1

EGERTON UNIVERSITY

DEPARTMENT OF INDUSTRIAL AND ENERGY ENGINEERING

COURSE: ITEC 112: THERMO-FLUIDS

LECTURE NOTES

Lecturer: Dr. L. K. Langat

Technologist: Mr. D. Chirchir

SEPTEMBER, 2011

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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

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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

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line, as shown in below. The points P, Q, and R represent the

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

When a liquid at boiling point is heated further at constant

pressurhe te additional heat supplied changes the phase of the

substance from liquid to vapour; during this change of phase the

pressure and temperature remain constant. The heat supplied is

called the latent heat of vaporization. It is found that the higher

the pressure then the smaller is the amount of latent heat required.

There is a definite value of specific volume of the vapour at

anyone pressure, at the point at which vaporization is complete,

hence a series of points such as p', Q', and R' can be plotted and

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.

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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

QQ

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

QQ

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.