FLUID POWER!
TODAY
• Learn some fluid power basics
• Hands-on project with pneumatic
components
• Some possibilities for your FIRST robot
What is fluid power?
•
Uses:
•
•
•
•
Heavy equipment
Construction industry
Off-road vehicles
Manufacturing
WHY FLUID POWER?
STRONG!
LIGHT!
EASY!
Fluid Power is Unique Unparalleled torque, power
and bandwidth for the same weight or volume.
Example: Power/Weight (kW/kg)
Pneumatic Motor
0.3-0.4
Hydraulic Motor
0.5-1.0
Electric Motor
0.03-0.1
Fluid power weight advantage = 10:1
Reference: I. L. Krivts and G. V. Krejnin, Pneumatic Actuating Systems for
Automatic Equipment, Taylor and Francis, 2006.
CCEFP
CENTER FOR COMPACT AND
EFFICIENT FLUID POWER
•100 mpg automobile
•Efficient off-road equipment
•Compact and portable
CCEFP
CCEFP testbeds
TB1: Excavator
TB2*: Injection
molding machine
Existing
FP applications
FP enabled
breakthroughs
in transportation
TB3: small Urban
Vehicle (sUV)
TB5*: FP assisted
hand tools
TB4: Compact
Rescue Crawler
* Reduced or
delayed funding
CCEFP
New industries
& applications
TB6:
FP assisted
orthoses &
prostheses
Master Pneumatic
National Tube Supply Company
Ralph Rivera
HIGH COUNTRY TEK
Member of the Schaeffler Group
Pneumatics compared to hydraulics
• No problems of a spills
• Compressibility stores energy
– Available for your use
– Dangerous if excessive
volumes or pressures
• Difficult to control precisely
• Fluid is readily available
– Should be filtered, dry
• Usually lower forces
Safety Must Always
Be Considered!
Pressure of an “ideal” Gas
Pressure P
P V  mR T
• Pressure of a gas is due to
the force of gas molecules
bouncing off the walls.
• Pressure increases when
molecules are moving
faster, heavier, or if there
are more molecules.
• Molecules move faster
when they are hot.
• mR depends on molecule.
Getting Work out of Air
• Work is force acting
over a distance, ftlbs.
• Put air in a container
under pressure
• Allow part of the
container to expand
• The expanding part
does work
How much energy is in a tank filled
with compressed air?
Assume constant temperature:
PV  mRT  constant
Energy:
E  PV ln(P / Patm )
P = pressure in tank (absolute)
V = volume of tank
Patm= atmospheric pressure = 101,325 Pa or 14.7 psi
How much energy is in a small air
tank?
• Tank Volume = 150 ml or 9.154 in3
• Pressure = 413,700 Pa or 60 psi (over Patm)
• Patm = 101,325 Pa or 14.7 psi
Answer:
Energy = PV ln(P/Patm)
= 0.15 x 515025 x ln(74.7/14.7) = 125 kJ
Challenge question: How high could the instructor
be lifted using the energy in one tank?
How much energy in your tank
can you use?
• Line losses:
Pressure drop proportional to flow
• Throttling losses:
Pressure drop proportional to flow squared
• Cylinder friction:
Coulomb plus viscous friction, depends on
seals
Force available
• Pressure x Area = Force
• Area = pi x Bore2 / 4
• For example cylinder:
– Bore = 10 mm = .394 in.  Area = .122 in2
– Force = PxA = 60 psi x .122 in2 = 7 lbs
Pressure P
Area AP
Force F
CYLINDER FORCE AT 60 PSI
800
700
fORCE (LBS)
600
500
400
300
200
100
0
0
1
2
3
BORE (IN.)
4
5
The Effect of Different Areas
Pressure P
Area AR
Area AP
Pressure P
How much force F is
necessary to hold the rod still?
Force F
Pressure
Patmosphere
Pneumatic components seen in the
FIRST Robotics competition
LET'S BUILD!