HYDRAUUC DEVICESA Module on Hydraulics and Equilibrium

SUNY at Binghamton

Malcolm Goldberg, Westchester Community CollegeJohn P. Ouderkirk, State University of New York Agricultural and Technical College at Canton

Bruce B. Marsh, State University of New York at Albany

Carl R. Stannard, State University of New York at BinghamtonBruce B. Marsh, State University of New York at Albany

Project Co- Directors

NEW YORKST. LOUIS

DALLASSAN FRANCISCO

AUCKLANDDUSSELDORF

JOHANNESBURGKUALA LUMPUR

LONDONMEXICO

MONTREALNEW DELHI

PANAMAPARIS

sAO PAULOSINGAPORE

SYDNEYTOKYO

TORONTO

The Physics of Technology modules were produced by the Tech Physics Project, which wasfunded by grants from the National Science Foundation. The work was coordinated by theAmerican Institute of Physics. In the early planning stages, the Tech Physics Project receiveda small grant for exploratory work from the Exxon Educational Foundation.

The modules were coordinated, edited, and printed copy produced by the staff at IndianaState University at Terre Haute. The staff involved in the project included:

Philip DiLavore .Julius Sigler . . .Mary Lu McFall.B. W. BarricklowStacy Garrett . .Elsie Green. . . .Donald Emmons

· Editor· . . . . . . Rewrite Editor· Copy and Layout Editor

.. Illustrator· Compositor· Compositor·Technical Proofreader

In the early days of the Tech Physics Project A. A. Strassenburg, then Director of the AlPOffice of Education, coordinated the module quality-<:ontrol and advisory functions of theNational Steering Committee. In 1972 Philip DiLavore became Project Coordinator andalso assumed the responsibilities of editing and producing the final page copy for the mod-ules.

The National Steering Committee appointed by the American Institute of Physics hasplayed an important role in the development and review of these modules. Member of thiscommittee are:

J. David Gavenda, Chairman, University of Texas, AustinD. Murray Alexander, DeAnza CollegeLewis Fibel, Virginia Polytechnic Institute & State UniversityKenneth Ford, New Mexico Institute of Mining and TechnologyJames Heinselman, Los Angeles City CollegeAlan Holden, Bell Telephone LabsGeorge Kesler, Engineering ConsultantTheodore Pohrte, Dallas County Community College DistrictCharles Shoup, Cabot CorporationLouis Wertman, New York City Community College

This module was written and tested at the State University of New York at Binghamton.In addition to the Project Co-Directors, the professional staff includes Arnold Benton, theAmerican Institute of Physics; Malcolm Goldberg, Westchester Community College; Gio-vanni Impeduglia, Staten Island Community College; Gabriel Kousourou, QueensboroughCommunity College; Ludwig P. Lange, Broome Community College; John F. Ouderkirk,SUNY ATC, Canton; and Arnold A. Strassenburg, SUNY, Stony Brook.

The authors wish to express their appreciation for the help of many people in bringing thismodule to final form. The criticisms of various reviewers and the cooperation of field-testteachers have been most helpful. Several members of the staff of the State University ofNew York at Binghamton also deserve special recognition for their contributions. Theyare: Felix K. Chen, Duane A. Damiano, Eugene C. Deci, Barry R. DeGraff, Max Eibert,Steve I. Finette, Chung-Yu Fu, Jeffrey W. Gray, Leonard R. Kelly, Stevan H. Leiden,Debra M. Simes, Laurie A. Sperling, William Standish, and Howard M. Zendle.

Copyright © 1975 by the Research Foundation of the State University of New York. Allrights reserved. Printed in the United States of America. No part of this publication may bereproduced, stored in a retrieval system, or transmitted, in any form or by any means,electronic, mechanical, photocopying, recording, or otherwise, without the prior writtenpermission of the publisher.

Except for the rights to material reserved by others, the publisher and copyright ownerhereby grant permission to domestic persons of the United States and Canada for use of thiswork without charge in the English language in the United States and Canada after JanuaryI, 1982. For conditions of use and permission to use materials contained herein for foreignpublication or publications in other than the English language, apply to the AmericanInstitute of Physics, 335 East 45th Street, New York, N.Y. 10017.

Section AHydraulic DevicesExperiment A-I. DevicesProperties of Liquids ..Experiment A-2. HydrometrySummary . . . . .Problems

2258

· 12· 14· 14

Section B . . . . . . . .Pressure . . . . . .Experiment B-1. Pressure Dependence on DepthPressure versus DepthTall Buildings . . . . . . . . . . . . .Archimedes' Principle .Derivation of Archimedes' PrincipleExperiment B-2. Measuring DensityQuestions for Experiment B-2Atmospheric Pressure . . . . . . . .Gauge Pressure . . . . . . . . . . . .Experiment B-3. A Closed Hydraulic System .Closed Systems-Pascal's LawFriction .SummaryProblems

· 15· 15· 16· 18· 19· 19.20.22.24.25.26.26.27.29.32.32

Section C-1 . .Fluids in Motion: HydrodynamicsExperiment C-1. The Bernoulli Effect.Bernoulli's Principle . . . . . . . . . . .Derivation of the Bernoulli Effect (Optional)

.34

.34

.35

.37

.38

Section C-2 .Other Hydraulic Devices .Pressure Measuring Devices (Gauges)Pressure Measurements .Experiment C-2. Blood Pressure MeasurementsHydraulically Actuated Controls .SummaryProblems .

.41

.41

.42

.44

.48· 51· 52.52

WorksheetsExperiment A-I 54

Experiment A-2Experiment B-1Experiment B-2Experiment B-3

· .55· .57· . 58

..... 60

Hydraulic

Hydraulic devices are used in many waysin our technical society. Many of thesedevices were developed in the nineteenthcentury, before electrical power was available.In those days the engineering work ofteninvolved applications of water and steampower. Hydraulic devices were an outgrowthof the technological needs of the times.

Today we still find many hydraulic(fluid-power) devices in our homes, hospitals,and industrial plants. Fluid-powered devicesare relatively cheap, easy to maintain, andhighly reliable; they are still very much a partof our technology. The basic physical prin-ciples of hydraulic devices are now well-known, so most hydraulic engineering todayinvolves only the improvement of materials,methods of manufacture, and the develop-ment of new applications of fluid-operateddevices. A few years ago an entirely new useof fluid technology resulted from the develop-ment of fluidic amplifiers, which sense andswitch like their electronic counterparts. Theyare extremely reliable and rugged, so they areused in applications which would be too harshfor other kinds of amplifiers.

In this module you will study the basicphysics of fluids in order to understand some

Devicessimple hydraulic devices. The goals of themodule are outlined here. Read them beforestarting on the module, and refer to them asyou work through the module.

The main goal of this module is to helpyou learn the basic principles and concepts ofhydraulics. When you have completed themodule, you should have a knowledge andunderstanding of the following:

3. The relationship between force andpressure.

4. The way in which pressure increases withdepth in a liquid.

5. The way in which pressure is transmittedthrough fluids.

7. The mechanical advantage and efficiencyof hydraulic jacks.

8. The buoyant forces exerted on objectsby liquids.

9. The way in which pressure depends onthe rate of flow of a fluid.

A treatment of the properties of liquidsis an important component of this module.You will study several devices in which thestatic (stationary) and dynamic (moving)properties of liquids are used to performuseful functions. The devices to be studiedinclude automobile brakes, the hydraulic jack,the toilet tank, the siphon, the aspirator, thehydrometer, pressure gauges, and thesphygmomanometer.

In an automobile brake system, thebrake fluid transmits pedal pressure to a brakeshoe (or a pad for a disc brake). The brakeshoe ru bs against the brake drum and pro-duces a frictional force to slow the auto. A2-ton car, or a 10-ton truck, can be stoppedby applying a force of a few pounds to thebrake pedal. Why is such a small applied forceable to stop a speeding car? What happens ifthere is air in the brake line? You will learn theanswers to these and other questions as youproceed through the module.

The toilet tank uses a system of levers torelease the water and a float-lever system toshut off the water supply after the tank hasrefilled. Did you ever watch the operation of

MASTERBRAKECYLINDER

the tank? What happens to shut off the waterfilling the tank?

The aspirator is a pump whose operationdepends on the properties of a flowing liquid.You will see this device in operation in thelaboratory .

The hydraulic jack has the effect of"magnifying" a force, so that a heavy object,such as a car, can be lifted with ease.

A siphon is a simple device (usually justa piece of tubing) which enables a liquid toflow from one container, over a high point,then down into a lower container, withoutthe aid of a pump. It can be used to drawliquid from the lower part of a containerwhen the surface is contaminated with scum,oil, or other unwanted matter. You might alsouse it to "borrow" gasoline from the tank of acar.

Figures I through 6 will serve as part ofthe basis for our discussions of fluid mechan-ics. You may need to look back at theseillustrations as you work through the module.

So far we have used many technicalterms without explaining their meaning. Someyou already know, others are new to you. Wewill explain most of the terms to you as wecome to them in the module.

WHEELCYLINDER

BRAKEPEDAL

DISCBRAKE

DRUMBRAKE

Figure 2. The carburetor is a device that makes useof the physical laws which apply to stationary fluids(hydrostatics) and moving fluids (hydrodynamics).

BALL COCKASSEMBLY

WATER SUPPLY-.J LOUTLET

Your instructor will provide you with aportable hydraulic jack that has a base plateand a load-lifting plate added to it. Thisdevice is a I.S-ton-capacity jack of the typeused to lift a car. Figure 7 shows this manual-ly operated miniature garage jack.

Operate the jack; work it up and down.Use it to lift one student standing on the liftplatform. Measure the force needed to movethe handle to lift the student. Use a springbalance and read the dial as the handle movesdownward. Fill in the worksheet for thisexperiment, which is found at the back of themodule.

1. How many times greater is the liftedweight than the force applied to thehandle? How shall we describe thisdevice? Inexpensive? Reliable? Liftsheavy loads with a small applied force?Multiplies force?

2. Carefully look over the jack, then-describe briefly what you see.

Did you notice the two cylinders? Theshaft from the smaller cylinder isattached to the bottom of the handle.The shaft from the larger cylinder trans-mits the lifting force to the object beinglifted. Look for these now if you didn'tfind them before.

Which cylinder is longer? The jackrequires many full strokes of the handleto extend the shaft of the longer cylin-der its full travel. Why?

3. What do you notice about the handle? Isit a lever? What is the approximatedistance from the pivot to the otherend? What is the approximate distancefrom the pivot to the shaft of the smallercylinder? How many times bigger is thefirst length than the second length? Doyou know the significance of this num-ber? If not, you will soon.

What's inside the jack? Can you guess?What reason do you have for youranswer?

This jack weighs less than 6 Ib, costsabout $12, and can lift 3000lb with about100-lb force applied to the handle. How doesit work? The answer is described in thismodule. We will explain the behavior of fluidsin terms of physical laws. When you under-stand those laws, the. operation of thishydraulic jack and many other devices will beclear.

The hydraulic jack illustrates that asimple two-piston hydraulic system can beused for force multiplication.

Using the two-piston demonstrationsystem (Figure 8) with valves and a pressuregauge (0-100 Ib/in2

):

I. Demonstrate to yourself that the fluid isalmost incompressible.

2. Observe that simple pressure-activateddevices can be operated to control sig-

nals similar to the oil pressure light orbrake lights in a car.

3. Push down on the smaller-diameterpiston until you develop 801b/in2 pres-sure in the system (you will have to closethe valve between the gauge and thelarger piston to build that pressure). Now,after adjusting the valves properly, pushdown on the larger-diameter piston todevelop an 80 Ib/in 2 pressure in thesystem. Which piston requires a largerforce to produce this pressure? Can youexplain why?

4. (Optional) Have a test of strength. Letthe instructor press down on the largepiston. Push down on the small piston-see who can push his piston all the waydown.

5. (Optional) Place a 5-lb weight on thesmall piston. What pressure is developedin the system? Now do it with a I G-Ib

and a 15-lb weight. Try the same weightson the larger piston. What pressures areproduced? Can you explain the differ-ences between the two sets of values?

6. In a hydraulic system, the amount offluid stays the same if there is no leak. Ifthe fluid is incompressible, then thevolume it occupies also is constant whenthe pressure changes. This may seemquite obvious, but if the system werepneumatic (air operated), the volumewould change as the pressure changed,since air is easily compressed. Using anempty syringe, demonstrate that the airinside is highly compressible by sealingthe exit port with your finger whileoperating the plunger. You should be

able to produce a substantial decrease involume. What fraction of the originalvolume is the compressed volume? Whenyou release the plunger, it should returnalmost to the starting point (keeping theexit port sealed). If it doesn't quitereturn, try to explain why and checkyour answer with your instructor.

Examine the other hydraulic devicesdisplayed in the lab. As you think about whateach one is for and how it operates, someunanswered questions may come to mind. Foreach device write down one or two questionsyou would like to have answered. Be sure tooperate the aspirator ~nd the siphon.

The behavior of liquids has top billing inthis module. It is important that we establishseveral of the basic properties of liquids.Some of these liquid properties will be veryfamiliar, others will be new and interesting.

1. A liquid conforms to the shape of itscontainer. This fact is one of the maindifferences between liquids and solids.

2. A liquid seeks its own level. When aliquid is poured into an open vessel, thesurface of the liquid will reach the samelevel in all parts of the vessel.

--- ------------------0--- ---------------------- --------------------------------------------------------------------------------------------------------------------------------------------------~--------------------------------------------------------------------------------------------------~-------------------------------------------------------

Figure 9. A liquid reaches the same level in all partsof an open vessel.

3. Liquids are incompressible. No matterhow hard you squeeze, you cannotnoticeably change the volume of asample of liquid. (There is a tiny bit ofchange at high pressures, but it can beneglected.) This is the most importantdifference between a liquid and a: gas. Agas is easily compressed into a smallervolume. Because of its incompressibility,the ratio of the mass to the volume is adistinctive property of a given liquid.This useful and easily measured propertyis called the mass density. The Greek

letter p (rho) is the commonly usedsymbol for mass density:

M'p=-V

where M = mass of the sample in kilo-grams (kg) and V = the volume of thesample in cubic meters (m3). Mass den-sity is measured in kg/m 3. Since this is alarge unit, mass density is more oftendescribed in g/cm3. 1 g/cm3 = 103

kg/m3.

In the English system of units it is morecommon to use weight density. which isthe ratio of weight to volume.

WD=-V

where D is the weight density, W is theweight, and V is the volume. Typicalunits used for weight density are Ib/ft3or ton/yd3. Table I lists density values forseveral common substances. Severalsolids are included, even though thismodule emphasizes liquids.

A word of caution: the density of amaterial depends on its temperature.Because the volume of a substanceusually changes when its temperaturechanges, while the mass and weightremain unchanged, the density alsochanges. To be precise, one shouldalways give temperature data along withthe density.

Weight density and mass density arerelated because the weight of an object isrelated to its mass:

where g is the acceleration due to gravity(9.8 m/s2 or 32 ft/s2

). Therefore, theweight density is simply related to themass density by

MD =-g =pg

V

Mass Density Weight Density

LIQUIDS (g/cm3) (kg/m3) (lb/ft3 )

Water 1.00 1,000 62.4

Alcohol 0.79 790 49.4

Maple Syrup . 1.4* 1,400* 87.5*

Ethylene Glycol (Antifreeze) 1.12 1,120 70.0

Glycerine 1.26 -1,260 78.8

Turpentine 0.87 870 54.4

Mercury 13.5 13,500 843.8

Olive Oil 0.91 910 56.9

SOLIDS

Iron 7.9 7,900 493.8

Copper 8.9 8,900 556.3

Aluminum 2.7 2,700 168.8

Gold 19.4 19,400 1206.3

Glass 2.6* 2,600* 162.5*

Bone 1.8* 1,800* 112.5*

Diamond 3.2* 3,200* 200.0*

Wood 0.7* 700* 43.8*

Example 1. A sample of metal has a volumeof 0.10 m3 and a mass of 790 kg. Determineits density, and check Table I to see whatmaterial it might be.

Solution. Because mass is specified, we canmost readily determine mass density.

Mp=-

V

790 kg= 30.1 m

From the table we see that this is the densityof iron.

Example 2. How many tons of water arethere in a swimming pool 15ft wide, 30 ftlong, and 4 ft deep?

Solution. The weight can be determined fromthe volume of the pool and the weight densityof water:

WD=-V

The volume of the pool can be determinedfrom the dimensions:

From Table I, the weight density of water is62.41b/fe.

W = (62.41b/ft3) (1800 ft3)

Sometimes a material is described by itsspecific gravity. The specific gravity is anumber which compares the density of thesubstance to the density of water.

S Of· . S density of substancepeCI ICgraVity = ° Go =--------density of water

Specific gravity is a dimensionless number.Since the density of water is 1.0 g/cm3, wecan easily calculate the density of any sub-stance if we know its specific gravity.

Example 3. By law, maple syrup must meetminimum density standards. The specificgravity of a sample of maple syrup is 1.412.What is its density?

density of syrupS.G. = ------

density of water

density of syrup1.412 = / 31.0gcm

density of syrup = 1.412 g/cm3

One practical use of specific gravity is inthe measurement of the "proof' of alcoholicsolutions, as Table II shows. Notice that thetemperature is specified.

There are many other practical uses formeasurements of specific gravity. For exam-ple, since the specific gravity of the acid in acar battery depends on the condition ofcharge, the battery can be tested simply bymeasuring that specific gravity. Likewise, thepercentage of antifreeze in the cooling systemcan be determined through a specific-gravitymeasurement. The specific gravity of a liquidcan be measured easily by using a simpleinstrument called a hydrometer.

Alcohol Strength and Specific Gravity,As Measured in the United States

Percent by volume Proof Specific gravity

0.0 0.0 1.0000

5.0 10.0 0.9928

10.0 20.0 0.9866

15.0 30.0 0.9810

20.0 40.0 0.9759

30.0 60.0 0.9653

40.0 80.0 0.9517

50.0 100.0 0.9342

60.0 120.0 0.9133

70.0 140.0 0.8899

A. A hydrometer is a simple device formeasuring the specific gravity of a liquid.Hydrometers are used in the chemical indus-try for process-control and testing. They areused in the maple syrup, wine, whiskey, andmilk industries for quality control tests.

In its simplest form, the hydrometer is arod of wood, or a hollow glass tube, weightedslightly at one end so that it floats in anupright position. A scale is usually marked onit. Each hydrometer, of course, has a certainfixed mass. When floated in a beaker of water,it will sink to a certain depth. The water lineis marked 1.0 on the hydrometer, since this isthe specific gravity of water. If the hydrom-eter is placed into any other liquid and itcomes to rest with the liquid surface at the1.0 line, then that liquid also has a specificgravity of 1.0 and a density of 1 g/cm3

.

Question 1. Suppose you have a beaker ofliquid which is more dense than water. Thatis, it has a higher specific gravity than water.Each cubic centimeter has mass greater thanone gram. Will the hydrometer sink deeperinto the new liquid than into water?

Question 2. Which hydrometer of Figure 11has the right scale? Why?

1. Obtain a beaker, a hydrometer, andseveral different liquids from your in-structor, and measure the specific gravityof each liquid. Compare your answers tothe known specific gravity of the liquidssupplied. You may have to look up theknown specific gravity in a book oftables, such as the Chemical Rubber

Company's Handbook of Chemistry andPhysics.

2. Your instructor may also ask you toidentify unknown liquids by their speci-fic gravities.

3. Heat some water in a beaker. With ahydrometer in the beaker, slowly addand dissolve sugar into solution. Whatrange of specific gravity do you mea-sure? Compare your results to those in ahandbook of physics or some otherreference source for density values (thehandbook may list sucrose, which is thechemist's name for sugar).

Question 3. The average human body has aspecific gravity of 0.9S. Should it float infresh water? How would it float differently insea water? What should happen when afloating person takes a deep breath or exhalesdeeply?

B. The freezing point of pure water is 32°F.Adding impurities lowers the freezing-pointtemperature. Usually ethylene glycol is addedas antifreeze to accomplish this. A tableshowing mixture ratios for water and anti-freeze to provide protection is often suppliedon the antifreeze container. If one wishes tocheck the antifreeze content already in aradiator, he can use a hydrometer and look upthe freezing point that corresponds to thedetermined specific gravity. The cheapestdo-it-yourself hydrometers do not have acontinuous specific gravity scale. Instead,they utilize four balls which have differentdensities. Each one just floats when thespecific gravity of the solution matches thatof the ball. The number of balls which floatindicates the specific gravity, thus the freezingpoint of the radiator fluid. Typically, the ballsare designed to float at concentrations ofantifreeze which produce freezing points of+30, + 15, zero, and -ISoF.

1. Using an automotive hydrometer, a grad-uated cylinder, some ethylene glycol, andwater, determine the density of each ofthe balls in the inexpensive hydrometer.When a ball just floats, its density isequal to the density of the antifreeze +water mixture. The density of the mix-ture then corresponds to the freezingtemperature indicated on the hydrom-eter scale. To start put SO cm3 of waterinto a 12S-cm3 graduated cylinder. Addethylene glycol in S-cm3 increments. Inthe table at the back of the module,record the volume of the solution. Testafter each increment of antifreeze to seehow many balls float. Note the mixturewhen each ball floats.

The density of the liquid at any incre-ment of antifreeze is given by:

total massPmix = total volume

Vw, the volume of water, was set atSO cm3

VA, the volume of antifreeze, is in-creased in S-cm3 increments

PA density of antifreeze1.12Sg/cm3

2. Calculate solution density (and ball den-sity) when each ball just barely floats.

3. Compare the mixture found experi-mentally for each degree of protectionwith the antifreeze-to-water ratio sug-gested on a container of ethylene glycolpermanent antifreeze for the same pro-tection.

A liquid conforms to the shape of its con-tainer.

Mass density P IS the ratio of mass M tovolume V for a substance:

MP=-

V

Weight density D is the ratio of weight W tovolume V for a substance:

WD=-V

density of a substanceS.G. =--------density of water

If you understand the module so far,then you should be able to do the followingproblems. Be sure to show the correct unitswith each numerical answer.

1. A block of wood weighs 160 Ib and hasdimensions of 2 ft X 2 ft X 1 ft. Calcu-late the weight density of the block inIb/fe.

3. A vat of maple syrup contains 1600gallons of syrup. Calculate the weight ofthe syrup in the vat. (7.5 gal = 1 fe)

4. The specific gravity of a metal alloy is2.62. Calculate the mass density and theweight density of the alloy.

5. The specific gravity of an oil sample is0.85. What is the volume of 40,000 Ib ofthis oil?

6. A jet plane loads 14,000 Ib of fuel,specific gravity 0.9. How many gallonsof fuel does the plane load? (7.5 gal =1 fe)

7. A government agent confiscates 6 kg of asmuggled narcotic. If the substance justfills a 5-liter container, what is its den-sity in g/cm3? (lliter = 103 cm3

)

Pressure is a ratio of force to area. Inliquids it is usually easier and more meaning-ful to measure pressures than forces. If a forceF acts over an area A and is perpendicular(normal) to the surface, the pressure is:

Fp=-A

Example 4. A cylinder, whose base has anarea of 5 in 2, contains 10 lb of water. What isthe pressure on the bottom due to the water?

Solution. We can apply the definition ofpressure, noting that the force is simply theweight of the water and that it is normal tothe bottom of the cylinder.

Fp=-A

Example S. An automobile radiator cap isdesigned to release steam if the pressurereaches 5 lb/in2. At that pressure, how muchforce acts on the cap if it is sealing an opening1.5 in in diameter?

Fp=-A

Since A = 1rd2/4, where d is the diameter ofthe circle,

22 1rd

F = (5lb/in )4

Example 6. A stack of books whose totalmass is 10 kg rests on a table top. The bottombook has an area of 300 cm2. Determine thepressure in N/m2 exerted on the table.

Solution. The force is the weight of theobject:

W=F=Mg

F = (l0 kg) (9.8 m/s2)

= 98 kgom/s2 = 98 N

p=~= 98 N _ 2A 300 cm2 - .33 N/cm

However, we want to calculate the pressure inN/m2, so we must convert from square centi-meters to square meters.

.33 N 104 cm2

P=--2- X ---cm m2

Although one commonly encounters pressurespecified in English units Ob/in 2), the SI unitis the metric unit (N/m2), which is called aPascal (Pa).

Nl-2=lPa

m

Pressure in a liquid is related to thedepth below the surface, as any swimmer willreadily tell you. This fact is the design basis ofmany water distribution systems. Towns andvillages pump their water up into high watertowers to assure adequate pressure every-where in their area. In cities, tall buildingshave their own water towers raised on plat-forms above the roof. Why is that done?Often, automatic fire sprinkler systems have aseparate water tank on the roof to assureadequate pressure on all floors. Nuclear powerreactors have high tanks of water to assureemergency reactor core cooling should anaccident occur. You may have observed thatin your own house the pressure upstairs is lessthan the pressure downstairs.

In this experiment you will use a simu-lated water stand-pipe to determine the pres-sure as a function of depth for the water inthe pipe.

In many engineering investigations it istoo difficult or too expensive to build andtest a real system. The engineer may build amodel or a simulator. In this context, a modelmeans a smaller, scaled replica of the realsystem. By a simulator one means a smallsystem that will behave similarly to the realsystem even if it is not an exactly scaled-downcopy. This experiment will be done using asimple simulation of a building water system,as shown in Figure I 2, to show you theeffects of pressure on depth. You mightconsider each port to correspond to a floor. Areal system would be about twelve times ashigh, or approximately 60 ft tall. Real pres-sures would be about twelve times what youwill measure.

Here you will measure the pressure as afunction of depth by employing the springpush balance as shown in Figures 13 and 14.Be sure the stand-pipe is full to the top beforebeginning.

rPUSHER PLATErBALANCE

H +t+H++++t++ft+ ~

LOIAPHRAGM

,,,~ ,

!

1. Carefully measure the force necessary toflatten the flexible diaphragm at eachsurface and record your answers in theworksheet at the end of the module. Ithelps to have one person align the pusherplate while another person reads thecompression balance.

Note that the balance is calibrated inmass units (grams). To convert to forceunits (dynes) one should multiply by g,the acceleration due to gravity. Thus

F Mgp=-=-

A A

2. The area of the pusher plate is deter-mined by measuring the diameter andusing the formula:

rrd2A=-

4

Calculate the pressure P for each port.Measure and record the distance h to thecenter of each port from the watersurface.

3. Plot on graph paper P versus h. Use a fullpage. Draw the straight line which bestfits your data points. (If you don't knowhow, ask your instructor for help.) Ifany point seems very far from the line,recheck your data. The pusher plate mayhave been hitting the metal wall of theport.

4. The slope of a line on a graph is the rise(the vertical distance between twopoints) divided by the run (the hori-zontal distance between the same twopoints). By taking measurements fromyour graph between any two points onthe line (pressures PI and P2 and depthshI and h2 respectively) find the slope ofthe graph. The worksheet at the back ofthe module shows you how to do thisstep by step. You will use this valuelater.

We will now develop an expression forthe way pressure varies with depth and see ifit corresponds to the results of your experi-ment. Figure 15 represents a column of fluidof depth h and cross-sectional area A.

The column of fluid rests on an area A. so thepressure at the base of the column due to thefluid is

F W DV DAhp=-=-=-=-A A A A

In words, the pressure beneath the sur-face of a liquid is equal to the product of

weight density (D) and depth (h). D is theslope of the P versus h graph. How does theslope you measured compare with theaccepted value of the weight density of water(D = 1 g/cm3)? Since D = pg, we can write

When using either Equation (6) or Equa-tion (7), one must be careful to be consistentwith the units. The following examples illus-trate this point.

Example 7. Maple syrup has a density of1.4 g/cm3

• Find the pressure in a large tank ofsyrup at a depth of 1 m below the surface.

Solution. Because mass density is specified,Equation (7) is appropriate. A consistent setof units is obtained if the depth is expressedin centimeters.

P = pgh

= (1.5 g/cm3) (980 cm/s2

) (l00 cm)

25 g.cm/s

= 1.47 X 10 2cm

= 1.47 X 105 dY~cm

Example 8. Determine the pressure on a diverwhen he is 30 ft below the surface of a freshwater lake.

Solution. The weight density of water is 62.4Ib/fe.

= (62.41b/fe) (30 ft)

= 1870 Ib/ft2

1. An interesting feature of Equations (6)and (7) is that there is nothing in eitherequation to specify the shape (geometry)

of the liquid column. In fact, if a seriesof differently shaped containers are allattached to pressure gauges and filled tothe same level with the same liquid, thepressures are seen to be the same. Thatis, the pressure at any given depth is thesame no matter what the shape of the

.container.

2. In addition, the pressure does notdepend upon the orientation of a solidsurface on which it might act. No matterwhich way the surface is tilted, thepressure at a given depth is the same.

3. Finally, no forces act in a directionparallel to a solid surface which is underthe surface of a stationary liquid. Thatis, the forces produced by pressure arealways perpendicular (normal) to thesurface.

In Experiment B-1 you probably noticedthat the pressure at the lowest port was aboutfour times the pressure at the highest port.Does this agree with your own experience intall buildings? If real buildings worked exactlythis way, some very high pressures would befound on the lower floors. For example, theEmpire State Building is 102 stories (about1100 ft) high. If the water pressure at theupper observation deck was 151b/in2

, then inthe lobby the pressure would be about1530 Ib/in2! Imagine how thick the pipe wallsmight have to be to contain that. If the waterpressure at the top floor is adequate, engi-neers install a device, called a pressure regu-lator, at each floor. It keeps the pressure fromgetting too high at that floor. A two- orthree-story house doesn't usually have pres-sure regulators, so you can often observepressure differences there.

Question 4. The water pressure in the topfloor of a factory is 60 Ib/in2

• How highabove the floor must the top of the watertower be to produce this pressure?

We have spent some time discussing theproperties of liquids-now we shall concernourselves with the behavior of objects that areplaced in liquids, i.e., things that sink or float.

Why does a 40,00Q-ton steel ship float?Exactly how big must the diving tanks on aresearch submarine be so that it can explorethe bottom of the ocean? Today many riverbarges are made from poured concrete.Doesn't concrete sink when you throw it inwater? Why does a heavy rock that you liftfrom the bottom of a lake feel heavier as itclears the water surface? The ball float in atoilet tank and the float in your car's carbu-retor experience forces that keep them at thesurface of a liquid. What causes those forces?

In Experiment A-2 did you observe thata hydrometer that floats at some level in oneliquid floats deeper in a second liquid andperhaps sinks to the bottom in a third liquid?In some way, the density of the liquiddetermines whether an object will float orsink. The Greek philosopher Archimedes hada clear understanding of the buoyancy (float-ability) of objects over 2000 years ago. Hisdiscoveries have influenced technology for 20centuries. The following is a statement ofArchimedes' Principle:

An object placed in a liquid experiencesan upward force equal to the weight ofthe liquid it displaces.

An illustration of this is shown in Figure16. Each cubic foot of liquid that is displacedcauses the metal cylinder to apparently "lose"50 Ib of weight. The buoyant force is equal tothe weight of the displaced fluid. (This liquidisn't water, which weighs 62.41b/fe .)

When the metal cylinder is suspendedthis way, the spring balance reads its apparentweight. The apparent weight is the trueweight minus the buoyant force exerted bythe liquid. By Archimedes' Principle, thatbuoyant force is equal to the weight of thedisplaced liquid.

Figure 16 shows the displaced liquidspilling out of the full container as the metalcylinder is lowered into it. Note, however, that

I ft3 SUBMERGED

OF DISPLACEDLIQUID

the displaced fluid doesn't have to leave thecontainer in order for the cylinder to experi-ence a buoyant force. "Displaced" means thatthe liquid is moved aside to make room for theobject. If the cylinder were lowered into apartially full container, the displaced liquidwould just rise in the container.

If a 125-lb piece of wood floats on thesurface of a lake, it displaces 2 ft3 (125 Ib) ofwater. Its apparent weight is zero. In general,an object floats if it displaces a weight ofliquid equal to its own weight. An additionaldownward force is necessary to completelysubmerge a floating object.

The size of a large ship is usuallycharacterized by its deadweight displacement,in tons. A 125,00D-ton displacement oiltanker displaces 125,000 tons of sea waterwhen it is fully loaded. The tanker floatswhen loaded because it weighs the same as do125,000 tons of sea water.

DERIVATION OF ARCHIMEDES'PRINCIPLE

We can show that the upward or buoy-ant force on a submerged object is equal tothe weight of the displaced liquid by analyz-ing Figure 17. Recall that a liquid exerts anormal (perpendicular) force on any surfacein the liquid.

When a cylinder, such as the one inFigure 17, is submerged in a liquid, thepressure on the top surface is:

FIP=Dhl =-A

where D is the weight density, A is thecross-sectional area of the cylinder, and FI isthe total force on the top surface. The forceon the top of the submerged cylinder is thus

Similarly, the force on the bottom is F2Dh2A. Since h2 is greater than hi, F2 isgreater than FI • There are also forces actingon the sides of the cylinder, but for each suchforce on one side there is an oppositelydirected matching force on the other side.Thus all the horizontal forces cancel out andthe net force on the cylinder is

The buoyant force, FB, is the net forceof the liquid on the object. The volume of thedisplaced fluid is V = A Llh. Therefore,FB = D V, which is the weight of the dis-placed fluid. We have derived Archimedes'Principle for a solid whose volume is easy tocalculate. The same analysis can be applied toother shapes as well, although it is moredifficult to analyze complicated shapes.

The shut-off valve in a toilet tank illus-trates an application of Archimedes' Principle.It relies on the buoyant force on a floatingball to shut off the valve.

Example 9. A cubic foot of iron is totallyimmersed in water.

Solution. a. Archimedes' Principle states thatthe buoyant force is equal to the weight ofthe displaced fluid. One cubic foot of water isdisplaced. It weighs 62.4 lb. Therefore, thebuoyant force is 62.4 lb.

b. The apparent weight is the trueweight minus the buoyant force. The trueweight is the product of the weight densityof the iron and the volume.

W = DV

= (494 }b/ft3) (l fe)

= 494 lb

Apparent weight = 494 lb - 62.4 lb431.6 lb.

Example 10. A wood float has a density of600 kg/m 3

. Its mass is 2 kg.

a. Determine how much of its volume issubmerged when it floats in water.

b. Determine the force required to pullit below the surface.

places an amount of water equal to its ownweight, W = Mg. Therefore, the mass of thedisplaced water is 2 kg. The volume of dis-placed water can be determined from themass density of water.

M 2 kgV=-= 3

P 1 g/cm2000 gI g/cm3

= 2000 cm3

= 2 X 10-3 m3

b. The force required to pull the floatbelow the surface is the difference betweenthe buoyant force on the completely sub-merged float and the true weight of the float.First, using the density of the wood, deter-mine the volume of the wood block, which isalso the volume of water displaced when thewood is completely submerged.

v = M = 2 kg 3 = 3.3 X 10-3m3p 600 kg/m

The buoyant force equals the weight of thedisplaced water.

FB = pgV

= (l03 kg/m3) X (9.8 m/s2

) X

The weight of the float, Mg, is 19.6 N.Subtracting, the required force is 12.7 N.

Now you should be ready for Experi-ment B-2, an experiment which usesArchimedes' Principle in the determination ofdensities.

The simplest way to determine the densi-ty of an object is to measure its mass andvolume. One can calculate the volume if theobject has a regular shape, as do a sphere, acube, and a cylinder. The density is thencalculated by taking the ratio of the mass tothe volume:

If the object is irregularly shaped, it maybe difficult or impossible to find the volumeby measuring its dimensions.

Here is a method that can be used withsmall objects which are irregularly shaped.

1. Determine the mass of the object with agram balance.

2. Partially fill a graduated cylinder withwater and record the volume of water incubic centimeters.

3. Place the object in the cylinder. If it ismore dense than the water, it will sinkand displace a volume of water equal toits own volume so that the water rises inthe cylinder. If it is less dense thanwater, you will have to hold it just under

the water with a stiff wire which doesnot displace much water.

4. Record the volume (water plus object) inthe graduated cylinder. The differencebetween the two readings of the waterlevel is the volume of the object.

5. Take the mass-to-volume ratio to obtainthe density, Px, for each object.

Perform this experiment for each of theobjects provided by your instructor.

Your measurement of mass is quiteaccurate because the balance is a sensitivedevice. You should be able to determine anobject's mass to the nearest 0.1 g. However,your measurement of volume using the gradu-ated cylinder is a relatively crude measure-ment; the graduated cylinder does not give aprecise measure of volume. Thus, using thismethod to measure volume isn't a very goodway to determine density. It is better tomeasure volume more accurately. One way todo this is by using the gram balance.

Using Archimedes' Principle, a balancecan be used as illustrated in Figures 18and 19.

1. With the balance supported over thetable, hang the object in question fromthe balance arm, as shown in Figure 18,and find its mass in air, MA.

2. Now suspend the object from the bal-ance arm so that it is submerged in abeaker of water, as in Figure 19. Becareful that it doesn't touch the sides.Record the apparent mass in water, Mw.

The difference between these two values(MA - Mw) is related to the apparentloss in weight, which is in turn equal tothe buoyant force exerted by the water.By Archimedes' Principle, that force isjust equal to the weight of the waterdisplaced. Thus, the weight of the waterdisplaced is just the weight of (MA -Mw), in grams (Le., MWater = MA -Mw). How many cubic centimeters ofwater are there in (MA - Mw) grams ofwater? Use the density relationship andthe density of water (l g/cm3

):

MWater (MA - Mw) 1 / 3PWater = = = g cm

V V

Thus, the volume (in cm 3) of the object

is numerically equal to the apparent lossin mass represented by (MA - Mw). Bymaking two readings of mass you haveobtained the volume of the object to amuch higher accuracy than by using thegraduated cylinder. Using the expressionabove for the volume, you can nowcompute the density, Px, of the object:

MAPx =---- X (l g/cm3

)MA -Mw

3. Using this method, repeat the determi-nation of Px for each of the objectspreviously used.

If you wish to use this latter method tofind the density of an object which floats, youmust use a sinker to pull it under the water.

1. Measure the mass in air, M A, of theobject by hanging it from the balancearm as before.

2. Suspend a heavy metal sinker from theobject, as shown in Figure 20.

3. Submerge the sinker, as in Figure 20,and measure and record the apparentmass of the object in air plus the sinkerin the water(MA +Msw).

LEADSINKER

4. Let the sinker pull the object downbeneath the surface of the water. Be surethe sinker doesn't touch bottom. Nowmeasure the apparent mass of the objectand the sinker, with both in the water(Mw +Msw)·

v= (MA - Mw)/(1 g/cm3)

= [(MA +Msw)- (Mw + Msw)]/(1 g/cm3)

Px = MA = MA (1 g/cm3)

V [(MA+Msw)-(Mw+Msw)]

5. Calculate the densities of the objectswhich float.

The use of a gram balance to determinethe volume of an object demonstrates twopractical technical principles. First, there isoften more than one way to make a satis-factory measurement. Second, different mea-suring techniques have different accuracies,which usually depend upon the instrumentused.

Archimedes developed his principle inorder to solve a very practical technicalproblem. The king wished to purchase a newgold crown. Before he paid the craftsman, hewanted to be sure the crown was pure gold.He asked Archimedes to determine if thecrown was pure gold or a mixture of metals.Archimedes measured the buoyant force onthe crown in water to provide the answer.Non-destructive testing is a widely used engi-

neering technique. Surely, this is one of theearliest examples recorded.

1. A partially filled fish tank rests on abalance. A fish is put into the tank. Thefish doesn't touch the sides. Does thebalance indicate more weight? Explainyour answer.

2. In Experiment B-2 we ignored the buoy-ant force of air on the objects. Showthat this was a reasonable procedure tofollow. (Density of air at standard con-ditions = .00129 g/cm 3

)

3. Suggest a method other than using ahydrometer to measure the density of anunknown liquid. Write out a detailedprocedure. Obtain a liquid sample anddetermine its density and specific gravi-ty. Check your answer with a hydrom-eter. List the most important sources oferror carefully.

4. A partially inflated rubber inner tube isjust barely floating on the surface of apond. It is pulled down lOft by aswimmer and released. Will it rise orsink? Explain.

5. A 1 fe block of cement has an apparentweight of 330 Ib when it is submerged ina lake. What is the buoyant force on it?

You have seen that pressure in a liquiddepends on both its density and the depth.The same is true for gases. We live at thebottom of an "ocean" of air, and the pressureof the atmosphere is significant. However, thedensity of the atmosphere varies with alti-tude, so Equation (7) can't be used tocompute air pressure. Figure 21 shows that acolumn of air I fe in cross-sectional area, andextending to the top of the atmosphere, weighsover a ton! The pressure at the bottom of thiscolumn can be calculated.

2120lbP(atmosphere) =PA = 2

I ft2120lb144 in2

You have probably encountered this valuebefore. A column of water would have to benearly 34 ft (10m) high and a column ofmercury would need to be 2Y2ft (76 cm) high

to exert the same pressure. As you know fromweather forecasts, the actual pressure exertedby the atmosphere varies considerably. Thispressure is a typical value at sea level and iscalled a standard atmosphere. Atmosphericpressure is often expressed in terms of theheight of a column of liquid that produces thesame pressure. A mercury barometer, like thatin Figure 22, measures atmospheric pressurein terms of the height of a column of mercurywhich is balanced by air pressure.

Since the air pressure decreases as oneascends above sea level, aneroid barometers,calibrated for altitude, are used as altimetersin airplanes. You will study both mercury andaneroid barometers in Section C.

There are a great many units of pressurecurrently in use. Table III gives a standardatmosphere (l atm) in terms of several ofthem.

Table III.A Standard Atmosphere in

Several Systems of Units

I atm = 14.7 Ib/in2

= 760 torr(formerly millimeters of mercury)

= 29.9 inches of mercury6 2= 1.013 X 10 dyn/cm

= 1.013 X 105 Pa (1 Pa = 1 N/m2)

Does air pressure have anything to dowith the pressure in liquids? The answer is yesand no. Atmospheric pressure acts on anyexposed surface of a liquid. At a depth hbelow the liquid surface, the total pressure Pis the sum of atmospheric pressure and thepressure due to the liquid. The pressure dueto the liquid is Dh. The total pressure is then

P = Dh + PA = hD + 14.71b/in2

This is usually called an absolute pressure.Usually liquid pressure problems require onlythe pressure due to the liquid. Such pressuresare called gauge pressures, because mostgauges read the pressure above atmosphericpressure. For example, a tire pressure gaugereads gauge pressure; inflating a tire "to301b" means raising its pressure to 301b/in2

above atmospheric pressure.Note that in previous sections of the

module we have been referring to gaugepressure (Dh) most of the time. How wouldthe inclusion of atmospheric pressure affectthe answers of the earlier examples?

The following example should help toclarify the difference between gauge pressureand absolute pressure.

Example 11. Suppose you have two pressuregauges, one of which reads absolute pressureand the other gauge pressure. What is thereading on each gauge at the surface of a lakeand at a depth of 15 ft?

Solution. At the surface the absolute pressureis one atmosphere, or 14.71b/in2

• Thereforeat the surface

The gauge pressure (pressure due to thewater) at a depth of 15 ft can readily bedetermined. Adding one atmosphere to itgives the absolute pressure. Thus at 15 ft

PGauge = Dh = (62.41b/fe) X (15 ft)

= 93.61b/fe

P = 14.71b/in2 + 6.51b/in2

= 21.21b/in2

EXPERIMENT B-3. A Closed HydraulicSystem

In this experiment you will investigatethe properties of a two-piston, closed hydrau-lic system, using two medical syringes whichare coupled together. In a hydraulic system,matter is conserved (if there isn't a leak).Since a liquid is incompressible, the totalvolume of liquid in a closed system alsoremains constant. This may seem obvious, butif the system were pneumatic (air-operated),the enclosed volume would change, since air iseasily compressed.

1. Use the syringes as a two-piston system,as shown in Figure 23. Demonstrate thatmatter is conserved as oil is transferredfrom one cylinder to the other. Measurethe diameters of the two syringeplungers. Note that the syringes aremarked for volume displacement in cc(cm3

).

2. Using a range of slotted weights and thetwo-syringe hydraulic system, hangweights up to 600 g so that the forceacts on the smaller piston. Using a springpush balance to measure the outputforce, determine the output force aftereach weight is applied to the small-diameter cylinder. Record the outputforce, then gently tap the mountingboard to break the frictional forcesimpeding the motion of the pistons.(This frictional force is termed thebreakaway force in hydraulic technol-ogy.) Record the new value of theoutput force.

""~~''''''''''''''-----=l ====------.vI/-~----~~----- -- ---- -- -~i~""""'" Az '1

3. Compute the pressure at the face of eachplunger.

4. Calculate the ratio of output force (aftertapping) to input force for each weight.

While many characteristics of liquids canbe explained by studying free-standing liquidsin open containers, at least one importantproperty relates to liquids in closed systems-systems which are not open to the atmo-sphere. Automobile brake systems andhydraulic jacks are examples of closedsystems. The behavior of closed systems wasstudied by Pascal, an eighteenth centuryscientist. His conclusion is known as Pascal'sLaw:

In a closed system filled with a fluid,pressure applied to the fluid in one partis transmitted equally to all parts of thesystem.

Frictional effects may have partiallymasked this effect in Experiment B-3, butyour results for the pressures on the twoplungers should have been close.

The hydraulic press and hydraulic jackare devices which work through the applica-tion of Pascal's Law. A discussion of thesedevices should lead to an understanding ofthis important principle of physics. Figure 24shows a simplified version of the hydraulicjack. The basic parts are:

b. A large piston which supports theload.

c. A small piston which applies theinput force.

Let's start with the system in balance.(Pressure differences due to differences inheight are small compared to the total pres-sure in the jack; therefore, we will ignorethem.) The applied force on the small pistonproduces a pressure given by

Pascal's Law states that this same pressure willbe transmitted to the large piston.

::::.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.:.::.:.:.:.:.:.:.:.:.:.:.:.:.-.:.:.:.-.:.:.:.:.:.:.:.: ............................••;........

Az 20 inz=::

AI = I in2: :: ::

: : :::::-: :: :

~rrrrrttttttttttttttttttR!Etttttttttt~ttttIttfrrrrrrrrrrrrrFigure 24. When the large piston moves up 1 in, the small piston must move down 20 in, and 1 Ib appliedto the small piston produces 20 Ib of force on the large one.

F2 rrd; /4---2-F1 rrd2/4

In other words, the force is multiplied by theratio of the squares of the diameters. Itsounds great to multiply force, but you maybe sure there is a catch somewhere. You can'tget something for nothing. Let's look at Sl ,the distance the small piston must move inorder to raise the large piston up a distanceS2' To move the large piston up S2 requires avolume of oil given by

Because the oil is incompressible, this must beequal to the volume of oil moved by the smallpiston:

Thus, the smaller piston moves a longerdistance than the larger one. In fact, in thisideal case the ratio of distances moved is justthe ratio of output to input forces. Forexample, in the system of Figure 24 the smallpiston moves 20 times as far as the largepiston and the force is multiplied 20 times.Rewriting Equation (10) shows that the prod-uct of force times distance is the same for thetwo pistons:

The product of a force and the distancethrough which the force is applied is thephysical definition of the term work. (This isa very special use of the term, quite differentfrom our everyday meaning. A displacementparallel to the direction of the applied forcemust occur if physical work is to be done.)Thus, Equation (II a) means that work-inequals work-out ideally:

In other words, the hydraulic jackdoesn't reduce the amount of work one mustdo; it reduces the force needed but increasesthe distance through which it must beapplied. If there is no friction (as we assumedin the above derivation), the work outputequals the work input.

Experience tells you that hydraulic jacksdon't raise a load with a single stroke of thesmall piston. Real jacks use repeated strokesof the small piston to do this lifting; Figure25 shows this refinement of the jack. SectionsA and B of the jack are the same as thoseshown in Figure 24. Section C is a reservoir ofoil at atmospheric pressure. 0 and E areone-way valves. If the pressure on the right ofvalve 0 is greater than that on the left, theball closes the valve. Conversely, if the pres-sure on the left is greater, the ball is forcedout of the cup and the valve opens. Thus the

VALVED

101_

VALVEE

liquid will flow only from left to rightthrough this valve. Valve E works exactly thesame way.

When the piston in the small cylinder ispushed downward, valve 0 is forced shut andvalve E is opened. This admits liquid to thelarge cylinder and causes the load to rise. Ifthe small piston is raised, the valves reversepositions and liquid is drawn from the reser-voir, refilling the cylinder at B. Add a lever(handle) to B and you have a hydraulic jack.(How do you get the load down?)

In garages and gas stations many hydrau-lic lifts employ a combination of a gas (air)and a liquid (oil). Multiple use is made of theair compressor, which also inflates tires, runspneumatic tools, and helps to signal incomingbusiness. In the automobile lift, the directpressure from the small piston is replaced byair pressure from the compressor. For exam-ple, in an automobile lift, if the compressoroutput is 40 Ib/in 2

, and the lift piston is 12 inin diameter(l13 in2 area), the lift can support4520 lb. See Figure 26.

We started the discussion of the hydrau-lic jack by neglecting friction; now let's put itback. With friction in the system, some workis lost and the output work W2 no longerequals the input work WI. We define theefficiency E:

W2E=-X 100%WI

This is the percentage of the work inputwhich is converted into useful output.

Efficiency is often defined in terms ofmechanical advantages. The actual mechanicaladvantage (AMA) includes frictional effects,and it is expressed in terms of the appliedforce FI and the actual output force 4 F2

(load):

AMA =F2

FI

This is just the ratio you computed inExperiment B-3.

The ideal mechanical advantage (lMA)ignores friction, and is usually expressed interms of distances or dimensions. It is thevalue that F2 IF. would have in the absenceof friction. Since Equation (10) was derivedon the basis of no friction, we can use it todetermine the ideal mechanical advantage forthe jack:

Now the efficiency can be expressed in termsof mechanical advantages:

Example 12. A hydraulic jack has a small-piston diameter of I in and a large-pistondiameter of 9 in. It is found that, to lift a200Q-Ib load, a force of 30 Ib must be appliedto the small piston. Determine

d; (9 in)2IMA=-=--=81

d~ (l in)2

AMA=_F2= _20_0_0_I_b= 67F. 30lb

AMA 67E = IMA X 100% ~81x 100%

If this jack were operated with a leveredhandle which has a mechanical advantage of10 (lever arms in the ratio 10: 1), the appliedforce on the end of the handle would be only3 lb. (Mechanical pivots are almost frictionlessif good bearings are used.) Even if the frictionwere so high that the efficiency of themechanical lever was only 50%, a 6-lb forceinput would produce the required 30 lbneeded to operate the hydraulic lift. With thiscombination of mechanical and hydraulicforce multiplication, even a small child couldlift a I-ton load.

Power steering units for automobiles,large hydraulic hoists, cherry-picker cranes,and dump-truck hoists are essentially like ahydraulic jack in which the lever and smallpiston are replaced by a pump which forcesliquid into the large cylinder. Technologistshave certainly put Pascal's Law to good use.One example is shown in Figure 27.

Figure 27. Hydraulic systems do much of the heavywork in our technological world.

Pressure is the ratio of the normal force to thearea over which the force acts and is usuallyexpressed in Pa (N/m2

), Ib/in2, or dyn/cm2

•

The pressure at any point a distance h belowthe surface of a liquid is equal to the productof weight density (D) and depth (h).

Archimedes' Principle states that, when anobject is placed in a liquid, it is buoyed up bya force equal to the weight of the displacedliquid.

The absolute pressure is equal to the gaugepressure plus the atmospheric pressure.

Pascal's Law states that the pressure appliedto an enclosed system is transmitted equallyto all parts of the system.

The hydraulic jack and the hydraulic pressboth operate according to Pascal's Law. Theactual mechanical advantage of these devicesis:

F2 Output forceAMA=-=----

F1 Input force

The ideal mechanical advantage of thesedevices is:

d2

IMA= -f where d = piston diameterdl

Workout AMAE=---X 100%=--X 100%

Work in IMA

A submarine's hull can withstand a waterpressure of 20 ton/fe. What is its safediving limit? (The weight density of seawater is about 641b/ft3 .)

2. An ice cube floats in a glass brim full ofwater. When the ice cube melts, does theglass overflow? State a precise reason foryour answer. Perform a simple experi-ment to prove your results.

3. Gin has a density of about 0.90 (depend-ing on the proof rating). Will an ice cubefloat higher or lower in a glass of ginthan in a glass of water? Will the ice cubefloat in pure alcohol (200 proof)? Testyour answers on a few friends.

4. A student reading a physics text isfloating on a raft in her swimming pool.In frustration she throws her text intothe pool from the raft. Does the waterlevel rise in the pool? Explain youranswer.

5. Using your data from Experiment A-I,calculate the actual mechanical advan-tage of the automotive jack.

6. The small piston of a hydraulic jack hasa cross-sectional area of 2 in 2

• The largepiston has cross-sectional area of 40 in2

•

What is the ideal mechanical advantageof this jack? How many 4-in strokes ofthe small piston are required to raise theload on the large cylinder 100 in?

7. The surface of a vinyl floor covering hasa breaking strength of 250 Ib/in2

• Will a105-lb girl wearing shoes having heelswith an area of 0.35 in2 break thesurface? State clearly any assumptionsthat you make. (In the days of "spike"heels, this was a serious problem in manypublic buildings.)

8. A block of aluminum 2 in X 6 in X I infloats on its large face in a pool ofmercury.

a. What pressure does the weight ofthe aluminum exert on themercury?

b. What is the buoyant force on thealuminum?

c. How much mercury is displaced bythe aluminum?

9. A hydraulic cylinder whose diameter is2.5 cm is coupled to a cylinder ofdiameter 12.5 cm. What force would berequired at the smaller cylinder to pro-vide an output force of 1100 Ib if thesystem operates with 88% efficiency?

10. A window on a spacecraft has a diameterof 12 in. The spacecraft is pressurized at0.8 atm. What pressure must the windowglass withstand in Ib/in 2? What totalforce is exerted on the window?

II. The face-mask window of a skin diverhas an area of 100 in 2

• The pressureinside is I atm. At what depth will thewater pressure be ten times the insidepressure? What will the total force be onthe outside of the window?

12. From your answers to questions 10 andII, can you explain why it is easier to goin to outer space than to go to thebottom of the ocean (46,000 ft)?

13. A naval depth charge is set to go offwhen its pressure sensor detectsISO Ib/in 2 pressure. If the density of seawater is 641b/ft3, at what depth will thecharge explode?

14. A Polaris submarine crew wishes tolaunch a missile from a depth of 80 ft.

To what pressure must the missile firingtube be pressurized to open the protec-tive door?

IS. An Eskimo tries to sneak up on a seal bystepping onto a circular ice floe 3 ft indiameter and 18 in thick. The floe is insalt water (D = 64 Ib/fe). If the Eskimoweighs 140 Ib, and the specific gravity ofsalt ice is 0.90, will the ice floe supporthim?

16. A 23-lb bicycle has a 137-lb rider on it.The weight is evenly distributed to bothwheels. If the pressure in the tires is80 Ib/in2

, how much tire area contactsthe road?

17. As the eye of a tornado passes over theroof of a house, the pressure outside theroof suddenly drops to 27 in of mercury.If the windows and doors are all closed,the inside pressure does not changerapidly. Calculate the pressure differ-ential in Ib/in2 if the inside pressure re-mains at 30 in of mercury. If the house is30 ft X 50 ft, calculate the total forcelifting the roof as the tornado passes.What should one do with the windowswhen one hears a tornado warning forthe area?

18. A boy weighs 185 lb. When he dives intoa pool, he displaces 22 gal of water. His110-lb girlfriend displaces 13 gal. Whosebody is more dense?

The study of the behavior of fluids inmotion is called hydrodynamics. Fluids inmotion display properties that differ dramat-ically from fluids at rest.

One of these properties is frictional drag.Real liquids such as water, beer, and oilexperience a frictional force, called drag, asthey flow through pipes or over surfaces. Theresult is that the force exerted by a movingliquid on the walls of the pipe has a compo-nent parallel to the pipe. Static fluids alwaysexert forces perpendicular (normal) to thewalls of the container. The frictional effectcauses a loss of mechanical energy and a lossof pressure as the fluid flows through the

pipe. This effect is easily seen in commongarden hoses. The longer the hose, the less thepressure at the end of the hose.

A detailed study of friction is beyondthe scope of this module, and the followingdiscussions assume that the effect can beignored. Frictional effects only cause a changein numerical values; they do not change themain features of the fluid behavior.

A second difference between static fluidsand flowing fluids is illustrated by Experi-ment C-l. Differences in the velocity of afluid produce differences in pressure in thefluid. After the experiment, an optional sub-section explains why this pressure differenceexists, and derives an equation which is usedto compute the difference.

In this experiment you will see severalexamples of the way in which the motion of afluid, air, affects its pressure. You will not berequired to record data in the usual way, butyou should write down what you see in eachcase. Sketches may be helpful.

1. Hold a piece of notebook paper in frontof your mouth as in Figure 28. Whathappens to the paper when you blowacross its upper surface?

2. Stick a pin through the center of a 3 inX 5 in index card. Insert the pin intothe center hole of an ordinary householdspool, so that the card covers the end ofthe spool. Blow through the other end ofthe spool. What happens?

3. Fold a 3 in X 5 in index card as shown inFigure 29. Place the card on the edge ofa table and blow through the "tunnel"formed by the card and the table top.What happens? Can you blow the cardoff the table in this way?

4. Attach an air supply hose to the nozzleprovided. Turn ,on the air and place aPing-Pong ball in the air flow as in Figure .30. Describe what happens as the airstream is slowly tilted away from thevertical position.

1AIR

5. Replace the nozzle with a funnel. Turnthe air on and place a Ping-Pong ball inthe throat of the inverted funnel (seeFigure 31). Release the ball and describewhat happens.

AIR IN

j

6. Tape a piece of thread to a Ping-Pongball and suspend it near a stream ofwater from a faucet. What happens?

7. Tape some thread to a second Ping-Pongball and hang the two balls so that theyare the same distance below a horizontal

support and about 5 cm apart. Blowgently between them. What happens?Describe the result of blowing harder.

8. Fill the V-shaped glass tube about halffull of water. Blow across one open endof the tube and ask a classmate todescribe the results. Be careful to blowhorizontally and not down into the tube.

9. Observe the behavior of the atomizerwhen you squeeze the bulb. Do you seeany connection between the behavior ofthis device and the effects observed insteps I through 8?

10. Examine the aspirator carefully. Thenhook up the aspirator supply hose to awater outlet. When you are sure that thedrain hose is properly placed, turn onthe water and hold your finger over theopen arm of the "T." What happens?

Think about the observations you havejust completed. Can you identify anycommon element of these effects? If so,you may be able to guess the physicalreason for the behavior of each of thenine "devices."

Did you guess that the pressure exertedby a fluid is affected by the motion of thatfluid? The greater the fluid speed, the lesspressure it exerts. This important physicalfact is called Bernoulli's principle.

The Bernoulli principle can be derivedby applying the principle of energy conser-vation to a small volume of the fluid as itmoves. Such an analysis, which we do later inan optional subsection, leads to a generalequation which is valid for incompressiblefluids. Referring to Figure 32, the Bernoulliequation is

• ~VI

IIh1 i •. V2

~"2*Figure 32.

PI' P2 = absolute pressure in the pipe atpoints 1 and 2.

hI, h2 = heights of pipe, measured from somearbitrary level.

In many cases of practical interest, the twosections of the "stream" are at equal height(hI = h2) and the Bernoulli equation reducesto

Note that this agrees with what you observedin your experiments: the point at which thefluid speed is greater is where the pressure islower. The following example illustrates theuse of this equation in a real device.

Example 13. Figure 33 illustrates an aspira-tor. Suppose the water speed into the aspira-tor is 6.66 m/s and water speed just past theconstriction is 20.5 m/s. If the input pres-sure is 2 X 105 N/m2 (1.97 atm), find the"pumping" pressure. (The density of water is1000 kg/m3

.)

VI ~ v2• ~PI / P2

i(Figure 33.

. Solution. We can apply Bernoulli's equationto the points 1 and 2 before and after theconstriction. The pumping pressure P 2 is thepressure following the constriction. We knowthe following quantities:

V2 = 20.5 m/s

PI = 2 X 105 N/m2

p = 103 kg/m3

Since the heights in the two parts of theaspirator are the same, we can use the simplerform of the Bernoulli equation:

= (2 X 105 + .22 X 105-

2.10 X 105) N/m2

= .12 X 105 N/m2

DERN ATION OF THE BERNOULLIEFFECT (OPTIONAL)

Figure 34 shows a tank of water with anoutlet at the bottom. Water leaves the bottomat a speed V2 and at atmospheric pressure PA.As you know, the greater the depth of thewater, the faster it flows from the outlet. Weseek first a mathematical expression for thedependence of the water velocity on itsdepth. Such an expression can be obtained byapplying the principle of conservation ofmechanical energy, but to do so we mustmake a number of simplifying assumptions.For example, we shall assume that there is nofriction and that the flow is perfectly smooth(laminar).

hi

T .•... ~... _/----+~-h

1 .-- - - - ...•- -"'h2 ,--)..--I I II I I

V2 PAFigure 34.

Another idealized assumption we makeis that VI is the velocity of the water floweverywhere in the main part of the container.Then a mass M of water in the container has akinetic energy Y2 MV;. If that same massleaves the container, its kinetic energy as itleaves is Y2 Mv;. Thus there has been anincrease in the kinetic energy of the mass Mgiven by

We assume that energy is conserved.Therefore, this increase in kinetic energy musthave come from somewhere. Where? The levelof water in the container drops somewhat,thereby decreasing the gravitational potentialenergy available. The gain in kinetic energymust be equal to this loss in potential energy.The decrease in height is exactly what wouldoccur if the mass M were removed from thetop of the container and placed at thebottom. Therefore, the change in gravitationalpotential energy is negative and is given by:

The law of conservation of energy states thatthe total energy remains constant or, in otherwords, that the change in total energy is zero:

This tells us how fast the fluid leaves thecontainer, expressed in terms of the depth h.A somewhat similar situation is discussed inthe following example.

Example 14. Water falls from the top of apower dam to the river below, a drop of 80 m(260 ft). What kinetic energy does each kg ofwater have as it reaches the river bed? Whatvelocity does the water have?

Solution. The kinetic energy gained justequals the potential energy lost as water falls80m.

tJ.KE =Mgh

= (l kg) (9.8 m/s2) (80 m)

Assuming that VI = 0 (the water has nokinetic energy as it dribbles over the top ofthe dam):

The process whereby falling water con-verts gravitational potential energy to kineticenergy is the basis of hydroelectric powergeneration. The falling water collides with theblades of a turbine, converting the water'skinetic energy to rotational energy whichdrives the generators. The same principle haspowered waterwheels for centuries.

Returning to the container of Figure 34,there is another way of looking at thesituation. Certainly, it is the pressure whichcauses the water to be pushed out of thebottom. The surface of the water is atatmospheric pressure PA. The difference be-tween this and the pressure in the water at thebottom of the container PB is

Notice that PA, the atmospheric pressure,is also the pressure in the region where thewater is flowing out of the bottom of thecontainer. Now let PA = P2 and PB = PI' sothat PI is the pressure in the region where thevelocity is VI and P2 is the pressure in theregion where the velocity is v2• We can nowwrite the preceding equation as:

Now we have a way of calculating thevelocity in terms of the pressure difference, orvice versa. It doesn't matter what causes thepressure difference. Equation (19) is true forany case of a liquid in continuous flowwithout friction. The pressure PI does nothave to be due to the weight of a column of

liquid, nor does P2 have to be atmosphericpressure. Although Equation (19) was derivedfor the situation shown in Figure 34, it is alsotrue for the situation shown in Figure 35,where three different regions are involved.

l fPz -VI- Vz v3 -

PI f l P3

Figure 35.

An interesting conclusion can be reachedby inspecting Equation (19): the pressure issmaller where the velocity is greater. This isexactly the Bernoulli effect that you observedin Experiment C-l.

We now consider the case in which thepipe carrying the fluid is not all at the sameheight, as shown in Figure 36. In this situa-tion, there is a pressure difference due to theheight difference, in addition to that due tothe differences in velocities. We can add thedynamic pressure difference (due to a velocitydifference) to the static pressure difference(due to a height difference) to get the netpressure difference.

Note: if the pipe diameter does not change,v2 must equal VI. Why? We can rearrange theterms in this equation to get the Bernoulliequation:

Since the subscripts refer to any twopoints in the pipe, the sum of the three termshas the same value for all points along thepipe. That is, Bernoulli's equation can also bewritten as:

P = absolute pressure (due to some impressedforce)

pgh = static pressure due to a depth h ofliquid

V2 pv2 = dynamic pressure due to fluidvelocity v

The Bernoulli equation is valid only ifcertain conditions are met. We have alreadyassumed that the fluid is incompressible andthat there is no friction between the fluid andthe walls of the pipes. In addition, we mustassume that the flow is perfectly smooth, orlaminar. Turbulent flow, involving eddies orvortices, is not adequately described by theBernoulli equation.

Except for rather low velocity flow, theBernoulli equation does not work well forgases, which are highly compressible. How-ever, as you saw in Experiment C-I, the basicfact of increased velocity resulting in de-creased pressure still holds.

\ fP2 -VI - V2 V3 -

PI f l P3 SUCTION •Figure 37. A venturi tube. PORT

Consider a tube with a constrictedthroat, as in Figure 37. The constricted throatis called a venturi. If PI is comparatively largeand the venturi has a very small diameter, thevelocity V2 can be quite large. Therefore P2 isless than PI or P3' In fact, P 2 can easilybecome less than the atmospheric pressureoutside this system, PA.

If we now connect a tube to the con-stricted throat, gas or liquid is drawn into themain stream because of the pressure differ-ence (P A - P2) (Figure 38). Such a device iscalled an aspirator. Chemistry students oftenuse an aspirator connected to an invertedfunnel to exhaust the noxious fumes theyfrequently generate in the laboratory. If a gas,such as air, is forced through a venturi, thepressure drop at the constriction can be usedto draw liquids into the gas stream. Atom-izers, paint sprayers, aerosol cans, and garden

sprayers thus are also aspirators. The carbura-tor throat in your car is a venturi which drawsgasoline from the bowl while mixing it withair.

Figure 39A shows a typical aspiratorpump. (Figure 39B is a photograph of anaspirator.) Water flows in the top and outthe bottom (into a sink). The horizontal armcan be connected to any container one wishesto pump.

Figure 39A and B. A schematic and a photo of anaspirator.

A simple but useful hydraulic device isthe siphon, which is simply a tube or hoseconnecting two containers of liquid, as shownin Figure 40. If the tube is filled with fluid,the siphoning effect causes fluid to flow fromthe higher container to the lower container.

To understand why the effect occurs,first imagine that a closed valve is inserted atthe top of the hose (point C in Figure 40).The pressure Pion the left side of the valve is

Thus, with h2 greater than hi , the pressure onthe left side of the valve is greater than thepressure on the right. Then if the valve isopened, the fluid will flow to the right.

You will find a problem if you try tosiphon water over a wall more than 34 ft high;it won't work. Atmospheric pressure cansupport a column of water 34 ft high, but nohigher. Thus, if the wall is higher than that, an

attempt to siphon over it would fail. If onestarted with the hose full of water, the levelwould merely drop until its height was 34 ftin each arm, with a vacuum (P ~ 0) in themiddle section. Then each arm would act as awater barometer. The system would sit therewith nothing happening.

If you have never used a siphon, youwould enjoy trying one out.

PRESSURE MEASURING DEVICES(GAUGES)

Perhaps the least complicated deviceused to measure static pressure is an open-tube manometer, shown in Figure 41. Thefluid whose pressure is to be measured exertsa pressure PM on one surface of the liquid inthe tube. The atmosphere exerts a pressurePA on the other surface of the liquid. If thetwo surfaces are at the same level, the twopressures are equal. If the two surfaces are avertical distance h apart, the difference inpressure between the two arms is just that dueto the column of liquid of height h. That is,PM - PA = pgh, where p is the density of theliquid in the tube.

If this result is not obvious, perhaps thefollowing discussion will help. The pressure at

FLUIDAT

PRESSUREPM

the bottom of the left part of the tube is PM+ pgh1 and the pressure at the bottom of theright side of the tube is PA + pgh2• At thebottom of the manometer, these two pres-sures are equal. Why? Then

The gauge pressure on the manometer is justPM - PA = pgh. The manometer scale may becalibrated in any convenient units.

Example 15. The fluid in an open-tubemanometer is mercury (p = 13.6 g/cm3

).

When it is attached to a cylinder of gas, the"pressure head" h produced is 10 cm. Deter-mine the gauge pressure and the absolutepressure in N/m2

•

Solution. The gauge pressure is the pressuredue to the O.I-m head of mercury.

PG =pgh

= (13.6 X 103 kg/m3) (9.8 m/s2

) (0.1 m)

= 1.34 X 104 N/m2

The absolute pressure is greater than thegauge pressure by one atmosphere.

=PG + 1.01 X 105 N/m2

=(l.34X 104 + 1.01 X 1Q5)N/m2

= 1.14 X 105 N/m2

The barometer, which was mentionedearlier, is an example of a closed-tubemanometer. (See Figure 42.) The weight of acolumn of mercury is balanced by the forceexerted on the open mercury surface by the

ATMOSPHERICPRESSURE

pressure of the atmosphere. The space at theclosed end of the tube exerts essentially zeropressure on the mercury column, so PA =pgh. At one atmosphere of pressure, theheight of the mercury column is

PA 1.01 X 105 N/m2

h =- = 4 3 2pg 1.36 X 10 kg/m X 9.8 m/s

The height of the column of mercury ina barometer in millimeters. of mercury orinches of mercury is most often used to

specify atmospheric pressure. You may haveheard a weather report saying that the barom-eter reading is, for example, 29.92 in. Thismeans that the measured value of atmosphericpressure at the weather station is 29.92 inchesof mercury. Actually the measurement proba-bly was not made with a mercury barometer.Instead, a more compact instrumen t called ananeroid barometer was likely used.

The aneroid barometer is another hy-draulic device. The small metal can (seeFigure 43) contains air at a pressure muchlower than atmospheric pressure. The flexiblelid of the can bends slightly in response tosmall changes in atmospheric pressure. Thisbending is measured by a mechanical systemattached to a scale. The scale is usuallycalibrated in inches of mercury by comparingthe response of the instrument to that of amercury barometer. Since the value of atmo-spheric pressure 'decreases as one's altitudeincreases, the scale of an aneroid barometercan also be calibrated to read altitude. Theinstrument is then called an altimeter.

Manometers are awkward to use, soanother type of gauge called a Bourdon gaugeis frequently used in technical applications.The Bourdon-type gauge (Figure 44) is basi-cally a curved tube sealed at one end. Pressureexerted through the open end of the tubetends to make the tube, which is fastened atits open end, straighten out. The movement

of the free end causes a pointer to move overa scale. A familiar party favor, Figure 45,helps to illustrate the principle.

For moving liquids, we may measure anyof three different pressures. One is the staticpressure which would exist if the liquid werenot moving. A second is the dynamic pres-sure, given by Y2 pv2• The third is the totalpressure which includes both the static anddynamic pressures.

A Bourdon gauge or a manometer can beused to measure either static pressure or totalpressure. Which pressure is measured dependson the way the instrument is connected to thepipe.

It is possible (see Figure 46A) to connecta gauge to the pipe in such a way that itspresence does not significantly disturb thefluid flow. If the gauge is a manometer, thepressure difference across the two arms of themanometer is

where PM is the density of the liquid in themanometer, Pl is the pressure in the pipe, andPAis atmospheric pressure. Thus the pressuremeasured by the gauge is the differencebetween the atmospheric pressure and thepressure in the flowing fluid. The pressure inthe fluid is then:

or it is the difference between atmosphericpressure and the pressure measured by thegauge.

~ pv=o

Alternatively, one can mount the gaugeas shown in Figure 46B. The open probefacing into the flow is called a Pitot tube(pronounced pea-tow). The pressure in thetube increases until it offsets the pressure atthe opening. In such a case, a "dead" point ofno fluid flow is formed at the opening of thetube. One can apply Bernoulli's equation tothe fluid to find the pressure differencebetween the opening of the tube and a pointwell removed from the probe opening

2Pl + Y2 PLVL = P + 0 (23)

where Pl is the pressure in the flowing fluid,VL is the fluid velocity, PL is the fluiddensity, and P is the pressure at the openingof the tube. As above, the manometer dis-plays the difference between atmosphericpressure and the pressure at the opening atthe Pitot tube. That is

Equation (23) suggests a way in whichthe velocity of a liquid can be measured. Byusing the two gauges of Figure 46, Pl and Pcan be measured. Then the velocity-dependent part of the pressure (dynamicpressure) is the difference of the tworeadings:

Example 16. An arrangement such as that inFigure 46 is used to measure velocity of waterin a pipe. If the difference between the twopressures is 2 Ib/in2

, what is the velocity ofthe water?

Solution. We may apply Equation 25 using D=pg:

V=j2~V2 (32 ft/s2) (2 Ib/in2)

62.4 Ib/ft3

2 (32 ft/s2) (288 Ib/ft2)62.4 lb/fe

A different device, also called a Pitottube, combines the two measurements ofFigure 46 into a single manometer. Thus itgives a direct reading of the pressure differ-ence for moving and stationary parts of thefluid. As shown in Figure 47, the pressure inside A of the manometer is the pressure of thestationary fluid at the inner tube. The pres-sure in side B is the pressure of the fluidmoving past the side holes in the Pitot tube.

------

The Pitot tube works well for measuringthe velocity of gases, provided that thevelocity is not so high that compression

46 effects become important. Airspeed indicators

use such a device to measure airplane speedsrelative to the air. Figure 48 is a photographof a typical Pitot tube.

Incidentally, if the flow is out of asystem, either through the end of a pipe orfrom a hole in a container, then the velocitycan be experimentally determined withoutusing a pressure measurement. One has tomeasure the volume that escapes in a periodof time. The volume which escapes can becomputed using Figure 49. If the water flowsat a speed v, then'in a time t it would fill animaginary cylinder whose base is A, the areaof the hole whose length is vt. Thus thevolume of outflow is V = Avt, or solving forthe velocity:

Vv=-At

------ -" '"

Example 17. A vandal shoots a hole in awater tower with a 22-caliber bullet (diameter= .25 inches). If 0.43 fe of water escapes in aminute, calculate the outflow velocity.

2

A = 3.14 (.25) = .05 in2 = 3.6 X 10-4 fe4

v = vLAt

V (0.43 fe)v ---L - At - (3.6 X 104 fe) (60 s)

One familiar pressure measurement isarterial blood pressure, using a sphygmo-manometer (sfig-mo-man-o-miter). This device(shown in Figure 50) enables a doctor ornurse to make an approximate measurementof the pressure in an artery without havingaccess to the fluid (blood). It provides anindirect measure of the gauge pressure in thecirculatory system. For medical purposes,easily reproducible information about abnor-mally high blood pressure or sudden changesin blood pressure is more important thanextreme precision. Blood pressure actuallydepends in a complex way on heart musclecontraction, circulatory system constriction,location in the body, blood flow velocity, andmany other factors. Difficulty with anyoneof a number of bodily functions may show upin an abnormal blood pressure.

Blood pressure measurement is an excel-lent example of a situation where the precisephysics is complex, but a good understandingcan come from an analogy with simplerhydraulic systems, such as those studiedelsewhere in this module. Physicists oftenanalyze complex systems in terms of simpler"models" they are able to understand moreclearly.

PRESSUREGAUGE

Blood pressure varies between a maxi-mum and a minimum value. The peak value,called systolic pressure, corresponds to thepumping action or contraction of the heart.Depending on the rate of heart contraction(pulse rate), the systolic pressure occurs about70 times each minute (or once every1/70 min). Figure 51 is a sketch of bloodpressure (torr) versus time for a normalindividual at rest. Notice that the time be-tween peaks gives the pulse rate. The mini-mum pressure is called the diastolic pressure,and is never zero. These two values ofpressure (maximum and minimum) are thevariables recorded for a blood pressure mea-surement. A typical value for a healthy adultmight be 130/80, which means that thesystolic (maximum) pressure is 130 torr andthe diastolic (minimum) pressure is 80 torr.The values found for a given person dependon many factors, including fatigue, excite-ment, anxiety, recent food, cigarettes oralcohol, and others. The most useful data areobtained under controlled conditions, whenthe person is at rest for a period of time priorto measurement. Often, repeated measure-ments are taken to improve accuracy.

RELEASEI VALVE

+ ,BULB

":: 150~o--IJJIOO0:::Jenen 50IJJ0::Cl..

Obtain a sphygmomanometer and identi-fy the following parts (see Figure 50):

1. An inflatable cuff to wrap around theupper arm.

2. A pressure gauge (mercury manometeror Bourdon type) which indicates thepressure in the cuff in torr (or theequivalent older unit, millimeters ofmercury).

4. A valve to release air from the inflatedcuff.

A stethoscope is required to use thesphygmomanometer for blood-pressure mea-surements. Blood pressure is measured bywrapping the inflatable cuff around the upperarm an inch or two above the elbow. The cuffis inflated to a pressure which is sufficient toconstrict the main artery and thus stop theblood flow. Since the sphygmomanometercuff, the skin on the arm, and the wall of theartery are all flexible, the pressure producedinside the cuff is transmitted to the inside ofthe artery. When the cuff pressure is greaterthan the arterial pressure, the artery is col-lapsed, stopping the flow of blood. Remem-ber, however, that the arterial pressure varies

1/70TIME (min)

with time between a maximum and a mini-mum value.

During measure men t, the cuff pressure isslowly decreased (by letting air out of thecuff) until the pressure in the cuff, Pc, justequals (or is slightly less than) the systolicarterial pressure, Ph. At this cuff pressuresome blood is forced through the constrictedartery at each heart beat. The blood passingthe constriction makes sound which can bedetected with a stethoscope placed on thearm, just below the cuff (inside the elbow).At higher cuff pressures no sound is heardbecause no blood gets through. Thus, thedetermination of the systolic pressure corre-sponds to the cuff pressure at which fluidsounds are first heard, as the cuff pressure isslowly decreased from a value high enough tostop all blood flow.

By continuing to reduce the cuff pres-sure, the flow restriction is diminished. How-ever, the stethoscope continues to pick upsounds as long as any constriction of theartery exists. Eventually the cuff pressureequals the minimum blood pressure and theconstriction disappears. With no resistance toflow, the blood passes through without asound. Thus, the cuff pressure when thesounds disappear is the diastolic pressure. Thiscuff pressure corresponds to the minimumblood pressure in an unconstricted artery.

Working in pairs, measure each other'sblood pressure. Take repeated readings anddon't expect great accuracy, since some prac-

tice is required. Look for left arm-right armdifferences, the effect of exercise, or othervariables.

Question. If you want to measure the heartpressure, you must make the measurementwith the upper arm at chest level. Why is this

necessary? Think about the simpler hydraulicsystems you have studied. (What affects pres-sure in a fluid?) If you think you know theanswer (or your instructor tells you), deviseand perform a simple experiment to testwhether or not this requirement is reallyimportant in blood pressure measurement.

HYDRAULICALLY ACTUATEDCONTROLS

Liquid pressure and fluid flow are usedto control mechanisms found in many com-mon devices. Some of these devices aredescribed below. You probably can think ofother examples.

Figure 3 illustrated the float-level con-trol in a toilet tank. As the float at one end ofthe lever rises with the water level, the otherend of the lever presses a rubber-tipped pistonagainst a valve seat, and this shuts off thewater supply.

A carburetor float control (Figure 52) isquite similar. As the level of gasoline in thebowl of the carburetor drops, the float-levermechanism opens the inlet from the fuelpump. What happens if the float is set toohigh or too low?

GASINLET

FLOATPIVOT

PIN

An automobile radiator cap and a pres-sure cooker safety valve are safety deviceswhich will open at a preset pressure. Thepressure at which the radiator cap opens isdetermined by the stiffness of the spring,shown in Figure 53. In addition, pressurecookers have a pressure-regulation valve, as inFigure 54. The operating pressure of thecooker is determined by the weight of thevalve. When the pressure becomes greatenough, the weight is lifted and jiggles a bit,releasing some of the steam, thus keeping thepressure approximately constant.

The automobile brake-light switch andthe oil-pressure indicator switch are goodexamples of pressure-activated electricalswitches (see Figure 55). Pressure-activatedswitches of this type are connected to lightsor buzzers to warn of high or low pressureconditions in hydraulic systems.

TOBATTERY

BRASS PLATE

RUBBERDIAPHRAGM

These devices are specific examples of atype of technology called fluidics. This tech-nology uses controlled fluid flow to performthe same control functions as do electronicdevices. Fluidic devices are much slower thanelectronic devices, but they have one great

advantage. Because they are part of a sealedsystem, they are very reliable in dirty environ-ments. Thus, they are often used in industrialprocesses such as metal refining and fabrica-t ion. In hospitals, portable respiratorytherapy units use fluidics to control air oroxygen flow in assisted breathing applica-tions. This use of pressure and flow controleliminates the need for a possibly dangerouselectrical system in the apparatus.

Bernoulli's principle states that the fastera fluid flows, the lower the pressure. Fornon-turbulent flow of an incompressible non-viscous fluid, the Bernoulli equation applies:

The aspirator is based on Bernoulli'sprinciple.

The barometer, the manometer, theBourdon gauge, and the Pitot tube are all usedfor pressure measurements.

The sphygmomanometer is used to mea-sure blood pressure by determining the maxi-mum and minimum pressures required toconstrict an artery.

Atmospheric pressure and the variationof pressure with depth are the basic ideasbehind the operation of a siphon.

1. The water tank of a building is filled to adepth of 4 ft. The air above the water iscompressed to a pressure of 10 psi whenthe tank has 4 ft of water in it.

a. What are the absolute and gaugepressures at the surface and at thebottom of the water?

b. What is the velocity of water flowout of an open pipe at the bottomof the tank? (Assume that the tankis so large that for all practicalpurposes v = 0 in the tank.)

2. A river is 60 m wide and 3 m deep as itgoes over a waterfall. The stream isflowing at 2 m. If the falls are 90 m high,how much energy would be availableeach second for hydroelectric powergeneration, if all the water could bechannelled through a turbine?

3. A chocolate bar manufacturer wants torun liquid chocolate uniformly onto amoving conveyor belt. The belt moves ata speed of 8 ft/s. What height of choco-late in the storage tank will maintain theproper outflow onto the moving belt?(You must make the assumption that theflow is frictionless.)

4. A swimming pool has a hole of area Y2in2 at the bottom. Water freely runs out.How many gallons per hour must besupplied to maintain the water level at4 ft above the bottom?

5. A small town (area 3.5 square miles) hasa single l6-in-diameter storm sewer pipewhich drains rain water from the townto a river at the bottom of a canyon60 ft below the town's level. If 1.3 in ofrain falls in 2 hand 40% is lost to groundabsorption and evaporation, how longwill it take for all the rainfall run-off tobe cleared through the sewer?

6. A stream of water with a I-in diameter isprojected with a velocity of 20 ft/sagainst a wall. What pressure would agauge indicate at the point of impact?

7. Water pressure in a house is 30 Ib/in2• A

leaky faucet drips I gal per h. What isthe effective area of the leak?

8. A test chamber at a submarine designlaboratory operates to simulate the sub'smaximum speed by flowing water past afixed model of the submarine. The pres-sure difference measured with a Pitottube is 20 Ib/in2

•

a. Calculate the water flow velocity(submarine's maximum speed).

b. If the test tank flows 100,000 gal in90 s and the test chamber is 3 fe incross-sectional area, recalculate theflow velocity and compare youranswer to the result in part a.

WeightForce

C. Questions you would like answeredabout the hydraulic devices in the lab:

Known MeasuredSpecific Specific

Liquid Gravity Gravity

Specific IdentificationLiquid Gravity of Liquid

A

B

C

D

E

3. Sugar SolutionRange of Specific Gravity.

Volume H2O Volume Antifreeze Number of Balls(cm3

)3 Which Float Pmix (g/cm3

)(em)

50 0

50 5

50 10

50 15

50 20

50 25

50 30

3. Comparison of experimental values for freezing points of antifreeze solution to freezingpoint indicated by container label.

Protection Percentage Antifreeze Percentage Antifreezeto CF) Experimental Value Value from Container

+300

+150

00

-150

EXPERIMENT B-t. Pressure Dependenceon Depth

Spring Balance Force DepthPort Reading (g) (dyn) (cm)

I

2

3

4

a. From your graph, find the values ofpressure and depth at two points(use two points on the line youdraw, not two actual data points):

PI =2(dyn/cm )

P2 = 2(dyn/cm )

hI = (cm)

h2 = (cm)

rise P2 - PISlope = - = ---run h2 - hI

Mass of Object Volume of Water Volume of Volume of Density of(g) in Graduated Water plus Object Object

Cylinder Object 3 (g/cm3)(cm )

3 3(cm ) (cm )

Object ObjectNo. I No.2

Mass in airMA (g)

Apparent massin waterMw (g)

Volume(MA - Mw)/(l g/cm3

) (cm3)

DensityPx (g/cm3

)

Object ObjectNo. I No.2

Mass of object in airMA (g)

Apparent massObject in air plus sinker in water(MA + Msw) (g)

Apparent massBoth object and sinker in water(Mw + Msw )(g)

Volume of object[(MA +Msw)- (Mw +Msw)]/(l g/cm3

)

DensityPx (g/cm3

)

Input Force Output Force Output Force Pressure at Pressure at Ratio ofMg (dynes) Before Tapping After Tapping Large Plunger Small Plunger Fout/Fin

Average ratio