Fluid Mechanics.pdf

Complete and detailed discussion of the formulas and principles Over 800 solved and supplementary problems AUseful tables and conversion factors • With index for easy access to topics in the book Includes recent board examination questions

-~-------~~~---~--~~

- -- - -

--

-

TABLE OF CONTENTS Preface .. ........................................................ ................................. vii Dedication .................................................................................... viii

CHAPTER 1 Properties of Fluid ..................... ............................................ :....... 1 Types of Fluid ........................................... .......................................... 1

To my mother Iluminada, my wife Imelda, and our Children Kim Deunice, Ken Dainiel, and Kq.rla Denise

Mass Density .............................................. ........................................ 2 SpL'Cific Volume ................................................................................. 3 Unit Weight or Specific Weight ...................................................... 3 Specific Gravity ................................................................................. 4 Viscosity ............................................................................, ................ 4 Kinentatic Viscosity ..................................................................... 5 Surface Tension ....................................................~ ............................ 6 Capillarity .......................................................................................... 7 Compressibility ..................................................... ............................ 8 Pressure Disturbances ...................................................................... 9 Property Changes in Ideal Gas ...................._. .................................. 9 \ 'upor Pressure ................................................................................ 10 SOLVFD PROBLEMS .......................................................... 11to23 SUPPLEMJ:NTARY PROBLEMS ...................................... 24 to 26

CHAPTER 2 Principles of Hydrostatics .......................................................... 27 ·unit Pressure ................... ................................................................ 27 Pascal's La \V . ................. ................................ .... ................ .... ........... 27 Absolute and Gage Pressures .................... .................................... 29 Variations in Pressure.. .. .............. ............................................ 31 Pressure below La> ers of Different Liquids .... .. ..................... 32 Pressure Hea9 .. ............ ............... ............. ........................ 33 Manometerc:; ............ ... ........ . .: ............. .. .............................. 34 SOLVFn PROBI l Ms...... .. ........................................35 to 68 "UPPL'EMENT ARY PROBLEMS ................................... 69 to 72

-

-- - - - - - - -- - - - -- ----

-

II

-

-

TABLE OF CONTENTS

~

r ABLE OF CONTENTS

111

CHAPTER 3

HAPTER 5

Total Hydrostatic Force on Surfaces ......................................... 73

11mlamentals of Fluid Flow ................... ................................. 241

Total Hydrostatic Force on Plane Surface ................................... 73 Properties of Common Geometric Shapes .............................. 76 Total Hydrostatic Force on Curved Surface ..: ............................. 78 Dams ............. :............ :...................................................................... 81 Types of DanlS ............................................................................ 81 Analysis of Gravity Dams ......................................................... 84 Buoyancy ................................... ....................................................... 88 Archimedes' Principles ............................................................. 88 Statical Stability of Floating Bodies .............................................. 90 Stress on Thin-Walled Pressure Vessels ...................................... 96 Cylindrical Tank ......................................................................... 96 Spherical Shell ............................................................................ 98 Wood Stave Pipes ....................................................................... 98 SOLVED PROBLEMS ........................................................ 99to195 SUPPLEMENTARY PROBLEMS .................................. 196 to 200

CHAPTER 4 Relative Equilibrium of Liquids ............. ............................ .'.... 201 Rectilinear Translation ................................................................. 201 Horizontal Motion ................................................................... 201 Inclined Motion ........................................................................ 202 Vertical Motion ................... ..................................................... 203 Rotation .......................................................................................... 203 Volume of Paraboloid ............................... ....................... ........ 205 Liquid Surface Conditions ...................................................... 206 SOLVED PROBLEMS ...................................................... 210 to 240

I)ischarge ......................................................................··················· 241

Definition of Terms ....................................................................... 241 I~nergy and Head .......................................................................... 244 Power and Efficiency .................................................... :............... 245 Bernoulli's Energy Theorem ........................................................ 246 Energy and Hydraulic Grade Lines ........................................... 248 SOLVED PROBLEMS ...................................................... 250 to 273 SUPPLEMENT ARY PROBLEMS .................................. 274 to 276

CHAPTER 6 I l11id Flow Measurement ........ ..... ............................................ 277

. c oeff"ic1en . ts ...................................................... ·.. ·············· 277 D ev1ce . · M . D evn::es ·. ....... ......................................... . 2/l) H ead 1ost m easurmg

Orifice ..................................................................................... ········ 281 c ond"1tions . . Values of H f or Vanous ...................................... -'18'.1Contraction of the Jet ............................................................... 284 Orifice under Low Heads ......................................................... 285 Venturi Meter ...................................................................... ·· ···· ·· ·· 285 Nozzle ............................................................................................. 287 Pi tot Tube ............................................................................. ·· ···· ·· ·· 288 Gates ............................................................................................... 290 Tubes ................................................................................................ 291 Unsteady Flow (Orifice) ............................................................... 294 Weir ................................................................................................. 297 Classification of Weirs ............................................................. 297 Rectangular Weir ......... ,............................................................ 298 Contracted Rectangular Weirs ...... .......... :......................... 301 Triangular Weirs ...................................................................... 301 Trapezoidal Weirs .................................................................... 304 Cipolletti Weir ..................................................................... 30-l Suttro Weir ................................................................................ 305

JV

TABLE OF CONTENTS

Submerged Weir ....................................................................... 305 Unsteady Flow .......... ,............................................................... 306 SOLVED PROBLEMS ...................................................... 307 to 371 SUPPLEMENT ARY PROBLEMS ................................... 372 to 374

CHAPTER 7 Fluid Flow in Pipes.'. .'. ...:··················· ········································· 375 Definitions .......................................................................... ;........... 375 Reynolds Number ......................................................................... 376 Velocity Distribution in Pipes ..................................................... 377 Shearing Stress in Pipes ............................................................... 379 Head Losses in Pipe Flow ............................................................ 381 Major Head Loss ...................................................................... 381 Darcy-Weisbach Formula................................................... 381 Value of/.......................................................................... 382 Moody Diagram .............................................................. 384 Manning Formula ............................................................... 385 Haz'e n Williams Formula ................................................... 386 Minor Head Loss ...................................................................... 387 Sudden Enlargement .......................................................... 388 Gradual Enlargement ......................................................... 388 Sudden Contraction ............................................................ 388 Bends and Standard Fittings.............................................. 390 Pipe Discharging from Reservoir ............................................... 390 Pipe Connecting Two Reservoirs ................................................ 391 Pipes in Series and Parallel.. .............. ., ......................................... 392 Equivalent Pipe ............................................................................. '394 Reservoir Problems ....................................................................... 394 Pipe Networks ............................................................................... 398 SOLVED PROBLEMS ..................................................... .400 to 476 SUPPLEMENTARY PROBLEMS ..................................477 to 480

f ABLE OF CONTENTS

v

HAPTER 8

1 48 pen Channe1 . .... .. .... .. .. .. ... ... . .. ... . ..... ................. ........................ , 481 Specific Energy ........................ .............................. ..........·············· 482 Chezy Formu1a ····· .....:..... ... ....... ... ... .. ... .. .. ... .... ... .. ......................... 483 Kutter and Gungmllet Formula ........................................... .. . F ormula .................................... .. .............................. 483 M anrung . FormuIa ... ··· .. ··· ......... ·· ............. ·......... .............................. 483 Bazm 484 . ............. ·· .... ·· ·· ·... ··· ···· ... ··· ·· ····· .. ..................... . Powell E quat ion 85 · . Flow ........................... ._............ ........................................ 4485 Uruform Boundary Shear Stress ................................................................. : . 486 NormaIDep th ...................................... ....................................... · 486 Most Efficient Sections ................ ··· ······ .. ··...... ·· ·· ·· ···· .. ··· ·· ··· ···· ·· .. ·· Proportions for Most Efficient Sections .. ····..... ············.. ······· 4 ~; RectanguIar Secti.o n ................... ·.. ··.. ··········......................... 4 89 T rapezo1'da ISec ti'on ..................................... ......................... 437 4 s ti' .......................... . 1 Triangu ar ec on ..................... ·· .. ············ . Iar Sec ti'ens........................................ ............................... 490 Cuen Velocity Distribution in Open Channel ..................................... 491 . of Flow ................................... ........................... 491 Alternate Stages 92 Frou d e N umb er.............................................. .......................... 492 · · ID ep th ...... ...... ... ....... .. ... ..... ..... .... .. ................................. 4495 Cntica Non-Uniform or Varied Flow ..................................................... I Iydraulic Jump ............................................................................. 497 500 Flow around Channel Bends ....··· ... ·......................... ·· ... ··· .... ·· .... · SOLVED PROBLEMS ..................................................... .501to547 SUPPLEMENT ARY PROBLEMs ................................. .547 to 550 L

vi

TABLE OF CONTENTS

CHAPTER9 Hydrodynamics ..............................................................~ .......... 551 Force against Fixed Flat Plates .................................................... 551 Force against Fixed Curved Vanes ............................................. 553 Force against Moving Vanes ....................................................... 554 Work Done on Moving Vanes ................................................ 555 Force Developed on Closed Conduit ......................................... 556 Drag and Lift. ........................................................'......................... 557 Terminal Velocity ....................................... :............................. 559 Water Hammer .............................................................................. 560 SOLVED PROBLEMS ...................................................... 563 to 597 SUPPLEMENTARY PROBLEMS .................................. 597 to 598

APPENDIX Properties of Fluids and Conversion Factors ........................ 599 Table A - 1: Viscosity and Density of Water at 1 ahn .............. 599 Table A - 2: Viscosity and Density of Air at 1 ahn ................... 600 Tabl~ A - 3: Properties of Common Liquids at 1 ahn & 20°C .. 601 Table A - 4: Properties of Common Gases at 1 ahn &20°C ..... 601 Table A - 5: Surface Tension, Vapor Pressure, and Sound Speed of Water ........................................... 602 Table A - 6: Properties of Standard Ahnosphere ..................... 603 Table A - 7: Conversism Factors from BG to SI Units .............. 604 Table A - 8: Other Conversion Factors ...................................... 605

INDEX I- IV

FLUID MECHANICS & HYDRAULICS

CHAPTER ONE Properties of Fluids

1

Chapter 1 Properties of Fluids FLUID MECHANICS & HYDRAULICS

Fluid Mechanics is a physical science dealing with the action of fluids at rest or in motion, and with applications and devices in engineering using fluids. Fluid mechanics can be subdivided into two major areas, fluid statics, which -deals with fluids at rest, and fluid dynamics, concerned with fluids in motion. The term lzydrodynamics is applied to the flow of liquids or to low-velocity gas flows where the gas can be considered as being essentially incompressible. Hydraulics deals with the application of fluid mechanics to engineering devices involving liquids, usually water or oil. Hydraulics deals with such problems as the flow of fluids through pipes or in open channels, the design of storage dams, pumps, and water turbines, and with other devices for the control or use of liquids, such as nozzles, valves, jets, and flowmeters.

TYPES OF FLUID

Fluids are generally divided into two categories: ideal fluids and real fluids. Ideal fluids . •Assumed to have no viscosity (and hence, no resistance to shear) • Incompressible •Have uniform velocity when flowing • No friction between moving layers of fluid • No eddy currents or turbulence Real fluids • Exhibit infinite viscosities •Non-uniform velocity distribution when flowing • Compressible • Experience friction and turbulence in flow

CHAPTER ONE Properties of Fluids

2

FLUID MECHANICS & HYDRAULICS

Real fluids are further divided into Newtonia1t fluids and 11011-Newtonian fluids. Most fluid problems assul11€ real fluids with Newtonian characteristics for convenience. This assumption is appropriate for water, air, gases, steam, and o lher simple fluids like alcohol, gasoline, acid solutions, etc. However, slurries, pastes, gels, suspensions may not behave according to simple fluid relationships. Fluids

Ideal Fluids

CHAPTER ONE Properties of Fluids

FLUID MECHANl(:S · & HYDRAULICS

where:

3

p =absolute pressure of gas in Pa R =gas constant Joule/ kg-°K For air: R = 287 J/kg - °K R = l,716 lb-ft/slug-0 R T = absolute temperature in °Kelvin °K = °C + 273 0 R = °F + 460 Table 1 - 1: Approximate Room-Temperature Densities of Common Fluids

pin kg/m3

Fluid

Real Fluids

Newtonian Fluids

Non-Newtonian Fluids

IPseudoplastic Fluids ~

Delatant Fluids

I

-

Bingham Fluids

1.29 1.20 790 602 720 1,260 13,600 1,000

AirJSTP) AirJ..21°F, a ltml Alcohol Ammonia Gasoline G!Y_cerin Mercl.!.!Y_ Water

Figure 1 - 1: Types of fluid

SPECIFIC VOLUME, Vs Specific volume, V,, is the volume occupied by a unit mass of fluid.

MASS DENSITY, p (RHO) The density of a fluid is its mass per unit of volume. p=

Units: English Metric SI

slugs/ft3 gram/cm3 kg/m3

m~ss of fluid,

M

volume, V

Note:

Pslugs

=

1 V,= -

Eq. 1-1

Plbm/ g

UNIT WEIGHT OR SPECIFIC WEIGHT, y Specific weight or unit weight, y, is the weight of a unit volume of a fluid.

f
y=

weightoffluid, W volume, V

y=pg p=L

RT

Eq. 1-3

p

Eq. 1- 2

Eq. 1- 4 Eq. 1- 5

4

CHAPTER ONE Properties of Fluids

Units: English Metric SI

FLUID MECHANICS & HYDRAULICS

CHAPTER ONE Properties of Fluids

FLUID MECHANICS & HYDRAULICS

the upper plate will adhere to it CU}d will move with the same velocity U while the fluid irr contact with the fixed plate will have a zero velocity. For small values of U and y, the velocity gradient can be assumed to be a straight line and F varies as A, U and y as:

lb/ ft3 dyne/cm3 N/m3 or kN/m3

AU

F

Foc - - or y A SPECIFIC GRAVITY Specific gravity, s, is a dimensionless ratio of a fluid's density to some standard reference density. For liquids and solids, the reference density is water at 4° C (39.2° F).

but

s=--

Eq. 1- 6

Pwater

In gases, the standard reference to calculate the specific gravity is the density of air. Pgas

s=--

u y

U

oc -

y

dV dy

(from the figure)

!_ =Shearin g stress, t A . t

P!iqui
I

oc

d V or -r = k dV dy dy

where the constant of proportionality k is called the d ynamic of absolute viscosity denoted asµ. dV t=µdy

Eq. 1- 7

µ=

Pair

For water at 4°C: y = 62.4 lb/ft3 = 9.81 kN/m3 p = l.94slugs/ft3=1000 kg/m3 s = 1.0

Area =-A

'

dV /dy

Eq. 1- 8

where: ·( = shear stress in lb/ ft2 or Pa µ=absolute viscosity in lb sec/ft2 (poises) or Pa-sec. y =distance between the plates in ft or m LT = velocity in ft/ s or m/ s

VISCOSITY,µ (MU) The property of a fluid which determines the amount of its resistance to shearing forces. A perfect fluid would have no viscosity.

Consider lwo large, parallel plates at a small distance y .apart, the space between them being filled with a fluid. Consider the upper plate to be subject to a force F so as to move with a constant velocity U. The fluid in contact with

5

u

KINEMATIC VISCOSITY v (NU)

Kinematic viscosity is the ratio of the dynamic viscosity of the fluid, µ, to its mass density, p.

F

--------~

moving plate

v = .!:'.. p

y

fixed plate

where: µ=absolute viscosity in Pa-sec. p =density in kg/m3 ·

Eq. 1- 9

~ -~~------

6

-

CHAPTER ONE

FLUID MECHANICS & HYDRAULICS

Prope rtie s of Fluids

English Metric S.l.

Absolu te, µ lb-sec/ ft2 J!'lu_g/ ft-se<:l_ dyne-s/cm 2 iE_oistl Pa-s _(.N-s/ m:2_

Note: 1 poise= 1 dyne·S/ cm 2 = 0.1 Pa-sec 1 stoke= 0.0001 m2/s

CHAPTERONJ;:

FLUID MECHANICS & HYDRAULICS

Properties of Fluids

Kinem a tic, v

d

·1

ft2/ sec cm2/s J!'tokaj_ m2/s

(1 d yne= 10-s N)

(b) Cohesion > adhesion

(a) Adhesion > cohesion

SURFACE T ENSION cr (SIGMA)

fhe membrane of "skin" that seems to form on the free surface of a fluid is du e to the intermolecular cohesive forces, and is known as surface tension. SurfaCl' tension is the r eason that insects are able to sit on water and a needle is ,1ble to float on it. Surface tension also causes bubbles and droplets to take on a spherical shape, since any o ther shape would have more surface area per unit volume.

Capillarity (Capillary action) is the name given to the behavior of the liquid in a thin-bore tube. The rise or fall or a fluid in a capillary tube is caused by surface tension and depends on the relative magnitudes of the cohesion of the liquid and the adhesion of the liquid to the walls of the containing vessel. Liquids rise in tubes they wet (adhesion> cohesion) and fall in tubes they do not wet (cohesion > adhesion). Capillary is important when using tubes smaller than about 3/8 inch (9.5 mm) in diameter. 4crcose h=--yd

Pressure inside a Droplet of Liquid:

4cr

· I~------------p_=_-_d_ __ where: cr =surface tension in N/ m d = diameter of the d roplet in m p = gage pressure in Pa

7

Capillarity

Table 1 - 2: Common Units of Viscosity

gyslem

-

_ _ _ _ _ _E_q_ . 1_-_1_0--'

Eq. 1-11

For complete wetting, as with water on clean glass, the angle e is 0° . Hence the formula becomes 4cr h=yd where: IL = capillary rise or depression in m y =unit weight in N/m3 d = diameter of the tube in m cr = surface tension in Pa

Eq. 1-12

., CHAPTER ONE Properties of Fluids

8

FLUID MECHANICS & HYDRAULICS

Table 1 - 3: Contact Angles, 0

Materials

Es = stress = 6p strain D. V

Angle, 0

mercury-glass water-paraffin water-silver kerosene-glass g.!Y_cerin:-glass water-glass eth_}'l alcohol-glass· ·

140° 107° 90° 26° 19°

oo oo

CHAPTER ONE Properties of Fluids

I UID MECHANICS ' HYDRAULICS

1

9 Eq. 1-15

v dp orEB=---dV /V

Eq. 1-16

'

PRESSURE DISTURBANCES I '11•ssure disturbances imposed on a fluid move in waves. The velocity or COMPRESSIBIUTY,

.•lerity of pressure wave (also known as acoustical or sonic velocity) is pressed as:

J3

Cr)111pressibility (also known as the coefficie11I of co1111iressibi/1t11) is th<' fractional chc:1nge in the volume of a fluid per unit changl' in pres~ure in ,, constantcmperature process.

c=lf =frh

Eq. 1-17

L\V

J3= _v_ D.p

dV /V orP=--dp

Eq. 1 -13 PROPERTY CHANGES IN IDEAL GAS

For any ideal gas experiencing any process, the equation of state is given by: Eq. 1 - 14 Eq. 1-18

where: D. V = change in volume V =original volume

6p •change in pressure dV/ V =change in volume (usually in perc:l·nt)

When temperature is held constant, Eq. 1 - 18 reduces to (Boyle's Law) Eq. 1-19 When temperature is held constant (isothermal condition), Eq. 1 - 18 reduces to (Charle's Law)

BULK MODULUS OF ELASTICITY, E8

The bulk modulus of elasticity of the fluid expr-..">.,C'S the rnmpr1.'Ssibilitv of the· fluid. It is the ratio of the change m unit pres~ure to the w rrespondmg volume change per unit of volume.

Eq. 1-20

!

...-

--

10

-

CHAPTER ONE Properties of Fluids

FLUID MECHANICS & HYDRAULICS

-

or

(~)k = E1.. =Constant V2

and

Eq. 1- 21

Fluid

Eq. 1- 22

turp_entine water

mere~

P1

~~ = (:~

rk

11

Table 1 - 4: Typical Vapor Pressures

For Adiabatic or Isentropic Conditions (no heat exchanged)

pi Vik= p2 V2k

CHAPTER ONE Properties of Fluids

FLUID MECHANICS & HYDRAULICS

et~alcohol

k-1

ether butane Freon-12 __E!O_r_ane ammonia

Eq. 1- 23

where: p1 = initial absolute pressure of gas p2 = final absolute pressure of gas Vi = initial volume of gas V2 = final volume of gas r, =initial absolute temperature of gas in °K (°K = °C + 273) T2 = final absolute temperature of gas in °K k = ratio of the specific heat at constant pressure to the specific heat at constant volume. Also known as adiabatic exponent.

VAPOR PRESSURE

Molecular activity in a liquid will allow some of the molecules to escape the liquid surface. Molecules of the vapor also condense back into the liquid. The vaporization and condensation at constant temperature are equilibrium processes. The equilibrium pressure exerted by these free molecules is known as the vapor pressure or saturation pressure.

!solved Pro~lems Problem 1-1 A reservoir of glycerin has a mass of 1,200 kg and a volume of 0.952 cu. m. Find its (a) weight, W, (b) unit weight, y, (c) mass density, p, and (d) specific gravity (s).

Solution (a) Weight, W =Mg = (1,200)(9.81) Weight, W= 11,772N or11.772 kN I (b)

Some liquids, such as propane, butane, ammonia, and Freon, have significant vapor pressure at normal temperatures. Liquids near their boiling point or that vaporizes easily are said to volatile liquids. Other liquids such as mercury, have insignificant vapor pressures at the same temperature. Liquids with low vapor pressure are used in accurate barometers. The tendency toward vaporization is dependent on the temperature of the liquid. Boiling occurs when the liquid temperature is increased to .the point that the vapor pressure is equal to the local ambient (surrounding) pressure. Thus, a liquid's boiling temperature depends on the local ambient pressure, as well as the liquid's tendency to vaporize.

kPa, 20°C 0.000173 0.0534 2.34 5.86 58.9 218 584 855 888

. h t, y = -w . Umtwe1g

v

11.772 0.952 Unit weight, y = 12.366 kN/m3

(c)

. M Dens1ty, p=V . 1200 Density, p = 0. 952 Density, p =1,260.5 kw'm3

-

12 (d)

CHAPTER ONE Properties of Fluids

FLUID MECHANICS & HYDRAULICS

. s= Pgty S pec1'fic gravity, Pwater 'fj . 1,260.5 SpeCI c gravity, s = - - 1,000 Specific gravity, s = 1.26

Problem 1- 2 The specific gravity of certain oil is 0.82. Calculate its (a) specific weight, in lb/ft3 and kN/m3, and (b) mass density in slugs/ft3 and kg/m3 .

CHAPTER ONE Properties of Fluids

13

Solution (a)

W = mg = 22(9.75) W=214.5N

(b)

Since the mass of an object is absolute, its mass will still be 22 kg.

Problem 1- 5 What is the weight of a 45-kg boulder if it is brought to a place where the acceleration due to gravity is 395 m/s per minute? Solution

Solution (a)

FLUID MECHANICS & HYDRAULICS

Specific weight, y = Ywater x s Specific weight, y = 62.4 x 0.82 = 51.168 lb/fP Specific weight, y = 9.81 x 0.82 = 8.044 kN/m3

(b) ' Density, p = Pwater x S Density, p = 1.94 x 0.82

= 1.59 slugs/fP Density, p = 1000 x 0.82 = 820 kglmJ

W=Mg

_ m/s lmin g- 395 - - x - min 60sec

g = 6.583 m/ s2

w = 45(6.583) W= 296.25 N

A liter of water weighs about 9.75 N. Compute its mass in kilograms.

Problem 1- 6 If the specific volume of a certain gas is 0.7848 m 3 /kg, what is its specific weight?

Solution

Solution

Problem 1- 3

Mass= W g 9.75 M ass= - 9.81 Mass= 0.994 kg

Problem 1- 4 If an object has a mass of 22 kg at sea level, (a) what will be its weight at a · point where the acceleration due to gravity g = 9.75 m/s2? (b) What will be its mass at that point?

V,= -1

p

p= -

1

vs

1.

= --

0.7848

p = 1.2742 kg/m3

Specific weight, y = p x g = 1.2742 x 9.81 Specific weight, y = 12.5 N/m3

CHAPTER ONE Properties of Fluids

14

FLUID MECHANICS & HYDRAULICS

CHAPTER ONE Properties of Fluids

FLUID MECHANICS & HYDRAULICS

15

Solution

Problem 1- 7

What is the specific weight of air at 480 kPa absolute and 21°C?

Density, p = .l

g

Solution

13.7 9.81 = 1.397 kg/m3

y=p xg

P-

p where R = 287 J/kg-°K RT 480 x 10 3 287(21 + 273) p = 5.689 kg

Density p = .1!._

'

RT

3

1.397 = (205+101 .325) x 10 R(32 + 273) Gas constant, R = 718.87 J/kg - °K

Note: Potm - l 01.325 kPa

y = 5.689 x 9.81

y = 55.81 N/m3 Problem 1 - 10 Problem 1- 8

\ ir is kept at a pressure of 200 kPa absolute and a tempt.>r~t11rP nt ~0°C in .~ 500-liter container. What is the mass of air?

Fin? the mass density of helium at a temperature of 4 °C and a pressure of 184 kPa gage, if atmospheric pressure is 101.92 kPa. (R = 2079 J/kg • °K)

Solution p

Solution

Density p =

'

RT

200x 10 3 287(30 + 273) p = 2.3 kg/m3

LRT

p = pg,1ge + p .1tm = 184 + 101.92 p = 285.92 kPa

Mass= p x V =23x 2illL . 1000

T = 4 + 273 = 277°K D

= .1!._

Mass= 1.15 kg

285.92x 10 3 . ensity, p = 2,079(277)

Density, p = 0. 4965 kglm3

Problem 1 - 11

Problem 1- 9

\t 32°C and 205 kPa gage, the specific weight of a certain gas was 13.7 N/m3 • I >etermine the gas constant of this gas.

.

'

A cylindrical tank 80 cm in diameter and 90 cm high is filled with a liquid The tank and the liquid weighed 420 kg. The weight of the empty tank is 40 kg. What is the unit weight of the liquid in kN/ m3.

CHAPTER ONE

16

Properties of Fluids

FLUID MECHANICS & HYDRAULICS

CHAPTER ONE

I UID MECHANICS HYDRAULICS

Properties of Fluids

17

.olution

Solution M

p=-

.

v

P=

420 - 40 = 840 k /m3 f(0.8[2 (0.90) g

.

I 1

dP

EB= - - dV /V

dp

p2 - p1 p1=0 dp = p2

y=pg

= 840(9.81) = 8240.4 N/m3

=

dV = V2 - V1 dV = -0.6% V = -0.006V

y = 8.24 kN/m3

'Problem 1 - 12

Pi 22 £B- 0.006V /V = .

A lead cube has a total mass of 80 kg. What is the length of its side? Sp. gr. of lead = 11.3.

pi= 13.2MPa

p2 = 0.0132 GPa

Solution Let L be the length of side of the cube:

M=p V 80 = (1000 x 11.3) v L = 0.192 m = 19.2 cm

•1roblem 1 - 15

\l,tter in a hydraulic press, i)tltially at 137 kPa absolute, is subjected to a .•rcssure of 116,280 kPa absolute. Using E8 = 2.5 GPa, determine the lll'rcentage decrease in the volume of water.

'iolution Problem 1 - 13 A liquid compressed in a container has a volume of 1 liter at a pressure of 1 MPa and a volume of 0.995 liter at a pressure of 2 MPa. The bulk modulus of elasticity (EB) of the liquid is:

Solution EB= __ d_P_ =---2_-_1__ dV /V (0.995-1)/1

EB= __:!E__ dV /V

2.5 x 109 = - (116,280-137) x 10 dV /V

3

dV = -0.04'65

v V = 4.65°/c, decrease dV

EB= 200 MPa

Problem 1 - 16 Problem 1 - 14

What pressure is required to reduce the volume of water by 0.6 percent? Bulk modulus of elasticity of water, EB= 2.2 ~Pa.

If 9 m 3 of an ideal gas at 24 °C and 150 kPaa is compressed to 2 m3 , (a) what 1s the resulting pressure assuming isothermal conditions. (b) What would have been the pressure and temperature if the process is isentropic. Use k = 1.3

.·

CHAPTER ONE Properties of Fluids

18

FLUID MECHANICS & HYDRAULICS

Two large plane surfaces are 25 mm apart and the space between them 1s fill~d with a liquid of viscosity µ = 0.958 Pa-s. Assuming the velocity gradient to ht> a straight line, what force is required to pull a very thin plate qf 0.37 m 2 area at a constant speed of 0.3 m/ s if the plate is 8.4 mm from one of the surfaces?

For isothermal condition: V, = p2 V2

p1

150(9) = p2 (2) pi = 675 kPa abs

(b)

19

Problem 1 - 18

Solution (a)

CHAPTER ONE Properties of Fluids

FLUID MECHANICS

& HYDRAULICS

Solution

For isentropic process: p1 Vik= pi Vik

F = F, +Fi

150(9) 1.3 = p 2 (2) L3 pi = 1,060 kPa abs

2 (El_)(k-1)/

T

=

r,

u /y F/A µ=-u /y

k

p, T2

-~\

T

~l=--

16.6

I I

:

~mm

I

\ \

\

F1 \

.....

8.4

F= µUA =

y

( 1,060 )(1.3-l)/1.3

24 + 273 150 = 466.4°K or 193.4°C

T2

F1 =

0.958(0.3)(0.37) 0.0166 = 6.4 N

Fi=

0.958(0.3)(0.37) O.OOS4 = 12.66 N

Problem 1 - 17

If the viscosity of water at 70 °C is 0.00402 poise and its specific gravity is 0.978 determine its absolute viscosity in Pa - s and its kinematic viscosity in mi/s and in stokes.

,

µ

poise

0.1 Pa-s

x ----

lpoise

= 0.000402 Pa - s

Kinematic viscosity: µ

0.000402 (1000 x 0.978)

v= -= - - - - -

p

v = 4.11 x 10-7 m 2/s

lstoke

v = 4.11 x10·7 mi/s x - - - - 0.0001 m 2 /s v = 4.11 x 10-3

F=19.06N

Problem 1 • 19

Solution Absolute viscosity: µ = 0.00402

F = 6.4 t 12.66

stoke

•

•

A cylinder of 125 mm radius rotates concentrically inside a fixed cylinder of 130 mm radius. Both cylinders are 300 mm long. Determine the viscosity of the liquid which fills the space between the cylinders if a torque of 0.88 N-m is

required to maintain an angular velocity of 2n radians/sec velocity gradient to be a straight line

Assume thE'

CHAPTER ONE Properties of Fluids

20

FLUID MECHANICS & HYDRAULICS

Solution fixed

1

~y=0.005m D.._

µ=--

u /y

U=rw LI= 0.125(2tt) U = 0.785 m/s

u~5

F=tA

,

fixed cylinder

Torque= F(0.125) Torque= -rA (0.125)

,,

0.88 = 1 [2rr(0.125)(0.3)] (0.125) 1 = 29.88 Pa

L

e-

IF,

u

= 1 A = ~1-A I y

= 0.3 m

+ f- liquid

I

0.005

21

176.58 sin 15° = 0.0814~ (0.3) 0.003 LI= 5.614 m/s (11 = 5.614 m/s

i I I I I I

CHAPTER ONE Properties of Fluids

fLF, = O] Wsin (:) - Fs = 0 Fs = Wsin (:) F, = 176.58 sin 15°

rotating cylinder

y = 0.005 m

29.88 µ = 0.785/0.005 µ = 0.19 Pa-s

FLUID MECHANICS & HYDRAULICS

0.125

0.13 m

Problem 1 - 21

l!stimate the height to which water will rise in a capillary tube of diameter 3 ll\m. Use cr = 0.0728 N/m and y = 9810 N/m3 for water

·01ution Problem 1 - 20

An 18-kg slab slides down a 15° inclined plane on a 3-mm-thick film of oil with viscosity µ = 0.0814 Pa-sec. If the contact area is 0.3 m2, find the terminal velocity of the slab. Neglect air resistance.

Solution W = 18(9.81) = 176.58 N

Note: (:) = 90° for water in clean tube 4 Capillary rise, h = cr yd

4(0.0728) Capillary rise, h = --'-----'9810(0.003) Capillary rise, Ii = 0.0099 m = 9.9 mm

.. y I I

Problem 1 - 22

I I

v=

0.003 m

.........

plane

Terminal velocity is attained when the s um of all forces in the direction of motion is zero.

I stimate the capillary depression for mercury in a glass capillary tube 2 mm in .l1ameter. Use cr = 0.514 N/m and(:)= 140°

'•olution 4(0.514)(cos 140°) 4cr cos El Capillary rise, Ji = - - yd (9810 x 13.6)(0.002) Capillary rise, '11 = -0.0059 m (the negative sign indicates capillary depression) Capillary depression, '1 = 5.9 mm

CHAPTER ONE Properties of Fluids

22

FLUID MECHANICS & , HYDRAULICS

CHAPTER ONE Properties vf Fluids

11 UID MECHANICS ~ HYDRAULICS

23

Problem 1 - 23

I 1oblem 1- 26

What is the value of the surface tension of a small drop of water 0.3 mm in diameter which is in contact with air if the pressure within the droplet is 561 Pa?

'lOnar transmitter operates at 2 impulses per second. If the device is held tl' of fresh water (£8 = 2.04 x 1Q9 Pa) and the echo i'> received midwa\ I ·tween impulses, how deep is the water1 1111• surface

Solution

olution 4o

p=-

I

he velocity of the pressure wave (sound wave) is

d·

561=~

0.0003 cr = 0.042 N/m

c=

ff

('=

..----2. 04x10 9 = 1,428 m/ s ,____

Sonar transmitter

1000 Problem 1 - 24

An atomizer forms water droplets 45 µm in diameter. Determine the excess pressure within these droplets using cr = 0.0712 N/m.

Solution

•11\Ce the echo is received 111dway between impulses, then •he total time of travel of sound. ¥2(0.5) = 114 sec and the total listance covered is 2h, then;

~Sound wave

~

0

9,

h

0

0

1

0

0

0

Bottom

4cr p= d 4 - (0.07l 2 ) = 6,329 Pa p - 45x 10-6

211 =

cI

2lt = 1,428(114) /1=178.5 m

Problem 1 - 27 Problem 1 - 25

Distilled water stands in a glass tube of 9 mm diameter at a height of 24 mm. What is the true static height? Use cr = 0.0742 N/m.

'\t what pressure will 80 °C water boil? (Vapor pressure of water at 80°C = 47 4 kPai

olution Solution h

=

Water will boil if the atmospheric pressure equals che vapor 4crcose yd where e = 0° for water in glass tube

h=

4(0.0742 ) 0 00336 33 9810(0.009) = · m = · 6 mm

True static height = 24 - 3.36 True static height = 20.64 mm

I

fherefore water at 80 °C will boil at 47.4 kP;;

~· ~sst.rt-

24

CHAPTER ONE

Properties of Fluids

FLUID MECHANICS & HYDRAULICS

!supplementary Problems Problem 1 - 28

What would be the weight of 1 3-kg mass on a planet where the acceleration due to gravity is 10 m/s27 Ans: 30 N

FLUID MECHANICS & HYDRAULICS

CHAPTER ONE

Properties of Fluids

25

Problem 1 - 33

(a) If 12 m3 of nitrogen at 30°C and 125 kPa abs is permitted to expand isothermally to 30 m3, what ~s the resulting pressure? (b) What would the pressure and temperature have been if the process had been isentropic? Ans: (a) 50 kPa abs (b) 34.7 kPa abs; -63°C

Problem 1 - 29

Problem 1 ... 3:4

A vertical cylindrical tank with a diameter of 12 m and a depth of 4 mis filled with water to the top with water at 20°C. If the water. is heated to 50°C, how much water will spill over? Unit weight o~ water at 20°C and 50°C is 9.79 kN/m3 and 9.69 kN/m3, r.espectively. Ans: 4.7m 3

A square block weighing 1.1 kN and 250 mm on an edge slides down an incline on a film of oil 6.0 µm thick. Assuming a linear velocity profile in the oil and neglecting air resistance, what is the terminal velocity of the block? fhe viscosity of oil is 7 mPa-s. Angle of inclination is 20°. Ans: 5.16 m / s

Problem 1 - 30

Problem 1 - 35

A rigid steel container is partially filled with a liquid at 15 atm. The volume of the liquid is 1.23200 L. At a pressure of 30 atm, the volume of the liquid is 1.23100 L. Find the average bulk modulus of elasticity of the liquid over the given range of pressure if the temperature after compression is ·allowed to return to its initial value. What is the coefficient of compressibility? Ans: ER= 1.872 GPa; P = 0.534 GPa·1

Benzene at 20°C has a viscosity of 0.000651 Pa-s. What shear stress is required lo deform this fluid at a strain rate of 4900 s·1? Ans: r = 3.19 Pa

Problem 1 - 31

Calculate the density of water vapor at 350 kPa abs and 20°C if its gas constant is 0.462 kPa-m3/kg- 0 K. Ans: 2.59 kg/ m3

Problem 1 - 36

A shaft 70 mm in diameter is being pushed at a speed of 400 mm/ s through a bearing sleeve 70.2 mm in diameter and 250 mm long. The clearance, assumed uniform, is filled with oil at 20°C with v = 0.005 m 2 /s and sp. gr. = 0.9. Find the force exerted by the oil in the shaft. · Ans: 987 N

Problem 1 - 37 Problem 1 - 32

Air is kept at a pressure of 200 kPa and a temperature of 30°C in a 500-L c0ntainer. What is the mass of the air? Ans: 1.15 kg

Two clean parallel glass plates, separated by a distance d = 1.5 mm, are dipped in a bath of water. How far does the water rise due to capillary action, if cr = 0.0730 N/m? Ans: 9.94 mm

26

CHAPTER ONE

Properties of Fluids

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO

FLUID MECHANICS & HYDRAULICS

Principles of Hydrostatics

27

Problem 1 - 38

hnd the angle the surface tension film leaves the glass for a vertical tube immersed in water if t~e diameter is 0.25 inch and the capillary rise is 0.08 inch. Use cr = 0.005 lb/ ft. Ans: 64.3°

Chapter 2 Principles of Hydrostatics

Problem 1 - 39 I

What force is required to lift a th(n wire ring 6 cm in diameter from a water surface at 20°C? (cr of water at 20°C = 0.0728 N/m). Neglect the weight of the ring. Ans: 0.0274 N

UNIT PRESSURE OR PRESSURE, p

Pressure is the force per unit area exerted by a liquid or gas on a body or surface, with the force acting at right angles to the surface uniformly in all directions.

p=

Force,F Area, A

Eq. 2-1

In the English system, pressure is usually measured in pounds per square inch (psi); in international usage, in kilograms per square centimeters (kg/ cm2), or in atmospheres; and in the international metric system (SI), in Newtons per square meter (Pascal). The unit atmosphere (atm) is defined as a pressure of 1.03323 kg/ cm2 (14.696 lb/in2), which, in terms of the conventional mercury baron;i.eter, corresponds to 760 mm (29.921 in) of mercury. The unit kilopascal (kPa) is defined as a pressure of 0.0102 kg/ cm2 (0.145 lb/ sq in).

PASCAL'S LAW

P11sc11/'s law, developed by French mathematician Blaisr Pascal, states that the pressure on a fluid is equal in all directions and in all parts of the container. In Figure 2 - l, as liquid flows into the large container at the bottom, pressure pushes the liquid equally up into the tubes above the container. The liquid rises to the same level in all of the tubes, regardless of the shape or angle of the tube.

28

CHAPTER TWO Principles of Hydrostatics

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO Principles of Hydrostatics

I l.UID MECHANICS ·'• HYDRAULICS

29

/\KSOLUTE AND GAGE PRESSURES #! -

- - __, .,.

«1 ge Pressure (Relative Pressure)

Figure 2 - 1: Illustration of Pascal's Law

The laws of fluid mechanics are observable in many everydp.y situation~. For example, the pressure exerted by water at the bottom of a p'Jnd will be the same as the pressure exerted by water at the bottom of a much narrower pipe, provided depth remains constant. If a longer pipe filled with water is tilted so that it reaches a maximum height of 15 m, its water will exert the same pressure as the other examples (left of Figure 2 - 2). Fluids can flow up _as well as down in devices such as siphons (right of Figure 2 - 2). Hydrostatic force causes water in the siphon to flow up and over the edge until the bucket is empty or the suction is broken. A siphon is particularly useful for emptying con tainers that should not be tipped.

' ..1ge pressures are pressures above or below the atmosphere and can be 111«'asured by pressure gauges or manometers. For small pressure differences, a U1t1hc manometer is used. It consists of a U-shaped tube with one end connected to llu• container and the other open to the atmosphere. Filled with a liquid, such as '11\ler, oil, or mercury, the difference in the liquid surface levels in the two 111,mometer legs indicates the pressure difference from local atmospheric 11nditlons. For higher pressure differences, a Bourdon gauge, named after the I r~nch inventor Eugene Bourdon, is used. This consists of a hollow metal tube , 1th an oval cross section, bent in the shape of a hook. One end of the tube is losed, the other open and connected to the measUJ"ement region.

tmospheric Pressure & Vacuum

\/111ospheric Pressure is the pressure at any one point on the earth's surface from the •veight of the air above it. A vacuum is a space that has all matter removed from it. II is impossible to create a perfect vacuum in the laboratory; no matter how 1dvanced a vacuum system is, some molecules are always present in the vacuum irea. Even remote regions of outer space have a small amount of gas. A vacuum 1 ,m also be described as a region of space where the pressure is less than the normal atmospheric pressure of 760 mm (29.9 in) of mercury. Under Norrnal conditions at sea level: P•tm = 2166 lb/ft2 = 14.7 psi = 29.9 inches of mercury (hg) =760rnmHg = 101.325 kPa

Absolute Pressure Figure 2 - 2: Illustration of Pascal's Law

Absolute pressure is the pressure above absolute zero (1111c1111m) Pabs = pg.1ge

+ p.1hn

Eq. 2- 2

Note: • Absolute zero is attained if all air is removed. It is the lowest possible pressure attainable. • Absolute pressure can never be negative. • The smallest gage pressure is equal to the negative of the ambient atmospheric pressu1•'

30

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO Principles of Hydrostatics

CHAPTER TWO Principles of Hydrostatics

I I IJID MECHANICS 1IYDRAULICS

31

Vl\IUATIONS IN PRESSURE . 11·. id er any two points (1 & 2), whose difference in elevation is II, to lie m the Standard atmosphere = 101.325 abs

60 gage

-40 gage

60 abs

58.675 gage

l•• I,

of an elementary prism having a cross-sectional area a and a length of L

l1u ,. this prism is at rest, all forces acting upon it must be in equilibrium. Free liquid surface •••• S,1 ••••••••••••••

Current atmosphere = 100 abs

Absolute zero or ·100 gage

P1 & P2 are gage pressures

=·101.325 gage

All pressure units in kPa Figure 2 - 3: Relationship between absolute and gage pressures

Note: Unless otherwise specified in this book, the term pressure signifies gage pressure.

MERCURY BAROMETER A mercury barometer is an accurate and relatively simple way to measure changes in atmospheric pressure. At sea level, the weight of the atmosphere forces mercury 760 mm (29.9 in) up a calibrated glass tube. Higher elevations yield lower readings because the atmosphere is less dense there, and the thinner air exerts less pressure on the mercury.

vacuum 760 mm

Mercury

Figure 2 - 4: Forces acting on elementary prism tfol 1•: Free Liquid Surface refers to liquid surface subject to zero gage pressure or with

atmospheric pressure only. "- at Sea Level

ANEROID BAROMETER In an aneroid barometer, a partially evacuated metal drum expands or contracts in response to changes in air pressure. A series of levers and springs translates the up and down movement of the drum top into the circular motion of the pointers along the aneroid barometer's face.

Ith reference to Figure 2 - 4 : W =yV W = y (nL)

[LF:r = O] F2 - F1 = W sin 0 p2 a - pi a= y (aL) sin p2 - pi = y L sin 0

e but L sin 0 = h

Eq. 2- 3

the difference in pressure between any two points in a homogenequs f11111I 11 rest is equal to the product of the unit weight of the fluid (y) to the vertical disl11111 (/1) between the points. l'herefore;

32

CHAPTERlWO

FLUID MECHANICS & HYDRAULICS

Principles of Hydrostatics

Also:

1I

CHAPTERlWO

IJID MECHANICS I IYDRAULICS

Principles of Hydrostatics

33

11~ider the tank shown to be filled with liquids of different densities and 1th air at the top under a gage pressure of pA, the pressure at the bottom of llu t.mkis: 1

Eq. 2-4

pi =pi +wh

fhis means that nny chnnge in pressure nt point 1 would cnuse an equal change at pornl 2. Therefore; a pressure applied at any point in a liquid at rest is transmitted equally and undiminished to every other point in the liquid.

Let us assume that point 0 in Figure 2 - 4 lie on the free liquid surface, then th,e gage pressure pi is zero and Eq. 2 - 4 becomes:

Pbottom =LY IL+ P = Y1 fti + Y2 '12 + Y3 h3 +PA

Eq. 2- 7

l'HESSURE HEAD

l 1t''iSUre head is the height "h" of a column of homogeneous liquid of unit tght y that will produce an intensity of pressure p.

Eq. 2-5

p=wh

lz =

This means that the pressure at any point "lz" belmo a free liquid surface is equnl to the product of the unit weight of the fluid (y) and h.

E.. y

Eq. 2- 8

1o Convert Pressure head (height} of liquid A to liquid B

Consider that points 0 and 6 in Figure 2 - 4 lie on the same elevation, such that h = O;. then Eq. 2 - 4 becomes:

Eq. 2- 9 Eq. 2- 6 This means that the pressure along the same horizontal plane in a homogeneous fluid

at rrst are equal.

Io convert pressure head (height} of any liquid to water, just multiply its h lght by its specific gravity lzwiller

Pressure below Layers of Different Liquids

Air, pressure = PA h1

Liquid 1

0 0

Liquid 2

h3

Liquid 3

0

0

= hti<Jllid X SJiqulJ

Eq. 2-10

34

CHAPTER TWO Principles of Hydrostatics

FLUID MECHANICS & HYDRAULICS

1I

CHAPTER TWO

lJIO MECHANICS I fYDRAULICS

Principles of Hydrostatics

35

l1 ps in Solving Manometer Problems:

MANOMETER

A mm10111eter is a tube, usually bent in a form of a U, containing a liquid of known specific gravity, the.surface of which moves proportionally to changes of pressure. ft is used to measure pressure

Types of Manometer

Open Type - has an atmospheric surface in one leg and is capable of measuring gage pressures. Differential Type - without an atmospheric surface and capable of measuring onJy differences of pressure.

I Decide on the fluid in feet or meter, of which the heads are to h1 expressed, (water is most advisable). ' Starting from an end point, number in order, the interface of diffNl'nl fluids. · Identify points of equal pressure (taking into account that for a homogeneous fluid at rest, the pressure along the same horizontal plane are equal). Label these points with the same number. I. Proceed from level to level, adding (if going down) or subtracting (if going up) pressure heads .as the elevation decreases or increa~e~, respectively with due regard for the specific gravity of the fluids.

Piezometer - The simplest form of open manometer. It is a tube tapped into a wall of a container or conduit for the purpose of measuring pressure. The fluid in the container or conduit rises in this tube to form a free surface

olved Problems

Limitations of Piezometer:

• Large pressures in the lighter liquids require long tubes • Gas pressures can not be measured because gas can not form a free surface

1•roblem 2 - 1

depth of liquid of 1 m causes a pressure of 7 kPa, what is the specific 111vity of the liquid?

11 ' '

•.olution

Pressure, p =y h 7 = (9.81 x s) (1)

s = 0.714

7 Specific Gravity

Problem 2- 2 (a) Open manometer

(b) Differential manometer

,Yhat is the pressure 12.5 m below the ocean? Use sp. gr. · olutlon

p=yh p = (9.81 xl.03)(12.5) p = 126.3 kPa

(c) Piezometer

= 1.03 for salt water.

CHAPTER TWO

36

Principles of Hydrostatics

FLUID MECHANICS & HYDRAULICS

I 1 llll MECHANICS t fYl)RAULICS

CHAPTER TWO

Principles of Hydrostatics

37

1bl"m 2- 5

Problem 2 - 3

If the pressure 23 meter below a liquid is 338.445 kPa, determine its unit weight y, mass density p , and specific gravity s

1111 pressure in the air space above an oil (s = 0.75) surface in a closed tank is

Solution

l11tlon

(11)

ld'a absolute, what is the gage pressure 2 m below the surface?

P=

Unit weight, y p=yh 138.445 = y (23) y = 14.715 kN/m 3

Psurface =

Note: Patm

= 101.325 kPa

p = 13.675 + (9.81x0.75)(2) p = 28.39 kPa

'··) Mass density, p p=

+ Yh 115 - 101.325 Psurface = 13.675 kPa gage

Psurface

l g

I • •1blem 2 - 6

14.715x10 3 p= 9.81 p = 1,500 kglm 3

·1

I the absolute pressure in kPa at a depth of 10 m below the free surface of

II if sp. gr. 0.75 if the barometric reading is 752 rnmHg. olution

(c)

Specific gravity, s

Pnl>s = Pn1111 + pgago·

~ = Pflu1d

P•tm = Y111 h111

= (9.81 x 13.6)(0.752)

Pwate1

1,500 ,=-1,000 '= 1.5

p,1tm

Pnbs =

100.329 + (9.81 x 0.75)(10)

p..1•• = 173.9 kPa

Problem 2 - 4 If the pressure at a point below this point?

= 100.329 kPa

1•1 oblem 2 - 7 in

the ocean is 60 kPa, what is the pressure 27 meters

pressure gage 6 m above the bottom of the tank containing a liquid reads 90 1 I .1 Another gage height 4 m reads 103 kPa. Determine the specific weight of 11,, liquid.

Solution

fhe difference in pressure between any two pomts liquid is p2 - pi = y h p2 = p1 + yh = 60 + (9.81xl.03)(27) f12 = 332.82 kPa

/

in

a

•Olution

p2 - p1

0

0

= yh

103 - 90 = y(2) y = 6.5 kN/m 3

0

0

0

~0

CHAPTER TWO Principles of Hydrostatics

38

FLUID MECHANICS & HYDRAULICS

Principles of Hydrostatics

39

Solution

Problem 2 - 8 ·

Since the density of the mud varies with depth, the pressure should be solved by integration

An open tank contains 5.8 m of water covered with 3.2 m of kerosene (Y = 8 kN/ m3). Find the pressure at the interface and at the bottom of the tank.

dp = y dh . dp = (10 + 0.5 h)dh

Solution (a) Press ure at the interface

,,

f

PA= Yk hk = (8)(3.2) PA= 25.6 kPa

dp

Kerosene

(b) Press ure at the bottom PB =2: yh = Yw h,,, + Yk hk = 9.81(5.8) + 8(3.2) p~ =

CHAPTER TWO

FLUID MECHANICS & HYDRAULICS

82.498 kPa

r

=

0

=8 kN/m

3

J1o + O.Sh)dh 0

0

p = 10h + 0.25h 2

Water

y

s

5 ]0

= [10(5) + 0.25(5)2] - 0 . p = 56.25 k Pa

=9.81 kN/m3 B

~roblem 2 - 11 Problem 2 - 9 If atmospheric pressure is 95) kPa and the gage attached to the tank r eads 188 mmHg vacuum, find the absolute pressure within the tank. Solution P••V< =

In tho figµre shown, if the atmospheric ltl'l.i~!lure

is 101.03 kPa and the ab.soh.ite ~m11~nure at the bottom of the tank is ~ ~ t.:\ kPa, what is the specific gravity ~tf

p.,,,,, + pgag.-

11live oil?

pg<1g1; = Ymercury hmerc ury

= (9.81x 13.6) (0.188) = 25.08 kPa vacu um

Ps·•s~ = -25.08

I

kPa

p.... = % .7 + (-25.08) I

p,,1,, = 70.62 kPa abs

Problem 2 - 10 The weight density of a mud is given by y = 10 + O.Sh, where y is in k N/ m 3 and his in meters. Determine the pressure, in kPa, at a depth of 5 m

!hhttlon I 1llf1.
,,, .. .Eyh] hm + Yo 'h,, + Yw hw + Yoil hon 'L30.27"' (9.81 x 13.6)(0.4) + (9.81 ... s)(2. 9) + 9.81 (2.5) + (9.81 x 0.89)(1.5)

(J • Ym ~

• 1.38

,.

CHAPTER TWO

40

FLUID MECHANICS & HYDRAULICS

Principles of Hydrostatics

CHAPTER TWO

FLUID MECHANICS

Principles of Hydrosta tics

& HYDRAULICS

Problem 2 - 12

Problem 2 - 14

lf air had a constant specific weight of 12.2 N/ml and were incompressible, w hat would be the height of the atmosphere if the atmospheric pressure (sea level) is 102 kPa?

Compute the barometric pressure in kPa at an altitude of 1,200 m if the pressure at sea level is 101.3 kPa. Assum e isothermal conditions a 21°C. Use R = 287 Joule /kg-°K.

Solution

Solution

Height of atmosp he re, Ir =

For gases:

!!.

I

y

dp = -pgdh

102x 10 3 12.2 Height o f atmosphere, h = = 8,360.66 m

p=

I

_e_ RT

p 287(21 +273)

Problem 2 - 13 (CE Board May 1994)

'\

41

p

Assuming specific weight of air to be constant at 12 N/ m3, w hat is the approximate height of Mount Banahaw if a mercury barometer at the base of the mountain reads 654 mm and at the same instant, another barometer at the top of the mou ntain reads 480 mm

= 0.00001185 p

= -(0.00001185 p)(9.81) dh dp = 0.0001163 dh

dp

p

fd~J

I'

Solution

101 .3xl0

Air

hm= 480 mm

y = 12 N/rn3

In p

= -0.0~01163

1

1

= -0.0001163 11

In p - In (101.3 x 103) lnp = 11.386 p = e11.386 p = 88,080 Pa

Pbo• - Ptop = y h 1 (y,., 11,.,)bottom - ('1111 h m)top = (y h).,. (9,810 x 13.6)(0.654) - (9,810 x 13.6)(0.48) h = 1,934.53 m

= 12 h

1r

0

I'

J101 .3 x10

f

1200

1200

J0

= - 0.0001163(1200 -

0)

CHAPTER TWO Principles of Hyd rostatics

42

FLUID MECHANICS & HYDRAULICS

mm of mercury to (rr) oil of sp. gr. 0.82 and (b) wate r.

Solution (n)

'1011

S mercury

43

Principles of Hydrostatics ''

Problem 2 - 18 (CE November 1998)

Prob.lem 2 - 15

Convert 760

CHAPTER TWO

FLUID MECHANICS.

& HYDRAULICS

Piston A has a cross-section of 1,200 sq. cm while that of piston B is 950 sq. cm. with the latter higher than piston A by 1.75 m . If the intervening passages are filled with oil whose specific gravity is 0.8, what is the difference in pressure between A and B.

= ltmNr ury - - -

Solution 13.6 = 0.76 0.82 /1011 =

(b)

B

PA - PB= Yo ho

= ( 9,810 x 0.8)(1.75)

12.605 m of oil

PA - ps = 13,734 Pa A

fi w•lcr = izmcrr ury Smercury

= 0.76(13.6) llwatN =

10.34 m of water

·~

Oil

s = 0.8 1200 OTI2

950 cm2

Problem 2 - 16 (CE Board May 1994)

A barometer reads 760 nunHg and a pressure gage attached to a tank reads 850 cm of o il (sp. gr. 0.80). What is the absolute pressu re in the tank in kPa? Solution p,ob; = p,11111

+ P g•gc

= (9.81 x 13.6)(0.76) + (9.81 x 0.8)(8.5) p.1,,, =

168.1 kPa abs

Problem 2 - 19

In the figure shown, determine the weight W that can be carried by the 1.5 kN force acting on the piston.

1.5 kN

Problem 2 - 17

A hydraulic press is used to raise an 80-kN cargo truck. If oil of sp. gr. 0.82 acts on the piston under a pressure of 10 MPa, what diameter of piston is required? Solution . Since th e pressure w1der the piston is uniform: Force = pressure x Area 80,000

= (10 x 10~)

Solution Sin ce points 1 and 2 lie on the same elevation, pi = pi

*

D = 0.1 m = 100 mm

02 I

1.5

w

t(0.03) 2

t(0.3) 2

W= 150 kN

1.5 kN 2

Oil, s .=.•0J12

44

CHAPTER TWO

FLUID MECHANICS & HYDRAULICS

Principles of Hydrostatics

~LUID MECHANICS ,_ HYDRAULICS

CHAP TER TWO

Prin ciples of Hydrostatics

Solution

Problem 2 - 20

A drum 700 min in d iameter and filled with water has a vertical pipe, 20 mm in diameter, attached to the top . How many New tons of water must be poured into the pipe to exert a fo rce of 6500 Non the top of the drum? Solution Force on the top: F= p x Area 6500 = p x (7002 - 202)

=

45

Cylinder W =44kN A = 0.323 m'

-r- - o

~·

t

Plunger, a 0.00323 m·

h = 4. 6 m

=

G_ l

h

p 0.016904 MPa p = 16,904 Pa

20mm 0

[p = y Ir]

l

Oil ,

Top

16,904 = 9810 ,, h = 1.723 m

s = 0.78

Oil, s

= 0.78

IP2 - p1 = y,, it]

F

F

pi = - = n 0.00323 pi = 309.6 F (kPa)

Weight = y x Volume = 9810 x t (0.02)2(1.723)

Area on top

Weight = 5.31 N

w

Pz = 700 mm 0

fl

44 =

o.32..1

p2 = 136.22 kPa

Problem 2 - 21

The figure shown shows a setup with a vessel containing a plunger and a cylinder. What force F is required to balance the weight of the cylinder if the weight of the plunger is negligible?

L36.22 - 309.6 F = (9.81 =326 N

x

0.78)(4 .6)

F = 0.326 kN

~roblem 2 - 22

Cylinder W=44 kN A= 0.323 m2

1lw hydraulic press show n is filled with oil w ith sp. gr. 0.82. Neglecting the

li'1'1 Hht of the two pis tons, w hat force F on the handle is required to support

Ill' 10 kN weight?

f 1

4.6 m

Plunger,

a =0.00323 m2 '100 mm

25 mm 0

Oil, s = 0.78

Oil, s = 0.78 oil

F

CHAPTER nxtO Principles of Hydrostatics

46

FLUID MECHANICS & HYDRAULICS

~Ince

Solution

Since points I and 2 lie on the samt> e le11ation, the n;

P1

flLUID MECHANIC::S ' HYDRAULICS

=

P2

_!)_

=

Ai

CHAPTER nxtO Pr.inciples of Hydrostatics

47

the gage reads "FULL" then the reading is equivalent to?,() cm of gasolin~·

Reading (pressure head) when the tank contain water= (y + 2 o1;a) cm of gasolinl'

!i_

y+ 2....L =30 0.68

rhen.

A2 10

11 =27.06 cm 2

f(0.075) Fi= 1.11 kN

oil

~l'Oblem 2 - 24 (CE Board November 2000)

[L Mo= OJ f(0.425) = f2(0.025) F(0.425) = 1.11 (0.025) r = 0.0654 kN F = 65.4 N

F

,,,l'or,,,the tank shown in the Figure, /1i = 3m and lh = 4 m

()ptPrmmE' tht> va luE>

FBD 'of the· lever arm

Problem 2 - 23

T'he fuel gage for a gasol ine (sp. gr. = 0.68) tank in a car reads proportional to its bottom gage. If the tank is 30 cm deep an accide ntally contaminated with 2 cm of water, how many centime ters of gaso line does the tank actually contain when the gage e rroneously read s "FULL" 1 .S olution

2

Summing-up pressure head from 1 to 3 in meters of water

+ /12(0.84) - x = P3 y y

El.

0 + 0.84 h2 - (4 - 3) = 0 lt2= 1.19 m

vent

ry

llc1lutlon

=

c°*~-----~ Water

"Full"

r

30 cm

.)

s = 0.84 2

2.

Gasoline, s = 0.68

l '------..r "Full"

x

0

0

n1

'l. Water

"

0

0

0

0

Vent

Gasoline, s = 0.68

Oil

ni

Water

"

CHAPTER TWO

48

FLUID MECHANICS & HYDRAULICS

Principles of Hydrostatics

Problem 2 - 25 (CE Board May 1992)

CHAPTER TWO

FLUID MECHANICS & HYDRAULICS

49

Principles of Hydrostatics

Problem 2 • 26

In tht> figurt> sho wn. what is the static pressure

in

kPa

in

the air cham~r?

For the manome ter shown, dl'termine the pressure at the <'lmter of the pipe.

I

1 rn

Air

Oil

Ii

s = 0.80

Mercury, s = 13.55

1.5 rn

L

/~~;· ~ ,...... ~,

.., . :

Sm

'

. \ .. '

"

\

. '

.............: .~Oil, s = 0.80

Solution

l olution

The pressure m the air spact> equ als the pressure on the surfact> of oil, pi

~.um-up

2m

r-

= y,., h,, = 9.81 (2)

2m

1•2 = 19.62 kPa

p2.- p1 = y,, 11.i. 19.62 - p3 = (9.81 • 0.80)(4) P• = -11.77 kPa

0

. 011 s = 0.80

6

6

Water

2 - 4(0.80) = P3

+

() + 2-~ 2 =

..El. 9.81

1•• = -11.77 kPa I

•

y·

4m

pressure head from

3 in meters o f water:

+ 1(13.55) + 1.5(0.8) =

0+14.75

12. y

=·Ji y

Another solution Sum-up pressure head trom I to ?o in me ters o t water

!!..!.

tll

l!..L

,,, = (l (12

I

'·

3m

l

P'?> y /IJ

= 14.75 m

of water

= 14.75(9.81)

111 =

144.7 kPa

/

'

0

\.. ................

. '

'

.~Oil, s = 0.80

CHAPTER TWO Prin cipl es of Hy drostatics

50

FLUID MECHANICS & HYDRAULICS

CHA PTER TWO Principles o f Hydrostatics

FLUID MECH AN ICS

& HYDRA ULICS

Problem 2 - 27 (CE Board Novembe'r 2001}

flroblem 2 - 28 ( CE May 1993)

Determine the value of yin the manometer shown in the Figure.

In the fi gure shown, when the funnel is empty the water surfoce 11 '1t point A and the mercury of •P· gr. 13.55 shows a deflection of 1~ cm. Determine the new 1l,1flcction of mercury when the tunnel is filled w ith water to B.

-i:-

lm

Air, 5 KPa

3m

s = 0.8

t

i

lm

Oil

Water

51

B

-..-

15 cm

.I.

-"-

...L 0.5 m

Mercury = 13 .55

s

lolution

1-

Solution

E..d_ + 3(0.8) + 1.5 - y(13.6) = y

-

PB

5

9.81

.

,:--

Summing-up pressure head from A to B in meters of water: '

·+ 3.9-13.6y= -

where pe = 0

y

ElL y

lm

t

3m

i

lm

...L

Air, 5 KPa A

30cm 0

j

T 80 cm

Oil

~

s =0.8 Water

__j_

2' 2

Lt

1_

1

l' -

l'

t+ R 0.15

~

0.15

y = 0.324 m

y+x

Mercury s = 13 .55

Figure (b): Level at B

Figure (a): Level at A

'iolve for yin Figure (a): Sum-up pressure head from A to 2 in meters of water:

El!_ +y - 0.15(13.55)= El_ y

0 + y - 2.03 = 0 y = 2.03 m

y

CHAPTER TWO Principles of Hydrostatics

52

FLUID MECHANICS & HYDRAULICS

In Figure (b): When the funnel is filled with water to B, point 1 w ill move down to 1' with the same value as point 2 moving up to 2'

E!.

..

y

+ 0.8 +

E1_ + y(l3.6) - x = E.!E... y

y

=~

Eq. (1)

In Figure (b):

y + x - (x + 0.15 + x)(13.55) = EL

0 + 0.80 + 2.03 + x - 27.lx - 2.03 26.l x = 0.80 .x = 0.031 m = 3.1 cm

53

Principles of Hydrostatics

Sum-up pressure head from 2 to 111 in meters of water:

13.6y - x

Sum-up pressure head from B to 2':

CHAPTER TWO

FLUID MECHANICS & HYDRAULICS

y

Sum-up pressure head from 2' tom' in ~eters of water:

=0

EL y

+ (0.2 sin 0 +

y + 0.2)(13.6) - (x + 0.2) =

EE:... y

0 + 2.72 sin 0 + 13.6y + 2.72 - x·..: 0.2 = ~ 13.6y - x = 8.183 - 2.72 sin 0

New reading, R = 15+2x=15 + 2(3.1) New reading, R = 21.2 cm

[13.6y - x = 13.6y - x] 8.183 - 2.72 sin 0 =

Problem 2 - 29

Water

The pressure at point m in the figure shown was increased from 70 kPa to 105 kPa. This causes the top level of .mercury to move 20 mm· in the sloping tube. What is the inclination, 0?

Eq. (2)

;3

1

sin 0 = 0.3852 0 = 22.66°

Problem 2 - 30

A closed cylindrical tank contains 2 m of water, 3 m of oil (s = 0.82) and the air 1\hove oil has a pressure of 30 kPa. If an· open mercury manometer at the hut tom of the tank has 1 m of water, determine the defle.c tion of mercury. Solution

Solution

Sum-up pressure head from 1 to 4 in meters of water: pair

y

L

&

L

+ 2.46 + 3 - 13.6y = 0

y= 0.626 m

0.2m

1

+ 3(0.82) + 2 + 1 - y(l3.6) =

Tl'

!!..!. y

3m

-*4-

2m

Oil, s

=0.82

Water 4 ...,..

lm

3

Figure ( a)

In Figure (11):

Figure (b)

3

y

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO Principles of Hydrostatics

54

l'ht• U-tube s ho wn 1s 10 mm m diameter crnd con tain s me rcury . If 12 ml of w a te r is poured into the right- ha nd leg, w ha t are the ultimate heig h ts in the two legs?

lOmm v

El y

·I E E

E E

-

0

0

Mercury

L L

Solution Volu mt? of water =

.!.

1N }2 /.i = 12 cm·'

Note:

~ ~1 0

_J

l20 mm

l ml

_j

= l cm3

Ir= 15.28 cm= 152.8 m m

Since the q ua nti ty of me rcury before and afte r water 1s poured re m ain the same, th e n;

E E

~

0

N

Mercury

L L

120 mm Figure (a)

_J _J

x= 114.38mm Ultimate heigh ts in each leg: Right-hand leg, lzR = h + x = 152.8 + 114.38 Right-han d leg, llR = 267.18 mm Left-hand leg, lzL = R + x = 11.24 + 114.38 Left-hand leg, lzL = 125.62 mm

1•l11:r.ometer columns E, F, and t < .ind (b) the deflection of the 111..rcury in th~ U-tube 11 11111ometer neglecting_ the \~ 1·lght of air.

rr

I

In Eq. (2): 11.24 + 2t = 240

Gage

11111· a gage reading of -17.1 kPa, tl..t1•rmjne the (n) elevations of Ili11 Liq uids in the open

10 mm. ·

E E

y

ftroblem 2 - 32

120(3) = R + x + 120 + .1 R + 2x = 240 -7 Eq . (1)

r

= .£.1-

N

~

b)

+' 152.8 - R(l3.6)

R =11.24mm

r·

(St't' fig11rt'

55

ln Figure (b): Su nuning-up pressure head from 1to3 in mm of water:

Pr oblem 2 - 31

'-;olv tng tor Ii.

CHAPTER TWO Princip les of Hyd rostatics

FLUID MECHANICS & HYDRAULICS •

02_ Air

El. 15 m

-

~·

s = 0.70 El. 12 m

~

Water

~t

El. 8 m

s= 1.6

U_ L

120 mm

_j

Figure (b} .

E

]h

El. 4m

y

Mercury

.

F

G

56

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO Principles of Hydrostat10

Solution

Gage G

"

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO

Principles of Hydrostatics

57

Column G Sum-up pressure head from 1 tog in meters of water;

El.

+ 3(0.7) + 4(1) - '13(1.6) =

y

11 ..

!£. y

i.~; + 2.1 + 4 - l.6h3 = 0 1

El. LS m

e· s

Sm

-t-~- :2_m_

= 2.72 m Surface elevation = 8 + 113 Surface elevation = 8 + 2.72 = 10.72 m 113

= 0.70 2 g

.Water

~m

3

3

Deflection of mercury Sum-up pressure head from 1 to 5 in meters of water;

El.

s"' 1.6

+ 3(0.7) + 4 + 4 - '4(13.6)

y

p1 I

!_

= p... = - 17

8.4 m - - - 4

Column E

i:;um-up pressure head from I t0 1' m metes of water. + lt1(0.7)

= Er_

ftroblem 2 - 33

1\1\ npen manometer attached to a pipe shows a deflection of 150 mmHg with 1111' lower level of mercury 450 mm below the centerline of the pipe carrying 11111111,r. Calculate the pressure at the centerline of the pipe.

lctlutlon

y

y

9~~/

9~~1 + 10.l - 13.6/14 '14 = 0.614m

I kPa

Mercurv. s = 13.o

El

y

1

H

4m

= J!2..

/~1(0 .7) = 0 h1 = 2.5 m Surface elevation= 15 - h, Surface elevation = 15 - 2.5 = 12.5 m +

- · -·-·-·-·- ·-·-·7·-·-·-·- ·~l._. Water 450 mm

Column F

<;um-up pressure head trom I to fin meters ot water. Pi

-

• W>.7\ - h2(1\

=

_PJ

llH1111 up pressure head from 1 to

lt1 meters of water; 111

y _!Z.J. ... ) Q.Rl

112

-·

1 - Ji, ·

= ll

= 0.357 m

5urface elevation = l2 + fi2 Surface elevation= 12 + 0357 = 1Z.357 m

+ 0.45- 0.15(13.6) = E1y

J!J_ + 0.45 - 2.04 = 0 1 UH /'I

15.6 kPa

Mercury

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO Principl es of Hydrostati cs

58

r-1m0-,

Problem 2 - 34

r:or th e configuration shown, calculate the weight of the piston if the pressure gage reading is 70 kPa.

Piston

Il m

CHAPTER TWO Principles o f Hydrostatics

FLUID MECHANICS & HYDRAULICS

"Gage liquld =mercury, 11=0.1 m Sum-up pressure head from 1 to 4 in meters _o f wa ter;

Air 0

y

y

y

El - E.!. y

y

1.5 m

g Water

y

El.. - Ei.

4

0

E.!. + x +It - /r(13.6) - x -1.5 = Ei.

Oil

s = 0.86

g

-1(0.86)

~ 9.81

0

0

0

0

x

= 1.5 - 0.1 + 0.1(13.6)

Water 0

'?, 0

= 2.76 m of water

3

h 2

Stun-up pressure head from A to B in meters of water;

y

Air

t

0

0

Solution

fu

59

- 0.86 =

r- 1m0-,

I

= }!.§_ y

Piston

.22._ 9.81

Weigh t = FA =PA x Area

=78.44 x

1m

Weight

B

Gage liquid

reading, h = ? Sum-up pressure head from 1 to 4 in meters of wateri.

El..

Piston

y

A Oil s = 0.86

p,i = 78.44 kPa

Pa= 70 kPa

(11) Gage liquid= carbon tetrachloride

+ x +It - /r(l.59) - x - 1.5 =

El.. - Ei. y

y

Ei. y

= 1.5 + 0.59/r '

FA= PA x Area

where

El.. - P4 y

= 2.76 m -:) from (a)

y

2.76 = 1.5 + 0.59h 11=2.136 m

t (1)2

Weight = 61.61 kN

Problem 2 - 35

lti the figure shown, deter mine

Two vessels are connected to a differential manometer using mercury, the connecting tubing being filled with water. The higher p ressure vessel is 1.5 m lower in elevation than the other. (a) If the mercury reading is 100 mm, what is the pressure head difference in meters of wa ter? (b) If carbon tetrachloride (~ = 1.59) were used instead of merc4ry, what would be the manometer rt..1d ing for the same pressure difference?

lht1 height Ii of water and the

..

Air, p

••H'' reading at A when the !'ltK(ll ute pressure at B is 290

0

i

0

W11.

B

=175 kPa abs Water

A

'

CHAPTER TWO Principles of Hy drostatics

60

FLUID MECHANICS & HYDRA ULICS

FLUID MECHANICS & HYDRAULICS

Solution

CHA PTER TWO Principles of Hydrostatics

61

Sum-up press ure (gage) head from 1 to 4 in meters of water;

Sum-up absolute pressure head from B to 2 in meters of water;

Air, p

I

~ - 0.7(13.6) - h = E1.. y

y

El..

=175 kPa abs

y

2

h = 2.203 m

Then, x + y = 28.42 m

700 mm

L_

y

+0.9x- 16.51 =O 9.81 ' x = 13.81 m

t-~·~/

175 .122. 9.8l - 9 .52 - Il -- 9.8T

= f.i_

~

oi Water

h

+ x(0.9) + 1.3(0.9) -1.3(13.6)

I.:-'fE==:'...=::===

B

=t..:.l

A

Problem 2 - 38

Sum-up absolute 'pressure head from B to A in meters of water;

Oil,

the manometer setup shown, 1h•tcrmine the difference in pressure l•l'lween A and B.

s = 0.85

l'llr

£..§_ - 0.7(13.6) + 0.7 = ~ y y

Water 680 mm

1

.i_

.l2!!. - 9.52 + 0.7 = Li_

9.81 p11 = 203.5 kPa abs 9.81

1700 mm

l Problem 2 - 37 In the figure shown, the atmospheric pressure is 101 kPa, the gage reading at A is 40 kPa, and the vapor pressure of alcohol is 12 kPa absolute. Compute x + y.

Alcohol vapor

lolution \ + 0.68 = y + 1.7 \ - y = 1.02 m -7 Eq. (1)

.._ A Air

x

Alcohol

s =0.90

_l_ 1.3 m

Sum-up pressure head from A to B In meters of water;

Ell_ - x - 0.68(0.85) + y = ~ y

y

Mercury

Oil, s

Ell_ - E.!_ = x - y + 0.578 -7 Eq. (2) y

y

Solution Sum-up absolute pressure head from 1 to 2 in meters of water;

p = 12 kPa abs

.._ A Air

El.. - y(0.9) = E1.. y

y

_g_

40+101 - 0.9tj = 9.81 9.81 y = 14.61 m

Alcohol

s = 0.90

"ubstitute x - y = 1.02 in Eq. (1) to Eq. (2):

Ell_ - ~ = 1.02 + 0.578 y

y

PA - Pe = 1.598 9.81 /IA - pa = 15.68 kPa

=0.85

,,,

11111111111.uuUUAMMllll

62

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO Principles of Hydrostatics

CHAPTER TWO Principles of Hydrostatics

FLUID MECHANICS

& HYDRf.ULICS fl!roblem 2 - 40

Problem 2 - 39

/\ difforential manori1etc r is attached to a pipl:.' <1S <; hown. Calcul ate ti1l.' . p r<'ssure diffe rence betwl'<>n p(iin ts !\ , a nd 8.

100 mm _i__

Mercury

·- -o!I·-·- _...: -·-·-·· Oil, s = 0.90

It' the fig ure sh own, the 1foflcction of m ercu ry is initially a1>() rnm. If the pressm e a t A is 1111 rcased by 40 kPa, w hile 11111intaining the p ressure at B 1'11 11 ~ tant, wha t will be th e new n w rur y deflection?

Solution Mercvry

lOO mm

1

y 0.25 m

·-·-0!3·- - ·-·-·-·- -· ·-

E

Oil, 's = 0.90

'° 0

Sum-up· pressure head from A to Bin meters of water;

2.1 m

fu - y(0.9)- 0.1(13.6) + 0.1(0.9) + y(0.9) = I!.!l. .y

'.

E

y

"'?

EA - fl.!_ = 0.1 (13.6) - 0.1(0.9) y

y

·pi, .-yu = 1.27m 9.81

p,i - pn = 12.46 kPa

Figure (b)

Figure (a)

111Pi1•,11rc11, su m-up pressu re head from A to Bin meters of water; 111 '

y

-

0.6 - 0.25(l3.6) + 0.25 + 2.'I =

t.A - EfL y

y

= 1.65 m of wa te r

EfL y .

63

64

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO

Principles of Hydrostatics

In Figu re b, pA' = p.~ + 40 Sum-up pressure head from A' to B in meters of wa ter;

CHAPTER TWO

FLUID MECHANICS & HYDRAULICS

65

Principles of Hydrostatics

Solution Kerosene, s = 0.82

EK_ (0.6 - .x) - (0.25 + 2.x)l3.6 + (2.35 + x) = .E£. y

y

p11 + y

Ell.

y .

40

+

- 0.6 +

~ 9.81

Ell. - .!!..!!.. y

y

x - 3.4 - 27.2x + 2.35 + x = Jj_ y

-1.65 - 25.2x =

= 25.2 x - 2.423

El!... y

But

Ell. - .E£. y

y

= 1.65

1.65 = 25.2 x - 2.423 x = 0.162rn=162 mm

'ium-up pressure head from A to B in meters of water;

!!.i.l.

New mercury deflection = 250 + 2x = 250 + 2(162) New n\erCury d eflection = 574 mm

y

+ 0.2(0.88) - 0.09(13.6) - 0.31 (0.82) + 0.25 - 0.1(0.0012) =

y

f!.!.l. - .f!..!}_, y

Problem 2 - 41 In the figure shown, determine the difference in pressure between poin ts A and B. Kerosene, s =0.82

E.!L

y

= 1.0523 m of water ,

PA - pa= 9.81(1.0523) = 10.32 kPa

•roblem 2 - 42 (CE Board) AA1~111ning

normal barometric pressure, how deep in the ocean is the point

hW li111• fm air bubble, upo n reaching the surface, has six times its volume than 1

l!h111 l11tthebottom?

T

·

llltlutlon

E E

Applying Boyle's Law isothermal condition)

(11 11'lt11ning

0

..,. 0

150 mm

200mm

90 mm

t

11'1 V,

=

p2 V2]

,,, = 101.3 + 9.81(1.03)h /It= 101.3 + 10.104 /J

Vi= V /'2= 101.3+0=101.3 Water

V1•6V \1\11.3

f 10.104/i)V = 101.3 (6 V) 111llJ.l/t • 101.3(6) -101.3 It r;o.13 m

'

66

CHAPTER TWO

FLUID MECHANICS & HYDRAULICS

Principles of Hydrostatics

Problem 2 - 43

A vertical tube, 3 m long, w ith one end closed is inserted vertically, w ith the open end dow n, into a tank of water to such a depth that an open manometer connected to the upper end of the tube reads 150 mm of mercury. Neglecting vapor pressure and ass.urning no rmal conditions, how far is the lower end of the tube below the water surface in the tank? Solution

FLUID MECHANICS & HYDRAULICS

CHAPTER TWO

Principles of Hydrostatics

67

Since the p ressure in air inside the tube is u niform . then Pr= p1. = 20.0124 kPa Pc= Yw h . 20.0124 = 9.81h; Ii = 2.04 m Then;

x = /1 + y = 2.04 + 0.495 r = 2.535 m

Area= A

Pfoblem 2 - 44

A bottle consisting of a cylinder 15 cm in diameter and 25 cm high, has a neck Which is 5 cm diameter and 25 cm long. The bottle is inserted vertically in water, with the op,en end down, such that the neck is completely filled with Wl\ter. Find the depth to which the open end is submerged . Assu me nor mal lt.1rometric pressu re and neglect vapor pressure

3m

x y

Solution

LS cm 0

l\ pplying Boyle's Law P1 Vi= pi V1 Applying Boyle's Law: p1Vi= p2 V2 Before the tube was inserted; . Absol ute pressure of air inside, p1= 101.3 Vo lume of air inside, Vi = 3A I· When the tube was inserted; Abso lu te pressure of air inside, p2 =101.3 + 9.81(13.6)(0.15) Absolute pressure of air inside, p2 = 121.31 kPa Volume of air inside the tube, .V2 = (3 - y)A

[p1 V1 = p2 Vz] 101.3 (3 A) = 121.31 [ (3 - y) A] 3 - y = 2.505 y = 0.495 m

From the manometer show n; P1> = y,,, /1,,, = (9.81 x 13.6)(0.15) p1. = 20.0124 kPa

''• t11re the bottle was inserted Volume of air: v, = (15)2 (25) + t (5)2 (25)

t

V, = 4,908.74 cm '

l\bsolute pressu re m ai r· p1= 101 .325

lt'V h1111 the bottle is inserted ·

~

~·

T d Wat er

W•Wc

V2 = t (15) 2 (25) V2 = 4,417.9 cm' in air: ,,, - 101.325 + 9.81 h

fl11·~~ure

11

'I

::=:::--

25 cm@

Vnlume of air:

lr

I'

111 • p2 V2J 10 1.325(4,908.74) = (1 01.325 + 9.81 h)(4,417 9) 10 1.325 + 9.81 11 = 11 2.58 It 1.15 cm /1 + 25 = 26.15 cm

x

CHAPTER TWO

68

Principles of Hydrostatics

FLUID MECHANICS & HYDRAULICS

Problem 2 - 45

/\ bicycle tire is inflated at sea level, where the atmospheric pressure is 101.3 kPaa and the temperature is 21 °C, to 445 kPa. Assuming the tire does not expand, what is the gage pressure within the tire on the top of a mountain where the altitude is 6,000 m, atmospheric pressure is 47.22 kPaa, and the temperature is 5 °C.

FLUID MECHANICS

CHAPTER TWO Principles of Hydrostatics

& HYDRAULICS

69

[supplementary Problems Problem 2 ·- 46

A we~ther repo rt indicates the barometric pressure is 28.54 inches of mercury. What 1s the atmospheric pressure in pounds per square inch? Aus: 14.02 psi

Solution Problem 2 - 47

P1V1 = P2V2

The tube shown is filled with oil. Determine the pressure heads at Band moters of water. At sea level: Absolute pressure of air: p1 = 101.3 + 445 Absolute pressure, pi== 546.3 kPaa Volume of air, V1 = V Absolute temperature of ajr. T 1 = 21 + 273 = 294 °K

B

2.2 m

i--

On the top of the mountain: · Absolute pressure of air, pi= 47.22 + p Since the tire did not expand, volume of air, V2 = V Absolute temperature of air, T 2 = 5 + 273 = 278 °K ( P1 V1

T1

= P2 V2

E.!L = -2.38 m y

E£ Air

y

c

= -0.51 m

0.6m

_L_ Oil, s

J

Oil, s

T2

546.3(V) = (47.22 + p}V 294 278 47.22 + p = 516.57 p = 469.35 kPa

Ans:

c in

=o.asl--
=0.85

1101 t~\C tank shown in the figure, compute the pressure at points 8, C, D,·and £ ~.,kl

11.

Neglect the unit weight of air.

A11s: /1B Air

= 4.9; pc = po= 4.9; PE= 21.64

B

0.4m 0.4

D

m

Oil

s

0.5 m

=0.90

c l m E

70

CHAPTER TWO Principles of Hydrostatics

CHAPTER TWO Principles of Hydrostatics

FLUID MECHANICS & HYDRAULICS

71

Problem 2 - 49 A glass U-tube open to the atmosphere at both ends is shown. If the U-tube rontains oil a nd water. d etermine the specific gravity o f the oil Ans: 0.86

!fh,• pressure pii = 13.4 kPa gage and Pl = 13.42 kPa gage.

Assume the cleaning

"~\lid is prevented from moving to the top of the tank. Use unit weight of W11t,•r = 9.79 kN /m3. (n) Determine the pressure p.1 in kPa, (b) the value of /1 in

1

·r

,,.. <"ylindrical tank conta ins wa ter at a height of 55 mm, as show n. Inside is a ~1111111 open cylii1drical tank containing cleaning fl u id (s.g. = 0.8) al a h eight /r.

lllll\, and (c) the value of y in millime ters. A11s: (11) 12.88; (b) 10.2; (c) 101

E

Air Water Water

55 mm

0

0

0

0

Problem 2 - SO Kerosene

A g lass 12 cm tall filled with water 1s inverted. fhe bottom 1s open. What 1~

the p ressure at the closed end ? Barometric pressure is 101.325 kPa. Ans: 100.15 kPaa

-V- D y

Mercury (s.g. = 13.6)

Prob!em 2 - 51

In Figure 13. in whic h fluid will a pressure of 700 kPa firs t be ac hieved? Ans: glycerm

A d1fforen tial manometer shown is measuring the differen ce in pressure two W1tkr pipes. The indica ting liquid is mercllry (specific gravity = 13.6), /i 1 is 675 ~1111. lt,,, 1 is 225 mm, and !t,,, 2 is 300 m m. What is the pressure differen tial

Po= 90 kPa

ethyl alcohol

60 m

" = 7~3.3 kg/m3

0 11

water

~ 979 kg/m 3 glycerin

,, = 1236 kg/m3

Ans: 89.32 kPa

10 m

,, =899.6 kg/m 3 ,,

1 11lvPPn the two pipes.

5m ,......,"'""".....,_....,,,,,.=,,_..,,...,..~

.-._......w.o-...--..;i;.r

i.-..;....;......

sm

hml

1

72

FLU ID MECH ANICS & H YDRAULICS

CHAPTER TWO

Principles of Hydro ~tatics

Problem 2 - 54 .A. force of 460 N is exerted on lever AB as shown. The end B is connected to

cl

piston w h ich fits into a cylinder having a d iameter of 60 mm . What force Fn r1cts on the larger piston, if the vo lume between C and D is filled with water? Ans: 15.83 kN 460 N A

220 mm

CHAPTER THREE

FLUID M CHANICS

& HYDRAULICS

Total Hydrostatic Force on Surfaces

73

Chapter 3 Total Hydrostatic Force on Surfaces TOTAL HYDROSTATIC FORCE ON PLANE SURFACES

Ir

120 mm

the pressure ov er a p lane area js uniform, as in th e case o f a horizo n tal uurface subme rged in a liquid or a p lane s urface ins ide a gas chamber, the II 1t.il hydros tatic fore~ (or total pressure) is giv en by:

D

•

B

L \~ l11·re pis

Problem 2 - SS

An open tube open tube 1s attached to a tank as shown. If water rises to a height of 800 mm in the tube, what a re the pressures PA a nd PB of the air above water? Neglect capillary effects in the tube Ans: PA= 3.92 kPa; pn = 4.90 kPa

PA

Pe

k?>

A

B

~

to;;"'°mm Water

..

r-

30.0 tm ;;

..

"

~

1 E E

0 0 00

Water ~

l

F = pA

Eq. 3-1

the urufor m p ressure and A is the area.

111 the case of a n inclined or vertical plan e submerged ·in a liq uid, the to tal 1 11•i.sure can be found by the following fo rmula:

74 ,

CHAPTER THREE Total Hydrostatic

~orce

on Surfaces

FLUID MECHANICS & HYDRAULICS

Consider the plane surface shown inclined at an angle a with the horizontal To get the total force F, consider a differential element of area dA . Since this element is horizo ntal the pressure is uniform over this area, then;

rw10MEC~ ry" =

p = yli p = yy sine

dF = y y sin0 dA

f

F = y sin 0

75

f

ydF dF ='Y y s in 0 rlA f = y sin 0 Ay

w here

f

y sin e Ay"y,, =

yd A

From calculus, J ydA

CHAPTER THREE Hydrostatic Force o n Surfaces

h1 hgure 3 - 1, taking moment of fo rce about S, (the intersection o f the ~ 111lungalion of the plane area and the liq uid surface),

dF = pdA

w here

To~al

4!, t IYDRAULICS

y sin 0 Ay y1,

Ay

F = y sin 0 Aj/

I

y('Y y s.in 0 di\)

f

y 2 dA

= y sin 0

From calculus,

F = y( y sin 0) A

f

2

(moment of inertia about S)

y rlA =ls

./

From the figure, Then,

y sin 8 =

/1

yp= F = ylr A

I AV

Eq. 3 - 4

Eq. 3 - 2

lly transfer formula of m oment of inertia: -2

Since y his the unit pressure 'at the centroid of the plane area, Peg, the formul a may also be .expressed as:

ls= 13 +A Y

l g+

F = Pc&A

!Ji· =

Eq. 3 - 3

AY 2

AY

y,.=

Eq. 3 - 2 is convenient to use if the plane is submerged in a single Jjquid and without gage pressure at the surface of the liquid. However, if the plane is submerged under layers of different liquids or if the gage pressure at the liquid surface is not zero, Eq . 3 - 3 is easier to apply . See Problem 3 - 15

lg

f +--=

Eq. 3 -5

AY

+ e, from Figure 3 - 1, then

,.

,, Eccentricity, e =

1

1

I

~

AY

Eq. 3- 6

I ,1ble 3 - 1 in Page 76 for the properties of common plane sections.

76

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

Total Hydrostatic Force on Surfaces

TABLE 3 - 1: Properties of Common plane sections

v,

a

------f9 ,-: Ye

-

Ix = -

I=-

12

36

8'

3 IT nb I, = - 8

Arca = bd lid~

/ = -

'

(

S'

rlb 3 3

"y

3

y

lid~ = -

3

Area =

1• IT 02 IT 04

'I• IT r 2 ;

IT•r4

4

64

~,=

x 4r

,,IT

IT

r4

y

Ellipse

'y

1

~r

~11

;r ,•

I=/= -

•

'"

Area=

1,, = 0.11

8 ,...

(

Hll

1Tllb

~·

= --

4

1rba~ I=-r.• . 4

~bh 3

r4

5

(O+ 1/2 sin 20)

Segment of arc 1

+~ i·-

,

Ye

'r/·~

iY /

h

j_ x

b

3

x

-

2

2 ' I, = - I ii>' 15

l+--b--+I y. = -

h

Area =

:L2fl=~T ! i

r';

•

3-o-

Xe

112 11

cg - y

I

Spandrel

I

'16

= 0.055b11'

Qb x,= -ii

4

y

18 ,

Parabolic segment

1y

y

I

-8-

r4

ly = -

16 18, = lgy = 0.0551"' '

16 I.~' = 0.055nb-'

1,, = "4 (fl - 'h sin 2G)

I=/=-

Semicircle

Art-.i •

x

Jrl:ln 3 / = --

'

Area = 'h , 2 (20) = ,2 0 2 r sinO

x, =y, = -;;-

1

fv, .:c /'(llc __ c - -

m1b3 / = --

nb~

ij)·

v•I

=

a

Arcil = V. 1111b 411 4b x,= y, = 3tr 31T

12

gy

LL1

m ·l

'

Sector of a circle

db ( =-

12

!8, = 0.11

gy -

'~ Arca •

'~-

·x

I - 71'1m3

/ = -

Quarter circle

Circle

af

I

Arca = 12 mib 4b y, = 3;r

y. =Ii/?>

Area= •12b/J b/1 3 blr'

a

1

I<- b/2 ->f.-- b/2 --1
1Y

IY

~h

.. •

I

Quarter ellipse

Y1

-------1 '--~

Total Hydrostatic Force on Surfaces

Half ellipse

Rectangle

Triangle

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

1 Arca= - -bli 11 + 1 1 II +1 x, = --b; y,= - - I r II+ 2 411 +2

.x

i~-"· ' , i

Xe '

Length of arc = r(20) = 21-Q rsinB f r= - 0 -

When e = 90° (semicircle) 2r

x( == -

1t

77

CHAPTER THREE

78

Total Hydrostatic Force on Surfaces

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

Total Hydrostatic Force on Surfaces FH=prgA

Eq. 3- 7

or F11 = y Ii A

Eq. 3- 8

Fv=y V

Eq. 3- 9

TOTAL HYDROSTATIC FORCE ON CURVED SURFACES CASE I : FLUID IS ABOVE THE CURVED SURFACE.

tan 8

---

... __

~i

79

= Fv/ F11

Eq. 3-10

where: A =vertical projection of submerged curve (plane area) Pc&= pressure at the cenb·ojd of A

0

cg of volume

N1>Le: The procedure used in solving FH is the same are that presented in Page 73.

FLUID BELOW AND ABOVE THE CURVED SURFACE

CASE II: FLUID IS BELOW THE CURVED SURFACE

Volume= V cg of volume

cg of volume 0

f

Curved surface Net vertical force

Net vertical projectlon of area

' 'l"!"l!l~IUMM++WlllllllllllUlllUIUUll

'

CHAPTER THREE

80

Total Hydrostatic Force on Surfaces

FLUID MECHANICS & HYDRAULICS

PLUID MECHANICS

a. t-IYDRAULICS

CHAPTER THREE

Total Hydrostatic Force on Surfaces

81

DAMS Dams are structures that block the flow of a river, stream, or other waterway. Some dams divert the flow of river water into a pipeline, canal, or ch annel. Others raise the level of inland waterways to make them navigable by ships and barges. Many dams harness the energy of falHng water to generate electric power. Dams also hold water for drinking and crop irrigation, and provide flood con trol.

PORPOSE OF A DAM

Dams 1. 2. 3. 4. 5.

are built for the following purposes: Irrigation and drinking.water Power s upply (hydroelectric) Navigatio n Flood control Multi purposes

"'uru

3 - 3: Boat Passing through canal Lock. Canal locks are a series of gates designed I ..1111w a boat or ship to pass from one level of water to another. Here, after a boat has t1l 111•d lhe lock and all gates ar~ secured, the downstream sluices open and water flows H•~1qh them. When the water level is equal on either side of the downstream gate, water 111h nowlng through the sluices; the downstream gate opens, and the boat continues on at I 11rw water level.


Gent!:rator

au water.

Turbine

- ---~---...... I

I·

-~··~--.;-~---~---......=o;i..-

Dr aft tube

Figure 3 - 2: Section of a dam used for hydroelectric

lhnt is, they hold back the water by the sheer force of their weight pushing downward. To do this, gravity dams must consist of a mass ~~I heavy that the water in a reservoir cannot push the dam downstream or tip it over . They are much thicker at the base than the lt1p-a shape that r eflects the distribution of the forces of the water "ti•linst the dam. As water becomes deeper, it exerts more horizontal pressure on the dam. Gravity dams are relatively thin near the surface 1d the reservoir, where the water pressure is light. A thick base 1'1111bles the dam · to 44withstand the more intense water pressure at llw bottom of the reservoir.

82

CHAPTER THREE

Total Hydrostatic Force on Surfaces

FLUID MECHANICS & HYDRAULICS

LUID MECHANICS HYDRAULICS ·

CHAPTER T H REE

Total Hydrostatic Force on Surfaces

83

4 A~tttress da111 consists of a wall, or face, s upported by several bu t resses on the downstream s ide. The vast majority of butlress d . s are made of concrete that i~ reinfor~ed wi lh steel. Buttresses are p1cally spaced across the dam site every 6 to 30 m (20 to 100 ft), depending upon the size and design of the dam. Buttress dams are sometimes called hollow dams because the buttresses do not form a solid wall stretching across a river valley.

Figure 3 - 4: Gravity dam

2 An embankmeut dam is a gravity dam formed out of loose rock, earth, or a combination of these materials. The ups tream and downstream slopes of embankment dams are flatter than those of concrete gravity dams. In essence, they more closely ma tch the natural slope of a pile of rocks or earth 3 Arclt dams are concrete or masonry structures that curve upstream mto a reservoir, stretching from one wall of a river canyon to the other. This design, based on the same principles as the architectural arch and vault. transfers some water pressure onto the walls of the canyon. Arch dams require a relatively narrow river canyon with solid rock walls capable of withstanding a significant amount of horizontal thrust. These dams do not need to be as massive as gravity dams because the canyon walls carry part of the pressure exerted by the reservoir

Fi gur e 3 - 6 : Buttress dam

Fig ure 3 - 7 : Multiple arch dam Figure 3 - 5: Arch dam

111111111ttllfffftttlttfffflllllNliMlllWll1111-M111

84

CHAPTER TH REE

Total Hydrostatic Force on Surfa ces

FLUID MECHANICS & HYDRAULICS

FLUID MECHANICS ' HYDRAULICS

Downstream Side

Upstream Side

(Tallwater)

.---r~.~w ~ .s~·~""'"'.'I~ W I I

ltY2

W3 =Yr

=Yr Vi;

v3

F=yliA 2. Wind Pressure

jW•

Projection of ~

the submerged face of dam

~1 =Ye V1;

Weight of water in the upstream side (if ariy) /: W4 = yV4 3. Weight or permanent structures on the dam 4. Hydrostatic Uplift U1 =y V,, 1 U2 = Y V112 8. Horizontal Force 1. Total Hyd rostatic Force acting at the vertical projection of the su~mergecl portion of the dam,

Headwater

Vertical

85

ertical for ces Weight of the dam

ANALYSIS OF GRAVITY DAM

A dam is subjected to hydrostatic forces due to wa ter which is raised on its upstream side. These forces cause the da m to. slide horizontally on its fo undation and overturn it about its downstream edge or toe. These tendencies are resisted by friction on the base of the dam and gravitational forces which causes a moment opposite to the overturning moment. The dam may also be prevented from sliding by keying its base

CHAPTER THREE

Total Hydrostatic Force on Surfaces

.3. Wave Action 4. Floating Bodies 5. Earthquake Load

h

F

Ill. Solve for the Reaction A. Vertical Reaction, Ry Ry= L.F,, Ry = W1 +. W2 + W3 + W4 - U1 - Ui I

Uplift Pressure Diagram ~

Toe

··r·····+··r·· ·· I

8. Horizontal Reaction, R, l~r

=Hi,

R,

=p

.

IV. Moment abou.t the Toe Ry

x

Figure 3 - 8: Typical section' of a gravity dam showing the possible forces acting

Steps of Solution With reference to Figure 3 - 8, for purposes of illustration, an assumption w,1 made in the shape of the uplift pressure diagram. I. Consider 1 unit (1 m) length of dam (perpendicular to the sketch) 11 Determine all the forces acting:

A. Righting Moment, RM (rotation towards the upstream side) RM = W , x, + W2 x2 + W3 X3 + W4 x~

8. Ove~urning Moment, OM (rotation towards the downstream side) OM - p y + U1 Z1 + Ui Z2 . Location of Rv ( x)

x

= RM-OM Ry

Eq. 3-11

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

Total Hydrostatic Force o n Surfaces

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

Total Hydrostat ic Force o n Surfaces

87

where: y = unit weight of water= 9.81 kN/m1 (or1000 kg/ m3) y, = unit weight of concrete y, = 2.4y (usually taken as 23.5 kN / m1)

Rv ( 1±6e) , where e $ 8/6 q = ---· .

Factors of Safety

Factor of safety against sliding, FSs:

~

B

Eq. 3 - 14

8

Oll': Use (+) to get the stress at point where Ry is neares t. Jn the diagram

uiwn a~ove, use (+)to get q-r and (-)to get 'q11 . A nega,tive stress indicates

µR y FSs = - - >1 Rx

Eq. 3-12

m\r ress1ve stress and a positive stress indicates tensile stress

h,u• soil cannot carry any tensile stress, the result of Eq. 3 - 14 is invalid if the jMh·'I~

Factor of safety agains t overturning, F:;n:

is positive. This will happen if e > 8/6. Should

thi~

happen, Eq. 3 - 15

lllrlll he used . RM OM

B/2

FSo = - - >1

where: p =coefficient of friction between the base of the dam and the foundation

Foundation Pressure Fore~

B/3

B/6

.(

B/3

1

}(II"'

B/6 i

k

>1

~~

~--·

P Mc q=-- ± -

= 8(1) = 8

M = R,,e I= 1(8)3

R,1

q=-13 ±

I (R ~ 0

e)(B/2) 8 3 /12

lm

/2(11)(q,)(l) )q,.

x

a

_ 2Ry

q, - -3x

QH

12 c = B/2

1

/(,, = 1/2(3

Heel ~---.-...--!"""f'-r"~"f-T""'lt"'"l''-ri

P = Rv !\

,, .. 3x

: I

: B/6

From combined axial and bending stress fo rmula: A

B/3

q.

\' =a/3

1 Middle Third '

I

I

cg+ ~I(

\:

• Ry

i

B/2 B

B/2

I :\

Eq. 3-15

88

CHAPTER THREE

Total Hydro sta tic Fo rce on Surfaces

FLU ID MECHANICS & HYDRAULICS

, LUID MECHAN ICS .. HYDRAULICS

C HAPTER THREE

Total Hyd rostatic Force on Surfaces

89

where: /

BUOYANCY

y = 1fnit weight. of the fl uid

ARCHIMEDES' PRINCIPLE A principle discovered by the Greek scientist Archi1~edes that states that "m1y body immersed i11 n fluid is acted upon by n11 upwnrd force (buoynnt force) eq11nl to till' wei8i1t of tlle displaced fluid".

..

This principle, a lso known as the law of l1ydrostntics, applies to both floating and submerged bod ies, and to all fluids. Consider the body shown in Figure 3 - 9 immersed in a fluid of unit weighty The horizontal components of the force acting on the body are a ll in equilibrium, since the vertical projection of the body in opposite sides is th1· same. The upper 'face of the body is s ubject to a vertical downward forc1• which is equal to the weight of the fluid above il, and the lower face is subjecl to an upward force equal to the weight of real or· imaginary liquid above It The net upward force acting on the body is the buoyant force.

V[

· volume displaced . Vo'. ume of the body below the liq uid surface

I1 ~"('"'.pi ob/ems 111 b11oyn11cy, 1de11tifi; t/ie forces nct111g and npply ro11t!it1ons of stntir 11111lrlm11111: >: fH = 0 :!: Fv = 0 :!:M=O ~t lt1>1nogeneous solid body of volume V "floating" i.n a homogeneous fluid at •1111 .

VD

= sp.gr.of body

--"---=-----"-

sp.gr.ofliquid

v=

Ybody

--

v

Eq. 3 - 17

Y1tquid

1111• l1udy of height H has a constant horizontal cross-sectional area such as t lh •d cylinders, blocks, etc.:

V0

=Volz -

Vol,

Vo

l

BF=

Fv2- Fv1

D = sp .gr.ofbody H

sp.gr.of liquid

=

Ytxxty

H

Eq. 3 - 18

YliquiJ

Figure 3 - 9 : Forces acting on a submerged body

BF= Fvi-Fvi

I 11d y is of unjform vertical cross-sectional area A, the area submerged A,

= y(Voh) - y(Voh) BF= y(Voh- Voh)

BF =y Vo

As= sp.gr. of body A == Ybody A sp. gr . ofliquid y liquid

Eq.3- 19

90

CHAPTER THREE Total Hydrostatic Force on Surfaces

FLUID MECHANICS & HYDRAULICS

, LUID MECHANICS

~' HYDRAULICS

CHAPTER THREE Total Hydrostatic Force on Surfaces

91

STATICAL STABILITY OF FLOATING BODIES A floating body 1s acted upo n by two equal opposmg torces fhese are, th~ body's weight W (acting at its center of gravity) and its buoyant force Bf (acting at the center of buoyancv that is located at the center of gravitv of thE> d isplaced liqu id ) · When these torces are colhnear as shown m Figu re 3 · 10 (a), 11 floats 111 an upright posilion. However, when the body tilts due to w ind or wave action, the cente r of buoyancy shifts to its new position as shown in Figure 3 · 10 (b) and the two forces, whic h . are no longer coUinear, produces a couple equ a l to W(x). T he body will not ovf?rtum if this couple makes the body rota te 'toward~ its original position as shown in Figure 3 · 10 (b), and w ill overturn if tlw ~ituation is as shown in Figure 3 · 10 (c). fhe point of intersection between the axis of the body and lhe lme o t action ot the buoyant force is called the 111etace11ter. The distance from the rnetacenter (M) to the center of gravity (G) of the body is called the metacentric height (MG). It can be seen that a body is stable if Mis above G as s hown in Figu re;\ 10 (b), and unstable if M is below G as shown in Figure 3 · 10 (c) If M coincides with G. the bodv is said to be 1usl ~lnhlr•

Figure 3 · 10 (c): Unstable position Figure 3 • 10: Forces on a floating body

~IGHTING MOMENT AND OVERTURNIN.G MOMENT

RM or OM= W(x)

BF

Figure 3 10 (a): Upright position

Figure 3 · 10 ( b): Stable pos1t1on

Eq. 3- 20

weight of the body buoyant fo rce (always equal to W for a floating body) 1 t' nter of gravity of the body renter of buoyancy in the upright position (t 1•ntroid of the displaced liq uid ) 1111 renter o f buoyancy in the tilted p osition \ 1 volume displaced I mctacenter, the point of intersection between the line of action nl the buoyant force and the axis of the body 11•nter of g ravity of the wedges (imm ersion and emersion) h111 izontal distance between the cg' s of the wedges volu me of the wedge of immersion .rngle of tilting f.1111 1 distance from M to Bo I 1111 •• 1hstance from G to Bo ~It 1 11\utacentric height, distance from M to G

11111111111111111111.UllllllllHtMt*MMHIU

92

CHAPTER THREE

Total Hydrostatic Force on Surfaces

FLUID MECHANICS & HYDRAULICS

Metacentric he ight, MG = MBo± GBo

Eq. 3- 21

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE Total Hydrostatic Force on Surfaces

93

M~ rnent du to shifting of BF= m oment due to shifting of wedge BF ( = F (s)

Use ~-) If G 1s above LJ<;e (+) 1f G 1s below

F=y Vo F=yv

8<• R1'

z = MBo sin e Note

M 15

always above S,

y v D MBo sin e = y v s

MBo= __v_s_ VALUE OF MB 0 l"ht> stabili ty o t tht> body depends on the c1m o unt ot the righting momen t

wh ic h in tu rn 1s depende nt on the metacentric height MG. When the body tilts. tht> center of buoyan cy shifts to a new position (Bo'). This shifting also cau se~ the wedge p ' to shift to a new position o The moment due to the shifting ol the huovant forc-e flF(zJ i~ must equal to moment due to wedge shift Ffs )

Eq. 3 - 22

V0 sine

INITIALVALUE OF MB0

ti r ~mall values of e, (6

""0 ore = 0):

,

~--

111111..110~ ~.·~; Wedge, volume

s

~ Rolhn9 Waterline

Section

=v

, Figure 3 - 11: Rectangular body

hli•1 11 body in the shape of a rectangular II

111Figure3_11;

wedge, v = 112(B/ 2)[(B/ 2) tan e]L 11111111• of wedge, v = LB2 tan e

111111111• c1f

t

I 11111111 1 values of e, 5 "' ~ B

II para elepiped length L as

llil 1 i'lillillll1illflfl""ltt~llllltiilillllll11itilllt..llllllHIH

94

.C HA PTER TH REE

Total Hydrostatic Fo rce o n Surfaces

14-----s--~

s/2--+

s/2

But for small values of e, sin e"' tan e

..L L8 3

~

_12_ _

I

Vo But

95

Total Hydrostatic Force on Surfaces R RECTANGULAR SECTION

vs

MB. = - - V0 sine .l. LB 2 tan e x l. B MB.= s . 3 V0 sine MB.=

CHAPTER THREE

FLU ID MEC HA NICS & HYDRAULICS

-f.i LB3 is the moment of inertia of the waterline section, J

(B/2)sece

x=s/2 __.

1

o cg

M

( B/2)

Note: This formula can be applied to any section.

Since the metacentric heig ht MG is dependent with MBo, the stability of a floating body therefore depends on the moment of inertia of the waterlim section. Lt can also be seen that the body is more stable in pitching than in rolling because the moment of inertia in pitching is greater than that in rolling

co~-'-

Centroid of wedge

h-m. =

vs V 0 sine

Vo= BDL

where L is the lengtti perpendicular to the figure

II= 1/2(B/2)[(B/2)

II=

tan e)L

tLB2 tan e

x From geometry, x =

<'en troid of triangle,

MOMENT

The righting or overturning moment on a floating body is:

T-

(B/2) tan l-l

812

x

RM or OM= W x = W (MG sin 0)

+ X2 + X3 3 = O + ( B I 2) sec e + ( B / 2) cos e 3

x=

x1

.!!_ ( 1- +cose)= .!!_ ( 1+cos2e) 6

!. 2

=

x = .!!_ ( 1 + cos 6

2

cose

~ .. .!!. ( 1 + cos 2 e I 3 l cose J

e)

case

6

l

cose

96

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

Total Hydrostatic Force on Surfaces

~LB' tan•Hl l+,::~ 9 )] MB.=

nsider a pipe of diameter D and lckness t be s ubjected to a net pressure 1'o determine the tangential stress in t pipe wall, let us cut a section of length along the diam eter. The forces acting on 11 section are th e total pressure F due to •t lntem aJ pressure and this is to be l11tcd by T which is the total stress of • pipe wal.l.

(BDL)sin8. 2

MB .. =

LB 3 sin8 l + cos 8 - ·x 24 cos8 cos8 BDLsin8

8 2 1 + cos 2 e MB = - - - - " 24D cos 2 8 2

MB,, = 24D B

MB 0

=

MB.

=

(

£

cosl 2 ~ + i lJ

F = 2T

8

F= pA

but sec2 e = 1 + tan2 8

= pDs

T= Sr Awall T= Sr (s x t)

11Ds = 2 x

(Sr (s x

t)J

8 (2 tan 0) - + - -·2

2

= ---~2+ tan2 8) = - 12(2)D

Projection of curved area

p:r,, =OJ

24D

£24D [(1 + tanl 8) + l ]

12D

2

2

2

B2- ( 1+-tan MB = 0 120 2

Tangential stress,

eJ

s.,. =

pD 21

Eq. 3- ~6

t

tl11h•rn1ine the longitudinal stress, let us 1111' 1•ylinder acr oss its length as shown.

0/'11

-- -\\ ~

OJ

I' T

STRESS ON THIN-WALLED PRESSURE VESSELS

f= pA /' r= pf 0 2

F

'f' = SLAwall

THIN-WALLED CYUNDRICAL TANK A tank or pipe carrying a fluid or gas under a pressure is subjected to teruill forces, w hich resist bursting, developed across longitudinal and transver secti ons.

97

Applying equilibrium cond ition;

(seci 0 + 1)

2

MB 0

I

CHAPTER THREE

Total Hydrostatic Force on Surfaces

JL

---

~t

Awall = 1tDL 1· .. s,_ 1tDt

I' 1J) l = SL rcDt Longitudinal stress, SL = pD 4t p = internal pressure - exter nal pr essure

Eq. 3- 27

Eq. 3 - 28

Total Hyd rostatic Force on Surfaces

SPHERICAL SHELL

If a spherical tan k of d iameter D and thickness I contains gas under a p ressun of p. the stress at the wall can be expressed as·

.

'

a

.D '

CHAPTER THREE

FLUID MEC HAN ICS & HYDRAULICS

CHAPTER THREE

98

l

~

Total Hydrostatic Force on Surfaces

99

olved Problems vertical rec tangular plane of heigh t d and base bis submerged in a liquid With its top edge at the liquid surface. Determine the total force F acting on Ill' side and its location from the liquid surface .

F =yhA

ii= d/2

Wall s tress, S = pD

4t

A= bd

F = y(d/2)(brl) F = 1h y b d2 SPACING OF HOOPS OF A WOOD STAVE PIPE

1

e=

_g_

Ay

y = Ti = rl/2 .l. bd 3 12

I!=

.

(bd)(d I 2)

I'= d/6

.

y,, = h + e ljp = l/1 r

25 1 A,, Spacing, S = --'--pD w here: S, = allowable tensile s tress of the hoop A1, =cross-sectional area of the hoop p = in ternal p ressure in the pipe 0 =diameter of the pipe

Pressure diagram (triangular prism)

d/2 + d/6

= 2d/3

llsing the pressure d iagram: F =Volu me of pressu re diagram F = 1/z(yd)(d)(b) = 112 y b d2 The location of F is at the centroid of the pressure diagram. I or rectangular surface (inclined or vertical) submerged in a fluid with top edge nushed on the liquid surface, the center of pressure from the bottom is 1/3 of its

l11•lght.

II liillillllllfllli 111111111111

111

,

CHAPTER THREE

100

Total Hydrostatic Force on Surfaces

FLUI D MECHA NICS & HYD RAULICS

CHAPTER THREE

Total Hydrostatic Force o n Surfaces

Problem 3 - 2 A vertical triangular surface of height d and horizontal base width b 1~ submerged in a liq uid with its vertex at the liquid s urface. Determine the total force acti n g on one side and its location from the liquid surface

F=yh A F = y(r)(rr r2) F=7tyr"

r

Solution

.

Ig

t>=-

Ay

(' = y /1 A

fr = 1...3 d

.lm.4

e=

A = hbd 1

r = y x ~ d x 112nd

F=

101

4

(7tr 2 )(r)

= r/4

YP = r + e y,,=r+r/4

t ybd2

yp

=Sr/4

Pressure diagram

(cylindrical wedge)

Ig

1'=-

Using the pressure diagram for this case is quie t complicated . With the

Ay

l'l' sh.own, its volume can be computed by integration. Hence, pressu re •~mm 1s easy to use only if the area is rectangular, with one side horizontal.

ii = fr = 2d/3 .l.. bd 3

•' = I'.=

3b (f bd)(2d /3)

Pressure diagram (pyramid)

d/12

ifio= /i

1;,. =

+

Otw ~nde of the gate and its location from the bo ttom.

I'

-t d + d /1 2 = 3d/4

Us ing the pressure diagram. ('=Volu m e of pressure diagram

* ( =*

F=

A1>.i-e

(b

x

/' .. yh A fl ,,;, 1.5 + 2 = 3.5 m

~ heigh t yd)(d)

=t

2m

, ,. 9.81(3.5)[(1 .5)(3))

ybd2

~ is located at the centroid of the diagram. w hic h is

from the base

V1 rtira l rectangular gate 1.5 m wide and 3 m high is submerged in wate r

lh Iii. ~op edge 2 m below the water surface. Find the.total pressu re actit~g

ii= 3.5 m

154.51 kN 1 /4

of the altitud1 ' r'

3m

.!.L Ay 3

A vertica circular gate o r radius r is submerged in a liquid with its top ed g~·1I flushed on the liquid surface. Determine the magnitud e and locati on of lh• tota l force actin g on one side of the gate

r

fi"(l.5)(3) ' (1.5 x 3)(3.5) = 0.214 m

v"' ·1:5 - e 1/ ... '1.5 - 0.214 t/ • l .286 m

cg e--t---e..... cpe--T-,--....-

y

t

102

Total Hydrostatic Force on Surfaces

Using the pressure diagram: F =Volume of p ressure diagram

F = (Sy ;

21

x

3) (1.5)

CHAPTER TH REE

FLUID MECHANI CS & HYDRAULICS

CHAPTER THREE

~-.·: .::. ....

.....

..

F = 15.75y F = 15.75(9.81)

103

Total Hydrostatic Force o n Surfaces

F= yhA

il h

= 2+

t (3)

= 3m =

2m

y

h=3m

F = (9.81(0.82))(3)[1/2(1.5)(3)] F= 54.3 kN

F = 154.51 k N 2y

*

('1.5)(3)3 e -- -1g -- -='-----

Ay

[t (l.5)(3)](3)

e.= 0.167 m y1, = ii + e y,. = 3.167 m from the oil surface

3y

2y

Location of F:

Ai= 2y(3) = 6y A1 = V2(3y)(3) = 4.Sy A = A I + Ai = 10.5y

Oil

s = 0.82

blem 3 - 6 (CE Board May 1994)

Pressure diagram (trapezoidal prism)

v1·rtical rectangular p late is submerged half in oil (sp. gr. = 0.8) and half in 1111·1· such that its top edge is flushed with th e oil surface. What is the ratio of ~II' l11rce exerted by water acting on the lower half to that by oil acting on the pJ11· r half?

\

[Ay =Lay] 10.5y y = 6y(l .5) + 4.Sy(l) y =1.286 m (much complicated to get than using the formula)

I '11rrc on upper h alf: l 'o =Yo h A

l'o = (Yw x 0.8)(d/ 4)[b(d/2)] l'o =0.1Yw.b d2 Problem 3 - 5

·A v~ticaJ trian gular gate w ith top base horizontal and 1.5 wide is 3 m high. 11 is subm erged in oil h av ing sp. gr. of 0.82 with its top base submerged to n depth of 2 m. Determine th e magnitude and location of the total hydrostatlr press ure acting o n one side of the gate.

I 111 rt' on lower h alf: I 1v = p cg2 x A flcg2

= Yo ho + Yw It,,. = (Ytu X 0.8}(d/2} + y (d/ 4}

/Jcg2

= 0.65 Yw d

Pcg2

10

I 1v 111 (0.65 Y11• d)[b(d/2))

I 1v,. 0.325 Yw b d2

Fw Fo

0.325y wbd 2

---'-"'-- = 3.25

O.ly 111 bd 2

FLUID MECHANICS & HYDRAULICS

CHAP TER THRE E · Tot al Hydrostat ic For ce on Surfaces

104

CHAPTER THREE Total Hydrosta ti c Force on Surfaces

blem 3 - 8 (CE Board May 1992)

Pr oblem 3 - 7 (CE Board May 1994}

A vertical ci.rcular gate in a tunnel 8 m in diameter has oil (sp. gr. 0.8) on 01w side and air on the other side. If oil is 12 m above the invert and the air pressure is 40 kPa, where will a single support be located (above the invert <'I the tunnel) to hold the gate in position?

dosed cylindrical tank 2 m in diameter and 8 m deep with axis ver tical lllains 6 m deep of oil (sp. gr. = 0.8). The air above the liquid surface has a "Hsure of 0.8 kg/cm 2. Determine the total normal force in kg acting on the •II at its location from the bottom of the tank.

Solution

Oil; s = 0.8

Sm Air; p = 40 kPa

12 m

Fu

J

Oil;

s=0.8

!+---.- -·-·- ·- ·-·-·- ·-.. 8 m

z-y

,.__:::::-~_.i*

Yo11 h A F01 1 = (9.81 x 0.80)(8) x

t (8)

4

-Y

O

Z

~ Foi1 =

2

4 rn

Invert

hinge

Fo,1 = 3,156 kN

e=

I

_g_

Ay

~ (8)4 I! =

t (8)2 (8)

1'1 • P•ir A fluir = 0.8 kg/ cm2 = 8,000 kg/ m2 I 1 8,000(2rr x 2) = 32,000n kg l/1 • 6 + 1 = 7 m •

I 1 .. p,8 A = 0.5 m

z = 4 - e = 3.5 m F,ur = Pair Ar = 40. x F.1, = 2,011 kN

t

105

/ltn = (1000 x 0.8)(3) + 8,000 l'i~ =10,400 kg/m2 I 1 '" I0,400(2rr x 6) = 124,800n kg 1

'11lve for e: F2 =Yo ii A

(8)2

124,800n = (1000 x 0.8) ii (2rr x 6)

ii • y = 13 m

The support must be located at point 0 where the moment d ue to I and F011 is zero. Since F0 u > F.1,, 0 must be below Fon·

Ig

fU1o= 0) Fo11(z - y) = F.11(4 - y) (3,156)(3.5 - y) = 2,011(4 - y) 1.569(3.5 - y) = 4 - y 5.493 - 1.569y = 4 - y y = 2.62 m

~ (2n)(6) 3

,.. - = -=---Ay (2rrx 6)(13) r • 0.23077 m Vt .. •I - e "" 2.77 m

I

Ill

II

I

F~ • 156,8001C kg

7 Total normal force

106

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE Total Hydrostatic Force on Surfaces

Fy = F1 y1 + F2 y2 (156,800it) y = (32,000it)(7) + (124,800it)(2.77) y = 3.63 m -7 Location of F from the bottom

(l: Mhinge = 0 ] F z = 40(1) F = y h A = 9.81 F = 14.715h

Using the pressure diagram: n{2) = 2n m

8000

CHAPTER THREE Total Hydrostatic Force on Surfaces

e= -

h

I8

Ay

h (1 )(1.5) -

Ti (1.5)(1) 3

1

(1.5 x 1 )h

12h

e= -=----

6

z = 0.5 + e = 0.5 +

w.s.

•••

1± h

!.Sm

L=

p-1

I "

h

where y =

r

107

~

r-""''

B

l m

40 kN

12/i

800(6)

=4800

8000

1 14.715h (o.5+ _) = 40 . 12h

Pressure Diagram

0.5 h + 0.08333 = 2.718

P 1 = 8000(8)(27!) = 128,000it kg P2 = 112(4,800)(6)(2it) = 28,800it kg

P = P 1 +P2 =156,SOOJt kg

h =5.27 m = h

-7 Total normal force

[P y = P1 y1 + P2 y2] (156,800it) y = (128,0007!)(4) + (28,8007!)(2) y = 3.63 m -7 Location of P from the bottom

lie .ti circular gate is submerged in a liquid so that its top edge is flushed ~ lh•• liquid surface. ~11 tlrnt

Find the rati o of the total force acting on the lower acting on the upper half.

Problem 3- 9

In the figure sh ow n, stop B will break if the force on it reaches 40 kN. Find the critical water depth. The len gth of the gate perpendicular to the sketch is 1.5 m

+ 0.5 = 5.77 m -7 critical water depth

F F,

w.s.

H11t10 = -1.. L

= 1.Sm

f

~

x

yh2 A2 y h 1 A1

/\I

A2

lfotlc•

,,,~2

Millie•

l.424r 0.5756r

lm

stop

o.sJs6r

,,.....;

N

I1111 lo

h

B

F1

Fi

x x = 4r/311

= 2.475

CHAPTER THREE

108

Total Hyd rostat ic Force on Surfaces

F LUID MECHA NICS & HYDRAULICS

CHAPTER· TH REE

Total Hydrosta t ic Force o n Surfaces

Problem 3 - 11

A 30 m long dam, retains 9 m of water as shown in the figure. Find the total resultant force acting on the ·dam and the location of the center of pressure from the bottom.

e=

-a (30)(10.392)

3

_--!=-.:...~:__--'----

L = 30m

hA h = 3.5 + 2/3 h =4.167m

F=y

h

A = 1/2(1)(2.61) A= 1.305 m2 F = (9810 x 0.83)(4.167)(1.305) F = 44,277 N P =44.277 kN

2/3 2m

j

inclined, · circular h• with water on one Id•• is shown in the Determine the resultant force

t

2m

(30_:< 10.392)(4.5 /sin 60°)

e = 1.732m 0

0

y = 1/2(10.392) -1.732 y = 3.464 m or

y=

0

0

t (10.392) = 3.464 m

Problem 3 - 12

The isosceles triangle gate shown in the figure is hinged at A and weighs 1500 N. What is the total hydrostatic force acting on one side of the gate in kiloNewton?

y /1 A

/1

,,

2 + 0.5 sin 60° 2.433

'IHI (2.433) f (1 )2 IH7116 kN

Oil (s = 0.83)

10 9

CHAPTER THREE Total Hydrostatic Force on Surfa ces

110

FLUID MECHANICS & HYDRAULICS

Problem 3 - 14 The gate in the figure shown is 1.5 m wide, hinged at point A, and rests against a smooth wall at B. Compute (n) the total force on the gate due to seawater, (b) the reaction at B, and (c) the reaction at hinge A. Neglect the

weight of the gate.

w.s.

CHAPTER THREE Total Hyd rosta t ic Force on Surfaces

e = _I8_ =

Ay

3

t12(1.5)(3.6) (1.5 x 3.6)(7.21)

e = 0.15 m x = 1.8- 0.15 x = 1.65 m (:EMA= 0) F(x) - R8 (2) = 0 218.25(1.65) = 2Ra Ru= 180 k N

Seawater

s=

1.03

(:Efa =OJ

0

RA11

o·

*o 0

RA11

+ F sin e - Ra = 0

= 180 - 218.25 sin 33.69°

RA,, = 58.94 kN (:E F" = 0)

2m

p cos e = 0 = 218.25 COS 33.69°

RAv -

l

RAo

RM= 181.6kN RA= RA=

2 2 JRA v + RAH

= J(l81.6) 2 + (58.94) 2

190.9 kN

Solution

w.s.

I

d2 = 32 + 22 d=3.6m tan e = 2/3 e = 33.69° -

Y = -

Tr sine

4

1/ = · sin33.69° ij = 7.21 m

(11

Sm

l

F = y Jr A F = (9.81 x 1 .03)(4)[(1.5)(3.6)] F= 218.25 k N

...............

h =4 m

!J

Y:..---1-- 2 m

IC11r mine the magnitude I lnl·ation of the total h11.,latic force acting on 1 111 x 4 m gate shown Uw llgure.

~

lm

t-

1.5 m

::::H:: :w! :~:~:~Hf:~:::::::: Oil, s = 0.80 Water Glycerin, s

3m

l m

F

= 1.26

1 11

CHAPTER TH REE

112

Total Hydrostatic Force on Surfaces

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

113

Total Hydrostatic Force on Surfaces roblem 3 - 16 ( CE November 1997)

Solution ~

1m

Oil, s

f-

= 0.80

Water

1. S m

t

Glycerin, s

= 1.26

3m

>etermine the magnitude of he force on the inclined gate .5 m by 0.5 m shown in the ttgure 001. The tank of Wllter is completely closed •nd the pressure gage at the ~l
.. .. ....... .. ..... .. . .......... .... ........... . . . . . . . . . . ... . . . ' ' .... ..

3m

•

Figure 001

Water

f

2m

F =peg A

'i:.ylr + p (9.8lxl.26)(3) + (9.81)(1.5) + (9.81x0.80)(1) + 32 peg = 91.645 kPa

peg =

peg=

F = 91.645(2.x 4) F = 733.16 kN

/' .. peg

A

172 - Per. Solving for e: Solve for Ji and

y:

F=yhA

733.16 = (9.81xl .26) Ji (2 x 4)

= 7.414 m y = h /sin 60° = 7.414 / y = 8.561 m e= !.J._ = fi-(2)(4)3 Ay (2 x 4)(8.561)

=ylr

Water

3m

90000 - Pee = 9,800(2'. 65) p eg = 64030 Pa ,. 64030 (0.5 x 1.5) I' • 48,022.5 N

.,.,E

h

l.O

sin 60°

.....:i

2

1'~--­ P2 = 90,000 Pa

e=0.166m

z = 2 - e = 1.844 m Therefore, F is located 1.844 from the bottom of th e gate.

f K•'"' shown in the figure is hinged at A and rests on a smooth floor at B. Iii iti•ll• is 3 m square and oil of having sp. gr. of 0.82 stands to a height of 1.5

~h11v•• the hinge A. The air above the oil surface is under a pressure of 7 kPa tllt 11hnosphere. If the gate weighs 5 kN, determine the vertical force r

Ot11•d to open it.

114

Total Hydrostatic Force on Surfaces

: ::.:..:..:.::. ::.:.:~i~; ~:~:1: ~~. ~.::.::.: :.: >> ... . . . . .

..

Total Hydrostatic Force on Surfaces

1 1s

,,,II\ pins 20 rrun in ctiameter are used for suppor ting flashboards at the crest

.........

.

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

f masonry dams. Tests sh ow that the yield point of iron to be 310 MPa (1'1C treme fiber stress). Neglecting ·the dynamic effect of water on flas hboards !IM\d assuming static conditions, what is the proper spacing, S, of the iron pins, L1 that the flashboards 600 mm high will yield when water flows 150 mm deep v~·r the top of the flashboards.

.

T

Oil

s = 0.82 Hinge

F

1

B

Floor

Solution p

ti =0.45

= pq; A p (g

~lsm

~

.... ·.· .. ·.·.·.· ·.

= P•ir + w.,11.,

P•&= 7 + 9.81(0.82)(2.56) P<& = 27.59 kPa

::::: :::::::::~i~~:P::~ :7:k~~:::::::::::::: ::

p = 27.59 [(3)(3)] . p = 248.34 kN

·±I

Oil 5 = 0.82 p

P=yhA 248.34 = (9.81 x0.82) h (3 x 3)

Da

F

fr = 3.43 m -

Ii - 3.43 Y = sin 45° - sin 45° y =4.85 m

Floor

fi-(3)(3)3 e=- Ay (3 x 3)(4.85) l g· -

e = 0.155 m x = 1.5 + e x = 1.655 m

[LMA= 0 ]

"

.

P(x) + W(l.06) - F(2.12) = 0

2.12F = 248.34(1.655) + 5(1.06) F= 196.37 kN

3 sin 45° = 2.12 m

Moment capacity of one iron p in (20 mm 0): 1r,, = M c/I] 310 =

M(2f) 6i(20) 4

M = 243,473.43 N-mm M = 0.24347 kN-m M11111l•n t caused by F (considering Sm width of flashboard): Mr .. Fx y = h A . where A = 0.6 S == 9.81(0.45)[0.6 SJ

r y r

0.6m

A

_J

CHAPTER THREE

116

FLUID MECHANICS & HYDRAULICS

Total Hydrostatic Force on Surfaces

.CHAPTER THREE

Total Hydro~tatic Force on Surfaces

117

F = 2.6495

y = 0.3 - ,. l'= -

Ix

/\y

nCS)(0.6) :=

I 20 °C, gage A in the figure reads 290 kPa absolute. The tank is 2 m wide r1wndicular to the figure. Assume atmospheric p ressure to be 1 bar. Sp. gr. lllt'rcury = 13.6. Determine the total press ure acting on side CD.

1

r

(0.6 S)(0.45)

-r-,___ ___,___ c

= 0.067 m y = 0.3 - 0.067 = 0.233 m <'

1m

M 1 = fxy= M

Aif: 175 kPa abs

_j__ i ·~-~-...,==-

2.649 5 x.0.233.= 0.24347 5 = 0.394 m = 394 m m

h . I

Problem 3 - 19

_L_ 70 cm

ws

The semi-circ ula r gate shown in Figu re 28 is hinge d at B. Determine the force F required to hold the gate in position .

Water

Gage A

.._

•-=-==-=_=,.... Mercury

r-------"'-~ D

"'

Gage B

F

h h

=

y = 10 - 1.698

=

y = 8.302 ft

lv111g for h: I' I - L Yh + Plop

f

Solution

4 ft

8

1

l_~~~

'1()

Ii

•'

I lid force o n side CD: (Note: 1 bar = 100 kPa)

P=yh /\ P = 62.4(8.302)[ 1h rc(4)2] P = 13,019.89 lbs lg

ws

175-100 I'!> kPa

e=--= . AY

e=

lg= 0.1098 r lg = 0.1098(4)~ /~ = 28.11 f~ 28.11

---=---!rc(4}2(8.302)

6 ft

F

P(b) = F(4) 13,019.89(1.5633) = F(4) F = 5088.5 lbs

h = y

.w--i'--t----:~--..

e

4ft

l_~~~

b = 1.698 - 0.1347=1.5633 ft

=OJ

11,81(2.9) ~H.449 kPa

--,f,.--- .....

e = 0.1347 ft [LMs

= (9.81 x 13.6)(0.70) + (9.8l)!t + 175 2.2 m

B

~~~----~

,,, (3. 9)(2) .''>(3. 9)(2) 't.'15kN

P1

c

lm

Air: 175 kPa abs

i2.2"'

-

Water

J

-

E

_L 0.7 m

/11(2.9)(2) I :(28.449)(2.9)(2) llJ5kN I 1

4r/31t = 1.698 ft

-i-

E <"'i °' Fi

Mercury

fj

D P2

P1

2m

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

118

Total Hydrostatic Force on Surface s

Problem 3 - 21 :

yh A

The funnel s hown in the figure 1s full of water. The volume of the· upper part is 90 liters and the lower part is 74 liters. Wha t is the for ce te nd ing to push the plug o ut?

1 19

Total Hydrostatic Fo rce on Surfaces

~

1·.· ::

9.81 (2.6)(J .6 x 1.2) 48.97 kN

~

1

:

...

AY

1.8 m

:

1.2(1.6)' _ 1_2_

2.6

m : :

(I .6x 1.2)(2.6)

A

• 0.082 m

"'

e

t

Solution

Since the plug area in contact with water is ho rizontal, the pressure all owr it is unifor m. The shape of the container does not affect the pressure on th1 plug Force= p x A Force= 9,810(3)( ~) 100 Force= 1353.78 N

Problem 3 - 22

In the fi gure shown, the gate AB rotates about an axis through B. The gate width is 1.2 meters. A torqu e T is applied to the shaft through B. Determine the torque T to keep the gate closed.

1

water

F

z

~5.16kN-m

1.6 m

i

)i

0.8m

T

I

B

(CE Board)

11hi< .1I box, l .5 m on each edge, has its base horizontal and is half-filled U1w.itcr. The remainder of the box is filled with air under a gage pressure 1 l..l 'a. O ne of he vertical sides is hinged at the top and is free to swing 111rd To what dep th can the top of this box be submerged in an o pen body Jr11~ h water witho ut allowing any water to enter?

h

Water

1.8 m

1

1.5 m

1

r

0.75

x

1.25

LI)

..-< x

1.6 m

LI)

Water

FJ

..-<

0.2~ ~-_._-~ ·----''--

8

1. 5 m x 1.5 m

9.81(0.75) : 7.36 kPa

82 kPa

120

CHAPTER THREE

Total Hydrostatic Force on Surfaces

FLUID MECHAN IC~ & HYDRAULICS

CHAPTER THREE

Total Hydrostatic Force on Surfaces

[r M11111ge = Oj F, (x) - F1 (0.75) • F2 (1.25) = 0 Fi = p.,.A Fi = 82[ (1.5)(1.S)J = 184.5 kN

7 Eq. (1)

w.s.

T~--

\~I

the magnitude and location of l11rre exerted by water on one hi n( the vertical annular disk

4m

L

F2 = 112(7.36)(0.75)(1.5) F1 = 4.14 kN Fi= y /1 A

F3 = 9 .81 Ii [(1 .5)(1 .5)] F, = 22.0711 I=

0.75

+ I'

1.5(1.~ e= -

lg

12 [(1 .5)(1.5)]'1

Ay

0.1875

r = -- -

h

I

= Q.75 +

0.1~75 11

/' .. yli A /' • 9.81(4}[rr(l.5)2- rr(1)2] I .. 154.1 kN I ,oration of F: I

.!!. (1.5)4

p=-g-=

Ay

_ 11 (1)4

4

n[(l.5)

2

4

-

(1) 2 ]{ 4)

p= 0.203 m

In Equation (1) : 01875 . 22.07 11 (0.75 + _ .___ ). 184.5(0.75) . 4.14(,1.25) = 0

y1, = 4 + .0.203 = 4.203 m below the w.s.

11

16.55;; + 4.138 . 138.375 . 5.175 = 0 16.55 ;;

= 139412

Ii = 8.42 m " = ;; . 0.75 11 = 7.67 m

H••k in the figure shown hl!I 5 kN for each meter '1,1111111 the paper. Its center of . lty 1•1 0.5 m from the left face 0 r1 1n above the lower face. I Ir 1!11 the gate just to come 4 lh1• vertical position.

Water h

121

122

CHAPTER THREE Total Hydrosta tic Fo rce on Surfaces

FLUID MECHANICS & HYDRAULICS

CHAPTER Tf-f REE Total Hydrostatic Force on Surfaces

Solution

Considering·l m length f 1 = 1h (9.81/r)(h)(l ) f 1 = 4.905 /12 kN F2 = 9.8lh(l.5)(1) F2 = 14.715h kN

[rMo =OJ F1(h/3) + W (0.6) - F2(1.5/2) = 0 4.905Jr2 (/z/3) + 5(0.6) - 14.715/i (0.75) = 0 l .6351r3 - 11.04/z + 3 = 0 '

p=yy

0.6·1

~. W= SkN h

1' y

rlA = 2x dy

lly syuared property of parabola:

•cg

22

x2

·-=-

h/3 . 9.Blh 9.Blh

•

0

y

1.5

12

= fy 2m

\'= 2~y/3

Fz

Solve lz by trial and error h = 0.2748 m

3

4m

fll' 'YY f2 (2~y /3) dy] '" • 2.31yy3/2 dy

ell' = 2.31y fy 1 dy

I

Problem 3 - 26

3 2

I~

In Problem 3 - 25, find lz when the force against the" stop" is a m aximum. Solution

3

0

•

[I:Mo = 0) F1(h/3) + W (0.6) + P(l.5)- P2(1.5/2) = 0 4.905'12 (lr/3) + 5(0.6) + P(l.5) - 14.715/i (0.75) = 0 p = 1.Q9h3 - 7.358/1 + 3 dP - = 3.27 /12 - 7.358 = 0 d/1 h2= 2.25 h = 1.5 m

I • !.31y [

0.6·1 ~. W=S kN h

•cg h/3

I

2.31 (9.81)

t [3'1'-

3

I

9,Blh 9.Bl h

111,

=

f<;)dP 0

f f 3

Fz

111 .3 )Jp =

Problem 3 - 27

Determine the force due to water acting on one side and its location on the parabolic gate shown using integration.

I.

141.3 k N

r

0

~ y~

y(2.31yy 312 dy)

0

3

v,

0.1604

y 5 12 dy

0

l/1·

0.1604 [<2/7)y 7 1 2

r

J:

,,,, • 0. 1604 (2/ 7) 37/2_0712 1 2.14 m below the w .s.

111,

()'I' J

123

CHAPTER THREE

124

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

Total Hydrostatic Force on Surfaces

Tota l Hydrostatic Force on Surfaces n>blem 3 - 29 (CE Board)

Problem 3 - 28

~ dc1m is triangular in cross-section with the upstream face vertical. Water is ~u1hed with the top. The dam is 8 m high and 6 m wide at the base and 11ly,hs 2.4 tons per cubic meter. The coefficient of friction between the base rict the foundation is 0.8. Determine (a) the maximum and minimum unit rll'~~ure on the foundation, and the (b) factors of safety against overturning r1cl against sliding. ·

In the figure shown, f\nd the w idth b · of the concrete dam necessary to prevent the dam from sliding. The specific gravity of concrete is 2.4 and the coefficient of friction between the base of the da m and the foundation is 0.4. Use 1.5 as the factor of safety against sliding. ls the d am also safe from overturning ?

lution <;p gr.

Scone= y cone Yw 2.4x 1000 'ip. gr. of cone, Scone= = 2.4 1000

Solution Consider 1 m length of dam Wc=Yc V,. W e= y(2.4)[~)(6)(1)] W, = 14.4 by

Of CO'nC,

1111•ader 1 m length of dam

W=y, V "' (yx2.4) [! (6)(8)(1)}

w.s

w.. 57.6 y

wherey =unit wt. of wate1

F=yh A

F = y(2.25)f(4.5)(1)] F= 10.125y 4.5

Rx= F = 10.125y Ry= W, Ry= 14.4 by

m

I

F

.:r

·L.:.:J ~

yh A y(4)(8 x 1) I • 32y

w.s.

P = 32y

R,

Rv W= 57.6y = 57.6y(4)

RM= 230.4y

1.5 = 0.4(14.4by) 10.125y b = 2.637 m

1IM= P(8/3)

= 32y(8/3) 1IM • 85.33y

FS~= RM

RM-OM

OM

Ry

FS = W, (b/2)

°

lm

' NM= W(4)

FS, = µRy Rx

'

125

230.4y - 85.33y = 2.519 m < 8/2 57.6y

F(l .5)

FS. = 14.4(2.637)y(2.637 I 2) 10.125y(l .5)

=

3.3 > 1 (Safe)

r 11

1112 ..

\

·

:x

2.519 = 0.481 m < B/6

8/6 = 1 m

IRv

126

,, .

FLUm MECHANICS & HYDRAULICS

CHAPTER THREE Total Hydrostatic Force on Surfaces

CHAPTER THREE

Total Hydrostatic Force on Surfaces

Figure:

-R,, ( 1 ±6e) -

R

FLUID MECHANICS ' HYDRAULICS

B

57 .6~.81 ) [ 1 ± 6(0.:81) J

•( •

Using(+) Using (-)"

!-~ •.

l/ t

= - 139.47 kPa

r/H = - 48.88

p/{ IJ

0.8(57.6y)

R,

32y

=-·

kPa

\.- - - - - - - - - - - l l : 2 m ~

7"''

soil pressure at the toe

-)
P'"''"'" ,,

FS.. = 1.44

_ RM _ 230.4y Fs,- OM - 85.33y

'

F.'>. = 2.7

Problem 3 - 30 (CE Board May 1992)

lutlon

A gravity dam of trapez01dal cross-section w ith one tace vertical and horizontal base 1s 22 m high and has a thickness of 4 m at the top. Wa te1 upstream stands 2 m below the crest of the dam fhe specific gravity ol

Neglecting hydrostatic uplift: I Consider 1 in length of dam

masonry is 2.4 A Neglecting hydrostatic uplift: I Find the base width B of the dam so that the resultant force will cut the extreme edge of the middle third near the toe. 2 Compute the factors o f safety aga inst slid ing a nd overturning

8

Use ~l = 0.5. Co nstdering uplift pressure to vary uniformly from fu ll hydrostatu pressure at the heel to zero at the toe: . I Find the base width B of the dam so that the resulta nt fo rce wi ll act ell the extremity of the middle third near the toe. 2 Com pute the maximum and minimum compressive c;tresses achnr against the base of the dam

II

Forces Wi =Ye Vi = (y x 2.4)[(4)(22)(1)] W1=211.2y

W2 = (y x 2.4)[ 112 (B-4)(20)(1)] W2 = 24By - 96y F = Yh A F= 200w

=y (10)[(20)(1)]

Reaction

R,.. = r.f, = p R., = 200y Ry= 'EFy = W1 + W2 = 211.2y + 24By - 9f1y

Rv = 24By + 115.2y

127

128

CHAPTER THREE Tota l Hydrostatic Force on Surfaces

CHAPTER THREE Total Hydrostatic Force on Surfaces

FLUID MECHANICS & HYDRAULICS

8=

- 44.8 ± ~(44.8) 2 -4(8)(- 1499.73) 2(8)

Lm

8=11.175 m h = 10 m

'

Factors of Safety: • Factor of safety agai nst sliding:

I I I I I

· 20 m

W,

F

I

FS, = µR y

l

I I I I I

20/3 4

I

Heel

R., = (0.5)[24(11.175)y + 1.15.2y J 200y

FS, = 0.9585

.

Factor of safety against overturnin g:

RM

Uplift

FS11 = -

0M

pressure diagram

= 16(11.175) 2 y + 83.2(11.175)y -166.4y

X = B/3

1333.33y 26/ 3--~

..._, ,- - - - B

- - - --..i

IV Mo me nt a bo ut the toe

RM= W1 (B - 2) + W2[ =

f{M

C'onsidering hydrostatic uplift:

i (B - 4) ]

21'1.2y(B - 2) + (24By - 96y) [ { (B - 4) J

= 211 .2By - 422.4y + 16811' - 128By +

25fry

= I 68 2y + 83 .2By - 166.4y

OM = F(20/ 3) OM

V

FS., = 2.07

= 200y(20 /3) = 1333.3Jy

Location of /~ R. r =RM - OM Smee the res ulta nt torce w ill pass thro u gh the ex tre me ed ge o f the middl e thirds near the toe. x = B/3. Th en . (248y + 115. 2y)(8/ 3) = 168~ + 83.28'! - 166.4y - 1 ::n3.3~y 882y + 38 .4By = 168~y + 83 2By 1499 73y 8/F + 44.81l . 1499 73 = 0

Uplift fo rce, U

= 112 (20y)(B)(1) = 10811•

Rv = W1 + W2- U · = 248y + 115.2y - 1081 Rv = 14By + 115.2y RM= W1(8 - 2) + W 2[

t (8 - 4)]

RM= 1682y + 83.28y - 166.4y OM= F(20/3) + U(2B/3) = 200y(20/3) + 108y (28/ 3) c>M = 6.67827 + 1333.33y f{ v x =RM-OM (148y + 115.2y)(B/3) = 16827 + 83.2By - 166.4y - (6.6782y + 1333.33y) .J.6682 + 44.88 - 1499.73 = 0

8=

- 44.8 ± ~(44.8) 2 -

4(4.66)(-1499.73)

2(4.66)

B = 13.766 m

129

FLU ID MECH AN ICS & HYDRAULICS

CHAPTER THREE

130

Total Hyd ro static Force on Surface.s

, LUID MECHANICS & HYDRAULICS

CHAPTER THREE

Total Hydrosta t ic Force on Surfaces

13 1

Foundation stress

-r

= B/3

f

r = 13.766/3 = 4.59 m ,, = 8/2 - x = 2.2943 m •I

=-

I

T=

176.58 kN

II=

t (6) = 2 m

I

R: (l ± 6;)

Ry= 14(13.766)(9.81) R11 = 3020.73 kN

=y,,,/i A

=9.81(3)(6 x 1)

I(

2m

)I

W,=y, v, +

115.2(9.81)

= • 3,020.73 [ ± 6(2.2943) 1 I 13.766 13.766

\

1

,,., = - 438.87 kPa

l/N = 0 kPa

Problem 3 - 3 1 (CE Board May 2002)

fhe section of a concrete gravity dam shown in the figure. The depth of water at the upstream side 1s 6 m. Neglect hydrostatic uplift and use unit weight of concrete equal to 23.5 kN/m3. Coefficient of friction between the base of the dam and the fou ndation is 0.6 Determine the following: (a) factor of safe.ty against sliding, (b} the facto r of safety against overturning, and (c} the overturning moment acting against the dam in kN-m

'iJ

= 23.5f2(8){1)]

2m

I'

'I

------.

W1 =376kN

0

0

E

INJ = y,. V2

CD

= 23.5[1/2{2)(8)(1)] 1% = 188 kN

0

0

ID

= 4 - 1/2(2) = 3

'~

m

N,

=F= 176.58 kN

(2/3)(2) = 1.333 m

'"

Nv"" W1+W2=376+188 Nv = 564 kN Rx

0

176.58

0

=

1.916

0

NM = w, X1 + W2 X2

E

NM

0

0

,= 376(3) + 188(1.333) =1378.604 kN-m

0

OM= Fx y

I,

4m

.I

W1

0

0

E

F II

/ ,!i, .. 0.6(564)

0

0

0

Ii.:.= µR,,

0

=-

• 176.58(2) I IM • 353.16 kN-m

,

~,.

Is,,

-) over turning moment

RM -OM 1378.604 353.16

=3.904

I(

4m

)I

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

132

Total Hydrostatic Force on Surfaces

Problem 3 - 32 (CE Board November 2001)

fhe section of a gravity dam is as show n in the figure. Assume hydrostatic upli ft to vary uniformly from full hydrostatic uplift from the heel to zero at the toe De te rmine the total reaction per unit le ngth at the base of the dam Use sp. gr of concrete = 2 4

3

lil.81 kN/m and that of concrete is 23.54 kN / m3 Assuming uplift pressure Yaries linearly from 7 · h d m El. 52m

~10


b.

c::,.

0

D

0

b.

.::,. 0

<:::::..

Solution

l~=J R ' 2 +R I/ 2

= 62.4(30)(60

x

<

30'

·I·

10'

1)

, 6'

I

24'

l'I\ll>:tm um y rostatic pressu re at the heel to zero at tlw location of the drain, dll'termi ne the (a) location of Ow resultant force, (b) factor If Nafety agains t sliding if l~ll·fficient of friction is 0.75, (1) foctor of safety against l)Yt•rturning, (d) the stress at lhr heel and at the toe, a nd (e) th_, unit horizontal shearing trl"s~ at the base

.

W2 = y, V2 60 + 40 (62.4 x 2.4) -2- (6)( 1)

60'

\1\/2 = 44,928 lbs

vv, = y, v, = (62.4 x 2.4)

1

12(24)(40)(1)

w, = 71,884.8 lbs W4

= y.,,

V4 = (62.4) 112(30)(60)

U = y.,. Vu = (62.4)

112(60)(70)(1 )

= 224,640 + 44,928 + R,, = 266.572.8 lb~

'~.

J(112 .~20)

! . - - 26 m _____.; mI

2 •

r

= 131,040 l b~

71 .884 8

(266.572 8) 2

+

52 m

'>Om

L

lm

w, = 56,160 lbs

r~ ,.

7

- - - - - - - - - - JI - - - -

w.s.

2

w, = 224.640 l b~ w,. =

som

~10

= W1 + W1 + W,. U w, = y, v, 10 + 40 w, = (62.4 x 2.4) - (60)(1)

.

El.

3

R. = 112,320 lbs R.

50(9.81)

Heel

Toe

= 490.5

56160 . 131 .04(1

=

133

f'hl! section of the masonry dam is as shown. Th e specific weight of water is 3

0

<:::::..o

I

CHAPTER THREE

Total Hydrostatic Force on Surfaces

Problem 3 - 33 (CE Board May 1986) 10'

Consider 1- toot le ngth of dam R, = F = y h A

FLUID MECHANICS & HYDRAULICS

28CJ.26CJ 1b s per foot

38.2 m

----.i

134

II

CHAPTER THREE

Total Hydrostatic Force on Surfaces

FS

Forces

" OM FS,, = 1.81

W1 = 23.54 11/2(5.2)(52)(1 )} ·W2

= 3, 183 kN

= 23.54 [(7)(52)(1) = 8,569 kN

W3 = 23.54 [1/2(26)(52)(1)}

= 15,913 kN

IV

683,900.12 377,758

Foundation pressure

e=B/2~ =

x

38.2/2 - 13.2 = 5.9 m < 8/6

F = yh A = 9 .81(25)(50(1)]

q-- - 23,201.3

= 12,263 kN

Reaction R, = F = 12263 kN

Rv

= W1 + W2 +

Rv

= 3,183 + 8,569 + 15,913 + ·1,226.3 - 5,690 = 23,201.3 kN

Moment RM= W1(34.73) + W2(29.5) + W3(17.33) + W4(36.53) = 3,183(34.73) + 8,569(29.5) + 15,913(17.33) + 1,226.3(36.53) RM = 683,900.12 kN-m

Location of Ry

Ry

38.2

Stress

~t

[1 ± - -

6(5.9)] 38.2

the toe, (use "+");

q1 = -1,170.21 kPa

W3 + W4 - U

x = RM - OM

23,201.3 x = 683,900.12 - 377,758 \: = 13.2 m (11)

=

RM

~1 = . R; (1 ± 6; )

OM = F(50/3) + U(30.47) = 12,263(50/3) + 5,690(30.47) OM = 377,758 kN-m V

=

= 9.81[1/z(5)(50)(1)] = 1,226.3 kN U = 1/2(490.5)(23.2)(1) = 5,690 kN W4

135

Total Hydrostatic Force on Surfaces

Cons ide r 1 m le ngth of dam

w, =y, v,

111

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

f he resultant force 1s 13 .2 m from the toe

(bl FS, = µRy

R,

FS, = 0.75(23,201.3) = 1.42 12,263

Stress at the heel, (use "·") q,, = -44.52 kPa 11•) Unit horizontal shearing stress, S,

S = ~ = 12,263 = 321 kPa ' Ahasi· 38.2{1)

w.s.

h•• ~ubmerged curve AB is one t.111rler of a circle of radius 2 m \ii 111 located on the lower Htll'r of a tank as shown. The ni-;th of the tank perpendicular . I tlw sketch is 4 m . Find the ~ld~n ltude and location of the nrl1.ontal and vertical I 1111ponents of the total force •11111y, on AB

4m 0

0

0

#

2m 0 - ·I

i

12m I

A

-

B -

-

2m

CHAPTER THREE

136

Total Hydrostatic Force on Surfaces

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

Total Hydrostatic Force on Surfaces

Solution F11 = y ii A FH = 9.81(5)l(4)(2)]. FH = 392.4 k N I/= 1 + ~

t'= -

I~

h

Ay

y =5 m

~~:M!>Joi, ::_OJ x - F11 y =

3

4(2) / 12

l' = -- - -

4m

[(4)(2)](5)

w.s.

hu·c unit pressure is always normal to the \lrfoce and a normal to the circle passes ~rough its center, then the total force F holl also pass through the center of the ltdc 0, hence the moment about 0 due to. 'or due to FH and Fv is zero.

·~37.13

"=0.067 m

0

x = 392.4(1.067)

T' = 0.9578 m e

I/= 1 + 0.067

l

I/= 1.067 m

Note: This is true to all cylindrical or spherical surfaces. A

Therefore; f

11

is acting 1.067 m below B

tvblem 3 - 35 (CE Board)

w.s.

he• crest ga te shown F11 = We1ghl ,,11cn F1

= yVABCD

V ~ Ren= 4(A) A= A,+ A2 A 1 = (4)(2) = 8 1112 A1 = Y4 rr(2) 2 = 3.14

m 2

A= 8 + 3.1 4 A= 11.14 m 2 VA8CO

= 4(11 .14) = 44.56 m:'

F11 = 9.81(44.56) F11 = 437.13 kN

tllffists of a cylindrical urfl\Ce of which AB is the 1111 supported by a rurtural frame hinged at , The length of the gate it) m . Compute the 111g11itude and location of ill horizontal and vertical r11ponents of the total on AB.

0

:.. 0

•••ure

Sm

lutlon .

Loca tion 1f Fv A

x=A

1 X1

+

c

A2X2

r1=1 m t2

t1

4r 4(2) = -311 37! = 0.849 m = -

11.14 x = 8(1) \' = 0.957 m

+

3.14(0.849)

Therefore; fv is acting 0.957 to the right of A

E ~ oc! 0

~

____

!

Fi

= 4.33m

! !

...__~

+---FH~~j

·~

.... Q

y

L

=10 m

A

10cos60° = Sm

137

138

CHAPTER THREE Total Hydrostatic Force on Surfaces

CHAPTER THREE Total Hydrostatic Force on Surfaces

FLUID MECHANICS

& HYDRAU LIC~

FH = yh A F11 = 9.81(4.33)[10(8.66))

( 1111sidering 1 meter length:

FH = 3679 kN

111 yh A

y=

I 11

1 (8.66) = 2.887 m

9.81(3)(6

J \ ' Y Vs A~

Fv = y V.~s<..

= V AOllC -

V AB('

=

VAac

= 125.9 m 3

= A sector - A trianglc

2 A - n( 6) ( 60°) -·1/2(6) 2 sin60°

V A08

S+,].'° (8.66) X 10 -

1)

I 11 176.58 KN

Therefore; FH is acting 2.887 m above 0

V ABC

X

139

V2(1Q)2

s360° A, = 3.26 m 2

[60°1ifuo] 10 X

I,, • 9.81(3.26 x 1) I 1• • 31.98 KN J JpH2 + Pv2

Fv = 9 .81(125.90) Fv = 1235 kN

Mo ment about 0 d ue to FH and f~ = 0 Fv (x) = FH (y) 1235 x = 3679(2.877) x = 8.57 m

J(176.58) 2 + (31.98),

~

2

I' 179.45 KN

lem 3 - 37

Therefore; Fv is acting 8.57 m to from 0

Ii 11l.1te

the magnitude of the ult.ml pressure on a 1-ft-wide strip 11 11c>rn.icircular taintor gate shown

Problem 3 - 36 (CE May 1999)

5' A

I lv,11 rc-12.

Calculate the magnitude of the resultant fo rce per meter length due to water qcting on the .radial tainter ga te shown in Figure 021 .

Figure-12 Figure 021

Peg

A

(62.4 x 2.5)(5 x 1) = 780 lbs

I, y VABC I 1• 62.4 x [ f (5)2(1)]

= 1225 lbs

J(FH )2 + (Fv )2 1·

~ (780) 2 + (1225) 2 = 1452 lbs

B

CHAPTER THREE

140

Total Hydrostatic

~orce

on Surfaces

FLUID MECHANIO & HYDRAULICS

CHAPTER THREE

Total Hydrostatic Force on Surfaces

141

blem 3 - 39 (CE Board November 1993)

Problem 3 - 38

Determine the magnitude of the horizontal and vertical components of the total force per meter length ~cting on the three-quarter cylinder gate shown

the figure shown,. the 1.20 m •meter cylinder, 1.20 m long is IC!d upon by water on the left and ll h11ving sp. gr. of 0.80 on the right. ll."rmine the components of the 1 lion at B if the cylinder weighs 62 kN.

2m

1.2m

8

f'H1

= y Ti A = 9.81(1.2)(1.2 x 1.2)

flll

= 16.95 kN

f'H1

Solution

l'v1 = YVi l'v1 = 9.81[1/2 7t (0.6)2(1.2)]

h2 = 4 m

·2:) =

FH2 0

Im

I 'VJ = 6.657 kN

(

=y h A= (9.81 x 0.8)(0.6)(1.2 x 1.2) /'112 =6.78 kN / 112

Im

l'v2 = yV2 = (9.81 x 0.8)[Y2 TC (0.6)2(1.2)] l'v2 = 5.32 kN

Im

f" .. ll]

FH = y h A FH = 9.81 (3)[(1)(2)] FH = 58.86 kN

1'111 -

FH2 - RsH = 0

/{1111=16.95 - 6.78 1~1111=10.17 kN 2m

2m

· ~'· =Qt Fv = yVol Fv = 9.81[4(2)(1) .+ 0.75ln(2)2(1)l Fv = 170.94 kN

II 1\ OJ N11v+ Fvi+Fvi - W=O l~" v

.. 19.62 - 6.657 - 5.32

N"v• 7.64 kN

l.2m

CHAPTER THREE Total Hydrostatic Force on Surfaces

142

.CHAPTER THREE Total Hydrostatic Force on Surfaces

FLUl.D MECHANIC\ & HYDRAULIC~

Problem 3 - 40

f2 = (9.81 x 0.82)(0.00628)

An inverted conical plug 400 mm diameter and.300 mm long closes a 200 mn1 diameter circula r hole at the bottom of a tank containi.ng 600 mm of oil h avi111•. sp. gr. of0.82. Dete rmine the total vertical force acting on the plug.

f2 = 0.0505 kN

143

fv = F1 - f2 Fv = 0.114 - 0.0505 Fv = 0.0635 kN = 63.5 N downward

Solution

~

0\ diameter ho~izontal cylinder 2 m long plugs a lm by 2m rectangular ~~ i1t the bottom of a tank. With what force is the cylinder pressed against lKiltom of th: tank due to the 4-m depth of water?

E

I~--

0.4 m

F

- - - ' > 'I

1- o.2m ~

h1 • 2 x

j 0.45

I

~I ·

! •

hi

i

m

h,

I

i

! •Oil s = 0.82

o.1s.m

Fi =yV, F1 = .(9.81 x 0.82)[n(0.1) 2 (0.45}] F1= 0.114 kN

F2 = yV2 V2 = VFrustum -

(1 cos 30°) ).732m ~ 4 - lti 2.268 m

'1 •·VyV,=Ai x 2 1

Area, Ai= Area of rectangle DEFG -A4 n(1) 2 (60°) Area of segment, A 4 = - 1/z(l) (1 ) sin 60° . ' 360° Area of segment, A 4 = 0.09059 m2 Area, A 1 =1(2.268) - 0.09059 Area, A 1 = 2.1774 m2 V1 =2.1774(2) = 4.355 m 3 , I - 9.81(4.355)

V cylinder

5

V2 = n(O.l } [(0.2) 2

3 V2 = 0.00628 m 3

~ (0.2)(0.1) + (0.1) 2 ] - n(0.1)2(0.15) '

''

112.72 kN

144

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

Total Hydrostatic Force on Surfaces

I

CHAPTER THREE

LUID MECHANICS HYDRAULICS

Tota l Hydrostatic Force on Surfaces

145

f ,, =y h A

1t(l ) 2 (120°) 1 - /2(1)(1) sin 120° 3600 Area of segment, A2 = 0.614 m 2 V2 = 0.614(2) = 1.228 m3 F2 = f 3 = 9.81(1.228) F2= f 3 = 12.05 kN

r,, = 9.81(6.12)[(4.24)(1] r,, = 254.56 kN

Area of segment A 2 = I

f v= y

V shadc
V shadcd

= ~Asemicirde + A1r• pezoid} X 1

V shadcd =

Net force = F1 - F2 - f3 Net force= 42.72 -12.05 - 12.05 Net force = 18.62 kN ·

Vs1iadc<1

lt1t(3} 2

+~(4.24)](1)

=40.l m3

r

v = 9.81(40.1) f v = 393.38 kN

Problem 3 - 42

In the figu re shown, determine the horizontal and vertical components of the · total force acting on the cylinder per m of its length.

w.s. Water

4m

Solution

lutlon 2m

A

/11 y h A lm Net Vertical Projection

,,

2m

t

, ,, I II

9.81(1)(2.5 x 2) 49.05 kN

I \I

'Y VABC

II I,

9.81[(2 x 2) - 0.251t(2)2 ](2.5) 21.05 kN

- ·-·- ·- ·- ·- ·- ·1C i i 1 2m I

FH )I

2/ 3 ""--------' ---"'--~----1--,µ,

2.Sm

0

Fv

..----z-- ~

B

146

Tota l Hyd rosta tic Force o n Surfaces

Solve for z and x Since the surface 1s circul ar. 2":M11 = 0 d ue Fv(z) = FH(2/3) 21.05(z) = 49.05(2/3) z = 1.55 m x = 2 -. z = 2 - 1 .55 t = 0.45 m l2":Mu =OJ FH(2/3) + Fv (x)- F(2) = 0 2F = 49.05(2/3) + 21.05(0.45) r = 21.09 k N

to

FH and f' v

3m

l Solution Open

4m

h 3m

3m

l

= Pcgo A = (9.81 x 0.80)(7 - 1.273) x lf2n:(3)2

t

l1111 rcs due to water: l'11w = pcgw A /'11w= [(9.81x0.8)(7)+9.81(1.273)] x 1/m(3)2 l'11w = 953.19 kN

4m

3m

I'110 1'110

l'vo = Y• Vo Vo= VoJume of imaginary oil above the surface v. = Volume of half cylinder - Volum e of% sphere v. = 1/211:(3)2(7) - % 11:(3)3 v. = 70.686 m3 l'vo = (9.81 x 0.80)(70.686) l'vo = 554.74 kN Open

h

Total Hydrosta tic Fo rce on Surfaces

l"ito = 635.4 kN

Problem 3 - 44

he cylindrica l tank shown has a he mispherical e nd cap. Compu te the horizontal and vertical components of the total force due to oil and water acting on the he misphere

CH APTER TH REE

FLUID MECHA NICS & HYDRAULICS

CHAPTER THREE

"., Oil, s = O.B.O

/ ''

''

l'vw =Weight of re al and imaginary oil above the surface +weight of real water above the surface l'vw= (9.81x0.8)x 1/m(3)2(7) +.9.81 x %f it{3)3 l'vw = 1,054.01

Water

1l>l.11 horizontal force, FH = FHo + FHw lot.ii horizontal force, FH = 635.4·+ 953.19 I ot.11 horiion tal force, FH ,;, 1,588.59 kN ~ 101111 ve rtical force, Fv = Fvw- Fvo ~ 111,11 vertical force, Fv = 1,054.01 - 554.74 .,-11!.11 vertical force, Fv = 499.27 k N

Another w ay to solve for the total vertical force, Fv: "· Oil, s = 0.80

/ 'v = weight of fluid within the hemisphere l'v = Yo Vo + "110 Vw /\, = (9.81x0.8)[ ~x ~7t (3)3)] + 9.81 [ ~ x ~7t (3)3)] l'v =499.27 k N

147

CHAPTER THREE

148

CHAPTER THREE

FLUI D MECHANIC\ & HYDRAULICS

To ta l Hydrostatic Force on Surfaces

Total Hydrostatic Force on Surfaces

149

Problem 3 - 45

Pressurized water fills the tank s hown in the figure hydrostatic force ~I surface

Compute the m I

lt•rmine the force required to open the quarter-cylinder gate shown lght of the gate is 50 kN acting 1.2 m to the right of()

The

Hem1sphencal surface

F

Solution 4

C0.,vert 100 kPa to its t!4 uivalent pressure head, 11"1 11,'l =

~('1•

the gate has circular surface, total water pressure passes ~ugh point 0 which is also the tlon of the hinge, therefore the n1i•nt d ue to water . pressure ~Ill the hinge is zero.

E.

y

= 100

h

9.81 Ji,'l = 10 .194 m " I

Ii= 10.194 - 5 11 = 5.194 m

1.2 m

F

~ SOkN E l/l

vi

11 • 0) J'(2.5) = 50(1.2) + Fr(O) f •24 kN

F = Weight of imaginary water above the hemispherical surface F = y.,, V•. V,,. = Volume of cylinder+ Volume of hemisphere V,,. = n(2)2(5. 194) +

V,,, = 82.025 m-1 I- = 9.81(82.025) F = 804.7 kN

t x 4 n(2)~

h11111lc;phericaJ dome shown is filled with oil (s = 0.9) and is attached to the It liy eight diametrically opposed bolts. What force in each bolt is required holcl the dome down, if the dome weighs 50 kN?

CHAPTER THREE

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

150

Total Hydrostatic Force on Surfaces

Total Hydrostatic Foice on Surfaces

151

I + F.,, - F0 ,1 = (I I •

F...

Fo11 •

Fnil = YV011 i\lx>ve the '"""t'

Dome '/~ £~ "".,._ .

Oil s = 0.9

' 2m \

{4=. ·-· r-"'--~oll '

F'n11 =

Y

2m

s. 0.8

63.91 kN

3m

F.,, =Pair A F.1r = 20 [ f (1.61)2 J = 40.72 kN , •• 63.91 . 40.72 f' 23.19 kN

Solution FV =

011

= (9.81x0.8)[n(0.805)l(5) . t n'(o.805)2(3)l

rml

Vun;.)~&n.r) 011 .tbo".e tht.> Jon\\

Fv = (9.81x0.9){n(2) 2 (8) · ~7t(2)3 1

F1· = 739.66 kN

'00 mm diameter steel pipe 12 mm thick carries water under a head of 50 m wnlt'r. Determine the stress in the steel

W = F1 - 739.66 - 50 Ft..111 8 Fho11 = 86.2 kN

8F1"'''

+

lutlon

rs,= po 1 21

s, = 9.81(50)(300) 2(12) S, = 6.13 MPa

Problem 3 - 48

Determine the force F required to ho ld the cone shown. Neglect the weight of the cone

Oil s =o.s

2m

3m

= 6131 .25

kPa

lem 3 - so r111ine the required thickness of a 450 mm diameter steel pipe to carry a pressure of 5500 kPa if the allowable working stress of steel is 124 M.Pa.

h11111n

pD) 21

l~1

124 )( 1000 = P(45 0) 21 I

"'9.98 mm say 10 mm

J5 2

_>i:,tA'PTER TAREE-...

/

Total Hydrostatic Force on Surfa ces

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

153

Total Hydrostatic Force on Surfaces

Problem 3 - 51

Pipe diame te r, D = 6 m = 6000 mm Maximum pressure the tank (at bottom), p = y0 11 h p = 9.81(0.8)(7) = 54.936 kPa

Determ ine the s tress at the walls of a 200 mm diameter pipe, 10 m m thick u nder a press u re of 150 m of water and submerged to a depth of 20 m in salt water

2(110 x 10 3 )(300) 54.936(6000) S = 200.23 mm say 200 mm

s=

Solutien

i.S, = pD I 21 P = P.in>1<1o

- P o u1.>1de

p = 9.81(150) - 9.81(1.03)(20) p = 1269.4 kPa = 1.269 M Pa -.;, =

1.209(200) 2(1 0)

=

12.69 MPa

11:1.n-:valled ha~l~w sp here 3.5 m

in diam e te r ho lds h elium gas a t 1700 kPa. )1 hr1rune the m1rum um wall thick ness of the s phe re if its allowable stress is MPa.

Problem 3 - 52 A 100-mm-10 steel pipe has a 6 mm wall thickness. For an allowable tens1h

.;tress of 80 MPa. w hat maximum pressure can the pipe withs tand ? Solution

I-"·=

pD 21

80 f'

Wall stress, S,

= -pD 41

60,000 = 1,700(3.5 x 1000) 4t I = 24.79 mm

I

= p(lOO) 2(6)

=<J.6 MPa = 9,600 kPa

~.. rllca.l cylin.d _rical ta1;1k is 2 m eters in diarrieter an d 3 meters h igh. lts sides l11•ld m position by means of two steel hoops, one a t the top and the other

lhc: bottom. If th e tank is filled with water to a depth of 2.1 m, dete rmine h 11~ile stress in each hoop.

'

Problem 3 - 53

A wood e n storage va t is 6 m in dia me ter and is fi lled with 7 m of oil, s = 0 ~ f he wood staves are bound by flat steel ba nds, 50 m m wide by 6 m m th11 ~ wh ose allowable te nsile s tress is 110 MPa. What is the required spacin g of Iii ba n d~ near the bottom of the va t. neglecting any initial stress 7

2m

,--

Solution

..,pacing ot hoops, .'>

=

?.:, 1 A 1, pO

Allowable te nsile stress of hoo ps, :,,= 110 MPa C ross-section al area of hoo ps. A i. = 50(6) = '.\00 mm 1

-..../ .

r.,., ./-

-i-,

3m

2. l rrj

3

w~ te

2.1

J:' '

.....

i,..

2 T;

CHAPTER THREE

FLUID MECHANIO & HYDRAULICS

CHAPTER THREE

154

Total Hydrostatic Force on Surfaces

Total Hydrostatic Force on Surfaces

155

p: M1op = 01 2T2(3) = F(2. 3) r~ = 0.38331

(ylindrical container 8 m high and 3 m in diameter is reinforced w ith two

ltl>ps 1 mete r from each end. When it is filled with water, what is the tension •till'h hoop due to water ?

7 Eq . (1 l

f = y /J A

r

9.81 ( 221 )[ (2) (2 1lJ =43.20 kN

In Eq. (1) Ti= 0. 3833(43.26) r2 = 16.58 kN . (te nsion in the bo ttom hoop)

12:

F11

3m

Tt---· ·-.

= OI

2T2 + 2T1= I 2T1 = F · 27 z 2T1 = 43. 26 • 2(16.58) r, = 5.05 kN (tension in the top h oop)

8l1m 6m . . ....Y'!~~e!.... .. ~

m

Problem 3 - 56

A vertical cylindrical tank, open at the top, is filled with a liquid . Its sides t1n held m position by means of two steel hoops, one at the top an d the other 111 the bo tto m Dete rmine the ratio of the stress in the upper hoop to that in lh lowe r hoop Solution

fl'l111p hoop = OJ 2'1'2(6) = F (13/3) /';.., 13F/36 ,., .. 13(941.76)/36 /', • 340.08 kN

1·

Raho

= r, I ri

ILM"'" = Ol 2T2(lt) = F(2'1/3) = F/3

ri

IE.M1x>11om = 01

I h

Rah o

l

hoop =

OJ

J'/'1(6) = F(S/3) /'1 SF/36 = 5(941.76)/36 I 1 130.8 kN

2 T1

= F/ 6 F/ 6 =- = 0.5

t\·l1.111om

Liq Id

2T1(/1) = F(/t/ 3)

T,

I .. yh A

I • 9.81(8/2)(8(3)] I • 941 .76 kN

l

2h/3

F /3

j"

F

h/3 2 T,

•m 3 - 58 (CE Board November 1982~

lhhlr lrcll tank with its axis vertical is 1 meter in diameter and 6 m high. It 1l l11gt lher. by two steel hoops, one at the top and the other at the bottom. ll1p1ids A, 8, and C having sp. gr. of 1.0, 2.0, and 3.0, respectively fill this -1i1d 1 having a d epth of 1.20 m. On the surface of A there is atmospheric 11111 Pind the tensile stress in each h oop if each h as a cross-sectional area ~1 I 111u1i . 1

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

156

Total Hyd rosta t ic Force on Surfaces

CHAPTER THREE

Tot al Hydrostatic Force on Surfaces

T2 = 3.6(9810)

Solution 2T 1

p1 =

o

Stress 52 = '

Ltqutd A

= 1.0

5

=35316 N

I1.. = 35316 A2

1250

Stress, 52 = 28.25 MPa

. 1.2 m

157

7 stress in bottom hoop

p. r11 =OJ Liquid B

s = 2.0

Liquid C

s =3.0

3.2

1.2 m

1.2 m

LIQuld B

1.2

l.JQUld C

l.(

+ Y4/I ~

p1 = 0 + (y

x

1 )(1 2) = 1.2y

r' = p2 + yµh~ p~

= 1 .2y +

11pcn cylindrical tank of 1.86 m 2 cross-sectional area and 3.05 m high l111ns 2,831 liters of water. Into it is lowered another smaller tank of the I'll' height but of 0.93 m2 cross-section in the inverted position, allowing its 11 1•nd to rest on the bottom of the bigger tank. Determine the maximum 11111 per vertical millimeters on the sides of the bigger tank. Neglect the ~rwss of the metal forming the inner tank and assume normal barometric

(yx2)(1 2) :. 3 .6y

p4 = p1 + ydr, p4 = 3.6y + (yx3)(l .2) = 7.2y F, = 1/2(p2)(l .2)(1 ) F, = 112(1 .2y)(l 2)(1 ) = 0 .72y

F2 =

Ti= 1.44(9810) T1 = 14126.4 N _ T _ 14,126.4 Stress, S1 - -1 - - - A1 1,250 Stress, 51 =11.3 MPa 7 stress in top hoop

pi= 0 p2 = p I

2T1 + 2T2 = F1 + f2 + f3 + f4 + Fs 2T1 = 0.72y + l.44y + l .44y + 4.32y + 2.16y - 2(3.6y) T1 = l .44y

p2(1.2)(1 )

F2 = 1.2y(l 2 )(1 )

= l.44y

F1 = 112(p1 pz)(l .2)(1) F1 = 112(3 tw r~ =

p,p

I 2y)(l 2)( l l

= I 44y

T.-_,__

2)(1)

F,

= 3 6y(l

F,

= 'li(p4

2)(1)

= -U21

pi)(l .2)(1) F~ = 112 (7 2y . ~.6y)(1 2)(1)

I-: I I

2

1.86 m

2

0.93 m

-::;-!

j l

3.05

l 1

= 2.1611•

t--

P2 V2

t--

305

ltM1op =OJ

F 3 2)

3 6(2T2) = F1(0.8) + F1(1 .8) + F;(2) + F4(3) + s( .

7:2T2 = 0.72y(0.8) + 1 44y(1.. 8) + 1.44y (2) + 4.32y(3) + 2.16y(3.2)

T2 = 3 .6y

----.....IL O Before lowering

I:.

. 6 After lowering

FLUID MECHANIC\ & H YDRAULICS

CHAPTER THREE

158

Total Hydrostatic Force o n Surfaces

In Figure 6 1.86b + 0.93/1 = 2.831 2/? + 11 = 3.044 ~ Eq. (1) b = 1.522 - 0.5/i

n iceberg having specific gravity of 0.92 is floating on salt water of sp. gr. 03. If the volume of ice above the water surface is 1000 cu. m., what is the tlt1l volume of the ice?

V = total volume of ice

[p1 V1 = p2 V2] p1 = 101.325 kPa (atmospheric pressure)

Vo= volume displaced

V1 = 0.93(3.05) = 2.8365 m 3 p2 = 101.325 + 9.81/J V2 = 0.93(3.05 - b)

Vo= V-1000 Wice= Yicc V = (9.81x0.92)( V) Wice= 9.0252 V

101.325(2.8365) = (101.325 + 9.811!}[0.93(3.05 - b)] 309.04 = 309.04 - 101.325b + 29.92/t - 9.81blt 29.92/i - 101.325b - 9.81blt = 0 29.92/i -101.325(1.522- 0.5/!) - 9.81(1.522 - 0.5/t)h = 0 29.92/J - 154.22 + 50.66/1 - 14.93/i + 4.9051!2 = 0 4.905172 + 65.651! - 154.22 = 0 -65.65±~..-(6-5.-65-)2---4-(4-.9-0_§-).(---15-4.-22-) 2( 4.905f

= 2.039

BF = Y>c•water Vo BF = (9.81xl0.3)(V - 1000) BF= 10.1043(V - 1000)

= 1.03

l~fv =OJ

W,,. =BF 9.0252 v = 10.1043(V - 1000) 1.0791 = 10104.3 V = 9,364 cu. m .

sbody

The maximum tensile stress occurs at th e bottom of the tank.

p = yH = 9.81(2.542)

Tension, T:

Seawater, s

"olher Solutio n : 1'11r hom ogeneous solid body floating on a homogeneous liquid:

b = 0.5027 m H = b + It = 2.542 m

p = 24.937 kPa = 0.024937 MPa

159

(CE Board November 1977)

.

Volume of water= (1.86 - 0.93)(b + /1) + 0.93b = ~-:d

h=

CHAPT ER THREE

Total Hydrosta tic Force o n Surfaces

Vo = -

-

Ybocty

Vbo
Sliquid

Yliquid

0 92 · v v - 1000 = 1.03

T

0.106796V = 1000 V = 9,364 cu. m.

2T = pD(l)

f

02=1.86 m 2

D = 1.539m=1,539 mrn 2r = 0.024937(1,539)(1) T=l9.2 N

blam 3 - 61 (CE Board May 2003, Nov 2002, May 2000, Nov 1992) ~1111 k of wood 0.60 m x 0.60 m x /1 meters in dimension was thrown into the •~ 1 1111d floats with 0.18 m projecting above the water surface. The same block

thrown into a container of .a liquid having a specific gravity of 0.90 and it 0.14 m. projecting above the surface. Determine the following: (11) the value of /1, (Ii) Che specific gravity of the block, and (1l tl w weight of the block. •~wi th

160

FLUID MECHANICS & HYDRAULICS

CHAPTER THREE

Total Hydrostatic Force on Surfaces

Solution In Water: Draft =

CHAPTER THREE

. y,tOl1l'

wstonc

=-v-SlOne

swood h

Y.iono =

Swaler

Ir - 0.18 =

5

~ 0.0163

wood 1 Ir

Swootl 11 =Ji - 0.18 -7 Eq. (1)

161

Total.Hydrostatic Force on Surfaces

=

28,204

N/m~

(CE Board May 1993) In Water (S

=1)

ll1>dy having asp. gr. of 0.7 fl oats on a liquid of sp. g r. 0.8. The vo lume of \1• body above.the liquid surface is what percen t of its total vol ume?

, In ano ther liquid 5 Draft= wood h

Sbody

S11quiJ

Vo= --V1iody Sliquid

/1 II - 0 .14 =. Swood 0.9 Swood It= 0.9/i - 0.126 -7 Eq. (2}

Vo= 8:~ V1io.iy = 0.875V1ioJy In other liquid (S = 0.9)

[Swood It= Swood Ii] . ,, - 0.18 = 0.9/i - 0.126 Ii= 0.54 m -7 height of the block Substitute Ii to Eq. (1): Swood(0.54) = 0.54 - 0.18 Swuod = 0.667 -7 Specific gravity of wood

Since the volume of the body disp laced (below the liquid surface) is 0.875 147.5% of its total volume, then the volume of the body above the liquid 1foce is 12.5'Yc• of its total volum e.

(CE November 1997)

bl1H'k of wood 0.20 m thick is floating in sea water. The specific gravity of while that of seawater is 1.03. Find the minim.um area of a block l~h h will support a man weighing 80 kg. ~11d is 0.65

Weight of block= Ywood V1i1ock Weight of block= (9.81 x 0.667)[(0.6 x 0.6)(0.54)] Weight of block= 1.272 kN ~

WMAN

Problem 3 - 62

A stone weighs 460 N in air. When submerged in water, it weighs 300 N Find the volume and specific gravity of the stone.

Solution Weight of stone= 460 N Weight of stone in water= 300 N Buoyant force, BF = 460 - 300 = 160 N

[BF= Ywdler Vstone] 160 = 9810 (Vsione) V5 10110 = 0.0163 cu. m.

\

= 80 kg

Wwooo

i t

~ 0.2m

t l~Fv =

O] BF=

Wm•n + Wwood Ys"' Vwood = Wnian +"!wood Vwood (1000 X 1.03) Vwood = 80 +(1000 X 0.65) Vwoocl Vwood = 0.2105 m 3 = Area x 0.2 Area = 1.05 square meter

I

162

,

FLUID MECHANIC\ & HYDRAULIC\

CHAPTER THREE

Total Hydrostatic Force on Surfaces

CHAPTER THREE

Total Hydrostatic Force on Surfaces

163

blem 3 - 67

Problem 3 - 65 (CE November 1997}

A cube of wood (s.g. = 0.60) has 9-in sides. Compute Lhe magnitude a11d direction of the force required to hold the wood completely submerged 11 water. Solution·

uniform block of steel (s = 7.85)

Ill float at a mercmy-water

Water

\tc•rface as shown in Figure 27. hat is the ratio of the distances '' and 11 b11 for this condition?

b

3

Weight of wood= (62.4 x 0.60) ({2 ) = 15.795 lbs Buoyant fo rce w hen complete ly s ubmerged in wa ter:

BF= 62.4 ( 192 )3 = 26.325 lbs Req uired force= 23.325 - 15.795 Required fo rce= 10.53 lbs downward

Figure 27

Water

Problem 3 - 66 (CE Nov 2000)

T he block shown in Figure 04 weighs 35,000 lbs. Find the value of /1.

.

Oil s = 0.8

9 ft

Figure 04

Solution

From the figure s hown: 12' x 12'

[l:Fv = O] Bf1 + BF2 = 35,000

21,565.44 + BF2 =35,000 Bf2 = 13,434.56 lbs

BF2 =Yw Vo 13,434.56 = 62.4 ((12)(1 2) Ii] It = 1.495 ft

w

12' x 12'

Water

BF, = Yo11 Vo Bf1 = (62.4x0.8)(12x12x3) BF1 = 21,565.44 lbs

BF1

I A be the horizontal cross-sectional Mercury (S = 13.~) ~.. of the block. 111'1 + BF2 = W Y11• Vo... +y,,, Vo,,, = y, V I) 8l(A x a)+ (9.81 x 13.56)(A x b) = (9.81 x 7.85)[A(a + b)] II I 13.56b = 7.85a + 7.85b ~.7 1 b = 6.85 a 41/b = 0.834

BF2

35,000 lbs blem 3 - 68 (CE May 1998)

Oil s = 0.8 Water

•~I

kg steel plate is attached to one end of a 0.1 m x 0.3 m x 1.20 m wooden "· what is the length of the pole above water? Use s.g. of wood of 0.50. t1h•t't buoyant fo~ce on steel ·

CHAPTER THREE

164

Solution Neglecti ng the buoyant force on steel: BFwooJ = Wstccl + Wwood 1000(0. l x 0.3 x y) = 5 + "1000(0.5)[0.1 x 0.3 x 1.2] y = 0.77 111

165

Total Hydrostatic Force on Surfaces

llOden buoy (s.g. = 0.62) is 50 mm by 50 mm by 3 m long i;. made to float water (s.g. = 1.025). How many N of steel (s.g. = 7.85) should be ·lwd to the bottom to make the buoy float with exactly 450 mm exposed 1• the wate r s urface? ·

••'I.\ h

E

"= 1.2 - !/ h 1.2 .- 0.77

CHAPTER THREE

FLUID ME CHANIC & HYDRAULIC\

,

Total Hydrostatic Force o n Surfaces

N

,...;

0.05 m

Utlon y

=

~

.. 0

~J',,.,,, + Bfwood - Wwood - W
Ii = 0.43 Dl

II f~1..,1 = Ysw

0

iD

0.45 m

V s1cc1

llfstc>cl = (9810 X1.025}Vsteel Hf,l<'e1=10,055.25 V,1,,.,1 N Hfwood

= Ysw Vo

Hfwood

= (9810 X 1.025}[(0.@5)2(2.55)]

Problem 3 - 69 If a 5-kg s teel plate is attached to o ne end of a 0.1 m x 0.3 m x 1.20 m wood• 11 1 pole, what is the len~th of the pole above water? Use sp. gr. of wood of 0 '

llf'wood

= 64.1 N

Wwood

= (9810

and th a t of steel 7.85.

Wwood

= 45.62 N

= (1000 X 0.5)(0.1 Wwood = 18 Kg

X

0.3

X

1 .2)

BFw = 1000(0.1 x 0.3 x d) BFvv = 30 d BFs = 1000 Vs Ws = (1000 x 7.85) Vs = 5 Vs = 0.000637 m 3 BFs = 0.637 Kg W wood + Ws1..-1 = BFs + BFw 18 + 5 = 0.637 + 30d d = 0.745 m

t

V wood X

0.62}[0.05}2{3)]

Vstc't!I.

IV,,,,,,= (9810 X 7.85) I v.lt'\'I = 77008.5 Vsteel

w,,,.,, = 5 kg.

x = 1.2 - d x = 0.455 m

= Ywood

IV,i.-cl = Ystccl

Solution Wwood

~Vwood

2.55 m

Vstccl BFst..1

(H)'i'i.25 Vstt'CI + 64.1 - 45.62 - 77008.5 Vstl'CI = 0 1 11,25 = 18.48 ' Y1 I 0.000276 m3

v.,.,.,,

1

9810(7.85)(0.00276) 21.255 N

..

d I

m 3-71 1• of lead (sp. gr. 11.3) is tied to a 130 cc of cork whose s pecific gravity is I l"'y Ooat just s ubme rged in water. What is the weight of the lead?

166

FLUID MECHANIO &HYDRAULIO

CHAPTER THREE Total Hydrostatic Force on Surfaces

, LUID MECHANICS &HYDRAULICS

CHAPTER THREE Total Hydrostatic Force on Surfaces

167

Solution

[l:Fv =OJ We+ WL = BFc + Bh We= Ye Ve We = (1 x 0.25)(130) We = 32.5 grams BFe=Ym Ve BFe= (1)(130) BFc = 130 grm

w.s.

We

t

Cork

BFc Water

WL=YL VL

BFc

Wt= (1 x 11.3) Vt Wt= 11.3 VL BF1. = y.,, Vt BFL = (1 ) vt =

Lead

BFc WL

BFL

vL

32.5+11.3 Vt = 130 + Vt VL = 9.47 cc

.

(a) Lead is fastened outside the cylinder

WL = 11.3(9.47) WL = 106.97 grams

) I.cad is fastened outside BFc = Yw Vo BFc = 9.81[ f (1)2(1 .S)J

Problem 3 - 72 (CE November 1993)

A hallow cylinder 1 m in diameter and 2 m high weighs 3825 N. (a) I h11 many kN of lead weighing 110 kN/m\ m ust be fastened to the outside bolli111 of the cylinder to make it float with 1.5 ~submerged in water? (b) How n11111 kN of lead if it is placed inside the cylinder? \ \ \

I

I

BFe = 11.56 kN

Bft = Yw Vt Bft = 9.81 Vt Wt = YLVt= 110VL

I>.: Fv = OJ BFc + Bft = We + WL n .56 + 9.81vL=3.825 + novl Vt = 0.0772 m 3 WL =110(0.0772) = ~.49 kN

) l.c•ad is inside the cylinder ll:l'v = OJ WL +We= Bfc

Wt + 3.825=11.56 Wt= 7.735kN

(b) Lead is placed inside the cylinder

I .

FLUID MECHANICS & H YDRAULICS

CHAPTER THREE

168

Total Hydrostatic Force on Surfaces

LUID MECHANICS HYDRAULICS

Problem 3 - 73 A sto ne cube 280 nun on each side and weighing 425 N is lowered into a tank containing a layer of wat.e r 1.50 m thick over a layer of mercury. Determine the position of the block when it has reached equilibrium.

CHAPTER THREE

Total Hydrostatic Force on Surfaces

s.g.

= Q.8

TT

Solution W= 425 N

16 9

s.g. = 0.7

2.2 ft

BFM= YM VDM BFM = (9,810 x 13 .6)[0.282(x)] BF,v1= 10,459.81 x

~k

0.28 m

s.g.

=1.6

;

:----__;.;::;.,;__J

BFw= Y1v Vmv BF1\ 1 = 9,810((0.28)2(0.28 - x)] BFw = 769.1(0.28 - x)

s.g.=1.4

I

2.2ft ~

I". .. (62.4 x 1.4)((2.2)2(2.2 - h)] + (62.4 x 0.8)((2 2)2(/t)] ~ I • (62.4 x 2.22] (3.08 _ l .4h + 0 _8h)

p:Fv= O] BFM+ BFw= W 10,459.81 x+769.1(0.28 - x) = 425 9690.71x = 209.652 x = 0.0216 m x = 21.6 m m

~ (62.4 x 1.6)((2.2)2(1.1)] + (62.4 x 0.7)((2.2)2(1.1)] n

Therefore; the b l ock will float wit h 2il.6 mm below t h e mercu ry su rface.

Problem 3 - 74 A cube 2.2 feet on an edge has its lower half of s.g. = 1.6 and upper half of s.g. = 0.7. It rests in a two-layer fluid, w ith lower s .g. = 1.4 and upper s.g. 0.8. Determine the height /1 of the top of the cu be above the interface. See Figure33.

\

IT' l

l

s.g. = 0.7

:.g.

G

'"' 162.4 "' W]x 2.221 (3.08 - 1.4/z + 0.8/z) = f62.4 x 2.22)(2 53) l.08 - 1.4h + 0.8'1=1.76 + 0.77 . ,, .. 0.917 ft

~mm . diam eter solid cylinder is 95 mm hi h and . . . r•ll'd m a liquid (y = 8.175 k N/ m3) cont.u! . weighing 3.75 N ts M11 diameter of 125 mm Befo . . ed m a tall metal cylinder · r e 1mmers1on th li ·d , e qm was 75 mm deep. hAl lt•vel will the solid cylinder floa t?

= 1.6

~i......-----'t:-s.g. = 1.4

[62.4 x 2.22)(2.53)

I m 3 - 75 (CE May 1997 )

. rJ. I

2.2 ft

·

2.2 ft Figure 33

--+'

170

CHAPTER THREE

FLUID MECH ANIC.~ & HYDRAULIC.\

CHAPTER THREE

Total Hydrostatic Force on Surfaces

Total Hydrostatic Force on Surfaces blem 3-76

Solution

woo~en bea.m of sp. g r. 0.64 is 150 nun by 150 nun and is hinged at A, as own in the Figure. At w hat angle 8 will the beam float in water?

lOO..mm0

E E E

,.... "'

17 1

E E

y

. "',....

=8.175 kN/m3

...,,.

T

u

I•

l m

•'

}

l

' 125 mm 0

(a} Before immersion

· (b) After immersion (5 - 0.5x) cos e

Solve for the draft Din figure (b): [BF= W} YL Vo =W (8,175) Vo = 3.75 Vo= 0.0004587 m3

= 458,716 mrn3

{ (100)2 x D = 458,716

Draft, D = 58.4 mm

Weight of beam, W= Yre•m Vream Weight of beam, W = (9,810 x 0.64}[(0.15)2(5)] Weight of oeam, W = 706.32 kN

\

'

\

\

\

When the solid cylinder is immersed, the liquid in lthe tall cylinder 11 du. e to volume of liquid displaced. Theref~re, ~he vo~u".'e of li1111 displaced equals the total volume of real and imaginary liqw d abov1 11 original level

V•bo•e orig. level = V0

t (125)2(x) = 458,716

Buoyant force, BF= Ywator Vo 1\uoyant fo rce, BF = 9,810[(0.15)2 x] Buoyant force, BF= 220.725x

IJ. M;, = O] Uf(5 - 0.5x) cos 8 = W(2.5 cos 8) 220.725x(5 - O.Sx) = 706.32(2.5) ~l - 0.5x2 = 8 l).5x2 - Sx + 8 = 0

-(-5)±~(-5) 2 -4(0.5)(8)

x =37.38 mm From Figure b: 75+x=D+y y = 75 + 37.38 - 58.4 y= 53.98mm T h e refore; the solid cylinder will float w ith its b ottom 53.976 mm abov1• I bottom of the h allow cylinder.

2(0.5)

'

2m

1 5- x 0 19.47° . ~•fl

0 = - - = 1 /3

FLUID MECHANIC\ &HYDRAULIO

CHAPTER THREE

172

Total Hydrostatic Force on Surfaces

Problem 3 - 77 (CE Board May 2003)

CHAPTER THREE

Total Hydrostatic Force on Surfaces

173

lem 3 - 78 (CE Board November 1993)

From the figure below, it is shown that the gate is 1.0 m wide and is hinged 11 the bottom of the gate. Compute the Uowing: (a) the hydrostatic force in kN actin on the gate, . (b) the location of the center of pressu of the gate from the hmge, (r) the minimum volume of concrete (u ·t weight= 23.6 kN/m3) needed'" keep the gate in closed position.

Ill going from salt water (sp. gr. = 1.03) to fresh water (sp. gr. = 1.0) sinks rm and after burning 72,730 kg of coal rises up by 15.24 cm. Find the h1al displacement of the boat in sea water in kN.

0.5 m

I

,-!

I~0

0

0

0

0

0

Stopper

r -.

Sea water i

I

~

~

0

0

L

0 + 0.0762 I

2m

Hinge

~

Figure (a)

Solution

F = y h A =9.81(1)(2 x 1) F= 19.62 kN

y=

t (2) = 0.667 m

[LMA= 0) Fx y = T x 2.5 19.62(0.667) = 2.ST T= 5.232 kN

t

~T..__,,B

L

0.5 m

T

I

2m

From the FBD of the concrete block:

D + 0.0762 - 0.1524 = D - 0.0762 r -1

1

i

Fresh water i

c::> c.g c.p.

]v eA

[Uv =OJ

Figure (b)

Figure (c)

~\'I' to assume that the boat have a co~stant cross-sectional area ilh•1 surface and use 'Ywater

= 1000 kg/ m3

T+BF=W BF= Yw Vconc = 9.81 Vconc W = Yconc Vconc = 23.6 Vconc 5.232 + 9.81 Vconc = 23.6 Vconc Vco'nc = 0.3796 m 3

y,,. Vn = W IV (1000 x 1.03)1A(D)] IV

'f030AD

7 Eq. (1)

A below

..

174

CHAPTER THREE

Total Hydrostatic Force on Surfaces

CHAPTER THREE

FLUID MECHANIC\ & HYDRAUlll \

Total Hydrostatic Force on Surfaces

175

on

In Figure (b): BF2=W Yrw Vo= W W= lOOOfA(D+0.0762)] W = 1000A(D + 0.0762)

~

Eq. (2)

In Figure (c) : BF3 =W-72730 ~ lOOO[A(D - 0.0762)] = W - 7273

~

Eq. (3)

Pnr any floating body; Buoyant force= Weight

From Eq. (1) and Eq. (2): [W = W] '1030AD = lOOOA(D + 0.0762) 10300 = 10000 + 76.2 D = 2.54 m (draft in sea water)

Solving for displacement in sea water: Yse•water Vo1 = W (64) Vo1 = 24,000 x 2,240 Vo1 = 840,000 ff3 Solving for displacement in fresh waterYrresh water V02 = W (62.2)(Vo2) = 24,000 x 2240 Vo2 = 864,308.68 ft3 Figure (a)

l c•t /1 be the difference in the drafts in fresh & seawater: Vo2 - Vo1= Area x Ji

Ir= 864,308.68 - 840,000

32,poo

From Eq. (1) W = 1030A(2.54) W=2616.2A

lr=0.76 ft 1)111ft in fresh water, 0

From Eq. (3) 1000A(2.54 - 0.0762) = 2616.2A - 72730 2463.8A = 2616.2A - 72730 A= 477.23 m 2

i.

Therefore: = 2616.2(477.23) W= 1,248,529 kg (9.81/1000) W= 12248 kN

w

= 34 + 0.76 = 34.76 ft

(CE Board November 1995)

hJc•r an arbitrary shaped body w.ith a submerged volume Vs and a Pr- What is the net vertical force on t.01ly due to hydrostatic fo rces? ·

lly p~, submerged in a fluid of density

ro·••.. Yt v~ yf = Pt X g

Problem 3 - 79

A ship having a displacement of 24,000 tons and a draft of 34 feet _in 1•1 enters a harbor of fresh water. If the horizontal section of the ship 111 · 32,000 sq · ft, what depth of fresh water is required to flout I waterline is ship? Assume that·marine ton is 2,240 lb and that sea water and fresh \ weight 64 pcf and 62.2 pcf, respectively.

I,.,, .. Prg v~

• m3-81 I

h11rkal balloon, 9 min diameter is filled with helium gas pressurized .to ,1t a temperature of 20°C, and anchored by a rope to the ground fl: ling the dead weight of the balloon, determine the tension in the rope "• 212 m/°K for helium gas and Yair ? 11.76 N/m3

kl111

176

CHAPTER THREE

Total Hydrostatic Force on Surfaces

FLUID MECHANIC\ & HYDRAULIC\

111x10

-

3

p -----RT - 212 (273 + 20)

Yhehu"' =

w

1.787N/111~

l ~Mo

4

Vhalloon

= 3 rr(9 / 2) 3

Vh.tlloon

= 381. 7 In~

= Yhehum

=OJ

w•. (1.5 cos 9) - BF((L/2) cos 9] = o

[Hv = 01 BF - W- T= 0 BF= Yair V1>.111oun BF= 11.76(381 .7) = 4488.8 N W

177

W,., = 45.62 N (from Figure 3 - I) /!f = Ysw Vn ll f = 9810(1.025)[(0.05)2LJ l! f = 25.138L

Solution Yhohum -

CHAPTER THREE

Total Hydrosta tic Force on Surfaces

45.62(1.5) - 25.138L(L/2) l2.57 L2 = 68.43 L = 2.33 m T

=

o

sin 9 = 2/L sin 9 = 2/2.33 0 =59°

V1>,11loon

W=l.787 (381.7) = 682.1 N 4488.8 - 682.1 = T T= 3806.7 N

Mht tircular cone is 100 mm in diameter and 200 nun high an<..~ weighs Problem 3 - 82 The buoy in Figure 3 - 1 has 80 N of steel weigh t attached. The buoy lw lodged against a rock 2 m deep. Compute the angle 0 with the horizontal .11 which the buoy will IE7an, assuming the rock exerts no m oment on the buoy Solution

lw 11•quired downward vertical force 1s I

Bf-W Hf= Y1iq111d Vco"'·

A

Hr = (9,81 Ox 0.8) [(n I 3 )(0. 1/ 2)2(0.2 ll /lr' = 4.11 N

4.1 1 - 1.6 2.51 N 1lui, fo rce F= 2.51 N becomes

matter how deep further ''submerged

111 110

111

(L/2) cos e

1.5 cos 0

l .,b

How much force is required to push the cone (vertex downward) ll h<>dy of liquid having sp. gr. of 0.8, so that its base is exactly at the jtl•', llow much additional force is required to p ush the base 10 mm W tlll' surface? 1111

F

11

FLUID MECHANll \ . & HYORAULll \

CHAPTER THREE . Total Hydrostatic Force on Surfaces

178

CHAPTER THREE Total Hydrostatic Force on Surfaces

LUID MECHANICS HYDRAULICS

179

lution

Problem 3 - 84

F

To what depth will a 2-m-diameter log, 4 m long and of sp. gr. 0.425 sink 111 fresh water Glycerin

Solution For a homogeneous solid body floating on a homogeneous liquid:

s

i

= lOON

=1.3

5 body

V0 = - - V1,.,.11

.'

Sliquid

As L = A~

0.425 A

l.0

= 0.425nr2

L

(shaded area)

l.l•t V =volume of wood In water:

From geometery: As

[:Efv =OJ

= A >eetor - A1r1angl•

0.425rrr2 = Vi r e, e, - sin e = 2.67 2

1h

r2

BF1 - W- F= 0 9810V-W = 40 V= 40+ W -7 Eq. (1 l 9810

sin e

Solve e by trial and error: Try 0= 170° 170°(rr/180°) - sin170° = 2.76

In glycerin:

(:.t:2.67}

f ry 0 = 166° 166°(rr/180°} - sin166° = 2.655 · Try 0 = 166.44° 166.44°(11/180°) - sin166.44° = 2.67

f:EFv =OJ

(;t:2.67) O.K.

II= r- y "= 1 - (1) cos (0/2) It = 1 - (1) cos (166.44° /2) /1=0.882 m

Problem 3 - 85 A block of wood requires a force of 40 N to keep it immersed in water .11111 force of 100 N to keep it immersed in glycerin (sp. gr. = 1.3). Find the w1·11 I and sp. gr. of the wood.

BF2-W- F = O (9,810 x 1.3) v -

w = 100

w]

(9,810 x 1.3) [ 40+ 9810 52+1.3W- W= 100 W=160N

- W= 100

flmm Eq. (1): V= 40+160

9810 V= 0.0204 m 1 . l W= 160Um.t we1g1t,y= V 0.0204 Unit weight, y = 7843 N/m3

c.;p, gr., 5 = Y woo
=;

7843 9810

'

CHAPTER THREE

180

Tota l Hydrostatic Force on Surfaces

Problem 3 - 86

FLUID MECHANIC!. & HYDRAU LIC\

CHAPTER THREE

Total Hydrostatic Fo rce on Surfaces Since the volume of oil remain unchanged;

·

Vot1 (initia l)= V oil (fina l)

A rectangular tank of ipternal width of

(0.5)(5)(1.25) = (0.5)(5)('7') - 0.1274 Ii'= 1.301 m •

5 111, as shown, contains oil of sp. gr. = 0.8 and water. (a) Find the depth of oil, h. (b) If a 1000-N block of wood is floated in the oil, what is the rise in free surface of the water in contact with air?

l\i; shown in Figure b, if the oil-water interface drops by a distance of y, the In~ surface of water will rise by y/2, since the .cross-sectional area of the 1lgh t comprutment is twice that of the left comparh11ent. 3m

oil

Um-up pressure head fro m surface to water surface in 111 of water: 0 + 1.301(0.8) + (3 - y)- 4 - y/2 = 0 1.0408 - 1 - 3y/2 = 0 3y/2 = 0.0408 y/2 = 0.0136 m. or 13.6 mm

Solution

T h

t

3m

l

...rr

F-"

'l'lwrefore; the free surface of water will r ise 13.6 mm .

0 OH

s = 0.8

-

6 M -v

-

Water ~

T 4m

O.Sm

1

lm

lm

3-y

l_L-_ _ _ _,

Figure (a)

.

open cylindrical tank 350 m m in diameter and 1.8 m high is inserted th ;illy into a body of water with the open end down and floats with a 1300 t l11C'k of concrete (sp. gr. = 2.4) s uspended at its lower end. Neglecting the tihl of the cylinder, to what depth will the open end be submerged in &Id

Figure (b)

(a) Depth of oiJ: (Refer to Figure a) Sum-up press ure head from oil s urface 0 to water surface 6 in m of wall'1

utlon

\' OJ ~ 1'..int

+ Bfcy1 - W = 0 ~ Eq. (1)

B f conc

= Ywa ler V conc

El + h(0.8) +3 - 4 = E3..

Vto ne-

0 + 0.8/i -1 = 0 /1= 1.25 m

Y cone 1300 V,onc= - - - 9810(2.4)

y

y

(b) Rise of the water surface: (Refer to Figure b)

-

Vconc

x .

w conc '

-

-

BF<>i

1.8m Y.

= 0.0552 m 3

BF = W

llfconc

= 9810(0.0552)

Vo= W (9810 x 0.8) Vo = 1000 Vo= 0.1274 m ~

Hfconc

= 541.7 N

Yoil

181

lll'cyl = YwaLer Vo lll'cyl = 9810( T(0.35)2h] Hf cy1=943.83 h

Water concrete block

182

CHAPTER THREE

Total Hyd rostatic Force on Surfaces

FLUI D MECHANIC\ & HYDRAULIC\

LUID MECHANICS HYDRAULICS

Fro m Eq. (1) 541.7 + 943.83/t -1300 = 0 Ir = 0.803 m

CHAPTER THREE

Total Hydrostatic Force on Surfaces

Wi = 205 kg W1 =205 kg

0.6m 0

Applying Boyles Law (taking pa1m = 101.325 kPa) Before il1sertion : Abso lute pressure in air, pi = 101.325 kPa Volum e of air inside .the cylinder, V1 = t (0.35)2(0.18)

0.6m 0

0.84 I.Sm

i' 0.96

w, L

y

2.1 m

Volume of air i.nside the cylinder, Vi = 0.0173 m 3 After inser tion: Absolu te p ressure in air, Jh = 101.325 + ylt Absolute pr essure in air, p2 = 101.325 + 9.81(0.803) Absolute pressure in air, p2 = 109.2 kPa Volume of air inside the cylinder, V2 = f (0.35)2 x

183

Chain

t"· ! t"•

_:L__

1.Bm

i

D

H H' L'

Volume of a ir inside the cylinder, V2,;,, 0.0962x Figure a: High Tide

[p1 V1 = pi Vi] 101.325(0.173) = 109.2(0.0962x) x= 1.67 m x- h + y = 1.8 y = 1.8 -1.67 + 0.803 y= 0.933 m The refore, the open end is su bmerged 0.933 m belo\v the wa ter surface,

Problem 3 - 88 (CE Board)

A cylindrical buoy 600 nun in d iameter and 1.8 m high weighs 205 k g. It moored in salt water to a· 12 111 length of chain weighing 12 kg per m c1I 11 length. At high tide, the height of buoy ·protruding above water surfa11 0.84. What co uld be the le ngth of protru sion of the buoy if the tide dro111 · 2.1 Density of steel is 7,790 kg/ m 3. Use density of water= 1000 k g/ m I

m?

Figure b: Low Tide

4 h1ht of chain

= 12 kg/m 11•1ity of s teel = 7,790 kg/m 3 Cllurne of steel (ch ain) = 12/ 7790 c1l11me of s teel (ch ain) = 0.00154 m3 p er me ter len gth

I l1~ure a: p.rv = O] BF1 + BF2 - W1 - W2 = 0 BF1 = Ysw Vo = (1000x1.03)[

f (0.6)2(0.96)]

BF, = 279.58 kg BF2 = Ysw Vchain = (1000 x l .03)[0.00154(L)J BF2 = 1.586L

W2= 12L 279.58 + l.586L - 205 - 12L = 0 I. = 7.16 m I l1•pth of water, H.= L + 0.96 I h•pth of w a ter, H = 8.12 m

•

184

CHAPTER THREE

Total Hydrostatic Force on Surfaces

FLUID MECHANIC\ & HYDRAULIC\

CHAPTER THREE

Total Hydrostatic Force on Surfaces

185

ln Figure b: Depth of water, H' = H - 2.1 Depth of water, H' = 6.02 m Draft, D = H' - L' Draft, D = 6.02 - L'

[l::Fv = O] BF'1 + BF'2 - W1 - W'2 = 0 BF'1 =('1000x1.03) )[ f (0.6)2D]

BF'1 = 291.23 (6.02 - L') BF' 1 = 1753.18 - 291.23L' BF'2 = (1000 x l.03)[0.00154(L')J BF'2=1.586L'

W'2 = 12i.' 1753.18 - 291 .23L' +.l.586L' - 205 -12L' = 0 L'=S.13m

W= Y1>.11 Vb.11 W = (9810x0.42)f1t(0.15)3

4.Jrr

W= 58.25 N

"1111yant Force: Jjf =Yw&ll!r

V tMll

l1r = (9810)

-

t 1t(0.15)3

/IF= 138.69 N

t)c•plh of pool: W()rk done by W= Work done by BF I V(4.3 + h) = BF(h) 118.25(4.3 + /1) = 138.69'1 /1 - 3.11 m

D = 6.02 - 5.13 = 0.89 y

= 1.8- D

y = 1.8- 0.89 y = 0.91 m (le11glh of protmsion)

~,lt1Hneter weighs 0.0214 N and has a stem at the upper end which is 2.79

111 di ameter. How m uch deeper will it floa t in oil (sp. gr. 1111 (sp. gr. = 0.821)?

= 0.78)

Problem 3 - 89 (CE Board)

A wooden spherical ball with specific gravity of 0.42 and a diarr~eter or :1ni1 nun is dropped from a height of 4.3 m above the surface of water .Ill a pool 1 unknown depth. The ball barely touched the bottom o f the pool befon ·t began to float. Determine the depth of the pool.

lllH IO x 0.821)Vo. = 0.0214 v,,,, • 2.657 x 1Q·6 m3 I '1,,, 2,657 mm3

("; /\Vr ~--

w ('>BIO x 0.78)Vo0 = 0.0214 V11,, 2.797 x 10-9 m 3 V11,, 2,797 mm ~ 1

V,,,.- Vo" ... J.,797 - 2,657 = 140 mm3 ~ (2.79)2 /1 = 140

Alcohol, s = 0.821

Oil, s = 0.78

that in

186

Problem 3 - 91

horizontal, what is its maximum height for stable equilibrium in the lght position?

I."

LxL

0 ·82

D=

D r afl '

1

L

Dra ft, D = 0.82L ...L (L)(L/

J

MB,.= V0

llJj_ r. '"

M• • G ;:---- • Bo

I---

I•• The body is stable when M is •vt•G and unstable if M is below

Df2

= __,1.::..2_ __

MB,, = 0.102 L

GB.,= L/ 2 - 0 / 2 GB,. = 0.09L

L

I

Since MB 0 > G.B 0 , M is above G The body is stable.

10 cm • 10 cm

With sm aller value of H, the llll'cnter M will become higher · ~r1 l; making it much stable. ~·m I l increases, M will move WI\ closer to G making it less bhi. Hence, the maximum ht for stable equilibrium is 11111 M coincides with G, or MB 0 =

~

(LxL)(0.82L)

r 1----1

H

l1 D

0/2

n1the figure: (, /l,, = H/2 - D/2 Draft, D = ~:~ H = 0.621H ( .tl" = 0.5H - 0.621H/2 (, /i,,"' 0.189H

Solution sp.gr. wood H sp. gr. oil

-z.......=;.....---

Draft, D = ~:~~ {1200) = 847 m m I

MB., = V D

MB.

t (300)

4

= --'-----

n(300)2 (847)

MB. = 26.56 mm GB 0 = 600 - V2(847) GB"= 176.5

r

=300 mm

: f;--1---• -

D

Vo

(lO}(lO)D 100 13.419 MB. = - - - - - - 12{0.621H) H

G

• Bo

LUi -----

Since MB 0 < GB0 , the metacenter is below G Therefore, the body is unstable.

IMll.= _, J 1~ (10)(10)3 MB. = - = - - - -

1

OR:

MB. =~ (1+ tan 28 l 12D

MB. =

l

102

12(0.621H}

•Bo

2

where8=0°

)

(1 + 0) = 13.419 . H

!'---

J2

_j_ Seawater, s = 1.03

I

1

•G

i:

Waterline Section

Problem 3 - 92 A solid wood cylinder of specific gravity 0.6 is 600 mm in diameter and l mm high. If placed vertically in oil (sp. gr. = 0.85), would it be stable?

Draft, D =

187

block of wood (sp. gr.= 0.64) is in the shape of a rectangular parallelepiped Ing a 10-cm square base. ff the block floats in salt water with its square

A plastic cube of side L and sp. gr. 0.82 is placed vertically in water. cube stable?

Solution The body is stable if M is above G.

CHAPTER THREE Total Hydrostatic Force on Surfaces

FLUID MECHANll \ & HYDRAUlll \

CH APTER THREE Total Hydr osta t ic Force on Surfaces

CHAPTER THREE

188

Total Hydrostatic Force on Surfaces

FLUID MECHANIC\ & HYDRA ULIC~

·.

[MBo = GBo]

CHAPTER THREE

Total Hydrostatic Force on Surfaces

lnltial metacentric height, MG= MB 0 - GB0 Initial metacentric height, MG = 77.49 - 117.45 lrn!ial metacenb·ic height, MG= -39.96 mm

13.419 = 0.189H H H = 8.43 cm

Problem 3 - 94

A wood cone, 700 mm diameter and 1,000 mm high floats in w~ter with rti; vertex down. lf the _specific gravity of the wood is 0.60, would 1t be stabl1• Determine also its initial metacen tric heigh t.

~d.ingular scow 9 m wide, 15 m long, and 3.6 m high has a draft in sea ~ 1· of 2.4 m.

Its center of gravity is 2.7 m above the bottom of the scow M!rm ine the follo:vving: · lirl The initial metacentric heigh t,

(hJ :n~e righting or_ overturning m0men t when rs JUS~ at the pomt of submergence.

Solution Vwood

= ~ 1t(350)2{1000}

Vwooct

= 128,281,700 m 3

Vo=

0 60 Vwond · 1 •

Vn = 0.6

E E

Ve>= 0.6 (128,281.700) = 76,969,020 mm-1

vwuu.J

~o

= (1

0.6Vwood

1+

2

tan 2

f Metacenter, M

El]

750

l

r

oo]

(9)2 [ tan.2 - 1 +--12(2.4)

2

2.8125 m 2.7 - 1.2 = 1.5 n1

= 350 843.4 1000 x = 295.2 mm I Vo

MB=•

Waterline Section

u •1.5 11•13°

4

--=---76,969,020

=

77.49 mm

From the Figure: GBo = 750 - 30/4 GB 0 = 750 - 3(843.4)/ 4 GB.= 117.45 mm Since MB,. < GB,., M is below G and the cone is UNSTABLE.

2

[ + -tan20] H- 1

120

2

~[1+(1.2/4.5) 2 ]

12(2.4) .''> I m

2.7m

·--~~- -·-·-·-·-=! I

o

B = 9m

_x_

3.4 (295.2)

D=2.4m

lll.11rnetacenb·ic height, MG= MB 0 - GB() llnl metacen tric height, MG= 2.8125 - 1.5 llnl 111ctacentr.ic height'. MG= 1.3125 m

0 = 843.4 mm

L = lSm

1

where El= 0°

§

By similar solids: V11

£_[ 120

0

Vn

(1000 );1 D .

the scow tilts until one side

~r1111,1J metacentric height: 1000/4 . ~·

V wood

VW()(ld =

18 9

2

l.2m

J

3.6m

190

\CHAPTER THREE Total Hydrostatic Force on Surfaces

CHAPTER THREE Total Hydrostatic Force on Surfaces

FLUID MECHANll & HYDRAUlll

f.lung longitudinal axis (rolling): ll=lOm

Metacentric height, MG = MB" - GB" Metacentric height, MG= 2.91-1.5 = 1.41 m Since MG > MB,., the moment is righting moment. Righting moment, RM= W (MG sin 9) W= BF =yVD W = (9.81 x 1.03)[9(15}(2.4)] = 3,273.8 kN Righting moment, RM= 3,273.8((1.41) sin 14.93°] Righting moment, RM= 1,189.3 kN-m

MB 0

= ~[1+tan 12D

0

29

2

]

where 9

= 0°

102

(1 + 0) = 5.45 m 12(1.53) Metacentric height, MG= 5.45 - 3.235 Metacentric height, MG = 2.215 m (the barge is stable in rolling}

MB.

=

~long transverse direction (pitching):

Problem 3 - 96

A barge floating

191

II== 30m

fresh water has the torm ot a parallelepiped h11\1 d imensions 10 m x 30 m by 3 m. It weighs 4,500 kN when loaded with u 1• of g rav ity alo ng its vertical axis 4 m from the bottom. Find the metan•11t height about its longest c1nd shortest centerline, and deternune whether 111 the barge is stable in

Solution

~[1+ tan 120 2

2

MB 0 =

9

]

w here 9 = 0°

30 2 (1 + 0) = 49.02 m 12(1.53) Metacentric height, MG= 49.02 - 3.235 Metacentric height, MG= 45.785 m (the barge is stable in pitching)

MB. =

em 3 .:. 97 (CE August 1973)

·f:J

t- · 3m

Rolllnu

I.

30 m

y V0 = W

CBu=4-0/2 GB,.= 4 - 1.53/2 GB, = 3.235 m

II 1\1ad roller weighing 20 sho~t tons, floats on fresh water with a d raft of

lm:tcrs and has its center of gravity located along its vertical axis at a II 1.50 meters above its bottom. Compu,te the horizontal distance out to Mld1 from the centerline of the barge through which the crane could swing ~() ton load which it had lifted from the center of the deck, and tip the 1 with the 20-meter edge just touching the water surface? 1

Solve for the d raft. D [BF= W] 9.81 [10 x 30 x D = 1.53 m

barge, 20 m long, 8 meters wide, and 2 me.ters high loaded at its center

IUll'

OJ

= 4,500

192

FLUID MECHANll \ & HYDRAULIC \

CHA PTE R TH REE

Total Hydrostat ic Force o n Surfaces

CHAPTER THREE

Total Hydrostatic Force on Surfaces

z = d sine z = 0.552 sin 11.31° = 0.108 m

's olution 1 short ton = 2000 lb =

193

900 kg

WR = (20 x 900) 9.81 W,,.= 176.58 kN

11.. =

120

G4

r

el E

..,,

N

!t_.[1

-

0.5 m

- - - -w,- - - - - - . Bo

1i

--r

1.5 m

0.6 m Bf

Bl-= yVo BF= 9.81 [8 x 1.2 x 20] BF= 1,883.52 kN = W ,

Weight of barge, WI!= BF - w,. Weight of barge, Wµ = 1,883.52 - 176.58 Weight of barge. WH = 1,706.94 kN ri lted position

+ tan 2 2

e]

Mfl ... _ j!

8_2_ [1+ tan211.31o] =4.533 .m 12(1.2) 2

I

(;

4.533 - 0.848 = 3.685 m

MG sine .\685 sin 11.31° = 0.723 m f1. =O J (/110 x = WR(L + z)

1,1183.52(0.723) = (176.58)(L + 0.108) I 7.604 m ~ Horizontal distance from the center of the deck

h·n barge of rectangular cross-section is 8 m wide, 4 m high, ·and 16 m long. 1111~porting

in seawater (s = 1.03) a total load of 1,500 kN including its own I 11nd cargo. If a weight of 75 kN (included in the 1;500-kN) is shifted a 'I' of 2.5 m to one side, it will cause the barge to go down 450 rrun in tht> •if 1111111ersion and also rise 450 mm in the corresponding wedge of emer~ion IM1w• floats vertically (on an even keel) before the shifting of the weight. uh• how far above the waterline is the center of gravity of the loaded barge.

n

tan e =

QJ!!l. 4.0

e = 11.31 ° Solve for the new position of G in the tilted position: W1(0.5) = Wa(d) 1,883.52(0.5) = 1.706.94(d} d = 0.552 m

l .,. I li E

~~--i- · - -

-

-

-

-

- ·-+-·

Bo i

BF Sm

J

.,. -

CHAPTER THREE

194

CHAPTER THREE

Total Hydrostatic Force on Surfaces

Total Hydrostatic Force on Surfaces

Solve for the draft, 0: BF=W (9.81 x 1.03)[8 x 16 x D] = 1,500 D=l.16 m

waterline section of a 1,500-kN barge is as shown. Its cent~r of gravity 1s

m above the center of buoyancy. Compute the initial metacentric height Inst rolling.

In the Tilted position

'

E1

~· 11 :

~

~:

- · -·- ·-·-·=~-·-·-·-·-·-·- · -·- -·-·-·-·- ·-·-·~·- · -·- ·-·-·-· 60 kN/m

- 4m

\ tan e =

0 ·45 ,4

8m

e = 6.42° 2

e]

tan MB= -a- [ 1 + -• 120 2

MB 0

=

j

(MB 0 = -

82 ( 1 + tan2 6.420) -- 4 .63 m \2(1 .16) 2

f

=

B F(c)

c = MB. sin 9 I° = 4.63 sin 6.42° c=0.518m

a = 3.42 sin 6.42° + 2.5 cos 6.42° a = 2.867 m b = (11 + 0.58) sin 6.42°

1,425((11 + 0.58) sin 6.42°] + 75(2.867) = 1,500(0.518) 11 = 2.947 m --7 distance of G from the w.s.

~i,

12m

~I-

]

=lrec:L>ngle + 11ria ngle + f sem1-circle

Ac12)(8)3 +

1\

(6)(4P x 2 +

t (4)4

I= 676.53 m4

[BF = VV) 9.81 Vo = 1,500 Vo = 152.9 m~

MB. = 676.53

[~Mao=

O] 1,425(b) + 75(a)

1

Vo

r= 2

195

= 4.425 m

152.9

IMG = MB. - GB0 ] MG = 4.425 - 1.5 MG= 2.925 m --7 initial metacentric height

6m]

tJ)

196

FLUID MECHANIC\ & HYDRAU!-JC\

CHAPTER THREE

Total Hydrosta tic Force on Surfaces

!supplementary Problems

CHAPTER THREE

Total Hydrosta tic Force on Surfaces

197

r in a tank is pressurized to 80 cmHg. Determine the total force per meter

lh on panel AB.

Problem 3 - 100

Ans: 482kN

A vertical rectangular gale 2 m wide and 1.2 m high has water on one s1d1 wi th surface 3 m above its top. Determine the magnitude of the to l.i 1 2m

hyd rostatic fo1:ce acting on the gate and its distance from tht> w;:iter surface Ans:

~

80 cm

- M .t> kl\.. y 1• ; 3.o.i 11

Water 4m Hg

Problem 3 - 101

B

A vertical semi-circular area of radius r ts submerged in a liquid with 11 diameter in the liquid surface. How far is the center of pressure from lh· liquid surface?

Problem 3 - 102

\

3m

t..=::========i A \

111\ure shown, the 8-ft-diameter cylinder, 3 feet long weighs 5~0 lbs and bottom of a tank that is 3 feet Jong. Water and oil are poured into II .ind right-hand of the tank to depths 2 feet and 4 feet, respectively. 111111c the magnitudes of the horizontal and vertical componen ts of the lh.11 will keep the cylinder touching the tank at A. Ans: FH = 749 lb ~ Fv = 2,134 lbs w

1111 lhe

An open vat holding oil (s = 0.80) is 8 m long and 4 m deep and h,1• trapezoidal cross-section 3 m wide at the bottom and 5 m wide a t the 1111 Determine the following: (n) the weight of oil, (b) the force on the bottom of Iii vat, and (c) the force on the trapezoidal end pa~el. Ans: (a) 1002 kN; (b) 752 I (c) 230 I ' Problem 3 - 103

Freshly poured concrete approximates a fluid with sp . gr. of 2.40. The figure shown .a wall poured between wooden fo rms which are connected by six bolts. Neglecting end effects, compute the fo rce in the lower bolts. Ans: 19,170 lbs

l Oin

Water

Oil, s

=0.75

'3 - 106 11h· lhl~ hydrostatic force and its location on semi-cylindrical indentation

h11wn. Consider only 1 meter length of"cylinder perpendicular to the l 1h1w. A ns: FH = 109.5 kN@ 1.349 m below D Fv = 20.5 kN@ 0.531 m to the left of B

198

CHAPTER THREE Total Hydrostatic Force on Surfaces

CHAPTER THREE Total Hydrostatic Force on Surfaces

199

.

4m

H

2m

!

r 2m

El. 14

Oil, s = 0.85

3m

El. 3

lm '-~~~~~--~~....

El. 3

El. 0

__-L

Problem 3 - 107

The 1-m diameter solid cylinder shown is 8 m lon g perpendicular to the 111 and rests in static equil ibrium against a frictionless wall at 0. Determi111 I unit weigh t of the cylinder.

Ans: ex = 10.5°

Water

Problem 3 - 108

The section of a concrete dam is shown in the figure. Concrete weighl kN per cubic meter and water weighs 9,790 N per cubic n1eter. Coefficl••l•I friction betweeh the dam and fo undation is 0.55. Determine the fact111 ~ safety against sliding and against overturning, and also the soil pressun• 111 heel and toe. Assume hydrostatic uplift varies uniformly from full hydr11 1 head at the heel of the darn to zero at the toe. Consider 1 m length of da111 Ans: FSo = 2.20; FSs I qhe<·I = 85.2 kPa; q,"" = 300 '~

200

FLUID MECHANI( \ & HYDRAULf( \

CHAPTER THREE Total Hydrostatic Force on Surfaces

CHAPTER FO U R Relat ive Equilibrium of Liquids

201

Problem 3 - 110

Two spheres, each 1.3 m in diameter, weigh 5 kN and 13 kN, respectiv1 !\ They are connected with a short rope and placed in water. What is the tensh•11 in the rope and what portion of the lighter sphere protrudes from the w ater 1 Ans: T = 1.74 kN; 40 I Problem 3 - 111

A block weighing 125 pcf is 1 ft square and 9 inches deep floats on a strattlh .t liq uid composed of a 7-in layer of wa ter above a layer of mercury. 1111 Determine the position of the bottom of the block. (b) If a downward vert11 I force 'o f 260 lb is applied to the center of mass of this block, w hat is the '" 1 posi tion of the bottom of the block? A11s: (a) 0.8 " below m en 1111 (b) 4.67" below mern111 .

I

hapter 4 lative Equilibrium l iquids mas~ may have no relatiw ~111 between ~ach o ther yet the mass itself may be in motion If a mass of I~ moving with a constant speed (unifor m velocity), the conditions are 111111.! as ~uid s ta~cs (as d iscussed in previous chapte rs). But if the bod y 1'Jw led to acceleration (whether translation or rotation). speci
lw•t ccrtam conditions, the par ticles of a fluid

I

Problem 3 - 112

Wou ld a wooden cylinder (sp. gr. = 0.61) 660 mm in diameter and 1.3 m be stable if placed vertically in oil (sp. gr . = 0.85)?

I

(p r 1

Problem 3 - 113

A rectangula r scow 7 ft by 18 ft by 32 ft long loaded with garbage has a drr1lt of 5 feet in water. Its center of gravity is 2 ft above the waterline. [s the Slt1I stable? What is the initial met~centric height?

ILINEAR TRANSLATION (MOVING VESSEL)

dt•r a mass of fluid moving with a linear acceleration 11 as shv v,, 11 rn thE:Consideri_n~ ~ particle _in the surface, the forces acting are the weigh I Ms ,111d the fictit10 us mertia force (reversed effecti ve force, REF) wh ich If> IP Ma, and the reaction N which must be normal to the surface

'lt'I

a

Problem 3 - 114

A cube of dimension Land sp. gr. 0.82 floats horizontally in water. stable?

> REF = Ma