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Training Handbook



Basic Aircraft Maintenance Electrical Avionics Module 2 - Physics



Manual No. : BCT-0012/A2



For Training Purpose Only



Rev. 0 : Aug 19, 2015



Basic Aircraft Maintenance Training Manual Module 2 – Physics



Module 2 Physics for



Basic Aircraft Maintenance



Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Module 2 Chapters 2.1 Matter 2.2 Mechanics 2.3 Thermodynamics 2.4 Optics (Light) 2.5 Wave Motion and Sound



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Module 2 Physics 2.1 Matter



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Knowledge Levels — Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows: LEVEL 1 A familiarisation with the principal elements of the subject. Objectives: The applicant should be familiar with the basic elements of the subject. The applicant should be able to give a simple description of the whole subject, using common words and examples. The applicant should be able to use typical terms. LEVEL 2 A general knowledge of the theoretical and practical aspects of the subject. An ability to apply that knowledge. Objectives: The applicant should be able to understand the theoretical fundamentals of the subject. The applicant should be able to give a general description of the subject using, as appropriate, typical examples. The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. The applicant should be able to read and understand sketches, drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using detailed procedures. LEVEL 3 A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: The applicant should know the theory of the subject and interrelationships with other subjects. The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. The applicant should understand and be able to use mathematical formulae related to the subject. The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Table of Contents Module 2.1 – Matter ______________________________________________________________________________________________ 5 The Nature of Matter ______________________________________________________________________________________________ 5 The Components of Atoms _________________________________________________________________________________________ 6 Periodic Table of the Elements ______________________________________________________________________________________ 7 Chemical Definitions ______________________________________________________________________________________________ 8 The Electronic Structure of Atoms ____________________________________________________________________________________ 11 Chemical Bonding ________________________________________________________________________________________________ 19 States of Matter __________________________________________________________________________________________________ 29 Changes between States ___________________________________________________________________________________________ 30



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Module 2.1 Enabling Objectives



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Objective



Reference



Matter Nature of matter: the chemical elements, structure of atoms, molecules Chemical compounds States: solid, liquid and gaseous Changes between states



2.1



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Module 2.1 - 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics The Components of Atoms Atoms are the smallest particles of matter whose properties we study in Chemistry. However from experiments done in the late 19th and early 20th century it was deduced that atoms were made up of three fundamental sub-atomic particles (table 1.1)



Particle



Relative mass



Electrical charge



Comments



Neutron



1



0 (zero)



In the nucleus



Proton



1



+1 (positive)



In the nucleus



Electron



1/l850



-1 (negative)



Arranged in energy levels or shells around the nucleus



Table 1.1: The sub-atomic components of atoms Figure 1.1 gives some idea on the structure of an atom.



Figure 1.1 : The structure of an atom



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Periodic Table of the Elements



Figure 1.2: 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) Manual No. : BCT-0012/A2 Electrical Avionics



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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. 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), Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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). Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



are 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.



are the two isotopes of helium with mass numbers of 3 and 4, with 1 and 2 neutrons respectively but both have 2 protons.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



are 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. Ionization 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. The 1st shell (K) has a maximum of 2 electrons The 2nd shell (L) has a maximum of 8 electrons The 3rd shell (M) has a maximum of 18 electrons The 4th shell (N) has a maximum of 32 electrons Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 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 electrons held in the orbit is given as 2n2. 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 1.3 : The atomic structure of Helium and Neon Figure 1.4: Electron shell (orbit) designation



Examples: diagram, symbol or name of element (Atomic Number = number of electrons in a neutral atom), shorthand electron arrangement



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Basic Aircraft Maintenance Training Manual Module 2 – Physics On Period 1



Figure 1.5 : Electron arrangement of Hydrogen and Helium



On Period 2



Figure 1.6 : Electron arrangement of Lithium, Berylium, Boron, Carbon, Nitrogen, Oxygen, Fluorine and Neon



On Period 3



Figure 1.7 : Electron arrangement of Sodium, Magnesium, Aluminium, Silicon, Phosphorus, Sulphur, Chlorine and Argon



On Period 4



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Figure 1.8 : 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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Table 1.2: Electrons in outer shells of some common elements



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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



0



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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 H20. Finally, Fluorine and Chlorine - 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. 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. Atomsthat are one electron short of a full shell are more reactive Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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



Atomic No. 1 2 3 4 5 6 7 8 9 10 11 12 13



Element Hydrogen Helium Lithium Beryllium Boron Carbon Nitrogen Oxygen Fluorine Neon Sodium Magnesium Aluminium



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



L



1 2 3 4 5 6 7 8 8 8 8



Electrons per Shell M N O



1 2 3



P



Q



Atomic No. 53 54 55 56 57 58 59 60 61 62 63 64 65



Element Iodine Xenon Cesium Barium Lanthanum Cerium Praseodymium Neodymium Promethium Samarium Europium Gadolinium Terbium



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K 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



Electrons per Shell M N O 18 8 7 18 18 8 18 8 8 18 8 8 18 8 9 18 19 9 18 20 9 18 21 9 18 22 9 18 23 9 18 24 9 18 25 9 18 26 9



P



Q



1 2 2 2 2 2 2 2 2 2 2



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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



Silicon Phosphorus Sulphur Chlorine Argon Potassium Calcium Scandium Titanium Vanadium Chromium Manganese Iron Cobalt Nickel Copper Zinc Gallium Germanium Arsenic Selenium Bromine K ton Rubidium Strontium Yttrium Zirconium Niobium Molybdenum Technetium Ruthenium Rhodium Palladium Silver



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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



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



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



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



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



1 2 2 2



0 1



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 Protactinium Uranium Neptunium Plutonium Amerium Curium Berkelium Californium Einsteinium



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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



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



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



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



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



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



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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 48 49 50 51 52



Cadmium Indium Tin Antimony Tellurium



2 2 2 2 2



8 8 8 8 8



18 18 18 18 18



18 18 18 18 18



2 3 4 5 6



100 101 102 103



Fermium Mendelevium Nobelium Lawrencium



2 2 2 2



8 8 8 8



8 18 18 18



32 32 32 32



29 30 31 32



9 9 9 9



2 2 2 2



Table 1.3: Electrons per shell



Chemical Bonding Adhesion and Cohesion '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 1.9: (Atomic Number) and electron arrangement



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 nonmetallic elements in a molecule. or IONIC BONDING - By one atom transferring electrons to another atom. The atom losingelectrons 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 combine to form



where both atoms have a pseudo helium structure of 2 outer electrons around each atom.



Example 2 - 2 chlorine atoms (2.8.7) form the molecule of the element chlorine Cl2



combine to form



where both atoms have a pseudo neon or argon structure



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



combine to form



where hydrogen is electronically like helium and chlorine like neon or argon



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 H20



combine to form



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so that the hydrogen atoms are electronically like helium and the oxygen atom



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becomes like neon and 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



combine form



like neon or argon. The molecule can be shown as



so that the hydrogen atoms are electronically like helium and nitrogen atom become



with the 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



combine to form



so that hydrogen atoms are electronically like helium and the nitrogen



atoms 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 The molecules has one double covalent bond Manual No. : BCT-0012/A2 Electrical Avionics



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Example 8- One atom of carbon (2.4) combines with two atoms of oxygen (2.6) to form carbon dioxide CO2 The molecules can be shown as



with two carbon = oxygen double covalent bonds.



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



with one carbon = carbon double bond and four carbon-hydrogen single covalent bonds.



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 andboiling 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 poorconductors 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Figure 1.10 : A plane of Carbon atoms from 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 Group6orGroup7 (The Halogens) Example 1 - A Group 1 metal + a Group 7 non-metal e.g. sodium + chlorine →sodium chloride NaCI or ionic formula Na+CIIn 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)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Example 2 - A Group 2 metal + a Group 7 non-metal e.g. magnesium + chlorine →magnesium chloride MgCI2 or ionic formula Mg2+(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) + 2Cl (2.8.7) →Mg2+ (2.8) 2Cl (2.8.8)



(* 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 or ionic formula AI3+(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) →Al3+ (2.8) 3F-(2.8.8)



Example 4 - A Group 1 metal + a Group 6 non-metal e.g. potassium + oxygen →potassium oxide K2O or ionic formula (K+)2O2In 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) Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 2K (2.8.8.1) + O (2.6) → 2K+ (2.8) O2-(2.8)



Example 5 - A Group 2 metal + a Group 6 non-metal e.g. calcium + oxygen →calcium oxide CaO or ionic formula Ca2+O2In 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) Ca (2.8.8.2) + O (2.6) →Ca2+ (2.8.8) O2-(2.8)



Example 6 - A Group 3 metal + a Group 6 non-metal e.g. aluminium + oxygen →aluminium oxide Al2O3or ionic formula (Al3+)2 (O2-)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) →2AI3+(2.8) 3O2- (2.8)



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Basic Aircraft Maintenance Training Manual 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 1.11 : Sodium Chloride lattice structure



Bonding in Metals The crystal lattice of metals consists of ions, NOT atoms. The outer electrons (-) from the original metal 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Figure 1.12: ‘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. Non-metallic 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 Changes of state are physical changes in matter. They are reversible changes that do not involve changes in matter’s chemical makeup or chemical properties. Common changes of states include melting, freezing, sublimation, deposition, condensation, and vaporization. Energy is always involved in changes of state. Matter either loses or absorbs energy when it changes from one state to another. For example, when matter changes from a liquid to a solid, it loses energy. The opposite happens when matter changes from a solid to a liquid. For a solid to change to a liquid, matter must absorb energy from its surroundings. When energy is added, the particles of matter move more slowly. The amount of energy in matter can be measured with a thermometer. That’s because a thermometer measures temperature and temperature is the average kinetic energy of the particles of matter. a.



Changes between solids and liquids: The process in which a liquid changes to a solid is called freezing. Energy is taken away during freezing. The temperature at which a liquid changes to a solid is its freezing point. The process in which a solid changes to a liquid is called melting. Energy is added during melting, and the melting point is the temperature at which a solid changes to a liquid.



b.



Changes between liquids and gases: If water gets hot enough, it starts to boil. Bubbles of water vapor form in boiling water. This happens as particles of liquid water gain enough energy to completely overcome the force of attraction between them and change to the gaseous state. The bubbles rise



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Basic Aircraft Maintenance Training Manual Module 2 – Physics through the water and escape from the pot as steam. The process in which a liquid boils and changes to a gas is called vaporization. The temperature at which a liquid boils is its boiling point. A liquid can also change to a gas without boiling. This process is called evaporation. It occurs when particles at the exposed surface of a liquid and escape into the air. This happens faster at warmer temperatures. Energy is added during evaporation and vaporization. The process in which a gas changes to a liquid is called condensation. Energy is removed during condensation, changing a gas to a liquid.



Figure 1.13. Changes in the state of matter



c.



Changes between solids and gases: Solids that change to gases generally first pass through the liquid state. However, sometimes solids change directly to gases and skip the liquid state. The reverse can also occur. Sometimes gases change directly to solids. The process in which a solid changes directly to a



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Basic Aircraft Maintenance Training Manual Module 2 – Physics gas is called sublimation. Energy is added during sublimation. It occurs when the particles of a solid absorb enough energy to completely overcome the force of attraction between them. Dry ice (solid carbon dioxide, CO2) is an example of a solid that undergoes sublimation. The opposite of sublimation is deposition. Energy is taken away during sublimation. This is the process in which a gas changes directly to a solid without going through the liquid state. It occurs when gas particles become very cold. For example, when water vapor in the air contacts a very cold windowpane, the water vapor may change to tiny ice crystals on the glass (frost).



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Module 2 Physics 2.2 Mechanics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Knowledge Levels — Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows: LEVEL 1 A familiarisation with the principal elements of the subject. Objectives: The applicant should be familiar with the basic elements of the subject. The applicant should be able to give a simple description of the whole subject, using common words and examples. The applicant should be able to use typical terms. LEVEL 2 A general knowledge of the theoretical and practical aspects of the subject. An ability to apply that knowledge. Objectives: The applicant should be able to understand the theoretical fundamentals of the subject. The applicant should be able to give a general description of the subject using, as appropriate, typical examples. The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. The applicant should be able to read and understand sketches, drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using detailed procedures. LEVEL 3 A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: The applicant should know the theory of the subject and interrelationships with other subjects. The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. The applicant should understand and be able to use mathematical formulae related to the subject. The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Table of Contents Module 2.2 Mechanics ____________________________________________________________________________________________ 6 Introduction of Measurements ___________________________________________________________________________________ 6 Statics ________________________________________________________________________________________________________ 7 Force ______________________________________________________________________________________________________ 7 Weight ____________________________________________________________________________________________________ 7 Moment and Couples ________________________________________________________________________________________ 9 Vectors ____________________________________________________________________________________________________ 12 Centre of Gravity ____________________________________________________________________________________________ 12 Stress, Strain, and Hooke’s Law _________________________________________________________________________________ 13 Young Modulus _____________________________________________________________________________________________ 17 Bulk Modulus _______________________________________________________________________________________________ 20 Nature and Properties of Solids, Liquids and Gas ___________________________________________________________________ 22 Pressure and buoyancy in fluids ________________________________________________________________________________ 25 Kinetics _______________________________________________________________________________________________________ 39 Linear Motion _______________________________________________________________________________________________ 39 Rotational Motion ____________________________________________________________________________________________ 47 Periodic Motion______________________________________________________________________________________________ 55 Simple Machines and the Principle of Work _______________________________________________________________________ 64 Dynamics ______________________________________________________________________________________________________ 77 Newton’s First Laws __________________________________________________________________________________________ 77 Newton’s Second Laws ________________________________________________________________________________________ 78 Newton’s Third Laws__________________________________________________________________________________________ 79 Work, Energy and Power ______________________________________________________________________________________ 83 Motion in a Circle ____________________________________________________________________________________________ 90 Momentum _________________________________________________________________________________________________ 94 Torque _____________________________________________________________________________________________________ 100 Extensions _________________________________________________________________________________________________ 101 Gyroscope __________________________________________________________________________________________________ 105 Fluid Dynamics _________________________________________________________________________________________________ 123 Density and Specific Gravity ____________________________________________________________________________________ 125 Viscosity____________________________________________________________________________________________________ 128 Drag and Streamlining ________________________________________________________________________________________ 129 Streamlining ________________________________________________________________________________________________ 132 Bernoulli’s Principle __________________________________________________________________________________________ 133 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Module 2.2 Enabling Objectives Objective



Reference



Level



Mechanics Statics Forces, moments and couples, representation as vectors Centre of gravity Elements of theory of stress, strain and elasticity: tension, compression, shear and torsion compression, shear and torsion



2.2 2.2.1



2



Nature and properties of solid, fluid and gas Pressure and buoyancy in liquids (barometers) Kinetics 2.2.2 Linear movement: uniform motion in a straight line, motion under constant acceleration (motion under gravity); Rotational movement: uniform circular motion (centrifugal/ centripetal forces); Periodic motion: pendular movement; Simple theory of vibration, harmonics and resonance; Velocity ratio, mechanical advantage and efficiency Dynamics 2.2.3 Mass (a) Force, inertia, work, power, energy (potential, kinetic and total (b) energy), heat, efficiency Momentum, conservation of momentum Impulse Gyroscopic principles Friction: nature and effects, coefficient of friction (rolling resistance) Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Fluid dynamics Specific gravity and density Viscosity, fluid resistance, effects of streamlining; Effects of compressibility on fluids Static, dynamic and total pressure: Bernoulli’s Theorem, venturi



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2.2.4 (a)



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2.2. Mechanics Introduction of Measurements The three basic quantities which require units of measurement are mass (weight), length (distance), and time. The metric system is used as basic measuring units, the centimeter to measure length, the gram to measure mass – the quantity of matter in a body, and the second to measure time. The English system uses different units for the measurement of mass and length. The pound is the unit of weight; the foot is used to measure length. The second is used to measure time as in the metric system. The units of one system can be converted to units in the other system by referring to a table similar to that shown in Table 2.1. Metric System



English Systems



Centimeter Length (distance)



Weight (mass)



Time



1 cm = 10 mm 1 dm = 10 cm 1 m = 100 cm 1 km = 1000 m Gram 1 g = 1000 mg 1 kg = 1000 g Second (same as English system)



Foot 1 foot = 12 inches 1 yard = 3 ft. 1 mile = 5,280 ft.



Equivalents 1 in = 2.54 cm 1 ft. = 30.5 cm 1 m = 39.37 in 1 km = 0.62 mile



Pound 1 lb. = 16 oz. 1 ton = 2,000 lbs.



1 lb. = 453.6 gr 1 kg = 2.2 lbs. Time same for both system



Table 2.1. Comparison of metrics and English systems of measurement



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 2.2.1. Statics 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 (lb.) in the English system and the Newton (N) in the metric system. The relationship between the metric and English units is given by the conversion factor: 1 lb. = 4.45 N. Weight Before getting into weight, we all know that weight is very dependent with mass. Mass is a measure of the quantity of matter in an object. In other words, how many molecules are in the object, or how many atoms, or to be more specific, how many protons, neutrons, and electrons. The mass of an object does not change regardless of where you take it in the universe, and it does not change with a change of state. The only way to change the mass of an object is to add or take away atoms. The acceleration due to gravity here on earth is 32.2 feet per second per second (32.2 fps/s). An object weighing 32.2 pounds (lbs.) here on earth is said to have a mass of 1 slug. A slug is a quantity of mass that will accelerate at a rate of 1 ft./s 2 when a force of 1 pound is applied. In other words, under standard atmospheric conditions (gravity equal to 32.2) a mass of one slug is equal to 32.2 lb. Weight is a measure of the pull of gravity acting on the mass of an object. The more mass an object has, the more it will weigh under the earth’s force of gravity. Because it is not possible for the mass of an object to go away, the only way for an object to be weightless is for gravity to go away. We view astronauts on the space shuttle and it appears that they are weightless. Even though the shuttle is quite a few miles above the surface of the earth, the force of gravity has not gone away, and the astronauts are not weightless. The astronauts and the space shuttle are in a state of free fall, so relative to the shuttle the astronauts appear to be weightless. Mathematically, weight can be stated as follows: Weight = Mass × Gravity



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



What is the mass of a body having a weight of 45 N? (4.59 kg)



2.



What is the weight of a body having a mass of 23 kg? (225 N)



3.



What is the mass of a body having a weight of 350 Ibs.? (10.9 slugs)



4.



What is the weight of a body having a mass of 23.6 slugs? (755 lbs.)



5.



What is the weight (in Ibs.) of the corn flakes in a box where the mass is listed as 680 g? (1.45 lbs.)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Moment and Couples Moment The moment (or torque) of a force about a turning point is the force multiplied by the perpendicular distance to the force from the turning point. Moment = F d 



F = the force in Newton (N) or pounds (lbs.) for English system







d = perpendicular distance from the point of force in meter (m) or feet (ft.) for English system.



Example; A 10 N force acts at a perpendicular distance of 0.50 m from the turning point. What is the moment of the force?



Figure 2.1. Example of simple moment



Then, in order to find the moment of the force we use the formula; Moment = F x d =10 x 0.5 = 5.0 Nm The principle of moments. ” When an object is in equilibrium the sum of the anticlockwise moments about a turning point must be equal to the sum of the clockwise moments.” sum of anticlockwise moments = sum clockwise moments



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Example;



Figure 2.2. Moment of force with two different forces.



sum of anticlockwise moments = sum clockwise moments F1 x d1 = F2 x d2 Other than that, we can see another example of moment with three different forces with three different distances that worked on an object.



Figure 2.3. Moment of force with three different forces.



sum of anticlockwise moments = sum clockwise moments F1 x d1 = (F2 x d2) + (F3 x d3)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Couples A couple is two equal forces which act in opposite directs on an object but not through the same point so they produce a turning effect. The moment of a couple is called the torque.The moment (or torque) of a couple is calculated by: Moment of couple = F x s 



F = the force in Newton (N) or pounds (lbs.) for English system







s = perpendicular distance between the forces in meter (m) or feet (ft.) for English system.



Figure 2.4. Moment of Couple.



Even though the expressions are similar for both the moment of a force and of a couple, the physics behind are different. Only one force is taken into account, though there are two forces in the couple. The turning effect of one force is countered by the other. Therefore, only the difference in the distance from the considered point accounts for the net turning effect. Hence, the moment of the couple is a constant for any point on the plain of the couple. Whenever a force is applied to create a turning effect, in reality a torque is produced by a couple. For example, consider using a wrench to unscrew a bolt. When the force is applied to the end of the wrench arm, a force with the same magnitude is created at the bolt, which is the pivot in this case. These two equal and opposite forces create a couple, and the couple is generating the torque required to turn the bolt.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Difference between Moment and Couple •



Moment of force is the measure of turning effect of a force about a point. A couple consists of two equal and opposite forces acting with two different but parallel lines of action. Each force has its own moment.



•



Moment of a force is dependent on the distance from the pivot and the magnitude of the force while the moment of a couple is the net effect of the two moments of the forces. Moment of a couple is independent of the location of the point considered. It is constant throughout the plane.



Vectors Quantities like length, mass, and temperature that have magnitude (size) but not direction are called scalars. While quantities like velocity, acceleration, and force that have both magnitude and direction, are called vectors. Geometrically, we can picture a vector as a directed line segment, whose length is the magnitude of the vector and with an arrow indicating the direction. The direction of the vector is from its tail to its head.



Figure 2.5. The direction of a vector line segment.



Centre of Gravity Gravity is a force which is always present and is a pulling force in the direction of the center of the earth. The center of gravity is a geometric property of any object. The center of gravity is the average location of the weight of an object, or the point from which the object will balance.



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Figure 2.6. Centre of Gravity



Why do we need to learn about CG? To ensure the aircraft is safe to fly because the center-of-gravity must fall within specified limits established by the manufacturer. C of G range – C of G limits are specified longitudinal (forward and aft) and/or lateral (left and right) limits within which the aircraft's center of gravity must be located during flight. To evenly load the aircraft with equipment, passengers, baggage, cargo, fuel, etc. So that C of G range will not be exceeded to prevent aircraft unstable during flight. C of G in flight will change as the changes of fuel quantity, passengers’ movement, etc. Stress, Strain, and Hooke’s Law Introduction Structural integrity is a major factor in aircraft design and construction. No production airplane 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 airplane 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, Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. The five basic structural stresses to which aircraft are subject are: a.



Tension 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 airplane forward. The wings, tail section and fuselage, however resists that movement because of the airflow around the,. The result is a stretching effect on the airframe. Bracing wires in an aircraft are usually in tension.



Figure 2.7. Tension Stress



b.



Compression Compression is a force that tries to crush an object. An excellent example of compression is when a sheet metal airplane is assembled using the fastener known as a rivet. The rivet passes through a hole drilled in the pieces of aluminum, and then a rivet gun on one side and a bucking bar on the other apply a force. This applied force tries to crush the rivet and makes it expand to fill the hole and securely hold the aluminum pieces together. The ability of material to meet compression requirements is measured in pounds per square inch (PSI).



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Figure 2.8. Compression Stress



c.



Torsion Torsion is a twisting force. Because aluminium is used almost exclusively from 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 measure of strength in pounds per 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.



Figure 2.9. Torsion Stress



Figure 2.10 Turbofan engine, torque creating torsion in the shaft.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics d.



Shear When a shear stress is applied to an object, the force tries to cut or slice through, like a knife cutting through butter (see figure 2.12). For example of the application of shear stress, A clevis bolt, which is often used to secure a cable to a part of the airframe, has shear stress acting on it. As shown in Figure 2.11, a fork fitting is secured to the end of the cable, and the fork attaches to an eye on the airframe with the clevis bolt. When the cable is put under tension, the fork tries to slide off the eye by cutting through the clevis bolt. This bolt would be designed to take very high shear loads.



Figure 2.11. Clevis bolt, red arrows show opposing forces trying to shear the bolt.



Figure 2.12. Shear Stress



e.



Bending An airplane in flight experiences a bending force (shown by figure 2.14 below) on the wing as aerodynamic lift tries to raise the wing. This force of lift actually causes the skin on the top of the wing to compress and the skin on the bottom of the wing to be under tension. When the airplane is on the ground sitting on its landing gear, the force of gravity tries to bend the wing downward, subjecting the bottom of the wing to compression and the top of the wing to tension. [Figure 2.13] During the testing that occurs prior to FAA certification, an airplane manufacturer intentionally bends the wing up and down to make sure it can take the stress without failing.



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Figure 2.13 Airplane on the ground, wing under tension and compression



Figure 2.14. Bending Stress



Young Modulus All real “rigid” bodies are to some extent elastic, which means that we can change their dimensions slightly by pulling, pushing, twisting, or compressing them. To get a feeling for the orders of magnitude involved, consider a vertical steel rod 1 m long and 1 cm in diameter attached to a factory ceiling. If you hang a subcompact car from the free end of such a rod, the rod will stretch but only by about 0.5 mm, or 0.05%. Furthermore, the rod will return to its original length when the car is removed. If you hang two cars from the rod, the rod will be permanently stretched and will not recover its original length when you remove the load. If you hang three cars from the rod, the rod will break. Just before rupture, the elongation of the rod will be less than 0.2%. Although deformations of this size seem small, they are important in engineering practice (Whether a wing under load will stay on an airplane is obviously important.) Figure 2.15. shows two ways in which a solid might change its dimensions when forces act on it. In Fig. 2.15a, a cylinder is stretched. In Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Fig. 2.15b, a cylinder is deformed by a force perpendicular to its long axis, much as we might deform a pack of cards or a book. What the two deformation types have in common is that a stress, or deforming force per unit area, produces a strain, or unit deformation. In Fig. 2.15, tensile stress (associated with stretching) is illustrated in (a), shearing stress in (b). The stresses and the strains take different forms in the two situations of Fig. 2.15, but—over the range of engineering usefulness— stress and strain are proportional to each other.The constant of proportionality is called a modulus of elasticity, so that Stress = Modulus x Strain



Figure 2.15 (a) A cylinder subject to tensile stress stretches by an amount L. (b) A cylinder subject to shearing stress deforms by an amount of x, somewhat like a pack of playing cards would. All the deformations shown are greatly exaggerated.



In a standard test of tensile properties, the tensile stress on a test cylinder (like that in Fig. 2.16) is slowly increased from zero to the point at which the cylinder fractures, and the strain is carefully measured and plotted.



Figure 2.16. A test specimen used to determine a stress–strain curve . The change L that occurs in a certain length Lis measured in a tensile stress–strain test.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics The result is a graph of stress versus strain like that in Fig. 2.17. For a substantial range of applied stresses, the stress–strain relation is linear, and the specimen recovers its original dimensions when the stress is removed. If the stress is increased beyond the yield strength of the specimen, the specimen becomes permanently deformed, 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). If the stress continues to increase, where the force and the deformation are not proportional, but rather a small amount of increase in force produces a large amount of deformation. In this region, the specimen eventually ruptures as it begins to ‘neck down’, at a stress called the ultimate strength, where finally the specimen actually breaks.



Figure 2.17. A stress–strain curve for a steel test specimen, such as that of Fig. 2.16. The specimen deforms permanently when the stress is equal to the yield strength of the specimen’s material. It ruptures when the stress is equal to the ultimate strength of the material.



The point at which the elastic region ends is called the elastic limit, or the proportional limit. In actuality, these 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics longer holds). Although these two points are slightly different, we will treat them as the same in this course. For simple tension or compression, the stress on the object is defined as F/A, where F is the magnitude of the force applied perpendicularly to an area A on the object.



The strain, or unit deformation, is then the dimensionless quantity L/L, the fractional (or sometimes percentage) change in a length of the specimen. If the specimen is a long rod and the stress does not exceed the yield strength, then not only the entire rod but also every section of it experiences the same strain when a given stress is applied.



Because the strain is dimensionless, the modulus in previous formula above, has the same dimensions as the stress—namely, force per unit area. The modulus for tensile and compressive stresses is called the Young’s modulus and is represented in engineering practice by the symbol E.



Bulk Modulus The stress is also exists in the fluid pressure “p” on the object, where pressure is a force per unit area. The strain is V/V, where V is the original volume of the specimen and V is the absolute value of the change in volume. The corresponding modulus, with symbol B, is called the bulk modulus of the material. The object is said to be under hydraulic compression, and the pressure can be called the hydraulic stress.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



A steel bolt with a cross-sectional area of 0.1 in2 and a length of 6.0" is subjected to a force of 580 Ibs. What is the increase in length of the bolt? (0.0012 in) (Hint: Find the stress. Then use Young's Modulus of steel to find the strain. From the strain find the extension)



2.



An iron body of volume 145 in3 is subjected to a pressure of 500 Ib/in2. What is the decrease in volume of this body? (0.005 in3)



3.



A copper rod has a cross-sectional area of 0.04 in2 and a length of 24". What longitudinal force must be applied to cause this rod to stretch by 0.0024 in? (64 lbs)



4.



An aluminium brace inside a wing of a plane has a cross-sectional area of 0.2 in2. What is the greatest longitudinal force that can be applied to the brace without causing the brace to be permanently deformed? (3800 lbs)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Nature and properties of solid, fluid 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 2.18. Atom arrangement in a solid



The properties of 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 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



Figure 2.19. Atom arrangement in a liquid



The 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 move 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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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



Figure 2.20. Atom arrangement in a gas.



The properties of a gas Gases have a low density (‘light’) because the particles are so spaced out in the container (density = mass : 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 particles 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 forces. 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Pressure and buoyancy in fluids Introduction What is a fluid? A fluid, in contrast to a solid, is a substance that can flow. Fluids conform to the boundaries of any container in which we put them. They do so because a fluid cannot sustain a force that is tangential to its surface. (In the more formal language, a fluid is a substance that flows because it cannot withstand a shearing stress. It can, however, exert a force in the direction perpendicular to its surface.) Some materials, such as pitch, take a long time to conform to the boundaries of a container, but they do so eventually; thus, we classify even those materials as fluids. 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./in2 ) or Newton per square meter (N/m2). Pressure may be 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 P=F/A Since area equals force divided by pressure, the formula is written A=F/P



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Figure 2.21. 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 and A.



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 = r, where A is the area, 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 7 1/2 miles. This layer is called the troposphere. 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 millimeters. 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 2.23). 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 2.23. Aneroid Barometer



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Gauge Pressure When an instrument, such as an oil pressure gauge, fuel pressure gauge, or hydraulic system pressure gauge, displays pressure which is over and above ambient, the reading is referred to as gauge pressure (psig). This can be seen on the fuel pressure gauge shown in Figure 2.24. When the oil, fuel, or hydraulic pump is not turning, and there is no pressure being created, the gauge will read zero.



Figure 2.24. Psig read on a fuel pressure gauge.



Absolute Pressure A gauge that includes atmospheric pressure in its reading is measuring what is known as absolute pressure, or psia. Absolute pressure is equal to gauge pressure plus atmospheric pressure. If someone hooked up a psia indicating instrument to an engine’s oil system, the gauge would read atmospheric pressure when the engine was not running. Since this would not make good sense to the typical operator, psia gauges are not used in this type of application. For the manifold pressure on a piston engine, a psia gauge does make good sense. Manifold pressure on a piston engine can read anywhere from less than atmospheric pressure if the engine is not supercharged, to more than atmospheric if it is supercharged. The only gauge that has the flexibility to show this variety of readings is the absolute pressure gauge. Figure 2.25. shows a manifold pressure gauge, with a readout that ranges from 10 "Hg to 35 "Hg. Remember that 29.92 "Hg is standard day atmospheric.



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Figure 2.25. Manifold pressure gauge indicatingabsolute pressure.



Differential Pressure Differential pressure, or psid, is the difference between pressures being read at two different locations within a system. For example, in a turbine engine oil system the pressure is read as it enters the oil filter, and also as it leaves the filter. These two readings are sent to a transmitter which powers a light located on the flight deck. Across anything that poses a resistance to flow, like an oil filter, there will be a drop in pressure. If the filter starts to clog, the pressure drop will become greater, eventually causing the advisory light on the flight deck to come on. Figure 2.26 shows a differential pressure gauge for the pressurization system on a Boeing 737. In this case, the difference in pressure is between the inside and the outside of the airplane. If the pressure difference becomes too great, the structure of the airplane could become overstressed.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 2.27, view A). This happens because the bar is rigid. The direction of the blow almost entirely determines the direction of the transmitted force.



Figure 2.27. Forces acting on solids and liquids.



When a force is applied to the end of a column of confined liquid (figure 2.27, 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 2.28, the outward push of the water is equal in every direction.



Figure 2.28. Flat and water filled water hoses.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Pascal’s Law When you squeeze one end of a tube to get toothpaste out the other end, you are watching Pascal’s principle in action. The principle was first stated clearly in 1652 by Blaise Pascal (for whom the unit of pressure is named): “A change in the pressure applied to an enclosed incompressible fluid is transmitted undiminished to every portion of the fluid and to the walls of its container.” Pascal’s Principle and the Hydraulic Lever Figure 2.29 shows how Pascal’s principle can be made the basis of a hydraulic lever. In operation, let an external force of magnitude Fi be directed downward on the left-hand (or input) piston, whose surface area is Ai. An incompressible liquid in the device then produces an upward force of magnitude Fo on the right-hand (or output) piston, whose surface area is Ao. To keep the system in equilibrium, there must be a downward force of magnitude Fo on the output piston from an external load (not shown). The force Fi applied on the left and the downward force Fo from the load on the right produce a change p in the pressure of the liquid that is given by



So



The equation above shows that the output force Fo on the load must be greater than the input force Fi if Ao Ai, as is the case in Figure 2.29. If we move the input piston downward a distance di, the output piston moves upward a distance do, such that the same volume V of the incompressible liquid is displaced at both pistons.



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Figure 2.29. A hydraulic arrangement that can be used to magnify a force .The work done is, however, not magnified and is the same for both the input and output forces.



Then Which we can write as



This shows that if, Ao



Ai (as in Figure 2.29), the output piston moves a smaller distance than the input piston moves. From those



equations above we can write the output work as



This shows that the work W done on the piston by the applied force is equal to the work W done by the output piston in lifting the load placed in it. So the advantage of a hydraulic lever is: With a hydraulic lever, a given force applied over a given distance can be transformed to a greater force applied over a smaller distance. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics The product of force and distance remains unchanged so that the same work is done. However, there is often tremendous advantage in being able to exert the larger force. Most of us, for example, cannot lift an automobile directly but can with a hydraulic jack, even though we have to pump the handle farther than the automobile rises and in a series of small strokes. Archimedes Principle It 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. 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, keeldown 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 example a body of weight density 63.4 Ibs./ft3 would float in ocean water (D = 64.4 Ibs./ft3) and sink in lake water (D = 62.4 Ibs./ft3). 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 Iow body will float. In order to understand Archimedes' Principle as applied to floating bodies, let us consider a submarine and imagine that a block of wood of weight density 48.3 Ibs./ft3 and volume 2 ft3 is thrust out of the hatch of a submarine into the ocean water. We know intuitively that this block of wood will rise to the ocean surface. The weight of the block is (48.3 Ibs./ft3) (2 ft3) = 96.6 Ibs. As long as the block is below the water surface (while it is rising to the top), it displaces 2 ft.3 of ocean water.



Figure 2.30. Archimedes’ Principle



We know that: BF = weight of displaced ocean water = (64.4 Ibs./ft.3) (2 ft.3) BF = 128.8 Ibs. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 above the water surface. It follows that 75% of 2 ft. 3 (= 1.5 ft.3) will be below the water surface. When this occurs, the BF on the block is (64.4 Ibs./ft.3)(1.5 ft.3) equals 96.6 Ibs. Note again that the BF equals the weight of the block while the block is floating. In the preceding example, note that the ratio of the weight density of the block (48.3 Ibs./ft. 3) to the weight density of the ocean water (64.4 Ibs./ft.3) 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 Iow a block will float in a given liquid.



Figure 2.31. A Father and a son are fishing.



In dealing with bodies that float, it is important to note that boats, made of materials denser 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 2.31. Note that some of the boat (shown with dotted lines) 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 Ibs./ft.3) or 499 Ibs. Figure 2.31 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. This is realistic (father 200 Ibs., son 75 Ibs., lunch 25 Ibs., fishing gear 49 Ibs.). Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics One final comment should be made regarding submarines. Submarines cruising at a definite depth in ocean water have a total weight density equal to the weight density of ocean water, 64.4 Ibs./ft.3. This means that the total weight of the submarine (metal shell, air, crew, load, ballast, etc.) divided by the total volume is 64.4 Ibs./ft.3 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 : A block of oak (0 = 45 Ibs./ft.3) is placed in a tank of benzene (0 = 54.9 Ibs Ift.3). 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. Example: The bag of a balloon is a sphere of radius 25 m filled with hydrogen of weight density 0.882 N/m3. What total weight (in Newton) of fabric, car, and contents can be lifted by this balloon in air of weight density 12.6 N/m3? We first calculate the volume of the spherical balloon by recalling that the volume of a sphere is given by:



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Basic Aircraft Maintenance Training Manual Module 2 – Physics (0.882 N/m3) (65,450 m3) = 57,700 N. The weight of the displaced air is: (12.6 N/m3) (65,450 m3) = 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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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



A solid aluminium object (Daluminium = 169 lbs/ft3) of volume 250 ft3 is resting on the ocean floor (Docean water = 64.4 lbs/ft3). A salvage crew plans to raise this object. What force will be needed? (26,200 lbs)



2.



A solid steel (Dsolid steel = 487 lbs/ft3) body of volume 125 ft3 is to be raised by a salvaging crew to the surface of a lake (D water = 62.4 lbs/ft3). What force will be needed? (53,100 lbs)



3.



What percentage of an iceberg (Dice = 575 lbs/ft3)is below the surface of the ocean (Docean water = 64.4 lbs/ft3).? (89%)



4.



A canoe is floating in such a way that it displaces 6 ft3 of lake water (Dwater = 62.4 lbs/ft3). If the canoe weighs 100 Ibs., what is the weight of its contents? (274 lbs)



5.



A balloon is spherical in shape and has a radius of 20 ft. It is filled with helium (weight density 0.01 Ib/ft 3) and is floating in air (weight density 0.08 Ib/ft3 ). What is the weight of the balloon (fabric, crew and contents etc.)? (2240 lbs)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 2.2.2. 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 V0 = 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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) In using this formula, a may be either positive or negative. If V is less than V0, then our value of a turns out to be a negative number. 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 – V0 After transposing, we can write: (3)



V = V0 + 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) In this equation, we can substitute for V (= V0 + at) using the value in equation (3).



After a bit of algebra, we obtain: (5) Equation (4) can be written, after cross-multiplication: 2s = (V0 + v)t



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Basic Aircraft Maintenance Training Manual Module 2 – Physics We can now multiply this equation by equation (2). After cancelling time (t) on the right: 2as = (V – V0) (V + V0) 2as = V2 – V02 The final form of this formula is: V2 = V02 + 2as



(6)



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. ii.



V = V0 + at



iii. iv.



V2 = V02 + 2as



v. When a body in straight line motion is not changing speed, or in cases where we are interested only in 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Examples: 1.



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?



Notice that the unit of acceleration has the square of a time unit in its denominator. 2.



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. V0 = 50 ft./sec. V = 75 ftIsec. s = 125ft. a=? Formula iv involves these four quantities. Note that i, ii, and iii do not involve these exact four quantities. Formula iv is the one to use. First it should be solved for the unknown, a. V2 = V02 + 2as V2 - V02 = 2as



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Accelerated Motion of a "Freely Falling" Body Common experience indicates that falling bodies accelerate or increase in speed as they fall. Close to the surface of the earth this "acceleration of a freely falling body" has been measured to be about 32 ft./sec.2 in the English system and 9.8 m/sec.2 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 of the acceleration will be either 9.8 m/sec 2 or 32 ft./sec2 if the body is rising and therefore decreasing its speed the values of the acceleration will be - 9.8 m/sec2 or – 32 ft/sec2. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual 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. Examples: 1.



A body started from rest and has been falling freely for 3 sec. At what speed is it falling? u = 0, t = 3 sec, a = 32 ft/sec2 v = ? We will use Formula ii. V = V0 + at V = 96 ft/sec



2.



A body started at rest and has been falling freely for 3 sec. How far has it fallen? u = 0, t = 3 sec, a = 32 ft/sec2 , s = ? We will use Formula iii.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 3.



A body is thrown upward with an initial speed of 120 ft./sec. How high does it rise? u = 120 ft/sec, v = 0, a = -32 ft/sec2, s = ? We will use Formula iv. V2 = V02 + 2as



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



A car on the motorway is accelerating at 25 ft/sec2. If it started from rest and has been accelerating for 5 sec., how far has it travelled during this time of acceleration? (312 ft)



2.



A truck had an initial velocity of 40 ft/sec. It accelerated at 10 ft/sec2 and reached a final velocity of 60 ft/sec. How far did this truck travel while it was accelerating? (100 ft)



3.



A car slowed down from 80 ft/sec. to 40 ft/sec. while travelling a distance of 100 ft. What was its acceleration? (-24 ft/sec2)



4.



A car, originally travelling at 25 ft/sec, increases its speed at a rate of 5 ft/sec2 for a period of 6 sec. What was its final speed? (55 ft/sec)



5.



A car has an initial velocity of 40 ft/sec. It slows down at a rate of 5 ft/sec 2 and covers a distance of 60 ft. while slowing down. What is its final velocity? (32 ft/sec)



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? (78 m)



7.



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



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? (100 ft or 30.6 m)



9.



A car starts with an initial velocity of 30 ft/sec. and accelerates for 5 sec. at 4 ft/sec2. How far has it travelled during this time? (200 ft)



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? (15 sec) (Hint: convert MPH to ft./sec. first) 11. A Grumman Tomcat, powered by two Pratt & Whitney turbofan engines, has a maximum acceleration during take-off of 20 ft/sec2. What velocity can it achieve by the end of a 1000 foot take-off run? (200 ft/sec)



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Basic Aircraft Maintenance Training Manual 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. 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 = 2 radians 2 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 2.32. As the point moves about the circle in a counter-clockwise sense, a line drawn between the point and the center 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. If the point moves with constant speed it also has a constant angular velocity. That is, the line drawn from the point to the center 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,



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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 α 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,



Figure 2.32. A point moving in a circle



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.



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V = V0 + at



V2 = V02 + 2as



Examples: 1.



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 α.



We now substitute our known values.



2.



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



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3.



A propeller starts at 1,000 rev./min. and accelerates at 100 rev./min.2 through 2,000 revolutions. What is its final angular velocity?



Radian Measure In figure 2.33 's' is the length along the path. We would like to relate this distance to the size of the central angle (θ) 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. The equation that relates s to θ and R is a very simple one if we limit the angular unit to radians.



Figure 2.33. s, R, and θ



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Basic Aircraft Maintenance Training Manual Module 2 – Physics This equation is: s= Rθ We see that this equation is true if we look at figure 2.33. We note, by measuring, that the equation is satisfied. We also see that it would not be true if the angle 9 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 ( ). s=R θ Let us divide both sides of this equation by the time, ) during which the motion occurred.



We can write: V = Rω 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 ( t). We can write: v=R ω Next we divide left and right members by t



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 (α). We can write: A = Rα Let us summarize the three important equations we have derived: s = Rθ v = 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. α must be in rad./min2 or rad./sec2 Note that the radian is called a "dimensionless" unit. We put it in or take it out for clarity. Examples: 1.



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:



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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 2.



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 = 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"/sec. 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



3.



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



We note that the unit is rad./sec.2 because the equation that we have used requires radian measure. To obtain a in rev./sec.2, we must use the standard conversion factor.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



A propeller starts from rest and accelerates at 120 rev/sec2 for 4 seconds. What is its final angular velocity in rev/sec? In rev/min? (480 rev/sec or 28,800 rev/min)



2.



A rotating turntable starts from rest and accelerates at 5 rev/min2 for 3 min. Through how many revolutions has it turned? (22.5 rev)



3.



A helicopter main rotor starts from an initial angular velocity of 2 rev/min and accelerates at 60 rev/min 2 while turning through 400 revolutions. What is its final angular velocity? (219 rev/min)



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? (89,000 ft)



5.



A plane is circling an airport in a circle of radius 5,000 ft. How far has it travelled after 4 revolutions? (23.8 miles)



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? (1/6 rad/s, 5/π rev/min)



7.



A race car is moving on a circular racetrack of radius 800 ft. It is accelerating at a rate of 10 ft/sec2 What is its angular acceleration in rev/sec2? (1/160π rev/sec2)



8.



A helicopter tail rotor starts with an initial angular velocity of 15 rev/sec and decelerates at a rate of 2.00 rev/sec 2 until it comes to rest. Through how many revolutions has the rotor turned while it comes to rest? (56.3 rev)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 9.8 m/s2 or 32 ft/s2



Figure 2.34. 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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:



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 2.35. Sinusoidal motion of a spring / mass system



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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:



Where:



k = the stiffness of the spring in N/m, or lb./in. m = the mass of the oscillating body



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, . 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 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 2.36. 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 =vT. Then using f=1/T gives the standard wave relationship This is a general wave relationship which applies to sound and light waves, other electromagnetic waves, and waves in mechanical media. 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). Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 minimize unwanted vibrations. The study of sound and vibration are closely related. Sounds, 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. 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 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 collapse of road bridges in gale force winds, as experienced at Tacoma Bridge on November 7th , 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 airplane 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 of 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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? (0.6 Hz, 16.8 sec)



2.



A pendulum has a mass of 0.05 slugs. It takes 15 seconds to oscillate 10 times. What is its length? (0.6 m)



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? (0.36 Hz, 2.8 sec)



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? (0.05 slugs)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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: W o = Fo D o W i = Fi D i 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 “Wf” to represent work lost because of friction. Wi = Wo + Wf 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 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. Manual No. : BCT-0012/A2 Electrical Avionics



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The ideal mechanical advantage of a machine can always be determined by measurements made on the machine itself. The efficiency (Eft) of a machine is the ratio of the output work to the input work.



The efficiency can be expressed as a decimal or as a percentage. For example, if the efficiency is calculated as 0.78, we can express 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: 1.



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?



2.



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?



3.



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



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4.



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



Method 2



We will next consider six simple machines. In each of these cases the IMA is expressed, not as the ratio of 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 2.37. 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 "Fo". Manual No. : BCT-0012/A2 Electrical Avionics



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Figure 2.37. Simple Lever System.



In figure 2.38. note that the input force acts through a distance (Di) and the load is raised a distance (D o).



Figure 2.38. 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 (L i) and the distance from the output end to the fulcrum is called the output lever arm (Lo). Recall that:



However, figure 2.38. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



Figure 2.39. First class Lever.



2nd Class: The load is between the fulcrum and the applied force. Examples are the nutcracker and wheelbarrow.



Figure 2.40. Second class lever.



3rd Class: The applied force is between the load and the fulcrum. An example is ice tongs.



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Figure 2.41. Third class lever.



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 (Di) is pulled down by a worker, the load will be raised a distance (Do). 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 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".



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Figure 2.44. Simple pulley systems.



In figure 2.44 (B), there is a single movable pulley. A study of the diagram shows that Di is always twice Do. 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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics The Wheel and Axis



Figure 2.45. Simple 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 FI 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, (Do).



The Inclined Plane In the inclined plane shown in figure 2.46 we note that the worker slides the load up the incline. The input distance (Di) 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 (Do) equals h also. Manual No. : BCT-0012/A2 Electrical Avionics



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Figure 2.46. Inclined plane



The Screw Jack The pitch of the screw (p) is the distance between adjacent threads (see figure 2.47). As the handle is turned through one revolution, a distance given by 2 r ft., the load is raised a distance of one pitch. Therefore, we have the relation:



Figure 2.47. The screw jack



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 a hydraulic press is shown in figure 2.48. 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 Ao). 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). A hydraulic press is filled with some fluid (gas or liquid). This fluid exerts a common fluid pressure throughout the device.



Figure 2.48. The hydraulic press



As the smaller piston moves downward, a distance (di) the larger piston moves upward a distance (do). 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: r2 di = R2 do



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Basic Aircraft Maintenance Training Manual Module 2 – Physics We can cancel the common factor (it) and rearrange the equation. We obtain: di / do = R2 / r2 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: IMA = R2 / r2 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 anyone 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. Examples: 1.



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 = 3.5 ft / 0.27 ft = 13.0 AMA = (Eff) (IMA) = (0.60) (13.0) = 7.8 Fo = (AMA) (Fi) = (7.8) (75 lbs) = 585 lbs.



2.



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 / sin32º = 1.89, AMA = 90 lbs / 50 lbs = 1.8 Eff = AMA / IMA = 1.8 / 1.89 = 0.95 = 95%



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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? (3)



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)



3.



What is the efficiency of a machine having an IMA of 7 and an AMA of 5? (71%)



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? (80 %)



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? (33.3 lbs)



6.



The large piston of an hydraulic press has area 1.5 ft2. and the small piston has area 0.30 ft2. Assume that the machine is 100% efficient. What load can be raised by a force of 75 Ibs.? (375 lbs)



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? (90.9%)



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? (836 lbs)



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2.2.3. Dynamics 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 continue 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 shoveling 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. 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 be 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 2.49 below.



Figure 2.49. The four forces acting on an airplane



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. 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 and 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



Figure 2.50. Gravitational force on apple that falls from the tree



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. Manual No. : BCT-0012/A2 Electrical Avionics



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rd



Figure 2.51. 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 2.52. 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 opposite force on the propeller. This reaction force of the air on the propeller provides the thrust for a propeller driven plane.



Figure 2.52. Action of a propeller. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics The acceleration of the air mass is:



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



Both of the velocities (V0 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 V0 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:



Recall that we have a symbol for "change in", this means that we can write the above formula as:



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?



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 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 = (35 lbs.) (50 ft.) cos 28º = 1,550 ft.lbs. 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 180º angle. The cosine of 180º 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 = (15 Ibs.)(2 ft.) cos 180º W = (15Ibs.)(2 ft.)(-1) Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics W = -30 ft.lbs. Energy 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).



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). W = Fs = (ma) (



Now we note that the speed obtained by the body during the time of acceleration is given by the equation: Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics V= at Therefore, we can substitute v for at in the equation above.



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.



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) 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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 increases 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. Examples: 1.



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



(4 slug) (32 ft/sec ) (6 ft.) = (4 slug) (32 ft/sec2) (2 ft.) + 1/2 (4 slug) v2 768 = 256 + 2v2 512 = 2v2 256 = v2 V = 16 ft/s. 2.



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 (10 kg) (9.8 m/sec2) (300 m) = (10 kg) V2 2,940 m2/sec2 = 1/2 V2



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 5,880 m2/sec2 = V2 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" anybody 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:



In symbols:



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 ft. lbs. / sec = 33,000 ft. lbs. / sec Example: 1.



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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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=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-Ibs./sec. is used. Thus:



Alternatively, we can take the speed of the aircraft in MPH and use the conversion 1 HP = 375 mi.lbs./hr. to obtain:



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.



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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



How much work is done by a person in raising a 45 lb. 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. (3370 ft.lb)



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? (15,000,000 J)



3.



A father has his 45 lb. son on his shoulders. He lowers the child slowly to the ground, a distance of 6 ft. How much work does the father do? (-270 ft.lb. Note the negative sign!)



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? (4800 ft.lb; 19,200 ft.lb.)



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? (39.2 J or 40 of g = 10 m/s2)



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? (70 ft/sec)



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 2000 N. How far will the aircraft travel before it comes to rest? (900 m)



8.



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? (30,000 J)



9.



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? (3 MN)



10. An aircraft engine weighing 12,000 N is lifted by a 3.6 kW motor a distance of 10m. What time was needed? (33 sec) 11. 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? (100 ft.lb./sec 0.182 HP) 12. 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) (32.5 sec)



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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 2.53) 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. 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).



Figure 2.53. Centripetal force exerted by a hammer thrower.



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:



Newton's Second Law connects acceleration and force, by Force = Mass x Acceleration. Thus, we can write the equation:



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 2.2. Each set of units, pound, slug, ft./sec.2) in the English system, or (Newton, kilogram, m/sec.2) in the metric system is said to be consistent in the following sense: a force of 1 lb. when applied to a mass of 1 slug gives it an acceleration of 1 ft./sec2. UNITS



ENGLISH



METRIC



Force



pound (lb.)



newton (N)



Mass



slug



kilogram (kg)



Acceleration



ft./sec.



2



m./sec.2



Table 2.2. Units of Force, Mass, and Acceleration. Similarly, a force of 1 Newton applied to a mass of 1 kilogram causes it to accelerate at 1 m/sec2 Using Newton's second law, we can write: 1 Newton = 1 kilogram m/sec2 And 1 pound = 1 slug ft./sec2 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,



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Examples: Find the acceleration of a 3 slug object acted upon by a net force of 1.5 Ibs.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



Find the mass of an object which accelerates at 5 m/sec2 when acted on by a net force of one Newton. (0.2 kg)



2.



Find the acceleration of a 3 slug object experiencing a net force of 12 Ibs. (4 ft/sec2)



3.



Find the net force on a 5 slug object which is accelerating at 3ft/sec2. (15 lbs)



4.



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



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. (5600 N)



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? (976 lbs)



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. (16050 N)



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. (16 ft/sec2)



9.



What centripetal force is needed to keep a 3 slug ball moving in a circular path of radius 2 feet and speed 4 ft/sec.? (24 lbs)



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? (16.67 ft/sec2)



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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 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, .i1) divided by the time (t). Now let us visualize two bodies of masses, m1 and m2 on a one dimensional track. If these two bodies collide, we have four different velocities to consider. We will name these velocities very carefully. V1' = the velocity of body one before the collision V1”= 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:



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. A boy and a man are both on ice skates on a pond. The mass of the boy is 20 slug and the mass of the man is 80 slug. They push on each other and move in opposite directions. If the recoil velocity of the boy is 80 ft/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 ft/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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Example: A truck of mass 1,550 slug is moving east at 60 ft/sec. A car of mass 1,250 slug is travelling west at 90 ft/sec. The vehicles collide and stick together after impact. What is the velocity of the combined mass after the collision has occurred? V1' = 60 ft/sec. m1 = 1,550 slug V2' = -90 ft/sec. m2 = 1,250 slug We will not include units in our substitution. However, we will note that the velocity, when we obtain it, will be in ft/sec. (1,550) (60) + (1,250) (-90) = (1,550 + 1,250)V" -19,500 = 2,800 V" V' = -6.96 ft/sec. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 ft/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. 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. 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: Conservation of Energy (1)



1



/2 m1V1’2 + 1/2 m2V2’2 = 1/2 m1V1”2 + 1/2 m2V2”2



(2)



m1V1’ + m2V2’ = m1V1” + m2V2”



(3)



m1V1’2 + m2V2’2 = m1V1” 2 + m2V2” 2



Conservation of Momentum 1



Divide (1) by /2 Now in both (2) and (3), we will transpose some terms: (4) Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics (5)



m1V1’ - m1V1” = + m2V2”- m2V2’



(6)



m1 (V1’ - V1”) = m2 (V2” - V2’)



(7)



m1 (V1’2 - V1”2) = m2 (V2”2 - V2’2)



(8)



m1 (V1’ - V1”) (V1’ + V1”) = m2 (V2” - V2’) (V2” + V2’)



Factorise (4) and (5):



In (7), Factor again: Divide (8) by (6):



After cancelling common factors, we obtain: V1’ + V1” = V2” + V2’ Again we transpose terms: (9)



V1’ – V2’ = V2” – V1”



In words, this equation says that the relative velocity of the balls before the collision is equal to the relative velocity of the balls after the collision. Equation (9) has been obtained algebraically from two equations, the conservation of momentum and the conservation of energy. We use equations (2), the conservation of momentum equation, and equation (9), called the relative velocity equation, to solve for the velocities of the two bodies after an elastic collision. We will rewrite these two important equations for future reference:



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(2)



m1V1’ + m2V2’ = m1V1” + m2V2”



1.



V1’ – V2’ = V2” – V1”



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 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? (3 m/sec)



2.



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 ftlsec. With what speed does the astronaut recoil? (0.6 ft/sec)



3.



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? (27 m/sec East)



4.



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? (2.4 m/sec East)



5.



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? (67.7 MPH)



6.



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? (The 3 kg ball is moving left at 1 m/sec, and the 2 kg ball is moving right at 4 m/sec)



7.



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? (The 2kg ball is moving right at 3 m/sec, and the 1 kg ball is moving right at 6 m/sec)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Torque Consider the diagrams 2.54. 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 (τ) 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: τ= 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. , lb. in. or the Nm.



Figure 2.54. Force acting at a distance creates torque.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Extensions Figure 2.55 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:



Where: L = distance between the driving tang and the center of the handle. X = length of extension spanner between centres.



Figure 2.55. 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) 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, which is



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b) 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 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 2.56, and the scale reading in this instance will be equal to specified torque x.



Figure 2.56. A torque wrench fitted with an extensionspanner 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 90º to the axis of the wrench.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Gyroscopic Principles Introduction A gyroscopic system is one where a universally mounted heavy metal wheel or rotor has three planes of freedom, as shown in the figure 2.57 below.



Figure 2.57. Gyroscopic simple system



The aligned axes provide: Spinning Freedom : Rotation about the spin axis (XX1) Tilting Freedom : Rotation of the spin axis in the vertical plane (YY1) Veering Freedom : Rotation of the spin axis in the horizontal plane (ZZ1) Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Freedom of movement within the three planes is obtained by mounting the rotor in two concentrically pivoted rings, called inner and outer gimbal rings. The gimbal system mounts so that with the gyro in its normal operating position, all of its axes run mutually at right angles to each other, and intersect at the centre of the rotor. A gyro with its axis of rotation in the horizontal plane is a horizontal gyro, and a gyro with its axis of rotation in the vertical plane is a vertical gyro. The gyro's plane of rotation contains the sensitive axes, rotation about which causes the gyro to precess (described later). Principle of Construction A gyroscope consists of a weighted wheel or rotor, which spins at high speed (8000-24 000 rpm) and is mounted in a series of hinged mounting rings, called gimbals, as shown on the next page. A gyro has 3 axes of freedom, one of which is its spin axis, and is able to move relative to the mounting base around one or both of the remaining axes. Ignoring the spin axis, one degree of freedom exists when the gyro can rotate around only one axis, and two degrees of freedom exist if it is free to move around both axes. A gyro with two degrees of freedom is known as a Free or Space Gyro. Gyroscopic Properties When the rotor spins at high speed, the device becomes a true gyroscope, and possesses the following fundamental properties: Rigidity In Space The gyro tries to remain pointing in the same direction or position in space, even when its mounting base is tilted or rotated (i.e. the gyro has a tendency to maintain its plane of spin, dependent on its speed, mass, and radius about which the mass is displaced [Newton's 1" Law]). The greater the rigidity the more difficult it is to move the rotor away from its plane of spin , unless an external force acts on it. For example, if a spinning bicycle wheel was to fall over its spin axis, it must be rotated through 90°, and unless an external force is applied, it does not do so as long as it maintains a reasonable speed.



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Where:



I = Moment of Inertia F= External Force S = Speed of Wheel (rpm)



Figure 2.58. Rigidity in uniform wheel and dished wheel.



Rigidity is proportional to the rotor's rpm, and its moment of inertia, which increases if the rotor has a large radius, with the bulk of its mass distributed around its rim.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Precession When an external force is applied to the spinning rotor via the gimbal assembly, the gyro does not move in the direction of the force, but in a direction perpendicular to that of the applied force. For example, if a force is applied to the spin axis, the gyro does not move in the direction of the applied force, but instead rotates due to a force applied 90° later in the plane of rotation, as shown below in figure 2.59.



Figure 2.59. Precession.



Similarly, if a downward push force is applied to the inner gimbal of a gyro system, the gyro precesses about its outer gimbal pivot point.



Figure 2.60. Precession in details.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics The strength and direction of the applied force, the moment of inertia of the rotor, and the angular velocity of the rotor, all affect the amount of precession. It follows that, the larger the force the greater the rate of precession, and the higher the rigidity, the lower the rate of precession. Thus in order to precess a gyro its rigidity must first be overcome.



Precession continues while applying the force, until the plane of rotation is in line with the direction of the applied force. At that point, the force no longer disturbs the plane of rotation, so that there is no further resistance to the force, and thus no further precession. The axis about which a force or torque is applied to a gyro is termed the input axis, and the one about which precession takes place is termed the output axis. Most gyros rotate at a constant rpm, particularly where precession plays a part in the operation of a system or indicator. If the direction of the rotor or applied force reverses, then the direction in which precession occurs also reverses. Types of Gyroscopes The number of degrees of freedom permitted by each type of gyroscope determines its usage, but for use in aeroplanes, they must exhibit two essential reference datums. The first is a reference against which pitch and roll attitude changes are detected, and the second is a directional reference against which change about the vertical axis is detected.



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Figure 2.61. Gyroscopes application in airplane.



The following types of gyroscopes exist: SPACE (OR FREE) GYRO This is a gyro having freedom to move in all three planes. It consists of two concentrically pivoted rings, called inner and outer gimbal rings. The three planes relate to the three axes of the aeroplane (i.e. fore and aft or roll axis, lateral or pitch axis, and the normal or yaw axis). Furthermore, there is no means of external control over this type of gyro, a feature that distinguishes it from a tied or Earth gyro. This type of gyro would have no practical use in an aeroplane instrument where the gyro is required to be set to, and maintain a given direction.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics TIED (OR DISPLACEMENT) GYRO



Figure 2.62. Displacement Gyro



The precession sensor on the Y axis provides a measure of the angular displacement of the rotor around the Z axis. In a vehicle guidance system, where the Z frame is rigidly attached to the vehicle body and turns with the vehicle, the rotor will precess or tilt about the Y axis as the vehicle turns. After every turn the angular tilt of the rotor will indicate the cumulative deviation from the desired course, only returning to zero as the vehicle returns to its original pre-set direction. This type of gyro is a space gyro, which has a means of external control, and has freedom of movement about all three planes. This type is for use as a directional gyro (e.g. in the Direction Indicator). EARTH GYRO This type of gyro is a tied gyro, where the controlling force is the gravity of the Earth. This type is for use in gyro horizon, or artificial horizon instruments.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics RATE GYRO



Figure 2.63. Rate Gyro



The rate sensor measures precession force and provides measure of the rate of turn. Regarding rate of turn, in a vehicle guidance system, when the Z frame is rigidly attached to the vehicle body, and turns with the vehicle, the precession force is opposed by a spring mechanism which can be used to indicate the magnitude of the force. On completion of the turn, the springs return the rotor to its stable position with zero tilt (precession) but the movement of the rotor on the Z axis is constrained by its rigid connection to the vehicle body. Thus after the completion of every turn, the rotor remains aligned with the body pointing in the new direction but with zero tilt. This is a gyro having one plane of freedom only, its plane of rotation being 90° removed from its plane of freedom. This type of gyro is used in measuring the rate of turn, and employs restraining springs (e.g. in the turn and balance indicator or turn coordinator). Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics RATE INTEGRATING GYRO This type of gyro is similar to the rate gyro, having a single degree of freedom, except it uses the viscosity of a fiuid (viscous restraint) to damp the precessional rotation about its output axis, instead of restraining springs. The main function of this type of gyro is to detect turning about its input axis by precessing about its output axis. This type is used on inertial navigation stabilized platforms. SOLID STATE (RING LASER) GYRO These are not gyros in the true sense, but they behave like gyros, and sense the angular rate of motion about a single axis. They consist of a solid block of temperature stable glass within which there is a cavity or laser path, filled with a lasing medium such as helium-neon. Some are triangular in shape (Honeywell), whilst others have four sides (Litton). They both have small tunnels drilled in them, with reflecting mirrors sited at each corner. Two beams of high-energy laser light pass in opposite directions around the sealed cavity and initially travel at the same speed. Any rotation of the gyro in the laser plane results in a change in the path lengths of each beam, and a control element measures the resultant frequency shift of the beams. The frequency differential is directly proportional to the angular turning rate. Power Sources for Gyroscopes Conventional gyroscopes in aeroplanes are either air (vacuum) driven, or electrically driven. In some aeroplanes, all of the gyros are either vacuum or electrically driven, whilst in others the vacuum (suction) system provides the power necessary to run the attitude and heading indicators, whilst an electrical system runs the slip and turn indicator. The electrically driven gyroscopic instruments use alternating or direct current for power. 1. Air Driven Gyroscopes This type is widely used on small aeroplanes, and is found on some large aeroplanes in order to power stand-by, or emergency instruments. In the air driven gyro, the gyro system is contained in an airtight case, the air having been removed from the casing via a suction pump or venturi, thus creating a partial vacuum inside. The vacuum source is normally an engine driven Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics vacuum pump, which is controlled to a value between 4.5-5.0 inches of Hg, but in very simple aeroplanes, this may be achieved using a venturi tube attached to the outside of the fuselage. Air under atmospheric pressure enters the instrument casing via a filter, and flows through a shroud, which encases the rotor.



Figure 2.64. Air Driven Gyroscopes.



The shroud acts as the inner gimbal and has a small opening in it that directs the airflow fed to it via an inlet port onto buckets cut in the rotor periphery. The shroud also has an exhaust port to allow air to escape to the outer case. The pipeline from the inner gimbal is fed through the inner gimbal axis along the outer gimbal and through its axis to a filter system, which covers a hole in the outer case. When the vacuum is applied to the system, the pressure in the outer case drops. Replacement air enters the system from outside the case and passes through the filter before being directed onto the rotor buckets. The rotor rotates at approximately 10000 rpm under normal operation. The air escapes from the rotor shroud, although it can be used to provide a controlling force (tie) to the gyro. 2. Electrical Gyroscope The majority of gyroscopes used in aeroplanes today are electrically driven, and normally use AC current, although a 24volt DC supply feeds some. The AC powered gyro is preferable, since it avoids the use of commutators and brush gear, which Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics require frequent servicing. The gyroscope itself is a 3-phase squirrel induction motor, constructed to obtain the maximum gyroscopic effect. To achieve this, the rotor does not rotate inside the field coils, as it does in conventional motors, but instead is positioned around the outside of the field coils. This method of construction ensures that the rotor mass is concentrated as near its periphery as possible, thus increasing gyro rigidity. A 115 volt, 3-phase, 400 hertz AC system feeds the gyro stator, and the rotor rotates at approximately 24 000 rpm. AC gyros are capable of higher rotational speeds than the DC ones, and are therefore the favoured option in instruments where high rigidity is required. DC powered gyros can also be powered from the aeroplane emergency electrical supply. Gyro Wander Any deviation of the gyro spin axis from its set direction is known as gyro wander, and is classified as follows: REAL WANDER Any physical deviation of the gyro spin axis is called real wander. A gyro should not wander away from its preset direction, but various forces act on the rotating mass of a gyro and cause it to precess. For example bearing friction, which is always present at the spin axis. If this friction is symmetrical, it merely slows down the rotor, but if it is asymmetrical, it causes the gyro to precess. Similarly, any friction in the gimbal bearings causes the gyro to precess. Wear on the gyro may result in movement of the C of G, which may also result in a precessing force. Such errors are not constant or predictable, and cannot be calibrated, nor can corrections be applied to nullify this error. APPARENT WANDER In this case, the gyro spin axis does not physically wander away from its pre-set direction, but to an observer it appears to have changed its direction. This is because the gyro maintains its direction with respect to a fixed point in space, whereas the observer rotates with the Earth, thus with the passage of time the gyro appears to have changed direction, with reference to an Earth datum.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Apparent wander, is also made up of horizontal components called drift, and vertical components called topple. The rate of drift and topple depends upon the latitude and can vary from zero to a maximum of 15.040 per hour (the rate at which the Earth rotates). Depending on whether a gyro has a vertical or horizontal spin axis, the rotation of the Earth also has a different effect. HORIZONTAL AXIS GYRO The figure 2.65 below shows a horizontal spin axis gyro positioned at the North Pole.



Figure 2.65. Horizontal Axis Gyro.



It shows an observer initially at position A, where the gyro is set so that its spin axis is directly in line with him. Six hours later, the Earth having rotated through 90°, the observer now views the gyro from position B. The observer does not however appreciate his own motion, and the gyro spin axis appears to have moved clockwise in the horizontal plane through 90°. Twelve hours later the gyro spin axis appears to have moved through 180°, and finally after twenty-four hours, with the observer back in the original position, the gyro spin axis again appears as it was when first aligned. The apparent motion in the horizontal plane is known as gyro drift. If a horizontal spin axis gyro has its axis aligned in a north/south direction along the equator, during the Earth rotation, the gyro spin axis continues to remain aligned with the local meridian. This occurs because at the equator, all of the meridians are parallel to one another, and a gyro aligned with a meridian thus remains with that meridian over a 24-hour period. This means that the gyros neither drift nor topple Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics when aligned in this manner. If the horizontal spin axis gyro is positioned at the poles, it drifts through 360° in 24 hours (maximum drift). For example, the rate of drift at the poles is the same as the angular velocity of the Earth, at 15.04° per hour, whilst at the equator the same gyro with its spin axis aligned with the local meridian has zero drift due to Earth rotation. Drift at intermediate latitudes = 15.04° x sin Latitudeº per hour.



Figure 2.66. A horizontal spin axis gyro with its spin axis aligned in an east/west direction along the equator, when observed at point A.



In this case, after six hours duration, and the Earth having rotated through 90°, the observer again views the gyro from position B, where it appears to have turned into a vertical axis gyro. This apparent change in its vertical plane is topple, and is the maximum value at the equator, but zero at the poles, due to the Earth 's rotation. Topple at intermediate latitudes = 15.04° x cos Latitudeº per hour.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics VERTICAL AXIS GYRO Figure 2.67 (a) below shows a vertical spin axis gyro, which is positioned at the North Pole, where the rotation of the Earth beneath the gyro spin axis has no effect on the gyro (i.e. the gyro axis appears to neither drift nor topple).



Figure 2.67. Vertical Spin Axis Gyro



Diagram (b) above shows a gyro at the equator, with its spin axis vertical to the observer, when viewed at position A. After six hours, and the Earth having rotated through 90°, the observer at point B will view the gyro as a horizontal axis gyro. After another six hours, the spin axis again appears vertical. This apparent change in the gyros vertical axis is known as gyro topple. At the equator gyro topple is 360° in 24 hours, whilst at the poles it is zero. From the above, notice that the vertical gyro suffers from topple but does not suffer from drift: Drift = Vertical gyro unaffected Topple = 15.04° cos latitudeº per hour Gyro drift and topple may be summarized as: Horizontal axis gyro: at the poles: maximum drift and no topple at the equator: no drift and maximum topple Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Vertical axis gyro:



at the poles: no drift and no topple at the equator: no drift and maximum topple



TRANSPORT WANDER This is an additional form of apparent topple/drift, which principally occurs when the gyro is placed on a platform, such as an aeroplane, that is flying in an east or west direction. The gyro carries in space in the same way as the Earth does, and this result in transport wander. Transport Drift = Rate of change of longitudeº per hour x sin latitudeº per hour Transport Topple = Rate of change of longitudeº per hour x cos latitudeº per hour. 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



Static Friction Static friction (or 'starting' friction) is the force between two objects that are not moving relative to each other. For example, static Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics friction can prevent an object from sliding down a sloped surface. The coefficient of static friction, typically denoted as μs, 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 μk, 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 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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Calculating Friction In all cases, the friction equation is the same. F=μN The symbol "μ" (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 2.3. 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 2.3. We note that the coefficients of sliding friction are less than the coefficients of starting friction. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 0.02 Roller bearings 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 = 100 Ibs. F=μN Force to start sliding motion = (0.15) (100 lbs.) = 15 lbs. Force to keep body sliding = (0.09) (100 lbs.) = 9 lbs.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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? (1,700 lbs)



2.



It is necessary to slide a 200 lb. 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? (140 lbs, 100 lbs)



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2.2.4. Fluid Dynamics Introduction 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 g/m3 (grams per cubic metre) of water vapour when saturated. If air at this temperature contains only 4.7 g/m3 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 (e.g., density of water vapour in a given parcel), usually expressed in grams per cubic meter. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Actual vapour pressure: The partial pressure exerted by the water vapour present in a parcel. Water in a gaseous state (Le. 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. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Density and Specific Gravity 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, ( ).



Other algebraic forms of this same equation are:



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 2.4 is a table of densities. You can refer to this table when you solve the problems dealing with mass, weight, and volume. Densities of Liquids and Solids Liquids



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3



3



Kg/m



Slug/ft



Water



1000



1.940



Sea Water



1030



Benzine



Metals



3



3



Kg/m



Slug/ft



Aluminium



2700



5.25



2.00



Cast Iron



7200



14.0



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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Mercury



13600



26.3



Non-Metals



Steel



7800



15.1



Tungsten



19000



37.0



Zinc



7140



13.9



Brass



8700



16.9



Ice



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 2.4. Comparison of densities - Liquids and solids.



Specific Gravity The term "specific gravity" is closely related to the idea of density. The definition is as follows:



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 2.4). If we use the metric units (kg/m3) we obtain:



If we use the English units (slug/ft3) we obtain:



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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



What is the specific gravity of kerosene? (0.8)



2.



What is the specific gravity of aluminium? (2.7)



3.



What is the specific gravity of ice? (0.922)



4.



What is the specific gravity of glass? (2.6)



5.



What is the weight of 85 gallons of kerosene? (544 lbs. or 3029 N, Hint : Calculate weight of water, 1 litre= 1 kg, or 1 pint = 1lb, then convert to kerosene by multiplying by its specific gravity of 0.8)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 super fluids) 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 nominal viscosity of 1.0 x 10-3 Pa's and motor oil has a nominal apparent viscosity of 250 x 10-3 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics The fluids without a constant viscosity are called Non-Newtonian fluids. Their viscosity cannot be described by a single number. NonNewtonian 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. 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. 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), Profile drag includes form drag, skin friction, and interference drag Lift-induced drag (also known as vortex drag or induced drag),. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. 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. 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, B = 6 πη r Where: r is the Stokes radius of the particle, and η is the fluid viscosity. Manual No. : BCT-0012/A2 Electrical Avionics



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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. Aerodynamic drag = Cd ½ ρ V2 A Where Cd = drag coefficient (dimensionless) ρ = fluid density (slug/ft or kg/m3) V = Velocity of object (ft/s or m/s) A = projected frontal area (ft2 or m2) 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. 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 2.68 (below) shows a flat plate with the fluid coming from the right and stopping at the plate. The 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.



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Figure 2.68. Effect of airflow on a flat plate.



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 2.69 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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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Figure 2.69. 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 2.70). A pressure higher than the freestream (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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 heavier-than-air aircraft.



Figure 2.70. A combined Pitot tube and static taps.



The Venturi Tube in Bernoulli’s Principle. 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. Bernoulli’s principle was originally stated to explain the action of a liquid flowing through the varying cross-sectional areas of tubes. In Figure 2.71 a tube is shown in which the cross-sectional area gradually decreases to a minimum diameter in its center section. A tube Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics constructed in this manner is called a “venturi,” or “venturi tube.” Where the cross-sectional area is decreasing, the passageway is referred to as a converging duct. As the passageway starts to spread out, it is referred to as a diverging duct. As a liquid (fluid) flows through the venturi tube, the gauges at points “A,” “B,” and “C” are positioned to register the velocity and the static pressure of the liquid. The venturi in Figure 2.71 can be used to illustrate Bernoulli’s principle, which states that: The static pressure of a fluid (liquid or gas) decreases at points where the velocity of the fluid increases, provided no energy is added to nor taken away from the fluid. The velocity of the air is kinetic energy and the static pressure of the air is potential energy.



Figure 2.71. Bernoulli’s principle and a venturi.



In the wide section of the venturi (points A and C of Figure 2.71), the liquid moves at low velocity, producing a high static pressure, as indicated by the pressure gauge. As the tube narrows in the center, it must contain the same volume of fluid as the two end areas. In this narrow section, the liquid moves at a higher velocity, producing a lower pressure than that at points A and C, as indicated by the velocity gauge reading high and the pressure gauge reading low. A good application for the use of the venturi principle is in a float-type carburetor. As the air flows through the carburetor on its way to the engine, it goes through a venturi, where the static pressure is reduced. The fuel in the carburetor, which is under a higher pressure, flows into the lower pressure venturi area and mixes with the air. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Bernoulli’s principle is extremely important in understanding how some of the systems used in aviation work, including how the wing of an airplane generates lift or why the inlet duct of a turbine engine on a subsonic airplane is diverging in shape. The wing on a slow moving airplane has a curved top surface and a relatively flat bottom surface. The curved top surface acts like half of the converging shaped middle of a venturi. As the air flows over the top of the wing, the air speeds up, and its static pressure decreases. The static pressure on the bottom of the wing is now greater than the pressure on the top, and this pressure differences create the lift on the wing. Application of Bernoulli's Principle to Aerofoil Section. The relative wind direction is the direction of the airflow with respect to the wing and is opposite to the path of flight (figure 2.72).



Figure 2.72. Relative Wind



The chord line of a wing is a straight line connecting the leading edge of a wing to its trailing edge (figure 2.73).



Figure 2.73. Chord Line.



Angle of Attack is the angle between the chord line of a wing and the relative wind direction (figure 2.74).



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Figure 2.74. Angle of Attack.



Figure 2.75 shows the cross section of a wing at rest and subject to atmospheric pressure which on the average is 14.7 Ibs/in. 2 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 2.75 Pressure forces on airfoil.



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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. 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 2.76, the following symbols apply: P1 = pressure on the upper surface of the wing P2 = pressure on the lower surface of the wing Va = relative wind velocity V1 = Wind velocity over upper surface V2 = Wind velocity over lower surface ρ = density of the air



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Figure 2.76. Velocities and pressures above and below an aerofoil.



We apply Bernoulli's principle P1 + ½ ρ V12 = P2 + ½ ρ V22 We note that the ones refer to the upper surface and the twos apply to the lower surface of the wing. P1 + ½ ρ V12 = P2 + ½ ρ V22 ½ ρ V12 - ½ ρ V22 = P2 - P1 ½ ρ (V12 - V22)= P2 - P1 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: P = ½ ρ (V12 - V22)



and



Lift (or Force) = P x Area



Lift = P x Area (Note: In some questions, the weight of the aircraft will be quoted. Thus, if the aircraft is flying straight and level, the Lift = Weight).



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



An aeroplane having wing area 500 ft2 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 200 ft/sec. The density of the air is 0.0025 slug/ft3. What is the lift? (75,000 lbs)



2.



An aeroplane having wing area 400 ft2 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 230 ft/sec. The density of the air is 0.0025 slug/ft 3. What is the weight of the aeroplane? (2,350 lbs)



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 ftlsec. The density of the air is 0.001 slug/ft3 The weight of the aeroplane is 12,800 Ibs. What is the wing area? (1000 ft3)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics



Module 2 Physics 2.3 Thermodynamics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Knowledge Levels — Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows: LEVEL 1 A familiarisation with the principal elements of the subject. Objectives: The applicant should be familiar with the basic elements of the subject. The applicant should be able to give a simple description of the whole subject, using common words and examples. The applicant should be able to use typical terms. LEVEL 2 A general knowledge of the theoretical and practical aspects of the subject. An ability to apply that knowledge. Objectives: The applicant should be able to understand the theoretical fundamentals of the subject. The applicant should be able to give a general description of the subject using, as appropriate, typical examples. The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. The applicant should be able to read and understand sketches, drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using detailed procedures. LEVEL 3 A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: The applicant should know the theory of the subject and interrelationships with other subjects. The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. The applicant should understand and be able to use mathematical formulae related to the subject. The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.



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Table of Contents Module 2.3 Thermodynamics ______________________________________________________________________________________ Temperature __________________________________________________________________________________________________ Heat _________________________________________________________________________________________________________ Heat Transfer __________________________________________________________________________________________________ Thermal Expansion _____________________________________________________________________________________________ The Gas Laws __________________________________________________________________________________________________ The General (Ideal) Gas Law ______________________________________________________________________________________ First and Second Law ofThermodynamics ___________________________________________________________________________ An Adiabatic Process ___________________________________________________________________________________________



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Module 2.3 Enabling Objectives Objective



Reference



Level



Thermodynamics



2.3



Temperature: thermometers and temperature scales:



(a)



2



(b)



2



Celsius, Fahrenheit and Kelvin; Heat definition



Heat capacity, specific heat Heat transfer: convection, radiation and conduction; Volumetric expansion First and second law of thermodynamics Gases: ideal gases laws; specific heat at constant volume and constant pressure, work done by expanding gas Isothermal, adiabatic expansion and compression, engine cycles, constant volume and constant pressure, refrigerators and heat pumps Latent heats of fusion and evaporation, thermal energy, heat of combustion



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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 3.1, 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:



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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics These formulas are: K=C+273



and



R= F + 460



Boiling



Freezing



Absolute



Point of



Point of



Zero



Water



Water



Centigrade



100º



0º



-273º



Kelvin



373



273



0



Fahrenheit



212º



32º



-460º



Rankin



672º



492º



0º



Table 3.1. Comparisons of boiling points, freezing points and absolute zero in different units. Notes: • Kelvin has no 0 sign in-front of the K. • The accurate conversion factor for ºC to K is +273.15



Figure 3.1. Temperature scale comparison.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



Change 20°C to degrees F. (68ºF)



2.



Change -15°C to degrees F. (5ºF)



3.



Change 86°F to degrees C. (30ºC)



4.



Change _4°F to degrees C. (-20ºC)



5.



Change 100°F to degrees R. (560ºR)



6.



Change 450ºR to degrees F. (-10ºF)



7.



Change 100°C to degrees K. (373 K)



8.



Change 383 K to degrees C. (110ºC)



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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 is 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. 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 lb. of fuel, or Btu per lb. 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 lb. = 2.2 kg, 1 Btu/lb. = 2326 J/kg. 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=wC T Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics In this equation: Q = heat gained or lost (Btu) w = weight of the body (lb.) C = the specific heat of the substance Btu/lb. ºf T = the temperature change (ºF or ºR) Q=mC T



(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/kgºC) T = the temperature change (ºC 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. Table 3.2 shows various specific heats of substances in English and Metric units.



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Table 3.2. Specific Heat Capacities of some common substances. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Specific Heat 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.



For example: Specific Heat Capacity of Dry Air At Constant Pressure



Cp



1004 J / K kg



At Constant Volume



Cv



717 J / K kg



Therefore



Examples: 1.



How much heat must be supplied to raise the temperature of a 32 lb. aluminium fitting from 60°F to 90°F? Q = w C T Q = ( 0.212 Btu / lb.ft.) (32 lbs.) (30ºF) Q = 204 Btu.



2.



How much heat is given up as 100 Ibs. of sea water cools from 90°F to 50°F? Q = w C T Q = ( 0.93 Btu / lb.ft.) (100 lbs.) (40ºF) Q = 3720 Btu



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Heat Transfers 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.



Figure 3.2. Heat Transfer – Convection



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 solids-but 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 colder because they conduct heat away from your hand. You perceive the heat that is leaving your hand as cold.



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Figure 3.3. Heat transfer– conduction.



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 3.4. Heat transfer – radiation.



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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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



T



= the temperature difference in ºF



L



= the thickness of the insulation in inches



k



= the thermal conductivity of the material from



which the insulation is made.



THERMAL CONDUCTIVITIES (Btu-IN./FT.2-HR.-ºF) Air



0.17



Corkboard



0.30



Cotton



0.54



Fiberboard



0.42



Foam Plastic



0.30



Glass Wool



0.27



Table 3.3. Thermal conductivities of some common materials.



Example: An outside wall of a house has total cross-sectional area of 2,000 ft.2 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?



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



How much heat must be supplied to raise the temperature of 67 Ibs. of ethyl alcohol from 32°F to 76°F? (2,064 Btu)



2.



How much heat is given up as 780 Ibs. of steel cool from 90°F to 45°F? (3,860 Btu)



3.



If 1 lb. of vodka (alcohol) at 90°F is mixed with 0.2 lb. of water at 40°F what is the final temperature? (79°F)



4.



If 3 Ibs. of hot water at 200ºF are poured into a 1.5 Ibs. aluminium container at 40ºF, what is the final temperature? (180°F)



5.



A house has an outside wall area of 3,000 ft2 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? (13,500 Btu/hr)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Thermal Expansion 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: L = α L0 T In this formula, L0 = the original length of the rod α = the coefficient of linear expansion L = the change in length of the rod T = the change in temperature Area Expansion Two-dimensional solid bodies also experience thermal area expansion. The formula is as follows: A = 2α A0 T In this formula, A0 = the original area of the body α = the coefficient of linear expansion A =the change in area of the body T = the change in temperature Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics COEFFICIENTS OF LINEAR EXPANSION (α) SUBSTANCE



Per Fº



Aluminum



13 x 10-6



Brass



10 x 10-6



Concrete (varles)



5 x 10-6



Copper



9.4 x 10-6



Glass (Pyrex)



1.6 x 10-6



Ice



28 x 10-6



Iron



6.6 x 10-6



Lead



16 x 10-6



Steel



11 x 10-6 Table 3.4. Coefficients of Linear Expansion



Volume Expansion Three-dimensional solid bodies experience volume expansion. V = 3α V0 T In this formula, V0 = the original volume of the body α = the coefficient of linear expansion V =the change in volume of the body T = the change in temperature



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. COEFFICIENTS OF VOLUME EXPANSION (β) LIQUIDS



Per Fº



Ethyl Alcohol



0.60 x 10-3



Methyl Alcohol



0.66 x 10-3



Benzene



0.69 x 10-3



Gasoline



0.58 x 10-3



Table 3.5. 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Problems 1.



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 35ºF? (0.0585 ft)



2.



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? (0.099 ft.)



3.



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 of and the highest recorded temperature is 115 of. How much space should the builders leave between each section? (0.20 ft)



4.



The volume of an aluminium tank is 200 gallons on a day when the temperature is 30 of. 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? (3.7 gallons)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics The Gas Laws 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 3.5). 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.



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Figure 3.5. Boyle’s Law.



Example: A cylinder fitted with a piston contains gas at a pressure of 35.5 Ibs./in2 as indicated by a gauge mounted to the outside of the cylinder. The atmospheric pressure is 14.5 Ibs./in2 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./in2 P1 = 50 lbs./in2 V2 = ¼V1 P1V1 = P2V2 (50 lbs./in2) (V1) = P2 (¼V1) Solving for P2 gives, P2 = 200 lbs./in2 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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Let us again consider a sample of gas containing a definite number of molecules. We 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 3.6). 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 3.6. 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: Manual No. : BCT-0012/A2 Electrical Avionics



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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 3.7). Using absolute pressures and temperatures the following simple relationship is obtained:



This equation is referred to as Gay-Lussac's Law.



Figure 3.7. Gay Lussac’s Law.



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Basic Aircraft Maintenance Training Manual 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 tyre does not change and the atmospheric pressure is 14.7 Ibs./in.2 We must first convert to absolute temperatures and pressures. P1 = 50.0 Ibs./in2+ 14.7Ibs./in2= 64.7 lbs./in2 T1 = 460 + 58ºF = 518ºR T2 = 460 + 86ºF = 546ºR Substituting these values into Gay-Lussac's Law gives:



Solving for P2, we obtain P2 = 68.2 lbs./in2. Finally, the new gauge pressure is obtained by subtracting the atmospheric pressure from P2. 68.2 Ibs./in2 - 14.7 Ibs./in2 = 53.5 Ibs./in2 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:



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



Figure 3.8. Ideal Gas Law.



Example: A tank of helium gas has a gauge pressure of 50.2 Ibslin2 and a temperature of 45°F. A piston decreases the volume of the gas to 68% of its original volume and the temperature drops to 10°F. What is the new gauge pressure? Assume normal atmospheric pressure. We must change both temperatures to absolute units. We must change the original gauge pressure to absolute pressure. We remember that when the final pressure is obtained it will be in absolute units. We also note that V2 = 0.68 V1.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics We transfer V2 from the numerator on the right to the denominator on the left. We also transfer T 2 from the denominator on the right to the numerator on the left. In this way, we solve our formula for P2.



Next we substitute our known values:



Alternate Form of the General (Ideal) Gas Law The general gas law tells us that for a fixed quantity of gas, the expression PVIT is constant. Since PVIT 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 3.6 gives values of R for various gases. Values of the Gas Constant, R, for Some Common Gases Pa m3 / kg K



Ft.lbs / slug ºR



Air



287



1,710



Carbon Dioxide



189



1,130



2,077



12,380



Nitrogen



297



1,770



Oxygen



260



1,550



Water Vapour



462



2,760



Helium



Table 3.6. Gas Constants, R.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics We can write:



PV = mRT If we divide both sides of this equation by V, we obtain:



We remember that the density of any substance is given by:



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:



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 known as Standard Temperature and Pressure. These are 0ºC and 1 atmosphere (273.15 K and 760 mmHg). Example: Find the density of air if the temperature is 80°F and the absolute pressure is 2,150 Ibs./ft2



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Basic Aircraft Maintenance Training Manual 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 240 kPa when the volume is 0.15 m3. What will the volume be when the absolute pressure of the gas is changed to 80 kPa while the temperature is held constant? (0.45 m3)



2.



A quantity of gas is contained in a cylinder fitted with a piston. The gauge pressure of the gas in the cylinder is 335 Ibs/in2 when the volume occupied by the gas is 72in3 What is the gauge pressure when the volume is decreased to 60 in3? Assume atmospheric pressure to be 151bs/in2, and assume that the temperature is held constant. (405 lb./in2)



3.



A sample of nitrogen is held at an absolute pressure of 1.50 atmospheres and a volume of 7.80 m3. A piston gradually reduces the volume to 6.30 m3. The temperature does not change. What is the new absolute pressure in atmospheres? (1.86 atmosphere)



4.



A volume of 1.35 m3 of air at 17ºC is heated to 42ºC while its pressure is held constant. What is the volume of the gas at this elevated temperature. (3.26 m3)



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? (0.007 slug/ft3)



6.



The air pressure and density at a point on the wing of a Boeing 747 flying at altitude are 70 kPa, and 0.9 kg/m 3 respectively. What is the temperature at this point on the wing in degrees Centigrade? (-2ºC)



7.



The Goodyear non-rigid airship, the Mayflower, has a volume of 4000 m3 and is filled with helium to an absolute pressure of 100 kPa. The temperature is 27ºC. Find the density and total mass of the helium in the ship. (0.16 kg/m3 , 640 kg)



8.



At an altitude of 8,000 ft. the absolute temperature of air is 500ºR and the absolute pressure is 1600 Ibs/ft2 What is the density of air at this altitude? (0.00188 slug/ft3)



9.



A tank of carbon dioxide is maintained at an absolute pressure of 5,830 Ibs/ft2 and a temperature of 70°F. What is the density of this carbon dioxide? (0.01 slug/ft3)



10. A quantity of air occupying 0.9 ft3 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? (0.3 ft3)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics First and Second Law of Thermodynamics 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 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 air-lines 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 3.9. 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 3.9. An internal combustion engine converts chemical energy into heat energy and then into mechanical work.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Hence our Work formula is now: Work = Pressure x Area x Distance (where ‘Distance’ is the movement of piston) If we simplify the diagram as shown in Figure 3.9, 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. Our formula for work then becomes: Work = Pressure x Change in Volume Work = P x V ( indicates a ‘change in’ the quantity of 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 Work = PV 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. First and Second Law of Thermodynamics The first law of thermodynamics, also known as Law of Conservation of Energy, states that energy can neither be created nor destroyed; it can only be transferred or changed from one form to another. For example, turning on a light would seem to produce energy; however, it is electrical energy that is converted (as shown in figure 3.10 below) A way of expressing the first law of thermodynamics is that any change in the internal energy (∆E) of a system is given by the sum of the heat (q) that flows across its boundaries and the work (w) done on the system by the surroundings: ΔE=q+w



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Basic Aircraft Maintenance Training Manual Module 2 – Physics This law says that there are two kinds of processes, heat and work, that can lead to a change in the internal energy of a system. Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the world outside the system. In other words, energy cannot be created or destroyed. If heat flows into a system or the surroundings to do work on it, the internal energy increases and the sign of q or w is positive. Conversely, heat flow out of the system or work done by the system will be at the expense of the internal energy, and will therefore be negative. The second law of thermodynamics explains that it is impossible to have a cyclic (repeating) process that converts heat completely into work. It is also impossible to have a process that transfers heat from cool objects to warm objects without using work. The second part of the law is more obvious. A cold body can't heat up a warm body. Heat naturally wants to flow from warmer to cooler areas. Heat wants to flow and spread out to areas with less heat. If heat is going to move from cooler to warmer areas, it is going against what is “natural”, so the system must put in some work for it to happen.



Figure 3.10. (a) The first law of thermodynamics.



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Figure 3.10. (b) The second law of thermodynamics.



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 of fusion, 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 vaporization 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 3.7.



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Table 3.7. Latent heats, freezing points and boiling points of some common.



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 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics How do you make 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. 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). 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 (the simple latent heat diagram is shown on figure 3.11. below)



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Figure 3.11. Latent heat diagram.



Refrigerant 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. Some of them are as follows:



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Table 3.8. Some common refrigerants.



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Figure 3.12. Main components of a refrigeration system



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? It is the removal and relocation of heat. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Air-source 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 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. How heat pumps work.



Figure 3.13. Heat Pump Cycle.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Stage 1 The heat transfer medium inside the evaporator (the ‘refrigerant’) is colder than the heat source (the air or ground). As the glycol (GSHP) or outside air (ASHP) Passes across the first heat exchanger (the evaporator) the liquid refrigerant absorbs the low temperature heat energy and evaporates. Stage 2 The vapour then flows to the compressor and is compressed. When compressed the pressure of this vapour is increased and the temperature of the vapour rises, effectively concentrating the heat, elevating the temperature. Stage 3 The hot vapour flows to the second heat exchanger (the condenser) where the heat is transferred into the heating distribution system. As this heat energy is transferred, vapour condenses back into a liquid. Stage 4 The liquid refrigerant then passes through an expansion valve reducing its pressure, this process is the exact reverse of the compression function in stage 2 and temperature decreases ready to start the whole cycle once again.



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Module 2 Physics 2.4 Optics (Light)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Knowledge Levels — Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows: LEVEL 1 A familiarisation with the principal elements of the subject. Objectives: The applicant should be familiar with the basic elements of the subject. The applicant should be able to give a simple description of the whole subject, using common words and examples. The applicant should be able to use typical terms. LEVEL 2 A general knowledge of the theoretical and practical aspects of the subject. An ability to apply that knowledge. Objectives: The applicant should be able to understand the theoretical fundamentals of the subject. The applicant should be able to give a general description of the subject using, as appropriate, typical examples. The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. The applicant should be able to read and understand sketches, drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using detailed procedures. LEVEL 3 A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: The applicant should know the theory of the subject and interrelationships with other subjects. The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. The applicant should understand and be able to use mathematical formulae related to the subject. The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.



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Table of Contents Module 2.4 Optics (Light) _________________________________________________________________________________________ Wavelength, Frequency and Speed ________________________________________________________________________________ Light Waves in Matter ___________________________________________________________________________________________ Lenses________________________________________________________________________________________________________ Fibre Optics ___________________________________________________________________________________________________ Fibre Optic Data Links ___________________________________________________________________________________________ The Transmission of Signals ______________________________________________________________________________________



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Module 2.4 Enabling Objectives Objective



Reference



Optics (Light)



2.4



Level



2



Nature of Light; speed of light Laws of reflection and refraction; reflection at plane surfaces, reflection by spherical mirrors, refraction, lenses Fibre optics



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2.4



Optics



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 times and has the value, to three significant digits, 3.00 x 108 m/sec. The wavelength of visible light is usually measured in a unit called the Angstrom (Å). 1 Å = 10-10m 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 Å are not visible and are called "ultraviolet" and wavelengths longer than 7,000 Å 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 Å. Colours gradually change as the wavelength changes. 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=fλ Violet



4500 Å



Blue



4800 Å



Green



5200 Å



Yellow



5800 Å



Orange



6000 Å



Red



6400 Å



Table 4.1. The wavelengths of various colours of light measured in Å.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics In table 4.2, there are listed 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 decreases in such a way that the product equals the speed of light (3 x 108 m/sec.). Freq



λ



Freq



λ



(Hz)



(m)



(Hz)



(m)



10



23



10



22



10



21



10



20



10



19



10



18



10



17



10



Cosmic



10



-14



10



13



Rays



10



-13



10



12



10



-12



10



11



10



-11



10



10



10



-10



10



9



10



-9



10



8



10



-8



10



7



16



10



-7



10



6



10



15



10



-6



10



5



10



14



10



-5



10



4



Gamma Rays



X-Rays



Ultraviolet



Visible



Infrared



Microwave / Radar



10



-4



10



-3



10



-2



10



-1



1 TV / FM Radio Shortwave Radio AM Radio Maritime Comm



10



1



10



2



10



3



10



4



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. Manual No. : BCT-0012/A2 Electrical Avionics



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Figure 4.1. The Electrical (E) and magnetic (B) components of the electromagnetic wave.



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 depends 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 λm 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).



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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.



Refractive Index Varies with Wavelength The 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. 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.



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Figure 4.3. Refractive index varies with wavelength.



The Law of Refraction When light travels from one medium to another, it generally bends, or refracts. The law of refraction gives us a way of predicting the amount of bend. This law is more complicated than that for reflection, but an understanding of refraction will be necessary for our future discussion of lenses and their applications. The law of refraction is also known as Snell's Law, named for Willobrord Snell, who discovered the law in 1621 Like with reflection, refraction also involves the angles that the incident ray and the refracted ray make with the normal to the surface at the point of refraction. Unlike reflection, refraction also depends on the media through which the light rays are travelling. This dependence is made explicit in Snell's Law via refractive indices, numbers which are constant for given



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Snell's Law is given in the following diagram.



Figure 4.4. Snell’s Law diagram.



As in reflection, we measure the angles from the normal to the surface, at the point of contact. The constants n are the indices of refraction for the corresponding media. 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Figure 4.4. Dispersion and chromatic aberration.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Lenses The Anatomy of Lenses If a piece of glass or other transparent material takes on the appropriate shape, it is possible that parallel incident rays would either converge to a point or appear to be diverging from a point. A piece of glass that has such a shape is referred to as a lens.



Figure 4.5. A set of prisms acting as a converging and diverging lens.



A lens is merely a carefully ground or molded piece of transparent material that refracts light rays in such as way as to form an image. Lenses can be thought of as a series of tiny refracting prisms, each of which refracts light to produce their own image. When these prisms act together, they produce a bright image focused at a point. Type of Lenses There are a variety of types of lenses. Lenses differ from one another in terms of their shape and the materials from which they are made. Our focus will be upon lenses that are symmetrical across their horizontal axis - known as the principal axis. In this unit, we will categorize lenses as converging lenses and diverging lenses. A converging lens is a lens that converges rays of light that are traveling parallel to its principal axis. Converging lenses can be identified by their shape; they are relatively thick across their middle and thin at their upper and lower edges. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics A diverging lens is a lens that diverges rays of light that are traveling parallel to its principal axis. Diverging lenses can also be identified by their shape; they are relatively thin across their middle and thick at their upper and lower edges.



Figure 4.6. Converging and Diverging Lenses



Converging Lens Ray Diagrams In this section we will investigate the method for drawing ray diagrams for objects placed at various locations in front of a double convex lens. To draw these ray diagrams, we will have to recall the three rules of refraction for a double convex lens: Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens. Any incident ray traveling through the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis. An incident ray that passes through the center of the lens will in effect continue in the same direction that it had when it entered the lens. Manual No. : BCT-0012/A2 Electrical Avionics



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Figure 4.7. Image Formation by a Converging Lens



In this diagram, five incident rays are drawn along with their corresponding refracted rays. Each ray intersects at the image location and then travels to the eye of an observer. Every observer would observe the same image location and every light ray would follow the Snell's Law of refraction. Yet only two of these rays would be needed to determine the image location since it only requires two rays to find the intersection point. Of the five incident rays drawn, three of them correspond to the incident rays described by our three rules of refraction for converging lenses. We will use these three rays through the remainder of this lesson, merely because they are the easiest rays to draw. Certainly two rays would be all that is necessary; yet the third ray will provide a check of the accuracy of our process. Step by step methods for drawing ray diagrams The method of drawing ray diagrams for double convex lens is described below. The description is applied to the task of drawing a ray diagram for an object located beyond the 2F point of a double convex lens. 1.



Pick a point on the top of the object and draw three incident rays traveling towards the lens.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Draw the third incident ray such that it travels directly to the exact center of the lens. Place arrowheads upon the rays to indicate their direction of travel.



Figure 4.6. Step 1.



2.



Once these incident rays strike the lens, refract them according to the three rules of refraction for converging lenses. The ray that passes through the focal point on the way to the lens will refract and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray that traveled parallel to the principal axis on the way to the lens will refract and travel through the focal point. And the ray that traveled to the exact center of the lens will continue in the same direction. Place arrowheads upon the rays to indicate their direction of travel. Extend the rays past their point of intersection.



Figure 4.7. Step 2.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 3.



Mark the image of the top of the object. The image point of the top of the object is the point where the three refracted rays intersect. All three rays should intersect at exactly the same point. This point is merely the point where all light from the top of the object would intersect upon refracting through the lens. Of course, the rest of the object has an image as well and it can be found by applying the same three steps to another chosen point.



Figure 4.8. Step 3.



4.



Repeat the process for the bottom of the object One goal of a ray diagram is to determine the location, size, orientation, and type of image that is formed by the double convex lens. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the mirror as the image of the top of the object. At this point the entire image can be filled in.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Object – Image Relations Case 1: The object is located beyond 2F When the object is located at a location beyond the 2F point, the image will always be located somewhere in between the 2F point and the focal point (F) on the other side of the lens. Regardless of exactly where the object is located, the image will be located in this specified region. In this case, the image will be an inverted image. That is to say, if the object is right side up, then the image is upside down. In this case, the image is reduced in size; in other words, the image dimensions are smaller than the object dimensions. If the object is a six-foot tall person, then the image is less than six feet tall. There is a thing called the magnification, which is the ratio of the height of the object to the height of the image. In this case, the magnification is a number with an absolute value less than 1. Finally, the image is a real image. Light rays actually converge at the image location. If a sheet of paper were placed at the image location, the actual replica or likeness of the object would appear projected upon the sheet of paper.



Figure 4.10. Case 1.



Case 2: The object is located at 2F When the object is located at the 2F point, the image will also be located at the 2F point on the other side of the lens. In this case, the image will be inverted (i.e., a right side up object results in an upside-down image). The image dimensions are equal to the object dimensions. A six-foot tall person would have an image that is six feet tall; the absolute value of the magnification is exactly 1. Finally, the image is a real image. Light rays actually converge at the image location. As such, the image of the object could be projected upon a sheet of paper.



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Figure 4.11. Case 2.



Case 3: The object is located between 2F and F When the object is located in front of the 2F point, the image will be located beyond the 2F point on the other side of the lens. Regardless of exactly where the object is located between 2F and F, the image will be located in the specified region. In this case, the image will be inverted (i.e., a right side up object results in an upside-down image). The image dimensions are larger than the object dimensions. A six-foot tall person would have an image that is larger than six feet tall. The absolute value of the magnification is greater than 1. Finally, the image is a real image. Light rays actually converge at the image location. As such, the image of the object could be projected upon a sheet of paper.



Figure 4.12. Case 3.



Case 4: The object is located at F When the object is located at the focal point, no image is formed. The refracted rays neither converge nor diverge. After refracting, the light rays are traveling parallel to each other and cannot produce an image.



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Figure 4.13. Case 4.



Case 5: The object is located in front of F When the object is located at a location in front of the focal point, the image will always be located somewhere on the same side of the lens as the object. Regardless of exactly where in front of F the object is located, the image will always be located on the object's side of the lens and somewhere further from the lens. The image is located behind the object. In this case, the image will be an upright image. That is to say, if the object is right side up, then the image will also be right side up. In this case, the image is enlarged; in other words, the image dimensions are greater than the object dimensions. A six-foot tall person would have an image that is larger than six feet tall. The magnification is greater than 1. Finally, the image is a virtual image. Light rays diverge upon refraction; for this reason, the image location can only be found by extending the refracted rays backwards on the object's side the lens. The point of their intersection is the virtual image location. It would appear to any observer as though light from the object were diverging from this location. Any attempt to project such an image upon a sheet of paper would fail since light does not actually pass through the image location.



Figure 4.14. Case 5.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 t is the total length of the lens. Diverging Lens Ray Diagrams Three simple rules of refraction for double concave lenses: Any incident ray traveling parallel to the principal axis of a diverging lens will refract through the lens and travel in line with the focal point (i.e., in a direction such that its extension will pass through the focal point). Any incident ray traveling towards the focal point on the way to the lens will refract through the lens and travel parallel to the principal axis. An incident ray that passes through the center of the lens will in effect continue in the same direction that it had when it entered the lens. These three rules will be used to construct ray diagrams. A ray diagram is a tool used to determine the location, size, orientation, and type of image formed by a lens. Step by step methods for drawing ray diagrams The method of drawing ray diagrams for a double concave lens is described below. 1.



Pick a point on the top of the object and draw three incident rays traveling towards the lens.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Using a straight edge, accurately draw one ray so that it travels towards the focal point on the opposite side of the lens; this ray will strike the lens before reaching the focal point; stop the ray at the point of incidence with the lens. Draw the second ray such that it travels exactly parallel to the principal axis. Draw the third ray to the exact center of the lens. Place arrowheads upon the rays to indicate their direction of travel.



Figure 4.15. Step 1.



2.



Once these incident rays strike the lens, refract them according to the three rules of refraction for double concave lenses. The ray that travels towards the focal point will refract through the lens and travel parallel to the principal axis. Use a straight edge to accurately draw its path. The ray that traveled parallel to the principal axis on the way to the lens will refract and travel in a direction such that its extension passes through the focal point on the object's side of the lens. Align a straight edge with the point of incidence and the focal point, and draw the second refracted ray. The ray that traveled to the exact center of the lens will continue to travel in the same direction. Place arrowheads upon the rays to indicate their direction of travel. The three rays should be diverging upon refraction.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 3.



Locate and mark the image of the top of the object. The image point of the top of the object is the point where the three refracted rays intersect. Since the three refracted rays are diverging, they must be extended behind the lens in order to intersect. Using a straight edge, extend each of the rays using dashed lines. Draw the extensions until they intersect. All three extensions should intersect at the same location. The point of intersection is the image point of the top of the object. The three refracted rays would appear to diverge from this point. This is merely the point where all light from the top of the object would appear to diverge from after refracting through the double concave lens.



Figure 4.17. Step 3.



4.



Repeat the process for the bottom of the object. The goal of a ray diagram is to determine the location, size, orientation, and type of image that is formed by the double concave lens. Typically, this requires determining where the image of the upper and lower extreme of the object is located and then tracing the entire image. After completing the first three steps, only the image location of the top extreme of the object has been found. Thus, the process must be repeated for the point on the bottom of the object. If the bottom of the object lies upon the principal axis (as it does in this example), then the image of this point will also lie upon the principal axis and be the same distance from the lens as the image of the top of the object. At this point the complete image can be filled in.



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Figure 4.18. Step 4.



Object – Image Relations Ray diagrams were constructed in order to determine the location, size, orientation, and type of image formed by double concave lenses (i.e., diverging lenses). The ray diagram constructed earlier for a diverging lens revealed that the image of the object was virtual, upright, reduced in size and located on the same side of the lens as the object. Here are ray diagrams of diverging lenses.



Figure 4.19. Ray Diagram of Diverging Lens.



The diagrams above show that in each case, the image is located on the object' side of the lens a virtual image an upright image reduced in size (i.e., smaller than the object) Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Unlike converging lenses, diverging lenses always produce images that share these characteristics. The location of the object does not affect the characteristics of the image. As such, the characteristics of the images formed by diverging lenses are easily predictable. Another characteristic of the images of objects formed by diverging lenses pertains to how a variation in object distance affects the image distance and size. The diagram below shows five different object locations (drawn and labeled in red) and their corresponding image locations (drawn and labeled in blue).



Figure 4.20. Five different object locations with its formed images by diverging lenses.



The diagram shows that as the object distance is decreased, the image distance is decreased and the image size is increased. So as an object approaches the lens, its virtual image on the same side of the lens approaches the lens as well; and at the same time, the image becomes larger. 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. The principal focus of the lens is 10 cm from the lens, so its focal length is -10 cm. The lens formula gives: Manual No. : BCT-0012/A2 Electrical Avionics



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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. 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



•



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 4.21 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics as data input.



Figure 4.21. Parts of 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 (LEOs) 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 4.21 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. 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: Good 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Freedom from short circuit and sparks Reduced size and weight cables Environmental Protection Resistant to radiation and corrosion Resistant to temperature variations 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Figure 4.22. The Electromagnetic Spectrum, with wavelength 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. 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 means that if the light frequency doubled, the losses would increase by a factor of 24 or 16 times. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 4.23. 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 n1. 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 4.23. Basic structure of an optical fibre.



The cladding layer is made of a dielectric material with an index of refraction n2. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Adds mechanical strength For extra protection, the cladding is enclosed in an additional layer called the coating or buffer. 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. 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 t 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. 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Bound rays propagate through the fibre by total internal reflection. Unbound rays are refracted out of the fibre core. Figure 4.24 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 n1. The cladding of a step-index has an index of refraction n2, that is lower than n1. Figure 4.24 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. 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.



Figure 4.24. Bound and unbound rays in a step-index fibre.



How a light ray is launched into a fibre is shown in figure 4.25. The incident ray I 1 enters the fibre at the angle θa. I1 is refracted upon entering the fibre and is transmitted to the corecladding interface. The ray then strikes the core-cladding interface at the critical angle (θc). I1 is totally reflected back into the core and continues to propagate along the fibre. The incident ray I 2 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. I2 strikes the core-cladding interface at an angle less than the critical angle (θc). I2 is refracted into the cladding and is eventually lost. The light ray incident on the fibre core must be within the cone of acceptance defined by the angle θa shown In figure 4.26. Angle θa is defined as the acceptance angle. The acceptance angle (θa) is the maximum angle to the axis of the fibre that light Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics entering the fibre is propagated. The value of the angle of acceptance (θa) depends on fibre properties and transmission conditions.



Figure 4.25. How a light ray enters an optical fibre.



Figure 4.26. 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 4.26 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 n2. 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: Manual No. : BCT-0012/A2 Electrical Avionics



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Since the medium next to the fibre at the launching point is normally air, no 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. Skew Rays A possible path of propagation of skew rays is shown in figure 4.27. Figure 4.27, 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.



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Figure 4.27. 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. Large portions 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 (λ) of the plane wave is given by:



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where c is the speed of light in a vacuum, fis the frequency of the light, and n is the index of refraction of the plane-wave medium. Figure 4.28 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.



Figure 4.28. 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 4.29. 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 4.29. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 4.29. 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 4.29. 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. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. 4.30). Some dispersion occurs in all types of fibres.



Figure 4.30. The spreading of a light pulse.



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 as multimode fibres. Single mode fibres are also manufactured by following the same fabrication process as multimode fibres. 1. 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 (A). 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. Signal loss depends on the operational wavelength (A). 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. 2. Multi mode 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 ).1m 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 (LEOs). Single mode fibres typically must use laser diodes. LEOs are cheaper, less complex, and last longer. LEOs 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 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. Attenuation reduces the amount of optical power transmitted by the fibre. Attenuation controls the distance an optical signal (pulse) can travel. 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. An error occurs when the receiver is unable to distinguish between input pulses caused by the spreading of each pulse. 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 Imperfections in the atomic structure induce absorption by the presence of missing molecules or oxygen defects The intrinsic or basic material-material properties 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. The extrinsic (presence of impurities) material-material properties 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics ions from one energy level to another. Scattering 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. Light is then partially scattered in all directions. In commercial fibres operating between 700-nm and 1600-nm wavelength, the main source of loss is called Rayleigh scattering - the main loss mechanism between the ultraviolet and infrared regions, it 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 proportional to the fourth power of the wavelength (1 / λ4). 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.



Figure 4.31. Rayleigh and Mie Scattering.



Losses Bending Loss Bending the fibre also causes attenuation. Bending loss is classified according to the bend radius of curvature: Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Microbend (small microscopic bends of the fibre axis that occur mainly when a fibre is cabled) loss 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. Macrobend (bends having a large radius of curvature relative to the fibre diameter) loss. 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 highorder 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 1. Intramodal Dispersion lntramodal, or chromatic, dispersion depends primarily on fibre materials. lntramodal dispersion occurs because different colours of light travel through different materials and different waveguide structures at different speeds.There are two types of intramodal dispersion: a) Material dispersion. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. b) Waveguide dispersion. Waveguide dispersion occurs because the mode propagation constant (P) 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. 2. Intermodal Dispersion lntermodal 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 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. 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. 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.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 (EM I) or High Intensity Radiated Fields (HIRF) will also give rise to electrical noise in the non-fibre 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 o. 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 O. The signal is then generated as a perfect copy of the original signal.



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Module 2 Physics 2.5 Wave Motion and Sound



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Knowledge Levels — Category A, B1, B2 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1 and B2 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each applicable subject. Category C applicants must meet either the category B1 or the category B2 basic knowledge levels. The knowledge level indicators are defined as follows: LEVEL 1 A familiarisation with the principal elements of the subject. Objectives: The applicant should be familiar with the basic elements of the subject. The applicant should be able to give a simple description of the whole subject, using common words and examples. The applicant should be able to use typical terms. LEVEL 2 A general knowledge of the theoretical and practical aspects of the subject. An ability to apply that knowledge. Objectives: The applicant should be able to understand the theoretical fundamentals of the subject. The applicant should be able to give a general description of the subject using, as appropriate, typical examples. The applicant should be able to use mathematical formulae in conjunction with physical laws describing the subject. The applicant should be able to read and understand sketches, drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using detailed procedures. LEVEL 3 A detailed knowledge of the theoretical and practical aspects of the subject. A capacity to combine and apply the separate elements of knowledge in a logical and comprehensive manner. Objectives: The applicant should know the theory of the subject and interrelationships with other subjects. The applicant should be able to give a detailed description of the subject using theoretical fundamentals and specific examples. The applicant should understand and be able to use mathematical formulae related to the subject. The applicant should be able to read, understand and prepare sketches, simple drawings and schematics describing the subject. The applicant should be able to apply his knowledge in a practical manner using manufacturer's instructions. The applicant should be able to interpret results from various sources and measurements and apply corrective action where appropriate.



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Table of Contents Module 2.5 Wave Motion and Sound _______________________________________________________________________________ Resonance ____________________________________________________________________________________________________ Sound ________________________________________________________________________________________________________ Beats ________________________________________________________________________________________________________ Supersonic Speed and Mach Number ______________________________________________________________________________ The Doppler Effect ______________________________________________________________________________________________



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Module 2.5 Enabling Objectives Objective



Reference



Wave Motion and Sound



2.5



Level



2



Wave motion; mechanical waves, sinusoidal wave motion, interference phenomena, standing waves Sound: speed of sound, production of sound, intensity, pitch and quality, Doppler effect



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2.5 Wave Motion and Sound 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 surface at the point (P), the pattern is moving to the right in the diagram below. 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).



Figure 5.2. Waveform dimensions



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



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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:



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:



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=vT Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics From this equation, we have:



And finally: fλ=v v=fλ Examples: 1.



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.



2.



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?



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 is 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics We can write a general relation as follows:



The vibration where n = 1 is called the fundamental mode of vibration of the body. The other vibrations are called overtone vibrations. Everybody 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 relatively large amplitude. We say that the body is vibrating in resonance with the impressed frequency.



Figure 5.4. Normal modes of vibration.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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? (5 Hz) 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 ft/sec) 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? (2 ft)



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Basic Aircraft Maintenance Training Manual Module 2 – Physics Sounds 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 cord 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". 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 0ºC. 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 atmospheric pressure that day is 14.7 Ibs/in2, 256 times each second the air pressure is slightly above 14.7 Ibs/in2 and 256 times each second the pressure is slightly below 14.7 Ibs/in2 it should be emphasized that "slightly" means very small. The human ear is a remarkably sensitive instrument. It can detect air pressure variations as small as about 0,000000005 Ibs./in.2 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:



In this equation IL is in decibels. The intensity, (Io) is the intensity of the "threshold of hearing", the softest sound that the average Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics human ear can detect. Also in the equation, I is the intensity of the sound we are measuring. We note that: Io = 10-12 Watts / m3 We also review that the log 10n = n. 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 ad 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. 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 anybody 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Figure 5.5. Constructive and destructive interference.



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



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Figure 5.6. Noise cancelling headphones.



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 displacementof the forward wave is 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 displacementof 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 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 twowaves. 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. Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 5.7. Stationary waves and harmonics.



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics 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 5.8. 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 (vo) 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: Manual No. : BCT-0012/A2 Electrical Avionics



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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 instant the source of sound was approaching your ear and after this instant, the source of sound is receding from your ear.



Figure 5.9. 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 5.10. 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 Manual No. : BCT-0012/A2 Electrical Avionics



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Basic Aircraft Maintenance Training Manual Module 2 – Physics approach each other. Also, the ear hears a lower frequency if the source and observer recede from each other.



Figure 5.10. Doppler effect caused by a stationary sound source and moving ear.



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