at constant volume is equal to the heat added at
P
constant pressure. [Mean effective pressure is the
ɤ
3 P3 = P4 4 PV = C
height of a rectangle having the same length and area
as the cycle plotted on a p-v diagram.]
2 5

1
The Stirling cycle

V2 V1 V This has an efficiency equal to that of the Carnot cycle but

has a higher work ratio. It consists of the following
The heat is supplied in two parts, the first part at constant
processes
volume and the remainder at constant pressure, hence the
Process 1-2 = reversible constant volume heating
Process 2-3 = isothermal expansion
 33

Process 3-4 = reversible constant volume cooling Process 2-3 = isothermal expansion
Process 4-1 = isothermal compression Process 3-4 = reversible constant pressure cooling
Process 4-1 = isothermal compression
p
2
p
T2= T3
1 1 2

T2= T3
3 T1= T4
T1= T4
4
3
v 4

v
Heat supplied,

p  Once again, because the two constant pressure processes

Q23  RT2 ln  2 
 p3  are bounded by the same temperature limits;

Heat rejected, mCp(T3-T2) = mCp(T4-T1)

then the process of regeneration is again possible. By
p 
Q41  RT1 ln  1  including the process of regeneration the Ericsson cycle
 p4 
becomes a reversible cycle and has the highest thermal
Thus
efficiency possible which = (T3 – T1)/T3
p 
RT1 ln  2 
  1  p3  (3.10) Problems
 p 
RT2 ln  1 
 p4  1. What is the highest cycle efficiency possible for a heat
engine operating between 800 and 15oC? [73.2%]
For the constant volume process 1-2,
2. Two reversible heat engines operate in series between a
p 2 T2
 source at 527oC and a sink at 17oC. If the engines have
p1 T1
equal efficiencies and the first rejects 400 kJ to the
p 3 T3 T2
and for process 3-4,   second, calculate:
p 4 T4 T1
(a) the temperature at which heat is supplied to the
p2 p
Therefore  3 second engine;
p1 p4
(b) the heat taken from the source;
p p
and 2  1 (c) the work done by each engine.
p3 p4
Assume that each engine operates on the Carnot cycle.
Hence
[208.7oC; 664.4 kJ; 264.4 kJ; 159.2 kJ]
T1
  1 (3.11) 3. In a Carnot cycle operating between 307 and 17 oC the
T2
maximum and minimum pressures are 62.4 bar and
= the Carnot efficiency
1.04 bar. Calculate the cycle efficiency and the work
The Ericsson cycle ratio. Assume air to be the working fluid. [50%; 0.286]

This also has an efficiency equal to that of the Carnot cycle 4. A closed-cycle gas turbine unit operating with

but has a higher work ratio. It consists of the following maximum and minimum temperatures of 760 and 20 oC

processes has a pressure ratio of 7/1. Calculate the ideal cycle

Process 1-2 = reversible constant pressure heating efficiency and the work ratio. [42.7%; 0.505]
 34

Air Compressors
5. In an air standard Otto cycle the maximum and minimum The function of a compressor is to take a definite quantity
temperatures are 1400 and 15oC. The heat supplied per kg of fluid and deliver it at a required pressure. The most
of air is 800 kJ. Calculate the compression ratio and the efficient machine is one which will accomplish this with e
cycle efficiency. Calculate also the ratio of maximum to minimum input of mechanical work. There are two general
minimum pressures in the cycle. [5.27/1; 48.5%; 30.65/1] types of compressors;

1. Rotary compressors
6. A four-cylinder petrol engine has a swept volume of 2000
cm3, and the clearance volume in each cylinder is 60 cm 3. 2. Reciprocating compressors
Calculate the air standard cycle efficiency. If the Reciprocating compressors have a low mass flow rate and
introduction conditions are 1 bar and 24oC, and the
high pressure ratios whereas rotary compressors have hig
maximum cycle temperature is 1400oC, calculate the mean
mass flow rate and low pressure ratios.
effective pressure based on the air standard cycle.
[59.1%; 5.28 bar]
Reciprocating compressors

7. Calculate the cycle efficiency and mean effective pressure The mechanism involved is the basic piston con-rod, crank
of an air standard diesel cycle with a compression ratio of and cylinder arrangement.
15/1, and maximum and minimum cycle temperatures of
1650oC and 15oC respectively. The maximum cycle
pressure is 45 bar. [59.1%; 8.38 bar]

8. In a dual-combustion cycle the maximum temperature is

2000oC and the maximum pressure is 70 bar. Calculate the
cycle efficiency and the mean effective pressure when the
pressure and temperature at the start of compression are 1
bar and 17oC respectively. The compression ratio is 18/1.
[63.6%; 10.46 bar]

9. An air standard dual-combustion cycle has a mean As piston moves down, the inlet valve is open and the

effective pressure of 10 bar. The minimum pressure and delivery valve is closed. A fresh charge of air is taken into

temperature are 1 bar and 17oC respectively, and the the cylinder. As the piston moves up, pressure in the

compression ratio is 16/1. Calculate the maximum cycle cylinder builds up, the inlet valve is closed. When pressure

temperature when the cycle efficiency is 60%. The is slightly in excess of that of the air on the outside of the

maximum cycle pressure is 60 bar. [1959oC] delivery valve, the delivery valve opens (by differential
pressures) and compressed air is delivered

Assuming air to be a perfect gas and that the compressor

operates at zero clearance the theoretical pV diagram is
shown as below
 35

= i.p + f.p
P
Therefore the mechanical efficiency of the machine is given by;
c b
P2
ɳ = Indicated.Work  Indicated.Power
Shaft.Work Shaft.Power
d
P1 a
to determine the power input required, the efficiency of the
driving motor must be taken into account in addition to the
V2 V1 V mechanical efficiency of the compressor

d-a -induction stroke. Mass of air in the cylinder increases Motor and drive efficiency = Shaft.Power
from zero at d to that required to fill the cylinder at a. Input.Power

The temperature,T1 remains constant for this process

mRT2  T1 
From Indicated Work = n
and there is no heat exchange with the surrounding n 1

a-b compression stroke ( reversible polytropic pvγ = n T 

We can rewrite; Indicated Work = mRT1  2  1
Constant) n 1  T1 

b-c delivery stroke at constatnt temperature T2 and constant n 1

But = T2   P2 
n
pressure P2. and PV = MRT so that
T1  P1 1 
The general form of the compression a-b is the reversible
 n 1

polytropic given by PVn = Constant Indicated Work = n  P2  n 
mRT1    1
n 1  1 1 
P 
The net work done in the cycle is given by the area  
enclosed by the pv diagram and is work done on the gas.
 n 1

Indicated Work = n  P2  n 
P1V1    1
In this section work done on the gas will be considered as
n 1  1 1 
P 
positive work  

Indicated work = area abcd = area (abef + bcoe –adof) Where V1 is the volume indicated per unit time

= P2Vb  P1Va  P2Vb  P1Va

n 1

= P V  P V  1  1 Condition for Minimum Work

 n 1
2 b 1 a


= P V  P V  1  n  1  P
 n 1 
2 b 1 a
P2 c b1 b b2

= n P2Vb  P1Va 
n 1
P1Va = mRT1 and P2Vb = mRT2
P1 a
mRT2  T1 
Therefore work input per cycle = n d
n 1 e f
o ve 2 v1 v
The actual work input to the compressor is larger than the indicated
work, due to the work necessary to overcome losses due to friction. The work done on the air in a compressor is given by the
area of the indicator diagram, and the work will be
Shaft work = indicated work + friction work
minimum when the area is minimum. The height of the
 36

diagram is fixed by the required pressure ratio and the Isothermal.work

Isothermal efficiency =
length of the line da is fixed by the cylinder volume. The Indicated.Work
only process that can influence the area of the diagram is
the compression stroke (line ab)
Reciprocating compressors including clearance
Possible processes
Clearance is necessary in a compressor to give mechanical
Line ab1 - is according to the law pv = constant freedom to the working parts and allow the necessary
(isothermal process) space for valve operations.

Line ab2 – is according to the law pvɤ = Constant

P
(Isentropic process) P2 c T2 b

Line ab – is according to the law pvn = constant (Polytropic

Process)
T1
P1 a
Isothermal process is the desirable process between a and b d
giving minimum work to be done on the air. Therefore
V
cooling of the gas is always provided either by air or by Vc Swept Volume

water to keep the temperature in the cylinder as constant as

possible.
When the delivery stroke bc is compete the clearance
Indicated work when the gas is compressed isothermally = volume, vc, is full of gas at a pressure P2 and temperature
area ab1cd T2. As the piston proceeds on the next induction stroke the

Area ab1cd = Area ab1ef + Area bc1oe – Area adof air expands behind the piston until the pressure P 1 is
reached. As soon as the pressure reaches P 1 the induction
Area ab1ef = P
P2Vb1 ln 2 (isothermal process) of fresh air will begin and continue until the end of the
P1
stroke at a
P2 ab – compression stroke
Area a b1c d = P2Vb ln  P2Vb1  P1V a
1
P1
bc – delivery stroke
Also P1Va  P2Vb (isothermal process
1
da – induction stroke

Therefore indicated work = area (abcd) = P2Vb ln P2 Clearance reduces induction volume Vs to (Va – Vd)
1
P1
ṁa = ṁb, ṁc = ṁd
P
= P1V a ln 2 Indicated work = area abcd
P1
= area (abef) – area (cefd)
but P1Va = mRT1 so that
= n ma RT2  T1   n md R(T2  T1 )
. .

P2 n 1 n 1
Indicated work = mRT1 ln
R(ma  md )T2  T1 
P1 = n . .

n 1
Where m and Va are the mass and the volume of the work
= n m RT2  T1 
.

induced respectively. If these parameters are per unit time n 1

then the equation gives the isothermal power. Where ṁ = ṁa - ṁd ie mass induced per unit time
 37

n 1 . Pv s
ms 
But = T2   P2 
n
and PV = mRT so that RT
T1  P1 1 
. Pv
m v
n . T   v   RT 
Indicated Work = m RT1  2  1 . pvs vs
n 1  T1  ms
RT
 n 1
 or volume induced = va- vd
n .  P2  n 
 m RT1    1
n 1  1 1 
P 
  P
P2 c T2 b
 n 1

Indicated Work = n  P2  n

P1V    1
n 1  P1 1  
  T1
P1 a
 n 1
 d
 P2  n 
P1 Va  Vd 
n
   1
n 1  P1 1   V
  Vc Swept Volume

The mass delivered per unit time can be increased by

designing the machine to be double acting i.e gas is dealt va-vd = vs + vc - vd
with on both sides of the piston, the induction stroke of one P 
1
n

and vd   2  Vc
side being the compression stroke of the other side  P1 
1
P 
n

Chamber 1  volume.induced  vs  vc  vc  2 
 P1 
Piston
 P  n 
1

Chamber 2 vs  vc  2   1
v  P1  
v  
vs vs

 P 
1

v  2 n

Volumetric Efficiency v  1 c    1
 1 
vs  P 

Clearance volume reduces the induced volume to a value
less than that of the swept volume. Volumetric efficiency is
c‖ b‖
defined as

The.mass.of .air.induced
v  c‘ b‘
Mass.of .air.which.would. fill.the.swept.volume
P
Free air delivery (FAD), is the volume delivered measured P2 c T2 b
at the pressure and temperature of the atmosphere in which
the machine operates.
T1
If FAD is V, at pressure p, and temperature T, then the P1 a
d d‘ d‖
mass of air induced (delivered) is given by,

.
V
Pv
m Vc Swept Volume
RT

Mass required to fill the swept volume (Va –Vd) > (Va –Vd‘) > (Va –Vd‖)
 38

Multi stage compression  n 1

  n 1

n  pi  n  n  p2  n 
 m RT1    1  m RT1    1
As the required pressure ratio for a single stage compressor n 1   p1   n  1   p i  
   
increases, the volumetric efficiency decreases (i.e volume induced
per cycle decreases). The volumetric efficiency of the compressor It is assumed that inter-cooling is complete and therefore the
can be improved by carrying out the compression in two stages. temperature at the beginning of each stage is T1
After the first stage of compression, the fluid is passed into a
 n 1 n 1

smaller cylinder in which the gas is compressed to the required n  pi  n  p2  n 
Total.Work  m RT1    1     1
final pressure. For minimum work to be done, the gas from the n 1  1 
p  i
p 
 
first stage of compression is cooled as it passes from one cylinder
to the other by passing it through an intercooler. If P1T1 and P2 are fixed, then the optimum value which makes the
work a minimum can be obtained by equating
H2O in H2O out
P1T1 d
PiTi PiTi (Work)  0
P2T2 in in in dpi
in
  n 1 n 1

d  n  pi  n  p2  n 
 mRT1       2  0
    
Intercooler dpi  n  1  1 
p  i
p 
  

 n 1 n 1


d  pi  n   p2  n  2  0
 p  p  
 1   i
dpi 
High Pressure Low Pressure 
Stage Stage
 n 1
n 1 n 1
n 1


d  1  n  p n  p n   1  n  2  0

 p  i 2 p  
 1   i
P dpi 

P2
c' b'
n 1
 n 1  n 1  1 n 
n 1  1  1 n 1 n n  1 
  pi n 
    p2  pi n 
0
Pi c T2 b n  p1  n
d' a'  1 2 n 
1 n 1 n 1
n 1 n 1 n n  
 p1  pi n  p2  pi n 
0
T1 n n
P1 a
d 1 n 1 n 1  1 2 n 
n 1 n n 1 n  
 p1  pi n  p2  pi n 

vc vs
V n n
 n1  1 n1  12 n 
Vc Swept Volume    
So that , p1  n 
 pi n  p2 n  pi n 

d - a - induction stroke of 1st stage 1

n 1 n 1
pi n
b - c – delivery stroke of 1st stage  1 2 n
 p2 n  p1 n
n
pi
d‘- a‘ – induction stroke of 2nd stage
2 n2 n1
b‘-c‘ – delivery stroke of 2nd stage  pi n

 p2  p1 n

 pi2  p1 p2
Ideal intermediate pressure for a two stage compressor
pi p2
Intermediate pressure pi influences the work to be done on the gas So that pi  p1 p2 and 
p1 pi
and its distribution between the two stages.
Therefore for the work to be a minimum the pressure ratios for
Total work = Low Pressure work + High Pressure Work the two stages has to be equal
 39

 n 1 n 1
 3. A single acting two stage air compressor runs at 300
n  pi  n  p2  n 
Total.Minimum.Work  m RT1    1     1 rev/min and compresses 8.5m3/min at 1 atmosphere
n 1   p1   p i  
 
and 15oC to 40bar. Calculate
But we have seen that pi  p2 a. The optimum pressure for each stage
p1 pi
b. The theoretical power consumption for each stage
 n 1 n 1

n  pi  n  pi  n  if the compression in each stage is polytropic with
Total.Minimum.Work  m RT1    1     1
n 1   p1   p1  
  n=1.3 and intercooling is complete i.e to a
 n 1
 temperature of 15oC
2n  p  n 
 m RT1  i   1
n 1  1 
p  c. The swept volumes if the volumetric efficiencies of
 
In terms of the overall pressure ratio, the low pressure and high pressure stages are
pi p1 p2 p2 0.90 and 0.85 respectively
 
p1 pi p1 d. The heat rejected into the cylinder cooling jackets
Substituting for pi we have;
p1
and into the intercooler. [6.283, 32.86 kW,
 n 1
 0.0315m3, 0.0053m3, 6.39kW per cylinder,
2n  p2  n

Total.mimimum.work  m RT1  
  1 26.88kW for the intercooler]
n 1  p1  
  4. Air is to be compressed in a single stage reciprocating
 n 1

2n  p2  2 n  compressor from 1.0 13 bar and 15°C to 7 bar.
 m RT1    1
n 1  p1  
  Calculate the indicated power required for a free air
Problems delivery of 0.3m3/min when the compression process
1. A single stage reciprocating compressor takes 1m3 of is:
air per minute at 1.013 bar and 15oC and delivers it at a) Isentropic [1.31 kW].
1.35
7 bar. Assuming the law of compression is PV = b) Reversible isothermal [0.98 kW].
Constant and that the clearance volume is negligible, c) Polytropic, with n = 1.25 [1.196 kW].
calculate the indicated power. If the compressor is 5. A single — acting compressor is required to deliver
driven at 300rev/min, determine the cylinder bore air at 70 bar from an induction pressure of 1 bar, at
required assuming a stroke to bore ratio of 1.5:1. the rate of 2.4 m3/min measured at free air conditions
Calculate the power of the motor required to drive the of 1.013 bar and 15°C. the temperature at the end of
compressor if the mechanical efficiency of the the induction stroke is 32°C. Calculate the indicated
compressor is 85% and that of the motor transmission power required if the compression is carried out in
is 90%. [4.23 kW, 141.5 mm, 5.53kW, 77%] two stages with an ideal intermediate pressure and
2. A single stage double acting air compressor is complete intercooling. The index of compression and
3
required to deliver 14m of air per minute measured at expansion for both stages is 1.25. What is the saving
o
1.013bar and 15 C. The delivery pressure is 7 bar and in power over single stage compression? If the
the speed is 300 r.p.m. take the clearance volume as clearance volume is 3% of the swept volume in each
5% of the swept volume with a compression index of cylinder, calculate the swept volumes of the cylinders.
n=1.3. Calculate the swept volume of the cylinder, the The speed of the compressor is 750 rev/mm. if the
delivery temperature and the indicated power. mechanical efficiency of the compressor is 85%,
3
[0.0281m , 450K, 57.58kW] calculate the power output in kW of the motor
required. [22.7 kW, 6 kW, 0.00396 m3, 0.000474
m3, 26.75 kW]
 40

CHAPTER FOUR 3600

ssc 
Vapour Power Cycles W

Introduction Carnot cycle

Characteristics of power cycles - It consists of two reversible isothermal processes at T a

and Tb respectively, connected by two reversible
- The working fluid is a condensable vapour which is in
adiabatic (isentropic) processes.
liquid phase during part of the cycle
- When the working fluid is a condensable vapour, the two
- The cycle consists of a succession of steady-flow
isothermal processes are easily obtained by heating and
processes, with each process carried out in a separate
cooling at constant pressure while the fluid is a wet
component specifically designed for that purpose.
vapour.
- Each component constitutes an open system, and
- The processes are:
all the components are connected in series so that as the
1-2: Saturated water is evaporated at constant pressure to
fluid circulates through the power plant each fluid
form saturated steam; heat added is Q12 = h2 – h1
element passes through a cycle of mechanical and
2-3: Saturated steam is expanded isentropically in a
thermodynamic states.
turbine; work done is W23 = h2 – h3
- To simplify the analysis, it is assumed that the change in
3-4: Wet steam is partially condensed at constant
kinetic and potential energy of the fluid between entry
pressure to state 4 where s4 = s1; heat rejected is Q34
and exit in each component is negligible compared to the
= h4 – h3
change in enthalpy. This implies that the energy equation
4-1: Steam is compressed isentropically in a compressor;
can be written as: Q – W = h2 – h1
work required is W41 = h4 – h1
- The working fluid is usually steam because it is cheap
and chemically stable but any condensable vapour may
Turbine T
be used. Boiler

Criteria of performance Cooling 1 2

water
(a) Ideal cycle efficiency – this is the efficiency of a cycle
4 3
when all the processes are assumed to be reversible.
(b) Actual cycle efficiency – this is the efficiency of a S
Compressor Condenser
cycle when process efficiencies are introduced
(c) Efficiency ratio – this is the ratio of the actual cycle
Example 1
efficiency to the ideal cycle efficiency.
Calculate
PRD the heat and work transfers, cycle efficiency,
(d) Work ratio (rw) – this is the ratio of the net work to the 571E Carnot cycle 3- 01
work ratio and steam consumption of a Carnot cycle using
positive work done in the cycle. (It is a measure of the
steam between pressures of 30 and 0.04 bar.
cycle‘s sensitivity to irreversibilities since
Solution
irreversibilities decrease the positive work and increase
From tables, at 30 bar
the negative work.)
T1 = T2 = 507.0 K
(e) Specific steam consumption (ssc) – this is the mass
h1 = hf = 1008 kJ/kg,
flow of steam required per unit of power output. It is
h2 = hg = 2803 kJ/kg
usually expressed in kg/kW h and if the numerical
Putting s4 = s1 and s3 = s2, then at the condenser pressure of
value of net work output per unit mass of flow is W
0.04 bar,
(kJ/kg) the ssc can be found from
T3 = T4 = 302.2 K
 41

x3 = 0.716, x4 = 0.276 
W

483
 0.277
Q12 1741
Hence from h = hf + xhfg,
h3 = 121 + x32433 = 1863 kJ/kg The specific steam consumption is
3600
h4 = 121 + x42433 = 793 kJ/kg ssc   7.45
483
The turbine work is W23 = h2 – h3 = 940 kJ/kg
The compressor work is W42 = h4 – h1 = -215 kJ/kg
Rankine cycle
The heat transfer in the boiler is Q12 = h2 – h1 = 1795 kJ/kg
The heat transfer in the condenser is Q34 = h4 – h3 = -1070 Unsuperheated cycle

kJ/kg - There are two reasons why the Carnot cycle is not used
The net work from the cycle is thus W = W 23 + W41 = 725 in practice
kJ/kg (This is also equal to Q12 + Q34) (a) It has a low work ratio.
The cycle efficiency is (b) It is difficult to control the condensation process so
W 725 that it is stopped at state 4, and then carry out the
   0.404
Q12 1795
compression of a very wet vapour efficiently. The
Since this is a Carnot cycle, this must also be given by liquid tends to separate out from the vapour and the
T T 507.0  302.2
 a b   0.404 compressor would have to deal with a non-
Ta 507.0
homogeneous mixture.
The work ratio is
- On the other hand, it is comparatively easy to condense
W 725
rw    0.771 the vapour completely and compress the liquid to boiler
W23 940
pressure in a small feed pump
The specific steam consumption is
- The resulting cycle is known as the Rankine cycle.
3600
ssc   4.97 kg/kW h
725

Turbine T
Boiler
Example 2
Cooling 1 2
Recalculate Example 1 with isentropic efficiencies of 0.80 water
5
for the compression and expansion process, to estimate the
4 3
actual cycle efficiency and steam consumption.
S
Solution Condenser
Pump
The actual turbine work is
W23 = h2 – h3 = 0.80(h2 – h3‘) = 752 kJ/kg
The actual compressor work is Example 3
PRD the cycle efficiency, work ratio, and the steam
Calculate
W41  h4  h1 
h4  h1
 269 kJ/kg 571E Simple Rankine cycle 3- 01
0.80 consumption of a Rankine cycle working between
The net work is therefore pressures of 30 and 0.04 bar.
W = W23 + W41 = 483 kJ/kg Estimate the actual cycle efficiency and steam
The enthalpy at state 1 is consumption when the isentropic efficiencies of the
h1 = h4 – W41 = 1062 kJ/kg expansion and compression processes are each 0.80.
Hence the heat transfer in the boiler is
Q12 = h2 – h1 = 1741 kJ/kg Solution
The thermal efficiency is (a) Ideal cycle
 42

As in Example 1, Superheater
T 2'
h2 = 2803 kJ/kg and h3 = 1863 kJ/kg
Turbine 1 2
The turbine work is as before Boiler
W23 = h2 – h3 = 940 kJ/kg 5

4 3
The compression work is Condenser Cooling
water S
W45 = h4 – h5 = vf(p4 – p5) Pump
= 0.001(0.04 – 30) x 100 = -3 kJ/kg
- It is evident that the average temperature at which heat is
Since h4 = hf = 121 kJ/kg then h5 = 124 kJ/kg.
supplied is increased by superheating and hence the ideal
The heat supplied is PRD
Rankine cycle with
cycle
571E
efficiency is increased.
superheater 3- 01
Q52 = h2 – h5 = 2803 – 124 = 2679 kJ/kg
Example 4
Thus
A steam power plant operates between a boiler pressure of
W 940  3
   0.350 42 bar and a condenser pressure of 0.035 bar. Calculate the
Q52 2679
cycle efficiency and specific steam consumption when the
W 940  3
rw    0.997
W23 940 steam is superheated to 500oC.
3600 Solution
ssc   3.84 kg/kW h
940  3
From tables, by interpolation, at 42 bar:
(b) Actual cycle h5 = 3442.6 kJ/kg and s5 = s6 = 7.066 kJ/kg K
The actual expansion work is Now s6 = s1 + x6sfg therefore 0.391 + x68.13 = 7.066
W23 = 0.80 x 940 = 752 kJ/kg i.e. x6 = 0.821
The actual compression work is Also h6 = h1 + x6hfg = 112 + (0.821 x 2438) = 2113 kJ/kg
3 From tables:
W41    4 kJ/kg (which is negligibly small)
0.80
h1 = 112 kJ/kg
The enthalpy at state 5 now becomes
Then W56 = h6 – h5 = 3442.6 – 2113 = 1329.6 kJ/kg
h5 = h4 – W45 = 125 kJ/kg
Neglecting the feed-pump term,
The heat supplied is
heat supplied Q25  h5 – h1 = 3442.6 – 112 = 3330.6
Q52 = h2 – h5 = 2803 – 125 = 2678 kJ/kg
kJ/kg
Therefore
W56 1329.6
752  4    0.399
  0.279 Q25 3330.6
2678
3600 3600
3600 ssc    2.71 kg/kW h
ssc   4.81 kg/kW h W56 1329.6
752  4

Rankine cycle with superheat

- By placing in the combustion chamber a separate bank of Reheat cycle

tubes (the superheater) leading saturated steam away - With the reheat cycle the expansion takes place in two
from the boiler, it is possible to raise the steam turbines.
temperature without at the same time raising the boiler - The steam expands in the high-pressure turbine to some
pressure. intermediate pressure, and is then passed back to yet
- This gives the Rankine cycle with superheat. another bank of tubes in the boiler where it is reheated at
constant pressure, usually to the original superheat
4
temperature.
 43

- It then expands in the low-pressure turbine to the Regenerative cycle

condenser pressure. One feed heater
SH - In a practical regenerative cycle, steam is bled off the
T
2' 2'''
HPT LPT Turbine turbine at some intermediate pressure during the
1
Boiler 2 expansion and mixed with feed water, which has been
C
H 5 2''
CW pumped to the same pressure.
Pump 4 3
- The mixing process is carried out in a feed water heater
S
SH
H
- Superheater
- Heater and the arrangement is shown below. Only one feed
LPT - Low pressure turbine
HPT - High pressure turbine
CW
C
- Cooling water
- Condenser
heater is shown but several could be used.

PRD
Rankine cycle with
Example
571E 5 superheater and reheat 3- 01 Boiler 2

Find the ideal cycle efficiency and steam consumption of a

1
reheat cycle operating between pressures of 30 and 0.04 y kg 4
o
Feed
bar, with a superheat temperature of 450 C. Assume that pump Condenser
3
the first expansion is carried out to the point where the (1-y) kg
5
steam is dry saturated and that the steam is reheated to the 1 kg
7 6
original superheat temperature. The feed pump term may Feed
be neglected. heater
Feed
Solution T 2 pump

From tables 1 kg
h2  h1 = 121 kJ/kg, h3 = 3343 kJ/kg, s5 = 7.082 kJ/kg K 1
y kg
To find the intermediate reheat pressure p 6, get from the
7 3
saturation table the pressure at which sg = s5. This gives
p6 = p7 = 2.3 bar, and hence h6 = 2713 kJ/kg 6
(1-y) kg
From the superheat table 5 4

h7 = 3381 kJ/kg and s7 = 8.310 kJ/kg K

- The steam expands from condition 2 through the turbine.
s
At the turbine outlet
- At the pressure corresponding to point 3, a quantity of
x8 = 0.980 and h8 = 2505 kJ/kg
steam, say y kg per kg of steam supplied to the boiler, is
The total heat transferred to the steam in the boiler is
bled off for heating purposes.
Q25 + Q67 = (h5 – h2) + (h7 – h6)
- The rest of the steam (1 - y) kg, completes the expansion
= 3222 + 668 = 3890 kJ/kg
and is exhausted at state 4.
The total turbine work is
- This amount of steam is then condensed to state 5 and
W56 + W78 = (h5 – h6) + (h7 – h8)
pumped to the same pressure as the bleed steam (i.e. p 6 =
= 630 + 876 = 1506 kJ/kg
p7 = p3).
Thus
- The bleed steam and the feed water are mixed in the feed
W56  W78 1506
   0.387
Q25  Q67 3890 heater, and the quantity of bled steam, y kg, is such that
3600 3600 after mixing and being pumped in a second feed pump,
ssc    2.39 kg/kW h
W56  W78 1506 the condition is as defined by state 1.
 44

- The heat to be supplied in the boiler is then given by (h 2 = 876 kJ per kg of steam delivered to the
– h1) kJ/kg of steam; this is the heat supplied between boiler.
temperatures T1 and T2. Therefore,
- The bleed temperature to obtain maximum efficiency for W23  W34 876
   0.396
Q12 2216
a regenerative cycle is approximately the mean of the
3600 3600
saturation temperatures corresponding to p 2 and p5. ssc    4.11 kg/kW h
W23  W34 876

Example 6
Example 7
Find the cycle efficiency and specific steam consumption
Find the cycle efficiency and specific steam consumption
of a regenerative cycle with one feed heater, if the steam
of a regenerative cycle with one feed heater, if the steam
leaves the boiler dry saturated at 42 bar and is condensed at
leaves the boiler dry saturated at 30 bar and is condensed at
0.035 bar. Neglect the feed pump work.
0.04 bar. Neglect the feed pump work.
Solution
Solution
At 42 bar, T1 = T2 = 253.2oC and at 0.035 bar, T 5 = 26.7oC.
At 30 bar, T2 = 233.8oC and at 0.04 bar, T4 = 29.0oC.
253.2  26.7 o
Therefore T3   140 C
2 T2  T4 233.8  29.0 o
Therefore T6    131.4 C
2 2
Selecting the nearest saturation pressure from the tables
Hence p3 = 2.8 bar.
gives the bleed pressure p3 as 3.5 bar (i.e. T3 = 138.9oC).
From tables, h7 = 551 kJ/kg; h5 = 121 kJ/kg; and s2 = s3 =
To determine the fraction y, consider the adiabatic mixing
s4 = 6.186 kJ/kg K.
process at the feed heater, in which y kg of steam of
Thus x3 = 0.846, h3 = 2388 kJ/kg, x4 = 0.716, h4 = 1863
enthalpy h3, mix with (1-y) kg of water of enthalpy h6, to
kJ/kg
give 1 kg of water of enthalpy h7. The feed pump term may
551  121
be neglected (i.e. h6 = h5). Therefore Hence y  0.1897 kg
2388  121
yh3 + (1-y)h5 = h7. Heat supplied in boiler Q12 = (h2 - h1) = 2803 - 551 = 2252
i.e. kJ/kg
h h
y 7 5 Total work output = (2803 - 2388) + (1 - 0.1897)(2388 -
h3  h5
1863)
Now, h7 = 584 kJ/kg; h5 = 112 kJ/kg; and s2 = s3 = s4 =
= 840 kJ per kg of steam delivered to the boiler.
6.049 kJ/kg K.
6.049  1.727 6.049  0.391 Therefore,
x3   0.829 and x4   0.696
5.214 8.130 W23  W34 840
   0.373
Hence: Q12 2252

h3 = hf3 + x3hfg3 = 584 + (0.829 x 2148) = 2364 kJ/kg 3600 3600

ssc    4.29 kg/kW h
W23  W34 840
and
h4 = hf4 + x4hfg4 = 112 + (0.696 x 2438) = 1808 kJ/kg
584  112 Several feed heaters
Therefore y   0.21 kg
2364  112
- The thermal efficiency increases with addition of further
Heat supplied in boiler = (h2 - h7) = 2800 - 584 = 2216
heaters, but the capital expenditure is also increased
kJ/kg
considerably since a feed pump is required at each feed
Total work output = W23 + W34
heater.
= (2800 - 2364) + (1 - 0.21)(2364 - 1808)
 45

- Because of the number of feed pumps required, the Solution

heating of the feed water by mixing is dispensed with, From tables:
and closed heaters are used. h3 = 112 kJ/kg; h1 = 3445.8 kJ/kg; s1 = 7.089 kJ/kg K =
- The feed water is passed at boiler pressure through the s2
feed heaters 2 and 1 in series. Thus
- An amount of bleed steam y1, is passed to feed heater 1, x2 = 0.824 and h2 = 2117 kJ/kg
and the feed water receives heat from it by the transfer of For the first stage of expansion, 1-7, s7 = s1 = 7.089 kJ/kg
heat through the separating tubes. K, and from tables at 10 bar sg < 7.089 kJ/kg K, hence the
- The condensed steam is then throttled to the next feed steam is superheated at state 7. By interpolation between
heater which is also supplied with a second quantity of 250 and 300oC at 10 bar,
bleed steam, y2, and a lower temperature heating of the  7.089  6.926 
h7  2944   3052  2944 = 3032.9 kJ/kg
feed water is carried out.  7.124  6.926 

- When the final feed heating has been accomplished, the For the throttling process, 11-12,

condensed steam is then fed to the condenser. h6 = h11 = h12 = 763 kJ/kg
- The temperature differences between successive heaters For the second stage expansion, 7-8, s7 = s8 = s1 = 7.089

are constant, and the heating process at each is kJ/kg K, and from tables at 1.1 bar sg > 7.089 kJ/kg K,

considered to be complete (i.e. the feed water leaves the hence the steam is wet at state 8. Therefore,

feed heater at the temperature of the bleed steam supplied 1.333 + (x8 x 5.994) = 7.089
to it). and so x8 = 0.961 and h8 = 2591 kJ/kg
For the throttling process, 9-10,

Example 8 h5 = h9 = h10 = 429 kJ/kg

In a regenerative cycle employing two closed feed heaters, Applying an energy balance to the first feed heater,

the steam is supplied to the turbine at 40 bar and 450 oC and remembering that there is no work or heat transfer,
is exhausted to the condenser at 0.035 bar. The y1h7 + h5 = y1h11 + h6

intermediate bleed pressures are obtained such that the So

saturation temperature intervals are approximately equal, h6  h5 763  429

y1    0.147
h7  h11 3032.6  763
giving pressures of 10 and 1.1 bar. Calculate the amount of
Similarly for the second heater, taking h4 = h3,
steam bled at each stage, the work output of the plant in
y2h8 + y1h12 + h4 = h5 + (y1 + y2)h9
kJ/kg of boiler steam and the thermal efficiency of the
i.e.
plant. Assume ideal processes where required.
y2(h8 – h9) + y1h12 + h4 = h5 + y1h9
T 1
y2(2591–429)+(0.147x763)+112 = 429+(0.147x429)
Therefore y2 = 0.124

1 kg
7
The heat supplied to the boiler, Q1, per kg of boiler steam

y1 kg 1-y1 kg
is
6 11
Q1 = h1 – h6 = 3445 – 763 = 2682 kJ/kg
5 9 y2 kg 8 The work output, neglecting pump work, is given by
4
(1-y) kg 1-y1-y2 kg W = (h1 – h7) + (1 – y1)(h7 – h8) + (1-y1-y2)(h8-h2)
3 10 2
= (3445 - 3032.9) + (1 - 0.147)(3032.9 - 2591)

s
 46

+ (1 - 0.147 - 0.124)(2591 - 2117) enthalpies in the plant can now be found. Before, during
= 1134.5 kJ/kg and after expansion in the turbine these are:

Then  
W 1134.5
  0.423
h4 = 3343 kJ/kg, h5 = 3049 kJ/kg,
Q1 2682
h6 = 2769 kJ/kg, h7 = 2458 kJ/kg,
h8 = 2133 kJ/kg
Example 9
Calculate the ideal cycle efficiency and specific steam
In finding the enthalpies in the feed line, the following
consumption of a regenerative cycle using three closed
assumptions will be made. (a) The feed pump term is
heaters. The steam leaves the boiler at 30 bar superheated
negligible, i.e. h9  h10. (b) In throttling the condensed
to 450oC, and the condenser pressure is 0.04 bar. Choose
bled steam, which is a process of equal initial and final
the bleed pressures so that the difference between the
enthalpy, the state after throttling lies approximately on the
saturation temperature corresponding to 30 bar and that
saturation line; e.g. in throttling from 14 to 15, h14 = h15
corresponding to 0.04 bar is divided into approximately
will be identical with hf corresponding to pressure p6. (c)
equal steps. (Such a choice of bleed pressures makes the
The enthalpy of the compressed liquid in the feed line is
efficiency of the ideal cycle approximately maximum.)
approximately equal to that of saturated liquid at the same
temperature, e.g. h12  h15.

With these assumptions, then

h10 = h18 = h19 = h9 = 121 kJ/kg

h11 = h16 = h17 = 336 kJ/kg
h12 = h14 = h15 = 551 kJ/kg
h1 = h13 = 781 kJ/kg
To determine the correct amounts of steam to be bled for
each heater per kg of steam leaving the boiler, an energy
equation can be written down for each heater.
1st heater: 1h1 + yah14 – yah5 – 1h12 = 0
h13  h15
ya   0.0921
h5  h15

2nd heater: 1h12 + (ya + yb)h16 – ybh6 – 1h11 – yah15 = 0

h15  h17
yb  1  ya   0.0809
h6  h17

3rd heater: 1h11 + (ya + yb + yc)h18 - ych7 - (ya + yb)h17 - 1h10

=0
h17  h9
yc  1  ya  yb   0.0761
h7  h9
Solution
(Always start with the highest pressure heater so as to have
Making the temperature differences (T 2-T13), (T13-T15),
only one unknown at each stage of calculation.)
(T15-T17), (T17-T9), approximately equal, the bleed
Now the heat and work transfers for the cycle can be
pressures become 11, 2.8 and 0.48 bar. All the relevant
calculated.
 47

The heat added in the boiler is: o Indicated thermal efficiency is given by:
Q1,4 = (h4 – h1) = 2562 kJ/kg indicated power i. p.
IT  
The heat rejected in the condenser is energy in the fuel m f  Q net ,v

Q8,9 = (h8 – h9) + (ya + yb + yc)(h19 – h8) = 1511 kJ/kg b. p  BT

Note: M  
The work done in the turbine is i.p IT
W4,8 = (h4 – h5) + (1 – ya)(h5 – h6) + (1 – ya – yb)(h6 –
Efficiency of steam boilers
h7) + (1 – ya – yb – yc)(h7 – h8) = 1051 kJ/kg
- This is the heat supplied to the steam in the boiler
[As a check, W4,8 must also be given by
expressed as a percentage of the chemical energy of the
W4,8 = Q1,4 - Q8,9 = 1051 kJ/kg]
fuel which is available on combustion, i.e.,
Thus
h1  enthalpyof the feedwater
Boiler efficiency =

W4,8

1051
 0.410 and ssc 
3600
 3.43 kg/kW h m f  GCV or NCV 
Q1,4 2562 1051
Where h1 is the enthalpy of the steam entering the turbine
and mf is the mass of fuel burned per kg of steam delivered
Engine trials from the boiler
- Indicated power = rate of work done by the gas on the GCV and NCV are the gross (higher) and net (lower)
piston as evaluated from an indicator diagram obtained calorific values of the fuel.
from the engine. An indicator diagram is a p-v diagram Steam turbines
of the engine cycle.
- A steam turbine is a power unit which produces power
- Brake power = power output of an engine as measured
from a continuous supply of steam, the steam being
by a dynamometer. The difference between indicated
delivered to the turbine at a high pressure and exhausted
power and brake power is the friction power, i.e.
to the condenser at a low pressure.
friction power is the power required to overcome the
- A back pressure turbine is a turbine exhausting into a
frictional resistance of the engine parts.
condenser at a relatively high pressure so that the
- Mechanical efficiency is given by
rejected heat is employed usefully, i.e., the exhaust steam
brake power
M  is used for some heating process, and the turbine work
indicated power
may be a by-product.
- Thermal efficiency of a heat engine is the ratio of the
- In a reaction turbine, radial tubes, which are connected
net work done in the cycle to the gross heat supplied in
to a vertical supply tube, are free to rotate. The end of
the cycle. Two ways of reporting thermal efficiency:
each tube is shaped as a nozzle and the steam from the
o Brake thermal efficiency is given by
supply tube passes along the radial tubes and then
brake power b. p.
 BT   expands through the nozzles to atmosphere in a
energy in the fuel m f  Q net ,v
tangential direction. There is an increase in velocity of
where:
the steam relative to the rotating tube, and hence there is
mf = mass of fuel consumed per unit time
a reaction on the tube, which makes it rotate.
Qnet,v = lower calorific value of the fuel
- In an impulse turbine, blades are attached to a wheel,
which is free to rotate. A jet of steam acts on the blades
forcing them to move and in the process rotate the wheel.
- A pressure compounding turbine (The Rateau turbine)
is a turbine with a series of simple impulse stages. (A
 48

stage refers to one expansion through a row of fixed

blades or nozzles, and a row of moving blades.)
- The isentropic efficiency of a turbine (also known as the
overall efficiency) is given by:
h uo
O 
h Io

where:
hIo = the isentropic overall enthalpy drop for the
turbine between p1 and p2.
huo = actual overall enthalpy drop for the turbine
between p1 and p2.

Supersaturation

-When a superheated vapour expands isentropically and

slowly, condensation within the vapour begins to form
when the saturated line is vapour line reached. As the
expansion continues below this line into the wet region,
the condensation proceeds gradually and the dryness
fraction of the steam becomes progressively less. Above
the saturated vapour line, the vapour is said to be
superheated; below the saturated vapour line, it is said to
be supersaturated or supercooled. The temperature of a
supercooled vapour is always less than the saturation
temperature corresponding to its pressure.

Degree of supercooling

- This is the difference between the actual temperature of a

supercooled vapour and the saturation temperature
corresponding to its pressure.