|
Pneumatic actuators; both linear and rotary, must “know” where,
when and what times logic level of intensity is needed to act on the
machinery. When a signal is sent from electrical processor to
either the directional control valve or pressure control valves to
the actuator will result in thus accomplishing the work. Single
acting actuators convert pneumatic energy into mechanical energy to
produce work. This is usually only carried out during the forward
stroke movement of the actuator. The actuators force (lbs) which is
exerted via the piston and the piston rod which increases in
accordance with the piston area expressed in (in2) and the air
pressure in the actuators chamber. Usually the force to return the
piston is a mechanical spring; this is normally accomplished when
air is exhausted from the actuator. However, in some pneumatic
applications when using pneumatic single-acting actuators the spring
may be place on the piston end to allow the actuator to snap into
place, example shot pins, when the air is shut off. This usually
occurs when emergence “E”-stops application are energized on the
machinery.
Double-acting actuators are similar to their cousin the
single-acting actuators. The major difference between the too are
there is “no” spring, the double-acting actuators have two ports
which may be used alternatively as either the supply or exhaust
ports. Double-acting actuators maybe controlled by two positions or
three positions two-way, three-way, four-way or four-way five ported
directional control valves. Linear actuator seals should always be
capable of withstanding pneumatic industrial applications
requirements. Seals must be able to provide the following
resilience, strength, low frication, wear resistance, temperature
changes and the ability to form a tight seal.
Other common types of linear actuators used in pneumatic
applications are; single-acting, double-acting, ram, double rod,
tandem, duplex, telescoping and booster (air/oil).
Pneumatic actuators are made-up of the following basic components
in their construction; head and cap ends, cylinder tubes, pistons,
piston rods, rod glands or referred as bearings, some form of seals
and a means of holding the actuators together. One of the most
common is the tie-rod version; they may be threaded together or have
no means of taking them apart. When considering which type to use,
remember that the actuator style is given by the application.
Mounting styles of linear actuators one should consider, either
the NFPA mounting styles or the ISO mounting styles. Both are
similar but the ISO mounting configuration does not have as many
styles to choose from and is becoming the global standard.
Linear actuators have many methods of applying their
applications. Pneumatic application engineers must considerer
applications regarding horizontal, inclined, vertical, wedge,
sliding surfaces, block and tackle, jib crane, lever- jib crane,
bent- lever, first- class levers, second-class, third-class lever
and toggle joint applications.
Pneumatic actuator sizing is very important for the pneumatic
application. Make sure that all parameters are known before
choosing the pneumatic actuators. Pneumatic engineers need to look
at the push and pull forces being applied to the pneumatic
actuators. Rod size, slide loading and bending moments will lead to
the actuator buckling. Cushions should be considered for the
maximum deceleration when using actuators. Cushion capabilities
for pneumatic actuator designers, have many variables to examine.
One must always consider the compressibility of the gas, weight,
velocity friction pressure and of course the load. If cushioning is
not applied properly the following major issues will occur: the
piston could impact the end cap, oscillating, movement in the
opposite direction and stopping in place are just a few examples.
Finally, if cushion issues of bouncing keep occurring, one should
consider making adjustments to both the cushions and/ or loading
when designing the pneumatic circuit. Examples of possible
solutions are meter-out, or meter-in and external shock absorber
being used.
Examples, for pneumatic actuators a 3.5-inches actuator is being
used to give a force output of 500-lbs. What is the pressure
required? P-(pressure Psi) = F-(force lbs) / area-(in2)
Therefore;
3.5 x 3.5 x.7854 = 9.62-in2, 500-lbs / 9.62-in2
= 51.97- Psi or 52-Psi needed to overcome the cylinder.
What is the total volume of the actuator if the cylinder with a
bore of 3.5-inches and a Stroke of 10-inches
9.62-in2 x 10-inches stroke = 96.2- in3 of volume.
A double acting air cylinder with a 2-inch bore and a 20-inch
stroke reciprocates at 200- cycles/ min using 100-Psig of air. What
is scfm flow-rate of air to the cylinder? Please ignore the piston
rod cross- sectional area and assuming the temperature remains
constant.
2-inch bore = 3.14-in2, 20-inch stroke, Pressure =
100-Psig
Cycles = 200 cycles /min
Q2 = 3.14-in2 x
(20x2) x 200 / 1728 = 25120 / 1728 = 14.54-ft3/ min
Q1 = P2 x Q2 / P1 = 114.7 x 14.54 / 14.7 =
113.69- scfm
Assuming a total deceleration force of 1200-lbs is on the
stationary end of a pneumatic cylinder with a 3-in bore and a 1.5-in
cylinder rod, please compute the pneumatic pressure at the rod-end
of the cylinder?
4-in bore = 12.57-in2, Rod diameter = 1.5-in = 1.18-in2
Area of effect = A-piston –A-rod = 12.57-in2 -1.18-in2 = 11.39-in2
P = F/ A
P= 1200-lbs / 11.39-in2 =
Pressure = 105.36-Psi
Calculate the cylinder force required to move a load, which is
traveling at a 22-degree angle with the horizontal, the load weight
is 1500-lbs, and the coefficient of friction with the sliding
surface is 0.7. Therefore:
WF (load weight creating friction) = 1500-lbs x cos-220- (.927) =
1390.5-lbs
WG (weight to be lifted against gravity) =1500-lbs x sine-220-
(.375) = 562.5-lbs
FF (friction force) = 1500-lbs x 0.7- (coefficient of friction) =
1050-lbs
TF (total force) = 1390.5-(friction) + 562.5-lbs-(gravity lift) =
1953-lbs of force |
|
Rotary actuators are use to convert potential pneumatic energy into
mechanical energy. The major difference between its cousin the
linear actuator is that rotary actuators or commonly called air
motors, is the fact that they produce torque (in/lbs or ft/lbs).
Pneumatic rotary actuators are broken down into five types: rotary
vane, gear, turbine and piston these are broken down into (radial
and axial) styles, rack and pinion for rotary motion.
When making a pneumatic rotary actuator selection, one should be
considering three distinct factors; HP (horsepower) vs. RPM
(revolution per minute), Torque (in/lbs) vs. RPM and air consumption
vs. RPM. Before making the choosing a pneumatic rotary actuators
one should consult the pneumatic manufacturer. I have been told
over the years that a good rule of thumb to follow when selecting
rotary actuators pertaining to HP (horsepower) is that the air motor
should deliver a required horsepower of 65% of the pneumatic line
pressure being supplied. Rotary actuators the torque required are
usually achieved at about 2/3 of the pneumatic operating system
pressure. In most cases rotary actuators are not chosen for their
efficiencies, but for their power, speed and torque.
Rotary actuators can have some major concerns that could lead to
failures in the field, contamination, misalignment, corrosion and
most of all lack of proper lubrication. These will lead to bad
symptoms such as low torque and sluggish operation.
So let’s look at some examples of formulas required to operate
rotary actuators and how to solve the equitation’s.
HP = T-(in/lbs) x RPM / 63025
RPM = HP x 63025 / T- (in/lbs)
T –(in/lbs) = HP x CIR / 2-pi
CIR = T-(in/lbs) x 2-pi / RPM
A pneumatic rotary actuator has a system pressure of
500-(in/lbs), and is delivering 3-HP at speed of 2000-RPM, what is
rotary actuator displacement (CIR)?
CIR = 500-in/lbs x 6.28 / 2000
CIR= 1.57-in3/rev/min
15- HP is being supply by the rotary actuator, and has a CIR of
5.5-in3/rev. What would be the torque (in/lbs) that would be
produce?
T-(in/lbs) = HP x CIR / 2/pi
T-(in/lbs) = 15-HP x 5.5-CIR / 6.28
T-(in/lbs) =13.14 –in/lbs
If a pneumatic rotary actuator is produce 20-HP and has a torque
capability of 500-in/lbs. What is the RPM that the pneumatic
actuator will produce?
RPM = HP x 63,025 / T-(in/lbs)
RPM = 20-HP x 63,025 / 500-(in/lbs)
RPM = 2521-rev/min
Mathematical problems pertaining to pneumatic actuators are
illustrations of how to achieve the different functions in the
pneumatic process. Remember, that any type of pneumatic actuators
(linear or rotary) being used are the work horse of the pneumatic
system to accomplish work. When designing the pneumatic systems
please make sure the applications and components being used are the
correct ones for the pneumatic system.
Symbology:
Single Acting Actuator
Double Acting Actuator
Oscillating Motor / Rotary Actuator
Air motor/ Bi directional Rotary Actuator
We would be pleased to hear about your
unusual applications or experience with pneumatic cylinders. Also
for suggestions about your requirements or possible assistance
contact us through the Fluid Power Journal.
|
