dr esmaeel system 3 hydraulic.pdf

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CHAPTER (1) INTRODUCTION

CHAPTRE ONE

INTRODUCTION

1.1 IntroductionFluid power means using pressurized oil or air

to accomplish work. Most hydraulic systems use

petroleum oils, but often synthetic or water-base

fluids are used where there is a lire hazard.

Pneumatic systems use air to power actuators, but

unlike hydraulic systems that return oil at low

pressure to the reservoir, the air from pneumatic

systems is exhausted to atmosphere after doing work.

The oil in a hydraulic system exhibits the

characteristics of a solid. This provides a rigidmedium to transfer power through the system.

Conversely, air

used in pneumatic systems is spongy, and additional

controls must be provided if actuator speed and

stiffness are to be regulated.

A fluid power system accomplishes two main

purposes. First, it provides substantial force to move

actuators in locations some distance from the power

source, where the two are connected by pipes, tubes,

or hoses. For example, a hydraulic pump mounted to

the engine in one area can be connected to hydraulic

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motors or cylinders located 100 feet or more away.

This is a decided advantage over systems using

gears, shafts, and chains, particularly as the location

of output actuators becomes less accessible. Second,

fluid power systems accomplish highly accurate and

precise movement of the actuator with relative ease.

This is particularly important in such applications as

the machine tool industry where tolerances are often

specified to one ten thousandth of an inch and must

be repeated during several million cycles.

1.2 Historical PerspectiveThe modem era in fluid power began around the

turn of the century. Hydraulic applications weremade in the main armament system of the USS

Virginia as early as 1906, where a variable speed

hydrostatic transmission was installed to drive the

main guns. Since that time, the marine industry has

applied fluid power to cargo handling and winch

systems, controllable-pitch propellers, submarine

control systems, shipboard aircraft elevators, aircraft

and missile launch systems, and radar-sonar drives.

Fig.1-1 illustrates a combination cargo-passenger

ship. A modern application of hydraulics to marine

vessels is the hydraulically powered propeller drive

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shown, which is used as a lateral thruster for close

quarter maneuvers for large ships, or the main drive

for smaller vessels.

1.3 CapabilitiesHydraulics and pneumatics have almost unlimited

applications in the production of goods and services in nearly

all sectors of the economy. Several industries are dependent

upon the capabilities that fluid power affords. Among these

are agriculture, aerospace and aviation, construction,

defense, manufacturing and machine tool, marine, material

handling, mining, transportation, undersea technology, and

public utilities, including communications transmission

systems.

(a)

(b)

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Figure1-1 (a) cargo-passenger ship;

(b)Hydraulically powered propeller (courtesy of

Rexroth).

The world's need for food and fiber production has

caused unprecedented leadership in agriculture equipment

development, and particularly in applying hydraulics to solve

a variety of problems. Fig.1-2 illustrates a modern wheeldrive tractor that features extensive use of fluid power. In a

typical tractor application, a 25 to 30 gal/min eight-cylinder

variable volume radial piston pump supplies fluid to a closed-

center load-sensitive circuit. The pump unloads to a

minimum standby pressure to reduce power consumption

when demand is low. Hydraulics power the rear and front

main drive and power take-off (pto) clutches, wet disc brakes,

remote valves, implement hitch, draft sensing, power shift

transmission, differential lock, and hydrostatic steering. A

pressurized reservoir is used to supplement the How from the

charge pump during maximum demand from large bore

cylinders.

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Figure 1-2 modern wheel tractor (courtesy of John

Deere Company).

Figure 1-3 Self-leveling combine (courtesy of John

Deere Company).

Other applications to agriculture include combines,

forage harvesters, back-hoes, chemical sprayers, and

fertilizer spreaders. Fig.1.3. illustrates a self-leveling

combine harvesting grain on a hillside. Notice the

hydraulic cylinder on the lower side extending the

suspension to keep the operator station level while the

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combine head follows the contour of the ground to

harvest the crop.

Combines and other equipment make extensive use

of hydraulic wheel motors to assist in marginal tractive

conditions. Wheel motors consist of a hydraulic motor

mounted integrally with a wheel and tire assembly.

Braking is usually designed into the system. All that is

required to make the wheel motor functional is mounting

to the suspension and connection to hydraulic lines that

lead back to the main power control valve. Auxiliary

power drive wheel motors have several advantages to the

agricultural industry. For example, they can be used to

maintain and improve the turning ability of regular row

crop tractors, allow for a high design that maintainsunder-axle crop clearances, assist in adjusting front axle

width, and provide on-the-go engagement and

disengagement in forward as well as reverse.

Fig.1-4 illustrates applications of hydraulic power

to the aviation and aerospace industries. Hydraulic pumps

running off the main engines and electric motors power

cylinders to operate the landing gear, flaps, rudder,

control surfaces, payload bay doors, and complex

mechanical arm.

Another sector of our economy that has benefited

from the brute power of hydraulics and pneumatics is the

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construction industry. Crawler tractors, road graders,

bucket loaders, trenchers, backhoes, hydraulic shovels,

and pan scrapers are just a few of the many applications.

Fig.1-5 shows a large off-the-road truck with many

hydraulic and pneumatic components, including

telescoping dump cylinders, steering cylinders, and

brakes being loaded by a hydraulic power shovel. Notice

the-six large hydraulic cylinders that operate the boom

and clamshell bucket.

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Figure 1-4 Aerospace and aviation applications of fluid

power (courtesy of Fluid Power Educational Foundation).

The manufacturing and machine tool industry is

dependent on hydraulic power to provide the force and

close tolerance necessary in controlling production.

Fig.1-6 illustrates large press for manufacturing thick-

walled pipe used to transport

crude oil, gas, and even water. Thick-walled pipe is

usually produced in limited quantities because of the

force limits of roll bending. The cutaway drawing shows

how hydraulic cylinders are used to control manipulator

feed tables that position sheet stock in the press while the

blade inset forms each half of the pipe radius with up to

100 individual bends. This machine has the capacity to

form pipes 40 ft long with an inside diameter to 5 ft and

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wall thickness of 2.5 in. A pipe this size weighs as much

as 25 tons.

Figure 1-5 Off-road truck and loader (courtesy of

Aeroquip).

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Figre1-6 Functional principle of pipe bending

press.

1.4 The Use of UnitsEnglish and SI metric units are used throughout this

book. The basic units used from the English system are

length, force, and time. Derived units include area,

pressure, velocity, volume, and flow rate.

The units of length are the foot and inch (ft, in). The

units of force are the pound and ounce (lbf, oz). The same

units are used for weight. The units for time are the

minute and second (min, sec). Derived units are made up

of basic units. Area is derived in square feet and square

inches (ft2, in.

2), pressure is in pounds per square inch

(lbf/in.2

), velocity is in feet per minute or feet per second(ft/min, ft/sec), and volume is in cubic feet or cubic

inches (ft3, in.

1). One gallon equals 231 in

3.

The units for distance in the SI metric system most

often used are the meter and centimeter (m, cm). The unit

of force is the Newton (N), and the unit of time is the

second (sec). Area is derived in square meters and square

centimeters (m2,cm

2). The unit of pressure is derived in

Pascal in honor of the French physicist. A Pascal equals

one "Newton per square meter (Pa,N/m2). This is

somewhat confusing because the Newton also is Isaac

Newton. Because the Pascal is small, the kilo Pascal

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(kPa), which is 1000 Pascal, and mega Pascal (MPa),

which is one million Pascal, are used often in hydraulics.

Velocity in the SI metric system is derived in meters per second

(m/s). Volume is derived in cubic meters or cubic centimeters

(m3,cm

3).

Flow rate is a derived unit, but it is not the same in

hydraulics as in pneumatics. In hydraulics, flow rate in a

liquid measure. In the English system the units are

gallons per minute (gal/min), and in SI metrics liters per

minute (l/min). In pneumatics, flow is a volumetric

measure: cubic feet per minute or cubic feet per second,

in the English system (ft3/min, ft3/sec), and cubic meters

per minute or cubic meters per second (m3/min,m

3/sec)

in the SI metric system .In most cases the units should be included with the

numbers when calculations are made. This is because

without the units, the answer would be incomplete .And

unless the units are included with each number in the

calculation; there would be no easy way to check

through the problem to see if errors have been made. The

use of units also is important when conversions are made

from one system to another, from liquid to volumetric

measures, or from larger to smaller units. Conversation

factors for derived units are difficult to remember so if

one can work through the conversion with the units

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themselves, it usually is easy to find the conversion

between base unit like inches, centimeters, pounds, and

Newtons.

1.5 Dimensional Calculation

When analyzing hydraulics or other mechanical

system involving dimensions such as length, area, volume,

and time, it is good practice to apply dimensional analysis to

their solutions. This technique assures the proper use of

conversion factors and virtually eliminates the need to

memorize formulas. With all dimensions properly arranged in

the statement of a problem, the solution consists merely of

performing the indicated operations of the numbers and

combining exponents of like bases. This method allowscancellation of all dimensions not needed, while keeping

those called for in the answer.

Although formulas are not needed in the solution of

dimensional problems, they are generally shown in this text

in order to ease your transition from formula to dimensional

calculations. But, by all means, do not continue to use a

memorized formula as a crutch. It is hoped that before you

finish this book you can leave your 'little black book" of for-

mulas at home.

Example (1):

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How far will a (3-in. ) piston move if it receives (30 in . ) of

fluid?

2 3

Solution:

Let's examine the problem. The answer we need is the

distance traveled (in.).

We are given the volume of oil supplied to the cylinder

(in .) and the area of the cylinder (in .). By dividing in. by

in. we get in. /in. = in. = in . = in.

3 2 3

2 3 2 3-2 1

The formula for distance traveled is:

PistonAreaVolumestrokeceDis

1)(tan =

2

2

.3

1.30

inin =

.10.3

30 23 inin ==

Note: The formula for distance could have been statedas:

Area

VolumestrokeceDis =)(tan

While dividing volume by area will in fact give usdistance, it is better generally to keep everything in

the multiplying mode. That is, multiply by (1/area)

instead of dividing by area.

First, combine exponents of like bases in. /in. =

in. = in. =

3 2

3-2 1

in. Since the answer has the proper

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dimension you will know that you have the

dimensions properly placed. Next, perform the

indicated operations of the numbers (30/3 = 10). The

total answer is the combination of the two answers,

or 10 in.

A general shortcut used for combining like bases

is to apply strikethroughs to show divisions or

cancellations of exponents from bottom to top or top

to bottom of the fraction. For example:

.10.3

.30tan2

3 inin

inceDis ==1

This is read "Inches cubed divided by inches

squared equals inches; and 30 divided by 3 equals

ten."To use the distance formula, without the

dimensions, is to apply the "rote method," which

generally leads to trouble and errors. There is no way

that in. and in. can be combined to provide in. other

than to divide in. by in. .

3 2

3 2If the dimensions had been

arranged differently in the solution, say, in. /in. , the

answer would have been in. = in. . Then we

would have realized immediately that the dimensions

have to be rearranged in order to obtain the proper

answer.

2 3

2-3 -1

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With the use of dimensional calculations you are

constantly checking your answer as you cancel

unwanted dimensions. You are learning from the

procedure instead of blindly applying a memorized

formula.

1.6 Review of Dimensions

Fig.1-7 provides a review of some common

dimensions.

Figure1-7 Dimensions

Table (1-1) Common conversion fractions using U.S. units.

U.S. Units Conversion Fractions Descriptions

Displacement128.6

=rad

rev

rev=revolution

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(revolution and

radians)

Rev=6.28 rad

128.6

=rev

rad

sec/1046.0sec60

min28.6

minrad

rev

radrev=

1sec/1046.0

min/

=rad

rev

1min/

sec/1046.0=

rev

rad

rad = radian

sec =seconds

min = minutes

Flow

gal/min = 231

in. /min3

1min//1

min/.2313

=gal

in

1min/.231

min/13

=in

gal

in. /min =

cubic inches

per minute

3

gal/min=

gallons per

minute

Length

(feet and

inches)

12 in.=1 ft

11/.12 =ftin

1.12/1 =inft

in. = inches

ft = feet

Power

(horsepower

and kilowatts)

hp = 0.746 kW

1746.0

=kW

hp

1746.0

=hp

kW

kW = kilowattshp=horsepower

Pressure[lb/in. and in.

of mercury (in.

Hg)] 29.92 in.

Hs. = 14.7

lb/in.

2

2

1.92.29

./7.14 2

=Hgin

inlb

1./7.14

.92.292=

inlb

Hgin

lb = pounds

in.=inches

Hg = mercury

Time 11min60

=h h = hourssec = seconds

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(hours,

minutes, and

seconds)

60 min = 1 h

1min60

1=

h

1min1

sec=

60

1sec60

min1 =

min = minute

Volume

(in. and

gallons)

3

231 in. = 1 gal3

11

.231 3=

gal

in

1.231

13=

in

gal

in.3 = cubic

inches

gal = gallons

1.7 CONVERSION FRACTIONS

When the value of the numerator (top) of a

fraction is equal to the value of its denominator(bottom), the absolute value of the fraction is one,

and because multiplying by l does not change the

absolute value of an expression, dimensions can be

converted at will with the use of the conversion

fractions given in Tables 1-1, 1-2, and 1-3.

respectively, show prefixes for forming SI units arid

some handy formulas.

Table 1-2 Common conversion fractions using SI

units.

SI Units Conversion

Fractions

Descriptions

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Pressure

(Pascals,

kilopascals,&bars*)

Pa =N/m2

1000 Pa=kPa

1 bar=100kPa

Pa/(N/m )=12

(N/m )/Pa=12

1000 Pa/kPa=1

kPa/1000Pa=1

bar/100 kPa=1

100 kPa/bar=1

Pa = Pascal

N = Newton

(force)

m 2 = square

meters

kPa = kilopascal

bar=approximately

1 atmosphere

Area

( m , dm , cm ,

mm )

2 2 2

2

One square meter

equals 100 square

decimeters:

m /100 dm =12 2

100 dm /m =12 2

m /10000 cm =12 2

10000 cm /m =12 2

m /1000000

mm =1

2

2

1000000 mm /m

=1

2 2

dm2 = square

decimeter

cm2 = square

centimeter

mm2= square

millimeter

Volume

(Liter, kiloliter, m3and cm3)

m3 = 1000L

liter=1000 cm3

m /1000L=13

1000L/ m =13

L/1000 cm =13

1000 cm3/L=1

m = cubic meter3

L = liter

1000 L = kiloliter

Cm = cubic

centimeter

3

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* One bar is approximately the average atmospheric pressure

at the earth's surface (or 14.5 lb/in. )2

Table 1-3 Common conversion fractions using both U.S.

and SI units (converting from one system to the other).

Units Conversion Fractions Descriptions

Flow

(gal/min and

L/min)

1 gal/min =

3.7854 L/min

1min/7854.3

min/ =gal

1min/

min/7854.3=

gal

L

gal/min =

gallons per

minute

L/min =

liters per

minute

Force orWeight

(pounds,

newtons, and

kilograms)

1 lb* ==

4.44822 N

144822.4

=N

lb

144822.4

=lb

N

lb = pound

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1 kilogram(weight or

force) =

9.806654453

newtons

(weight or

force)

kgf= 9.8066 N

1 kilogram

(force)# = 2.2

lb

18066.9

=N

kg f

18066.9

=fkg

N

12.2

=lb

kgf

12.2

=fkg

lb

kg =

kilogram

mass or

kilogram

weight

Length

(meters and

inches)

1 m = 39.37 in.

1/.37.39 =min

1.37.39/1 =inm

in. = inches

m = meter

Pressure

(kPa, and bar

versus lb/in.2)

6.895kPa= 1

lb/in.2

1 bar = 14.5

lb/in.2

1./

895.62

=inlb

kPa

1895.6

./ 2=

kPa

inlb

1./5.142=

inlb

bar

1./5.14 2

=bar

inlb

kPa =

kilopascals

lb/in. =

pounds persquare inch

2

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Square

Measures

(m2and in.

2)

1 m2

= 1550

in.2

1.1550

2

2

=m

in

1.1550

12

2

=in

m

in. = square

inch

2

m = square

meter

2

Torque

(lb-in.) =

0.112979 N-m

Volume

(L and gal)

3.7854 L = gal

Volume

(m3 and in.3)

1 m3

= 61,024

in.3

17854.3

=L

gal

17854.3

=gal

L

1.024.613

3=

m

in

1.024.61

13

3

=in

m

L = liter

gal =

gallons

in. = cubic

inches

3

m = cubic

meter

3

Weight(see Force above)

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* some expert use lbf here to be consistent with kgf versus

Newton. This is not necessary, however, because "lb" always

means force or weight (never mass).

# The term kgf is used here to indicate that we are using

kilograms of weight or force and not of mass.

1.8 How Fluid Power Works

Fluid power works in accordance with laws

governing behavior of the fluid itself. The two most

common fluids used are air in pneumatic systems and

oil in hydraulic systems. The beginning of an

understanding about how fluid power works is

attributed to Blas Pascal (1650), who discovered

that pressure exerted by a confined fluid acts

undiminished the same in all directions at right

angles to the inside wall of the container. This isknown as Pascal's Law and is often thought of as the

foundation of the discipline. The law can be extended

to include transmission and multiplication of force,

as shown in Fig.1-8.

Fluid power works by applying a force against a

movable area. In a typical fluid power system, a fluid

power pump or compressor delivers fluid through the

control valve to a system actuator, such as a cylinder,

through lines at high pressure. Unused fluid in

hydraulic systems is returned to a reservoir at low

pressure for cooling, storage, and later use. Air used

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to power pneumatic actuators is not returned to the

receiver. Instead it is exhausted to atmosphere.

Fig.1-9illustrates a simulated hydraulic system used

for lowering and raising the nose wheel of an aircraft

landing gear. Movement of the control valve to lower

the wheel causes fluid to be delivered at high

pressure from the pump to the blank end of the

cylinder. To raise the wheel, the control valve directs

fluid to the rod end of the cylinder. In both cases

fluid power works by applying a force against the

area of one side of the movable piston in the cylinder

or the other.

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Figure 1-8 Transmission and multiplication of fluid

power force.

Figure1-9 How fluid power works.

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Examining the definition and example closer

Force = Pressure Area

All we need to remember is that units of force,

pressure, and area must be consistent. This means that in

the English system of measurement weight or force is in

pounds (lbf), area is in square inches (in. ), and pressure

is in lbf/in. . In SI units, force is in Newtons

2

2(N), area is

in square meters (m ), and pressure is in N/m , which is

given the name Pascals (Pa) alter the famous French

physicist. Because the Pa is a small unit, gauges usually

read pressure in kPa, where:

2 2

1KPa = 1000 N/m2

Following are derivations for the most commonly used

conversions.

1 lb=4.448 N

and

1N=0.2248lbf

1 in=0.254 m

and

1 m= 39.37 in

Converting units of pressure

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1 lbf/in. 4.448 222 N/lbf 1550 in /m =6895

N/m =6.895 kpa

2 2 2

2

or

1 lbf/in. = 6.895 kPa2

2222 /000145.0448222.4/11550/1/1 inlbfNlbfinmmN =

and

1 kPa = 0.145 lbf/in2

Table (1-4) gives constants for conversions among

various units used for pressure around the world. Only thelbf/in. and Pa will be used here. The Pa is the recognized

international unit of pressure, but the kgf/cm and bar are still

used in some countries.

2

2

When applying fluid power to a system, the force

required at the output is used to determine the pressure of the

system and the cross section area of the cylinder. System

pressures are also determined from the strength of

components, cost factors, and safely precautions. When the

system pressure is known and the force that the system must

apply is specified, the area of the cylinder then can be

computed by solving the formula for this value.

essure

ForceArea

Pr=

Table (1-4) units of pressure and conversion Factors

Convert

to

lb /in.f2

kPa kg /cmf2

bar

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Convert

from

lb /in.f2

1 6.895 0.07 0.069kPa 0.1 45 1 0.0101 0.010

kg /cmf2

14.22 98.07 1 0.98

bars 14.5 100 1.02 1

Pascal (Pa) = 1 N/m2

Conversions are made by multiplying the units in the

left margin by the conversion factors in the boxes to arrive at

the units across the top. Example: .kPainlbf 6895895.6./10002 =

Example 2Pressure = 1500 lbf/in

2

Area = (0.7854)(4 in. ) = 12.57 in.2 2

Force = AP = (1500 lbf/in )( 12.57 in ) = 18,855lbf

2 2

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Figure1-10 Force, pressure, and area relationships.

To simplify calculations using this formula, the desired

value can be determined using Fig. 1-10. Bycovering the desired value, the relationship between the other

two is given in the proper order. That is, force equals

pressure times area, pressure equals force divided by area,

and area equals force divided by pressure. Look at

the example that follows the figure, which asks for the

solution of the force that would, be applied to the cylinder

rod if the cylinder has a bore of 4 in. and the pressure in the

cylinder is 1500 lbf/in.2. In hydraulic and pneumatic

applications, bore is the same as the inside diameter of the

cylinder. Many helpful devices such as this have been

developed to assist the fluid power mechanic and technician

to make calculations.

1.9 Head PressureThe first pressure of which we should be aware,

even before the pump is started, is the pressure caused by

the weight of the fluid in the system, or the fluid density

[weight per unit volume (lb/in. )] and height of the fluid

(ft) above the test point. In this book we call this the

3

head pressure.

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Fig.1-11 shows three separate containers of fluids

depicting the effects of fluid height, volume, and

container shape on head pressure.

1.9.1 Effect of Fluid Height on Head Pressure

Fig.1-11a shows a square container with 1 ft of oil

(1 ft by 1 ft by 1 ft). The weight of the oil is

approximately 58 lb. Since the area of the tank bottom is

(12 in.12 in. = 144 in. ), the weight of the oil above

each square inch of the tank bottom is ( 58 lb/144 = 0.4

lb). Therefore, the head pressure is (58 lb/144 in = 0.4

lb/in . If the oil level were reduced to a height of (6 in.),

the weight of the oil would now be (29 lb) and the head

pressure would be (29 lb/144 in = 0.2 lb/in .)

3

2

2

2)

2 2So, the

head pressure is halved as the height of the fluid is

halved.

Filled with oil to a level of (6 in.) into the tube. The

weight of the oil in the tank is (29 lb) and the oil inside the

tube weighs (0.2 lb.)

The head pressure at the bottom of the tube is (0.2

lb/in.2). The pressure difference from the top of the tank to

the bottom (sometimes called psid) is (29 lb/144 in2

= 0.2

lb/in.). The head pressure at the bottom of the tube acts on the

entire area of the fluid at the top of the tank. So, the pressure

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at the bottom of the tank is the sum of the pressure difference

of the tank plus the pressure at the top of the tank or

0.2 lb/in2

+ 0.2 lb/in2

= 0.4 lb/in2

This is the same head pressure that we saw at the

bottom of the tanks of both Figures 1-11a and 1-11b.So, the

head pressure is proportional to the height of the fluid

regardless of the shape of the container.

1.9.2 Significant Head Pressure

Head pressures are always present in hydraulic systems,

but frequently they can be ignored as insignificant when

dealing with higher system pressures. There are three cases,

however, for which head pressures are quite significant: (1)

where the other pressures involved are very low, such as at

the inlet of a pump; (2) where the height of the fluid is great,

such as from the bottom of the ocean; or (3) where the area

affected by the head pressure is very large.

1.9.3 Destructive Head Pressure

Fig.1-12 shows a drawing depicting a home basement

with a standpipe screwed into the floor drain. This procedure

has been tried by home owners in the mistaken belief that it

would be a good method of keeping the basement dry should

the water in the storm sewer try to back up into the basement.

With weeping tile installed under the basement floor for

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normal drainage, the underside of the floor will be exposed to

the backup water, and will be acted on by the head pressure

created by the standpipe.

Assume 3 ft of water in the standpipe (measured from

the bottom of the floor). For this example, let's assume that

the head pressure of water is the same as that of oil:

The head pressure at the bottom of the pipe would be :

In other words, with 0.4 lb/in.2

of pressure per foot of water,

the pressure for 3 ft would be 3 x 0.4 lb/in.2= 1.2 lb/in.

2

22

./2.1/4.0

3 inlbft

inlbft =

Note: Observe that the unit of measure ' ft" has a

strikethrough at each location. This means that the "ft" on the

top of the fraction cancels the "ft" on the bottom.

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Figure1-12 Basement standpipe.

One might think that 1.2 lb/in.2

is insufficient pressure

with which to be concerned. But consider the force pushing

upward from the bottom of the floor. The area of the base-

ment floor is:

2400.8612

3012

20 inft

inft

ft

inft =

The force pushing the floor upward would be

lbinin

lb680.103400.86

2.1 22

=

or

tonsooolb

tonlb 85.51

2680.103 =

This force could break the basement floor.

Note:

ft

in12and

lb

ton

2000are conversion fractions. The

numerator (on top) is equal to the denominator (on the

bottom), which gives the fraction an absolute value of one.

Multiplying by these fractions changes the dimensions but

not the absolute value of an equation.

Of course, the basement would have been flooded had

the standpipe not been screwed into the drain. Flooding the

basement would have saved the floor, however, because the

weight of the floodwater in the basement would have pushed

down on the floor with the same magnitude at which the head

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pressure force was pushing upward, thus preventing damage

to the floor.

1.9.4 Use of Head Pressure for Level Control

NASA's Crawler transporter was designed to move a

missile and its launch tower from the assembly building to

the launch site on Merrit Island. This giant structure is

maintained perpendicular during travel by a leveling device,

which senses the difference of fluid head pressure at the fourcorners of the Crawler compared to the tank "bubble" located

at the center of the Crawler.

1.10 PASCAL'S LAW

Fig.1-13a shows a jug of wine filled to a height of 1 ft.

As usual, the head pressure graduates down the height of the

fluid to 0.2 lb/in. at 6 in. and 0.4 lb/in. at the bottom of the

jug. The area of the jug bottom is 50 in. , and the force of the

bottom caused by the head p

2 2

2

ressure is .

Note that "in

lbininlb 20.50./4.0 22 =

2" has been canceled, in both places, by

strikethroughs. The jug will have been designed to carry this

weight with, perhaps, a 100% safety margin.

Fig.1-13b shows the same jug with a 1 in.2

cork inserted

so that it touches the wine (no air pocket). The calculations

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show what happens if the cork is pushed downward with a

force of 10 lb.

Pascal's law states (in effect): Pressure applied to a

confined fluid is transmitted in all directions, and

acts with equal force on equal areas, and at right

angles to them. This law, as illustrated at the beginning of

this chapter, allows the applied force to be conducted around

corners and through irregular passages to the desired

destination.

The 10 lb of force per the 1 in.2

cork is transmitted to

each and every 1 in.2

of the jug. So every square inch will see

an additional 10 lb of force added to the head pressure force

already there. Now the bottom of the jug will see an

additional force of

lbinin

500.50 22

=lb10

or a total of 520 lb, counting the head pressure. This force,

which exceeds the safety margins of most jugs, could

break the jug.

Note: The head pressure is generally left out of thistype of calculation because the 20 lb of head force issmall compared to the applied force transmitted fromthe cork to the bottom of the jug.

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Figure1-13 (a) Vine jug head pressure; (b) Vine

jug head pressure with Jed force.

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Study Questions and Problems

1-List ten hydraulic applications and ten pneumaticapplications.

2-List the components on one hydraulic application andgive their specifications.

3-Compute the area and volume for a cylinder with a 2-in.bore and a 12-in. stroke.

4- Compute the bore of a cylinder which must exert10,000lbf with a pressure limitation of 1200 lbf/in .

2

5-Compute the force available from a hydraulic cylinderwith a bore of 75 mm under a pressure of 10 MPa.

6- What bore would be necessary on a cylinder operating

at 15 MPa to exert a force of 65 kN?

7- How much fluid is needed to stroke a cylinder with a

100 mm bore and 0.50 in stroke?

8- If a hydraulic cylinder has a rod diameter half the size

of the bore, what will be the difference in force available if

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the pressure applied alternately to each end remains

constant? (Clue: Try two convenient values such as 4 in.

for the piston and 2 in. for the rod diameter.)

9- A house weighing 30 tons is to be lifted by 4

hydrnulic_ram jacks. If the maximum lift pressure

available from the pump is 5000) lbf/in , what is the

theoretical diameter of the ram?

2

10- Two independent hydraulic jacks, A and B, lift a

10.000 lbf steel beam from opposite ends. If Jack A has a

bore of 4 in. and Jack B has a bore of 3 in), what would be

the theoretical pressure in each jack when the beam is in

the raised position?

11- How is a cylinder actuator engineered to provide the

same force extending and retracting?

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