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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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CHAPTER ONE
Introduction
Energy exists in many forms from the energy locked up in matter
itself to the intense radiant heat emitted by the sun, and between
these limits energy sources are available such as chemical energy
fuels and the potential energy of large water masses evaporated
by the sun. in order for this energy to serve man‘s needs
(purpose) means must be found to transform the energy into a
convenient form.
Thermodynamics is made up of two words ―Thermo‖ and
―Dynamics‖. These words come from Greek and have the
meanings: ―Thermo‖ = hot or heat; ―Dynamics‖ = the study of
matter in motion. Thus ―Thermodynamics‖ can be defined as the
study of heat related to matter in motion, i.e, it is the science of
the of the relationship between heat, work and the properties of
systems. This means that it a science of the transformation of
energy, and the accompanying change in the state of matter. It is a
physical theory of great generality affecting practically every
phase of human experience. It is based on two master concepts,
energy and entropy, and two great principles, the first and second
law of thermodynamics.
The two branches of thermodynamics of interest are: (1) classical
thermodynamics – which concerns itself with the phenomena that
occur in real macroscopic systems, e.g. changes in volume when
a gas is heated at constant pressure; and (2) applied (or
engineering) thermodynamics – which is concerned with work
producing or utilizing machines, e.g. engines, turbines or
compressors together with the working substances used in such
machines. [Other branches of thermodynamics are kinetic theory
and statistical thermodynamics, physical & chemical
thermodynamics.] Thus, Applied thermodynamics is the science
of the relationship between heat work and the properties of
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.
The principles of thermodynamics apply in whole or in part to all
engineering activities because every engineering operation
involves an interaction between energy and matter. They are used
in the design, development and analysis of all power producing
systems, e.g. internal combustion engines, gas turbines, steam
power plants, nuclear power plants, refrigeration systems, air
conditioning systems, propulsion systems (for rockets, missiles,
aircrafts, ships and land vehicles), energy conversion devices
(e.g. fuel cells), energy sources, heat exchangers, production of
low temperatures, etc.
Like all sciences, the basis of thermodynamics is experimental
observation and these findings have been formalized into certain
basic laws which are known as the Zeroth, First, Second (and
Third) laws of thermodynamics.
A Heat Engine is the name given to a system, which by operating
in a cyclic manner produces net work from a supply of heat.
The laws of thermodynamics are natural hypotheses based on
observations of the world in which we live. For example, heat
and work are mutually convertible forms of energy and this is the
basis of the first law of thermodynamics. Like a river never flows
uphill unaided, heat never flows from an object at a low
temperature to one at a higher temperature unaided. This is the
basis of the second law of thermodynamics, which can be used to
show that a heat engine cannot convert all the heat supplied to it
into mechanical work but must always reject some at a lower
temperature.
Thermodynamic System, Heat and Work.
A Thermodynamic System is a collection of matter within
prescribed and identifiable boundaries. The boundaries may be
flexible (e.g the fluid in the cylinder of a reciprocating engine
during the expansion stroke) or fixed and either real or imaginary.
The region outside the boundary is known as the surrounding
There are three types of thermodynamic systems namely; a closed
system, an open system and an isolated system
1. Closed system: A closed system allows energy
transfer across the boundaries, but the mass is
fixed ; for example the fluid expanding behind a
piston of an engine.
System
Surrounding Boundary
Piston
3
2. Open System: An open system is one in which
there is mass transfer across the boundaries. For
example the fluid in a turbine at any instant.
3. An Isolated System: This is a system which can
neither exchange mass nor energy with its
surrounding e.g.
Heat – Heat is a form of energy, which is transferred from a body
at a higher temperature to another body at a lower temperature by
virtue of the temperature difference between the bodies. Heat
may be said to be a transitory characteristic of a body. For
example, if two systems A and B show no change in their
observable characteristics when they are brought into contact with
one another then they are in thermal equilibrium. However if a
third body C at a higher temperature is brought into contact with
A there will be heat transfer from C to A and hence an decrease in
the intrinsic energy of C and an increase in the intrinsic energy of
A until they are at equilibrium.
This principle of thermal equilibrium is called the Zeroth law of
thermodynamics. The possibility of devising a means of
measuring temperatures rests upon this principle.
Work: Work is the product of force and the distance moved in the
direction of the force. Work can be done on the system by the
surrounding or vice versa. Work is measured in Nm or Nm/kg for
a unit mass of fluid. Note that heat and work are transitory
energies and should not be confused with the intrinsic energy
possessed by a system.
Properties of the working fluid
In practice, the matter contained within the boundaries of a
thermodynamic system can either be liquid, vapour or gas, and is
known as the working fluid. At any instant the state of the
working fluid may be defined by certain characteristics. These
characteristics are referred to as thermodynamic properties of the
working fluid.
To define the state of a thermodynamic system, only two
thermodynamic properties are required. Any such property must
be measurable and have a unique numerical value when the fluid
is in any particular state. The value of a property must be
independent of the process through which the fluid has passed in
reaching that state. This implies that a change in the value of a
property depends only on the initial and final states of the system.
Some of the common properties are pressure, volume,
temperature, internal energy, enthalpy and entropy. Thus, the
state of a system may be represented by a point on a diagram of
system properties as shown in fig 1.2.
Pressure: Pressure of a system is the force exerted by the system
on a unit area of its boundaries. Pressure is measured in N/m3 or
bar (1bar = 105 N/m2). Gauge pressure is the pressure read on a
measuring gauge. Absolute pressure is the sum of the gauge
pressure and the atmospheric pressure. Vacuum pressure is the
pressure below the atmospheric pressure. A barometer is an
instrument used to measure atmospheric pressure using, for
example, the height of column of liquid. Standard atmosphere
(101325 N/m2) gives a height of 760 mm of mercury or 10.326 m
of water.
To measure differences in pressure a U-tube manometer can be
used. When the pressure difference is small a liquid of a lower
density can be used and when the difference is even smaller, an
inclined manometer can be used. When one end of the
manometer is open to the atmosphere, the pressure measured is
known as gauge pressure, which is the pressure above or below
the atmospheric pressure.
The actual pressure (also known as absolute pressure) is given by:
Absolute pressure = atmosphere pressure ± gauge pressure. When
gauge pressures are too big such that the mercury manometer
becomes too long, the Bourdon pressure gauge is used.
Specific Volume: This is the volume occupied by a unit mass of
the system and is denoted by, v, while V denotes the absolute
volume. Specific volume is measured in m3/kg.
1
1 1
1
2 2
2
2
P
T
P
T
v
T v
S
4
Temperature: Though a familiar term, its exact definition is
difficult. It is the degree of hotness of a body or system and is
measured using a thermometer. A thermometer employs the
thermometric properties of a substance such as mercury and
alcohol or an electrical resistance as in the case of a
thermocouple. The temperature scales are calibrated in degrees
Fahrenheit (oF) or degrees centigrade (oC). A thermodynamic
scale provides a basis for absolute temperature measurement in
Kelvin, K. Normally capital T is used for absolute temperature
while small t is used for other temperatures. Two bodies are said
to have equal temperatures if there is no net heat transfer between
them when they are brought into contact with one another; they
are then said to be in thermal equilibrium.
The possibility of devising a means of measuring temperature
rests upon the principle of thermal equilibrium which is stated in
the Zeroth law of thermodynamics: If two bodies are separately in
thermal equilibrium with a third body then they must be in
thermal equilibrium with each other.
The temperatures of any group of bodies may be compared by
bringing a particular system known as a thermometer into contact
with each in turn. The thermometer must posses an easily
observable characteristic known as a thermometric property, e.g.
pressure of a gas in a closed vessel; the length of a column of
mercury in a capillary tube; the resistance of a platinum wire, etc.
To report temperature, a scale must be devised so that any device
used to measure temperature will record the same value when
used in the same conditions.
Scale of temperature
The two widely used temperature scales are Fahrenheit and
Celsius; the Celsius scale in more common. The lower fixed point
on Celsius scale is the temperature of the melting of pure ice (ice
point) at standard atmosphere = 0oC. The upper fixed point is the
temperature at which pure water boils (steam point) at standard
atmosphere = 100oC.
If X = value of any thermometric property, then on the Celsius
scale,
Temperature T oC = 0100
0
XX
XX T
Where
XT = value of property at temperature T
Xo = value of property at 0oC
X100 = value of property at 100oC
The second law of thermodynamics shows that there is the
possibility of an absolute zero of temperature which suggests an
absolute temperature scale. An absolute zero of temperature
would be the lowest temperature possible and this would
therefore be a more reasonable temperature to adopt as the zero
for a temperature scale.
The absolute thermodynamic temperature scale is called the
Kelvin scale and on this scale the unit of temperature is Kelvin
(K) (not oK). The zero on the Celsius scale is defined as 00C =
273.15K. The temperatures on the absolute scale are related to the
temperatures on the Celsius scale by K = oC + 273.15, e.g. 373.15
K = 100oC.
On the absolute scale, there is only one fixed point which is
known as the standard fixed point and is usually chosen as the
triple point of water, the temperature at which ice, liquid water
and water vapour coexist in equilibrium.
The triple point of water has been chosen to be 273.16 K exactly
and so if X =value of any thermometric property, then on the
absolute scale;
Temperature θ K = 273.16 tpX
X
Where
Xθ = value of thermometric property at temperature θ K
Xtp = value of thermometric property at temperature 273.16 K
Reversibility
When a fluid undergoes a reversible process both the fluid and
it‘s surrounding can be restored back to their original state.
A reversible process is one in which a system changes in such a
way that at any instant during the process, the state point can be
located on a diagram of system properties. A reversible process
between two states can be drawn as a line on a diagram of
properties as shown below. The fluid undergoing the process
passes through a continuous series of equilibrium states
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Criteria for reversibility
A process is said to be reversible when;
1. The process is frictionless This implies that the fluid
itself must have no internal friction and there must be
no mechanical friction (eg. between cylinder and
piston)
2. The difference in pressure between the fluid and its
surroundings during the process is infinitesimally
small. This means that the process must take place
infinitely slowly since the force to accelerate the
boundaries of the system is infinitely small.
3. The difference in temperature between the fluid and its
surroundings during the process is infinitesimally
small. This means that the heat supplied or rejected to
or from the fluid must be transferred infinitely slowly
In practice, all process are irreversible and are usually represented
by a dotted line joining the end state to indicate that the
intermediate states are indeterminate.
Consider an ideal frictionless fluid contained in a cylinder behind
a piston shown below
Assume ideal conditions for a reversible process and that the
pressure and temperature of the fluid are uniform. Let the area of
cross-section of the piston be A and the pressure of the fluid at
any instance be P.
Let the piston move under the action of the force exerted by the
fluid pressure, a distance dl to the right.
The work done by the fluid on the piston is given by;
Work done = Force × Distance Moved
= (PA) × dl
But Adl = dV
Work done = PdV
Where dV is a small increase in volume
For a unit mass;
Work done – Pdv
Where v is the specific volume.
Hence when a reversible process takes place between states 1 and
2, the work done by a unit mass of fluid is given by;
Work done W = 2
1Pdv
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)
V(m3)
P
dV
1
2
When a fluid undergoes a series of processes and finally returns
to its initial state, then it is said to have undergone a
thermodynamic cycle.
1
2
P
(N/m2)
V (m/s)
1
2
P
(N/m2)
V (m/s)
Fluid
Pressure
F
6
Example 1.
A fluid of volume 0.05m3 is contained behind a piston at a
pressure of 10 bar. After a reversible expansion at constant
pressure, the final volume is 0.2 m3. Calculate the work done by
the fluid. [150 KJ].
Example 2
A fluid at a pressure of 3 bar, and with a specific volume of 0.18
m3/Kg, contained in a cylinder behind a piston expands reversibly
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]
When a compression process takes place reversibly, the work
done on the fluid is given by the shaded area.
P
( N/m 2 )
V ( m 3 )
P
dV
2
1
Work done on the fluid = 1
2Pdv
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
positive). Conversely a process from right to left is one in which
the fluid has work done on the fluid by the surrounding (i.e. W is
negative).
A reversible cycle consisting of four reversible processes 1 to 2, 2
to 3, 3 to 4, and 4 to1 is shown below.
1
2
3
4
P
V
The net work done is equal to the shaded area
Example 3
1 Kg of a certain fluid is contained in a cylinder at an initial
pressure of 20 bar. The fluid is allowed to expand reversibly
behind a piston according to a law pv2 = constant until the
volume is doubled. The fluid is then cooled reversibly at constant
pressure until the piston regains the original position; heat is then
supplied reversibly with the piston firmly locked in position until
the pressure rises to the original value. Calculate the net work
done by the fluid for an initial volume of 0.05m3. [25 KJ]
Example 4
A certain fluid at 10 bar is contained in a cylinder behind a
piston, the initial volume being 0.05m3. Calculate the work
done by the fluid when it expands reversibly;
a. At a constant pressure to a final volume of 0.2m3
b. According to a linear law to a final volume of 0.2m3 and a
final pressure of 2 bar.
c. According to a law pv = constant to a final volume of
0.1m3
d. According to a law pv3 = constant to a final volume of 0.
06m3
e. According to a law
v
b
v
ap
2 to a final
volume of 0.1m3 and a final pressure of 1 bar if a and b
are constants.
Sketch all the process on a p-v diagram.
[150 KJ; 90 KJ; 34.7KJ; 7.64K; 19.2KJ ]
The First Law of Thermodynamics
The hypothesis that states that energy can neither be created or
destroyed is basically principle of conservation of energy The
First Law of Thermodynamics is merely one statement of this
general principle with particular reference to heat energy- and
mechanical energy (i.e. work).
The first law of thermodynamics which is a statement of the
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
by the system on the surroundings.
00
dWdQ
Where 0
is the sum for a complete cycle.
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
is denoted by the symbol u for unit mass or U for mass m of the
fluid. It is a property which depends on the pressure and
temperature of the fluid.
If the final internal energy of the system is greater than the initial
internal energy, then the difference between the net heat supplied
and the net work output is the increase in the internal energy.
Thus Q=(U2 –U1)+ w
For 1 kg,
Q=(u2 –u1)+ W
This equation is known as the Non-Flow Energy Equation
(NFEE). Its differential form can be written as,
dQ = du + dW
For a reversible non – flow process, 2
1
pdvW
For small quantities, this expression becomes,
dW = pdv
Substituting this in 3 gives
dQ=du+ pdv
Therefore,
2
1
12 )( pdvuuQ
The Steady-Flow Energy Equation, SFEE
In most practical problems, the rate at which a fluid flows through
a machine Or piece of apparatus is assumed constant. This type of
flow is known as Steady Flow.
Consider a steady flow of one kg of fluid through a piece of
apparatus in figure below.
Let Q be the heat supplied to the system per kg of fluid and W be
the work done by the fluid as it passes through the apparatus.
Then at inlet: At outlet:
Internal energy" = u. Internal energy = u2
Flow work = P1V1 Flow work = P2V2
Kinetic energy (K.E.) = ½ C12 Kinetic energy (K.E.) = ½ C2
2
Potential energy (P.E) = Z1g Potential energy (P.E) = Z2g
Heat supplied = Q Work Done = W
For steady flow, energy entering must be exactly equal to energy
leaving the system.
The sum of internal energy and the pv term is known as enthalpy
and is denoted by the symbol h.
Thus enthalpy, h = u + pv
However, enthalpy for any mass, m, other than the unit mass is
denoted H and is given by H = mh
Therefore, neglecting changes in elevation (i.e. Z1g = Z2g) the
equation becomes,
WvpC
uQvpC
u 22
2
2211
2
11
22
The rate of mass flow in the steady flow system can be
determined from the continuity of mass equation.
v
CAm
where C is the velocity of fluid, A the area of cross section and v
the specific volume.
Steam As A Working Fluid
The matter contained within the boundaries of a thermodynamic
system is defined as the working fluid. The working fluid may be
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
same liquid when it is boiling at a low pressure. A series of
boiling points plotted on a p-v diagram will appear as a sloping
8
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
joined to form a line.
When the two curves already drawn are extended to higher pres-
sures they form a continuous curve, thus forming a loop. The
pressure at which the turning point occurs is called the critical
pressure and the turning point itself is called the critical point
(point C on the diagram). It can be seen that at the critical point
the latent heat is zero. The substance existing at a state point
inside the loop consists of a mixture of liquid and dry vapour and
is known as a wet vapour.
A saturation state is defined as a state at which a change of phase
may occur without change of pressure or temperature. Hence the
boiling points P, Q, and R are saturation states, and a series of
such boiling points joined up is called the saturated liquid line.
Similarly the points P', Q', and R', at which the liquid is
completely changed into vapour, are saturation states, and a series
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
some of it into a vapour, and it is no longer a liquid but is now a
wet vapour. Similarly when a substance just on the saturated
vapour line is cooled slightly, droplets of liquid will begin to
form, and the saturated vapour becomes a wet vapour. A saturated
vapour is usually called dry saturated to emphasize the fact that no
liquid is present in the vapour in this state.
Lines of constant temperature, called isothermals, can be plotted
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
between the actual temperature of the superheated vapour and the
saturation temperature at the pressure of the vapour is called the
degree of superheat. For example, the vapour at point S is
superheated at pQ and T3, and the degree of superheat is T3 - T2•
The condition or quality of a wet vapour is most frequently
defined by its dryness traction, and when this is known as well as
the pressure or temperature then the state of the wet vapour is
fully defined.
Dryness fraction, x = The mass of dry vapour in 1 kg of the
mixture
(Sometimes a wetness fraction is defined as the mass of liquid in
1 kg of the mixture, i.e. Wetness fraction = (1-x.)
Note that for a dry saturated vapour, x = 1; and that for a
saturated liquid, x = 0.
The use of vapour tables
Tables of properties of different working fluids are available (e.g.
those by Mayhew and Rogers) which can be used to determine
the state properties during any thermodynamic process. In such
tables, the properties are designated with subscripts f for saturated
liquid; g for dry saturated; and hfg for the intermediate stages.
Superheat tables contain properties for superheated vapour.
9
Saturation state properties
The saturation pressures and corresponding saturation
temperatures of steam are tabulated in parallel columns in the first
table, for pressures ranging from 0·006112 bar to the critical
pressure of 221·2 bar. The. specific volume, internal energy,
enthalpy, and entropy are also tabulated for the dry saturated
vapour at each pressure and corresponding saturation
temperature. The suffix g is used to denote the dry saturated state.
The following are some of the formulae used to calculate the
properties of a wet vapour:
(i) Volume, v, is given by
v = vf (1-x) + x vg
Where x is the dryness fraction, vf volume of the liquid and vg the
volume of the dry saturated vapour.
For most practical problems, volume of the liquid is usually
negligibly small compared with the volume of dry saturated
vapour.
Therefore, v = xvg
u = (l-x)uf + xug = uf + x(ug - uf)
(iii.) Similarly enthalpy, h, is given by
h = (l-x)hf + xhg
= hf + x(hg-hf) = hf +xhfg
The change in specific enthalpy from hf to hg is given the symbol
hfg•
When saturated water is changed to dry saturated vapour, from
equation
Q=(u2-u1) + W = (ug – uf) + W
Also W is. represented by the area under the horizontal line on the
P-V diagram,
W= (vg – vf)p
So that Q = (ug-uf) + p(vg-vf)
= (ug + pvg) – (uf + pvf)
But h = u + pv
Therefore Q = hg – hf = hfg
Thus hfg is the heat required to change a saturated liquid to a dry
saturated vapour and is called the latent heat.
In the case of steam tables, the internal energy of saturated liquid
is taken to be zero at the Triple point (Le. at 0·01 °e and
0·006112 bar). Then since, from equation 2.7, h=u+pv, we have,
h at 0·01 °e and 0.006112 bar = 3
5
10
0010002.010006112.00
(where Vf at O'01°e is 0·0010002 m3/kg)
i.e h = 6.112 × 10-04 kJ/kg
This is negligibly small and hence the zero for enthalpy may be
taken as 0.01oC.
Note that at the other end of the pressure range tabulated in the
first table the pressure of 221·2 bar is the critical pressure,
374·15°e is the critical temperature, and the latent heat, hfg, is
zero.
Properties of wet vapour
For a wet vapour the total volume of the mixture is given by the
volume of liquid present plus the volume of dry vapour present.
Properties of superheated vapour
For steam in the superheat region temperature and pressure are
independent properties. When the temperature and pressure are
given for superheated steam then the state is defined and all the
other properties can be found. For example, steam at 2 bar and
200°C is superheated since the saturation temperature at 2 bar is
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
tables of properties of superheated steam range in pressure from
0·006112 bar to the critical pressure of 221·2 bar, and there is an
additional table of supercritical pressures up to 1000 bar. At each
pressure there is a range of temperatures up to high degrees of
superheat, and the values of specific volume, internal energy,
enthalpy, and entropy are tabulated at each pressure and
temperature for pressures up to and including 70 bar; above this
pressure the internal energy is not tabulated. For reference the
saturation temperature is inserted in brackets under each pressure
in the superheat tables and values of Vg, Ug, hg and Sg are also given.
Interpolation
For properties which are not tabulated exactly in the tables it is
necessary to interpolate between the values tabulated.
The Perfect gas and the Characteristic gas equation
In practice, there is no perfect gas, however many gases tend
towards a perfect condition. A perfect gas is an imaginary ideal
gas which obeys the law,
RtConsT
PV tan
Where p and V are the pressure and volume ofthe gas
10
respectively and R is the specific gas constant
PV=RT
For gas occupying mass m kg, occupying volume V m3 the
equation becomes,
PV=mRT
Another form of the characteristic equation is derived from the
kilogrammole. The kilogramme-mole is defined as a quantity of a
gas equivalent to M kg of the gas (M is the molecular weight of
the gas.) Thus for oxygen gas whose molecular weight is 32, 1
kg-mole is equivalent to 32 kg of oxygen.
From the of the kilogram-mole, for m kg of a gas
m=nM
Where n is the number of moles
Hence, pV= nMRT or nT
pVMR
However, V/n is the same for all gases at the same pressure, p,
and temperature, T (Avogadro's hypothesis,). Hence the quantity
pV/nT is constant for all gases. This constant denoted by the
symbol Ro' is called the universal gas constant.
pV=nRoT
From MR=Ro' R= Ro/M
From empirical information, the volume of 1 mole of any perfect
gas at pressure of 1× 105 N/m2 and temperature 0oC is
approximately 22.71 m3• Therefore Ro is given by;
moleKNmnT
pVRo /14.8314
15.2731
71.22101 5
Relationship between specific heats
Specific heat is generally defined as
dQ=mCpdT
For a perfect gas, the specific heat at constant pressure, cp' and the
specific heat at constant volume, cy' can be assumed to be
constant at all pressures and temperatures.
Hence for a reversible non-flow process at constant pressure,
Q = mCp(T2-T1)
And for a reversible non-flow process at constant volume,
Q = mCv(T2-T1)
It can be shown that for any process for a perfect gas, between
states 1 and 2
U2-U1 = mCv(T2-T1)
In a constant pressure process, the work done by the fluid is
given by,
W = p(V2-V1) = mR(T2-T1)
Substituting this in the N.F.E.E we have
Q= (U2-U1)+W = mCv(T2-T1) + mR(T2-T1) =m(Cv+R)(T2-T1)
But Q = mCp(T2-T1), hence
Q = mCp(T2-T1) = m(Cv+R)(T2-T1)
So that Cp = Cv + R
And R = Cp - Cv
Also the ratio of the specific heats is
v
p
C
C
Also from Cp – Cv = R
vv
p
C
R
C
C1
vC
R 1 so that
1
RCv
And Cp = ɤCv = 1
R
Applications of the First Law of Thermodynamics
To appreciate the applications of the concepts relating to
the first law of thermodynamics we now consider
processes which are approximated to in practice.
Reversible Non-flow processes
Constant volume process
In a constant volume process the working substance is
contained in a rigid vessel, hence the boundaries of the
system are immovable and no work can be done on or by
the system, other than paddlewheel work input. It will be
assumed that' constant volume' implies zero work-unless
stated otherwise.
From the non-flow energy equation,
Q = (u2-u1)+ W
11
Since no work is done, we therefore have
Q = u2-u1
or for mass, m, of the working substance, Q = U2-U1
All the heat supplied in a constant volume process goes to
increasing the internal energy.
Fig a Fig b
A constant volume process for a vapour is shown on a p-v
diagram is shown in fig. (a). The initial and final states
have been chosen to be in the wet region and superheat
region respectively. In fig.(b) a constant volume process is
shown on a p-v diagram for a perfect gas. For a perfect gas
we have,
Q = U2-U1 = mcV(T2 - T1)
Constant pressure process
In a constant volume process the boundary of the system is
inflexible and thus, the pressure rises when heat is
supplied. Hence for a constant pressure process the
boundary must move against an external resistance as heat
is supplied; for instance a fluid in a cylinder behind a
piston can be made to undergo a constant pressure process.
Since the piston is pushed through a certain distance by the
force exerted by the fluid, then work is done by the fluid
on its surroundings.
Fig ( c) Fig ( d)
From the non-flow energy equation, 2.2,
Hence for a reversible constant pressure process
Now enthalpy, h = u +pv, hence,
or for mass, m, of a fluid,
A constant pressure process for a vapour is shown on a p-v
diagram below. The initial and final states have been
chosen to be in the wet region and the superheat region
respectively. In fig. (d) a constant pressure process for a
perfect gas is shown on a p-v diagram.
For a perfect gas we have
Q = mCp(T2 - T1)
Constant temperature or isothermal process
A process at constant temperature is called an isothermal
process. When a fluid in a cylinder behind a piston
expands from a high pressure to a low pressure there is a
P1=P2
v1 V2
P
V(m3)
P1=P2
v1 V2
P
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
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).
Fig (e) Fig ( f)
From state 1 to state A the pressure remains at PI, since in
the wet region the pressure and temperature are the
corresponding saturation values. It can be seen therefore
that an isothermal process for wet steam is also at constant
pressure and heat supplied from state 1 to state A per kg of
steam =hA –h1 In the superheat region the pressure falls to
P2 as shown
Fig. (f) shows an isothermal process for a perfect gas on a
p-v diagram. The equation of the process is pv=constant,
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
to state 2 isothermally and reversibly is given by the
shaded area on fig.(f) shown.
But C= PV = P1V1 = P2V2
Also since P1V1 = P2V2 then
2
1
1
2
P
P
v
v
For a mss m of the gas
But for a unit mass of gas, P1 V1 = RT
Hence, we have, W = RT
2
1lnP
P. or for mass, m, of the gas
W = mRT
2
1lnP
P
For a perfect gas from Joule's law, we have, U=CvT
Hence for an isothermal process for a perfect gas, since T2
= T1 then U2-Ul = 0
i.e. the internal energy remains constant in an isothermal
process for a perfect gas.
From the non-flow energy equation for an isothermal
process for a perfect gas.
Hence in an isothermal process for a perfect gas the heat
flow is equivalent to the work done.
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
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
Also for a reversible process dW = p dv, hence for an
adiabatic process
u—cvT or du_—c,dT
Dividing through by T to give a form that can be
integrated,
we have T= (pv)/R, therefore substituting,
Dividing through by cv
but
We therefore have a simple relationship between P and v
for any perfect gas undergoing a reversible adiabatic
process, each perfect gas having its own value of γ
From, pv=RT, we have,
Therefore for a reversible adiabatic process for a perfect
gas between states 1 and 2 we have:
The work done in an adiabatic process is; W = (u2 – u1).
And for a perfect gas the gain in internal energy is;
14
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,
Therefore, since pvγ = constant, c, then
Polytropic processes
Many processes in practice approximate to a reversible law
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,
For a process in which pvn = constant, we have p = c/v
n,
where c is a constant.
But, c = PVn = P1V1
n = P2V2 and
n
v
v
P
P
1
2
2
1
For a perfect gas;
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.
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
and
From the non-flow energy equation, the heat flow during
15
the process is given by,
Hence substituting,
In an expansion, work is done by the gas, and hence the
term W is positive. Thus when the polytropic index n is
less than γ, in an expansion, then the right-hand side of the
equation is positive (i.e. heat is supplied during the
process). Conversely, when n is greater than γ in an
expansion, then heat is rejected by the gas. Similarly, the
work done in a compression process is negative, therefore
when n is less than γ, in compression, heat is rejected; and
when n is greater than γ, in compression, heat must be
supplied to the gas during the process.
N.B γ for all perfect gases has a value greater than unity.
In a polytropic process the index n depends only on the
heat and work quantities during the process. The various
processes considered earlier are special cases of the
polytropic process for a perfect gas. For example,
This is illustrated on a p-v diagram in fig.below.
Thus,
Similarly, I to A‘ is constant pressure heating; 1 to B‘ is
isothermal expansion; 1 to C‘ is reversible adiabatic
expansion; 1 to D‘ is constant volume cooling. Note that,
since y is always greater than unity, then process 1 to C
must lie between processes 1 to B and 1 to D, similarly,
process 1 to C‘ must lie between processes 1 to B‘ and 1 to
D‘.
For a vapour a generalization such as the above is not
possible.
One important process for a vapour should be mentioned
here. A vapour may undergo a process according to a law
pv = constant. In this case, since the characteristic equation
of state, pv = RT, does not apply to a vapour, then the
process is not isothermal. Tables must be used to find the
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
negligible. Certain processes cannot be assumed to be
16
internally reversible, and the important cases will now be
briefly discussed.
Unresisted, or free, expansion
This process was mentioned in Section 1.5 in order to
show that in an irreversible process the work done is not
given by J P dv. Consider two vessels A and B,
interconnected by a short pipe with a valve X, and
perfectly thermally insulated. Initially let the vessel A be
filled with a fluid at a certain pressure, and let B be
completely evacuated. When the valve X is opened the
fluid in A will expand rapidly to fill both vessels A and B.
the pressure finally will be lower than the initial pressure in
vessel A. This is known as an unresisted expansion or free
expansion
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
Now in this process no work is done on or by the fluid,
since the boundary of the system does not move. No heat
flows into or from the fluid since the system is lagged.
The process is therefore adiabatic, but irreversible.
u2 – u1= 0 or u2 = ul
In a free expansion therefore the internal energy initially
equals the internal energy finally
For a perfect gas, we have, u = CvT
Therefore for a free expansion of a perfect gas T1 = T2
That is, for a perfect gas undergoing a free expansion the
initial temperature is equal to the final temperature.
Throttling
A flow of fluid is said to be throttled when there is some
restriction to the flow, when the velocities before and after
the restriction are either equal or negligibly small, and
when there is a negligible heat loss to the surroundings.
The restriction to flow can be a partly open valve, an
orifice, or any other sudden reduction in the cross-section
of the flow.
An example of throttling is shown in fig below. The fluid,
flowing steadily along a well-lagged pipe, passes through
an orifice at section X. Since the pipe is well lagged it can
be assumed that no heat flows to or from the fluid.
Applying the steady flow equation between any two
sections of the flow,
Now since Q=O, and W=O, then,
22
2
22
2
11
Ch
Ch
When the velocities Cl and C2 are small, or when Cl is very
nearly equal to C2, then the kinetic energy terms may be
neglected. (Note that sections 1 and 2 can be chosen well
upstream and well downstream of the disturbance to the
flow, so that this latter assumption is justified.)
Therefore for a throttling process, the enthalpy initially is
equal to the enthalpy finally.
The process is adiabatic, but is highly irreversible because
of the eddying of the fluid round the orifice at X. Between
17
sections 1 and X the enthalpy drops and the kinetic energy
increases as the fluid accelerates through the orifice.
Between sections X and 2 the enthalpy increases as the
kinetic energy is destroyed by fluid eddies.
For a perfect gas, from, h=CpT, therefore,
CpTl = CpT2 or Tl = T2
For throttling of a perfect gas, therefore, the temperature
initially equals the temperature finally and no work is done
during the process. For a vapour, throttling can be used as
a means of finding the dryness fraction of wet steam.
Adiabatic mixing
The mixing of two streams of fluid is quite common in
engineering practice, and can usually be assumed to
occur adiabatically. Consider two streams of a fluid
mixing as shown in fig. below. Let the streams have
mass flow rates m1 and m2, and temperatures T1 and
T2• Let the resulting mixed stream have a temperature
Ta. There is no heat flow to or from the. fluid, and no
work is done, hence neglecting changes in kinetic
energy from the flow equation, we have,
For a perfect gas, from h=CpT, hence,
The mixing process is highly irreversible due to
the large amount of eddying and churning of the
fluid that takes place.
Reversible flow processes
Although flow processes in practice are usually highly
irreversible it is sometimes convenient to assume that a
flow process is reversible in order to provide an ideal
comparison. Some work is done on or by the gas by virtue
of the forces acting between the moving gas and its
surroundings. For example, for a reversible adiabatic flow
process for a perfect gas,
WC
hQC
h 22
2
22
2
11
Then since Q = 0
2
2
1
2
221
CChhW
Also since the process is assumed to be reversible, then for
a perfect gas PVY = constant. This equation can be used to
fix the end states.
NB: Even if the kinetic energies terms are negligibly
small, the work done in a reversible adiabatic flow process
between two states is not equal to the work done in a
reversible adiabatic non-flow process between the same
states. Note that the kinetic energy change is small
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.
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
steady flow process, but varies with time. This happens
when;
The rate of mass flow crossing the boundary of a
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
- The steady flow energy equation is:
Q W h h C C g z z 2 112 2
212
2 1
where Q and W are the heat and work
transfers per unit mass flowing through
the system.
- As in previous section, analysis will be done while
considering the fluid to be either steam or air.
- Irreversibility due to viscous friction is not neglected
unless the velocity of flow is very small.
- In many cases it is necessary to calculate the unknown
quantity assuming the process to be reversible, and
multiply the result by a process efficiency.
- In other cases irreversibility is dealt with by treating the
process as a polytropic process.
Boiler and condenser
For any fluid
W = 0
Thus Q h h C C 2 112 2
212
- The velocity in these devices is usually so small that the
effect of friction can be neglected and the process
regarded as internally reversible.
- This implies that there is no pressure drop due to friction,
and the pressure can be assumed constant throughout the
system.
Example 1: 1500 kg of steam are to be produced per hour
at a pressure of 30 bar with 100 K of superheat. The feed
water is supplied to the boiler at a temperature of 40oC.
Find the rate at which heat must be supplied, assuming
typical values for the velocity at inlet and outlet: 2 m/s in
the feed pipe and 45 m/s in the steam main.
Nozzle and diffuser
- For any fluid
Q = 0 (adiabatic process)
W = 0
Thus 12 2
212
1 2C C h h
- When the fluid is a perfect gas
12 2
212
1 2C C c T Tp
- The state of the fluid is usually known at the inlet of
these devices but only one property e.g. pressure, is
known at the outlet.
- To fix the final thermodynamic state of the fluid, and
hence determine h2, it is necessary to assume the process
to be reversible; the flow is then isentropic and the fact
that s1 = s2 can be used.
- Since the velocity of flow is very high in these devices,
the effect of friction cannot be neglected. One way of
accounting for this is to assume the process to be
reversible and then multiply the result by a process
efficiency to obtain a more realistic estimate.
- The value of the process efficiency must be determined
from tests carried out on a similar device.
- For nozzles the process efficiency is given by
N
C
C
22
2
2
Where C2 is the actual outlet velocity and C‘2 is the outlet
velocity which would have been achieved had the final
pressure been reached isentropically.
- For a diffuser the process efficiency is given by:
D
p p
p p
2 1
2 1
Where p2 is the actual outlet pressure and p‘2 is the
pressure which would result from an isentropic process
leading to the same outlet velocity.
Example 2: A fluid expands from 3 bar to 1 bar in a
nozzle. The initial velocity is 90 m/s, the initial
temperature is 150oC, and from experiments on similar
nozzles it has been found that the isentropic efficiency is
likely to be 0.95. Find the final velocity when the fluid is:
(a) steam; (b) air.
19
Turbine and rotary compressor
For any fluid
Q = 0 (adiabatic process)
W = (h2 - h1) (Since the velocity at inlet is
approximately the same as that at outlet.)
When the fluid is a perfect gas
W = cp(T1 - T2)
The process efficiencies of turbines and rotary compressors
are given by:
T
T T
T T
1 2
1 2
and C
T T
T T
2 1
2 1
respectively
Example 3: A fluid enters a turbine at the rate of 14 kg/s
with an initial pressure and temperature of 3 bar and
150oC. If the final pressure is 1 bar and the isentropic
efficiency of the turbine is 0.85, find the power developed
and the change of entropy between inlet and outlet when
the fluid is: (a) steam; (b) air.
Throttling
- For any fluid
h1 = h2
- When the fluid is a perfect gas
T1 = T2
- A throttling calorimeter is used to determine the dryness
fraction of a wet vapour. In this device, a sample of the
vapour (which is not too wet) is throttled in a well lagged
chamber such that the final state is in the superheat
region in which pressure and temperature are
independent. (Pressure and temperature are not
independent properties in the wet region.)
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.
Reciprocating compressor (or expander)
- The process is approximately isothermal and low
velocities can be assumed.
- Thus for any fluid
Q – W = (h2 – h1)
- When the fluid is a perfect gas enthalpy is a function of
temperature and so h2 = h1 and Q = W. Also since T is
constant, 1
212 ln
p
pRss and hence
1
2lnp
pRTQW
Example 5: A fluid, initially at 155.5oC and 1 bar, is
compressed reversibly and isothermally in a steady-flow
process to a state where the specific volume is 0.28 m3/kg.
Find the heat transferred and work done, per kg of fluid
when the fluid is: (a) steam (b) air.
Multi-stream steady-flow process
Provided the streams do not react chemically with one
another, the steady flow energy equation can be applied to
give:
inout
gzChmgzChmWQ 2
212
21
and conservation of mass gives:
inout
mm
Example 6: Steam is to be condensed by direct injection of
cold water. The steam enters the condenser at a rate of 450
kg/h with a dryness fraction of 0.9 and a pressure of 1 atm.
The estimated heat loss from the condenser to the
surroundings is 8500 kJ/h. If the cold water enters with a
temperature of 15oC, and the mixture of condensate and
cooling water is to leave at 95oC, determine the rate of
flow of cooling water required.
20
CHAPTER TWO
Second Law of Thermodynamics
Cycle efficiency
For any closed system taken through a cycle, the First
Law of thermodynamics can be expressed symbolically as:
Q1 - Q2 = W (2.1)
Where
Q1 = heat supplied from a heat reservoir
Q2 = heat rejected to a heat sink
W = net work done by the system during the cycle
A system operating in a cycle and producing a net
quantity of work from a supply of heat is called a heat
engine. The greater the proportion of heat supply converted
into work, the better is the engine. Consequently, the cycle
efficiency of a heat engine is defined as:
work done
heat supplied
W
Q
Q Q
Q
Q
Q1
1 2
1
2
1
1 (2.2)
There is nothing implicit in the First Law to say that
some proportion of the heat supplied to an engine must be
rejected, and therefore that the cycle efficiency cannot be
unity; all that the First Law states is that net work cannot
be produced during a cycle without some supply of heat.
The Second Law is an expression of the fact that some heat
must always be rejected during the cycle, and therefore that
the cycle efficiency is always less than unity.
Statements of the Second Law
Kelvin-Planck: It is impossible to construct a system
(device) which will operate in a cycle, extract heat from
a reservoir, and do an equivalent amount of work on the
surroundings.
Clausius: It is impossible to construct a system (device)
which will operate in a cycle and transfer heat from a
cooler to a hotter body without work being done on the
system by the surroundings.
Others:
It is impossible for a heat engine to produce a net work
output in a complete cycle if it exchanges heat only with
a single energy reservoir.
It is impossible to construct a device that operating in a
cycle will produce no effect other than the transfer of
heat from a cooler to a hotter body.
Consequences of the Second Law
1. If a system is taken through a cycle and produces
work, it must be exchanging heat with at least two
reservoirs at different temperatures.
2. If a system is taken through a cycle while exchanging
heat with only one reservoir, the work done must be
either zero or negative.
3. Since heat can never be converted continuously and
completely into work, whereas work can always be
converted continuously and completely into heat,
work is a more valuable form of energy transfer than
heat.
Reversibility and irreversibility
When a fluid undergoes a reversible process, both the
fluid and its surroundings can always be restored to their
original states. If either following phenomena is present
during a process it cannot be reversible:
- friction, and
- heat transfer across a finite temperature difference.
Since at least one of these is always present in some
degree, no real process can be reversible.
A cycle is reversible if it consists only of reversible
processes; if any of the processes in a cycle are
irreversible, the whole cycle is irreversible. The efficiency
of an engine in which irreversible processes occur must
always be less than that of the hypothetical reversible
engine, i.e. it is impossible to construct an engine operating
between only two reservoirs which will have a higher
efficiency than a reversible engine operating between the
same reservoirs.
Carnot cycle
The original concept of a reversible cycle is due to
Carnot who thought of a particular cycle, called the Carnot
cycle, which is composed of four processes:
1 A reversible isothermal heat addition from a source
at temperature TH.
21
2 A reversible adiabatic process in which work is
done by the system.
3 A reversible isothermal heat rejection to a sink at
temperature TL.
4 A reversible adiabatic process in which work is
done on the system.
Thermodynamic temperature scale
All reversible engines operating between the same two
reservoirs have the same efficiency. This efficiency must
depend upon the only feature that is common to them all,
viz. the temperatures of the reservoirs. This efficiency is
called the Carnot efficiency.
Recall that
1 2
1
Q
Q (2.3)
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
T TQ
Q 0
0
(2.4)
The value T is then fully and uniquely determined by:
(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
reservoirs.
If Q is measured for several sinks at different
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
the fraction 1/T0 of the interval from T = 0 to T0. The
temperature scale defined in this way is called the
thermodynamic scale, because it is dependent solely on the
laws of thermodynamics and not upon the properties of any
particular substance. It is an absolute scale because it
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
zero, and therefore it can be concluded that absolute zero is
a conceptual limit and not a temperature that can ever be
reached in practice.
Consider a series of reversible engines shown below, each
operating between only two reservoirs and each producing
the same quantity of work. Each sink is a source for the
following engine, the heat entering a reservoir being equal
to the heat leaving it.
If the temperatures of the reservoirs are defined in the way
suggested by equation (1.1), then
and the efficiencies of the engines therefore become
But since
It follows that
Simplifying we have
22
Thus the differences between the temperatures of
successive reservoirs are the same and can provide
intervals or units of temperature. T0 may be as high as
we wish to make it, and the intervals may be made as
small as desired by increasing the number of engines in
the series. The series of reservoirs could theoretically
be used as a standard set of temperatures with which to
calibrate any practical thermometer.
We define the unit of temperature by choosing two
fixed points such as the ice point of water Ti and the
steam point Ts and define the number of degrees, i.e,
Ts = Ti = 100
This is equivalent to placing 100 reversible engines in
series between the fixed points.
When used in conjunction with the equation
i
s
i
s
Q
Q
T
T ,
this arbitrary choice defines a linear thermodynamic
scale.
Ti = 273.16 K, Where K – Kelvin
Conversion of t to T is given by T = t + 273
Engines operating between more than two reservoirs
In many practical cycles the heat is received and rejected during
processes which involve a continuous change in the temperature
of the fluid. At any instant during a heating and cooling process,
heat must be exchanged between the system and a source or sink
which differs only infinitesimally in temperature from the fluid in
the system.
Characteristics of engines operating between only two
reservoirs are:
(a) All reversible engines, operating between a source at
temperature T1 and a sink at temperature T2, have an
efficiency equal to (T1 - T2)/T1. [This is the efficiency
of a Carnot cycle.]
(b) For a given value of T2, the efficiency increases with
T1. Since the lowest possible temperature of a practical
infinite sink is fixed within close limits, i.e. the
temperature of the atmosphere or the sea, it may be said
that a given quantity of heat is more useful for
producing work the higher the temperature of the
source from which it is received.
Entropy
Definition
This is a property of a closed system such that a change
in its value is equal to
dQ
T1
2
(2.5)
for any reversible process undergone by the system
between state 1 and state 2.
It is denoted by S and so
dQ
TS S
rev
1
2
2 1 (2.6)
or in differential form
dSdQ
T rev
(2.7)
It is an extensive property (like internal energy or
enthalpy) which may be calculated from specific entropies
based on a unit mass of the system so that
S = ms (2.8)
Characteristics of entropy
- If the process undergone by a system is a reversible
adiabatic one (dQ = 0), the entropy change will be zero,
and this is called an isentropic process; if the process is
irreversible and adiabatic, then the entropy must
increase.
- The entropy of any closed system which is thermally
isolated from the surroundings either increases or, if the
process undergone by the system is reversible, remains
constant.
- For a reversible isothermal process,
Q = TS (2.9)
Determination of values of entropy
The property entropy arises as a consequence of the
Second Law, in much the same way as the property
23
internal energy arises from the First Law. There is,
however, an important difference. The change in internal
energy can be found directly from a knowledge of the heat
and work crossing the boundary during any non-flow
process undergone by a closed system. The change in
entropy, on the other hand, can be found from a knowledge
of the quantity of heat transferred only during a reversible
non-flow process.
General equations
- As with internal energy, only changes of entropy are
normally of interest. The entropy at any arbitrary
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
function of other thermodynamic properties which can
be measured in experiments involving real processes.
- Two important relations of this kind can be obtained by
combining the equations expressing the First and
Second Laws. Thus the First Law yields the equation
(dQ)rev = du + p dv (2.10)
and combining this with Eq. (2.7) gives
T ds = du + p dv (2.11)
- 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.
For any reversible process
-
This equation is analogous to 2
1pdvW for any
reversible process
Thus, as there is a diagram on which areas represent work
done in a reversible process, there is also a diagram on
which areas represent heat flow in a reversible process.
These diagrams are the pv and the T-s diagrams
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,
represents work done; for a reversible process 1-2 in fig.
2.2b, the shaded area
T ds, represents heat flow.
- .
Fig 2.2 a P-v diagram Fig 2.2b T-s diagram
(a) Entropy Change for a vapour
Thermodynamic processes can be represented on T-s
diagrams in a way that is dictated by the working fluid. In
its simplest form the T-s diagram consists of a series of
constant pressure lines and saturation curve.
Where
AB – represents the liquid state
B – Saturated liquid state
BC – Represents wet vapour
24
C – Represents dry saturated steam
CD- superheated steam
- For all vapour substances values of specific entropy
may be tabulated along with enthalpy, specific volume,
and other thermodynamic properties of interest.
- 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
represent heat supplied.
Entropy change for perfect gas
- For a perfect gas, enthalpy and internal energy are
functions of temperature alone (i.e. du = cvdT and dh =
cpdT respectively) and pv = RT and so Eq. (2.11) can be
written as
ds cdT
TR
dv
vv (2.13)
For constant specific heats, Eq. (2.13) may be
integrated between two end states to give:
s s cT
TR
v
vv2 1
2
1
2
1
ln ln (2.14)
- Alternatively, Eq. (2.12) may be written as
ds cdT
TR
dp
pp (2.15)
and integrated to give
s s cT
TR
p
pp2 1
2
1
2
1
ln ln (2.16)
Reversible processes on the T-s diagram
We now consider various reversible processes in relation to
the T-s diagram. The constant volume and constant
pressure processes have already been represented on the T-
s diagram.
Reversible isothermal process
A reversible isothermal process will appear as a straight
line on a T-s diagram, and the area under the line must
represent the heat flow during the process.
The figure above represents a reversible isothermal
expansion of wet steam into the superheat region. The
shaded area represents the heat supplied during the
process,
i.e. Heat supplied = T(s2 —s1)
Note that the absolute temperature must be used. The
temperature tabulated in steam tables is t°C, and care must
be taken to convert this into T K.
Isothermal process for a perfect gas
A reversible isothermal process for a perfect gas is shown
on a T-s diagram in fig below. The shaded area represents
the heat supplied during the process,
i.e. Q = T(s2—s1)
For a perfect gas undergoing an isothermal process it is
possible to evaluate s2—s1. From the non-flow for a
reversible process we have,
25
From previous derivation
Note that this result is the same as that derived earlier
Reversible adiabatic process (or isentropic process)
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
into the wet region is shown in fig. below.
Using the fact that the entropy remains constant, the end
states can be found easily from tables.
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
reversible process,
dQ = du + pdv
Also for unit mass of a perfect gas from Joule‘s law du = cv
dT, and pv = RT.
Also
Hence between any two States 1 and 2
This can be illustrated on a T-s diagram as in fig. below
Since in the process T2 < T1, then it is more convenient to
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,
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
effect replaced the process by two simpler processes; from
1 to A and then from A to 2. It is clear from figure that
s2 - s1 = (SA—S1) - (SA-S2)
Any two processes can be chosen to replace a polytropic
process in order to find the entropy change. For example,
going from 1 to B and then from B to 2 as in figure above,
we have
= (SB—s1)—(SB—s1)
At constant temperature between P1 and P2‘
and at constant pressure between T1 and T2 we have
Hence
There are obviously a large number of possible equations
for the change of entropy in a polytropic process. Each
case can be dealt with by sketching the T-s diagram and
replacing the process by two other simpler reversible
processes, as in foregoing figure.
Problems
1. 1 kg of steam at 20 bar, dryness fraction 0.9 is heated
reversibly at constant pressure to a temperature of
300oC. Calculate the heat supplied, the change of
entropy, and show the process on a T-s diagram,
indicating the area which represents the heat flow.
[415 kJ/kg; 0.8173 kJ/kg K]
2. 0.05 kg of steam at 10 bar, dryness fraction 0.84, is
heated reversibly in a rigid vessel until the pressure is
20 bar. Calculate the change of entropy and the heat
supplied. Show the area which represents heat flow on
a T-s diagram. [0.0704 kJ/kg K; 36.85 kJ]
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
change of entropy and the heat supplied. Sketch the
process on a T-s diagram. Take the isentropic index, ,
for nitrogen as 1.4 and assume that nitrogen is a
perfect gas. [0.00125 kJ/K; 0.404 kJ]
4. 1 kg of steam undergoes a reversible isothermal
process from 20 bar and 250oC to a pressure of 30 bar.
Calculate the heat flow, stating whether it is supplied
or rejected, and sketch the process on a T-s diagram.
[- 135 kJ/kg]
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.
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
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
volume becomes 0.25 m3. Calculate the change of
entropy and mark the initial and final states on a T-s
diagram. [0.083 kJ/kg K]
7. Steam at 15 bar is throttled to 1 bar and a temperature
of 150oC. Calculate the initial dryness fraction and the
change of specific entropy. Sketch the process on a T-s
diagram. [0.992; 1.202 kJ/kg K]
8. A turbine is supplied with steam at 40 bar, 400oC,
which expands through the turbine in steady flow to an
exit pressure of 0.2 bar, and a dryness fraction of 0.93.
The inlet velocity is negligible, but the steam leaves at
high velocity through a duct of 0.14 m2 cross-sectional
area. If the mass flow is 3 kg/s, and the mechanical
efficiency is 90%, calculate the power output of the
turbine. Show that the process is irreversible and
calculate the change of specific entropy. Heat losses
from the turbine are negligible.
[2048 W; 0.643 kJ/kg K]
27
CHAPTER THREE
Gas Power Cycles
Introduction
Many work producing devices (engines) use or utilize a
working fluid that is always air e.g the petrol engine, diesel
engine, gas turbine etc. In these engines there is a change
of the composition of the working fluid, because during
combustion it changes from air and fuel to combustion
products. For this reason, such engines are called internal
combustion engines.
The working fluid does not go through a complete
thermodynamic cycle (even though the engine operates in a
mechanical cycle and thus the internal combustion engine
operates on the so-called open cycle. In order to analyze
internal combustion engines, closed cycles that closely
approximate to the open cycles are derived. One such
approach is the air standard cycle.
At intake – air + fuel
Compression – Air and Fuel are compressed
Power stroke – combustion products
Exhaust Stroke
The air standard cycle is based on several assumptions
1. A fixed amount (Mass) of air is the working fluid
throughout the entire cycle. Thus there is no inlet
or outlet (exhaust) process
2. Air is always an ideal gas
3. the cycle is completed by heat transfer from the
surrounding
4. All processes are internally reversible
5. Air has a constant specific heat
6. The source of heat supply and the sink for heat
rejection are assumed to be external to the air.
The air standard cycle
- Cycles in which the fuel is burned directly in the
working fluid are not heat engines in the true meaning
of the term since the system is not reduced to its initial
state.
- The working fluid undergoes a chemical change by
combustion and the resulting products are exhausted to
the atmosphere.
- In practice such cycles are used frequently and are
called internal-combustion cycles; the fuel is burned
directly in the working fluid, which is normally air.
- By supplying fuel inside the cylinder, higher
temperatures for the working fluid can be attained; the
maximum temperature of all cycles is limited by the
metallurgical limit of the material used and the
efficiency of the cooling system.
- Examples of internal combustion cycles are the open
cycle gas turbine unit, the petrol engine, the diesel or
oil engine, and the gas engine.
o In the open cycle gas turbine the working fluid
flows at a steady rate from one component to
another round the cycle.
o In the petrol engine a mixture of air and petrol is
drawn into the cylinder, compressed by the
piston, then ignited by an electric spark. The hot
gases expand, pushing the piston back, and are
then swept out to exhaust, and the cycle
recommences with the introduction of a fresh
charge of petrol and air.
o In the diesel or oil engine, the oil is sprayed
under pressure into the compressed air at the end
of the compression stroke, and the combustion is
spontaneous due to the high temperature of the
air after compression.
Connecting
Rod Crank
Inlet Exhaust
Piston
Cylinder
28
o In a gas engine a mixture of gas and air is
induced into the cylinder, compressed and then
ignited as in the petrol engine by an electric
spark.
- The cycle can be represented on any diagram of
properties, and is usually drawn on the p-v diagram,
since this allows a more direct comparison to be made
with the actual engine machine cycle.
- Note that an air standard cycle on a p-v diagram is a
true thermodynamic cycle, whereas a record of the
pressure variations in an engine cylinder against piston
displacement is a machine cycle.
The air standard cycle enables us to examine the influence
of a number of variables in performance
Thermal efficiency = cievedHeatNet
DoneWorkNet
Re..
.).(
OutputWorkGross
DoneWorkNetRatioWork
..
...
A cycle which has a good thermal efficiency and a good
work ratio suggests good overall efficiency potential in a
practical power producing plant
Specific Fuel Consumption =)(.
)/(...
kwoutputpower
hkgusedfuelofMass
Low specific fuel consumption indicates a better energy
conversion. For reciprocating engines, a means of
comparison between cycles can be made based on ―mean
effective pressure‖ This is the theoretical pressure which if
it was maintained throughout the volume changes of the
cycle would give the same output of the work as that
obtained from the cycle.
Indicated work = area of cycle = ɠW
Stroke volume of the diagram = V1-V2
Mean effective Pressure (Pm) =
21 VV
W
A cycle with a higher mean effective pressure will indicate
that it has better work characteristics than a cycle with a
lower mean effective pressure.
We will now consider the following cycles starting with
the Carnot cycle.
Carnot cycle
An ideal theoretical cycle which is the most efficient
conceivable is the Carnot cycle. This cycle forms a
reference point for the determination of the efficiencies of
the other power cycles. By calculating the thermal
efficiency, it is possible to establish the maximum possible
efficiency between the temperature limits taken (corollary
II of the 2nd
law)
This cycle consists of two isothermal processes joined by
two adiabatic/isentropic processes. It is most conveniently
represented on a T-s and p-v diagrams as follows:
Process 1-2 = isothermal heat supply.
Process 2-3 = isentropic expansion from T2 to T3
Process 3-4 = isothermal heat rejection
Process 4-1 = isentropic compression from T4 to T1
The cycle is completely independent of the working
substance used.
The cycle efficiency is given by
2 1
3 4 T3=T4
T1=T2
T
A B S
P
V
2
1 P1
P2
P4
P3
A V4 B V2
3 4
Pm
P
V2 V1 V
Shaded Area =ɠW
Area of the cycle =ɠW
29
suppliedheat
rejectedheat suppliedheat
suppliedheat
outputnet work
AB
ABAB
ssT
ssTssT
1
21
AB
AB
ssT
ssTT
1
21
1
21T
T
There is no attempt to use the Carnot cycle with gas as
working substance in practice because of two reasons
1. The pressure of the gas changes continuously from p4
to p1 during the isothermal heat supply, and from p2
to p3 during the isothermal heat rejection. But in
practice it is much more convenient to heat a gas at
approximately constant pressure or at constant
volume.
2. The Carnot cycle, despite its high thermal efficiency,
has a small work ratio. [Work ratio is the ratio of the
net work output (area 12341) to the gross work output
of the system (area 123BA1); the work done on the
gas is given by 341AB3.]
Examples
3.1. What is the highest possible theoretical efficiency of a
heat engine operating with a hot reservoir of furnace
gases at 2000oC when the cooling water is available
at 10oC?
3.2. A hot reservoir at 800oC and a cold reservoir at 15
oC
are available. Calculate the thermal efficiency and the
work ratio of a Carnot cycle using air as the working
fluid, if the maximum and minimum pressures in the
cycle are 210 bar and 1 bar.
Joule Cycle
This is also known as Brayton or Constant Pressure cycle
and forms the basis for the closed cycle gas turbine unit. In
this cycle the heat supply and heat rejection processes
occur reversibly at constant pressure. The expansion and
compression processes are isentropic.
Neglecting velocity changes and applying the steady-flow
energy equation to each part of the cycle gives:
Work input to compressor = (h2 – h1) = cp(T2 – T1)
Work output from turbine = (h3 – h4) = cp(T3 – T4)
Heat supplied in heater = (h3 – h2) = cp(T3 – T2)
Heat rejected in cooler = (h4 – h1) = cp(T4 – T1)
Thus
23
1423
TTc
TTcTTc
p
pp
23
141TT
TT
(3.2)
Since process 1 to 2 and 3 to 4 are isentropic between the
same pressures P2 and P1, then
/1
/1
1
2
4
3
1
2
pr
p
p
T
T
T
T
where rp is the pressure ratio, p2/p1.
i.e. /143
prTT and /112
prTT
so 14/1
23 TTrTT p
B
T2
A
T4
1
2
3
s
4
T
p1 1
2 3
v
4
p
p2 T3
T1
p2
p1
Heat
rejected
Cooler
Heat
supplied
Compressor Turbine
Net work
output
Heater
4 1
2 3
30
Hence substituting in the expression for the efficiency
gives
/1/114
14 111
pp rrTT
TT (3.3)
Thus for the Joule cycle the cycle efficiency depends only
on the pressure ratio.
The work ratio (rw) is:
output work gross
outputnet work wr
43
1243
TTc
TTcTTc
p
pp
43
211TT
TT
(3.4)
Now, as previously
/1
4
3
1
2 prT
T
T
T
therefore
/112
prTT and /1
34
pr
TT
Hence substituting
/13
/11
/11
11
p
p
wrT
rTr
/1
3
11 prT
T (3.5)
Thus the work ratio depends not only on the pressure ratio
but also on the ratio of the minimum and maximum
temperatures. For a given inlet temperature, T1, the
maximum temperature, T3, must be made as high as
possible for a high work ratio.
For an open-cycle gas turbine unit the actual cycle is not
such a good approximation to the ideal Joule cycle, since
fuel is burned with the air, and a fresh charge is
continuously induced into the compressor. The ideal cycle
nevertheless provides a good basis for comparison, and in
many calculations for the ideal open-cycle gas turbine the
effects of the mass of fuel and the charge in the working
fluid are neglected.
Example
3.3. In a gas turbine unit, air is drawn at 1.02 bar and 15oC,
and is compressed to 6.12 bar. Calculate the thermal
efficiency and the work ratio of the ideal cycle, when
the maximum cycle temperature is limited to 800oC.
The Otto cycle
This is the ideal air standard cycle for the petrol engine, the
gas engine and the high speed oil engine. It consists of the
following processes:
Process 1-2 = isentropic compression
Process 2-3 = reversible constant volume heating
Process 3-4 = isentropic expansion
Process 4-1 = reversible constant volume heating
To give a direct comparison with an actual engine the ratio
of the specific volumes, v1/v2, is taken to be the same as the
compression ratio of the actual engine, i.e.
Compression ratio,
2
1
v
vrv
volumeclearance
volumeclearance meswept volu
The heat supplied at constant volume between T2 and T3 is
given by:
Q1 = cv(T3 – T2)
Similarly the heat rejected per unit mass at constant
volume between T4 and T1 is given by
Q2 = cv(T4 – T1)
Processes 1 to 2 and 3 to 4 are isentropic and therefore
there is no heat flow.
Thus
23
1423
TTc
TTcTTc
v
vv
v
3
1
4
2
P
V2 V1
2
3
4
1
S1 S3
T
31
23
141TT
TT
(3.6)
Now for processes 1 to 2 and 3 to 4, which are isentropic,
1
1
3
4
1
2
1
4
3
1
2
vrv
v
v
v
T
T
T
T
Then 143
vrTT and 1
12
vrTT
Hence substituting gives
114
141
vrTT
TT
1
11
vr
(3.7)
Thus the thermal efficiency of the Otto cycle depends only
on the compression ratio, rv.
Example
3.4. Calculate the ideal air standard cycle efficiency based
on the Otto cycle for a petrol engine with a cylinder
bore of 50 mm, a stroke of 75 mm and a clearance
volume of 21.3 cm3.
The Diesel Cycle
This is ideal air standard cycle for the original diesel
engine and consists of the following processes:
Process 1-2 = isentropic compression
Process 2-3 = reversible constant pressure heating
Process 3-4 = isentropic expansion
Process 4-1 = reversible constant volume cooling
Heat supplied
Q1 = cp(T3 – T2)
Heat rejected
Q2 = cv(T4 – T1)
There is no heat flow in processes 1-2 and 3-4 since they
are isentropic.
By substituting in the equation of thermal efficiency, i.e.,
1
21
Q
At constant pressure process 2-3
Q1 =CpdT = Cp(T3 – T2)
At constant Volume process 4-1
Q2 = CvdT = Cv(T4 –T1)
)(
)(1
23
14
TTC
TTC
p
v
diesel
but
v
p
C
C
)(
)(1
23
14
TT
TTdiesel
Process 3-4 is isentropic expansion
From the characteristic gas equation
4
44
3
33
T
VP
T
VP
and 4433 VPVP
1
4
2
2
3
1
4
3
3
4
4
3
3
4
V
V
V
V
V
V
V
V
V
V
T
T
If we denote
2
3
V
Vrc
= Cut off ratio, and
2
1
V
Vrv
= Compression ratio
and given that V1 = V4
Then 1
3
4
v
c
r
r
T
T So that 1
34
v
c
r
rTT
Also 1
2
1
1
2
V
V
T
T so that 1
1
1
2
112
vrTV
VTT
For constant pressure process 2-3
3
33
2
22
T
VP
T
VP and P2 = P3
3
3
2
2
T
V
T
V So that
crV
V
T
T
2
3
2
3
Therefore T3 = T2 rc = cv rrT 1
1
And c
v
c
cv rTr
rrrTT 1
1
14 1
v
3
1
4
2
V2 V1
2
3
4
1
S1 S3
T PVɤ = C
P
32
Substituting for T2 T3 and T4 in the equation for diesel
efficiency
)(
)(1
23
14
TT
TTdiesel
=
1
1
1
1
11 )(1
vvc
c
rTrrT
TrT
=
1.
)1(1
1
1
1
cv
c
rrT
rT
= 1.
)1(1
1
cv
c
rr
r
(3.8)
Hence the efficiency of a diesel engine depends on the cut-
off ratio as well as the compression ratio.
Example
3.5. A diesel engine has an inlet temperature and pressure
of 15oC and 1 bar respectively. The compression ratio
is 12/1 and the maximum cycle temperature is 1100oC.
Calculate the air standard thermal efficiency based on
the diesel cycle.
The dual-combustion cycle
This is also known as the limited-pressure or mixed cycle
and is the ideal air standard cycle of modern diesel and oil
engines. It consists of the following processes:
Process 1-2 = isentropic compression
Process 2-3 is reversible constant volume heating
Process 3-4 = reversible constant pressure heating
Process 4-5 = isentropic expansion
Process 5-1 = reversible constant volume cooling
The heat is supplied in two parts, the first part at constant
volume and the remainder at constant pressure, hence the
name ‗dual-combustion‘. In order to get the thermal
efficiency, three factors are necessary. These are:
- The compression ratio, rv = v1/v2,
- The ratio of pressure, rp = p3/p2, and
- The ratio of volumes, = v4/v3.
Then it can be shown that
111
11
vpp
p
rrr
r (3.9)
Thus the thermal efficiency of a dual-combustion cycle
depends not only on the compression ratio but also on the
relative amounts of heat supplied at constant volume and at
constant pressure.
Equation (3.9) is much too cumbersome to use, and the
best method of calculating thermal efficiency is to evaluate
each temperature round the cycle and then get the total
heat supplied (Q1) and the total heat rejected (Q2) as:
Q1 = cv(T3 – T2) + cp(T4 – T3)
Q2 = cv(T5 – T1)
Note that when rp = 1 (i.e. p3 = p2), then Eq (3.9) reduces
to the thermal efficiency of the diesel cycle.
Example
3.6. An oil engine takes in air at 1.01 bar, 20oC and the
maximum cycle pressure is 69 bar. The compressor
ratio is 18/1. Calculate the air standard thermal
efficiency and the mean effective pressure based on
the dual-combustion cycle. Assume that the heat
added at constant volume is equal to the heat added at
constant pressure. [Mean effective pressure is the
height of a rectangle having the same length and area
as the cycle plotted on a p-v diagram.]
The Stirling cycle
This has an efficiency equal to that of the Carnot cycle but
has a higher work ratio. It consists of the following
processes
Process 1-2 = reversible constant volume heating
Process 2-3 = isothermal expansion
5
4
1
V1
2
3 P3 = P4
V2
P
V
PVɤ = C
33
Process 3-4 = reversible constant volume cooling
Process 4-1 = isothermal compression
Heat supplied,
3
2223 ln
p
pRTQ
Heat rejected,
4
1141 ln
p
pRTQ
Thus
4
12
3
21
ln
ln
1
p
pRT
p
pRT
(3.10)
For the constant volume process 1-2,
1
2
1
2
T
T
p
p
and for process 3-4, 1
2
4
3
4
3
T
T
T
T
p
p
Therefore 4
3
1
2
p
p
p
p
and 4
1
3
2
p
p
p
p
Hence
2
11T
T (3.11)
= the Carnot efficiency
The Ericsson cycle
This also has an efficiency equal to that of the Carnot cycle
but has a higher work ratio. It consists of the following
processes
Process 1-2 = reversible constant pressure heating
Process 2-3 = isothermal expansion
Process 3-4 = reversible constant pressure cooling
Process 4-1 = isothermal compression
Once again, because the two constant pressure processes
are bounded by the same temperature limits;
mCp(T3-T2) = mCp(T4-T1)
then the process of regeneration is again possible. By
including the process of regeneration the Ericsson cycle
becomes a reversible cycle and has the highest thermal
efficiency possible which = (T3 – T1)/T3
Problems
1. What is the highest cycle efficiency possible for a heat
engine operating between 800 and 15oC? [73.2%]
2. Two reversible heat engines operate in series between a
source at 527oC and a sink at 17
oC. If the engines have
equal efficiencies and the first rejects 400 kJ to the
second, calculate:
(a) the temperature at which heat is supplied to the
second engine;
(b) the heat taken from the source;
(c) the work done by each engine.
Assume that each engine operates on the Carnot cycle.
[208.7oC; 664.4 kJ; 264.4 kJ; 159.2 kJ]
3. In a Carnot cycle operating between 307 and 17oC the
maximum and minimum pressures are 62.4 bar and
1.04 bar. Calculate the cycle efficiency and the work
ratio. Assume air to be the working fluid. [50%; 0.286]
4. A closed-cycle gas turbine unit operating with
maximum and minimum temperatures of 760 and 20oC
has a pressure ratio of 7/1. Calculate the ideal cycle
efficiency and the work ratio. [42.7%; 0.505]
1
2
3
v
4
p
T2= T3
T1= T4
1 2
3
v
4
p
T2= T3
T1= T4
34
5. In an air standard Otto cycle the maximum and minimum
temperatures are 1400 and 15oC. The heat supplied per kg
of air is 800 kJ. Calculate the compression ratio and the
cycle efficiency. Calculate also the ratio of maximum to
minimum pressures in the cycle. [5.27/1; 48.5%; 30.65/1]
6. A four-cylinder petrol engine has a swept volume of 2000
cm3, and the clearance volume in each cylinder is 60 cm
3.
Calculate the air standard cycle efficiency. If the
introduction conditions are 1 bar and 24oC, and the
maximum cycle temperature is 1400oC, calculate the mean
effective pressure based on the air standard cycle.
[59.1%; 5.28 bar]
7. Calculate the cycle efficiency and mean effective pressure
of an air standard diesel cycle with a compression ratio of
15/1, and maximum and minimum cycle temperatures of
1650oC and 15
oC 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
effective pressure of 10 bar. The minimum pressure and
temperature are 1 bar and 17oC respectively, and the
compression ratio is 16/1. Calculate the maximum cycle
temperature when the cycle efficiency is 60%. The
maximum cycle pressure is 60 bar. [1959oC]
Air Compressors
The function of a compressor is to take a definite quantity
of fluid and deliver it at a required pressure. The most
efficient machine is one which will accomplish this with e
minimum input of mechanical work. There are two general
types of compressors;
1. Rotary compressors
2. Reciprocating compressors
Reciprocating compressors have a low mass flow rate and
high pressure ratios whereas rotary compressors have hig
mass flow rate and low pressure ratios.
Reciprocating compressors
The mechanism involved is the basic piston con-rod, crank
and cylinder arrangement.
As piston moves down, the inlet valve is open and the
delivery valve is closed. A fresh charge of air is taken into
the cylinder. As the piston moves up, pressure in the
cylinder builds up, the inlet valve is closed. When pressure
is slightly in excess of that of the air on the outside of the
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
d-a -induction stroke. Mass of air in the cylinder increases
from zero at d to that required to fill the cylinder at a.
The temperature,T1 remains constant for this process
and there is no heat exchange with the surrounding
a-b compression stroke ( reversible polytropic pvγ =
Constant)
b-c delivery stroke at constatnt temperature T2 and constant
pressure P2.
The general form of the compression a-b is the reversible
polytropic given by PVn = Constant
The net work done in the cycle is given by the area
enclosed by the pv diagram and is work done on the gas.
In this section work done on the gas will be considered as
positive work
Indicated work = area abcd = area (abef + bcoe –adof)
= ab
ab VPVPn
VPVP12
12
1
=
1
1
112
nVPVP ab
=
1
1112
n
nVPVP ab
= ab VPVPn
n12
1
P1Va = mRT1 and P2Vb = mRT2
Therefore work input per cycle = 121
TTmRn
n
The actual work input to the compressor is larger than the indicated
work, due to the work necessary to overcome losses due to friction.
Shaft work = indicated work + friction work
= i.p + f.p
Therefore the mechanical efficiency of the machine is given by;
ɳ = PowerShaft
PowerIndicated
WorkShaft
WorkIndicated
.
.
.
.
to determine the power input required, the efficiency of the
driving motor must be taken into account in addition to the
mechanical efficiency of the compressor
Motor and drive efficiency = PowerInput
PowerShaft
.
.
From Indicated Work = 121
TTmRn
n
We can rewrite; Indicated Work =
1
1 1
21
T
TmRT
n
n
But = n
n
P
P
T
T
1
11
2
1
2
and PV = MRT so that
Indicated Work =
11
1
11
21
n
n
P
PmRT
n
n
Indicated Work =
11
1
11
211
n
n
P
PVP
n
n
Where V1 is the volume indicated per unit time
Condition for Minimum Work
The work done on the air in a compressor is given by the
area of the indicator diagram, and the work will be
minimum when the area is minimum. The height of the
P
b
a P1
P2 c
V2 V1 V
d
b2 b1
P
P2 c b
a P1
e
e v2 v1 o
d
f
v
36
diagram is fixed by the required pressure ratio and the
length of the line da is fixed by the cylinder volume. The
only process that can influence the area of the diagram is
the compression stroke (line ab)
Possible processes
Line ab1 - is according to the law pv = constant
(isothermal process)
Line ab2 – is according to the law pvɤ = Constant
(Isentropic process)
Line ab – is according to the law pvn = constant (Polytropic
Process)
Isothermal process is the desirable process between a and b
giving minimum work to be done on the air. Therefore
cooling of the gas is always provided either by air or by
water to keep the temperature in the cylinder as constant as
possible.
Indicated work when the gas is compressed isothermally =
area ab1cd
Area ab1cd = Area ab1ef + Area bc1oe – Area adof
Area ab1ef =
1
22 ln
1 P
PVP b
(isothermal process)
Area a b1c d = abb VPVP
P
PVP 12
1
2
2 11ln
Also 121 ba VPVP (isothermal process
Therefore indicated work = area (abcd) =
1
22 ln
1 P
PVP b
=
1
2
1 lnP
PVP a
but P1Va = mRT1 so that
Indicated work =
1
2
1 lnP
PmRT
Where m and Va are the mass and the volume of the work
induced respectively. If these parameters are per unit time
then the equation gives the isothermal power.
Isothermal efficiency = WorkIndicated
workIsothermal
.
.
Reciprocating compressors including clearance
Clearance is necessary in a compressor to give mechanical
freedom to the working parts and allow the necessary
space for valve operations.
When the delivery stroke bc is compete the clearance
volume, vc, is full of gas at a pressure P2 and temperature
T2. As the piston proceeds on the next induction stroke the
air expands behind the piston until the pressure P1 is
reached. As soon as the pressure reaches P1 the induction
of fresh air will begin and continue until the end of the
stroke at a
ab – compression stroke
bc – delivery stroke
da – induction stroke
Clearance reduces induction volume Vs to (Va – Vd)
ṁa = ṁb, ṁc = ṁd
Indicated work = area abcd
= area (abef) – area (cefd)
= )(11
12
.
12
.
TTRmn
nTTRm
n
nda
= 12
..
)(1
TTmmRn
nda
= 12
.
1TTRm
n
n
Where ṁ = ṁa - ṁd ie mass induced per unit time
b
d
c
a
T1
P2
Swept Volume Vc
P
V
T2
P1
37
But = n
n
P
P
T
T
1
11
2
1
2
and PV = mRT so that
Indicated Work =
1
1 1
21
.
T
TRTm
n
n
11
1
11
21
. n
n
P
PRTm
n
n
Indicated Work =
11
1
11
21
n
n
P
PVP
n
n
11
1
11
21
n
n
daP
PVVP
n
n
The mass delivered per unit time can be increased by
designing the machine to be double acting i.e gas is dealt
with on both sides of the piston, the induction stroke of one
side being the compression stroke of the other side
Volumetric Efficiency
Clearance volume reduces the induced volume to a value
less than that of the swept volume. Volumetric efficiency is
defined as
volumesweptthefillwouldwhichairofMass
inducedairofmassThev
........
....
Free air delivery (FAD), is the volume delivered measured
at the pressure and temperature of the atmosphere in which
the machine operates.
If FAD is V, at pressure p, and temperature T, then the
mass of air induced (delivered) is given by,
RT
Pvm .
Mass required to fill the swept volume
RT
Pvm s
s .
sss
vv
v
RT
pvRT
Pv
m
m
.
.
or volume induced = va- vd
va-vd = vs + vc - vd
and cd V
P
Pv
n1
1
2
n
P
Pvvvinducedvolume ccs
1
1
2.
s
cs
s
vv
P
Pvv
v
v
n
1
1
1
2
11
1
1
2n
s
c
vP
P
v
v
(Va –Vd) > (Va –Vd‘) > (Va –Vd‖)
b
d‘
c
a
T1
P2
Swept Volume Vc
P
V
T2
P1
c‘ b‘
b‖
d‖ d
b
d
c
a
T1
P2
Swept Volume Vc
P
V
T2
P1
Chamber 2
Chamber 1
Piston
c‖
38
Multi stage compression
As the required pressure ratio for a single stage compressor
increases, the volumetric efficiency decreases (i.e volume induced
per cycle decreases). The volumetric efficiency of the compressor
can be improved by carrying out the compression in two stages.
After the first stage of compression, the fluid is passed into a
smaller cylinder in which the gas is compressed to the required
final pressure. For minimum work to be done, the gas from the
first stage of compression is cooled as it passes from one cylinder
to the other by passing it through an intercooler.
d - a - induction stroke of 1st stage
b - c – delivery stroke of 1st stage
d‘- a‘ – induction stroke of 2nd stage
b‘-c‘ – delivery stroke of 2nd stage
Ideal intermediate pressure for a two stage compressor
Intermediate pressure pi influences the work to be done on the gas
and its distribution between the two stages.
Total work = Low Pressure work + High Pressure Work
11
1
1
1
n
n
i
p
pRTm
n
n
11
1
21
n
n
ip
pRTm
n
n
It is assumed that inter-cooling is complete and therefore the
temperature at the beginning of each stage is T1
111
.
1
2
1
1
1
n
n
i
n
n
i
p
p
p
pRTm
n
nWorkTotal
If P1T1 and P2 are fixed, then the optimum value which makes the
work a minimum can be obtained by equating
0)( Workdp
d
i
021
1
2
1
1
1
n
n
i
n
n
i
i p
p
p
pRTm
n
n
dp
d
02
1
2
1
1
n
n
i
n
n
i
i p
p
p
p
dp
d
0211
11
2
11
1
n
n
i
n
n
n
n
i
n
n
i ppp
pdp
d
0111 1
11
2
1
1
11
n
n
in
nn
n
n
n
i ppn
n
pp
n
n
011
211
2
11
1
n
n
in
n
ni
n
n
ppn
npp
n
n
n
n
in
n
ni
n
n
ppn
npp
n
n211
2
11
1
11
So that ,
n
n
in
n
ni
n
n
pppp
211
2
11
1
n
n
n
n
n
n
i
ni pp
p
p1
1
1
221
1
n
n
n
n
i ppp
1
12
22
21
2 pppi
So that 21 pppi and
i
i
p
p
p
p 2
1
Therefore for the work to be a minimum the pressure ratios for
the two stages has to be equal
b
d
c
a
T1
Pi
Swept Volume Vc
P
V
T2
P1
P2
vs vc
b' c'
a' d'
H2O in H2O out
P2T2 in
PiTi
in
PiTi
in
High Pressure Stage
Low Pressure Stage
Intercooler
P1T1
in
39
111
..
1
2
1
1
1
n
n
i
n
n
i
p
p
p
pRTm
n
nWorkMinimumTotal
But we have seen that
i
i
p
p
p
p 2
1
111
..
1
1
1
1
1
n
n
in
n
i
p
p
p
pRTm
n
nWorkMinimumTotal
11
21
1
1
n
n
i
p
pRTm
n
n
In terms of the overall pressure ratio,
1
221
1 p
p
p
pp
p
p
i
i
Substituting for
1p
pi we have;
11
2..
1
1
21
n
n
p
pRTm
n
nworkmimimumTotal
11
2 2
1
1
21
n
n
p
pRTm
n
n
Problems
1. A single stage reciprocating compressor takes 1m3 of
air per minute at 1.013 bar and 15oC and delivers it at
7 bar. Assuming the law of compression is PV1.35
=
Constant and that the clearance volume is negligible,
calculate the indicated power. If the compressor is
driven at 300rev/min, determine the cylinder bore
required assuming a stroke to bore ratio of 1.5:1.
Calculate the power of the motor required to drive the
compressor if the mechanical efficiency of the
compressor is 85% and that of the motor transmission
is 90%. [4.23 kW, 141.5 mm, 5.53kW, 77%]
2. A single stage double acting air compressor is
required to deliver 14m3 of air per minute measured at
1.013bar and 15oC. The delivery pressure is 7 bar and
the speed is 300 r.p.m. take the clearance volume as
5% of the swept volume with a compression index of
n=1.3. Calculate the swept volume of the cylinder, the
delivery temperature and the indicated power.
[0.0281m3, 450K, 57.58kW]
3. A single acting two stage air compressor runs at 300
rev/min and compresses 8.5m3/min at 1 atmosphere
and 15oC to 40bar. Calculate
a. The optimum pressure for each stage
b. The theoretical power consumption for each stage
if the compression in each stage is polytropic with
n=1.3 and intercooling is complete i.e to a
temperature of 15oC
c. The swept volumes if the volumetric efficiencies of
the low pressure and high pressure stages are
0.90 and 0.85 respectively
d. The heat rejected into the cylinder cooling jackets
and into the intercooler. [6.283, 32.86 kW,
0.0315m3, 0.0053m
3, 6.39kW per cylinder,
26.88kW for the intercooler]
4. Air is to be compressed in a single stage reciprocating
compressor from 1.0 13 bar and 15°C to 7 bar.
Calculate the indicated power required for a free air
delivery of 0.3m3/min when the compression process
is:
a) Isentropic [1.31 kW].
b) Reversible isothermal [0.98 kW].
c) Polytropic, with n = 1.25 [1.196 kW].
5. A single — acting compressor is required to deliver
air at 70 bar from an induction pressure of 1 bar, at
the rate of 2.4 m3/min measured at free air conditions
of 1.013 bar and 15°C. the temperature at the end of
the induction stroke is 32°C. Calculate the indicated
power required if the compression is carried out in
two stages with an ideal intermediate pressure and
complete intercooling. The index of compression and
expansion for both stages is 1.25. What is the saving
in power over single stage compression? If the
clearance volume is 3% of the swept volume in each
cylinder, calculate the swept volumes of the cylinders.
The speed of the compressor is 750 rev/mm. if the
mechanical efficiency of the compressor is 85%,
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
Vapour Power Cycles
Introduction
Characteristics of power cycles
- The working fluid is a condensable vapour which is in
liquid phase during part of the cycle
- The cycle consists of a succession of steady-flow
processes, with each process carried out in a separate
component specifically designed for that purpose.
- Each component constitutes an open system, and
all the components are connected in series so that as the
fluid circulates through the power plant each fluid
element passes through a cycle of mechanical and
thermodynamic states.
- To simplify the analysis, it is assumed that the change in
kinetic and potential energy of the fluid between entry
and exit in each component is negligible compared to the
change in enthalpy. This implies that the energy equation
can be written as: Q – W = h2 – h1
- The working fluid is usually steam because it is cheap
and chemically stable but any condensable vapour may
be used.
Criteria of performance
(a) Ideal cycle efficiency – this is the efficiency of a cycle
when all the processes are assumed to be reversible.
(b) Actual cycle efficiency – this is the efficiency of a
cycle when process efficiencies are introduced
(c) Efficiency ratio – this is the ratio of the actual cycle
efficiency to the ideal cycle efficiency.
(d) Work ratio (rw) – this is the ratio of the net work to the
positive work done in the cycle. (It is a measure of the
cycle‘s sensitivity to irreversibilities since
irreversibilities decrease the positive work and increase
the negative work.)
(e) Specific steam consumption (ssc) – this is the mass
flow of steam required per unit of power output. It is
usually expressed in kg/kW h and if the numerical
value of net work output per unit mass of flow is W
(kJ/kg) the ssc can be found from
Wssc
3600
Carnot cycle
- It consists of two reversible isothermal processes at Ta
and Tb respectively, connected by two reversible
adiabatic (isentropic) processes.
- When the working fluid is a condensable vapour, the two
isothermal processes are easily obtained by heating and
cooling at constant pressure while the fluid is a wet
vapour.
- The processes are:
1-2: Saturated water is evaporated at constant pressure to
form saturated steam; heat added is Q12 = h2 – h1
2-3: Saturated steam is expanded isentropically in a
turbine; work done is W23 = h2 – h3
3-4: Wet steam is partially condensed at constant
pressure to state 4 where s4 = s1; heat rejected is Q34
= h4 – h3
4-1: Steam is compressed isentropically in a compressor;
work required is W41 = h4 – h1
Boiler
Compressor
Turbine
Condenser
Coolingwater
T
1 2
34
S
3- 01PRD
571E Carnot cycle
Example 1
Calculate the heat and work transfers, cycle efficiency,
work ratio and steam consumption of a Carnot cycle using
steam between pressures of 30 and 0.04 bar.
Solution
From tables, at 30 bar
T1 = T2 = 507.0 K
h1 = hf = 1008 kJ/kg,
h2 = hg = 2803 kJ/kg
Putting s4 = s1 and s3 = s2, then at the condenser pressure of
0.04 bar,
T3 = T4 = 302.2 K
41
x3 = 0.716, x4 = 0.276
Hence from h = hf + xhfg,
h3 = 121 + x32433 = 1863 kJ/kg
h4 = 121 + x42433 = 793 kJ/kg
The turbine work is W23 = h2 – h3 = 940 kJ/kg
The compressor work is W42 = h4 – h1 = -215 kJ/kg
The heat transfer in the boiler is Q12 = h2 – h1 = 1795 kJ/kg
The heat transfer in the condenser is Q34 = h4 – h3 = -1070
kJ/kg
The net work from the cycle is thus W = W23 + W41 = 725
kJ/kg (This is also equal to Q12 + Q34)
The cycle efficiency is
404.01795
725
12
Q
W
Since this is a Carnot cycle, this must also be given by
404.00.507
2.3020.507
a
ba
T
TT
The work ratio is
771.0940
725
23
W
Wrw
The specific steam consumption is
97.4725
3600ssc kg/kW h
Example 2
Recalculate Example 1 with isentropic efficiencies of 0.80
for the compression and expansion process, to estimate the
actual cycle efficiency and steam consumption.
Solution
The actual turbine work is
W23 = h2 – h3 = 0.80(h2 – h3‘) = 752 kJ/kg
The actual compressor work is
26980.0
141441
hhhhW kJ/kg
The net work is therefore
W = W23 + W41 = 483 kJ/kg
The enthalpy at state 1 is
h1 = h4 – W41 = 1062 kJ/kg
Hence the heat transfer in the boiler is
Q12 = h2 – h1 = 1741 kJ/kg
The thermal efficiency is
277.01741
483
12
Q
W
The specific steam consumption is
45.7483
3600ssc
Rankine cycle
Unsuperheated cycle
- There are two reasons why the Carnot cycle is not used
in practice
(a) It has a low work ratio.
(b) It is difficult to control the condensation process so
that it is stopped at state 4, and then carry out the
compression of a very wet vapour efficiently. The
liquid tends to separate out from the vapour and the
compressor would have to deal with a non-
homogeneous mixture.
- On the other hand, it is comparatively easy to condense
the vapour completely and compress the liquid to boiler
pressure in a small feed pump
- The resulting cycle is known as the Rankine cycle.
Boiler
Pump
Turbine
Condenser
Coolingwater
T
1 2
34
5
S
3- 01PRD
571E Simple Rankine cycle
Example 3
Calculate the cycle efficiency, work ratio, and the steam
consumption of a Rankine cycle working between
pressures of 30 and 0.04 bar.
Estimate the actual cycle efficiency and steam
consumption when the isentropic efficiencies of the
expansion and compression processes are each 0.80.
Solution
(a) Ideal cycle
42
As in Example 1,
h2 = 2803 kJ/kg and h3 = 1863 kJ/kg
The turbine work is as before
W23 = h2 – h3 = 940 kJ/kg
The compression work is
W45 = h4 – h5 = vf(p4 – p5)
= 0.001(0.04 – 30) x 100 = -3 kJ/kg
Since h4 = hf = 121 kJ/kg then h5 = 124 kJ/kg.
The heat supplied is
Q52 = h2 – h5 = 2803 – 124 = 2679 kJ/kg
Thus
350.02679
3940
52
Q
W
997.0940
3940
23
W
Wrw
84.33940
3600
ssc kg/kW h
(b) Actual cycle
The actual expansion work is
W23 = 0.80 x 940 = 752 kJ/kg
The actual compression work is
480.0
341 W kJ/kg (which is negligibly small)
The enthalpy at state 5 now becomes
h5 = h4 – W45 = 125 kJ/kg
The heat supplied is
Q52 = h2 – h5 = 2803 – 125 = 2678 kJ/kg
Therefore
279.02678
4752
81.44752
3600
ssc kg/kW h
Rankine cycle with superheat
- By placing in the combustion chamber a separate bank of
tubes (the superheater) leading saturated steam away
from the boiler, it is possible to raise the steam
temperature without at the same time raising the boiler
pressure.
- This gives the Rankine cycle with superheat.
Boiler
Superheater
Pump
Turbine
Condenser Coolingwater
T
1 2
2'
34
5
S
3- 01PRD
571E
Rankine cycle withsuperheater
- It is evident that the average temperature at which heat is
supplied is increased by superheating and hence the ideal
cycle efficiency is increased.
Example 4
A steam power plant operates between a boiler pressure of
42 bar and a condenser pressure of 0.035 bar. Calculate the
cycle efficiency and specific steam consumption when the
steam is superheated to 500oC.
Solution
From tables, by interpolation, at 42 bar:
h5 = 3442.6 kJ/kg and s5 = s6 = 7.066 kJ/kg K
Now s6 = s1 + x6sfg therefore 0.391 + x68.13 = 7.066
i.e. x6 = 0.821
Also h6 = h1 + x6hfg = 112 + (0.821 x 2438) = 2113 kJ/kg
From tables:
h1 = 112 kJ/kg
Then W56 = h6 – h5 = 3442.6 – 2113 = 1329.6 kJ/kg
Neglecting the feed-pump term,
heat supplied Q25 h5 – h1 = 3442.6 – 112 = 3330.6
kJ/kg
399.06.3330
6.1329
25
56 Q
W
71.26.1329
36003600
56
W
ssc kg/kW h
Reheat cycle
- With the reheat cycle the expansion takes place in two
turbines.
- The steam expands in the high-pressure turbine to some
intermediate pressure, and is then passed back to yet
another bank of tubes in the boiler where it is reheated at
constant pressure, usually to the original superheat
temperature.
4
43
- It then expands in the low-pressure turbine to the
condenser pressure.
Boiler
SH
SH - SuperheaterH - Heater
LPT - Low pressure turbineHPT - High pressure turbineCW - Cooling water
C - Condenser
H
Pump
LPTHPT Turbine
C
CW
T
1
2
2''
2' 2'''
34
5
S
3- 01PRD
571E
Rankine cycle withsuperheater and reheat
Example 5
Find the ideal cycle efficiency and steam consumption of a
reheat cycle operating between pressures of 30 and 0.04
bar, with a superheat temperature of 450oC. Assume that
the first expansion is carried out to the point where the
steam is dry saturated and that the steam is reheated to the
original superheat temperature. The feed pump term may
be neglected.
Solution
From tables
h2 h1 = 121 kJ/kg, h3 = 3343 kJ/kg, s5 = 7.082 kJ/kg K
To find the intermediate reheat pressure p6, get from the
saturation table the pressure at which sg = s5. This gives
p6 = p7 = 2.3 bar, and hence h6 = 2713 kJ/kg
From the superheat table
h7 = 3381 kJ/kg and s7 = 8.310 kJ/kg K
At the turbine outlet
x8 = 0.980 and h8 = 2505 kJ/kg
The total heat transferred to the steam in the boiler is
Q25 + Q67 = (h5 – h2) + (h7 – h6)
= 3222 + 668 = 3890 kJ/kg
The total turbine work is
W56 + W78 = (h5 – h6) + (h7 – h8)
= 630 + 876 = 1506 kJ/kg
Thus
387.03890
1506
6725
7856
WW
39.21506
36003600
7856
WW
ssc kg/kW h
Regenerative cycle
One feed heater
- In a practical regenerative cycle, steam is bled off the
turbine at some intermediate pressure during the
expansion and mixed with feed water, which has been
pumped to the same pressure.
- The mixing process is carried out in a feed water heater
and the arrangement is shown below. Only one feed
heater is shown but several could be used.
- The steam expands from condition 2 through the turbine.
- At the pressure corresponding to point 3, a quantity of
steam, say y kg per kg of steam supplied to the boiler, is
bled off for heating purposes.
- The rest of the steam (1 - y) kg, completes the expansion
and is exhausted at state 4.
- This amount of steam is then condensed to state 5 and
pumped to the same pressure as the bleed steam (i.e. p6 =
p7 = p3).
- The bleed steam and the feed water are mixed in the feed
heater, and the quantity of bled steam, y kg, is such that
after mixing and being pumped in a second feed pump,
the condition is as defined by state 1.
Boiler
Feed
pump
Feed
pump
Condenser
y kg
7 6
5
4
3
2
1
(1-y) kg
Feed
heater
1
2
3
4 5
6
7 y kg
(1-y) kg
1 kg
s
T
1 kg
44
- The heat to be supplied in the boiler is then given by (h2
– h1) kJ/kg of steam; this is the heat supplied between
temperatures T1 and T2.
- The bleed temperature to obtain maximum efficiency for
a regenerative cycle is approximately the mean of the
saturation temperatures corresponding to p2 and p5.
Example 6
Find the cycle efficiency and specific steam consumption
of a regenerative cycle with one feed heater, if the steam
leaves the boiler dry saturated at 42 bar and is condensed at
0.035 bar. Neglect the feed pump work.
Solution
At 42 bar, T1 = T2 = 253.2oC and at 0.035 bar, T5 = 26.7
oC.
Therefore 1402
7.262.2533
T
oC
Selecting the nearest saturation pressure from the tables
gives the bleed pressure p3 as 3.5 bar (i.e. T3 = 138.9oC).
To determine the fraction y, consider the adiabatic mixing
process at the feed heater, in which y kg of steam of
enthalpy h3, mix with (1-y) kg of water of enthalpy h6, to
give 1 kg of water of enthalpy h7. The feed pump term may
be neglected (i.e. h6 = h5). Therefore
yh3 + (1-y)h5 = h7.
i.e.
53
57
hh
hhy
Now, h7 = 584 kJ/kg; h5 = 112 kJ/kg; and s2 = s3 = s4 =
6.049 kJ/kg K.
829.0214.5
727.1049.63
x and 696.0
130.8
391.0049.64
x
Hence:
h3 = hf3 + x3hfg3 = 584 + (0.829 x 2148) = 2364 kJ/kg
and
h4 = hf4 + x4hfg4 = 112 + (0.696 x 2438) = 1808 kJ/kg
Therefore 21.01122364
112584
y kg
Heat supplied in boiler = (h2 - h7) = 2800 - 584 = 2216
kJ/kg
Total work output = W23 + W34
= (2800 - 2364) + (1 - 0.21)(2364 - 1808)
= 876 kJ per kg of steam delivered to the
boiler.
Therefore,
396.02216
876
12
3423
Q
WW
11.4876
36003600
3423
WW
ssc kg/kW h
Example 7
Find the cycle efficiency and specific steam consumption
of a regenerative cycle with one feed heater, if the steam
leaves the boiler dry saturated at 30 bar and is condensed at
0.04 bar. Neglect the feed pump work.
Solution
At 30 bar, T2 = 233.8oC and at 0.04 bar, T4 = 29.0
oC.
Therefore 4.1312
0.298.233
2
426
TTT
oC
Hence p3 = 2.8 bar.
From tables, h7 = 551 kJ/kg; h5 = 121 kJ/kg; and s2 = s3 =
s4 = 6.186 kJ/kg K.
Thus x3 = 0.846, h3 = 2388 kJ/kg, x4 = 0.716, h4 = 1863
kJ/kg
Hence 1897.01212388
121551
y kg
Heat supplied in boiler Q12 = (h2 - h1) = 2803 - 551 = 2252
kJ/kg
Total work output = (2803 - 2388) + (1 - 0.1897)(2388 -
1863)
= 840 kJ per kg of steam delivered to the boiler.
Therefore,
373.02252
840
12
3423
Q
WW
29.4840
36003600
3423
WW
ssc kg/kW h
Several feed heaters
- The thermal efficiency increases with addition of further
heaters, but the capital expenditure is also increased
considerably since a feed pump is required at each feed
heater.
45
- Because of the number of feed pumps required, the
heating of the feed water by mixing is dispensed with,
and closed heaters are used.
- The feed water is passed at boiler pressure through the
feed heaters 2 and 1 in series.
- An amount of bleed steam y1, is passed to feed heater 1,
and the feed water receives heat from it by the transfer of
heat through the separating tubes.
- The condensed steam is then throttled to the next feed
heater which is also supplied with a second quantity of
bleed steam, y2, and a lower temperature heating of the
feed water is carried out.
- When the final feed heating has been accomplished, the
condensed steam is then fed to the condenser.
- The temperature differences between successive heaters
are constant, and the heating process at each is
considered to be complete (i.e. the feed water leaves the
feed heater at the temperature of the bleed steam supplied
to it).
Example 8
In a regenerative cycle employing two closed feed heaters,
the steam is supplied to the turbine at 40 bar and 450oC and
is exhausted to the condenser at 0.035 bar. The
intermediate bleed pressures are obtained such that the
saturation temperature intervals are approximately equal,
giving pressures of 10 and 1.1 bar. Calculate the amount of
steam bled at each stage, the work output of the plant in
kJ/kg of boiler steam and the thermal efficiency of the
plant. Assume ideal processes where required.
Solution
From tables:
h3 = 112 kJ/kg; h1 = 3445.8 kJ/kg; s1 = 7.089 kJ/kg K =
s2
Thus
x2 = 0.824 and h2 = 2117 kJ/kg
For the first stage of expansion, 1-7, s7 = s1 = 7.089 kJ/kg
K, and from tables at 10 bar sg < 7.089 kJ/kg K, hence the
steam is superheated at state 7. By interpolation between
250 and 300oC at 10 bar,
29443052926.6124.7
926.6089.729447
h = 3032.9 kJ/kg
For the throttling process, 11-12,
h6 = h11 = h12 = 763 kJ/kg
For the second stage expansion, 7-8, s7 = s8 = s1 = 7.089
kJ/kg K, and from tables at 1.1 bar sg > 7.089 kJ/kg K,
hence the steam is wet at state 8. Therefore,
1.333 + (x8 x 5.994) = 7.089
and so x8 = 0.961 and h8 = 2591 kJ/kg
For the throttling process, 9-10,
h5 = h9 = h10 = 429 kJ/kg
Applying an energy balance to the first feed heater,
remembering that there is no work or heat transfer,
y1h7 + h5 = y1h11 + h6
So
147.07636.3032
429763
117
561
hh
hhy
Similarly for the second heater, taking h4 = h3,
y2h8 + y1h12 + h4 = h5 + (y1 + y2)h9
i.e.
y2(h8 – h9) + y1h12 + h4 = h5 + y1h9
y2(2591–429)+(0.147x763)+112 = 429+(0.147x429)
Therefore y2 = 0.124
The heat supplied to the boiler, Q1, per kg of boiler steam
is
Q1 = h1 – h6 = 3445 – 763 = 2682 kJ/kg
The work output, neglecting pump work, is given by
W = (h1 – h7) + (1 – y1)(h7 – h8) + (1-y1-y2)(h8-h2)
= (3445 - 3032.9) + (1 - 0.147)(3032.9 - 2591)
6
1
10 2 3
4
11 y1 kg
(1-y) kg
1 kg
s
T
7
8 5 9 y2 kg
1-y1 kg
1-y1-y2 kg
46
+ (1 - 0.147 - 0.124)(2591 - 2117)
= 1134.5 kJ/kg
Then 423.02682
5.1134
1
Q
W
Example 9
Calculate the ideal cycle efficiency and specific steam
consumption of a regenerative cycle using three closed
heaters. The steam leaves the boiler at 30 bar superheated
to 450oC, and the condenser pressure is 0.04 bar. Choose
the bleed pressures so that the difference between the
saturation temperature corresponding to 30 bar and that
corresponding to 0.04 bar is divided into approximately
equal steps. (Such a choice of bleed pressures makes the
efficiency of the ideal cycle approximately maximum.)
Solution
Making the temperature differences (T2-T13), (T13-T15),
(T15-T17), (T17-T9), approximately equal, the bleed
pressures become 11, 2.8 and 0.48 bar. All the relevant
enthalpies in the plant can now be found. Before, during
and after expansion in the turbine these are:
h4 = 3343 kJ/kg, h5 = 3049 kJ/kg,
h6 = 2769 kJ/kg, h7 = 2458 kJ/kg,
h8 = 2133 kJ/kg
In finding the enthalpies in the feed line, the following
assumptions will be made. (a) The feed pump term is
negligible, i.e. h9 h10. (b) In throttling the condensed
bled steam, which is a process of equal initial and final
enthalpy, the state after throttling lies approximately on the
saturation line; e.g. in throttling from 14 to 15, h14 = h15
will be identical with hf corresponding to pressure p6. (c)
The enthalpy of the compressed liquid in the feed line is
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
0921.0155
1513
hh
hhya
2nd
heater: 1h12 + (ya + yb)h16 – ybh6 – 1h11 – yah15 = 0
0809.01176
1715
hh
hhyy ab
3rd
heater: 1h11 + (ya + yb + yc)h18 - ych7 - (ya + yb)h17 - 1h10
= 0
0761.0197
917
hh
hhyyy bac
(Always start with the highest pressure heater so as to have
only one unknown at each stage of calculation.)
Now the heat and work transfers for the cycle can be
calculated.
47
The heat added in the boiler is:
Q1,4 = (h4 – h1) = 2562 kJ/kg
The heat rejected in the condenser is
Q8,9 = (h8 – h9) + (ya + yb + yc)(h19 – h8) = 1511 kJ/kg
The work done in the turbine is
W4,8 = (h4 – h5) + (1 – ya)(h5 – h6) + (1 – ya – yb)(h6 –
h7) + (1 – ya – yb – yc)(h7 – h8) = 1051 kJ/kg
[As a check, W4,8 must also be given by
W4,8 = Q1,4 - Q8,9 = 1051 kJ/kg]
Thus
410.02562
1051
4,1
8,4
Q
W and 43.3
1051
3600ssc kg/kW h
Engine trials
- Indicated power = rate of work done by the gas on the
piston as evaluated from an indicator diagram obtained
from the engine. An indicator diagram is a p-v diagram
of the engine cycle.
- Brake power = power output of an engine as measured
by a dynamometer. The difference between indicated
power and brake power is the friction power, i.e.
friction power is the power required to overcome the
frictional resistance of the engine parts.
- Mechanical efficiency is given by
M
brake power
indicated
power
- Thermal efficiency of a heat engine is the ratio of the
net work done in the cycle to the gross heat supplied in
the cycle. Two ways of reporting thermal efficiency:
o Brake thermal efficiency is given by
BTf net v
brake power b p
m Q
energy in the fuel
. .
,
where:
mf = mass of fuel consumed per unit time
Qnet,v = lower calorific value of the fuel
o Indicated thermal efficiency is given by:
ITf net v
indicated power i p
m Q
energy in the fuel
. .
,
Note:
M
BT
IT
b p
i
.
.p
Efficiency of steam boilers
- This is the heat supplied to the steam in the boiler
expressed as a percentage of the chemical energy of the
fuel which is available on combustion, i.e.,
Boiler efficiency = NCVor GCV
feedwater theofenthalpy 1
fm
h
Where h1 is the enthalpy of the steam entering the turbine
and mf is the mass of fuel burned per kg of steam delivered
from the boiler
GCV and NCV are the gross (higher) and net (lower)
calorific values of the fuel.
Steam turbines
- A steam turbine is a power unit which produces power
from a continuous supply of steam, the steam being
delivered to the turbine at a high pressure and exhausted
to the condenser at a low pressure.
- A back pressure turbine is a turbine exhausting into a
condenser at a relatively high pressure so that the
rejected heat is employed usefully, i.e., the exhaust steam
is used for some heating process, and the turbine work
may be a by-product.
- In a reaction turbine, radial tubes, which are connected
to a vertical supply tube, are free to rotate. The end of
each tube is shaped as a nozzle and the steam from the
supply tube passes along the radial tubes and then
expands through the nozzles to atmosphere in a
tangential direction. There is an increase in velocity of
the steam relative to the rotating tube, and hence there is
a reaction on the tube, which makes it rotate.
- 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:
Ouo
Io
h
h
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.