Mechanical Technology: Airplane Gear Landing System - Report Example

Mechanical engineering Mechanical engineering Introduction For an airplane gear landing system it is necessaryto have strong struts as to ensure that they don’t fail when the craft lands, it is therefore necessary to carry out buckling analysis, force analysis, deflection analysis and material yielding analysis. In the determination of the gear landing loads the following diagram has been developed to help in the evaluation Fig 1 the plane and the major forces acting when it lands Taking the wheels to be springy, the two main condition for the landing gear are given by Fig 2 showing the forces acting when a plane lands Here, the landing unit is represented using springs. The two diagrams above show the resolution of forces acting on the struts Defection analysis When the wheel are touching the ground, the kinetic energy is given by Where W is the weight in pounds V is the velocity in feet per second G is the gravitational acceleration (feet/S2) During the landing process, a small force lifts the plane upwards. The lift coefficient is usually 0.4 to 0.5. For a wing lift factor (P), the lift force is given by; Lift = weight of plane (1-P) For a lift coefficient equal to 0.4 to 0.5, P is given by 0.667. The potential energy at landing is given by Potential energy = weight of plane (1-P) × X=0 Where X is the landing gear stroke (deflection that occur on struts when the airplane lands) When the plane lands, at the bottom of the stroke, the velocity is 0 so the kinetic energy is also 0, the force exerted by the tires and the landing gear upwards is given by; F=XK where X is the deflection and K is the spring constant Since the landing gear is springy. The energy by the spring is given b The reaction force, which is a constant, is given by The principle of the conservation of energy states that no energy is created or destroyed, hence E= PE+KE Where: KE = W×V2/2g PE = W (1-P) X KX2/2 - W×V2/2g=0 The deflection (X) is given by; Where W is the weight of plane V is the velocity K is the spring constant G is the gravitational acceleration This deflection can be countered by (a) the plane tire (b) the landing gear Experimentally, the spring constant of 6.00×6 tire is given by Ktire = 5500/3.6=1.528 lb/in or 18.333 lb/ft The deflection of the cantilever leg is thus given by Y= R Z3/(3EI ) For the rectangular crossectional the moment of inertia is given by I= t3 w/12 1(b) ways of improving the design (i)Having the section being changed from rectangular section to a streamline crossectional (ii)Using the computer and software to optimize the design instead of relying on manual calculation. (iii)Using the pneumatic system instead of a spring loaded strut 1(c) safety factors were not included in the analysis. There is need to develop, methods of analyzing the deflection and buckling while using the safety factor. 1. Variation of friction factor with the Reynolds number Theory The head loss expressed as of a fluid in motion due to the friction is calculated using the Darcy-Weisbach equation and is given by; (1) Where: is the length of the pipe? The Darcy friction factor and is a measure of the shear stress exerted by the turbulent flow on the pipe wall and can be expressed in the dimensionless form as where is the shear stress, is the density of the liquid and is the mean velocity. The Hagen poiseuille equation for laminar flow is given by; (2) Where; h1 is the head loss V is the average velocity D is the diameter g is the gravitational acceleration l is the length If the flow in the pipe is laminar flow, then as shown in equation (2), the velocity varies inversely to the pipe diameter. The friction factor (f) varies as shown in the equation (3). (3) Thus the friction factor varies proportionately to the viscosity ( ) and inversely proportional to velocity (V), the pipe diameter (D) and the density of the fluid. As stipulated in the equation (3), the friction factor does not depend on the pipe roughness in laminar flow. Turbulent flow: for the turbulent flow, the friction factor becomes complicated and can only be represented by a graph. This is because the friction factor is expressed as a function of the Reynolds number and the pipe roughness. Nikuradse showed that the pipe roughness affects the friction factor. This is given by (4) Where the absolute roughness and D is is always the diameter The relationship between the friction factor and the Reynolds number for a rough pipe is given by (5) This equation is used only for high turbulent flow. Calculations As defined in the theory part the variation of the friction factor against the Pipe roughness is represented graphically as shown below Fig 3 showing the moody diagram (source: The engineering toolbox.2009.moody diagram. [Online]. Available at http://www.engineeringtoolbox.com/moody-diagram-d_618.html . Accessed 28 august 2009 ) From the diagram, the plot is for the friction factor against the Reynolds number. However, due to the various pipe roughness. Several curves are drawn to depict different pipes. For laminar flow we have only one curves, however for turbulent flow, more than one curve are drawn due to the different pipe roughness. The diagram showing the regimes of flow is indicated in the figure below Fig 4 Source (Finnemore, E. John, Joseph B. Franzini. Fluid Mechanics: with Engineering Applications. Tenth Edition) The left end shows the laminar flow, in between, we have the critical zone and to the right side we have the turbulent flow (very rough pipes). The approximate Reynolds number for the critical zone as shown by the moody chart is between 2300 Re to 4000 Re. Mathematically represented as 2300 < Re < 4000 Question 1(ii) Theory The Reynolds number is used to determine if the flow in pipes is laminar, turbulent or in the transition zone. The Reynolds number is computed using the equation (6) Where is the Reynolds number which is a ratio of the inertial forces to the viscous forces. V is the velocity of the fluid D is the diameter Is the fluid density µ is the dynamic viscosity is the kinematic viscosity Q is the volumetric flow rate A is the crossectional area Calculation The variables given are: Density of oil = 950 Kg/m3 Diameter of pipe = 0.25 m Velocity or the volumetric flow rate = 0.6 m3/s Kinematic viscosity = 2×10-6m2/s The crossectional area of a pipe is given by Q = 0.6m3 D= 0.25m Since we have Q, D, v and A, the last part of equation (6) can be used to evaluate the Reynolds number. =1528662.42 The Reynolds number is greater than 4000, hence the flow is turbulent Definition of Mass flow rate The mass flow rate is the mass of a substance passing through a given surface per unit time. The formula for computation of mass flow rate is given by Where is the mass flow rate Is the density Is the viscosity A Is the crossectional area Question 1(iii) Determine the loss in energy when there is sudden expansion Theory When a fluid flows in different crossectional area through a pipe, a sudden expansion or contraction occurs, the borda-carnot equation is used to compute the mechanical energy loss due to the sudden change. The Borda- Carnot equation is expressed as; The loss coefficient is an empirical formula and this can be determined using different Coefficient, in this case it is determined by; [1-(d1/d2)2]2 If the loss coefficient is 1, then Hence The energy loss is then equal to The volumetric flow rate is given by Where (A) is the area and (v) is the velocity Calculations The loss coefficient is given by: [1-(d1/d2)2]2 = The initial velocity before expansion is given by; The initial crossectional area is given by The fluid velocity before the expansion is thus given by The area after expansion is given by The oil density is given by 950 Kg/m3 The energy loss is given by = =35136.1 joules Lost energy =35.136 Kj Specific lost energy Volume flowing at a given time = 0.12m3/s Mass flowing Therefore = 0.12 ×950 =114 Kg (mass flowing per second) The specific energy loss = 35.136 Kj/114= 308.2 Kj/kg QUESTION 2. THE HYDROGEN FUEL CELL 2(i) Sketch of the hydrogen fuel cell Fig 5: the hydrogen fuel cell Mode of operation The hydrogen is channeled to the anode located at one side of the cell; oxygen is channeled on the other side of the cell as shown in figure The platinum anode causes the hydrogen atom to split into it respective ions, that is the positive ions (protons) and the negative ions (electrons). The cell reactions are; The polymer electrolyte membrane causes the positively charged ions to travel to the cathode while the negatively charged ions flow through the external circuit. These actions results to the creation of a potential difference hence result to the flow of an electric current. At the cathode the oxygen ions combine with the hydrogen ions to form water which is discharged as the waste. At the cathode, the reaction is given by; The overall cell reactions are given by; QUESTION 3 3(i) automatically controlled dew point sensor The dew point temperature is that temperature where water begins to condense. The dew point is mainly affected by pressure and the temperature. The more negative the dew point, the more safe it is to operate and also the more dry the air. At dew point, the air reaches the saturation conditions and water precipitates. The dew point sensors are used in many applications such as dryers and also in preventing the condensation at ceiling surfaces of chilled water supply. The dew point sensor has a sensor element whose resistance varies with changes in humidity. This element changes resistance. Equipment is fitted with controllers which are used for either switching the system ON or OFF. An example is the chilled water supply diagram shown in figure (6), The dew point sensor senses the humidity of the environment; this causes the resistance of the element to change. The change in electrical resistance causes a corresponding change in electrical current. This current is the manipulated and amplified in the signal conditioner. The signal is then applied to an actuator or relays that switches OFF/ON the pump unit. 3(ii) sling hygrometer A hygrometer is equipment that is used to measure the relative humidity. A simple and manual method is used for the measurement of humidity. This equipment is called a psychometric and is made up of two thermometers, the dry bulb and the wet bulb thermometer. The evaporation in the wet bulb lowers the temperature so that the wet bulb shows the lower temperature while dry bulb measures the ambient temperature. The difference in temperature is computed from the two measurements. The relative humidity can be compute by locating the intersection of the dry and wet bulb temperatures on a psychometric chart. Fig 7. The sling hygrometer A sling hygrometer is used for measuring the relative humidity 3(iii) Wet bulb temperature = 180C Ambient temperature =250C The relative humidity from the psychometric chart is given by the intersection of the two temperatures. Relative humidity = 50 The moisture content =0.010 Specific volume = 0.85 m3/kg Due point temperature = 140C Question 4 4(i) stoichiometric condition The stoichiometric condition is the condition that is arrived at, when the air fuel mixture, present in a combustion camber is enough to completely burn the available fuel. This is an ideal situation and is not arrived at due to the short time for combustion. An example is the air fuel mixture in a gasoline engine, the stoichiometric mixture of air to fuel (air: fuel) is 14.7:1. A mixture less than this ratio, say 10:1 is considered a rich mixture while a mixture greater than this ratio say 17:1 is considered a lean mixture. The air –fuel ratio is given as Where AFR is the air –fuel ratio. And is the mass of air, and is the mass of fuel. Calculation of the air fuel ratio for Heptane The chemical formula for Heptane; C7 H16 The balanced reaction equation with oxygen is given by; The atomic weights for the different elements are given by; Carbon; 12.01 Oxygen; 16 Hydrogen: 1.008 The molecular weight of one (1) Heptane molecule is given by; =100.198 The molecular weight of one (1) oxygen molecule is given by The oxygen –fuel ratio is given by 352:100.198 Simplifying the ratio we get Oxygen: fuel 3.5:1 However in ordinary air conditions, the oxygen level is 23% and since the air aspirated by the engine has the other components of air. The amount of air required is; Thus the stoichiometric air-fuel ratio is given by 15.22:1 for Heptane Question 5 5(i) the dehumidifier The dehumidifier is an appliance that is used to reduce the humidity in the air. High humidity in the air results to the growth of mould and mildew and also exposes human being to great health risks. The relative humidity required at any room is 30% to 50%. Any excess above this level is considered harmful to human beings. High humidity acts a good breeding ground for moths, fleas, cockroaches and many more insects. Fig 8: the dehumidifier 5(ii) The P-H diagram for a refrigerator Refrigeration is the process of removing heat from a given compartment and exhausting it at a place where it’s not objectionable. The main parts of refegeration system are The compressor; used to compress the working fluid. This compression heats up the gas The coil; lets the hot working fluid loose heat, upon dissipating the heat the working fluid cools back to a liquid The expansion valve; this is a small hole that has high pressure on one side and low pressure on the other side. It allows for the throttling process The working fluid: expands and cools causing heat transfer process. The compartment being cooled. Fig 9: refrigerator system From the diagram, the refrigeration cycle has the following main steps The refrigerant gas enters the compressor where it is compressed at constant entropy. It leaves the compressor as superheated vapor. The superheated vapor flows through the condenser where it cools. This removes the super heat. Additional heat is removed when the vapor condenses to liquid at constant temperature and pressure. The refrigerant then undergoes the throttling process through the expansion valve. The sudden reduction of pressure causes flash evaporation resulting to a mixture of liquid and vapor. This mixture goes through the evaporation chamber where it is completely vaporized by the warm air in the cabinet causing cooling and the process starts over again. Fig 10 the refrigeration cycle The different points in the refrigeration cycle are; Points 1-2: the saturated vapor is compressed resulting to its heating to the superheat. Points 2-3: the superheat is removed in the condenser Points 3-4: the vapor changes to liquid in the condenser Points 4-5: the throttling process results to the flashing of the liquid to form liquid and vapor Points 5-1: evaporation takes place and the liquid + vapor are converted to vapor causing cooling References Garde, R. J. (1997), Fluid Mechanics Through Problems, New Age Publishers Burstall, Aubrey F. (1965). A History of Mechanical Engineering. The MIT Press Stoecker and Jones.2004. Refrigeration and Air Conditioning, New Delhi: Tata-McGraw Hill Publishers Andrew D., Carl H., Alfred F.2003. Modern Refrigeration and Air Conditioning (18th Edition ed.). Goodheart-Wilcox Publishing. Aircraft design. 2003. Landing gear loads. [Online]. Available at http://www.aircraftdesigns.com/landing-gear-loads.html Read More