Module 2.1 PDF - Thakur Institute of Aviation Technology

THAKUR INSTITUTE OF AVIATION TECHNOLOGY TRAINING NOTES FORWORD UNCONTROLLED COPY  ITISIMPORTANTTONOTE THAT THE INFORMATION IN THIS BOOK IS OF STUDY/ TRAINING PURPOSES ONLY AND NO REVISION SERVICE WILL BE PROVIDED TO THE HOLDER....

THAKUR INSTITUTE OF AVIATION TECHNOLOGY TRAINING NOTES FORWORD UNCONTROLLED COPY  ITISIMPORTANTTONOTE THAT THE INFORMATION IN THIS BOOK IS OF STUDY/ TRAINING PURPOSES ONLY AND NO REVISION SERVICE WILL BE PROVIDED TO THE HOLDER.  WHEN CARRYING OUT APROCEDURE/ WORK ONAIRCRAFT/ AIRCRAFT EQUIPMENT YOU MUSTALWAYS REFER TOTHE RELEVANT AIRCRAFT MAINTENANCE MANUAL OREQUIPMENT MANUFACTURER'S HANDBOOK.  FOR HEALTH ANDSAFETY IN THE WORKPLACE YOU SHOULD FOLLOW THE REGULATIONS/ GUIDELINES AS SPECIFIED BYTHE EQUIPMENT MANUFACTURER, YOUR COMPANY, NATIONAL SAFETY AUTHORITIES AND NATIONAL GOVERNMENTS. JULY 2024 Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever: i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Thakur Institute of Aviation Technology. TIAT MODULE 2.1 Page 1- 1  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Knowledge Levels – Category A, B1, B2, B3 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1, B2 and B3 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each application 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 familiarization 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. TIAT MODULE 2.1 Page 1- 2  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Module 2.1: Matter Certification Statement These Study Notes comply with the syllabus of EASA Regulation (EC) No.1321/2014 Annex III (Part-66) Appendix I, as amended by Regulation (EC) No.989/2023, and the associated Knowledge Levels as specified below: Level Objective EASA 66 Reference B1 B2 2.1 Matter 2 2 Nature of matter: the chemical elements, structure of atoms, molecules; Chemical compounds; States: solid, liquid and gaseous; Changes between states Changes between states. TIAT MODULE 2.1 Page 1- 3  THAKUR INSTITUTE OF AVIATION TECHNOLOGY 2.1 MATTER and chilling are both applied to a gas, it assumes a liquid state. Liquid air, which is a mixture of oxygen and nitrogen, is produced Nature of Matter in this manner. Although matter is the most basic of all things related to the field of CHEMICAL ELEMENTS physics and the material world, it is the hardest to define, since It cannot be rigidly defined, this chapter will point out those THE PERIODIC TABLE OF THE ELEMENTS characteristics which are easily recognizable. Hanging on the wall of every chemistry laboratory, and Matter itself cannot be destroyed, but it can be changed from one emblazoned on many a chemist‘s favorite mug or T-shirt, is one state into another state by chemical or physical means, it is usually of chemistry’s most important basic tools, the periodic table of the considered in terms of the energy it contains, absorbs, or gives off. elements (Figure 1.1). This table is like the map of the world on Under certain controlled conditions, it can be made to aid man in the wall of every geography classroom. If a geography instructor his everyday life. points to a country on the map, its location alone will tell you what the climate would be like and perhaps some of the Matter is any substance that occupies space and has weight. There characteristics of the culture. Likewise, you may not be familiar are three states of matter: (1) Solids (2) liquids and (3) gases. Solids with the element potassium, but we shall see that the position of have a definite volume and a definite shape, liquids have a definite its symbol, K, on the periodic table, tells us that this element is volume, but they take the shape of the containing vessel. Gases very similar to sodium and that it will react with the element have neither a definite volume nor a definite shape. Gases not only chlorine to form a substance that is very similar to table salt. The take the shape of the containing vessel, but they expand and fill the elements are organized on the periodic table in a way that makes vessel, no matter what its volume. Water is a good example of it easy to find important information about them. You will quickly matter changing from one state to another at high temperature it is come to appreciate how useful the table is when you know just a in the gaseous state known as steam. At moderate temperatures it is few of the details of its arrangements. a liquid, and at low temperatures it becomes ice, a solid state. In this example, the temperature is the dominant factor in determining the state that the substance assumes. Pressure i.e. another important factor that will effect changes in the state of matter. At pressures lower than atmospheric, water will boil and thus change into steam at temperatures lower than 2120F. Pressure is a critical factor in changing some gases to liquids or solids. Normally, when pressure TIAT MODULE 2.1 Page 1- 4  THAKUR INSTITUTE OF AVIATION TECHNOLOGY In the United States, there are two common conventions for numbering the columns (Figure 1.1) Check with your instructor to find out which numbering system you are expected to know. Groups 1 to 18: The vertical columns can be numbered from 1 to 18. This text will use this numbering convention most often. Groups A and B: Some of the groups are also commonly described with a number and the letter A or B. For example, sometimes the group headed by N will be called group 15 and sometimes group 5A. The group headed by Zn can be called 12 or 2B. Because this convention is useful and is common, you will see it used in this text also. Some chemists use Roman numerals with the A- and B-group convention. In short, the group headed by N can be 15, 5A, or VA. The group headed by Zn can be 12, 2B, or IIB. The periodic table is arranged in such a way that elements in the The groups in the first two and last two columns are the ones that same vertical column have similar characteristics. Therefore, it is have names as well as numbers. You should learn these names; often useful to refer to all the elements in a given column as a they are used often in chemistry. group or family. Each group has a number, and some have a group name. For example, the last column on the right is group Most of the elements are classified as metals, which mean they 18, and the elements in this column are called noble gases, the have the following characteristics. elements in group 1 are called alkali and in group 17, are called halogen. TIAT MODULE 2.1 Page 1- 5  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Metals have a shiny metallic luster. Metals conduct heat well and in the solid form conduct electric currents. Metals are malleable, which means they are capable of being extended or shaped by the blows of a hammer. For example, gold (Au) can be hammered into very thin sheets without breaking There is more variation in the characteristics of the nonmetal elements. Some of them are gases at room temperature and pressure, some are solids, and one is a liquid. They have different colors and different textures. The definitive quality shared by all nonmetals is that they do not have the characteristics mentioned above for metals. For example, sulfur is a dull yellow solid that does not conduct heat or electric currents well and is not malleable. It shatters into pieces when hit with a hammer.The electrons in insulators are tightly bound. A few of the elements have some but not all of the characteristics of metals. These elements are classified as metalloids or semimetals. Authorities disagree to some extent concerning which elements belong in this category, but the elements in yellow boxes in the image on the left are commonly classified as metalloids. The portion of the periodic table that contains the metallic TIAT MODULE 2.1 Page 1- 6  THAKUR INSTITUTE OF AVIATION TECHNOLOGY elements is shown here in gray, and the portion that contains the The gap represents the proper location of the first row of the inner nonmetallic elements is shown in light blue. The stair-step line transition metals—that is, lanthanum, La, which is element that starts between B and Al on the periodic table and descends number 57, through ytterbium, Yb, which is element 70. These between Al and Si, Si and Ge, and so on separates the metallic elements belong in the sixth period. Similarly, the second row of inner transition metals, the elements actinium, Ac, through elements from the nonmetallic elements. The metals are below nobelium, No, belong in the seventh period between radium, Ra, and to the left of this line and the nonmetals are above and to the and lawrencium, Lr. At room temperature (20°C) and normal right of it. Most of the elements that have two sides of their box pressures, most of the elements are solid, two of them are liquid forming part of the stair-step line are metalloids. (Hg and Br), and eleven are gas (H, N, O, F, Cl, and the noble gases). Aluminum and polonium are usually considered metals. It is STRUCTURE OF ATOMS, MOLECULES: often useful to refer to whole blocks of elements on the periodic table. The elements in groups 1, 2, and 13 through 18 (the ―A‖ THE ATOM groups) are sometimes called the representative elements. They The atom is the smallest part of the element that retains the same are also called the main-group elements. The elements in groups 3 chemical characteristics of the element itself. (You will be better through 12 (the ―B‖ groups) are often called the transition prepared to understand descriptions of the elements chemical metals. The 28 elements at the bottom of the Objective 16 table characteristics after reading more of this book. For now, it is enough are called inner transition metals. The horizontal rows on the to know that the chemical characteristics of an element include how it periodic table are called periods. There are seven periods in all. combines with other elements to form more complex substances). For The first period contains only two elements, hydrogen1, H, and our purposes, we can think of the atom as a sphere with a diameter of helium, He. about 10-10 meters. This is about a million times smaller than the diameter of the period at the end of this sentence. If the atoms in your The second period contains eight elements: lithium, Li, through body were an inch in diameter, you would have to worry about neon, Ne. The fourth period consists of eighteen elements: bumping your head on the moon. Because atoms are so small, there potassium, K, through krypton, Kr. are a tremendous number of them in even a small sample of an element. A ½-carat diamond 21 contains about 5 × 10 atoms of Note that the sixth period begins with cesium, Cs, which is carbon. If these atoms, tiny as they are, were arranged in a straight element number 55, and barium, Ba, which is number 56, and line with each one touching its neighbors, the line would stretch from then there is a gap which is followed by lutetium, Lu, element 71. here to the sun.If we could look inside the gold atom, we would find TIAT MODULE 2.1 Page 1- 7  THAKUR INSTITUTE OF AVIATION TECHNOLOGY that it is composed of three types of particles: protons, neutrons, and disregard the question of what electrons are and how they move electrons. For example, every gold atom in nature has 79 protons, 79 and focus our attention only on the negative charge that they electrons, and 118 neutrons. Gold is different from phosphorus, generate. We can visualize each electron as generating a cloud of because natural phosphorus atoms have 15 protons, 15 electrons, and negative charge that surrounds the nucleus. Most of the carbon 16 neutrons. atoms in a diamond in a necklace have 6 protons, 6 neutrons, and 6 electrons. The protons (positive charge particles) and neutrons The particles within the atom are extremely tiny. A penny weighs about are in the nucleus, which is surrounded by cloud of negative 2.5 grams, and a neutron, which is the most massive of the particles in charge created by the 6 electrons. For now, we will continue to the atom, weighs only 1.6750×10-24 grams.The protons have about the picture the electron clouds of all the atoms has spherical Carbon same mass as the neutrons, but the electrons have about 2000 times less atom 6 protons 6 neutrons (in most carbon atoms) 6 electrons (in mass. Because the masses of the particles are so small, a more uncharged atom) Particle Charge Mass e− electron −1.000549 convenient unit of measurement has been devised for them. An atomic u(9.1096 × 10-28 g) proton +1.00728 u (1.6726 × 10-24 g) neutron mass unit (also called the unified mass unit) is 1/12 the mass of a 0 1.00867 u (1.6750 × 10−24 g) Cloud representing the −6charge carbon atom that has 6 protons, 6 neutrons, and 6 electrons. The from six electrons. According to electron configuration, electrons modern abbreviation for atomic mass unit is u, but amu is commonly are arranged in different shell or orbit. The shells are K, L, M, N, used. Protons have a positive charge, electrons have a negative charge, O -- or in numbers 1, 2, 3, 4, 5 ----. Each shell (1 to 4) can hold and neutrons have no charge. Charge, a fundamental property of matter, maximum electrons 2, 8, 18, 32 – and can use the formula 2n2 is difficult to describe. Most definitions focus less on what it is than on where n is the orbit number. what it does. For example, we know that objects of opposite charge attract each other, and objects of the same charge repel each other. An electron has a charge that is opposite but equal in magnitude to the ATOMIC NUMBER charge of a proton. We arbitrarily assign the electron a charge of −1, so the charge of a proton is considered to be + 1 We know that protons are present in the nucleus of an atom. It is THE ELECTRON the number of protons of an atom, which determines its atomic number. It is denoted by Z. All atoms of an element have the We do not think that electrons are spherical particles orbiting same atomic number, Z. In fact, elements are defined by the around the nucleus like planets around the sun. Scientists agree number of protons they possess. For hydrogen, Z = 1, because in that electrons are outside the nucleus, but how to describe what hydrogen atom, only one protons present in the nucleus. they are doing out there or even what they are turns out to be Similarly, for carbon, Z = 6. Therefore, the atomic number is difficult tasks. Until the nature of electrons is described, we will TIAT MODULE 2.1 Page 1- 8  THAKUR INSTITUTE OF AVIATION TECHNOLOGY defined as the total number of protons present in the nucleus of an CHEMICAL COMPOUND atom. Compounds can be classified into two types, molecular compounds and salts. In molecular compounds, the atom binds each other through covalent bonds. In salts, it is held together with ionic bonds. These are the two types of bonds out of which every compound is made of. MASS NUMBER Chemical compounds can generally be classified into two broad groups: After studying the properties of the subatomic particles of an atom, we molecular compounds and ionic compounds. Molecular compounds can conclude that mass of an atom is practically due to protons and involve atoms joined by covalent bonds and can be represented by a neutrons alone. These are present in the nucleus of an atom. Hence variety of formulas. Ionic compounds are composed of ions joined by protons and neutrons are also called nucleons. Therefore, the mass of an ionic bonding, and their formulas are generally written using oxidation atom resides in its nucleus. For example, mass of carbon is 12 u states. because it has 6 protons and 6 neutrons, 6 u + 6 u = 12 u. Similarly, the Molecular compounds are composed of atoms that are held together by mass of aluminum is 27 u (13 protons+14neutrons). The mass number covalent bonds. These bonds are formed when electrons are shared is defined as the sum of the total number of protons and neutrons between two atoms. present in the nucleus of an atom. In the notation for an atom, the atomic number, mass number and symbol of the element are to be written as: Mass Number. Isotopes In nature, a number of atoms of COVALENT BONDS some elements have been identified, which have the same atomic number but different mass numbers. For example, take the case of Covalent chemical bonds involve the sharing of a pair of valence Hydrogen atom, it has three atomic species, namely protium (1H1), electrons (in the outer most shell) by two atoms, in contrast to the deuterium (2H1 or D) and tritium (3H1 or T). The atomic number of each transfer of electrons in ionic bonds. Such bonds lead to stable one is 1, but the mass number is 1, 2and 3, respectively. Other such molecules if they share electrons in such a way as to create a noble gas examples are (i) carbon, 12C6, 13C6, and 14C6, (ii) chlorine, 35Cl17and configuration for each atom. Hydrogen gas forms the simplest covalent 37 Cl17, etc. On the basis of these examples due to more or less neutrons, bond in the diatomic hydrogen molecule. The halogens such as chlorine we can say that there are three isotopes of hydrogen atom, namely also exist as diatomic gases by forming covalent bonds. The nitrogen protium, deuterium and tritium. and isotopes are defined as the atoms of the same element, having the same atomic number but different mass numbers due TIAT MODULE 2.1 Page 1- 9  THAKUR INSTITUTE OF AVIATION TECHNOLOGY negative in order to produce a noble gas electron configuration, the bond is called an ionic bond. Typical of ionic bonds are those in the alkali halides such as sodium chloride (NaCl) Ionic bonding can be visualized with the aid of Lewis diagrams. IONIC BONDS Oxygen which makes up the bulk of the atmosphere also exhibits covalent bonding in forming diatomic molecules. Covalent Properties of Ionic and Covalent Compounds bonding can be visualized with the aid of Lewis diagrams. Intermolecular Forces Physical Properties & Bond Types Physical properties of substances are affected by the forces In chemical bonds, atoms can either transfer or share their valence between particles. electrons. In the extreme case where one or more atoms lose Greater attraction between the molecules → more energy is electrons, becomes positive and other atoms gain them, becomes required to overcome the attractive forces between molecules. TIAT MODULE 2.1 Page 1- 10  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Increased melting point or boiling point. Cannot conduct electricity as solid, liquid, or when dissolved– molecules will not separate into ions. Solubility depends on ―like dissolves like.‖ and forces between Particles, Interionic Forces. Will dissolve in liquids with similar molecular polarity– ―Like dissolves like‖ Non-polar Covalent Bonds Due to the attraction of oppositely charged ions. There is a greater attraction between ions of greater charge. Some atoms have similar attraction for electrons, so the shared pair(s) will spend equal time around each atom– There are no Strongest forces between particles found in ionic compounds. ―positive‖ or ―negative‖ sides. Chemical compound: When the two or more different atoms or This is referred to as a non-polar covalent Bond. elements contain in a substance or chemically bonded together is Attraction between Polar Compounds. called chemical compound. General Properties of Ionic 2H2 + O2 = 2H2 O Compounds:. High melting/boiling points– Solids at room temperature. Many are soluble in water– Ions separate when dissolved in water. Liquids & Solutions → Good Conductors– Ions are separated. Solids → Poor conductors– No separation of ions General Properties of Covalent Compounds: Lower Melting Points than ionic compounds– Intermolecular Forces aren‘t as strong as inter ionic forces. TIAT MODULE 2.1 Page 1- 11  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Differences between Polar and Non-polar Covalent Compounds Polar Compounds will attract each other SLIGHTLY. Non-polar compounds have no attraction for each other. Polar Compounds will have slightly higher melting points, but not as high as ionic compounds. Dipole Forces Polar molecules are attracted to each other via the ”negative” end of one molecule being attracted to the ”positive” end of another Forces between Particles molecule. Not as strong as ionic forces.  Strongest intermolecular bonds,– Bonds between molecules, not ions. Hydrogen bonding arises where hydrogen atoms are bonded to highly electronegative elements such as F, O, and N. TIAT MODULE 2.1 Page 1- 12  THAKUR INSTITUTE OF AVIATION TECHNOLOGY HYDROGEN BONDING STATES OF MATTER Solid – A state or phase of matter that has a definite shape and a definite volume. Liquid – A state or phase of matter that has a definite volume but no definite shape. Gas – A state or phase of matter that has no definite shape or volume. Phase change – A change of matter from one phase to another. HYDROGEN BONDING Describes the properties of a solid, liquid and a gas; explains The effect of heating or cooling on the states of matter The matter General Properties of Covalent Compounds that Exhibit around you is made up of moving molecules in the form of Solids, Hydrogen Bonding liquids, and gases. The form that matter takes depends on its shape and volume. Solids have a definite shape and a definite Lower Melting Points than ionic compounds, but higher than volume. The molecules in a solid are packed very closely together non-polar or polar covalent compounds – Intermolecular Forces and has maximum density.Although the molecules in a solid are aren‘t as strong as interionic forces, but Hydrogen bonding is moving, they are so close to one another that they don‘t look like stronger than other intermolecular forces. they are moving at all but vibrate due to temperature. They basically vibrate in their place. An ice cube, the table, the floor, Cannot conduct electricity as solid, liquid, or when dissolved – and a metal bar are all examples of solids. A liquid has a definite Still no ions. volume but no definite shape. Liquid molecules are also packed closely together, but there is enough space between the molecules Will dissolve in liquids with similar molecular polarity – ―Like to allow them to slide past each other. This sliding action prevents dissolves like. the liquid from taking a shape of its own. As a result, liquids take the shape of the container that holds them. Some examples of TIAT MODULE 2.1 Page 1- 13  THAKUR INSTITUTE OF AVIATION TECHNOLOGY liquids are water, milk, bleach, blood, and mercury (liquid metal). and turn into water vapor, a gas. Phase changes also result when A gas has no definite shape and no definite volume. The thermal energy is removed from matter. When a gas such as water molecules in a gas are spaced farther apart, and they move very vapor is cooled, it condenses or changes into water, a liquid. The rapidly. Gas molecules move to fill up the size of their container removal of thermal energy allows the molecules to slow their like when you blow up a balloon and can be compressed. Some kinetic motion. As the molecules slow down, they pack more examples of gases are water vapor, air, carbon dioxide, ozone, and closely together and change the state of matter. Another phase oxygen. change occurs when a liquid changes to a solid, or freezes. Freezing is the result of removing thermal energy from a liquid. Example: As the molecules slow down, they pack closer together and form a solid, such as when water turns to ice. Usually, the addition or removal of thermal energy from a substance will result in a change in the average kinetic motion of the molecules, or temperature, of the substance. During a phase change, however, the thermal energy being used to change the state of matter will not cause a change in the temperature of the substance.The energy(latent heat) being added or removed from the system is being used to break or form the molecular bonds holding the molecules together. As a result, the temperature will remain A steady, or reach a plateau, during a phase change. phase change is a change in the state of matter from one form to another. There are four types of phase changes: melting, boiling Sublimation: It is a process of changing matter from solid state to or vaporizing, condensing, and freezing. An example of melting is gaseous state without attending into liquid state. when a solid changes to a liquid. When thermal energy is added to an ice cube, its molecules begin to move faster. This creates more example: Naphtha, camphor etc. space between the molecules and allows them to slip past each other changing the ice into water.00 C ice gives more cooling effect as it contains less thermal energy(latent heat released) and 00 C water contains more thermal energy(latent heat absorbed). If more thermal energy is added, the molecules in the water further increase their motion. This causes the water to boil or vaporize TIAT MODULE 2.1 Page 1- 14  THAKUR INSTITUTE OF AVIATION TECHNOLOGY CHANGES BETWEEN THE STATES Figure 1.10 TIAT MODULE 2.1 Page 1- 15  THAKUR INSTITUTE OF AVIATION TECHNOLOGY TRAINING NOTES FORWORD UNCONTROLLED COPY  ITISIMPORTANTTONOTE THAT THE INFORMATION IN THIS BOOK IS OF STUDY/ TRAINING PURPOSES ONLY AND NO REVISION SERVICE WILL BE PROVIDED TO THE HOLDER.  WHEN CARRYING OUT APROCEDURE/ WORK ONAIRCRAFT/ AIRCRAFT EQUIPMENT YOU MUSTALWAYS REFER TOTHE RELEVANT AIRCRAFT MAINTENANCE MANUAL REQUIPMENT MANUFACTURER'S HANDBOOK.  FOR HEALTH ANDSAFETY IN THE WORKPLACE YOU SHOULD FOLLOW THE REGULATIONS/ GUIDELINES AS SPECIFIED BYTHE EQUIPMENT MANUFACTURER, YOUR COMPANY, NATIONAL SAFETY AUTHORITIES AND NATIONAL GOVERNMENTS. Copyright Notice © Copyright. All worldwide rights reserved. No part of this publication may be reproduced, stored in a retrieval system or transmitted in any form by any other means whatsoever : i.e. photocopy, electronic, mechanical recording or otherwise without the prior written permission of Thakur Insti``tute of Aviation Technology. TIAT MODULE 2.2 Page 1- 1  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Knowledge Levels – Category A, B1, B2, B3 and C Aircraft Maintenance Licence Basic knowledge for categories A, B1, B2 and B3 are indicated by the allocation of knowledge levels indicators (1, 2 or 3) against each application 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 familiarization 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. TIAT MODULE 2.2 Page 1- 2  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Module 2.2: Mechanics Certification Statement These Study Notes comply with the syllabus of EASA Regulation (EC) No.1321/2014 Annex III (Part-66) Appendix I, as amended by Regulation (EC) No.989/2023, and the associated Knowledge Levels as specified below: EASA 66 Level Objective Reference B1 B2 Mechanics : 2.2 2.2.1 Statics 2 2 Forces, moments and couples, representation as vectors; Centre of gravity; Elements of theory of stress, strain and elasticity: tension, compression, shear and torsion; Nature and properties of solid, fluid and gas; Pressure and buoyancy in liquids (barometers). 2.2.2 Kinetics 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: pendulum movement; Simple theory of vibration, harmonics and resonance; Velocity ratio, mechanical advantage and efficiency. TIAT MODULE 2.2 Page 1- 3  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Level EASA 66 Objective Reference B2 B1 2.2.3 Dynamics 2 2 a. Mass; Force, inertia, work, power, energy (potential, kinetic and total energy), heat, efficiency; b. Momentum, conservation of momentum; 2 2 Impulse; Gyroscopic principles; Friction: nature and effects, coefficient of friction (rolling resistance). 2.2.4 Fluid dynamics 2 2 a. Specific gravity and density; b. Viscosity, fluid resistance, effects of streamlining; 2 2 Effects of compressibility on fluids; Static, dynamic and total pressure: Bernoulli’s Theorem, venture TIAT MODULE 2.2 Page 1- 4  THAKUR INSTITUTE OF AVIATION TECHNOLOGY 2.2 MECHANICS a contact force. If two surfaces are not in contact, they can't exert a normal force on each other. 2.2.1 Statics  Kinetic Friction Kinetic friction is defined as a force that acts between moving FORCES, MOMENTS AND COUPLES surfaces. A body moving on the surface experiences a force in the opposite direction of its movement. The magnitude of the Force is a push or a pull that changes or tends to change the state of rest force will depend on the coefficient of kinetic friction between or uniform motion of an object or changes the direction or shape of an the two materials. object. It causes objects to accelerate. SI unit is Newton. Types of Forces  Gravity.  Static Friction.  Tension. Static friction is a force that keeps an object at rest. Static  Normal (Contact) Force. friction definition can be written as: The friction experienced  Kinetic Friction. when individuals try to move a stationary object on a surface,  Static Friction. without actually triggering any relative motion between the body and the surface on which it is on.  Air Resistance (Drag) Elastic (Spring) Force  Air Resistance (Drag)  Gravity. Air resistance is a force that is caused due to air when an object moves through it. This force acts in the opposite direction to a The gravitational force is a force that attracts any two objects with body passing through the air. Air resistance exerts a frictional mass. We call the gravitational force attractive because it always tries force against the moving body. to pull masses together, it never pushes them apart.  Elastic (Spring) Force  Tension. Elastic Force The force that allows some materials to return to Tension is defined as the force transmitted through a rope, string or its original shape after being stretched or compressed wire when pulled by forces acting from opposite sides. The tension force is directed over the length of the wire and pulls energy equally on the bodies at the ends. MOMENTS  Normal (Contact) Force The normal force is the force that surfaces exert to prevent solid The moment (or torque) of a force about a turning point is the force objects from passing through each other. Normal force is multiplied by the perpendicular distance to the force from the turning point. TIAT MODULE 2.2 Page 1- 5  THAKUR INSTITUTE OF AVIATION TECHNOLOGY 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 are measured in Newtonmeters (Nm). moments.‖ Sum of anticlockwise moments = Sum of clockwise Moment = F d moments  F = the force in Newton (N)  d = perpendicular distance in metres (m) Example Example; A 10N force acts at a perpendicular distance of 0.50m from the turning point. What is the moment of the force? Fig.2.1 Fig 2.2 Sum of anticlockwise moments = Sum of clockwise moments F1 x d1 = F2 x d2 OR Moment=Fd =10 x 0.50 = 5.0 Nm EXAMPLE : 50 lbs force is applied on a bolt by a 8 inches torque wrench. If a 2 inch extension is needed to reach the bolt head. what will be the actual reading? Hints: let the force = F therefore, 50 × 8 = (8 + 2) × F, F = 400 ÷ 10 =40 lbs inch Fig 2.3 Sum of anticlockwise moments = Sum of clockwise moments THE PRINCIPLE OF MOMENTS F1 x d1 = (F2 x d2) + (F3 x d3) TIAT MODULE 2.2 Page 1- 6  THAKUR INSTITUTE OF AVIATION TECHNOLOGY COUPLES Example: Moment of Couple = 50 Nm, control wheel dia. =0.8m, find the magnitude of one of the forces? 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. Moment of couple =Fs = force × distance The moment (or torque) of a couple is calculated by multiplying the 50Nm = force × 0.8m. Therefore, force = 50 ÷ 0.8 = 62.5 N. size of one of the force (F) by the perpendicular distance between the two forces (s). EQUILIBRIUM OF COPLANAR FORCES E.g. a steering wheel in a car; If a body is acted on by a number of coplanar forces and is in Equilibrium (i.e. there is rest or uncelebrated motion) then following conditions must apply. The components of the forces in both of any two directions (usually taken at right angles) must balance. The sum of the clockwise moments about any point equals the sum of the anticlockwise moments about the same point. Fig. 2.5 Fig.2.4 TIAT MODULE 2.2 Page 1- 7  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Example Coplanar forces acting on a body Figure 2.7 shows a beam with negligible mass and a total length of 8m The first statement above is a consequence of there being no pivoted about point F. Three forces A, B and C are shown acting translational motion in any direction and the second follows since there perpendicular to the beam. Calculate the additional force must be is no rotation of the body. If a body is in equilibrium the forces and applied to the beam at D to maintain equilibrium and state whether it moments must both balance. acts up or down. CWM = ACWM Example Figure 2.6 shows beam with negligible mass pivoted about Forces up = Forces down point B with perpendicular forces of 50N and 125N acting at either Neglecting force D at this point in time. end. The 50N force produces an anti-clockwise moment of 50 x 3 = 150 Nm about point B and force 12 produces a clockwise moment of 125 x Y = 125Y Nm. CWM = (1000 × 1) + (250 × 3.5) = 1875 Nm ACWM = (500 × 3) = 1500 Nm If ACWM = CWM then force D must be acting in an anticlockwise direction, and the moment produced by D must be equal to 1875 – 1500 = 375 Nm The value of force D = moment ÷ distance from pivot point = 375 ÷ 5 = 75 N Force D is equal to 75 N and acts upwards on the beam. Fig 2.6 Simple beam problem The beam is assumed to be in equilibrium  Clockwise moments = Anti-clockwise moments Fig 2.7 Simple beam problem 125Y = 150 Y = 1.2 m TIAT MODULE 2.2 Page 1- 8  THAKUR INSTITUTE OF AVIATION TECHNOLOGY REPRESENTATION AS VECTORS Scalar and Vector quantities Quantities are thought of as being either scalar or vector. A scalar quantity has only magnitude and is completely specified by a number and a unit. Examples of scalars are speed; distance, time, mass, Fig 2.9 volume, temperature and frequency. The magnitude of a scalar can be thought of as being represented by a simple scale (Fig 2.8). VECTOR ADDITION: TRIANGLE METHOD The total effect or resultant of a number of forces acting on a body may Fig 2.8 be determined by vector addition. A vector quantity has both magnitude and direction. Examples of vectors are; force, velocity, acceleration, momentum, displacement and field strength. Forces for example give rise to all changes in motion: a force is needed to start a stationary object moving, to change its direction of motion, and to stop it. From this example it can be seen that we need to know Fig.2.10 the direction of the force as well as its magnitude to determine what its effects will be. Triangle method of application of a vector is important a space diagram may be used. To There many ways of representing vectors such a force, either add vector B to vector A, draw B so that its tail is at the head of A. The algebraically or graphically, one common method is by the use of vector sum A + B is the resultant R that joins the tail of A and the head vector diagrams (Fig 2.9). The arrow is used to represent magnitude of B. The order in which A and B are added is not significant, so that and the angle is used to represent direction. If the point A+B=B+A TIAT MODULE 2.2 Page 1- 9  THAKUR INSTITUTE OF AVIATION TECHNOLOGY ADDITION: GRAPHICAL METHOD 1. Set a suitable scale representing magnitude. 2. Set a reference or datum direction. 3. Draw lines representing the vectors being careful to preserve their correct lengths and directions. 4. The resultant is drawn from the tail of the first vector to the head of the last. Fig.2.11 VECTOR ADDITION: POLYGON METHOD Exactly the same procedure is followed when more than two vectors of the same kind are to be added. Consider three forces A, B and C (Fig 2.12), A and B can be added to produce resultant (A + B). If force C is then added a new resultant (A + B + C) is produced. A + B + C Fig. 2.14 VECTOR ADDITION: TRIGONOMETRIC METHOD It is easy to apply trigonometry to find the resultant R of two vectors A and B that are perpendicular to each other (Fig 2.15). Fig.2.12 This procedure can be repeated again and again. Fig 2.15 Trigonometric method Fig 2.13 Polygon method TIAT MODULE 2.2 Page 1- 10  THAKUR INSTITUTE OF AVIATION TECHNOLOGY The magnitude of the resultant is given by the Pythagorean Theorem vectors. If vectors A and B are together equal to C, then vector C is as: equivalent to the two vectors A and B (Fig 2.16). A + B = C R  √𝐴2 + 𝐵2 C The direction, angle  between R and A may be found from: A 𝐵 𝐵 tan  𝜃 = tan-1 B 𝐴 𝐴 Fig. 2.16 Resolving vectors Example 2.1 A Calculate the resultant of two forces A and B, force B = 3N and acting due north and force A = 4N acting due east. When a vector is replaced by two or more others, the process is called resolving the vector. The new vectors are known as components of the initial vector. The components into which a vector is resolved are nearly always chosen to be perpendicular to one another. Figure 2.17 shows a force F FX Using Pythagoras: R = √𝐴2 + 𝐵 2 R = √32 + 42  Magnitude = 5N Fig. 2.17 -1 𝐵 -1 3   tan 𝜃  tan 𝐴 4 Resolving vectors = 37° 4 This force can be resolved into two component vectors Fx and Fy To express R in terms of north we find the value of  = 90° -  = 90° - where: Fx = horizontal component of F 37° = 53° Fy = vertical component of F The magnitudes of these components are: RESOLVING VECTOR Fx = F cos Just as two or more vectors can be added to yield a single resultant vector, so it is possible to break up a single vector into two or more Fy = F sin  TIAT MODULE 2.2 Page 1- 11  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Calculate the magnitude and direction of the single resultant force 6. VECTOR ADDITION: COMPONENT METHOD. Resolve the initial vectors into components in the x, y directions. When vectors are to be added are not perpendicular, the method of Fx = 500 Cos 120° + 1200 Cos 40° + 900 Cos 250° = 361.435 addition by components can be used. Fy = 500 Sin 120° + 1200 Sin 40° + 900 Sin 250° = 358.634 1. Resolve the initial vectors into components in the x, y, z directions. The magnitude of F = √(361.435)2 + (358.634)2 = 509.169N 𝐹𝑦 2. Add the components in the x direction to give Rx, add the The direction of F = tan-1 𝐹𝑥 components in the y direction to give Ry, add the components in the z direction to give Rz, = 358.634 tan-1 361.435 E.g. Rx = Ax + Bx + Cx + …… = 44.770 Ry = Ay + By + Cy + …… Rz = Az + Bz + Cz + …… Applying CAST both Fx and Fy are positive so convention the direction will be 44.77° from the positive x-axis. 3. Calculate the magnitude of the resultant R from its components Rx, Ry, Rz by using Pythagorean Theorem. COPLANAR FORCES Forces whose line of action act in the same plane (usually the x and y R √𝑅𝑋2 + 𝑅𝑦2 + 𝑅𝑧2 plane) are said to be coplanar. Note: If the vectors being added all lie in the same plane, only two Note: In general, however, three mutually perpendicular components components need to be considered. are required to completely describe the magnitude and direction of a Example : vector quantity, conventionally labeled x, y, z axes. Three coplanar forces act at a single point: Force A = 500 N at 120° F1 = 10N Force B = 1200 N at 40° Force C = 900 N at 250° TIAT MODULE 2.2 Page 1- 12  THAKUR INSTITUTE OF AVIATION TECHNOLOGY If the two forces F1 and F2 are equal in magnitude and opposite in direction (i.e. F1 and F2 = 0), then the object is in translational equilibrium (Fig 2.18). Fig.2.18 Fig 2.20 When two equal but opposite forces are present, whose lines of action are not coincident they will create a rotation termed a couple. A CENTRE OF GRAVITY moment of a couple is equal to the magnitude of a force F, multiplied by the distance between them. Centre of gravity, in physics, imaginary single point in a body of matter where, for convenience in certain calculations, the total mass or weight of the body may be thought to be concentrated. The concept is If in addition the two forces act along a common line of action then the sometimes useful in designing static structures (e.g., buildings and object is also in rotational equilibrium (Fig 2.19). bridges) or in predicting the behavior of a moving body when it is acted on by gravity in a uniform gravitational field the Centre of gravity is identical to the centre of mass, a term preferred by physicists. The two F1 = 10 N do not always coincide, however. For example, the Moon‘s centre of mass is very close to its geometric centre (it is not exact because the Moon is not a perfect uniform sphere), but its centre of gravity is slightly displaced toward the Earth because of the stronger gravitational force on the Moon’s near side. The location of a body’s center of gravity may coincide with the Fig 2.19 Forces in rotational equilibrium geometric centre of the body, especially in a symmetrically shaped object composed of homogeneous material. An asymmetrical object composed of a variety of materials with different masses, however, is If the vector sum of three forces is zero (F1+ F2 + F3 = 0), then the likely to have a centre of gravity located at some distance from its object is in translational equilibrium. If in addition the lines of action of geometric centre. In some cases, such as hollow bodies or irregularly the three forces pass through a common point then the object is in shaped objects, the centre of gravity (or centre of mass) may occur in rotational equilibrium as well. Such a system of forces is called space at a point external to the physical material—e.g., in the centre of concurrent (Fig 2.20). a tennis ball or between the legs of a chair. TIAT MODULE 2.2 Page 1- 13  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Published tables and handbooks list the centre of gravity for most Stress ( pronounced “sigma”) is the force or load (in N) acting on unit common geometric shapes. For a triangular metal plate such as that cross-sectional area (1 m²). depicted in the figure, the calculation would involve a summation of Area of a square is (side) x2 and area of rectangle is length × breadth. the moments of the weights of all the particles that make up the metal The units of stress is the Pascal (Pa) which equals 1 Nm-2 plate about point A. By equating this sum to the plate’s weight W, multiplied by the unknown distance from the centre of gravity G to AC, the position of G relative to AC can be determined. The summation of the moments can be obtained easily and precisely by means of integral calculus. The centre of gravity of anybody can also be determined by a The terms stress and strain are used when referring to deforming forces simple physical procedure. When an object is suspended from any and the deformation they produce. Components fail due to being over- single point, it’s centre of gravity lies directly beneath that point. stressed, not over-loaded. Stresses can occur in differing forms, dependent on the manner of application of the external force. There are five different types of stress in mechanical bodies. These are tension, compression, torsion, bending and shear. Strain The relative change in the size or shape of a body due to applied stress is called strain. Strain  (pronounced epsilon) is the extension of unit length (m). Strain is a unit and has no units, but can be expressed as Fig. 2.21 percentage. Strain can be tensile or compressive in nature STRESS, STRAIN AND ELASTIC TENSION STRESS The mechanical properties of a material are concerned with its behavior under the action of external forces. This is a matter of importance to the engineers, when selecting a material for a particular job. Information about mechanical properties may be obtained by observing the behavior of materials when placed under load. This allows engineers to analyses the external forces and then makes deductions about the internal forces or stresses that are produced internally. TIAT MODULE 2.2 Page 1- 14  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Fig 2.22 Examples of Tensile and Compressive Strain Example A steel rod 20 mm diameter and 1m in length carries a load of 45 kN. This causes an extension of 1.8mm. Calculate the stress and strain in the steel rod. Fig 2.23 Graph of stress against strain Thereafter there is a slight increase in strain with increased stress until a point (L) is reached. This is the elastic limit; up to this point the deformation of the specimen is elastic, i.e. when the stress is removed the specimen returns to its original length. Beyond point L there is permanent deformation when the stress is removed, i.e. the material has ceased to be elastic and has become strained or plastic. In the plastic region individual materials behavior varies, however at To show strain as a percentage you multiply by 100 point (B) there is a sudden increase in strain with further increases in  Strain as a percentage = 0.0018 x 100 = 0.18%. stress - this is the yield point. Point (C) represents the material ultimate tensile strength and point (D) represents the specimens fracture point. Elasticity as the elastic modulus. Within the region of elastic deformation strain is Elasticity is the property of certain materials that enables them to found to be proportional to stress. return to their original dimensions after an applied stress has been Stress  Strain = Constant removed. Figure 2.23 shows the graph of a stress when applied to a specimen. The strain increases in proportion (OA) until a certain point This constant is called Young‘s Modulus and is denoted by the letter E. called the limit of proportionality is reached (A). This is in accordance E = Longitudinal stress  Longitudinal strain with Hooke‘s Law. TIAT MODULE 2.2 Page 1- 15  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Tension or Tensile Stress Tensile stress describes the effect of a force that tends to pull an object The value of Young‘s Modulus is given by the gradient of a stress – apart. Flexible steel cable used in aircraft control systems is an example strain graph and depends on the nature of the material and not on the of a component that is in designed to withstand tensile loads (Fig 2.25). dimensions of the sample being tested. Steel cable is easily bent and has little opposition to other types of stress, but, when subjected to a purely tensile load, it performs exceptionally well. Example Calculate Young‘s modulus of the steel rod used in example 2.8. Young‘s Modulus =E = Longitudinal stress  Longitudinal strain Taking the results for stress and strain from example 2.8: Stress  = 143 MN m-2 Strain = 0.0018 Young‘s Modulus E = (143 x 106)  (0.0018) =7.94 x 1010 Nm-2 Note: The units of Young‘s Modulus of elasticity are the same as those COMPRESSION OR COMPRESSIVE STRESS for stress Nm-2 Compression is the resistance to an external force that tries to force an object together. Aircraft shock struts, rivets are driven with a compressive force. When compression stress is applied to a rivet, the rivet firstly expands until it fills the hole and then the external part of the shank spreads to form a second head, which holds the sheets of metal tightly together (Fig 2.26) Approximate values of E for some common metals Fig. 2.24 TIAT MODULE 2.2 Page 1- 16  THAKUR INSTITUTE OF AVIATION TECHNOLOGY TORSION An aircraft wing acts as a cantilever beam, with the wing supported at A torsional stress is applied to a material when it is twisted. Torsion is the fuselage attachment point. When the aircraft is on the ground the a combination of both tension and compression. For example, when an force of gravity causes the wing to bend in a similar manner to the object is subjected to torsional stress, tensional stresses operate beam shown in Fig. 2.28. In this case, the top of the wing is subjected diagonally across the object whilst compression stresses act at right to tensile stress whilst the lower skin experiences compression stress. angles to the tension stress (Fig 2.27). In flight, the force of lift tries to bend an aircraft's wing upward. When this happens the skin on the top of the wing is subjected to a compressive force, whilst the skin below the wing is pulled by a tension force (Fig 2.29). An engine crankshaft is a component whose primary stress is torsion. The pistons pushing down on the connecting rods rotate the crankshaft against the opposition, or resistance of the propeller. The resulting stresses attempt to twist the crankshaft. Bending Stress If a beam is anchored at one end and a load applied at the other end, the Shear stress beam will bend in the direction of the applied load (Fig.2.28). A shear stress attempts to slice, (or shear) a body apart. The upper diagram shows a clevis bolt in an aircraft control system, which is designed to withstand shear loads. They are made of high-strength steel and are fitted with a thin nut that is held in place with a split pin. Whenever a control cable moves, shear forces are applied to the bolt. However, when no force is present, the clevis bolt is free to turn in its hole. The lower diagram shows two sheets of metal held together with a rivet. If a tensile load is applied to the sheets (as would happen to the top skin of an aircraft wing, when the aircraft is on the ground), the rivet is subjected to a shear load. TIAT MODULE 2.2 Page 1- 17  THAKUR INSTITUTE OF AVIATION TECHNOLOGY Properties of Gas: 1) Gases have neither a fixed shape nor a fixed volume. 2) Gases can be compressed easily. 3) Gases have the least density among the three. 4) Intermolecular forces of attraction are weakest. 5) The space between the gas particles is large. Pressure When a force acts perpendicular to a surface, the pressure exerted is the ratio between the magnitude of the force and the area of the surface: The diagram shows two sheets of metal held together with a rivet. If a Where pressure can be expressed as Area = 𝐴 = 𝜋𝑟 2 tensile load is applied to the sheets (as would happen to the top skin of an aircraft wing, when the aircraft is on the ground), the rivet is Other units of pressure are often used: subjected to a shear load 1 bar = 105 Pa = 1013 millibar NATURE AND PROPERTRIES OF SOLID,LIQUID AND GAS 1 atm = 1.013 x 105 Pa  14.7 lb in-2 Properties of Solids: Absolute pressure is pressure measured on a gauge reading zero at zero 1) Solid has a fixed shape and a fixed volume. pressure rather than at atmospheric pressure. Gauge pressure is 2) Solid cannot be compressed. measured on a gauge that reads zero at atmospheric pressure. 3) Solids have a high density.  Absolute pressure = gauge pressure + atmospheric pressure 4) Force of attraction between the particles in a solid is very strong. 5) The space between the particles of solids is negligible. Example Properties of Liquids: Applying the above formula: 1) Liquid has a fixed volume but no fixed shape. 2) Liquids can be slightly compressed. large pressure is required to A tyre whose gauge pressure is 2 bar contains air at an absolute compress them. pressure of about 3 bar, since sea-level atmospheric pressure is about 1 3) Liquids have lesser densities than solids. bar. 4) Intermolecular forces of attraction is weaker than solids. Pressure in a fluid Pressure is a useful quantity where fluids (gases and 5) They have considerable space between the particles. liquids) are concerned because of the following properties of fluids: TIAT MODULE 2.2 Page 1- 18  THAKUR INSTITUTE OF AVIATION TECHNOLOGY 1. The forces that a fluid exerts on the walls of its container, and Fig 2.32 Fluid pressure exerted due its weight above those that the walls exert of the fluid, always act perpendicular to the walls (Fig 2.31). Weight W = mg Where mass m = volume × density = height x cross-sectional area × density = h × A ×  Downward force W = h××g ×A acting on A ℎ.𝑝.𝑔.𝐴 𝒇𝒐𝒓𝒄𝒆 𝑭 = 𝐴 P= =𝑨 𝒂𝒓𝒆𝒂 Therefore the pressure at a depth h in a fluid of density  due to the weight of the fluid above is The S.I. unit of pressure is the Pascal where 1 Pa = 1 Nm-2 P = h..g 2. The force exerted by the pressure in a fluid is the same in all Hence the total pressure at that depth is directions at a given depth. P = Pext. + h..g An external pressure exerted on a fluid is transmitted uniformly When a body of a fluid is in an open container, the atmosphere exerts throughout the fluid. This does not mean that pressures in a fluid are an external pressure on it. the same everywhere, because the weight of the fluid itself exerts pressures that increase with increasing depth. Example Accumulator charging: When any fluid is pumped in any pre- charged cylinder or accumulator, the gauge pressure of the In figure 2.32, the pressure acting on XX 1 is due to the weight of the cylinder/accumulator cannot be more than the supply (source) pressure. fluid acting downwards Example: 1500 psi hydraulic pressure is pumped into an accumulator which contains 1000 psi, after charging the gauge will read 1500 psi only. HYDRAULIC PRESS The hydraulic press is a basic machine which uses the fact that an external pressure exerted on a fluid is transmitted uniformly throughout the fluid. TIAT MODULE 2.2 Page 1- 19  THAKUR INSTITUTE OF AVIATION TECHNOLOGY BUOYANCY IN LIQUIED ARCHIMEDES PRINCIPLE An object immersed in a fluid is acted upon by an upward force that arises because pressure in a fluid increase with depth. Hence the upward force on the bottom of the object is more than the downward force on top of it. The difference between the two is called the buoyant force and is equal to the weight of a body of the fluid whose volume is the same as that of the object. This is Archimedes principle: the buoyant force on a submerged object is equal to the weight of the fluid the objects displaces. If the buoyant force is less than weight of the object itself, the object SIMPLE HYDRAULIC PRESS sinks; if the buoyant force equals the weight of the object, the object Figure shows a force F1 applied to a piston with area A1 containing a floats in equilibrium at any depth in the fluid. If the buoyant force is 𝐹 fluid. The pressure on the fluid given by 𝐴1 is transmitted through the more than the weight of the object, the object floats with part of its 1 pipe to the larger piston. The force of the Fluid on the larger piston is volume above the surface. when a block of ice floats in a glass of 𝐹1 𝐴 water, level of water will rise but after melting the level remains same given by F2 = 𝐴 2 as volume of ice decreases when changes to liquid or water. 1 Since the area A2 of the larger piston is greater than the area A1 the Example smaller piston, then F2 is greater than F1. So a small effort F1 can be A block is weighed in air then immersed fully in water and reweighed used to move a much greater load (Fig. 2.34). The readings obtained were 2.4 N in air and 2.0 N in water. However, the effort must travel further than the load when the forces Given that the density of water is 1000 kgm-3, calculate the density of move and do work. the block. Example A hydraulic press has an input cylinder 2 cm in diameter and an output cylinder 12 cm in diameter. Assuming the press is 100 % efficient calculate the force exerted by the output piston when a force of 80 N is applied to the input piston