Module 2

Part 66 Syllabus Module 2 – Physics For B1 & B2 Categories

Part 66 Training Syllabus Module 2 Physics Table of Contents Matter.................................................................................................................................................................. 5 The Nature of Matter ....................................................................................................................................... 5 The Components of Atoms ............................................................................................................................. 5 Periodic Table of the Elements ....................................................................................................................... 6 Chemical Definitions ....................................................................................................................................... 6 Elements ..................................................................................................................................................... 6 Mixtures ....................................................................................................................................................... 6 Compounds ................................................................................................................................................. 7 Atomic Number............................................................................................................................................ 7 Mass Number .............................................................................................................................................. 7 Molecules .................................................................................................................................................... 7 Isotopes ....................................................................................................................................................... 7 lonization ..................................................................................................................................................... 8 The Electronic Structure of Atoms .................................................................................................................. 8 Valency ......................................................................................................................................................10 Chemical Bonding .........................................................................................................................................13 Cohesion and Adhesion ............................................................................................................................13 Noble Gases ..............................................................................................................................................13 Covalent and Ionic Bonding ......................................................................................................................13

Covalent Bonding .................................................................................................................................13 The Properties of Small Covalent Molecules ............................................................................................15 Large Covalent Molecules and their Properties ........................................................................................15 Ionic Bonding ................................................................................................................................................15 The properties of Ionic Compounds ..............................................................................................................17 States of Matter .............................................................................................................................................17 Solids .........................................................................................................................................................17 Liquids .......................................................................................................................................................18 Gases ........................................................................................................................................................18 Plasma ......................................................................................................................................................18 Changes between States ..............................................................................................................................18 Mass .................................................................................................................................................................19 Force.................................................................................................................................................................19 Weight...............................................................................................................................................................19 Distinction between Mass and Weight .............................................................................................................19 Stress, Strain and Hooke's Law .......................................................................................................................22 Introduction .......................................................................................................................................................22 Types of Structural Stress ................................................................................................................................22 Stress, Strain and Young's Modulus ................................................................................................................23 Related Definitions ...........................................................................................................................................25 Bulk Modulus ....................................................................................................................................................25 Poisson's Ratio .................................................................................................................................................25 Cantilever..........................................................................................................................................................25 Materials Behaviour ..........................................................................................................................................26 Elastic ...............................................................................................................................................................26 Brittle.................................................................................................................................................................26 Ductile ...............................................................................................................................................................26 Viscous .............................................................................................................................................................26 Nature and Properties of Solids, Liquids and Gas ...........................................................................................28 Solid ..................................................................................................................................................................28 Liquid ................................................................................................................................................................28 Gas ...................................................................................................................................................................29 Changes of State ..............................................................................................................................................29 Evaporation and Boiling (liquid to gas) .............................................................................................................29 Condensing (gas to liquid) ................................................................................................................................30 Melting (solid to liquid) ......................................................................................................................................30 Freezing (liquid to solid) ...................................................................................................................................30 Summary ..........................................................................................................................................................31 Pressure and Force ..........................................................................................................................................31 Computing Force, Pressure, and Area .............................................................................................................31 Atmospheric Pressure ......................................................................................................................................31 Transmission of Forces Through Liquids ......................................................................................................32 Pressure and Force in Fluid Power Systems................................................................................................34 Part 66 Module 2 Page 1

Part 66 Training Syllabus Module 2 Physics The Hydraulic Ram Principle........................................................................................................................ 35 Differential Areas ...................................................................................................................................... 36 Volume and Distance Factors .................................................................................................................. 37 Relationship between Force, Pressure, and Head ...................................................................................... 37 Static and Dynamic Factors ......................................................................................................................... 37 Operation of Hydraulic Components ............................................................................................................ 38 Hydraulic Jack .............................................................................................................................................. 38 Hydraulic Brakes ...................................................................................................................................... 39 Accumulators ............................................................................................................................................ 40 Barometers ................................................................................................................................................... 42 Mercury barometers ................................................................................................................................. 42 Aneroid barometers .................................................................................................................................. 43 Buoyancy...................................................................................................................................................... 43 Archimedes Principle ................................................................................................................................ 43 Archimedes' Principle Applied to Bodies that Float .................................................................................. 44 Archimedes' Principles as Applied to Airships and Balloons ................................................................... 45 Kinetics ......................................................................................................................................................... 48 Linear Motion ............................................................................................................................................ 48 The Equations of Motion........................................................................................................................... 48 Accelerated Motion of a "Freely Falling" Body ............................................................................................. 50 Rotational Motion ......................................................................................................................................... 53 Degrees and Radians .................................................................................................................................. 53 Periodic Motion............................................................................................................................................. 58 Mass and Spring .......................................................................................................................................... 59 Simple Harmonic Motion (SHM)................................................................................................................... 59 Properties of SHM ........................................................................................................................................ 60 Vibration ....................................................................................................................................................... 60 Types of vibration ......................................................................................................................................... 61 Resonance ................................................................................................................................................... 61 Examples of Resonance .............................................................................................................................. 61 What Causes Resonance? .......................................................................................................................... 61 Design Implications of Resonance ............................................................................................................... 62 Harmonics .................................................................................................................................................... 62 Simple Machines and the Principle of Work ................................................................................................ 63 General Theory of All Machines ................................................................................................................... 63 The Lever ..................................................................................................................................................... 65 The Wheel and Axle ..................................................................................................................................... 67 The Screw Jack ............................................................................................................................................ 67 Dynamics...................................................................................................................................................... 70 Newton's First Law ....................................................................................................................................... 70 Newton's Second Law .................................................................................................................................. 70 Newton's Third Law ...................................................................................................................................... 72 Every action has an equal and opposite reaction ........................................................................................ 72 Motion in a Circle ......................................................................................................................................... 76 Units of Force ............................................................................................................................................... 77 Friction .......................................................................................................................................................... 79 Static Friction................................................................................................................................................ 79 Rolling Friction.............................................................................................................................................. 79 Kinetic Friction .............................................................................................................................................. 79 Calculating Friction ....................................................................................................................................... 80 Work, Energy and Power ............................................................................................................................. 81 Gravitational Potential Energy...................................................................................................................... 82 Power ........................................................................................................................................................... 83 Alternate Form for Power ............................................................................................................................. 84 Momentum ................................................................................................................................................... 86 Momentum - mV ........................................................................................................................................... 86 Conservation of Momentum ......................................................................................................................... 86 Recoil Problems ........................................................................................................................................... 86 Collision Problems ....................................................................................................................................... 87 Inelastic Collisions ........................................................................................................................................ 87 Elastic Collisions .......................................................................................................................................... 88 Torque .......................................................................................................................................................... 92 Extensions .................................................................................................................................................... 93 Couples ........................................................................................................................................................ 94 Part 66 Module 2 Page 2

Part 66 Training Syllabus Module 2 Physics The Gyroscope .............................................................................................................................................94 Apparent Drift (or Wander)............................................................................................................................95 Transport Drift (or Wander) ...........................................................................................................................96 Fluid Dynamics .............................................................................................................................................97 Cabin Altitude ................................................................................................................................................98 Humidity ......................................................................................................................................................101 Definitions ...................................................................................................................................................101 Density and Specific Gravity .......................................................................................................................102 Specific Gravity ...........................................................................................................................................104 Compressibility in Fluids .............................................................................................................................104 Viscosity ......................................................................................................................................................105 Viscosity Measurement ...............................................................................................................................105 Units of Measure .........................................................................................................................................106 Kinematic viscosity......................................................................................................................................106 Viscosity of air .............................................................................................................................................106 Viscosity of water ........................................................................................................................................106 Drag and Streamlining ................................................................................................................................106 Stokes's Drag ..............................................................................................................................................107 Viscous resistance = - bv ............................................................................................................................107 Drag Coefficient ..........................................................................................................................................107 Streamlining ................................................................................................................................................108 Bernoulli's Principle .....................................................................................................................................109 The Venturi Tube ........................................................................................................................................111 Application of Bernoulli's Principle to Aerofoil Sections ..............................................................................112 Temperature ...............................................................................................................................................115 The Gas Laws .............................................................................................................................................116 Boyle's Law .................................................................................................................................................117 Charles' Law ...............................................................................................................................................117 Gay-Lussac's Law .......................................................................................................................................118 The General (Ideal) Gas Law......................................................................................................................119 Alternate Form of the General (Ideal) Gas Law ..........................................................................................120 Application of the General Gas Law to Compressors .................................................................................121 Thermal Expansion .....................................................................................................................................122 Linear Expansion ........................................................................................................................................123 Area Expansion ...........................................................................................................................................123 Volume Expansion ......................................................................................................................................124 Expansion of Liquids and Gases ................................................................................................................124 The Interesting Case of Water ....................................................................................................................124 Heat.............................................................................................................................................................126 Heat Exchange ...........................................................................................................................................129 Latent Heat of Fusion and Vaporisation .....................................................................................................130 Further Discussion on Latent Heat .............................................................................................................131 Heat Transfer ..............................................................................................................................................133 Refrigeration and Heat Pumps ...................................................................................................................134 How do things get colder? ..........................................................................................................................134 The States of Matter ...................................................................................................................................135 The Magic of Latent Heat............................................................................................................................135 So what is a refrigerant? .............................................................................................................................136 So is water a refrigerant? ............................................................................................................................137 Heat Pumps ................................................................................................................................................138 Thermodynamics and the 1st and 2nd Laws ..............................................................................................139 Boyle's Law .................................................................................................................................................139 Charles' Law ...............................................................................................................................................139 Gay Lussac's Law .......................................................................................................................................139 An Adiabatic Process ..................................................................................................................................139 Thermodynamic Work .................................................................................................................................140 Work = PV ...................................................................................................................................................141 Internal Energy ............................................................................................................................................141 Enthalpy ......................................................................................................................................................141 The First Law of Thermodynamics .............................................................................................................141 The Second Law of Thermodynamics ........................................................................................................142 Wavelength, Frequency and Speed ...........................................................................................................144 The Speed of Light in Various Substances.................................................................................................146 Light Waves in Matter .................................................................................................................................147 Part 66 Module 2 Page 3

Part 66 Training Syllabus Module 2 Physics Refraction and the speed of light waves .................................................................................................... 147 Refractive Index Varies with Wavelengthe speed of light in a given transparent medium is also likely to vary with frequency - the refractive index is different for different frequencies. The diagram below shows how the refractive index of fused quartz and crown glass varies with the vacuum wavelength of radiation between short ultraviolet wavelengths ( ~200nm) and near infrared ( ~750nm). Fused quartz is widely used in optical devices as it is transparent over a wide range of wavelengths. ........................................ 148 Dispersion and Chromatic Aberration ........................................................................................................ 148 Lenses ........................................................................................................................................................ 149 Predicting the image .................................................................................................................................. 151 Drawing to find the image .......................................................................................................................... 151 Finding the image by formula ..................................................................................................................... 152 The Diverging Lens .................................................................................................................................... 153 The sign convention ................................................................................................................................... 153 The Power of a Lens .................................................................................................................................. 154 Lens Power = ........................................................................................................................................ 154 Mirrors ........................................................................................................................................................ 154 The angle of incidence equals the angle of reflection. ............................................................................... 154 Front-silvered mirrors ................................................................................................................................. 156 Reflecting Prisms ....................................................................................................................................... 156 Concave Mirrors ......................................................................................................................................... 157 Convex Mirrors ........................................................................................................................................... 158 Fibre Optics ................................................................................................................................................ 158 Fibre Optic Data Links ................................................................................................................................ 158 History of Fibre Optic Technology .............................................................................................................. 159 Advantages and Disadvantages of Fibre Optics ........................................................................................ 160 Frequency and Bandwidth ......................................................................................................................... 161 Basic Structure of an Optical Fibre ............................................................................................................ 162 Propagation of Light along a Fibre ............................................................................................................. 163 Ray Theory ................................................................................................................................................. 163 Mode Theory .............................................................................................................................................. 165 Optical Fibre Types .................................................................................................................................... 168 Single Mode Fibres .................................................................................................................................... 169 Multimode Fibres ........................................................................................................................................ 169 Properties of Optical Fibre Transmission ................................................................................................... 169 Attenuation ................................................................................................................................................. 170 Dispersion .................................................................................................................................................. 173 Intermodal Dispersion ................................................................................................................................ 173 The Transmission of Signals ...................................................................................................................... 174 Analogue Transmission .............................................................................................................................. 174 Digital Transmission ................................................................................................................................... 174 Optical Fibres and Cables .......................................................................................................................... 174 Optical Fibre and Cable Design ................................................................................................................. 174 Optical Fibres ............................................................................................................................................. 175 Multimode Step-Index Fibres ..................................................................................................................... 176 Multimode Graded-lndex Fibres ................................................................................................................. 177 Optical Time-Domain Reflectometry .......................................................................................................... 180 Power Meter ............................................................................................................................................... 182 Wave Motion .............................................................................................................................................. 183 Progressive and Stationary Waves ............................................................................................................ 183 The Wave Formula ..................................................................................................................................... 183 Resonance ................................................................................................................................................. 186 Sound ......................................................................................................................................................... 186 Intensity of Sound ...................................................................................................................................... 187 Sound Waves and Resonant Vibrations .................................................................................................... 188 Constructive and Destructive Interference ................................................................................................. 188 Noise Cancelling Headphones ................................................................................................................... 188 Reflected Waves ........................................................................................................................................ 189 Producing Stationary Waves ...................................................................................................................... 190 Beats .......................................................................................................................................................... 191 Supersonic Speed and Mach Number ....................................................................................................... 191 The Doppler Effect ..................................................................................................................................... 191

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Part 66 Training Syllabus Module 2 Physics

Matter Matter The Nature of Matter Scientists for a long time suspected that all substances were composed of small particles which they called atoms. However, it wasn't until the beginning of this century that the existence of atoms was demonstrated to everyone's satisfaction. The size of the atom was found to be so small that a few hundred million, if placed side by side in a row, would form a line less than an inch long. All atoms are, crudely speaking, the same size and can be thought to consist of two main parts. The outer part is composed of 1 or more orbits of electrons. These orbits makes up most of the volume of the atom yet contributes practically nothing to its substance. The other part, located at the centre, is extremely small compared to the atom as a whole, yet essentially all of the real substance of the atom can be attributed to this small speck. We call this speck the nucleus. Further investigation revealed that the nucleus is actually composed of two kinds of particles of roughly equal size and substance packed closely together. These nuclear particles are the proton and neutron. When we refer to the amount of material or substance in an object, we are really talking about the number of protons and neutrons in that object. Also, what we perceive as the mass of an object is related directly to the number of protons and neutrons contained it. The simplest atom is hydrogen which has a single proton for a nucleus. An atom of lead, on the other hand, has 82 protons and 125 neutrons in its nucleus and so has 207 (125 + 82) times as much material or substance as an atom of hydrogen. The size of an atom bears no simple relation to the number of particles in its nucleus. A sodium atom, for example, with 11 protons and 12 neutrons is approximately the same size as an atom of mercury with 80 protons and 121 neutrons. In general, we can say that the size of an atom is determined by its electron orbits, its substance is determined by the total number of protons and neutrons in its nucleus.

The Components of Atoms Atoms are the smallest particles of matter whose properties we study in Chemistry. However from th th experiments done in the late 19 and early 20 century it was deduced that atoms were made up of three fundamental sub-atomic particles (Table 0-1). Particle

Relative mass

Electrical charge

Neutron

1

0 (zero)

Proton

1

+1 (positive)

Electron

Comments In the nucleus

In the nucleus Arranged in energy levels or shells around 1/1850 -1 (negative) the nucleus Table 0-1 The sub-atomic components of atoms

Figure 1.1 gives some idea on the structure of an atom.

Figure 0-1 The structure of an atom.

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Part 66 Training Syllabus Module 2 Physics

Periodic Table of the Elements

Figure 0-2 The Periodic Table of the Elements The elements are laid out in order of Atomic Number. Hydrogen 1, H, does not readily fit into any Group. A Group is a vertical column of like elements e.g. Group IA, The Alkali Metals (Li, Na, K etc.), Group VIIB, The Halogens (F, Cl, Br, I etc.) and Group VIII (or 0), The Noble Gases (He, Ne, Ar etc.). The Group number equals the number of electrons in the outer shell (e.g. chlorine's electron arrangement is 2.8.7, the second element down, in Group 7). A Period is a horizontal row of elements with a variety of properties. The Period number equals the number of shells (1-7).

Chemical Definitions Elements Pure substances, made up of atoms with the same number of protons. Note that an element:  consists of only one kind of atom,  cannot be broken down into a simpler type of matter by either physical or chemical means, and  can exist as either atoms (e.g. argon) or molecules (e.g., nitrogen).

Mixtures Mixtures are of pure substances. Mixtures have the properties of the different substances that make it up. Mixtures melt at a range of temperatures and are easy to separate. Note that a mixture:  consists of two or more different elements and/or compounds physically intermingled,  can be separated into its components by physical means, and  often retains many of the properties of its components.

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Part 66 Training Syllabus Module 2 Physics

Compounds Pure substances made up more than 1 element which have been joined together by a chemical reaction therefore the atoms are difficult to separate. The properties of a compound are different from the atoms that make it up. Splitting of a compound is called chemical analysis. Note that a compound:  consists of atoms of two or more different elements bound together,  can be broken down into a simpler type of matter (elements) by chemical means (but not by physical means),  has properties that are different from its component elements, and  always contains the same ratio of its component atoms.

Atomic Number The atomic number (also known as the proton number) is the number of protons found in the nucleus of an atom. It is traditionally represented by the symbol Z. The atomic number uniquely identifies a chemical element. In an atom of neutral charge, atomic number is equal to the number of electrons.

Mass Number The mass number (A), also called atomic mass number or nucleon number, is the number of protons and neutrons (also defined as a less commonly known term, nucleons) in an atomic nucleus. The mass number is unique for each isotope of an element and is written either after the element name or as a superscript to the left of an element's symbol. For example: Carbon-12 (12C) has 6 protons and 6 neutrons. The full isotope symbol would also have the atomic number (Z) as a subscript to the left of the element symbol directly below the mass number, thus:

The difference between the mass number and the atomic number gives the number of neutrons (N) in a given nucleus: N=A-Z. For example: Carbon-14 is created from Nitrogen-14 with seven protons (p) and seven neutrons via a cosmic ray interaction which transmutes 1 proton into 1 neutron. Thus the atomic number decreases by 1 (Z: 76) and the mass number remains the same (A = 14), however the number of neutrons increases by 1 (n: 78). Before: Nitrogen-14 (7p, 7n) After: Carbon-14 (6p, 8n).

Molecules A pure substance which results when two or more atoms of a single element share electrons, for example O2. It can also more loosely refer to a compound, which is a combination of two or more atoms of two or more different elements, for example: H2O. Atoms combine to form more complex structures which we call molecules. Like building blocks, these molecules organize to form all of the materials, solid, liquid and gas, which we encounter in our daily lives. Solids and liquids are materials in which the molecules attract one another so strongly that their relative motion is severely restricted. In a gas, the freedom of motion of the molecules is only slightly influenced by their mutual attraction. This is why gases fill the entire space to which they are confined, They spread out unconstrained until they encounter the walls of their container.

Isotopes Isotopes are atoms of the same element with different numbers of neutrons. This gives each isotope of the element a different mass or nucleon number but being the same element they have the same atomic or proton number. There are small physical differences between the isotopes e.g. the heavier isotope has a greater density and boiling point. However, because they have the same number of protons they have the same electronic structure and are identical chemically. Examples are illustrated below. Do not assume the word isotope means it is radioactive, this depends on the stability of the nucleus i.e. unstable atoms might be referred to as radioisotopes.

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Part 66 Training Syllabus Module 2 Physics

Figure 0-3 The three isotopes of hydrogen Figure 1-3 shows the three isotopes of hydrogen. They are called hydrogen, deuterium, and tritium respectively. How do we distinguish between them? They each have one single proton (Z = 1), but differ in the number of their neutrons. Hydrogen has no neutron, deuterium has one, and tritium has two neutrons. The isotopes of hydrogen have, respectively, mass numbers of one, two, and three. Hydrogen-1 is the most common, there is a trace of hydrogen-2 naturally but hydrogen-3 is very unstable and is used in atomic fusion weapons.

Figure 0-4 The two isotopes of helium Figure 1-4 shows the two isotopes of helium with mass numbers of 3 and 4, with 1 and 2 neutrons respectively but both have 2 protons. Helium-3 is formed in the Sun by the initial nuclear fusion process. Helium-4 is also formed in the Sun and as a product of radioactive alpha decay of an unstable nucleus. An alpha particle is a helium nucleus, it picks up two electrons and becomes the atoms of the gas helium.

Figure 0-5 The two isotopes of sodium Figure 1-5 shows the two isotopes of sodium with mass numbers of 23 and 24, with 12 and 13 neutrons respectively but both have 11 protons. Sodium-23 is quite stable e.g. in common salt (NaCI, sodium chloride) but sodium-24 is a radio-isotope and is a gamma emitter used in medicine as a radioactive tracer e.g. to examine organs and the blood system.

lonization When the atom loses electrons or gains electrons in this process of electron exchange, it is said to be ionised. For ionisation to take place, there must be a transfer of energy which results in a change in the internal energy of the atom. An atom having more than its normal amount of electrons acquires a negative charge, and is called a negative ion (or 'anion'). The atom that gives up some of its normal electrons is left with less negative charges than positive charges and is called a positive ion (or 'cation'). Thus, ionisation is the process by which an atom loses or gains electrons.  Cation - A cation is a positively charged ion. Metals typically form cations.  Anion - An anion is a negatively charged ion. Non-metals typically form anions.

The Electronic Structure of Atoms The electrons are arranged in energy levels or shells around the nucleus and with increasing distance from the nucleus. The shells are lettered from the innermost shell outwards from K to Q. There are rules about the maximum number of electrons allowed in each shell.    

st

The 1 shell (K) has a maximum of 2 electrons nd The 2 shell (L) has a maximum of 8 electrons rd The 3 shell (M) has a maximum of 18 electrons th The 4 shell (N) has a maximum of 32 electrons

Our knowledge about the structure of atoms depends on the mathematical formulations predicted by Neils Bohr. He suggested that electrons are distributed in orbits and the number of electrons held in the orbit

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Part 66 Training Syllabus Module 2 Physics depends on the number of the orbit. The orbits are counted outwards from the nucleus. Higher the orbit number, farther are the electrons in that orbit from the nucleus. If the orbit number is "n", then the maximum 2 electrons held in the orbit is given as 2n . The first orbit has n=1, and will hold maximum of 2 electrons, the second orbit has n=2 and is capable of holding a total of 8 electrons; similarly the third orbit will be able to contain 18 electrons and so on. Electrons within an atom have definite energies. The electrons closest to the nucleus (n=1) are most tightly bound; the reason is because of stronger electrostatic attraction with the nucleus. Electrons in the highest orbit are least tightly bound. Electrons in the same orbit have same energies. The electron orbits are also called as electron energy levels or shells. Electronic shells are known as K shell, L shell, M shell, N shell corresponding to orbit number n=1, 2, 3 and 4 respectively. Higher number orbits are assigned shell names in alphabetical order after N.

Figure 0-6 The atomic structure of Helium and Neon

Figure 0-7 Electron shell (orbit) designation Examples: diagram, symbol or name of element (Atomic Number = number of electrons in a neutral atom), shorthand electron arrangement:

On Period 1

Figure 0-8 Electron arrangement of Hydrogen and Helium

On Period 2

Figure 0-9 Electron arrangement of Lithium, Carbon and Neon

On Period 3 to Figure 0-10 Electron arrangement of Sodium, Chlorine and Argon

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Part 66 Training Syllabus Module 2 Physics

On Period 4

Figure 0-11 Electron arrangement of Potassium and Calcium

Valency Hydrogen is the simplest element. It has one electron. Its outer shell only holds two electrons. Let us use Hydrogen as a standard to see how other atoms combine with it. Table 1-2 lists the simplest compound of selected elements with Hydrogen. Valency can be simply defined as the number of Hydrogen atoms that an element can combine with. In the above table, Helium, Neon and Argon have a valency of 0. They do not normally form compounds. Lithium, Sodium and Potassium have a valency of 1 because they combine with one Hydrogen atom. Beryllium, Magnesium and Calcium all have a valency of 2: they combine with two Hydrogen atoms. Note that the valences of all these atoms are equal to the number of outer electrons that these elements have. Boron and Aluminium combine with three Hydrogen atoms - their valences are 3 - and they have three outer electrons. Carbon and Silicon combine with four Hydrogen atoms. The valency of these elements is 4. It will come as no surprise that they both have four outer electrons. Any element with 4 electrons in its outer shell is known as a semiconductor Atom Symbol Outer Shell Compound Helium He Full None Lithium Li 1 LiH Beryllium Be 2 BeH2 Boron B 3 BH3 Carbon C 4 CH4 Nitrogen N 5 NH3 Oxygen Ο 6 H2O Fluorine F 7 HF Neon Ne Full None Sodium Na 1 NaH Magnesium Mg 2 MgH2 Aluminium Al 3 AIH3 Silicon Si 4 SiH4 Phosphorus P 5 PH3 Sulphur S 6 H2S Chlorine Cl 7 HCI Argon Ar Full None Potassium K 1 KH Calcium Ca 2 CaH2 Table 0-2 Electrons in outer shells of some common elements What about Nitrogen and Phosphorus? They have five outer electrons. But they normally only combine with three Hydrogen atoms. Their valences are 3. Note that 3 is 5 less that 8. These atoms are three electrons short of a full shell. Please note that both Nitrogen and Phosphorus can also have a valency of 5. Some atoms are capable of having more than one valency. That will confuse the issue so we will talk of normal valency. Now to Oxygen and Sulphur. Both have six outer electrons. Six is two short of a full shell. Their normal valences are 2 and they combine with two atoms of Hydrogen. Water is H 2O. Finally, Fluorine and Chlorine – have seven outer electrons. This is one short of a full shell. They both combine with a single Hydrogen atom and their normal valences are 1.

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Part 66 Training Syllabus Module 2 Physics As a side note, Chlorine can also have valences of 3, 5 and 7. The reasons are well beyond the scope of these notes. The rules above can be summarised as follows: The normal valency of an atom is equal to the number of outer electrons if that number is four or less. Otherwise, the valency is equal to 8 minus the number of outer electrons. The atoms with full electron shells (Helium, Neon, Argon) are chemically inert forming few compounds. The atoms don't even interact with each other very much. These elements are gases with very low boiling points. The atoms with a single outer electron or a single missing electron are all highly reactive. Sodium is more reactive than Magnesium. Chlorine is more reactive than Oxygen. Generally speaking, the closer an atom is to having a full electron shell, the more reactive it is. Atoms with one outer electron are more reactive than those with two outer electrons, etc. Atoms that are one electron short of a full shell are more reactive than those that are two short. Atoms with only a few electrons in its outer shell are good electrical conductors. Atoms with 8, or close to 8 electrons in its outer shell are poor conductors (or good insulators). This is why atoms with 4 electrons in its outer shell are semi-conductors. When a semiconductor (such as silicon or germanium) atom bonds with another similar atom, it does so covalently. Each atom shares one electron with 4 neighbour atoms. Thus all its electrons are used up in what becomes a solid lattice of semiconductor atoms. The solid material has therefore no free electrons (and no holes for electrons to fit into). The following names are given to ions of the specific number of electron bindings (valence):      

1 electron binding - monovalent 2 electron binding - divalent 3 electron binding - trivalent 4 electron binding - tetravalent 5 electron binding - pentavalent 6 electron binding – hexavalent

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Part 66 Training Syllabus Module 2 Physics  Atomic No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52

Electrons per Shell

Element Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminium Silicon Phosphorus Sulphur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine Krypton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver Cadmium indium Tin Antimony Tellurium

K 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

L

1 2 3 4 5 6 7 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

M

1 2 3 4 5 6 7 8 8 8 9 10 11 13 13 14 15 16 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18

N

1 2 2 2 2 1 2 2 2 2 1 2 3 4 5 6 7 8 8 8 9 10 12 13 14 15 16 18 18 18 18 18 18 18

Ο

1 2 2 2

1 2 3 4 5 6

P

Q

Atomic No. 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103

Electrons per Shell

Element iodine Xenon Caesium Barium Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium Dysprosium Holmium Erbium Thulium Ytterbium Lutetium Halnium Tantalum Tungsten Rhenium Osmium iridium Platinum Gold Mercury Thallium Lead Bismuth Polonium Asatine Radon Francium Radium Actinium Thorium Proactinium Uranium Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium

K 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

L 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8 8

M 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 18 8 8 8 8 8 8 8 8 8 8 8 8 18 18 18 18 18 18 18 8 8 8 8 8 18 18 18

N 8 18 8 8 8 19 20 21 22 23 24 25 26 27 28 29 30 31 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32 32

Ο 7 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9 10 11 12 13 14 15 16 18 18 18 18 18 18 18 18 18 18 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32

Table 0-3 Electrons per shell

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P

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 2 3 4 5 6 7 8 8 8 9 9 9 9 9 9 9 9 9 9 9 9 9 9 9

Q

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

Part 66 Training Syllabus Module 2 Physics

Chemical Bonding Cohesion and Adhesion 'Cohesion' is the intermolecular force between liquid particle types (for example, it is what makes water molecules stick together, or 'cohere', to make a rain drop). 'Adhesion' is the intermolecular force between dissimilar atoms (for example, it is what makes the rain drops 'adhere' to a washing line). These types of bonding are temporary. Atomic bonding refers to the permanent bonding between atoms which holds all materials together.

Noble Gases Some atoms are very reluctant to combine with other atoms and exist in the air around us as single atoms. These are the Noble Gases and have very stable electron arrangements e.g. 2, 2.8 and 2.8.8 and are shown in the diagrams below.

Figure 0-12 (Atomic Number) and electron arrangement

Covalent and Ionic Bonding All other atoms therefore, bond to become electronically more stable, that is to become like Noble Gases in electron arrangement. Atoms can do this in two ways:  COVALENT BONDING - sharing electrons to form molecules with covalent bonds, the bond is usually formed between two non-metallic elements in a molecule. or  IONIC BONDING - By one atom transferring electrons to another atom. The atom losing electrons forms a positive ion and is usually a metal. The atom gaining electrons forms a negative ion and is usually a non-metallic element. The types of bonding and the resulting properties of the elements or compounds are described in detail below. In all the electronic diagrams ONLY the outer electrons are shown.

Covalent Bonding Covalent bonds are formed by atoms sharing electrons to form molecules. This type of bond usually formed between two non-metallic elements. The molecules might be that of an element i.e. one type of atom only OR from different elements chemically combined to form a compound. The covalent bonding is caused by the mutual electrical attraction between the two positive nuclei of the two atoms of the bond, and the electrons between them. One single covalent bond is a sharing of 1 pair of electrons, two pairs of shared electrons between the same two atoms gives a double bond and it is possible for two atoms to share 3 pairs of electrons and give a triple bond. The Bonding in Small Covalent Molecules The simplest molecules are formed from two atoms and examples of their formation are shown below. The electrons are shown as dots and crosses to indicate which atom the electrons come from, though all electrons are the same. The diagrams may only show the outer electron arrangements for atoms that use two or more electron shells. Examples of simple covalent molecules are... Example 1 - 2 hydrogen atoms (1) form the molecule of the element hydrogen H2: and combine to form electrons around each atom.

where both atoms have a pseudo helium structure of 2 outer

Example 2 -2 chlorine atoms (2.8.7) form the molecule of the element chlorine:

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Part 66 Training Syllabus Module 2 Physics

and combine to form outer electrons around each atom.

where both atoms have a pseudo neon or argon structure of 8

Example 3 - 1 atom of hydrogen (1) combines with 1 atom of chlorine (2.8.7) to form the molecule of the compound hydrogen chloride HCI: and combine to form neon or argon.

where hydrogen is electronically like helium and chlorine like

Example 4 - 2 atoms of hydrogen (1) combine with 1 atom of oxygen (2.6) to form the molecule of the compound we call water H2O and and combine to form so that the hydrogen atoms are electronically like helium and the oxygen atom becomes like neon or argon. The molecule can be shown as:

with two hydrogen - oxygen single covalent bonds. Example 5 - 3 atoms of hydrogen (1) combine with 1 atom of nitrogen (2.5) to form the molecule of the compound we call ammonia NH3. Three of and one combine to form so that the hydrogen atoms are electronically like helium and the nitrogen atom becomes like neon or argon. The molecule can be shown as:

with three nitrogen - hydrogen single covalent bonds. Example 6 - 4 atoms of hydrogen (1) combine with 1 atom of carbon (2.4) to form the molecule of the compound we call methane CH4. Four of and one of combine to form so that the hydrogen atoms are electronically like helium and the nitrogen atom becomes like neon or argon. The molecule can be shown as

with four carbon - hydrogen single covalent bonds. All the bonds in the above examples are single covalent bonds. Below are three examples 7-9, where there is a double bond in the molecule, in order that the atoms have stable Noble Gas outer electron arrangements around each atom. Example 7 - Two atoms of oxygen (2.6) combine to form the molecules of the element oxygen O2.

The molecule has one double covalent bond

.

Example 8 - One atom of carbon (2.4) combines with two atoms of oxygen (2.6) to form carbon dioxide CO2. .

The molecule can be shown as

with two carbon = oxygen double covalent bonds.

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Part 66 Training Syllabus Module 2 Physics Example 9 - Two atoms of carbon (2.4) combine with four atoms of hydrogen (1) to form ethane C2H4.

The molecule can be shown as hydrogen single covalent bonds.

with one carbon = carbon double bond and four carbon -

The Properties of Small Covalent Molecules The electrical forces of attraction between atoms in a molecule are strong and most molecules do not change on heating. However the forces between molecules are weak and easily weakened further on heating. Consequently small covalent molecules have low melting and boiling points. They are also poor conductors of electricity because there are no free electrons or ions in any state to carry electric charge. Most small molecules will dissolve in a solvent to form a solution.

Large Covalent Molecules and their Properties It is possible for many atoms to link up to form a giant covalent structure. This produces a very strong 3-dimensional covalent bond network. This illustrated by carbon in the form of diamond. Carbon can form four single bonds to four other atoms etc. etc. This type of structure is thermally very stable and they have high melting and boiling points. They are usually poor conductors of electricity because the electrons are not usually free to move as they can in metallic structures. Also because of the strength of the bonding in the structure they are often very hard and will not dissolve in solvents like water. Figure 1-12: A plane of Carbon atoms from a diamond crystal.

Ionic Bonding Ionic bonds are formed by one atom transferring electrons to another atom to form ions. Ions are atoms, or groups of atoms, which have lost or gained electrons. The atom losing electrons forms a positive ion (a cation) and is usually a metal. The overall charge on the ion is positive due to excess positive nuclear charge (protons do NOT change in chemical reactions). The atom gaining electrons forms a negative ion (an anion) and is usually a non-metallic element. The overall charge on the ion is negative because of the gain, and therefore excess, of negative electrons. The examples below combining a metal from Groups 1 (Alkali Metals), 2 or 3, with a non-metal from Group 6 or Group 7 (The Halogens). Example 1 - A Group 1 metal + a Group 7 non-metal e.g. sodium + chlorine  sodium chloride NaCI or + ionic formula Na CI . In terms of electron arrangement, the sodium donates its outer electron to a chlorine atom forming a single positive sodium ion and a single negative chloride ion. The atoms have become stable ions, because electronically, sodium becomes like neon and chlorine like argon. +

-

Na (2.8.1) + Cl (2.8.7)  Na (2.8) Cl (2.8.8)

One Na combines with one Cl to form:

Example 2 - A Group 2 metal + a Group 7 non-metal e.g. magnesium + chlorine  magnesium chloride 2+ MgCI2 or ionic formula Mg (CI )2. In terms of electron arrangement, the magnesium donates its two outer electrons to two chlorine atoms forming a double positive magnesium ion and two single negative chloride ions. The atoms have become stable ions, because electronically, magnesium becomes like neon and chlorine like argon. Mg (2.8.2) + 2CI (2.8.7)  Mg

2+

-

(2.8) 2Cl (2.8.8)

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One Mg combines with two Cl to form:

* *NOTE: You can draw two separate chloride ions, but in these examples a number subscript has been used, as in ordinary chemical formula. Example 3 - A Group 3 metal + a Group 7 non-metal e.g. aluminium + fluorine  aluminium fluoride AIF3 3+ or ionic formula AI (F )3. In terms of electron arrangement, the aluminium donates its three outer electrons to three fluorine atoms forming a triple positive aluminium ion and three single negative fluoride ions. The atoms have become stable ions, because electronically, aluminium becomes like neon and also fluorine. Al (2.8.3) + 3F (2.8.7)  AI

One

combines with three

3+

-

(2.8) 3F (2.8)

to form

Example 4 - A Group 1 metal + a Group 6 non-metal e.g. potassium + oxygen  potassium oxide K2O or + 2ionic formula (K )2O . In terms of electron arrangement, the two potassium atoms donates their outer electrons to one oxygen atom. This results in two single positive potassium ions to one double negative oxide ion. All the ions have the stable electronic structures 2.8.8 (argon like) or 2.8 (neon like) +

2-

2K (2.8.8.1) + O (2.6)  2K (2.8.8) O (2.8)

Two

combine with one

to form

Example 5 - A Group 2 metal + a Group 6 non-metal e.g. calcium + oxygen  calcium oxide CaO or ionic 2+ 2formula Ca O . In terms of electron arrangement, one calcium atom donates its two outer electrons to one oxygen atom. This results in a double positive calcium ion to one double negative oxide ion. All the ions have the stable electronic structures 2.8.8 (argon like) or 2.8 (neon like): 2+

Ca (2.8.8.2) + O (2.6) => Ca

ONE

combines with ONE

2-

(2.8.8) O (2.8)

to form

Example 6 - A Group 3 metal + a Group 6 non-metal e.g. aluminium + oxygen  aluminium oxide Al2O3 or 3+ 2ionic formula (AI )2(O )3. In terms of electron arrangement, two aluminium atoms donate their three outer electrons to three oxygen atoms. This results in two triple positive aluminium ions to three double negative oxide ions. All the ions have the stable electronic structure of neon 2.8 2AI (2.8.3) + 3O (2.6) => 2AI

Two

combine with three

3+

2-

(2.8) 3O (2.8)

to form

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Part 66 Training Syllabus Module 2 Physics

The properties of Ionic Compounds  The ions in an ionic solid are arranged in an orderly way in a giant ionic lattice shown in the diagram on the left. The ionic bond is the strong electrical attraction between the positive and negative ions next to each other in the lattice. Salts and metal oxides are typical ionic compounds.  This strong bonding force makes the structure hard (if brittle) and have high melting and boiling points. Unlike covalent molecules, ALL ionic compounds are crystalline solids at room temperature.  Many ionic compounds are soluble in water, but not all.  The solid crystals DO NOT conduct electricity because the ions are not free to move to carry an electric current. However, if the ionic compound is melted or dissolved in water, the liquid will now conduct electricity, as the ion particles are now free. Figure 0-13 Sodium Chloride lattice structure The crystal lattice of metals consists of ions, NOT atoms. The outer electrons (-) from the original metal 1 atoms are free to move around between the positive metal ions formed (+). These free or 'delocalised electrons are the 'electronic glue' holding the particles together. There is a strong electrical force of attraction between these mobile electrons and the immobile positive metal ions - this is the metallic bond.

Figure 0-14 'Electron cloud' formation of Ionic (or Metallic) Bonding  This strong bonding generally results in dense, strong materials with high melting and boiling points.  Metals are good conductors of electricity because these 'free' electrons carry the charge of an electric current when a potential difference (voltage!) is applied across a piece of metal.  Metals are also good conductors of heat. This is also due to the free moving electrons. Nonmetallic solids conduct heat energy by hotter more strongly vibrating atoms, knocking against cooler less strongly vibrating atoms to pass the particle kinetic energy on. In metals, as well as this effect, the 'hot' high kinetic energy electrons move around freely to transfer the particle kinetic energy more efficiently to 'cooler' atoms.  Typical metals also have a silvery surface but remember this may be easily tarnished by corrosive oxidation in air and water.

States of Matter Solids A solid object is characterized by its resistance to deformation and changes of volume. At the microscopic scale, a solid has these properties:  The atoms or molecules that comprise the solid are packed closely together.  These constituent elements have fixed positions in space relative to each other. This accounts for the solid's rigidity. In mineralogy and crystallography, a crystal structure is a unique arrangement of atoms in a crystal. A crystal structure is composed of a unit cell, a set of atoms arranged in a particular way; which is periodically repeated in three dimensions on a lattice. The spacing between unit cells in various directions is called its lattice parameters.  If sufficient force is applied, its lattice atomic structure can be disrupted, causing permanent deformation.

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Part 66 Training Syllabus Module 2 Physics Because any solid has some thermal energy, its atoms vibrate. However, this movement is very small, and cannot be observed or felt under ordinary conditions.

Liquids A liquid's shape is confined to, but not determined by, the container it fills. That is to say, liquid particles (normally molecules or clusters of molecules) are free to move within the volume, but they form a discrete surface that may not necessarily be the same as the vessel. The same cannot be said about a gas; it can also be considered a fluid, but it must conform to the shape of the container entirely.

Gases Gases consist of freely moving atoms or molecules without a definite shape and without a definite volume. Compared to the solid and liquid states of matter a gas has lower density and a lower viscosity. The volume of a gas will change with changes in temperature or pressure, as described by the ideal gas law. A gas also has the characteristic that it will diffuse readily, spreading apart in order to uniformly fill the space of any container.

Plasma A plasma is typically an ionized gas. Plasma is considered to be a distinct state of matter, apart from gases, because of its unique properties. 'Ionized' refers to presence of one or more free electrons, which are not bound to an atom or molecule. The free electric charges make the plasma electrically conductive so that it responds strongly to electromagnetic fields. Plasma typically takes the form of neutral gas-like clouds (e.g. stars) or charged ion beams, but may also include dust and grains (called dusty plasmas). They are typically formed by heating and ionizing a gas, stripping electrons away from atoms, thereby enabling the positive and negative charges to move more freely.

Changes between States Solids can melt and become liquids, and liquids can boil to become gases. Likewise, gases can condense to become liquids, and liquids can freeze to become solids. Sometimes solids can become gases without ever becoming liquids. This is called subliming.

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Part 66 Training Syllabus Module 2 Physics

Mechanics Mass In physics, the term for what we have up to now referred to as the amount of substance or matter is "mass". A natural unit for mass is the mass of a proton or neutron. This unit has a special name, the "atomic mass unit" (amu). This unit is useful in those sciences which deal with atomic and nuclear matter. In measuring the mass of objects which we encounter daily, this unit is much too small and therefore very inconvenient. For example, the mass of a bowling ball expressed in amu's would be about 4,390,000,000,000,000,000,000,000,000. One kilogram equals 602,000,000,000,000,000,000,000,000 amu. Since one amu is the mass of a proton or neutron we know immediately that a kilogram of anything has this combined number of protons and neutrons contained in it. The kilogram is the SI unit of mass. In the English system, the standard unit of mass is the slug. The conversion is: 1 slug = 14.59 kg = 8,789,000,000,000,000,000,000,000,000 amu We will use the conveniently sized units, the slug in the English system and the kilogram in the metric system, for all of the problems that we will do in this course. Note that the above conversion, 1 slug = 14.59 kilogram, is listed with your conversion factors in the table of conversion factors (Table 1-1).

Force The physicist uses the word "force" to describe any push or pull. A force is one kind of vector. A vector is a quantity that has both size and direction. A force has a certain magnitude or size. Also, a force is always in a certain direction. To completely describe a force, it is necessary to specify both the size of the push or pull and its direction. The units in which force are measured are the pound (Ib.) in the English system and the Newton (N) in the metric system. The Newton is named for Sir Isaac Newton, a famous British physicist who lived in the 17th century. The relationship between the metric and English units is given by the conversion factor: 1 Ib. = 4.448 N

Weight A weight is one kind of force. It is defined as the gravitational pull of the earth on a given body. The direction of this force is toward the geometrical centre of the earth.

Distinction between Mass and Weight The physicist very carefully distinguishes between "mass" and "weight". As we have seen, mass is the quantity of matter, determined by the number of protons and neutrons in the body, and weight is a measure of the gravitational pull of the earth on this quantity of matter. It may seem that this is an unimportant distinction. However, there is one important difference. The mass of an object is the same wherever this object is in the universe. The mass of a stone is the same if the stone is on the earth, on Mars, in a space ship, or some place in the Milky Way Galaxy. If the stone is not on the earth but is in a space station orbiting the earth some distance from the earth's surface, the weight of this stone is different from its weight on the earth's surface. If the stone is on the planet Mars, we speak of its "weight on Mars", the gravitational pull of Mars on the stone. As you have probably figured out, the greater the mass of an object on the surface of the earth, the greater is the weight of this object. These two quantities are approximately proportional to each other as long as the body remains on the surface of the earth. The word "approximately" in the previous sentence refers to the fact that the pull of the earth on a body of a given mass varies slightly with the position of the body on the earth's surface. For example, a body that weighs 57.3 Ibs. at the North Pole would weigh 57.0 Ibs. at a place on the equator. This occurs because a body at either pole is slightly closer to the centre of the earth than it is

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Part 66 Training Syllabus Module 2 Physics at the equator. Thus, the pull of the earth on the body is greater at the poles and slightly smaller at other places on the earth. However, we usually neglect this slight difference. Physicists and engineers measure masses of bodies in slugs or kilograms and weights in pounds or Newtons. The equation relating mass and weight is: w = mg In this equation, g has a definite numerical value. We will use the following relations: g = 32

or g = 9.8

There is a great source of confusion in British marketing practices. For example, we often see on a packet of sugar the information regarding the contents: 1 kg or 2.2 Ibs We note that 2.2 Ibs. equals 1 kg. We have just learned that 2.2 Ibs. is the weight of the sugar and that 1 kg is the mass of the sugar. In other words, British packaging practices list the weight of the product if we deal with the English system and list the mass of the product if we are in the metric system. For example, suppose the weight of a piece of cheese is marked 32 oz. and we wish to know the number of grams. First we convert the weight in ounces to 2 Ibs. Then we convert from pounds to Newtons. W = 2lbs x

= 8.9 N

Next, we use the relation: w = mg or m = Therefore, we write: m=

= 0.908 kg = 908 grams

Note that we can convert from pounds to Newtons since both are units of weight and we can convert from kilograms to slugs since both are units of mass. However, if we want to find a mass if we know a weight or a weight if we know a mass we must use the equation: m=

or w = mg

In summary, let us note that mass is a measure of the quantity of matter - ultimately, a measure of the number of protons and neutrons in the body and weight is the force with which the earth pulls on a body. These are related but not identical concepts. The units of mass are slugs and kilograms. The units of weight are pounds and Newtons. A mass can be changed from slugs to kilograms and vice versa. A weight can be changed from Newtons to pounds or vice versa. However, one cannot say that one pound equals 454 grams. The only correct statement is that a body having a weight of one pound has a mass of 454 grams. The equation relating mass and weight is: m=

or w = mg

Problems 1. What is the mass of a body having a weight of 45 N? 2. What is the weight of a body having a mass of 23 kg? 3. What is the mass of a body having a weight of 350 Ibs.? 4. What is the weight of a body having a mass of 23.6 slugs? 5. What is the weight (in Ibs.) of the corn flakes in a box where the mass is listed as 680 g? 6. What is the mass in grams of 2.5 Ibs. of bologna?

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Part 66 Training Syllabus Module 2 Physics

Answers All answers are to 3 significant figures. 1. 4.59kg 2. 225 N 3. 10.9 slugs 4. 755 Ibs. 5. 1.45 Ibs. 6. 1140g

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Part 66 Training Syllabus Module 2 Physics

Stress, Strain and Hooke's Law Introduction Structural integrity is a major factor in aircraft design and construction. No production aeroplane leaves the ground before undergoing extensive analysis of how it will fly, the stresses it will tolerate and its maximum safe capability. Every aircraft is subject to structural stress. Stress acts on an aeroplane whether on the ground or in flight. Stress is defined as a load applied to a unit area of material. Stress produces a deflection or deformation in the material called strain. Stress is always accompanied by strain. Current production general aviation aircraft are constructed of various materials, the primary being aluminium alloys. Rivets, bolts, screws and special bonding adhesives are used to hold the sheet metal in place. Regardless of the method of attachment of the material, every part of the fuselage must carry a load, or resist a stress placed on it. Design of interior supporting and forming pieces, and the outside metal skin all have a role to play in assuring an overall safe structure capable of withstanding expected loads and stresses. The stress a particular part must withstand is carefully calculated by engineers. Also, the material a part is made from is extremely important and is selected by designers based on its known properties. Aluminium alloy is the primary material for the exterior skin on modern aircraft. This material possesses a good strength to weight ratio, is easy to form, resists corrosion, and is relatively inexpensive.

Types of Structural Stress The five basic structural stresses to which aircraft are subject are: 1. 2. 3. 4. 5.

Tension Compression Torsion Shear Bending

While there are many other ways to describe the actual stresses which an aircraft undergoes in normal (or abnormal) operation, they are special arrangements of these basic ones.  Tension - is the stress acting against another force that is trying to pull something apart. For example, while in straight and level flight the engine power and propeller are pulling the aeroplane forward. The wings, tail section and fuselage, however, resist that movement because of the airflow around them. The result is a stretching effect on the airframe. Bracing wires in an aircraft are usually in tension.  Compression - is a squeezing or crushing force that tries to make parts smaller. Anti-compression design resists an inward or crushing force applied to a piece or assembly. Aircraft wings are subjected to compression stresses. The ability of a material to meet compression requirements is measured in pounds per square inch (PSI).  Torsion - is a twisting force. Because aluminium is used almost exclusively for the outside, and, to a large extent, inside fabrication of parts and covering, its tensile strength (capability of being stretched) under torsion is very important. Tensile strength refers to the measure of strength in pounds per square inch (PSI) of the metal. Torque (also a twisting force) works against torsion. The torsional strength of a material is its ability to resist torque. While in flight, the engine power and propeller twist the forward fuselage. The force, however, is resisted by the assemblies of the fuselage. The airframe is subjected to variable torsional stresses during turns and other manoeuvres.  Shear - stress tends to slide one piece of material over another. Consider the aircraft fuselage. The aluminium skin panels are riveted to one another. Shear forces try to make the rivets fail under flight loads; therefore, selection of rivets with adequate shear resistance is critical. Bolts and other fasteners are often loaded in shear, an example being bolts that fasten the wing to the spar or carrythrough structure. Although other forces may also be present, shear forces try to rip the bolt in two. Generally, shear strength is less than tensile or compressive strength in a particular material.

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Part 66 Training Syllabus Module 2 Physics  Bending - is a combination of two forces, compression and tension. During bending stress, the material on the inside of the bend is compressed and the outside material is stretched in tension. An example of this is the G-loading an aeroplane structure experiences during manoeuvring. During an abrupt pull-up, the aeroplane's wing spars, wing skin and fuselage undergo positive loading and the upper surfaces are subject to compression, while the lower wing skin experiences tension loads. There are many other areas of the airframe structure that experience bending forces during normal flight. An aircraft structure in flight is subjected to many and varying stresses due to the varying loads that may be imposed. The designer's problem is trying to anticipate the possible stresses that the structure will have to endure, and to build it sufficiently strong to withstand these. The problem is complicated by the fact that an aeroplane structure must be light as well as strong.

Stress, Strain and Young's Modulus What is known as Axial (or Normal) Stress, is defined as the force perpendicular to the cross sectional area of the member divided by the cross sectional area. Or: Stress =

2

2

(units Ib/in or N/m )

In figure 2-1, a solid rod of length L, is under simple tension due to force F, as shown. If we divide that axial force, F, by the cross sectional area of the rod (A), this would be the axial stress in the member. Axial stress is the equivalent of pressure in a gas or liquid. As you remember, pressure is the force/unit area. So axial stress is really the 'pressure' in a solid member. Now the question becomes, how much 'pressure' can a material bear before it fails.

Figure 0-1Tensile Stress In fact, if we look at a metal rod in simple tension as shown in figure 2.1, we see that there will be an elongation (or deformation) due to the tension. If we then graph the tension (force) verses the deformation we obtain a result as shown in figure 2.2.

Figure 0-2 Force-Extension diagram

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Part 66 Training Syllabus Module 2 Physics In figure 2.2, we see that, if our metal rod is tested by increasing the tension in the rod, the deformation increases. In the first region the deformation increases in proportion to the force. That is, if the amount of force is doubled, the amount of deformation is doubled. This is a form of Hooke's Law and could be written this way: F ∞ k (deformation), where k is a constant depending on the material (and is sometimes called the spring constant). After enough force has been applied the material enters the plastic region - where the force and the deformation are not proportional, but rather a small amount of increase in force produces a large 1 amount of deformation. In this region, the rod often begins to 'neck down , that is, the diameter becomes smaller as the rod is about to fail. Finally the rod actually breaks. The point at which the Elastic Region ends is called the elastic limit, or the proportional limit. In actuality, these two points are not quite the same. The Elastic Limit is the point at which permanent deformation occurs, that is, after the elastic limit, if the force is taken off the sample, it will not return to its original size and shape, permanent deformation has occurred. The Proportional Limit is the point at which the deformation is no longer directly proportional to the applied force (Hooke's Law no longer holds). Although these two points are slightly different, we will treat them as the same in this course. Next, rather than examining the applied force and resulting deformation, we will instead graph the axial stress verses the axial strain (figure 2-2). We have defined the axial stress earlier. The axial strain is defined as the fractional change in length or: Strain = (deformation of member) divided by the (original length of member)

LO

LO + ΓL ΓL

Figure 0-3 Axial force in a member of length LO causing deformation (extension) of ΓL We may write: Strain =

=

where LO is the original length of the member. Strain has no units - since its length divided by length, however it is sometimes expressed as 'in/in (or inches per inch)' in some texts. As we see from figure 2.4, the Stress verses Strain graph has the same shape and regions as the force verses deformation graph in figure 2.2. In the elastic (linear) region, since stress is directly proportional to strain, the ratio of stress/strain will be a constant (and actually equal to the slope of the linear portion of the graph). This constant is known as Young's Modulus, and is usually symbolized by an E or Y. We will use E for Young's modulus. We may now write: Young's Modulus (E) = (This is another form of Hooke's Law.)

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Part 66 Training Syllabus Module 2 Physics

Figure 0-4 Stress-Strain graph The value of Young's modulus - which is a measure of the amount of force needed to produce a unit deformation - depends on the material. 6

2

6

2

Young's Modulus for Steel is 30 x 10 Ib/in , for Aluminium E = 10 x 10 Ib/in , and for Brass E = 6 2 15x10 lb/in . To summarize our stress/strain/Hooke's Law relationships up to this point, we have: Stress =

2

2

(units Ib/in or N/m )

Strain = Young's Modulus (E) =

Related Definitions Bulk Modulus The bulk modulus gives the change in volume of a solid substance as the pressure on it is changed. The formula for bulk modulus is very similar to that for Young's Modulus: Bulk Modulus (B) =

=

= Pressure x

Some examples of Bulk Modulus for different materials are given on the next page.

Poisson's Ratio As a member is stressed in tension, its length increases (axial strain) and its width decreases (transverse strain). Poisson's Ratio is the ratio of transverse strain to the axial strain in a stressed member.

Cantilever Figure 2.5 illustrates a cantilever structure. The beam is under bending stress (which is greatest at the root end) and shear stress (which is constant along the beam). FORCE

The Tailplane is a one piece cantilever structure, mounted high on the vertical stabilizer to keep it clear of the jet efflux.

Figure 0-5 A cantilever structure

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Part 66 Training Syllabus Module 2 Physics

Materials Behaviour Elastic Material deforms under stress but returns to its original size and shape when the stress is released. There is no permanent deformation. Some elastic strain, like in a rubber band, can be large, but in metals it is usually small.

Brittle Material deforms by fracturing. Glass is typically brittle.

Ductile Material deforms without breaking. Metals and most plastics are ductile.

Viscous Materials that deform steadily under stress. Purely viscous materials like liquids deform under even the smallest stress. Even metals may behave like viscous materials under high temperature and pressure. This is known as creep and affects plastics far more than metals

YOUNG MODULUS, E MATERIAL Aluminum Brass Copper Glass Iron Steel Ethyl Alcohol Oil Water Mercury

2

(LB./IN. ) 6 10 x 10 6 13 x 10 6 16 x 10 6 7.8 x 10 6 13 x 10 6 29 x 10

BULK MODULUS, B 2 (LB./IN. ) 6 10 x 10 6 8.5 x 10 6 17 x 10 6 5.2 x 10 6 1.45 x 10 6 23 x 10 6 0.16 x 10 6 0.25 x 10 6 0.31 x 10 6 4.0 x 10

ELASTIC LIMIT MATERIAL Aluminum Brass Copper Iron Annealed Steel Spring Steel

2

(LB./IN. ) 4 1.9 x 10 4 5.5 x 10 4 2.3 x 10 4 2.4 x 10 4 3.6 x 10 4 6.0 x 10

ULTIMATE STRESS 2 (LB./IN. ) 4 2.1 x 10 4 6.6 x 10 4 4.9 x 10 4 4.7 x 10 4 7.1 x 10 4 10 x 10

Table 0-1 Elastic Limit and Ultimate Stress of some common materials Table 0-2 Young's Modulus and Bulk Modulus of some common materia EXAMPLE: 4 2 4 2 The elastic limit for copper is 2.3 x 10 Ib/in and the ultimate strength is 4.9 x 10 Ib/in . Suppose that a 2 copper rod has a cross-sectional area of 0.5 in . A force of 11,500 Ibs. applied longitudinally to this rod would just be within the elastic limit. A force of 12,000 Ibs. would deform the rod in such a way that it would not return to it original size after the force is removed. A force of 24,500 Ibs. would cause the rod to rupture.

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Part 66 Training Syllabus Module 2 Physics Problems 2 1. A steel bolt with a cross-sectional area of 0.1 in and a length of 6.0" is subjected to a force of 580 Ibs. What is the increase in length of the bolt? (Hint: Find the stress. Then use Young's Modulus of steel to find the strain. From the strain find the extension) 3 2 2. An iron body of volume 145 in is subjected to a pressure of 500 Ib/in . What is the decrease in volume of this body? 2 3. A copper rod has a cross-sectional area of 0.04 in and a length of 24". What longitudinal force must be applied to cause this rod to stretch by 0.0024 in? 2 4. An aluminium brace inside a wing of a plane has a cross-sectional area of 0.2 in . What is the greatest longitudinal force that can be applied to the brace without causing the brace to be permanently deformed? Answers 1. 0.0012 in. 2. 0.05 in3 3. 64 Ib. 4. 3800 Ib.

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Nature and Properties of Solids, Liquids and Gas All matter exists in one of three states - solid, liquid or gas. The following notes characterize the three states:

Solid  The greatest forces of attraction are between the particles in a solid and they pack together in a neat and ordered arrangement.  The particles are too strongly held together to allow movement from place to place but the particles vibrate about there position in the structure.  With increase in temperature, the particles vibrate faster and more strongly as they gain kinetic energy.

Figure 0-6 Atom arrangement in a solid The properties of a Solid  Solids have the greatest density ('heaviest') because the particles are closest together.  Solids cannot flow freely like gases or liquids because the particles are strongly held in fixed positions.  Solids have a fixed surface and volume (at a particular temperature) because of the strong particle attraction.  Solids are extremely difficult to compress because there is no real 'empty' space between the particles.  Solids will expand a little on heating but nothing like as much as liquids because of the greater particle attraction restricting the expansion (contract on cooling). The expansion is caused by the increased strength of particle vibration.

Liquid  Much greater forces of attraction between the particles in a liquid compared to gases, but not quite as much as in solids.  Particles quite close together but still arranged at random throughout the container, there is a little close range order as you can get clumps of particles clinging together temporarily.  Particles moving rapidly in all directions but more frequently colliding with each other than in gases.  With increase in temperature, the particles move faster as they gain kinetic energy.

Properties of a Liquid  Liquids have a much greater density than gases ('heavier') because the particles are much closer together.  Liquids flow freely despite the forces of attraction between the particles but liquids are not as 'fluid' as gases.  Liquids have a surface, and a fixed volume (at a particular temperature) because of the increased particle attraction, but the shape is not fixed and is merely that of the container itself.  Liquids are not readily compressed because of the lack of 'empty' space between the particles.  Liquids will expand on heating (contract on cooling) but nothing like as much as gases because of the greater particle attraction restricting the expansion. When heated, the liquid particles gain kinetic energy and hit the sides of the container more frequently, and more significantly, they hit with a greater force, so in a sealed container the pressure produced can be considerable.

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Gas  Almost no forces of attraction between the particles which are completely free of each other.  Particles widely spaced and scattered at random throughout the container so there is no order in the system.  Particles moving rapidly in all directions, frequently colliding with each other and the side of the container.  With increase in temperature, the particles move faster as they gain kinetic energy.

Properties of a Gas  Gases have a low density ('light') because the particles are so spaced out in the container (density = mass -5- volume).  Gases flow freely because there are no effective forces of attraction between the particles.  Gases have no surface, and no fixed shape or volume, and because of lack of particle attraction, they spread out and fill any container.  Gases are readily compressed because of the 'empty' space between the particles.  If the 'container' volume can change, gases readily expand on heating because of the lack of particle attraction, and readily contract on cooling. On heating, gas particles gain kinetic energy and hit the sides of the container more frequently, and more significantly, they hit with a greater force. Depending on the container situation, either or both of the pressure or volume will increase (reverse on cooling).  The natural rapid and random movement of the particles means that gases readily 'spread' or diffuse. Diffusion is fastest in gases where there is more space for them to move and the rate of diffusion increases with increase in temperature.

Figure 2-7 States of matter

Changes of State We can use the diagrams shown below, to explain changes of state and the energy changes involved.

Figure 2-8 Gas to liquid transformation

Evaporation and Boiling (liquid to gas) In evaporation and boiling the highest kinetic energy molecules can 'escape' from the attractive forces of the other liquid particles. The particles lose any order and become completely free. Energy is needed to overcome the attractive forces in the liquid and is taken in from the surroundings. This means heat is taken in (endothermic). Boiling is rapid evaporation at a fixed temperature called the boiling point and requires

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Part 66 Training Syllabus Module 2 Physics continuous addition of heat. Evaporation takes place more slowly at any temperature between the melting point and boiling point and results in the liquid becoming cooler.

Condensing (gas to liquid) On cooling, gas particles lose kinetic energy and eventually become attracted together to form a liquid. There is an increase in order as the particles are much closer together and can form clumps of molecules. The process requires heat to be lost to the surroundings i.e. heat given out, so condensation is exothermic.

Figure 2.9: Liquid to solid transformation.

Melting (solid to liquid) When a solid is heated the particles vibrate more strongly and the particle attractive forces are weakened. Eventually, at the melting point, the attractive forces are too weak to hold the structure together and the solid melts. The particles become free to move around and lose their ordered arrangement. Energy is needed to overcome the attractive forces, so heat is taken in from the surroundings and melting is an endothermic process.

Freezing (liquid to solid) On cooling, liquid particles lose kinetic energy and become more strongly attracted to each other. Eventually at the freezing point the forces of attraction are sufficient to remove any remaining freedom and the particles come together to form the ordered solid arrangement. Since heat must be removed to the surroundings freezing is an exothermic process.

Figure 2-9 Phase changes

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Summary

Figure 2-10 Solid, liquid and gas summaries

Pressure and Force The terms force and pressure are used extensively in the study of fluids. It is essential that we distinguish between the terms. Force means a total push or pull. It is the push or pull exerted against the total area of a particular surface and is expressed in pounds or grams. Pressure means the amount of push or pull (force) applied to each unit area of the surface and is expressed in pounds per square inch (Ib/in ) or Newtons per 2 square meter (N/m ). Pressure maybe exerted in one direction, in several directions, or in all directions.

Computing Force, Pressure, and Area A formula is used in computing force, pressure, and area in fluid power systems. In this formula, P refers to pressure, F indicates force, and A represents area. Force equals pressure times area. Thus, the formula is written: F=PxA Pressure equals force divided by area. By rearranging the formula, this statement may be condensed into

Since area equals force divided by pressure, the formula is written:

Figure 2-11 The Pressure, Force and Area equation finder Figure 2.11 illustrates a memory device for recalling the different variations of this formula. Any letter in the triangle may be expressed as the product or quotient of the other two, depending on its position within the triangle. For example, to find area, consider the letter A as being set off to itself, followed by an equal sign. Now look at the other two letters. The letter F is above the letter P; therefore, NOTE: Sometimes the area may not be expressed in square units. If the surface is rectangular, you can determine its area by multiplying its length (say, in inches) by its width (also in inches). The majority of areas you will consider in these calculations are circular in shape. Either the radius or the diameter may be given, but you must know the radius in inches to find the area. The radius is one-half the diameter. To determine the area, use the formula for finding the area of a circle. This is written A = 2 2 πr , where A is the area, π is 3.1416 (3.14 or 22/7 for most calculations), and r indicates the radius squared.

Atmospheric Pressure Recall that the atmosphere is the entire mass of air that surrounds the earth. While it extends upward for about 500 miles, the section of primary interest is the portion that rests on the earth's surface and extends upward for about 7½ miles. This layer is called the troposphere.

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Part 66 Training Syllabus Module 2 Physics If a column of air 1-inch square extending all the way to the "top" of the atmosphere could be weighed, this column of air would weigh approximately 14.7 pounds at sea level. Thus, atmospheric pressure at sea level is approximately 14.7 PSI. As one ascends, the atmospheric pressure decreases by approximately 1.0 PSI for every 2,343 feet. However, below sea level, in excavations and depressions, atmospheric pressure increases. Pressures under water differ from those under air only because the weight of the water must be added to the pressure of the air. Atmospheric pressure can be measured by any of several methods. The common laboratory method uses the mercury column barometer. The height of the mercury column serves as an indicator of atmospheric pressure. At sea level and at a temperature of 20° Celsius (C), the height of the mercury column is 29.92 inches, or 760 millimetres. This represents a pressure of approximately 14.7 PSI. The 30-inch column is used as a reference standard. Another device used to measure atmospheric pressure is the aneroid barometer. The aneroid barometer uses the change in shape of an evacuated metal cell to measure variations in atmospheric pressure (figure 2.12). The thin metal of the aneroid cell moves in or out with the variation of pressure on its external surface. This movement is transmitted through a system of levers to a pointer, which indicates the pressure. The atmospheric pressure does not vary uniformly with altitude. It changes more rapidly at lower altitudes because of the compressibility of the air, which causes the air layers close to the earth's surface to be compressed by the air masses above them. This effect, however, is partially counteracted by the contraction of the upper layers due to cooling. The cooling tends to increase the density of the air.

Figure 2-12 Aneroid barometer Atmospheric pressures are quite large, but in most instances practically the same pressure is present on all sides of objects so that no single surface is subjected to a great load. Atmospheric pressure acting on the surface of a liquid is transmitted equally throughout the liquid to the walls of the container, but is balanced by the same atmospheric pressure acting on the outer walls of the container. In figure 2.13, atmospheric pressure acting on the surface of one piston is balanced by the same pressure acting on the surface of the other piston. The different areas of the two surfaces make no difference, since for a unit of area, pressures are balanced.

Transmission of Forces Through Liquids When the end of a solid bar is struck, the main force of the blow is carried straight through the bar to the other end (figure 2.14, view A). This happens because the bar is rigid. The direction of the blow almost entirely determines the direction of the transmitted force.

Figure 2-13 Forces acting on liquids

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Figure 2-14 Forces acting on solids (A) and liquids (B) When a force is applied to the end of a column of confined liquid (figure 2.14, view B), it is transmitted straight through to the other end and also equally and undiminished in every direction throughout the column—forward, backward, and sideways—so that the containing vessel is literally filled with pressure. An example of this distribution of force is illustrated in figure 2.15. The outward push of the water is equal in every direction. So far we have explained the effects of atmospheric pressure on liquids and how external forces are distributed through liquids. Let us now focus our attention on forces generated by the weight of liquids themselves. To do this, we must first discuss density, specific gravity, and Pascal's law.

Figure 2-15 Flat and water filled water hoses

Pascal's Law The foundation of modern hydraulics was established when Pascal discovered that pressure in a fluid acts equally in all directions. This pressure acts at right angles to the containing surfaces. If some type of pressure gauge, with an exposed face, is placed beneath the surface of a liquid (figure 2.16) at a specific depth and pointed in different directions, the pressure will read the same. Thus, we can say that pressure in a liquid is independent of direction. Figure 2-16 Pascals Law Pressure due to the weight of a liquid, at any level, depends on the depth of the fluid from the surface. If the exposed face of the pressure gauges, are moved closer to the surface of the liquid, the indicated pressure will be less. When the depth is doubled, the indicated pressure is doubled. Thus the pressure in a liquid is directly proportional to the depth. Consider a container with vertical sides (fig. 2.17) that is 1 foot long and 1 foot wide. Let it be filled with water 1 foot deep, providing 1 cubic foot of water. We learned earlier in this chapter that 1 cubic foot of water weighs 62.4 pounds. Using this information and the equation, P = F/A, we can calculate the pressure on the bottom of the container.

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= 62.4 Ib/ft

2

Since there are 144 square inches in 1 square foot, 0.433 Ib/in

2

This can be stated as follows: the weight of a column of water 1 foot high, having a cross-sectional area of 1 square inch, is 0.433 pound. If the depth of the column is tripled, the weight of the column will be 3 x 0.433, 2 or 1.299 pounds, and the pressure at the bottom will be 1 .299 Ib/in (PSI), since pressure equals the force divided by the area. Thus, the pressure at any depth in a liquid is equal to the weight of the column of liquid at that depth divided by the cross-sectional area of the column at that depth. The volume of a liquid that produces the pressure is referred to as the fluid head of the liquid. The pressure of a liquid due to its fluid head is also dependent on the density of the liquid. If we let A equal any cross-sectional area of a liquid column and h equal the depth of the column, the volume becomes Ah. Using the equation for density, D = Weight (W) / Volume (V), the weight of the liquid above area A is equal to AhD.

W = Ahd Water Weight = 62.4 Ib 2 Water Pressure = 62.4 Ib/ft

0.433 Ib/in

2

Figure 2-17 A body of water Since pressure is equal to the force per unit area, set A equal to 1. Then the formula pressure becomes P = hD It is essential that h and D be expressed in similar units. That is, if D is expressed in pounds per cubic foot, the value of h must be expressed in feet. If the desired pressure is to be expressed in pounds per square inch, the pressure formula becomes:

Pressure and Force in Fluid Power Systems Pascal was also the first to prove by experiment that the shape and volume of a container in no way alters pressure. Thus in figure 2-17, if the pressure due to the weight of the liquid at a point on horizontal line H is 8 PSI, the pressure is 8 PSI everywhere at level H in the system. The equation P=F/A also shows that the pressure is independent of the shape and volume of a container. If there is a resistance on the output piston and the input piston is pushed downward, a pressure is created through the fluid, which acts equally at right angles to surfaces in all parts of the container. If force 1 is 100 pounds and the area of the input piston is 10 square inches, then the pressure in the fluid is 10 PSI.

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Surface A HEAD H

Figure 2-18: Hydrostatic pressure in different shaped containers

Figure 2-19: Forces, pistons and pressure Recall that, according to Pascal's law, any force applied to a confined fluid is transmitted in all directions throughout the fluid. NOTE: Fluid pressure cannot be created without resistance to flow. In this case, resistance is provided by the equipment to which the output piston is attached. The force of resistance acts against the top of the output piston. The pressure created in the system by the input piston pushes on the underside of the output piston with a force of 10 pounds on each square inch.

Figure 2-20: Pressure due to forces on pistons In this case, the fluid column has a uniform cross section, so the area of the output piston is the same as the area of the input piston, or 10 square inches. Therefore, the upward force on the output piston is 100 pounds (10 PSI x 10 sq. in.), the same as the force applied to the input piston. All that was accomplished in this system was to transmit the 100-pound force around the bend. However, this principle underlies practically all mechanical applications of fluid power. At this point you should note that since Pascal's law is independent of the shape of the container, it is not necessary that the tube connecting the two pistons have the same cross-sectional area of the pistons. A connection of any size, shape, or length will do, as long as an unobstructed passage is provided. Therefore, the system shown in figure 2-20, with a relatively small, bent pipe connecting two cylinders, will act exactly the same as the system shown in figure 2-19.

The Hydraulic Ram Principle Consider the situation in figure 2-21, where the input piston is much smaller than the output piston. Assume that the area of the input piston is 2 square inches. With a resistant force on the output piston a downward force of 20 pounds acting on the input piston creates a pressure 10 PSI in the fluid.

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Figure 2-21: The hydraulic ram principle Although this force is much smaller than the force applied in figures 2-19 and 2-20, the pressure is the same. This is because the force is applied to a smaller area. This pressure of 10 PSI acts on all parts of the fluid container, including the bottom of the output piston. The upward force on the output piston is 200 pounds (10 pounds of pressure on each square inch). In this case, the original force has been multiplied tenfold while using the same pressure in the fluid as before. In any system with these dimensions, the ratio of output force to input force is always ten to one, regardless of the applied force. For example, if the applied force of the input piston is 50 pounds, the pressure in the system will be 25 PSI. This will support a resistant force of 500 pounds on the output piston. The system works the same in reverse. If we change the applied force and place a 200-pound force on the output piston (figure 2-21), making it the input piston, the output force on the input piston will be one-tenth the input force, or 20 pounds. (Sometimes such results are desired.) Therefore, if two pistons are used in a fluid power system, the force acting on each piston is directly proportional to its area, and the magnitude of each force is the product of the pressure and the area of each piston. Note the white arrows at the bottom of figure 2-21 that indicate up and down movement. The movement they represent will be explained later in the discussion of volume and distance factors.

Differential Areas

Figure 2-22: Pressures on unequal areas Consider the special situation shown in figure 2-22. Here, a single piston (1) in a cylinder (2) has a piston rod (3) attached to one of its sides. The piston rod extends out of one end of the cylinder. Fluid under pressure is admitted equally to both ends of the cylinder. The opposed faces of the piston (1) behave like two pistons acting against each other. The area of one face is the full cross-sectional area of the cylinder, say 6 square inches, while the area of the other face is the area of the cylinder minus the area of the piston rod, which is 2 square inches. This leaves an effective area of 4 square inches on the right face of the piston. The pressure on both faces is the same, in this case, 20 PSI. Applying the rule just stated, the force pushing the piston to the right is its area times the pressure, or 120 pounds (20 x 6). Likewise, the force pushing the piston to the left is its area times the pressure, or 80 pounds (20 x 4). Therefore, there is a net unbalanced force of 40 pounds acting to the right, and the piston will move in that direction. The net effect is the same as if the piston and the cylinder had the same cross-sectional area as the piston rod.

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Volume and Distance Factors You have learned that if a force is applied to a system and the cross-sectional areas of the input and output pistons are equal, as in figures 2-19 and 2-20, the force on the input piston will support an equal resistant force on the output piston. The pressure of the liquid at this point is equal to the force applied to the input piston divided by the piston's area. Let us now look at what happens when a force greater than the resistance is applied to the input piston. In the system illustrated in figure 2-21, assume that the resistance force on the output piston is 100 PSI. If a force slightly greater than 100 pounds is applied to the input piston, the pressure in the system will be slightly greater than 10 PSI. This increase in pressure will overcome the resistance force on the output piston. Assume that the input piston is forced downward 1 inch. The movement displaces 10 cubic inches of fluid. The fluid must go somewhere. Since the system is closed and the fluid is practically incompressible, the fluid will move to the right side of the system. Because the output piston also has a cross-sectional area of 10 square inches, it will move 1 inch upward to accommodate the 10 cubic inches of fluid. You may generalize this by saying that if two pistons in a closed system have equal cross-sectional areas and one piston is pushed and moved, the other piston will move the same distance, though in the opposite direction. This is because a decrease in volume in one part of the system is balanced by one equal increase in volume in another part of the system. Apply this reasoning to the system in figure 2-21. If the input piston is pushed down a distance of 1 inch, the volume of fluid in the left cylinder will decrease by 2 cubic inches. At the same time, the volume in the right cylinder will increase by 2 cubic inches. Since the diameter of the right cylinder cannot change, the piston must move upward to allow the volume to increase. The piston will move a distance equal to the volume increase divided by the surface area of the piston (equal to the surface area of the cylinder). In this example, the piston will move one-tenth of an inch (2 cu. in. + 20 sq. in.). This leads to the second basic rule for a fluid power system that contains two pistons: The distances the pistons move are inversely proportional to the areas of the pistons. Or more simply, if one piston is smaller than the other, the smaller piston must move a greater distance than the larger piston any time the pistons move.

Relationship between Force, Pressure, and Head In dealing with fluids, forces are usually considered in relation to the areas over which they are applied. As previously discussed, a force acting over a unit area is a pressure, and pressure can alternately be stated in pounds per square inch or in terms of head, which is the vertical height of the column of fluid whose weight would produce that pressure. In most of the applications of fluid power, applied forces greatly outweigh all other forces, and the fluid is entirely confined. Under these circumstances it is customary to think of the forces involved in terms of pressures. Since the term head is encountered frequently in the study of fluid power, it is necessary to understand what it means and how it is related to pressure and force. At this point you need to review some terms in general use. "Gravity head", when it is important enough to be considered, is sometimes referred to as simply "head". The effect of atmospheric pressure is referred to as "atmospheric pressure" (Atmospheric pressure is frequently and improperly referred to as suction). Inertia effect, because it is always directly related to velocity, is usually called "velocity head"; and friction, because it represents a loss of pressure or head, is usually referred to as "friction head".

Static and Dynamic Factors Gravity, applied forces, and atmospheric pressure are static factors that apply equally to fluids at rest or in motion, while inertia and friction are dynamic factors that apply only to fluids in motion. The mathematical sum of gravity, applied force, and atmospheric pressure is the static pressure obtained at any one point in a fluid at any given time. Static pressure exists in addition to any dynamic factors that may also be present at the same time. Remember, Pascal's law states that a pressure set up in a fluid acts equally in all directions and at right angles to the containing surfaces. This covers the situation only for fluids at rest or practically at rest. It is true only for the factors making up static head. Obviously, when velocity becomes a factor it must have a direction, and as previously explained, the force related to the velocity must also have a direction, so that Pascal's law alone does not apply to the dynamic factors of fluid power. The dynamic factors of inertia and friction are related to the static factors. Velocity head and friction head are obtained at the expense of static head. However, a portion of the velocity head can always be reconverted to

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Part 66 Training Syllabus Module 2 Physics static head. Force, which can be produced by pressure or head when dealing with fluids, is necessary to start a body moving if it is at rest, and is present in some form when the motion of the body is arrested, therefore, whenever a fluid is given velocity, some part of its original static head is used to impart this velocity, which then exists as velocity head.

Operation of Hydraulic Components To transmit and control power through pressurized fluids, an arrangement of inter-connected components is required. Such an arrangement is commonly referred to as a system. The number and arrangement of the components vary from system to system, depending on the particular application. In many applications, one main system supplies power to several subsystems, which are sometimes referred to as circuits. The complete system may be a small compact unit; more often, however, the components are located at widely separated points for convenient control and operation of the system. The basic components of a fluid power system are essentially the same, regardless of whether the system uses a hydraulic or a pneumatic medium. There are five basic components used in a system. These basic components are as follows:     

Reservoir or receiver Pump or compressor Lines (pipe, tubing, or flexible hose) Directional control valve Actuating device

Several applications of fluid power require only a simple system; that is, a system which uses only a few components in addition to the five basic components. A few of these applications are presented in the following paragraphs. We will explain the operation of these systems briefly at this time so you will know the purpose of each component and can better understand how hydraulics is used in the operation of these systems.

Hydraulic Jack The hydraulic jack is perhaps one of the simplest forms of a fluid power system. By moving the handle of a small device, an individual can lift a load weighing several tons. A small initial force exerted on the handle is transmitted by a fluid to a much larger area. To understand this better, study figure 2-23 right. The small input piston has an area of 5 square inches and is directly connected to a large cylinder with an output piston having an area of 250 square inches. The top of this piston forms a lift platform. If a force of 25 pounds is applied to the input piston, it produces a pressure of 5 PSI in the fluid, that is, of course, if a sufficient amount of resistant force is acting against the top of the output piston. Disregarding friction loss, this pressure acting on the 250 square inch area of the output piston will support a resistance force of 1,250 pounds. In other words, this pressure could overcome a force of slightly under 1,250 pounds. An input force of 25 pounds has been transformed into a working force of more than half a ton; however, for this to be true, the distance travelled by the input piston must be 50 times greater than the distance travelled by the output piston. Thus, for every inch that the input piston moves, the output piston will move only one-fiftieth of an inch. This would be ideal if the output piston needed to move only a short distance. However, in most instances, the output piston would have to be capable of moving a greater distance to serve a practical application. The device shown in figure 2-23 is not capable of moving the output piston farther than that shown; therefore, some other means must be used to raise the output piston to a greater height.

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Figure 2-23: A simple hydraulic jack The output piston can be raised higher and maintained at this height if additional components are installed as shown in figure 2-24. In this illustration the jack is designed so that it can be raised, lowered, or held at a constant height. These results are attained by introducing a number of valves and also a reserve supply of fluid to be used in the system. Notice that this system contains the five basic components the reservoir; cylinder 1, which serves as a pump; valve 3, which serves as a directional control valve; cylinder 2, which serves as the actuating device; and lines to transmit the fluid to and from the different components. Figure 2-24 (A): A hydraulic jack with valves In addition, this system contains two valves, 1 and 2, whose functions are explained in the following discussion. As the input piston is raised (fig. 2-24, view A), valve 1 is closed by the back pressure from the weight of the output piston. At the same time, valve 2 is opened by the head of the fluid in the reservoir. This forces fluid into cylinder 1. Figure 224 (B): A hydraulic jack with valves When the input piston is lowered (fig. 2-24, view B), a pressure is developed in cylinder 1. When this pressure exceeds the head in the reservoir, it closes valve 2. When it exceeds the back pressure from the output piston, it opens valve 1, forcing fluid into the pipeline. The pressure from cylinder 1 is thus transmitted into cylinder 2, where it acts to raise the output piston with its attached lift platform. When the input piston is again raised, the pressure in cylinder 1 drops below that in cylinder 2, causing valve 1 to close. This prevents the return of fluid and holds the output piston with its attached lift platform at its new level. During this stroke, valve 2 opens again allowing a new supply of fluid into cylinder 1 for the next power (downward) stroke of the input piston. Thus, by repeated strokes of the input piston, the lift platform can be progressively raised. To lower the lift platform, valve 3 is opened, and the fluid from cylinder 2 is returned to the reservoir.

Hydraulic Brakes The hydraulic brake system used in the automobile is a multiple piston system. A multiple piston system allows forces to be transmitted to two or more pistons in the manner indicated in figure 2-25. Note that the pressure set up by the force applied to the input piston (1) is transmitted undiminished to both output pistons (2 and 3), and that the resultant force on each piston is proportional to its area. The multiplication of forces from the input piston to each output piston is the same as that explained earlier. The hydraulic brake system from the master cylinders to the wheel cylinders on most automobiles operates in a way similar to the system illustrated in figure 2-26. Figure 2-25: Principle of hydraulic brake

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Part 66 Training Syllabus Module 2 Physics When the brake pedal is depressed, the pressure on the brake pedal moves the piston within the master cylinder, forcing the brake fluid from the master cylinder through the tubing and flexible hose to the wheel cylinders. The wheel cylinders contain two opposed output pistons, each of which is attached to a brake shoe fitted inside the brake drum.

Figure 2-26: Hydraulic brake system Each output piston pushes the attached brake shoe against the wall of the brake drum, thus retarding the rotation of the wheel. When pressure on the pedal is released, the springs on the brake shoes return the wheel cylinder pistons to their released positions. This action forces the displaced brake fluid back through the flexible hose and tubing to the master cylinder. The force applied to the brake pedal produces a proportional force on each of the output pistons, which in turn apply the brake shoes frictionally to the turning wheels to retard rotation. As previously mentioned, the hydraulic brake system on most automobiles operates in a similar way, as shown in figure 2-26. It is beyond the scope of this manual to discuss the various brake systems.

Accumulators An accumulator is a pressure storage reservoir in which hydraulic fluid is stored under pressure from an external source. The storage of fluid under pressure serves several purposes in hydraulic systems. In some aircraft hydraulic systems it is necessary to maintain the system pressure within a specific pressure range for long periods of time. It is very difficult to maintain a closed system without some leakage, either external or internal. Even a small leak can cause a decrease in pressure. By using an accumulator, leakage can be compensated for and the system pressure can be maintained within an acceptable range for long periods of time. Accumulators also damp out fluctuations in pressure due to the operation of services such as control surfaces and landing gear. They can supply extra pressure when all the hydraulic services are being operated at one time (flaps, control surfaces, landing gear etc.) and when the hydraulic pump is unable to cope. They can also be used in an emergency when all other hydraulic power pressure supplies (pumps etc) have failed. Thus a large modern aircraft can be controlled on accumulator power alone, for up to an hour. Accumulators also compensate for thermal expansion and contraction of the liquid due to variations in temperature.

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Part 66 Training Syllabus Module 2 Physics

Figure 2-27: Hydraulic Accumulator

The accumulator consists of an air chamber, which is charged with air or nitrogen. This is called the precharge pressure and is usually about 1000 PSI. This pressure is measured when there is no hydraulic pressure. The air chamber is the under side of the piston shown in figure 2-27. With no hydraulic pressure, the air/nitrogen pressure will push the piston to the top of the accumulator. A pressure gauge may be attached to the accumulator to indicate the air/nitrogen pressure. When the hydraulic pumps are switched on, the hydraulic pressure (acting on top of the piston, in opposition to the air/nitrogen pressure) begins to rise. When the hydraulic pressure exceeds the air/nitrogen pre-charge pressure (1000 PSI), the piston will begin to move down and further compress the air/nitrogen pressure. At all times that the hydraulic pressure is above the air/nitrogen pre-charge pressure of 1000 PSI, the air/nitrogen and the hydraulic pressures are equal. Thus when the hydraulic pressure has reached its working level of 3000 PSI, the air/nitrogen pressure is also 3000 PSI.

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Part 66 Training Syllabus Module 2 Physics

It is the additional pressure supplied to the air/nitrogen by the hydraulic pressure, which can be used to feed back the pressure to the hydraulic fluid if the hydraulic fluid pressure falls below that of the air/nitrogen. However, when the air/nitrogen gauge indicates 1000 PSI, the hydraulic pressure is zero, since the air/nitrogen has expanded back to its original pre-charge pressure. Problems 1. Calculate the pressure on a gas when a force of 3100 N is exerted on a piston of diameter 2 cm 2. Calculate the force exerted when a pressure of 1 bar acts on a piston of diameter 8 cm which has a piston rod of diameter 2 cm taking some of the piston area. 3. The piston face area in the hydraulic jack shown above is 0.3 sq.in. The rod cross sectional area is 0.1 sq.in. Calculate the force and direction the ram rod will move if a pressure of 12 PSI enters equally into both sides of the cylinder chamber.

4.

5.

A brake master cylinder has a piston diameter of 0.4 ins. It feeds pressure to 4 identical wheel cylinders, each having just one piston of diameter 2 ins. What is the force on one wheel brake when the driver applies a force of 80 Ibs to the master cylinder? An hydraulic accumulator is charged with nitrogen to 600 PSI. The hydraulic pump is then switched on and it feeds 3000 PSI to the other side of the accumulator piston. What will be the new pressure on the nitrogen side of the accumulator?

Answers 1. 10MPa 2. 450 N 3. 1.2lbs, right 4. 2000 Ibs 5. 3000 PSI

Barometers A barometer is an instrument used to measure atmospheric pressure. It can measure the pressure exerted by the atmosphere by using water, air, or mercury.

Mercury barometers A standard mercury barometer has a glass tube of about 30 inches (about 76 cm) in height, closed at one end, with an open mercury-filled reservoir at the base. Mercury in the tube adjusts until the weight of the mercury column balances the atmospheric force exerted on the reservoir. High atmospheric pressure places more force on the reservoir, forcing mercury higher in the column. Low pressure allows the mercury to drop to a lower level in the column by lowering the force placed on the reservoir. Since higher temperature at the instrument will reduce the density of the mercury the scale for reading the height of the mercury is adjusted to compensate for this effect. The standard temperature for reading a mercury barometer is 0°C (32°F). A correction factor is read from a graph and applied to the reading for temperatures above 0°C. The barometer over-reads at higher temperatures. Figure 2-28:Mercury Barometer The mercury barometer's design gives rise to the expression of atmospheric pressure in inches or millimetres: the pressure is quoted as the level of the mercury's height in the vertical column. 1 atmosphere is equivalent to about 29.9 inches, or 760 millimetres, of mercury. Barometers of this type normally measure atmospheric pressures between 28 and 31 inches of mercury. The reading from a barometer (in mm.hg or in.hg) can be converted into Pascals or PSI by using the formula Pressure = pgh (if using Metric units) Or

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Part 66 Training Syllabus Module 2 Physics Pressure = ph (if using English units) Where: Pressure = the converted pressure (Pa or PSI) 3 3 p = density of the mercury (13,600 kg/m or 62.4 lbf./ft ) h = height of mercury (m or ft)

Aneroid barometers An aneroid barometer uses a small, flexible metal vacuum chamber called an aneroid cell. This aneroid capsule (cell) is made from an alloy of beryllium and copper. The evacuated capsule (or usually more capsules) is prevented from collapsing by a strong spindle spring. Small changes in external air pressure cause the cell to expand or contract. This expansion and contraction drives mechanical levers such that the tiny movements of the capsule are amplified and displayed on the face of the aneroid barometer. Many models include a manually set needle which is used to mark the current measurement so a change can be seen. In addition, the mechanism is made deliberately 'stiff so that tapping the barometer reveals whether the pressure is rising or falling as the pointer moves. They are used for measuring atmospheric pressure. Fig.2-29: Aneroid Barometers

Buoyancy Archimedes Principle Archimedes was a Greek philosopher and mathematician who lived about 250BC. There is a story (maybe even true) about Archimedes that every physics student should hear. It goes as follows: The king who ruled Greece at that time asked his royal metalworkers to make him a gold crown. When the crown was delivered it was indeed beautiful. However, the king suspected that the crown was not pure gold. He did not want to destroy the crown but he wanted to know if he had been cheated. What he needed was some type of non-destructive evaluation (NDE dates back many years!). He asked Archimedes to solve his problem. Archimedes pondered the question. The density (mass / volume) of gold was well known. He knew of course how to determine the weight and mass of the crown by simple weighing. However, since the crown did not have a regular shape it was impossible to determine the volume by a mathematical calculation. The solution came to Archimedes one day when his servant filled his bathtub too full. As Archimedes stepped into his bath, he noticed that a volume of water equal to his volume overflowed! With a flash of insight he ran through Athens, stark naked, shouting "Eureka, Eureka, I have the solution!" The experiment was performed, the king was notified that his crown was not pure gold and the royal metal workers lost their lives. The point of the above story is that a body submerged in a liquid displaces a volume of water equal to its own volume. A corollary is that a body that floats in a liquid displaces a volume of liquid less than its volume since some portion of the body is above the water level.

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Part 66 Training Syllabus Module 2 Physics Archimedes' principle, states that a body immersed in a fluid is buoyed up by a force equal to the weight of the displaced fluid. The principle applies to both floating and submerged bodies and to all fluids, i.e., liquids and gases. It explains not only the buoyancy of ships and other vessels in water but also the rise of a balloon in the air and the apparent loss of weight of objects underwater. In determining whether a given body will float in a given fluid, both weight and volume must be considered; that is, the relative density, or weight per unit of volume, of the body compared to the fluid determines the buoyant force. If the body is less dense than the fluid, it will float or, in the case of a balloon, it will rise. If the body is denser than the fluid, it will sink. Figure 2-30: Archimedes' principle Relative density also determines the proportion of a floating body that will be submerged in a fluid. If the body is two thirds as dense as the fluid, then two thirds of its volume will be submerged, displacing in the process a volume of fluid whose weight is equal to the entire weight of the body. In the case of a submerged body, the apparent weight of the body is equal to its weight in air less the weight of an equal volume of fluid. The fluid most often encountered in applications of Archimedes' principle is water, and the specific gravity of a substance is a convenient measure of its relative density compared to water. In calculating the buoyant force on a body, however, one must also take into account the shape and position of the body. A steel rowboat placed on end into the water will sink because the density of steel is much greater than that of water. However, in its normal, keel-down position, the effective volume of the boat includes all the air inside it, so that its average density is then less than that of water, and as a result it will float.

Archimedes' Principle Applied to Bodies that Float A body will float in any liquid that has a weight density greater than the weight density of the body. For 3 3 example a body of weight density 63.4 Ibs./ft. would float in ocean water (D = 64.4 Ibs./ft. ) and sink in lake 3 water (D = 62.4 Ibs./ft. ). When bodies float they can float "high" or float "low". The ratio of the weight density of the floating body relative to the weight density of the liquid determines exactly how high or low a body will float. In order to understand Archimedes' Principle as applied to floating bodies, let us consider a submarine and 3 3 imagine that a block of wood of weight density 48.3 Ibs./ft. and volume 2 ft. is thrust out of the hatch of a submarine into the ocean water. We know intuitively that this block o1 wood will rise to the ocean surface. 3

3

The weight of the block is (48.3 Ibs./ft. ) (2 ft. ) = 96.6 Ibs. As long as the block is below the water surface 3 (while it is rising to the top), it displaces 2 ft. of ocean water. We know that:

3

3

BF = weight of displaced ocean water = (64.4 Ibs./ft. ) (2 ft. ) BF = 128.8 Ibs.

Figure 2-31: Archimedes' principle

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Part 66 Training Syllabus Module 2 Physics We can see why the block rises. How far will the block rise? it will rise until the BF exactly equals its weight. In our example it will rise until the BF has been reduced to 96.6 Ibs. (the weight of the block). The BF will be reduced as the block emerges from the water. In our example, it will rise until 25% of the block's volume is 3 3 above the water surface. It follows that 75% of 2 ft. (= 1.5 ft. ) will be below the water surface. When this 3 3 occurs, the BF on the block is (64.4 lbs./ft. )(1.5 ft. ) equals 96.6 Ibs. Note again that the BF equals the weight of the block while the block is floating. 3

In the preceding example, note that the ratio of the weight density of the block (48.3 Ibs./ft. ) to the weight 3 density of the ocean water (64.4 Ibs./ft. ) was 0.75. We recall that 75% of the floating block was under water. This is generally true and makes a much easier procedure to determine how low a block will float in a given liquid. In dealing with bodies that float, it is important to note that boats, made of materials more dense than water, are shaped in such a way that the total weight density is less than water. In order to understand this, consider the rowing boat with contents (people, lunch, fishing gear, etc.) shown in figure 2-32. Note that some of the boat is below the water surface. Suppose that the row boat floats in such a way that it displaces 8 cu ft. of lake water. The weight of the displaced water is 8 ft. (62.4 3 Ibs./ft. ) or 499 Ibs. Therefore, the BF is 499 Ibs. The boat and contents must weigh 499 Ibs. to float at this level. If the boat weighs 150 Ibs. the contents must weigh 349 Ibs. Figure 2-32: This is realistic (father 200 Ibs., son 75 Ibs., lunch 25 Ibs., fishing gear 49 Ibs.). One final comment should be made regarding submarines. Submarines cruising at a definite depth in ocean 3 water have a total weight density equal to the weight density of ocean water, 64.4 Ibs./ft. . This means that the total weight of the submarine (metal shell, air, crew, load, ballast, etc.) divided by the total volume is 64.4 3 Ibs./ft. . The ballast used in submarines is ocean water. These vessels can take on water or pump out water. If the submarine wants to descend, it takes on water. If it wants to rise toward the surface it pumps out water.

EXAMPLE: 3 3 A block of oak (D = 45 Ibs./ft. ) is placed in a tank of benzene (D = 54.9 Ibs /ft. ). The oak floats since its weight density is less that the weight density of the benzene. What percentage of the oak will be below the surface of the benzene? We find the ratio of the two weight densities.

We conclude that 82% of the oak block will be below the surface of the benzene.

Archimedes' Principles as Applied to Airships and Balloons In all of the above materials, we have talked about Archimedes' principle as if it applied only to liquids. Since most of our experience with this principle is with liquids, it seemed easier to do this at first. However, it must now be emphasized that buoyant forces exist also with gases. The obvious example is that of a hot air balloon or a lighter-than-air aircraft.

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: The bag of a balloon is a sphere of radius 25 m filled with hydrogen of weight density 0.882 N/rn3. What total weight (in Newtons) of fabric, car, and contents can be lifted by this balloon in air of weight density 12.6 N/rn3? We first calculate the volume of the spherical balloon by recalling that the volume of a sphere is given by:

The weight of the hydrogen is found from the formula D V = w: 3

3

(0.882 N/m ) (65,450 m ) = 57,700 N The weight of the displaced air is: 3

3

(12.6 N/m ) (65,450 m ) = 824,700 N Since the weight of the displaced air is the BF we can say that: BF = 824,700 N This BF must hold up the hydrogen, fabric, car, and contents. It follows that fabric, car, and contents weighing 767,000 N can be lifted by this balloon. Note that this number was obtained by subtracting 57,700 N from 824,700 N. Usually balloons are not filled with hydrogen since hydrogen is explosive. Of course, since hydrogen is the lightest of all gases it is the most efficient. However, the danger of explosion outweighs this advantage. The next lightest gas is helium of weight density 1.74 N/m. Usually, balloons are filled with this gas.

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Part 66 Training Syllabus Module 2 Physics

WEIGHT DENSITIES AT 68°F N/m

3

LBS./FT.

3

LIQUIDS Water

9,807

62.4

Ocean Water

10,100

64.4

Benzene

8,620

54.9

Carbon Tetrachloride

15,630

99.5

Ethyl Alcohol

7,740

49.3

Gasoline

6,670

42.5

Kerosene

7,850

49.9

Lubricating OH

8,830

56.2

Methyl Alcohol

7,770

49.4

Sulphuric Acid, 100%

17,960

114.3

Turpentine

8,560

54.5

Aluminum

26,500

169

Cast Iron

70,600

449

Copper

87,200

555

Gold

189,300

1,205

Lead

111,200

708

Magnesium

17,100

109

Nickel

86,600

553

Silver

103,000

656

Steel

76,500

487

Tungsten

186,000

1,190

Zinc

70,000

446

Brass or Bronze

85,300

543

Ice

9,040

57.5

Concrete

22,600

144

Earth, Packed

14,700

94

Glass

25,500

160

Granite

26,500

169

Balsa

1,270

8

Pine

4,700

30

Maple

6,300

40

Oak

7,100

45

SOLID METALS

NONMETALIC SOLIDS

WOODS

Table 2-3: Weight densities for common materials

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Part 66 Training Syllabus Module 2 Physics Problems 3 1. A solid aluminium object of volume 250 ft is resting on the ocean floor. A salvage crew plans to raise this object. What force will be needed? 3 2. A solid steel body of volume 125 ft is to be raised by a salvaging crew to the surface of a lake. What force will be needed? 3. What percentage of an iceberg is below the surface of the ocean? 4. A canoe is floating in such a way that it displaces 6 cu.ft. of lake water. If the canoe weighs 100 Ibs., what is the weight of its contents? 5. A balloon is spherical in shape and has a radius of 20 ft. It is filled with helium (weight density 0.01 3 3 Ib/ft ) and is floating in air (weight density 0.08 Ib/ft ). What is the weight of the balloon (fabric, crew and contents etc.)? Answers 1. 26,200 Ibs. 2. 53,100 Ibs 3. 89% 4. 274 Ibs 5. 2240 Ibs

Kinetics Linear Motion When a body is moving in a straight line with constant speed it is not accelerating. We say, in this case, that it is moving with constant velocity. If a body's velocity is not constant, it is accelerating. A body accelerates if it is changing its speed and/or its direction. When we discuss a body's straight-line motion, then we do not have any change in direction. In this special case, any acceleration is due to a change in speed.

The Equations of Motion In all of the following discussion, certain symbols will be used. These symbols are summarized below: Vav = average velocity t = time u = initial velocity v = final velocity a = acceleration s = distance covered* *NOTE: That 's' is the traditional notation for distance in almost all physics textbooks. This choice reduces confusion with the symbol d for derivative, a concept from calculus. There is a formula dealing with the motion of a body that you have used for many years. In school, you probably memorized the formula in these words: distance = rate (or speed) x time Using our above symbols, we could write: (1)

s = Vavt

Note that for the rate, we have used the average speed. We all know that even though sometimes speed changes, we can always talk about the average speed. Thus, if we travel at an average speed of 50 MPH for 6 hours, we cover 300 miles. Now we must extend our treatment of motion to include the concept of acceleration. Acceleration (for straight-line motion) is the rate of change of speed in time. We define acceleration (for straight-line motion) in the following manner: (2)

a=

In using this formula, a may be either positive or negative. If v is less than u, then our value of a turns out to be a negative number.

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: A truck is initially travelling at a speed of 50 ft./sec. The driver applies his brakes for 15 sec. The final speed of the car is 20 ft./sec. What is the acceleration? a= a= a=

=-2 ft/sec/sec 2

a = -2 ft/sec Notice that the unit of acceleration has the square of a time unit in its denominator.

A little thought will convince you that an acceleration is positive if the body is increasing speed and negative when the body is decreasing its speed. If we cross-multiply in formula (2) we obtain: at = v - u After transposing, we can write: (3)

v = u + at

If an automobile is on an expressway and the driver is increasing speed smoothly and regularly, we note that his average speed is the average of his initial and final speed. The equation can be written: = If this value of Vav is substituted into equation (1), we have: (4)

s=

In this equation, we can substitute for v (= u + at) using the value in equation (3). s=

=

After a bit of algebra, we obtain: (5)

1

2

s = ut + /2 at

Equation (4) can be written, after cross-multiplication: 2s = (u + v)t We can now multiply this equation by equation (2). After cancelling time (t) on the right:

Or

2as = (v - u) (v + u) 2as = -

The final form of this formula is: (6)

2

2

v = u + 2as

These equations are very important. They enable us to deal with all kinds of motion problems where the body is in straight line motion and is changing its speed. These formulas will be summarized below. They will be numbered with Roman numerals and can be referred to by these numbers when used in the problem exercises. i. s= t ii. v = u + at iii. 1 2 s = ut + /2 at 2 2 iv. v = u + 2as

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Part 66 Training Syllabus Module 2 Physics

When a body in straight line motion is not changing speed, or in cases where we are interested only m the average speed, the formula is more simple. s = Vavt Formulas i through iv are used in many practical physics problems. Note that each one involves four quantities. When a problem is given to you to solve, be sure to determine which of these three quantities are given to you, and which quantity is to be found. Choose the formula which involves these four quantities. If the formula is not solved for the unknown quantity solve for this quantity algebraically. Finally substitute the known quantities and solve for the unknown quantity. An example should clarify the above procedure. EXAMPLE: An automobile has an initial speed of 50 ft./sec. and a final speed of 75 ft./sec. While it is undergoing this change of speed, it travels a distance of 125 ft. What is its acceleration? In attacking this problem it is wise to write down exactly what is known and what is unknown. u = 50ft./sec. V= 75 ft/sec. S = 125ft. a=? Formula iv involves these four quantities. Note that i, ii, and ill do not involve these exact four quantities. Formula iv is the one to use. First it should be solved for the unknown, a. 2 2 v = u + 2as 2 2 v - u = 2as a= a= a= a=12.5 ft/

Accelerated Motion of a "Freely Falling" Body Common experience indicates that falling bodies accelerate or increase in speed as they fall. Close to the 2 surface of the earth this "acceleration of a freely falling body" has been measured to be about 32 ft./sec. in 2 the English system and 9.8 m/sec. in the metric system. The "about" in the preceding sentence indicates that this quantity varies somewhat over the face of our earth. The values given are average values. When we use the words "freely falling", we mean that we are neglecting the effects of air resistance (as if we were in a vacuum). Of course, there is always air resistance, so how can we neglect it? When a body is falling with a great speed, air resistance can certainly not be neglected. To use the acceleration formulas in these cases would give us results that are not valid. However, if a body is falling close to the surface of the earth, the acceleration formulas do give us valid results if the height from which it falls is not too great. Some numerical data should clarify the preceding statements. If a compact body, such as a stone, is dropped (not thrown) from a height of 324 ft. above the surface of the earth, it will take about 4.5 sec. for the body to reach the ground. It will have obtained a speed of 144 ft/sec. (98 MPH). At this speed, the effects of air resistance are still quite negligible. Above this speed (98 MPH), the effects of air resistance are not negligible. Therefore, we can conclude that the fall of a body from a height of 324 ft. or less (or equivalently during a time of 4.5 sec. or less) can be handled quite accurately with the ordinary acceleration formulas. The value 2 2 of the acceleration will be either 9.8 m/sec or 32 ft./sec if the body is rising and therefore decreasing its 2 2 speed the values of the acceleration will be - 9.8 m/sec or - 32 ft/sec

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Part 66 Training Syllabus Module 2 Physics If a body falls from a height greater than about 324 ft. above the surface of the earth, the air resistance becomes very important. As we have said, a height of 324 ft. corresponds to a fall of 4.5 sec. When the time of fall increases to about 8 seconds, the speed of fall has increased to about 115 MPH. When the time of fall is between 4.5 sec. and 8 sec. the speed increases in a non-linear manner from 98 MPH to 115 MPH. As the time of fall increases beyond 8 seconds the speed of fall remains constant at about 115 MPH. This speed of fall is called the "terminal velocity". All of the above data indicates that it is possible to use the acceleration formulas with accurate results for many applications dealing with falling bodies. We will limit our applications to cases where the formulas are valid: heights less than 324 ft. and times of fall less than 4.5 seconds. EXAMPLE A body started from rest and has been falling freely for 3 sec. At what speed is it falling? u = o,

2

t = 3 sec, a = 32 ft/sec , v = ?

We will use Formula ii. v = u + at v=0+( 32

) (3 sec)

v=96ft/sec

EXAMPLE : A body started at rest and has been falling freely for 3 sec. How far has it fallen? 2 u = 0, t = 3 sec, a = 32 ft/sec , s = ? We will use Formula iii. 1

2

s = ut + /2 at 2 s = (0)(3 sec) + (32 ) (3 sec ) s = 144 ft EXAMPLE: A body is thrown upward with an initial speed of 120 ft./sec. How high does it rise? 2 u = 120 ft/sec, v = 0, a = -32 ft/sec , s = ? We will use Formula iv.

2

2

v = u + 2as s= s= s=225 s = 225 ft.

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Part 66 Training Syllabus Module 2 Physics Problems 2 1. A car on the motorway is accelerating at 25ft/sec . If it started from rest and has been accelerating for 5 sec., how far has it travelled during this time of acceleration? 2

2.

A truck had an initial velocity of 40ft/sec. It accelerated at 10ft/sec and reached a final velocity of 60ft/sec. How far did this truck travel while it was accelerating?

3.

A car slowed down from 80 ft/sec, to 40 ft/sec, while travelling a distance of 100 ft. What was its acceleration?

4.

A car, originally travelling at 25 ft/sec, increases its speed at a rate of 5 ft/sec for a period of 6 sec. What was its final speed?

5.

A car has an initial velocity of 40 ft/sec. It slows down at a rate of 5 ft/sec and covers a distance of 60 ft. while slowing down. What is its final velocity?

6.

A stone is dropped from a high building and falls freely for 4 sec. How far (in meters) has it fallen during this time?

7.

A stone is thrown upward with an initial velocity of 64 ft/sec. How high does it rise?

8.

A ball is dropped from a bridge into the river below and 2.5 sec. after the ball is dropped a splash is heard in the water below. How high is the bridge?

9.

A car starts with an initial velocity of 30 ft/sec, and accelerates for 5 sec. at 4 ft/sec . How far has it travelled during this time?

10.

A Cessna Agcarryall has a take-off run of 900 feet, at the end of which its speed is 80 MPH. How much time does the run take? (Hint: convert MPH to ft./sec. first)

2

2

2

11. A Grumman Tomcat, powered by two Pratt & Whitney turbofan engines, has a maximum 2 acceleration during take-off of 20 ft/sec . What velocity can it achieve by the end of a 1000 foot takeoff run? Answers 1.

312ft.

2.

100ft.

3.

-24 ft/sec

4.

55ft/sec.

5.

32 ft/sec.

6.

78m

7.

64 ft.

8.

100ft. or 30.6m

9.

200 ft.

10.

15 sec.

11.

200 ft/sec

2

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Part 66 Training Syllabus Module 2 Physics

Rotational Motion Introduction Previously we discussed constant speed and accelerated motion in a straight line and derived four important formulas which will be reviewed below. In this chapter we will consider motion which takes place on a circular path. Such motion is very common in our complex society and we need to understand more about motion in curved paths.

Degrees and Radians Before we begin our discussion, we need to define a new unit for measuring angles, the radian (see figure 2.35). A radian is defined as the central angle subtending a length of arc equal to the radius of the circle. A radian is approximately equal to 57.3°. The conversion factors for angle units are: 1 revolution = 360° 1 revolution = 2π radians 2π radians = 360° 1 radian = 57.3° Now let us consider a body (represented by a point) moving in a circular path. An initial reference line is shown in figure 2.36. As the point moves about the circle in a counter-clockwise sense, a line drawn between the point and the centre of the circle continuously sweeps out an angle. This angle can be measured in revolutions, radians or degrees. We call this angle the angular displacement of the point and use the Greek letter theta (θ) to represent this angular displacement. Figure 2.35 If the point moves with constant speed it also has a constant angular velocity. That is, the line drawn from the point to the centre of the circle sweeps out a definite number of revolutions, radians, or degrees each second or minute. The symbol used to represent angular velocity is the Greek letter omega (ω). Angular velocity can be expressed in different units, such as,

It is also possible that the point is not moving with constant angular velocity. It may be increasing or decreasing its angular velocity. When a CD starts rotating in a CD drive the angular velocity increases until it reaches a constant value. After the reject button is pushed the angular velocity decreases until the CD comes to rest. In both of the above cases we say that the point has an angular acceleration. The Greek letter alpha (α) is used for angular acceleration. Note that a is positive if the angular velocity is increasing and negative if the angular velocity is decreasing. Angular acceleration can also be expressed in different units,

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Part 66 Training Syllabus Module 2 Physics Now as a body moves in a circular path four similar equations hold as in the case of a body moving in a straight-line path. Both sets of equations will be shown below. It is important to re-memorize the equations for straight-line motion. In this way the other four equations will also be known, since they are exactly analogous.

Figure 2.36: A point moving in a circle θ=

s= v=u+at s= ut+ ½ at 2

2

v =u +2as

2

2

θ=ω1t+ ½ αt 2

2

ω2 =ω1 + 2 αθ

Study these equations carefully and note that the set to the right, the "rotational analogy" are easily remembered if the left set is well known. We recall that the subscripts "u" and "v" indicate "initial" and "final". These four rotational equations help us to solve many practical problems dealing with rotating bodies. EXAMPLE: A rotating machine part increases in angular velocity from 3 rev./min. to 35 rev./min. In 3.5 minutes. What is its angular acceleration? We use the following equation and solve it for a. ω2 = ω1+αt =α We now substitute our known values. 2 α= = 9.14 rev /min

EXAMPLE: 2

A propeller starts from an angular velocity of 900 rev./min. and accelerates at 100 rev./min. for 5 minutes. Through how many revolutions has it turned? 2 θ = ω1t+ ½ αt 2

θ=(900rev./min.)(5min) + ½ (100rev./min. )(5min.)

2

θ = 5,750 revolutions EXAMPLE: 2 A propeller starts at 1,000 rev./min. and accelerates at 100 rev./min. through 2,000 revolutions. What is its final angular velocity? 2 2 ω2 = ω1 +2αθ 2 2 2 ω2 = (1,000 rev./min.) +2 (100 rev./min. )(2,000 rev.) ω2 =1,180 rev./min.

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Part 66 Training Syllabus Module 2 Physics

We note that there is an acceleration of the body "in the path", called the tangential acceleration. The body is increasing or decreasing its speed, or traversing the circle. We recall also that when a body moves in a circle there 2 is also a centripetal acceleration, V /R, that is always directed toward the centre of the circular path. Thus when a body is increasing speed as it moves in a circular path there are two acceleration vectors, one tangential to the path, and the other directed to the centre of the path (centripetal acceleration). In figure 2.37, the body is increasing speed in the counter-clockwise sense. The directions of the two acceleration vectors are shown.

Figure 2.37: Tangential acceleration (at) and Centripetal acceleration (ac) Radian Measure In figure 2.38, 's' is the length along the path. We would like to relate this distance to the size of the central angle (6) and the radius (R) of the circular path. In our preceding discussion, the angle (θ) was measured in any of three different units, degrees, revolutions, or radians.

Figure 2.38: s, R and θ The equation that relates s to θ and R is a very simple one if we limit the angular unit to radians. This equation is: S = Rθ We see that this equation is true if we look at figure 2.38. We note, by measuring, that the equation is satisfied. We also see that it would not be true if the angle θ was in revolutions or degrees We now have a new problem to deal with in our treatment of rotational motion. There is a limit to the units that may be used in this equation. We repeat that, for this equation, we must use radian measure. Also, any equation that is derived from s = Rθ will have this same restriction. Suppose that a body moves a small distance along the path and sweeps out a small central angle. The usual mathematical notation for a very small quantity is the use of the Greek letter Delta (Γ). Γs = R Γθ Let us divide both sides of this equation by the time, (Γt) during which the motion occurred. =R We can write: ν= Rω

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If this velocity in the path is changing, there is also a change in the angular velocity. Assume that this change occurs in the small time interval (At). We can write: Γv = R Γω Next we divide left and right members by Γt. =R The tangential acceleration (a) in the left side is the rate at which a body moving in a circular path is picking up speed in the path. It is equal to the radius times the angular acceleration (a). We can write: a = Rα Let us summarize the three important equations we have derived: s = Rθ ν = Rω a = Rα All three of these equations require the use of radian measure. This means that: θ must be in radians ω must be in rad/min. or rad./sec. 2 2 α must be in rad./min or rad./sec Note that the radian is called a "dimensionless" unit. We put it in or take it out for clarity. EXAMPLES: A car is moving on a circular racetrack of radius 150 ft. It sweeps out an angle of 2000. How far has it travelled? We note that: θ = 200° x

= 3.49 rads. s = Rθ s= (150 ft.) (3.49 rad.) s = 523 ft.

3.36 rev./min

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: A race car is travelling at a speed of 176 ft./sec. (120 MPH) around a circular racetrack of radius 500 ft. What is the angular velocity of this car in rev./min.? Use the equation: v = Rω or ω= = ω = 0.352 rad./sec. Note that we knew that the unit of our answer is rad./sec. and not rev./sec. since the equation we used always is in radian measure. The units in the right side of the second equation above actually come out as "nothing'Vsec. We put in the radian unit in the numerator for clarity. in order to find our answer in rev./min. we use the proper conversion factors. ω=

3.36 rev/min.

EXAMPLE: A race car is moving on a circular racetrack of radius 4,000 ft.. It is increasing its speed at a rate of 15 2 2 ft./sec. What is its angular acceleration rev./sec. ? We use the equation: a = Rα α=

= 2

α = 0.00375 rad./sec. 2 We note that the unit is rad./sec. because the equation that we have used requires radian measure. 2

To obtain a in rev./sec. , we must use the standard conversion factor. α= a - 0.000597 rev ./sec.

2

Problems 2

1.

A propeller starts from rest and accelerates at 120 rev/sec for 4 seconds. What is its final angular velocity in rev/sec? In rev/min?

2.

A rotating turntable starts from rest and accelerates at 5 rev/min for 3 min. Through how many revolutions has it turned?

3.

A helicopter main rotor starts from an initial angular velocity of 2 rev/min and accelerates at 60 2 rev/min while turning through 400 revolutions. What is its final angular velocity?

4.

A plane is circling O'Hare in a circular pattern of radius 15,000 ft. It sweeps out an angle of 340°? How far has it travelled?

5.

A plane is circling an airport in a circle of radius 5,000 ft. How far has it travelled after 4 revolutions?

6.

A race car is moving on a circular track of radius 600 ft. It is travelling at a speed of 100 ft/s. What is its angular velocity in rev/min?

2

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Part 66 Training Syllabus Module 2 Physics 7.

A race car is moving on a circular racetrack of radius 800 ft. It is accelerating at a rate of 10 2 2 ft/sec What is its angular acceleration in rev/sec ?

8.

A helicopter tail rotor starts with an initial angular velocity of 15 rev/sec and decelerates at a rate 2 of 2.00 rev/sec until it comes to rest. Through how many revolutions has the rotor turned while it comes to rest?

Answers 1. 480 rev/sec.

28,800 rev/min.

2. 22.5 rev. 3. 219 rev/min. 4. 89,000 ft.

5. 23.8 miles 6. 1/6 rad/s, 7.

1/16071 rev/sec

5ln rev/min 2

8. 56.3 rev.

Periodic Motion Simple Pendulum A simple pendulum is one which can be considered to be a point mass suspended from a string or rod of negligible mass. It is a resonant system with a single resonant frequency. For small amplitudes, the period of such a pendulum can be approximated by: Where: L = the length of the pendulum is m, or ft g = the magnitude of acceleration due to gravity = 9.81 2 2 m/s or 32 ft/s Note: The Natural Frequency of Oscillation is independent of the mass of the pendulum, and of the amount of initial displacement

Figure 2.39: A simple pendulum This expression for period is reasonably accurate for angles of a few degrees, but the treatment of the large amplitude pendulum is much more complex It is interesting to note that the pendulum will oscillate at only one frequency, regardless of how far the pendulum is initially displaced, or for how long the pendulum is left to oscillate. The only factor that changes, is the linear velocity of the mass. This fixed frequency is known as the Natural Frequency of Oscillation. If we consider only the horizontal motion of the mass and neglect its vertical motion as it swings (an assumption which can be made if the string is long compared to the amplitude of swing), then the periodic motion is said to be Simple Harmonic Motion (SHM). Time period (T) and frequency (f) can also be related to each other by the formulae: T=

or f =

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Mass and Spring When a mass is acted upon by an elastic force which tends to bring it back to its equilibrium position, and when that force is proportional to the distance from equilibrium (e.g., doubles when the distance from equilibrium doubles - a Hooke's Law force), then the object will undergo periodic motion when released. A mass on a spring is the standard example of such periodic motion. If the displacement of the mass is plotted as a function of time, it will trace out a pure sine wave. The motion of the medium in a travelling wave is also simple harmonic motion as the wave passes a given point in the medium.

Figure 2.40: Sinusoidal motion of a spring/mass system It is interesting to note that the spring/mass system will oscillate at only one frequency, regardless of how far the mass is initially displaced, or for how long the system is left to oscillate. The only factor that changes, is the linear velocity of the mass. The fixed frequency is known as the Natural Frequency of Oscillation, and can be calculated from the formula: fn=

√

Where: k = the stiffness of the spring in N/m, or Ib/in m = the mass of the oscillating body Note: The Natural Frequency of Oscillation is independent of the magnitude of gravity, and of the amount of initial displacement

Simple Harmonic Motion (SHM) What is SHM Motion which repeats itself precisely and can be described with the following terms:  Period: the time required to complete a full cycle, T in seconds.  Frequency: the number of cycles per second, f in Hertz (Hz)  Amplitude: the maximum displacement from equilibrium, A and if the periodic motion is in the form of a travelling wave, one needs also  Velocity of propagation: v  Wavelength: repeat distance of wave, λ. Simple harmonic motion is the motion of a simple harmonic oscillator (such as a pendulum or spring/mass system), a motion that is neither driven nor damped. The motion is periodic, as it repeats itself at standard intervals in a specific manner - described as being sinusoidal, with constant amplitude. It is characterized by its amplitude, its period which is the time for a single oscillation, its frequency which is the number of cycles

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Part 66 Training Syllabus Module 2 Physics per second, and its phase, which determines the starting point on the sine wave. The period, and its inverse the frequency, are constants determined by the overall system, while the amplitude and phase are determined by the initial conditions (position and velocity) of that system. A single frequency travelling wave will take the form of a sine wave. A snapshot of the wave in space at an instant of time can be used to show the relationship of the wave properties frequency, wavelength and propagation velocity.

Figure 2.41 : The sinusoidal waveform terminology The motion relationship "distance = velocity x time" is the key to the basic wave relationship. With the wavelength as distance, this relationship becomes λ=vT. Then using f= relationship

gives the standard wave

This is a general wave relationship which applies to sound and light waves, other electromagnetic waves, and waves in mechanical media. ν= f λ

Properties of SHM Considering the motion of a mass on the end of a spring, or the horizontal motion of a pendulum, the following properties can be observed:  The velocity of the body is always changing. It is maximum at the undisturbed position (centre of its motion) and zero at the extremities of its motion (maximum displacement position)  The acceleration of the body is always changing. It is maximum at the extremities of its motion (maximum displacement position) and zero at its undisturbed position (centre of motion). In other words, when its velocity is zero, its acceleration is a maximum, and when its acceleration is zero, its velocity is a maximum.

Vibration Vibration refers to mechanical oscillations about an equilibrium point. The oscillations may be periodic such as the motion of a pendulum or random such as the movement of a tire on a gravel road. Vibration is occasionally desirable. For example the motion of a tuning fork, the reed in a woodwind instrument or harmonica, or the cone of a loudspeaker is desirable vibration, necessary for the correct functioning of the various devices. More often, vibration is undesirable, wasting energy and creating unwanted sound - noise. For example, the vibrational motions of engines, electric motors, or any mechanical device in operation are typically unwanted. Such vibrations can be caused by imbalances in the rotating parts, uneven friction, the meshing of gear teeth, etc. Careful designs usually minimise unwanted vibrations. The study of sound and vibration are closely related. Sound, pressure waves, are generated by vibrating structures (e.g. vocal cords) and pressure waves can generate vibration of structures (e.g. ear drum). Hence, when trying to reduce noise it is often a problem in trying to reduce vibration.

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Types of vibration Free vibration occurs when a mechanical system is set off with an initial input and then allowed to vibrate freely. Examples of this type of vibration are pulling a child back on a swing and then letting go or hitting a tuning fork and letting it ring. The mechanical system will then vibrate at one or more of its natural frequencies and damp down to zero. Forced vibration is when an alternating force or motion is applied to a mechanical system. Examples of this type of vibration include a shaking washing machine due to an imbalance, transportation vibration (caused by truck engine, springs, road, etc), or the vibration of a building during an earthquake. In forced vibration the frequency of the vibration is the frequency of the force or motion applied, but the magnitude of the vibration is strongly dependent on the mechanical system itself.

Resonance What is Resonance? Resonance is the phenomenon of producing large amplitude of vibrations by a small periodic driving force. It is the tendency of a system to oscillate at maximum amplitude at a certain frequency. This frequency is known as the system's resonance frequency (or resonant frequency). When damping is small, the resonance frequency is approximately equal to the natural frequency of the system, which is the frequency of free vibrations. Under resonance condition the energy supplied by the driving force is sufficient enough to overcome friction.

Examples of Resonance One familiar example is a playground swing, which is a crude pendulum. When pushing someone in a swing, pushes that are timed with the correct interval between them (the resonant frequency), will make the swing go higher and higher (maximum amplitude), while attempting to push the swing at a faster or slower rate will result in much smaller arcs. Other examples:  acoustic resonances of musical instruments  the oscillations of the balance wheel in a mechanical watch  electrical resonance of tuned circuits in radios that allow individual stations to be picked up  the shattering of crystal glasses when exposed to a strong enough sound that causes the glass to resonate. A resonator, whether mechanical, acoustic, or electrical, will probably have more than one resonance frequency (especially harmonics of the strongest resonance). It will be easy to vibrate at those frequencies, and more difficult to vibrate at other frequencies. It will "pick out" its resonance frequency from a complex excitation, such as an impulse or a wideband noise excitation. In effect, it is filtering out all frequencies other than its resonance.

What Causes Resonance? Resonance is simple to understand if you view the spring and mass as energy storage elements - the mass storing kinetic energy and the spring storing potential energy. When the mass and spring have no force acting on them they transfer energy back forth at a rate equal to the natural frequency. In other words, if energy is to be efficiently pumped into the mass and spring the energy source needs to feed the energy in at a rate equal to the natural frequency. Applying a force to the mass and spring is similar to pushing a child on swing - you need to push at the correct moment if you want the swing to get higher and higher. As in the case of the swing, the force applied does not necessarily have to be high to get large motions. The pushes just need to keep adding energy into the system. A damper, instead of storing energy dissipates energy. Since the damping force is proportional to the velocity, the more the motion the more the damper dissipates the energy. Therefore a point will come when the energy dissipated by the damper will equal the energy being fed in by the force. At this point, the system has reached its maximum amplitude and will continue to vibrate at this amplitude as long as the force applied

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Part 66 Training Syllabus Module 2 Physics stays the same. If no damping exists, there is nothing to dissipate the energy and therefore theoretically the motion will continue to grow to infinity. Such catastrophic resonance can be witnessed frequently, in, for example, the failure of complete aircraft wing structures during control surface "flutter", failure of helicopter structural components, and even the th collapse of road bridges in gale force winds, as experienced at Tacoma Bridge on November 7 ,1940.

Design Implications of Resonance Designers of aircraft must be seriously concerned about the phenomenon of resonant frequency because if a certain component of an aeroplane or helicopter is caused to vibrate at its resonant frequency the amplitude of the vibration can become very large and the component will destroy itself by vibration. Let us examine the case of a helicopter which has a tail boom with a natural or resonant frequency of 1 Hz. That is, if you were to strike the boom with your fist it would oscillate once each second. The normal rotational speed of the rotor is 400 RPM and the helicopter has 3 blades on its main rotor. Each time a rotor blade moves over the tail boom the blade is going to cause a downward pulse of air to strike the tail boom. The designer must determine the speed at which the pulses will be equal to the resonant frequency of the boom. One cycle per second is equivalent to 60 cycles/minute. Since each of the three blades causes a pulse each revolution, there will be 3 x 60 or 180 pulses/minute. Therefore a rotor speed of 180 RPM would be critical and the pilot would be warned against operating at that speed. Since the boom also has a secondary, or overtone, resonant frequency of twice the fundamental resonant frequency, 360 RPM would also have to be avoided but would not be as critical as 180 RPM. The third frequency of concern would be 3 x 180 or 540, but that is above the rotor operating speed, so is not a problem. The natural frequency of vibration is also an extremely important consideration in designing the wings, horizontal and vertical stabilizers of an aircraft. The designer must be certain that the resonant frequency when the surface is bent is different from that resonant frequency when it is twisted. If that is not the case, an aerodynamic interaction with the elasticity of the surface can result in "flutter" which can cause the surface to fracture in a fraction of a second after it begins.

Harmonics The harmonic of an oscillation is a component frequency of the oscillation that is a multiple of its natural frequency (known as the fundamental frequency). For example, if the fundamental frequency is f, the harmonics have frequency 2f, 3f, 4f, etc. The harmonics have the property that they are all periodic at the input frequency. Thus, if an oscillating body (e.g. a spring/mass system) can be oscillated by an excitation input of frequency equal to its natural frequency (the 'fundamental frequency'), it will also be oscillated at frequencies that are harmonics of that natural frequency. Problems 1.

A pendulum has a length of 0.7m. What is its frequency of oscillation, and how long will it take to oscillate 10 times?

2.

A pendulum has a mass of 0.05 slugs. It takes 15 seconds to oscillate 10 times. What is its length?

3.

A mass of 0.4 kg oscillates freely on the end of a spring. The stiffness of the spring is 2 N/m. What is its natural frequency of oscillation and its time period?

4.

A ball on the end of a spring bounces such that it nearly hits the floor 30 times in a minute. The spring has a stiffness of 0.5 Ib./in. What is the value of the mass of the ball?

Answers 1. 0.6 Hz, 16.8 seconds 2.

0.9 m

3.

0.36 Hz, 2.8 seconds

4.

0.05 slugs

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Simple Machines and the Principle of Work The definition of work is as follows: W = FD cos θ The symbol for "distance" has been switched from S to D, to emphasize that we are dealing with distances in our treatment of simple machines. The angle (θ) in this definition is the angle between the direction of the force vector and the direction of the displacement vector. In this chapter, we will assume that in all the cases we will study the force and displacement vectors act in the same direction. This implies that the angle (θ) is a 0° angle and since the cosine of a 0° angle equals one, the equation for work becomes the simple equation: W=FD In this chapter, we will study six simple machines:      

The lever The pulley The wheel and axle The inclined plane The screw The hydraulic press

General Theory of All Machines In discussing machines, we will assume that there is an object on which work is to be done. We will call this object the load. In most cases, it is required that the load be raised a certain distance in a gravitational field. For example, we wish to put cement blocks originally on the ground into the bed of a truck. A machine is a device for doing this work. The input work is, by definition, the work done by the worker, that is, the force applied by the worker multiplied by the distance through which the worker's force acts. The output work is, by definition, the force that actually acts on the load multiplied by the distance the load is raised. We note that one way to do work is to do it directly. For example, it is possible for the worker to raise each cement block directly to the truck bed. This is possible but can be difficult if each block weighs, say, 175 Ibs. In this case it would be better to use a machine since a machine usually decreases the force supplied by the worker and increases the distance through which his force acts. In the equations which follow, the subscript "o" will stand for output and the subscript "i" will indicate input. We will use the following defining equations: Wo=FοDο W i=FiW f It is important to realize that there is no perfect machine. In our real world, on our earth, there is always some friction. We always have, at least, air resistance. In addition, there is friction due to the nooks and crannies that we would see if we inspected the surfaces of our machine parts with a high-powered microscope. Because of the constant presence of friction the input work is always greater than the output work. Some of the input work is not useful work but serves to produce sound energy (a squeak), light energy (a spark), or heat energy. We will use the symbol "W f" to represent work lost because of friction. W i=W o+W f We define two kinds of "mechanical advantage". The actual mechanical advantage (AMA) is the ratio of the output force to the input force. This actual mechanical advantage tells us how much easier it is for the

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Part 66 Training Syllabus Module 2 Physics worker. The ideal mechanical advantage (IMA) is the mechanical advantage that would exist if there were no friction in the machine. It is the ratio of input distance to the output distance. AMA=

IMA =

The ideal mechanical advantage of a machine can always be determined by measurements made on the machine itself. The efficiency (Eff) of a machine is the ratio of the output work to the input work. Eff= Eff= The efficiency can be expressed as a decimal or as a percentage. For example, if the efficiency is calculated as 0.78, we can expressed it as 78%. One final point should be made regarding efficiency. There is no machine that is 100% efficient. We always have some friction. However, sometimes we assume that there is no friction and that the machine is perfect or ideal! if a problem says that the efficiency is 100%, we are doing a make-believe problem. This kind of a problem is not meaningless, however, because it tells us the best that this machine can do. In this ideal case the AMA equals the IMA. EXAMPLES: A worker is able to raise a body weighing 300 Ibs. by applying a force of 75 Ibs. What is the AMA of the machine that he is using? AMA=

=

=4

A worker applied his force through a distance of 15 ft. The load is raised a distance of 2.5 ft. What is the IMA of the machine that he used? IMA=

=

The actual mechanical advantage of a machine is 8 and the efficiency of this machine is 78%. What is the ideal mechanical advantage? IMA =

=

=10.3

A worker uses a machine to raise a load of 500 Ibs. a distance of 2 ft. He does this by applying a force of 100 Ibs. through a distance of 12 ft. What was the efficiency of the machine? Method 1: AMA= =5 IMA = Eff=

=6 = = 83%

Method 2: W 0 = (500 lbs)(2 ft) = 1,000 ft.lbs. W i= (100 lbs)(12 ft) = 1,200 ft.lbs. Eff=

=

= 0.83 = 83%

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Part 66 Training Syllabus Module 2 Physics We will next consider six simple machines. In each of these cases the IMA is expressed, not as the ratio Di/Do , but in some other manner. We will study the geometry of each of these simple machines to determine how to express the IMA in some simple equation.

The Lever Consider the diagram in figure 2.42. Note that the lever always pivots about some point called the fulcrum. The input force (F1) is downward force and in our diagram, is applied at the right end of the lever. This input force gives rise to an upward force at the left end in our diagram. This upward force causes the load to be raised and is called "F0". Figure 2.42: Simple lever system In figure 2.43. note that the input force acts through a distance (Di) and the load is raised a distance (Do).

Figure 2.43: Distances moved in a simple lever system The distance from the input end of the lever to the fulcrum is called the input lever arm (Lf) and the distance from the output end to the fulcrum is called the output lever arm (L0). Recall that: IMA= However, figure 2.43 shows that the ratios of lever arms and distances are equal: = Since it is much easier to measure lever arms that the distances of rotation, we always use the ratio on the right hand side of the above equation to express the IMA of a lever. (Lever) There are three classes of levers:  1st Class: The fulcrum is between the load and the applied force. Examples are the claw hammer, scissors, and crowbar  2nd Class: The load is between the fulcrum and the applied force. Examples are the nutcracker and wheelbarrow.  3rd Class: The applied force is between the load and the fulcrum. An example is ice tongs. In a third class lever, the IMA is less than one. There is no force advantage. However, there is a speed advantage. The work can be done in less time. The Pulley Some pulleys are firmly attached to an overhead support while other pulleys move up or down with the load. We will refer to pulleys as "fixed" or "movable". In figure 2.44 (A), we have shown a single fixed pulley. If a length of pulley cord (Dj) is pulled down by a worker, the load will be raised a distance (D0). We see from the diagram that these distances equal each other. Therefore we conclude that the IMA of this type of pulley is one. For example, it would take 100 Ibs. of force to raise a 100 Ibs. load. The advantage of using this type of pulley is that the worker is able to pull

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Part 66 Training Syllabus Module 2 Physics down on the pulley cord and in this way an upward force is applied to the load. We say that a single fixed pulley is a "direction changer". FI

Figure 2.44: Simple pulley systems In figure 2.44 (B), there is a single movable pulley. A study of the diagram shows that D i is always twice D0. For example, if the load is to be raised 2 ft. the worker must pull in 4 ft. of cord. Note also that there are 2 strands supporting the load. The IMA of a single movable pulley is 2. In figure 2.44 (C), there is a single movable pulley and a single fixed pulley. The fixed pulley again serves to change the direction of the input force. The IMA is still 2. Note also that there are again 2 strands supporting the load. We conclude that the IMA of a pulley equals the number of strands supporting the load. (Pulley) IMA = the number of strands supporting the load Several other examples of various types of pulley blocks are shown in figure 2.45.

Figure 2.45: More complex pulley systems

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The Wheel and Axle

Figure 2.46: Wheel and axle Note that one cord is wrapped around the axle of radius (r). The load is attached to this cord. Another cord is wrapped around the wheel of radius (A). The worker applies his force to this second cord. Both wheel and axle turn together. This means that if the wheel rotates through one revolution the axle also turns through one revolution. Let us suppose that the worker pulls in a length of cord equal to one circumference of the wheel (D1) The load will be raised a distance equal one circumference of the axle, (D0). IMA= (Wheel and Axle)

= IMA=

The Inclinked Plane

Figure 2.47: Inclined plane In the inclined plane shown in figure 2.47 we note that the worker slides the load up the incline. The input distance (Dj) is therefore equal to the length of the incline (L). The effect of this is that the load is raised a distance (h). This means that the output distance (D0) equals h also. IMA=

=

=

We note that the sine of the angle of inclination (θ) is also h/L. Therefore, we can write the expression for the IMA as follows: (Inclined Plane)

IMA=

The Screw Jack

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Part 66 Training Syllabus Module 2 Physics The pitch of the screw (p) is the distance between adjacent threads (see figure 2.48). As the handle is turned through one revolution, a distance given by 2 TI r ft., the load is raised a distance of one pitch. Therefore, we have the relation: (Screw Jack)

ΙΜΑ =

Figure 2.48: The screw jack A screw Jack has a great deal of friction. Therefore its efficiency is usually very low. However, the distance through which the input force acts in comparison to the pitch is usually very large. This gives a screw jack a large mechanical advantage. The Hydraulic Press A cross section of an hydraulic press is shown in figure 2.49. The small rectangles are cross sections of the circular input and output pistons. Usually, we talk about the areas of the input and output pistons (Ai and A0). We note that the smaller of the two pistons is the input piston (radius = r) and, of course, the larger piston is the output piston (radius = R). An hydraulic press is filled with some fluid (gas or liquid). This fluid exerts a common fluid pressure throughout the device.

Figure 2.49: The hydraulic press As the smaller piston moves downward a distance (d i) the larger piston moves upward a distance (d0). We recall that the volume of a cylindrical shape is equal to the area of the circular base x the height. Also, a volume of fluid is transferred from the input (left) cylinder to the output (right) cylinder. The volume of fluid is constant since the pressure is constant. Therefore, we can write the equation: 2

2

π r di = π R do We can cancel the common factor (it) and rearrange the equation. We obtain:

The left member of this equation is, by definition, the IMA. Therefore, the IMA is also equal to the right member of this equation. Thus, we can finally say that: (Hydraulic Press)

IMA =

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We have obtained equations for the IMA of each of the six simple machines. We will do an example of a typical problem dealing with machines. Note that any one of the six could be chosen as an example. In the problems that follow the example, be sure to use the correct formula for the IMA.

EXAMPLE: The radius of the wheel in a windlass (wheel and axle) is 3.5 ft. and the radius of the axle is 0.27 ft. The efficiency of the machine is 60%. What load can be lifted by this machine by using a force of 75 Ibs.? IMA=

=13.0

AMA = (Eff) (IMA) AMA = (0.60) (13.0) = 7.8 F0=(AMA)(Fi) F0=(7.8)(75lbs.) F0=585lbs. EXAMPLE: An inclined plane has a 32° angle of incline. A force of 50 Ibs. Is required to slide a 90 Ibs. load up the incline. What is the efficiency of this machine? IMA=

= 1.89

AMA=

= 1.8

Eff=

=

= 95%

Problems 1. It takes a force of 80 Ibs. to raise a body that weighs 240 Ibs. What is the actual mechanical advantage of the machine that was used? 2. A load is raised a distance of 6 ft. by a force acting through a distance of 18 ft. What is the ideal mechanical advantage of the machine that was used? 3. What is the efficiency of a machine having an IMA of 7 and an AMA of 5? 4. A load weighing 120 Ibs. is raised a distance of 4 ft. by a machine. The worker using the machine exerts a force of 50 Ibs. through a distance of 12 ft. What was the efficiency of the machine? 5. The radius of the wheel of a windlass is 4.0 ft. and the radius of the axle is 0.2 ft. The machine is 75% efficient. What force must be exerted to raise a load of 500 Ibs. with this machine? 2

6. The large piston of an hydraulic press has area 1.5 ft , and the small piston has area 2 0.30 ft . Assume that the machine is 100% efficient. What load can be raised by a force of 75 Ibs.? 7. A pulley system has four strands supporting the load. A force of 55 Ibs. is needed to raise a load of 200 Ibs. What is the efficiency of this pulley system? 8. A light aircraft has a hydraulic braking system. Each rudder pedal is connected to a master cylinder which provides braking for one of the main landing gear wheels. Each master cylinder has a radius of 1/4-inch. The cylinder on the wheel has a radius of 1.0 inch. If the system is 95% efficient and the pilot exerts a force of 55 Ibs. on the pedal, how much force is exerted on the brake disc by the wheel cylinder?

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Answers 1. 2. 3. 4. 5. 6. 7. 8.

3 3 71% 80% 33.3 Ibs. 375 Ibs. 90.9% 836 Ibs.

Dynamics Newton's Laws The rapid advance in aviation in the first half of the last century can be attributed in large part to a science of motion which was presented to the world three centuries ago by Sir Isaac Newton, a British physicist. Published in 1686, Newton's treatise on motion, The Principia, showed how all observed motions could be explained on the basis of three laws. The applications of these laws have led to great technological advances in the aerodynamics, structure, and power plant of aircraft. It is safe to say that any future improvements in the performance of aircraft will be based on these laws. This chapter will be devoted to Newton's laws, examining some of their applications in aviation.

Newton's First Law The old magician's trick of pulling a cloth out from under a full table setting is not only a reflection of the magician's skill but also an affirmation of a natural tendency which dishes and silverware share with all matter. This natural tendency for objects at rest to remain at rest can be attested to by any child who ever tried kicking a large rock out of his path. It is also a well known fact that once a gun is fired, the command "stop" has no effect on the bullet. Only the intervention of some object can stop or deflect it from its course. This characteristic of matter to persist in its state of rest or continue in whatever state of motion it happens to be in is called inertia. This property is the basis of a principle of motion which was first enunciated by Galileo in the early part of the 17th century and later adopted by Newton as his first law of motion. The first law of motion is called the law of inertia. It can be summarized: A body at rest remains at rest and a body in motion continues to move at constant velocity unless acted upon by an unbalanced external force. The importance of the law of inertia is that it tells us what to expect in the absence of forces, either rest (no motion) or straight line motion at constant speed. A passenger's uncomfortable experience of being thrown forward when an aircraft comes to a sudden stop at the terminal is an example of this principle in action. A more violent example is the collision of a vehicle with a stationary object. The vehicle is often brought to an abrupt stop. Unconstrained passengers continue to move with the velocity they had just prior to the collision only to be brought to rest (all too frequently with tragic consequences) by surfaces within the vehicle (dashboards, windshields, etc.). A less dramatic example of Newton's first law comes from the invigorating activity of shovelling snow. Scooping up a shovel full of snow, a person swings the shovel and then brings it to a sudden stop. The snow having acquired the velocity of the shovel continues its motion leaving the shovel and going off onto the snow pile.

Newton's Second Law A Learjet accelerates down the runway a distance of 3,000 feet, takes off and begins its climb at 6,000 feet per minute quickly reaching a cruising altitude of 35,000 feet, where it levels off at a speed of 260 knots. Subsequently, the plane may have to perform a variety of manoeuvres involving changes in heading, elevation, and speed. Every aspect of the aeroplane's motion is governed by the external forces acting on its wings, fuselage, control surfaces and power plant. The skilled pilot using his controls continually adjusts these forces to make the plane perform as desired.

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Part 66 Training Syllabus Module 2 Physics The interplay between force and motion is the subject of Newton's second law. An understanding of this law not only provides insight into the flight of a plane, but allows us to analyse the motion of any object. While Newton's first law tells us that uniform velocity is to be expected when an object moves in the absence of external forces, the second law states that to have a change in speed or direction an unbalanced force must act on the object. Using acceleration to describe the change in motion of an object, the second law can he expressed Fnet = ma In words, the second law states that a net or unbalanced force acting on an object equals the mass of the object times the acceleration of that object. Here, the net force is the total force acting on the object, obtained by adding vectorially all of the forces influencing the object. The mass is a scalar quantity. However, both the net force and the acceleration are vector quantities. Mathematically, this means that they must always point in the same direction. That is, at each instant the acceleration is in the same direction as the net force. Before we consider cases where the net force acting on a body is not zero, it is most important to understand that sometimes the net force acting on a body is zero. The vector sum of the forces acting on the body in the x-direction is zero and the vector sum of the forces acting on the body in the y-direction is also zero. In this case we say that the body is in equilibrium. From the law, net force equals mass times acceleration, we know that since the net force is zero the acceleration is also zero. Zero acceleration means that the velocity of the body in not changing in direction or in magnitude. This means that the body is moving in a straight line with constant speed or it has the constant speed, zero (it is at rest). If we observe that a body is at rest we know that all of the forces on this body are balanced. Similarly, if a body is moving in a straight line with constant speed, all of the forces acting on this body are balanced. For example, if a plane is travelling on a straight stretch of runway at constant speed, there are four forces acting on this plane: the earth is pulling down on the plane (its weight), the earth is pushing up on the plane (the normal force), the engine is giving a forward thrust to the plane, and frictional forces (air resistance, tires on runway, etc.) are acting backward. This is illustrated in figure 2.50. Next, we must consider some examples where the net force acting on a body is not zero. The body is accelerating. The body is experiencing a change in its direction or in its speed or both. As a first example, a plane accelerating down a runway gets a change in velocity in the direction of its motion. This is the same direction as the thrust provided by the power plant. In figure 2.51, note that the thrust is greater than the frictional forces. The net forward force is the thrust minus the friction. It is this net forward force that results in the acceleration of the plane.

Figure 2.50: The four forces acting on an aeroplane

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Figure 2.51: Thrust, drag and acceleration forces

Newton's Third Law Newton's third law is sometimes referred to as the law of action and reaction. This law focuses on the fact that forces, the pushes and pulls responsible for both the stability of structures as well as the acceleration of an object, arise from the interaction of two objects. A push, for example, must involve two objects, the object being pushed and the object doing the pushing.

Every action has an equal and opposite reaction The third law states that no matter what the circumstance, when one object exerts a force an a second object the second must exert an exactly equal and oppositely directed force on the first. An apple hanging from a tree is pulled by the earth with a force which we call its weight. Newton's third law tells us that the apple must pull back on the earth with an exactly equal force. The weight of the apple is a force on the apple by the earth, directed downward. The force which the apple exerts back on the earth, is a pull on the earth directed upward. Another force acting on the apple is the upward pull exerted by the branch. The law of action and reaction tells us that the apple must be pulling down on the branch with the same magnitude of force. People are often confused by this principle because it implies, for instance, that in a tug of war the winning team pulls no harder than the losing team. Equally enigmatic is how a horse and wagon manage to move forward if the wagon pulls back on the horse with the same force the horse pulls forward on the wagon. We can understand the results of the tug of war by realizing that the motion of the winning team (or losing team) is not determined exclusively by the pull of the other team, but also the force which the ground generates on the team members feet when they "dig in". Recall, it is the net force, the sum of all of the acting forces which determines the motion of an object. The results of a "tug of war" can be quite different if the "winning team", no matter how big and strong, is standing on ice while the "losing team" is able to establish good solid footing on rough terrain. Similarly, the horse moves forward because the reaction force which the ground exerts in the forward direction on its hooves is greater than the backward pull it receives from the wagon. By focusing now on the wagon, we see that it moves forward because the forward pull of the horse is greater than the backward pull of friction between its wheels and the ground.

Figure 2.52: Gravitational force between two objects

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rd

Figure 2.53: Equal and opposite forces of Newton's 3 Law One of the main difficulties people have with the third law comes from not realizing that the action and reaction forces act on different objects and therefore can never cancel. Another difficulty comes from forgetting that the motion of an object is determined by the sum of all of the forces acting on that object. In canoeing or rowing, a paddle is used to push water backward. The water reacts back on the paddle generating a forward force which propels the boat. Consider now a propeller as shown in figure 2.55. The plane of rotation of the propeller is assumed to be perpendicular to the plane of the paper. The flow of air is from left to right. We can imagine the action of the propeller is to take a mass (m) of air on the left and accelerate it from some initial velocity (u) to a final velocity (v) to the right of the propeller. The acceleration of this air mass requires a force which is provided by the propeller. The air mass, in turn, reacts with an equal and

Figure 2.54: Equal and opposite forces of an oar

Figure 2.55: Action of a propeller opposite force on the propeller. This reaction force of the air on the propeller provides the thrust for a propeller driven plane. The acceleration of the air mass is:

a= Substituting this into Newton's second law, we find for the net force on the air mass: F=m

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Both of the velocities (u and v) are the velocities relative to the plane of rotation of the propeller. The time (t) is the time involved in accelerating the air mass from u to v. By Newton's third law, the thrust, which is the force the air mass exerts back on the propeller, is equal in magnitude to F. Therefore, the thrust (T) is given by: T=m Recall that we have a symbol for "change in", this means that we can write the above formula as: T=m The velocities of the air mass are relative to the plane, and therefore change as the plane's speed changes. Also the time involved in accelerating the air mass changes with the speed of the plane. This causes considerable variation in the thrust provided by a propeller. EXAMPLE: Each second a propeller accelerates an air mass of 12.2 slugs from rest to a velocity of 137 ft./sec. How much thrust is provided? T=(12.2 Slugs) T=1,670lbs.

In contrast to the reciprocating engine driven propeller which imparts a small change in velocity to a relatively large mass of air, a turbojet induces a large change in velocity to a relatively small mass of air. Here, the sole action of the Jet engine is considered to be the intake of a mass of air at some velocity (u) and its exhaust at a higher velocity (v). Figure 2.50 is a sketch of a turbojet engine. The velocity (u) in the figure denotes the relative intake velocity and v denotes the exhaust gas velocity. The thrust formula which was obtained above for the propeller will now be applied to a Jet engine. The thrust formula above can be rewritten: T= v - u T = Gross thrust - Ram drag The gross thrust is provided by the exhaust gases. The ram drag of the incoming air is due to the speed of the aeroplane. The effect of the ram drag is to reduce the thrust provided by the engine as the speed of the plane increases.

Figure 2.57: Principle of the turbojet engine

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EXAMPLE: The Pratt & Whitney J60 has a mass air flow of 23 kg/sec. During a static test (initial velocity 0) the exhaust velocity was measured to be 580 m/sec. Determine the thrust produced. Note that the ram drag is zero since v1 is zero. Therefore, the thrust is equal simply to the gross thrust. T= v Substituting the given values we have:

T=

(580 m/sec) = 13,300N

EXAMPLE: What would the thrust have been if the J60 of the previous example had been in a plane moving at 250 knots? Assume the same mass flow and exhaust velocity. Note that the ram drag is not zero in this case. In order to calculate this ram drag we must use the formula: Ram drag = u Before substituting, we must express the initial velocity in m/sec. u= 250 knots x ram drag=

x

= 129 m/sec

129m/sec

ram drag = 2,970N

T = gross thrust - ram drag T= 13,300 N-2,970 N T=10,300N EXAMPLE: During a static test (initial velocity = zero), a Pratt & Whitney J75 produced a thrust of 16,000 Ibs. with an air mass flow of 8.23 slugs/sec. Determine the exhaust gas velocity of the engine. Since u is zero, the ram drag is zero and T = gross thrust. T= v We solve for the final velocity: v=

v= v = 1,940 ft./sec.

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Part 66 Training Syllabus Module 2 Physics The air intake velocity of a turbojet will be approximately equal to the airspeed of the plane. Let us again examine the thrust formula. T= (v-u) It can be seen that the thrust may be increased in two ways, either by increasing the air mass flow through the engine (M/t) or increasing the exhaust gas velocity (v). EXAMPLE: A French Dassault Falcon 30 is powered by two Lycoming ALF 502 turbofan engines. Flying at sea level with a velocity of 154 m/sec. the air intake velocity is 154 m/sec. and the air exhaust velocity is 224 m/sec. The airflow through each engine is 109 kg per second. Determine the thrust of each engine.

EXAMPLE: A Lockheed Jet Star is equipped with four Pratt & Whitney JT12 engines. Cruising at 220 knots, each engine was found to be providing 1,420 Ibs. of thrust. If the airflow through each engine was 1.55 slug/sec., what was the exhaust gas velocity? U = 220 knots = 371 ft/sec. = v-u v= +u v=

+ 371 ft /sec v = 1,290ft./sec.

Motion in a Circle A ball whirled in a circle experiences an acceleration toward the centre of the circle. This can be proven by considering that the ball is continually changing direction as it moves in a circle. Newton's first law tells us that the ball would prefer to follow a straight path, and that for it to deviate from a straight path, a force must be applied to it. It is a direct result of Newton's first law that a hammer thrower (Figure 2.58) must continually pull towards the centre of rotation, applying his full weight to make the hammer accelerate continually towards the centre of rotation. As soon as the athlete stops applying the force towards the centre (i.e. releases the hammer) the hammer travels in a straight line, at a tangent to the circle. Figure 2.58: Centripetal force exerted by a hammer thrower This acceleration is in the same direction as the force which makes it move in a circle. This force is called centripetal force (from the Latin meaning centre-seeking) Since we have a constant change in the direction of the motion of the hammer, we have a constant acceleration. This is called centripetal acceleration and can be calculated by the square of the velocity divided by the radius of the circular path, thus:

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Part 66 Training Syllabus Module 2 Physics Centripetal acceleration = Newton's Second Law connects acceleration and force, by Force = Mass x Acceleration. Thus, we can write the equation: Fnet= mac or Fnet=

Units of Force The units which we will use in our discussion of Newton's laws are the same as the units used in the formula relating weight to mass (w = mg). These units are reviewed and summarized in Table 2.4. 2

2

Each set of units, pound, slug, ft./sec. ) in the English system, or (Newton, kilogram, m/sec. ) in the metric system is said to be consistent in the following sense: a force of 1 Ib. when applied to a mass of 1 slug 2 gives it an acceleration of 1 ft./sec . UNITS

ENGLISH

METRIC

Force

pound (Ib.)

newton (N)

Mass

slug

kilogram (kg)

Acceleration

ft/sec.

2

m/sec.

8

Table 2.4: Units of Force, Mass and Acceleration 2

Similarly, a force of 1 Newton applied to a mass of 1 kilogram causes it to accelerate at 1 m/sec . Using Newton's second law, we can write: 1 Newton = 1 kilogram m/sec and 2 1 pound = 1 slug ft./sec

2

We note that Newton's second law is correctly written as: Fnet= ma However, we often assume that the force acting on mass (m) is the net force. Thus, we usually write the second law simply as: F= ma or, for circular motion, F=

(Centripetal Force )

Newton's second law when applied to bodies moving in a circular path states that the force directed toward the centre of the path must equal the mass of the body times the square of the speed of the body divided by the radius of the path. This force is called the centripetal (centre-seeking) force. EXAMPLE: Find the acceleration of a 3 slug object acted upon by a net force of 1.5 Ibs. a= a= a=0.5 ft/sec

2

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: 2 A mass of 6 kilograms accelerates at 5 rn/sec. Find the force which is acting on this object. F = ma 2 F = (6 kg) (5 m/sec ) = 30 N

Problems 1.

2

Find the mass of an object which accelerates at 5 m/sec when acted on by a net force of one Newton.

2.

Find the acceleration of a 3 slug object experiencing a net force of 12 Ibs.

3.

Find the net force on a 5 slug object which is accelerating at 3ft/sec

4.

A Learjet Model 24 of mass 6,000 kg is observed to accelerate at the start of its takeoff at 4 2 m/sec . What is the net forward force acting on the plane at this time?

5.

During a static test, a Continental engine driving a two blade constant speed propeller was found to accelerate each second a mass of 140 kg from rest to a velocity of 40 m/sec. Determine the thrust on the propeller.

6.

A Piper Archer ii has an Avco Lycoming engine driving a two blade propeller. Each second 8 slugs of air are given a change in velocity of 122 ft/sec. How much thrust is generated on the propeller?

7.

The Garrett TFE 731 turbofan engine which powers the Rockwell Saberliner 65 under static testing has an exhaust gas velocity of 321 m/sec and an airflow of 50 kg/sec. Find the static thrust of the engine.

8.

A plane weighs 36,000 Ibs. The forward thrust on the plane is 20,000 Ibs. and the frictional forces (drag) add up to 2,000 Ibs. What is the acceleration of this plane? Hint: Be sure to find the mass of the plane from its weight.

9.

What centripetal force is needed to keep a 3 slug ball moving in a circular path of radius 2 feet and speed 4ft/sec.?

10.

A boy is swinging a stone at the end of a string. The stone is moving in a circular path. The speed of the stone is 5 ft/sec, and the radius of the path is 1.5 ft. What is the centripetal acceleration of the stone?

2

Answers 1.

0.2kg

2.

4 ft./sec

3. 4.

15 Ib . 24,000 N

5.

5600 N

6.

976 Ib.

7.

16,050 N

8.

16 ft/sec

9.

24 Ib.

10.

16.67 ft/sec

2

2

2

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Friction When a body rests on a horizontal surface or is dragged or rolled on such a surface there is always contact between the lower body surface and the horizontal surface. This contact results in friction. Friction is work done as the surfaces rub against each other. This work heats the surfaces and always results in wasted work. We need to define a force known as the normal force. A body resting on a horizontal surface experiences two forces, the downward force due to the gravitational pull of the earth on this body (the weight of the body), and the upward push of the surface itself on the body (the normal force). The weight (w) and the normal force (N) are equal to each other. There are three kinds of friction: 1. Static friction 2. Sliding friction 3. Rolling friction

Figure 2.58: Weight and its reaction

Static Friction Static friction (or 'starting' friction) is the force between two objects that are not moving relative to each other. For example, static friction can prevent an object from sliding down a sloped surface. The coefficient of static friction, typically denoted as JL/S, is usually higher than the coefficient of kinetic friction. The initial force to get an object moving is often dominated by static friction. Another important example of static friction is the force that prevents a car wheel from slipping as it rolls on the ground. Even though the wheel is in motion, the patch of the tire in contact with the ground is stationary relative to the ground, so it is static rather than kinetic friction. The maximum value of static friction, when motion is impending, is sometimes referred to as limiting friction, although this term is not used universally

Rolling Friction Rolling friction is the frictional force associated with the rotational movement of a wheel or other circular objects along a surface. Generally the frictional force of rolling friction is less than that associated with kinetic friction. One of the most common examples of rolling friction is the movement of motor vehicle tyres on a road, a process which generates heat and sound as byproducts.

Kinetic Friction Kinetic (or dynamic) friction occurs when two objects are moving relative to each other and rub together (like a sled on the ground). The coefficient of kinetic friction is typically denoted as u k, and is usually less than the coefficient of static friction. Since friction is exerted in a direction that opposes movement, kinetic friction usually does negative work, typically slowing something down. There are exceptions, however, if the surface itself is under acceleration. One can see this by placing a heavy box on a rug, then pulling on the rug quickly. In this case, the box slides backwards relative to the rug, but moves forward relative to the floor. Thus, the kinetic friction between the box and rug accelerates the box in the same direction that the box moves, doing positive work. Examples of kinetic friction:  Sliding friction is when two objects are rubbing against each other. Putting a book flat on a desk and moving it around is an example of sliding friction  Fluid friction is the friction between a solid object as it moves through a liquid or a gas. The drag of air on an aeroplane or of water on a swimmer are two examples of fluid friction.

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Calculating Friction In all cases, the friction equation is the same. The symbol >" (the Greek letter mu) is called the coefficient of friction. Every pair of flat surfaces has two different coefficients of friction: Coefficients of

Friction

Material

μstart

μslide

Steel on Steel

0.15

0.09

Steel on ice

0.03

0.01

Leather on Wood

0.5

0.4

Oak on Oak

0.5

0.3

Rubber on Dry Concrete

1.0

0.7

Rubber on Wet Concrete

0.7

0.5

Table 2.5: Some examples of Coefficients of Friction The coefficient of starting friction — μstart The coefficient of sliding friction —μslide Some values for the coefficients of starting and sliding friction are given in Table 2.5. We note that the coefficients of sliding friction are less than the coefficients of starting friction. This means that the force needed to start a body sliding is greater than the force needed to keep a body sliding with constant speed. When we deal with a body that rolls over a flat surface, we have another coefficient of friction to consider: the coefficient of rolling friction. The coefficients of rolling friction (μroll) are very small. Therefore, rolling friction is much smaller than either starting or sliding friction. Some values are: Rubber tires on dry concrete Roller bearings

0.02 0.001 to 0.003

EXAMPLE: A steel body weighing 100 Ibs. is resting on a horizontal steel surface. How many pounds of force are necessary to start the body sliding? What force is necessary to keep this body sliding at constant speed? W=N = 100lbs. Force to start sliding motion = (0.15)(100 Ibs.) =15 Ibs. Force to keep body sliding = (0.09) (100 Ibs.) = 9 Ibs.

Problems 1.

An aircraft with a weight of 85,000 Ibs. is towed over a concrete surface. What force must the towing vehicle exert to keep the aircraft rolling?

2.

It is necessary to slide a 200 Ib. refrigerator with rubber feet over a wet concrete surface. What force is necessary to start the motion? What force is necessary to keep the motion going?

Answers 1. 1,700lbs.

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2.

140 Ibs. / 100lbs.

Work, Energy and Power Work Work is done on a body when a force acts through a distance. The definition of work involves the force acting on the body (F) the distance through which this force acts (S) and the angle (θ) between the force vector and the distance vector. The definition of work is: W = FS cos θ Very often the force vector and the distance vector act in the same direction. In this case, the angle (θ) is a zero degree angle. If you check on your calculator, you will find that the cosine of a zero degree angle is equal to one. This simplifies things in this case because then work is simply equal to the product of force times distance. The unit of work in the English system is the foot-lb. Note that the two units are multiplied by each other. Students tend to write ft./lb. This is incorrect. The unit is not feet divided by pounds. In the metric system, the unit is the Newton-meter or the Joule (J). Note that the Newton-meter has a name, the Joule. The foot-lb. has no special name. EXAMPLE: A puck lies on a horizontal air table. The air table reduces the friction between the puck and the table to almost zero since the puck rides on a film of air. A player exerts a force of 70 Ibs. on this puck through a distance of 0.5 feet, and he is careful that his force is in the same direction as the distance through the force is applied. The player has done 35 ft-lbs. of work on the puck.

EXAMPLE: A book weighing 8 pounds is raised a vertical distance by a student demonstrating work. The book is raised 2 feet. The student has done 16 ft.lbs. of work.

EXAMPLE: A sled is dragged over a horizontal snowy surface by means of a rope attached to the front of the sled. The rope makes an angle of 28° with the horizontal. The sled is displaced a distance a 50 ft. The worker exerts a force of 35 pounds. How much work does the worker do? We use the formula: W = FS cos θ W = (35 Ibs.) (50 ft.) cos 28° W= 1550 ft. Ibs.

Sometimes the force and the displacement are in the opposite directions. This situation gives rise to negative work. Note, in this case, the angle between the force and the displacement is a 180° angle. The cosine of 180° is negative one. One example of negative work occurs when a body is lowered in a gravitational field. If a student carefully lowers a book weighing 15 pounds through a distance of 2 feet, we note that the displacement vector points downward and the force vector point upwards W = FS cos θ W=(15lbs.)(2ft.)cos180° W=(15lbs.)(2ft.)(-1) W=-30ft.lbs. Energy

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Part 66 Training Syllabus Module 2 Physics The concept of energy is one of the most important concepts in all of physical science. We often hear of energy sources, alternate energy, shortage of energy, conservation of energy, light energy, heat energy, electrical energy, sound energy, etc. What is the exact meaning of this word, energy? Sometimes energy is defined as the "capacity to do work". This definition is only a partial definition. However, it has the advantage of immediately relating the concept of energy to the concept of work. These two ideas are intimately related to each other. Energy is a quality that a body has after work has been done on this body. Once work has been done on a body of mass (m) this body has energy. The body can then do work on other bodies. Consider the following situation: A body of mass (m) was resting on a horizontal air table. A player exerted a horizontal force (F) on this mass through a distance (s). Since the angle between the force and the displacement was a zero degree angle, the work done on this body was simply Fs. At the instant the player removed his hand from the body we note two facts. The body accelerated while the force (F) was acting on the body and the body has acquired a velocity (v) during this time of acceleration (a). The body has moved through a distance (s) in time (t). 1

2

s = /2 at

Also note that the force (F) is related to the acceleration by the relation: F = ma We now look again at this body at the instant the force (F) has ceased acting. We note that work (W) has been done on this body and that the body moves with speed (v). 1

2

W=Fs = (ma)( /2at ) 1

W = /2m(at)

2

Now we note that the speed obtained by the body during the time of acceleration is given by the equation: V = at Therefore, we can substitute v for at in the equation above. 1

2

W = /2 mV

The equation we have obtained is the defining equation for a quantity known as kinetic energy. Usually, we use the symbol "KE" for kinetic energy. 1

2

KE = /2 mV

After the work has been done on the mass (m) it moves off on the frictionless air table with this kinetic energy. This body now is capable of doing work on other bodies that it contacts. For example, it probably will strike the edge of the table. When this happens this kinetic energy will be changed into other types of energy such as sound energy or heat energy. We note that the initial kinetic energy of the mass (m) was zero. This is true because the body was initially as rest. We can say that the work done on the body is equal to the change in the energy of the body.

Gravitational Potential Energy Another equally important situation where an agent easily can do work on a body (and thus give the body energy) occurs when the agent raises a body vertically in a gravitational field, at the surface of the earth. In this case, the work done on the body again equals the force applied multiplied by the distance the body is raised. W=Fs W = (weight of body) (distance raised)

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Part 66 Training Syllabus Module 2 Physics We recall that w = mg. Also since the distance is a vertical distance we use the symbol "h" for height. In our discussion we will assume that the symbol "h" always represents the vertical distance of the body above the surface of the earth. Therefore, we write: W= mgh Again we have a case where an agent did work on a body and the body has acquired "energy". This type of energy is known as gravitational potential energy, however, we usually symbolize it as "PE". PE = mgh If we neglect air resistance (which results in loss of energy to heat), we note that there is a conservation of kinetic and potential energy of a body moving in a gravitational field. As a body falls from a height (h) and moves closer to the surface of the earth, its potential energy decreases and its kinetic energy in creases while it is falling. Therefore, there is an easy way of finding the speed of a falling body during any instant of its fall. The units for energy are the same as the units for work, the Joule (J) in the metric system and the foot-pound in the English system. EXAMPLE: A body of mass 4 slugs is held by an agent at a distance of 6 ft. above the surface of the earth. The agent drops the body. What is the speed of the body when it is on the way down and at a distance of 2 feet above the earth's surface? We note that the initial potential energy is equal to the sum of the kinetic and potential energies on the way down (wd). PE = PEwd + KEwd 2 2 1 2 (4 slug) (32 ft/sec ) (6 ft.) = (4 slug) (32 ft/sec ) (2 ft.) + /2 (4 slug) v

EXAMPLE: A body of mass, 10 kg, falls to the earth from a height of 300 m above the surface of the earth. What is the speed of this body just before it touches ground? PEi = KEf 2 2 (10 kg) (9.8 m/sec ) (300 m) = (10 kg) V 2 2 1 2 2,940 m /sec = /2 V 2 2 2 5,880 m /sec = V V = 76.7 m/sec

The kinetic energy that the body has just before it reaches the ground immediately changes to sound energy and heat energy on impact. It may also "squash" any body in its path or make an indentation in the earth this is strain energy (energy to deform).

Power Power is the rate of doing work. The more rapidly a piece of work can be done by a person or a machine the greater is the power of that person or machine. We define power by the following equation: Power=

=

In symbols:

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Part 66 Training Syllabus Module 2 Physics P= In the English system the unit of power is the horsepower and in the metric system the unit is the Watt. Conversion factors exist giving information regarding these units. 1 Horsepower =550

= 33,000

1,000 Watts = 1 kilowatt EXAMPLE: An aircraft engine weighing 4,000 Ibs. is hoisted a vertical distance of 9 feet to install it in an aircraft. The time taken for this piece of work was 5 minutes. What power was necessary? Give the answer in ft.lb./sec. and in horsepower. P=

=

P=120 ft.lbs./sec.x P = 0.218 HP

EXAMPLE: An elevator cab weighs 6,000 N. It is lifted by a 5 kW motor. What time is needed for the cab to ascend a distance of 40 m? t= t= t = 48 seconds

Alternate Form for Power We can put our formula for power in another form by recognizing that - is speed (v). This leads to the formula: P= =F P = Fv This form is particularly useful for obtaining an expression for the power output of a turbine engine. These engines are ordinarily rated in terms of the thrust which they produce. To obtain an expression for their power output it is necessary to multiply their thrust by the speed of the plane. This thrust power, which is usually expressed in units of horsepower (THP, thrust horsepower), can be obtained by multiplying the thrust in pounds by the speed in ft./sec. and dividing by 550 where the conversion 1 HP = 550 ft-lbs./sec. is used. Thus: THP= Alternatively, we can take the speed of the aircraft in MPH and use the conversion 1 HP = 375 mi.lbs./hr. to obtain:

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Part 66 Training Syllabus Module 2 Physics THP = EXAMPLE: A gas turbine engine is producing 5,500 Ibs. of thrust while the plane in which the engine is installed is travelling 450 MPH. Determine the THP. THP=

= 6,600 HP

It is important to note that while the thrust of a gas turbine engine may not vary much over a particular range of aircraft speeds, the power must be recalculated each time the plane changes its speed. Problems 1.

How much work is done by a person in raising a 45 Ib. bucket of water from the bottom of a well that is 75 ft. deep? Assume the speed of the bucket as it is lifted is constant.

2.

A tugboat exerts a constant force of 5,000 N on a ship moving at constant speed through a harbour. How much work does the tugboat do on the ship in a distance of 3 km?

3.

A father has his 45 Ib. son on his shoulders. He lowers the child slowly to the ground, a distance of 6ft. How much work does the father do?

4.

A 6 slug body has a speed of 40 ft/sec. What is its kinetic energy? If its speed is doubled, what is its kinetic energy?

5.

A 2 kg ball hangs at the end of a string 1 m in length from the ceiling of a ground level room. The height of the room is 3 m. What is the potential energy of the ball?

6.

A body of mass 3 slug is a distance of 77 ft. above the earth's surface and is held there by an agent. The agent drops the body. What is the speed of the body just before it hits ground?

7.

An aircraft of mass 4 tonnes lands at 30 m/s and the pilot immediately applies the brakes hard. The brakes apply a retarding force of 2000N. How far will the aircraft travel before it comes to rest. A pile driver of mass 1000 kg, hits a post 3 m below it. It moves the post 10 mm. What is the kinetic energy of the pile driver? A pile driver of mass 1000 kg, hits a post 3 m below it. It moves the post 10 mm. With what force does it hit the post when it hits the post? An aircraft engine weighing 12,000 N is lifted by a 3.6 kW motor a distance of 10m. What time was needed? A hand-powered hoist is used to lift an aircraft engine weighing 3,000 Ibs. a vertical distance of 8 ft. If the worker required 4 minutes to do this job, what horsepower was developed by the mechanic? How long does it take a 5 kW motor to raise a load weighing 6,000 Ibs. a vertical distance of 20 ft.? (Hint: convert KW to ft.lb./sec first)

8. 9. 10. 11.

12.

Answers 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

3370ft.lb. 15,000,000 J -270 ft.lb. (Note the negative sign!) 4800 ft.lb; 19,200 ft.lb. 2 39.2J(or40ofg = 10m/s ) 70 ft./sec. 900 m 30,000 J 3MN 33 sec. 100ft.lb./sec. 0.182 HP 32.5 sec.

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Part 66 Training Syllabus Module 2 Physics

Momentum Definition of Momentum Momentum is a vector quantity defined as the product of mass times velocity. Note that velocity (V) is also a vector quantity. We write the defining equation as: Momentum=mV

Momentum - mV Momentum is a very important quantity when we are dealing with collisions, because it is conserved in all such cases.

Conservation of Momentum In a collision, there are always at least two bodies that collide. We will deal only with collisions of two bodies. We will also limit our discussion to collisions occurring in one dimension. Such collisions are called "head-on" collisions. At this time, we need to recall two of Newton's laws. We need Newton's second law, F = ma, and Newton's third law, which tells us that if two bodies collide, the force that the first body exerts on the second body is equal in magnitude and opposite in direction to the force that the second body exerts on the first body. Also recall that the acceleration (a) equals the change in the velocity (symbolized by the Greek letter Delta, Γ) divided by the time (t). Now let us visualize two bodies of masses, mi and m 2 on a one dimensional track. If these two bodies collide, we have four different velocities to consider. We will name these velocities very carefully. Vi' = the velocity of body one before the collision Vi"= the velocity of body one after the collision V2' = the velocity of body two before the collision V2" = the velocity of body two after the collision From Newton's two laws, we can conclude that: m1

= -m2

After cancelling ‘t’ , we obtain: m1(V1’’ – V1’ ) = - m2(V2’’ – V2’) If we remove the parentheses, transpose terms, and switch left and right parts we obtain: m1V1’ + m2V2’= m1V1’’+m2V2’’ The equation tells us that the total momentum before the collision is equal to the total momentum after the collision. Sometimes we say simply that "momentum is conserved".

Recoil Problems The simplest example of the conservation of momentum is in recoil problems.

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: A boy and a man are both on ice skates on a pond. The mass of the boy is 20 kg and the mass of the man is 80 kg. They push on each other and move in opposite directions. If the recoil velocity of the boy is 80 m/sec., what is the recoil velocity of the man? First we note that both the man and the boy are at rest before the collision occurs. m1V1’ + m2V2' = m1V1" + m2V2" (20) (0) + (80) (0) = (20) (80) + (80)V2" 0 =1,600 + 80V2" -1,600 = 80V2" V2" = - 20 m/sec. The negative sign indicates that the man recoils in the opposite direction from the boy.

Collision Problems Whenever two bodies collide, momentum is always conserved. This is simply the result of applying Newton's second and third laws as we have done in the preceding discussion. Sometimes kinetic energy is also conserved in a collision. This happens when the bodies are so hard that there is very little deformation of the bodies in the actual collision process. Billiard balls are a good example. These collisions are known as elastic collisions. We will derive a formula for determining the velocities of the bodies after the collision has occurred. Another type of collision that we will discuss is the perfectly inelastic collision. In this type of collision, the bodies are deformed so much that they actually stick together after the collision. An example would be the collision of two masses of putty. We will also do some problems for this type of collision.

Inelastic Collisions We use the conservation of momentum for dealing with this type of collision. As we have said, the colliding bodies stick together after impact. Therefore, the equation is simply: m1V1’ + m2V2’ = (m1 + m2) V’’ Note that we use the symbol V" for the common velocity of the two bodies (which are now one body) after the collision. It is important to include the signs of the velocities of the bodies in setting up momentum equations. As usual, we use a positive sign for east and a negative sign for west, a positive sign for north and a negative sign for south.

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EXAMPLE: A truck of mass 1,550 kg is moving east at 60 m/sec. A car of mass 1,250 kg is travelling west at 90 m/sec. The vehicles collide and stick together after impact. What is the velocity of the combined mass after the collision has occurred? V1' = 60 m/sec. m1 = 1,550 kg V2' =-90 m/sec. m2 = 1,250 kg M1V1‘ + m2V2' = (m1 + m2) V" We will not include units in our substitution. However, we will note that the velocity, when we obtain it, will be in m/sec. (1,550) (60) + (1,250) (-90) = (1,550 + 1,250)V" -19,500 = 2,800V" V" = -6.96 m/sec. Since the calculated velocity has a negative sign, we conclude that the combined mass is travelling west after the impact has occurred. Our answer is that the wreckage starts moving west with a speed of 6.96 m/sec. Sometimes the principle of conservation of momentum in the case of an inelastic collision can be used by the police to determine the speed of a vehicle engaged in a head-on collision.

Sometimes the principle of conservation of momentum in the case of inelastic collision can be used by the police to determine the speed of a vehicle engaged in a head -on collision. Suppose that a large truck with a weight of 12,000 Ibs. (mass of 375 slugs) travelling east with an unknown velocity enters into a head-on collision with a smaller truck of weight 6.400 Ibs. (mass of 200 slugs) initially travelling west with a speed of 30 MPH (44 ft./sec.). The trucks stick together in the collision and marks on the highway indicate that the wreckage travelled a distance of 120 feet east. The condition of the roadway (amount of friction) indicates that the wreckage would travel for a time of 4 sec. Determine the initial speed of the large truck. The equation:

s=

t

can be used to determine the initial velocity of the wreckage. Note that the final velocity of the wreckage is zero. U=

=

= 60 ft./sec

Next, we can use the conservation of momentum equation to determine the velocity of the large truck at the instant of the impact. We will use the symbol V to represent this velocity. (375 slugs) (V) + (200 slugs) (- 44 ft./sec.) = (575 slugs) (60 ft./sec.) 375V = 43,300 V = 115 ft./sec. V = 78.4 MPH

Elastic Collisions Elastic collisions are collisions that occur between bodies that deform very little in the collision. Therefore we assume that no energy is lost. An example of such a collision is the collision between pool balls.

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Part 66 Training Syllabus Module 2 Physics In elastic collisions, both kinetic energy and momentum are conserved. In an ordinary elastic collision problem, we know the masses and the velocities of two bodies that will collide. We want to predict, by a mathematical calculation, the velocities the bodies will have after the collision has occurred, the two unknowns. If we write the two conservation equations, we have two equations in these two unknowns. It is possible to solve these two equations for these two unknowns. However, one of the conservation equations, the energy equation, is a "second order" equation. A "second order" equation contains the squares of the unknowns. This makes the solution more difficult. Instead, we will use an algebraic trick! The two conservation equations can be solved together producing a third equation. This third equation and the

momentum conservation equation provide the two first order equations that we will use in solving elastic collision problems. We will write the two conservation equations:

Again we transpose terms:

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In using these two equations, the two unknowns are usually VV and V 2", the velocities of the two bodies after the collision has occurred. The known quantities are usually the two masses and the velocities of the bodies before the collision. Also be careful to include the signs of the velocities. If you forget to do this, you will always end up with incorrect results.

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: A billiard ball of mass 2 kg is moving east at 3 m/sec. and undergoes an elastic collision with another billiard ball of mass 3 kg moving west at 4 m/sec. Find the velocities of the two balls after the collision. m1 = 2

Vi'= 3 (east)

m2=3

V2' = -4(west)

Substitute in equation (2): (2)(3) +(3)(-4) =2V1" + 3V2" (10)

-6 = 2V1" + 3V2"

Substitute inequation (9): 3 - (-4) = V2" – V1 " (11)

7= V2"-V1"

Rewrite equations (10) and (11) putting the unknowns in the left members and in order. (10)

2V1" + 3V2"=-6

(11)

-V1’’ + V2" = 7

We now have two equations and two unknowns. There are several methods of solving such a system of equations. We will use the method of addition. In this method we multiply either or both of the equations by constants to make the coefficient of one of the unknowns in one of the equations a positive number and to make the coefficient of this same unknown in the other equation a negative number of the same magnitude. We then add the two equations to eliminate one of the unknowns. We then solve for the other unknown by substituting in either equation. We will multiply (11) by the number 2. (12)

-2V1" + 2V2" = 14

Add (10) and (12): 5V2" = 8 V2" = 1.6m/sec. Substitute this value back into (11): -V1" + 1.6 = 7 V1’’ = -5.4m/sec. We note that we interpret a positive sign for the velocity as motion east and a negative sign as motion west. Our final result is that the 2 kg ball is moving west with a speed of 5.4 m/sec after the collision and the 3 kg ball is moving east with a speed of 1.6 m/sec. after the collision.

Problems 1. 1.A gun of mass 5 kg fires a bullet of mass 20 grams. The velocity of the bullet after firing, is 750 m/sec. What is the recoil velocity of the gun?

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Part 66 Training Syllabus Module 2 Physics 2.

3.

4.

5.

6.

7.

An astronaut on a space walk has a mass of 5 slugs and is at rest relative to the space station. She is working with a tool having a mass of 0.5 slug. She accidentally throws this tool away from herself with a speed of 6 ft/sec. With what speed does the astronaut recoil? An automobile having mass 1,500kg is travelling east on an expressway at 30m/sec. It overtakes a truck of mass 2,000kg also travelling east and moving with a speed of 25 m/sec. The automobile rear-ends the truck. The vehicles become locked together in this collision and continue east. What is the velocity of this combined mass? Two balls of putty become one mass of putty in a collision. The first, of mass 6 kg, was originally moving east at 10 m/sec., and the second, of mass 4 kg was originally moving west at 9 m/sec. What is the velocity of the total mass after the collision has occurred? Due to a controller's error two aircraft are directed to land in opposite directions on the same runway in a fog. A Cessna 150 of mass 50 slugs and a Beechcraft Bonanza of mass 80 slugs undergo a direct head-on collision. The Beech-craft Bonanza was originally travelling north at a speed of 30 MPH and the Cessna was travelling south. The wreckage travels a distance of 20 ft. south during a time of 3.6 sec. What was the original speed of the Cessna? A 3 kg ball is moving right with a speed of 3 m/sec. before a collision with a 2 kg ball originally moving left at 2 m/sec. What are the directions and speeds of the two balls after the collision? A 2 kg ball moving right at 5 m/sec. overtakes and impacts a 1 kg ball also moving right at 2 m/sec. What are the speeds and directions of the two balls after the impact?

Answers 1. 2. 3. 4. 5. 6. 1.

3 m/sec. 0.6 ft/sec. 27 m/sec. East 2.4 m/sec. East 67.7 MPH The 3 kg ball is moving left at 1 m/sec. and the 2 kg ball is moving right at 4 m/sec. The 2 kg ball is moving right at 3 m/sec. and the 1kg ball is moving right at 6 m/sec.

Torque Consider the diagrams 2.59 shown below. We define torque as the force (F) applied to a body that is pivoted at a point (0) multiplied by the distance from the pivot point to the place where the force is applied and multiplied by the sin of the angle between r and F. We will use the Greek letter Tau (T) for torque. The distance mentioned in the preceding sentence is called the lever arm and symbolized by the letter r. The defining equation is: Τ=

r F sin θ

In the diagram , we note that θ = 90°. This is by far the most common case. Since sin 90° = 1, this common case reduces to the more simple equation: T= r

F

However, it must be remembered that in those cases where θ is not 90°, the full equation must be used. Note also that the unit for torque is the Ib.ft., Ib.in. or the Nm.

Figure 2.59: Force acting at a distance creates torque

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Extensions Figure 2.60 shows a typical beam type torque wrench which has an extension spanner attached. If this combination is used to torque load a fastener then the following formula should be used to calculate the wrench scale reading which corresponds to the specified torque value: Scale reading = specified torque x Where

L = distance between the driving tang and the centre of the handle X = length of extension spanner between centres

Figure 2.60: A torque wrench fitted with an extension spanner A simple way of calculating the scale reading required without using the formula is set out in the following example, for which the specified torque loading is 300 Ib in and the lengths of the wrench and spanner are 10 and 5 inches respectively. (a)

which is (b)

Force required on wrench handle to produce a torque of 300 Ib in is 300 Ib in divided by the distance between nut and wrench handle, = 20lb Scale reading when force on handle is 20 Ib is, 20 Ib x 10 in 200 Ib in.

Force must therefore be applied to the wrench handle until a reading of 200 Ib in is shown on the wrench 1 scale, and this will represent a 300 Ib in torque load applied to the nut. With the 'break type wrench, the adjustment must be preset at 200 Ib in. NOTE: For the purpose of conversion, 1 lb.in. = 115 kg cm or 0.113 N.m. When using an extension spanner with a torque wrench, the spanner and wrench should be as nearly as possible in line. If it is necessary to diverge by more than 15° from a straight line (due, for example, to intervening structure), then the direct distance (D) between the nut and wrench handle must be substituted for 'L + X' in the formula for calculating wrench scale reading. This is shown in figure 2.61, and the scale reading in this instance will be equal to specified torque x

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Figure 2.61: A torque wrench fitted with an extension spanner positioned out-of-line with the wrench Whenever a torque wrench is used, it must be confirmed that the specified torque and the wrench scale are in the same units; if not, then the specified torque should be converted, by calculation, to the units shown on the wrench scale, and any measurements taken in appropriate units. When applying torque the wrench handle should be lightly gripped and force applied smoothly at 90° to the axis of the wrench.

Couples A 'couple' is a pair of forces of magnitude F that are equal and opposite but applied at points separated by a distance d perpendicular to the forces. The combined moment of the forces produces a torque Fd on the object on which they act. An example is the cutting of an internal thread with a tap and tap wrench. The force applied at one end of the wrench handle, multiplied by the distance to the centre of rotation is just half of the torque felt at the tap itself, since there is an equal torque applied at the other wrench handle. Torque applied by a couple = one of the forces (F) x distance to centre of rotation (r) x 2 = one of the forces (F) x distance between the forces (d) = Fd Another example is the forces applied to a car steering wheel. Problems 1. 2. 3. 4.

Answers 1. 2. 3. 4.

Calculate the torque applied to a nut and bolt by a 12 in. spanner when a force of 12 Ib. is applied perpendicular at the end of the spanner. How much force is required to torque a nut and bolt to 50 Nm with a wrench 0.5 m long? A nut is to be torqued to 50 in.lb. A torque wrench of 17 in is used with an extension of 3 in. What setting should the torque wrench be adjusted to? A ships wheel has a couple applied to it by the captain of 60 Nm. The diameter of the wheel is 0.8 m. What is the force applied on just one side of the wheel?

144in.lb. 100 N 42.5 in.lb. 75 N

The Gyroscope Gyros are fascinating to study and a great deal of material is available on them. For the most part, we will be concerned with only two of the properties of spinning gyros. The first is the tendency of a spinning gyro to remain fixed in space if it is not acted upon by outside forces such as bearing friction. This is the property of rigidity. Rigidity is used to measure position in position gyros such as the HSI (gyro compass) and ADI (artificial horizon).

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The other property of a spinning gyro that concerns us is its right angle obstinacy. It never goes in the direction that you push it, but off to one side. Figure 2.62 illustrates this obstinate characteristic.

’

Figure 2.62: Force and resultant movement obstinacy - 'Precession

Whichever way you apply the force to the axis of a gyro, it will move in a direction 90° (in direction of rotation) to the force. The speed at which it moves is proportional to the force applied. This action is called precession. The force of procession is used in rate gyros, such as those in a turn and slip indicator, where the speed of turn is measured by the force that the processing gyro exerts on a spring.

Apparent Drift (or Wander) Figure 2.63 illustrates the behavior of a gyro. A perfect gyro would be one without any external forces acting upon it, mounted in a perfect suspension system that would give it complete freedom of movement in all three axes. All the gyros in this figure are perfect gyros. Such gyros are called free gyros. Only four gyros are represented - A, B, C, and D. The other gyro symbols shown illustrate the various positions of B, C, and D as the earth rotates.

Gyro A has its spin axis parallel with the spin axis of the earth, sitting on top of the North Pole. It could maintain that position indefinitely. Gyro B has its spin axis parallel to the earth's spin axis, and is located above the equator. The other gyros in its group represent Gyro B as it would appear at different times of the day. If we were to look at Gyro B sitting on a table in front of us, we would see that the upper end of its spin axis is pointing off toward the north star. As time goes on and the earth turns 360°, we would not see any change in its attitude on the table. Its spin axis would always point toward the north star. Gyro C is situated on the equator. The other gyros in its group represent Gyro C as it would appear at different times of the day. Let's say that we have the Gyro C in front of us on a table. Its spin axis is parallel to the earth's surface. As time goes on and the earth rotates, we would see its spin axis gradually tilting upward at one end until, six hours later (90° of earth rotation), we would see it perpendicular to the earth's surface, illustrated by the gyro shown to the right of the earth. Six hours later (behind the earth out of sight in this drawing) the spin axis would once again be parallel to the earth, but with the end which was first pointing east now pointing west.

Figure 2.63: Apparent Drift

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Part 66 Training Syllabus Module 2 Physics Another six hours later, the spin axis would once again be perpendicular, but this time the opposite end of the axis would be another six hours later. When we get to the same time of day at which we started, the gyro will again be occupying its original position. Gyro D and its group illustrate another changing aspect of a gyro, in different positions as viewed from the earth's surface at different times of day. These perfect gyros illustrate what any gyro tries to do but cannot because of its orientation of the spin axis always in the same direction in space

Transport Drift (or Wander) The outer ring of gyros in Figure 2.64 demonstrates that a completely free gyro in an aircraft circling the earth would be perpendicular to the earth's surface at only two points. The gyros drawn in the aircrafts are continuously being corrected to a vertical position as the aircraft moves around the surface of the earth. The corrections are gentle and slow, since the amount of correction needed in a ten minute period, for example, is small The gyro is relatively very stable during the pitch and roll maneuvers of the aircraft. Such a gyro is called an earth gyro or tied gyro. The gyro's stable position with respect to the movements of the aircraft makes it possible for the pilot to know the actual attitude of his aircraft, nose up or down, and wings level or not. This is quite important to him when all he can see out of the window is a gray fog. The aircraft attitude information derived from the gyros is also used by such systems as the autopilot, radar antenna stabilization, flight recorders and flight directors.

Figure 2.64: Transport Drift

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Figure 2.65: Horizon indication and Compass indication (Position Gyros)

Figure 2.66: Turn Rate indication (Rate Gyro)

Fluid Dynamics The Atmosphere On November 21, 1783, a hot air balloon carrying Marquis d'Arlandes, flew 5 miles across the skies of Paris opening up new possibilities in travel and a fresh interest in our atmosphere. It wasn't however, until heavierthan-air flight became a reality that a detailed understanding of the medium enveloping our globe became essential. The atmosphere is a mixture of gases which we call air. Dry air is composed of approximately 21% oxygen, 78% nitrogen, and 1% carbon dioxide. These percentages remain fairly constant as we ascend in altitude. However, the density of air decreases. This drop in density with altitude has great significance in aviation as it not only places limits on the attainable altitudes, but also the powerplant performance of an aircraft. The mapping out of our atmosphere, that is, determining its density, pressure, and temperature at different altitudes, required the effort of many individuals working over many years. The fruit of this labour is a vast quantity of data which has led to the definition of a standard atmosphere. The standard atmosphere, a term coined by Willis Ray Gregg in 1922, is a compilation of mean annual atmospheric properties. Since our atmosphere undergoes seasonal variations in properties such as temperature, a mean or average value is used. Tables 2.6 and 2.7 are two tables of values for the standard atmosphere. The first table (table 2.6) gives values in English units and the second (table 2.7) in metric units. It must be kept in mind that the numbers in these tables are annual averages which can be useful for reference purposes but do not indicate the actual atmospheric conditions existing at any particular moment.

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: Using the Gas Law and the temperature and pressure at an altitude of 12,000 meters listed in the Standard Atmosphere Table (table 2.5), verify that the density of air at this altitude is 0.312 3 kg/m . We will use the equation

ρ= ρ= =0.312 kg/m

3

Cabin Altitude Cabin altitude is a term used to express cabin pressure in terms of equivalent altitude above sea level. For example, a cabin altitude of 6,000 feet means that the pressure inside the aircraft cabin is the same as the atmospheric pressure at an altitude of 6,000 feet. Looking at the Standard Atmosphere Table (table 2.6), the 2 2 pressure is found to be 1,696 Ibs/ft which upon division by 144 gives the pressure in Ib/in to be 11.78 2 Ibs/in At a cabin altitude of 8,000 feet, the passengers and crew can ride in relative comfort without any special oxygen supply. Planes which fly at much higher altitudes than 8,000 feet must be furnished with a special atmosphere control system. It is highly advantageous to fly at high altitudes both for economy of fuel consumption, and the smooth air high above the level of turbulent weather systems. At these high altitudes, the pressure outside the plane can be significantly lower than the cabin pressure. 2

2

At 8,000 ft., the Standard Atmosphere Table tells us that the air pressure is 1,572 Ibs/ft or 10.92 Ibs/in This is the pressure that is normally maintained in the cabin even though the plane is flying at a higher altitude. Suppose the plane is flying at an altitude of 40,000 ft. At this altitude the pressure (from the table 5-1) is 393 2 2 Ibs/ft or 2.73 Ibs/in This means that for a cabin altitude of 8,000 ft. for a plane flying at 40,000 ft., there is a 2 2 net outward pressure of 8.19 Ibs/in This number was obtained by subtracting 2.73 Ibs/in from 10.92. For a 2 Learjet with a pressurized area of 45,000 in , we are dealing with a bursting force of over 368,000 Ibs. (8.19 x 45 thousand), in addition to being able to withstand this much force, a safety factor of 1.33 is generally used by design engineers. Therefore, the pressurized portion of the fuselage must be constructed to have an ultimate strength of over 460 thousand pounds or about 230 tons. The challenge of finding lightweight materials which can withstand these large forces is great. In the description of an aircraft's air conditioning and pressurization system, a differential pressure is given. The differential pressure is the maximum difference between cabin pressure and atmospheric pressure which the pressurization system can sustain. For example, the air-cycle air-conditioning system of a Boeing 2 747 can maintain a pressure differential of 8.9 Ibs/in This means that the system can maintain a cabin 2 pressure 8.9 Ibs/in greater than the atmospheric pressure surrounding the plane. This also means that there is an upper limit imposed by the pressurization system on the altitude at which the plane can fly.

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Standard Atmosphere — English Units 2

3

Altitude (ft.)

Temperature (°R)

Pressure (Ib./Ft )

Density (Slug/Ft )

0 500

519 517

2,116 2,078

0.002377 0.002342

1,000

515

2,041

0.002308

2,000

512

1,968

0.002241

3,000

508

1,897

0.002175

4,000

504

1,828

0.002111

5,000

501

1,761

0.002048

6,000

497

1,696

0.001987

7,000

494

1,633

0.001927

8,000

490

1,572

0.001869

9,000

487

1,513

0.001811

10,000

484

1,456

0.001756

15,000

465

1,195

0.001496

20,000

447

973

0.001267

25,000

430

786

0.001066

30,000

412

630

0.000891

35,000

394

499

0.000738

40,000

390

393

0.000585

45,000

390

309

0.000462

50,000

390

244

0.000364

55,000

390

192

0.000287

60,000

390

151

0.000226

65,000

390

119

0.000178

Table 2.6: Quantities within the Standard Atmosphere (English Units)

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Standard Atmosphere — Metric Units Altitude

Temperature

Pressure

Density

(M)

(K)

(kPa)

(kg/m )

0

288

101.3

1.225

100

288

100.1

1.213

200

287

98.9

1.202

300

286

97.8

1.190

400

286

96.6

1.179

500

285

95.5

1.167

600

284

94.3

1.156

700

284

93.2

1.145

800

283

92.1

1.134

900

282

91.0

1.123

1,000

282

89.9

1.112

1,500

278

84.6

1.058

2,000

275

79.5

1.007

2,500

272

74.7

0.957

3,000

269

70.1

0.909

3,500

265

65.8

0.863

4000 4,500

262 259

61 7 57.8

0819 0.777

5,000

256

54.0

0.736

5,500

252

50.5

0.697

6,000

249

47.2

0.660

6,500

246

44.1

0.624

7,000

243

41.1

0.590

7,500

239

38.3

0.557

8,000

236

35.7

0.526

8,500

233

33.2

0.496

9,000

230

30.8

0.467

9,500

227

28.6

0.439

10,000

223

26.5

0.414

12,000

217

19.4

0.312

14,000

217

14.2

0.228

16,000

217

10.4

0.166

18,000

217

7.57

0.122

20,000

217

5.53

0.0889

22,000

217

4.04

0.0650

3

Table 2.7: Quantities within the Standard Atmosphere (Metric Units)

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Humidity Some water in the form of invisible vapour is intermixed with the air throughout the atmosphere. It is the condensation of this vapour which gives rise to most weather phenomena: clouds, rain, snow, dew, frost and fog. There is a limit to how much water vapour the air can hold and this limit varies with temperature. When the air contains the maximum amount of vapour possible for a particular temperature, the air is said to be saturated. Warm air can hold more vapour than cold air. In general the air is not saturated, containing only a fraction of the possible water vapour. The amount of vapour in the air can be measured in a number of ways. The humidity of a packet of air is usually denoted by the mass of vapour contained within it, or the pressure that the water vapour exerts. This is the absolute humidity of air. Relative humidity is measured by comparing the actual mass of vapour in the air to the mass of vapour in saturated air at the same temperature. For example, air at 10°C contains 9.4 3 g/m (grams per cubic metre) of water vapour when saturated. If air at this temperature contains only 4.7 3 g/m of water vapour, then the relative humidity is 50%. When unsaturated air is cooled, relative humidity increases. Eventually it reaches a temperature at which it is saturated. Relative humidity is 100%. Further cooling leads to condensation of the excess water vapour. The temperature at which condensation sets in is called the dew point. The dew point, and other measures of humidity can be calculated from readings taken by a hygrometer. A hygrometer has two thermometers, one dry bulb or standard air temperature thermometer, and one wet bulb thermometer. The wet bulb thermometer is an ordinary thermometer which has the bulb covered with a muslin bag, kept moist via an absorbent wick dipped into water. Evaporation of water from the muslin lowers the temperature of the thermometer. The difference between wet and dry bulb temperatures is used to calculate the various measures of humidity.

Definitions Absolute humidity: The mass of water vapour in a given volume of air (i.e., density of water vapour in a given parcel), usually expressed in grams per cubic meter Actual vapour pressure: The partial pressure exerted by the water vapour present in a parcel. Water in a gaseous state (i.e. water vapour) exerts a pressure just like the atmospheric air. Vapour pressure is also measured in Millibars. Condensation: The phase change of a gas to a liquid. In the atmosphere, the change of water vapour to liquid water. Dewpoint: the temperature air would have to be cooled to in order for saturation to occur. The dewpoint temperature assumes there is no change in air pressure or moisture content of the air. Dry bulb temperature: The actual air temperature. See wet bulb temperature below. Freezing: The phase change of liquid water into ice. Evaporation: The phase change of liquid water into water vapour. Melting: The phase change of ice into liquid water. Mixing ratio: The mass of water vapour in a parcel divided by the mass of the dry air in the parcel (not including water vapour). Relative humidity: The amount of water vapour actually in the air divided by the amount of water vapour the air can hold. Relative humidity is expressed as a percentage and can be computed in a variety of ways. One way is to divide the actual vapour pressure by the saturation vapour pressure and then multiply by 100 to convert to a percent. Saturation of air: The condition under which the amount of water vapour in the air is the maximum possible at the existing temperature and pressure. Condensation or sublimation will begin if the temperature falls or water vapour is added to the air. Saturation vapour pressure: The maximum partial pressure that water vapour molecules would exert if the air were saturated with vapour at a given temperature. Saturation vapour pressure is directly proportional to the temperature.

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Part 66 Training Syllabus Module 2 Physics Specific humidity: The mass of water vapour in a parcel divided by the total mass of the air in the parcel (including water vapour). Sublimation: In meteorology, the phase change of water vapour in the air directly into ice or the change of ice directly into water vapour. Chemists, and sometimes meteorologists, refer to the vapour to solid phase change as "deposition." Wet bulb temperature: The lowest temperature that can be obtained by evaporating water into the air at constant pressure. The name comes from the technique of putting a wet cloth over the bulb of a mercury thermometer and then blowing air over the cloth until the water evaporates. Since evaporation takes up heat, the thermometer will cool to a lower temperature than a thermometer with a dry bulb at the same time and place. Wet bulb temperatures can be used along with the dry bulb temperature to calculate dew point or relative humidity. Problems 1. Verify, that using the Gas Law (p = P/RT) and the temperature and pressure from the Standard 3 Atmosphere Table, at an altitude of 65,000 ft., the density of air is 0.000178 slug/ft. 2. A pressurized Cessna Centurion II has a cabin pressurization system which can maintain a 2 pressure differential of 3.45 Ibs/in What is the maximum altitude at which the plane can fly and still maintain a cabin altitude of 8,000 feet? 2 2 3. (Hint: convert the of 3.45 Ibs/in to Ibs/ft and compare with the Standard Atmosphere table) What is the maximum altitude at which this same Cessna plane can fly and maintain a cabin altitude of 6,000 ft? Answers 2. About 18,000 ft 3. About 14,000 ft

Density and Specific Gravity Density The density of a material is defined as the mass of a sample of the material divided by the volume of the same sample. The symbol used for density is the Greek letter rho, (p). ρ= Other algebraic forms of this same equation are: m= ρV and V= Density is a very important and useful concept. If a body is made of a certain kind of material its density is known. If the weight of the body is also known, it is possible to determine the volume of this body. Similarly, if the kind of material and volume are known it is possible to determine the weight of the body. Table 2.8 is a table of densities. You can refer to this table when you solve the problems dealing with mass, weight, and volume.

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Densities of Liquids and Solids Liquids

Kg/m

Water

1000

Sea Water

3

Slug/ft

3

3

Metals

Kg/m

1.940

Aluminium

2700

5.25

1030

2.00

Cast Iron

7200

14.0

Benzine

879

1.71

Copper

8890

17.3

Alcohol

789

1.53

Gold

19300

37.5

Gasoline

680

1.32

Lead

11340

22.0

Kerosene

800

1.55

Nickel

8850

17.2

Sulphuric Acid

1831

3.55

Silver

10500

20.4

Mercury

13600

26.3

Steel

7800

15.1

Tungsten

19000

37.0

Zinc

7140

13.9

Brass

8700

16.9

Non-Metals

Slug/ft

Ice (32°F, 0°C)

922

1.79

Concrete

2300

4.48

Glass

2,600

4.97

Woods

Granite

2700

5.25

Balsa

130

0.25

Pine

480

0.93

Maple

640

1.24

Oak

720

1.4

Ebony

1200

2.33

Table 2.8: Comparison of densities - Liquids and solids EXAMPLE: An order has been placed for 120 gal. of lubricating oil. How much will this oil weigh? V=120gal x

= 16.0

Ρ=1.75 The density of the lubricating oil has been obtained from table 2-1. m=ρV m=

=

W=mg=(28 slugs)(

= 4.69 slug ) = 896 lbs

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3

Part 66 Training Syllabus Module 2 Physics EXAMPLE: An order has been placed for 150 Ibs. of turpentine. How many gallons of turpentine will be delivered? m=

=

V=

= 4.69 slugs

=

= 2.78 ft 3

V=2.178 ft x

3

= 20.8 gal

Specific Gravity The term "specific gravity" is closely related to the idea of density- The definition is as follows: Specific Gravity = The calculation will give the same result (for a given substance) no matter what units are used. The example below will calculate the specific gravity of sulphuric acid (see table 2.8). 3

If we use the metric units (kg/m ) we obtain: Specific Gravity =

= 1.83

3

If we use the English units (slug/ft ) we obtain: Specific Gravity =

= 1.83

The specific gravity number (1.83) is unitless. It tells us that, for sulphuric acid, the density is 1.83 times as dense as water. Problems 1. 2. 3. 4. 5. Answers 1. 2. 3. 4. 5.

What is the specific gravity of kerosene? What is the specific gravity of aluminium? What is the specific gravity of ice? What is the specific gravity of glass? What is the weight of 85 gallons of kerosene?

0.8 2.7 0.922 2.6 544 Ibs. or 3029 N (Hint: Calculate weight of water, 1 litre = 1kg, or 1 pint = 1 Ib, then convert to kerosene by multiplying by its specific gravity of 0.8)

Compressibility in Fluids Fluids are defined as any substance which flows readily. Gases and liquids are such substances Gases A gas is relatively easy to compress, and the effects of which have already been discussed in the chapter on Pressure and Force.

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Part 66 Training Syllabus Module 2 Physics Liquids Many people think that a liquid is incompressible. However, liquids are, like any material, to a certain amount compressible. In calculations, the amount of compressibility of liquid is considered to be 1 volume-% per 100 bar. This means that for example when there is liquid supplied to a 200 litre oil drum which already is completely filled with liquid, the pressure increases with 100 bar for each 2 litre of extra supplied liquid. When we supply 3 litre of extra oil the pressure increases with 150 bar. The compressibility of liquid plays a key role in for example fast hydraulic systems like servo-systems of a flight simulator. To obtain a maximum dynamic performance, the compressibility should be as little as possible. This is achieved by mounting the control valves directly on the hydraulic motor or cylinder. In that case the amount of liquid between the control valve and the motor/cylinder is minimised. In some situations, the compressibility of liquids is made use of in design. A 'liquid spring' for example, is the principle of a particular type of landing gear leg, which uses no gas. The leg is completely filled with oil. The oil is compressed under the extremely large forces encountered on landing, and the shock of the landing is absorbed by the compressibility of the liquid.

Viscosity Viscosity is a measure of the resistance of a fluid to being deformed by either shear stress or extensional stress. It is commonly perceived as "thickness", or resistance to flow. Viscosity describes a fluid's internal resistance to flow and may be thought of as a measure of fluid friction. Thus, water is "thin", having a lower viscosity, while vegetable oil is "thick" having a higher viscosity. All real fluids (except superfluids) have some resistance to stress, but a fluid which has no resistance to shear stress is known as an ideal fluid or inviscid fluid. The study of viscosity is known as rheology. Viscosity coefficients When looking at a value for viscosity, the number that one most often sees is the coefficient of viscosity. There are several different viscosity coefficients depending on the nature of applied stress and nature of the fluid.    

Dynamic viscosity determines the dynamics of an incompressible fluid; Kinematic viscosity is the dynamic viscosity divided by the density; Volume viscosity determines the dynamics of a compressible fluid; Bulk viscosity is the same as volume viscosity

Shear viscosity and dynamic viscosity are much better known than the others. That is why they are often referred to as simply viscosity. Simply put, this quantity is the ratio between the pressure exerted on the surface of a fluid, in the lateral or horizontal direction, to the change in velocity of the fluid as you move down in the fluid (this is what is referred to as a velocity gradient). For example, at "room temperature", water has a -3 -3 . nominal viscosity of 1.0 x 10 Pa-s and motor oil has a nominal apparent viscosity of 250 x 10 Pa s.

Viscosity Measurement Dynamic viscosity is measured with various types of viscometer. Close temperature control of the fluid is essential to accurate measurements, particularly in materials like lubricants, whose viscosity can double with a change of only 5°C. For some fluids, it is a constant over a wide range of shear rates. These are Newtonian fluids. The fluids without a constant viscosity are called Non-Newtonian fluids. Their viscosity cannot be described by a single number. Non-Newtonian fluids exhibit a variety of different correlations between shear stress and shear rate. One of the most common instruments for measuring kinematic viscosity is the glass capillary viscometer. In paint industries, viscosity is commonly measured with a Zahn cup, in which the efflux time is determined and given to customers. The efflux time can also be converted to kinematic viscosities (cSt) through the conversion equations. Also used in paint, a Stormer viscometer uses load-based rotation in order to determine viscosity. The viscosity is reported in Krebs units (KU), which are unique to Stormer viscometers.

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Vibrating viscometers can also be used to measure viscosity. These models use vibration rather than rotation to measure viscosity.

Units of Measure Viscosity (dynamic/absolute viscosity) Dynamic viscosity and absolute viscosity are synonymous. The symbol for viscosity is the Greek symbol eta (η), and dynamic viscosity is also commonly referred to using the Greek symbol mu (μ). The SI physical unit . of dynamic viscosity is the pascal-second (Pa s), which is identical to 1 kg/(m s) (kilogram per metre-second). If a fluid with a viscosity of one Pa-s is placed between two plates, and one plate is pushed sideways with a shear stress of one pascal, it moves a distance equal to the thickness of the layer between the plates in one second.

Kinematic viscosity In many situations, we are concerned with the ratio of the viscous force to the inertial force, the latter characterised by the fluid density p. This ratio is characterised by the kinematic viscosity (v), defined as follows: ν= where μ is the (dynamic) viscosity, and p is the density. 2.

Kinematic viscosity (Greek symbol: v) has SI units (m / s). The cgs physical unit for kinematic viscosity is the stokes (abbreviated S or St), named after George Gabriel Stokes. It is sometimes expressed in terms of centistokes (cS or cSt). In U.S. usage, stoke is sometimes used as the singular form. 2

2

1 stokes = 100 centistokes = 1 cm /s = 0.0001 m /s. 2 1 centistokes = 1 mm /s

Viscosity of air

-5

The viscosity of air depends mostly on the temperature. At 15.0 °C, the viscosity of air is 1.78x10 kg/(m-s).

Viscosity of water The viscosity of water is 8.90 x 10

-4

.

Pa s at about 25 °C.

Drag and Streamlining In fluid dynamics, drag (sometimes called resistance) is the force that resists the movement of a solid object through a fluid (a liquid or gas). Drag is made up of friction forces, which act in a direction parallel to the object's surface (primarily along its sides, as friction forces at the front and back cancel themselves out), plus pressure forces, which act in a direction perpendicular to the object's surface. For a solid object moving through a fluid or gas, the drag is the sum of all the aerodynamic or hydrodynamic forces in the direction of the external fluid flow. (Forces perpendicular to this direction are considered lift). It therefore acts to oppose the motion of the object, and in a powered vehicle it is overcome by thrust. Types of drag are generally divided into three categories: profile drag (called 'parasitic' drag in USA), liftinduced drag (also known as vortex drag or induced drag), and wave drag. Profile drag includes form drag, skin friction, and interference drag. Lift-induced drag is only relevant when wings or a lifting body are present, and is therefore usually discussed only in the aviation perspective of drag. Wave drag occurs when a solid object is moving through a fluid at or near the speed of sound in that fluid. The overall drag of an object is characterized by a dimensionless number called the drag coefficient, and is calculated using the drag equation. Assuming a constant drag coefficient, drag will vary as the square of velocity. Thus, the resultant power needed to overcome this drag will vary as the cube of velocity. The standard equation for drag is one half the coefficient of drag multiplied by the fluid density, the cross sectional area of the specified item, and the square of the velocity. Wind resistance or air resistance is a layman's term used to describe drag. Its use is often vague, and is usually used in a relative sense (e.g., A badminton shuttlecock has more wind resistance than a squash ball).

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Stokes's Drag The equation for viscous resistance or linear drag is appropriate for small objects or particles moving through a fluid at relatively slow speeds where there is no turbulence. In this case, the force of drag is approximately proportional to velocity, but opposite in direction. The equation for viscous resistance is:

Viscous resistance = - bv where: b is a constant that depends on the properties of the fluid and the dimensions of the object, and v is the velocity of the object. For the special case of small spherical objects moving slowly through a viscous fluid, George Gabriel Stokes derived an expression for the drag constant, Β= 6π η r where: r is the Stokes radius of the particle, and n is the fluid viscosity. For example, consider a small sphere with radius r = 0.5 micrometre (diameter =1.0 urn) moving through -3 water at a velocity v of 10 um/s. Using 10 Pa-s as the dynamic viscosity of water in SI units, we find a drag force of 0.28 pN. This is about the drag force that a bacterium experiences as it swims through water.

Drag Coefficient The drag coefficient (Cd) is a dimensionless quantity that describes a characteristic amount of aerodynamic drag caused by fluid flow, used in the drag equation. Two objects of the same frontal area moving at the same speed through a fluid will experience a drag force proportional to their Cd numbers. Coefficients for rough unstreamlined objects can be 1 or more, for smooth objects much less. 1

2

Aerodynamic drag = Cd /z p V A Where Cd = drag coefficient (dimensionless) 3 3 ρ = fluid density (slug/ft or kg/m ) V = Velocity of object (ft/s or m/s) 2 2 A = projected frontal area (ft or m ) The drag equation is essentially a statement that, under certain conditions, the drag force on any object is approximately proportional to the square of its velocity through the fluid.

Figure 2.67: Effect of airflow on a flat plate A Cd equal to 1 would be obtained in a case where all of the fluid approaching the object is brought to rest, building up stagnation pressure over the whole front surface. Figure 2.67 (top) shows a flat plate with the

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Part 66 Training Syllabus Module 2 Physics fluid coming from the right and stopping at the plate. The graph to the left of it shows equal pressure across the surface. In a real flat plate the fluid must turn around the sides, and full stagnation pressure is found only at the centre, dropping off toward the edges as in the lower figure and graph. The Cd of a real flat plate would be less than 1, except that there will be a negative pressure (relative to ambient) on the back surface. Some examples of Cd Object

cd

A smooth brick

2.1

A bicycle plus rider

0.9

A rough sphere A smooth sphere A flat plate parallel to the flow A bullet A man (upright position) A flat plate perpendicular to flow A skier Wires and cables

0.4 0.1 0.001 0.295 1.0-1.3 1.28 1.0-1.1 1.0-1.3

Drag coefficients of some complete aircraft Aircraft type

cd

Cessna 172/182

0.027

Cessna 310 Learjet 24 Boeing 747 X-15

0.027 0.022 0.031 0.95

Streamlining Streamlining is the shaping of an object, such as an aircraft body or wing, to reduce the amount of drag or resistance to motion through a stream of air. A curved shape allows air to flow smoothly around it. A flat shape fights air flow and causes more drag or resistance. Streamlining reduces the amount of resistance and increases lift. To produce less resistance, the front of the object should be well rounded and the body should gradually curve back from the midsection to a tapered rear section. Figure 2.68 shows how the drag of a flat plate can be reduced if its shape is changed to a sphere, and more still if it is streamlined with fairings.

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Figure 2.68: Streamlining of an object reduces its drag

Bernoulli's Principle Basic Definitions Before we begin our discussion of the lift and drag on an aircraft wing, the following definitions must be understood. The Pitot tube (named after Henri Pitot in 1732) measures a fluid velocity by converting the kinetic energy of the flow into potential energy. The conversion takes place at the stagnation point, located at the Pitot tube entrance (Figure 2.69). A pressure higher than the free-stream (i.e. dynamic pressure) results from this conversion. This static pressure is measured at the static taps (known as static ports or vents). The static pressure is not affected by the speed of the aircraft, but is dependant upon the surrounding atmospheric static pressure. Pitot Pressure is the sum of static and dynamic pressures, thus:

Pitot Pressure = Static Pressure + Dynamic Pressure

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Figure 2.69: A combined Pitot tube and static taps Bernoulli's Principle applies the ideas of work and energy and the conservation of energy to a mass of fluid (liquid or gas). Since it is not as easy to think of a mass of fluid as it is to think of a discrete body, the derivation of this principle requires some thought and effort. It is worth the thought and effort, however, since this principle is the basic principle of the flight of heavierthan-air aircraft. We review that the density of a fluid (p) is related to the mass and volume of the sample of fluid by the relation: m = ρV V = AL

Figure 2.70: A volume of fluid

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Figure 2.71: Fluid flowing through a tube of increasing cross sectional area The fluid flows from a region where the cross-sectional area is less (labelled with 1's in the diagram) to a region where the cross-sectional area is greater (labelled with 2's in the diagram). We assume that the volume of fluid in the left cylindrical shape of fluid (labelled with 1's) is equal to the volume of fluid in the right cylindrical shape of fluid (labelled with 2's). Hence, the volume flow rate in any part of the tube is constant, regardless of the tube diameter. And, since the density of the fluid is constant (and air flowing at subsonic speed is considered incompressible) than the mass flow rate is also constant, regardless of the diameter of the tube.

The Venturi Tube A venturi tube is a tube constructed in such a way that the cross sectional area of the tube changes from a larger area to a smaller area and finally back to the same larger area. As a fluid flows through this tube the velocity of the fluid changes from a lower velocity to a higher velocity and finally back to the same lower velocity. We note that, if the rate (volume per second) of fluid flow is to remain constant, the fluid must flow faster when it is flowing through the smaller area.

A diagram of a venturi tube is shown in figure 2.72.

Figure 2.72: A venturi tube The height of the fluid column in the vertical tubes at the three places shown in the diagram, is an indication of the fluid pressure. As we expect from Bernoulli's Principle, the pressure is greater where the velocity is lower and vice versa. Venturi tubes in different shapes and sizes are often used in aircraft systems.

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Part 66 Training Syllabus Module 2 Physics If we consider the types of energy involved in the flowing fluid, we find that there are three types - potential (gravitational), pressure and kinetic energies. Now if we consider only two positions in the venturi -the wide part (marked '1') and the narrow part (the throat, marked '2'), and consider the conservation of energy principle, we have: Potential Energy at 1 +

Potential Energy at 2 +

=

Pressure Energy at 1 + Kinetic energy at 1

Pressure Energy at 2 + Kinetic Energy at 2

The above is assumed since the total energy in the fluid cannot change, only transferred from one form to another. This then, is the basis for Bernoulli's Formula. Since the venturi in this case is horizontal, there in no change in potential energy, and so the potential energies can be cancelled from the formula thus: Pressure Energy at 1 + Kinetic energy at 1

Pressure Energy at 2 + Kinetic energy at 2

=

2

Since Kinetic Energy is ½mV where m = mass of fluid, and V = velocity of fluid and Pressure energy is m

where P = pressure, ρ = density of the fluid

Thus: m

2

2

+ ½ mV1 = m

+ ½ mV2

Note that the mass, m, has no suffix, since the mass flow rate is constant regardless of the area of the flow. The density, p, is also a constant since the fluid is considered incompressible (even air, providing its velocity is subsonic). Cancelling the mass, m, from each equation, and multiplying each term by the density, p, gives us P1 +

1

2

/2ρV1

=

P2 +

1

2

/2ρV2

This is the standard mathematical form of the Bernoulli's Equation. It can be rearranged to give the pressure difference (for example between the upper and lower surfaces of a wing) thus: 2

1

2

P1 - P2 =

½ ρ v2 - /2 p V1

P1 - P2 =

1/2 ρ (V2 – V1 )

Factorising gives: 2

2

Application of Bernoulli's Principle to Aerofoil Sections The relative wind direction is the direction of the airflow with respect to the wing and is opposite to the path of flight (figure 2.73). The chord line of a wing is a straight line connecting the leading edge of a wing to its trailing edge (figure 2.74).

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Figure 2.73: Relative wind

Figure 2.74: Chord line Angle of Attack is the angle between the chord line of a wing and the relative wind direction (figure 2.75). Figure 2.76 shows the cross section of a wing at rest and subject to atmospheric pressure which on the 2 average is 14.7 Ibs./in.

Figure 2.75: Angle of attack ANGLE OF ATTACK A force of 14.7 Ibs. can be imagined as acting perpendicular to every square inch of the wing. The resultant of these 14.7 Ibs. force vectors is zero and therefore has no effect on the dynamics of the plane.

Figure 2.76: Pressure forces on an aerofoil It is the motion of air past the wing that alters the pressure pattern. Whether the wing is in motion through the air or the air is flowing past a stationary wing the result is the same. For example, if a plane is moving through stationary air at a speed of 200 MPH, the effect is the same (as far as the plane and air are concerned) as if the plane were stationary and the air was moving with velocity 200 MPH past the plane.

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Part 66 Training Syllabus Module 2 Physics There is a thin layer of air in direct contact with the wing surface, which, due to skin friction, is actually stationary (relative to the wing). This is called the boundary layer. In these discussions we will disregard the boundary layer and assume that the airflow is unaffected by friction. As air streams past the wing of a plane, the speed of the air past the upper surface of the wing is greater than the speed of the air past the lower surface of the wing. These exact speeds are determined by the shape of the wing and the angle of attack. For example, if the speed of the relative wind (equal to the speed of the plane) is 200 MPH, the speed of the air past the upper surface of the wing may be 210 MPH and the speed of air past the lower surface of the wing may be 195 MPH. As indicated above, the exact values for a given case depend on the shape of the wing and the angle of attack. In this example, we could say that the speed past the upper surface of the wing is [1.05 (200 MPH)] and the speed past the lower surface if the wing is [0.975 (200 MPH)]. In figure 2.77, the following symbols apply P1 P2 V0 V1 V2 ρ

= = = = = =

pressure on the upper surface of the wing pressure on the lower surface of the wing relative wind velocity Wind velocity over upper surface Wind velocity over lower surface density of the air

We apply Bernoulli's principle

1

Figure 2.77: Velocities and pressures above and below an aerofoil 2

2

P1 + /2 ρ V1 = P2 + ½ ρ V2

We note that the ones refer to the upper surface and the twos apply to the lower surface of the wing. 1

2

1

2

/2 p V1 = P2 + /2 ρ V2 2 1 2 /2 ρ V1 - /2 ρ V2 = P2 – P1 1 2 2 /2 ρ (V1 - V2 ) = P2 – P1

P1 +

1

When finding the lift on a wing, the pressure difference between the upper and lower surfaces is found from the above equation, and, since Force = Pressure x Area, simply multiply the calculated pressure difference by the area of the wing, thus: ΓP =

1

2

2

/2 ρ (V2 – V1 ),

and

Lift (or Force) = ΓP x Area

Lift = ΔP xArea (Note: In some questions, the weight of the aircraft will be quoted. Thus, if the aircraft is flying straight and level, the Lift = Weight). Problems 2 1. An aeroplane having wing area 500 ft is moving at 300 ft/sec. The speed of the air moving past the top surface of the wing is 400 ft/sec, and the speed of the air past the bottom surface of the wing is 3 200 ft/sec. The density of the air is 0.0025 slug/ft . What is the lift? 2 2. An aeroplane having wing area 400 ft is cruising at 230 ft/sec. The speed of the air moving past the top surface of the wing is 240 ft/sec and the speed of the air past the bottom surface of the wing is 3 230 ft/sec. The density of the air is 0.0025 slug/ft . What is the weight of the aeroplane? 3. An aeroplane is cruising at 310 ft/sec. The speed of the air moving past the top surface of the wing is 340 ft/sec and the speed of the air past the bottom surface of the wing is 300 ft/sec. The density of 3 the air is 0.001 slug/ft The weight of the aeroplane is 12,800 Ibs. What is the wing area? Answers 1. 75,000 Ibs. 2. 2350 Ibs. 2 3. 1000ft

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Module 2.3 Thermodynamics Temperature Temperature Scales Our common notion of hot and cold has its precise expression in the concept of temperature. As objects are heated their molecules move faster. In a solid the molecules vibrate more rapidly. In liquids and gases the molecules move all over in the container at a faster rate of speed. These variations in speed of the molecules cause objects to expand when they are heated. This expansion can be used to construct instruments called thermometers. The ordinary mercury thermometer uses the expansion of a volume of mercury contained in a bulb to indicate temperature. A number of temperature scales are currently in use. The Fahrenheit scale is the one we have used most extensively. On this scale the freezing point of water is 32° and its boiling point is 212°. The metric scale is the Celsius or centigrade scale. On this scale the freezing point of water is zero and the boiling point is 100°. In theory, if we cool any substance enough, we can cause all molecular motion to cease. We call this lowest possible temperature "absolute zero". Ordinary gases like air would be rock solid at this temperature. Low temperature physicists have never been able to reach this extremely low temperature in their laboratories. However, they have come close—down to a fraction of a centigrade degree. Absolute zero is a limiting temperature which can never be reached. Two other temperature scales are used by engineers and experimental scientists. In both of these scales the zero of the scale is placed at absolute zero, the coldest possible temperature. These scales are the metric Kelvin scale and the English Rankin scale. In Table 3-3, the four temperature scales are compared. There are formulas that enable us to change from a centigrade reading to a Fahrenheit reading and vice versa. These formulas are: C= (F-32)

and

F= C + 32

Note that there are parentheses in the first formula but not in the second formula. Be careful! There are also formulas that change from a centigrade reading to a Kelvin reading and from a Fahrenheit reading to a Rankin reading. These formulas are very important to us at this time since we will have to use absolute temperatures in the gas laws. These formulas are: K = C + 273

and

R = F + 460

Boiling Point of Water

Freezing Point of Water

Absolute Zero

Centigrade

100°

0°

-273°

Kelvin

373

273

0

Fahrenheit

212°

32°

-460°

Ran kin

672°

492°

0°

Table 3.1: Comparisons of boiling points, freezing points and absolute zero in different units Notes:  Kelvin has no ° sign in-front of the K.  The accurate conversion factor for °C to K is +273.15

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Part 66 Training Syllabus Module 2 Physics : ,

Figure 3.1: Temperature scale comparison Problems 1. Change 20°C to degrees F. 2. Change -15 °C to degrees F. 3. Change 86°F to degrees C. 4. Change -4°F to degrees C. 5. Change 100°F to degrees R. 6. Change 450°R to degrees F. 7. Change 100°C to degrees K. 8. Change 383 K to degrees C. 9. Gas turbine engine performance is very sensitive to variations in the temperature of the air. All engines are rated with the air at a standard temperature of 59°F. What is the equivalent Centigrade temperature? 10. On some large commercial turbojet engines, the temperature at the front end of the combustion section is approximately 400°C. What is this temperature on the Fahrenheit scale? 11. As air enters the combustion chamber of a turbojet fuel is added and the temperature is raised to about 3,500°F in the hottest part of the flame. What is this temperature on the Centigrade scale? Answers 1. 68°F 2.

5°F

3.

30°C

4.

-20°C

5.

560°R

6.

-10°F

7.

373 K

8.

110°C

9.

15°C

10.

752°F

11.

1,930°C

The Gas Laws We will next discuss the volume and density of gases under varying conditions of temperature and pressure. Three gas laws, named after the scientists that discovered them, will be considered.

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Boyle's Law A cylinder containing gas is fitted with a light piston. This cylinder contains a certain mass of gas and therefore a certain number of molecules of gas. The gas has a definite absolute temperature. This temperature is a measure of the average speed of the gas molecules in the sample. Some of the molecules are moving faster and some are moving slower. The average speed determines the temperature. If the temperature of the gas remains constant and the volume of the gas sample is decreased, the molecules, still moving with the same average speed, are "squashed" into a smaller space (see figure 3.2). The result is that the sides of the container experience more collisions per unit time. This results in an increase in the absolute pressure the molecules exert on the walls of the container. Note that a decrease in volume produces an increase in absolute pressure. This is characteristic of an inverse proportion. We write the equation as:

= If we cross multiply in the above equation we reach the form in which Boyle's Law is usually written: P1V1 = P2V2 Here P1 and P2 are the absolute pressures corresponding to the volumes V1 and V2 respectively. In working with Boyle's Law, it must always be remembered to use absolute pressures. Figure 3.2: Boyle's Law EXAMPLE: 2

A cylinder fitted with a piston contains gas at a pressure of 35.5 Ibs./in as indicated by a gauge mounted to 2 the outside of the cylinder. The atmospheric pressure is 14.5 Ibs./in if the piston is forced down reducing the volume in the cylinder to one fourth of its original volume while holding the temperature of the gas constant, determine the new reading on the pressure gauge. P1 = (35.5 + 14.5) Ibs./in 2 P1 = 50 ibs./in V2= ¼ V1 P1V1 = P2V2

2

2

(50 Ibs./in ) (V1) = P2 ( ¼ V1 ) Solving for P2 gives, 2

P2 = 200 ibs./in absolute We still must express this new pressure as a gauge pressure since the problem asked for the new reading on the pressure gauge. Our final answer is: 2

P2 = (200 - 14.5) Ibs./in = 186 Ibs./in

2

Charles' Law Toward the end of the 18th century, investigations carried out by French physicists, Jacques Alexandre Charles and Joseph Louis Gay-Lussac led to the discovery of a relation between the volume and absolute temperature of gases under conditions of constant pressure.

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Part 66 Training Syllabus Module 2 Physics Let us again consider a sample of gas containing a definite number of molecules. W e stipulate that the pressure on this sample of gas will remain constant. If the pressure is to remain constant, an increase in absolute temperature must be accompanied by a corresponding increase in volume (see figure 3.3). We say that the volume is directly proportional to the absolute temperature, provided that the pressure remains constant. We write the equation as: The absolute temperatures must be either Kelvin, or Rankin degrees.

Figure 3.3: Charles' Law EXAMPLE: A quantity of air occupies a volume of one cubic foot on a day when the temperature is 15°F. What will be the volume of this quantity of air when the temperature increases to 85°F, and the pressure stays the same? = Note that we have changed the temperatures from degrees Fahrenheit to degrees Rankin, because we must express the temperatures in absolute units. Cross multiplying, we obtain: V2=

= 1.15

Failure to convert to absolute temperatures will always lead to incorrect answers when working with the gas laws!

Gay-Lussac's Law This third gas law relates the absolute pressure to the absolute temperature of a gas when its volume is held constant. Again we consider a certain number of molecules of gas in a closed container where the volume of the gas is held constant. If we increase the absolute temperature of the gas, the average speed of the molecules increases. As these molecules strike the walls of the container they exert a greater pressure since they are moving faster (see figure 3.4). Using absolute pressures and temperatures the following simple relationship is obtained: Figure 3.4: Gay Lussac's Law This equation is referred to as Gay-Lussac's Law.

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: The tyre of a bicycle is filled with air to a gauge pressure of 50.0 Ibs./in. at 58°F. What is the gauge pressure in the tyre on a day when the temperature rises to 86° F? Assume that the volume of the 2 tyre does not change and the atmospheric pressure is 14.7 Ibs.fin. We must first convert to absolute temperatures and pressures. 2

2

P1 = 50.0 Ibs./in + 14.7 Ibs./in = 64.7 ibs./in T1 = 460 + 58°F = 518°R T2 = 460 + 86°F = 546°R

2

Substituting these values into Gay-Lussac's Law gives: = 2

Solving for P2, we obtain P2= 68.2 ibs./in . Finally, the new gauge pressure is obtained by subtracting the atmospheric pressure from P2. 2 2 2 68.2 Ibs./in -14.7 Ibs./in = 53.5 Ibs./in

The General (Ideal) Gas Law The three properties, pressure, temperature, and volume are interrelated for a fixed mass (number of molecules) of gas in such a way that if two of them change in value the third can immediately be determined. Combining the three gas laws the following general gas law can be written: = Note that this equation gives us the three gas laws that we have studied. If the temperature of the gas remains constant, we can cancel the temperatures in the denominators and obtain: P1V1=P2V2 Boyle's Law If the pressure remains constant, we can cancel the pressures in the numerators and obtain: =

Charles' Law

If the volume remains constant, we can cancel the volumes in the numerators and obtain: =

Gay-Lussac's Law

EXAMPLE: 2 A tank of helium gas has a gauge pressure of 50.2 Ibs/in and a temperature of 45°F. A piston decreases thevolume of the gas to 68% of its original volume and the temperature drops to 10°F. What is the new gaugepressure? Assume normal atmospheric pressure. We must change both temperatures to absolute units. We must change the original gauge pressure toabsolute pressure. We remember that when the final pressure is obtained it will be in absolute units. We alsonote that V2 = 0.68 V1.

= We transfer V2 from the numerator on the right to the denominator on the left. We also transfer T 2 from thedenominator on the right to the numerator on the left. In this way, we solve our formula for P 2. Next we substitute our known values:

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Part 66 Training Syllabus Module 2 Physics

P2 = 2

P2= 88.8 Ibs/in Absolute 2 P2= 74.1 Ibs/in (New Gauge Pressure)

= P2 Alternate Form of the General (Ideal) Gas Law The general gas law tells us that for a fixed quantity of gas, the expression PV/T is constant. Since PV/T is a constant for a fixed mass of gas, we can set this expression equal to the product of the mass (m) of the gas and what is referred to as a gas constant (R). This gas constant (R) varies according to the type of gas. Table 3.2 gives values of R for various gases. We can write: =mR PV=mRT If we divide both sides of this equation by V. we obtain: P= We remember that the density of any substance is given by: ρ= Values of the Gas Constant, R, for Some Common Gases 3 Pa m /kg K Air Carbon Dioxide Helium Nitrogen Oxygen Water Vapour

287 189 2,077 297 260 462 Table 3.2: Gas Constants, R

ft.lbs/slug °R 1,710 1,130 12,380 1,770 1,550 2,760

Therefore we can write: P = ρRT The most important application of this formula enables us to obtain the density of any particular kind of gas if we know its absolute pressure and absolute temperature. We write the equation in the form: ρ= Note: When comparing the density of one type of gas to another, we need to use equal temperatures and pressures for each gas (since, as the above equation shows, density changes with pressure and temperature changes). The temperature and pressure we use for this is as Standard Temperature and Pressure . These are 0°C and 1 atmosphere (273.15 K and 760 mmHg).

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: Find the density of air if the temperature is 80°F and the absolute pressure is 2,150 Ibs./ft ρ=

=

= 0.00233 slug/ft

2

3

Application of the General Gas Law to Compressors We can apply the general gas law to the flow of air through the compressor of a turbojet engine. The function of the compressor is to provide a large quantity of high pressure air to the limited space of the combustion chamber. The reason for this is that the energy released in the combustion chamber is proportional to the mass of air consumed. The pressure of the air when it leaves the compressor is called the compressor discharge pressure (CDP) and the ratio of this to the compressor inlet pressure (CIP) is the compression ratio. That is, Compression Ratio = Note that the compression ratio can also be expressed as: Compression Ratio = where the 1 's refer to the inlet pressure and the 2's to the discharge pressure. Air entering a compressor having a compression ratio of 12.5:1 at a pressure of 14.7 PSIA will leave with a pressure of: (12.5)(14.7)= 184PSIA If however, the temperature of the air is increased too much in the compression process the volume of a quantity of air entering the combustion chamber will not be reduced significantly and the compressor efficiency will be low. EXAMPLE: A quantity of air occupying 1 cu.ft. at pressure of 14.7 PSIA and a temperature of 59°F enters the compressor of a turbojet engine having a compression ratio of 12.5:1 and is discharged at a temperature of 2,000°F. With what volume will this quantity of air enter the combustion chamber? Solving our general equation:

= = V1( ) (

V2 =

)

Therefore, we can substitute our given values:

V2=(1ft3)(

)(

)

V2=0.379 FT.3 EXAMPLE: With what volume would the quantity of air of the previous problem enter the combustion chamber if the discharge temperature of the compressor were 750°F instead of 2,000°F? 3

V2(1 ft ) ( V2=0.187ft.

)(

)

3

3

We see that the volume of the original cubic foot of air is less (0.187 ft. ) when the temperature is 750°F 3 than it is (0.379 ft. ) when the temperature is 2,000°F.

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Part 66 Training Syllabus Module 2 Physics

Problems 1. A quantity of gas is contained in a cylinder fitted with a piston. The absolute pressure of the gas is 3 240 kPa when the volume is 0.15 m . What will the volume be when the absolute pressure of the gas is changed to 80 kPa while the temperature is held constant? 2. A quantity of gas is contained in a cylinder fitted with a piston. The gauge pressure of the gas in the 2 3 cylinder is 335 Ibs/in when the volume occupied by the gas is 72in What is the gauge pressure 3 2 when the volume is decreased to 60 in ? Assume atmospheric pressure to be 15 Ibs/in , and assume that the temperature is held constant. 3 3. A sample of nitrogen is held at an absolute pressure of 1.50 atmospheres and a volume of 7.80 m . 3 A piston gradually reduces the volume to 6.30 m . The temperature does not change. What is the new absolute pressure in atmospheres? 3 4. A volume of 1.35 m of air at 17°C is heated to 427°C while its pressure is held constant. What is the volume of the gas at this elevated temperature. 2 5. A tank of carbon dioxide is maintained at an absolute pressure of 5,000 Ibs/ft . The temperature is 190°F. What is the density of this carbon dioxide? 6. The air pressure and density at a point on the wing of a Boeing 747 flying at altitude are 70 kPa, and 3 0.9 kg/m respectively. What is the temperature at this point on the wing in degrees Centigrade? 3 7. The Goodyear non-rigid airship, the Mayflower, has a volume of 4000 m and is filled with helium to an absolute pressure of 100 kPa. The temperature is 27°C. Find the density and total mass of the helium in the ship. 8. At an altitude of 8,000 ft. the absolute temperature of air is 500°R and the absolute pressure is 1600 2 Ibs/ft . What is the density of air at this altitude? 2 9. A tank of carbon dioxide is maintained at an absolute pressure of 5,830 Ibs/ft and a temperature of 70°F. What is the density of this carbon dioxide? 3 10. A quantity of air occupying 0.9 ft at a pressure of 15 PSIA and a temperature of 40°F enters the compressor of a turbojet engine having a compression ratio of 13:1 and is discharged at a temperature of 1,540°F. With what volume will this quantity of air enter the combustion chamber? Answers 3 1. 0.45m 2 2. 405 Ib./in 3. 1.86 atmosphere 3 4. m 3 5. 0.007 slug/ft 6. -2°C 3 7. 0.16kg/m ,640kg 3 8. 188 slug/ft 3 9. slug/ft 3 10. 0.3ft

Thermal Expansion The temperature of a body is a measure of the average kinetic energy of the molecules of that body. It follows that molecules of warm liquids and gases move around faster in their containers than molecules of cool liquids and gases. As a solid is heated its molecules vibrate faster about their equilibrium positions. As a result of this increased motion of molecules as they are heated, solids and liquids expand as the temperature is raised. We're going to return to the idea of how temperature is related to molecular motion later, but first let's look at what happens to materials when they change temperature. Let's say that you've got a jar lid that's stuck and you want to get it off. How can you do this? One common way is by running the jar under hot water so that the jar lid expands and can come off the jar. Thinking back to our model of solids, if temperature is a measure of how fast things are moving, when a solid heats up, the molecules vibrate about their normal positions. At higher temperature, they vibrate more and the material actually grows in size. When a material is cooled, the molecules don't move as much and the material shrinks. If we look at a long strip of metal, with length LO , we might want to find out what its change in length is under certain conditions. This is important, for instance, in building roads that must undergo temperature extremes. Experimentally, we find that the change in length is directly related to the change in temperature and to the initial length of the bar. The dependence on the initial length of the bar comes about because there are that many more molecules moving, so the change in length will be greater than that of a shorter bar.

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But let's think back to the jar. When you heat the lid, you're also heating the glass, too. Doesn't the glass also expand? The answer is that it does, but it expands less than the material from which the lid is made. This means that we somehow have to account for the fact that different materials expand or contract by different amounts under the same temperature change. Let's try to arrange these materials on a scale. The way we account for the different rates of different materials in out equation is via the coefficient of linear expansion, α. α has units of /°C (pronounced 'per degree Celsius').

Linear Expansion A rod of a substance will increase its length for a given temperature change. The increase in length depends on the original length of the rod, the temperature change, and the material of the rod. The increase in size of the object comes about by the fact that an increase in temperature results in a n increase in kinetic energy of the molecules or atoms which make up the material. Increasing the movement of the molecules forces it to occupy more space. We define alpha (α), the coefficient of linear expansion. Tables of values of alpha for various substances are found in handbooks of physics. The formula is: ΔL =αLOΔT In this formula, LO = α = ΓL = ΓT =

the original length of the rod the coefficient of linear expansion the change in length of the rod the change in temperature

Area Expansion Two-dimensional solid bodies also experience thermal area expansion. The formula is as follows: ΔA =2αA0ΔT In this formula, A0 α ΓA ΓT

= = = =

the original area of the body the coefficient of linear expansion the change in area of the body the change in Temperature COEFFICIENTS OF

LINEAR EXPANSION (α)

SUBSTANCE

perF°

Aluminum

13 x 10

Brass

10 x 10

Concrete (varies)

5 x 10

Copper

9.4 x 10

Glass (Pyrex)

1.6 x 10

Ice

28 x 10

Iron

6.6 x 10

Lead

16 x 10

Steel

11 x 10

-6 -6

-6 -6 -6 -6 -6

-6 -6

Table 3.3: Coefficients of Linear Expansion

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Part 66 Training Syllabus Module 2 Physics

Volume Expansion Three-dimensional solid bodies experience volume expansion. ΔV =3αV0ΔT In this formula: V0 α ΓV ΓT

= the original volume of the body = the coefficient of linear expansion = the change in volume of the body = the change in temperature

Expansion of Liquids and Gases We have a minor problem with our expression for the thermal expansion of solids, which is that it only works for solids. Neither liquids nor gases have a fixed shape when left on their own. The expression also fails if you have to consider the expansion of a solid in all directions, β is called the coefficient of volume expansion. For solids, β is approximately equal to 3α. This is true only when the change in volume is small compared to the original volume. The problem is that for liquids and gases, β is very large and this formula sometimes won't work. Liquids also experience thermal expansion. We introduce beta (P), the coefficient of volume expansion. There are also tables of the coefficients of volume expansion. ΓV =βV0ΓT

Table 3.4: Coefficients of Volume Expansion Generally, liquids expand more than solids, and gases much more than liquids, for any given change in temperature. This is because the molecules of liquids are not tied to each other and have more room and freedom to vibrate than do the molecules or atoms in solids. The molecules of gases of course, are completely free to move, and thus will move much more vigorously when heated than either solids or liquids

The Interesting Case of Water Most materials expand when heated and contract when cooled. Water is an exception. Between 0°C and 4°C, water actually expands when cooled. Above this range, it behaves normally. Water therefore has its greatest density at 4°C. This turns out to be quite important for things that live underwater. In the winter, you notice that the top of a pond always freezes first. As the temperature decreases, there is a temperature gradient in the water. The top will be cooler than the bottom because it is in contact with the cold air. When the water on the top of the lake reaches 4°C, it becomes denser and sinks to the bottom of the lake, being replaced by warmer water from the bottom. The water that is now on top cools to 4°C, and so on, until the whole lake is at 4°C. The surface water cools even more, but now it is less dense than the water below it, so it stays on the top of the lake and turns to ice (which is even less dense than cold water). If the ice sank instead of floating, the lake would freeze all the way through and pretty much everything inside would die. The layer of ice additionally acts as an insulator, keeping the rest of the water away from the surface and the colder environment.

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: A steel rail of length 140 ft. Is laid down when the temperature is 20°F. What is the increase in length of this rail when the temperature is 95°F? ΓL = αLoΓT -6 ΓL =(11x10 /°F) (140 ft.) (75 °F) ΓL = 0.116ft.

EXAMPLE: 3 An aluminium tank has volume 35 ft. What is the increase in volume of this tank when the temperature increases from SOT to 90°F? It should be noted that a solid block of a substance increases in volume as the body is heated. Also, a container has a bigger volume as the temperature of the container increases. ΓV = 3αV0ΓT -6 ο 3 ΓV = 3 (13 x 10 / F) (35 ft. ) (60 °F) 3 ΓV =0.0819 ft.

EXAMPLE: The manager of an airport accepts delivery of 1,000 gallons of avgas on a cool evening when the temperature is 35°F. This avgas completely fills a 1,000 gallon aluminium tank. A warm front moves in the next morning and the temperature rises to 95°F. How much avgas will overflow? If the avgas costs £1.25/gal. what is the loss to the airport? For the gasoline: ΓV =βV0ΓT 3 ΓV = (0.58 x 1 0' /°F) (1,000 gal.) (60 °F) ΓV = 34.8 gal. For the tank: ΓV = ρV0ΓT -6 ο ΓV = 3(13 x 10 / F) (1,000 gal.) (60 °F) ΓV =2.3 gal. The new volume of the avgas is 1,034.8 gal. and the new volume of the tank is 1,002.3 gal. We note that 32.5 gallons of avgas will overflow! Loss £1.25/gal. x 32.5 gal. = £40.63.

EXAMPLE: A motorist puts 20.1 gallons of petrol in his gas tank on a hot summer day when the temperature is 95°F. He uses 0.1 gal. in driving home. The temperature falls to 45°F that evening after a cool front has moved into the area. How many gallons are in his tank the next morning when he leaves for work? ΓV =ρV0ΓT -6 ΓV = (0.58 x 10 /°F) (20 gal.) (50 °F) ΓV = 0.58 gal. There are 19.42 gallons of petrol in his tank the next morning!

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Part 66 Training Syllabus Module 2 Physics

Problems 1. 2. 3.

4.

Answers 1. 2. 3. 4.

A 90 ft. aluminium rail is put in place on a hot summer day when the temperature is 85°F. What is the decrease in length of this rail when the temperature is 35 °F? A 150 ft. steel rail is put in place when the temperature is 35°F. What is the increase in length of this rail when the temperature is 95°F? A concrete bridge is laid down in sections with some space between sections to allow for expansion. The length of one section is 250 ft. The lowest recorded temperature in the area is - 45 °F and the highest recorded temperature is 115 °F. How much space should the builders leave between each section? The volume of an aluminium tank is 200 gallons on a day when the temperature is 30 °F. It is completely filled with gasoline from a supply truck. The temperature rises to 70°F when a warm front moves in. How many gallons of gasoline overflow?

0.0585ft 0.099 ft. 0.20ft 3.7 gallons

Heat We recall that temperature is a measure of the average kinetic energy of molecules or atoms in a gas, and therefore the average velocity, of the molecules of the substance whose temperature is being measured. Heat is a measure of the total energy of molecular motion. The more molecules that are moving, the greater is the heat energy. Let us compare a teaspoon of water at 100°F with a cup of water at 50°F. The molecules of water in the teaspoon are moving faster than the molecules of water in the cup. However, since we have so many more molecules in the cup, the heat energy in the cup is greater than the heat energy in the teaspoon. If the teaspoon of water is placed on a large block of ice and the cup of water is also placed on this block of ice, the cup of water at 50°F would melt more ice than the teaspoon of water at 100°F. There are definite units for measuring heat energy. The units are the Btu (British thermal unit) and the metric units, the large Calorie (written with a capital "C") and the small calorie. The definitions are: 1 British thermal unit (Btu) = the amount of heat needed to raise the temperature of 1 Ib of water 1°F 1 Calorie = the amount of heat needed to raise the temperature of 1 kilogram of water 1°C (Note: 1 Calorie = 4186 J, 1 Btu = 0.252 Cal.) 1 calorie

= the amount of heat needed to raise the temperature of 1 gram of water 1°C

1 Celsius Heat Unit (CHU) = the amount of heat needed to raise the temperature of 1 Ib of water 1°C (Note: The CHU is a mix of English and Metric units and is rarely used) When we talk about the heat content of fuel (which must be burned to be released) -commonly called the heat of combustion, we talk about Calories per Ib. of fuel, or Btu per Ib. of fuel, or Joules per kg of fuel. Since 1 Btu = 252 calories, and 1 calorie = 4.186 Joules, there are 1055 Joules in 1 Btu. And since 1 Ib. = 2.2 kg, 1 Btu/lb. = 2326 J/kg. We note that the Calorie is the famous dietary Calorie. The body stores excess food as fat and we measure the Calories in a certain foodstuff by burning these foodstuffs and measuring the heat produced! In the solution of heat problems, we will limit our discussion to the English system, since this is the system that is most often encountered in our society.

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Part 66 Training Syllabus Module 2 Physics As heat is added to a body its temperature increases. However, the same amount of heat added to a piece of aluminium and a piece of copper will not produce the same temperature change. Aluminium and copper have different "specific heats". The important equation is the following: Q = Wcδt

(when using English units)

In this equation: Q = heat gained or lost (Btu) w = weight of the body (Ib.) C = the specific heat of the substance Btu/lb.°F ΓT = the temperature change (°F or °R) Q = mCΔT

(when using Metric units)

In this equation: Q = heat gained or lost (J) m = mass of the body (kg) C = the specific heat of the substance (J/kg°C) ΓT = the temperature change (°C or K) It is important to note that this equation deals with substances that are not changing their states of matter. Another equation will deal with heat added or lost as a body changes from one state (solid, liquid, or gas) to another. Since there are two equations, (depending on whether you are using English or Metric units) there are also two sets of Specific Heat Capacity constants. Table 3.5 shows various specific heats of substances in English and Metric units.

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Part 66 Training Syllabus Module 2 Physics

SPECIFIC HEAT CAPACITIES J/kgK

Btu/lb°F

Liquids Acetic acid

2130

0.51

Alcohol Ammonia Paraffin Petroleum Turpentine Water, fresh

2930 470 2140 2090 1980 4190

0.70 0.11 0.51 0.50 0.33 1.00

Water, sea 4°C

3940

0.93

Metals Aluminium

912

0.212

Antimony Copper Gold Iron Lead Mercury Nickel Platinum Silver Tin

214 389 130 460 130 138 452 134 234 230

0.051 0.093 0.031 0.110 0.031 0.033 0.108 0.032 0.056 0.055

Zinc

393

0.094

Solids Asbestos

84

0.20

Ashes 84 0.20 Asphalt 80 0.19 Brick 92 0.22 Carbon 71 0.17 Coal 1310 0.314 Coke 850 0.203 Concrete 1130 0.27 Cork 2030 0.485 Glass 840 0.20 Granite 750 0.18 Graphite 710 0.17 Ice 2110 0.504 Wood 2300 - 2700 0.55 - 0.65 Table 3.5: Specific Heat Capacities of some common substances Gases have a different specific heat capacity depending upon whether they are held at constant volume or constant pressure. Therefore all gases, as well as having a specific heat capacity, also has a value of its ratio of CP to Cv. γ= For example:

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Part 66 Training Syllabus Module 2 Physics Specific Heat Capacity of Dry Air: at Constant Pressure, at Constant Volume,

CP Cv

1004 J / K kg 717 J / K kg

Therefore γair = Also Cp - Cv = R

= 1 .4

(the ideal gas constant we saw in Chapter 4).

Since both R and γ are always positive values, and greater than 1 , CP is always greater than Cv. EXAMPLE: How much heat must be supplied to raise the temperature of a 32 Ib. aluminium fitting from 60°F to 90°F? Q = wCΓT Q =(

(imperial) )(32lbs.) (30 F) o

Q = 204 Btu.

EXAMPLE: How much heat is given up as 100 Ibs. of sea water cools from 90°F to 50°F? Q = wCΓT (imperial) o Q=( )(100 lbs)(40 F) Q = 3720 Btu.

Heat Exchange When hot bodies and cool bodies are mixed heat exchange occurs. The heat lost by the hot body equals the heat gained by the cold body: Heat Lost = Heat Gained On each side of this equation there is a wCΓT term. In writing an expression for ΓT, we always express this change as the larger temperature minus the smaller temperature. EXAMPLE: If 5,000 Ibs. of sea water at 100°F are mixed with 7,000 Ibs. of ordinary water at 40°F, what is the final temperature of the mixture? We note that, if the final temperature is T, the temperature 100° is more than T and the temperature 40° is less than T. Therefore the temperature change of the sea water is (100 -T) and the temperature change of the ordinary water is (T - 40). Heat Lost = Heat Gained In setting up the wCΓT left and right members of the above equation, we will not include the units. However we will note that the weights must be in Ibs. and the temperature changes in Fahrenheit degrees. (0.93) (5,000) (100 - T) = (1.00) (7,000) (T - 40) 465,000 - 4,650 T = 7,000 T - 280,000 745,000 =11,650 T T = 63.9°F

Cooling and Heating Curves When a substance changes phase, that is it goes from either a solid to a liquid or liquid to gas, the energy, it requires energy to do so. The potential energy stored in the inter-atomic forces between molecules needs to be overcome by the kinetic energy of the motion of the particles before the substance can change phase.

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Part 66 Training Syllabus Module 2 Physics If we measure the temperature of the substance which is initially solid as we heat it we produce a graph like Figure 3.5.

Figure 3.5: Temperature change with time. Phase changes are indicated by flat regions where heat energy used to overcome attractive forces between molecules Starting a point A, the substance is in its solid phase, heating it brings the temperature up to its melting point but the material is still a solid at point B. As it is heated further, the energy from the heat source goes into breaking the bonds holding the atoms in place. This takes place from B to C. At point C all of the solid phase has been transformed into the liquid phase. Once again, as energy is added the energy goes into the kinetic energy of the particles raising the temperature, (C to D). At point D the temperature has reached its boiling point but it is still in the liquid phase. From points D to E thermal energy is overcoming the bonds and the particles have enough kinetic energy to escape from the liquid. The substance is entering the gas phase. Beyond E, further heating under pressure can raise the temperature still further is how a pressure cooker works. Note the temperature stays constant during the state changes of melting and boiling. Note: Since 'fusion' (meaning 'to melt') is the opposite of 'solidification', the Latent Heat of Fusion is the same as the Latent Heat of Solidification. Also, since Vaporisation' is the opposite of 'condensation', the Latent Heat of Vaporisation is the same as the Latent Heat of Condensation. A heating curve summarises the changes: solid => liquid => gas The principle of latent heat (especially of vaporization) is what is behind the operation of the fridge and air conditioning system, water injection of gas turbine engines, and the cooling effect you feel when you perspire. That principle is that if you make a fluid vaporize, it extracts heat (latent heat) to cause it to vaporize, but the fluid does not change temperature.

Latent Heat of Fusion and Vaporisation The energy required to change the phase of a substance is known as a latent heat. The word latent means hidden. When the phase change is from solid to liquid we must use the latent heat effusion, and when the phase change is from liquid to a gas, we must use the latent heat of vaporisation. The latent heat energy required is given by the formula: Q=mL where m is the mass of the substance and L is the specific latent heat of fusion or vaporisation which measures the heat energy to change 1 kg of a solid into a liquid. Some values of Specific Latent Heats of Fusion and Vaporisation are shown in table 3.6.

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Specific Substance latent heat of fusion kJ/kg

Freezing Temperature °C

Specific latent heat of vaporisation kJ/kg

Boiling Temperature °C

Water

334

0

2258

100

Ethanol

109

-114

838

78

Chloroform

74

-64

254

62

Mercury

11

-39

294

357

Sulphur

54

115

1406

445

Hydrogen

60

-259

449

-253

Oxygen

14

-219

213

-183

Nitrogen

25

-210

199

-196

Table 3.6: Latent heats, freezing points and boiling points of some common substances

Further Discussion on Latent Heat Each time water changes physical state, energy is involved. In the vapour state, the water molecules are very energetic. The molecules are not bonded with each other, but move around as single molecules. Water vapour is invisible to us, but we can feel its effect to some extent, and water vapour in the atmosphere is a very important factor in weather and climate. In the liquid state, the individual molecules have less energy, and some bonds form, break, then re-form. At the surface of liquid water, molecules are continually moving back and forth from the liquid state to the vapour state. At a given temperature, there will be an equilibrium between the number of molecules leaving the liquid, and the number of molecules returning. In solid water-ice-the molecules are locked together in a crystal structure: a framework. They are not moving around, and they contain less energy. How do you make water evaporate? Here is a bowl of water. Make the water evaporate. Go ahead. How did you make the water evaporate? Probably you added heat. You might have put it out in the sun, or possibly put it over a fire. To make water evaporate, you put energy into it. The individual molecules in the water absorb that energy, and get so energetic that they break the hydrogen bonds connecting them to other water molecules. They become molecules of water vapour. Evaporation is the change of state from liquid to vapour. In the process of evaporation, the molecule absorbs energy. This energy is latent heat. Latent means hidden, so latent heat is "hidden" in the water molecule-we can't feel it, but it is there. Wherever that individual molecule of water vapour goes, it takes that latent heat with it. To get the molecule of water vapour to become liquid again, we have to take the energy away, that is, we have to cool it down so that it condenses (condensation is the change from the vapour state to the liquid state). When water condenses, it releases latent heat. Now, how do you make ice melt? Here is a block of ice, water in the solid state. Make it melt. Go ahead. Again, you probably melted the ice by adding energy. The additional energy was absorbed by the individual molecules of water, which became so energetic that they broke some of the hydrogen bonds holding the ice crystal together, and became liquid (that is, the ice melted). This energy is also latent heat, and each molecule of the liquid water is holding that latent heat. To change the liquid water back to ice, you have to take that latent heat away, or in other words, cool the water. Water could change directly from the frozen state to the vapour state without passing through the liquid state first. This process is called sublimation. Water can also change from the vapour state to the frozen state without passing through the liquid state. This is usually called deposition, and is what you see when frost forms on grass or windows on a cold night. (Sometimes the term sublimation is used when water changes state in either direction, that is, from solid to vapour, or vapour to solid).

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Part 66 Training Syllabus Module 2 Physics The really important thing to remember is that each time water changes state, energy is absorbed or released. This energy is latent heat. Latent heat is the energy absorbed or released when a substance changes its physical state. Latent heat is absorbed upon evaporation, and released upon condensation to liquid (as in clouds). Latent heat is also absorbed when water melts, and released when it freezes. How much heat does it take to get water to change state? if the water is at a temperature of 100 degrees C (that is, the boiling point, or 212 degrees F) it takes an additional 540 calories of heat to convert one gram of water from the liquid state to the vapour state. When the vapour converts to the liquid state, 540 calories of energy will be released per gram of water. If you are converting solid water (ice) to liquid water at 0 degrees C, it will require about 80 calories of heat to melt one gram of ice, and the 80 calories will be released when the liquid water is frozen to the solid state.

Figure 3.6: Ice, water and water-vapour Water does not have to be at the boiling point to evaporate, if you don't believe this, set a pan of water out in the sun and watch it slowly disappear. The sun's heat is not boiling the water, but it is evaporating it. in a given amount of water at a given temperature, some molecules of water will have more energy than others, so some molecules will be able to evaporate, while others remain in the liquid state. The lower the temperature of the water, the more energy is required for evaporation, if the water is liquid at a temperature of 0 degrees C, the latent heat of vaporization is 597 cal/g, compared to 540 cal/g at 100 degrees C. in between, at 50 degrees C, an input of 569 cal/g would be required for evaporation. It will take a total of about 720 calories per gram to sublimate water, that is change it directly from ice at 0 degrees C, to vapour at 100 degrees C: this includes 80 calories from latent heat effusion (melting) + 100 calories to raise the temperature of the water 100 degrees C + 540 calories to make the liquid water evaporate (latent heat of vaporization). Similarly, about 720 calories per gram will be released when water is changed directly from vapour to ice, the process called deposition. Methods of Heat Transfer Heat can be transferred from one place to another by one or more of the following processes: Convection is the transfer of heat by the actual movement of the warmed matter. Heat leaves a coffee cup as the currents of steam and air rise. Convection is the transfer of heat energy in a gas or liquid by movement of currents. Think of air and water currents! (it can also happen is some solids, like sand.) The heat moves with the fluid. Consider this: convection is responsible for making macaroni rise and fall in a pot of heated water. The warmer portions of the water are less dense and therefore, they rise. Meanwhile, the cooler portions of the water fall because they are denser. Conduction is the transfer of energy through matter from particle to particle. It is the transfer and distribution of heat energy from atom to atom within a substance. For example, a spoon in a cup of hot soup becomes warmer because the heat from the soup is conducted along the spoon. Conduction is most effective in solidsbut it can happen in fluids. Have you ever noticed that metals tend to feel cold? Believe it or not, they are not colder! They only feel

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Part 66 Training Syllabus Module 2 Physics colder because they conduct heat away from your hand. You perceive the heat that is leaving your hand as cold. convection water. Figure 3.7: Heat - convection Radiation: Electromagnetic waves that directly transport ENERGY through space. Sunlight is a form of radiation that is radiated through space to our planet at the speed of light without the aid of fluids or solids. The energy travels through nothingness! Just think of it! The sun transfers heat through 93 million miles of space. Because there are no solids (like a huge spoon) touching the sun and our planet, conduction is not responsible for bringing heat to Earth. Since there are no fluids (like air and water) in space, convection is not responsible for transferring the heat. Thus, radiation brings heat to our planet. Figure 3.8: Heat – conduction

Heat Transfer We know that heat flows through insulating materials from the warm side to the cool side. It is possible to predict how many Btu will flow through a given insulator in a given amount of time. The equation is:

The equation is less difficult than it seems at first. We will carefully define each symbol. Q = heat flow in Btu t = time in hours A = the surface area of the insulation in square feet ΓT = the temperature difference in°F L = the thickness of theinsulation in inches k = the thermal conductivity of the material from which the insulation is made Table 3.7: Thermal conductivities of some common materials material from which the insulation is made

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EXAMPLE: 2

An outside wall of a house has total cross-sectional area of 2,000 ft. The thickness of the fibreboard insulation is 3 inches. The inside temperature is 70°F and the outside temperature is 20°F. What is the heat loss per hour through this outside wall?

=

(

)

= 14,000 Btu / hr

Problems 1. How much heat must be supplied to raise the temperature of 67 Ibs. of ethyl alcohol from 32°F to 76°F? 2. How much heat is given up as 780 Ibs. of steel cool from 90°F to 45°F? 3. If 1 Ib. of vodka (alcohol) at 90°F is mixed with 0.2 Ib. of water at 40°F what is the final temperature? 4. If 3 Ibs. of hot water at 200 °F are poured into a 1.5 Ibs. aluminium container at 40 °F, what is the final temperature? 2 5. A house has an outside wall area of 3,000 ft These walls are insulated with corkboard 4 in. thick. The inside temperature is 75°F and the outside temperature is 15°F. What is the heat loss per hour through these outside walls?

Answers 1. 2,064 Btu 2. 3,860 Btu 3. 79°F 4. 180°F 5. 13,500 Btu/hr

Refrigeration and Heat Pumps First of all, did you know that there is no such thing as cold? You can describe something as cold and everyone will know what you mean, but cold really only means that something contains less heat than something else. All there really is, is greater and lesser amounts of heat. The definition of refrigeration is The Removal and Relocation of Heat. So if something is to be refrigerated, it is to have heat removed from it. If you have a warm can of pop at say 80 degrees Fahrenheit and you would prefer to drink it at 40 degrees, you could place it in your fridge for a while, heat would somehow be removed from it, and you could eventually enjoy a less warm pop. (oh, all right, a cold pop.) But lets say you placed that 40 degree pop in the freezer for a while and when you removed it, it was at 35 degrees. Even "cold" objects have heat content that can be reduced to a state of "less heat content". The limit to this process would be to remove all heat from an object. This would occur if an object was cooled to Absolute Zero which is -273° C or -460° F. They come close to creating this temperature under laboratory conditions and strange things like electrical superconductivity occur

How do things get colder? Things get cold because they lose heat by one or more of the following methods:  Radiation  Conduction  Convection

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Figure 3.9: Summary of radiation, conduction and convection The latter two are used extensively in the design of refrigeration equipment. If you place two objects together so that they remain touching, and one is hot and one is cold, heat will flow from the hot object into the cold object. This is called conduction. This is an easy concept to grasp and is rather like gravitational potential, where a ball will try to roll down an inclined plane. If you were to fan a hot plate of food it would cool somewhat. Some of the heat from the food would be carried away by the air molecules. When heat is transferred by a substance in the gaseous state the process is called convection. And if you kicked a glowing hot ember away from a bonfire, and you watched it glowing dimmer and dimmer, it is cooling itself by radiating heat away. Note that an object doesn't have to be glowing in order to radiate heat, all things use combinations of these methods to come to equilibrium with their surroundings. So you can see that in order to refrigerate something, we must find a way to expose our object to something that is colder than itself and nature will take over from there. We are getting closer to talking about the actual mechanics of a refrigerating system, but there are some other important concepts to discuss first.

The States of Matter They are of course; solid, liquid and gas. It is important to note that heat must be added to a substance to make it change state from solid to liquid and from liquid to a gas. It is just as important to note that heat must be removed from a substance to make it change state from a gas to a liquid and from a liquid to a solid.

The Magic of Latent Heat Long ago it was found that we needed a way to quantify heat. Something more precise than "less heat" or "more heat" or "a great deal of heat" was required. This was a fairly easy task to accomplish. They took 1 Lb. of water and heated it 1 degree Fahrenheit. The amount of heat that was required to do this was called 1 BTU (British Thermal Unit). The refrigeration industry has long since utilized this definition. You can for example purchase a 6000 BTU/H window air conditioner. This would be a unit that is capable of relocating 6000 BTU's of heat per hour. A larger unit capable of 12,000 BTU/H could also be called a One Ton unit. There are 12,000 BTU's in 1 Ton. o

Figure 3.10: Raising 1lb of by water 1 F To raise the temperature of 1 LB of water from 40 degrees to 41 degrees would take 1 BTU. To raise the temperature of 1 LB of water from 177 degrees to 178 degrees would also take 1 BTU. However, if you tried raising the temperature of water from 212 degrees to 213 degrees you would not be able to do it. Water boils at 212 degrees and would prefer to change into a gas rather than let you get it any hotter. Something of utmost importance occurs at the boiling point of a substance. If you did a little experiment and added 1 BTU of heat at a time to 1 LB of water, you would notice that the water temperature would increase by 1 degree each time. That is until you reached 212 degrees. Then something changes. You would keep adding BTU's, but the water would not get any hotter! it would change state into a gas and it would take 970 BTU's to vaporize that pound of water. This is called the Latent Heat of Vaporization and in the case of water it is 970 BTU's per pound. So what! you say. When are you going to tell me how the refrigeration effect works? Well hang in there, you have just learned about 3/4 of what you need to know to understand the process. What keeps that beaker of water from boiling when it is at room temperature? if you say it's because it is not hot enough, sorry but you are wrong. The only thing that keeps it from boiling is the pressure of the air molecules pressing down on the

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Part 66 Training Syllabus Module 2 Physics surface of the water. When you heat that water to 212 degrees and then continue to add heat, what you are doing is supplying sufficient energy to the water molecules to overcome the pressure of the air and allow them to escape from the liquid state. If you took that beaker of water to outer space where there is no air pressure the water would flash into a vapour. If you took that beaker of water to the top of Mt. Everest where there is much less air pressure, you would find that much less heat would be needed to boil the water, (it would boil at a lower temperature than 212 degrees). So water boils at 212 degrees at normal atmospheric pressure. Lower the pressure and you lower the boiling point. Therefore we should be able to place that beaker of water under a bell jar and have a vacuum pump extract the air from within the bell jar and watch the water come to a boil even at room temperature. This is indeed the case! A liquid requires heat to be added to it in order for it to overcome the air pressure pressing down on its' surface if it is to evaporate into a gas. We just learned that if the pressure above the liquids surface is reduced it will evaporate easier. We could look at it from a slightly different angle and say that when a liquid evaporates it absorbs heat from the surrounding area. So, finding some fluid that evaporates at a handier boiling point than water (i.e. lower) was one of the first steps required for the development of mechanical refrigeration. Chemical Engineers spent years experimenting before they came up with the perfect chemicals for the job. They developed a family of hydroflourocarbon refrigerants which had extremely low boiling points. These chemicals would boil at temperatures below 0 degrees Fahrenheit at atmospheric pressure. So finally, we can begin to describe the mechanical refrigeration process.

Figure 3.11: Main components of a refrigeration system There are 4 main components in a mechanical refrigeration system. Any components beyond these basic 4 are called accessories. The compressor is a vapour compression pump which uses pistons or some other method to compress the refrigerant gas and send it on it's way to the condenser. The condenser is a heat exchanger which removes heat from the hot compressed gas and allows it to condense into a liquid. The liquid refrigerant is then routed to the metering device. This device restricts the flow by forcing the refrigerant to go through a small hole which causes a pressure drop. And what did we say happens to a liquid when the pressure drops? if you said it lowers the boiling point and makes it easier to evaporate, then you are correct. And what happens when a liquid evaporates? Didn't we agree that the liquid will absorb heat from the surrounding area? This is indeed the case and you now know how refrigeration works. This component where the evaporation takes place is called the evaporator. The refrigerant is then routed back to the compressor to complete the cycle. The refrigerant is used over and over again absorbing heat from one area and relocating it to another. Remember the definition of refrigeration? (The removal and relocation of heat.)

So what is a refrigerant? Remember that a refrigerant, in order to cool the space, must evaporate at the temperature of the space. So in the case of an air conditioning unit, we need a chemical which will evaporate (boil) at the temperature of the room you are trying to cool. In the case of a fridge, you need a chemical which will boil at the temperature of the inside of the fridge. There are hundreds of chemicals which will do this. Several of the commonest are as follows:

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Refrigerant Formula

Boiling temperature °C

Ammonia

NH3

R12

C CI2F2 Chlorodi-29.8 fluoromethane

R11

CCI3F

-33

8.9

Properties Penetrating odour, soluble in water. harmless in concentration up to 1/30%, non flammable, explosive Little odour, colourless as gas or liquid, non flammable, non corrosive of ordinary metals, stable Non flammable, non corrosive non toxic, stable Little odour, colourless as gas or liquid, non toxic, non irritating, non flammable, non corrosive, stable

Applications

Large industrial plants

Small plants reciprocating compressors.

Commercial plants with centrifugal compressors. Packaged air conditioning units where size of equipment and economy are important.

R22

CH Cl F2

R500

C CI2 F2 (73.8%) CH3CH F2 -33 (26.2%)

Similar to R12

R502

CCIF2(48.8%) CCI F2- -45.6 CF3(51.2%)

Non flammable, non Capacity toxic, non corrosive, to R22. stable

R134a

F3CCH2F Tetrafluoroethane

-40.8

with

Offers approx. 20% more refrigeration capacity than R12 for same compressor. comparable

Totally replaces other Non flammable, non Freon types for auto -26.6 toxic, non corrosive, and aircraft stable applications Table 3.8: Some common refrigerants

The range of refrigerants beginning with the 'R' prefix (R12, R22 etc.) are complex compounds of 1 fluorocarbons or hydroflourocarbons collectively known as 'Freon . Other chemicals could be used and have been used in the past. Methyl bromide is a refrigerant but has almost completely been phased out for safety and environmental reasons. Carbon dioxide (dry ice) has also been used as a refrigerant (infact at one time Carbon Dioxide and Ammonia were the only two refrigerants in use) but is no longer used because other complex formulas are much more efficient.

So is water a refrigerant? Remember that a refrigerant, in order to cool the space, must evaporate at the temperature of the space. Unless the room you are trying to cool, or the inside of your fridge, is 10OoC or more, water will not work as a refrigerant at normal atmospheric pressure. You would have to reduce the pressure to almost nothing before water will boil at 0°C (0.089 PSI to be exact). This is impractical. However, if you want to cool something which is very hot (i.e. over 100oC) like the rods of a nuclear reactor, then pouring water on it will do the job. In turning to steam (evaporating) the rods are cooled. This is how cooling towers work - water is poured onto the hot object at the base of the cooling tower. The water turns to steam and rises (because water vapour is lighter than air) and expands, which forces it to cool (Charles' Law). The cooled steam condenses into water which runs down the inside of the cooling tower and back onto the hot object where the process repeats. So you have water, as a refrigerant, cooling an object in much the same way as Freon or Ammonia cools the interior of your car, office or fridge.

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Part 66 Training Syllabus Module 2 Physics But also, water can vaporise at temperatures below 100°C, thus missing out the boiling process. Hence water which exudes from the pores of your skin when you are hot (sweat) evaporates, and in changing state absorbs the heat from your skin, and cools you. Again water is a refrigerant. However, to be efficient, it usually helps to have a breeze over the skin to help the evaporation process.

Heat Pumps A heat pump is a machine or device that moves heat from one location (the 'source') to another location (the 'sink’ or 'heat sink’), using work. Most heat pump technology moves heat from a low temperature heat source to a higher temperature heat sink. Common examples are food refrigerators and freezers and air conditioners and reversible-cycle heat pumps for providing thermal comfort. Heat pumps can be thought of as a heat engine which is operating in reverse. One common type of heat pump works by exploiting the physical properties of an evaporating and condensing a refrigerant. In heating, ventilation, and cooling (HVAC) applications, a heat pump normally refers to a vapour-compression refrigeration device that includes a reversing valve and optimized heat exchangers so that the direction of heat flow may be reversed. Most commonly, heat pumps draw heat from the air or from the ground. Airsource heat pumps do not work well when temperatures fall below around -5°C (23°F). According to the second law of thermodynamics heat cannot spontaneously flow from a colder location to a hotter area; work is required to achieve this. Heat pumps differ in how they apply this work to move heat, but they can essentially be thought of as heat engines operating in reverse. A heat engine allows energy to flow 1 from a hot 'source' to a cold heat 'sink , extracting a fraction of it as work in the process. Conversely, a heat pump requires work to move thermal energy from a cold source to a warmer heat sink. Since the heat pump uses a certain amount of work to move the heat, the amount of energy deposited at the hot side is greater than the energy taken from the cold side by an amount equal to the work required. Conversely, for a heat engine, the amount of energy taken from the hot side is greater than the amount of energy deposited in the cold heat sink since some of the heat has been converted to work.

Figure 3.12. A heat pump's vapour-compression refrigeration cycle: 1) condenser, 2) expansion valve, 3) evaporator, 4) compressor. The working fluid, in its gaseous state, is pressurized and circulated through the system by a compressor. On the discharge side of the compressor, the now hot and highly pressurized gas is cooled in a heat exchanger called a condenser until it condenses into a high pressure, moderate temperature liquid. The condensed refrigerant then passes through a pressure-lowering device like an expansion valve, capillary tube, or possibly a work-extracting device such as a turbine. This device then passes the low pressure, barely liquid (saturated vapour) refrigerant to another heat exchanger, the evaporator where the refrigerant evaporates into a gas via heat absorption. The refrigerant then returns to the compressor and the cycle is repeated. In such a system it is essential that the refrigerant reaches a sufficiently high temperature when compressed, since the second law of thermodynamics prevents heat from flowing from a cold fluid to a hot heat sink. Similarly, the fluid must reach a sufficiently low temperature when allowed to expand, or heat cannot flow from the cold region into the fluid. In particular, the pressure difference must be great enough for the fluid to condense at the hot side and still evaporate in the lower pressure region at the cold side. The greater the temperature difference, the greater the required pressure difference, and consequently more energy is needed to compress the fluid. Thus as with all heat pumps, the energy efficiency (amount of heat moved per

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Part 66 Training Syllabus Module 2 Physics unit of input work required) decreases with increasing temperature difference. Thus a ground-source heat pump, which has a very small temperature differential, is relatively efficient. (Figures of 75% and above are quoted.) Due to the variations required in temperatures and pressures, many different refrigerants are available. Refrigerators, air conditioners, and some heating systems are common applications that use this technology. In HVAC applications, a heat pump normally refers to a vapour-compression refrigeration device that includes a reversing valve and optimized heat exchangers so that the direction of heat flow may be reversed. The reversing valve switches the direction of refrigerant through the cycle and therefore the heat pump may deliver either heating or cooling to a building. In the cooler climates the default setting of the reversing valve is heating. The default setting in warmer climates is cooling. Because the two heat exchangers, the condenser and evaporator, must swap functions, they are optimized to perform adequately in both modes. As such, the efficiency of a reversible heat pump is typically slightly less than two separately-optimized machines.

Thermodynamics and the 1st and 2nd Laws We have already seen, the three thermodynamic processes. Here is a brief recap:

Boyle's Law Pressure and volume are inversely proportional to each other, providing temperature remains constant (i.e. if you decrease the volume of a gas, its absolute pressure will increase by the same proportion). This is an ISOTHERMAL process (ISO means 'equal' and THERMAL means 'temperature')

Charles' Law Volume and temperature are directly proportional to each other, providing the pressure remains constant (i.e. if you increase the absolute temperature of an unrestrained volume of gas, its volume will increase by the same proportion). This is an ISOBARIC process (BARIC means 'pressure')

Gay Lussac's Law Pressure and Temperature are directly proportional to each other, providing the volume remains constant (i.e. if you increase the absolute temperature of a confined gas, its absolute pressure will increase by the same proportion). This is an ISOCHORIC process (CHORIC means 'volume' or 'size') Note that all three processes involve absolute pressures and temperatures, those pressures which include atmospheric pressure, and temperatures measured in Kelvin or degrees Rankin.

An Adiabatic Process Another definition to know is that of ADIABATIC. That is: When a change in the volume and pressure of the contents of a system takes place without exchange of heat between the system and its surroundings. Although there is no flow of heat, the system temperature can change as heat energy can be converted to mechanical work and vice versa. It is easy to appreciate how Charles' Law and Gay Lussac's Law are NOT adiabatic processes, since heat clearly must cross the boundary of the system (in both systems heat is added). It makes no difference to the definition of adiabatic if the heat is added in the form of fuel, which is subsequently ignited, or the gas is heated from an external heat source such as a flame. A little more difficult to determine, is whether Boyle's Law is an adiabatic process or not. It is true that no heat is added whilst the gas is compressed, but the natural tendency is for the gas to rise in temperature as it is compressed (as Charles law tells us). The key then, to performing a compression under Boyles's Law is

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Part 66 Training Syllabus Module 2 Physics to compress the gas so slowly, that any heat which builds up leaks away before it has a chance to show itself in the form of a temperature rise. The fact that heat is leaking away (through the casing of the system) precludes it from being an adiabatic process. The definition of adiabatic does not differentiate between heat being added (as in Charles' and Gay Lussac's Laws), conventionally considered positive, and heat removed (as in Boyle's Law), conventionally considered negative. One example of a process which is adiabatic, is that of an air compressor. Inside the cylinder of an air compressor, as the volume is rabidly reduced, the air is compressed so quickly, that any rise in temperature does not result in any loss of heat through the cylinder walls - the air being removed from the cylinder before the heat has a chance to escape. However, if you put your hands on the air compressor that supplies the airlines around your hangar, you will almost certainly find it to be hot. The conclusion is then that an adiabatic process exists mainly in theoretical terms only, and is very difficult to achieve in practice.

Thermodynamic Work Most thermodynamic processes are carried out in order to do work - the combustion of fuel inside your car engine, or inside an aero engine for example. But as we saw previously, to do work, there must be some movement, and a force in the same direction as that movement. Consider a piston inside a cylinder as shown in Figure 3.13. After both valves are closed, the gas is ignited and it expands. This expansion pushes the piston down the cylinder, with a considerable force. Hence, work is done and can be calculated by Work = Force x Distance But the force on the piston is given by Force = Pressure x Area Where the Pressure is the pressure of the gas in the cylinder, and the Area is the piston face area. Figure 3.13: An internal combustion engine converts chemical energy into heat energy and then into mechanical work Hence our Work formula is now Work = Pressure x Area x Distance

where 'distance' is the movement of the piston

If we simplify the diagram as shown in Figure 3.13, you will see that the distance that the piston moves, multiplied by the Area of the piston, equates to the increase in volume of the gas as it burns.

Figure 3.14: A simplified heat engine Our formula for work then becomes: Work = Pressure x Change in Volume i.e.

Work = P ΔV

Where 'Γ' (delta) indicates a 'change in' the quantity V

This is the standard formula for Work done in a thermodynamic process (it is only applicable to processes where the volume changes, if the volume does not change, no work is done). It applies regardless of the direction of work, so if an external force is applied to the piston, the volume of gas in the cylinder will reduce. The work required to reduce the volume is given by the same formula. Thermodynamicists often simplify the formula to

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Work = PV It makes certain assumptions, for example, that the pressure remains constant as the gas expands. In a piston engine, this is not the case, as the pressure rises to a peak as the gas is burned, then reduces as the piston is displaced. It is however, a fair assumption to make, in the case of a gas turbine engine, as the gas in the combustion chamber is burned, and expands at constant pressure. Even in the case of the piston engine however, the above formula can still be used in conjunction with some complex calculus (which is not part of the EASA Part-66 syllabus, and way beyond the scope of these notes).

Internal Energy Whenever a gas increases in temperature, the molecules of the gas vibrate and move around more vigorously. It is this increased kinetic energy that actually produces the increased pressure as they collide with each other and with the walls of the container in which it is confined. This kinetic energy is directly proportional to the absolute temperature in the gas. Thermodynamicists call the sum of this kinetic and potential energies in the gas, "Internal Energy", symbol U, and is measured in Joules. When the temperature of a gas is increased, it will increase in volume (and provide work) or increase in internal kinetic and potential (due to a rise in pressure) energies - that is, increase Internal Energy, or more likely, a combination of both volume and internal energy changes.

Enthalpy The combination of internal energy, and pressure and volume are the most likely things to change whenever a gas is heated. Since this happens so often in thermodynamics, it is all grouped together and given its own name and unit. It is called "Enthalpy" and given a symbol H. It is simply the sum of the internal energy plus the pressure volume product. Thus

Enthalpy

H

=

U

+

PV

Remember that PV is the work done when a gas expands when it is heated. Infact, Enthalpy is the amount of energy in a gas which is capable of doing work.

The First Law of Thermodynamics If you have understood this chapter so far, you will be pleased to know that we have already covered the First Law of Thermodynamics. It is best thought of by thinking of the air/fuel mixture that enters your car engine cylinder. It enters at about 15°C and standard atmospheric pressure. When it is ignited to release all its internal chemical energy as heat energy, it will expand, push the piston with a certain pressure, and provide work. The chemical energy of the fuel (before it is ignited) is measured in J/kg, of the order of around 10 MJ/kg. You would be forgiven for thinking that all the heat energy (symbol Q) released on burning can be transformed into work (Symbol W), thus: Heat energy added to system (Q) = Work done by system (W) However, now imagine that you walk around the car and put your hand over the exhaust. Two things you will notice; one is that it is considerably hotter than it went into the engine, and two; that it is considerably higher pressure than when it went into the engine. This is because not all of the heat energy added to the gas can be turned into useful work, some of the heat energy added to the gas will leave via the exhaust in the form of Internal Energy (symbol U). Our formula is now: Heat energy added to the system (Q) = Work done by the system (W) plus a change in Internal Energy (U) Or in its simplest terms Q = W+U st

This is the mathematical version of the 1 Law of Thermodynamics.

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Part 66 Training Syllabus Module 2 Physics It is typical for chemistry texts to write the first law as ΓU = Q - W. It is the same law, of course -the thermodynamic expression of the conservation of energy principle. It is just that W is defined as the work done on the system instead of work done by the system. In the context of physics, the common scenario is one of adding heat to a volume of gas and using the expansion of that gas to do work, as in the pushing down of a piston in an internal combustion engine. In the context of chemical reactions and process, it may be more common to deal with situations where work is done on the system rather than by it. Engine designers of course try to minimise the internal energy (U) rise which is wasted through the exhaust, and maximise the Work output, thus making most efficient use of the chemical energy locked away in the nd petrol. But as we will see in the 2 Law of Thermodynamics, an engine of 100% thermal efficiency (where all the heat energy is transformed into work) is not possible, theoretically as well as practically. st

During the mid-eighteenth century, when the 1 law of thermodynamics was discovered, it was thought possible that an engine could be produced which ran off the heat energy (internal energy) in the exhaust of another engine. Indeed, it seemed possible theoretically to run an engine off its own exhaust. This led to much research into such a machine, which would have been the invention of perpetual motion. This same research led only to the conclusion that such a machine, and with it, perpetual motion, was impossible, and a new law which described this impossibility, the Second Law of Thermodynamics...

The Second Law of Thermodynamics The First Law of Thermodynamics is really a prelude to the second. It states that the total energy output (as that produced by a machine) is equal to the amount of heat supplied. Generally, energy can neither be created nor destroyed, so the sum of mass and energy is always conserved. Physicists attempting to transform heat into work with full efficacy quickly learned that always some heat would escape into the surrounding environment, eternally doomed to be wasted energy (recall that energy can not be destroyed). Being obsolete, this energy can never be converted into anything useful again. One physicist noted for significant experiments in this field is the Frenchman, Sadi Carnot. His ideal engine, so properly titled the 'Carnot Engine,' would theoretically have a work output equal to that of its heat input (thus not losing any energy in the process). However, he fell into a similar trap as in the first law, and failed to conduct his experiments as would naturally occur. Realizing his error, he concluded (after further experimentation) that no device could completely make the desired conversion, without losing at least some energy to the environment. Carnot created an equation he employed to prove this statement, and which is currently used to show the thermodynamic efficiency of a heat machine: efficiency = 1 (the efficiency of a heat machine is equal to one minus the low operating temperature of the machine in degrees Kelvin, divided by the high operating temperate of the machine in degrees Kelvin). For a machine to attain 100% efficiency, temperatures of absolute zero would have to be incorporated. Reaching absolute zero is later proved impossible by the Third Law of Thermodynamics (which would surface in the late 19th century). The irrevocable loss of some energy to the environment was associated with an increase of disorder in that system. Scientists wishing to further penetrate the realm of chaos needed a variable that could be used to calculate disorder. Thanks to mid-nineteenth century physicist, R.J.E. Clausius, this Pandemonium could be 1 measured in terms of a quantity named entropy (the variable S). Entropy acts as a function of the state of a system - where a high amount of entropy translates to higher chaos within the system, and low entropy signals a highly ordered state. Like Carnot, Clausius worked out a general equation, his being devoted to the measurement of entropy change over a period of time: (change)S = Q / T (the change in entropy is equal to the amount of heat added to the system [by an invertible process] divided by the temperature in degrees Kelvin). The beauty of this 1

Specifically is a measure of randomness or disorder in a system. Darnell Ebbing, in the textbook General Chemistry, very usefully suggests thinking of a deck of cards. A new pack fresh out of the box, arranged by suit and in sequence from ace to king, can be said to be in its ordered state. Shuffle the cards and you put them into a disordered state. Entropy is a way of measuring just how disordered that state is and of determining the likelihood of particular outcomes with further shuffles. Of course, if you wish to have any observations published in a respectable journal you will need also to understand additional concepts such as thermal non-uniformities, lattice distances, and stoichometric relationships, but that's the general idea.

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Part 66 Training Syllabus Module 2 Physics equation is that it can be used to compute the entropic change of any exchange in nature, not solely limited to machines. This development brought thermodynamics out of the industrial workplace, and opened the possibility for further studies into the tendencies of natural order (and lack therefore of), eventually extending to the universe as a whole. Applying this knowledge to nature, physicists found that the total entropy change (change in S) always increases for every naturally occurring event (within a closed system) that could be then observed. Thus, they theorized, disorder must be continually augmenting evenly throughout the universe. When you put ice into a hot cup of tea (aristocrats of the Victorian era were constantly thinking of tea), heat will flow from the hot tea to the cold ice and melt the ice in the beloved beverage. Then, once the energy in the cup is evenly distributed, the cooled tea would reach a maximum state of entropy. This situation represents a standard increase in disorder, believed to be perpetually occurring throughout the entire universe. The second law of thermodynamics (also known as 'the entropy law' or 'law of entropy') was formulated in the middle of the nineteenth century by Clausius and Thomson following Carnot's earlier observation that, like the fall or flow of a stream that turns a mill wheel, it is the "fall" or flow of heat from higher to lower temperatures that motivates a steam engine. The key insight was that the world is inherently active, and that whenever an energy distribution is out of equilibrium a potential or thermodynamic "force" (the gradient of a potential) exists that the world acts spontaneously to dissipate or minimize. All real-world change or dynamics is seen to follow, or be motivated, by this law. So whereas the first law expresses that which remains the same, or is time-symmetric, in all real-world processes the second law expresses that which changes and motivates the change, the fundamental time-asymmetry, in all real-world processes. Clausius coined the term "entropy" to refer to the dissipated potential and the second law, in its most general form, states that the world acts spontaneously to minimize potentials (or equivalently maximize entropy), and with this, active end-directedness or time-asymmetry was, for the first time, given a universal physical basis. The balance equation of the second law, expressed as S > 0, says that in all natural processes the entropy of the world always increases, and thus whereas with the first law there is no time, and the past, present, and future are indistinguishable, the second law, with its one-way flow, introduces the basis for telling the difference. The active nature of the second law is intuitively easy to grasp and empirically demonstrate. If a glass of hot liquid, for example, as shown in Figure 16-3, is placed in a colder room a potential exists and a flow of heat is spontaneously produced from the cup to the room until it is minimized (or the entropy is maximized) at which point the temperatures are the same and all flows stop. The glass of liquid at temperature T’ is placed in a room at temperature T’’ such that the disequilibrium produces a field potential that results in a flow of energy in the form of heat from the glass to the room so as to drain the potential until it is minimized (the entropy is maximized) at which time thermodynamic equilibrium is reached and all flows stop. This refers to the conservation of energy in that the flow from the glass equals the flow of heat into the room. Figure 3.15: Heat transfer from a hot space to a less-hot space The two laws have far reaching ramifications in physics and the real world. One such is that no heat engine can ever be 100% efficient, unless the exhaust temperature (known as the 'sink' temperature) is absolute zero. Absolute zero is both theoretically and practically impossible to achieve. Another ramification is the fact that no thermodynamics process can be reversed, precisely due to the first law which states that not all the heat energy supplied can be turned into work. The closest one can come to a fully reversible process is forcing a flow through a constricted pipe. Ideal means no boundary layer losses. As the flow moves through the constriction, the pressure, temperature and velocity change, but these variables return to their original values downstream of the constriction. The state of the gas returns to its original conditions and the change of entropy of the system is zero. Aerodynamicists call such a process an isentropic process. Isentropic means constant entropy. The second law states that if the physical process is irreversible, the combined entropy of the system and the environment must increase. The final entropy must be greater than the initial entropy for an irreversible process: Summary

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Part 66 Training Syllabus Module 2 Physics Thermodynamics is the study of the inter-relation between heat, work and internal energy of a system. The British scientist and author C.P. Snow had an excellent way of remembering the three laws (including rd the 3 law): 1. You cannot win (that is, you cannot get something for nothing, because matter and energy are conserved). 2. You cannot break even (you cannot return to the same energy state, because there is always an increase in disorder; entropy always increases). 3. You cannot get out of the game (because absolute zero is unattainable).

Module 2.4 Optics (Light) Wavelength, Frequency and Speed Light is a form of electromagnetic radiation. There is a certain band of frequency of electromagnetic radiation that affects the retina of the human eye. We call this band of radiation "visible light". Sometimes the word "light" means only visible light and sometimes the word "light" is used as a generic word to mean any kind of electromagnetic radiation. Electromagnetic radiation is a type of wave. As in the case of all wave motion, the wave moves with a definite speed (c) called the speed of light. The speed of light has been measured many 8 times and has the value, to three significant digits, 3.00 x 10 m/sec. The wavelength of visible light is usually measured in a unit called the Angstrom (A). -10 1A=10 m Various colours of visible light have characteristic wavelengths. Table 4.1 is a list of some colours and their approximate wavelengths. Wavelengths of electromagnetic radiation shorter than 4,000 A are not visible and are called "ultraviolet" and wavelengths longer than 7,000 A are also not visible and are called "infrared". We also note that "colours" such as "blue-green" also exist. The wavelength would be about 5,000 A. Colours gradually change as the wavelength changes. Table 4.1: The wavelengths of various colours of light measured in A As in the case of all wave motion, the speed of electromagnetic radiation equals the frequency times the wavelength. Therefore, for light, we have the relation: c=f λ In table 4.2. we list some common types of electromagnetic radiation with ranges of frequency and wavelength. Note that the above equation is always satisfied. As the frequency increases, the wavelength 8 decreases in such a way that the product equals the speed of light (3 x 10 m/sec.).

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Table 4.2: Radiation types - their frequencies and wavelengths EM radiation consists of two different waves; an electrical field (E field) and a magnetic field (B field) moving at 90 degrees to each other (Figure 4.1). We see that both the electric field lines and magnetic field lines vary sinusoidally. The electric field lines lie in a plane that is perpendicular to the plane of the magnetic field lines. All light radiation, or electromagnetic (EM) radiation, consists of these patterns of electric and magnetic field lines moving in free space (vacuum) with speed (c) or in some other transparent medium. We note that the frequencies and wavelengths of the various types of EM radiation vary greatly. EXAMPLE: 18 (a) The frequency of an x-ray is 5 x 10 Hz. What is the wavelength of this radiation? c=f λ 10

λ= =

= 0.6x10 m 11

= 6x10 m

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(c)

Figure 4.1: The Electrical (E) and magnetic (B) components of the electromagnetic wave An FM radio wave has a wavelength of 6 m. What is the frequency in megahertz? f= = 6

f = 50 x 10 Hz = 50 megahertz

The Speed of Light in Various Substances

8

We have already stated that the speed of light in vacuum (free space) is, to three significant digits, 3.00 x 10 m/sec. The speed of light is less in various transparent substances. We define the "index of refraction" (n) as the ratio of the speed of light in vacuum to the speed of light in the substance (v). N=

=

From this equation, we see that we can find the speed (v) in various transparent substances by using the relation:

V= EXAMPLE: Find the speed of light in water. v= 8 v = 2.25 x 10 m/sec

Table 4.3: Indices of refraction for various substances Problems 1. What is the wavelength in meters of an FM radio wave having a frequency of 90MHz? 10 2. What is the frequency of an x-ray having a wavelength of 4 x 10 m? 3. What is the frequency in kilohertz of an AM radio transmission if the wavelength is 500m? (1 kHz =1,000 Hz) 4. Find the speed of light in crown glass.

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Part 66 Training Syllabus Module 2 Physics 5. 6. Answers 1. 2. 3. 4. 5. 6.

8

The speed of light in carbon tetrachloride is measured to be 2.05 x 10 m/sec. What is the index of refraction of carbon tetrachloride? Find the speed of light in a diamond.

3.33m 17 7.5x10 Hz 600 kilohertz 8 2x10 m/sec. 1.46 8 1.24x 10 m/sec

Light Waves in Matter The speed of light waves is a maximum in a vacuum and less in materials which are 'transparent' to the waves. As a general rule, electromagnetic waves cannot travel at all through 'opaque' materials containing free electrons (e.g. metals) as the waves lose so much energy to the electrons. Bound charged particles (including electrons) may also absorb energy from the waves, but do so at definite frequencies or bands of frequencies that depend on the atoms or molecules of the medium.

Refraction and the speed of light waves We have noted that light waves travel more slowly in a transparent medium than in a vacuum. The diagram below shows what happens as waves enter a transparent medium in which their speed is cm. Their frequency stays the same but their wavelength gets less such that: Cm = f λ m This change in speed also causes the refraction effect — the wave-fronts change direction when they enter or leave the surface of the material at other than 90° (at an angle to the normal). The diagram shows a set of parallel wavefronts of single frequency radiation entering a transparent medium. As the leading edge enters the medium, the wave slows down but the 'outside' section of the front does not, so it catches up on the inside section. Inside the medium the distance between successive fronts is smaller and the direction of travel of the wave has changed. Snell's law of refraction follows directly from this effect.

Figure 4.2: Refraction of light as it travels from a material of low to a high refractive index

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Refractive Index Varies with Wavelengthe speed of light in a given transparent medium is also likely to vary with frequency - the refractive index is different for different frequencies. The diagram below shows how the refractive index of fused quartz and crown glass varies with the vacuum wavelength of radiation between short ultraviolet wavelengths ( ~200nm) and near infrared ( ~750nm). Fused quartz is widely used in optical devices as it is transparent over a wide range of wavelengths.

Figure 4.3: Refractive index varies with wavelength Note that glass has a higher refractive index for light of shorter wavelengths (higher frequencies) than for longer wavelengths: light of shorter wavelength is refracted more. This is why prisms produce a spectrum from white light, with blue light deviated more than red light.

Dispersion and Chromatic Aberration Dispersion is a serious problem that the makers of optical instruments with lenses have to solve. Dispersion means that red light is brought to a focus further away from a positive lens than blue light is. This blurs images, an effect called chromatic aberration. Newton solved the problem for telescopes by designing one in which the light was focused by a curved mirror. An achromatic lens can be made - a combined double-lens using two different types of glass (e.g. crown and flint glass). One lens is positive and stronger than the other, negative, lens. The overall combination is positive, but the negative lens is made from a more dispersive type of glass so that the total dispersion of the combination can be made very small.

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Figure 4.4: Dispersion and chromatic aberration

Lenses The Converging Lens How a Converging Lens Forms an image We are familiar with convex glass lenses. They have equal surfaces that we can think of as parts of a sphere, and are called spherical (or simple) lenses. Figure 4.5 shows what happens to the wave-fronts of light waves when they pass through a simple glass lens. A lens like this, which alters a plane wavefront to make the light waves pass through a point, or focus, is called a converging or positive lens. in the diagrams you can see that the wavelengths of the light are shorter after they pass across the air— glass boundary. This is because the speed of the waves is less in glass than in air, and the waves don't go as far in each period of the wave motion. This drop in speed causes refraction, meaning that the light changes direction wherever the wavefront is not parallel to the boundary. Wave diagrams like those in the diagram are hard to draw, and they also hide some of the features of the light paths. Here, it makes sense to revert to an old but very useful model of light which assumes that light travels in straight lines. Where this model breaks down (as it will!), we shall use the more advanced wave model. The following diagrams, (a) and (b) show just some of the light rays we could draw. The rays show the direction of the waves, which means that the rays are at right angles to the wavefront. When a ray hits the glass surface, it is refracted as shown. Notice in (b) that the light ray hitting the centre of the lens does so at 90° and that light rays increasingly far from the centre of the lens hit the surface at smaller and smaller angles. We will return to this soon.

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Figure 4.5: Parallel and point-source light rays being 'focussed' by a lens Figure 4.6 shows light passing through a plane surface, such as the surface of a glass block. The line at right angles to the boundary is the normal, and we can see that the angle i made with the normal by the incident ray is greater than the angler of the refracted ray. At the same time, as i increases, so r increases, while remaining less than i see (b).

Figure 4.6: Incident (i) and Refracted (r) light angles The light rays are obeying Snell's law, the law of refraction: =

=n

The constant n is the refractive index (air to glass for glass lenses).

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Part 66 Training Syllabus Module 2 Physics Figure 4.7 which shows rays from a point-source passing through a lens. The radius lines of the curved surface are shown extended beyond the surface as normal lines. We can see again that the law of refraction applies to the rays as they cross the air—glass boundary. The spherical geometry of the simple lens makes the rays converge (but not accurately) to one point or focus. For parallel rays (that form a plane wavefront), this point is called the principal focus. The distance of this point from the centre of the lens is called the focal length of the lens, CF in figure 4.8. A lens has two principal focuses, one on each side of the lens, at equal distances from it, F and F. normals Figure 4.7: Point-source light passing through a lens In practice, the focus for simple spherical lenses is only at a point if the diameter of the lens is very small compared with the radius of curvature of its surfaces. Otherwise, the lens forms a partly blurred image. A lens defect like this, caused either by the shape or the material of the lens, is called an aberration. For a lens with spherical geometry, it is called spherical aberration. Removing aberrations is technically difficult, which explains why good optical instruments, such as camera lenses, are expensive.

Figure 4.8: Object and image

Predicting the image The diagram above shows how a simple lens forms an image of an object. We can predict the position and size of an image either by drawing or by using formulae.

Drawing to find the image We can draw any number of rays to help us find the position and size of an image, but the three rays that are most helpful to draw are shown above, with the labels we use when drawing ray diagrams. Figure 4.8 shows an upright object OX close to a lens. In ray diagrams, the rays are assumed to change direction at a line that represents a plane in the centre of the lens. The object could be anything, but by convention it is drawn simply as an arrow. Any ray that is parallel to the principal axis of the lens is refracted to pass through the principal focus, F. In this way, we can predict the direction of the ray labeled 1. We use the same idea for ray 2. Going through F', it emerges from the lens parallel to the principal axis. The third useful line, for ray 3, goes straight through the centre of the lens — it does not deviate. But see the close-up drawing. At the centre of the lens, the two faces are parallel to each other. If the lens is thin compared with the distances of object and image, the slight sideways displacement of the ray is not significant. All three rays pass through the same point, Y They all started at X, and it is clear that a screen placed at Y would catch them all together again: Y is a focused image of X. Similarly, any point on OX would give a focused image of itself somewhere between i and Y. IY is a real image.

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Part 66 Training Syllabus Module 2 Physics These construction rays can be drawn to scale to find the position and relative size of the image of any object placed in front of the lens. Objects viewed through a lens, and their images, are usually much smaller than the distances they are from the lens. In such cases, it is best to make the vertical scale larger than the horizontal one.

Figure 4.9: Virtual and real images, depending on whether the object is in-front of, or behind, the focal point Note that the image may be inverted, or not, real or virtual, depending on whether the object is placed behind or in-front of the focal point.

Finding the image by formula It is usually quicker and more accurate to find the position and size of an image by using the lens formula: + = where u is the distance of the object from the lens centre, the object distance; v is the distance of the image from the lens centre, the image distance; and f is the focal length of the lens. The following diagram gives the geometry required to prove this formula for a thin converging lens. You can now find the position and size of any image simply by inserting the other given values in the formula.

Figure 4.10: components of the lens formula

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The Diverging Lens Figure 4.11 to the right shows a lens with concave spherical surfaces and what happens to parallel light rays (that is, a plane wavefront coming from a distant object) when they pass through it. The rays are refracted so that they seem to diverge from a single point, F. This point is the principal focus of a diverging lens. As light doesn't actually come from the point, or pass through it, it is a virtual.

Figure 4.11: Parallel light entering a divergent lens Figure 4.12 shows how a diverging lens forms an image of an object. As for a converging lens, you would see the image by looking at the object through the lens. But light doesn't actually pass back through the diverging lens to the image, so you cannot catch the image on a screen. It is therefore a virtual image.

Figure 4.12: Convergent lens forming a Virtual images

The sign convention The lens formula works for all simple optical devices. But we have to know whether the images, principal focuses - and even objects - are real or virtual. Where they are virtual, the convention is to give negative values to distances measured from them to the lens or mirror. For example, the principal focus of a diverging lens is virtual, so its focal length is given a negative sign. Suppose we place a real object 20 cm from the diverging lens as shown in the figure 4.12. The principal focus of the lens is 10 cm from the lens, so its focal length is -10 cm. The lens formula gives: + = =-

–

=

v= -

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Part 66 Training Syllabus Module 2 Physics So the distance of the image from the lens centre is 6.7 cm, and the negative sign tells us that the image is also virtual.

The Power of a Lens In optics, lenses are usually described in terms of their power. The more powerful the lens, the closer to the lens is the image that the lens forms of a distant object. The power of a lens is defined as the reciprocal of its focal length measured in metres:

Lens Power = The unit of power is called the dioptre, symbol D, and so a lens of focal length +10 cm (0.1 m) has a power of +10 D. A diverging lens of focal length -5 cm (0.05 m) has a power of -20 D. This way of describing lenses lets us work out what happens when two lenses are used together. The combined power of the lenses is simply the sum of the power of each lens, bearing in mind their signs, as shown in figure 4.13.

Figure 4.13: Adding and subtracting lens powers

Mirrors Plane mirrors are the most common optical devices we come across. The ordinary household mirror is a sheet of flat glass, 'silvered' on the back with a layer of metal paint (the metal is usually aluminium), which is then protected by a coat of ordinary paint. Light passes through the glass and is reflected by the silvering. A plane wavefront is reflected as a plane by a flat mirror, and the geometry results in the well-known rule:

The angle of incidence equals the angle of reflection. The angles are shown in the diagram to the right, and are measured from the normal line drawn at right angles to the surface. As shown in the figure 4.14, the image of an object reflected in a plane mirror is behind the mirror and as far from the mirror as the object is.

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Figure 4.14: Source and image in a mirror Figure 4.15: Incident angle (i) and reflected angle (r) Notice that light does not pass through the mirror, and the image is really an optical illusion. It isn't really 'there' — you couldn't, for example, catch it on a photographic film placed at the image position. The image is therefore a virtual image. Compare this image with the one produced by the convex lens which is formed from rays of light and can be caught on a photographic film. Our brain perceives ('sees') virtual images by making them into 'real' ones using another optical device — our eyes. Figure 4.16: Mirrors produce a 'virtual' image

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An ordinary glass mirror produces a blurred image because of multiple reflections — see figure 4.17. Such a not suitable for optical instruments which need to give images. There are two main solutions to this problem.

mirror is sharp

Figure 4.17: Reason for the 'blurred' image of a mirror

Front-silvered mirrors These have the metallic layer on the front surface. Very fine metal particles are deposited on glass either from solution or from metal vapour in a vacuum. The large mirrors used in reflecting telescopes are made in this way. The metal surface is easily damaged, by touching or corrosion, and has to be treated with care.

Reflecting Prisms A glass (or clear plastic) reflecting prism provides a much cheaper and more practical plane mirror by using the effect of total internal reflection. Figure 4.17 shows light rays which are partly internally reflected. Figure 4.18 shows how total internal reflection occurs when the angle of incidence of a ray exceeds the critical angle c for the medium. When a ray reaches the inside of a plane boundary (say, between air and glass) at an angle of incidence greater than the critical angle, it cannot pass through the surface.

Figure 4.18: Stages of Total Internal Reflection Total internal reflection occurs at angles of (internal) incidence greater than c, the critical angle. By Snell's taw: = n or n =

(since sin 90 = 1)

A prism is often used as a mirror. Angle a must be greater than the critical angle c of the material used:

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Figure 4.19: A reflecting prism

Concave Mirrors Spherical mirrors can be thought of as a portion of a sphere which was sliced away and then silvered on one of the sides to form a reflecting surface. Concave mirrors were silvered on the inside of the sphere and convex mirrors were silvered on the outside of the sphere. If a concave mirror is thought of as beinga slice of a sphere, then there would be a line passing through the centre of the sphere and attaching to the mirror in the exact centre of the mirror. This line is known as the principal axis. The point in the centre of sphere from which the mirror was sliced is known as the centre of curvature and is denoted by the letter C in the diagram below. The point on the mirror's surface where the principal axis meets the mirror is known as the vertex and is denoted by the letter A in the diagram below. The vertex is the geometric centre of the mirror. Midway between the vertex and the centre of curvature is a point known as the focal point; the focal point is denoted by the letter F in the diagram below. The distance from the vertex to the centre of curvature is known as the Figure 4.20: Diamensions of a concave mirror radius of curvature (abbreviated by "R"). The radius of curvature is the radius of the sphere from which the mirror was cut. Finally, the distance from the mirror to the focal point is known as the focal length (abbreviated by "f"). Since the focal point is the midpoint of the line segment adjoining the vertex and the centre of curvature, the focal length would be one-half the radius of curvature. The focal point is the point in space at which light incident towards the mirror and travelling parallel to the principal axis will meet after reflection. The diagram at the right depicts this collected by a principle. In fact, if some light from the Sun was concave mirror, then it would converge at the focal point. Because the Sun is such a large distance from the Earth, any light rays from the Sun which strike the mirror will essentially be travelling parallel to the principal axis. As such, this light should reflect through the focal point. Perhaps you remember the outdoors demonstration in which a pencil was engulfed in flames in a matter of seconds when placed at the focal point of the demonstration mirror. In the demonstration, whatever light from the Sun which hit the mirror was focused at the point where the pencil was. To the surprise of many, the heat was sufficient to ignite the pencil.

Figure 4.21: Light rays pass through the focal point

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Figure 4.22: The two images of a concave mirror depend on where the object is placed Note that the image may be inverted or not, depending on whether the object is placed outside or inside the focal point. They are virtual images in both cases.

Convex Mirrors Convex mirrors are used to shrink the image, and therefore fit more detail into the mirror (such as car rearview mirrors). The focal point is at the half radius point (but behind the mirror. The image is virtual.

Figure 4.23: Line diagram for a convex mirror

Figure 4.24: The virtual image of a convex mirror

Fibre Optics Definition of Fibre Optics Fibre optics uses light to send information (data). More formally, fibre optics is the branch of optical technology concerned with the transmission of radiant power (light energy) through fibres.

Fibre Optic Data Links A fibre optic data link sends input data through fibre optic components and provides this data as output information. It has the following three basic functions:  To convert an electrical input signal to an optical signal  To send the optical signal over an optical fibre

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Part 66 Training Syllabus Module 2 Physics  To convert the optical signal back to an electrical signal A fibre optic data link consists of three parts - transmitter, optical fibre, and receiver. Figure 4.25 is an illustration of a fibre optic data-link connection. The transmitter, optical fibre, and receiver perform the basic functions of the fibre optic data link. Each part of the data link is responsible for the successful transfer of the data signal. A fibre optic data link needs a transmitter that can effectively convert an electrical input signal to an optical signal and launch the data-containing light down the optical fibre. A fibre optic data link also needs a receiver that can effectively transform this optical signal back into its original form. This means that the electrical signal provided as data output should exactly match the electrical signal provided as data input.

Figure 4.25: Parts of a fibre optic data link. The transmitter converts the input signal to an optical signal suitable for transmission. The transmitter consists of two parts, an interface circuit and a source drive circuit. The transmitter's drive circuit converts the electrical signals to an optical signal. It does this by varying the current flow through the light source. The two types of optical sources are light-emitting diodes (LEDs) and laser diodes. The optical source launches the optical signal into the fibre. The optical signal will become progressively weakened and distorted because of scattering, absorption, and dispersion mechanisms in the fibre waveguides. The receiver converts the optical signal exiting the fibre back into an electrical signal. The receiver consists of two parts, the optical detector and the signal-conditioning circuits. An optical detector detects the optical signal. The signal-conditioning circuit conditions the detector output so that the receiver output matches the original input to the transmitter. The receiver should amplify and process the optical signal without introducing noise or signal distortion. Noise is any disturbance that obscures or reduces the quality of the signal. Noise effects and limitations of the signal-conditioning circuits cause the distortion of the receiver's electrical output signal. An optical detector can be either a semiconductor positive-intrinsic-negative (PIN) diode or an avalanche photodiode (APD). A fibre optic data link also includes passive components other than an optical fibre. Figure 4.25 does not show the optical connections used to complete the construction of the fibre optic data link. Passive components used to make fibre connections affect the performance of the data link. These components can also prevent the link from operating. Fibre optic components used to make the optical connections include optical splices, connectors, and couplers. Proof of link performance is an integral part of the design, fabrication, and installation of any fibre optic system. Various measurement techniques are used to test individual parts of a data link. Each data link part is tested to be sure the link is operating properly.

History of Fibre Optic Technology People have used light to transmit information for hundreds of years. However, it was not until the 1960s, with the invention of the laser, that widespread interest in optical (light) systems for data communications began. The invention of the laser prompted researchers to study the potential of fibre optics for data communications, sensing, and other applications. Laser systems could send a much larger amount of data than telephone, microwave, and other electrical systems. The first experiment with the laser involved letting the laser beam transmit freely through the air. Researchers also conducted experiments letting the laser beam transmit through different types of waveguides. Glass fibres, gas-filled pipes, and tubes with focusing lenses are examples of optical waveguides. Glass fibres soon became the preferred medium for fibre optic research.

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Part 66 Training Syllabus Module 2 Physics Initially, the very large losses in the optical fibres prevented coaxial cables from being replaced. "Loss" is the decrease in the amount of light reaching the end of the fibre. Early fibres had losses around 1,000 dB/km making them impractical for communications use. In 1969, several scientists concluded that impurities in the fibre material caused the signal loss in optical fibres. The basic fibre material did not prevent the light signal from reaching the end of the fibre. These researchers believed it was possible to reduce the losses in optical fibres by removing the impurities. By removing the impurities, construction of low-loss optical fibres was possible. There are two basic types of optical fibres, multimode fibres and single mode fibres. In 1970, Corning Glass Works made a multimode fibre with losses under 20 dB/km. This same company, in 1972, made a high silica-core multimode optical fibre with 4dB/km minimum attenuation (loss). Currently, multimode fibres can have losses as low as 0.5 dB/km at wavelengths around 1300 nm. Single mode fibres are available with losses lower than 0.25 dB/km at wavelengths around 1500 nm. Developments in semiconductor technology, which provided the necessary light sources and detectors, furthered the development of fibre optics. Conventional light sources, such as lamps or lasers, were not easily used in fibre optic systems. These light sources tended to be too large and required lens systems to launch light into the fibre. In 1971, Bell Laboratories developed a small area light-emitting diode (LED). This light source was suitable for low-loss coupling to optical fibres. Researchers could then perform source-tofibre jointing easily and repeatedly. Early semiconductor sources had operating lifetimes of only a few hours. However, by 1973, projected lifetimes of lasers advanced from a few hours to greater than 1,000 hours. By 1977, projected lifetimes of lasers advanced to greater than 7,000 hours. By 1979, these devices were available with projected lifetimes of more than 100,000 hours. In addition, researchers also continued to develop new fibre optic parts. The types of new parts developed included low-loss fibres and fibre cables, splices, and connectors. These parts permitted demonstration and research on complete fibre optic systems. Advances in fibre optics have permitted the introduction of fibre optics into present applications. These applications are mostly in the telephone long-haul systems, but are growing to include cable television, computer networks, video systems, and data links. Research should increase system performance and provide solutions to existing problems in conventional applications. The impressive results from early research show there are many advantages offered by fibre optic systems.

Advantages and Disadvantages of Fibre Optics Fibre optic systems have many attractive features that are superior to electrical systems. These include improved system performance, immunity to electrical noise, signal security, and improved safety and electrical isolation. Other advantages include reduced size and weight, environmental protection, and overall system economy. The following list details the main advantages of fibre optic systems. Advantages of Fibre Optics  System Performance  Greatly increased bandwidth and capacity  Lower signal attenuation (loss)  Immunity to Electrical Noise  Immune to noise (electromagnetic interference [EMI] and radio-frequency interference [RFI])  No crosstalk  Lower bit error rates  Signal Security  Difficult to tap  Nonconductive (does not radiate signals) - Electrical Isolation  No common ground required  Freedom from short circuit and sparks  Reduced size and weight cables  Environmental Protection  Resistant to radiation and corrosion  Resistant to temperature variations

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Part 66 Training Syllabus Module 2 Physics  Improved ruggedness and flexibility  Less restrictive in harsh environments Silica is the principal, abundant, and inexpensive material (source is sand) Despite the many advantages of fibre optic systems, there are some disadvantages. Because of the relative newness of the technology, fibre optic components are expensive. Fibre optic transmitters and receivers are still relatively expensive compared to electrical interfaces. The lack of standardization in the industry has also limited the acceptance of fibre optics. Many industries are more comfortable with the use of electrical systems and are reluctant to switch to fibre optics. However, industry researchers are eliminating these disadvantages. The cost to install fibre optic systems is falling because of an increase in the use of fibre optic technology. Published articles, conferences, and lectures on fibre optics have begun to educate managers and technicians. As the technology matures, the use of fibre optics will increase because of its many advantages over electrical systems.

Frequency and Bandwidth Bandwidth is defined as the amount of information that can be transmitted at one time. In the early days of radio transmission when the information transmitted was mostly restricted to Morse code and speech, low frequencies were (long waves) were used. The range of frequencies available to be transmitted (which determines the bandwidth) was very low. This inevitably restricted us to low speed data transmission.

Figure 4.26: The Electromagnetic Spectrum, with wavelengths measured in Angstrom (10

-10

m)

As time went by, we required a wider bandwidth to send more complex information and to improve the speed of transmission. To do this, we had to increase the frequency of the radio signal used. The usable bandwidth is limited by the frequency used - the higher the frequency, the greater the bandwidth.

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Part 66 Training Syllabus Module 2 Physics When television was developed we again had the requirement of a wider bandwidth and we responded in the same way - by increasing the frequency. And so it went on. More bandwidth needed? Use higher frequency. For something like sixty years this became an established response - we had found the answer! Until fibre optics blew it all away. The early experiments showed that visible light transmission was possible and we explored the visible spectrum for the best light frequency to use. The promise of fibre optics was the possibility of increased transmission rates. The old solution pointed to the use of the highest frequency but here we met a real problem. We found that the transmission losses were increasing very quickly. In fact the losses increased by the fourth power. This 4 means that if the light frequency doubled, the losses would increase by a factor of 2 or 16 times. We quickly appreciated that it was not worth pursuing higher and higher frequencies in order to obtain higher bandwidths if it meant that we could only transmit the data over a very short distance. The bandwidth of a light based system was so high that a relatively low frequency could be tolerated in order to get lower losses and hence more transmission range. So we explored the lower frequency or the red end of the visible spectrum and then even further down into the infrared. And that is where we are at the present time. Infrared light covers a fairly wide range of wavelengths and is generally used for all fibre optic communications. Visible light is normally used for very short range transmissions using plastic fibre.

Basic Structure of an Optical Fibre The basic structure of an optical fibre consists of three parts; the core, the cladding, and the coating or buffer. The basic structure of an optical fibre is shown in figure 4.27. The core is a cylindrical rod of dielectric material. Dielectric material conducts no electricity. Light propagates mainly along the core of the fibre. The core is generally made of glass. The core is described as having a radius of (a) and an index of refraction PH. The core is surrounded by a layer of material called the cladding. Even though light will propagate along the fibre core without the layer of cladding material, the cladding does perform some necessary functions.

Figure 4.27: Basic structure of an optical fibre.

The cladding layer is made of a dielectric material with an index of refraction n 2. The index of refraction of the cladding material is less than that of the core material. The cladding is generally made of glass or plastic. The cladding performs the following functions:    

Reduces loss of light from the core into the surrounding air Reduces scattering loss at the surface of the core Protects the fibre from absorbing surface contaminants Adds mechanical strength

For extra protection, the cladding is enclosed in an additional layer called the coating or buffer.

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Part 66 Training Syllabus Module 2 Physics The coating or buffer is a layer of material used to protect an optical fibre from physical damage. The material used for a buffer is a type of plastic. The buffer is elastic in nature and prevents abrasions. The buffer also prevents the optical fibre from scattering losses caused by microbends. Microbends occur when an optical fibre is placed on a rough and distorted surface. Microbends are discussed later in this chapter.

Propagation of Light along a Fibre The concept of light propagation, the transmission of light along an optical fibre, can be described by two theories. According to the first theory, light is described as a simple ray. This theory is the ray theory, or geometrical optics, approach. The advantage of the ray approach is that you get a clearer picture of the propagation of light along a fibre. The ray theory is used to approximate the light acceptance and guiding properties of optical fibres. According to the second theory, light is described as an electromagnetic wave. This theory is the mode theory, or wave representation, approach. The mode theory describes the behaviour of light within an optical fibre. The mode theory is useful in describing the optical fibre properties of absorption, attenuation, and dispersion. These fibre properties are discussed later in this chapter.

Ray Theory Two types of rays can propagate along an optical fibre. The first type is called meridional rays. Meridional rays are rays that pass through the axis of the optical fibre. Meridional rays are used to illustrate the basic transmission properties of optical fibres. The second type is called skew rays. Skew rays are rays that travel through an optical fibre without passing through its axis. Meridional Rays. - Meridional rays can be classified as bound or unbound rays. Bound rays remain in the core and propagate along the axis of the fibre. Bound rays propagate through the fibre by total internal reflection. Unbound rays are refracted out of the fibre core. Figure 4.28 shows a possible path taken by bound and unbound rays in a step-index fibre. The core of the step-index fibre has an index of refraction n-i. The cladding of a step-index has an index of refraction na, that is lower than n-i. Figure 4.28 assumes the core-cladding interface is perfect. However, imperfections at the core-cladding interface will cause part of the bound rays to be refracted out of the core into the cladding. The light rays refracted into the cladding will eventually escape from the fibre. In general, meridional rays follow the laws of reflection and refraction.

Figure 4.28: Bound and unbound rays in a step-index fibre. It is known that bound rays propagate in fibres due to total internal reflection, but how do these light rays enter the fibre? Rays that enter the fibre must intersect the core-cladding interface at an angle greater than the critical angle (θC). Only those rays that enter the fibre and strike the interface at these angles will propagate along the fibre. How a light ray is launched into a fibre is shown in figure 4.29. The incident ray h enters the fibre at the angle θa. Ι1 is refracted upon entering the fibre and is transmitted to the core-cladding interface. The ray then strikes the core-cladding interface at the critical angle (θC). Ι1 is totally reflected back into the core and continues to propagate along the fibre. The incident ray I2 enters the fibre at an angle greater than θa. Again, I2 is refracted upon entering the fibre and is transmitted to the core-cladding interface, la strikes the corecladding interface at an angle less than the critical angle (θC). Ι2 is refracted into the cladding and is eventually lost. The light ray

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Part 66 Training Syllabus Module 2 Physics incident on the fibre core must be within the cone of acceptance defined by the angle 0a shown in figure 4.30. Angle 9a is defined as the acceptance angle. The acceptance angle (6a) is the maximum angle to the axis of the fibre that light entering the fibre is propagated. The value of the angle of acceptance (0 a) depends on fibre properties and transmission conditions.

Figure 4.29. - How a light ray enters an optical fibre.

Figure 4.30: Fibre acceptance angle. The acceptance angle is related to the refractive indices of the core, cladding, and medium surrounding the fibre. This relationship is called the numerical aperture of the fibre. The numerical aperture (NA) is a measurement of the ability of an optical fibre to capture light. The NA is also used to define the cone of acceptance of an optical fibre. Figure 4.30 illustrates the relationship between the acceptance angle and the refractive indices. The index of refraction of the fibre core is n1. The index of refraction of the fibre cladding is n 2. The index of refraction of the surrounding medium is n0. By using Snell's law and basic trigonometric relationships, the NA of the fibre is given by: 2

2

ΝΑ=n0 x sinθa = (n1 – n2 ) Since the medium next to the fibre at the launching point is normally air, n 0 is equal to 1.00. The NA is then simply equal to sin θa. The NA is a convenient way to measure the light-gathering ability of an optical fibre. It is used to measure source-to-fibre power-coupling efficiencies. A high NA indicates a high source-to-fibre coupling efficiency. Typical values of NA range from 0.20 to 0.29 for glass fibres. Plastic fibres generally have a higher NA. An NA for plastic fibres can be higher than 0.50. In addition, the NA is commonly used to specify multimode fibres. However, for small core diameters, such as in single mode fibres, the ray theory breaks down. Ray theory describes only the direction a plane wave takes in a fibre. Ray theory eliminates any properties of the plane wave that interfere with the transmission of light along a fibre. In reality, plane waves interfere with each other. Therefore, only certain types of rays are able to propagate in an optical fibre. Optical fibres can support only a specific number of guided modes. In small core fibres, the number of modes supported is one or only a few modes. Mode theory is used to describe the types of plane waves able to propagate along an optical fibre.

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Part 66 Training Syllabus Module 2 Physics Skew Rays. - A possible path of propagation of skew rays is shown in figure 4.31. Figure 4.31, view A, provides an angled view and view B provides a front view. Skew rays propagate without passing through the centre axis of the fibre. The acceptance angle for skew rays is larger than the acceptance angle of meridional rays. This condition explains why skew rays outnumber meridional rays. Skew rays are often used in the calculation of light acceptance in an optical fibre. The addition of skew rays increases the amount of light capacity of a fibre. In large NA fibres, the increase may be significant.

Figure 4.31: Skew ray propagation: A. Angled view; B. Front view. The addition of skew rays also increases the amount of loss in a fibre. Skew rays tend to propagate near the edge of the fibre core. A large portion of the number of skew rays that are trapped in the fibre core are considered to be leaky rays. Leaky rays are predicted to be totally reflected at the core-cladding boundary. However, these rays are partially refracted because of the curved nature of the fibre boundary. Mode theory is also used to describe this type of leaky ray loss.

Mode Theory The mode theory, along with the ray theory, is used to describe the propagation of light along an optical fibre. The mode theory is used to describe the properties of light that ray theory is unable to explain. The mode theory uses electromagnetic wave behaviour to describe the propagation of light along a fibre. A set of guided electromagnetic waves is called the modes of the fibre. Plane Waves. - The mode theory suggests that a light wave can be represented as a plane wave. A plane wave is described by its direction, amplitude, and wavelength of propagation. A plane wave is a wave whose surfaces of constant phase are infinite parallel planes normal to the direction of propagation. The planes having the same phase are called the wavefronts. The wavelength (A) of the plane wave is given by: Wavelength(λ)= where c is the speed of light in a vacuum, f is the frequency of the light, and n is the index of refraction of the plane-wave medium. Figure 4.32 shows the direction and wavefronts of plane-wave propagation. Plane waves, or wavefronts, propagate along the fibre similar to light rays. However, not all wavefronts incident on the fibre at angles less than or equal to the critical angle of light acceptance propagate along the fibre. Wavefronts may undergo a change in phase that prevents the successful transfer of light along the fibre.

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Figure 4.32: Plane-wave propagation. Wavefronts are required to remain in phase for light to be transmitted along the fibre. Consider the wavefronts incident on the core of an optical fibre as shown in figure 4.33. Only those wavefronts incident on the fibre at angles less than or equal to the critical angle may propagate along the fibre. The wavefront undergoes a gradual phase change as it travels down the fibre. Phase changes also occur when the wavefront is reflected. The wavefront must remain in phase after the wavefront transverses the fibre twice and is reflected twice. The distance transversed is shown between point A and point B on figure 4.33. The reflected waves at point A and point B are in phase if the total amount of phase collected is an integer multiple of 2π radian. If propagating wavefronts are not in phase, they eventually disappear. Wavefronts disappear because of destructive interference. The wavefronts that are in phase interfere with the wavefronts that are out of phase. This interference is the reason why only a finite number of modes can propagate along the fibre.

Figure 4.33: Wavefront propagation along an optical fibre. The plane waves repeat as they travel along the fibre axis. The direction the plane waves travel is assumed to be the z direction as shown in figure 4.33. The plane waves repeat at a distance equal to λ/sin θ. Plane waves also repeat at a periodic frequency β=2π sin θ/λ The quantity β is defined as the propagation constant along the fibre axis. As the wavelength (λ) changes, the value of the propagation constant must also change. For a given mode, a change in wavelength can prevent the mode from propagating along the fibre. The mode is no longer bound to the fibre. The mode is said to be cut off. Modes that are bound at one wavelength may not exist at longer wavelengths. The wavelength at which a mode ceases to be bound is called the cut-off wavelength for that mode. However, an optical fibre is always able to propagate at least one mode. This mode is referred to as the fundamental mode of the fibre. The fundamental mode can never be cut off. The wavelength that prevents the next higher mode from propagating is called the cut-off wavelength of the fibre. An optical fibre that operates above the cut-off wavelength (at a longer wavelength) is called a single mode fibre. An optical fibre that operates below the cut-off wavelength is called a multimode fibre. Single mode and multimode optical fibres are discussed later in this chapter.

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In a fibre, the propagation constant of a plane wave is a function of the wave's wavelength and mode. The change in the propagation constant for different waves is called dispersion. The change in the propagation constant for different wavelengths is called chromatic dispersion. The change in propagation constant for different modes is called modal dispersion. These dispersions cause the light pulse to spread as it goes down the fibre (fig. 4.34). Some dispersion occurs in all types of fibres. Dispersion is discussed later in this chapter.

Figure 4.34: The spreading of a light pulse. Modes. - A set of guided electromagnetic waves is called the modes of an optical fibre. Maxwell's equations describe electromagnetic waves or modes as having two components. The two components are the electric field, E(x, y, z), and the magnetic field, H(x, y, z). The electric field, E, and the magnetic field, H, are at right angles to each other. Modes travelling in an optical fibre are said to be transverse. The transverse modes, shown in figure 4.35, propagate along the axis of the fibre. The mode field patterns shown in figure 4.35 are said to be transverse electric (TE). In TE modes, the electric field is perpendicular to the direction of propagation. The magnetic field is in the direction of propagation. Another type of transverse mode is the transverse magnetic (TM) mode. TM modes are opposite to TE modes. In TM modes, the magnetic field is perpendicular to the direction of propagation. The electric field is in the direction of propagation. Figure 4.35 shows only TE modes.

Figure 4.35: Transverse electric (TE) mode field patterns. The TE mode field patterns shown in figure 4.35 indicate the order of each mode. The order of each mode is indicated by the number of field maxima within the core of the fibre. For example, TE0 has one field maxima. The electric field is a maximum at the centre of the waveguide and decays toward the core-cladding boundary. TE0 is considered the fundamental mode or the lowest order standing wave. As the number of field maxima increases, the order of the mode is higher. Generally, modes with more than a few (5-10) field maxima are referred to as high-order modes. The order of the mode is also determined by the angle the wavefront makes with the axis of the fibre. Figure 4.36 illustrates light rays as they travel down the fibre. These light rays indicate the direction of the wavefronts. High-order modes cross the axis of the fibre at steeper angles. Low-order and high-order modes are shown in figure 4.36.

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Figure 4.36: Low-order and high-order modes. Before we progress, let us refer back to figure 4.30. Notice that the modes are not confined to the core of the fibre. The modes extend partially into the cladding material. Low-order modes penetrate the cladding only slightly. In low-order modes, the electric and magnetic fields are concentrated near the centre of the fibre. However, high-order modes penetrate further into the cladding material. In high-order modes, the electrical and magnetic fields are distributed more toward the outer edges of the fibre. This penetration of low-order and high-order modes into the cladding region indicates that some portion is refracted out of the core. The refracted modes may become trapped in the cladding due to the dimension of the cladding region. The modes trapped in the cladding region are called cladding modes. As the core and the cladding modes travel along the fibre, mode coupling occurs. Mode coupling is the exchange of power between two modes. Mode coupling to the cladding results in the loss of power from the core modes. In addition to bound and refracted modes, there are leaky modes. Leaky modes are similar to leaky rays. Leaky modes lose power as they propagate along the fibre. For a mode to remain within the core, the mode must meet certain boundary conditions. A mode remains bound if the propagation constant β meets the following boundary condition:

where n1 and n2 are the index of refraction for the core and the cladding, respectively. When the propagation constant becomes smaller than 2πn2/λ power leaks out of the core and into the cladding. Generally, modes leaked into the cladding are lost in a few centimetres. However, leaky modes can carry a large amount of power in short fibres. Normalized Frequency. - Electromagnetic waves bound to an optical fibre are described by the fibre's normalized frequency. The normalized frequency determines how many modes a fibre can support. Normalized frequency is a dimensionless quantity. Normalized frequency is also related to the fibre's cut-off wavelength. Normalized frequency (V) is defined as: 2

2

V= (n1 – n2 ) where n1 is the core index of refraction, n2 is the cladding index of refraction, a is the core diameter, and λ is the wavelength of light in air. The number of modes that can exist in a fibre is a function of V. As the value of V increases, the number of modes supported by the fibre increases. Optical fibres, single mode and multimode, can support a different number of modes. The number of modes supported by single mode and multimode fibre types is discussed later in this chapter.

Optical Fibre Types Optical fibres are characterized by their structure and by their properties of transmission. Basically, optical fibres are classified into two types. The first type is single mode fibres. The second type is multimode fibres. As each name implies, optical fibres are classified by the number of modes that propagate along the fibre. As previously explained, the structure of the fibre can permit or restrict modes from propagating in a fibre. The basic structural difference is the core size. Single mode fibres are manufactured with the same materials

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Part 66 Training Syllabus Module 2 Physics as multimode fibres. Single mode fibres are also manufactured by following the same fabrication process as multimode fibres.

Single Mode Fibres The core size of single mode fibres is small. The core size (diameter) is typically around 8 to 10 micrometers (μm). A fibre core of this size allows only the fundamental or lowest order mode to propagate around a 1300 nanometre (nm) wavelength. Single mode fibres propagate only one mode, because the core size approaches the operational wavelength (λ). The value of the normalized frequency parameter (V) relates core size with mode propagation. In single mode fibres, V is less than or equal to 2.405. When V ≤ 2.405, single mode fibres propagate the fundamental mode down the fibre core, while high-order modes are lost in the cladding. For low V values (£1.0), most of the power is propagated in the cladding material. Power transmitted by the cladding is easily lost at fibre bends. The value of V should remain near the 2.405 level. Single mode fibres have a lower signal loss and a higher information capacity (bandwidth) than multimode fibres. Single mode fibres are capable of transferring higher amounts of data due to low fibre dispersion. Basically, dispersion is the spreading of light as light propagates along a fibre. Dispersion mechanisms in single mode fibres are discussed in more detail later in this chapter. Signal loss depends on the operational wavelength (λ). In single mode fibres, the wavelength can increase or decrease the losses caused by fibre bending. Single mode fibres operating at wavelengths larger than the cut-off wavelength lose more power at fibre bends. They lose power because light radiates into the cladding, which is lost at fibre bends. In general, single mode fibres are considered to be low-loss fibres, which increase system bandwidth and length.

Multimode Fibres As their name implies, multimode fibres propagate more than one mode. Multimode fibres can propagate over 100 modes. The number of modes propagated depends on the core size and numerical aperture (NA). As the core size and NA increase, the number of modes increases. Typical values of fibre core size and NA are 50 to 100 (am and 0.20 to 0.29, respectively. A large core size and a higher NA have several advantages. Light is launched into a multimode fibre with more ease. The higher NA and the larger core size make it easier to make fibre connections. During fibre splicing, core-to-core alignment becomes less critical. Another advantage is that multimode fibres permit the use of light-emitting diodes (LEDs). Single mode fibres typically must use laser diodes. LEDs are cheaper, less complex, and last longer. LEDs are preferred for most applications. Multimode fibres also have some disadvantages. As the number of modes increases, the effect of modal dispersion increases. Modal dispersion (intermodal dispersion) means that modes arrive at the fibre end at slightly different times. This time difference causes the light pulse to spread. Modal dispersion affects system bandwidth. Fibre manufacturers adjust the core diameter, NA, and index profile properties of multimode fibres to maximize system bandwidth.

Properties of Optical Fibre Transmission The principles behind the transfer of light along an optical fibre were discussed earlier in this section. You learned that propagation of light depended on the nature of light and the structure of the optical fibre. However, our discussion did not describe how optical fibres affect system performance. In this case, system performance deals with signal loss and bandwidth. Signal loss and system bandwidth describe the amount of data transmitted over a specified length of fibre. Many optical fibre properties increase signal loss and reduce system bandwidth. The most important properties that affect system performance are fibre attenuation and dispersion. Attenuation reduces the amount of optical power transmitted by the fibre. Attenuation controls the distance an optical signal (pulse) can travel as shown in figure 4.37. Once the power of an optical pulse is reduced to a point where the receiver is unable to detect the pulse, an error occurs. Attenuation is mainly a result of light absorption, scattering, and bending losses. Dispersion spreads the optical pulse as it travels along the fibre. This spreading of the signal pulse reduces the system bandwidth or the information-carrying capacity of the fibre. Dispersion limits how fast information is transferred as shown in figure 4.37. An error occurs when the receiver is unable to distinguish between input pulses caused by the

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Part 66 Training Syllabus Module 2 Physics spreading of each pulse. The effects of attenuation and dispersion increase as the pulse travels the length of the fibre as shown in figure 4.38.

Figure 4.37. - Fibre transmission properties.

Figure 4.38: Pulse spreading and power loss along an optical fibre. In addition to fibre attenuation and dispersion, other optical fibre properties affect system performance. Fibre properties, such as modal noise, pulse broadening, and polarization, can reduce system performance. Modal noise, pulse broadening, and polarization are too complex to discuss as introductory level material. However, you should be aware that attenuation and dispersion are not the only fibre properties that affect performance.

Attenuation Attenuation in an optical fibre is caused by absorption, scattering, and bending losses. Attenuation is the loss of optical power as light travels along the fibre. Signal attenuation is defined as the ratio of optical input power (Pj) to the optical output power (P0). Optical input power is the power injected into the fibre from an optical source. Optical output power is the power received at the fibre end or optical detector. The following equation defines signal attenuation as a unit of length: Attenuation = ( )log10( ) Signal attenuation is a log relationship. Length (L) is expressed in kilometres. Therefore, the unit of attenuation is decibels/kilometre (dB/km). As previously stated, attenuation is caused by absorption, scattering, and bending losses. Each mechanism of loss is influenced by material-material properties and fibre structure. However, loss is also present at fibre connections. Fibre connector, splice, and coupler losses are discussed later. The present discussion remains relative to optical fibre attenuation properties. Absorption - Absorption is a major cause of signal loss in an optical fibre. Absorption is defined as the portion of attenuation resulting from the conversion of optical power into another energy form, such as heat. Absorption in optical fibres is explained by three factors:  Imperfections in the atomic structure of the fibre material  The intrinsic or basic material-material properties  The extrinsic (presence of impurities) material-material properties Imperfections in the atomic structure induce absorption by the presence of missing molecules or oxygen defects. Absorption is also induced by the diffusion of hydrogen molecules into the glass fibre. Since intrinsic and extrinsic material properties are the main cause of absorption, they are discussed further.

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Part 66 Training Syllabus Module 2 Physics Intrinsic Absorption - Intrinsic absorption is caused by basic material-material properties. If an optical fibre were absolutely pure, with no imperfections or impurities, then all absorption would be intrinsic. Intrinsic absorption sets the minimal level of absorption. In fibre optics, silica (pure glass) fibres are used predominately. Silica fibres are used because of their low intrinsic material absorption at the wavelengths of operation. In silica glass, the wavelengths of operation range from 700 nanometres (nm) to 1600 nm. Figure 4.39 shows the level of attenuation at the wavelengths of operation. This wavelength of operation is between two intrinsic absorption regions. The first region is the ultraviolet region (below 400-nm wavelength). The second region is the infrared region (above 2000-nm wavelength).

Figure 4.39: Fibre losses. Intrinsic absorption in the ultraviolet region is caused by electronic absorption bands. Basically, absorption occurs when a light particle (photon) interacts with an electron and excites it to a higher energy level. The tail of the ultraviolet absorption band is shown in figure 4.39. The main cause of intrinsic absorption in the infrared region is the characteristic vibration frequency of atomic bonds. In silica glass, absorption is caused by the vibration of silicon-oxygen (Si-O) bonds. The interaction between the vibrating bond and the electromagnetic field of the optical signal causes intrinsic absorption. Light energy is transferred from the electromagnetic field to the bond. The tail of the infrared absorption band is shown in figure 4.39. Extrinsic Absorption - Extrinsic absorption is caused by impurities introduced into the fibre material. Trace metal impurities, such as iron, nickel, and chromium, are introduced into the fibre during fabrication. Extrinsic absorption is caused by the electronic transition of these metal ions from one energy level to another. -

Extrinsic absorption also occurs when hydroxyl ions (OH ) are introduced into the fibre. Water in silica glass forms a silicon-hydroxyl (Si-OH) bond. This bond has a fundamental absorption at 2700 nm. However, the harmonics or overtones of the fundamental absorption occur in the region of operation. These harmonics increase extrinsic absorption at 1383 nm, 1250 nm, and 950 nm. Figure 4.39 shows the presence of the three OH harmonics. The level of the OH harmonic absorption is also indicated. These absorption peaks define three regions or windows of preferred operation. The first window is centred at 850 nm. The second window is centred at 1300 nm. The third window is centred at 1550 nm. Fibre optic systems operate at wavelengths defined by one of these windows. Visible light has a wavelength between 400 and 750 nm. Therefore all three of these wavelengths used in fibreoptic data transmission are within the infrared range. The amount of water (OH ) impurities present in a fibre should be less than a few parts per billion. Fibre attenuation caused by extrinsic absorption is affected by the level of impurities (OH ) present in the fibre. If the amount of impurities in a fibre is reduced, then fibre attenuation is reduced. Scattering. - Basically, scattering losses are caused by the interaction of light with density fluctuations within a fibre. Density changes are produced when optical fibres are manufactured. During manufacturing, regions of higher and lower molecular density areas, relative to the average density of the fibre, are created. Light travelling through the fibre interacts with the density areas as shown in figure 4.40. Light is then partially scattered in all directions.

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Figure 4.40: Light scattering . In commercial fibres operating between 700-nm and 1600-nm wavelength, the main source of loss is called Rayleigh scattering. Rayleigh scattering is the main loss mechanism between the ultraviolet and infrared regions as shown in figure 4.39. Rayleigh scattering occurs when the size of the density fluctuation (fibre defect) is less than one-tenth of the operating wavelength of light. Loss caused by Rayleigh scattering is 4 proportional to the fourth power of the wavelength (1/ λ ). As the wavelength increases, the loss caused by Rayleigh scattering decreases. If the size of the defect is greater than one-tenth of the wavelength of light, the scattering mechanism is called Mie scattering. Mie scattering, caused by these large defects in the fibre core, scatters light out of the fibre core. However, in commercial fibres, the effects of Mie scattering are insignificant. Optical fibres are manufactured with very few large defects. Bending Loss - Bending the fibre also causes attenuation. Bending loss is classified according to the bend radius of curvature: microbend loss or macrobend loss. Microbends are small microscopic bends of the fibre axis that occur mainly when a fibre is cabled. Macrobends are bends having a large radius of curvature relative to the fibre diameter. Microbend and macrobend losses are very important loss mechanisms. Fibre loss caused by microbending can still occur even if the fibre is cabled correctly. During installation, if fibres are bent too sharply, macrobend losses will occur. Microbend losses are caused by small discontinuities or imperfections in the fibre. Uneven coating applications and improper cabling procedures increase microbend loss. External forces are also a source of microbends. An external force deforms the cabled jacket surrounding the fibre but causes only a small bend in the fibre. Microbends change the path that propagating modes take, as shown in figure 4.41. Microbend loss increases attenuation because low-order modes become coupled with high-order modes that are naturally leaky

Figure 4.41: Microbend loss.

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Part 66 Training Syllabus Module 2 Physics Macrobend losses are observed when a fibre bend's radius of curvature is large compared to the fibre diameter. These bends become a great source of loss when the radius of curvature is less than several centimetres. Light propagating at the inner side of the bend travels a shorter distance than that on the outer side. To maintain the phase of the light wave, the mode phase velocity must increase. When the fibre bend is less than some critical radius, the mode phase velocity must increase to a speed greater than the speed of light. However, it is impossible to exceed the speed of light. This condition causes some of the light within the fibre to be converted to high-order modes. These high-order modes are then lost or radiated out of the fibre. Fibre sensitivity to bending losses can be reduced. If the refractive index of the core is increased, then fibre sensitivity decreases. Sensitivity also decreases as the diameter of the overall fibre increases. However, increases in the fibre core diameter increase fibre sensitivity. Fibres with larger core size propagate more modes. These additional modes tend to be more leaky.

Dispersion There are two different types of dispersion in optical fibres. The types are intramodal and intermodal dispersion. Intramodal, or chromatic, dispersion occurs in all types of fibres. Intermodal, or modal, dispersion occurs only in multimode fibres. Each type of dispersion mechanism leads to pulse spreading. As a pulse spreads, energy is overlapped. This condition is shown in figure 4.42. The spreading of the optical pulse as it travels along the fibre limits the information capacity of the fibre.

Figure 4.42: Pulse overlap. Intramodal Dispersion - Intramodal, or chromatic, dispersion depends primarily on fibre materials. There are two types of intramodal dispersion. The first type is material dispersion. The second type is waveguide dispersion. Intramodal dispersion occurs because different colours of light travel through different materials and different waveguide structures at different speeds. Material dispersion occurs because the spreading of a light pulse is dependent on the wavelengths' interaction with the refractive index of the fibre core. Different wavelengths travel at different speeds in the fibre material. Different wavelengths of a light pulse that enter a fibre at one time exit the fibre at different times. Material dispersion is a function of the source spectral width. The spectral width specifies the range of wavelengths that can propagate in the fibre. Material dispersion is less at longer wavelengths. Waveguide dispersion occurs because the mode propagation constant (β) is a function of the size of the fibre's core relative to the wavelength of operation. Waveguide dispersion also occurs because light propagates differently in the core than in the cladding. In multimode fibres, waveguide dispersion and material dispersion are basically separate properties. Multimode waveguide dispersion is generally small compared to material dispersion. Waveguide dispersion is usually neglected. However, in single mode fibres, material and waveguide dispersion are interrelated. The total dispersion present in single mode fibres may be minimized by trading material and waveguide properties depending on the wavelength of operation.

Intermodal Dispersion Intermodal or modal dispersion causes the input light pulse to spread. The input light pulse is made up of a group of modes. As the modes propagate along the fibre, light energy distributed among the modes is delayed by different amounts. The pulse spreads because each mode propagates along the fibre at different

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Part 66 Training Syllabus Module 2 Physics speeds. Since modes travel in different directions, some modes travel longer distances. Modal dispersion occurs because each mode travels a different distance over the same time span, as shown in figure 4.43. The modes of a light pulse that enter the fibre at one time exit the fibre at different times. This condition causes the light pulse to spread. As the length of the fibre increases, modal dispersion increases.

Figure 4.43: Distance travelled by each mode over the same time span. Modal dispersion is the dominant source of dispersion in multimode fibres. Modal dispersion does not exist in single mode fibres. Single mode fibres propagate only the fundamental mode. Therefore, single mode fibres exhibit the lowest amount of total dispersion. Single mode fibres also exhibit the highest possible bandwidth.

The Transmission of Signals Analogue Transmission This is the simplest method. Although it is not generally used on aircraft systems, it is discussed here as a background to digital techniques. The incoming information signal, speech, music, video etc. is used to control the power output from the LED or the laser. The light output is, as near as possible, a true copy of the electrical variations at the input. At the far end of the fibre, the receiver converts the light back into an electrical signal which is, hopefully, the same as the electrical signal. However, any nonlinearity of the characteristics of the transmitter or receiver will reduce the accuracy of the electrical/optical (E/O) and optical/electrical (O/E) conversions and give rise to distortion in the output signal. Another problem is noise. Since the receiver is receiving an analogue signal, it must be sensitive to any changes in amplitude. Any random fluctuations in light level caused by the light source, the fibre or the receiver will cause unwanted noise in the output signal. Electrical noise due to lightning, electromagnetic interference (EMI) or High Intensity Radiated Fields (HIRF) will also give rise to electrical noise in the nonfibre parts of the system.

Digital Transmission In a digital system, the information signal is represented by a sequence of on/off levels. The 'on' state is often referred to as logic 1 and the 'off state as logic 0. The 1 and 0 have no numerical significance and are just convenient ways to differentiate between the two states. The 'yes' and 'no' approach means that it ignores noise and distortion since all voltages above the threshold level are recognised as logic 1 state and all below this level as a logic 0. The signal is then generated as a perfect copy of the original signal.

Optical Fibres and Cables Optical Fibre and Cable Design Optical fibres are thin cylindrical dielectric (non-conductive) waveguides used to send light energy for communication. Optical fibres consist of three parts: the core, the cladding, and the coating or buffer. The choice of optical fibre materials and fibre design depends on operating conditions and intended application. Optical fibres are protected from the environment by incorporating the fibre into some type of cable structure. Cable strength members and outer jackets protect the fibre. Optical cable structure and material composition depend on the conditions of operation and the intended application.

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Optical Fibres Optical fibres are classified as either single mode or multimode fibres. Fibres are classified according to the number of modes that they can propagate. Single mode fibres can propagate only the fundamental mode. Multimode fibres can propagate hundreds of modes. However, the classification of an optical fibre depends on more than the number of modes that a fibre can propagate. An optical fibre's refractive index profile and core size further distinguish single mode and multimode fibres. The refractive index profile describes the value of refractive index as a function of radial distance at any fibre diameter. Fibre refractive index profiles classify single mode and multimode fibres as follows:    

Multimode step-index fibres Multimode graded-index fibres Single mode step-index fibres Single mode graded-index fibres

In a step-index fibre, the refractive index of the core is uniform and undergoes an abrupt change at the corecladding boundary. Step-index fibres obtain their name from this abrupt change called the step change in refractive index. In graded-index fibres, the refractive index of the core varies gradually as a function of radial distance from the fibre centre. Single mode and multimode fibres can have a step-index or graded-index refractive index profile. The performance of multimode graded-index fibres is usually superior to multimode step-index fibres. However, each type of multimode fibre can improve system design and operation depending on the intended application. Performance advantages for single mode graded-index fibres compared to single mode stepindex fibres are relatively small. Therefore, single mode fibre production is almost exclusively step-index. Figure 10-28 shows the refractive index profile for a multimode step-index fibre and a multimode gradedindex fibre. Figure 10-28 also shows the refractive index profile for a single mode step-index fibre. Since light propagates differently in each fibre type, figure 4.44 shows the propagation of light along each fibre.

Figure 4.44: The refractive index profiles and light propagation in multimode step-index, multimode gradedindex, and single mode step-index fibres. Previously you learned that fibre core size and material composition can affect system performance. A small change in core size and material composition affects fibre transmission properties, such as attenuation and dispersion. When selecting an optical fibre, the system designer decides which fibre core size and material composition is appropriate.

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Part 66 Training Syllabus Module 2 Physics Standard core sizes for multimode step-index fibres are 50 μm and 100 μm. Standard core sizes for multimode graded-index fibres are 50 μm, 62.5 μm, 85 μm, and 100 μm. Standard core sizes for single mode fibres are between 8 μm and 10 μm. In most cases, the material used in the preparation of optical fibres is high-quality glass (SiO2). This glass contains very low amounts of impurities, such as water or elements other than silica and oxygen. Using high-quality glass produces fibres with low losses. Small amounts of some elements other than silica and oxygen are added to the glass material to change its index of refraction. These elements are called material dopants. Silica doped with various materials forms the refractive index profile of the fibre core and material dopants are discussed in more detail later in this chapter. Glass is not the only material used in fabrication of optical fibres. Plastics are also used for core and cladding materials in some applications. A particular optical fibre design can improve fibre optic system performance. Each single mode or multimode, step-index or graded-index, glass or plastic, or large or small core fibre has an intended application. The system designer must choose an appropriate fibre design that optimizes system performance in his application.

Multimode Step-Index Fibres A multimode step-index fibre has a core of radius (a) and a constant refractive index n1. A cladding of slightly lower refractive index n2 surrounds the core. Figure 4.45 shows the refractive index profile n(r) for this type of fibre. n(r) is equal to ni at radial distances r < a (core). n(r) is equal to n2 at radial distances r a (cladding). Notice the step decrease in the value of refractive index at the core-cladding interface. This step decrease occurs at a radius equal to distance (a). The difference in the core and cladding refractive index is the parameter Γ : Δ is the relative refractive index difference. n(r)

Figure 4.45: The refractive index profile for multimode step-index fibres. The ability of the fibre to accept optical energy from a light source is related to A. A also relates to the numerical aperture by NA ~ n1√ The number of modes that multimode step-index fibres propagate depends on Γ and core radius (a) of the fibre. The number of propagating modes also depends on the wavelength (λ) of the transmitted light. In a typical multimode step-index fibre, there are hundreds of propagating modes. Most modes in multimode step-index fibres propagate far from cut-off. Modes that are cut off cease to be bound to the core of the fibre. Modes that are farther away from the cutoff wavelength concentrate most of their light energy into the fibre core. Modes that propagate close to cutoff have a greater percentage of their light energy propagate in the cladding. Since most modes propagate far from cut-off, the majority of light propagates in the fibre core. Therefore, in multimode step-index fibres, cladding properties, such as cladding diameter, have limited affect on mode (light) propagation.

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Part 66 Training Syllabus Module 2 Physics Multimode step-index fibres have relatively large core diameters and large numerical apertures. A large core size and a large numerical aperture make it easier to couple light from a light-emitting diode (LED) into the fibre. Multimode step-index fibre core size is typically 50 μm or 100 μm Unfortunately, multimode step-index fibres have limited bandwidth capabilities. Dispersion, mainly modal dispersion, limits the bandwidth or information-carrying capacity of the fibre. System designers consider each factor when selecting an appropriate fibre for each particular application. Multimode step-index fibre selection depends on system application and design. Short-haul, limited bandwidth, low-cost applications typically use multimode step-index fibres.

Multimode Graded-lndex Fibres A multimode graded-index fibre has a core of radius (a). Unlike step-index fibres, the value of the refractive index of the core (ni) varies according to the radial distance (r). The value of n1 decreases as the distance (r) from the centre of the fibre increases. The value of nι decreases until it approaches the value of the refractive index of the cladding (n2). The value of n1 must be higher than the value of n2 to allow for proper mode propagation. Like the step-index fibre, the value of n2 is constant and has a slightly lower value than the maximum value of n 1. The relative refractive index difference (A) is determined using the maximum value of n1 and the value of n2. Figure 10-30 shows a possible refractive index profile n(r) for a multimode graded-index fibre. Notice the parabolic refractive index profile of the core. The profile parameter (α) determines the shape of the core's profile. As the value of a increases, the shape of the core's profile changes from a triangular shape to step as shown in figure 4.46. Most multimode graded-index fibres have a parabolic refractive index profile. Multimode fibres with near parabolic graded-index profiles provide the best performance. Unless otherwise specified, when discussing multimode graded-index fibres, assume that the core's refractive index profile is parabolic (α =2).

Figure 4.46. - The refractive index profile for multimode graded-index fibres.

Figure 4.47: The refractive index profiles for different values of a. Light propagates in multimode graded-index fibres according to refraction and total internal reflection. The gradual decrease in the core's refractive index from the centre of the fibre causes the light rays to be refracted many times. The light rays become refracted or curved, which increases the angle of incidence at the next point of refraction. Total internal reflection occurs when the angle of incidence becomes larger than the critical angle of incidence. Figure 4.48 shows the process of refraction and total internal reflection of light in multimode graded-index fibres. Figure 4.48 also illustrates the boundaries of different values of core refractive index by dotted lines. Light rays may be reflected to the axis of the fibre before reaching the corecladding interface.

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Figure 4.48: Refractive index grading and light propagation in multimode graded-index fibres. The NA of a multimode graded-index fibre is at its maximum value at the fibre axis. This NA is the axial numerical aperture [NA(0)]. NA(0) is approximately equal to: n1√ However, the NA for graded-index fibres varies as a function of the radial distance (r). NA varies because of the refractive index grading in the fibre's core. The NA decreases from the maximum, NA(0), to zero at distances greater than the core-cladding boundary distance (r>a). The NA, relative refractive index difference (Γ), profile parameter (α), and normalized frequency (V) determine the number of propagating modes in multimode graded-index fibres. A multimode graded-index fibre with the same normalized frequency as a multimode step-index fibre will have approximately one-half as many propagating modes. However, multimode graded-index fibres typically have over one-hundred propagating modes. Multimode graded-index fibres accept less light than multimode step-index fibres with the same core Γ. However, graded-index fibres usually outperform the step-index fibres. The core's parabolic refractive index profile causes multimode graded-index fibres to have less modal dispersion. Figure 4.48 shows possible paths that light may take when propagating in multimode graded-index fibres. Light rays that travel farther from the fibre's axis travel a longer distance. Light rays that travel farther from the centre travel in core material with an average lower refractive index. Earlier, you learned that light travels faster in a material with a lower refractive index. Therefore, those light rays that travel the longer distance in the lower refractive index parts of the core travel at a greater average velocity. This means that the rays that travel farther from the fibre's axis will arrive at each point along the fibre at nearly the same time as the rays that travel close to the fibre's axis. The decrease in time difference between light rays reduces modal dispersion and increases multimode graded-index fibre bandwidth. The increased bandwidth allows the use of multimode graded-index fibres in most applications. Most present day applications that use multimode fibre use graded-index fibres. The basic design parameters are the fibre's core and cladding size and Γ. Standard multimode graded-index fibre core and cladding sizes are 50/125 μm, 62.5/125 μm, 85/125 μm, and 100/140 μm Each fibre design has a specific Γ that improves fibre performance. Typical values of Γ are around 0.01 to 0.02. Although no single multimode graded-index fibre design is appropriate for all applications, the 62.5/125 μm fibre with a Γ of 0.02 offers the best overall performance. A multimode graded-index fibre's source-to-fibre coupling efficiency and insensitivity to microbending and macrobending losses are its most distinguishing characteristics. The fibre core size and Γ affect the amount of power coupled into the core and loss caused by microbending and macrobending. Coupled power increases with both core diameter and Γ, while bending losses increase directly with core diameter and inversely with Γ. However, while these values favour high Γ s, a smaller Γ improves fibre bandwidth. In most applications, a multimode graded-index fibre with a core and cladding size of 62.5/125 μm offers the best combination of the following properties:     

Relatively high source-to-fibre coupling efficiency Low loss Low sensitivity to microbending and macrobending High bandwidth Expansion capability

For example, local area network (LAN) and aircraft applications use multimode graded-index fibres with a core and cladding size of 62.5/125 μm. In LAN-type environments, macrobend and microbend losses are hard to predict. Cable tension, bends, and local tie-downs increase macrobend and microbend losses. In

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Part 66 Training Syllabus Module 2 Physics aircraft applications, a cable-way may place physical restrictions, such as tight bends, on the fibre during cable plant installation. The good microbend and macrobend performance of 62.5/125 |4.m fibre permits installation of a rugged and robust cable. 62.5/125 (am multimode graded-index fibres allow for uncomplicated growth because of high fibre bandwidth capabilities for the expected short cable runs onboard aircraft. Single Mode Step-Index Fibres There are two basic types of single mode step-index fibres: matched clad and depressed clad. Matched cladding means that the fibre cladding consists of a single homogeneous layer of dielectric material. Depressed cladding means that the fibre cladding consists of two regions: the inner and outer cladding regions. Matched-clad and depressed-clad single mode step-index fibres have unique refractive index profiles. A matched-clad single mode step-index fibre has a core of radius (a) and a constant refractive index ni A cladding of slightly lower refractive index surrounds the core. The cladding has a refractive index n2 .Figure 4.49 shows the refractive index profile n(r) for the matched-clad single mode fibre.

Figure 4.49: Matched-clad refractive index profile. Figure 4.50 shows the refractive index profile n(r) for the depressed-clad single mode fibre. A depressedclad single mode step-index fibre has a core of radius (a) with a constant refractive index nι. A cladding, made of two regions, surrounds the core. An inner cladding region surrounds the core of the fibre and has a refractive index of n2. The inner cladding refractive index n3 is lower than the core's refractive index n 1. An outer cladding region surrounds the inner cladding region and has a higher refractive index na than the inner cladding region. However, the outer cladding refractive index n 2 is lower than the core's refractive index n3.

Figure 4.50: Depressed-clad refractive index profile. Single mode step-index fibres propagate only one mode, called the fundamental mode. Single mode operation occurs when the value of the fibre's normalized frequency is between 0 and 2.405 (0 V < 2.405). The value of V should remain near the 2.405 level. When the value of V is less than 1, single mode fibres carry a majority of the light power in the cladding material. The portion of light transmitted by the cladding material easily radiates out of the fibre. For example, light radiates out of the cladding material at fibre bends and splices. Single mode fibre cut-off wavelength is the smallest operating wavelength when single mode fibres propagate only the fundamental mode. At this wavelength, the 2nd-order mode becomes leaky and radiates

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Part 66 Training Syllabus Module 2 Physics out of the fibre core. As the operating wavelength becomes longer than the cut-off wavelength, the fundamental mode becomes increasingly leaky. The higher the operating wavelength is above the cut-off wavelength, the more power is transmitted through the fibre cladding. As the fundamental mode extends into the cladding material, it becomes increasingly sensitive to bending loss. Single mode fibre designs include claddings of sufficient thickness with low absorption and scattering properties to reduce attenuation of the fundamental mode. To increase performance and reduce losses caused by fibre bending and splicing, fibre manufacturers adjust the value of V. To adjust the value of V, they vary the core and cladding sizes and relative refractive index difference (Γ). A single mode step-index fibre has low attenuation and high bandwidth properties. Present applications for single mode fibres include long-haul, high-speed telecommunication systems. Future applications include single mode fibres for sensor systems. However, the current state of single mode technology makes installation of single mode systems expensive and difficult. Short cable runs, low to moderate bandwidth requirements, and high component cost make installation of single mode fibre aircraft systems impractical at this time. Therefore, field measurements may require two people. The main field measurement technique involves optical time-domain reflectometry. An optical time-domain reflectometer (OTDR) is recommended for conducting field measurements on installed optical fibres or links of 20 meters or more in length. An OTDR requires access to only one fibre end. An OTDR measures the attenuation of installed optical fibres as a function of length. It also identifies and evaluates optical connection losses along a cable link and locates any fibre breaks or faults. End users can also measure fibre attenuation and cable plant transmission loss using an optical power meter and a stabilized light source. End users use this measurement technique when optical time-domain reflectometry is not recommended. Measurements obtained with a stabilized light source and power meter are more accurate than those obtained with an OTDR. Measuring fibre attenuation and transmission loss using a power meter and light source requires access to both ends of the fibre or link. An optical loss test set (OLTS) combines the power meter and source functions into one physical unit.

Optical Time-Domain Reflectometry End users use optical time-domain reflectometry to characterize optical fibre and optical connection properties in the field. In optical time-domain reflectometry, an OTDR transmits an optical pulse through an installed optical fibre. The OTDR measures the fraction of light that is reflected back due to Rayleigh scattering and Fresnel reflection. By comparing the amount of light scattered back at different times, the OTDR can determine fibre and connection losses. When several fibres are connected to form an installed cable plant, the OTDR can characterize optical fibre and optical connection properties along the entire length of the cable plant. A fibre optic cable plant consists of optical fibre cables, connectors, splices, mounting panels, jumper cables, and other passive components. A cable plant does not include active components such as optical transmitters or receivers. The OTDR displays the backscattered and reflected optical signal as a function of length. The OTDR plots half the power in decibels (dB) versus half the distance. Plotting half the power in dB and half the distance corrects for round trip effects. By analyzing the OTDR plot, or trace, end users can measure fibre attenuation and transmission loss between any two points along the cable plant. End users can also measure insertion loss and reflectance of any optical connection. In addition, end users use the OTDR trace to locate fibre breaks or faults. Figure 4.51 shows an example OTDR trace of an installed cable plant. OTDR traces can have several common characteristics. An OTDR trace begins with an initial input pulse. This pulse is a result of Fresnel reflection occurring at the connection to the OTDR. Following this pulse, the OTDR trace is a gradual down sloping curve interrupted by abrupt shifts. Periods of gradual decline in the OTDR trace result from Rayleigh scattering as light travels along each fibre section of the cable plant. Periods of gradual decline are interrupted by abrupt shifts called point defects. A point defect is a temporary or permanent local deviation of the OTDR signal in the upward or downward direction. Point defects are caused by connectors, splices, or breaks along the fibre length. Point defects, or faults, can be reflective or non-reflective. An output pulse at the end of the OTDR trace indicates the end of the fibre cable plant. This output pulse results from Fresnel reflection occurring at the output end-end face

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Figure 4.51: OTDR trace of an installed cable plant. Attenuation - The accuracy of this test method depends on the user entering the appropriate source wavelength, pulse duration, and fibre length (test range) into the OTDR. In addition, the effective group index of the test fibre is required before the attenuation coefficient and accurate distances can be recorded. The group index (N) is provided by fibre manufacturers. By entering correct test parameters, OTDR fibre attenuation values will closely coincide with those measured by the cutback technique. Test personnel can connect the test fibre directly to the OTDR or to a dead-zone fibre. This dead-zone fibre is placed between the test fibre and OTDR to reduce the effect of the initial reflection at the OTDR on the fibre measurement. The dead-zone fibre is inserted because minimizing the reflection at a fibre joint is easier than reducing the reflection at the OTDR connection. Figure 10-88 illustrates the OTDR measurement points for measuring the attenuation of the test fibre using a dead-zone fibre. Fibre attenuation between two points along the test fibre is measured on gradual down sloping sections on the OTDR trace. There should be no point defects present along the portion of fibre being tested.

Figure 4.52: OTDR measurement points for measuring fibre attenuation using a dead-zone fibre. OTDRs are equipped with either manual or automatic cursors to locate points of interest along the trace. In figure 4.52, a cursor is positioned at a distance z0 on the rising edge of the reflection at the end of the deadzone fibre. Cursors are also positioned at distances z1 and z2. The cursor positioned at zi is just beyond the recovery from the reflection at the end of the dead-zone fibre. Since no point defects are present in figure 1088, the cursor positioned at z2 locates the end of the test fibre. Cursor z2 is positioned just before the output pulse resulting from Fresnel reflection occurring at the end of the test fibre.

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Power Meter Test personnel also use an optical power meter and stabilized light source to measure fibre attenuation and transmission loss in the field. Optical power meter measurements are recommended when the length of an installed optical fibre cable or cable plant is less than 50 meters. A test jumper is used to couple light from the stabilized source to one end of the optical fibre (or cable plant) under test. An additional test jumper is also used to connect the other end of the optical fibre (or cable plant) under test to the power meter. Optical power meter measurements may be conducted using an optical loss test set (OLTS). An OLTS combines the power meter and source functions into one physical unit. When making measurements, it does not matter whether the stabilized source and power meter are in one physical unit or two. Power meter measurements are conducted on individual optical fibre cables installed onboard the aircraft. The installed optical fibre cable must have connectors or terminations on both ends to make the measurement. If the installed optical fibre cable does not have connectors or terminations on both ends, an OTDR should be used to evaluate the cable. If the cable is too short for evaluation with an OTDR, cable continuity can be verified using a flashlight. Power meter measurements for cable assembly link loss require that test personnel clean all optical connections at test jumper interfaces before performing any measurement. Test personnel should use cotton wipes dampened with alcohol to clean connectors and blow dry before making connections. End users should also ensure that test equipment calibration is current. Power meter measurements connecting a test reference cable between the light source and power meter. The test reference cable has the same nominal fibre characteristics as the cable under test. The optical power present at the power meter is the reference power (Pi). Disconnect the test reference cable and connect the optical fibre cable under test between the light source and power meter using test jumpers. If possible, the test reference cable should be used as the input jumper cable for the test cable measurement. The test jumper fibre properties, such as core diameter and NA, should be nominally equal to the fibre properties of the cable being tested. The optical power present at the power meter is test power (P2). Test personnel use P-i and Pz to calculate the cable assembly link loss. The cable assembly link loss (BCA) of optical fibre installed with connectors or terminations on both ends is BCA=(P1=P2)dB The cable assembly link loss should always be less than the specified link loss for that particular link. Besides measuring individual cables, test personnel measure the transmission loss of installed fibre optic cable plants. The transmission loss of fibre optic cable plants is measured using method B (multimode fibre) or (single mode fibre). The procedure measures the internal loss of the cable plant between points A and B, plus two connection losses. Figure 4.53 (A) illustrates the method for measuring the reference power (P 1). Figure 4.53 (B) shows the final test configuration for measuring the cable plant test power (P 2).

Figure 4.53: Methods for measuring the reference power (P1). The procedure is exactly the same as described for measuring the link loss of an individual cable assembly. The total optical loss between any two termination points, including the end terminations, of the optical fibre cable plant link is measured. The measured cable plant link loss should always be less than the specified cable plant link loss. Test personnel should conduct cable assembly link loss, and cable plant transmission loss measurements in both directions and at each system operational wavelength. By performing

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Part 66 Training Syllabus Module 2 Physics these measurements in each direction, test personnel can better characterize cable and link losses. Unlike optical time-domain reflectometry, bi-directional readings are always possible when performing power meter measurements. In power meter measurements, by definition, end users have access to both ends of the cable or cable plant. Module 2.5 Wave Motion and Sound

Wave Motion Transverse and Longitudinal Waves Mechanical waves can be classified as transverse or longitudinal according to how they travel. Both types of wave can be demonstrated using a slinky (a long steel spring). The transverse wave occurs when the coils move at right angles to the direction of motion of the wave, with the motion along the length of the slinky. To produce a transverse wave, the slinky is rested on a flat surface and one end is moved from side to side, setting up the oscillation and hence the traveling wave.

Figure 5.1: Transverse and longitudinal waves produced on a slinky The end of the slinky can also be moved in and out along its axis. The coils undergo compression, followed by rarefaction when the coils open out. Displacement of the coils is now along the axis of the spring.

Progressive and Stationary Waves When we start a wave in the slinky, either transverse or longitudinal, we can watch it travel from one end to the other. Because it progresses along the slinky it is called a progressive wave. However, if the far end of the slinky is fixed, waves are reflected back. These can combine with the next waves which are traveling forwards. At the right combination of frequency and speed, the waves traveling in opposite directions can produce a stationary or standing wave. Here we shall consider both types.

The Wave Formula There are many types of waves: light waves, sound waves, radio waves, cosmic rays, x-rays, communication waves, waves on cords, etc. In our first discussion of waves, we will deal with that type which is called just "wave", that is, a water wave. Let us assume that a stone is thrown into the middle of a large, calm pond on a day when there is no wind. If there is a perpendicular plane surface cutting the water surface through a point where the stone hits the water, an observer would see the water surface disturbed in such a way that a curve would be visible. This curve would have a shape as shown in figure 5.2. In figure 5.2 it is important to note that the pattern of crests and troughs is moving. If the stone hits the water Figure 5.2: Waveform dimensions surface at the point (P), the pattern is moving to the right in the diagram above. Of course, the entire pattern is moving out from point (P) in all directions, but we are looking in only one direction. We should also note that the pattern is moving with a definite speed, called the wave speed (v).

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Part 66 Training Syllabus Module 2 Physics The amplitude (A) of the wave is the greatest displacement from the rest position. The amplitude is shown in figure 5.2. Another distance that we will need in our discussion of waves is the wavelength, λ. (Greek letter lambda). The wavelength is defined as the distance from one point on the wave pattern to the next point in a similar position. The distance from the top of a crest to the top of the next crest is a wavelength. Also the distance from the bottom of one trough to the bottom of the next trough is also the same distance, one wavelength. The distance λ, is also shown in the diagram. Let us next consider sinusoidal wave motion impressed on a very long flexible cord by an oscillating body. Assume that the oscillating body is a sphere attached to a vertical spring. After the spring has been oscillating for some time, the physical situation is as shown in figure 5.3. The frequency (f) of the oscillating body is defined as the number of complete oscillations in one second. Frequency is expressed in cycles/sec, or Hertz. The period (T) is defined as the time for one complete oscillation. It is expressed in seconds. Let us suppose that the oscillating body completes 6 oscillations in one second. It follows that the time for one oscillation is one-sixth of a second. In this case: F=6Hz and T=

Figure 5.3: A waveform produced on a piece of string by a mass oscillating on the end of a spring From this example we see that f and T are reciprocals of each other. T=

and f =

We next seek a relationship between wave speed (v), frequency (f), and wavelength (λ,). We note that the wave moves forward a distance of one wavelength in a time of one period. Of course, the wave moves with speed (v). Since the distance equals the speed times the time, we can write the equation: λ=νΤ From this equation, we have: ν And finally: fλ = ν

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Part 66 Training Syllabus Module 2 Physics EXAMPLE: (a) A body oscillates with a frequency of 8 Hz, and sends out a wave having a wavelength of 0.2 ft. What is the speed of the wave? V = (8 cycles/sec.) (0.2 ft.) = 1.6 ft./sec. (b) What is the wavelength of a wave moving with a speed of 5 ft./sec. If the frequency of the oscillating body which is the source of the wave is 12 Hz? λ=

=

= 0.417 ft

(c) An observer times the speed of a water wave to be 2 ft./sec. and notes that the wavelength is 0.5 ft. What is the frequency of the disturbance that gives rise to this wave? f= =

4 cycles/sec. = 4 Hz ν=f λ

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Resonance In the case of water waves and in the case of waves on a very long cord, we were able to neglect waves that were reflected back along the medium. We now must consider reflected waves. The most common example is the case of waves originating in a disturbance impressed on a cord or string of a definite length. Many musical instruments depend on such vibrations. If a sinusoidal disturbance is impressed on a very long cord a sinusoidal wave travels continuously along the cord. However, if the sinusoidal wave meets a fixed end, a reflected wave moves back along the cord. The wave patterns which are observed are called the normal modes of vibration of the cord. In figure 5.4. the length of the cord is L. The wavelength in the various modes of vibration are X. The n is the index of the mode. In the equations which follow, n has an integral value, that is n = 1 , 2, 3, 4. We can write a general relation as follows: λn = The vibration where n = 1 is called the fundamental mode of vibration of the body. The other vibrations are called overtone vibrations. Every body which can vibrate has a certain fundamental mode of vibration of a definite frequency. If this frequency is impressed on the body, it will vibrate with a relatively large amplitude. We say that the body is vibrating in resonance with the impressed frequency. Figure 5.4: Normal modes of vibration Problems 1. A water wave has a wavelength of 0.9 ft. and the wave speed is 4.5ft/sec. What is the frequency of the disturbance setting up this wave? 2. A wave on a cord is set up by a body oscillating at 12 Hz. The wavelength is 0.25ft. What is the wave speed? 3. A water wave is set up by a source oscillating at 12 Hz. The speed of the wave is 24ft/sec. What is the wavelength?

Answers 1. 5 Hz 2. 3ft/sec. 3. 2ft.

Sound Sound waves are usually defined as pressure waves in air or in some other material medium. Sound waves originate in some vibrating body such as the oscillation of a person's vocal cords or the periodic rotation of a plane's propeller. As the source of sound vibrates, the air surrounding the source is periodically compressed and rarefied (made less dense). This periodic change in the atmospheric pressure moves forward with a definite speed of propagation called the "speed of sound".

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The speed of sound in air is dependent on the temperature of the air. This is not surprising since the molecules of air move faster in their random motion if the temperature is higher. Thus we should expect these pressure waves to move somewhat more rapidly in warmer air. The speed of sound in air is approximately 331.5 m/s at 0°C At an air temperature of 20°C, the speed of sound increases to 344 m/s If an ear and its eardrum are in the vicinity of a sound wave, the air which strikes that eardrum has a periodically changing atmospheric pressure. If the frequency of the sound is middle C (256 Hz), and the 2 atmospheric pressure that day is 14.7 Ibs/in , 256 times each second the air pressure is slightly above 14.7 2 2 Ibs/in and 256 times each second the pressure is slightly below 14.7 Ibs/in it should be emphasized that "slightly" means very small. The human ear is a remarkably sensitive instrument. It can detect air pressure 2 variations as small as about 0,000000005 lbs./in. Sound travels faster in liquids, and even faster still, in solids.

Intensity of Sound For those working in the aviation industry it is important to understand something regarding the intensity of a sound wave. The intensity level (IL) of sound waves is measured in a unit called the decibel (after Alexander Graham Bell). The equation is: IL = 10 log n

We also review that the log 10 = n EXAMPLE: 2 The intensity of a given sound is 10~ Watts/rn . What is the intensity level (IL) in decibels? -5

IL=10 Log

12

= 10 log ( 10 )(10 ) 7 =10 log 10 IL= 10(7)= 70 db

INTENSITY LEVELS OF SOME COMMON SOUNDS SOUND Rocket Engine

2

INTENSITY (W/m ) 6 10

INTENSITY LEVEL (db) 180

3

150

-1

110

-3

90

-4

80

-5

70

-6

60

-8

40

-10

20

Jet Plane at Takeoff

10

Amplified Rock Music

10

Riveting

10

Elevated Train

10

Busy Street Traffic

10

Conversation in Home

10

"Quiet" Radio in Home

10

Whisper

10

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10

Hearing Threshold

10

-11

-12

10 0

It should be noted that 120 db is the "threshold of pain". Sound of this intensity is painful to the normal ear. If the ear is continuously subjected to sound of this intensity, ear damage and hearing loss can result. Those who work in the aviation industry should take precautionary measures and wear ear protectors. The intensity of sound decreases inversely with the square of the distance from the source of sound. Therefore, doubling the distance from a source of sound decreases the intensity to one-fourth of the previous value. A worker who is suddenly subjected to a very intense sound with unprotected ears should move as quickly as possible away from the sound of the source.

Table 5.1: Intensity levels of some common sounds

Sound Waves and Resonant Vibrations Intense sound waves can cause resonant vibrations in pieces of equipment. There is a fundamental mode of vibration and a set of overtone vibrations (multiples of the fundamental) for any body that can vibrate. The frequencies of these vibrations are all natural frequencies for the given body. Vibrations of moving parts of equipment are often caused by "sympathetic vibrations" to some impressed sound wave. The Italian tenor, Enrico Caruso, had a powerful voice. Wine glasses have a natural frequency of vibration. As an attention getter at parties. Caruso used to sing the resonant note of a wine glass and cause the glass to vibrate with such amplitude that it would shatter! Try it sometime!

Constructive and Destructive Interference When two sinusoidal waves superimpose, the resulting waveform depends on the frequency (or wavelength) amplitude and relative phase of the two waves. If the two waves have the same amplitude A and wavelength the resultant waveform will have an amplitude between 0 and 2A depending on whether the two waves are in phase or out of phase.

Figure 5.5: Constructive and destructive interference

Noise Cancelling Headphones Noise-cancelling headphones reduce unwanted ambient sounds (i.e., acoustic noise) by means of active noise control. Essentially, this involves using a microphone, placed near the ear, and electronic circuitry

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Part 66 Training Syllabus Module 2 Physics which generates an "anti-noise" sound wave with the opposite polarity of the sound wave arriving at the microphone. This results in destructive interference, which cancels out the noise within the enclosed volume of the headphone. Keeping noise low at the ear makes it possible to enjoy music without raising the volume unnecessarily. It can also help a passenger sleep in a noisy vehicle such as an airliner.

Figure 5.6: Noise cancelling headphones Another effect of constructive and destructive interference is the "dead zones" produced when two identical waves emanate from separate locations, as shown in figure 5.7. Here, water waves are progressing from two points, causing destructive interference where a peak from one source coincides with a trough from the other source, the effect being to cancel each other at those points. Striations (or "rays") of undisturbed water result.

Figure 5.7: Destructive interference causing "dead zones"

Reflected Waves Let us look in more detail at how to set up standing waves. We set off a short wave on a slinky which has been firmly fixed at its far end. Assume that the wave consists of one and a half wavelengths. The wave travels along the slinky until it reaches the far end. At this point, the wave can travel no further forwards and is reflected back. This means that the velocity has changed sign. In addition, the phase of the wave has changed. If the displacement of the forward wave is

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Part 66 Training Syllabus Module 2 Physics upwards at the instant of time when it reaches the far end, then its displacement is downwards on reflection. This makes sense. At the fixed end, the displacement of the incoming and outgoing waves sum to zero. This must be so because there can be no displacement of the string at the fixed point. The reflected wave is out 1 of phase by it. It passes back 'through the forward wave (think how ripples can pass through each other on the surface of a pond). Where the two waves overlap, the displacement of the slinky is the sum of the two waves. But, eventually, we see the reflected wave emerge complete and pass back along the slinky. The frequency, velocity and wavelength of the wave all remain the same in reflection. If no energy is lost at the far end, the amplitude of the reflected wave equals that of the incoming one. The phase difference of n which we have identified and is crucial to the setting up of standing waves. When waves pass through each other, the displacement at any point is the sum of the individual displacements of the two waves passing in opposite directions.

Producing Stationary Waves Stationary waves are set up in stringed instruments such as a guitar. What we see is the string vibrating from side to side. At the moment that the string is plucked, a progressive transverse wave is set up traveling out from that point. It meets the fixed end of the string and is reflected back. The amplitudes of the two waves add together as they meet. The string vibrates naturally at certain frequencies because it is fixed at both ends. When the outgoing and reflected waves are added together subject to this condition, a stationary wave is set up in the string. If the string is plucked centrally we get the fundamental mode (shape of wave). In this case, the string vibrates with maximum displacement at the central position (called the antinode) and the displacement falls away to zero at the two ends (called nodes).

Figure 5.8: Stationary waves and harmonics When a string on an instrument is plucked, vibrations, that is, waves, travel back and forth through the medium being reflected at each fixed end. Certain sized waves can survive on the medium. These certain sized waves will not cancel each other out as they reflect back upon themselves. These certain sized waves are called the harmonics of the vibration. They are standing waves. That is, they produce patterns which do not move. On a medium such as a violin string several harmonically related standing wave patterns are possible. The first four of them are illustrated above. It is important to understand that for any one given medium fixed at each end only certain sized waves can stand. We say, therefore, that the medium is tuned. The first pattern has the longest wavelength and is called the first harmonic. It is also called the fundamental. The second pattern, or second harmonic, has half the wavelength and twice the frequency of the first harmonic. This second harmonic is also called the first overtone. This can get confusing with the second member of the harmonic group being called the first member of the overtone group. The third harmonic, or pattern, has one third the wavelength and three times the frequency when compared to the first harmonic. This third harmonic is called the second overtone. The other harmonics follow the obvious pattern regarding wavelengths, frequencies, and overtone naming conventions described in the above paragraph.

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Depending upon how the string is plucked or bowed, different harmonics can be emphasized. In the above animation all harmonics have the same maximum amplitude. This is for purposes of illustration. Actually, the higher harmonics almost always have maximum amplitudes much less than the fundamental, or first harmonic. It is the fundamental frequency that determines the note that we hear. It is the upper harmonic structure that determines the timber of the instrument.

Beats Suppose we tune two strings of a guitar to vibrate at almost, but not quite, the same frequency. Plucked simultaneously, the volume of the sound produced by them appears to rise and fall continuously. This rise and fall has a fixed frequency called the beat frequency. What is happening is that the sound waves produced by the two guitar strings interfere and our ears detect the variation of the resultant intensity. Maximum intensity is heard when the waves add together (interfere constructively) and minimum intensity is heard when the waves cancel each other out (interfere destructively). We can see what is happening by adding together the two separate waves as shown in the diagram below. The resultant, obtained by the principle of superposition, is shown.

Figure 5.9: A beat created by two sound waves of similar (but not the same) frequencies

Supersonic Speed and Mach Number Jet planes can travel at speeds greater than the speed of sound. In this case, we have a source of sound, the plane, moving at a greater speed than the sound itself. The pressure waves of the sound all "pile up" and a very strong V-shaped pressure "bow-wave" is produced. A sonic boom results as this strong pressure ridge reaches the earth. The Mach number is the ratio of the speed of the plane (VQ) to the speed of sound (v). If a plane is travelling at 1.000 MPH and the local speed of sound is 750 MPH. the Mach number is calculated in the following way: Mach Number =

=

=

= 1.25

We say that the plane is travelling at Mach 1.25.

The Doppler Effect The "Doppler effect" is named after Christian Doppler (1803-1853), the American physicist who first named the effect. The effect is present for all wave motion. However, we will describe it for sound waves since it is most easily understood for a case where it can be observed (heard might be a better word). Whenever you have stood on a railway platform and a train blows its whistle as it approaches, passes, and recedes, you have heard the Doppler effect. In this case, the sound suddenly changes from a higher pitch (frequency) as the source of sound approaches to a lower pitch as the source of sound recedes from your ear at rest on the station platform. The change in pitch occurs at the instant the train passes. Before this

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Part 66 Training Syllabus Module 2 Physics instant the source of sound was approaching your ear and after this instant, the source of sound is receding from your ear.

Figure 5.10: Effect on frequency of a stationary and moving sound source There is another problem to be considered. Suppose that the source is at rest and the ear is moving. Consider the figure 5.11. As the ear moves to the left, it picks up more waves than it normally would if it were at rest. If the observer moves away from the source, the ear picks up less waves than it would if it were at rest. As a conclusion, note that the ear hears a higher frequency if source and observer approach each other. Also, the ear hears a lower frequency if the source and observer recede from each other.

Figure 5.11: Doppler effect caused by a stationary sound source and moving ear

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