Thermodynamics Module

Chapter two The laws of thermodynamic

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Chapter one Introduction

1.0 The word thermodynamics

Thermo – is the Greek word meaning hot or heat.

Dynamics -Is the study of matter in motion, from the Greek word meaning power.

1.1 What is thermodynamics as field of study?

Def (1) it is science in which the storage, the transformation and the transfer of energy are studied.

Energy is stored as internal energy, kinetic energy, potential energy and chemical energy. It is

transformed from one of these specifics to another and it is transferred across a boundary as either

heat or work. Thermodynamics is a system of mathematical equations that relate the

transformations and transfer of energy to the material properties such as temperature, pressure and

enthalpy.

Def (2) it is the science of the relationship between heat, work and the properties of systems, it is

concerned with means necessary to convert heat energy from available sources such as fossil fuels

into mechanical work.

1.2 Why do engineers study thermodynamics?

They are concerned with how much heat energy can be effectively converted into work and vice

versa in the case of refrigeration.

1.3 Historical background to the thermodynamics

Man used his muscles as the source of power at the beginning to carry, walk etc.

Later animals were used for carrying purposes and in ploughing fields.

Wind was the next source of power for driving vessels, milling, sawing etc.

Water was the next source of power but its disadvantage was that it was not always

available.

Steam engines were developed in the seventeenth century

Internal combustion engines were next to being developed.

Hydroelectric power was the next

Currently people are turning to solar power.

1.4 What is a thermodynamic system?

Is a definite quantity of matter contained within a closed surface. All matter and space external to

the system is called its surroundings. A system interacts with its surroundings by transferring

energy, not matter across its boundary. If a system does not exchange energy with its surrounding

it is an isolated system.

1.5 Types of energy

Internal energy- it is that energy which results from the random motion of the atoms and

molecules of a body. An internal energy increase does not always result in increase in temperature


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of the body e.g. during phase change the temperature does not change. Internal energy is a

property of the substance.

Heat- it is a transient quantity, it describes the energy transfer process through a system of

boundary resulting from temperature difference. Heat ceases to exist when the process of heat

transfer ceases.

Potential energy – If a fluid is at some height Z above a given datum level, then as result of its

mass it possess gravitational potential energy with respect to the datum.

Potential energy per unit mass =gZ…………………………………………1.1

Work- if a system exists in which a force at the boundary of the system is moved through a

distance then work is done by or on the system. Work is therefore a transient quantity, being

descriptive of that process by which a force is moved through a distance.

W = PV………………………………………1.2

Kinetic energy- is the energy that can be obtained from a moving body by slowing it down. The

maximum energy is obtained by slowing the body down to rest

k. e. =1/2mc2 ……………………………………….1.3.


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Chapter two

Laws of thermodynamics

2.0 Introduction

The laws of thermodynamics are really statements of thermodynamic behavior. They are natural

laws, which are based on observable phenomena. These are considered as law because they have

never been shown to be contradicted.

2.1 The Zeroth law of thermodynamics.

It states that if two bodies are separately in thermal equilibrium with a third body then they must

be in thermal equilibrium with each other. If bodies B and C are in thermal equilibrium with body

A then bodies B and C must be in thermal equilibrium with each other. By thermal equilibrium is

meant that there is no change of state and hence the zeroth law implies that the bodies A, Band C

will all be at the same temperature.

Thermal Equilibrium

It is observed that a higher temperature object, which is in contact with a lower temperature

object, will transfer heat to the lower temperature object. The objects will approach the same

temperature, and in the absence of loss to other objects, they will then maintain a constant

temperature. They are then said to be in thermal equilibrium. Thermal equilibrium is the subject of

the Zeroth Law of Thermodynamics.

The "zeroth law" states that if two systems are at the same time in

thermal equilibrium with a third system, they are in thermal

equilibrium with each other.

If A and C are in thermal equilibrium with B, then A is in thermal equilibrium with B. Practically

this means that all three are at the same temperature, and it forms the basis for comparison of

temperatures. It is so named because it logically precedes the First and Second Laws of

Thermodynamics.

Application

Measurement of temperature of a body using mercury in glass thermometer, when the

thermometer is steady, it is assumed that the mercury, the glass container and the body whose

temperature is being measured, are all at the same temperature and hence, are in thermal

equilibrium.

2.2 The first law of thermodynamics

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Statement 1

Net heat transfer - net work transfer = 0

Net heat transfer = net work transfer

§ Q = § W………………………………………………………………..2.1

Written symbolically § Q = § W § means the summation round the cycle.

The implications of the above statement are that there should be heat transfer for there to be work

done

Statement 2

If the heat transfer is not the same as work then some of the heat has been converted into internal

energy.

Q = U + W …………………………………………………………………2.2

where Q is the heat transfer,

U change of internal energy.

and W is the work transfer.(work done by the system)

Statement 3

The steady flow energy equation is a further statement of the first law of thermodynamics

U1 + p1v1 +2

1

2C

+ gZ1+Q = U2 +p2v2 +2

2

2C

+ gZ2 + W……………………….2.3

Where U = specific internal energy

p1v1 = specific flow work

C = velocity, it gives specific K.E.

Z = height above given datum level ( gives gravitational

potential Energy)

Q = heat transfer

W = work transfer

2.3 The second law of thermodynamics

It is a directional law in that it states that the heat transfer will occur down a temperature gradient

as a natural phenomenon. Heat transfer can occur up the gradient but not without the aid of

external energy e.g. in refrigeration and heat pumps. The second law of thermodynamics is a

general principle which places constraints upon the direction of heat transfer and the attainable

efficiencies of heat engines. In so doing, it goes beyond the limitations imposed by the first law of

thermodynamics.

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2.3.1 The concepts of the second law of thermodynamics

Rudolf Claussius (1822-1888)

It is impossible for a self-acting machine, unaided by an external agency, to convey heat

from a body at low temperature to one at a higher temperature.

Lord Kelvin (1824-1907)

We cannot transfer heat into work merely by cooling a body already below the temperature

of the coldest surrounding objects.

Marx Planck (1858-1947)

It is impossible to construct a system, which will operate in a cycle, extract heat from the

reservoir, and do an equivalent amount of work on the surroundings

Kelvin – Planck

It is impossible for heat engine to produce net work in a complete cycle if it exchanges

heat only with bodies at a single fixed temperature.

Second Law: Heat Engines

Second Law of Thermodynamics: It is impossible to extract an amount of heat QH from a hot

reservoir and use it all to do work W. Some amount of heat QC must be exhausted to a cold

reservoir. This precludes a perfect heat engine.

This is sometimes called the "first form" of the second law, and is referred to as the Kelvin-Planck

statement of the second law.

Second Law: Refrigerator

Second Law of Thermodynamics: It is not possible for heat to flow from a colder body to a

warmer body without any work having been done to accomplish this flow. Energy will not flow

spontaneously from a low temperature object to a higher temperature object. This precludes a

perfect refrigerator. The statements about refrigerators apply to air conditioners and heat pumps,

which embody the same principles.

This is the "second form" or Clausius statement of the second law.

Second Law: Entropy

Second Law of Thermodynamics: In any cyclic process the entropy will either increase or remain

the same.

Entropy:

a state variable whose change is defined for a reversible process at temperature T where Q is the

heat absorbed.

T

QS

Entropy: a measure of the amount of energy, which is unavailable to do work.

Entropy: a measure of the disorder of a system.

Entropy: a measure of the multiplicity of a system.

Since entropy gives information about the evolution of an isolated system with time, it is said to

give us the direction of "time's arrow" . If snapshots of a system at two different times shows one

state which is more disordered, then it could be implied that this state came later in time. For an

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isolated system, the natural course of events takes the system to a more disordered (higher

entropy) state.

2.3.2 Implications of the second law of thermodynamics.

Heat transfer will only occur, and will always naturally, when a temperature difference

exists, and always naturally down the temperature gradient.

If due to temperature difference, there is heat transfer availability, then work transfer is

always naturally down the temperature gradient.

Temperature can be elevated but not without the expenditure of external energy. Elevation

of temperature cannot occur unaided.

There is no possibility of work transfer if only a single heat energy source or reserve at

fixed temperature is available.

It is useful to note that if work transfer is supplied to a system then this can all be

transformed into heat energy. Examples of this are seen in the cases friction and the

generation of electricity. Heat energy cannot be all transformed into work transfer; there

will always be some loss. Thus work transfer appears to have a higher value than heat

transfer.

No contradiction with the first law of thermodynamics has been demonstrated.

From this second law it follows that inorder to run all the engines and devices in use at

present, and hence to maintain and develop modern industrial society, a source of supply

of suitable fuels is essential it is by burning and consuming these fuels that the various

working substances have their temperatures put up above that of their surroundings, that’s

enabling them to release energy by heat transfer in a natural manner according to the

second law of thermodynamics.

2.4 The third law of thermodynamics

The law suggests that the internal energy of a substance results from the random motion of

the atoms and the molecules which makeup the substance, the motion of the atoms is

associated with the temperature and it develops the idea of an absolute zero temperature

when all random motion ceases. The above considerations led to the development of the

third law of thermodynamics, which is the concept that at absolute zero of the temperature

the entropy of a perfect crystal of a substance is zero.

2.5 Conclusion

The laws of thermodynamics govern the way thermodynamic processes take place.

The zeroth law compares the temperature of two bodies. The first law is a conservation of energy

principle. The second law indicates the direction in which the heat would flow. The third law

gives the relationship between kinetic energy of particles and the temperature of the body.


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Chapter three

The properties of the thermodynamic media

3.0 Introduction

The molecules of a gas are in a rapid and random motion such that they fill completely the

container in which they are placed, the rapidly moving molecules continually bombard the surface

of the container and their effect is to produce a force normal to the surface, the number of impacts

per unit time is so large that it appears as if the force is continuous and uniform.

Varying any one of the gas properties such as pressure, temperature, volume, density will witness

the change in other properties.

Heat: is defined as thermal energy transferred across the boundaries of a system mainly because

of temperature difference between the two points. Specifically heat is transferred from a body of

higher temperature to a body in contact, which is at a lower temperature unless aided by

mechanical means to act otherwise.

Specific heat capacity of a body is the amount of heat required to raise one kg of a substance

through one degree.

Specific heat capacity t

Qc

in the limit as 0t , then

dt

dQC

The specific heat capacity is generally found to vary with temperature; they also vary with

pressure and volume. Cp denote the specific heat capacity at constant pressure and at constant

volume it is represented by Cv

For a solid the quantity of heat energy Q = mct

Q = mc(T1-T2)

For a gas of mass m heat added at constant pressure is given by

Q = mcp(T1-T2)……………………………………. .3.1

The heat added at constant volume Q = mcv(T1-T2)…….……………………… ………….3.2

Joule`s law states that the internal energy of a body/ perfect gas is a function of absolute

temperature only.

U = f (T)

But from the first law of thermodynamics the energy supplied to the gas is given by

Q = U + W

U — is internal energy ,

W — is the work done

But at constant volume W = 0

then Q = U

but also at constant volume Q = mCv T


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for one kg Q = Cv T

Q = Cv (T2-T1) + k

3.1 Relationship between specific heats

Q = U + W

= m Cv (T2 -T1) + W

but W = P(V2 - V1)

also PV = mRT

W =mR(T2 - T1)

Heat supplied to the system Q = mCv(T2 - T1) + mR(T2 - T1)

But also Q =mCp(T2-T1) for constant pressure

Comparing the two equations above it is deduced that

Cp = Cv + R

Cp - Cv = R……………………………………………………………3.3

3.2 Ratio of specific heats It has been proved that,

constantC

C

v

p

v

p

C

C …………………………………………..………………………….3. 4

is the expansion/ compression index

Cp = Cv

But from above Cp – Cv= R

Dividing through out by Cv

VV

V

V

P

C

R

C

C

C

C

- 1 =R/Cv

1-

RC

V ……………………………………………………………………………3.5


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3.3 Partial pressures:

Dalton’s and Gibbs law

The relationship between the partial pressures of the constituent gases is expressed by Dalton`s

law as follows:

The pressure of a mixture of a gas is equal to the sum of the partial pressures of the constituents.

The partial pressure of each constituent is that pressure which the gas would exert if it occupied

alone that volume occupied by the mixture.

The total pressure P= Partial pressure of gas1 + partial pressure of gas 2 + partial pressure gas 3

Thus P = P1 + P2 + P3

Thus from the characteristic gas equation Pv = mRT substituting for the pressure of each gas and

also for the mixture

3

333

2

222

1

111

V

TRm

V

TRm

V

TRm

V

mRT

Since it can be considered as though each gas occupy the same volume, then the volume is

common throughout. The temperature is also common.

MRt = T(m1R1 + m2R2 + m3R3 + etc)………………………………………………………….3. 6

Where m is the total mass of the mixture and R is the gas constant of the mixture.

3.4 The ideal gases

The various laws governing the pressure- volume-temperature relations of gases as discussed in

this section apply accurately only to a hypothetical ideal or perfect gas. A perfect gas is described

as one, in which there is no interaction between the molecules of the gas. The molecules of such a

gas are entirely free and independent of each other’s attraction forces. Hence none of the energy

transferred either to or from an ideal gas has an effect on the internal potential energy.

At high temperature and low pressure gases approach the ideal gas, at low pressure the density

and intermolecular forces become very small. Under this situation the gases obey the law

RT

PV

Where R is the characteristic gas constant whose units are kJ/kgK

Each perfect gas has a different gas constant

For unit mass P v = RT

For m kg Pv = mRT…

the above is known as the characteristic gas equation

for a substance of molecular weight M =m/n, where n is the number of moles, the equation

becomes PV = nMRT


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nT

PVMR

Avogadro`s hypothesis states that that the volume of one mole of any gas is the same as the

volume of any other gas at the same temperature and pressure

Thus V/n is the same for all gases at the same temperature and pressure

Thus Tn

PV is constant for all gases

Therefore Tn

PVRMR

0

Where R0 is the molar gas constant or universal gas constant

PV = n R0 T………………………………………………………………………3.7

R0 = 8314, 5J/kmol

3.5 Expansion and compression of ideal gases

Introduction

The state of a working fluid can be identified by its thermodynamic properties, which are

pressure, temperature, volume, internal energy, entropy etc. if two of the properties are chosen

they are enough to define the state of a system and this state is represented by a point situated on

the diagram of the properties shown below.

Fig 3. 1 Pressure volume diagram Fig 3.2 Pressure and Temperature diagram

1

2

3

p1

p2

p3

V1 V2 V2

1

2

3

p1

p2

p3

T1 T2 T2


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3.5.1 Reversibility

Definition – when a fluid undergoes a reversible process, both the fluid and its surrounding can

always be restored to their original state.

Conditions of reversibility: The process must be frictionless (no internal friction and mechanical friction).

The differences in pressure between the fluid and its surrounding during the process must

be infinitely small.

Difference between the fluid temperature and its surrounding must be infinitely small.

When a fluid undergoes a reversible process a series of state points can be joined up to

form a line on a diameter of properties.

The work done by a fluid during an irreversible process is given by the area under the P-V

diagram

P1

P2

V1 V2 Volume

pre

ssure

1

2

Fig3.3 Pressure– volume relationship

Work done is the area under graph 1-2 = v

v

1

2

Pdv

3.5.2 The isothermal process

Also known as the constant temperature process

When a gas in the cylinder is compressed from v1 to v2 the temperature of the gas will increase

but if the change takes place slowly there will be a very small change in temperature. In the latter,

when a lapse of time is allowed the temperature will return to its original value. And the process

will approximate an isothermal process.

When the volume of the gas is increased or decreased under such conditions that the temperature

remains constant the absolute pressure will vary inversely with the volume.


14

Boyles laws

It states that when the temperature of the body is maintained constant as its volume is varied, the

change in absolute pressure is inversely proportional to the volume

V

1P

PV = C

P1V1 = P2V2 …………………………………………………..3.8

P1

P2

V1

V2 Volume

pre

ssure

1

2

Vv

P

PV = C

Fig. P-v Diagram showing work done in an isothermal process

Work done during an isothermal process dvP

V2

V1

w

but V

CP therefore dv

V

C

V2

V1

w

W = C [ ln ] 2

1

v

vInCW

Then PV = C then 1

211

V

VlnVPW ………………………………3.9

Heat transfer during constant temperature process

Q = U + W


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But U =0 since there is no temperature change

Q = W

3.5.3 Constant pressure process

If the temperature of the gas is increased by the addition of heat while the gas is allowed to

expand and its pressure is maintained constant, then the volume of the gas will increase in

accordance with Charles law.

Charles law states that:

When energy is supplied to a gas under such conditions that the pressure of the gas is kept

constant, then the volume of the gas will increase in direct proportion to the change in absolute

temperature of the gas.

V T

V/T = constant

V1/T1 =V2/T2………………………………………………………….3.10

Force exerted on the piston = P A

Work done in pushing piston = PAL

But AL = V2- V1

P

V1 V2

L

Area (A) Piston

Fig 3.5 Work done at constant pressure

Work done on the piston = P(V2–V1) ……………………………………………….3.11

Heat transfer Q = U + W…………………………………...……………….3.12


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3.5.4 Constant volume process ( isochoric)

Assume that the gas is confined in a closed cylinder so that its volume cannot change as it is

heated or cooled, as the energy to the gas is varied, the absolute pressure will increase

proportionally to the increase in absolute temperature

Combining Charles law and Boyles law

2

22

1

11

T

VP

T

VP Known as the characteristic gas equation

For constant V, V1 =V2

2

2

1

1

T

P

T

P …………………………………………………………….……..3.13

Work done at constant volume W = 0

Heat transfer at constant volume Q = U + W

but W = 0

Q = U

= mC

= mCv(T2- T1)……………………………………….3.14

3.5.5 Adiabatic compression and expansion: Also known as the constant energy process

Def – an adiabatic process occurs when the compression or expansion of a gas takes place without

an heat flow out or into the system ( it occurs very fast)

If a gas expands or is compressed in a cylinder, which is well lagged, and the change takes place

very quickly the conditions are approximately adiabatic. An adiabatic process is a special case of a

polytropic process

Polytropic expansion or compression of gases

PVn = C

All expansions of the above form are polytropic

P1V1n = P2V2

n………………………...…………………3.15

Also PV/T =C then 2

22

1

11

T

VP

T

VP ………………………….………..…………………3.16

From 3.15

1

2

1

2

1

2

V

V

P

P

T

T…………………………………………………3.17


17

From 3.16 (V2/V1)n = P1/P2

V2/V1 = (P1/P2)1/n

= (P2/P1) –1/n

……………..…………………………………….3.18

substituting into 17, T2/T1 = P2/P1 (P2/P1) –1/n

T2/T1 =P2/P1 n-1/n

……………………….…………………………………………….………3.19

from 2 T2/T1 =P2V2/P1V1

From 1 P2/P1 =V1/V2 n

Then T2/T1 = (V1/V2)n x V2/V1

= (V1/V2)n x (V1/V2)

–1

T2/T1 = ( V1/V2)n-1

…………………………………………………………3.20

3.5.6. For non-flow processes Work done during a polytropic expansion

P1

P2

V1

V2 Volume

pre

ssure

1

2

Vv

P

PVn = C

Work done = area under the curve 1-2

dvP

V2

V1

w


18

but nV

CP therefore dv

V

C

V2

V1

nw

= C[ V1-n ]

v

v

1

2

1-n

=

n1

VPVP 1122

…………………………….……………………..………………..3.21

=

1n

VPVP 2211

……………………………………………………………………..3.22

ALSO Since PV = mRT W =

1n

TTmR 21

For non-flow processes

Work done during an adiabatic expansion

P1

P2

V1

V2 Volume

pre

ssu

re

1

2

Vv

P

PVγ = C

Work done = area under the curve 1-2

dvP

V2

V1

w


19

but

V

CP therefore dv

V

C

V2

V1

w

= C[ V1-]v

v

1

2

1-

=

1

VPVP 1122 ………………………..………………..3.23

=

1

VPVP 2211

……………………………………………………………………………………..3.24

ALSO Since PV = mRT W =

1

TTmR 21

Worked examples

QUESTION 1

A petrol engine cylinder has diameter of 9,5 cm and stroke 12,7 cm, clearance volume 230 cm3. If the

temperature at the beginning of the compression is 570C,

Find the temp at the end of compression, and the work done during the compression stroke if the law

of compression is pV1.3

= c. take the initial pressure as 100 kN/m2.

SOLUTION

V2 = VC = 230 X 10-6

V1 = 10 230 0.127095.04

6-2 X X X

= 1130 X 10-6

From 3.111

3.122 VPVP

3.1

2

12

V

V100P

3.1

2230

1130100P

P2

P1

V2=VC V1=VC+VS Volume

pre

ssu

re

2

1

VC VS


20

= 792.1kN/m2

Temperature

From 2

22

1

11

T

VP

T

VP

11

2212

VP

VPTT

1301100

2307921330T2

X

X

= 532K or 259oC

work done

W = 1n

VPVP 2211

W = 3.0

102301.792101130100 66 XXXX

= -0.23061kJ

= -230.61J

(–)indicates work done on the gas

QUESTION 2

1 m3 of air, initially at 110 kN/m

2 and 15

0C, is compressed according to the law pV

1.3 = c in a

cylinder to a final pressure of 1.4 MN/m2 . Taking R for air = 287 J/kgK and cp = 1005 J/kgK,

determine

a) the volume and temp of air at the end of compression;

b) the work done in compressing the air;

c) the change of internal energy;

d) the heat exchange through the cylinder walls, stating the direction of heat flow.

a) i)

3.1

1

2

112

P

PVV


21

3.1

1

21400

1101V

= 0.141m3

ii) n

1n

1

212

P

PTT

3.1

13.1

2110

1400288T

= 518 K or 245oC

b) work done

W = 1n

VPVP 2211

W = 3.0

141.014001110 XX

= -291.3kJ

C) U2 – U1 = mcV(T2 – T1)

But RT

PVm =

288287.0

1110

= 1.33

CP - CV = R

Thus Cv = 1.005 - 0.287

= 0.718 kJ/kg

U = 1.33 x 0.718(518 – 288)

= 219.77 kJ/kg

Heat flow Q = W + (U2 – U1)

= –291.3 + 219.6362

= –71.6638 kJ/kg

QUESTION 3

1 kg of a certain gas expands adiabatically in a closed system until its pressure is halved. During the

expansion the gas does 67 kJ of external work and its temp falls from 2400C to 145

0C. Calculate the

value of the adiabatic index and the characteristic constant of the gas.


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From

1

1

2

1

2

P

P

T

T

1

2

1

513

418

0.814814814 =

1

2

1

log 0.8 =

1 log

2

1

1 =

5.0log

8.0log

1 = 5.0log

8.0log

- 0.295γ = 1

0.7 = 1

= 1.419

Work done

W =

1

TTmR 21

67 =

1

TTR 21

R =

21 TT

167

R = 418513

419.067

= 0.2955kJ/kg

QUESTION 4

In a closed system, 1 kg of air initially at 100 kN/m2 and 27

0C is compressed adiabatically to 3

MN/m2 and then expanded isothermally back to its original volume. determine the excess of the work

done by the gas over the work done on the gas.Taking R for air = 287 J/kgK and = 1.4.


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Given data

m = 1 kg

P1 = 100 kN/m2

T1 = 300K

P2 = 3 000 kN/m2

R = 0.287 kJ/kgK

Final temperature—adiabatic process

1

1

212

P

PTT

4.1

4.0

2100

0003300T

K7858662.792T2

Adiabatic work

W =

1

TTmR 21

W =

14.1

793300287.01

W = –353.573859 kJ

Isothermal work

2

322

V

VlnVPW

But P2V2 = mRT2 2

32

V

VlnmRTW

Also 2

1

2

3

V

V

V

V

From adiabatic process

2

3

1

1

2

2

1

V

V

P

P

V

V

4.114.1

1

1

30100

0003

P2

P1

V2 V1=V1 Volume

pre

ssu

re

2

1

3P3


24

thus Work 4.11

2 30lnmRTW

4.11

30ln7858662.792287.01W

= 552.7663485 kJ

= 552.8 kJ

Excess of the work done

= 552.7663485 kJ –353.573859kJ

= 199.1924895kJ

Chapter 3

Tutorial questions

1. A volume of 0.5 m3 of gas is expanded in a cylinder from a pressure of 660 kN/m

2 and temp 165

0C

to a pressure of 120 kN/m2 according to the law pV

1.3 = c. Find:-

a) the final volume and temp of air ;

b) the work done by the air during the expansion;

c) the change of internal energy during expansion;

d) the heat exchange through the cylinder walls, stating the direction of heat flow.

Take Cv = 710 J/kg K and R for air = 287 J/kg K.

2. Air is expanded adiabatically from a pressure of 800 kN/m2 to

128 kN/m2. If the final temp is 57

0C, calculate the temp at the beginning of expansion, taking = 1.4.

3.The ratio of compression in a petrol engine is 9 to 1. Find the temp of the gas at the end of

compression if the temp at the beginning is 240C, assuming compression to follow the law pV

n =

constant where n = 1.36.

4.The vol. and temp of a gas at the beginning of expansion are 0.0056 m3 and 183

0C, at the end of

expansion the values are 0.0238 m3 and 22

0C respectively. Assuming expansion to follow the law

pVn = constant, find the value of n.

5. 7.08 litres of air at a pressure of 13.79 bar and temp 3350C are expanded according to the law

pV1.32

= constant, and the final pressure is 1.206 bar. Calculate

i) the vol at the end of expansion

ii) the work transfer from the air

iii) the temp at the end of expansion


25

iv) the mass of air in the system taking R for air = 287 J/kgK.

6. A perfect is compressed in a cylinder according to the law pV1.3

= constant. The initial condition

of the gas is 1.05 bar, 0.34 m3 and 17

0C. If the final pressure is 6.32 bar, calculate:-

a) the mass of gas in the cylinder;

b) the final volume;

c) the final temperature;

d) the work done to compress the gas;

e) the change in internal energy;

f) the transfer of heat between the gas and cylinder walls.

7. Assume a compression according to the law PV = CONSTANT,

i. Calculate the final volume when 1m3 of gas at 120 k N/m

2 is compressed to a

pressure of 960 k N /m2.

ii. Calculate the initial volume of gas at a pressure of 1.05 bar, which will occupy

a volume of 5.6m3

when it is compressed to a pressure of 42 bar.

iii. 0.25m3 of gas at temperature of 21

0C is heated at constant pressure to a

temperature of 3150C. Calculate the final volume.

8. A closed vessel contains air at a pressure of 140kN/m2 gauge and temperature 20

0C. Find the

final gauge pressure if the air is heated at constant volume to 400C. Take the atmospheric pressure

as 759mm Hg.

9.Taking the characteristic gas constant R for nitrogen as 0.297kJ/kgK., calculate:

i. The mass of 0.05m3 of nitrogen at 550kN/m

2 and 28

0C.

ii. the volume of 1kg of nitrogen at 1MN/m2 at 0

0C.

10.An air reservoir contains 20kg of air at 3200kgkN/m2 gauge and 16

0C. Calculate the new

pressure

and heat energy transfer if the air is heated to 350C. Neglect any expansion of the reservoir, take

the

R for air =0.287kJ/kgK, specific heat at constant volume Cv =0.718kJ/kgK, and atmospheric

pressure = 100kN/m2.

11. heat energy is transferred to 1.36kg of air which causes its temperature to increase from

400C to 468

0C. calculate for two separate cases of heat transfer at a). Constant volume b).

Constant pressure:

i. the quantity of heat energy transfer

ii. the external work done

iii. the increase in internal energy

Take Cv and Cp as 0.718 and 1.005kJ/kgK respectively.

12. A closed vessel of 500cm3 capacity contains a sample of flue gas at 1.015 bars and 20

0C. If

the analysis of the gas by volume is 10% CO2. 8% O2, and 82% n2 calculate the partial

pressure and mass of each constituent in the sample. R for CO2, O2 and N2 =0.189, 0.26

and 0.297kJ/kgK respectively.

13. Gas is expanded in an engine cylinder according to the law Pvn =C. at the beginning of

the expansion the pressure and the volume are 1750kN/m2 and 0.05m

3 respectively, and at

the end of expansion the respective values are 122.5kN/m2 and 0.375m

3. calculate the

value of n.


26

14. 0.113m3 of air at 8.25 bar is expanded in cylinder until the volume is 0.331m

3 . calculate

the final pressure and the work done if the expansion is ;

i. isothermal,

ii. adiabatic, taking =1.4

15. a gas is expanded in a cylinder behind a gas tight piston. At the beginning of the expansion

the pressure is 36bar, volume 0.125m3 and temperature 510

0C. at the end of expansion the

volume is1.5m3 and the temperature 40

0C. Taking R =0.284kJ/kgK and Cv =0.71kJ/kgK,

calculate;

i. the pressure at the end of expansion.

ii. The index of expansion

iii. The mass of gas in the cylinder,

iv. the change in internal energy,

v. work done by the gas

vi. heat transfer during expansion.


27

Chapter four

The first law of thermodynamics

4.0 Introduction:

Conservation of mass states that the energy can neither be created nor destroyed but can be

changed from one form to another. The total energy in a system remains constant hence

initial energy of the system plus energy entering the system is equal to the final energy of

the system plus energy leaving the system.

4.1 Energy forms in thermodynamics:

Gravitational potential energy, for unit mass of fluid potential energy =gz where z is some

height above a given datum.

Kinetic energy, k.e. = C2/2 where C is the velocity of flow of the fluid.

Internal energy U –all fluids store energy. The energy stored within the fluid results from

the internal motion of the atoms and molecules.

Heat received or rejected (Q)- in an system a fluid can have a direct reception or rejection

of heat through the system boundary. Heat received is positive heat rejected is negative.

External work done (W)

In any system a fluid can do external work or have external work done on it, transferred

through the system boundary. External work done by the system is positive, external work

done on the system is negative.

4.2 The-closed system: The substance enclosed by the system will possess internal energy U, there could also be

some other random energies, which are electrical, magnetic and surface tension but these

forms of energy will be neglected as being insignificantly small.

In a closed system the sum of all the energies possessed by the contained substance called

the total energy E:

E = U + P.e + K.e

Process carried out on a closed system:

E1 = initial energy of the contained substance

E2= final total energy of the contained substance

Q = heat transferred to or from the substance.

W = work transferred to or out of the system.

E1 +Q = E2 + W

4.3 The non-flow energy equation:

It was stated that the sum of all energies possessed by the contained substances was called

total energy E, where

E = U + P.E + K.E


28

If the substance is considered to be at rest then there is no turbulence. In this case the random

potential and kinetic energy will be zero. Thus for a substance at rest the contained energy will be

only internal energy then equation 1 becomes.

U1 + Q = U2 + W

Q = U2-U1 + W………………………………………………..4.1

4.4 The open system:

The purpose here is to examine the two flow open system in which an equal mass of fluid

per unit time is entering and leaving the system. This is known as the continuity of mass

flow.

Q

W

Input Output

Fig 4.1 energy balance in an open system

Entering

1. internal energy (U)

2. displacement or flow energy (PV)

3. kinetic energy (KE)

4. Gravitational potential ( PE)

Leaving

1. U2

2. KE2

3. P2V2

4. PE2

As the fluid enters the system let the total actual energy of the fluid in the system= E1

after passing through the system as the fluid leaves let the total energy of the fluid mass

remaining in the system = E2 in its passage through the system, let the fluid mass transfer heat

=Q. and the transfer work = W. for two flow open systems:

Fluid mass entering the system =U1 + P1V1 + KE1 + Q1

Energy of fluid mass leaving the region = U2 + P2V2 + PE2 + KE2


29

from the principle of conservation of energy

U1 + E1 +P1V1 + KE1 + PE1 + Q = U2 + E2 + P2V2 + KE2 + PE2 +W

But U + PV = H = enthalpy

E1 + H1 + PE1 + KE1 + Q = E2 + H2 + PE2 + KE2 + W

Q - W = (E2-E1) + (H2-H1) + (KE2-KE1) + (PE2-PE1)

4.5 The steady flow energy equation

In a stead flow system it is considered that the mass rate of the fluid throughout the system is

constant. It is further considered that the total fluid mass in the system remains constant

For unit mass through the system

U1 + P1V1 +2

C21 + Q + Z1 = U2 + P2V2 +

2

C22 2 + W + Z2 …………. ….4.2

`for most industrial applications the difference in height is negligible Z2-Z1=0

and also U + PV = H

H1 + 2

C21 + Q = H2 +

2

C22 + W………………. ….. . 4.3

W

Q

P1, V1, C1, U1

Entry

P2, V2, C2, U2

Exit

Z2

Z2

Datum

Fig 4.2 energy balance in a steady flow


30

Worked example 1

A turbine operating under steady flow conditions receives steam at the flowing state; pressure,

13.8 bar, specific volume 0.143m3/kg, specific internal energy 2590 kJ/kg,

velocity 30 m/s. the state of the steam leaving the turbine is as flows: pressure 0.35 bar,

specific volume 4.37m3, specific internal energy 2360 kJ/kg, velocity 90m/s. heat is

rejected to the surroundings at the rate of 0.25 kW and the rate of steam flow through

the turbine is 0.38 kg/s. calculate the power developed in the turbine.

Given data

Entering Leaving

P1 = 1 380 kN/m2 P2 = 35 kN/m

2

V1 = 0.143 m3/kg V2 = 4.37m

3/kg

U1 = 2 590 kJ/kg U2 = 2 360 kJ/kg

C1 = 30 m/s C2 = 90 m/s

= – 0.25 kJ/s

Q = = 38.0

25.0 – 0.657894736kJ/kg

W = (U1 – U2) + (P1V1 – P2V2) + 3

22

21

102

CC

+ Q

= 2 590 – 2 360 + 1 380 x 0.143 – 35 x 4.37 + 3

22

102

9030

– 0.658

= 270.1321kJ/kg

P = W = 270.1321 x 0.38

= 102.65 kW

Q• Q

•

Q•

Q•

m•

2.In the turbine of a gas turbine unit the gases

= – 0.25 kJ/s

Q =

= – 0.657894736kJ/kg

38.0

25.0

m•


31

Chapter four

Tutorial Questions

The first law of thermodynamics

1.A nozzle is a device for increasing the velocity of a steadily flowing

fluid . At the inlet to a certain nozzle the specific enthalpy of the fluid is 3025 kJ/kg

and the velocity is 60m/s. at exit from the nozzle the specific enthalpy is 2790 kJ/kg

the nozzle is horizontal and there is negligible heat loss from it. Calculate

The velocity of the fluid at exit.

The rate of flow of fluid when the inlet area is 0.1m2 and the specific volume at inlet is

0.19m3 /kg

The exit area of the nozzle when the specific volume at the exit is 0.5m3/kg

[688m/s; 31.6kg/s; 0.0229m2.] esatop

2.In the turbine of a gas turbine unit the gases flow through the turbine at 17 kg/s and the power

developed by the turbine is 14000kW . The specific enthalpies of the gases at

inlet and outlet are 1200kJ/kg and 360kJ/kg respectively, and the velocities of the

gases at inlet and outlet are 60m/s and 150m/s respectively. Calculate the rate at which

heat is rejected from the turbine. Find also the area of the inlet pipe given that the

specific volume of the gases at the inlet is 0.5m3/kg

[119.34KJ/s 0.142m3 ]


32

Chapter five

The second law of thermodynamics

5.0 Engine cycles and thermal efficiency

During a cycle there will be some heat transfer and work transfer to and from the substance. After

performing the cycle, if the substance returns to its original state then from the first law of

thermodynamics.

§ Q = § W

Thus for the cycle the net work transfer = net heat received- net heat rejected

Thermal efficiency receivedheatnet

doneworknet

Net work done = positive work – negative work

Specific steam consumption kWinoutputPower

hr/kgsteamofflowmass

Specific fuel consumption kWinoutputPower

uesdfuelofflowmass


33

5.1 The Carnot cycle for the gas

Carnot`s principle states that no engine can be more efficient than a reversible engine working

between the same limits of temperature

Carnot conceived of a cycle made up of thermodynamically reversible process.

Figs 5.1 P-V diagram for Carnot cycle

1-2 isothermal expansion

Work done =P1V1lnV2/V1 = mRT1lnV2/V1

for the isothermal process, Q = W

therefore heat received =mRT1lnV2/V1

2-3 adiabatic expansion

work done =P2V2-P3V3 = mR(T2-T3)

( – 1) ( – 1)

for the adiabatic process Q = 0, there is no heat transfer.


34

3-4 isothermal compression

Work done = P3V3ln V4/V3 = - P3V3ln V3/V4

= - mRT3lnV3/V4

for the isothermal process, Q = W

Therefore heat rejected = mRT3lnV3/V4

4-1 adiabatic compression

Work done = P4V4- P1V1 = -(P1V1-P4V4)

( – 1) ( – 1)

for the adiabatic Q = 0, no heat transfer during the process

This process returns the cycle to the starting point 1

Net work done per cycle = § W

= area under 1-2 + area under 2-3- area under 3-4 – area 4-1

= area 1 2 3 4

= area enclosed by cycle

§ W =mRT1lnV2/V1+ mRln(T2-T3) - mRT3lnV3/V4 – mR(T1-T4)

( – 1) ( – 1)

now T1 = T2 and T3 = T4 from the isothermals

therefore mR(T2-T3) = mR(T1-T4)

( – 1) ( – 1)

from the above equation

§ W = mRT1lnV2/V1-mRT3ln V3/V4

For the adiabatic 1-4

T1/T4 = (V4/V1) ( – 1)

For the adiabatic 2-3

T2/T3 = (V3/V2) ( – 1)

But T1 =T2 and T3 =T4

Therefore T1/T4 =T2/T3

Hence, from equations above

V4/V1 =V3/V2 or V2/V1 = V3/V4

Substituting equation 6 into equation 2

§ W =mRlnV2/V1(T1-T3)


35

Thermal efficiency = heat received – heat rejected

heat received

from the above analysis thermal efficiency = mRT1ln(V2/V1) –mRT3ln(V3/V4)

mRT1ln(V2/V1)

= mRln(V2/V1) (T1-T3)

mRln(V2/V1) T1

Since V2/V1 = V3/V4

Therefore thermal = T1 –T3

T1

= Max. absolute temp.– min. absolute temp.

Max. absolute temp .

5.1.1 Carnot efficiency using T-s diagram

Carnot showed the most efficient possible cycle is one with all the heat supplied at one fixed

temperature and then all the heat is rejected at a lower fixed temperature. The cycle therefore

consists of two isothermals joined with two adiabatic

1

23

4

A B

T1

T2

S

T

Fig 5.2 Carnot cycle on a T-s diagram

1-2 isentropic expansion from T1 to T2

2-3 isothermal heat rejection

3-4 isentropic compression from T2 to T1

4-1 isothermal heat supply

The cycle is completely independent of the substance used

The cycle efficiency = - W = Q/Q1

Q1


36

From the graph the gross heat supplied Q1 is given by the area 41BA4

Q1 =T1( sB-sA)

The net heat supplied Q is given by the area 41234

Q =(T1-T2) (sB-sA)

Hence the carnot cycle efficiency = T1-T2) (sB-sA)

T1( sB-sA)

= (T1-T2)

T1

5.20 Entropy

It is the property of a body, which is defined as Q/T

From the first law of thermodynamics the work transfer in a reversible process

W rev = PdV

The second law provides the corresponding expression for heat transfer which can be

written as

Q rev = T s , s is the change in entropy

Entropy is an abstract property it can not be measured but can only be calculated, entropy

like enthalpy can be treated in three stages of the formation of vapour from the liquid.

The energy of the liquid can be expressed as Q = mc

And for unit mass Q = cθ, for θ = T And for constant pressure Q = cpl T

Dividing by T Q/T = cpl T/T

S = cpl T/T

To calculate the total change in entropy between given temperatures

S = m cpl T/T

s2 –s1 =cpl ln (T2/T1) ………………………………….[5.1]

Change in entropy for a vapour

S = mcpv T/T

s2 –s1 = cpv ln (T2/T1)

s2 –sf = cpv ln (T2/T1)

s2 = sf + cpv ln (T2/T1)………………………………….[5.2]


37

Entropy of steam

s2 –sg =cps ln (T2/T1)

s2 = sg + cps ln (T2/T1)…………………………………………….[5.3]

5.2.1 Change of entropy for a gas

from the first law of thermodynamics

Q = U +W where U = cv T and W =P V

Q = cv T + P V Dividing by T

Q = cv T + P V T T T

From pv = RT then P/T = R/v put in above equation

Q = cv T + R V T T V

S = cv T + R V T V

S = cv T + R V T V

s2-s1 =cvln (T2/T1) +Rln( V2/V1)………………………………………[5.4]

but cp- cv =R

Substituting s2-s1 =cvln (T2/T1) +( cp- cv) ln( V2/V1)

s2-s1 =cvln (T2/T1) + cp ln( V2/V1) - cv ln( V2/V1)……………….. [5.5]

s2-s1 =cvln{ (T2/T1)(V1/V2) } + cp ln( V2/V1)

And from P1V1/T1 =P2V2/T2; T2V1 = P2/P1

T1V2

s2-s1 =cvln (P2/P1) + cp ln( V2/V1)……………………[5.6]

But also cv = cp-R put into equation [5.4]

s2-s1 =cPln (T2/T11) -R ln (T2/T11) + R ln( V2/V1)

s2-s1 =cPln (T2/T11) + R ln{( V2/V1)(T1/T2)}


38

From P1V1/T1 =P2V2/T2; T1V2 = P1/P2

T2V1

s2-s1 =cPln (T2/T11) +R ln (P1/P2) ……………………………………… [5.7 ]

5.2.2 Change of entropy at constant pressure

From equation [5.4]

s2-s1=cvln(T2/T1)+Rln(V2/V1)…………………………………………… [5.8]

from equation [5.6] p1=p2

s2-s1 = cp ln( V2/V1)…………………………………………………… [5.9]

from equation [5.7]

s2-s1 =cPln (T2/T11)…………………………………………………… [5.10]

5.2.3 Change of entropy at constant temperature

from equation [5.6]

s2-s1 =cvln (P2/P1) + cp ln( V2/V1)………………………………… [5.11]

from equation [5.4]

s2-s1 =Rln( V2/V1)…………………………………………………. . [5.12]

from equation [5.7]

s2-s1 = R ln (P1/P2) ………………………………………………… [5.13]

5.2.4 Change of entropy at constant volume

from equation [5.7]

s2-s1 =cP ln (T2/T11) +R ln (P1/P2)………………………………… [5.14]

from equation [7]

s2-s1 =R ln (P1/P2)…………………………………………………. [5.15]

from equation [4]

s2-s1 =cvln (T2/T1) ………………………………………………… [5.16]

5.2.5 Change of entropy during a polytropic process

Q = –n x polytropic work

–1

Q = –n x P V

–1

dividing by T

Q = –n x P V from pv = RT, P/T =R/V

T –1 T


39

s = –n x R V

–1 V

s2-s1 = –n x R ln(V2/V1) ………………………………………..[5.17]

–1

From cp –cv =R and cp/cv =

R = cv(–1) put into [5.17]

s2-s1 = –n x cv( –1) ln(V2/V1)

–1

s2-s1 = cv (–n ) ln(V2/V1) ……………………………….. [5.18]

but from P1V1n =P2V2

2

V2/V1 = (P1/P2)1/n

Putting into [5.18]

s2-s1 = cv (–n ) ln(P1/P2) 1/n

s2-s1 = cv ( –n ) ln(P1/P2) ……………………………………………[5.19]

n

from T2/T1 = (V2/V1)n-1

, V2/V1 =( T2/T1 )1/(n-1)

put into equation [18]

s2-s1 = cv ( –n ) ln( T2/T1 )1/(n-1)

s2-s1 = cv ( –n ) ln( T2/T1 ) …………………………………………[5.20]

n-1

Worked example A rigid vessel contains 0.5 kg of a perfect gas of specific heat at constant volume 1.1 kJ/kg K. A

stirring paddle is inserted into the vessel; and 11 kJ of work are done on the paddle by the stirrer

motor. Assuming that the vessel is well lagged and that the gas is initially at the temperature of the

surroundings, which are at 17 oC, calculate the effectiveness of the process.

Solution

Q + W = U2 +U1 but Q = 0 , W = 11 kJ

Giving

W = mCv(T2 – T1)

11 = 0.5 x 1.1 x (T2 – 290)

T2 = 310K = 37 oC


40

Change in entropy

s2 – s1 = Cvln(T2/T1) = 1.1 x ln(310/290)

= 0.07336 kJ/kg K

Increase in specific energy

= (U2 +U1) – To(s2 – s1)

= 1.1(310 – 290) – (290 x 0.07336)

= 0.7256 kJ/kg

therefore

effectiveness = gssurroundinofenergyinloss

systemaofenergyinincrease

effectiveness = 11

5.07256.0 = 0.033 or 3.3%

Tutorial questions

Second law of thermodynamics

1. The overall expansion ratio of a Carnot cycle is 15. The temperature limits of the cycle are

2600C and 21

0C. Determine;

The volume ratios of the isothermal and adiabatic processes

The thermal efficiency of the cycle

=1.4 [3.39; 44.8%] RJ

2. 0.23kg of gas is taken through a cycle whose temperature limits are 3000C and 50

0C if the

volume ratio of the isothermal processes is2.5:1 determine for the cycle,

the thermal efficiency

the net work done

the work ratio

Take =1.4 R =0.28kJ/kgK [43.6%; 14.75kJ; 0.2] RJ

3. Steam at a pressure of 1.9MN/m2 and with temperature of 250

0C is expanded isentropically

to a pressure of 0.3MN/m2. it is further expanded hyperbolically to a pressure of 0.12MN/m

2.

using tables determine;

the final condition of steam

the change of specific entropy during the hyperbolic process


41

[0.954; 0.5727kJ/kgK, an increase] RJ

4. A quantity of gas as an initial pressure, volume, and temperature of 1.1bar, 0.16m3 and 18

0C,

respectively. It is compressed isothermally to pressure of 6.9bar. Determine the change of

entropy. Take R =0.3kJ/kgK [-111kJ/K]

5. 1kg of dry saturated steam at 22bar is throttled to a pressure of 7bar, then expanded at constant

entropy to a pressure of 1.4bar. calculate;

the degree of superheat at 7bar

the increase in entropy

the dryness fraction at 1.4bar [15.80C, 0.4848kJ/kgK, 0.9217] REEDS


42

Chapter six

Gas engine cycles 6.0 Introduction

Gas engine cycles are ideal gas cycles the working media is assumed to be air. In studying this

chapter the students should be able to calculate the change in thermodynamic property at each

turning point. From which the heat supplied, heat rejected work done and the thermal efficiency

can be calculated.

6.1. Constant pressure cycle

Also known as the [isobaric] cycle, heat is added and rejected at constant pressure

Fig 6.1 P-v and T-s for constant pressure cycle

1-2 adiabatic compression according to the law PV = C

at point 1 assume that P1 V1 T1

1

1

2

2

1

V

V

T

T

1

2

112

V

VTT

= T1( rv)

-1

where 2

1

V

Vr

also 2211 VPVP

2

112

V

VPP


43

2-3 constant pressure heat addition

P3 = P2

2

2

3

3

T

V

T

V

T3 =T2 V3/V2 =V3/V2T1 rv(-1)

3-4 adiabatic expansion according to the law PV = C

i

4

3

3

4

V

V

T

T

4-1 constant pressure heat rejection

T4 =T1V4/V1

It follows that V4/V1 = V3/V2 =constant pressure ratios

This could also be obtained from the fact that since V1/V2 =V4/V3 = rv then

V4/V1 = V3/V2

Also P4V4

=P3V3

Then P4 =P3 (V3/V4) = P3/r

v

Net work done by the system= area under 2-3 + area under 3-4 – area under 4-1 – area1-2

= 1

VP-VP - )V-(VP -

1

)VP-V(P )V-(VP 1122

1414433

232

= 1

VP-VP-VP-VP)V-(VP )V-(VP 11224433

141232

1

T-T-T-TmR)T-(TmR )T-mR(T 1243

1423

1

T-T-T-T)T-(T )T-(TmR 1243

1243

1

11)T-(T )T-(TmR 1243 but PCR

1

)T-(T )T-(TmC 1243P

Work done = heat received – heat rejected

= mcp (T3-T2)- mcp(T4-T1)

= mcp (T3-T2)-(T4-T1)


44

Net work done =heat received x thermal efficiency

Thermal efficiency = 1- heat rejected

heat received

= 1-mcp(T4-T1)

mcp(T3-T2

= 1- (T4-T1)

(T3-T2)

6.2. Constant volume Also known as an otto cycle heat is added and rejected at constant volume

fig 6.2 T-s and P-V for constant volume cycle

1-2 adiabatic compression of the gas according to the law PV = C

assume that the thermodynamic properties at 1 are P1, V1, T1

T1/T2 = (V2/V1) -1

T2 = T1(V1/V2) -1

where, r v =V1/V2 , the adiabatic compression , volume ratio

=V4/V3, adiabatic expansion, volume ratio

also P1V1 = P2V2

P2 =P1(V1/V2)

=P1 rv

2- 3 constant volume heat addition

V3 =V2 since the volume remains constant

P3/T3 =P2/T2

T3 = T2 P3/P2 = T1P3/P2 r-1

v from 2

3-4 adiabatic expansion of the gas according to the law PV =C

T3/T4 =(V4/V3)-1

= r-1

v


45

T4 =T3/ r-1

v

= T1(P3/P2) r-1

v/ r-1

v

= T1(P3/P2) from 3

AlsoP4V4 =P3V3

P4 =P3 (V3/V4)

= P3/rv

4-1 constant volume heat rejection

and also from the constant volume process

P4/T4 = P1/T1

T4 = T1 (P4/P1) = T1P3/P2 from above

from the above then it follows that P4/P1 =P3/P2

Net work done = area under 3-4 – area under 1-2

= P3V3 –P4V4 – P2V2 –P1V1

-1 -1

= P3V3 –P4V4 – P2V2 –P1V1

-1

= m R {( T3-T4)-(T2-T1)} Since pv =mRT

-1

6. 3.Diesel cycle

heat is added at constant pressure and then released at constant volume

fig 6.3 P-V and T-s for the diesel cycle

1-2 adiabatic compression according to the law PV =C

2-3 constant pressure heat addition P2 =P3

3-4 adiabatic expansion according to the law PV =C

4-1 constant volume heat rejection V4 =V1

the cycle is sometimes referred to as the modified constant pressure cycle

assume that at point 1, P1,V1,T1, are known


46

1. P1,V1,T1

2. T1/T2 =(V2/V1) -1

T2 T1 (V2/V1) -1

Also P1V1 = P2V2

P2 = P1( V1/V2)

2. P3 =P2 pressure remains the same

V3/T3 =V2/T3

T3 =T2 V3/V2 =V3/V2T1 rv -1

= T1 rv -1

= V3/V2 cut off ratio

3. T4/T3 = (V3/V4) -1

T4 = T3(V3/V4) -1

Now V3/V4 =V3/V1 since V4 = V1 and

V3/V1 =V3 V2 =/rv

V2 V1

T4 =T3( /rv) -1

= T1 rv-1

-1

rv-1

T4 = T1 -1

= T1

and also P4V4 = P3V3

P4 = P3( V3/V4) =P3 (/ rv)

Work done during the process= area under2-3 + area under 3-4 –area under 1-2

=P2(V2-V3) +(P3V3 –P4V4) – P2V2 –P1V1

-1 -1

= P2(V2-V3) + (P3V3 –P4V4) –P2V2 –P1V1

-1

6.4. The dual combustion cycle

Heat is added at constant pressure and at constant volume hence the name dual. Heat is then

rejected at constant volume, it is a modified diesel cycle


47

1-2 isentropic compression

2-3 reversible constant volume heating

3-4 reversible constant pressure heat addition

4-5 isentropic expansion

4-6 reversible constant volume cooling

1

2

34

5

p

vv1

v2=v3

p3=p4

The heat is supplied in two parts first at constant volume and then at constant pressure

Heat supplied =mcv(T3-T2) + mcp(T4-T3)

Heat rejected = mcv(T5-T1)

Net work done = area under 3-4 + area under 4-5 – area under 1-2

= P3(V4-V3) +P4V4 –P5V5) – (P2V2 –P1V1)

( –1 ) ( –1)

= P3(V4-V3) +(P4V4 –P5V5) – (P2V2 –P1V1)

( –1)

= mR{T4-T3) +{(T4-T5)-T2-T1)}

( –1)

or net work = heat received – heat rejected

=mcv(T3-T2) +mcp(T4-T3) –mcv(T5-T1)

Thermal efficiency =1 - heat rejected

Heat received

=1 - mcv(T5-T1)

mcv(T3-T2) +mcp(T4-T3)

=1- (T5-T1)

(T3-T2) + (T4-T3)


48

6.5. Gas turbine cycles

combustion chamber

air compressor

2 3

turbine

coupling shaft

1 4

air intake exhaust

2s2

1

3

4s

4

p2

p2

T

s

fig 6.5 T-s diagram for gas turbine

The basic gas turbine unit is that which operates on an open cycle in which a rotary compressor is

coupled to a turbine. Air is drawn into the compressor, and after compression it is passed into the

combustion chamber where energy is supplied in the form of sprayed fuel. The resulting hot gases

expand through the turbine to the atmosphere.

1-2 represents irreversible adiabatic compression

2-3 constant pressure heat supply in the combustion chamber

3-4 irreversible adiabatic expansions

1-2s represents the ideal isentropic process between same pressures p1=p2

3-4s represents the ideal isentropic expansion between the same pressures

Assume that the kinetic energy through the system is negligible compared with enthalpy changes


49

Applying the flow equation to each part of the cycle for unit mass

Compressor work input =cpa(T2-T1)

For combustion chamber heat supplied =cpc(T3-T2)

For the turbine work output =cpt(T3-T4)

Net work output =cpt(T3-T4)- cpa(T2-T1)

The thermal efficiency = net work output

heat supplied

= cpt(T3-T4)- cpa(T2-T1)

cpc(T3-T4)

cpa is the specific heat capacity at constant pressure for entering compressor

cpc is the specific heat capacity at constant pressure for fuel in the combustion chamber

cpt is the specific heat capacity at constant pressure for combustion products in the turbine

isentropic efficiency of the compressor is defined as the ratio of work in put required in isentropic

compression between p1 and p2 to the actual work required

compressor isentropic efficiency c = cp(T2s –T1)

cp(T2-T1)

= (T2s –T1)

(T2-T1)

Isentropic efficiency of the turbine is defined as the ratio of the actual work to the isentropic work

output between the same pressures

Turbine isentropic efficiency = cp(T3-T4)

cp(T3-T4s)

= (T3-T4)

(T3-T4)

Worked Example

A closed cycle gas turbine unit operating with maximum and minimum temperatures of 760oC and

20oC has a pressure ratio of 7:1. Calculate the ideal cycle efficiency and the work ratio.

T2 = 293 x (7) 0.4/1.4

= 510.9 K

Then heat supplied

Q = Cp(T3 –T2)

= 1.005(1033 – 510.9)

= 52407 kJ/kg

T4 = 1033/(7) 0.4/1.4

= 592.4 K


50

Then heat rejected

Q = Cp(T4 –T1)

= 1.005(592.4 - 293)

= 300.9 kJ/kg

thus cycle efficiency = pliedsupHeat

rejectedHeatpliedsupHeat

= 7.524

8.223

7.524

9.3007.524

= 42.7%

The net work output is equal to the net heat supplied

i.e. net work output = 223.8 kJ/kg

Gross work output = Cp(T3 – T4)

= 1.005(1033 – 592.4)

= 442.8 kJ/kg

Work ratio = outputworkgross

outputworknet

= 8.442

8.223= 0.505

Worked example

In a dual combustion cycle the maximum temperature is 20000C and the maximum pressure is 70

bar. Calculate the cycle efficiency and the mean effective pressure when the pressure and

temperature at the beginning of the compression are 1 bar and 170C respectively. The compression

ratio is 18/1 (63.6%; 10.46bar) T.D. Eastop

Solution

P2 = P1x (18)1.4

= 1 x (18)1.4

= 57.2 bar

T2 = T1x (18)0.4

= 290 x (18)0.4

= 921.5 K

Also 2

3

2

3

P

P

T

T

T3 = K6.11292.57

70923

And 012.26.1129

2273

T

T

V

V

3

4

3

4

i.e. 945.8012.2

18

V

V

4

5


51

Then T5 =

4.0

5

44

V

VT

= 946.2 K

Heat supplied = Cv(T3 – T2) + Cp(T4 – T3)

= 0.718(1123 – 921.5) + 1.005(2273 -11296)

= 1293.8 kJ/kg

Heat rejected = Cv(T5 –T1)

= 0.718(946.2 - 290)

= 471.2 kJ/kg

Net heat supplied = net work output

= 1293.8 – 471.2

= 822.6 kJ/kg

then, cycle efficiency

= %6.63or636.0pliedsupheat

outputworknet

now kg/m8323.0101

290287

P

RTV 3

51

11

V2 –V1 = (.8323/18) – 0.8323 = 0.7861 m3/kg

mean effective pressure = 12 VV

1000outputworknet

= 5107861.0

10006.822

= 10.46 bar


52

Chapter six

Gas power cycles Tutorial Questions

1. The pressure, volume and temperature at the beginning of the compression of a constant

volume (otto) cycle are 105kN/m2, 0.002m

3 and 25

0C respectively. The maximum

temperature of the cycle is 1250 0C. The volume ratio of the cycle is 8:1. The cycle is

repeated 4000times/min. determine for the cycle,

the theoretical output in kilowatts;

the thermal efficiency;

the mean effective pressure;

the carnot efficiency within the same temperature limits

take cp =1.007kJ/kgK.cv =0.717kJ/kgK [54.3kW; 57%; 465.1kN/m2; 80.4% ] R.JOEL

2. 0.5 kg of air is taken through constant pressure cycle. The pressure and temperature at the

beginning of the adiabatic compression are 96.5kN/m2 and temperature 15

0C, respectively.

The pressure ratio o9f the compression is 6:1. Constant pressure heat addition occurs after the

adiabatic compression until the volume is doubled. Determine for the cycle,

the thermal efficiency;

the heat received

the net work done

the work ratio

the mean effective pressure

[40%; 241kJ; 96.4kJ; 0.466; 134.3kN/m2 ] R. JOEL

4. An engine working on an ideal Diesel cycle has a clearancevolume of 0.000 25m3. it has a

bore and stroke 0f 152.5mm and 200mm, respectively. At the beginning of the adiabatic

compression the air in the cyclinder has a pressure of 100kN/m2 and a temperature of 20

0C

respectively. The maximum temperature of the cycle is 10900C ,. Determine,

The temperature and pressure at the end of the adiabatic compression;

the temperature and pressure at the end of an adiabatic expansion

the thermal efficiency of the cycle

take =1.4 [6060C;4680kN/m

2; 269

0C;185kN/m

2 63.3% ] R. JOEL

5. A continuous combustion constant pressure gas turbine takes in air at a pressure of 93KN/m2

with a temperature of 20 0C .A rotary air compressor compresses the air to a pressure of 552

kN/m2 with an isentropic efficiency of 83%.The compressed air is passed to a combustion

chamber in which its temperature is increased to 8700C .From the combustion chamber the

high temperature air passes into a gas turbine in which it is expanded to 93kN/m2 with an

isentropic efficiency of 80%.For an air flow of 10 kg/s and neglecting the fuel mass as

small, determine:

(a) the net power output of the plant if the turbine is coupled to the

compressor;

(b) the thermal efficiency of the plant;

(c) the work ratio.

Take, =1.4 and cp =1.00kJ/kgK

[(a) 1310KW; (b) 21%; (c) 0.36]


53

Chapter seven

Vapour power cycles:


54

7.0 Introduction

The steam power plant is the plant, which is used to convert water into steam, whose kinetic

energy is used to drive the turbine. The turbine in turn is coupled to the generator, which is used to

generate electricity. The main components of a steam power plant are as flows:

1. the boiler

2. the turbine

3. the condenser

4. the feed pump

5. extraction pump

6. cooling tower

7. the economizer

fig 7.1 major components of a steam power plant

7.1 Vapours

Formation of steam

When water is put into a boiler it absorbs heat at constant pressure, steam bubbles are formed near

the heating surface and they rise through the water, however the water around is cold hence it

absorbs heat energy from the steam bubbles, which immediately collapse. Finally the steam

bubbles can escape from the boiling surface. The mass is in extreme state of turbulent called

boiling. it will be noted that as boiling temperature ceases to rise. The temperature remains

constant despite the increase in heat being added (saturation temperature). As the steam breaks

away from the water surface it will carry with it droplets of water. The steam with these droplets is

called wet steam. Further heating to the wet steam will convert the droplets tom dry steam, which

is then known as dry saturated steam. This makes the end of the constant temperature heat

addition. Further addition of heat to this steam result in the formation of superheated steam. And

there is a sudden increase of temperature.


55

7.1.1 Stages in the formation of steam

Temp,

tsup liquid water vapour

water/ steam superheated steam

boiling point dry saturated

tf

hf enhalpy of steam

hg

hsup

fig. 7.2 formation of steam from water

Stage 1

Warming phase, temperature rise to saturation temperature. The energy supplied, is the liquid

Enthalpy hf in kJ/kg

Stage 2

It starts with water at saturation temperature tf at the beginning and a dry saturated steam at

saturated temperature at the end. The energy required, is the enthalpy of evaporation, hfg

Stage 3

Formation of superheated steam, temperature rise as the energy is added, superheat enthalpy

Liquid enthalpy

For 1kg of water, the specific liquid enthalpy is written as hf, the accurate value of hf at any given

saturation temperature corresponding to any given constant pressure given from steam tables, also

hf = mc

Enthalpy of evaporation

Dryness fraction of steam x, quality of steam is the fraction of dry steam to the total mass of wet

steam

x = mass of dry steam present

total mass of water and steam


56

= ms

mw + ms

= v/vg …………………………………………………………………………….7.1

VfVg

c=0 c=1

fig 7.3 dryness fraction of steam

Specific enthalpy of wet steam

hwet = hf + x ( hg-hf) …………………………………………7.2

hwet = hf + x hfg

Enthalpy of superheated steam

hsup = hg + cp (tsup –tf ) cp specific heat capacity of superheated steam


57

Formation of the T-s or T-h diagrams

When experiments are performed at different pressures the temperature enthalpy diagrams will be

as shown in the first diagram. As the operating pressure increases the boiling temperature also

increases.

If points 2,6,10,13,14, 11,7, 3 are joined with a smooth curve the second diagram is formed. At

point c is that pressure which will ensure that the water suddenly changes into gas without passing

through the vapor phase, it is known as the critical pressure and temperature point.

Curve ac is the liquidous line, and cb is the gaseous line. The second graph is the one applied

when dealing with different cycles

7.2 The Rankine cycle

T

h

T

h

Vapor

phase

Gaseous

phase

Liquid

phase

c

12 3

4

5

6 7

8

9

10 11

1213 14

ab


58

Fig 7.4 Rankine cycle with superheat

1-2 water is being pumped into the boiler with volume V and pressure P2,

2-3 in the boiler water is converted into steam at pressure P2, with volume V3 which is then

fed into the turbine.

3-4 the steam is expanded frictionless adiabatic ally in the engine or turbine.

4-1 steam after expansion, is passed from the turbine into a condenser in the condenser the

volume of the steam is reduced.

2-3 shows constant pressure formation of steam in the boiler

3-4 isentropic expansion of steam in the turbine

4-1 constant pressure and temperature formation of steam.

1-2 pumping of water into the boiler at constant volume.

Heat supplied to unit mass of water in the boiler = h3 –h2

Specific work done in the turbine = h3 –h4

Feed pump work = ( p2- p1) v1

where v1 is the specific volume of water

net work done/cycle = (h3 –h2) -( p2- p1) v1

total energy of water entering the boiler at 2 h2 = liquid enthalpy at 1 + work done by pump

h2 = h1 +( p2- p1) v1

Heat transfer required in the boiler = ( h3 - h1 )+( p2- p1) v1

Thermal efficiency of the cycle = workdone per cycle

heat received per cycle

Thermal efficiency of rankine cycle = (h3-h4)- ( p2- p1) v1

( h3 - h1 )+( p2- p1) v1

if the feed pump work is neglected, then the efficiency

= (h3-h4) …………..7.3

( h3 - h1)

to calculate h4, from s3= s4 isentropic expansion in the turbine

then s4 = sf + x4 sfg x4 =( s4-sf)/sfg

therefore h4 = hf + x4 hfg


59

7.3 The reheat cycle:

The pressure ratio through the turbine can be considerable, therefore after partial expansion in

high pressure side of the turbine, the partially wet steam is fed back to the boiler were it is

reheated at constant pressure to a higher temperature it is then passed to the lower pressure

side of the turbine, the expansion which follows is largely dry and superheated.

The reheat cycle reduces erosion and corrosion by wet steam. It also results in higher thermal

efficiency.

fig7.5 reheat cycle

Work done in the turbine = ( h3-h4) +(h5-h6)

S3 = S4 Isentropic expansion in the high pressure side of the turbine

S5 =S6 Isentropic expansion in the low pressure side of the turbine

Work done by the pump = v1(P2-P1)

Heat supplied in the boiler = (h3- h2) + (h5 –h4)

Net work output = work done by turbine – work of pump

=( h3-h4) +(h5-h6) - v1(P2-P1)

also h2 = h1 + v1(P2-P1)

thermal efficiency = net work

heat supplied

= h3-h4) +(h5-h6) - v1(P2-P1)

(h3- h2) + (h5 –h4)

Neglecting work done by the pump

Thermal efficiency = h3-h4) +(h5-h6)

(h3- h2) + (h5 –h4)………………………….7.4

7.4 The regenerative cycle:

Also known as the feed heat

The steam is made to expand from condition 1 through the turbine. At the pressure corresponding

to point 5, aquantity of steam, say ykg per kg of steam supplied from the boiler, is bled off for

feed 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 and pumped to the same pressure as the feed


60

steam, y kg is such that, after the mixing and being in the second feed pump the condition is

defined by state 7

The heat to be supplied in the boiler is given by =h1-h7

to determine the bled pressure when one or more feed heaters are used this can be based on the

assumption that the bleed steam temperature to obtain maximum efficiency the arithmetic mean of

the temperatures at 4 and 7

t –bleed =(t4 + t7)/2

Fig 7.6 T-s diagram for regenerative cycle

m5 =y, m2 = (1-y)

h2(1-y) +yh5 =h6

y = h6-h2

h5-h2

also s3 = s5 = s4

heat supplied in the boiler = h3-h7

total work output -W = -W35 –W54

= h3-h5 +(1-y)(h5-h4)

cycle efficiency = h3-h5 +(1-y)(h5-h4) …………………………..7.5

(h3-h7)


61

Worked Example

In a regenerative cycle employing three closed feed heaters the steam is supplied to the turbine at

42 bar and 5000C and is exhausted to the condenser pressure at 0.035 bar. The bled steam for

feed heating is taken at a pressure of 15, 4, and 0.5 bar. Assuming ideal process and neglecting

pump work, calculate,

The fraction of the boiler steam bled at each stage;

the power output of the plant per unit mass flow rate of the boiler steam;

the cycle efficiency

T.D Eastop

i) using the tables and h-s chart:

h1 = 3443 kJ/kg, h2 = 3127 kJ/kg

h3 = 2815 kJ/kg h4 = 2458 kJ/kg

h5 = 2114 kJ/kg h6 = 112 kJ/kg

h7 = 340 kJ/kg h8 = 605 kJ/kg

h9 = 845 kJ/kg

Applying an energy balance to each feed heater assuming that the bleed steam fractions are y1, y2

and y3 as shown in the figure:

3127y1 + (1 - y1)605 = 845

i.e. y1 = 0.0952

2815y2 + (1 – 0.0952 - y2)340 = (1 – 0.0952)605


62

i.e. y2 = 0.0969

2155y3 + (1 – 0.0952 – 0.0969 – y3)112 = (1 – 0.0952 – 0.0969)340

i.e. y3 = 0.02

BOILER

42 bar

PPP

P

9 8 7

6

y 1kg/s

y 2kg/s

y 3kg

/s

15 bar

4 bar

0.5 b

ar

TURBINE

5

(1-y1-y2-y3)kg/s

CONDENSER

0.035bar

42 bar

500oC

1 kg/s

1

1 kg/s

23

4

ii) Power output per unit mass flow rate of boiler steam is given by

Power output = (h1 –h2) + (1 - y1)(h2 –h3) + (1 - y1 – y2)(h3 –h4) + (1 - y1 – y2 – y3)(h4–h5)

=(3443-3127)+(1-0.0952)(3127-2815)+(1-0.0952-0.0969)(2815-2458)+(1-0.0952-0.0969-

0.0902)(2585-2114)

i.e. Power output = 1 133.6 kW per kg/s

iii) heat supplied per unit mass of boiler steam

= h1 –h9 = 3443 – 845 = 2598 kJ/kg

then, cycle efficiency = 1134/2558 = 43.6%


63

Tutorial Questions

Vapor power cycle

1. Steam at a pressure of 30 bar and temperature of 2500C is fed into a steam turbine from the

boiler. In the turbine the steam is expanded isentropically to a pressure of 1 bar. The steam is then

exhausted into the condenser where it is condensed but not under cooled. The condensate is then

pumped back into the boiler, determine

the dryness fraction of the steam after expansion;

the Rankine efficiency

[0.823; 23.88] R.J

2. Steam at 1.7MN/m2 and with a temperature of 250

0C is fed into a steam engine in which it is

expanded adiabatically to arelease pressure of 0.35MN/m2. From this pressure it is released at

constant volume to a condenser pressure of 50kN/m2.the steam is then exhausted from the engine

into the condenser where it is condensed but not undercooled the condensate is then pumped back

into the boiler. The stem flow rate is 1500kg/h determine

the power output of the turbine

the Rankine efficiency

the Carnot efficiency for the same temperature limits of the cycle

[ 187kW; 17.4%; 32.25%]R.J

3. In a steam turbine plant, the initial pressure and temperature of the steam are 3MN/m2

and

3200C, respec tively and the exhaust pressure is 0.06NM/m

2 . after isentropic expansion to

1MN/m2 the steam is reheated isobarically to 250

0C and after further isentropic expansion in the

turbine to 0.4MN/m2 it is again reheated isobarically to a temperature of 200

0C. the steam is

finally expanded isenropically in the turbine to exhaust pressure. Determine,

thermal efficiency of the arrangement

[ 26.5%] RJ


64

Chapter eight

Refrigeration cycles

8.0 Introduction

In general refrigeration is defined as any process of heat removal, specifically it is defined as the

branch of science that deals with the process of reducing and maintaining the temperature of a

space or material below the temperature of the surrounding.

Fig. 8.1 T-s refrigeration cycles


65

8.1 Components of vapor refrigeration cycle

1.Compressor

Work is input by a compressor, which increases the pressure of the vapour and also the

temperature through isentropic compression in the ideal cycle.

2.The condenser.

From the compressor the vapour enters the condenser, where heat is extracted, resulting in

saturated liquid. The heat extracted is released to the surrounding.

3. Expansion valve

The pressure of the saturated vapour is then reduced through the expansion valve or the

Throttle valve. The expansion takes place at constant enthalpy.

4. Evaporator

From the expansion valve the refrigerant enters the evaporator, where it absorbs heat from

the refrigerated space. This was the actual refrigeration of the items takes place. The items

release the heat to the refrigerant, resulting in them getting cooler and the refrigerant

getting hotter.

8.2 The refrigeration cycle

1-2 isentropic compression in the compressor

s1 = s2 , for an isentropic compression

work done by the pump= h2- h1

to find h1; at pressure p1

s1 = sf + x1 sfg

x1 = (s1 - sf )/ sfg

then, h1 = hf + x1 hfg

to get h2, at pressure p2, assume the temperature at 2 to be t2 and the saturation

temperature at p2 to be tf , then,

the degree of superheat = t2 - tf

if (t2 - tf ) 15 K, choose the value h, under this column

if (t2 - tf ) > 15 K choose the h, value under the 30K value.

2-3 heat rejection in the condenser,

heat rejected = h2 – h3

h3 = hf at p2

3-4. Expansion of the refrigerant at constant enthalpy.

h3 = h4

also x4 =( h4 – hf1)/ hfg1


66

4-1 constant pressure heat addition to the refrigerant

Refrigerating effect = h1 – h4

Coeffient of performance C.O.P. = heat absorbed from the refrigerated space.

Heat energy equivalent of the energy supplied by comp.

= refrigerating effect ……………….. 8.1

heat of compression


67

8.3. The absorption refrigerating cycle

fig 8.2 diagrammatic presentation of absorption

it replaces the compressor in the conventional refrigerating cycle with an absorber, pump, heat

exchanger and a generator.

Saturated low-pressure refrigerant enters the absorber, where it is absorbed in aweak carrier

solution. Heat is released in the process. In the absorber the temperature is maintained low by

removing heat Q`A. The much stronger solution leaves the absorber and is pumped into the

higher condenser pressure using little power., it goes through the heat exchanger which

increases its temperature and enters the generator, the added heat boils off the refrigerant

which then passes off into the condenser. The remaining weak carrier solution is then returned


68

from the generator to the absorber. To be recharged with the refrigerants on its way to the

absorber the temperature of the carrier solution is reduced in the heat exchanger. And its

pressure is reduced in the regulating valve.

From the condenser the refrigerant is passed through the expansion valve where its pressure is

reduced before being fed into the evaporator where absorbs heat from the refrigerated space

and is then fed into the absorber.

The solutions normally used are that of ammonia in water or water and lithium bromide for air

conditioning.

The concentration fraction of liquid ammonia x` = mass of NH3 liquid

Mass of mixture

Concentration of vapour ammonia x`` = mass NH3 vapour

Mass of mixture

Enthalpy of liquid mixture = hL, enthalpy of vapour mixture =hv

fig 8.3 concentration versus temperature


69

8.4 Gas refrigeration cycle

fig. 8.4 T-s diagram for gas refrigeration cycle

1-2 the refrigerant absorbs heat from the refrigerated space at constant pressure

2-3 isentropic compression in the compressor

3-4 the gas releases heat at constant pressure

4-1 isentropic expansion in the turbine to a low exit temperature

The net work input to the plant W =W14 + W23

Applying steady – flow energy equation

W14 = -(h4- h1 ) and W23 =h3-h2

For a perfect gas

W 41 =cp(T4- T1) and W23 =cp(T3- T2)

The refrigerating affects Q1 = (h2-h1)

= cp(T2-T1)

The actual cycle would be as shown below


70

T

3

3s

4

2

1S 1

S

Fig. T-s diagram for gas turbine

The cycle would have W23 = cp(T3-T2)

W34 = -cp(T4-T1)

Isentropic efficiency of the compressor c = T3s –T2 ……………………………… 8.2

T3-T2

Isentropic efficiency of the turbine T = T4- T1 ………………… 8.3

T4- T1s

Then the refrigerating effect Q1 =cp (T2-T1)

Worked Example

In a refrigerating plant using R12 the vapour leaves the evaporator dry saturated at 1.826 bar and

is compressed to 7.449 bar. The temperature of the vapour leaving the compressor is 45ºC and the

liquid leaves the condenser at 25ºC and is throttled to the evaporator pressure calculate:

the refrigerating effect

the specific work input

the COPref

Solution

i) h1 = 180.5 kJ/kg

After compression the refrigerant is superhe

ated by (45 – 30) = 15K at 7.449 bar therefore

from tables,

h2 = 210.63 kJ/kg

also h3 = 59.7 kJ/kg = h4

the refrigerating effect = h1 – h4

= 180.67 – 59.7

= 121. 27 kJ/kg

T

S

2

3

41

45oC

30oC25oC

-15oC


71

ii) Work in put = h2 – h1 = 210.63 – 180.97 = 29.66 kJ/kg

iii) COPref = 66.29

27.121

put in Work

effect ingRefrigerat

= 4.09


72

Chapter eight

Refrigeration

Tutorial questions

1. A heat pump using ammonia as a refrigerant operates between saturation temperatures 6 and

38ºC. the refrigerant is compressed isentropically from dry saturation and there is 6K of

undercooling in the condenser, calculate

the COPhp

the mass flow of refrigerant per kilowatt power input

the heat available per kilowatt power input

( 8.77; 25.03kg/h; 8.77Kw) eastop

2. Freon 12 leaves the condenser of the refrigerating plant as saturated liquid at 5.673 bar. The

evaporator pressure is 1.509 bar and the refrigerant leaves the evaporator at this pressure and at a

temperature-50C. Calculate,

the dryness fraction of the refrigerant as it enters the evaporator

the refrigerating effect per kg


73

Chapter nine

Nozzles and Jet propulsion 9.0 Nozzles A nozzle is a duct of whose cross-sectional area varies with length such that when a fluid is

passed through it is either accelerated or decelerated.

Applications;

Steam and gas turbines

Jet engines

Rocket motors

Inflow measurements

General nozzle shapes

Convergent divergent nozzle, at the entry the area of the nozzle converges down to the

throat and then diverges up to the exit

Convergent nozzle, the cross-sectional area converges from entry down to the exit.

Diffuser, is the opposite of the above the fluid is decelerated and the pressure is increased.

entryexit

throat

convergent divergent

fig 9.1 divergent- convergent nozzle

9.2 Flow analysis

A1, c1

P1,V1,T1,

h1

A2, c2

P2,V2,T2h1

Neglecting change in potential energy and putting W =0 since no work is done in the nozzle, the

steady flow energy equation for the nozzle becomes

h1 + c12/2 +Q = h2+c2

2/2………………………………………………………………9.1

The time taken by the substance to pass through the nozzle is very small hence there is little time

for heat exchange. Hence Q =0


74

The above equation becomes

h1 + c12/2 = h2+c2

2/2

c22 - c1

2 = h1 - h2

2

but c2 >> c1

therefore c2 = 2(h1-h2)1/2

………………………………………………………………9.2

for gas,

h1-h2 =cp(T1-T2) = R(T1-T2)………………….……………………………….9.3

( -1)

Since cp = R…………………………………………………………………………9.4

( -1)

substituting equation 9. 3 into 9. 2

c= {2 R(T1-T2)}………………………………………………….9.5

(-1)

since Pv = mRT, then

c= {2 P1V1(1- P2V2)}………………………………………….9.6

(-1) P1V1

At any section of the nozzle

.

m v =AC

Where .

m = mass flow rate in kg/s

v =specific volume in m3/kg

A = the crosssectional area in m2

C = velocity in m/s

9.2 Steam turbines


75

INLET

Cai = is the absolute velocity at inlet, is the velocity of steam as is leaves the nozzle

The nozzles are inclined at an angle to the plane of rotation of the turbine blade

U = is the mean blade speed at the mean height of the blades

Cri = is the velocity of steam relative to the blades is obtained by compounding U and Cai.

It is inclined at angle to the plane of rotation, if the steam enters without shock then the

angle must be the inlet angle of the blades

Cwi = is the velocity of whirl at inlet is the component of Cai in the plane of rotation

Cfi = is the velocity of flow at inlet is the component of Cai along the axis of rotation

OUTLET

Cre = is the exit relative velocity, at an angle to the plane of rotation, which also the exit

angle of the blade

U= is still the mean velocity of the blades

Cae= is the absolute velocity at exit is obtained by compounding U and Cre

It is inclined at an angle to the plane of rotation.

Cwe= is the velocity of whirl at exit it is the component of Cae in the plane of rotation.

Cfe = is the velocity of flow at exit, is the component of Cae along the axis of rotation

9.3.1 The combination of inlet and outlet velocity triangles.

CfiCai

UCwi

Cri

CreCaeCfe

Cwe U

EXIT

INLET

Direction of

motion

Φβ

θ

θα

Fig 9.3 Velocity diagram for impulse turbine

Axis of rotation

Plane of rotation


76

-CweCwe

CfiCri

CaiCre

U

Cae

Cfe

Change of velocity of whirl

= -Cwe-Cwi= -(Cwe+Cwi)

A Bβ α Φ θ

Cfi-Cfe

fig 9.4 combination of inlet and outlet velocity triangles

The exit velocity triangle remains as originally shown. The inlet velocity triangle is drawn sa the

other half of the parallelogram of velocities the original triangle is shown dotted. Absolute

velocities start at A. For no friction Cri = Cre

Work done by the blades

The whirl velocities are the effective part in producing motion in the blades

From Newton`s

The force to change the velocity of whirl of steam

Force = rate of change of momentum

=mass x change of velocity change in V = – Cwe – Cwi

= .

m x[-(Cwi +Cwe)] m is the mass flow rate in kg/s

The negative sign shows that this force acts in opposite direction to that of rotation, then the actual

force being exerted on the blades

F = .

m (Cwi + Cwe)

Work done = force x distance

Power = force x mean velocity

= .

m U( Cwi + Cwe)

Kinetic energy of the steam supplied = C2

ai J/kg steam/s

2

Blade or diagram efficiency = work done by blade/kg steam

energy supplied /kg steam


77

= U(Cwi + Cwe)

C2ai

2

= 2U( Cwi + Cwe)

C2ai

The velocity of flow is that velocity which passes the steam across the blades, this also causes

end thrust along the turbine shaft.

End thrust force =mass rate x change of velocity

= .

m (Cfe –Cfi)

9.3.2 The blade height

The blade height at any particular section of turbine will be a function of the following

Mass flow rate in kg/s

specific volume of steam in m3/kg

the area through which the steam is passing in m2

the velocity of steam as it passes the section in m/s

For a given section we have

.

m v = AC

9.3.3 Blade height for impulse turbine

p

Cri

psi

nθ

θ

Cri

Cfi = Crisinθ

θ

(a)

(b)

Cfi=Cri Sin

For an impulse turbine blade row,let,


78

M=mass flow of steam,kg/s

V=specific volume of the steam,m3/Kg

N=number of blades covered by nozzles

P=pitch of blades,m(taken at mean blade height)

H=blade height,m

=blade inlet angle

Cri=relative velocity of steam at inlet,m/s

Cfi=velocity of flow at inlet,m/s

Then,from equatin (1) .

m v =AC

=N p sin

=NPH x Cri sin , which from equation (2)

=N P H Cfi

H =.

m v

NPCfi

NP= circumference at the mean blade diameter

=d

Where d = mean blade diameter, m.

(b) Reaction turbine

In this case, let,

D =mean blade diameter, m.

Then, since the reaction turbine has full admission, from equations (3) and (5)

H = .

m v

d Cfi

9.4.0 Jet propulsion

Aircraft propulsion is achieved by using a heat engine to drive an airscrew or propeller, or by

allowing a high-energy fluid to expand and leave the aircraft in a rear direction as a high velocity

jet. In the propeller type of aircraft engine the propeller takes a large mass flow and gives it a

moderate velocity backwards relative to the aircraft. In the jet engine the aircraft induces a

comparatively small airflow

and gives it high velocity backwards relative to the aircraft. In both

cases the rate of change of momentum of the air provides a reaction forward thrust which propels

the engine.

9.4.1 Turbojets

Ca turbojet Cj


79

Air craft velocity, Ca

Thrust per unit mass flow rate = Cj – Ca

Propulsive power, thrust power per unit mass flow rate = Ca( Cj – Ca)

The above is the rate at which work is done inorder to keep the air craft moving.

The net work output from the engine is given by the increase in kinetic energy

( Cj2 -Cj

2)/2

the work output can be divided into two:

It provides the thrust work as given by Ca( Cj – Ca)

Gives the air a kinetic energy (Cj-Ca)2/2

Ca( Cj – Ca) +(Cj-Ca)2/2 =CaCj -Ca

2 +Cj

2/2 +Ca

2/2 –2CjCa/2

Therefore workoutput = C2j –C

2a

2

the propulsive efficiency p = thrust work

rate at which work is done on the air in the air craft

= 2 Ca( Cj – Ca)

C2j –C

2a

= 2Ca

Cj + Ca

9.4.1The ram jet Is the simplest form of jet engine. In this engine the air is compressed by the conversion of

the kinetic energy of the atmospheric air relative to the aircraft, it is known as the ram

effect. Fuel is burnt in the compressed air stream at approximately constant pressure and

the hot gases are allowed to expand through the nozzle reaching a high velocity backwards

relative to the aircraft

Fuel ramjet velocity nozzle

Ca

1 2 3 4


80

air inlet exhaust

1 2 3 4

T PI

P2=P3

I 3

T2=T1 2

2S P1=P4

4S 4

1

S

The ramjet velocity is Ca, the air enters the diffuser with a kinetic of =C2

a/2 per unit mass

Assuming isentropic, using flow equation

h1 +C2a/2 = hi +Ci

2/2

cpT1 +C2a/2 = cp(Ti + C

2i/2) =cpToi

C2a/2 = cp (Toi –T1)

Toi –T1 = C2a/2 cp

The total pressure Poi is the pressure the air attains when the diffusion process is isentropic, when

the procee is irreversible, though still approximately adiabatic, then the total pressure attained

isPo2 which is less than Poi

The kinetic energy available, C2a/2 is the same whether or not the process is irreversible the

temperatures remain the same Toi =To2

To2 –To1 =Ca2/2cp

Intake efficiency = Tt2s –T1

Tt2 – T1

Worked Example


81

A convergent–divergent nozzle expands air at 6.89 bar and 427oC into a space at 1 bar. The throat

area of nozzle is 650 mm2 and the exit area is 975mm

2. the exit velocity is found to be 680m/s

when the inlet velocity is negligible. Assuming that friction in the convergent portion is

negligible, calculate:

i) the mass flow through the nozzle, stating whether the nozzle is underexpanding or

overexpanding;

ii) the nozzle efficiency and the coefficient of velocity

Solution

i) 5283.04.2

2

p

p 4.0

4.1

1

c

therefore,

pc = 6.89 x 0.583 x 3.64 bar

also,

Tc = K3.5832.1

700

2.1

T1

And, Cc = s/m4833.5832874.1

Vc = kg/m459.01064.3

3.583287 3

5

Therefore,

Mass flow rate =

610459.0

6504830.684 kg/s

To determine whether the nozzle is underexpanding or overexpanding it is necessary to

find the correct exit pressure

V2 = kg/m97.010684.0

975680 3

6

Also.

C2 = s/m680TT10052 21

21 TT = 230 K

2T = 700 – 230 = 470 K

therefore,

p2 = bar39.11097.0

4702875

since the actual back pressure is 1 bar the nozzle is underexpanding.

ii) For isentropic expansion between pressures of 6.89 bar and 1.39,

58.139.1

89.6

T

T 4.14.0

2

1

i.e. T2 = K44358.1

700


82

21 TT = 700 – 443 = 257 K

Then,

Nozzle efficiency = 100257

230 = 89.5%

And,

Coefficient of velocity = 985.0 = 0.946

Worked Example

A turbo air craft is traveling at 625 km/h in atmospheric conditions of 0.45 bar and –26oC. The

compressor pressure ratio is 8, the air mass flow rate is 45 kg/s, and the maximum allowable cycle

temperature is 800oC. The compressor, turbine, and jet pipe stagnation isentropic efficiencies are

0.85, o.89, and 0.9 respectively, the mechanical efficiency of the drive is 0.98, and the combustion

efficiency is 0.99. Assuming a convergent propulsion nozzle, a loss of stagnation pressure in the

combustion chamber of 0.2 bar, and a fuel of calorific value 43 300 kJ/kg, calculate:

i) the require nozzle exit area;

ii) the net thrust developed;

iii) the air-fuel ratio

iv) the specific fuel consumption.

For gasses in the turbine and propulsion nozzle take γ = 1.333 and Cp = 1.15 kJ/kg K;

for the combustion process assume an equivalent Cp value of 1.15 kJ/kg K.

Solution

i) Kinetic energy at inlet = 23600

1000925

21

= 33 kJ/kg

= cp(T01 – T0)

i.e. T01 = (-26 + 273) + (33/1.005) = 279.85 K

then, 247 + (279.85 - 247) x 0.9 = 276.6 K

therefore, p01 = 0.45 x

5.3

247

6.276

= 0.5585 bar

For the process in the compressor:

T02s = T01 x 80.4/1.4

= 1.8115 T01

i.e. T02s = 279.85 x 1.8115 = 506.9 K

then, T02 - T01 =

85.0

85.2799.506 = 267.16 K

i.e. T02 = 279.85 + 267.16 = 547 K


83

Also, p02 =8 x p01 = 8 x 0.6685 = 5.348 bar

Therefore, p03 = 5.348 – 0.2 = 5.148 bar

cpg(T03 – T04) = cpa(T02 – T01)/0.98

i.e.(T03 – T04) = 98.015.1

16.267005.1

= 238.2 K

therefore T04 1073 – 238.2 = 834.8 K

and, (T03 – T04s) = 89.0

8.8341073 = 267.7 K

i.e. T04s = 1073 – 267.7 = 805.3 K

then, 333.0

333.1

04

03

3.805

1073

p

p

= 3.154

i.e. p04 = 5.148/3.154 = 1.632 bar

For an isentropic nozzle,

333.0

333.1

04

c

333.2

2

p

p

= 0.5398

pc = 1.632 x 0.5398 = 0.881 bar

Since the back pressure is 0.45 bar the nozzle is underexpanding. The actual exit nozzle pressure,

p5 is slightly different from the above since there is friction in the nozzle.

800

-26

2s

0

2

1s

1

3

4

54s

5s

0.45 bar

Tem

per

ature

/(oC

)

Entropy/(S)


84

Assuming that since the nozzle is choked the actual temperature at exit is the critical temperature,

333.2

2

T

T

04

c = 0.8573

i.e. T5 = Tc = 834.8 x 0.8573 = 715.6 K

therefore

T5s = 9.0

6.7158.8348.834

= 702.4 K

And, 333.0

333.1

5

04

4.702

8.834

p

p

= 1.996

i.e. p5 =0.8176 bar

Then, 55

108176.0

6.715287v

= 2.5146 m

3/kg

And, 6.715287333.1c5 = 523.5m/s

Therefore

Nozzle exit area = 5.523

455146.2 = 0.216 m

2

ii) momentum thrust = a5 ccma

=

3600

109255.52345

3

=11 996 N

and, Pressure thrust = (0.8176 – 0.45) x 105 x 0.216

= 7940 N

i.e. total thrust = 11 996 + 7 940 = 19 936 N

= 19.94 kN

iii) Heat supplied = 45 x 1.15 x (1073 – 547)

= 27 220 kJ/kg

therefore, mass flow rate of fuel = 3004399.0

20027


85

= 0.635 kg/s

ie. Air-fuel ratio = 635.0

45 = 70.87

iv) specific fuel consumption = 936.19

635.0 = 0.0319 kg/kN s

Chapter nine


86

Tutorial questions

1. Air enters the nozzle at a pressure of 3 MN/m

2 and with a temperature of 400

0C. it leaves at

a pressure of 0.5 MN/m2. the exit area is 5000 mm

2 the expansion through the nozzle is

adiabatic according to the law PV =C. Determine

the mass flow through the nozzle

the throat area

the mach number at the exit

take =1.4 and R = 0.287 KJ/kgK [15.86 kg/s, 3390mm2, 1.55 ]

2. Air enters the nozzle with a pressure of 700kN/m2 and with a temperature of 180

0C. Exit

pressure is 100kN/m2. the law connecting pressure and specific volume during the expansion in

the nozzle is PV1.3

= C. Dtermine the velocity at exit from the nozzle takeCp =1.006kJ/kgK and

Cv = 0.717kJ/kgK [576m/s]

3. Steam leave as the nozzle of a single stage impulse turbine with a velocity of 1000m/s. the

nozzles are inclined at an angle of 24o to the direction of motion of the turbine blades. The mean

blade speed is 400m/s. and the blade inlet and exit angles are equal. The steam enters the blades

without shock and the flow over the blades is considered to be frictionless.

Determine i) the inlet angles of the blades

ii) the force exerted on the blade in the direction of their motion

iii) the power developed when the steam flow rate is at 4000kg/h

39o, 1.133kN, 453.2kW

Chapter ten

Effects of temperature change

10.Introduction


87

When matter is subjected to heat energy, the particles which make up the body absorbs the heat

energy and they start vibrating showing that the inter –atomic/ inter molecular forces holding the

particles will be weakened

Weakening of solid particles will result in the formation of liquid

Further addition of heat to liquid would result in the formation of gas

The degree of vibration of particles of a substance is a measure of its temperature

10.1 Thermal expansion and contraction

It is the increase in size or reduction in size of a body when heat is added or subtracted

Most substances increase in size when they are being subjected to heat energy

Their size is reduced when they are cooled, with the exception of water, which expands when it is

cooled to 4 oC

10.2. Linear Expansion It is the increase in length when a body is subjected to heat energy

Coefficient of Linear Expansion ()

It is the increase in length per unit length per degree rise in temperature

The change in length L of a solid bar when heated or cooled through a temperature change t is

given by the experimental relation L = lo t

t

1

Lo

-----------------------------------------------------------(10.1)

l = change in length

lo = original length

t = change in temperature

Final length L1 = l0 + l

By transposing in (1) for l ,

l = lo t

L1 = lo + lo t

= l0(1+ t)

but t = (1 – 2)

L1 = l0 { 1+ (2 – 1)}---------------------------------------------------------------(10.2)

10.3 Superficial expansion / area expansion ()

It is the increase in area per unit area per degree rise in temperature

= t

1

A

A

0

---------------------(10.3)


88

A1 = a1b1

a1 = a0 + a

= a0{1+ (2 – 1)} but for 1degree ;

a1 = a0(1+)

b1 = b0(1+ )

A1 = a1b1 = a0(1+) b0(1+ )

= a0b0(1+ )2

= a0b0(1+2 +

2)

but = small change

then 2 would be approaching 0

A1 = a0b0(1+ 2)

For temperature change of ( 2 -1)

A1 = a0b0{1+ 2 (2 -1 )}--------------------------------------------------(10.4)

10.4 Volumetric or cubical expansion

It is the increase in volume of an object when subjected to heat energy

Coefficient Of Cubical Expansion () It is the increase in volume per unit volume per degree rise in temperature

V1 = a1b1c1

But a1 = ao + a

b1 = bo + b

c1 = co + c

for change in length of one side a,

a1 = ao{1 + (2 – 1)} but for 1o

a1 = ao (1 + )

b1 = bo (1 + )

c1 = co (1 + )

V1 = a1b1c1

= ao (1 + ) bo (1 + ) co (1 + )

= aoboco(1 + )3

= aoboco{1 + 3 + 32 +

3}

A0

b1

a1a0

b0

A1


89

but is very small therefore 32 and

3 approach

zero

giving V1 = aoboco{1 + 3 }

for a change in temperature from 1 to 2

V1 = aoboco{1 + 3 (2– 1)}

V1 = Vo{1 + 3 (2 – 2)} but = 3

V1 = Vo{1 + (2 – 2)}----------------------------------------(10.5)

10.5 Change of phase

All matter exists in one of the three states viz. solid, liquid and gas.

Heating a solid enough will result in the formation of a liquid and further application of heat will

result in the formation of gas.

Enthalpy

It is the energy content of a given mass of a substance. Units are J/kg K.

Specific heat capacity

It is the energy, which is required to raise the temperature of 1 kg of a body by 1 degree.

Latent heat

It is that amount of energy which when added or subtracted from a body system will cause a phase

change at a constant temperature.

Sensible heat

It is the amount of energy which when supplied to a body will cause a temperature rise.

c1

co

b1

b2

a1

ao

Vo

V1


90

SOLID

SOLID

&

LIQUID

LIQUID

&

GASLIQUID

GAS

ENTHALPY

TE

MP

ER

AT

UR

E O

C

Latent

heat

of

fusion

Latent

heat of

Vapour-

isation

``

In general if an amount of heat Q produces a rise in temperature t in a body of mass m and

specific heat capacity c and the same heat produces the same rise in temperature in a mass mw of

water then mw is the water equivalent of the body, thus:

Q = cm t = cwmw t

i.e. equivalent mass of water, mw = mc

c

w

WORKED EXAMPLES

QUESTION 1

AN 80 mm Ø aluminium piston fits into a cast iron cylinder with uniform radial clearance of 0.75

mm at room temperature. Determine the percentage increase in area of the gap between piston and

cylinder at 130°c. For aluminium, =24x10-6

/K and for Cylinder, =10X10-6

/K.

Area of piston = r2 = x 40

2 = 5027mm

2

Area of cylinder = r2= x 40.75

2 =5217mm

2

Initial gap area = x 40.752 - 40

2

=190.3 mm2

Coefficient of superficial expansion for aluminium is,

= 2 = 2 x24 x10-6

/k

Change in temperature = (130oC -15

oC)= 115

oC


91

Increase in area of piston =2tAo

=2 x 24 x 10-6/k x 5027 x 115 = 27.8mm2

Increase in cylinder area =2 x 10 x 10-6/k x 52217 x 115

=12 mm2

Change in area of cylinder gap

=27.8 mm2 – 12 mm

2

=15.8mm2

Percentage increase =15.8/190.3 x 100

= 8.3 %

QUESTION 2

A steel tank of mass 50 kg contains 200 kg of water at 15 oC. A further 100 kg of water is poured

into the tank at 95 oC. If the specified heat capacity of steel is 0.44 kJ/kg K, calculate the final

temperature of the water in the tank.

Let t be the final temperature of the mixture.

Heat given up by the liquid poured in the tank

= mct

= 4.2 x 100 x (95 – t)kJ

Heat gained by steel tank = mc t

=.45 x 50 x (t -15)KJ

=22.5 (t – 15)KJ

Heat gained by water in tank = mc t

= 4.2 x 200 x (t – 15)

Equating heat given up by the water poured into the heat taken by the tank and water then =

4.2 x 100 x (95 – t) = (22.5 (t – 15)) + 4.l2 x 200 (t – 15)

= 862.5 (t – 15)

t = 41.3 oC.

QUESTION 3

Calculate the heat required to convert 8 Kg of ice -10 oC to water at 60

oC. The specific Heat

capacity of ice is 2.1 KJ/Kg K and its latent heat is 335KJ/Kg.

Heat required to raise temperature of the ice from -10 oC to melting point 0

oC

= mct

= 2.1 x 8 x 10

=168 KJ

Heat to melt the ice at constant temperature = mass x latent heat

= 8 x 335

=2680 KJ


92

Heat to raise temperature of 8 Kg of water from 0 oC to 60

oC

= mct

= 8 x4.2 x 60

= 20166 KJ

Total Heat required = 2016KJ + 2680KJ + 168KJ

= 4864 KJ

Chapter ten

Tutorial questions

1.A duralumin piston slides in a cast iron cylinder at 20 oC. At this temperature, there is a radial

clearance of 0.13m and the piston diameter is 100 mm. Determine the area of gap at 20 oC and

also when the temperature is raised to 120 oC. When the coefficient of linear expansion is 22.5 x

10 -6

/k. [20mm²]

2. How much heat energy would be required to convert 20g of ice at -15 oC into water at 40

oC

take the relative specific heat capacity of ice as 0.5 and the specific enthalpy of fusion as

335kJ/Kg [10.69 KJ]


93

3 .A glass flask of volume 200 cm ³ is first filled with mercury at 20ºC. How much mercury will

over flow when the temperature of the system is raised to 100ºC? The coefficient of cubical

expansion of glass is 1.2 x 10 -6

and mercury is 18 x 10 -5

. [27 cm ³]

4.A refrigerator produces ice by evaporation of to liquid ammonia. How much liquid ammonia

must be evaporated to convert 500g of water at 15ºC to ice at -5ºC. If no heat energy is

transmitted through the outside walls the refrigerator.

Specific enthalpy of fusion of ice is 335 KJ/Kg and specific heat capacity of ice is 2.1 KJ/Kg.

[204.25 KJ]


94

Chapter eleven

Heat transfer 11.0 Introduction

The laws of heat transfer find application in many engineering fields i.e. in process engineering,

manufacturing and in everyday life. Heat transfer is by virtue of temperature difference between

parts of material. There are three ways by which heat can be transferred. These are conduction,

convection and radiation.

11.1 Heat Transfer By Conduction

It is the main mode of heat transfer in solid materials. Heat transfer by conduction occurs by

virtue of temp difference between parts of a material. It also takes place in liquids and gasses.

11.2 Mechanism of Conduction

source

of

heat

particles Figure 11.1.

The particles, which are in contact with the source of heat, absorb heat energy through contact. As

the particles get heated, they start vibrating, thereby getting in contact with the adjacent particles

thus passing heat to the next particle until the energy is distributed within the material.

Conduction within a solid material is the transfer of internal energy. This energy is the kinetic

energy of the constituent molecules, atoms and particles of the material. This kinetic energy is

proportional to the absolute temperature. Metals are generally considered as good conductors of

heat. Materials such as glass, wood, wool, paper, and asbestos are poor conductors or insulators. A

good conductor of heat has a high thermal conductivity. Good conductors of heat are usually good

electrical conductors.

11.3 Fourier’s Rate Equation Of Conduction

Molecular collision leads to energy transfer to regions of low kinetic energy. Under steady state

conditions, a molecule will pass on the same amount of energy as it receives.

The figure shows a section of a rod through which heat is flowing.

If the temperature T2 and T1 of a section are kept for a sufficient long time the temperature at

points within the rod is found to decrease uniformly with distance from the hot to the cold face.

However at each point the temperature remains constant with time. This condition is the steady

state.


95

Experiments have shown that:-

The rate of heat flow Q at steady state is proportional to area A and temperature difference T2—T1

is inversely proportional to the change in length .

12 TTA

Q

12 TTkA

Q

TkAQ

A is the area of transfer, m2.

T1 is the inlet face temperature.

T2 is the outlet face temperature.

is the change in thickness.

k is the coefficient of thermal conductivity

of the material (W/m K) Figure 11.2.

– negative sign shows that the temperature is

decreasing with increase in distance.

11.4 Conduction through A Flat Plate

The transfer of heat by conduction is found to depend upon: -

The area through, which heat transfer, takes place. The temperature difference of the two faces through which the heat is passing.

The time taken for heat transfer.

The thickness of material through which the heat is passing.

The greater the area and temperature difference, the greater would be the heat transfer. The heat

transfer is, therefore, inversely proportional to the thickness of wall. Consider a flat plate, or wall

thickness and heat transfer area A. Let the temperatures of its faces be T1 and T2. This is shown

on figure below

Quantity of heat required

12 ttkA

A

T1

T2

Q


96

21 TTkA

Q ------------------------11.1

t1

t2

Direction of

heat transfer Q

Heat transfer

area A

Figure 11.3.

11.5 Conduction through A Composite Wall

Consider the composite wall shown in figure 4. In this case there are three layers.

t1

2 3

t2

k2k1 k3

t3

t4

1

Direction of

heat transfer Q

Heat transfer

area A

Figure 11.4.

If Q is passing through this wall, then Q is passing through each wall.


97

Thus

1

211 ttAkQ

……………………………………..1

2

322 ttAkQ

…….……………………..………..2

3

432 ttAkQ

………………………………..……..3

Transposing equations 1, 2 and 3 gives,

Ak

Qtt

1

121

………………………….…………………..4

Ak

Qtt

2

232

……………………………………………..5

Ak

Qtt

3

343

……………………………………………..6

Adding equations 4, 5 and 6 gives,

3

3

2

2

1

141

kkkA

Qtt ……………………………….7

From equation 7,

2

2

2

2

1

1

41

kkk

TTAQ

------------------------11.2

Therefore in general for n number of layers,

k

TTAQ 1n1 -----------------------11.3


98

Thus the heat transfer per second can be calculated. When this is known, by substitution in

equations 1, 2 and 3 the interface temperatures can be calculated.

11.6 Surface and Overall Heat Transfer

Now there must be a temperature difference a surface and its surroundings for heat transfer to

occur.

As before if heat transfer Q passes through wall, then Q passes through each layer of the wall.

Let ta1 = ambient temperature (temperature of surrounding) on inlet side

t1 = inlet surface temperature

t2 = interface temperature

t3 = exit face temperature

ta2 = temperature of surrounding on outlet side

t2

k1

t1

k2

t3

ta2Inlet surface

film

a1

Outlet surface

film

Direction of

heat transfer Q

21

Figure 11.5.

Q = Us1A(ta1-t1)

= Us2A(t3-ta2)

U is called the surface transfer coefficient and has units W/m2K

2a2

2

1

1

1a

2a1aU

1

kkU

1

A

Qtt


99

2a2

2

1

1

1a

2a1a

U

1

kkU

1

ttAQ

= UA(ta1-ta2)…………………………………………11.4

2a2

2

1

1

1a U

1

kkU

1

1U …………………………11.5

= Overall transfer coefficient sometimes called the U coefficient value.

11.7 Conduction through A Thin Cylinder

A thin cylinder may be one whose internal surface area is almost the same as its external surface

area. From heat transfer point of view, if this is the case, then the area through which the heat is

passing is always very nearly the same. For a thin cylinder of radius r, (either external or internal)

and thickness x, then the area of heat transfer for a length of cylinder L = 2rL.

21 TTkA

Q

21 ttrL2k

Q

Figure 11.6.

11.8 Conduction through A Thick Cylinder

Let L be the length of the cylinder. The internal surface area of a thick cylinder is much smaller

than the external surface area. The fig shows a thick cylinder of internal radius r1 and external

radius r2 with internal and external temperatures t2 and t2 respectively. Assume that the heat is

flowing from the inside to the outside. Consider the elementary cylinder of radius r and thickness

r, let the change across this elementary cylinder t.

t1

r1 t2L


100

t2

t1

r2

r1

r

Q Q

r

r

Figure 11.7.

Hence the heat transfer r

tkAQ

r

trL2kQ

Make t the subject,

rL2k

rQt

rr

1

L2k

Qt

2r

1r

2t

1t

1

212

r

rln

L2k

Qtt

1

212

r

rln

L2k

Qtt

1r2rln

ttL2kQ 21


101

k

ln

ttL2Q

12r

r21

-------------------------------------11.6

11.9 Heat Transfer through A Composite Cylinder

Similarly Q is given by,

321

41

k

ln

k

ln

k

ln

ttL2Q

34

23

12

rr

rr

rr

t1

t4

r1

r4

r2

r3

Qt2

t3

Figure 11.8.

For inner cylinder

t1— t2 = (Q/2Lk1) ln r2/r1 ……………………………………..1

For intermediate cylinder

T2— t3 = (Q/2Lk2) ln r3/r2 ……………………………………..2

For outer cylinder

T3— t4 = (Q/2Lk3) ln r4/r3 ……………………………………..3

Adding equations 1, 2 and 3 gives,

321

41k

ln

k

ln

k

ln

L2

Qtt 3

42

31

2r

rr

rr

r


102

321

41

k

ln

k

ln

k

ln

ttL2Q

34

23

12

rr

rr

rr

----------------------------------------------------------11.6

11.10 heat transfer by convection

It is the name given to gross motion of fluid itself so that fresh fluid is continuously available for

heating or cooling.

Mechanism

Air next to the source of heat is heated by conduction, due to its contact with the hot surface. The

heated mass of air will expand, increasing its volume and density decreasing. The cool air around

with higher density will displace the warm air, which is forced to rise. The cool air gets in touch

with hot surface and become warm by conduction. This process continues with heat getting

distributed within the fluid.

Types Of Convection Transfer

Natural convection — is the movement of fluid caused by differences in density

resulting from temperature differences.

Forced convection — is produced by mechanical means e.g. by fans.

Newton’s Law Of Cooling

The heat transfer from one fluid to another through a wall is understood by studying

how heat is transferred from one to a wall and vice versa.

Newton’s law of cooling states that:

Heat transfer from a solid surface of area A at temperature tw to a fluid of

temperature t is given by:

Q = hA(tw — t) ------------------------11.7

h is the heat transfer coefficient with units W/m2K. h depends on the properties of

the fluid and its velocity.

Transfer Of Heat From Fluid A To

Fluid B Through A Wall.

In fluid the temperature decreases

suddenly from tA to t1 and in fluid B,

temperature decreases rapidly from t2

to tB. this is due to the thin film close

to the wall where transfer is by

conduction.

k

Fluid A

Ta

t1

Direction of

heat transfer Q

Tbt2

Fluid B


103

X is thickness of wall.

K is thermal conductivity.

Figure 11.9.

Considering unit surface area

From fluid A to the wall the rate of heat transfer is,

1a

A

A ttk

Q

A

AA1aA

khwhere,tthQ

From wall to fluid B

b2

B

B ttk

Q

B

BBb2B

khwhere,tthQ

Transmission through the wall per unit area

21 ttk

Q

For steady state, heat transfer flowing from fluid A to the wall = heat flowing

through the wall = heat flowing through from wall to the wall.

Q = hA(tb — t1) = hB(t2 — tb) =

21 ttk

Expressing in terms of temperature,


104

(ta — t1) = Ah

Q

(t1 — t2) =

k

Q

(t2 — tb) = Bh

Q

Adding the corresponding sides,

(tA — tB) = Q + Q + Q

hA k hB

X

(tA — tB) = Q 1 + X + 1

hA k hB

Q = (tA — tB)_______

1 + X + 1

hA k hB --------------------------11.8

Q = U(tA — tB)

1 = 1 + X + 1

U hA k hB

U is the overall heat transfer coefficient has the same units as h.

11.11 transmission of heat by radiation


105

Thermal radiation consists of electromagnetic waves emitted due to the agitation of

the molecules of substance. These waves are similar to the light waves in that they

are propagated in a straight line at the speed of light. It requires no medium to be

transmitted. Heat energy from the sun to the earth is mainly transmitted by radiation

because the space between is largely vacuum.

Radiation striking a body (absorptivity )

Reflected from the body (reflectivity )

Transmitted through the wall (transmissivity )

The emissivity of a body radiating energy at a temperature t is equal to the to the

absorptivity (α) of the body when receiving energy from the source from

temperature t.

The Boltzmann’s law

States that the emissive power of the black body is directly proportional to the

fourth power of its absolute temperature.

EB = s AT4

Where s is Boltzmann’s constant = 5.67-8

Wm2/K

4

Energy absorbed by a grey body,

E = s T4

Where is the emissivity of the body.

The heat transfer from the body to its surroundings per square metre is given by,

Q = s(T14— T2

4) -----------------------------11.9

Worked Example

In a small furnace, the heat loss to the surroundings is to be kept down to 1 700 W/m

2. The

internal temperature of 150 mm firebrick wall which lines the furnace is 650oC, and the

temperature of the air surrounding the furnace is 25oC. Neglecting the temperature drop through

the steel casing, calculate the thickness of exterior lagging required. Take the thermal conductivity

of the firebrick 1W/mK, the thermal conductivity of the lagging as 1.2W/mK and the convection

heat transfer coefficient of the lagging surface as 20W/m2K. Radiation from the lagging surface


106

may be ignored. Estimate also the temperature of the outer surface of the lagging.

lm

Lagging

Steel

shell

Fire

brick

0.150m

25°C

650°C

Figure 11.10.

For conduction through the fire brick Q = 1700W/m2

1650kAQ

)1..(..........m/W

15.0

6501Q 21

For conduction through the lagging

)2..(..........m/W

2.1Q 221

For convection from the lagging surface

25hAQ 2

)3..(..........m/W2520Q 22

From (1)

15.01700650 1

From (2)

2.1

170021

From (3)

20

11700252

Adding:

2.1

05.015.0170025650

thus 05.015.01700

625

2.1

l = 0.201 m

For temperature of the lagging surface, from equation (3)

1700 = 20(2 – 25)


107

2 = 1100C

Worked Example

A steel pipe shown in Figure 1 carries steam at 260 oC. The atmospheric temperature is 15

0C. The

heat transfer coefficient for the inside and outside surfaces are 550 and 15W/m2K respectively,

and the thermal conductivities of steel, diatomaceous earth and asbestos felt are 50, 0.09, and

0.07W/mK respectively. Calculate:

The rate of heat loss by the steam per unit length of pipe

The temperature of the outside surface.

Diatomaceous Earth

Asbestos Felt

Ø 3

14

Ø 1

94

Ø 1

14 Ø

100

Steel

all dimensions are in milimetres

Figure 11.11.

Rate of heat loss Q

aafdesta

as

h

1

k

ln

k

ln

k

ln

h

1

ttL2Q

34

23

12

rr

rr

rr

15

1

07.0

ln

09.0

ln

50

ln

500

1

1526012Q

097.0157.0

057.0097.0

05.0057.0

Q = 116W

Rate of heat loss per meter length of pipe = 116W

0675.0

15t116Q 4


108

t4 = (116 x 0.0675) + 15

= 22.8oC

Temperature of outside surface = 22.8oC

Chapter eleven

Tutorials

1.Calculate the quantity of heat conducted per minute through a circular disc of diameter 127 mm

19mm thick hen the temperature drop across he thickness of the plate is 5oC. Take the coefficient

of thermal conductivity as 150W/mK. [30kJ]

2.A cold storage compartment is 4.5m long by 4m wide by 2.5m high. The four walls, ceiling and

floor are covered to a thickness of 150mm with insulating material, which has a coefficient of

thermal conductivity of 5.8 x 10-2

W/mK. Calculate the quantity of heat leaking through the

insulation per hour when the inside and the outside face temperatures of the material is 15oC and –

5oC respectively. [2.185MJ or 2185kJ]

3.One side of a refrigerated cold chamber is 6m long by 3.7m high and consist of 168mm

thickness of cork between outer and inner walls of wood. The outer wood wall is 13mm thick and

its outside face temperature is 20oC, the inner wood wall is 35mm thick and its inside face

temperature is –3oC.Taking the coefficient of thermal conductivity of the cork and wood as 0.042

and 0.2W/mK respectively, calculate:


109

(a) the heat transfer per second per square metre of surface area, [5.318J]

(b) the total heat transfer through the chamber side per hour, [425kJ]

(c) the interface temperatures. [19.2oC , -2.07

oC ]

4.Hot gasses at 280oC flow on one side of metal plate of 10mm thickness and air at 35

oC flows on

the other side. The heat transfer coefficient the gasses is 31.5 W/m2K and that of air is 32W/m

2K.

The coefficient of thermal conductivity of the metal plate is 50W/mK. Calculate:

(a) the overall heat transfer coefficient, [15.82W/m2K]

(b) the heat transfer from gasses to air per minute per square metre of plate area.

[232.7kJ ]

5.The wall of a cold room consists of a layer of cork sandwiched between outer and inner walls of

wood, the wood walls being each 30 mm thick. The inside atmosphere of the room is maintained

at –20oC when the external atmospheric temperature is 25

oC, and the heat loss through the wall is

42W/m2. Taking the thermal conductivity of cork and wood as 0,2W /m K and 0.05 W/mK

respectively, and the rate of heat transfer between each exposed wood surface and their respective

atmospheres as 15W/m2k, calculate;

(a) the temperatures of the exposed surfaces, [22.2oC , -17.2

oC]

(b) the temperatures of the interfaces, [15.9oC , -10.9

oC]

(c) the thickness of the cork. [31.9mm]

6.A flat circular plate is 500 mm diameter. Calculate the theoretical quantity of heat radiated per

hour when its temperature is 215oC and the temperature of its surrounds is 45

oC. Take the value of

the radiation constant as 5.67 X 10-11

kJ/m2sK

4

[1862kJ]

7.The steam drum of a water-tube boiler has hemispherical ends, the diameter is 1.22m and the

overall length is 6m. Under steaming conditions the temperature of the shell lagging was 230oC

and the temperature of the surrounds was 51oC. The temperature of the cleading after lagging was

69oC and the surrounds 27

oC. Assuming 75% of the total shell area to be lagged and taking the

radiation constant as 5.67X10-11

kJ/m2sK

4, estimate the saving in heat energy per hour due to

lagging. [167MJ]


110

Documents

Thermodynamics Module