Download Pressure and Hydraulic/Pneumatic

Pressure and Hydraulic/Pneumatic
Systems
Formula (Pascal’s Law)
P = Force
Area
Pressure: a measure of the distribution of force over a given area
Have you ever noticed how much air you can pump in to a bicycle tire? When you pump more air
into a tire, the particles in the air move closer together; the air becomes compressed. Air and other gasses
can be compressed because in a gas there are large empty spaces between the particles that can be filled up.
In a liquid (e.g., water) the particles are closer together and can not be compressed as much. This
can be noticed when doing our syringe experiment where the pneumatic system (syringes filled with air),
moved slower than the hydraulic system (syringes filled with water). This happens because as you press on
the master plunger, the movement first compresses the air, then it will move the slave plunger. Alternately,
when you push the master plunger on a hydraulic system, the slave plunger moved right away because the
water can not be compressed as easily.
The hydraulic system shown in figure 1 illustrates an application of Pascal’s Law. It contains two
pistons (syringes) that slide up and down inside a tube. One syringes plunger area is 1 m2 and the other is
25 m2. As you can see, a small force can be magnified many times in a hydraulic system. The amount of
force magnification equals to how many time the area of the larger piston is greater than the area of the
smaller piston.
We can calculate the mechanical advantage of the system in figure 1. The effort force of 100 N is applied to
the small siring. The resulting load force on the large piston is 2500 N. This system can lift objects that are
25 times heavier than the input or effort force.
We can also look at hydraulic and pneumatic systems in terms of mechanical advantage. The formula
should be familiar
Mechanical Advantage = Load Force
Effort Force
= 2500 N
100 N
= 25
An application of Pascal’s Law:
Small Piston
Area (A) = 1m2
F = 100N
Large Piston
Area (A) = 25m2
P = 100 Pa
= 100 N/m2
P=F
A
This formula can be rearranged:
F=AxP
F = 25 m2 x 100 N
m2
F = 2500 N
P=F
A
= 100N
1m2
= 100Pa
2
1m
Effort 100N
25 m2
Load 2500 N
Another way to look at this is by observing what the syringes do:
When you have a hydraulic system with a small syringe and a large syringe, when you
press the plunger on the small syringe, the surface area of the plunger (small syringe) is less and
therefore can only push a small amount of water. It is easier to push, but the large syringes
plunger will move less. There is greater mechanical advantage, but less movement (sounds
familiar)
Alternately, when you press the large syringe in this system, it is more difficult to press
as you are moving a large amount of water, but the small syringes plunger moves more. There is
less mechanical advantage, but the syringe moves more (now it should sound familiar).
Questions:
1. What is a hydraulic system?
2. What is a pneumatic system?
3. In your own words, explain why there is mechanical advantage when you press a small
syringe that is connected to a large syringe (use pictures).
Mechanical Efficiency
We have been working this year with mechanical advantage, but we have not taken one
thing into account with our calculations: Friction. Friction decreases our mechanical advantage
(e.g., the rope and pulley create friction). This friction causes our mechanical advantage to be
lower than expected and we need to keep this in mind when building machines. All systems will
have some type of friction decreasing the mechanical advantage.
Example:
Velocity Ratio = 12 m = 4
3m
Without Friction
M.A. = 3000 N
750 N
=4
With Friction
M.A. = 3000 N
900 N
= 3.3
How efficient is the system?
% Efficiency = Mechanical Advantage X 100
Velocity Ratio
= 3.3
4
= 82.5%
This means that the machine is 82.5% efficient. A machine that has 82.5% efficiency is a much
better machine that is only 50% efficient because it can do more work with less effort.
W=FxD
This formula is the basis for all mechanical systems.
For example:
Picking up a weight that is 10 N 5 m gives:
W=FxD
= 10 N x 5m
= 50 J
When using a mechanical system, with a M.A.= 2 (not taking friction into account) our
force needed is cut in half, but our distance doubles:
W=FxD
= 5 N x 10m
= 50 J
The work is the same. Bringing us back to the idea that trade distance for force, but the
work needed is the same.
Questions
1. Calculate the work needed to lift a 50 N weight 10 meters.
2. Calculate the % efficiency with a velocity ratio of 10 and a mechanical advantage
of 5. Is this an efficient machine? Explain.