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Sections in this Chapter: Section 1.1 - Introduction to Aircraft Hydraulics Section 1.2 - Hydraulics Systems Principles of Operation Section 1.3 - Hydraulic System Power Requirements ....
Aircraft HydraulicsSections in this Chapter:Section 1.1 - Introduction to Aircraft Hydraulics Section 1.2 - Hydraulics Systems Principles of Operation Section 1.3 - Hydraulic System Power Requirements Section 1.4 - Hydraulic Pressure Regulated Power System Section 1.5 - Aircraft Hydraulic System Reservoir Design Section 1.6 - CavitationSection 1.7 - Pressurized Reservoirs Section 1.8 - Aircraft Hydraulic System Power Pumps Section 1.9 - Hydraulic System Check Valves Section 1.10 - Pressure Control Section 1.11 - Hydraulics System Accumulators Section 1.12 - Pressure Regulation in Hydraulic Systems Section 1.13 - Hydraulic System Hand Pumps Section 1.14 - Flow Control Section 1.15 - Flow Conditions Section 1.16 - Flow Restrictors Section 1.17 - Synchronizing Circuits Section 1.18 - Types of Actuation Cylinders Section 1.19 - Hydraulic Motors & Variable Displacement Pumps Section 1.20 - Automatic Hydraulic Transmissions Section 1.21 - Variable Displacement Pump Power System & Open Center Type Power SystemSection 1.22 - Pressure Boosters, Pressure De-Boosters, & Hydraulic Fluids Section 1.23 - REVIEW EXERCISESection 1.1 - Introduction to Aircraft Hydraulics Aircraft Hydraulics DefinitionAircraft Hydraulics is a means of transmitting energy or power from one place to another efficiently.What is a hydraulics system?It is a system where liquid under pressure is used to transmit this energy. Hydraulics systems take engine power and converts it to hydraulic power by means of a hydraulic pump. This power can be distributed throughout the airplane by means of tubing that runs through the aircraft. Hydraulic power may be reconverted to mechanical power by means of an actuating cylinder, or turbine.(1) - A hydraulic pump converts mechanical power to hydraulic power(2) - An actuating cylinder converts hydraulic power to mechanical power(3) - Landing Gear(4) - Engine power (mechanical HP) If an electrical system were used instead of a hydraulic system, a generator would take the place of the pump and a motor would take the place of the actuating cylinder. Advantages of Hydraulic Systems (over other systems for aircraft use) 1. It is lighter in weight than alternate existing systems.2. It is dead beat, that is, there is an absence of sloppiness in its response to demands placed on the system.3. It is reliable; either it works or doesn't.4. It can be easily maintained. 5. It is not a shock hazard; it is not much of a fire hazard. 6. It can develop practically unlimited force or torque. Example: A gun turret must be able to change direction almost instantaneously. This is what is accomplished by this hydraulic system. In an electrical system, the rotating armature must come to full stop and then reverse direction or else the armature will burn out. This doesn't happen with a hydraulic system because there is no need for a motor in the hydraulic system. Example: In a landing gear the hydraulic motor can produce enough power to pull up the landing gear system without trouble even though air loads act on the system and the slip stream air is impinging against it. The actuating cylinder can change hydraulic power to linear or rotating motion. It has a reduction gear in it to reduce rotating motion to that amount which is needed. Previously, systems used to control motion by using steel cables connected by pulleys between the controlling mechanism (such as the pedals) and the controlled surface (such as the rudder). The cables were affected by expansion rates of the cables due to temperature changes. Hydraulic systems can control motion without worrying about the effect of temperature since it is a closed system (not open to the atmosphere) compared to a cable system. This means better control of the plane and less lag time between the pilot's movement to control the plane and the response by the control surface. Some Devices Operated by Hydraulic Systems in Aircraft1. Primary control boosters2. Retraction and extension of landing gear 3. Sweep back and forth of wings 4. Opening and closing doors and hatchways 5. Automatic pilot and gun turrets 6. Shock absorption systems and valve lifter systems 7. Dive, landing, speed and flap brakes 8. Pitch changing mechanism, spoilers on flaps 9. Bomb bay doors and bomb displacement gears Some Devices Operated by the Hydraulic Systems in Spacecraft1. Gimbeling of engines and thrust deflector vanes 2. Thrust reversers and launch mechanisms Section 1.2 - Hydraulics Systems Principles of Operation Introduction Pressures in hydraulic systems can be extremely high and normally are measured in thousands of pounds per square inch (psi) when using British units of measurement, or pascals (Newtons/square meter). Part of the hydraulic system is the actuating cylinder whose main function is to change hydraulic (fluid) power to mechanical (shaft) power. Inside the actuating cylinder is a piston whose motion is regulated by oil under pressure. The oil is in contact with both sides of the piston head but at different pressures. High pressure oil may be pumped into either side of the piston head. The diagram below shows an actuating cylinder controlled by a selector valve. The selector valve determines to which side of the actuating cylinder the high pressure oil (red colored side) is sent. The piston rod of the actuating cylinder is connected to the control surface, in this case, an elevator. Press to see Animation As the piston moves out, the elevator moves down. As the piston moves in, the elevator moves up. The selector valve directs the high pressure oil to the appropriate side of the piston head causing movement of the piston in the actuating cylinder. As the piston moves, the oil on the low pressure side (blue colored side) returns to the reservoir since return lines have no pressure! The differential in oil pressure causes movement of the piston. The force generated by this pressure difference can be sufficient to move the necessary loads. Each cylinder in the plane, boat, etc., is designed for what it must do. It can deliver the potential it was made for; no more, no less. Air loads generally determine the force needed in aircraft applications. For example, if a force of 40,000 pounds is required and the high pressure oil is pumped in at a pressure of 1000 psi, then the piston is designed to have a surface area of 40 square inches on which the oil acts. Hydraulic SystemA hydraulic system transmits power by means of fluid flow under pressure. The rate of flow of the oil through the system into the actuating cylinder will determine the speed with which the piston rod in the actuating cylinder extends or retracts. When the cylinder is installed on the aircraft, it is already filled with oil. This insures that no air bubbles are introduced into the hydraulic system, which can adversely affect the operation of the system.Pascal’s TheoryThe method by which fluid is used to create force was explained by Pascal. In a confined stationary liquid, neglecting the effect of gravity, pressure is distributed equally and undiminished in all directions; it acts perpendicular to the surface it touches. Because the actuating cylinder is not vented, the force delivered through the piston to the surface of the fluid is translated into a pressure on the surface of the fluid. The pressure (p) acting on the incompressible oil does work [(pressure) x (Area of piston) x (piston's stroke) = Work]. In the diagram below, the force acting on the right side piston does work and moves the fluid from the right cylinder to the left cylinder. The fluid movement into the left cylinder creates a pressure on the left piston's surface area. That in turn creates a force that moves the left piston up. Multiplication of ForcesPascal's Law states that the pressures in both cylinders are the same (p1=p2). Thus, given a force, F1, of 10 pounds (lbs) in the right cylinder acting on a piston area, A1, of 2 square inches (sq. in.) a pressure in the right cylinder, p1, of 5 pounds per square inch (lbs/sq. in. = psi) is produced. Now if A2 is given as 5 sq. in., then the force developed in the left cylinder is F2 = p2xA2, or 25 lbs. This is due to the fact that p1=p2. Thus Pascal's Law shows the way in which one can increase the output force for a given input force .regulate the areas of the pistons!Press to see Animation The only disadvantage is the size of the piston stroke involved. Let's say, piston 2 moves (up) 10 inches. For the previous problem the work done by piston 2 is F2 times the stroke of piston 2 (10 in. x 25 lbs). If no losses exist in the system due to friction, then work is conserved and piston 1 must do 250 in-lb of work. Therefore, the F1 must move down 25 inches (250 in-lbs/10 lb force)! To move piston 2 up, a volume of 50 cubic inches (cu. in.) of incompressible oil must be pumped in at 5 psi (since pressure times Volume is also another way to find work). The movements of the pistons are measured relative to the bottom of the cylinder with all the measurements computed to produce 100 % efficiency.How to Increase the Output Force of Cylinder 21. Increase the pressure generated. The disadvantage with this idea is that you must remove the old tubing and replace it with new tubing that can withstand the new loading.2. Increase the area of piston 2. That may be restricted by the size of the actuating cylinder you can place in the location slated for the cylinder.3. Increase the stroke of piston 1. This may also be restricted by the location of the actuating cylinder 1.How to Increase Input Force, F11. Increase the force by increasing the pressure.2. Increase the stroke of piston 1.3. Decrease the area of piston 2. Just to reinforce what was said before: the distance of piston movement for the piston in the output cylinder is determined by the volume of oil being pushed into the output cylinder.Brake System in a Car – Hydraulic SystemAn example of a hydraulic system that we deal with every day is the brake system in our cars. That system is an example of the material we have just discussed. Look at the picture given below. When the brake pedal is pressed down, the piston in the 1st cylinder goes down, pushing the oil through the tubing into the little wheel actuating cylinder near the brake shoes. The oil, in turn, pushes the little pistons out and this, in turn, pushes the shoes up against the brake drum causing the car wheel to be slowed to a stop.Section 1.3 - Hydraulic System Power RequirementsTypical ProblemSuppose you were asked to determine the mechanical horsepower (HP) required to retract a landing gear in a required time period. How would you do the calculations? Example: Given- Force Requirements = 5000 lb (this is the force that has to be moved) Distance moved = 2 ft (this is the distance you must move the force) Time required = 30 s (this is the time required to move that distance)Power is given as Force times velocity for a constant force (P=Fv). If the force is not constant, then you can use the average force over the time required. The velocity in this case is the average velocity, namely, the distance traveled over the time required. Therefore, Power=Force x distance / timeWe convert to horsepower (HP) using the conversion factor 550 equals 1 HP. Therefore, by multiplying 334 by [1 HP/ (550 )], we find that we need 0.61 HP. Thus, an actuating cylinder must then be mounted which can deliver 0.61 HP. The actuating cylinder for the retractable landing gear is mounted so that it can move in order that the piston rod in the actuating cylinder won't bend. A flexible hose to the oil pressure lines is put at the cylinder attachment so that it won't break during movement.Selection of an Actuating CylinderThe selection of the actuating cylinder depends upon two parameters: 1. Piston stroke - the distance that it must travel to do the job.2. Piston head area which must be large enough to develop the proper force with the pressure available. Flow Requirements to Accomplish TaskThe hydraulic system oil flow rate, Q, may be measured in gallons per minute (gpm). The flow rate required can be related to the volume of fluid required to be moved (in cubic inches-cu in) and the time required for the job (in minutes).The volume of fluid required to be moved is given by the input force times the piston stroke (in inches) divided by the system oil pressure. Remember that force divided by pressure is an area, and, multiplied by the piston stroke defines the volume moved. Therefore,Example: If the pressure in the system = 2000 psi, find Q of previous problem.Hydraulic HorsepowerThe hydraulic horsepower is the power provided by the hydraulic system. It is directly proportional to the rate of flow, the pressure, a constant and inversely proportional to the efficiency of the system. The coefficient equals 0.000583 and is the conversion factor between gallon-lbs/(minute-square inches) and horsepower. Therefore:Example: Find the hydraulic HP of the previous problem if the system has an efficiency of 1.F-111 sweep back problemLet's look at a typical problem. Find the hydraulic and mechanical HP required to vary the sweep back of an experimental F-111 wing, given the following data:Force required = 160,000 lb; Cross-sectional area of the actuating cylinder piston A = 32 square inFluid Pressure P = 5000 lb/sq. in. = 5000 psiPiston stroke D=30 inchesTime required for sweeping the wing T=75 seconds=1.25 minutesHydraulic HP is found by getting the flow rate, Q, in gpm, FIRSTNow having found Q, we can now find the Hydraulic HP, assuming an efficiency of 1, using The mechanical HP is found usingBy comparing both results, we can see that the hydraulic system will meet the requirements of the mechanical system HP.Note: In reality, the actual value is about 9.7 HP, but because systems that deliver the horsepower are rated in 1/4 HP increments, the results would be listed to the next 1/4 HP increment, namely 9.75 HPSection 1.4 - Hydraulic Pressure Regulated Power SystemIntroductionThe system in drawing below represents a pressure regulated power system comprised of two parts: 1) the power system, and 2) the actuating system part of the overall hydraulic system. This type of system was used in all aircraft between 1937 and 1945. The system had a pressure regulator which "knew" when to withdraw horsepower from the engine when it was needed. The concurrent blue-red system drawn in that manner because when the bypass part of the system was used, the blue-red part of the system acted as the return line to the reservoir and, thus, was "blue". However, the power system was used to produce hydraulic power, the blue-red part of the system was filled with high pressure oil and, therefore, was "red". Functions of Parts of the Power System1. Reservoir -- holds an extra supply of fluid for system from which oil was drawn when needed, or oil was returned to it when not needed.2. Accumulator -- absorbs pulsation within the hydraulic system and helps reduce "linehammer effects" (pulses that feel and sound like a hammer has hit the hydraulic tubes). It is an emergency source of power and it acts as another reservoir. 3. Filter -- removes impurities in the hydraulic system and in the reservoir. The reservoir has one big filter inside the tank. 4. Power Pump -- it changes mechanical horsepower (HP) to hydraulic HP.5. System Relief Valve -- relieves pressure on system as a safety.measure and takes over as a pressure regulator when pressure regulator fails. 6. Pressure Regulator -- as the name implies, regulates the pressure in the hydraulic system. When it senses a built-up in pressure in the lines to the selector valves, it acts so that the system automatically goes to bypass.Section 1.5 - Aircraft Hydraulic System Reservoir DesignFunctions of the Reservoir1. Provides air space for expansion of the oil due to temperature changes2. Holds a reserve supply of oil to account fora. thermal contraction of oil.b. normal leakage - oil is used to lubricate piston rods and cylinder seals. When the piston rod moves, it is scraped to remove impurities that might collect on the rod when returning into actuating cylinders. If many actuating cylinders are operating at the same time, then the amount of oil lost is greater. c. emergency supply of oil - this case occurs only when the hand pump is used. d. volume changes due to operational requirements - oil needed on side 2 of piston head is less than that needed on side 1 of cylinder piston (which occurs during actuation). . speed and flap brakes 8. Pitch changing mechanism, spoilers on flaps 9. Bomb bay doors and bomb displacement gears Some Devices Operated by the Hydraulic