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CHAPTER-1 INTRODUCTION
This project is made in order to make a pneumatic jack that could lift up a complete car or automobile when placed on the bottom of the car at its center. That is why named as center jack. Pneumatics is a branch of technology, which deals with the study and application of use of pressurized gas to affect mechanical motion.
Pneumatic systems are extensively used in industry, where factories are commonly plumbed with compressed air or other compressed inert gases. This is because a centrally-located and electrically-powered compressor that powers cylinders and other pneumatic devices through solenoid valves is often able to provide motive power in a cheaper, safer, more flexible, and more reliable way than a large number of electric motors and actuators.
Pneumatics also has applications in dentistry, construction, mining, and other areas.
Here, we are using pneumatic system to move the jack up with the air pressure input.
This centre jack is used in order to lift a car completely by putting force on the center of the bottom of the car while other jacks are used to lift some part of the car.
The best part of this project is to drive the complete system with the help of compressed air up completely without any part of the automobile touching the earth.
SPECIFICATIONS OF THE CENTER JACK:
Maximum weight carrying capacity: 2000Kgs
Maximum voltage of operation: 24V dc
Fuel to be used: Compressed air
Technology used: Pneumatic
Force applied:
Pneumatic systems in fixed installations such as factories use compressed air because a sustainable supply can be made by compressing atmospheric air. The air usually has moisture removed and a small quantity of oil added at the compressor, to avoid corrosion of mechanical components and to lubricate them.
Factory-plumbed, pneumatic-power users need not worry about poisonous leakages as the gas is commonly just air. Smaller or stand-alone systems can use other compressed gases which are an asphyxiation hazard, such as nitrogen - often referred to as OFN (oxygen-free nitrogen), when supplied in cylinders.
Any compressed gas other than air is an asphyxiation hazard - including nitrogen, which makes up 77% of air. Compressed oxygen (approx. 23% of air) would not asphyxiate, but it would be an extreme fire hazard, so is never used in pneumatically powered devices.
Portable pneumatic tools and small vehicles such as Robot Wars machines and other hobbyist applications are often powered by compressed carbon dioxide because containers designed to hold it such as soda stream canisters and fire extinguishers are readily available, and the phase change between liquid and gas makes it possible to obtain a larger volume of compressed gas from a lighter container than compressed air would allow. Carbon dioxide is an asphyxiant and can also be a freezing hazard when vented inappropriately.
Comparison to hydraulics
Both pneumatics and hydraulics are applications of fluid power. Pneumatics uses an easily compressible gas such as air or a suitable pure gas, while hydraulics uses relatively incompressible liquid media such as oil. Most industrial pneumatic applications use pressures of about 80 to 100 pounds per square inch (550 to 690 kPa). Hydraulics applications commonly use from 1,000 to 5,000 psi (6.9 to 34 MPa), but specialized applications may exceed 10,000 psi (69 MPa).
Advantages of pneumatics
• Simplicity of Design And Control
• Machines are easily designed using standard cylinders & other components. Control is as easy as it is simple ON - OFF type control.
• Reliability
• Pneumatic systems tend to have long operating lives and require very little maintenance.
• Because gas is compressible, the equipment is less likely to be damaged by shock. The gas in pneumatics absorbs excessive force, whereas the fluid of hydraulics directly transfers force.
• Storage
• Compressed Gas can be stored, allowing the use of machines when electrical power is lost.
• Safety
• Very low chance of fire (compared to hydraulic oil).
• Machines can be designed to be overload safe.
CHAPTR-2: ELEMENTS OF THE PNEUMATIC SYSTEM
A – Compressor: a pump which compresses air, raising it to a higher pressure, and delivers
it to the pneumatic system (sometimes, can also be used to generate a vacuum).
B – Check valve: one-way valve that allows pressurized air to enter the pneumatic system,
but prevents backflow (and loss of pressure) into the compressor when it is stopped.
C – Accumulator: stores compressed air, preventing surges in pressure and relieving the
duty cycle of the compressor.
D – Directional valve: controls the flow of pressurized air from the source to the selected
port. Some valves permit free exhaust from the port not selected. These valves can be
actuated either manually or electrically (the valves typically provided in the FIRST kits use
dual solenoids to change the direction of the valve, based on input signals from the control
system).
E – Actuator: converts energy stored in the compressed air into mechanical motion. A
linear piston is shown. Alternate tools include rotary actuators, air tools, expanding
bladders, etc.
Pneumatic System Design Notes
Constants:
1 mm = 0.0394 inches
1 square mm = 0.0016 square inches
1 litre = 0.0353 cubic feet
1 bar = 14.50 psi
As we design a pneumatic system of the type used in the FIRST competitions, we want to
know three things:
- how much force can an actuator apply?
- is that force sufficient to move the desired load?
- how fast can the load be moved?
To determine how much force an actuator can apply, we need to calculate the Theoretical
Force. For a pneumatic piston actuator, that is determined by multiplying the surface area
of the moving piston by the pressure applied. In other words, for a round piston:
Ft = p * D2/4 * P
Where D is the diameter of the piston and P is the
working pressure of the injected air. Note that on
the reverse stroke of the piston, the available surface
area of the piston is decreased by the area of the
piston rod. In that case:
Ft = p * (D2 – d2)/4 * P
Note this does not account for inefficiencies in the actuator due to friction between the
piston and the cylinder wall, the piston rod and the packing gland, stiction forces, etc. For
our purposes, these factors contribute to an approximate 5% loss in efficiency (i.e. the
practical force available from the piston is about 95% of the calculated force).
Next, we need to know how much force is required to move the object we want to move.
To determine the Required Force, we need to know the mass of the object to be moved,
the direction of motion relative to gravity, and the effects of any friction between the object
to be moved and whatever is supporting it. To calculate the required force, use:
Fr = G * (sin a + μ * cos a)
where G is the mass of the object to be moved, a is
the angle of inclination that the mass will move
(between –90 and 90 degrees), and μ is the
coefficient of friction between the moving object
and any supporting structure or surface (μ may
vary between 0.1 and 0.4 for sliding metal-onmetal
parts, or about 0.005 for iron rolling on iron as in a ball bearing, etc.).
The Load Ratio is the relationship between the force required to move the load and the
available force from the actuator. The load ratio is determined by:
Load ratio = (Required Forced / Theoretical Force) * 100 %
In theory, the load ratio must be 100% or lower to be able to perform the task. In practical
applications, the load ratio should be 85% or lower. Also note that if the actuator is able to
deliver more force than the minimum needed to move the load, then the excess force
delivered by the actuator is used to accelerate the load. In other words:
Acceleration Force (Fa) = Theoretical Force (Ft) – Required Force (Fr)
From the first lesson on forces and accelerations, we know that
Acceleration (A) = Force (Fa) / Mass (G)
Distance (d) = 1/2 * Acceleration (A) * Time (t)2
Knowing the acceleration of the object and the distance to be traveled (the stroke of the
piston), we can calculate the time required for the object to move from rest to the end of
the piston stroke (remember that the value of G is determined by the weight of the object
divided by acceleration due to gravity; 32 ft/sec2 for English units, or 9.8 m/sec 2 for
metric). Note that this theoretical acceleration is based on the assumption that there is an
instantaneous supply of compressed air, and there is no back pressure on the back side of
the piston. Each of these factors limits the practical acceleration of the load.
The effects of these limiting factors can be reduced by applying a number of strategies
when designing the pneumatic system. A very complex set of calculations can be used to
evaluate the different design parameters, or we can use a set of “Rules of Thumb” which
result in approaches that are close enough for our purposes. These are captured below.
“Rules of thumb” for pneumatic design:
- Larger tubing, valves and fittings are preferable to smaller (large diameter tubing adds
less resistance to the air flow than smaller tubing)
- For a given tubing size, the shorter the run, the better (long tubing runs add resistance
to the air flow from the source to the fittings and actuators)
- The straighter the tubing run, the better (bends and curves induce turbulence, which
slows the flow of air into the fittings and actuators)
- The fewer valves (and other fittings), the better (fittings add resistance to the air flow)
- The higher the air pressure, the better (for a given equivalent flow section, this gives a
higher flow rate)
- For bi-directional piston actuators, place the control valve as close to the actuator as
possible (this reduces the back pressure on the exhaust side of the piston)
Example and problems:
A 100 pound object needs to be lifted vertically one foot. You have a 2-inch diameter
piston actuator with a 12-inch stroke connected to a 50 psi source of pressurized air. The
mass is restricted by a set of guide rails, which limit it to vertical motion, and which use
roller bearings with a friction coefficient of 0.002. Can the object be lifted? How long
will it take?
The theoretical force of the piston is (! * 22/4 * 50) = 157 lb. With a 5% efficiency loss,
the available force is 149 lb (157 lb * 95%).
Since the mass is being lifted vertically, the inclination angle a is 90°. So, the required
force is 100 lb * (sin 90° + 0.002 * cos 90°) = 100 lb. This is well below the theoretical
force, so the object can be lifted. The load ratio is (100 lb / 149 lb)*100 = 67.1%
The acceleration force is (149 lb – 100 lb) = 49 lb. Therefore, the acceleration is (49 /
(100/32)) = 15.75 ft / sec2, or 189 inches / sec 2. With this acceleration, the object can be
raised 12 inches in about 0.35 seconds (12 = 1/2 * 189 * t2