msc energy notes

7/30/2019 Msc Energy Notes

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ENERGY MANAGEMENT T T AL-SHEMMERI P.1


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PREFACE

This book is concerned with the subject of Energy, as resources, use and the ways

it can be optimised. The need for this type of book is very well documented, and in

the following paragraph I will try to summarise that.

Coal fuelled the industrial revolution in the 18th and 19th century. It remained as the

prime fuel supplying steam engines used in road vehicles and rail road trucks. The

industrialised nations were in huge competition in their search for additional fuels

and the discovery of oil in the Middle East and elsewhere extended the use oil

opening the applications to a wider range. With the advent of the automobile,

airplanes and the spreading use of electricity, oil became the dominant fuel during

the twentieth century. The Arab-Israeli wars in the 1967 and 1972, the price of oil

increased from 5 to 45 US dollars per barrel, there was a shift away from oil. Coal

and nuclear became the fuels of choice for electricity generation and conservation

measures increased energy efficiency. The use of fossil fuels has continued to grow

and their share of the energy supply has increased.

In 2005, total worldwide energy consumption was 500 EJ (= 5 x 1020 J) (or 138,900

TWh) with 86.5% derived from the combustion of fossil fuels. The estimates of

remaining worldwide energy resources vary, with the remaining fossil fuels totalling

an estimated 0.4 YJ (1 YJ = 1024 J).

Political considerations over the security of supplies, environmental concerns

related to Global Warming and Sustainability will move the world's energy

consumption away from fossil fuels.

In order to move away from fossil fuels it is expected to create economic pressure

through Carbon trading and Green taxation. Some countries are taking action as a

result of the Kyoto Protocol, and further steps in this direction are proposed. For

example, the European Union Commission has proposed that the Energy Policy

should set a binding target of increasing the level of renewable energy in the EU's

overall mix from less than 7% today to 20% by 2020.

This book tackles the fundamental principles of thermodynamics into day-to-day

engineering concepts; it provides the tools to accurately measure process efficiency

and sustainability in the power and heating applications-helping engineers to

recognize why losses occur and how they can be reduced utilizing familiar

thermodynamic principles.


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CONTENTS

1 ENERGY & THE ENVIRONMENT ......................................................................... 7

1.1 Introduction...................................................................................................... 8

1.2 Forms of energy............................................................................................... 8

1.2.1 Mechanical energy................................................................................. 9

1.2.2 Electrical energy................................................................................... 10

1.2.3 Chemical energy .................................................................................. 11

1.2.4 Nuclear energy..................................................................................... 12

1.2.5 Thermal energy.................................................................................... 13

1.3 Energy conversion......................................................................................... 14

1.4 The burning question ..................................................................................... 18

1.4.1 Combustion of coal............................................................................... 19

1.4.2 Combustion of oil.................................................................................. 20

1.4.3 Combustion of natural gas ................................................................... 21

1.5 Environmental Impact from fossil fuels .......................................................... 23

1.6 Energy world-wide ......................................................................................... 24

1.7 Energy and the future! ................................................................................... 26

1.7.1 The dream scenario ............................................................................. 27

1.7.2 The Renewable scenario...................................................................... 27

1.8 Worked Examples.......................................................................................... 36

1.9 Tutorial Problems........................................................................................... 41

1.10 Case Study- Energy for the World ............................................................... 42

2 MODES OF HEAT TRANSFER ........................................................................... 46

2.1 Modes of Heat Transfer ................................................................................. 47

2.2 Fouriers Law of Thermal Conduction............................................................ 48

2.3 Film heat transfer between a solid and a fluid................................................ 49

2.4 Heat transfer through a composite wall separating two fluids........................ 50

2.5 Radial conduction through pipe wall .............................................................. 51

2.6 Heat exchange though a tube with convection on both sides ........................ 52

2.7 Composite tube with fluid on inner and outer surfaces .................................. 53

2.8 Heat transfer by convection ........................................................................... 55

2.9 Heat transfer by radiation............................................................................... 63

2.16 Worked Example ......................................................................................... 66

2.17 Tutorial Problems - Heat Transfer................................................................ 79

3 DESIGN OF HEAT EXCHANGERS..................................................................... 81

3.1 Mechanism of heat exchange ........................................................................ 82


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3.1.1 Double-pipe.......................................................................................... 83

3.1.2 Shell-and-Tube heat exchangers ......................................................... 84

3.1.3 Cross flow heat exchangers................................................................. 85

3.2 Overall heat transfer coefficient ..................................................................... 88

3.3 Analysis of heat exchangers.......................................................................... 89

3.3.1 The Logarithmic Mean Temperature Difference Method...................... 89

3.3.2 The F-Method for Analysis of Heat Exchangers................................... 91

3.3.3 The Effectiveness NTU Method for Analysis of Heat Exchangers..... 96

3.4 Optimisation of Heat Transfer Surfaces (Fins)............................................. 105

3.4.1 Fin Types .................................................................................................. 105

3.4.2 Theory of fins............................................................................................ 106

3.5 Worked Examples........................................................................................ 109

3.6 Tutorial Problems - Heat Exchangers .......................................................... 146


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LIST OF FIGURES

Figure 1.1 Energy conversions in a typical coal fired power plant ........................... 15

Figure 1.2 Evolution of energy consumption by mankind ........................................ 24

Figure 1.3 Worlds Energy consumption, past and forecast .................................... 28

Figure 1.4 Worlds CO2 emission, past and forecast .............................................. 28

Figure 1.5 Global warming, the evidence ................................................................ 29

Figure 1.6 Carbon Footprint Calculator example..................................................... 30

Figure 2.1 Conduction Heat Transfer ...................................................................... 48

Figure 2.2 Film convection concept......................................................................... 49

Figure 2.3 Conduction through multi-layers............................................................. 50

Figure 2.4 Radial conduction................................................................................... 51

Figure 2.5 Combined convection-conduction-convection heat Transfer.................. 52

Figure 2.6 Radial conduction through multi-layers .................................................. 53

Figure 3.1 Types of Double pipe heat exchangers .................................................. 83

Figure 3.2 Shell and Tube heat Exchanger............................................................. 84

Figure 3.3 Cross flow Heat Exchanger.................................................................... 85

Figure 3.4b Derivation of the Log Mean Temperature Difference (LMTD) .............. 90

Figure 3.5a Correction Factors for Shell and Tube heat Exchangers...................... 92

Figure 3.5b Correction Factors for Shell and Tube heat Exchangers...................... 93

Figure 3.5c Correction Factors for Crossflow heat Exchangers .............................. 94

Figure 3.5d Correction Factors for Crossflow heat Exchangers .............................. 95

Figure 3.6a NTU-effectiveness chart for Counter flow heat Exchangers................. 99

Figure 3.6b NTU-effectiveness chart for Parallel flow heat Exchangers................ 100

Figure 3.6c NTU-effectiveness chart for Shell and Tube heat Exchangers........... 101

Figure 3.6d NTU-effectiveness chart for Shell and Tube heat Exchangers........... 102

Figure 3.6e NTU-effectiveness chart for Cross flow heat Exchangers .................. 103

Figure 3.6f NTU-effectiveness chart for Cross flow heat Exchangers ................... 104

Figure 3.7 Efficiency of Straight Fins..................................................................... 107

Figure 3.8 Efficiency of Annular Fins of Rectangular Profile.................................. 108


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LIST OF TABLES

Table 1.1 Energy conversion matrix ........................................................................ 16

Table 1.2 Typical Power ratings .............................................................................. 17

Table 1.3 Energy content of fuels............................................................................ 22

Table 1.4 Environmental impacts of fossil fuels....................................................... 23

Table 1.5 worlds energy production, past present and future................................. 31

Table 1.6 Energy consumption for various countries............................................... 32

Table 1.7 Energy needs of a typical family.............................................................. 33

Table 1.8 Energy usages ( EJ ) by sector for the UK ............................................. 34

Table 1.9 Estimated energy reserves of the world .................................................. 35

Table 2.1 Thermal Conductivity of Some Common Materials.................................. 54

Table 2.2 Properties of air at 1.01325 bar ............................................................... 59

Table 2.3 Properties of typical engine oil................................................................. 60

Table 2.4 Properties of saturated water and steam................................................. 61

Table 2.5 Emissivity of some materials ................................................................... 64

Table 3.1 Typical Overall Heat-transfer Coefficients ............................................... 86

Table 3.2 Typical Fouling Factors ........................................................................... 87


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1 ENERGY & THE ENVIRONMENT

This Chapter helps the reader to develop an awareness of environmental issues

related to energy. It describes the principles of energy sources, utilisation and

conversions. Finally, it presents a global overview of energy.


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1.1 Introduction

It is necessary to appreciate that energy will be needed to modify the state of

working environment and keep it at a comfortable condition whether to provide

warm or cool air.

Heating the air is a simple process of increasing its thermal energy and

consequently raising its temperature. Heating of air can be achieved either by direct

or indirect means; examples such as coal or gas fires represents the direct forms of

heating normally used in houses; on the other hand, the electrical resistance

heating elements is an indirect method because electricity was produced

elsewhere. Hot water radiators also used to provide indirect heating of indoor air.

Similarly, the process of cooling the air or reducing its thermal energy is energy

driven. Cooling and heating require further process, the air after being treated ( i.e.

heated or cooled ) has to be delivered to the space where it is needed, and hence

an electric fan is usually used to circulate the air from the apparatus to the room.

The following sections will discuss the various forms of energy, and how energy can

be converted from one form to another which convenient for heating, cooling etc.

This chapter will demonstrate the environmental impact of using different fuels to

provide energy for heating or electricity.

1.2 Forms of energy

We associate energy with devices whose inputs are fuel based such as electrical

current, coal, oil or natural gas; resulting in outputs such as movement, heat or light.

Unit of energy is the Joule (J). The rate of producing energy is POWER which has

the unit of Joule per second or the Watt (W).

There are FIVE forms of Energy:

1.2.1 Mechanical Energy

1.2.2 Electrical Energy

1.2.3 Chemical Energy

1.2.4 Nuclear Energy

1.2.5 Thermal Energy


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1.2.1 Mech anic al energ y

This type of energy is associated with the ability to perform physical work.

There are two forms in which this energy is found; namely potential energy and

kinetic energy.

Potential energy

As the name implies is contained in a body due to its height above its surroundings,

examples such as the gravitational energy of the water behind a dam, and the

energy stored in batteries.

Potential Energy = mass x acceleration due to gravity (9.81) x height above datum

Or Ep = m x g x h (1.1)

The energy produced by one kilogram of water falling from a height of 100m above

ground is a potential energy, which can be calculated as follows:

Potential Energy = mass x acceleration due to gravity x height above datum

Ep = 1 x 9.81 x 100 = 981 J/kg

Kinetic energy

Kinetic energy is related to the movement of the body in question. Examples of KE

such as the flywheel effect and the energy of water flowing in a stream.

Kinetic Energy = mass x velocity squared

Or Ek = x m x v2 (1.2)

The water stream in a river flowing at a velocity of 2 m/s has a kinetic energy of:

Kinetic Energy = mass x velocity squared = x 1 x (2)2 = 2 J/kg


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1.2.2 Electric al energy

This type of energy as the name implies is associated with the electrons of

materials.

Electrical energy exists in two forms:

Electrostatic electricity

This type of electrical energy is produced by the accumulation of charge on the

plates of a capacitor.

Electromagnetic energy

This type is due the Inductive- field energy.


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1.2.3 Chemic al energ y

This type of is associated with the release of thermal energy due to a chemical

reaction of certain substances with oxygen. Burning wood, coal or gas is the main

source of energy we commonly use in heating and cooking.

Calculation of chemical energy

The energy liberated from the combustion of a given mass of fuel, with a known

calorific value in a combustion chamber of known efficiency is given by:

Chemical Energy = Mass of fuel x calorific value x efficiency of combustion

(1.3)

Typical coal has an energy value of 30 MJ/kg, which implies that during the

combustion of one kilogramme of coal, there will be a release of 30 megajoules of

thermal energy.

The energy contained in the food we take is another example of chemical energy.

Analyses of thermal energy liberated from stored chemical energy during the

combustion of coal, oil and natural gas will be discussed later in this chapter.


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1.2.4 Nuclear energ y

This energy is stored in the nucleus of matter, and is released as a result of

interactions within the atomic nucleus.

There are three nuclear reactions:

Radioactive decay:

In which one unstable nucleus (radioisotope) decays into a more stableconfiguration resulting in the release of matter and energy

Fission:

A heavy nucleus absorbs a neutron splitting it into two or more nuclei accompanied

by a release of energy. Uranium U235 has the ability to produce 70x109 J/kg

Einstein proposed the following equation to calculate the energy produced from

nuclear fissioning (i.e. conversion of matter into energy) :

E = m C2 (1.4)

Where m is the mass, and

C is the speed of light

This reaction forms the bases for current nuclear power generation plants.

Fusion:

Two light nuclei combine to produce a more stable configuration accompanied by

the release of energy. Heavy water (Deuterium) fusion reaction may produce

energy at the rate of 0.35x1012 J/kg.

This reaction is yet to be realised to produce electricity on commercial basis.


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1.2.5 Thermal ene rgy

Thermal energy is associated with intermolecular vibration resulting in heat and a

temperature rise above that of the surroundings. Thermal energy is calculated for

two different regimes:

When the substance in a pure phase, say if it is in a liquid, gas or solid, then

Thermal Energy = mass x specific heat x temperature difference (1.5)

During a change of phase, such as evaporation or condensation, it can be

calculated by:

Thermal Energy = mass x latent heat (1.6)

However, if there is a change of phase, say during the condensation of water

vapour into liquid, there is an additional amount of heat released while the

temperature remains constant during the change of phase. For 1 kg of water to be

heated at ambient pressure from 20 to 120 oC, the requirement is

Thermal energy = heating water (20-100) oC + evaporation at 100 oC + heating

vapour (100-120) oC

Thermal energy = 1x 4.219 x (100-20) + 1x2256.7 + 1x2.01x(120-100)

= 337.52 + 2256.7 + 40.2

= 2634.42 kJ

Note that the specific heat capacity for water at 1 atmosphere is

For liquid state Cpf = 4.219 kJ/kg

For vapour state Cpg = 2.010 kJ/kg

The latent heat of evaporation = 2256.7 kJ/kg


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1.3 Energy conversion

It is important to understand that losses are encountered during the transformation

of energy from the source into a useful form for a given application, for example

consider a power station using coal as a fuel, the following conversions take place:

Energy liberated by the combustion of coal, (chemical to thermal); in other words

not all coal is burnt completely.

Thermal energy of the combustion gases is used to raise steam (thermal to

thermal); some thermal energy is lost as the exhaust gases leaving the boiler have

a temperature above that of the ambient air outside the boiler.

High pressure and temperature steam turns the steam turbine (thermal to

mechanical). Not all energy of the steam is converted into rotational energy as the

steam leaving the turbine has a temperature well above that of the ambient.

The steam turbine drives the electrical generator (mechanical to electrical). Some

losses are dissipated through the mechanical connections between the turbine and

the electrical generator.

Electricity is used by customers for lighting / heating or to operate electric devices

such as radio, television, etc. electrical devices are designed to operate on an

optimum condition, efficiency of the operation will vary depending on its use, age

and maintenance.

The input energy to a coal-fired power station (Fig. 1.1) can be analysed in a simple

way by considering the energy flow diagram, which may look like this:

100 units of energy are stored in the coal fuel

95 units are converted into thermal energy entering the boiler

70 units are absorbed by the water, converting into steam

45 units are converted into mechanical energy at the steam turbine rotor

38 units are converted into electrical energy in the electrical generator

30 units go up to waste through the chimney-stack.

Most of the devices in common use have the ability to convert energy from one form

to another. Table 1.1 shows the 25 possibilities of energy transfer (the symbol, x,

indicates unrealised link at the present time).


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It is important to appreciate the amount of energy consumed in order to develop an

understanding of energy efficiency, Table 1-2 displays examples of some typical

energy ratings.

Figure 1.1 Energy conversions in a typical coal fired power plant

Numbers quoted represents the units of energy

Boiler

Steam

Coal100

Turbine

Electrical

Generator

Steam

70

Torque

45

Electricity

38Heat 95

Lighting Heating


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Table 1.1 Energy conversion matrix

From \ To Mechanical Electrical Thermal Chemical Nuclear

Mechanical Gear

Nutcracker

push

mower

Electric

generator

friction x x

Electrical Electric

motor

Light bulb Electric fire Electrolysis Particle

accelerator

Thermal Steam

turbine

Thermo-

couple

Heat

exchanger

x Fusion

reactor

Chemical jet engine

Rocket

Battery

fuel cell

Car engine

Boiler

Intermediat

e reaction

x

Nuclear x x Nuclear

reactor

x x


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Table 1.2 Typical Power ratings

Application Rating

Energy efficient light bulb 10 W

Electric Kettle 1500

(1.5kW)

Family car 100000

(100kW)

Coal fired power station 500000000

(500MW)

Electricity demand for a big city 10000000000

(10GW)


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1.4 The burning question

Although man discovered fire very long time ago, the fuel for cooking, heating and

light relied on burning just wood. By the 4 th century AD, the Romans sent an entire

fleet of ships to bring wood from France and North Africa.

It was not until the 17th century that coal was discovered, and was extensively used

during and after the Industrial revolution in Europe.

In the nineteenth century oil was discovered and was extensively used mainly by

the major industrialised world. There were times when the west enjoyed cheap oil

prices due to competition between oil producing countries. The major shock was feltin the 1973 war between the Arab and Israel and the subsequent tripling of oil

prices.

Although, the main fuels are coal, oil and natural gas, there other sources which can

substitute in some cases, such as dry wood, agricultural waste, and general

household waste. Table 1.3 presents the typical heating values of miscellaneous

substances that can be burnt in typical energy furnaces.

For all combustion processes, the fuel has to have a combination of any of all three

elements (Carbon, Hydrogen, and Sulphur) which when reacts with oxygen; they

produce thermal energy the amount is quoted as its calorific value.

Combustion of Carbon:

C + O2 CO2 + Energy

Carbon has energy content of 32793 kJ/kg

Combustion of Hydrogen:

H2 + 2O2 2H2O + Energy

Hydrogen has energy content of 142920 kJ/kg

Combustion of Sulphur:

S + O2 SO2 + Energy

Sulphur has energy content of 9300 kJ/kg


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1.4.1 Combu st ion o f coal

Typical coal in the UK has the following composition (percentage by weight):

Carbon 64.4

Hydrogen 4.4

Oxygen 4.4

Nitrogen 0.9

Sulphur 10

Ash 15

Total 100%

The heat release of burning one kilogram of coal can thus be estimated as the sum

of energy released from the proportions of carbon, Hydrogen and Sulphur contained

in coal. Hence, Calorific value of coal is calculated as follows:

Part due to combustion of carbon +32793x0.644 = 21118.7

Part due to combustion of Hydrogen +142920x0.044 = 6288.5

Part due to combustion of Sulphur +9300x0.1 = 930

Part lost in combustion of water content -2256.7x 0.044 = - 893.6

Calorific value of coal 2118.7+6288.5+930-893.6= 27443.6 kJ/kg

This value obtained on the assumption of 100% efficient and complete combustion

Emissions:

Production of CO2 from coal = 0.644 x(44/12) = 2.36 kg CO2 /kg coal

Production of CO2 per kWh = 2.36 x 3600 / 27443 = 0.31 kg/kWh

Production of CO2 per GJ energy = 2.36 x 106 / 27443 = 86 kg/GJ

Production of SO2 from coal = 0.009 x(64/32) = 0.018 kg SO2 /kg coal


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1.4.2 Combu st ion o f oi l

Typical oil in the UK has the following composition (percentage by weight):

Carbon 86

Hydrogen 12

Sulphur 2

-----------------------------------------

Total 100%

The heat release of burning one kilogram of oil can thus be estimated as the sum of

energy released from the proportions of carbon, Hydrogen and Sulphur contained in

oil. Hence Calorific value of oil is calculated as follows:

Part due to combustion of carbon +32793x0.86

Part due to combustion of Hydrogen +142920x0.12

Part due to combustion of Sulphur +9300x0.02

Part lost in combustion of water content -2256.7x(0.12x9)

Calorific value of oil = 28202 + 17150 +186 -2437

= 43101 kJ/kg

Emissions:

Production of CO2 from oil = 0.86 x(44/12) = 3.153 kg CO2 /kg oil


Production of CO2 per GJ energy = 3.153 x 106 / 43101 = 73.15 kg/GJ

Production of SO2 from oil = 0.02 x(64/32) = 0.040 kg SO2 /kg oil


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1.4.3 Combus t ion of n atural gas

Typical natural gas in the UK has the following composition (percentage by weight):

Carbon 75

Hydrogen 25

Sulphur 0

-----------------------------------------

Total 100%

The heat release of burning one kilogram of natural gas can thus be estimated as

the sum of energy released from the proportions of carbon, and Hydrogen

contained in natural gas. Hence Calorific value of natural gas is calculated as

follows:

Part due to combustion of carbon +32793x0.75

Part due to combustion of Hydrogen +142920x0.25

Part lost in combustion of water content -2256.7x(0.25x9)

Calorific value of natural gas = 24595 + 35730 -5077

= 55247 kJ/kg

Emissions:

Production of CO2 from natural gas = 0.75 x(44/12) = 2.75 kg CO2 /kg


Production of CO2 per GJ energy = 2.75 x 106 / 55247 = 49.77 kg/GJ

Production of SO2 from natural gas = ZERO


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Table 1.3 Energy content of fuels

SourceCalorific value

MJ/kg

Paper 15

Vegetable 7

Rag 16

Plastics 37

Dry wood (20% moisture

content max.)15

Municipal waste 9

Natural gas 55

Oil 42

Coal 26


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1.5 Environmental Impact from fossil fuels

Coal, Oil and Natural gas have their relative merits in terms of availability, price and

thermal performance. Table 1.4 below is constructed for comparison of the heat

capacity, CO2 and SO2 production by the three fossil fuels. The 4th column is of

particular importance in comparing all three fuels; it represents the quantity of

carbon dioxide emitted for every unit of energy produced.

Coal produces the highest amount of Carbon dioxide for a given output of energy;

then oil, then Natural gas which produces nearly half the emission of coal and a

third less than that of oil.

Table 1.4 Environmental impacts of fossil fuels

Fuel Calorific Value

MJ/kg

CO2

kg / kg fuel

CO2 / Energy

kg / GJ

SO2

kg / kg fuel

Coal 26 2.36 92 0.018

Oil 42 3.153 73 0.040

Natural

Gas

55 2.75 49.77 0


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1.6 Energy world-wide

The consumption of energy by humankind has evolved over the ages. It began with

the invention of fire, man relied on wood burning to cook and to provide warmth and

light for millions of years. As civilisation evolved, the needs for energy became

greater and other sources were sought. In the long search, man discovered coal.

Over the years coal provided much greater resource for energy, encouraging man

to push its use into further applications. Michael Faraday demonstrated in 1831 the

first electric current generator. Other discoveries needed

Figure 1.2 Evolution of energy consumption by mankind


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The industrial revolution in Europe:

A major leap in the nineteenth century was achieved by the discovery of oil in the

Middle East. This unfortunate discovery eventually led to TWO World wars as the

leading industrial nations attempted to dominate the world market and to secure the

energy supply for their huge manufacturing industries.

The oil crises due to the Arab-Israeli war in 1973 resulted in tripling of oil prices, this

was a major shock for non producing countries, particularly in Europe, on one hand

it has put tremendous increase on the energy consumers budget;

However, on the positive side, this was a major advantage to humankind and the

environment, it forced consumers to reduce the excessive consumption of energy, it

helped man to review manufacturing processes and attempt to increase energy

efficiency, and probably it pushed governments to search for newer sources of

energy. Substantial funds were allocated for the research into renewable resources

such as Hydropower, wind turbines and solar energy.

Energy consumption world-wide has continued to rise, it is estimated that in 1900,

the world consumption was around 30EJ and by 1950 it rose to 70 EJ; this reached

400EJ in 1992.

The continued increase in population and the associated increase in manufacturing

industry to cater for greater dependence of man on energy driven devices and the

culture of multi-car ownership has put even greater importance for energy.

It is interesting to examine the history of energy production at selected windows,

Table 1-5 shows the evolution of the worlds production of energy.

Table 1-6 indicates the energy consumption for selected countries; it is not

surprising that the United States of America has the highest consumption per

inhabitant of 6 kW compared with a mere 100 W per person in India.

On individual scale, consider a typical family of four, the annual energy consumption

is estimated to be around 200000 kWh, or 50000 kWh per person!

The energy usage by sector is shown in Table 1.8 (space heating is one the highest

users, even higher than the consumption by the Transport sector)


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1.7 Energy and the future!

Fossil fuel resource is finite; In the figure below is an estimate to the production of

oil since 1900, it is estimated that currently we are at the start of the expected drop

in production, and the prediction of this loss of production is very severe as can be

seen in the diagram. The rate of consumption is unsustainable, it has taken the

earth millions of years to form these fossil fuels, and it seems that we are about to

consume all in a matter of 200 300 years; see Figure 1.3

The issue is not one of "running out" so much as it is not having enough to keep our

current exuberance waste of energy running. It is a matter of responsibility and

obligation to keep some reserves for generations to come.

The second critical issue is that as energy will become scarce, there will be wars

over it, the world have already witnessed few major wars in the middle east, and to

the contrary of the politicians in the white house or white hall would say, these wars

are over the control of oil. The situation will deteriorate as the fuel supply is

reduced.

The top curve in Fig. 1.4 shows the total CO2 emitted worldwide, by far coal

represents the biggest part, and the trend is rising till the end of the 21st Century.

Table 1.10 gives one of the many estimates of the world energy reserves. It is clear

that by the end of the 21st century, fossil fuels will almost be exhausted; and even

before then the world will undoubtedly witness some serious crises.

A major environmental disaster is in the making, in fact global warming is a reality, it

is scientifically proven that there is a direct correlation between Energy

consumption/CO2 production and environmental ambient temperature (Figure 1.5).

Education plays a major role in preparing the next generations to be aware of the

need for that change, and there are many tools written to give a simple and

straightforward indication of the carbon footprint of the daily activities we do, see

Figure 1.6, which is taken from http://www.carbonify.com/carbon-calculator.htm.

The shift from fossil fuel to alternative sources such as renewable is not going to be

easy, it is going to take time and much effort and sacrifice in order to adjust.

As affordable oil is necessary to power any serious attempt at a switchover to

alternative sources of energy, these extreme prices will severely hamstring if not -

completely cripple - the ability of the market to handle these problems. The

economic fallout from high prices will almost certainly produce geopolitical tensions

(i.e. war) thereby further hampering the development of large-scale alternative

sources of energy. Worse still, in a global environment characterized by massive
http://www.carbonify.com/carbon-calculator.htmhttp://www.carbonify.com/carbon-calculator.htm


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energy-wars, the bulk of the world's financial capital is likely to be disproportionately

invested in weapons technologies over alternative energy technologies.

Albert Einstein put forward in 1905 his equivalence theory, and in 1938 Otto Hahn

discovered nuclear fission, however, it was not until 1957 that the first nuclear

electricity generation was achieved in the USA. Today France produces 20% of its

electricity by nuclear fission.

It is well known by now that although nuclear fission is a valuable source of energy,

it has its long term hazards during the production period and the burden of long

term storage of waste after use and the decommissioning of the power plant at the

end of its production life. There has to be a Solution in order to avoid such crises,

the way forward takes two possibilities:

1.7.1 The d ream s cenario

If by extreme good fortune, someone is able to realise nuclear fusion on realistic

magnitude, it is estimated that this form of energy will sustain life on earth for

millions of years to come. However, realising this may be quoted as someone

planning to go to heaven!

1.7.2 The Renewable scen ario

Renewable energy sources include solar, wind, hydro, geothermal and biomass. All

have been realised to a limited degree so far.

This is a more realistic Solution for the near future, Renewable energy is freely

available, it is pollution-free, but at the present time, it does not compete on

economic terms, with cheap fuels such as coal, oil or gas. This is because so far the

technology of harnessing renewable energy is relatively inefficient. However,

considerable international effort has been coordinated to increase the use of

renewable energy in order to reduce global warming caused by the combustion of

fossil fuels


28/149


Figure 1.3 Worlds Energy consumption, past and forecast

Figure 1.4 Worlds CO2 emission, past and forecast


29/149


Figure 1.5 Global warming, the evidence


30/149


Quantity Total Tons of CO2 annually

Small car

(40 mpg fuel economy)

1000

miles per month

3.54

Average/medium car(21 mpg fuel economy)

1000

miles per month

6.6

SUV/4 wheel drive

(15 mpg fuel economy)

1000

miles per month

9.42

Electricity usage1000

kwh/month

9

Natural/propane gas1000

cubic feet/month

0.1968

Fuel oil heating1000

gallons per month

133.74

Air Travel1000

miles per month

5.82

Train Travel1000

miles per month

2.7

General food & waste - how

many people in your

household?

1

no. of people

1.8

How many people in your

household have meat in

their diet?

1

no. of people

1.5

Total annual emissions 174.3168tons

Total monthly emissions 14.5264 tons

No. of trees to offset

per year871.58400000

trees

No. of trees to offset

per month72.632

trees

Figure 1.6 Carbon Footprint Calculator example


31/149


Table 1.5 worlds energy production, past present and future

Year Coal

EJ

Oil

EJ

Nat.

gas

EJ

RE

EJ

TOTAL

EJ

Population

millions

Energy

kW/capita

1860 3.8 EJ 0 0 0 3.8 1000 3.8

1900 20.8 0.8 0.3 0 21.9 1700 12.8

1920 35.8 3.8 0.9 0 40.5 1900 21.3

1940 42.1 11.2 3.1 0 56.4 2000 28.2

1960 60.0 40.2 17.9 10 128.1 2400 53.4

1972 66 115 46 26 253 2500 101.2

1985 115 216 77 33 441 3884 113.5

2000 170 195 143 56 564 5780 97.6

2020 259 106 125 100 590 8846 66.7


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Table 1.6 Energy consumption for various countries

Based on the US office of Science and Technology data 1961

Country GNP

$/capita

Energy

consumption

kW/capita

Energy

consumption

Relative to India

USA 1500 6 60

CANADA 1000 4.5 45

UK 750 3.8 38

FRANCE 750 2.0 20

BRASIL 225 0.8 8

INDIA 100 0.1 1


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Table 1.7 Energy needs of a typical family

Energy source Quantity and units kWh equivalent

Electricity 3400 kWh 10300

@ 30% efficiency

Natural gas 4500 m3 47500 ( 38MJ/m3)

Coal 10 Tonnes 83333 ( 30 MJ/kg)

Wood 1 tonne 3889 ( 14 MJ/kg)

Petrol 5000 Litre 55555 ( 40 MJ/Litre)

Total 200577


34/149


Table 1.8 Energy usages ( EJ ) by sector for the UK

(BP statistical review)

Sector Energy

EJ

Energy

%

Space and Process Heating 3.3 36.6

Transport 2.0 22.2

Food/ cooking 0.4 4.4

Lighting 0.3 3.3

Losses in conversion 3.0 33.3

Total 9 EJ 100%


35/149


Table 1.9 Estimated energy reserves of the world

SOURCE QUANTITY FORM

Earths daily receipt of

solar energy

1.49x1022 J Thermal /

electromagnetic

Hydropower 3x1012 J/s Mechanical

Tidal power 6.7x1010 J/s Mechanical

Geothermal 0.4x1021 J Thermal

Natural gas 11x1021 J Chemical

Petroleum 11.7x1021 J Chemical

Coal 200x10

21

J Chemical

Uranium 1800x1021 J Nuclear fission

Deuterium 6000000x1021 J Nuclear fusion


36/149


1.8 Worked Examples

Worked Example 1.1

Determine the free energy of water contained in a lake, if the lake size is 2000m3

and its water level is 300m above the axis of a water turbine.

Determine the velocity of the water in the supply pipe at the turbine inlet, if losses

due to friction in the pipe represent an equivalent of 50m energy head.

Solution:

The potential energy of the water behind the dam is calculated using equation 1.1:

Potential energy = m g h

Mass of water = volume of the lake x density of water = 2000 x 1000 = 2x106 kg

Potential Energy PE = 2x106 x 9.81 x300

= 5.886x109 J

Potential energy is converted into kinetic energy at the turbine inlet, less the energy

consumed in overcoming friction, hence:

Potential energy ( m g h ) = kinetic energy ( m x V2)

Hence V =2gh = 2x9.81x250

= 70 m/s

Note that the effective head used is 300 - 50 = 250 m to allow for losses.


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Worked Example 1.2

A car engine consumes 0.2 litre of fuel every minute; the petrol has a heating value

of 42000kJ/kg and a density of 800 kg/m3. If only 30% of the fuel is converted into

useful mechanical energy, determine the power of the car. Discuss the fate of the

wasted energy.

Solution:

Mass of fuel used = volume x density

= (0.2/1000) x 800

= 0.16 kg/minute

= 0.0026 kg/s

Chemical energy liberated from burning fuel is calculated using equation 1.3:

Useful energy = mass x calorific value x combustion efficiency

= 0.0026 x 42000000 x 0.30

= 33600 J/s (W)

= 33.6 kW

This is the useful part utilised to move the car.

The remaining energy content of fuel has been wasted as

Thermal energy carried away by the exhaust gases

Thermal energy carried away by the engine cooling system

Mechanical energy wasted as friction in various places.

Other minor forms such as sound and light generated due to excessive friction

between tyres and the road.


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Worked Example 1.3

Using Einsteins mass-energy equation ( E=mC2)

(a) Determine the energy produced when one milligram of a fissionable material

undergoes a nuclear chain reaction.

(b) What is the power rating of such a power plant if the consumption of Uranium

is 1 kg per day?

Assume speed of light C = 3x108 m/s

Solution:

(a) Using the famous Einsteins mass-energy equation, the Energy liberated

from nuclear fissioning is calculated:

E = m C2

= 10-6 x( 3x108)2

= 90x109 J

Note that the mass has to be in units of kilograms, and the speed of light in metres

per second.

(b) The power output from 1 kg, in 1 day = 24 x3600 = 86400 seconds

Power = E/time = m C2/time

= 1.0 x( 3x108)2/86400

= 1.04x109 kW


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Worked Example 1.4

(a) Calculate the thermal energy needed to raise the temperature of 20 litres of

water from 20 to 50oC.

(b) If the above process is completed in 4 minutes, what is the power of the

heater used?

Assume for water:

1 litre weights 1 kg and Specific heat capacity Cp = 4200 J/kgK

Solution:

(a) Using equation 1.5 to calculate the heat (thermal energy) required to raise

the temperature of water:

Thermal energy = Mass x specific heat x temperature difference

= 20 x 4200 x (50-20)

= 2520000 J

Note that 20 litres of water has a mass of 20 kg.

(b) By definition, power is the rate of energy consumed, hence:

Power = Energy / time

= 2520000 / (4x60)

= 10500 W or 10.5 kW


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Worked Example 1.5

Describe the energy conversion in each of the following units:

a motor car

a battery

a gas boiler

an electric fire

Solution:

Take for example the first part; a. the motor car converts fuel (chemical energy) into

torque (mechanical energy)

UNIT INPUT OUTPUT

Motor car chemical Mechanical

Battery chemical electrical

Boiler chemical thermal

Electric fire electric thermal


41/149


1.9 Tutorial Problems

1.1 Describe the energy conversion in each of the following units

an aeroplane

a push mower

a radio

Ans. ( chem/mech,mech/mech,elec/mech)

1.2 Calculate the thermal energy needed to raise the temperature of 40 litres of

water from 10 to 40 oC.

If the above process is completed in 5 minutes, what is the power of the

heater? Assume for water:

1 Litre weights 1 kg

Cp = 4200 J/kgK

Ans. (5040000J, 16.8kW)

1.3 Using Einsteins mass-energy equation ( E=mC2 ) Determine the energy

produced when a bar of Uranium of 25mm diameter, one meter long ( density 8400

kg/m3 ) undergoes a nuclear chain reaction.

Assume speed of light C = 3x108 m/s

Ans. (3.17x1017 Joules)

1.4 A car engine consumes 0.2 Litre of fuel every minute; the petrol has a

heating value of 45 MJ/kg and a density of 800 kg/m3. If only third of the fuel is

converted into useful mechanical energy, determine the power of the car.

Discuss the fate of the wasted energy.

Ans.(40 kW)

1.5 Determine the potential energy of water contained in a lake, if the lake size is

200m3 and its water level is 30m above the axis of a water turbine.

Determine the velocity of the water at the turbine inlet, if losses due to friction in the

pipe represent an equivalent of 5m energy head.

Ans.(22 m/s)


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1.10 Case Study- Energy for the World

Case Study Brief:

In this case study, the purpose is to examine the current world energy consumption

and investigate the various strategies for securing parallel energy demands from

fossil fuels.

Data: The annual world energy consumption for 1992 is shown below with the

associated CO2 release for the different fuels used.

Type of fuel World

consumption

EJ

CO2

kg/ GJ

Total CO2

x109 kg

% CO2

Oil 131 73

Coal 91 92

Natural Gas 75 50

Biomass 55 77

Hydro 24 5

Nuclear 22 12

Note: for hydro and nuclear plants, the figures correspond to construction of the

plant, and manufacture of components.

Determine

(a) The percentage contribution of CO2 released by each fuel based on the data

available for 1992.

Propose a scheme/s for reducing the total emission by 12%.

Discuss the merits of each proposal.


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Solution:

the table below demonstrate the effect of each fuel

Type of fuel World

consumption

EJ

CO2

kg / GJ

Total CO2

x109 kg

% CO2

Oil 131 73 9563 36.36

Coal 91 92 8372 31.83

Natural Gas 75 50 3750 14.25

Biomass

wood

55 77 4235 16.1

Hydro 24 5 120 0.45

Nuclear 22 12 264 1.00

Total CO2 emission = 26304 x 109 kg in year 1992


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Scheme 1 - Reduce emission of CO2 by 12 -Replace coal by Nuclear!

Type of fuel World

consumption

EJ

CO2

kg / GJ

Total CO2

x109 kg

% CO2

Oil 131 73 9563 50.26

Coal -- -- -- --

Natural Gas 75 50 3750 19.71

Biomass

wood

55 77 4235 22.26

Hydro 24 5 120 0.63

Nuclear 113 12 1356 7.13

Total = 19024 x 10

9

kg

% Reduction over 27% Great result.

Disadvantages - - - - - - - long term hazard of nuclear waste

Risk of contamination to personnel

Uncertainty in price of fuel, could be another


45/149


Scheme 2 - Reduce emission of CO2 by 12%, - Replace coal by Natural gas

Type of fuel World

consumption

EJ

CO2

kg / GJ

Total CO2

x109 kg

% CO2

Oil 131 73 9563 42.54

Coal -- -- -- --

Natural Gas 166 50 8300 36.92

Biomass

wood

55 77 4235 18.84

Hydro 24 5 120 0.53

Nuclear 22 12 264 1.17

Total = 22482 x 10

9

kg% Reduction over 14% Great result.

At what price!

OTHER POSSIBILITIES TRY:

The use of Renewable Energy such as

Hydro!; Waste to energy ( Biomass ); Wind Energy!; Solar Energy!


46/149


2 MODES OF HEAT TRANSFER

CONDUCTION, CONVECTION AND RADIATION

T1

T2

dx


47/149


2.1 Modes of Heat Transfer

Conduction, convection and radiation are the three basic mechanisms by which

heat is transferred from a hot source to cooler surroundings. Heat transfer between

two media can only take place when the two parties have a finite temperature

difference between them.

To give an example of heat transfer consider when you hold one end of a steel rod

while the other end is kept in the fire, before too long your end of the rod becomes

hot. There has been a flow of heat along the rod from the high temperature end.

This form of heat transfer through a solid material with no apparent movement of

the material is called conduction.

If you boil an egg in a pan on an electric cooker, the egg is cooked even through it

is not in direct contact with the cooker ring. This is explained as: the water at the

base of the pan. This water will move upwards as its density decreases with

heating and therefore is replaced by colder water coming down from upper layers,

and so on. This movement of the heated water results in heat being transferred

from the hot water to the egg. This mode of heat transfer is called convection.

The sun warms the earth, the heat energy transferred through vast distance in

space, mostly vacuum. This mode of transfer without involving any intervening

medium is called radiation.

Studying the different modes of heat transfer is necessary for two reasons

In order to optimise heat transfer, this may sound contradictory, but heat loss needs

to be reduced in certain applications such as buildings, while energy transfer is

encouraged in other applications such as heat exchangers eg car engine radiators.

As energy is commonly derived from fuel, saving energy such as insulating

buildings will result in a reduction of energy costs.


48/149


2.2 Fouriers Law of Thermal Conduction

Conduction heat transfer takes place through a solid medium of finite thickness and

surface area. Consider a solid material which is heated on one side while the

opposite side is subject to the ambient. The heated side is at a higher temperature

than the ambient, heat energises the molecules adjacent to the source of heat,

hence there will be a flow of energy from the hot surface to the cool surface, and

this is how thermal energy propagates by conduction.

Fourier found that the rate of heat transfer by conduction is proportional to:

the area of cross-section, A

the temperature difference, dT

the distance, dx

i.e. Q (A ,dx

dT) (2.1)

The constant of proportionally is known as the thermal conductivity of the material

through which the heat transfer is taking place

i.e. Q = - k A12

21

12

12

dx

dT

xx

TTAk

xx

TTAk

+=

= (2.2)

The minus sign appears due to the fact that the temperature is decreasing with

increased distance from the hot surface.

Figure 2.1 Conduction Heat Transfer

Q

Heat-flow

direction

T1

T2

dx


49/149


2.3 Film heat transfer between a solid and a fluid

Consider the situation where a heated solid surface is kept at a temperature higher

than the surrounding air or fluid. This situation is very common, for example

domestic central heating "radiators". In order to analyse such situation, examine the

situation described in Figure 2.2, imagine the conduction in the solid layer to

continue into the adjacent fluid layer ( gas, or liquid ); if it can be assumed that the

law of heat conduction to apply, then

Q = - k A

fs TT (2.3)

In the fluid thin film or layer of thickness the ratio of thermal conductivity and the

film thickness is better known as the convection heat transfer coefficient

Hence:

Q = h A (Ts Tf) (2.4)

Figure 2.2 Film convection concept

Ts

Tf


50/149


2.4 Heat transfer through a composite wall separating two fluids

In buildings where there is a finite difference between the inside and outside air

temperatures, the situation in terms of heat transfer constitute a mixture of

conduction through the solid envelop and convection on either sides of the walls,

roof, etc. Heat transfer between two fluids kept at Ta and Tb are separated by solid

layer/s "x distance/s apart, consists of two modes; convection from fluid a to the

solid boundary layer/s then subsequently convected to the second fluid.

A

Q= ha (Ta T1) convection from air to face a

=1

211

x

)T(Tk conduction across layer X1

=2

322

x


=3

433

x


= hb (T4 Tb) convection from air to face b

Since the heat flux ( Q/A ) is constant, then rewriting the five equations in terms of

temperatures and adding them will result in :

Ta Tb = A

Q

]h

1

k

x

k

x

k

x

h

1

[ b3

3

2

2

1

1

a ++++

A

Q=

)h

1

k

x

k

x

k

x

h

1(

)T(T

b3

3

2

2

1

1

a

ba

++++

(2.5)

or Q = A.U. (Ta Tb) (2.6)

Figure 2.3 Conduction through multi-layers

U is the Overall heat transfer coefficient:

]h

1

k

x

k

x

k

x

h

1[

1

b3

3

2

2

1

1

a

++++=U

X1 X2 X3

k1 k2 k3

TT

T

TT

Tha h


51/149


2.5 Radial conduction through pipe wall

Consider the flow of hot water through pipes in central heating network. The radial

conduction heat flow through concentric layers of the pipe must be constant but

since the surface area normal to flow increases with radius it follows that dT/dx must

decrease with radius. The temperature profile is shown in Figure 2.4.

At any radius r , the radial heat conduction is given by Q = -k Adx

dT

where the heat flow direction is along r , and the conduction area is A =2 r L

dr

dTkL2- rQ

= (2.5)

Integrating between the inner and out surfaces of the pipe, we have

=

2

1

2

1

r

dr

dTkL2- rQ

; The result of the integration yields:

)(

)(2

1

2

21

r

rn

TTLkQ

l

= (2.6)

Figure 2.4 Radial conduction

L

r1r2

T1T2


52/149


2.6 Heat exchange though a tube with convection on both sides

Consider a pipe carrying a fluid (liquid or gas) which is kept at a temperature higher

than the ambient. This situation encompasses radial conduction through the pipe

material, as well as convection on either side of the pipes surface.

With reference to Figure 2.5, the heat transfer through a pipe separating two fluids

(i.e. conduction and convection on inside and outside layer) is given by:

( )

( )

( )

b21

12

a1

ba

22

11

12

12

Lh2

1

k2

)/rn(r

Lh2

1

TT

.

.

)/

)(2

rLrr

Q

rewriting

TThAQ

TThAQ

rrn

TTkLQ

bb

aa

++

=

=

=

=

l

l

The heat transfer can be written in relation to the inside area as follows:

b

21121

a

bai

h

)/(r

k

)/rn(r*

h

1

)T(T*A

rrQ

++

=

l(2.7)

Figure 2.5 Combined convection-conduction-convection heat Transfer

T

T2

Tb

hb

Ta

k

r r2


53/149


2.7 Composite tube with fluid on inner and outer surfaces

Composite tube geometries are found in situations when a relatively hot or cold fluid

passes through an ambient with markedly different temperature. In other words,

composite tubes are required in situations where insulation is beneficial in order to

save energy. Consider Figure 2.6., heat transfer consist of four components,

convection on either side of the solid surfaces, and conduction through the pipe

material and the insulation material, hence

Q = 2 L r1 ha (Ta T0) inside film

= 2 L k1 (T0 T1)/ l n(r1/r0) first layer conduction

= 2 L k1 (T1 T2)/ l n(r2/r1) second layer conduction

= 2 Lk2 (T2 T3)/ l n(r3/r2) third layer conduction

= 2 L r3 hb (T3 Tb) outside film

Rearrange the above equations to get:

b33

23

2

12

1

01

a0

ba

Lh2

1

Lk2

)/rn(r

Lk2

)/rn(r

Lk2

)/rn(r

Lh2

1

TT

rr

Q

++++

=

lll(2.8)

T3

T1

T2

T0

Tar0

Tb

r1

r3

r2

Figure 2.6 Radial conduction through multi-layers


54/149


Table 2.1 Thermal Conductivity of Some Common Materials

Substance Thermal Conductivity

W/m K at 300 K

Air, still at 15oC 0.025

Aluminium, pure 237

Alloy 2024 T6 177

Alloy 195 (4.5%Cu) 168

Brick 0.6

Concrete 0.85

Copper Pure 401

Bronze 52

Brass 110

Cork 0.05

Felt 0.04

Glass 1.0

Glass fibre 0.04

Iron, pure 80.2

Steel

Carbon, steel A151 1010 63.9Carbon silicon 51.9

Carbon manganese silicon 41.0

Chromium steel 37.7

Stainless steel

A151 302

A151 304

A151 316

A151 347

15.1

14.9

13.4

14.2

Magnesium 156.0

Molybdenum 138

Nickel, pure 90.7

Wood 0.15

Zinc 116

Zirconium 22.7


55/149


2.8 Heat transfer by convection

This type of heat transfer is caused by the motion of a fluid past a solid boundary.

The heat loss from the solid surface to the fluid or vice versa is mainly by

convection. In general, the rate of heat transfer by convection is governed by

NEWTONS EQUATION:

Q = h . A . dT (2.9)

Where h is the convection heat transfer coefficient (units of W/m2 K)

A Surface area in contact with the fluid

dT Temperature difference between the solid surface and the

surrounding fluid.

There are two types of convection heat transfer:

Free or Natural convection. Example: a cup of hot coffee left in a room in the

absence of draughts. i.e. due to density gradients.

Forced or artificial convection. Example: a cup of hot coffee left in a room under a

ceiling fan.

The convection heat transfer coefficient is an empirical quantity determined by

experiments and dimensional analysis and is dependent upon the geometry and

properties of the situation.

A list of convection relationships is given below.

The procedure to determine h is to identify whether it is A free convection or B

forced convection; identify the geometry; determine whether it is laminar or turbulent

situation and hence an equation for the Nusselt number is identified. Since Nusselt

number (Nu) is a function of the heat transfer coefficient (h) as shown below, hence

h is calculated and the heat transfer by convection can be calculated.


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Convective heat transfer: experimental correlations

Nusselt No: Nu =k

hl

Prandtl No: Pr =

k

Cp

Reynolds No: Re =

Vl

Grashof No: Gr =2

23

Tgl

Where h = convective heat transfer coefficient

l = Characteristic dimension of a surface

T = surface temp. (Ts) fluid temp. (Tf)

g = gravitational acceleration = 9.81 m/s2

V = mean fluid velocity

Fluid properties are taken at the mean film temperature: Tm = (Tf+ Ts)/2.

k = thermal conductivity,

= dynamic viscosity,

= density,

Cp = specific heat at constant pressure

= 1/Tf coefficient of cubic expression for air

Tfshould be in degrees Kelvin


57/149


Free convection

Vertical flat plates and cylinders ( l - vertical length of surface):

(i) Laminar range 104 < (Gr Pr) < 109, Nu = 0.59 (Gr Pr)0.25

(ii) Turbulent range 109 < (Gr Pr) < 1012, Nu = 0.13 (Gr Pr)0.33

Outside horizontal cylinders ( l - outside diameter of the cylinder):

(i) Laminar range 104 < (Gr Pr) < 109, Nu = 0.53 (Gr Pr)0.25

(ii) Turbulent range 109 < (Gr Pr) < 1012, Nu = 0.13 (Gr Pr)0.33

Horizontal flat plate ( l - length of longer side)

Hot plate facing upwards or cold facing downwards

Laminar range 105 < (Gr Pr) < 2 x 107, Nu = 0.54 (Gr Pr)0.25

Turbulent 2 x 107 < (Gr Pr) < 3 x 1010, Nu = 0.14 (Gr Pr)0.33


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Forced convection

Flow along flat plates ( l = length of plate in the direction of flow):

Laminar range Re < 105, Nu = 0.664 Re0.5 Pr0.33

Correlation is suitable for 0.6 < Pr < 10 and heat is transferred over the whole of the

plate.

Turbulent range Re > 105, Nu = 0.037 Re0.8 Pr0.33

Correlation is suitable for Pr > 0.6 and heat is transferred over the whole of the

plate.

Flow inside tubes ( l = inside diameter of the tube):

Laminar flow Re < 2500, Nu = 4.1

Fully developed flow, heating or cooling.

Turbulent flow Re= > 2500, Nu = 0.023 Re0.8 Prn

Suitable for 0.7 < Pr < 100

Where n = 0.4 if fluid is being heated

n = 0.3 if fluid is being cooled

External flow over cylinders ( l - cylinder outside diameter)

Across isolated cylinders and tubes:

1 < Re < 4000, Nu = 0.43 + 0.53 Re0.50 Pr0.31

4000 < Re < 40000, Nu = 0.19 Re0.62 Pr0.31

40000 < Re < 400000, Nu = 0.027 Re0.81 Pr0.30

Across banks of cylinders or tubes

2000 < Re < 32000 Nu = b Re0.6 Pr0.33

Where b = 0.33 for staggered tubes and b = 0.26 for tubes in line.


59/149


Table 2.2 Properties of air at 1.01325 bar

Temperature

[K]

T

specific heat

K][J/kg

Cp

dynamic

viscosity

s][kg/m10

5

thermal

conductivity

K][W/m10

k2

Density

][kg/m

3

175

200

225

250

275

300

325

350

375

400

450

500

550

600

650

700

750

800

850

900

950

1000

1002.3

1002.5

1002.7

1003.1

1003.8

1004.9

1006.3

1008.2

1010.6

1013.5

1020.6

1029.5

1039.8

1051.1

1062.9

1075.0

1087.0

1098.7

1110.1

1120.1

1131.3

1141.1

1.182

1.329

1.467

1.599

1.725

1.846

1.962

2.075

2.181

2.286

2.485

2.670

2.849

3.017

3.178

3.332

3.482

3.624

3.763

3.897

4.026

4.153

1.593

1.809

2.020

2.227

2.428

2.624

2.816

3.003

3.186

3.365

3.710

4.041

4.357

4.661

4.954

5.236

5.509

5.774

6.030

6.276

6.520

6.754

2.017

1.765

1.569

1.412

1.284

1.177

1.086

1.009

0.9413

0.8824

0.7844

0.7060

0.6418

0.5883

0.5430

0.5043

0.4706

0.4412

0.4153

0.3922

0.3716

0.3530


60/149


Table 2.3 Properties of typical engine oil

Temperature

[C]

T

Specific Heat

K][J/kg

Cp

Kinematic

viscosity

s/m 2

x10-3

Thermal

conductivity

KW/m

k

Density

][kg/m

3

0

20

40

60

80

100

120

140

160

1796

1880

1964

2047

2131

2219

2307

2395

2483

4.28

0.900

0.200

0.084

0.037

0.020

0.012

0.008

0.006

0.147

0.145

0.144

0.140

0.138

0.137

0.135

0.133

0.132

899

888

876

864

852

840

829

817

806


61/149


Table 2.4 Properties of saturated water and steam

Note : Subscripts w Water; s Steam

Temp

C][

To

Pressur

e

[bar]

P

Specific

volume

/kg][m10

v

32

w

Specific

heat

K][J/kg

CC pspw

Dynamic

viscosity

s][kg/m10

6

sw

Thermal

conductivity

K][W/m10

kk

3

sw

0.01

5

10

15

20

25

30

35

40

45

50

55

60

65

70

75

80

85

90

95

100

105

110

115

120

125

130

135

140

0.006112

0.008719

0.01227

0.01704

0.02337

0.03166

0.04242

0.05622

0.07375

0.09582

0.1233

0.1574

0.1992

0.2501

0.3116

0.3855

0.4736

0.5780

0.7011

0.8453

1.01325

1.208

1.433

1.691

1.985

2.321

2.701

3.131

3.614

0.10002

0.10001

0.10003

0.10010

0.10018

0.10030

0.10044

0.10060

0.10079

0.10099

0.1012

0.1015

0.1017

0.1020

0.1023

0.1026

0.1029

0.1032

0.1036

0.1040

0.1044

0.1048

0.1052

0.1056

0.1060

0.1065

0.1070

0.1075

0.1080

4210 1860

4204 1860

4193 1860

4186 1870

4183 1870

4181 1880

4179 1880

4178 1880

4179 1890

4181 1890

4182 1900

4183 1900

4185 1910

4188 1920

4191 1930

4194 1940

4198 1950

4203 1960

4208 1970

4213 1990

4219 2010

4226 2030

4233 2050

4240 2070

4248 2090

4260 2120

4270 2150

4280 2180

4290 2210

1752 8.40

1501 8.66

1300 8.83

1136 9.00

1002 9.18

890 9.35

797 9.52

718 9.70

651 9.87

594 10.0

544 10.2

501 10.4

463 10.6

430 10.7

400 10.9

374 11.1

351 11.3

330 11.4

311 11.6

294 11.8

279 12.0

265 12.2

252 12.4

241 12.6

230 12.8

220 13.0

211 13.2

203 13.4

195 13.5

569 16.3

578 16.7

587 17.1

595 17.5

603 17.9

611 18.3

618 18.7

625 19.1

632 19.5

638 19.9

643 20.4

648 20.8

653 21.2

658 21.6

662 22.0

666 22.5

670 22.9

673 23.3

676 23.8

678 24.3

681 24.8

683 25.3

684 25.8

686 26.3

687 26.8

687 27.3

688 27.8

688 28.3

688 28.8


62/149


145

150

160

170

180

190

200

210

220

230

240

250

260

270

280

290

300

4.155

4.760

6.181

7.920

10.03

12.55

15.55

19.08

23.20

27.98

33.48

39.78

46.94

55.05

64.19

74.45

85.92

0.1085

0.1091

0.1102

0.1114

0.1128

0.1142

0.1157

0.1173

0.1190

0.1209

0.1229

0.1251

0.1276

0.1302

0.1332

0.1366

0.1404

4300 2250

4320 2290

4350 2380

4380 2490

4420 2620

4460 2760

4510 2910

4560 3070

4630 3250

4700 3450

4780 3680

4870 3940

4980 4220

5100 4550

5240 4980

5420 5460

5650 6180

188 13.7

181 13.9

169 14.2

159 14.6

149 15.0

141 15.3

134 15.7

127 16.0

121 16.3

116 16.7

111 17.1

107 17.5

103 17.9

99 18.3

96 18.8

93 19.3

90 19.8

687 29.4

687 30.0

684 31.3

681 32.6

676 34.1

671 35.7

665 37.5

657 39.4

648 41.5

639 43.9

628 46.5

616 49.5

603 52.8

589 56.6

574 61.0

558 66.0

541 72.0


63/149


2.9 Heat transfer by radiation

All bodies radiate thermal energy. If two bodies at different temperature and the

space between them is unoccupied, the hotter body will emit radiation and the

colder body will absorb part of this radiant energy. Thermal radiation like light,

travels at the same speed through a vacuum and this is the secret of our heat

energy received from the sun.

The net outcome of heat exchange by radiation between two bodies is given by

Stefan-Boltz equation.

Qr= e F12 A1 (T14 T24) (2.10)

The absolute temperatures (K) for hot surface (T1), and ambient air (T2)

A1 is the surface are of emitter.

= 5.67 x 10-8 W/m2 K4.

F12 is the shape factor, for small size surfaces emitting heat in a large space F12 = 1.

e is the emissivity for a hot surface radiating to atmosphere, If two surfaces (with

emissivity values e1 and e2) are exchanging radiant heat then effective emissivity is

calculated as:

e =

111

1

21

+ee

(2.11)

Thermal energy falling onto a surface may be absorbed, reflected and/or

transmitted depending on the nature of the surface. Some gases transmit nearly all

the radiant heat. A body that absorbs all the impinging radiant heat is called a

black body, a hypothetical conception in which the absorptivity is unity. The

nearest approach to a black body is obtained by a hollow vessel penetrated only by

a small pin hole through which radiant heat may pass to the inside; once inside, little

of this radiation is reflected back through the pin hole.

Black body has nothing to do with the colour, and a painted white surface may

absorb about the same total radiation as a painted black surface. However,

surfaces have a certain selectivity regarding reflection; over visible wave lengths, a

white painted surface reflects a large part, a black surface absorbs a large part of

the incident energy and so it is considered as a good heat sink

Since absorptivity is a surface property, accumulation of dust, corrosion etc, may

have a drastic effects on heat transfer rates by radiation.

Brightly polished metals are such good reflectors that most of the radiant heat may

be reflected.


64/149


Table 2.5 Emissivity of some materials

SURFACE AND CONDITION TEMPERATURE/OC EMISSIVITY

Aluminium

Unoxidised

Oxidised

Anodised

Brass

Unoxidised

Oxidised

Brick

Chromium

Unoxidised

Oxidised

Copper

Polished

Oxidised

Gold

Glass

Iron

Unoxidised

Oxidised

Cast, Unoxidised

Cast, Oxidised

Wrought, Dull

Wrought, Oxidised

Lacquer

White

Matt Black

Lead

Grey, Oxidised

Grey, Unoxidised

Red

Nickel

Unoxidised

0 100

0 200

50

25 100

0 600

20

100

70

0 250

0 250

0 250

20 90

100

100

100

200

25

350

20

80

20

100

100

100

0.03

0.10

0.72

0.04

0.60

0.93

0.08

0.91

0.02

0.60

0.02

0.88

0.05

0.74

0.21

0.64

0.94

0.94

0.80

0.97

0.28

0.05

0.93

0.06


65/149


Oxidised

Paint

Paper

Quartz

Silver

Steel

Polished

Oxidised

Tungsten

Unoxidised

Filament

Wood (Beech)

Zinc

Unoxidised

Grey, Oxidised

Cast

Galvanising

200

30

95

30 230

30 230

30 200

30 530

30

30 500

70

300

20

20

20

0.37

0.95

0.92

0.90

0.02

0.08

0.79

0.02

0.32

0.94

0.05

0.25

0.05

0.23


66/149


2.10 Worked Example

Worked Example 2.1

A composite wall is made up of external brickwork, a layer of fibre-glass thick. The

fibreglass is faced internally by an insulating board thick.

The coefficient of thermal conductivity and thickness for these materials are as

follows:

Brick Work 0.60 W/m K 110 mm

Fibre-Glass 0.04 W/m K 75 mm

Insulating Board 0.06 W/m K 25 mm

The surface heat transfer coefficient on the inside wall is 2.0 W/m2 K while that on

the outside wall is 3.0 W/m2 K. Determine the overall heat transfer coefficient for the

wall and the heat lost through such wall 6 m high and 10 m long. Take the internal

ambient temperature as 20oC and the external ambient temperature as 10oC.

Solution:

The overall heat transfer coefficient

U =

)

h

1

k

x

k

x

k

x

h

1(

1

o3

3

2

2

1

1

i

++++

=

)0.3

1

6.0

110.0

04.0

075.0

06.0

025.0

0.2

1(

1

++++

KmW2

/302.0

333.0183.0875.1417.05.0

1

=

++++=

The Heat Transfer Rate

Q = U A (Ti To)

= 0.302 x 6 x 10 x (20 10) = 181 W

X1 X2 X3

k1 k2 k3

TT

T

T T

Tha h


67/149


Worked Example 2.2

A cold store is used to keep food products at a temperature of -5oC while the

ambient temperature is 15oC. The convective heat transfer coefficients on the inside

and outside of the store are 10 and 20 W/m2K respectively. Thermal conductivity of

the wall is 0.8 W/mK and its thickness is 40cm.

(a) Calculate the heat transfer rate across the store, assuming its dimensions

are 5x3x2m.

(b) It is desired to save energy by adding an insulating material on the inside; the

insulator has a thermal conductivity of 0.05 W/mK. Determine the required thickness

of insulation required to provide a 20% reduction in the heat transfer rate.

Solution:

The Overall h.t.c.

KmW

hok

xU 2/538.1

20

1

8.0

4.0

10

1

1

1

hi

1

1 =++

=++

=

Surface Area of the store = 2*[5x3 + 5x2 + 3x2] = 62 m2

The heat transfer rate is Q = A U (Ti To)

= 62x 1.538 ( 15 - (-5))

= 1908 W

(b) The New heat transfer = 0.8 *Q1 = 1526 W

mmx

x

hok

x

k

xQ

inswall

new

8

20

1

05.08.0

4.0

10

1

20*62

1526

1][][

hi

1

dTA.

=

+++=

+++=


68/149


Worked Example 2.3

A steel pipe carries steam at 320 oC has internal diameter 24mm, 3 mm thick, and

wrapped with 10mm thick insulation. The thermal conductivity of the pipe material is

50, and the insulating material is 0.03 W/mK. The inside and outside convective

heat transfer coefficients are 100 and 10W/m2K. The outside air temperature is

20oC. Calculate:

the Heat transfer per meter length from the Un-insulated pipe

the heat transfer per meter length from the insulated pipe

Solution

i. The un-insulated pipe:

KmW

Ui

2

o

i

o1

12i

i

/11

10*015.0

01.0

50

)12/15ln(*012.0

100

1

1

r

r.

h

1

k

)/rn(rr

h

1

1

=++

=

+

+=

l

Q = A.U.dT = (2*0.015*1)*11*(320-20) =311 W/m

ii. The insulated pipe:

KmW

k

rrrUi

2

o

1

o2

231

1

121

i

/8.3

10*025.0

01.0

03.0

)15/25ln(*012.0

50

)12/15ln(*012.0

100

11

r

r.

h

1)/ln(*

k

)/rn(rr

h

1

1

=+++

=

++

+=

l

Q = A.U.T = (2*0.015*1)*3.8*(320-20) = 107 W/m

Quite a big difference, only one-third of the original heat loss, or a saving of 65%


69/149


Worked Example 2.4

(a) Estimate the film coefficients for flow of a liquid of density 800 kg/m,

viscosity 0.0008 kg/m/s and thermal conductivity 0.2 W/m/K and specific heat

capacity of 2000 J/kgK, which flows inside a tube at a mean velocity of 0.5 m/s:

tube internal diameter is 40mm

tube external diameter is 44mm.

(b) Estimate the overall heat transfer coefficient if heat is to be transferred

between these two fluid streams.

The following empirical relations are available:

Flow inside tubes

Laminar flow Re < 2500 Nu = 4.1

Turbulent flow Re => 2500 Nu = 0.023 Re0.8 Pr0.3

External flow over cylinders and tubes:

1 < Re < 4000 Nu = 0.43 + 0.53 Re0.50 Pr0.31

4000 < Re < 40000 Nu = 0.19 Re0.62 Pr0.31

Solution:

Inside flow:

82.0

2000*0008.0*Pr ===

k

Cp

000,200008.0

04.0*5.0*800D*V*Re ===

Turbulent flow

Nu = 0.023 Re0.8 Pr0.3 = 0.023*(20,000)0.8 *(8)0.3 = 118

hi = Nu.k/D = 118 x 0.2 / 0.04 = 590 W/m2/K

Outside flow:

000,220008.0

044.0*5.0*800D*V*Re ===

Turbulent flow

Nu = 0.19 Re0.62 Pr0.31= 0.19*(22000)0.62*(8)0.31=178

ho = Nu.k/D = 178 x 0.2 / 0.044 = 809 W/m2/K

(b) It will be reasonable to neglect wall resistance, so

U =1/[ 1/hi + 1/ho ]= 1/[1/590 + 1/809] = 341 W/m2/K.


70/149


Worked Example 2.5

Central heating panel (have emissivity of 0.85) is placed by a brick wall (have

emissivity of 0.93). The surface temperature of the heating panel is kept at 80 oC

while the brick surface is 20 oC.

(a) Determine the radiation heat exchange per unit area between the two

surfaces

(b) In order to reduce the radiation between the two surfaces, a sheet of

aluminium is placed parallel between the two surfaces. If Aluminium has emissivity

of 0.05, determine the percentage reduction in radiation.

Solution

(a) No foil, with A1 =1.0 = 5.67 x 10-8 W/m2 K4. F12 = 1.

799.0

193.0

1

85.0

1

1

111

1

21

=+

=+

=

ee

e

Hence Qr = e F12 A1 (T14 T24)

= 0.799*5.67*10-8*1*1*[(80+273)4-(20+273)4]

= 369.5 W/m2

(b) When the foil is used, the exchange of radiation is between the central heating

panel and the foil. Hence:

049.0

105.0

1

85.0

1

1

111

1

1

=+

=+

=

foilee

e

Hence

Qr = e F12 A1 (T14 T24)

= 0.049*5.67*10-8*1*1*[(80+273)4-(20+273)4]

= 22.9 W/m2

Hence the % reduction = 100*(369.5 -22.9)/369.5 = 94 %


71/149


Worked Example 2.6

An electric heater consists of a single horizontal bar 25 mm diameter and 0.3 m

long is used to maintain a room temperature at 17oC. If the heater surface

temperature is 537oC, calculate:-

(a) The convective heat transfer.

(b) The radiative heat transfer, if that the heater surface has an emissivity of 0.6.

(c) The total heat transfer.

Solution:

(a) To determine the convection heat transfer, follow the steps outlined earlier:

Step 1: determine mean temperature Tm =2

53717+= 277 + 273 = 550 K

Step 2: at this temperature find properties of air

Cp = 1.0398 kJ/kg K, = 2.849 x 10-5 kg/ms,

k = 4.357 x 10-5 kW/mK = 0.6418 kg/m3

Step 3: the situation is a free convection, hence calculate Gr & Pr

= 1/ Ta = 1/(273+17) = 3.448x10-3

Gr =25

332

2

32

)10849.2(

)17537(81.910448.3025.06418.0

gTd

=x

= 139478

Pr = 2

5

104.357

1039.8102.849

k

Cp

= = 0.68

Step 4: check Gr Pr = 139478 x 0.68 = 94845 < 109

Nu = 0.53 (Gr Pr)0.25 =0.53 (94845 < 109) 0.25 = 9.3

Nu =k

dh h = 9.3 x 4.357 x 10-2/0.025 = 16.21 W/m2K

Qconv = h A T = 16.21 x x 0.025 x 0.3 (537 17) = 198.6 W

(b) the radiative heat transfer is

msc energy notes

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