Hydraulics is the science of transmitting
force and/or motion through the medium of a confined liquid. In a
hydraulic device, power is transmitted by pushing on a confined
liquid. Figure 1-1 shows a simple hydraulic device. The transfer
of energy takes place because a quantity of liquid is subject to
pressure. To operate liquid-powered systems, the operator should
have a knowledge of the basic nature of liquids. This chapter
covers the properties of liquids and how they act under different
conditions.
1-1. Pressure and Force.
Pressure is force exerted against a specific area (force per unit
area) expressed in pounds per square inch (psi). Pressure can
cause an expansion, or resistance to compression, of a fluid that
is being squeezed. A fluid is any liquid or gas (vapor). Force is
anything that tends to produce or modify (push or pull) motion
and is expressed in pounds.
- a. Pressure. An example
of pressure is the air (gas) that fills an
automobile tire. As a tire is inflated, more air
is squeezed into it than it can hold. The air
inside a tire resists the squeezing by pushing
outward on the casing of the tire. The outward
push of the air is pressure. Equal pressure
throughout a confined area is a characteristic of
any pressurized fluid. For example, in an
inflated tire, the outward push of the air is
uniform throughout. If it were not, a tire would
be pushed into odd shapes because of its
elasticity.
- There is a major difference
between a gas and a liquid. Liquids are slightly
compressible (Figure 1-2). When a confined liquid
is pushed on, pressure builds up. The pressure is
still transmitted equally throughout the
container. The fluid's behavior makes it possible
to transmit a push through pipes, around corners,
and up and down. A hydraulic system uses a liquid
because its near incompressibility makes the
action instantaneous as long as the system is
full of liquid.
- Pressure can be created by
squeezing or pushing on a confined fluid only if
there is a resistance to flow. The two ways to
push on a fluid are by the action of a mechanical
pump or by the weight of the fluid. An example of
pressure due to a fluid's weight would be in an
ocean's depths. The water's weight creates the
pressure, which increases or decreases, depending
on the depth.
- By knowing the weight of a cubic
foot of water, you can calculate the pressure at
any depth. Figure 1-3 shows a column of water 1
foot square and 10 feet high, which equates to 10
cubic feet. (One cubic foot of water weighs 52.4
pounds.) The total weight of water in this column
is 624 pounds. The weight at the bottom covers
1,445 square inches (1 square foot). Each square
inch of the bottom is subject to 1/144 of the
total weight, or 4.33 pounds. Thus, the pressure
at this depth is 4.33 psi. You can also create an
equal pressure of 4.33 psi in a liquid using the
pump and figures shown in Figure 1-4.
- Before pressure, head was the
only way to express pressure measurement. It was
expressed as feet of water. Today, head is still
the vertical distance between two levels in a
fluid. In Figure 1-3, the head between the top
and bottom of the water is 10 feet, which is
equivalent to 4.33 psi. Therefore, each foot of
water is equal to 0.433 psi.
- The earth has an atmosphere of
air extending 50 miles up, and this air has
weight. This air creates a head of pressure that
is called atmospheric pressure. A column of air 1
square inch in cross section and the height of
the atmosphere would weigh 14.7 pounds at sea
level. Thus, the earth's atmospheric pressure is
14.7 psi at sea level. The role of atmospheric
pressure in most hydraulic systems is
significant. Figure 1-5 shows the interaction of
hydraulic and atmospheric pressures under the
three sets of conditions listed below:
- (1) Diagram A. In the diagram,
the tube is open at both ends. When it is placed
in a liquid, the liquid will rise, inside and
outside, in proportion to the amount of liquid
displaced by the submerged tube wall.
- (2) Diagram B. In the diagram,
ends of the tube are closed. When placed in a
liquid, the liquid level in the tube is forced
down because the air in the tube must occupy a
space. Therefore, the liquid is displaced. The
liquid level outside the tube rises in proportion
to the volume of the cylinder wall and the volume
of the trapped air below the original liquid
level. The atmospheric pressure (14.7 psi) on the
liquid outside the tube is not heavy enough to
force the liquid inside the tube upward against
the pressure of the trapped air, which is more
than 14.7 psi.
- (3) Diagram C. In the diagram,
the upper end of the tube is closed, but some of
the air has been removed from this tube so that
the pressure within the tube is less than 14.7
psi (a partial vacuum). A perfect vacuum would
exist if all pressure within the tube could be
eliminated, a condition that never happens.
Because the liquid outside the tube is subject to
full atmospheric pressure, the liquid is forced
up into the tube to satisfy the vacuum. How far
the liquid rises depends on the difference in air
pressure between the trapped air and the
atmosphere.
- b. Force. The
relationship of force, pressure, and area is as
follows:
F = PA
where-
F = force, in
pounds
P = pressure, in psi
A = area, in square inches
Example:
Figure 1-6 shows a pressure of 50 psi being
applied to an area of 100 square inches. The total force on the
area is-
F = PA
F = 50 x 100 = 5,000 pounds
1-2. Pascal's Law. Blaise
Pascal formulated the basic law of hydraulics in the mid 17th
century. He discovered that pressure exerted on a fluid acts
equally in all directions. His law states that pressure in a
confined fluid is transmitted undiminished in every direction and
acts with equal force on equal areas and at right angles to a
container's walls.
- Figure 1-7 shows the apparatus
that Pascal used to develop his law. It consisted
of two connected cylinders of different diameters
with a liquid trapped between them. Pascal found
that the weight of a small piston will balance
the weight of a larger piston as long as the
piston's areas are in proportion to the weights.
In the small cylinder, a force of 100 pounds on a
1-square-inch piston creates a pressure of 100
psi. According to Pascal's Law, this pressure is
transmitted undiminished in every direction. In
the larger cylinder, the 100 psi of pressure from
the small cylinder is transmitted to an area of 5
square inches, which results in a force of 500
pounds on the second piston. The force has been
multiplied 5 times-a mechanical advantage of 5 to
1. Using the same factors, you can determine the
distance the pistons move. For example, if the
small piston moves down 10 inches, the larger
piston will move up 2 inches. Use the following
to determine the distance:
where-
F1
= force of the small piston, in pounds
D1 =
distance the small piston moves, in
inches
D2 =
distance the larger piston moves, in
inches
F2 = force
of the larger piston, in pounds
Example: Determine D2
1-3. Flow. Flow is the
movement of a hydraulic fluid caused by a difference in the
pressure at two points. In a hydraulic system, flow is usually
produced by the action of a hydraulic pump-a device used to
continuously push on a hydraulic fluid. The two ways of measuring
flow are velocity and flow rate.
- a. Velocity. Velocity
is the average speed at which a fluid's particles
move past a given point, measured in feet per
second (fps). Velocity is an important
consideration in sizing the hydraulic lines that
carry a fluid between the components.
- b. Flow Rate. Flow rate
is the measure of how much volume of a liquid
passes a point in a given time, measured in
gallons per minute (GPM). Flow rate determines
the speed at which a load moves and, therefore,
is important when considering power.
1-4. Energy, Work, and Power.
Energy is the ability to do work and is expressed in foot-pound
(ft lb). The three forms of energy are potential, kinetic, and
heat. Work measures accomplishments; it requires motion to make a
force do work. Power is the rate of doing work or the rate of
energy transfer.
- a. Potential Energy.
Potential energy is energy due to position. An
object has potential energy in proportion to its
vertical distance above the earth's surface. For
example, water held back by a dam represents
potential energy because until it is released,
the water does not work. In hydraulics, potential
energy is a static factor. When force is applied
to a confined liquid, as shown in Figure 1-4,
potential energy is present because of the
static pressure of the liquid. Potential energy
of a moving liquid can be reduced by the heat
energy released. Potential energy can also be
reduced in a moving liquid when it transforms
into kinetic energy. A moving liquid can,
therefore, perform work as a result of its static
pressure and its momentum.
- b. Kinetic Energy.
Kinetic energy is the energy a body possesses
because of its motion. The greater the speed, the
greater the kinetic energy. When water is
released from a dam, it rushes out at a high
velocity jet, representing energy of
motion-kinetic energy. The amount of kinetic
energy in a moving liquid is directly
proportional to the square of its velocity.
Pressure caused by kinetic energy may be called
velocity pressure.
- c. Heat Energy and Friction.
Heat energy is the energy a body possesses
because of its heat. Kinetic energy and heat
energy are dynamic factors. Pascal's Law dealt
with static pressure and did not include the
friction factor. Friction is the resistance to
relative motion between two bodies. When liquid
flows in a hydraulic circuit, friction produces
heat. This causes some of the kinetic energy to
be lost in the form of heat energy.
- Although friction cannot be
eliminated entirely, it can be controlled to some
extent. The three main causes of excessive
friction in hydraulic systems are-
- In a liquid flowing through
straight piping at a low speed, the particles of
the liquid move in straight lines parallel to the
flow direction. Heat loss from friction is
minimal. This kind of flow is called laminar
flow. Figure 1-8, diagram A, shows laminar flow.
If the speed increases beyond a given point,
turbulent flow develops. Figure 1-8, diagram B,
shows turbulent flow.
- Figure 1-9 shows the difference
in head because of pressure drop due to friction.
Point B shows no flow resistance (free-flow
condition); the pressure at point B is zero. The
pressure at point C is at its maximum because of
the head at point A. As the liquid flows from
point C to point B, friction causes a pressure
drop from maximum pressure to zero pressure. This
is reflected in a succeedingly decreased head at
points D, E, and F.
- d. Relationship Between
Velocity and Pressure. Figure 1-10, page
1-8, explains Bernouilli's Principle, which
states that the static pressure of a moving
liquid varies inversely with its velocity; that
is, as velocity increases, static pressure
decreases. In the figure, the force on piston X
is sufficient to create a pressure of 100 psi on
chamber A. As piston X moves down, the liquid
that is forced out of chamber A must pass through
passage C to reach chamber B. The velocity
increases as it passes through C because the same
quantity of liquid must pass through a narrower
area in the same time. Some of the 100 psi static
pressure in chamber A is converted into velocity
energy in passage C so that a pressure gauge at
this point registers 90 psi. As the liquid passes
through C and reaches chamber B, velocity
decreases to its former rate, as indicated by the
static pressure reading of 100 psi, and some of
the kinetic energy is converted to potential
energy.
- Figure 1-11 shows the combined
effects of friction and velocity changes. As in
Figure 1-9 pressure drops from maximum at C to
zero at B. At D, velocity is increased, so the
pressure head decreases. At E, the head increases
as most of the kinetic energy is given up to
pressure energy because velocity is decreased. At
F, the head drops as velocity increases.
- e. Work. To do work in
a hydraulic system, flow must be present. Work,
therefore, exerts a force over a definite
distance. It is a measure of force multiplied by
distance.
- f. Power. The standard
unit of power is horsepower (hp). One hp is equal
to 550 ft lb of work every second. Use the
following equation to find power:
P = f x d/t
where-
P = power, in hp
f = force, in GPM
d = distance, in psi
t = time (1,714)
HOMEPAGE
