Mechanics
& Hydraulics
Rc~ ;i~eu
Editiorl
DIEGO INOCENCIO T. GILLESANIA
Civil EtJgilJeer
BSCE, LIT - Magna Cum Laude
5th Place, PICE National Students' Quiz, 1!>89
Awardee, Most Outstanding Student, 1989
3rd Place, CE Roard November 1989
Review Director & Reviewer in av SLtbjects
Gillesania Engineering Review Center
Reviewer in Mathematics and General Enzinccring Sciences
MERIT Philippines Review, Manila
Author of Various Engineering Books
Fluid Mechanics
and Hydraulics
Rel'isPd Edition
Copyright © 1997, 1999, 2003
by Diego l11ocencio Tapa11g Gil/esauia
.·lit ngbts r(lsen·ed No part of this book ma) be
reprod11ced, stored 111 a rC'triC'l•a/ S)'Stem or
transferred. 111 cmy.form or b)' em:)' medns,
u•itho111 the> prior perm issio11 of tbe ctllthor
The cardinal objective of this book is to provide reference lo
Engineering students taking-up Fluid Mechanics and.HydTaulics.
This may also serve as a guide to engi~eering students who will be
laking the licensure examination given by the PRC.
The book has 9 chapters. Each chapter presents the principles
and formulas involved, followed by solved problems and
supplementary problems. Each step in the solution is carefully
explained to ensure that it will be readily understood. Some
problems are even solved in several methods to give the reader a
choice on the type of solution he may adopt.
To provide the reader easy access to the different topics, the
book includes index.
Most of the materials in this book have been used in my
review classes. The choice of these materials was guided by their
effectiveness as tested in m1 classes.
lSBN 971-8614-28-1
Pnnted by:
GPP Gilfestmia Pri11ling Press
Ormoc Crty, Leyte
I wish to thank all my friends and relatives who inspired me
in writing my books and especially to my children and beloved
wife Imelda who is very supportive to me.
I will appreciate any errors pointed out and will welcome any
suggestion for further improvement.
Philippines
'
Cot>er desig11l~)' the a11thor
DIEGO INOCENCIO T. GILLESANIA
Cebu Cihj, Philippines
TABLE O F CONTENTS
Preface ........................................................................................... vii
Dedication .................................................................................... viii
CHAPTER 1
Properties of Fluid ........ ................................................................. 1
To my mother Iluminada,
my wife Imelda,
aud our Children Kim Deunice,
Ken Dainiel,
and Karla Deuise
Types of Fluid .........................................~ .......................................... 1
Mass Density ..........: .............................. ......... .................................... 2
Specific Volume ................................................................................. 3
Unit Weight or Specific Weight ...................................................... 3
Specific Gravity ................................................................................. 4
Viscosity ............................................................................................. 4
Kinematic Viscosity ..................................................................... 5
Surface Tension ....................................................~ ............................ 6
c,,pillarity .......................................................................................... 7
CnmprL'Ssibility ................................................................................. 8
Pressure Disturbances ...................................................................... 9
Proper tv Changes in Ideal Gas ....................................................... 9
\ 'apor Pressure .............................. .................................................. 10
SOLVFD PROBLEMS .......................................................... 11 to 23
SUPPLEME~TARY 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 Head... ............. .............. .. ...................................... 33
Manometerc; ............
............ . . ,.. ,......... .. .............................. 34
SOLVFD PROBI fMS..... ....
.. ..~ ..................................... 35 to 68
SUPPLEMENTARY PROBLEMS ................................... 69 to 72
II
TABLE OF CONTENTS
TABLE OF CONTENTS
iii
CHAPTER 3
CHAPTER 5
Total Hydrostatic Force on Surfaces ......................................... 73
Fundamentals of Fluid Flow .................................................... 241
Total Hydrostatic Force on Plane Surface ................................... 73
Properties of Common Geometric Shapes .............................. 76
l'otal Hydrostatic Force on Curved Surface ........ ........................ 78
Discharge ........................................................................................ 241
Definition of Terll.1S ............................................................. .......... 241
Energy and Head .......................................................................... 244
Power and Efficiency .................................................................... 245
Bernoulli's Energy Theorem .................~ ...................... ................ 246
Energy and Hydraulic Grade Lines ........................................... 248
SOLVED PROBLEMS .............................................. ........ 250 to 273
SUPPLEMENTARY PROBLEMS .................................. 274 to 27b
Dams ................................................................................................. 81
Types of Dan1S ............................................................................ 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 .............................. .......................... 99 to 195
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
CHAPTER6
Fluid Flow Measurement ......................................................... '277
Device Coefficients ....................................................... ................ '27":
Head lost in Measuring Devices ................................................. 2/ll
Orifice ............................................................................................. 281
Values of H for Various Conditions ...................................... 2R3
Col1h·action of the Jet .............................. ,........................ ........ 2~4
Orifice under Low I leads ........................................................ 285
Venturi Meter ................................................................................ 285
Nozzle ............................................................................................. 287
Pitot Tube ....................................................................................... 288
Gates .................................................... ........................................... 290
Tubes ................................ .............. ................................................. 291
Unsteady Flow (Orifice) ............................................................... 294
Weir ................. :~.......... .................................................................... 297
Classification of Weirs ............................................................. 21.}7
Rectangular Weir. ..................................................................... 2lJ8
Contracted Rectangular Weirs .......................................... 301
Triangular Weirs ...................................................................... 301
Trapezoidal Weirs ..................................................................... 304
Cipolletti Weir ............................................................ ~ ....... ~0-l
Suttro Weir ................................................................................ 305
IV
TABLE OF CONTENTS
Submerged Weir ....................................................................... 305
Unsteady Flow .......................................................................... 306
SOLVED PROBLEMS ...................................................... 307 to 371
SUPPLEMENTARY PROBLEMS .................................. 372 to 374
CHAPTER 7
Fluid Flow in Pipes .................................................................... 375
Definitions ................................................................. ..................... 375
Reynolds Number ......................................................................... 376
Velocity Distribution in Pipes ..................................................... 377
Shearing Sh·ess in Pipes .......... :.................................................... 379
Head Losses in Pipe Flow ............................................................ 381
Major Head Loss ........................, ............................................. 381
Darcy-Weisbach Formula ................................................... 381
Value of f .......................................................................... 382
Moody Diagram ............................................................. 384
Maru1ing Fonnula ............................................................... 385
I Iazcn 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
TABLE OF (:ONTENTS
v
CHAPTER 8
Open Channel ........................................................... ····..·.·····.. ··· 481
Specific Energy ........................................................... ··· .. ·············· 481
Chezy Formula ..................................................................... ·· ...... · 482
Kutter and Gunguillet Formula ..............._. ............................. 483
Manning Formula .................................................................... 483
Bazin Formula ..................................... :..................................... 483
Powell Equation ....................................................................... 484
Uniform Flow ....................................................................... ·· ·.. ·.. · 485
Boundary Shear Stress .................................................................. 485
Nor1nal Depth ................................................................................ 486
Most Efficient Sections ................................................................. 486
Proportions for Most Efficient Sections ..... ,.......................... 487
Rectangular Section............................................................. 487
Trapezoidal Section............................................................. 487
Triangular Section ............................................................. ·. 489
Circular Sections ....................................................................... 490
Velocity Distribution in Open Cha1mel ...................... ............... 491
Alternate Stages of Flow .............................................................. 491
Froude Number ........................................................................ 492
Crit{cal Depth............................................................................ 492
Non-Uniform or Varied Flow ..................................................... 495
Hydraulic Jump ............................................................................. 497
Flow around Channel Bends .................. .. ................................... 500
SOLVED PROBLEMS ...................................................... 501 to 547
SUPPLEMENTARY PROBLEMS .................................. 547 to 550
vi
TA BLE OF CONTENTS
CHAPTER 9
Hydrodynamics ......................................................................... 551
Force against Fixed Flat Plates .................................................... 551.
Force against Fixed Curved Vanes ............................................. 553
Force against Moving Vanes ....................................................... 554
Work Done on Movil<g Vanes ..................... :................ .......... 555
~orce Deve~oped on Closed Conduit .. : ........... ........................... 556
rag
and Lift.......................
................................. ........................ .. 557
Velocity
Terminal
w
..................................................................... 559
ater Hammer .............................................................................. 560
SOLVED PROBLEMS ...................................................... 563 to 597
SUPPLEMEN!ARY PROBLEMS .................................. 597 to 598
APPENDIX
Properties of Fluids and Conversion Factors ........ ...... .. ........ 599
Table A -1: Viscosity and Density of Water at 1 atm ...... ........ 599
Table A - 2: Viscosity and Density of Air at 1 atm ................... 600
Table A- 3: Properties of Common Liquids at 1 atm & 20°C .. 601
Table A- 4: Properties of Common Gases a t 1 atm & 20°C ..... 601
Table A- 5: Surface Tension, Vapor Pressure,
and Sound Speed of Water ........................................... 602
Table A - 6: Properties of Standard Atmosphere ..................... 603
Table A - 7: Conversion 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
Fluicl Mrclumics is a physical science dealing w ith the action of fluids at rest or
in motion, and with applications and devices in engineering using fluids.
Fluid mecharucs can be subdivided into two major areas, fluid strztics, which
deals with fluids at rest, and fluid dynamics, concerned with fluids in motion.
The term hydrodynamics is applied to the flow of liquids or to low-velocity gas
flows where the gas can be considered as being essentially incompressible.
IIydmulics deals with the application of fluid mechanics to engineering devices
involving liquids, usually water or oil. Hydra\.t!ics deals with such problems
as the Oow 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, jels, and flovvmeters.
TYPES OF FLUID
Fluids are generally divided jnto 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 cur~nts or turbulence
Rea/fluids
• Exhibit infillite viscosities
• Non-uniform velocity distribution when flowing
• Compressible
• Experience friction and turbulence in flow
2
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
CHAPTER ONE
Properties of Fluids
I'LUID MECHANICS
;, HYDRAULICS
Real fluids are further divided into Newtonian fluids and 11011-NL"lvtoninn fluids.
where:
3
p =absolute pressure ot gas in Pa
R = gas constant Joule/ kg-°K
Most fluid problems assume real fluids with Newtonian characteristics for
convenience. This assumption is appropriate for water, air, gases, steam, and
other simple fluids like alcohol, gasoline, acid solutions, etc. However,
shnTies, pastes, gels, suspensions may not behave according to simple fluid
relationships.
For air:
R = 287 J/kg- °K
R = 1,716 lb-ft/ slug-0 R
T =absolute temper ature in °Kelvin
°K = oc + 273
0
R = °F + 460
Table 1 - 1: Approximate Room-Temperature
Densities of Common Fluids
pin kgjm3
Fluid
IPseudoplastic Fluids}
1.29
1.20
790
602
720
1,260
13,600
1,000
Air (STP)
Air (21°F, a 1lm)
Alcohol
Ammonia
Gasoline
Glycerin
Mercury
Water
Bingham Fluids
Figure 1 - 1: Types of fluid
PECIFIC VOLUME, Vs
'lwc.ific 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
gramf cm3
mass of fluid, M
volume, V
Eq. 1 -1
Note: P slugs = Plbm/ g
For an ideal gas, its density can be found from the specific gas constant and
1deal gas law:
RT
Eq. 1 -3
=----------------------·~p--------------------~
UNIT WEIGHT OR SPECIFI! WEIGHT, y
'lpl'Lific weight or unit weight, y, is the weight of a unit volume of a fluid.
kgjm3
p = _.E_
V = ..:!._
[
Eq. 1-2
y=
weightoffluid, W
volume, V
Eg. 1-4
y=pg
Eq. 1-5
4
CHAPTER ONE
FLUID MECHANICS
& HYDRAULICS
Properties of Fluids
Units:
English
Metric
SI
CHAPTER ONE
r lUID MECHANLCS
Properties of Fluids
l· HYDRAULICS
5
thl• upper plate will adhere to it and will move \Nilh the same velocity U whik•
tlw fluid in contact with the fixC'd plate \\'til hc~ve a zero velocity. For small
v.11uL'S of U andy, the velocity gradiL•ntc.m hL' .1ssumed to be a straight line
md F varies as A, U and y as:
lb/ft3
dynejcm3
Njm3 or kNjm3
u
F
- All
1 o:: - - or - 0 : : -
y
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
y
A
u
!I
dV
dy
(from Lhe figure)
I_ =Shearing stress, T
A
< u: ~ or T = k dV
PJiquitl
s=---
Eq. 1-6
Pwater
In gases, the standard reference to calculate the specific gravity is the density
of air.
Pgas
s=--
Eq. 1-7
Pair
For water at 4°C:
y = 62.4lb/ft3 = 9.81 kNjm3
p = 1.94 slugsjft3 = 1000 kg/m3
s = 1.0
dy
dy
where the constant of proportionality k ts Lolled the dynamk of
absolute viscosity denoted as ).l.
tiV
·= ~t t!y
'
[- - - -~~ =---
Eq. 1-8
!IV /liy
whcr<':
'=shear stress in lb/ ft2 or Pa
).l =absolute viscosity in lb sec/ ft2 (poises) or Pa-scc.
y = distance between the plates in ft or m
U = velocity in ft/ s or m/ s
VISCOSITY, ).l (MU)
The property of a fluid which determines the amount of its resistance to
shearing forces. A perfect fluid would have no viscosity.
Consider two 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
Area=A
--------·U
KINEMATIC VISCOSITY v (NU}
Kinematic viscosity is the ratio of the dynamic viscosity of the fluid, ~1, to its
IH.lSS density,(>.
F
v= ~
p
moving plate
y
fixed plate
where:
J..l =absolute viscosity in Pa-sec
p = density in kg/ml
Eq. 1 -9
6
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
English
Metric
5.1.
Absolute, 1-1
lb-sec/ft2
(slug/ ft-sec)
dyne-s/cm2
(poise)
Pa-s
(N-s/ m2)
Note:
1 poise= 1 d yne·s/cm 2 = 0.1 Pa-sec
1 s toke= 0.0001 m2js
Kinematic, v
d
1- i
ft2/sec
cm2/s
(stoke)
m2js
(1 d yne= 10-s N)
(b) Cohes1on > adhes1on
(a) Adhesion• > cohesion
SURFACE TENSION cr (SIGMA)
rhe nw m brane of "skin" that seem s to form on the free surface of a fluid is
d ue to the intermolecular cohesive forces, and is kno wn as sw face tension.
'::iurfaCL' tension is the reason that insects are able to sit on water and a needle is
.1ble to fl oat on it. Surface tension also causes bubbles and drople ts to take on
a spherical s hape, since any other shape would have more surface area per
unit volume.
Pressure inside a Droplet of Liquid:
Utprlfarity (Cuprllury acli01r) is the name given to the behavior of the liquid in a
thtn bore tube. The rise or fall or a fluid in a capillary tube is caused by
s111 face tension and depends on the relative magnitudes of the cohesion of the
liquid and the adhesion of the liq uid to the walls of the containing vessel
l 1quids rise in tubes they wet (adhesion> cohec;ion) and fall in tu bes they do
rwt wet (cohesion > adheesion). Capillary is important w hen using tu bes
~""'ller tha n about 3/8 incJ1 (9.5 mm) in d iameter.
[
4cr
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
System
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
Eq. 1-10
4crcos0
h=--yd
Eq. 1- 11
_ _ _ ______.
For complete wetting, ms with water on clean glass, the angle 0 is 0° Hencl'
the formu la becomes
4cr
11=yd
where:
h =capilla ry rise or ·dep ression in m
y =unit weig ht in N'jm3
d =diameter of the ttube in m
cr = surface tension iin Pil
Eq. 1 - 12
8
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
CHAPTER ONE
Properties of Fluids
I I UID MECHANICS
· HYDRAULICS
Table 1 - 3: Contact Angles, 0
stress
strain
l\p
Eq. 1 -15
Eu= - - = - -
Materials
Angle, 9
mercury-glass
water-paraffin
water-silver
kerosene-glass
_gly_cerin-glass
water-glass
ethyl alcohol-glass
140°
107°
90°
26°
19°
9
/\V
v
tip
dV/V
Eq. 1- 16
orEs=----
oo
a~
PRESSURE DISTURBANCES
l'n";sure disturbances imposed on a fluid move in waves. The velocity ~r
l<'lt•rity of pressure wave (also known as acollslical or some veloctty) IS
t•xpressed as:
COMPRESSIBIUTY, ~
Cf)mpressibility (also known as the coefficient of COIItJoressibilt/.f) is th, fraction.1l
change in the volume of a fluid per unit chang, 1n pre~--ure in .a constantTemperature process.
l
c=~ =fe
_
Eq.1-17
___.
~-_:___:,___
t\V
13 = _
I'
llp
dV I \'
orP=--dp
Eq I -13
PROPERTY CHANGES IN IDEAL GAS
1or any ideal gas experiencing an y process, Lhc equation of state is g iven by:
Eq.l 14
Eq. 1 -18
where:
6. V =change in volume
V =original Vtllume
D.p =change in pressure
dV/ V = change in volume (uc;ually in per<.~ 1 t
When temperature is held constant, Eq. 1 - 18 reduces to (Boyle's Lnw)
Eq. 1 -19
When temperature is held constant (isothermal condition), Eq. 1 - 18 reduces
Lo (Ciwrle's Lnw)
~
BULK MODULUS OF ELASTICITY, E8
The bulk modulus ol dasticity of tht• fluid exp'"'"'t'<; the~ ·m ~·n·c;sibiht \ '' the
fluid. It is the ratio of the change rn unit pre~~u rt> to t1\l , orresponJ rng
volume change- per unit of volume.
Eg.1-20
10
CHAPTER ONE
FLUID MECHANICS
& HYDRAULICS
Properties of Fluids
or
Eq. 1-21
(~Jk = E2. =Constant
V2
Eq. 1- 22
P1
~--1
and
~: = ( ;~ J-k-
Properties of Fluids
11
Table 1 - 4: Typical Vapor Pressures
For Adiabatic or Isentropic Conditions (no heat exchanged)
p1 V1 k = p2 V2k
CHAPTER ONE
I I UJD MECHANICS
HYDRAULICS
Fluid
kPa,20°C
mercury
turpentine
water
e thyl alcohol
0.000173
0.0534
elhcr
butane
Freon-12
propane
ammonia
Eq. 1-23
where:
P• =initial absolute pressure of gas
pz = final absolute pressure of gas
V 1 = initial volume of gas
v2 =final volume of gas
T 1 = initial absolute temperature of gas in °K (°K = oc + 273)
T2 = finc:ll 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.
olved Problems
Ptoblem 1-1
ll ~·voir of glycerin has a mass of 1,200 kg and a volume
Molecular activity in a liquid will allow SOIT)e 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 vnpor pressure or safttrntion pressure.
olution
(11)
(l1)
Weight, W = M g
= (1,200)(9.81)
Weight, W = 11,772 Nor 11.772 kN
w
.
. h
Umtwe1g
t,y= V
11.772
0.952
Unit weight, y = i-2.366 kN/m 3
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.
of 0.952 cu. m
)llld 1ts (a) weight, W, (b) unit weight, y, (c) mass density, p, and (d) specific
I lVII Y (s).
VAPOR PRESSURE
2.34
5.86
58.9
2 18
584
855
888
(1)
.
M
0 ens1ty, p = V
.
1200
Density, p = _
0 952
Density, p = 1,260.5 kglrn3
12
(d)
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
S peel'fi c gravity,
. s= Pgiy
Pwatcr
Specific gravity, s =
rt UID MECHANICS
f. HYDRAULICS
13
olution
( t}
W = mg = 22(9.75)
W=214.5 N
(/J}
Since the mass of an object is absolute, its mass will still be 22 kg
~
1 2605
1,000
Specific gravity, s = 1.26
CHAPTER ONE
Properties of Fluids
Problem 1-2
Problem 1-5
The specific gravity of certain oil is 0.82. Calculate its (n) specific weight, in
lb/ft3 and kN/m3, and (b) mass density in slugs/ft3 and kg/m3.
Whttl is the weight of a 45-kg boulder if it is brought to a place where tht>
~t•lt>ration due to gravity is 395 m/s per minute>?
olution
Solution
(a)
Specific weight, y = Yw•ter x s
Specific weight, y = 62.4 x 0.82 = 51.168 lb/ft-1
Specific weight, y = 9.81 x 0.82 = 8.044 kN/ml
(b)
Density, p = Pw•ter 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
lmm
g- 395 - - x - min
60sec
g = 6.583 mjs2
w = 45(6.583)
W= 296.25 N
l'r oblem 1 - 6
Problem 1-3
A liter of water weighs about 9.75 N. Compute its mass in kilograms.
Solution
w
Mass=g
9.75
Mass=-9.81
Mass = 0.994 kg
Problem 1-4
If ~n object has a mass o.f 22 kg at sea level, (n) what will be its weight at a
pomt where th~ acceleration due to gravity g = 9.75 mjs2? (b) What will be its
mass at that pomt?
II lh~ specific volume of a certain gas is 0.7848 m 1/kg, what is ils specific
~light?
olution
1
V,=p
1
1
p=-=---
v,
0.7848
p = 1.2742 kgjm3
Specific weight, y = p x :5
= 1.2742 X 9.Hl
Specific weight, y = 12.5 Njm3
CHAPTER ONE
14
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
Problem 1-7
CHAPTER ONE
15
Properties of Fluids
Solution
What is the specific weight of air at 480 kPa absolute and 21 °C?
Solution
y=pxg
Jl
p= RT
FLUID MECHANICS
& HYDRAULICS
where R = 287 J/kg-°K
480 X 10 3
287(21 + 273)
p = 5.689 kg
Density, p = J....
g
13.7
9.81
= 1.397 kg/ m 1
Density, p = L
RT
.
= (205 + 101.325) X 10 3
1 397
R(32 + 273)
Gas constant, R = 718.87 J/kg- °K
Note: P otm - L01.325 kPa
y = 5.689 X 9.81
y = 55.81 NjmJ
Problem 1 - 10
Problem 1-8
Find the mass density of helium at a temperature of 4 oc and a pressure of 184
kPa gage, if atmospheric pressure is 101.92 kPa. (R = 2079 J/kg • °K)
0
(
m ,1
Solution
p= _L
Solution
RT
Density p = L
RT
I
P = P&••&• + p,,tm
= 184 + 101.92
p = 285.92 kPa
T = 4 + 273 = 277°K
D
\iris kept at a pressure of 200 kPa nbsolule nnd " tl'mpt>rMun' "f i()
500-liter container. What is the mass of air?
,
285.92 X 103
enslty, P = 2,079(277)
Density, p = 0. 4965 k!ifm3
Problem 1-9
\ t 32°C and 205 kPa gage, tlile specific weight of a certain gas was 13.7 N/m3 .
I ldermine the gas constant of this gas.
200 X 103
287(30 + 273)
p = 2.3 kg/m 3
Mass= p x V
-23x
500
.
l()l)(l
Mass = 1.15 kg
Problem 1 - 11
A cylindrical tank 80 em in diameter and 90 em 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 kNjml.
16
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
Solution
CHAPTER ONE
Properties of Fluids
I I UID MECHANICS
& HYDRAULICS
olution
M
dP
Es=---
p=-
v
y=pg
dV IV
dp = /)2- J'i
p1 = 0
dp = p2
= 840(9.81) = 8240.4 Nlm3
y = 8.24 kNim3
dV= V2- VI
dV = -0.6% V = -0.006V
p=
17
420 - 40
t (0.8) (0 ..90)
2
= 840 k I m3
g
Pz
Problem 1- 12
E8 -
A lead cube has a total mass of 80 kg. What is t11e length of its side? Sp. gr. of
lead= 11.3.
p2 = 0.0132 GPa
p2 = 13.2 MPa
- - 0.006V
Iv
= 2.2
Solution
Let L be the length of side of the cube:
M=pV
80 = (1000 X 11.3) LJ
L = 0.192 m = 19.2 em
Problem 1 - 15
W.ttcr in a hydraulic press, initially at 137 kPa absol ute, is subjected to a
pn•ssure of 116,280 kPa absolute. Using FH = 2.5 GPa, determine the
p•·rcentage decrease in the vol ume of water.
olution
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 (E8 ) of the liquid is:
•
dp
Ea=--dV IV
_ X
= _ (116,280 -137) X 10
2 5 109
dV IV
dV
Solution
dP
v
2-1
(0.995-1)11
Es=---- =-------dV IV
dV
-
v
'\
= -0.0465
= 4.65'\'u decrease
Es = 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, E8 = 2.2 GPa.
..
If 9 m3 of an ideal gas at 24 oc and 150 kPaa is compressed to 2 m 3, (n) what IS
the resulting pressure assuming isothermal conditions. (b) What would hav<:>
h0en the pressure and temperature if the process is i<;enh·opic_ Use k = 1.3
CHAPTER ONE
18
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
CHAPTER ONE
I llJID MECHANICS
liYDRAULICS
Prop erties of Fluids
19
Solution
(a)
(b)
For isothermal condition:
\, !.1rge p lane surfaces are 25 m m apart ,md the space between them ts filled
P1 V, = pz Vz
150(9) = p 2 (2)
llh ,, liquid of v iscosity 1-1 = 0.958 Pa-s. Ass uming the velocity gradient to be
lr ught line, what force is required to pull a very thin pla te of 0.37 m 2 area at
Pz = 675 kPa abs
unst,mt speed of 0.3 m /s if the plate is 8.4 mm h·om one of the surfaces?
For isentropic process:
Pl V, k = pz Vz k
150(9) u = p 2 (2) u
pz = 1,060 kPa abs
olutlon
/' = F, + F~
U/y
F/A
T
( ) (k- 1)/ k
_1_=12._
r,
=
(
f' = ~tllA
1,o6o)'l.3- 1l;u
oc is 0.00402 poise and its specific gravity is 0 978
dete~nune Its absolute viscosity in Pa - s and its kinematic viscos·ty
· . z;
1
If the v~sco~ity of water at 70
m m
Solution
I
I
v
'
-
F1
Fl
_ 0.958(0.3)(0.37) = _ N
64
F1 0.0166
Fz =
and m stokes.
8.4
I
I
I
I
I
y
Problem 1 - 17
0.958(0.3)(0.37)
0.0084
= 12.66 N
r = 6.4 + 12.66
F= 19.06 N
s
I roblem 1- 19
Absolute viscosity:
0.1 Pa-s
= 0.00402 poise x - -- 1poise
~· = 0.000402 Pa - s
Kinematic viscosity:
v = !: =
0.000402
p
(1000 x 0.978)
v = 4.11 x 1Q-7 m2fs
1 stoke
v = 4.11 x 10·7 m2fs x - - - --
0.0001 m 2 I s
v = 4.11 x 1Q-3 stoke
25 mm
u /y
24 + 273
150
Tz = 466.4°K or 193.4°C
~
16.6
~-~=--
p,
T2
1\
T
~t= - -
tylinder of 125 mm radius rota tes concentrically inside a fixed cylinder of
I 0 mm radius. Both cylinders a re 300 m m long. Determine the v iscosity of
lh• ltquid w hich fills the space between the cylind ers if a torque of 0.88 N- m IS
qu1red to maintain a n a ngular velocity of 27f radians / sec Assume tlw
l'IOl'Ity gradient to be a strai&h t line
CHAPTER ONE
20
FLUID MECHANICS
& HYDRAULICS
Properties of Fluids
r~r,
Solution
fixed
T
Jl= - -
Xy=0.005m
u /y
~~
U= rw
u = 0.125(27t)
U = 0.785 m /s
U = 0.785
F =tA
rotatmg
cylinder
I
I
I
I
I
I
I
-e
=OJ
Wsin e- F, = 0
f,=WsinO
r, = 176.58 sin 15°
176.58 sin 15° = 0.0814-U- (0.3)
0.003
ll = 5.614 m/s
11, = 5.614 m/s
Torque= f(0. 125)
Torque= tA (0.125)
0.005
21
y
y = 0.005 m
29.88
Jl = 0.785/0.005
Jl = 0.19 Pa-s
J>ropert ies of Fluids
IF, = T A = p ~ A I
fixed cylinder
0.88 = t [2n(0.125)(0.3)) (0.125)
T = 29.88 Pa
CHAPTER ONE
FLUID MECHAN ICS
& HYDRAULICS
L
+- -
=0.3 m
Problem 1 - 21
rstimalc the height to which wa ter will rise 111 a capll lilrv tube of diamctL'r ":\
mm Use a= 0.0728 N/m andy= 9810 N/m 1 for water
liquid
0.125
0.13 m
\
Problem 1 - 20
A n 18-kg sla b slides down a 15° inclined plan e on a 3-mm -thick film of oil
with viscosity p = 0.0814 Pa-sec. [f the contact a rea is 0.3 m2, find the termin al
velocity of the slab. Neglect air resista nce.
Solution
Solution
Note: 0 = 90° for water in dea n lube
40
Cupillary rise, 11 =
yrl
4(0.0728)
= _ . ! . . . . __ _:.._
'
981 0(0.003)
Capi llary rise, 11 = 0.0099 m = 9.9 mm
Ca pill~ry rise h
W = 18(9.81) = 176.58 N
.. y
I
Problem 1 - 22
Estimate the capillary depression for me rcury in a glass capillary tube 2 mm 111
diu mcler. Use o = 0.514 N/m ~nd 0 = 1-10°
I
I
I
y = 0.003 m
plane
..
Solution
X
Capillary rise, 11 =
4crcosU
yT/
Capillary rise, II = -0.0059 111
Terminal velocity is attained w hen the sum of all forces in the direction of
motion is zero .
4(0.51-+ )(cos 140°)
(9810 X 13.6)(0.002)
(the negative sign indicates capillary depress1on)
Capillary depression, h = 5.9 mm
CHAPTER ONE
Properties of Fluids
22
FLUID MECHANICS
& HYDRAULICS
Problem 1 - 23
~hat is the value of the surface tension of a small drop of water 0.3 nun in
diameter which is in contact with air if the pressure within the droplet is 561
Pa?
Solution
rl
h vt•I<Kity of the pressure wave (sound w,we) IS
Ff
561=~
0.0003
a= 0.042 N/m
1 1.11 ttme of travel of sound.
(0 5) 1/4 sec and the total
11111 L' covered is 2/1, then;
'it
11
4(0.0712)
= 6,329 Pa
45 X 10-6
Distilled water stands in a glass tube of 9 mm diameter at a height of 24 mm.
What is the h·ue static height? Use a= 0.0742 Nfm.
11
yrl
where 0 = oofor water in glass tube
II =
4(0.0742)
0 00336
m = 3·36 mm
981 0(0.009) = ·
True static height= 24- 3.36
True static height= 20.64 mm
0
0
0
0
<I
1,428(1/.J)
178.5 m
1 ,., h.ll pressure will80 oc water boil?
por pressure of water at 80°C = 47 4 kPa,
lution
11
= 4acos0
~
0
roblcm 1-27
1
Solution
'=-'
Bottom
4a
p=rl
Problem 1 - 25
.r.:,.
the echo is received
1 lwt~v between tmpulses, then
Solution
=
Sonar
transmitter
2.0-l X 109
= 1,428 m/s
1000
Problem 1 - 24
An a tomizer forms water droplets 45 J..!m in diameter. Determine the excess
pressure w ithin these droplets using a= 0.0712 Nfm.
23
ruhlcm 1-26
11 11 transmitter operates at 2 tmpube~ per second If the deviCe ts held to
urt,lCe of fresh water (E 8 = 2.0-l :-. lOq Pa) and the ech<' i~ received midwc\\
n imptllses, how deep is the water?
nlution
4a
p= -
P
CHAPTER ONE
Properties vf Fluids
lllJIIJ MECHANICS
I IVDRAULICS
r will boil if the atmosph~ric pressure equ,l!S 1hP vilpr,r , · ,--.~L• 1"'
1111 rC'IllTL' water at ~0 oc will boil at 47.4 kP;:
24
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
!supplementary Problems
FLUID MECHANICS
& HYDRAULICS
CHAPTER ONE
Properties of Fluids
25
l'roblem 1 - 33
Problem 1 - 28
1) If J2 m3 of nitrogen at 30°C and 125 kPa abs is permitted to expand
(b) What would the
pressure and temperature have been if the process had been isentropic?
Aus: (a) 50 kPa abs
(b) 34.7 kPa abs; -63°C
f~othct mally to 30 m3, what is the resulting pressure?
What would be the weight of 1 3 k
due to gravity is 10 mjs27
- g mass on a planet where the ~cceleration
A11s: 30 N
Problem 1 - 29
A vertical cylindrical tank with a diameter of
with water to the top with water at 2ooc If~ m and a depth of 4 m is filled
much water will s ill over?
.
. .
e water ts heated to sooc, how
kNjm3 and 9 69 kJ/ 3
. U.rut wetght of water at 2ooc and sooc is 9 79
·
m , respectively
·
Ans: 4.7 ml
Jlroblem 1 - 34
1\ SlJU,ue block weighing 1.1 kN and 250 mm on an edge slides down an
Incline on a film of oil 6.0 f..IID thick. Assuming a linear velocity profile in the
nl and neglecting air resistance, what is the terminal velocity of the block?
'f lw VIStosity of oil is 7 mPa-s. Angle of inclination is 20°.
AilS: 5.16 mjs
Problem 1 - 30
A rigid steel container is partially filled with a 1' .
the liquid is 1.23200 L At a r
tqutd at 15 atm. The volume of
1.23100 L. Find the a~erage b~~s:~~:tu;o atm, t~~ volume of the liquid is
given range of pressure if th t
of elastictty of the liquid over the
·
·
e emperature after com
return to its initial value Wh t. th
ff .
presswn ts allowed to
.
a IS e coe tcient of compressibility?
Ans: En = 1.872 CPa; f3 = 0.534 GPa·l
Problem 1 - 31
Calculate the density of water va or at 350 kP
.
o . .
ts 0.462 kPa-m3jkg-oK.
p
a abs and 20 C tf tts gas constant
Problem 1 - 35
II •tWt>nc at 20°C has a viscosity of 0.000651 Pa-s. What shear stress is required
(o dl'form this fluid at a strain rate of 4900 s·1?
Ans: t = 3.19 Pa
Problem 1 - 36
1\ sh,,ft 70 mm in diameter is being pushed at a speed of 400 mm/ s through a
lw.mng sleeve 70.2 mm in diameter and 250 mm long. The clearance, assumed
Ulllform, is filled with oil at 20°C with v = 0.005 m 2/s and sp. gr.= 0.9. Find
llu' fPrtl' exerted by the oil in the shaft.
Ans: 987 N
A11s: 2.59 kgjm3
Problem 1 - 32
Probl m 1-37
Air is kept at a pressure of 200 kPa
d
C0ntainer. What is the mass of th . ? an a temperature of 30oc in a 500-L
e arr.
''' il b.tth of water.
Ans: 1.15 kg
I wo cll•an parallel glass plates, separated by a distanced= 1.5 mm, are dipped
0 07311 Njm?
How far does the water rise due to capillary action, if cr =
,
Ans: 9.94 mm
26
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
CHAPTERlWO
rlUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
27
.
Problem 1 ~ 38
~ind ~e a~gle the ~urface tension film leaves the glass for a vertical tube
~mmheiUsed m water tf the diameter is 0.25 inch and the capillary rise is 0 08
me . se cr = 0.005 Ib/ft.
·
A11s: 64.3°
Chapter 2
Principles of Hydrostatics
Problem 1 ~ 39
What force isocrequired to lift a thin wire ring 6 em in di'ame ter f rom a water
f ace at 2o ? (cr of water at 20°C = 0 ·0728 N/ m ) · N eg1ect the weight
sur
·
ring.
of the
A11s: 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
.urface, with the force acting at right angles to the surface uniformly in all
d trections.
p=
Force,F
Area, A
Eq. 2-1
In the English system, pressure is usually measured in pounds per square inch
2
(psi); in international usage, in kilograms per square centimeters (kg/cm ), or
m atmospheres; and in the international metric system (SI), in Newtons per
(1uare meter (Pascal). The unit atmosphere (atm) is defined as a pressure of
I 03323 kg/cm2 (14.696 lb/in2), which, in terms of the conventional mercury
l>.~rometer, corresponds to 760 mm (29.921 in) of mercury. The unit kilopascal
(k Pa) is definid as a pressure of 0.0102 kg/ cm2 (0.145 lb/ sq in) .
.,ASCAL'S LAW
l'ascnl's law, developed by French mathematician Blaisr Pnscal, states that the
pressure on a fluid is equal in all directions and in all parts of the container. In
ltgure 2 - 1, as liquid flows into the large container at the bottom, pressure
pushes the liquid equally up into the tubes above the container. The liquid
ttscs to the same level in all of the tubes, regardless of the shape or angle of the
lube.
28
CHAPTER TWO
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
CHAPTER TWO
Principles of Hydrostatics
II UJD MECHANICS
I. HYDRAULICS
29
ABSOLUTE AND GAGE PRESSURES
Figure 2 - 1: Illustration of Pascal's Law
The laws of fluid mechanics are observable in many everydal/ situations. For
ex~mple, the pressure exerted by water at the bottom of a· p')nd 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
containers that should not be tipped.
, ge Pressure (Relative Pressure)
(., 'l;~ pressures are pressures above or below the atmosphere and can be
Olt ll'>ured by pressure gauges or manometers .. For small pressure differences, a Utul•• manometer is used. It consists of aU-shaped tube with one end connected to
llw container and the other open to the atmosphere. Filled with a liquid, such as
w.rh•r, oil, or mercury, the difference in the liquid surface levels in the two
rn.mometer legs . indicates the pressure difference from local atmospheric
,,,llhtions. For higher pressure differences, a Bourdon gauge, named after the
I '' rrch inventor Eugene Bourdon, is used. This consists of a hollow metal tube
rth an oval cross section, bent in the shape of a hook. One end of the tube is
l1.' .I'd, the other open and connected to the measurement region.
Atmospheric Pressure & Vacuum
luw~plteri.c Pressure is the pressure at any one point on the earth's surface from the
rght of the air above it. A vncuum is a space that has all matter removed from it.
II 1•. impossible to create a perfect vacuum in the laboratory; no matter how
'iv,mced a vacuum system is, some molecules are always present in the vacuum
Even remote regions of outer space have a small amount of gas. A vacuum
n .1Iso be described as a region of space where the pressure is Jess than the
t1 Irma! atmospheric pressure of 760 mm (29.9 in) of mercury.
llndt•r Normal conditions at sea level:
fl•tm = 2166lb/ft2
= 14.7 psi
= 29.9 inches of mercury (hg)
= 760mmHg
= 101.325 kPa
Figure 2 - 2: Illustration of Pascal's Law
Ab olute Pressure
All•olute pressure is the pressure above absolute zero (1>ncuum)
P•bs = Pc•s• + p.11m
Eq. 2-2
Ntl
• Absolute zero is attained if all air is removed. It Is the lowest possible pressure attarnable.
• Absolute pressure can never be negative.
• The smallest gage pressure is equal to the negative of the ambient atmospheric pressure.
30
CHAPTER TWO
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
CHAPTER TWO
II UID MECHANICS
& I fYDRAULICS
Pnnc1ples of Hydrostatics
31
VARIATIONS IN PRESSURE
vnsider any two points (1 & 2), whose difference in elevation is II, to lie m the
Standard
ndo; of an elementary prism having a cross-sectional area a and a length of L
atmosphere = 101.325 abs
ll\ll' this prism is at rest, all forces acting upon it must be in equilibrium.
-40 gage
Free liquid surface
Current atmosphere = 100 abs
s:z
...
160 abs
P1 & P2 are gage pressures
60 abs
Absolute zero= -101.325 gage
__....__ _ _ _ _......;t__--1!__ or ·100 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 Sl'gm'fi1es gage pressure.
MERCURY BAROMETER
~ mercury barometer is an accurate and relatively
Simple way to measure changes in atmospheric
pressure. At sea level, the weight of the atmosphere
forces mercu~ 760 mm (29.9 in) up a calibrated
glass tube. Higher elevations yield lower readings
b~cause ~he atmosphere is less dense there, and the
thmner air exerts less pressure on the mercury.
Figure 2 - 4: Forces acting on elementary prism
N ol!•: Free liqurd Surface refers to liquid surface subject to zero gage pressure or w1th
atmospheric pressure only.
' '- at sea Level
W 1th reference to Figure 2-4:
W=-y V
W= y (nL)
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 c1rcular motion of the
pointers along the aneroid
barometer's face.
[~F,
= 0]
F2 - F, = wsin e
p2 a - p, a = -y (aL) sin 9
p2 - p1 = y L sin e
but L sin 9 = f1
Eq. 2-3
llwrefore; tl1e difference in pressure between any two points in a homogeneous jluuf
rl/ rest is equnl to the product of the tmit weight of tile fluid (y) to the vertical distant
1/r) behoeen the points.
32
CHAPTER TWO
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
C
Also:
Eq. 2-4
P2 = Pt + wh
33
.der the tank shown to be filled with liquids of different densities an~
w~:lair at the top under a gage pressure of pA, the pressure at the bottom o
the tank is:
This means that nny change in pressure nt point ~ would cnuse nn equal change nt
point 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
the gage pressure p1 is zero and Eq. 2- 4 becomes:
CHAPTER TWO
Pnnciples of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
Eq. 2-7
pbouom
PRESSURE HEAD
.
.
.
Pressure head is the height "lr" of a column of homogeneous liqwd of urut
weighty that will produce an intensity of pressure p.
Eq. 2-5
p=wh
Eq. 2-8
It = E.
y
This means that the pressure at any point "lr" below a free liquid surface is equal to
the product of the unit weight of the fluid (y) and h.
To Convert Pressure head (height) of liquid A to liquid B
Consider that points 0 and f) in Figure 2 - 4 lie on the same elevation, such
that h = 0; then Eq. 2- 4 becomes:
Eq. 2-6
This means that the pressure along the same horizontal plane in a homogemous fluid
at rest are equal.
sA
PA
I - 1 ~
ha = h A
- or lla = h A - or ta - LA
s8
Pa
To convert pressure head (height) of any liquid to water, just multiply its
height by its specific gravity
lzwaler
Pressure below Layers of Different Liquids
Air, pressure = PA
h1
Liquid 1
0
0
hz
Liquid 2
h3
Uquid 3
0
0
'---------4..0ot Pbottom
Eq. 2-9
Ya
ltliquid X Stoquid
Eq: 2-10
34
CHAPTER TWO
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
t I tJID MECHANICS
I fYDRAULICS
CHAPTER TWO
Principles of Hydrostatics
35
MANOMETER
l J>S In Solving Manometer Problems:
A 11tn110111efer 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 It is used to measure pressure
I Decide on the fluid in feet or meter, of which the heads are to h1
l
pressed, (water is most advisable).
t.;tarting from an end point, number in order, the interface of diffNenl
fluids.
Identify points of equal pressure (taking into account that for a
homogeneous fluid at rest, the pressure along the same horizontal plane
1re equal). Label these points with the same number.
Proceed from level to level, adding (if going down) or subtracting (il
going up) pressure heads as the elevation decreases or incrf'il"l''>.
respectively with due regard for the specific gravity of the fluids.
Types of Manometer
Open Type - has an atmospheric surtace m one leg and rs capable ot
measuring gage pressures
Differential Type - without an atmospheric surface and capable of
measuring only differences of pressure
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 requtre long tubes
• Gas pressures can not be measured because gas can not form a free
surface
t•1ohlem 2- 1
II
d<'pth of liquid of 1 m causes a pressure of 7 kPa, what is the specific
r vrty of the liquid?
olutlon
Pressure, p = y 1r
7 = (9.81 x s) (1)
s = 0.714
-7 Specific Gravity
t 10blem 2-2
(a) Open manometer
(b) Drfferentral manometer
h 11 is the pressure 12.5 m belo~ the ocean? Use sp. gr. = 1.03 for salt water.
lution
p =y 1r
p = (9.81 x1.03)(12.5)
p = 126.3 kPa
(c) Piezometer
J
CHAPTER TWO
36
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
I LUID MECHANICS
/. HYDRAULICS
CHAPTER TWO
Principles of Hydrostatics
37
Problem 2-3
Problem 2-5
If the pressure 23 meter below a liquid is 338.445 kPa, determine its unit
weighty, mass density p . and specific gravity ~
II the pressure in the air space above an oil (s = 0.75) surface in a closed tank is
115 kPa absolute, what is the gage pressure 2m below the surface?
Solution
olution
Unit weight, y
(a)
p = yh
338.445 = y (23)
y = 14.715 kNfmJ
.., l Mass density, p
P = Psurlaco + Y h
Psurloce = 115- 101.325
P•urlacr = 13.675 kPa gage
Note: Patm = 101.325 kPa
p = 13.675 + (9.81x0.75)(2}
p = 28.39 kPa
p = 1.
g
14.715 X 10 3
p=
9.81
p = 1,500 kglml
Problem 2-6
Find the absolute pressure in kPa at a depth of 10m below the free surface of
oil of sp. gr. 0.75 if the barometric reading is 752 mmHg.
Solution
Specific gravity,~
(c)
Pn"' = Pnlm + Pgugr
~ = Pnuid
Palm= y,., h,
= (9.81 X 13.6){0.752)
Pwater
1,500
s= - 1,000
s = 1.5
P•tm = 100.329 kPa
Pnl'> = 100.329 + (9.81 X 0.75)(1 0)
P••los = 173.9 kPa
Problem 2-4
Problem 2-7
If the pressure at a point in the ocean 1s 60 kPa, what is the pressure 27 meters
below this point?
A pressure gage 6 m above the bottom of the tank containing a liquid reads 90
kPa. Another gage height 4 m reads 103 kPa. Determine the specific weight o(
the liquid
Solution
rhe difterence In pressure between any two points In a
liquid is P2 - Pt = y /1
P2 = Pt + yh
= 60 + (9.81xl.03)(27)
112 = 332.82 kPa
Solution
P2- PI= y h
103 - 90 = y(2) •
y= 6.5 kN/m 3
CHAPTER TWO
Principles of Hydrostatics
38
FLUID MECHANICS
&HYDRAULICS
CHAPTER TWO
Principles of Hydrostatics
I I UID MECHANICS
'• HYDRAULICS
39
olution
Problem 2-8
An open tank contains 5.8 m of water covered with 3.2 m of kerosene (Y = 8
kNjm3). Find the pressure at the interface and at the bottom of the tank.
Solution
Since the density of the mud varies with depth, the pressure
should be solved by integration
dp = y dh
dp = (10 + 0.5 h)dh
(a) Pressure at the interface
PA = Yk hk
= (8)(3.2)
PA = 25.6 kPa
,
fdp = foo +o.sh)dh
0
Kerosene
(b) Pressure at the bottom
PB :~ yh
= Yr" hw + Yk hk
= 9.81(5.8) + 8(3.2)
pR = 82.498 kPa
5
., = 8 kN/m 3
Water
y
=9.81 kN/m3
0
p = 10h + 0.25fr 2
5
]0
= [10(5) + 0.25(5)2] - 0
p = 56·.25 kPa
l11'0blem 2- 11
Problem 2-9
If atmospheric pressure is 95.7 kPa and the gage attached to the tank reads 188
mmHg vacuum, find the absolute pressure within the tank.
Solution
In I h<• figure shown, if the atmospheric
(' .ure is 101.03 kPa and the absolute
1'' • ure at the bottom of the tank is
1 ~ kPa, what is the specific gravity
I •livl' oil?
SAE Oil, s = 0.89
Water
Pubs = P•lllll + pgtrg•
P8"8' "' Ymereury hmercury
= (9.81 X 13.6)(0.188)
= 25.08 kPa vacuum
Olive, s =?
pgagr '" -25.08 kPa
Mercury, s = 13.6
P•bs = 95.7 + (-25.08)
1
1.5 m
t
2.5 m
+
2.9 m
-l
0.4 m
-L.
p.,,. = 70.62 kPa abs
Problem 2 - 10
The weight density of a mud is given by y = 10 + 0.5/t, where y is in kNjm3 and
I! is in meters. Determine the pressure, in kPa, at a depth of 5 m
nlutlon
1 ..,ge pressure at the bottom of the tank, p = 231.3- 101.03
<..tgc pressure at the bottom of the tank, p = 130.27 kP!"
II' L.ylt]
P -= Ym fzm + Yo ho + Yw hw + Y011 l1o11
130.27 = (9.81 X 1i.3.6)(0.4) + (9.81
s - 1.38
}
~}(2.9) T 11.81 (2.5) + (9.81 X 0.89}(1.5)
CHAPTER TWO
40
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Problem 2 - 12
CHAPTER TWO
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
41
If air had a constant specific weight of 12.2 Njm3 and were incompressible,
what would be the height of the atmosphere if the atmospheric pressure (sea
level) is 1 02 kPa?
Problem 2 - 14
Com ute the barometric pressure in kPa at an altitude ?~ t200 ~ if the
p
1 . 1 . 101 3 kPa Assume isothermal conditions a 21 C. Use
pressure at sea e" e ts
·
·
R = 287 Joule /kg-oK.
Solution
Solution
For gases:
Height of atmosphere, 11 = !!..
y
1dp = -pg dh
102 X 103
12.2
Height of atmosphere, 11 = = 8,360.66 m
1
p= ..L
RT
p
287(21 + 273)
p = 0.00001185 p
Problem 2- 13 (CE Board May 1994)
Assuming specific weight of air to be constant at 12 Njm3, what 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 mountain reads 480 mm.
dp = -(0.00001185 p)(9.81) dll
dp
p
= 0.0001163 dll
p
Solution
1200
Jd:
= -0.0001163
101.3x 10'~~~'
Air
y = 12 N/m3
hm
=480 mm
ln p
l
J
(y., hm)bottom - (Ym ltm)top = (y h)alr
(9,810 X 13.6)(0.654)- (9,810 X 13.6)(0.48) c 12 h
h • 1,934.53 m
0
I'
= - 0.0001163 h
101.3x103
1200
J
0
In p-In (101.3 x 1Q3) =- 0.0001163(1200- 0)
In p = 11.386
p-_ e11.386
p = 88,080 Pa
Pbot - Ptop • y II
Jdlt
CHAPTER TWO
42
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
CHAPTER TWO
flUID MECHANICSIIYDRAULICS
43
Principles of Hydrostatics
ltoblem 2- 18 {CE November 1998)
Problem 2 - 15
Convert 760 mm of mercury to (a) oil of sp. gr. 0.82 and (b) water.
1 l tn A has a cross-section of 1,200 sq. em while that of piston B is 950 sq. em.
Solution
lth the latter higher than piston A by 1.75 m. If the intervening passages are
II d with oil whose specific gravity is 0.8, what is the difference in pressure
I\\ Pl'n A and B.
5
(ll) Ifool -- 1lmrrrury mcrcunt·
Soil
13.6
= 0.76 0.82
/roll = 12.605 m of oil
lutlon
B
PB =Yo flo
Jl
= ( 9,8}0 X 0.8)(1.75)
I
i'A- PB = 13,734 Pa
A
(b)
frw.olor = hmrrruoy Sonercury
= 0.76(13.6)
!twa lor = 10.34 m of water
1.75 m
_j
Oil
s = 0.8
950 cm2
1200 cm 2
Problem 2 - 16 {CE Board May 1994)
A barometer reads 760 mmHg and a pressure gage attached to a tank reads
850 em of oil (sp. gr. 0.80). What is the absolute pressure in the tank in kPa?
Solution
p,oh>
=p,olon + P~·'t"
= (9.81 X 13.6)(0.76) + (9.81 X 0.8)(8.5)
p,,los = 168.1 kPa abs
ltoblem 2- 19
lu the figure shown,
I I r mine the weight W
lh 11 ,,,n be carried by the
I k N force a-cting on the
1l loll
1.5 kN
Oil, s = 0.82
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
requu·ed?
lutlon
Solution
Since the pressure under the piston is uniform:
Force = pressure x Area
80,000 = (10 X 10~) t D2
D = 0.1 m = 100 mm
~11\ce points 1 and 2 lie on the
ulll\C elevation, Pl = P2
1.5
2
4 (0.03)
-
*
w
(0.3)
2
IV 150 kN
2
Oil, s = 0.82
44
CHAPTERlWO
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Problem 2 - 20
CHAPTERlWO
~ LUID MECHANICS
Prrnciples of Hydrostatics
ft. HYDRAULICS
45
.,olution
A drum 700 mm in diameter and filled with water has a vertical pipe, 20 mm
in diameter, attached to the top. How many Newtons of water must lx>
poured into the pipe to exert a force of 6500 Non the top of the drum?
W = 44 kN
Cylinder
W
=44 kN
A = 0.323 m'
~r-- 0
Solution
Force on the top:
F= p x Area
6500 = p X t (7002 - 202)
Plunger,
I
p = 0.016904 MPa
p = 16,904 Pa
a =0.00323 m ·
h
1
[p =y h]
16,904 = 9810 ,,
h=1.723m
Top
IP2 - Pr = y,, Jz)
F
F
pr = -;; = 0.00323
Weight= y x Volume
= 9810 X
p1 = 309.6 F (kPa)
t (0.02)2(1.723)
Area on top
w
44
r 2 =A = 0.32...1
Weight= 5.31 N
700 mm 0
p2 = 136.22 kPa
Problem 2 - 21
The figure shown shows a setup with a vessel containing a p lunger and a
cylinder. What force F is required to balance the w eight of the cylinder if the
weight of the .plunger is negligible?
Problem 2 - 22
·
· h OJ·1 WI' th s p. gr_- 0.82 . . Neglecting tlw
fhe hydraulic press shown IS fJlled
Wit
rt
weight of the two pis tons, w h at fo rce F on the handle ts requtred to suppo
Cylinder
W = 44 kN
A = 0.323 m2
f
4.6 m
1
136.!2- 309.6 F = (9.81 X 0.71$)(4.6)
F = 0.326 kN = 326 N
the 1 0 kN weight?
Plunger,
a = 0.00323 m2
CHAPTER TWO
46
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Solution
111
ce the gage reads " F.ULL" then the re,Hling is equivalC'nt t0 10 em of gasolin•
Reading (pressure head) when the tank contmn
water= \Y
tv+ 2-1- ) em of gasolinP
0.68
_5_=!1_
A1
47
Pnnciples of Hydrostatics
• HYDRAULICS
,
Since point~ I and 2 he on the samt>
elevation, then,
PI= P2
CHAPTER TWO
I I UID MECHANICS
A2
y+
fhen;
10
2"5'h = 30
IJ = 27.06 em
t (0.075) 2
h = 1 11 kN
Problem 2 - 24 (CE Board November 2000}
I o1 the tank shown in the Figure, h1 = 3m and '" = 4 m DE>termine the value
iL Mo =OJ
F(0.425) = F2(0.025)
F(0.425) = l 11 (0 025)
F = 0.0654 kN
F = 65.4 N
F
1( /r~
-,..-
\1.
Oil
0
0
.n
s = o.s4g
h2
0
0
0
0
hI
FBD of the lever arm
~Water
0
h
Water
0
I
0
0
0
Problem 2 - 23
The fuel gage for a gasoline (sp. gr. = 0.68) tank in a car reads proportional to
1ts bottom gage. If the tank is 30 em deep an accidentally contaminated with 2
em of water, how many centimeters of gasoline does the tank actually contain
when the gage erroneously reads "FULL"?
olllution
'lum.ming- up pressure head
rrom 1 to 3 in meters of water
h
y
Solution
y
y
0 + 0.84 112 - (4 - 3) = 0
lt2 = 1.19 m
Air
r
3
+ /r 2(0.84) - x = P
Gasoline, s = 0.68
2cmL_~~========~~
T
Water
"Full"
1
30 em
r
Gasoline, s = 0.68
l
1...----------.j'
"Full"
0
g Water
0
CHAPTER TWO
Principles of Hydrostatics
48
FLUID MECHANICS
& HYDRAULICS
Problem :l- 25 (CE Board May 1992)
In the figure shown. what is the static pressure m kPa m the atr chamber7
OfAPTEn TWO
FWI& MEGHANICii
& HVDR'AUUC-s
Principle~ of HydtO'Statlia!;
(J:OOS: 1
Problem 2 - 26
nwvolt1 b1608 3J) ~~ - c H1"'•r~o,q
I
~,
For the manometershowr11
determine the pressure at the
center of the pipe.
IJI·<I
•
lm
Mercury, s = 13.55
t2
_/'loil, s == o.so
Solution
The pressure m the air space
equals the pressure on th e surface
of oil, p'
noi:h
-=il r
CJolutionJ
A
., >t
I
r>t\J,r, Jrl ·P ,,..
'
II
''
'Jl
•,um-up , ressure head {•om
1 to 3 in tlieters )lb watel1
'
. f!
II
I !I'• Tf
I
.\
0
- 1m
12 +1(13.~5) +l.S(O.S) = p
3
r-
P2 = Y11• It,.
= 9.81(2) .
P2 = 19.62 kPa
2m
P2 - P' = y, h,.
I 9.62 - Pl = (9.81 " 0.80)(4)
p, = -11.77 kPa
I
0 + 14!.75 =
y
I
J!J.
y
11 rr1 t
= 14.75'rn of water
11 - 14!71i(9:81)
1,, = 144.7 kPa
y
')
1.5 m
L
\
Another solution.
Sum-up pressure head from 1 to 3 in meters ot water
!?..!.. + 2 - 4(0.80) = p3
v
() + 2 - 3.2
y
y
= ..f!1_
9.81
I'• = -11.77 kPa
r•• l
...... \
.
'
-~Oil, s = 0.80
...... ··--"
CHAPTER TWO
Principles of Hydrostatics
50
FLUID MECHANICS
& HYDRAULICS
CHAPTER TWO
Pnnc1ples of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
Problem 2 - 27 (CE Board November 2001)
Problem 2- 28 (CE May 1993}
Determine the value of yin the manometer shown in the Figure.
In the figure shown, when the
funnel is empty the water surface
is at point A and the mercury of
sp. gr. 13.55 shows a deflection of
15 em.
Determine the new
deflection of mercury when the
funnel is filled with water to B.
-;y;-
lm
t
3m
t
Air, 5 KPa
~
Oil
s = 0.8
y
y)
51
Water
lm
L
0.5 m
Mercury
5 = 13 .55
Solution
Solution
-;y;-
Summing-up pressure head from
A to B in meters of water:
fu + 3(0.8) + 1.5- y(13.6) = ElL
y
y
PB
- 5 +3.9-13.6y=9.81
y
w here PB = 0
lm
Air, 5 KPa A
3m
Oil
5 = 0.8
t
t
lm
...L
rt
Water
-:r--
y+x
R 0.15
0.15
l
4
y = 0.324 m
Mercury
5 = 13 .55
Figure (b): Level at B
Figure (a): Level at A
Solve for yin Figure (a):
Sum-up pressure head from A to 2 in meters of water:
fu + y- 0.15(13.55) = E1_
y
0 + y- 2.03 = 0
y = 2.03 m
y
CHAPTER TWO
Pnnciples of Hydrostatics
CHAPTER TWO
Principles of Hydrostatics
52
In Figure (b):
When the funnel is filled with water to B, point 1 will move down to 1
with the same value as point 2 moving up to 2'
ElL + 0.8 + y ,+ x- (x + 0.15 + x)(13.55) = h
y
•llln-up pressure head from 2 to 111 in meters of water:
1!1._ + y(13.6) - X = E.!!!_
y
13.6y- X= :.~1
In hgure (b):
Sum-up pressure head from B to 2':
53
y
,Eq. (1)
•,urn· up pressure head from 2' to nz' in meters of water:
y
0 + 0.80 + 2.03 + x- 27.1x- 2.03 = 0
26.1 X= 0.80
x = 0.031 m = 3.1 em
1?1;_ + (0.2 sine + y + 0.2)(13.6)- (x + 0.2) = Pm'
y
0 + 2.72 sin 8 + 13.6y + 2.72- X - 0.2 =
13.6y- X= 8.183- 2.72 sine
New reading, R = 15 + 2x = 15 + 2(3.1)
New reading, R = 21.2 em
y
*
Eq. (2)
111 6y- x = 13.6y - x]
8.183- 2.72 sine = i~1
~
ly
sin e = 0.3852
Problem 2 - ~9
e = 22.66°
The pressure at point 111 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, 8?
J, "·I'd cylindrical tank contains 2m of water, 3m of oil (s = 0.82) and the .11r
I 1 ,. oil has a pressure of 30 kPa. If an open mercury manometer at tlw
•II• •111 of the tank has 1 m of water, determine the deflection of mercury.
lutlon
Solution
',urn-up pressure head from
I to 4 in meters of water:
p "" + 3(0.82) + 2 + 1 - y(13.6) = Ei_
y
& + 2.46 + 3 - 13.6y = 0
y= 0.626m
y
3m
1-
2m
-*-
lm
Figure (a)
In Figure (n):
Figure (b)
Oil, s =0.82
2
Water
y
CHAPTER TWO
54
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Problem 2- 31
rhe U-tube shown IS 10 mm In diameter
and contains mercury. If 12 ml of water is
poured into the right-hand leg, what are
the ultimate heights in the two legs?
CHAPTER TWO
I Ull) MECHANICS
II\ IJRAULICS •
55
Pnncrples of Hydrostatics
In Ftgure (b):
Summing-up pressure head from 1 to 3 in mm of water:
10 mm 1£
El + 152.8- R(13.6) = E1_
y
R=11.24mm
r
E
E
E
E
0
y
In Eq. (2~
11.24 + 2x = 240
x= 114.38 mm
0
N
N
......
~
Mercury
L '!:t:==:::!J _j
L
Solution
120mm
_j
IJitimate heights in each leg:
Right-hand leg, lrR = h + x
= 152.8 + 114.38
Right-hand leg, hR = 267.18 mm
Solvtng tor It . (SI'I' figu rl' b)
2
Volumeofwater= .!1.
( 10 ) 11= 12cm.l
~
10
Left-hand leg, hL = R + x
= 11.24 + 114.38
Left-hand leg, hL = 125.62 mm
Note: 1 ml = 1 cm3
/1 = 15.28 em= 152.8 mm
Since the quantity of mercury before and after water is poured
remain the same, then;
t 1ohlem 2- 32
120(3) = /~ + X + 120 + A.
I~ + 2x = 240
~ Eq. (1 )
10 mm ~,"'
lOmm V
r
I
E
E
E
E
0
0
....
N
N
......
Mercury
L
L
120mm
Figure (a)
_j
_j
Gage
I''' ,1 gage reading of -17.1 kPa,
[h,t
li
3
Water
2
2
. . . -- . .. -.-
Mercury
L
120 mm
Figure (b)
_j
n
l·
]
I 1 lllline the (If) elevations of
liquids in the open
1 lo ometer columns E, F, and
111d (b) the deflection of the
in
the
U-tube
neglecting the
<9
E
Air
El.15 m
=0.70
5
El.12 m
-
:=
Water
El. 8 m
s = 1.6
]h
El. 4 rn
~
Mercury
r
F
G
CHAPTERlWO
CHAPTER TWO
56
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Solution
E
3m
El + 3(0.7) + 4(1)- lt3(1.6) = !!.!_
y
e·
h,
2
h,
s = 0.70
-~- ~· ~2_m__
Water
4m
hl
1- ~-~~-
3
lt3=2.72m
Surface elevation = 8 + 113
Surface elevation= 8 + 2.72 = 10.72 m
Deflection of mercury
Sum-up pressure head from 1 to 5 in meters of water;
El + 3(0.7) + 4 + 4- h4(13.6) = h
y
y
-J.~il + 10.1 - 13.6114
H
114 = 0.614 m
P1 = p.,. = - I 7 1 kPa
5
j_ -"'<_m_
y
-i.~il + 2.1 + 4- 1.61!3 = 0
s =1.6
4m
57
G
"'"
-t- El.- -15m
---
Principles of Hydrostatics
Column G
Sum-up pressure head from 1 to g in meters of water;
Gage
'
FLUID MECHANICS
& HYDRAULICS
]h.
4
Mercury, s = 13.6
Column E
Sum-up pressure head from 1 toe m metes of water,
El + /rl(0.7) = f!..£._
y
olutlon
y
-.}~i
1
Problem 2 - 33
An open manometer attached to a pipe shows a deflection of 150 mmHg with
the lower level of mercury 450 mm below the centerline of the pipe carrying
w.1ter. Calculate the pressure at the centerline of the pipe.
+ /ri(0.7) = 0
·-·-·-·-·-·- -·-·-- -·-·-·~~-
It,= 2.5 m
Surface elevation = 15 _ h,
Surface elevation = 15 _ 2.5 = 12.5 m
Column F
Sum-up pressure head from 1 to fin meters of water;
El + 3(0.7) - h2(1) = !!..1_
y
1
9 ~,
y
1
+ 2.1 - h? = 0
lr2 = 0.357 m
Surface elevation = 12 + h2
Surface elevation = 12 + 0.357 = 12_357 m
Water
450 mm
um up pressure head from 1 to
In nwlers of water;
1' 1 + 0.45- 0.15(13.6) = 12_
y
y
111 + 0.45- 2.04 = 0
9 Hl
15.6 kPa
58
CHAPTER TWO
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Problem 2 - 34
For the configuration shown, calculate the
weight of the piston if the pressure gage
reading is 70 kPa.
CHAPTER TWO
I LUID MECHANICS
'- HYDRAULICS .
r-lm 0"1
'>olution
1) Gage liquid= mercury, 11 = 0.1 m
Piston
Sum-up pressure head from
1 to 4 in meters of water;
Air
0
1.5 m
g water
_ P4
y
y
4
0
P + x + l1 - 1!(13.6) -X- 1.5- _1
Oil
s = 0.86
t
.
0
~
El - b._ = 1.5-0.1 + 0.1(13.6)
y
59
Principles of Hydrostatics
Air
1
0
0
g
.
0
y
0
0
El _b._ = 2.76 m of water
y
y
Solution
Sum-up pressure head from
A to B in meters of water;
Pe = 70 kPa
f./}_ - 1 (0.86) = E.!.
B
y
y
..E_!l_ - 0.86 = ~
9.81
9.81
p, = 78.44 kPa
Weight= FA
=PAx Area
= 78.44 X f {1)2
Piston
(I) Gage liquid= carbon tetrachloride
Weight
reading, It = ?
Sum-up pressure head from 1 to 4 in meters of water;
P
_1 + X + /t - 1!(1.59) - X - 1.5
Piston
y
= -P4
y
El - b._ = 1.5 + 0.591!
Oil
y
s = 0.86
FA = PA x Area
y
where El - E..!. = 2.76 m ~ from (a)
y
y
7..76 = 1.5 + 0.59h
11 2.136m
Weight= 61.61 kN
11 roblem 2 - 36
Problem 2 - 35
Two vessels are connected to a differential manometer using mercury, the
connecting tubing being filled with water. The higher pressure 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 water? (b) If carbon tetrachloride
(; = 1.59) were used instead of mercury, what would be the manometer
re<~ding for the same pressure difference?
Water
Air, p = 175 kPa abs
In Ul figure shown, determine
lh• hl••ght h of water and the
1gl' reading at A when the
I olulL• pressure at B is 290
.
0
~
\1'
B
Water
CHAPTER of
TWO
Principles
Hydrostatics
61
60
:::::::-----~~~~~::::~------------------!&~H~
.Y~D~RA~U~L~IC~S _&_H__v_D_R_A_u_L_Ic_s____________________P_r_in_c_ip_l_cs_o_f_H_y_d_r_o_s_ta_t_ic_s_________
FLUID MECHANICS
Solution
FLUID MECHANICS
CHAPTER TWO
Sum-up pressure (gage) head from 1 to 4 in meters of water;
Sum-up absolute pressure head
from B to 2 in meters of water;
PB - 0.7(13.6) - h = __1_
p
y
El. + x(0.9) + 1.3(0.9)- 1.3(13.6) = E.:!.
Air, P = 175 kPa ab5
0
y
9,
~ + 0.9x- 16.51 = 0
Water
9.81
0
~Sol
- 9.52- II = ..1Z2.
·
9Bl
y
y
x = 13.81 m
h = 2.203 m
Then, x + y = 28.42 m
A
Problem 2 - 38
For the manometer setup shown,
determine the difference in pressure
between A and B.
~ - 9.52 + 0.7 = .E.L
9.81
Solution
'
= y + 1.7
x- y = 1.02 m
-7 Eq. (1)
X + 0.68
A
Air
Sum-up pressure head from A to B
in meters of water;
Alcohol
s =0.90
~ - X - 0.68(0.85) + y = !!.!_
y
Mercury
y
~ - E.!!_ = x- y + 0.578 -7 Eq. (2)
y
y
Solution
Substitute x- y = 1.02 in Eq. (1) to Eq. (2):
Sum-up absolute pressure head from
1 to 2 in meters of water;
El. - y(0.9) = .!!1_
y
y
40 + 101
12
-0.9y=981
.
9.81
y = 14.61 m
680mm
_j'__
1
Problem 2- 37
(
i1
Water
1700 mm
PA = 203.5 kPa abs
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.
Oil, 5 = 0.85
~ - !!.!_ = 1.02 + 0.578
y
Air
y
p A - p B = 1.598
9.81
PA - ps = 15.68 kPa
Alcohol
5 = 0.90
Mercury
CHAPTER TWO
Principles of Hydrostatics
62
CHAPTER TWO
t I UID MECHANICS
ltYDRAULICS
Problem 2 - 39
A difterentinl mcmomt:•ter i~
atlnclwd Inn pipe a'> <;hown
l.akulcltl'
tlw
prt''>'>UI"l'
differenn• bet\\'l'l'll plllnt~ 1\
and B.
Mercury
A
- - ·- - <> ·- - - -
-
B
~- · --
-·---
Pnnc1ples of Hydrostatics
I' I oblem 2 - 40
thl' figure shown, the
I ll•• lion of mercury is initially
11 mm. If the pressure at A is
,
t w;t•d
by 40 kPa, while
unt.tining the pressure at B
" t,u\t, what will be the new
"' ••ury deflection?
E
"!
Oil, s = 0.90
Solution
2
olution
y
I
0.25 m
---~<>----~9 -~§ _______ \
Otl, s = 0.90
Sum-up pressure head from A toBin meters ot water;
~ - y(0.9) - 0.1(13.6)+0.1(0.9)+y(0.9)= .E..!:_
y
y
....
....
~-.E..!:_ = 0.1(13.6)-0.1(0.9)
y
E
Ll1
E
Ll1
y
PA PH = 1.27 m
9.Sl
p 1- p11 = 12.46 kPa
Figure (b)
Figure (a)
In Figure a, sum-up pressure head from A to B in meters of water;
EA - o.6 - 0.25(13.6) + o.2s + 2.1 = E.!L
y
y
~ -· E.!L = 1.65 m of water
y
y
63
64
CHAPTER TWO
CHAPTER TWO
I I UID MECHANICS
1.. HYDRAULICS
Principles of Hydrostatics
In Figure b, p.,' = p 1 + 40
Sum-up pressu re head from A ' toBin meters of water;
65
Pnnc1plcs of Hydrostatics
olution
Kerosene, s = 0.82
p A - (0.6 - x) - (0.25 + 2T)13.6 + (2.35 + x ) = .£.£_
y
y
pA +
40
y
- 0.6 + X- 3.4 - 27.2x + 2.35 + X = 1.!.£_
y
~ + _iQ__ -1.65- 25.2x = 1!.!!_
y
9.81
E.!l. - PH = 25.2 X - 2.423
y
y
y
Rut ~ - .£.£_ = 1.65
·:
!
y
90 mm
.j.
-1
Mercury
1.65 = 25.2 X- 2.423
.\ = 0.162 m = 162 111111
Sum-up pressure head from A to B in meters of water,
~ + 0.2(0.88)- 0.09(13.6)- 0.31(0.82) + 0.25- 0.1(0.0012) =Eli.
New mercury deflection= 250 + 2.r = 250 + 2(162)
New mercury deflection = 574 mm
y
y
~ - E..!. = 1.0523 m of water
y
y
PA - po = 9.81 (1.0523) = 10.32 kPa
Problem 2 - 41
ln the figure shown, determine the difference in pressure bctvvcen points A
and B.
Kerosene, s = 0.82
11 toblem 2 - 42 {CE Board)
.~uming normal barometric pressure, how deep in the ocean is the point
here an air bubble, upon reaching the surface, has six times its volume than
'' had at the bottom?
olution
Applying Boyle's Law
(assuming Isothermal condition)
150 mm
Water
[p, V, = p2 V2]
p1 = 101.3 + 9.81(1.03)/J
p, = 101 .3 + 10.104 /•
V1 = v
p2 = 101.3 + 0 = 101 .3
v2 = 6V
(101.3 + 10.104/z) v = 101 .3 (6 V)
10.104/z = 101 .3(6) - 101.3
II= 50.13 m
CHAPTER TWO
Pnnciples of Hydrostatics
CHAPTER TWO
Principles of Hydrostatics
. 66
Problem 2 - 43
A vertical tube, 3 m long, with one end closed is inserted vertically, with
open end down, into a tank of water to such a depth that an open
connected to the upper end of the tube reads 150 mm of mercury.
vapor pressure and assuming normal conditions, how far is the lower end
the tube below the water surface in the tank?
67
,, , the pressure in air inside the tube 1s uniform.
h 11 I'· PI• = 20.0124 kPa
1'· Yw It
20.0124 = 9.811!; I! = 2.04 m
Ill II,
X= /t + y = 2.04 + 0.495
r = 2.535 m
Solution
lift 1 onsisting of a cylinder 15 em in djameter and 25 em high, has a neck
h I' ~ em diameter and 25 em long. The bottle is inserted vertically in
r ,, 1th the open end down, such that the neck is completely filled with
1 hnd the depth to which the open end is submerged. Assume normal
1111 I ric pressure and neglect vapor pressure.
3m
y
X
L5cm0
h•tlon
llyiny, Boyle's Law
It~,
p2 V,
Applying Boyle's Law:
P1 V1 = P2 v2
Before the tube was inserted;
Absolute pressure of air inside, p1 = 101.3
Volume of air inside, V1 = 3A
When the tube was inserted;
Absolute 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
[p, V, = p2 V2]
101.3 (3 A) = 121.31 [ (3- y) A ]
3- y = 2.505
y = 0.495 m
From the manometer shown;
PI• = y, h,
= (9.81 X 13.6)(0.15)
Ph = 20.0124 kPa.
, lht• bottle was mserted
V11lume of air.
VI = t (1!1)2 (25) + t (5)2 (2~)
V1 = 4,908.74 cm 1
I• .olute pressure tn air·
,,, = 101.325
11 the bottle is inserted ·
olume of air:
v2 =
(15)2 (25)
v2 = 4,417.9 em~
l't t•ssure in air:
1'2 = 101.325 + 9.81 h
t
G
I'
'I
:,
"L
-----JT
l
WatE
Water
x
"L~
1/'1 V, = p2 V2]
101.325(4,908.74) = (101.325 + 9.81 h)(4,417 9)
101.325 + 9.81 It = 112.58
I!= 1.15 em
1 =It+ 25 = 26.15 em
CHAPTER TWO
Principles of Hydrostatics
68
CHAPTER TWO
Principles of Hydrostatics
69
Problem 2 45
M
A bicycle tire is inflated at sea level, where the atmospheric pressure is 101
kPaa and the temperature is 21 °C, to 445 kPa. Assuming the tire does
expand, what is the gage pressure within the tire on the top of a 1uu•uuuu~
where the altitude is 6,000 m, atmospheric pressure is 47.22 kPaa, and
temperature is 5 oc.
tlur report indicates the barometric pressure is 28.54 inches of m ercury.
1 thl ,\tmospheric pressure in pounds per square inch?
A11s: 14.02 psi
Solution
P1 V1
=
P2V2
Determine the pressure heads at B and C in
At sea level:
Absolute pressure of air, p1 = 101.3 + 445
Absolute pressure, p 1 = = 546.3 kPaa
Volume of air, V1 = V
Absolute temperature of air. T1 = 21 + 273 = 294 °K
.
B
Ans: .EJL = -2.38 m
y
I
+--
2.2m
On the top of the mountain:
Absolute pressure of air, p2 = 47.22 + p
Since the tire did not expand, volume of air, V2 = V
Absolute temperature of. air, T2 = 5 + 273 = 278 °K
=
Air
===r
Pc = -0.51 m
y
c
0.6 m
l
?
Oil, 5 = O.ssrvt===~
r P1 V1 = P2 V2 1
T2
546.3(V) = (47.22 + p)V
294
278
47.22 + p = 516.57
p = 469.35 kPa
--
Oil, 5 =0.85
Tl
I •hi Ill 2M 48
the ,mk shown in the figure, compute the pressure at points B, C, 0, and£
I 1 Nl•glcct the unit weight of air.
A11s: p8 = 4.9; pc = po = 4.9; pF = 21.6-l
Air
Air
B
0.4 m
0
0.4 m
Oil
5
0.5 m
=0.90
c
1m
Water
E
70
CHAPTER TWO
Prrnciples of Hydrostatics
CHAPTER TWO
Principles of Hydrostatics
69
Problem 2 - 49
A glass U-tube open to the atmosphere at both ends 1s shown.
contains oil and water. determine the specific gravity of the oil
Ans: 0.8n
r
E
tther report indicates the barometric pressure is 28.54 inches of mercury.
II tlw abnospheric pressure in pounds per sguare inch?
Ans: 14.02 psi
1
E
"'
M
0
luh • 'ihown is filled with oil. Determine the pressure heads at Band C in
I t l>l water.
Water
B
Ans: E.!L = -2.38 m
y
r
2.2 m
+-
Problem 2 - 50
A glass 12 em tall filled with water is inverted . fhe bottom is open. What 1~
the pressure at the closed end? Barometric pressure is 101.325 kPa.
Ans: 100.15 kPad
=
Air
Pc = -0.51 m
=
y
c
0.6m
!
Oil, 5
= 0.85r
/'
Problem 2 - 51
"
In Figure 13, in which fluid will a pressure of 700 kPa first be achieved?
A11s: glycenn
1
Po= 90 kPa
ethyl alcohol
,, = 773.3 kg/m3
60 m
Oil
=0.85
Ill m 2- 48
tht t mk shown in the figure, compute the pressure at points B, C, 0, and [
I 1 Nt•glect the unit weight of air.
Ans: JIH= -1.9; P< = pv = 4.9; pr = 21 .6-l
Air
,, = 899.6 kg/m3
Oil, 5
B
Air
0.4 m
10m
D
0.4 m
Oil
water
1> = 979 kg/m3
5 m
" = 1236 kg/m
5 m
glycerin ,...-'"===-----3
5
0.5 m
=0.90
c
lm
Water
E
70
CHAPTER TWO
CHAPTER TWO
IIIII MECHANICS
Principles of Hydrostatics
Pnnciplcs of Hydrostatics
II VI >RAULICS
71
Problem 2 - 49
hi~ Ill 2- 52
A glass U-tube open to the atmosphere at both ends is shown
contains oil and water, determine the specific gravity of the oil
hmlrical tank contains water at a he ig ht of 55 mm, as shown. Inside is a
II open cylindrical tank containing cleaning Ouid (s.g. = 0.8) at a height II
pt••ssure PR =13.4 kPa gage and p, =13.42 kPa gngc. /\ssume the cleaning
I I'• prevented from moving to the top of the tc1nk. Use unit weight of
1 1 9.79 kN/ m~. (a) Determine the press ure p 1 in kPa, (b) the value of lr in
1 .md (c) the value of yin millimeters.
A11s: (11) 12.88; (b) 10.2; (c) 101
1
E
r---------------------~~ PA
Air
Water
Water
55 mm
0
0
0
0
Problem 2 - 50
Kerosene
A glass 12 em tall filled with water Is inverted. The bottom IS open. What
the pressure at the closed end? Barometric pressure is 101.325 kPa.
A11s: 100.15 k
Mercury
(s.g. 13.6)
Problem 2 - 51
=
In Figure 13, in which fluid will a pressure of 700 kPa first be achieved?
A11s: glycenn
11
ethyl alcohol
,, = 773.3 kg/m3
Po= 90 kPa
60 m
l•llblem 2 -.53
,IJrfcrential manometer shown is measuring the difference in pressure two
,h•r pipes. The indicating liquid is mercury (specific gravity= 13.6), h1 is 675
What is the pressure differential
1 ,, /r.,1 is 225 mm, and 11.,2 is 300 mm.
'ween the two pipes.
A11s: 89.32 kPa
hm1
Oil
,, = 899.6 kg/m 3
10m
water
,, = 979 kg/m3
,, =
5 m
glycenn ,......,.-===--------.
1236 kg/m3
5 m
72
CHAPTER TWO
Principles of Hydrostatics
CHAPTER THREE
Total Hydrostatic Force on Surfaces
73
Problem 2 - 54
A force of 460 N is exerted on lever AB as shown. The end B ts connected
piston which fits into a cylinder having a diame ter of 60 mm. What force
acts on the larger pis ton, if the volume between C and D is filled with water?
A us: 15.83
apter 3
I Hydrostatic Force
r Surfaces
Water
220 mm
I AL HYDROSTATIC FORCE ON PLANE SURFACES
III'.,SUre over a plane area is u niform, as in the case of a horizontal
uhmerged in a liquid or a p lane surface inside a gas chamber, the
h\ dm.tatic force (or total pressu re) is given by:
0
F=pA
Eq. 3 - 1
Problem 2 - 55
I' I• the uniform pressure and A is the area.
An open tube open tube ts attached to a tank as shown. If water rises to
height of 800 mm in the tube, what are the pressures PA and p8 of the air
water? Neglect capillary effects in the tube.
, ,t• of an inclined or vertical plane submerged in a liquid, the total
1111 , 1n be foun d by the following formula:
p.
A
De
~
8
0
E
E
lOO..mm
Water
r-
300 mm
1
~
Water
"='=
l
- 1: Forces on an Inclined plane
74
CHAPTER THREE
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Consider the plane surface shown inclined at an angle 8 with the h
To get the total force F, consider a differential element of area riA. Since th
element is horizontal the pressure is uniform over this area, then;
dr = p dA
where
1111:
'
-
1, taking moment of force about 5, (the intersection of the
lum of the plane area and the liquid surface),
I 'h
p = yh
J1 = y y sine
fydF
where
df = y y sinO dl\
Jrlf=ysin8
fydA
dF = y y sin 8 dA
F = y sin 8 Ajj
y rnOA]iyp= fy(yysin8dA)
FdA
From calculus,
75
Total Hydrostatic Force on Surfaces
Ay
111 0
F=ysin8 Ay
F = y( y sin 0) A
Ay yp = y sin 8
Jy dA
2
hom calculus, fy 2 dA = Is
From the figure, y sin 8 = h
Then,
Ay Yr -Is
(moment of inertia about S)
/
Is
Eq. 3-4
yp=~
AY
F= yh A
lr n.• fcr forrfmla of moment of inertia:
Since Y his the unit pressure at the centroid of the plane area, P<r.• the form
may also be expressed as:
J,
11
' •
Eq. 3 - 2 is convenient to use if the plane is submerged in a single liquid
without gage pressure at the surface of the liquid. However, if the plane
submerged under layers of different liquids or if the gage pressure at
liquid surface is not zero, Eq. 3 - 3 is easier to apply. See Problem 3- 15
18 +A Y2
I g +AY
AY
2
Eq. 3-5
11 t• 'lr
Y + e, from Figure 3 - 1, then
Eccentricity, e =
,.
I
~
AY
Eq. 3-6
1 tble 3- 1 in Page 76 for the properties of common plane sections.
76
CHAPTER THREE
Total Hydrostatic Force on Surfaces
CHAPTER THREE
Total Hydrostatic Force on Surfaces
TABLE 3 - 1: Properties of Common plane sections
Triangle
Quarter ellipse
Half ellipse
Rectangle
IY
IY
Yl
(
blcg
a
I
;
b~-
b.
af
a
X
Area= Y2 m1b
ol+b
:.
bd
Area= 112 b Jr
blr 3
/.Ia12
blr1
/ =,,
36
Circle
.
Area= bd
y. = l1/ ?.
1. = - -
3
db
3
1
, =3-
1=-
3
db 3
I=-
y
bd
12
I~-
t•
I
(.
I
I r
- - :_~ _r_
_. X
,,
D =2r
Area = trrl = If• IT L)2
nr
4
rr0 4
'·· = 1.. = - - = - 4
64
·C>
LfJ
Semicircle
y
1,, = - 4
rrba·1
lgy= - 4
Y' = ~~~
!!
I = 2_11/J'
I= ~bh'
15
y
7
4
Spandrel
Segment of arc
~ ~-----~----------+-----~--------~
Xc
I
:
,_ _ b
Area= '"' b
rr ab 3
,. = ~~~
·(
5
T
r,/v
IIV ,.'
:L?B=kx"T
a
-!r
Arcil = ~bh
3
ly
ib
-X
I = -(0+ 1hsin20)
16
I
3;r
Yci
h
',
r4
- - -·~· - -QCi·-·-·-·
16
ego Xc
I
I
I
,.4
.
I, = - (0 - 112 sm 29)
4
IT 1'4
•
l~v = 0.055bn'
iL
'"'''
J'b
x,= 3-0-
X
I, =ly= - -
1
An•a = 1/2 n r'; y, = -
I = !Tbn3
Parabolic segment
Area = 'h r2 (20) = r2 0
2 rsin O
4r
31T
Ellipse
I = mru3
16
J.~, = 0.055nb'
Xc
Area = 11• IT rl; x, = y, = -
11 , = In= o.o55r
y, = 31T
'
~+'
Quarter circle
yl
J
11 , = 0.11 nb~
4b
,\, = 31T
Sector of a circle
12
gy
4n
1Tbn3
l;ry=-8-
3
x
Area = 1/.1 m1b
4b
y,. = 31T
11'nb3
I,= - 8
77
Yc
-----+1
..!<.: o_ - .., .x
1'·,0
h
j_
I
i<
1
'•,
'---:
Xc '
X
1
Area= - - blr
11+1
1
11+1
r,= --b; y,= - - I I
11+2
411 +2
Length of arc = r(28) = 2r0
rsinO
.\, = - 0 When e = 90° (semicircle)
2r
x.=rr
CHAPTER THREE
Total Hydrostatic Force on Surfaces
CHAPTER THREE
Total Hydrosta tic Force on Surfaces
78
79
TOTAL HYDROSTATIC FORCE ON CURVED SURFACES
Fu = Prr. A
Eq. 3-7
CASE I : FLUID IS ABOVE THE CURVED SURFACE.
or FH = yh A
Eq. 3-8
Fv=y V
Eq. 3-9
tan 8 = Fv/ Fu
Eq. 3-10
.-.
... _______ _
~!
1K
0
Vt'rtical projection of submerged curve (plane area)
pressure at the centroid of A
cg of volume
1he procedure used in solving FH is the same are that presented in Page 73.
I HI: FLUID BELOW AND ABOVE THE CURVED SURFACE
tlj Fv2
r~:
CASE II: FLUID IS BELOW THE CURVED SURFACE
-
~;,H2>-·~·- ·--Fvt
-~~0
-- -·- ---
Volume= V
cg of
volume
cg of volume
0
?
FH
Curved surface
Net vertical force
t ' f'..
Fv
...... __ .,.. ...
Net vertical
projection of a~ea
80
CHAPTER THREE
Total Hydrostatic Force on Surfaces
IIJID MECHANICS
ll 'I'DRAULICS
CHAPTER THREE
Total Hydrost~trc Force on Surfaces
81
DAMS
Dams are structures that block the flow of a river, stream, or other tAT:>+o....., .. ,
Some dar:'s divert the flow of river water into a pipeline, canal, or
Others ratse the level of inland waterways to make them navigable by
and barges. Many dams harness the energy of falling water to generate .. ,.,,,...,..;,....
power. Dams also hold water for drinking and crop irrigation, and ,...,.,....,;,r~ ..
flood control.
PURPOSE OF A DAM
Dams are built for the following purposes:
1. Irrigation and drinking.water
2. Power supply (hydroelectric)
3. Navigation
4. Flood control
5. Multi purposes
1tc.~ure 3 - 3: Boat Passing through Canal Lock. Canal locks are a series of gates designed
to .1llow a boat or ship to pass from one level of water to another. Here, after a boat has
lrtered the lock and all gates are secured, the downstream sluices open and water flows
tlnough them. When the water level is equal on either side of the downstream gate, water
lops flowing through the sluices; the downstream gate opens, and the boat continues on at
1111' new water level.
TYPES OF DAMS
1. Gravitt) dams use only the force of gravity to resist water pressurethat is, they hold back the water by the s heer force of their weight
pushing downward . To do this, gravity dams must consist of a mass
so heavy that the water in a reservoir cannot push the dam
downstream or tip it over. They arc much thicker at the base than the
top- a shape that reflects the distribution of the forces of the water
against the dam. As water becomes deeper, it exerts more horizontal
pressure on the dam. Gravity dams are relatively thin near the surface
of the reservoir, where the water pressure is light. A thick base
enables the dam to 44withstand the more intense water pressure at
the bottom of the reservoir.
Figure 3 - 2: Section of a dam used for hydroelectric
82
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FlUID MECHANICS
& HYDRAULICS
I I UID MECHANICS
I IYPRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
83
\ buttress daur consists of a wall, or face, supported by several
buttresses on the downstream side. The vast majority of buttress
d \ms are made of concrete that is reinforced with steel. Buttresses are
typically spaced across the dam site every 6 to 30 m (20 to 100 ft),
dL•pending upon the size and design of the dam. Buttress dams are
~ometimes called hollow dams because the buttresses do not form a
~olid wall stretching across a river valley.
Figure 3 - 4: Gravrty dam
2 An embankment dam is a gravity dam formed out of loose rock,
earth, or a combination of these materials. The upstream and
downstream slopes of embankment dams are flatter than those of
concrete gravity dams. In essence, they more closely match the
natural slope of a pile of rocks or earth
3 Arclr dams are concrete or masonry structures that curve upstream into
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
rlat,htlt
Figure 3 - 6: Buttress dam
Figure 3 - 7: Multiple arch dam
Figure 3 - 5: Arch dam
84
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
ANALYSIS OF GRAVITY DAM
A dam is subjected to hydrostatic forces due to water which is raised on its
upstream side. These forces cause the dam to slide horizontally
foundatio n and ove rturn it about its downstream edge or toe.
tendencies are resisted by friction on the base of the dam and
forces which causes a moment opposite to the overturning mome nt.
may also be prevented from sliding by keying its base
Upstream Side
Downstream Side
(Taiiwater)
CHAPTER THREE
Total Hydrostatic Force on Su rfa ces
85
A. Ver tical forces
1. Weight of the dam
W1 = Yr V,; W2 = y, V2. W y, V3
2. Weight of water in the upstream side (if any)
w4 = yV4
3. Weight or per manent str].lctures on the dam
4. Hydrostatic Uplift
Lit =y Vut
U2 = y Vu2
B. I lorizontal Force
1. Total Hydrostatic Force acting at the vertical projection
of the submerged portion of the dam,
~~--~~~X~·~-~
----~-+~
FLUID MECHANICS
& HYDRAULICS
I
Vertical
Projection of --7
the submerged
face of dam h
2.
3.
4
5
F
f= yh A
Wind Pressure
Wave Action
Floating Bodies
Earthquake Load
111. Solve for the Reaction
A. Vertical Reaction, Rv
Rv = r.F,,
Rv = w, + W2 + W1 + w4 - u, - lh
Toe
B. Horizon tal Reaction, R,
R, = r.F,,
Diagram
R, = p
x
Rv
Figure 3 - 8: Typical section of a gravity dam showing the possible forces acting
Steps of Solution
With referen ce to Figure 3 - 8, for purposes of illus tration, an ass umption w as
made in the shape of the uplift pressure diagram .
IV. Moment about the Toe
A Righting Moment, RM (rotntwn towards tire 11p~trea111 sufe)
RM = WI X! + Wz Xz + W3 XJ + w4 X4
B. Overturning Moment, OM (rotatwn towards tlw downstream stde)
OM= P y + U, Z1 + Liz Zz
V Location of Ry (:X)
I. Co nsider 1 unit (1 m) length of dam (p erpendicular to the sketch )
II Deter mine all the forces acting:
_
X
RM-OM
=-
Eq.3-ll
CHAPTER THREE
86
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
where:
Y = uni~ wei?ht of water= 9.81 kN/m~ (or 1000 kgjml)
y, = umt we1ght of concrete
y, = 2.4y
(usually taken as 23.5 kN/m')
I LUID MECHANICS
1. HYDRAULICS
CHAPTER THREE
87
Total Hydrostatic Force on Surfaces
[~:_______________q__=_-_RB_Y_(_1_±_6_;_)_,_w_h_e_re_e_~__
-_14~
B_f6
_________
E_q_.3__
Factors of Safety
Factor of safety against sliding, FS~:
f.lRy
FSs= - - >1
Eq. 3-12
Rr
Note: Use (+) to get the stress at point where Ry is nearest. In the diagram
hown above, use (+) to get qT and (-) to get q11 . A negative stress indicates
t •>mpressive stress and a positive stress indicates tensile stress
·•nce soil cannot carry any tensile stress, the result of Eq. 3 - 14 is invalid if the
.tress is positive. This will happen if e > B/6. Should this happen, Eq. 3- 15
will be used. ·
Factor of safety against overturning, F!>p:
RM
FSo = - - >1
B/2
Eq. 3-13
OM
where:
f.l =coefficient of friction between the base of the dam and the foundation
when e > B/6
q.
Foundation Pressure
x = a/3.,
1 Middle Third •
Fore~ B/6
B/3
From combined axial and bending
stress formula:
B/3
I
I
I
I
1,
B/6 ''- B/6 ,:
a=3x
B/3
Ry = h(a)(q.)(l)
Rv = 112(3 x )q,
I
P
Me
q=-- ± A
X
a
:I
l
Heel
P = Rv
A= B(]J = B
Toe
Eq. 3-15
QH
M=Rve
- 1(8)
a/3
1
Qr
3
l--12
c=B/2.
I
I
q = _!.:!._ ± (R!, e)(B j 2)
B
8 3 /12
lm
I
j:
cg
B/2
f
Jl~
B
• Ry
B/2
I
:j
88
CHAPTER THREE
CHAPTER THREE
Total HydrostatiC Force on Surfaces
IIIJID MECHANICS
IIVDRAULJCS
Total Hydrostatic Force on Surfaces
here:
y =unit weight of the fluid
BUOYANCY .
.
89
.
Vn =volume displaced. Volume of the body below the hqu1d surface
ARCHIMEDES' PRINCIPLE
A principle discovered by the Greek scientist Archimedes that s tates that "a11y
body immersed in a fluid is acted upon by an upward force (buoyant force) equal to Hu•
v/oe problems i11 brwyn11 cy, irlentifiJ the forces nct111g 111rd apply corutitrorzs of sfntir
lllilhl'llllll:
weight of the displaced fluid".
.
L FH = 0
~ Fv
This principle, also known as the law of hydrostatics, applies to both floating
and submerged bodies, 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 all in
equilibrium, since the vertical projection of the body in opposite sides is the
same. The upper face of the body is subject to a vertical downward force
which is equal to the weight of the fluid above it, and the lower face is subject
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.
'
=0
LM=O
l 11 homogeneous solid body of volume V "floating" in a homogeneous fluid at
I
1
s..!..p~.g~r_.o_f_b_o_d...:.y_ V = Ybody V
Vo= sp.gr.ofliquid
Yhqutd
Eq. 3- 17
11 the body of height H has a constant horizontal cross-sectional area such as
. •tical cylinders, blocks, etc.:
)
Cross-sectiOnal area, A
Vo = Volz- Vol.
Vo
t
fvz
t
D= sp.gr.ofbody H= Ybouy H
sp.gr.ofliquid
YhqutJ
BF = fvz - fvt
Eq.3-18
Figure 3- 9: Forces acting on a submerged body
BF=FV2-FVJ
11 the body is of uniform vertical cross-sectional area A, the area submerged A.
= y(Voh)- y(Voh)
BF = y(Voh- Vol,)
BF = 'Y Vo
Eq. 3-16
l
A s=
-s.!.p....:.g~r_._of_b_o_d....:y-A = =
Ybody
A
Eq.3-19
-------------------s~p~.gr~.o_f_H~q~u-id______Y_uq~u-•d--------------~
90
CHAPTER THREE
I llJID MECHANICS
HYDRAULICS
Total Hydrostatic Force on Surfaces
CHAPTER THREE
91
Total Hydrostatic Force on Surfaces
STATICAL STABILITY OF FLOATING BODIES
A floating body ts acted upon by two equal oppostng forces. These are, tht
body's weight W (acting at its center of gravity) and its buoyant force Bl
(acting at the center of buoyancy that is located at the center of gravity of tht•
displaced liquid)
When these forces are collmear as shown m Figure 3 - I0 (a), 1t floats m an
upright position. However, when the body tilts due to wind or wave action.
the center of buoyancy shifts to its new position as shown in Figure 3 - 10 (b)
and the two forces, which are no longer collinear, produces a couple equal to
W(x). The body will not overturn if this couple makes the body rotate towards
its original position as shown in Figure 3 - 10 (b), and will overturn if the
situation is as shown in Figure 3- 10 (c).
rhe point of intersection between the ax1s of the body and the line of action ol
the buoyant force is called the metacenter. The distance from the metacenter
(M) to the center of gravity (G) 0f the body is called the metacentric height
(MG). It can be seen that a body is stable if M is above G as shown in Figure 3
· 10 (b), and unstable if M is below G as shown in Figure 3 - 10 (c) If M
coincides with G. the body is said to be jus/ slnhlt•
Rgure 3 - 10 (c): Unstable position
Figure 3- 10: Forces on a floatmg body
RIGHTING MOMENT AND OVERTURNING MOMENT
..
Wedge of
emerslon
BF
IX
Figure 3- 10 (a): Upright position
Figure 3 - 10 (b): Stable position
RMor OM W(x)
Eq. 3-20
ll EMENTS OF A FLOATING BODY:
W =weight of the body
BF = buoyant force (always equal to W for a floating body)
G =center of gravity of the body
Bo =center of buoyancy in the upright position
(centroid of the displaced liquid)
..
Bo' =center of buoyancy in the tilted position
Vo =volume displaced
.
.
M = metacenter, the point of intersection between the !me of action
of the buoyant force and the axis of the body
.
c =center of gravity of the wedges (irrunersion and emerswn)
5 =horizontal distance between the cg's of the wedges
11 =volume of the wedge of irrunersion
a = angle of tilting
MBo = distance from M to Bo
GBo = distance from G to Bo
MG = metacentric height, distance from M to G
92
CHAPTER THREE
3-21
93
Total Hydrostatic Force on Surfaces
I. HYDRAULICS
=MB.± GBo
Moment due to shifting of BF = moment due to shifting of wedge
BF (z) = F (s)
Use (-J If G 1s above Bu
Use(+) if G is below Bn
Note
CHAPTER THREE
I I UID MECHANICS
Total Hydrostatic Force on Surfaces
BF = y Vo
F=yv
z = MBo sin 8
~ IS always above B.
y Vo MBo sin 8 = y v s
vs
Vo sinO
Eq. 3-22
MBo= - - -
VALUE OF MB 0
fhe stability of the body depends on the amount of the rightmg mom nt
which in turn IS dependent on the metacentric height MG. When the body tilts.
the center of buoyancy shifts to a new position (Bo'). This shifting also causes
the wedge rl' to shift to a new position v. The moment due to the shifting of
the ouovant force BF(z) is must equal to moment clue to wedge shift F(s)
INITIAL VALUE OF MB0
I or small values of 8, (8 "' 0 or 8 = 0) :
I
I
!
I
c·G---~---
c:__:-----J
~
(B/2) tan
9~
Wedge, volume
1 - - --
s - ---+!
BF
B
=v
Figure 3 - 11: Rectangular body
Waterline Sectron
Consider a body in the shape of a rectangular parallelepiped length L as
~hown in Figure 3 - 11;
Volume of wedge, v = Y2(B/2)[(B/2) tan 8]L
Volume of wedge, v = LB2 tan 8
t
For small values of 8, s"'
t8
CHAPTER THREE
94
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
I I UID MECHANICS
95
Total Hydrostatic Force on Surfaces
I. HYDRAULICS
r OR RECTANGULAR SECTION
vs
MBa=--Vo sine
1 LB 2 tanexl.B
MBn = 8
3
V0 sine
l
But for small values of e, sine"' tan e
LB 3
MB = _12_ _
,,
v
F
0
But ~ LB~ is the moment of inertia of the waterline section, I
I
MB.=Vo
Eq. 3-23
(B/2) sec o
s/2-
l
r(B/2.) tan e
~
B/2
(B/2)<~
Note: Th1s formula can be applied to any sect1on.
Centroid of wedge
Since the metacentric height MG is dependent with MBo, the stability of a
floating body therefore depends on the moment of inertia of the waterline
section. It 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.
[I om Eq. 3- 22,
vs
v 0 sine
MB = - --
•
where L Is the length perpendicular to the figure
V0 = BDL
v = th(B/2)[(B/2) tan e]L
2 tan 8
v = lLB
8
Centroid of triangle, x
MOMENT
The righting or overturning moment on a floating body is:
RM or OM= W x = W (MG sin 9)
From geometry, x =
Xl +X2 +X3
3
0 + (B /2)sece + (B I 2)cos8
x=
3
Eq. 3-24
x = Ji ( -1-+cose)= Ji (1+cos2e)
6
2
.:_ =
2
x = !i ( 1 + cos e )
6
cose
2
s = !i ( 1 + cos e )
3
cose
cose
6
cose
96
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
~LB' tan a{~( 1 ·,::~'e)]
M8. =
(BDL)sin e
LB sine l+cos 2 8
---X---24 cose
cose
BDLsine
2
8
l + cos 2 8
M8.= - - - - - 24D cos 2 8
M8. =
M8. =
M8. =
12
~
( -- + 1)
24D cos e
82
240
82
240
Total Hydrostatic Force on Surfaces
• onsider a pipe of diameter D and
lhickness t be subjected to a net pressure
I' J'o determine the tangential stress in
I Ill' pipe wall, let us cut a section of length
1long the diameter. The forces acting on
I Ius section are the total pressure F due to
1hl' internal pressure and this is to be
', •sis ted by T which is the total stress of
1he pip~ wall.
3
M8,=
CHAPTER THREE
I LUID MECHANICS
1. HYDRAULICS
97
Projection of
curved area
Applying equilibrium condition;
[LFH = 0)
(sec2e + 1)
F= 2T
F=pA = pDs
but sec2 8 = 1 + tan2 8
·r= SrAw,,u
T = Sr (s x t)
[ (1 + tan 2e) + 1)
pDs = 2 x [Sr(s x t)]
2 ( 3_ + tan 2 e )
M8. = - 82
- - (2 + tan2 e) = ...!__
12(2)D
12D 2
2
MBo =~
12D
(1 + tan2 e)
2
pD
Tangential stress, Sr = 2t
Eq. 3-25
l'o determine tfle longitudinal stress, let us
cut the cylinder across its length as shown.
[LFH = 0]
F=T
F=pA
STRESS ON THIN-WALLED PRESSURE VESSELS
THIN-WALLED CYLINDRICAL TANK
A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile
forces, which resist bursting, developed across longitudinal and transverse
sections
Eq. 3- :2tJ
F= p
F
t oz
T= SL Awall
Awan = nDt
T= SL nDt
Longitudinal stress, SL = pD
4t
p = internal pressure - external pressure
Eq. 3-27
Eq. 3-28
98
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
SPHERICAL SHELl
Iffa spherical tank of diameter D and thickness I con tams gas under a pressure
o p. the stress at> the wall can be expressed as·
I I UID MECHANfCS
I. HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
99
I~olved Problems
\T
Problem 3 - 1
vertical rectangular plane of height d and base b is submerged in a liquid
Jth its top edge at the liquid surface. Determine the total force F acting on
1•m• side and its location from the liquid surface .
-
'>olution
0
D
.. l
Wall stress, S = pD
4t
Eq. 3- 29
F=yhA
h =d/2
A= bd
F = y(d/2)(bd)
F=%ybd2
SPACING OF HOOPS OF A WOOD STAVE PIPE
I
e= _g_
Ay
lj = h = d/2
e=
...1.. brl 3
12
(bd)(d 1 2)
e = d/6
YP = h + e
yp=df2+d/6
YP = 2d/3
Spacing, S = 251 A,,
pD
where.
S, = allowable tensile stress of the hoop
A,, ~ cross-sectional area of the hoop
p = mternal pressure in the pipe
D =diameter of the pipe
Eq. 3-30
Pressure diagram
(triangular prism)
Using the pressure diagram:
F =Volume of pressure diagram
F = 1/l(yd)(d)(b) = 1/2 y b d2
The location ofF is at the centroid of the pressure diagram.
Note: For rectangular surface (inclined or vertical) submerged in a fluid with top edge
flushed on the liquid surface, the center of pressure from the bottom is 1/3 of its
height.
CHAPTER THREE
Total Hydrostatic Force on Surfaces
100
II UID MECHANICS
I HYDRAULICS
Problem 3-2
CHAPTER THREE
Total Hydrostatic Force on Surfaces
101
olution
A vertical triangular surface of height d and horizontal base width b 1s
submerged in a liquid with its vertex at the liquid surface. Determine the tal
force F acting on one side and its location from the liquid surface.
F=yhA
F = y(r)(rr r2)
F=nyr3
Solution
I
e= _g_
F = yh A
Alj
h = l.3 d
lnr4
A = 1hbd
e=
F = y X l. d X 1!2bd
~
F=
t ybtF
I~:
,,=Ay
= r/4
Pressure diagram
(cylindrical wedge)
Using the pressure diagram for this case is. quiet c.omplicated. With the
lt.lpe shown, its volume can be computed by mt~grat10n.. Hen~e, pressure
d 1,1gram is easy to use only if the area is rectangular, w1th one s1de honzonta l.
ii = h = 2d/3
e=
4
(rrr 2 )(r)
y1, = r + e
y1, = r+ r/4
y1, = Sr/4
...Lbd 3
36
(t brl)(2d 13)
I'= d/12
Problem 3-4
Pressure diagram
(pyramid)
IIJ,= h +e
IJ,.= fd+d/12=3d/4
Using the pressure diagram.
F = Volume of pressure diagram
F = t A1..-. x height
f =
t (b yd)(d) = t ybcf2
X
F IS located at the centroid of the diagram, which 1s 1/4 of the altitude
from the base
·\ vertical rectangular gate 1.5 m wide and 3 m high is submerged in w~tcr
with its top edga;2 m below the water surface. Find the total pressure achng
on one side of the gate and its location from the bottom.
Solution
F=yil A
li = 1.5 + 2 = 3.5 m
F = 9.81(3.5)[(1.5)(3)]
F= 154.51 kN
"
2m
____: ~
3m
1
e= _g_
Alj
Problem 3- 3
A vertical circular gate or radius r is submerged in a liquid with its top edged
flushed on the liquid surface. Determine the magnitude and location of the
total force acting on one side of the gate
e=
ii = 3.5 m
cg e
cpe
e
i
y
1 .S) (3 ) 3 = 0.214 m
-lf(
(1.5 X 3)(3.5)
y=1.5-e
y = 1.5 - 0.214
y =1.286 m
___:t:.._
.j,
CHAPTER THREE
102
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Using the pressure diagram:
F =Volume of pressure diagram
5
F = ( Y;
X
(1.5)
2Y 3)
........... . .
F=y h A
, =2 +
::
,':
''
''
'
3m
L
I
I
6 i •
t (3}
2m
h =3m= y
'
1.5 m
h=3m
F = [9.81(0.82)](3)(1/2(1.5}(3) ]
F= 54.3kN
2y
Iii
103
Total Hydrostatic Force on Surfaces
~
~-
''
F = 15.75y
F = 15.75(9.81)
F = 154.51 kN
CHAPTER THREE
I IIJID MECHANICS
I IYDRAULICS
1g
e=-=
Ay
ll'rl
0
it; (1.5}(3) 3
[t (1.5)(3)](3}
3m
Oil
s = 0.82
e=0.167m
YP =It + e
y1, = 3.167 m from the oil surface
0
0
lm
3y
2y
Location of F:
A1 = 2y(3) = 6y
Az = '!2(3y)(3) = 4.5y
A =A, +A2 =10.5y
*
Pressure diagram
(trapezoidal prism)
[Ay = !:ay]
10.5y y = 6y(1.5) + 4.5y(1)
y = 1.286 m
(much complicated to get than using the formula)
~roblem 3-5
A vertical triangular gate with top base horizontal and 1.5 wide is 3 m high. It
is submerged in oil having sp. gr. uf 0.82 with its top base submerged to a
depth of 2 m. Determine the magnitude and location of the total hydrostatic
pressure acting on one side of the gate.
Problem 3- 6 (CE Board May 1994)
vertical rectangular plate is submerged half in oil (sp. gr. = 0.~) and ha.lf in
,\ter such that its top edge is flushed with the oil surface. What ts the ratio of
tile force exerted by water acting on the lower half to that by oil acting on the
upper half?
C)olution
Force on upper half:
Fo =Yo II A
Fo = (Yw x 0.8)(d/4}[b(d/2)]
Fo=0.1y,..bd2
.
Force on lower half:
fiV= Pcg2 X A
Pcr,2 =Yo ho + Yw It"'
Pcy,2 = (Yw X 0.8}(d/2} + Yw(d/ 4}
pcg2 = 0.65 Yw d
Fw = (0.65 y,.. d)[b(d/2)]
Fw = 0.325 Y11• b d2
.
Fw
Ratio=Fa
Ratio=
0.325y wbd
0.1ywbd 2
2
=
.
3 25
104
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
\
Problem 3-7 (CE Board May 1994)
0.81o
CHAPTER THREE
~ l UID MECHANICS
Total Hydrostatic Force on Surfaces
I-. HYDRAULICS
105
Problem 3 - 8 (CE Board May 1992)
A vertical circular gate in a tunnel 8 min diameter has oil (sp. gr.
one
side and air on the other side. If oil is 12 m above the invert and e air
pressure is 40 kPa, w~ere will a single support be located (above the in ert of
the tunnel) to hold the gate in position?
dosed cylindrical tank 2 m in diameter ,md 8 m deep with axis vertical
•mtains 6 m deep of oil (sp. gr. = 0.8) The air above the liquid surface has a
l'll'Ssure of 0.8 kg/cm2. Determine the total normal force in kg acting on the
\ .111 at its location from the bottom of the tank
Solution
'iolution
n.
-l+--2m0=4
Oil; s = 0.8
am
12m
Air; p = 40 kPa
L----1'4-- ~
Fa~r
t z.yt
z
!
Fool= Yoil h A
Foil = (9.81 X 0.80){8) X f (8)2
0
T
1
-·-·-·- -·-·-·-·· 8m0
4-y
r-
y
c.h1nge
4m
,rlnve
Foil= ~,156 kN
I
e= _g_
Ay
~=
.1!..(8)4
64
t (8)2(8)
-::::::
,....
lr ~·· ,.~::Bm
[
011;•=0.8
___.,.
.......
F1 = p.,,A
paor = 0.8 kg/ cm2 = 8,000 kg/ m2
F1 = 8,000(2rr ~ 2) = 32,000rr kg
y; = 6 + 1 = 7 m
F2 =Pes A
= o.5 m
z = 4- e = 3.5 m
Fair =Pair A, = 40 X f (8)2
F.1, = 2,011 kN
Th~ support must be located at point 0 where the moment due to F...
Pes= (1000 x 0.8)(3) + 8,000
peg= 10,400 kg/m 2
F2 = 1 0,400(2rr x 6) = 124,800rr kg
Solve fore:
F2 = Yoh A
124,800rr = (1000 x 0.8) h (2rr x 6)
h=lj=13m
and Foil is zero. Since Foil > F.,,, 0 must be below Fon.
(:EMo =0]
Fou(z- y) = F.;,(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 =262m
e= -
Ig
Alj
=
n(2rr)(6)
3
(2rrx6)(13)
e = 0.23077 m
y2 = 3- e = 2.77 m
F = F 1 + F2 = 156,8001t kg
-7 J'otal normal force
106
CHAPTER THREE
Total Hydrostatic Force on Surfaces
~=A~+~~
(156,800n) y = (32,000n)(7) + (124,800n)(2.77)
y = 3.63 m
FLUID MECHANICS
HYDRAULICS
f-LUID MECHANICS
&. HYDRAULICS
/
'">olution
CHAPTER THREE
[2: Mhmge = 0)
F z = 40(1)
-7 Location ofF from the bottom
F = y h A= 9.81 h (1)(1.5)
Using the pressure diagram:
F = 14.71511
n(2) = 2n m
8000
lg
2m
t'
=-
Ay
fi (1.5)(1) 3 =1t>=
6m
rr j: I
lJ±:
ii
where y = h
(1.5 x 1)h
107
Total Hydrostatic Force on Surfaces
1211
L•LSm
~ ~ r- hinge
h
1
z = 0.5 +I' = 0.5 + --=
8
lm
40 kN
12/r
800(6)
=4800
14.715 1i (o.5 + ~) = 40
12h
Pressure Diagram
0.5 h + 0.08333 = 2.718
h = 5.27 m = h + 0.5 = 5.77 m -7 critical water depth
Pt = 8000(8)(2n) = 128,000n kg
P2 =1/2(4,800)(6)(2n) =28,800n kg
P = Pt + Pz = 156,8007t kg
-7 Total normal force
Problem 3 - 10
[P y = pI Yt + p2 Y2]
(156,800n) y = (128,000n)(4) + (28,800n)(2)
y = 3.63 m -7 Location of P from the bottom
A vertical circular gate is submerged in a liquid so that its top edge is flushed
with the liquid surface. Find the ratio of the total force acting on the lower
half to that acting on the upper half.
Solution
Problem 3-9
In the figure shown, stop B will
break if the force on it reaches
40 kN. Find the critical water
depth. The length of the gate
perpendicular to the sketch is
1.5m
w.s.
Ratio= F2
F,
L = l.Sm
r- hinge
.
y lt2 A2
Ra tJ o = ~"---=y lt1 A1
At =A2
h
Ji2
Ratio= =-
h,
lm
1-1
B
'\__stop
Ratio =
1.424r
= 2.475
0.5756r
o.Js:l
108
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANI CS
& HYDRAULICS
Problem 3 - 11
CHAPTER THREE
Total Hydrostatic Force on Surfaces
.,olution
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.
F=y /i A
h = 3.5 + 2/3
9m
j
Solution
F=yh A
F = 9.81(4.5)[(30)(10.392)]
F= 13,763 kN
L =30m
h =4.167m
A = 1/2(1)(2.61)
A= 1.305 m 2
F = (9810 X 0.83)(4.167)(1.305)
F= 44,277 N
F= 44.277 kN
h
2m
j
Oil (s = 0.83)
Problem 3 - 13
An
I
e= _g_
inclined,
circular
)~ate with water on one
Ay
e=
I LUID MECHANICS
At HYDRAULICS
3
-f2 (30)(10.392)
(30 x 10.392)(4.5 j sin 60°)
e = 1.732 m
o;ide is shown in the
figure. Determine the
total resultant force
.tcting on the gate.
0
y = 112(10.392) - 1.732
0
0
y=3.464m
0
or
y = ! (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?
t
2m
Solution
F=y h A
1i = 2 + 0.5 sin 60°
h = 2.433
F = 9.81(2.433) t (1) 2
F= 18.746 kN
109
CHAPTER THREE
110
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Problem 3 - 14
(b) e = .!L_ =
The gate in the figure shown is 1.5 m wide, hinged at point A, and rests
against a smooth wall at B. Compute (a) 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
I LUID MECHANICS
1v HYDRAULICS
Alj
Total Hydrostatic Force on Surfaces
-f2(1.5)(3.6)3
(1.5x3.6)(7.21)
e = 0.15 m
x=1 .8-0.15
x = 1.65 m
{LMA= 0]
F(x) - Rs(2) = 0
218.25(1.65) = 2 Rs
Ra = 180 kN
Seawater
s = 1.03
{c) (LfH = 0)
RAI• + F sin 8 - Rs = 0
RAI• = 180 - 218.25 sin 33.69°
RAil= 58.94 kN
0
0
*0
0
Sm
[L F,, = 0]
2m
R....v - P COS 9 = 0
1
RAv = 218.25 COS 33.69°
RAil= 181.6 kN
RA = ~RA/ + RAH
2
2
= ~(181.6) + (58.94)
2
RA = 190.9 kN
Solution
w.s.
c[l
I
f"j
= 32 + 22
d=3.6 m
tan 9 = 2/3
9 = 33.69°
h=4m
Sm
•Y.:~
o ~
y = -lz -
-
r----+--2m
sine
-If=
·
4
sin 33.69°
11 = 7.21 m
_.!.__] -
f-A
RAh
~v
(•'
F = y !1 A
r = (9.81 x 1.03)(4)[(1.5)(3.6)]
f= 218.25 kN
3m--~
Problem 3 - lS
Determine the magnitude
md location of the total
hydrostatic force acting on
the 2 m x 4 m gate shown
111 the figure.
-;y;--
:::::::: !#fi:~ +~H%::: :::::
lm
Oil, s = 0.80
1.5 m
Water
t
Glycerin, s = 1.26
t-
3m
F
111
112
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
\
Solution
r- :::::::::~i~<p :~:~~:~~~:::<::
I LUID MECHANICS
I. HYDRAULICS
CHAPTER THREE
Vroblem 3- 16 {CE November 1997)
I >••termine the magnitude of
.... ... . . . .... . . . .
•
1m
Oil, s = 0.80
t-·t---------J
r
1.5 m
Water
Glycerin, s = 1.26
3m
F
113
Total Hydrostatic Force on Surfaces
\h!' force on the inclined gate
1 ~ m by 0.5 m shown in the
I tgure 001.
The tank of
1\ otter is completely closed
•nd the pressure gage at the
I Hlttom of the tank reads
'10,000 N/m2• Use 9,800
I Jfcu. m. for water.
•
•
••
•
•
•
•
•
•
•
•••
•
•
•
0
0
•••••••
••
•
•
•••
•
3m
Figure 001
F =Pee A
Peg= L.yh + p
Peg= (9.81x1.26)(3) + (9.81)(1.5) + (9.81x0.80)(1) + 32
Pes = 91.645 kPa
F = 91.645(2 X 4)
F = 733.16 kN
Solving for e:
Solve for ii and y :
F=yhA
~3.16 = (9.81x1.26) h (2 x 4)
It = 7.414m
Solution
F =Pes A
Pz- Pez = Y~
Water
3m
90000- peg= 9,800(2.65)
pez = 64030 Pa
F = 64030 (0.5 X 1.5)
F = 48,022.5 N
lj = ii /sin 60° = 7.414/ sin 60°
1j = 8.561 m
e = .!.!_ = "fi(2)(4)3
Ay (2 X 4)(8.561)
e = 0.156 m
z = 2 - e = 1.844 m
Therefore, F is located 1.844 from the bottom of the gate.
P2 = 90,000 Pa
Problem 3- 17
The gate shown in the figure is hinged at A and rests on a smooth floor at B.
l'he gate is 3m square and oil of having sp. gr. of 0.82 stands to a height of 1.5
m above the hinge A. The air above the oil surface is under a pressure of 7 kPa
above atmosphere. If the gate weighs 5 kN, determine the vertical force F
required to open it.
114
CHAPTER THREE
Total Hydrostatic Force on Surfaces
... . . . . . . . ... ...
:.:.:.:.::::}~G:P:~:7: ~~~.::
I LUID MECHANICS
1. HYDRAULICS
. . ' ..
:·:--:::::
T
Oil
5
= 0.82
Hinge
CHAPTER THREE
Total Hydrostatic Force on Surfaces
t•roblem 3 - 18 (CE Board)
Iron pins 20 m.m in diameter are used for supporting flashboards at the crest
nf masonry dams. Tests show that the yield point of iron to be 310 MPa
( xtreme fiber stress). Neglecting the dynamic effect of water on flashboards
111d assuming static conditions, what is the proper spacing, S, of the iron pins,
1 that the flash boards 600 mm high will yield when water flows 150 mm deep
rver the top of the flashboards.
'iolution
F
Floor
B
Solution
P=p,r.A
Peg = P•" + IP•.ft.,
P•K = 7 + 9.81(0.82)(2.56)
p,g = 27.59 kPa
:::::::::.::.:
:~ir;:p ·~ ::i:k~~:::::
...... . ...
p = 27.59 [(3)(3)]
P = 248.34 kN
:.: :.: -.: ·: ·:
Oil
5 = 0.82
p
P = yh A
~8.34 = (9.81 x0.82) Ti (3 x 3)
It
5m 45°
y =4.85 m
l'
F
= 3.43 m
y=-.-'-'-=~
sin 45°
= !.£. = }i (3)(3)3
Ay (3 x3)(4.85)
£'=0.155m
x=1.5+e
x = 1.655 m
(1:M~ = 0]
P(x) + W(1.06)- F(2.12) = 0
2.1 2F = 248.34(1.655) + 5(1.06)
F= 196.37kN
Floor
3 sin 45" = 2.12 m
115
Moment capacity of one iron pin (20 mm 0):
(Fb = Mcjl]
310 =
M(.f!)
if(20)4
M = 243,473.43 N-mm
M = 0.24347 kN-m
Moment caused by F (considering S m width of flashboard) :
Mr= Fx y
F=yhA
whereA=0.6S
F = 9.81(0.45)[0.6 S]
CHAPTER THREE
116
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
\
r= 2.M9S
y = 0.3- ('
I~
i~ (.S)(0.6) l
e= - ·- .=
Ay
(0.6 S)(O...l5)
CHAPTER THREE
I LUID MECHANICS
I. HYDRAULICS
117
Total Hydrostatic Force on Surfaces
Problem 3 - 20
\t 20 °C, gage A in the figure reads 290 kPa absolute. The tank is 2 m wide
Jll'rpendicular to the figure. Assume atmospheric pressure to be 1 bar. Sp. gr.
nf mercury= 13.6. Determine the total pressure acting on side CD.
e = 0.067 m
y = 0.3 - 0.067 = 0.233 m
,....--------, c
r
Air: 175 kPa abs
1m
M1 = Fxy=M
i-
2.649 5 X 0.233 = 0.24347
5 = 0.394 m = 394 mm
Water
h
j _ __ -Problem 3 - 19
Solution
li = y = 10 -1.698
li = y = 8.302 ft
P=yhA
P = 62.4(8.302)[1 m(4)2J
Gage A '
f
4ft
B
j_~~w.
lg = 0.1 098(4)~
Ig = 28.11 fp
28.11
e= ----:~-tn(4)2 (8.302)
e = 0.1347 ft
b = 1.698-0.1347 = 1.5633 ft
[LMB=OJ
P(b) = F(4)
13,019.89(1.5633) = F(4)
F =5088.5 I bs
~-----~o:~.---::\ -' Gage B
D
':>olving for fl:
PA = L Y II + Ptop
290 = (9.81 X 13.6)(0.70) + (9.81)/t + 175
It= 2.2m
fetal force on side CD: (Note: 1 bar = 100 kPa)
WS
~
AY
Ig = 0.1098 r
Mercury
Solution
P = 13,019.89lbs
I
e=
70cm
ws
The semi-circular gate shown
in Figure 28 is hinged at B.
Determine the force F required
to hold the gate in position.
F
p1 = 175-100
p1 = 75 kPa
p2 = 9.81(2.9)
P2 = 28.449 kPa
f1 = p,(3.9)(2)
F1 = 75(3.9)(2)
r, = 585 kN
t
1m
t- 2.2
Water
-
E
0\
EM
j_
0.7m
F2 = 1/2 P2(2.9)(2)
F2 = 1/2(28.449)(2.9)(2)
F2 = 82.5kN
F= F1 + F2
F= 667.5 kN
c
Air: 175 kPa abs
Fz
Mercury
D
Pz
PI
n
2m
118
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Problem 3 - 21
fLUID MECHANICS
1. HYDRAULICS
CHAPTER THREE
119
Total Hydrostatic Force on Surfaces
..olution
The funnel shown in the figure is
full of water. The volume of the
upper part is 90 liters and the
lower part is 74 liters. What is the
force tending to push the plug
out?
--l.b"'::c:=----l:::: ---~
f=y/r A
r = 9.81(2.6)(1.6 x 1.2)
r= 48.97kN
,,=
I
1
~F
~
1.8 m
AY
2.6m
1.2(1.6)'
I' = ---~...!1-=..2
____ _
·.·.
••••
::::
(1.6 X 1.2)(2.6)
£'= 0.082m
z = 0.8- e
z = 0.718 m
Solution
Since the plug area in contact with water is horizontal, the pressure all over
it is uniform. The shape of the container does not affect the pressure on the
plug.
Force= p x A
Force= 9,810(3)( ~)
100
Force = 1353.78 N
Problem 3 - 22
In the figure shown, the gate AB
rotates about an axis through B.
The gate width is 1.2 meters. A
torque T is applied to the shaft
through B. Determine the torque T
to keep the gate closed.
e
T=Fxz
T = 48.97 X 0.718
T = 35.16 kN-m
TF
A
1.6 m
j
-.
0.8 m
L T !('"" 1"\. 1
B
l•roblem 3- 23 {CE Board)
'\ cubical box, 1.5 m on each edge, has its base horizontal and is half-filled
with water. The remainder of the box is filled with air under a gage pressure
of 82 kPa. One of he vertical sides is hinged at the top and is free to swing
mward. To wl:iat-d.epth can the top of this box be submerged in' an open body
of fresh water without allowing any water to enter?
Solution
l7
: '::1
I
w.s. '<7
1
water
I
h
I
Water
I
1.8 m
•
•
•
0
•
•
•
•
•••••••
•
0
••••
•
•
•
•
0
••
•
0
••••
.. .. .. .. .. .. .. .. ..
..... . .. .. ... ......... ... .
..........
.........
1.6 m
n~
... ......
•
•
1.5 m
1.25 Fl
Water
0.2
1.5 m x 1.5 m
X
1
= 7.36 kPa
lJ')
....;
-'£·
F2!
:_'-;
9.81(0.75)
B
h='Y
FJ
82 kPa
X
Lf'!
....
120
CHAPTER THREE
Total Hydrostatic Force on Surfaces
[:E Mhinge = 0]
F3 (x)- F1 (0.75)- F2 (1.25) = 0
F, = p.,.A
F, = 82[(1.5)(1.5)] = 184.5 kN
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Problem 3 - 24
-7 Eq. (1)
w.s
.
• • ·SOZ" ••••
Find the magnitude and location of
the force exerted by water on one
side of the vertical annular disk
shown.
F2 = 1/2(7.36)(0.75)(1.5)
F2 = 4.14 kN
F3 = yh A
F3 = 9.81 h [(1.5)(1.5)]
F3 = 22.07h
Solution
x = 0.75 + e
1.5(1.5)
3
I
e = _g_ =
12
Alj
[(1.5)(1.5)]1!
0.1875
e=--lt
X= 0.75 +
0.1~75
h=7.67m
I
.!!.(1.5) 4 -.!!.(1) 4
4
4
Alj n[(1.5) 2 - (1) 2 ](4)
e = 0.203m
0 1 75
y1, = 4 + 0.203 = 4.203 m below the w.s.
· ~ ) -184.5(0.75)- 4.14(1.25) = 0
11
16.55 h + 4.138-138.375-5.175 = 0
16.55 h = 139.412
h = 8.42m
h = h -0.75
Location of F:
h
In Equation (1):
22.07 h (0.75 +
F=yh A
F = 9.81(4)[n(1.5)2 -n(1)2]
F= 154.1 kN
e=-g-=
Problem 3 - 25
The gate in the figure shown
weighs 5 kN for each meter
normal to the paper. Its center of
gravity is 0.5 m from the left face
and 0.6 m above the lower face.
Find 11 for the gate just to come
up to the vertical position.
121
Water
h
122
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Solution
Total Hydrostatic Force on Surfaces
Solution
\
Considering 1 m length
f1 = 112 (9.8111)(/1)(1)
r1 = 4.905112 kN
r2 = 9.81h(1.5) {1)
f2 = 14.715/1 kN
CHAPTER THREE
• LUID MECHANICS
I. HYDRAULICS
dF = p dA
0.6!
dA = 2xdy
~ W= 5kN
h
By squared property of parabola:
•cg
[LMo = 0]
x2
22
x2=
1-Y
-y
3
11/3
Fi(ll/3) + W (0.6)- F2(1.5/2) = 0
4.905/12 (11/3) + 5(0.6) -14.715/z (0.75) = 0
1.635/!J- 11.0411 + 3 = 0
t y
p=yy
X=
Solve lr by trial and error
II= 0.2748 m
2m
2~y /3
4m
dF = yy [2 (2~y /3) dy]
dF = 2.31yy3/2 dy
F
In Problem 3- 25, find It when the force againsl lhe "stop" is a maximum.
0
F 2.3l y [
[LMo = 0]
!!!= 3.27 ,,2- 7.358 = 0
dll
112 = 2.25
lr = 1.5 m
0
~
Solution
fl(/1/3) + W(0.6) + P(1 .5)- P2(1.5/2) = 0
4.9051!2 (h/3) + 5(0.6) + P(l.5)
- 14.71511 (0.75) = 0
p = 1.09111 - 7.35811 + 3
3
fdF = 2.31y fy 3 12 dy
Problem 3 - 26
0.6!
~ W = 5 kN
h
•cg
h/3
0
P
I
iy% ~ 2.31(9.81) %[3'''- O'l' I
F= 141.3 kN
Location:
1.5
3
3 2
141.3y,.= Jy(2.31yy 1 dy)
Problem 3 - 27
Determine the force due to water
acting on one side and its location on
the parabolic gate shown using
integration.
0
3
y,, = 0.1604 fy 5 12 dy
0
y,. = 0.1604 [<2/7)l' 2
J:
YP = 0.1604 (2/7) [ 37/ 2 - 07/ 2 ]
yp = 2.14 m below the w.s.
123
124
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Problem 3 - 28
CHAPTER THREE
II UID MECHANICS
Total Hydrostatic Force on Surfaces
I. HYDRAULICS
125
r•roblell) 3 - 29 (CE Board)
In the figure shown, find the
width 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 dam and the
foundation is 0.4. Use 1.5 as the
factor of safety against sliding. Is
the dam also safe from
overturning ?
' dam is triangular in cross-section with the upstream face vertical. Water is
The dam is 8 m high and 6 m wide at the base and
1 l'rghs 2.4 tons per cubic meter. The coefficient of friction between the base
111d the foundation is 0.8. Determine (a) the maximum and minimwn unit
prPssure on the foundation, and the (b) factors of safety against overturning
md against sliding.
llu~hed with the top.
\olution
Ycone
5p. gr. of cone, Sco'"'-_ -
Yw
2.4x 1000
Sp . gr. of cone, Scom =
= 2.4
1000
Solution
Consider 1 m length of dam
Wr =yr V.W, = y(2.4)[(b)(6)(1)]
= 14.4/ry
r onsider 1 m length of dam
w.
w.s
F=yhA
F = y(2.25)r(4.5)(1)]
F = 10.125y
R, = F = 10.125y
Ry=W..
Ry = 14.4 by
~tRy
4.5 m
F
~
T~
~~~
FSs= - Rx
1.5 = 0.4(14.4by)
10.125y
b = 2.637m
RM
FS.= - -
OM
FSa = W,(b/2)
F(1.5)
FS. = 14.4(2.637)y(2.637 I 2)
= 3.3 > 1 (Safe)
10.125y(1.5)
W=yc V
= (yx2.4) [!(6)(8)(1))
W= 57.6y
where y = unit wt. of water
F=yh A
= y(4)(8 X 1)
F=32y
Rx = P = 32y
Ry = W = 57.6-y
RM = W(4)
= 57.6y(4)
RM= 230.4y
OM= P(8/3)
= 32y(8/3)
OM= 85.33y
Toe
RM -OM
x= ---
B/6 = 1 m
Ry
i' = 230.4y- 85.33y = 2519 m <
57.6y
e = B/2- x
e=3-2.519=0.481 m<B/6
812
CHAPTER THREE
126
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
t LUJD MECHANICS
lot HYDRAULICS
Total Hydrostatic Force on Surfaces
Figure:
R, ( 11/3
6e )
,, = - R
2m
_ 57.6~.81) [I ± 6(0.:81)]
•/
u~mg (+)
Using(-)
</I = - 139.47 kPa
q,, - 4R.RR kPa
~
-? soil pressure at the toe
0
-7 ~oil pressure at the heel
0
0
0
pi<,
!-'> = _ _.1
..
R,
YJ'
ll.8(57.6y)
1-\. = 1.44
RM
I'>. = OM
20m
~~'~~~-·1
~2y
230.4y
R533y
/- ', = 2.7
B
Problem 3- 30 (CE Board May 1992)
dam of trapezotdal c ro~s-~el· tt on wtth one face vertical and
horizontal base ts 22 m high and has a thickness of 4 m at the top. Water
upstTeam stands 2 m below the crest of the dam
rhe ~pecifk gravity of
masonry is 2.4
A Neglecting hydrostc1ttc uplift·
I
Find the base width 8 of the dam so that the resu ltant force wtll cu t
the extreme edge of the midd le third near the toe
2 Compute the factors of safety against sli<i ing and overturntng
Usc~~ = 0.5
R Considering uplift pre~sure to va ry uniformh from full hvdrostatH
pressure at the heel to zero at the toe·
I
Find the base wid th 8 of the dam so that the resultant force w ill act at
the e>..tremity of the middle third ncar the toe
2 Compu te the maximum and minimum compre-;stve ~tresses ach ng
ngatnst the basC' of the <iam
f\ gravity
Solution
A. Neglecting hydrostatic uplift:
I.
Consider 1 m length of dam
II.
Forces
wl = Yc vl = (Y X2.4)[(4)(22)(1)]
w1 = 211.2y
w2 = (Y x 2.4}[ 1/2 (B-4)(20)(1)]
w2 = 24By - 9&y
F = y h A = y (10)[(20)(1)]
F = 200w
Ill
Reaction
Rx = LFx = P
R, = 200y
Ry = LFy = W1 + W2
= 211.2y + 24By- 96y
R,1 = 24By + 115.2y
127
CHAPTER THREE
128
Total Hydrostatic Force o n Surfaces
2m:;=--=--r-=--1m--
(
FLUID MECHANICS
& HYDRAULICS
J
~
Total Hydrostatic Force on Surfaces
2
- .t4.8
± J(44.8)
-!(8)(
B= _
__
.:...__ _ -_
__l-l'N.73)
_ __
2(8)
R = 11.175 m
h = 10m
20m
CHAPTER THREE
I I UID MECHANICS
J.. HYDRAULICS
Factors of Safety:
Factor of safety against sliding:
l_
J.!R
rs,= -'
R,
= (0.5)[24(11.175)y + 115.2y ]
200y
rs, = o.9585
Factor of safety against overturning:
Uplift
FS
RM
_,, = OM
pressure
dtagram
1
= 16(11.175) y + 83.2(11.175)y- lo6.4y
1333.33y
rs.. = 2.01
IV
B
Mo m ent about the toe
I~M = W,(B- 2) + W2 J
f (B - 4)]
= 2'11 .2y(8- 2) + (248y- 96y) [
Considering hydrostatic uplift:
+
Uplift f~rcl, U = 1h (20y)(8)(1) = lOB rl'
(8- 4)]
R,,= W, + W2- U
= 211 .2By - 422.4y + l682y - 128Ry + 256y
= 248y + 115.2y" 108y
I~M = 1682y + 83.2By - 166.4y
R,, = 148y + 115.2y
OM= F(20/3)
= 200y(20/3)
OM= 133333y
V
RM = W,(8- 2) + W2 [
•
RM = 1682y + 83.28y - 166.4y
l,ocat10n ot R
R. I= RM- OM
'>mce the resultant torce w tll pass through the extreme edge of
the mtddle thtrds near the toe, x = 8/3 Then,
(24By + 115.2y)(B /3) = l6B 2y + 83.28y 1on 4y
882y + 38.48y = 168'y + 83 2By 1499 71~
88 7 + 44 8R . 1499 71 = 0
-f (8- 4)]
1333 3::\y
OM = F(20/3) + l/(28/3)
= 200y(20/3) + 10By (28/3)
OM= 6.6782y + 1333.33y
Rv x = RM- OM
(148y + 115.2y)(B/3) = 1682y + 83.28y -166.4y- (6.678 2y + 1333.33y)
4.668 2 + 44.88 -1499.73 = 0
2 -- -4-(
- 44.8 ± ~r-(4-4-.8)-::4-.6-6)-(--1-49_9_.
73-)
8 = -----''-'----'----'----'----'2(4.66)
B =13.766m
129
130
CHAPTER THREE
1w 1tYDRAULICS
Foundation stress
r = B/3
r = 13.766/3 = 4.59 m
,. = 8/2- x = 2.2943 m
,, = -
I -y,.hA
9.81(3}(6 X 1)
I= 176.58 kN
1±
11 =
1 (6) =2 m
WI= y, VI
Ry = 14(13.766)(9.81) + 11 5.2(9.8'1)
Rv = 3020.73 k N
= 23.5(2(8)(1)]
~v,
= 376 kN
0
Problem 3 - 31 (CE Board May 2002)
= 4 - 112(2) = 3
,. 2
= (2/3)(2) = 1.333 m
0
E
0
0
\C)
IV2 = 188 kN
\1
~
0
CX)
= 23.5[V2(2)(8)(1)]
13.766
0
E
w, =y, Vz
,, =- 3,020.73[1 ± 6(2.2943)]
T'he section of a concrete gravity
dam shown in the figure. The
depth of water a t the upstream side
1s 6 m. Neglect hydrostatic uplift
dnd use unit weight of concrete
equal to 23.5 kN/m3 Coefficient of
friction be tween the base of the
dam and the fo undation is 0.6.
Determine the following: (a) factor
of safety against sliding, (b) the
factor of safety against
overturning, and (c) the
overturning moment acting agai ns l
the dam in kN-m
131
lution
:v ( ~)
13.766
</I = - 438.87 kPa
1/ H = 0 kPa
CHAPTER THREE
Total Hydrostatrc Force on Surfaces
f I UID MECHANICS
Total Hydrostatic Force on Surfaces
F
R, = F = 176.58 kN
R, = w, + w2 = 376 + 188
,. 2m .,
Rv = 564 kN
J..IR ,,
rs,=-
R\ .
0.6(56"4)
= 1.916
FS, = 176.58
'~~-=0
0
0
0
RM = w, x, + w2 x2
E
\C)
= 376(3) + 188(1.333)
0
RM = 1378.604 kN-m
0
0
0
OM= fxy
=176.58(2)
L
4m
.I
OM = 353.16 kN-m
~ overturning moment
RM
FS., = OM
FSo =
1378.604
353.16 = 3.904
4m
)I
134
CHAPTER THREE
FLUID MECHJ..\IJCS
Total Hydrostatic Force on Surfaces
& HYDRA~LICS
Consider 1 m length of dam
(c)
Forces
W1=y, V,
W1 = 23.54 [lfz(5.2)(52)(1)] = 3,183 kN
w2 = 23.54 [(7)(52)(1) = 8,569 kN
W, = 23.54 [lfz(26)(52)(1)) = 15,913 kN
W4 = 9.81[Vz(5)(50)(1)] = 1,226.3 kN
U = 1h {490.5)(23.2)(1) = 5,690 kN
F = y/i A= 9.81(25)[50(1)] = 12,263 kN
Ill
Reaction
R, = F = 12263 kN
Rv = WI + w2 + w3 + w4 - u
= 3,183 + 8,569 + 15,913 + 1,226.3- 5,690
Rv = 23,201.3 kN
IV
Moment
RM = W1(34.73) + Wz(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
OM= F(50j3) + U(30.47)
= 12,263(50/3) + 5,690(30.47)
OM = 377,758 kN-m
V
Location of Ry
Ry x = RM- OM
23,201.3 :X =683,900.12- 377,758
\' = 13.2 m
(11)
(11)
fhe resultant force IS 13.2 m from the toe
~~n.,
rs, = - ·
R,
FS, =
0.75(23,201.3)
12,263
Total Hydrostatic Force on Surfaces
135
FS = RM = 683,900.12
"
II
CHAPTER THREE
I I.UID MECHANICS
I. HYDRAULICS
OM
377,758
FS,. = 1.81
(d)
Foundation pressure
e=Bj2- x
(" = 38.2/2- 13.2 = 5.9 m < B/6
q=-
;(1±6;)
q = - 23,201.3 [1 ± 6(5.9) ]
38.2
38.2
Stress at the toe, (use"+");
q, =.-1,170.21 kPa
Stress at the heel, (use"-")
q11 = -44.52 kPa
(e) Unit horizontal shearing stress, 5,
S, = ~ = 12,263 = 321 kPa
A,.~,
38.2(1)
Problem 3 - 34
w.s.
I he submerged curve AB is one
quarter of a circle of radius 2 m
md is located on the lower
orner of a tank as shown. The
lt>ngth of the tank perpendicular
to the sketch is 4 m. Find the
magnitude and location of the
horizontal and vertical
, omponents of the total force
tr ting on AB .
4m
0
0
0
0
2m
0 ·- -
I
I
12m
= 1.42
A
B
-- -
2m
CHAPTER THREE
136
Solution
~~
FH = yll A
FH = 9.81(5)[(4)(2))
FH = 392.4 kN
-
If= l + l'
I
f=
t' =
_K
2m
Dr------1
~~
~
h=y=Sm
Ay
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Total Hydrostatic Force on Surfaces
\nother way of solving x:
w.s.
·,jnce unit pressure is always normal to the
.urface and a normal to the circle passes
through its center, then the total force F
.hall also pass through the center of the
1 1rcle 0, hence the moment about 0 due to
I or due to F11 and Fv is zero.
I
4(2) 3 /12
I ~Mo = 0]
4m
[(4)(2)](5)
I'= 0.067 m
If= 1 + 0.067
11 = 1.067 m
Fv X- FH y = 0
437.13 = 392.4{1.067)
:X= 0.9578 m
x
e
1
Note: This is true to all cylindrical or spherical surfaces.
A
Therefore; F,, is acting 1.067 m below B
F" = WergiltA 8co
F,. = yVABCD
V ABCD = 4(A)
A= A1 + A2
A1 = (4)(2) = 8 m 2
A2 = V. rr(2) 2 = 3.14 m2
A =8+3.14
A= 11 .14 m2
VABW = 4(11.14) = 44.56 m·'
F~ = 9.81(44.56)
F, = 437.13 kN
Problem~ ~ 35
(CE Board)
w.s.
l'he crest gate shown
l'Onsists of a cylindrical
'iurface of which AB is the
base supported by a
'>lructural frame hinged at
0. The length of the gate
1s 10 m.
Con:tpute the
magnitude and location of
the horizontal and vertical
components of the total
pressure on AB.
0
0
;...
<I
Sm
Solution
Location tf Fv
c
Ax =A,x1+A2x,
r, = l m
_ 4r _ 4(2)
\1--- --
3n
t2
11.14
3rr
= 0.849 m
x = 8(1) + 3.14(0.849)
r = 0.957 m
Th~refore; Fv is acting 0.957 to the right of A
ih
E
:8
oO
II
f6
0
!I
= 4.33m
i
F
!
---...J-.L-~
c:
'Vi
....
y
0
L =10m
A
10cos60° = Sm
138
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Total Hydrostatic Force on Surfaces
139
solution
FH =yh A
FH = 9.81(4.33)[10(8.66)]
FH = 3679 kN
!I =
CHAPTER THREE
I LUID MECHANICS
1.. HYDRAULICS
Considering 1 meter length:
FH=yh A
FH = 9.81(3)(6 X 1)
FH = 176.58 KN
t (8.66) = 2.887 m
Therefore; FH is acting 2.887 m above 0
Fv = y Vs
A s = Asector - Atnanglc
7t(6) 2 (60°)
As =
- 1/2(6)2 sin 60°
3600
As= 3.26 m 2
Fv = y V11u<.
VABC = VAOBC- V AOB
VAB( = 5 +210 (8.66) X 10 - 112(10)2 [60°"'ifool X 10
VAB< =125.9m3
Fv = 9.81(3.26 x 1)
Fv = 31.98 KN
Fv = 9.81(125.90)
F= ~PH +P/
2
F,· = 1235 kN
F = ~(176.58) 2 + (31.98)_
F=179.45KN
2
Moment about 0 due to FH and Fv = 0
Fv (x) = fu {y)
1235 X= 3679(2.877)
x = 8.57 m
Problem 3 - 37
Therefore; Fv is acting 8.57 m to from 0
,,lculate the magnitude of the
H'sultant pressure on a 1-ft-wide strip
nf a semicircular taintor gate shown
In Figure-12.
5'
1
Problem 3 - 36 (CE May 1999)
Calculate the magnitude of the
resultant force per meter length due
to water acting on the radial tainter
gate shown in Figure 021 .
Figure-12
Solution
Figure 021
A
FH =Pes A
FH = (62.4 x 2.5)(5 x 1) = 780 lbs
Fv=yVABC
Fv = 62.4 x [ f (5)2(1)] = 1225lbs
F= ~ (FH ) +(Fv )
2
2
F = ~ (780) 2 + (1225) 2 = 1452lbs
B
140
CHAPTER THREE
.. LUID MECHANICS
I>. HYDRAULICS
Total Hydrostatic Force on Surfaces
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Problem 3 - 38
Problem 3 - 39 (CE Board November 1993)
Determine the magnitude of
the horizontal and vertical
components of the total force
per meter length acting on the
three-quarter cylinder gate
In the figure shown, the 1.20 m
,fiameter cylinder, 1.20 m long is
.1rted upon by water on the left and
oil having sp. gr. of 0.80 on the right.
I>ctermine the components 'of the
"'•1Ction at B if the cylinder weighs
2m
shown
1~.62kN.
141
0.6m
1.2m
8
Solution
FHl =ylt A
FHJ = 9.81(1.2}(1.2 X 1.2)
Fill = 16.95 kN
Solution
h=4m
em
=
lm
FH = yh A
FH = 9.81(3)[(1)(2)]
FH = 58.86 kN
p:FH = 0]
FHt - F1-12- Ra11 = 0
RBH = 16.95 - 6.78
RBII = 10.17 kN
1m
i
Zm
2m
~-·-·-!·-·-·-·:i
1m!I
FH2 = y h A= (9.81 X 0.8)(0.6)(1.2 X 1.2)
FH2 = 6.78 kN
FV2 = yV2 = (9.81 X 0.8)[% 1t (0.6)2(1.2)]
FV2 = 5.32 kN
lm
i
Fv1 = yV1
Fv1 = 9.81(Y27t (0.6)2(1.2)]
FVJ = 6.657 kN
Fvt !
I
I
,·-·-·1m
i
l2m
.
I
0
Fv = yVol
Fv = 9.81(4(2}(1) + 0.75(n(2)2(1)]
Fv = 170.94 kN
F\n
=
?~
~
[:EFv = 0]
RBv+Fv1+FV2-W=O
R8 v = 19.62- 6.657- 5.32
RBv= 7.64 kN
0.6m
1.2m
FLUID~CS
CHAPTER THREE
142
Total Hydrostatic Force on Surfaces
& HYDRAULICS
Problem 3 - 40
An inverted conical plug 400 mm diameter and 300 mm long closes a 200 mm
diameter circular hole at the bottom of a tank containing 600 mm of oil having
sp. gr. of 0.82. Determine the total vertical force acting on the plug.
~0.4m~
Solution
I
0.6 m
l
CHAPTER THREE
Total Hydrostatic Force on Surfaces
143
F2 = (9.81 X 0.82)(0.00628)
F2 = 0.0505 kN
F,, = Ft- F2
Fu = 0.114- 0.0505
F" = 0.0635 kN = 63.5 N downward
Problem 3 - 41
'\ 2m diameter horizontal cylinder 2m long plugs a 1m by 2m rectangular
hole at the bottom of a tank. With what force is the cylinder pressed against
I he bottom of the tank due to the 4-m depth of water?
i
Oil
fLUID MECHANICS
& HYDRAULICS
0.15 m
s = 0.82
~
'iolution
E
Ft
F
0.15 m
l
~~--
1--
h2
E
0.4 m --~1
0.2 m
v
=:1
/IJ = 2 X (1 C9S 30°)
h1 = 1.732 m
F, =yV1
F, = (9.81 X 0.82)[7t(0.1)2(0.45)]
Ft = 0.114 kN
F2 = yV2
v2 = VFrustum- Vcyhnder
7t(0.15) r
]
V2 = - -l(0.2) 2 + (0.2)(0.1) + (0.1) 2 - 7t(0.1)2(0.15)
3
v2= 0.00628 m3
F3
c
H
112 = 4- h,
lz2 = 2.268 m
0.45 m
0.15 m
F2
Oil
s = 0.82
Ft = yV,
VI =AI X 2
Area, A1 =Area of rectangle DEFG- A4
7t(l) 2 (60°)
.
0
Area of segment, A 4 =
- 1/2(1)(1) sm 60
360°
Area of segment, A 4 = 0.09059 m 2
Area, A 1 = 1(2.268)- 0.09059
Area, A 1 = 2.1774 m 2
V1 = 2.1774(2) = 4.355 m 3
F, = 9.81(4.355)
Ft = 42.72kN
F2 = F3 = yV2
v2 = A2 x 2
ht
144
~!
&HY~~~:~S
CHAPTER THREE
FLUIDM
Total Hydrostatic Force on Surfaces
1t(1)2 (120°)
-%(1)(1) sin 120°
3600
Area of segment, A2 = 0.614 m2
v2 = 0.614(2) = 1.228 m3
F2 =F3 =9.81(1.228)
F2 = F3 =12.05 kN
Area of segment, A2 =
FLUID MECHANICS
& HYDRAULICS
145
FH = yll A
FH = 9.81(6.12}[(4.24)(1]
FH = 254.56 kN
FV = Y Vshaded
Vshaded
= ~Asemicircle + Atrapezmd} X 1
Vshadcd =
Net force = F1 - F2 - F3
Net force= 42.72-12.05 -12.05
Net force = 18.62 kN
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Vshactcd
ltn{3) 2 +~(4.24)]{1}
= 40.1 m 3
Fv = 9.81(40.1)
Fv = 393.38 kN
Problem 3 - 42
In
the
figure
shown,
determine the horizontal and
vertical components of the
total force acting on the
cylinder perm of its length.
Problem 3 - 43
w.s.
Water
4m
rhe gate shown is a
tluarter circle 2.5 m
wide. Find the force F
JUSt sufficient to prevent
rotation about hinge B.
Neglect the weight of
the gate.
Solution
w.s.
Solution
2m
A
FH = yh A
FH = 9.81(1)(2.5 X 2)
FH = 49.05 kN
lm
Net Vertical
Projection
2m
t
- ·-
- · - · - ·- · - I C
i
i
1
2m
I
FH
...
2/3
~--~~----..J-~--~~------~--,~8
Fv = y VABC
2.5m
Fv = 9.81({2 x 2}- 0.257t{2)2]{2.5)
Fv = 21.05 kN
. . · .· · .
0
Fv
1+-- z -->It~
146
~S
&HYDt~~:~S
CHAPTER THREE
FLUIDME
Total Hydrostatic Force on Surfaces
Solve for z and .\
Since the s urface IS cncu lar, i:Me> = 0 due to f , and 1- ·
Fv (z) = FH(2/3)
21.05(z) = 49.05(2/3)
"= 1.55 m
\ = 2 - z = 2 - I .55
\' = 0.45111
li:Mn "' OJ
Fu (2/3) + fv (x) - f(2) = 0
2f = 49.05(2/3) + 21.05(0.45)
F = 21.09 kN
Problem 3 - 44
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Forces due to oil:
FHo = PcgoA
FHo = (9.81 x 0.80)(7- 1.273) x 1f2rr(3)2
FHo = 635.4 kN
Fvo = y,, Vo
V0 =Volume of imaginary oil above the surface
Vo = Volume of half'cylinder - Volume of 1/4 sphere
V., = %rr(3)2(7) -
%! rr(3)
3
v.
= 70.686 m 3
Fvo = (9.81 x 0.80)(70.686)
Fvo = 554.74 kN
Forces due to water:
F1 O'V = Pcgw A
FHw = [(9.81 x 0.8)(7) + 9.81(1.273)] x %rr(3)2
FHw = 953.19 kN
Open
he cylindrical tank shown has a
hemispherical
end
cap
Compute the horizontal and
vertical components of the total
force due to o il and water acting
on the hemisphere
FLUID MECHANICS
& HYDRAULICS
4m
Fvw = Weight of real and imaginary oil above the surface
+ weight of real water above the surface
Fvw = (9.81 x 0.8)x 1/m(3)2 (7) + 9.81 x 1/4 rr(3)3
·•. 011, s = 0;80
~
t
'
Fvw = 1,054.01
Water
Solution
Total horizontal force, FH = FHo + F11w
Total horizontal force, FH = 635.4 + 953.19
Total horizontal force, FH = 1,588.59 kN 7
•
Open
T
4m
7m
•.. 011, s = 0.80
l
Total vertical force, Fv = F vw- Fvo
Total vertical force, Fv = 1,054.01- 554.74
Total vertical force, Fv = 499.27 kN
Another way to solve for the total vertical force, Fv:
Fv =weight of fluid within the hemisphere
Fv =Yo Vo + Yw Vw
Fv = (9.81x0.8)[ ~x ~7t (3)3)] + 9.81[ ~x ~rr (3)3)]
Fv = 499.27 kN
4( 3)/3n
147
CHAPTER THREE
148
FLUID MECHANICS
&. HYDRAULICS
Total Hydrostatic Force on Surfaces
Problem 3 - 45
CHAPTER THREE
Total Hydrostdtic Force on Surfaces
149
Problem 3 - 46
Pressurized water fills the tank shown m the figure
hydrostatic force acting on the hemispherical surface
~
/
Compute the nel
Determine the force required to open the quarter-cylinder gate shown
•vcight of the gate is 50 kN acting 1.2 m to the right of 0
The
~ Hem1sphencal
; surface
F
E
U"l
Lri
Solution
Ccvwert 100 kPa to tts
e4u1valent pressure
head, h,.q
Solution
It"<I = E
y
It
=
--------------r-;...-------
------------~-
LOO
9.8]
lt,'<1 = 10 1 94 m
''<1
Sm
'iince the gate has circular surface,
lhe total water pressure passes
lhrough point b which is also the
location of the hinge, therefore the
moment due to water pressure
tbout the hinge is zero.
E
U"l
Lri
It = 10.194 -5
Water
It= 5194 m
ll:Mo = 0)
F(2.5) = 50(1.2) + Fr(O)
F= 24 kN
F = Weight of tmaginary water above the hemtsphencal surface
F= y.,, v".
V". = Volume of cylinder+ Volume of hemisphere'
v,..= rr(2)2(5.194)+
4rr(2)'
+x
V,.. = 82.025 m-1
f = 9.81 (82.025)
F = 804.7 kN
Problem 3 - 47
1\ hemispherical dome shown is filled with oil (s = 0.9) and is attached to the
Jtoor by eight diametrically opposed bolts. What force in each bolt is required
lo hold the dome down, if the dome weighs 50 kN?
150
CHAPTER THREE
flUID MECHANICS
E. HYDRAUliCS
FLUID MECHANICS
& HYDRAUliCS
Total Hydrostatic Force on Surfaces
CHAPTER THREE
Total Hydrostatic Force on Surfaces
151
<;olution
Dom•
/
I
,J~~
~~ ~~.9~~
,
FOil =
y
v
Oil •"'". lh• <UI>•
Fo1l = (9.81x0.8)[n(0.805)l(5)
Oil
-! n(0.805)2(3))
'\\
\ Bolt
0..,
I
= 0]
F + F... - f.o11 = 0
r = F0 ,1- F•.,
fv
Fo1t = 63.91 kN
F," = P•" A
F.lr = 20 [ f (1.61)2 I= 40.72 kN
F = 63.91 - 40.72
Solution
F= 23.19 kN
F\ = y vlffidg!O..I\ HllttllO\~Ihl'liOrnl·
F1 = (9.8lx0.9)[n(2)2(8) -~n(2)3]
Problem 3 - 49
F1 = 739.66 kN
\ 300 mm diameter steel pipe 12 nun thick carries water under a head of 50 m
•f water. Determine the stress in the steel
8Ft..ll + W = f\
F _ 739.66-50
'~'"8
Solution
F'"'" = 86.2 kN
[S, = pD)
2l
S - 9.81(50)(300) = 6131.25 kPa
I2(12)
Problem 3 - 48
Determine
the
force
F
required to hold the cone
shown. Neglect the weight of
the cone
51 = 6.13 MPa
Air
Oil
2m
s = 0.8
3m
Problem 3 - 50
Determine the required thickness of a 450 mm diameter steel pipe to carry a
maximum pressure of 5500 kPa if the allowable working stress of steel is 124 MPa
c;olution
[S,= pD I
21
124 X 1000 = P( 4SQ)
21
I = 9.98 mm say 10 mm
152
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Problem 3- 51
Determine the stress at the walls of a 200 mm diameter pipe, 10 mm thick
under a pressure of 150m of water and submerged to a depth of 20m in salt
water
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
153
Total HydrostatiC Force on Surfaces
Pipe diameter, D = 6 m = 6000 mm
Maximum pressure the tank (at bottom), p = Yo•l h
p = 9.81(0.8)(7) = 54.936 kPa
3
S = 2(110 X 10 )(300)
54. 936(6000)
S = 200.23 mm say 200 mm
Solution
J.S, = pD I
21
P = Pu»ll1l• - Poutsotle
p = 9.81(150)- 9.81(1.03)(20)
p = 1269.4 kPa = 1.269 MPa
...;, = l .2o9(200) = 12.69 MPa
2(10)
Problem 3 - 54
t\ thin-walled hallow sphere 3.5 min diameter holds helium gas at 1700 kPa.
1>etermine the minimum wall thickness of the sphere if its allowable stress is
110 MPa.
Solution
Problem 3 - 52
A 100-mm-10 steel ptpe has a 6 mm wall thickness. For an allowable tensilt>
~tress of 80 MPa. what maximum pressure can the pipe withstand?
pD
Wall stress, S, = .
4t
OOO = 1,700(3.5 X 1000)
4t
t = 24.79 mm
60
Solution
'
J.S. = pD I
21
HO = p(100)
2(6)
p = Q.6 MPa = 9,600 kPa
Problem 3 - 55
'\ vertical cyl\fidrical tank is 2 meters in diameter and 3 meters high. [ts sides
1re held in position by means of two steel hoops, one at the top and the other
11 the bottom. If the tank is filled with water to a depth of 2.1 m, determine
the tensile stress in each hoop.
Problem 3 - 53
A wooden storage vat is 6 m in diameter and is filled with 7 m of oil, s = 0.8
fhe wood staves are bound by flat steel bands, 50 mm wide by 6 mm thick,
whose allowable tensile stress is 110 MPa. What is the required spacing of the
bands near the bottom of the vat. neglecting any initial stress?
«;olution
-....
Y""
-....-
Solution
'-;pacing ot hoops, S =
I~
2m
l.S 1 A 11
pD
Allowable tensile stress of hoops, .S, = 110 MPa
Cross-sectional area of hoops. A,, = 50(6) = ';00 mm 1
2.3
3m
2.1rr
2 T1
~
W te
F
I
0.77
r-
o- . -
3
2.1
"----
_;t__
2 T2
154
MECH~NICS
CHAPTER THREE
FLUID
& HYDRAl\LICS
Total Hydrostatic Force on Surfaces
\
p.: Mtop =OJ
2T2(3) = F(2.3)
r~ = 0.3833F
-7 Eq. (1)
F = yh A
r = 9.81 ( 1f )[ (2)(2.1)1= 43.26 k N
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total HydrostCltic Force on Surfaces
155
l,roblem 3 - 57
\ cylindrical container 8 m high and 3 m in diameter is reinforced with two
hoops 1 meter from each end. When it is filled with water, what is the tension
111 each hoop due to water?
'iolution
In Eq. (1)
r2 = 0.3833(43.26)
/'2 = 16.58 kN (tens10n m the bottom hoop)
p.:: F11- OJ
2T2 + 2T, = r
2fl = F- 2Tz
2[1 = 43.26- 2(16.58)
T1 = 5.05 kN (tensio n m the top hoop)
Problem 3 - 56
A vertical cylindrical tank, open at the top, is filled with a liquid. Its sides are
held in position by means of two steel hoops, one at the top and the other at
the bottom. Determine the ratio of the stress in the upper hoop to that in the
lower hoop
f=y lt A
F = 9.81(8/2)[8(3)]
f= 941.76 kN
[:EM top hoop = 0]
Solution
]:EMtup=O]
2T2(1t) = F(2h/3)
r2 = F/3
[:EMoouom= OJ
2T1(It) = F(h/3)
r
.,.-
2T2(6) = F (13/3)
T2 = 13F/36
D-+
..........
T2 = 13(941.76)/36
T2 = 340.08 kN
h
[:EMoottom hoop = 0]
Uq ~ld
l
2T1 (6) = F(S/3)
r, = sr/36 =5(941.76)/36
T1 = 130.8 kN
T, = F/6
Problem 3 - 58 (CE Board November 1982}
F/6
Ratto= - - = 0.5
F/3
F
h/3
h
A cylindrical tank with its axis vertical is 1 meter in diameter and 6 m high. It
I<; held together by two steel hoop s, one at the top and the other at the bottom.
lhree liquids A, B, and C having sp. gr. of 1.0, 2.0, and 3.0, respectively fill this
t.mk, each having a depth of 1.20 m . On the surface of A there is atmospheric
pressure. Find the tensile stress in each hoop if each has a cross-sectional area
of1250 mm2.
156
I I UID MECHANICS
HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
-
lm
~
r
L
3.6 m
Liquid A
2T,
Liquid B
= 2.0
.......
liquid
K
0
'
A2
1250
Stress, 5 2 = 28.25 MPa
-7 stress in bottom hoop
1.2 m
5 = 1.0
5
PJ
1.2 m
-
c
s =3.0
LIQUid 8
157
r2 = 3.6(9810) = 35316 N
r2 35316
Stress 52 = - = - -
Solution
r=
CHAPTER THREE
Total Hydrostatic Force on Surfaces
1.2
[UH = 0]
2T, + 2T2 = F, + F2 + F3 + F4 + Fs
2T1 = 0.72y + 1.44y + 1.44y + 4.32y + 2.16y- 2(3.6y)
r, = 1.44y
r, = 1.44(9810)
r, = 14126.4 N
1.2 m
......
p.
Sb·ess 5, = .Ii_ = 14,126.4
A,
1,250
I
2T1
PI= 0
P2 =PI+ y~Jt •
. P2 = 0 + (y x 1)(1 .2) = 1.2y
fl~ = P2 + y~lt~
p~ = 1.2y + (yx2)(1 2) = 3 6y
fl4 = PI + Yc he
P4 = 3.6y + (y><3)(1.2) ~ 7 2y
F1 = 1h (p2)(1 .2\(1)
F1 = 1h(1 2y)(1 2)(1) = 0 72y
. '· .,., F, = p2(l .2)(1)
. ' F~ = 1 2y(1 2)( I) = 1.44y
F, = 112(P• f12)(1 .2)(1)
F, = 112(3 o-.. I 2y)(1 2)( 1l = 1 44y
f 4 = p,(1 2)(11
F, = 3 6y(l 2)(1) = 4 32)
Fs = lfz(p4 · p,)(1 .2)(1)
F~ = lfz(7 2y · ~.6y)(1 2)(1) = 2. 16w
[I.Miop"' OJ
3.6(2T2) = F,(0.8) + F2(1.8) + F3(2) + F4 (3) + F5 (3.2)
7.2T, = 0 72y(0.8) + 1 44y(1.8) + 1.44y(2) + 4.32y(3) + 2.16y(3.2)
r, = 3 oy
Sb·css, 5 1 =11.3 MPa
-7 stress in top hoop
Problem 3 - 59
<\n open cylindrical tank of 1.86 m2 cross-sectional area and 3.05 m high
ontains 2,831 liters of water. Into it is lowered ano ther smaller tank of the
une height but of 0.93 m2 cross-section in the inverted position, allowing its
11'cn e11d to rest on Lhe bottom of the bigger tank. Determine the maximum
l•·nsion per vertical millimeters on the sides of the bigger tank. Neglect the
thickness of the metal forming the inner tank and assume normal barometric
pressure.
Solution
1.~
l
305
1
Pz
Vz
1---
305
L..----__.11
o Before lowering
f--
6 After lowering
158
CHAPTER THREE
Total Hydrostatic Force on Surfaces
In Figure 6
CHAPTER THREE
I I UID MECHANICS
.'. HYDRAULICS
Total Hydrostt~trc Force on Surfaces
159
Problem 3- 60 (CE Board November 1977)
Volume of water= {1.86- 0.9.3)(b +II)+ 0.93b = ~~0~
n iceberg hav ing specific gravity of 0.92 IS floating on salt water of sp. gr.
1.86b + 0.93/t = 2.831
2/J + h = 3.044
b = 1.522- 0.5/t
-7 Eq. (1)
!r\. If the volume of ice above the water s urface is 1000 cu. m., what is the
1~>1.11 volume of the ice?
olution
[p1 v1 =7'2 V2]
pr = 101.325 kPa (atmospheric pressure)
Let
vl = 0.93(3.05) = 2.8365 m 3
P2 = 101.325 + 9.81h
v2 = 0.93(3.05 - b)
Woce
V = total volume of ice
Vo =volume displaced
Vo = V -1000
W;w = Y~<<' V = (9.81x0.92)(V)
W;c~ = 9.0252 V
101.325(2.8365) = (101.325 + 9.81h)[0.93(3.05- b)]
309.04 = 309.04- 101.325b + 29.9211- 9.81blt
29.9211- 101.325b- 9.81bll = 0
29.9211- 101.325(1.522- 0.5h) - 9.81(1.522- 0.5/t)h = 0
29.92/t- 154.22 + 50.66/t- 14.93/t + 4.905h2 = 0
4.905Jt2 + 65.65/t- 154.22 = 0
~-----------------
h=
-65.65± ~(65.65)2 -4(4.905)(-154.22)
2(4.905)
= 2.039
b = 0.5027111
H = b + It = 2.512 m
The maximum tensile stress occurs at the bottom of the tank.
p = yH = 9.81(2.542)
p = 24.937 kPa = 0.024937 MPa
Tension, T:
2T= pD(1)
f 02 = 1.86 m2
0 = 1.539 m = 1,539 mm
2T = 0.024937(1,539)(1)
T=l9.2N
BF = Y"''"·•ter Vo
BF = (9.81xl0.3)(V -1000)
Bf = 10.1043(V- 1000)
.,..
0
0
0
[LFv = 0]
W;". = Bf
9.0252V = 10.1043(V -1000)
1.0791 = 10104.3
V = 9,364 cu. m .
Another Solution:
For homogeneous solid body floa ting on a homogeneou s liquid:
sbody
_ Ybod y
Vn = - - Vbody - - - Vt>ody
Sliquid
Y!iquid
0.92
then; V - 1000 = 1 .03
v
0.106796 v = 1000
V = 9,364 cu. m.
Problem 3- 61 (CE Board May 2003, Nov 2002, May 2000, Nov 1992)
1\ block of wood 0.60 m x 0.60 m x It meters in dimension was thrown into the
water and floats w ith 0.18 m projecting above the water surface. The same block
was tlu-own into a container of a liquid having a specific gravity of 0.90 and it
floats w ith 0.14 m projecting above the surface. Determine the following:
(n) the value uf It,
(b) the specific gravity of the block, and
(c) the weight of the block.
CHAPTER THREE
160
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
CHAPTER THREE
I LUID MECHANICS
I. HYDRAULICS
161
Tot<tl Hydrost.lttc Force on Surfaces
Solution
In Water:
Draft =
5
wood It
Swater
/r -0.18=
5
y,lono -- -460
- - -- 28,204 N/m 1
0.0163
wood h
1
Swnuo lr = It - 0.18 ~ Eq. (1)
l'roblem 3- 63 (CE Board May 1993)
In Water (S
= 1)
\ body having asp. gr. of 0.7 floats on a liquid of sp. gr . 0.8. The volume ot
tlw body above the liquid surface is what percent of its tota l volume?
In another liquid
Draft=
·s wood It
shquid
It - 0.14 = Swood It
0.9
Swood It= 0.9/t- 0.126 ~ Eq. (2)
~ ~0.14
LLJ)_j _
0.14
In other liquid (S = 0.9)
[Swoocl /r = Swood /r)
lr - 0.18 = 0.9h- 0.126
It = 0.54 rn
~ height of the block
S ubs titute lr to Eq . (1):
Swood(0.54) = 0.54- 0.18
Swood = 0.667
~Specific gravity of wood
Weig ht of block= Ywoo<t Vt>to<k
Weight of block= (9.81 x 0.667)[(0.6 x 0.6)(0.54)]
Weight of block= 1.272 kN
Solution
sbody
Vn = - - Vt"'.t"
5 Jtqllld
Vn = g:~ Vt,•.tr = 0.875Vt,•.t"
Since the volume of the body displaced (below the liqutd surface) is 0.875
••r 87.5'Yo of its total volume, then the volume of the body above the liquid
ll ttrface is 12.5''/., of its total volume.
Problem 3- 64 (CE November 1997)
\ block of wood 0.20 m thick is floating in sea water. T he specific g rav ity of
,,·ood is 0.65 whi le lhat of seawater is 1.03. Find lhc minimum area of a block
hich will suppo rt a ma n we ig hing 80 kg.
<,olution
w..,.N = ao kg ~
Problem 3 - 62
A s tone weighs 460 N in air. When s ubmerged in water, it weighs 300 N.
Find the volume and specific gravity of the s tone.
w~
[BF = Yw•t~r Vstonr]
160 = 9810 (Vstono)
Vstune = 0.0163 cu. m.
-lt0.2 m
t
Solution
Weight of stone = 460 N
Weight of s tone in water= 300 N
Buoyant force, Bf = 460- 300 = 160 N
t
[~fv = 0]
BF = Wman + Wwood
Ysm Vwood = Wm.m +')'wood \/wood
(1 000 X 1.03) Vwood - 80 + (1000 X 0.65) Vwom1
wood = 0.21 05 m 3 = Area X 0.2
Area = 1.05 square m eter
v
I
CHAPTER THREE
Total Hydrostatic Force on Surfaces
162
FLUID MECHANICS
& HYDRAULICS
Problem 3- 65 (CE November 1997)
Problem 3 - 67
A cube of wood (s.g. = 0.60) has 9-in sides. Compute the magnitude
direction of the force required to hold the wood completely submerged
water.
·~ill
Solution
CHAPTER THREE
Total Hydro\t.tttc Force on Surfaces
163
\ uniform block of steel (s = 7.85)
float at a mercury-water
111terface as shown in Figure 27.
What is the ratio of the distances
'a" and "b" for this condition?
Weight of wood= (62.4 x 0.60) (?zY = 15.795lbs
Buoyant force when completely submerged in water:
)
BF = 62.4 C~ 3 = 26.325lbs
Required force= 23.325 -15.795
Required force = 10.53 lbs downward
c.iolution
Problem 3 - 66 (CE Nov 2000)
The block shown in Figure 04
weighs 35,000 lbs. Find the
value of IL.
12' X 12'
_Jj
Oil
5 = 0.8
L-9ft_
Water
Figure 04
Solution
From the figure shown:
[I Fv =OJ
BF, + BF2 = 35,000
BF, = Yoat Vo
BF1 = (62.4x0.8)(12x12x3)
BF1 = 21,565.44 lbs
21,565.44 + BF2 = 35,000
BF2 = 13,434.56 lbs
Bf2 = Yw Vo
13,434.56 = 62.4 [(12)(12) 11]
It = 1.495 ft
12' X 12'
ll't A be the horizontal cross-sectional
trea of the block.
8ft+ Bf2 = W
Y11• Vn"' + Ym Vnm = y, V
9.81(/\ X a)+ (9.81 X 13.56)(/\ X /J) = (9.81 X7.85)[/\(rl + /1)]
ll + 13.56b = 7.8!'¥1 + 7.85/J
5.71 b =·6.85 (/
a/b = 0.834
35,000 lbs
Problem 3 - 68 (CE May 1998)
Oil
s = 0.8
Water
- ·- -·-·-· ·- - · -~·- ·
--~~t9ft
----BF2
3ft
h
1 a 5-kg steel plate is attached to one end of a 0.1 m x 0.3 m x 1.20 m wooden
pole, what is the length of the pole above water? Use s.g. of wood of 0.50.
-leglect buoyant force on steel
CHAPTER THREE
164
FLUID MECHANICS
& HYDRAULICS
To tal Hydrostatic Force on Surfaces
Solution
VJ .. .11
Neglecting the buoyant force on steel:
BFwood = Wsteel + Wwood
1QQQ(Q.1 X 0.3 X y) = 5 +
}QQQ(Q.5)[Q.1 X 0.3 X 1.2]
y = 0.77m
CHAPTER THREE
11roblem 3 - 70
wooden buoy (s.g. = 0.62) is 50 mm by 50 mm by 3 m long is made to float
sea water (s.g. = 1.025). How many N of steel (s.g. = 7.85) should be
ltached to the bottom to make the buoy float w ith exactly 450 mm exposed
1hove the water surface?
<"'!
.....
~olution
h· =0
BF''"''' + BFwoot!- Wwood- W,~.~., = 0
wwood
0.05 m
~D
/"f-
i
0.45 m
8 F''""' = Ysw V'""''
8 F,,.,, = (981 ox 1.025) v,,,~.,
8f,,,..,, = 10,055.25 V,....,, N
Problem 3 - 69
8fwoou = Ysw VD
8fwood = (9810 X 1.025)[(0.05)2(2.55))
BFwood = 64.1 N
If a 5-kg steel plate is attached to one end of a 0.1 m x 0.3 m x 1.20 m wooden
pole, what is the length of the pole above water? Use sp. gr. of wood of 0.50
and that of steel 7.85.
Wwood = Ywood Vwood
Wwood = {9810 X 0.62)[0.05)2(3))
Wwood = 45.62 N
Solution
w,,,..,, = y,""'' v,,,.,.,
w",.,., = (9810 x 7.85) v.'"'''
Wwood = (1000 X 0.5)(0.1 X 0.3 X 1.2)
Wwood = 18 Kg
w.....,, = 5 kg
BFw = 1000(0.1 x 0.3 x d)
BFw=30d
BF~
= 1000 Vs
Ws = (1000 x 7.85) Vs = 5
165
Tota l HydrostatiC Force on Surfaces
11
E
It = 1.2- y
It= 1.2- 0.77
It= 0.43 m
ILUJD MECHANICS
,._ HYDRAULICS
Wsteel = 77008.5 V,,,.,,,
-
1-
BFwood
t
i-
~
2.55 m
h.
l~ I)
Wsteel
10055.25 Vslct•l + 64.1- 45.62- 77008.5 V,,,.,., = 0
66953.25 v,,,.•, = 18.48
..r = 0.000276 m 3
v,,•
Wstccl = 9810{7.85)(0.00276)
W,1,.,.1 = 21.255 N
Vs = 0.000637 m3
BFs = 0.637 Kg
Problem 3 - 71
Wwood + Wstccl = BFs + BFw
18 + 5 = 0.637 + 30d
d = 0.745 m
X= 1.2- d
x = 0.455 m
\ piece of lead (sp. gr. 11.3) is tied to a 130 cc of cork whose specific gravity ts
11.25. They float just submerged in water. What is the weight of the lead?
166
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Solution
CHAPTER THREE
Total Hydrostatic Force on Surfaces
167
Solution
[l:Fv = 0]
We+ W, = BFc + 8ft
Wc=yc Vc
We= (1 x 0.25)(130)
We= 32.5 grams
w.s.
We
8Fc = Ym Vc
8 r c = (1 )(1 30)
8Fc = 130 grm
w, =yl v,
w, = (1 X 11.3) Vt
BFc
Wt = 11.3 VI
BFc
8fr = y,., Vt
8[1 = (1) Vt = VL
32.5 + 11.3 v, = 130 + v,
Vt = 9.47 cc
.
w, = 11.3(9.47)
W1. = 106.97 grams
Problem 3 - 72 (CE November 1993}
A hallow cylinder 1 m in diameter and 2 m hjgh weighs 3825 N. (a) How
many kN of lead weighing 110 kNjm3 must be fastened to the outside bottom
of the cylinder to make it float with 1.5 m submer ged in water? (b) How many
kN of lead if it is placed inside the cylinder ?
(a) Lead is fastened outside the cylinder
(a) Lead is fastened outside
Bfc = Yw Vo
8rc = 9.81r (1)2(1.5)]
t
Brc = 11.56 kN
Bft = y" Vt
8Ft= 9.81 V,
WI.= YL Vt = 110VL
(:E Fv = 0)
8Fc + 8Ft= We+ Wt
11.56 + 9.81 Vt = 3.825 + 110Vt
Vt = 0.0772 m 3
Wt = 110(0.0772) = 8.49 kN
(b) Lead is inside the cylinder
[:EFv = 0]
Wt + Wc=8Fc
Wt + 3.825 = 11.56
Wt = 7.735 kN
(b) Lead is placed inside the cylinder
168
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Problem 3 - 73
A s tone cube 280 nun on each side and weighing 425 N is lowered into~nk
containing a layer of water 1.50 111 thick over a layer of mercury. ~ermine
lhe position of the block when it has reached equilibrium.
Solution
W =425N
CHAPTER THREE
I I UID MECHANICS
Total Hydrostatic Force on Surfaces
1, HYDRAULICS
169
.,olution
s.g. = 0.8
IT
s.g. = 0.7
2.2 ft
BFAI = 'YA IVnM
BFA1 = (9,810 x 13.6)[0.282(x)]
BFAI = 10,459.81 X
w
~r=c-
BF,v='Yiv Vmv
BF1v = 9,810[(0.28)2(0.28- x)]
BFw = 769.1(0.28- x)
/
0.28m
I I
s.g.
~~---------~F
= 1.6
..·.Y
~~r
BFwt
Jl·~_.__
l --4•~VJ7M~rr
s.g. = 1.4
~2.2 ft ~l
/
[H, =OJ
BfM + BFw= W
BFM
RF = (62.4 X 1.4)[(2.2)2(2.2- h)] + (62.4 X 0.8)[(2.2) 2(11)]
BF = [62.4 X 2.22] (3.08 -1.4h + 0.817)
W= (62.4 X 1.6)[(2.2)2(1.1)] + (62.4 X 0.7)[(2.2)2(1.1)]
W = [62.4 X 2.22](2.53)
10,459.81 x + 769.1(0.28- x) = 425
9690.71x = 209.652
X= 0.0216111
x = 21.6mm
Therefore; the block will float with 21.6 mm below the mercury surface.
[BF = W]
[62.4 X 2.22) (3.08 - 1.4/7 + 0.8/t) = (62.4 X 2.22}(2.53)
3.08 - 1.4/t t 0.8/t = 1.76 + 0.77
It = 0.917 ft
Problem 3 - 74
A cube 2.2 feet on an
ed ge has its lower half of
s.g. = 1.6 and upper half
of s.g. = 0.7. It rests in a
two-layer fluid, with
lower s.g. = 1.4 and
upper
s.g.
0.8.
Determine the height It
of the to p of the cube
above the interface. See
Figure 33.
l'toblem 3- 75 (CE May 1997)
s.g. = 0.8
IT
100-nun diameter solid cylinder is 95 mm high and weighing 3.75 N is
Immersed in a liquid ('Y = 8.175 kNjm3) contained in a tall metal cylinder
b.1ving a diameter of 125 mm. Before immersion, the liquid was 75 mm d eep
t what level will the solid cylinder float?
s.g. = 0.7
2.2 ft
~
s.g. = 1.4
s.g. = 1.6
1--.
IL-------------~
. I
~ 2.2ft~
Figure 33
170
CHAPTER THREE
I I UID MECHANICS
HYDRAULI CS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Total HycJrostc1tic Force on Surfaces
171
I'• oblem 3 - 76
Solution
125 mm 0
vooden beam of sp. gr. 0.64 is 150 111111 by 150 mm and is hinged at/\, as
h '" n in the Figure. At what angle 0 will the beam float in water?
•.
olution
100. mm 0
.....,r
E
E
E
T
lm
E
E
~
"'....
1
Jv
125 mm 0
(a) Before immersion
(b) After Immersion
(5 - 0.5x) cos 0
Solve for the draft D in figuxe (b):
[Bf = W)
Yt Vo =W
(8,175) Vo = 3.75
Vn = 0.0004587 m3 = 458,716 mm 3
f (100)2 X D = 458,716
Draft, D = 58.4 mm
When the solid cylinder is immer~ed, the liquid in the tall cylinder
due to volume of liquid displaced. Therefore, the volume of
displaced equals the total volume of real and imaginary liquid above
original level
Vaboveorig.level = Vo
f (125)2(x) = 458,716
x =37.38 mm
Weight of beam, w = Ylx••m vbNm
Weight of beam, W = (9,810 x 0.64)[(0.15)2(5)]
Weight of beam, W = 706.32 kN
Buoyant force, BF = Y"•ter Vo
Buoyant force, B.P = 9,810[(0.15)2 x]
Buoyant force, BF = 220.725x
[L MA = 0]
BF(5- 0.5x) cos 8 = W(2.5 cos 8)
220.725x(5- 0.5x) = 706.32(2.5)
5x- 0.5x 2 = 8
0.5x2_ 5x + 8 = 0
- (-5)± ~(-5) 2 - 4(0.5)(8)
__..:.___:_.:..___:_
X = __:.__:,_!....:..__:__
2(0.5)
From Figuxe b:
75 +X= D + y
y = 75 + 37.38 - 58.4
y=53.98 mm
Therefore; the solid cylinder will float with its bottom 53.976 mm above
bottom of the hallow cylinder.
x= 2m
1
sin 0 = - - = 1/3
5-x
8 = 19.47°
I
/,12
CHAPTER THREE
Total Hydrostatic Force on Surfaces
_ / __:
Problem 3 - 77 (CE Board May 2003)
Frornthe figure below, it is shown that the gate is 1.0 m wide and is hinged
the oottom of the gate. Compute the following:
(ajthc hydrostatic force in kN acting on the gale,
(f)the location of the center of pressure of the gate from the hinge,
(i)thc minimum volume of concrete (unit weight= 23.6 kN/m3) needed
kPep the gate in closed position.
CHAPTER THREE
Total Hydro;t.lttc Force on Surfaces
I I UID MECHANICS
1. HYDRAULICS
173
t•roblem 3- 78 (CE Board November 1993)
boat going from salt water (sp. gr
1.03) Lo fresh water (sp. gr. = 1.0) sinks
, 7 em and after burning 72,730 kg of coal rises up by 15.24 em.
Find the
11ginal displacement of the boat in sea water in kN.
'>nlution
t.
f
0.5 m
I
L
0 + 0.0762
r-r
~
Sea water
~
~I
2m
~
Figure (b)
Figure (a)
Solution
f= Yh J\:: 9.81(1)(2 X 1)
f= 19.62 kN
y= t (2) = 0.667 m
I~MA = 0]
fxy = T X 2.5
19.62(0.667) = 2.5T
T=5.232 kN
ftOJU the FBD of the
crncrete block:
f
0.5 m
D + 0.0762-0.1524
=0- 0.0762
L
-::r--1
1
I
~
Figure (c)
2m
l~fv =O]
T+ BF= W
BF = Yw Vconc = 9.81 Yconc
W = Yconc Vconc = 23.6 Yconc
5.232 + 9.81 Yronc = 23.6 Vconc
Yron< = 0.3796 m1
1
Fresh water i
We have to assume that the boat have a constant cross-sectional area A below
the water surface and use Yw•tPr = 1000 kg/ m 3
In Figure (a):
BF1=W
Y•w Vn = W
W= (1000 X 1.03)[A(D)]
W= 1030AD
-7 Eq. (1)
1~ CHAPTER THREE
Total Hydrostatic Force on Surfaces
In Figure (b):
BF2= W
Yr" Vo = W
W = 1000[A(0+0.0762)J
W = 1 OOOA(O + 0.0762)
In Figure (c):
BF3 = W- '72730
1000[A(O- 0.0762)) = W- 72730
II UID MECHANICS
HYDRAULICS
CHAPTER THREE
Total Hydrostatrc force on Surfaces
175
ulution
For any floating body; Buoyant force= Weight
~olvi ng for displacement in sea water:
-7 Eq. (2)
y".• walcr v()l = w
(64) Vo 1 = 24,000 x 2,240
Vo1 = 840,000 ft3
-7 Eq. (3)
Solving fo r displacement in fresh water:
Yrrt•;h w,•ler Vnz = W
From Eq. (1) and Eq. (2):
[W =WJ
1030AO = 1000A(O + 0.0762)
10300 = 10000 + 76.2
0 = 2.54 m (draft in sea water)
From Eq. (1)
W= 1030A(2.54)
W= 2616.2A
(62.2)(V02) = 24,000 x 2240
V 02 = 864,308.68 ft:-1
~
~I
Figure (a)
Let II be the difference in the drafts in fresh & seawater:
v[l2- vlll = Area X "
" = 864,308.68-840,000
32,000
" = 0.76 ft
Draft in fresh water, 0 = 34 + 0.76 = 34.76 ft
From Eq. (3)
1000A(2.54 - 0.0762) = 2616.2A - 72730
. 2463.8A = 2616.2A - 72730
A= 477.23 m2
Therefore:
w = 2616.2(477.23)
w = 1,248,529 kg (9.81/1000)
W=12248 kN
l,roblem 3 - 80 (CE Board November 1995)
• onsid.er an arbitrary shaped body with a submerged volume Vs and a
,[,•nsity p, submerged in a fluid of density Pr· What is lhe net vertical force on
lht• body due to hydrostatic forces?
• olution
F,,.,t = Yt V~
yf = Prx g
Problem 3 - 79
A ship having a displacement of 24,000 tons and a draft of 34 feet in ocean
enters a harbor of fresh water. If the horizontal section of the ship at the
waterline is 32,000 sq. ft, what depth of fresh water is required to float the
ship? Assume that marine ton is 2,240 lb and that sea water and fresh water
weight 64 pcf and 62.2 pcf, respectively.
Fn,•t = PtgV~
Problem 3 - 81
\ spherical balloon, 9 m in diameter is filled with helium gas pressurized to
Ill kPa at a temperature of 20°C, and anchored by a rope to the ground.
Neglecting the dead weight of the balloon, determine the tension in the rope
UseR= 212 m;oK for helium gas andy,"= 11.76 N/m3
176
CHAPTER THREE
'• HYDRAULICS
Solution
111 X 10
P
Ytwilum = 1 . 787 N
w
I m·'
"
m·1
[Lr v =OJ
BF- W- T = 0
BF = Yaor Vt,.lloon
Bf = 11.76(381.7) = 4488.8 N
W = Yheilum Vt~o~lloon
W=1.787 (381.7) = 682.1 N
4488.8 - 682.1 = T
T= 3806.7 N
177
[2:Mn = Ol
W... (1.5 cos 0) - Bf[(L/2) cos 8] = 0
45.62(1.5)- 25.138L(L/2) = 0
12.57 L2 = 68.43
L = 2.33 m
V tMIIoon =::~n{9/2)3
V t,.ttuuu = 381.7
Total Hydro~tcltrc Force on Surfaces
W,. = 45.62 N
(from Figure 3 - I)
Bf = y," Vn
Rf = 9810(1.025)[(0.05)2LI
BJ' = 25.138L
3
- ----Yh t ' h-u m - -RT
212(273+20)
CHAPTER THREE
I LUIS rv1ECHAN JCS
Total Hydrostatic Force on Surfaces
T
sin 0 = 2/ I
sin 0 = 2/2.33
() =59°
L ~2
~
Problem 3 - 83
Problem 3 - 82
The b uoy in Figure 3 - 1 has 80 N of s teel weight a ttached. The buoy has
lodged agams t a rock 2 m deep . Compu te the angle 9 w ith the horizontal at
which the buoy w ill lean, assum ing the rock exer ts n o moment on the bu oy.
, right circular cone is 100 mm in diameter and 200 mm high and weighs 1.6
'\1 in air. How much force is required to push the cone (vertex downward)
nto a body of liquid having sp. gr. o f 0.8, so that its base is exactly at the
11rfacc? l!ow much additional force is required to push the base 10 mm
••·low the surface?
<;olution
Solution
J'he required d ow nward ve rtical force IS
A
r = Br- w
Br = y'"'""' v"'""
Bf = (9,810 x 0.8) [(n/3)(0.1 /2)2(0.2)1
BF=.f.l1N
F = 4.11 - 1.6
r = 2.51 N
1\lotc: This force f= 2.51 N becomes
onstant no matte r how deep fu r ther
thc cone is s ubmerged
1.5 cos 9
F
178
CHAPTER THREE
CHAPTER THREE
I LUID MECHANICS
"- HYDRAULICS
on Surfaces
Problem 3 - 84
Total HycJrostatic Force on Surfaces
'-\olution
To what depth will a
fresh water
F
iameter log, 4 m long and of sp. gr. 0.425 sink 1n
Solution
Water
~
=40N
Glycerin
s = 1.3
For a homogeneous solid body floating on a
homogeneous liquid:
_ 5 body
VrJ - - - Vt"'d'
5 hqUld
A\ I.= }~~ A L
A, = 0.425nr2 (shaded area)
0
Lcl V =volume of wood
From geometery:
In water:
[~f,, = 0]
As= A>l'tlor- AtrMnp,h•
0.4251tr2 = 1/2 r 2 0, - 1
Bf1- w- r = o
9810V- W = 40
12 r 2 sin 9
0, -sin 9 = 2.67
Solve 0 by trial and error:
Try 8 = 170°
170°(1t/180°)- sin170° = 2.76
V = 40 + W
9810
(¢2.67)
Try 0 = 166°
166°(1t/180°)- sin166° = 2.655
Try 6 = 166.44°
166.44°(n/180°) - sin166.44° = 2.67
(¢2.67)
O.K.
lt=r-y
"= 1 - (1) cos (0/2)
1 - (1) cos (166.44° /2)
h = 0.882m
"=
Problem 3 - 85
A block of wood requires a force of 40 N to keep it immersed in water and a
force of 100 N to keep it immersed in glycerin (sp. gr. = 1.3). Find the weight
and sp. gr. of the wood
F = lOON
-7 Eq. (l l
In glycerin:
fHv = 0]
BF2-W-f =O
(9,810 X 1.3)V- W= 100
40
w= 100
(9,810 x 1.3)[ +
9810
52+ 1.3W- W = 100
W= 160 N
w] -
From Eq. (1):
v = 40 + 160
9810
V = 0.0204 m 1
.
w 160
Unitwe•ght,y= - = - V
0.0204
Unit weighl, y = 7843 N/m3
Sp. gr., s = Ywood = 7843
Ywat~r
98"1 0
Sp. gr., s = 0.8
179
180
CHAPTER THREE
f I UID MECHANICS
Total Hydrosta tic F
IIYDRAULICS
on Surfaces
Vaal (anah•l) = Vail (final)
A rectangular tank of
5 m, as shown, contains oil o
0.8 and water. (a) Find the
h. (b) If a 1000-N block of ood is
floated in the oil, what is the rise in free
surface of the water in contact with air?
(0.5)(5)(1.25) = (0.5)(5)(1!') - 0.1274
h' = 1.301 m
1\.s shown in Figure b, if the oil-water interface drops by a distance of y, the
lree surface of water will rise by y/2, since the cross-sectional area of the
1ight compartment is twice that of the left compartment.
Solution
1000 N
t
3m
l
'n1m-up pressure head from oil surface to water surface in m 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 m.m
I herefore; the free surface of water w ill rise 13.6 mm.
SL o
h
181
O..,mce the volume of oil remain unchanged;
Problem 3 - 86
T-
CHAPTER THREE
Total Hydrostatic Force on Surfaces
=
6,....,
011
s- 0.8
l•r oblem 3 - 87
Water
O.Sm
11 open cylindrical tank 350 mm in diameter and 1.8 m high is inserted
• 1tically into a body of water with the open end down and floats with a 1300
I block of concrete (sp. gr. = 2.4) suspended at its lower end. Neglecting the
··1ght of the cylinder, to what depth will the open end be submerged in
.tier?
lm
Figure (b)
Figure (a)
olution
I v = 0]
(a) Depth of oil: (Refer to Figure a)
Sum-up pressure head from oil surface 0 to water surface 6 in m of water
fl.+ lt(0.8) + 3 - 4 = J2
y
0 + 0.811 - 1 = 0
h = 1.25 m
y
(b) Rise of the water surface: (Refer to Figure b)
BF=W
Yoil Vo = W
(9810 x 0.8) Vn = 1000
Vo = 0.1274 m3
RFcanc + Bfcyl- W
= 0 -7 Eq. (1)
= Ywalcr Vcanc
=
Vcone wconc
BFcanc
Ycone
1300
Vcanc = _:.:....;....:.,__
9810(2.4)
Vcanc = 0.0552 m 3
= 9810(0.0552)
Bfcanc = 541.7 N
0
Bfcanc
= Yw•ler Vo
Bfcyl = 9810[ { (0.35)2h)
Bfcyl
Bfcyl =943.83 It
Water
concrete
block
182
CHAPTER THREE
Total Hydrostatic
on Surfaces
CHAPTER THREE
I I UID MECHANICS
1. HYDRAULICS
Total Hydrostatic Force on Surfaces
183
ulution
From Eq. (1)
541.7 + 943.83lt - 1300 = 0
It= 0.803 m
w, = 205 kg
w, = 205 kg
0.6m 0
Applying Boyles Law (taking P•tm = 101.325 kPa)
Before insertion:
Absolute pressure in air, p1 = 101.325 kPa
Volume of air inside the cylinder, V1 = (0.35)2(0.18)
t
l.Bm
1
w,
L
2.1 m
..±.__
t BF,
0.96
Volume of air inside the cylinder, V1 = 0.0173 m3
After insertion:
Absolute pressure in air, p2 = 101.325 + ylt
Absolute pressure in air, pz = 101.325 + 9.81 (0.803)
Absolute pressure in air, pz = 109.2 kPa
Volume of air inside the cylinder, Vz = { (0.35)2 x
0.6m0
0.84
Chain
l
Lem
A cylindrical buoy 600 nm1 in diameter and 1.8 m high weighs 205 kg.
moored in salt water to a 12 m length of chain weighing 12 kg per m of
length. At high tide, the height of buoy protruding above water surface
0.84. What could be the length of protrusion of the buoy if the tide d
2.1 m? Density of steel is 7,790 kg/ m3. Use density of water= 1000 kg/m3.
tBF',
L'
~
Figure b: Low Tide
Wl'ight of chain = 12 kg/m
llt•nsity of steel= 7,790 kgjm3
ulume of steel (chain)= 12/7790
\ olume of s teej (chain) = 0.00154 1n3 per m eter length
l11 Figure n:
•
[IFv = 0]
BF1 + BF2- w1- w2 = o
Problem 3 - 88 (CE Board)
D
H'
t BF,
Figure a: High Tide
Therefore, the open end is submerged 0.933 m b elow the water surface.
i
H
Volume of air ins ide the cylinder, V2 = 0.0962.:r
[p1 V1 = p2 V2]
101.325(0.173) = 109.2(0.0962x)
x = 1.67 m
X- ft + y =
1.8
y = 1.8 -1.67 + 0.803
y = 0.933 m
y
BF1 = Y>w Vo
= (1000 X 1.03}[
t (0.6) (0.96))
2
BF1 = 279.58 kg
BF2 = Ysw V,hain
= (1000 X 1.03)[0.00154(L)]
BF2 = 1.586L
w2 = 12L
279.58 + 1.586L - 205 - 12L = 0
L = 7.16m
Depth of water, H = L + 0.96
Depth of water, H = 8.12 m
184
CHAPTER THREE
I LUID MECHANICS
Total Hydrostatic Force on Surfaces
I. HYDRAULICS
----
In Figure b:
-Depth of water, H' = H- 2.1
Depth of water, H' = 6.02 m
CHAPTER THREE
Total Hydrostatic Force on Surfaces
185
\olution
Weight of ball:
Draft, D = H'- L'
Draft, D = 6.02- L'
W= Yb•ll Vb.n
[L:Fv =OJ
BF', + BF'2- Wt- W'2 = 0
BF', = (1000 X 1.03) )[ (0.6)20]
W= 58.25 N
W = (9810 X 0.42}-t 7t(0.15)3
t
BF' 1 = 291.23 (6.02- L')
BF't = 1753.18- 291.23L'
BF'2 = (1 000 X 1.03)[0.00154(L')]
BF'2 = 1.586L'
W'2 = 12L'
1753.18- 291.23L' + 1.586L'- 205 -12L' = 0
L' = 5.13 m
D = 6.02-5.13 = 0.89
y = 1.8- D
y = 1.8-0.89
y = 0.91 m (le11gth of prolmsiou)
4.31T'
I
~t
Buoyant Force:
BF = Yw•ter Vb.n
BF = (9810) t 7t(0.15)3
BF = 138.69 N
Depth of pool:
Work done by W= Work done by BF
W(4.3 +It) = BF(h)
58.25(4.3 + It) = 138.69h
It= 3.11 m
l•roblem 3 - 90
\hydrometer weighs 0.0214 Nand has a stem at the upper end which is 2.79
How much deeper will it float in oil (sp. gr. = 0.78) that in
tlt·ohol (sp. gr. = 0.821)?
111111 in diameter.
\olution
Problem 3 - 89 (CE Board)
A wooden spherical ball with specific gravity of 0.42 and a diameter of 300
nun is dropped from a height of 4.3 m above the surface of water in a pool of
unknown depth. The ball barely touched the bottom o f the pool before it
began to float. Determine the depth of the pool.
In alcohol:
BF=W
(9810 x 0.821)Von = 0.0214
Von= 2.657 x 10-6m3
VOn = 2,657 mm3
In Oil:
BF=W
(9810 x 0.78)Vo. = 0.0214
Von= 2.797 x 10-6 m3
Von= 2,797 mm~
= Voo- Vo,.
= 2,797-2,657 = 140 mm3
~ Vo = t (2.79)2 lr = 140
~Vo
~Vo
Jr = 22.9 mm
A.lcohol, s • 0.821
Oil, s =0.78
186
CHAPTER THREE
~
Problem 3 - 9)
L• L
The body is stable if M is above G.
Draft' 0 = 0 ·182 L
Draft, 0 = 0.82L
-f2 (L)(L) 3
l
MB,,=-
Vo
Total Hydrostatic Force on Surfaces
187
l'roblem 3 - 93
A plastic cube of side L C)nel sp. gr. 0.82 is placed vertically in water.
cube stable?
Solution
CHAPTER THREE
I LUID MECHANICS
.•, HYDRAULICS
Total Hydrostatic Force on Surfaces
!~,
Me
• G
lli_
~
D/2
J,
~
MB,, = 0.102 L
GB. = L/2- 0/2
GB. = 0.09L
.,olution
Jote: The body is stable when M is
thove G and unstable if M is below
' , With smaller value of H, the
metacenter M will become higher
than G making it much stable.
When H increases, M will move
.Jnwn closer to G making it less
.t,\ble. Hence, the maximum
Iwight for stable equilibrium is
hen M coincides with G, or MB. =
LR.
-:r-• Bo
(LX L)(0.82L)
block of wood (sp. gr. = 0.64) is in the shape of a rectangular parallelepiped
h.tving a 10-cm square I:?ase. If the block floats in salt water with its square
J,,,sc horizontal, what is its maximum height for stable equilibrium in the
upright position?
L
Since MB. > GB., M is above G
The body is stable.
10 em • 10 em
r
H
,~
D
1l
• Bo
Df2
:!
I rom the figure:
A solid wood cylinder of specific gravity 0.6 is 600 mrn in diameter and
mm high. If placed vertically in oil (sp. gr. = 0.85), would it be stable?
Solution
Draft, 0 = sp.gr. wood H
sp.gr.oil
r = 300 mm
c
Draft, 0 = ~:~~ (1200) = 847 mm
Vn
f(300)
MB" = rc(300) 2 (847)
MB. = 26.56 mm
GB. = 600 - 12(847)
GB" = 176.5
1
Since M8 0 < G80 , the metacenter is below (,
Therefore, the body is unstable.
•G
"'
D/2
1
Vn
.....
]
3
r-----
~
4
GBo = 0.5H- O.ti21H/2
GBo = 0.189H
fMB. = -
{
MB.=-
GB.=H/2-0/2
Draft, 0 = ~:~ H = 0.621H
...!..12 (10)(10)
MB.= ...!.!:.....:.....__ _
(10)(10)0
13.419
100
MB.= - - - - - - H
12(0.621H)
•B.
MB.= ~ (1+ tan
12D
2
2
OR:
l
MB,, =
2
el where9=0°
J
10
(l + O) = 13.419
12(0.621H)
H
H~2
_1_
Seawater, s = 1.03
Waterline Section
Problem 3-92
~
•G
CHAPTER THREE
Total Hydrostatic Force on Surfaces
I WID MECHANICS
1.. HYDRAULICS
CHAPTER THREE
Total Hydrostatic ~on Surfaces
188
189
Initial metacentric height MG = MBo - GBo
Initial metacentric height, MG = 77.49 - 117.45
Initial metacentric height, MG = -39.96 rnrn
[MB .. = GBo]
13 419
= 0.189H
·
H
H = 8.43 em
'
Problem 3 - 94
A wood cone, 700 mm diameter an~ 1,000 mm high floats in water with its
vertex down. If the specific gravity of the wood is 0.60, would it be stable?
Determine also its·initial metacentric height.
Solution
*
1t(350)2 (1000)
Vwood = 128,281,700 m3
Vwood =
lution
(•t) Initial metacentric height:
vwood
- -0.60
VCl1-
MB 0 = ~[1 + tan2
Vo = 0.6 Vwood
120
E
Vo = 0.6 (128,281,700)
Vn = 76,969,020 mm'
E
~
1
Vwvorl = ( 1000 )
0
,
750
X
350
843.4
1000
x = 295.2 mm
1
bI Metacenter, M
1
:G:
I
j
12(2.4)
(I)
11 (295.2) 4
4
MB. = =
= 77.49 mm
Vn
76,969,020
From the Figure:
GB. = 750-30/4
GB. = 750- 3(843.4)/4
GB. = 117.15 mm
Since MB., < GB.,, M is below G and the cone ts UNSTABLE.
2.7m
Initial metacentric height, MG = MBo- GBo
Initial metacentric height, MG = 2.8125 - 1.5
Initial metacentric height, MG = 1.3125 m
I
Waterline Section
D=2.4m
MBo = 2.8125 m
GBo = 2.7- 1.2 = 1.5 m
I
X
oo]
2
MB 0 = _j2t__ [1 + tan 2
i
I
VWIIIJd
0.6Vwootl
0 =843.4 mm
2
a]
where e = 0°
0
0
By similar solids:
Vn
f'roblem ~ • 95
rectangular scow 9 m wide, 15 m long, and 3.6 m high has a draft in sea
"•Iter of 2.4 m. Its center of gravity is 2.7 m above the bottom of the scow.
I ll'lerrnine the following:
(a) The initial metacentric height,
(b) The righting or overturning moment when the scow tilts until one side
is just at the point of submergence.
tan a= .L1
4.5
0 = 14.93°
MB 0 = ~[1 + tan2
120
2
a]
MB 0 = ~[1 + (1.2/ 4.5)2]
12(2.4)
MBo= 2.91 m
2
L"' 15m
-+:------ J=1,.••
!
1.2m
190
CHAPTER THREE
Metacentric height, MG = MB"- GB"
Metacentric height, MG = 2.91 -1.5 = 1.41 m
Total Hydrostatic Force on Surfaces
191
Along longitudinal axis (rolling):
B=10m
~[1
+ tan e]
12D
2
2
Since MG > MB", the moment is righting moment.
Righting moment RM = W (MC sin 0)
MB =
o
W= BF = yV,
w = (9.81
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic-F-Cll.C.e...on Surfaces
where 6 =
o•
10 2
(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)
1.03)[9(15)(2.4)I = 3,273.8 kN
R1ghhng mo ment, RM = 3,273.8[(1 .41) sin 14.93°1
Righting moment RM = 1,189.3 kN-m
MB. =
X
Problem 3 - 96
A barge tloa ting tn tresh water has the torm ot a parallelepiped ha
dimensions 10m x 30m by 3 m. It wetghs 4,500 kN when loaded with
of gravity along its vertical axis 4 m from the bottom. Find the metacentn
height about its longest c1nd shortest centerline, and determine whether or not
the barge is s table
Solution
Along transverse direction (pitching):
B=30m
MB = ~[1 + tan
o
12D
2
2
e]
where 6 = oo
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)
MBo =
Problem 3 - 97 (CE August 1973)
·fo
Rolling
Solve for the draft, D
[BF= W]
y Vo= W
9.81 [10 X 30 X Dl = 4,500
0 = 1.53 m
CB.= 4-0/2
GB,, = 4- 1.53/2
c-;B" = 3.2..15 m
1\ crane barge, 20cQ1long, 8 meters wide, and 2 meters high loaded at its center
with a road roller weighing 20 short tons, floats on fresh water with a draft of
I 20 meters and has its center of gravity located along its vertical axis at a
point 1.50 meters above its bottom. Compute the horizontal distance out to
one side from the centerline of the barge through which the crane could swing
I he 20-ton load which it had lifted from the center of the -deck, and tip the
barge with the 20-meter edge just touching the water surface?
192
CHAPTER THREE
I I UIO MECHANICS
IIYDRAULICS
Total Hydrostati~r~o.-Surfaces
Solution
<.B. = 1.4 - d = 0.848 m
w~ = (20 x 900) 9.R1
2
2
tan MB.= - - [ 1+ 120
2
WR = 176.58 kN
el
"'
E
N
1!
193
z = d sin 8
z = 0.552 sin 11.3JO = 0.108 m
1 c;hort ton = 2000 lh
=· 900 kg
r
CHAPTER THREE
Total Hydrostatic Force on Surfaces
B
G~
0.5 m
- - - - - - - - -w,- - - - - - .
Bo
1.5 m
MB 0 =
8]
~ [1 + tan211.31o] = 4.533 m
12(1.2)
2
MG = 4.533 - 0.848 = 3.685 m
\ =MGsin8
BF
BF = yVll
BF = 9.81 [8 X 1.2 X 20]
BF = 1,883.52 kN = W,
Weight of barge, Wn = Br- WR
Weight of barge, w,, = 1,883.52- 176.5R
Weight of barge, W11 = 1.706.94 kN
\ = 3.685 sin 11.31 o = 0.723 m
[rMc= 0]
(BF) x = WR(L + z)
1,883.52(0.723) = (176.58)(L + 0.108)
L = 7.604 m 7 Horizontal distance from the center of the deck
,., oblem 3 - 98
Tilted posi tion
· wooden barge of rectangular cross-section is 8 m wide, 4 m high, and 16 m long.
It 1s transporting in seawater (s = 1.03) a total load of 1,500 kN i,ncluding its own
'"ight and cargo. If a weight of 75 kN (included in the 1,500-kN) is shifted il
·il'ltance of 2.5 m to one side, it will cause the barge to go down 450 mm in tht>
wrlge of immersion and also rise 450 mm in the corresponding wedge of e1ner~ion
I Ill' barge floats vertically (on an even keel) before the shifting of the weight.
ompute how far above the waterline is the center of gravity of the loaded barge.
olution
·80
tan 8 = 04.11
8 = 11.31 °
Solve for the new position of G m the tilted position:
Wr(O.S) = Wu(d)
1,883.52(0.5) = 1.706.94(d)
d = 0.552 m
r
CHAPTER THREE
194
CHAPTER THREE
t I UID MECHANICS
I IYDRAULICS
Total Hydrostatic l=n•·rp.-nn Surfaces
Solve for the draft, 0:
BF=W
(9.81 X 1.03)[8 X 16 X OJ = 1,500
D = 1.16 m
Total Hydrostatic Force on Surfaces
I'• oblem 3 - 99
,,. waterline section of a 1,500-kN barge IS as shown. Its center of gravity IS
m above the center of buoyancy. Compute the initial metacentric height
1\ unst rolling.
In the Tilted position:
60 kNfm
'>olution
tan e
= 0 ·45
4
Bm
e = 6.42°
82
12(1.16)
A
I= 676.53 m 4
(1 + tan226.42o) 4.63 m
=
[LMso =OJ
1,425(b) + 75(a) = BF(c)
c = MB. sin 8
c = 4.63 sin 6.42°
c = 0.518 m
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)
h = 2.947 m
1
[MB.= ]
Vo
I = 1rcocl•ngle + I ln•ngle + Iscmo-<'trrlc
I~
(12)(8)3 +
(6)(4)3 X 2 + t (4)•1
A
MB 0 = ~[1 + tan2 (:)]
120
2
MBo =
~distance of G from the w.s.
195
[BF= W)
9.81 Vo = 1,500
Vo = 152.9 m~
MB 0 = 676.53 = 4.425 m
152.9
[MG = MB. - GB.)
MG = 4.425 - 1.5
MG = 2.925 m ~ initial metacentric height
196
CHAPTER THREE
Total Hydrostatic Fo
lementa
II UID MECHANICS
• lfYDRAULICS
Surfaces
CHAPTER THREE
Total Hydrostatic Force on Surfaces
197
fl• oblem 3 - 104
\ Iter in a tank is pressurized to 80 cmHg. Determine the total force per meter
1dth on panel AB.
Ans:482kN
Problems
Problem 3 - 100
A vertical rectangular gate 2 m w1de and 1.2 m high has water on one sidt
with surface 3 m above its top. Determine the magnitude of the total
hydrostatic force acting on the gate and it..<t distance from th<> wr~ter surface
2m
Am;: 1- - M.o k!\1, y1• .=. 3.63 111
Water
4m
Problem 3 - 101
...,r-
A vertical semi-circular area of radius r IS submerged in a liquid with 1ts
diameter in the liquid surface. How far: is the center of pressure from tht.>
liquid surface?
Ill
80 em
-:r-
YHg
B
3m
A
Problem 3 - 105
11 the figure shown, the 8-ft-diameter cylinder, 3 feet long weighs 550 lbs and
Problem 3 - 102
1, ~ts on the bottom of a tank that is 3 feet long.
An open vat holding oil (s = 0.80) is 8 m long and 4 m deep and has a
trapezoidal cross-section 3 m wide at the bottom and 5 m wide at the top
Determine the following: (n) the weight of oil, (b) the force on the bottom of the
vat, and (c) the force on the trapezoidal end panel.
Ans: (n) 1002 kN; (b) 752 kN
(c) 230 kN
Water and oil are poured into
!Ill' left-and right-hand of the tank to depths 2 feet and 4 feet, respectively.
lll'termine the magnitudes of the horizontal and vertical components of the
force that will keep the cylinder touching the tank at A.
Ans: FH = 749 lb -7
Fv = 2,134lbs
,
w
Problem 3 - 103
Freshly poured concrete approximates a fluid with
sp. gr. of 2.40. The figure shown a wall poured
between wooden forms which are connected by six
bolts. Neglecting end effects, compute the force in
the lower bolts.
Ans: 19,170 lbs
Water
Concrete
Oil, s
= 0.75
Problem 3 - 106
c ompute the hydrostatic force and its location on semi-cylindrical indentation
/lCD shown. Consider only 1 m eter length of cylinder perpendicular to the
11gure below.
Ans: FH = 109.5 kN@ 1.349 m below D
Fv = 20.5 kN@ 0.531 m to the left of B
200
CHAPTER THREE
Total Hydrostatic Force on Surfaces
r liJID MECHANICS
IIYDRAULICS
CHAPTER FOUR
Relative Equilibrium of Liquids
201
Problem 3 - 110
Two spheres, each 1.3 m in diameter, weigh 5 kN and 13 kN, roo· .....,,,...;,,,,,
They are connected with a short rope and placed in water. What is the
in the rope and what portion of the lighter sphere protrudes from the water?
Ans: T = 1.74 kN; 40.1
Problem 3 - 111
A block weighing 125 pcf is 1 ft square and 9 inches deep floats on as
liquid composed of a 7-in layer of water above a layer of mercury.
Determine the position of the bottom of the block. (b) If a downward
force of 260 lb is applied to the center of mass of this block, what is the
position of the bottom of the block?
A11s: (a) 0.8" below
(b) 4.67" below
hapter 4
Relative Equilibrium
of Liquids
I 11der certain conditions, the particles of a fluid
mass may have no relative
lllnhon between each other yet the mass itself may be in motion. If a mass of
lhud is moving with a constant speed (uniform velocity), the conditions are
lh·· ~ame as in fluid statics (as discussed in previous chapters). But if the body
1 ubjected to acceleration (whether translation or rotation). special treahnent
1 ~t•quired, and this will be discussed in this chapter
Problem 3 - 112
Would a wooden cylinder (sp. gr. = 0.61) 660 nun in diameter and 1.3 m
be stable if placed vertically in oil .(sp. gr. = 0.85)?
HI CTILINEAR TRANSLATION {MOVING VESSEL)
tlorizontal Motion
Problem 3 - 113
A rectangular scow 7 ft by 18 ft by 32 ft long loaded with garbage has a
of 5 feet in water. Its center of gravity is 2ft above the waterline. Is the
stable? What is the initial metacentric height?
Ans:
Problem 3 - 114
• 1111sider a mass of flutd moving with a linear acceleration n as shown tn tlw
t 11•,ure. Considering a particle in the surface, the forces acting are the weight
I \1 Mg and the fictitious inertia force (reversed effective force, REF) which i~
•111al to Ma, and the reaction N which must be normal to the surface
a
A cube of dimension Land sp. gr. 0.82 floats horizontally in water. Is the cube
stable?
Ans: Stable
REF= Ma
202
CHAPTER FOUR
I I UJD MECHANICS
Relative
of Liquids
w
203
V• rtical Motion
From the force polygon sh
tan 9 = REF
Rclctttvc Equilibrium of Liquids
I. HYDRAULICS
nnsider a mass of fluid accelerated upwards or downwards with an
ll'leration of a as shown in the Figure. The forces acting at a point It below
llu· liquid surface are the weight of the liquid above the point, yV. the inertia
fntre, Mn, and the pressure force F = pA, then.
\
Ma
tan e = -
Mg
I I,,= 0]
tan e = !!:..
g
/'= Mn + yV
REF= Ma
M=pV= l.v
I
L:
g
Therefore; the surface and all planes of equal hydrostatic pressure must
tnclined at this angle e with the horizontal
f= l..va+yV
g
Volume, V = Alt
F= pA
Inclined Motion
Consider a mass of fJUJd being accelerated upwards at an inclination a
the horizontal so that n, = n cos a and nv = n sin a
pA = 1.. (Alt)a + y(A/t}
g
Area = A
F= p A
p = y/t{1 + n/ g)
REFv
W = yV
h
=M av
a
t
W = Mg
Eq. 4-3
~a..
t
'jol
Use(+) for upward motion and(-) for downward motion
M av
Note: a is positive for acceleration and negative for deceleration.
Mg
From the force polygon shown
MaH
tan e = _ _
___;_'--M g+ Mnv
ROTATION (ROTATING VESSELS)
When a liquid mass is rotated about a vertical axis at a constant angular speed
ro (in radians per second), every particle experiences a normal acceleration
llf
tan e = _!!jj_
g + av
2
uf a, which is equal to ~ = ro 2 x where xis the particle's distance from the
X
txis of rotation. This acceleration causes an inertia force (centrifugal force or
tan e =_!!jj_
g±ay
Use(+) sign for upward motion and (-)sign for downward motion.
Eq. 4-2
n•versed normal effective force) which is equal toM n, or
gw w x.
2
204
dy
rrom calculus, slope = - = tan 0
'J'
I•
I, di
dy
w2 X
-=tanS=--
dx
zos.
CHAPTER FOUfl
Relative Equilibrium of Liquids
PLIJID.'MECHANICS
t. HYDRAULICS
CHAPTER FOUR
IITOITIOL-100 JOA":f71U8 OIUOIJ
---
g
~F
ParabolOid
of revolution
Figure
J - 1 (a)
I
Figure 4 - 1 (b)
I nr cylindrical contain1r of radius r revolved about its ~ertfcal ~xis, the height
~'" W=Mg
+--X---+1 /
'
CF
Eq. 4-5
11
of paraboloicl is:
'
I
~
i'
Eq. 4-6
---..-----''- --~-o.
T
CF = (W/g) ,,,2 x
1
where will the ang~lar speed in radians per se(:ond.
•
I I
NOTE: 1 r m\~/30 rad/sec
Figure 4 - 1 (c)
111 Figure 4 J 1
Figure 4- 1: Paraboloid of revolut1on
1
1
/
\
:
v
1
(b), the relationship between any two points in the p,arabola
can
1
~~iven by ~squa'ted property of parabola):
,
1....._
¥
_..)
From the force polygon
tan e = CF
,, •r I
w
II
'
2
(WI g)w ·'
tan e = ..;___;_~--
w
Volume of Paraboloid of Revolu ion
w2 x
tanO = - g
Eq. 4 - 4
Where tan e is the slope of the paraboloid any point x from the axis of rotation
Eq. 4-7
206
CHAPTER FifSI,tJn
I I UID MECHANICS
CHAPTER FOUR
I
LIQUID SURFACE CO
For open cylindrical containers more than half-full of liquid, rotated
its vertical axis (I! > H/2):
~
~
y/2 < 0
y/2 =0
(No liquid spilled)
Liquid surface just
touching the top rim
(No liquid spilled)
Relative Equilibrium o(Ligui~$
HYDRAULICS
For closed cylindrical containers more than hc\lf-fuH of liCJ.uicl, rotated
about its vertical axis (11 > H/2):
y
y/2 < 0
y/2 > 0
y=H
(Some liquid spilled)
Vortex at the bottom
(Some liquid spilled)
If
I'
I
y>H
Vortex (imaginary) below
the bottom
(Some liquid spilled)
y/2 > 0
(with imaginary
paraboloid above)
!\lute: For closed vessels, there can
m•ver be any liliUid spilled, so the initial
\ ulume of liquid (before rotation) is
,,(ways equal to tile final volume of the
hquid (after rotation) or tile iuitinl volume
ofnir inside is equal to tile finn/ volume ofnir
msirlc. The volume ofnir rclntiou is more
ronvenient to usc in solving this type of
problem.
y2= (D/H)(y- K); K= H2/2D
...,/1/
y/2 = 0
( hquid surface JUst
touching the top nm)
y = H2/20
(vortex just touching the
bottom)
y > H2/2D
(Vortex below
the bottom)
208
CHAPTER FOUR
CHAPTER FOUR
I LUID MECHANICS
I. HYDRAULICS
Relative Equilibrium of liquids
209
Relative Equilibrium of Liquids
For closed cylindrical containers completely filled with liquid:
U-tube revolved about its own axis:
Note: the pressure head at any point in
the tube is the vertical distance from the
tube to the paraboloid. The pressure is
positive if the paraboloid is above the
point and negative if it is below the
point. The limiting pressure is absolute
zero.
ro= O
Without pressure at top
y/2[
With pressure at top
...
·-"'._..·-·- - ·- ....................
...... _
For pipes and tubes:
"'
T-~
~ ----------------y
I
I
"
I
I
I
I
I
I
I.
I
I
I
I
I
I
/
1+- X1 ~
y1 I
-:;),.-: /
~"""~I
//
]?Jr
±========;:Dd
f:
l
+-----0--Xz--L-------6~)1
Without initial pressure inside
With initial pressure inside
\
I
\
\
\
\
\
I
.
\
- ·\ - ·- ·- ·- ·-I
\
I
.
\
\
' ' ......
'
I
\
I
I
\
·- ~\· - · - · -·- · -
·- ·- - - 7 L .
I
\
\
I
I
\
I
\
\
'
I
' . . . _L . . "'
I
I
CHAPTER FOUR
210 Relative Equilibrium of Liquids
Solved Problems
CHAPTER FOUR
Rci<Hivc Equilibrium of Liquids
II UID MECHANICS
I. HYDRAULICS
211
{c) When n = 6 m/s2
ll
6
tan a=-=-g 9.81
e = 31.45°
Problem4- 1
An open recta ngu lar tank mounted o n a truck IS 5 m lo ng, 2m wide a nd 2.5
high is fi.led with water to a depth of 2 m (u) What maximum horizo
accelerahnn can be tmposed o n the tank without spilling any water and
determindhe accelerating force on the liquid mass? (r) If the accele ration
mcreasedtoo m /s 2• how mu ch wa ter is s pilled o ut?
'I
a= 6 m/s2
x = 2.5 cot31.45°
x = 4.0875 < 5m
Vlcfl = 1/2(4.0875)(2.5)(2)
Ytclt = 10.22 m 3
VorigioMI = (2){2)(5)
Yorrgonal = 20 m 3
Solution
(a)
Vsprllcd = Vougon•t- V1en
V<poll<•d = 20- 10.22
Vsr•ll•·d = 9.78 m 3
11roblem 4 - 2
!'he li!ure s how'> lht> wa ter levt:'l und t:'t ma xrmum 11 w ht:'n no wa te r 1~
~p tlledout
0·5 = () 2
tantl "' TI
tan~= :!.. = 0.2
g
tl =0.2(9.81)
\ closed h orizontal cylindrical tank 1.5 m in diameter and 4 m long is
nmpletely filled with gasoline (sp. gr. = 0.82) and accelerated horizontally at 3
111js2 . Find the total force acting at the rear wall a nd at the front wall of the
l.tnk. Find also the accelerating force on the fluid mass.
'iolution
tan 0 =!!..
g
u =1.962 m/s 2
y
3
-=--
{b)
4
i\ccelerating f'orce, F = Mn
Mass, M = p(Volume of liqutd)
Mass, M = 1000(5 x 2 x 2]
Mass, M = 20,000 kg
Accelerating Force, F = 20,000 x I 962
~ccele rating Force, F = 39,240 N
9.81
y =1.223m
1i = 1.223 + 0.75
fi = 1.973 m
F,w = y h A
Frc.r = (9.81 X 0.82)(1.973)[ T(1 .5)2]
F,..., = 28.05 kN
or.
F
=9.81 < 225 )[2.5(2) 1- 9.81 ( 125 )[1 .5(2) I
f= 39.24 kN
Frront = Y ll A
Frront = (9.81 X 0.82)(0.75) ( T (1.5)2]
Frronl =10.66 kN
a= 3 m/s2
212
CHAPTER FOUR
Relative Equilibrium of
CHAPTER FOUR
Relative Equilibrium of Liquids
Acceleratmg rorn•
213
Also:
r = Ma
VAir (original) = Vair (final)
4(0.2)(2) = (1/2)xz(2)
xz = 1.6
~ Eq. (2)
t = 1.6/x
~ Eq. (3)
Mas'>. M = p(Vol unH:' J
I = 1(1000 · OH2)j-J(l 5)l(4)1l (31
f = 17,390 N
f = 17.39 kN
.,ubstitute z and xz to Eq. (1)
4(1.6/x) -1.6 = 4.1x
-7 multiply by x
6.4- 1.6x = 4.1x2
4.1x2 + 1.6x- 6.4 = 0
,-------- 1.6±~(1.6)2 -4(4.1)(-6.4)
x=
= 1.0695 m
2(4.1)
z = 1.6/1.0695 = 1.496 m
or
f = fJ,,.fl, JJituut
f = 28.05 - 10.6h
I = 17.36 kN
Problem 4- 3
..X closed rectangular ta nk -1 m long, 2 m wrde, a nd 2 m h1gh IS filled w 1th
wa ter to a Jepth of 18m If the a llowable force a t the rear wa ll of the tank 1 ~
200 kN ho"' fast can 1t bE.> accelerated horizontally?
Solution
rrF-q· 4m-·-1
E
.._,
B
cg•
h
!Iv
o.z '
E
T
N"'!
o'
'~ ·:·~
-----,;-~-iookN
a • ? m/s'
.
______"":..<·f·
~
/
a
z
g
X
tane=-=a
1.496
= 1.0695
a =13.72 mfs2 (horizontal acceleration)
g
f'•oblem4 -4
"open tank 1.82 m square weighs 3,425 N and contains 0.91m of water. It is
, lt•d by an unbalanced force of 10,400 N parallel to a pair of sides. What is
tit•• force acting in the side with the smallest depth?
-.. ...J..:-~-d.,.
\olution
2m
P=yhA
Solve for a andy:
F = Ma = 10,400
M =Mwater + Mtank
M = 1,000[(1.82)(1.82)(0.91)] + 3,425/9.81
a
M = 3,363.42 kg
----->
10,400 = 3,363.42 x a
a = 3.092 m/ s2
F = yh A
200 = 9.81 h !2(2)J
lr =51 IT\
'I = h - I = 4. I m
By simila r triangles
4- ..\
.\
a
y
tanS=-=-g
0.91
---4.1
z
4z - l'Z = 4 h
7 Eq. (1)
1.82 m x 1.82 m
0.91 m
214
CHAPTER FOUR
Relative Equilibrium of Liquids
CHAPTER FOUR
Relative Equilibrium of liquids
215
AABE = 5.11 m 2
3.092 = _jf_
9.81
0.91
y = 0.29 m < 0.91 m (OK)
"= 0.91- y
/1 = 0.62m
VABC =5.11(1.5) = 7.665 m3
Vtcrt = 12.4635- 7.665
Vtcft = 4.7985 m 3
p = 9,810 (0.62/2) (0.62 X 1.82]
P = 3,432 N
V,pillcd = 8.6985- 4.7985
V;pillcd = 3.9 nt3
Problem 4-5
An open trapezoidal tank having a bottom width of 3 m is 2 m high, 1.5
wide, and has its sides inclined 60° with the horizontal. It is filled with
to a depth of 1.5 m. If the tank is accelerated horizontally along its length
4.5 m/s2, how much water is spilled out?.
J•roblem 4- 6 (CE Board)
vcssel3 min diame ter containing 2.4 m of water is being raised. (a) Find the
l"'ssure at the bottom of the vessel in kPa when the velocity is constant, and
(I•) find the pressure at the bottom of the vessel when it is accelerating 0.6 m/s2
t('Wards.
Solution
3 + 2(2 cot 60°) =5.309 m
olution
ror vertical motion:
p =ylt (1 ±~g)
lz =2.4 m
rr
4.5
tane =- = g
9.81
f.l = 24.64°
(a) When the 'lelocity is constant, a= 0, then
p = ylz
p = 9.81(2.4)
p = 23.544 kPa (pressure at the bottom)
Yspillcd = Vong.- Ytelt
3m
Vong = 3+42732 (1.5) X 1.5
Vonr. = 8.6985 m3
VABCH - VABE
VABW = 3+52309 (2)(1.5)
Vteft =
(b) When a= 0.6 m /s 2 (use"+" for upward motion)
p = 9.81(2.4)(1 + (1 + ;~t)
p = 24.984 kPa
VA Ben = 12.4635 m3
Problem 4-7
VAse= AAec(1.5)
AABC = V2 (AB)(AE)sin f.l'
a= 180°- 60°- 24.64°
a= 95.36°
AE =sin 60°
AE = sin60° = 4.618 m
AABE = 1/2(5.309)(4.618)sin 24.64°
\ vessel containing oil is accelerated on a plane inclined 15° with the
horizontal at 1.2 m/ s2. Determine the inclination of the oil surface when the
1110tion is (a) upwards, and (b) downwards.
216
CHAPTER FOUR
CHAPTER FOUR
t LUID MECHANICS
Relative Equilibrium of Liquids
Relative Equilibrium of Liquids
I. HYDRAULICS
Solution
217
(c) Downward motion with a positive acceleration (use"-" with a= +8 m/s2)
p = (9.81 X 0.8)(3) (1 - 9 1 )
1
tan e = __!!_t!__
g±av
a,= a cos a
a, = 1.2 cos 15°
p=4.34 kPa
(d) Dow:ward motion with a ~e~ative acceleration (use "-" with a= -8m/ s2)
p- (9.81 X 0.8)(3) (1- 9 .81 )
p =42.74 kPa
a11 = 1.159 m/s 2
av = a sin a
av = 1.2 sin 15°
n,, = 0.31 mjs2
Problem 4-9
{a) When the motion is upwards:
cylindrical water tank used in lifting water to the top of a tower is 1.5 m
lt1gh. If the pressure at the bottom of the tank is must not exceed 16 KPa, what
'"'' ximum vertical acceleration can be imposed in the cylinder when it is filled
' 1th water.
1.159
tan O= - - - - 9.81 + (+0.31)
e = 6.533°
\olution
{b) When the motion is downwards:
1.159
tan S = - - - - 9.81 - (+0.31)
e = 6.955°
p = y lr (1 + ajg)
16 = 9.81(1.5)(1 + a/9.81)
a = 0.857 mfs2
Problem 4-8
Problem 4 - 10
An open tank containing oil (sp. gr. = 0.8) is accererated vertically at 8 m/
Determine the pressure 3 m below the s urface if the motion is (a) upward
a positive acceleration, (b) upward with a negative acceleration, (c) down
with a positive acceleration, and (d) downward with a negative acceleration.
\n open cylindrical vessel having a height equal to its diameter is half-filled with
".1ter and revolved about its own vertical axis with a constant angular speed of
110 rpm. Find its minimum diameter so that there can be no liquid spilled.
Solution
that there's no liquid spilled, base of the
1·.traboloid must just coincide with the
upper rim of the cylinder. Since the cylinder
• •nitially half-full, the height of the
Jl<lraboloid is therefore equal to the height of
tlu• cylinder.
co2r2
l r = -2g
lr = H = D
w = 120 rpm x n/30
co = 4n rad / sec
lution
•J
The pressure at a depth h is given by, p =
yh(1 ± ~)
(a) Upward motion with a p ositive acceleration (use"+" with n = +8 mjs2)
p = (9.81 X 0.8)(3) (1 + 9~:J
p = 42.74 kPa
(b) Upward motion w ith a negative acceleration (use"+" with a = -8 mjs2)
p = (9.81 X0.8)(3) (1 + 9~:l)
p = 4.34 kPa
~ ro = 120rpm
CHAPTER FOUR
Relative Equilibrium of Liquids
218
I LUID MECHANICS
'• HYDRAULICS
CHAPTER FOUR
Relative Equilibrium of Liquids
(·ht) 2 (0 / 2)'2
0 = ..:____:__~____:_-
Vspill!'d
= Vatr (fin•l)- Yair (intlial)
2(9.81)
0 = 0.497 m or 497 mm
Vspillcd
= 1/21t(0.6)2(1.63) - n(0.6)2(0.7)
Vsplllro = 0.13 m 3 x 1000 lit/m3
Vspilled = 130 liters
Problem 4 - 11
An open cylindrical tank 1.6 m in diameter and 2 m high is full of wat('r
When rotated about its vertical a'<is at 30 rpm, what would be the slope of
water su rface at the rim of the tank?
Solution
Slope = tan 0
(t)
2X
Slope = - g
rev
2;r rad
1 min
w = 30 - - X - - - X - min
rev
60 'il'C
<•> = 1t rad / sec
l'roblem 4 - 13
r\n open cylindrical tank, 2m in diameter and 4 m high contains water to a depth
,,f 3m. !tis rotated about its own vertical axis with a constant angular speed w.
(a) If w = 3 rad/ sec., is there any liquid spilled?
(b) What maximum value of w (in rpm) can be imposed without spilling
any liquid?
.
(c) If w = 8 radjs, how much water is spilled out and to what depth wdl
the water stand when brought to rest?
(d) What angular speed w (in rpm) will just zero the depth of water at the
center of the tank?
(e) If w = 100 rpm, how much area at the bottom of the tank is uncovered?
~
.,olution
w2r2
Slope= (n) '2 (0. 8 ) = 0.805
9.81
/z=-2g
(11) w = 3 radjsec
lz=
Problem 4 - 12 (CE Board November 1978)
An open cylindrical vessel 1.2 m in diameter and 2.1 m high is 2/3 full
water. Compute the amount of water in liters that will be spilled out is
vessel is rotated about its vertical axis at a constant angular '>peed of 90 rpm.
Solution
I
ll = - 2g
h/2 = 0.23 < 1 m
:. no liquid is spilled out
2
(1)2 (1)2
2=-2(9.81)
2
(3n) (0.6)
2(9.81)
It= 1.63 m
11/ 2 = 0.815 > 0.7 m (some liquid spilled)
w = 6.26 rad/ sec x 3,0
(I) =
' r = 0.6 m
4
Ifmt
tt
r =1m
(II) The maximum w so that there is no liquid spilled is such that
h/2 = 1 m or It = 2m
II=-2g
<•> = 3n rad / s
=
2(9.81)
It= 0.46
w2r2
<•> = 90 rpm x n/30
11
(3)2 (1)2
~ <•>=90rpm
w 2r 2
219
59.78 rplJl
CHAPTER FOUR
Relative Equilibrium of liquids
220
11/2 = 1.63 m > I m
V,""'•·.t = V.111 <•"~·•II - V.,".'""'1.111
V~ur (lln.tH = VPM·''"''md
V.ttl (1111<11) =
5. 121 l1l I
221
= 100 rpm
By the squared property
of parabola:
:. some liquid spilled but the vortex of the
paraboloid is inside the tank since II < -+m.
v." cr."·•'> =
(l)
Area, A = 1t x 2
y = 5.58-4
y= 1.58 m
(c) (1) = 8 rad/seL
(8):!(1):!
II =
= 3.26 m
2(9.81)
1
'2 n:(1F (3.26)
CHAPTER FOUR
Relative Equilibrium of Liquids
II UID MECHANICS
I. HYDRAULICS
x2
r2
y
II
x= s:~
2
(1.58)
x2 = 0.283
r =1m
n:(l ) 2(1)
V,ltr (tnlllotl) = 3.1-!2 111 I
v.lr(ttttll,lll =
Area, A= n(0.283)
Area, A = 0.889 m2
v,,,u,,, = 5. 121 - 1. 1-!2
V, 1, 1h•d = 1.979 n1 1
l'•oblem 4- 14 (CE November 1993)
Another solution:
When the tan k is brought to rest, the
water level w ill rc•st c1t 11/2 from top.
11 open vertical cylindrical vessel, 2m in diameter and 4 m high is filled
with
.cter to the top. If rotated on its own v-ertical axis in order to discharge a
pc.mtity of water to uncover a circular area at the bottom of the vessel 1 m in
ll.uneter: (a) Determine the angular speed in rpm, and (b) how much water is
HI in the cylinder after rotation?
y=ll/2- I
y = 1.63-1
!I= 0.63 111
V,p1ll•·•• = n: r 2 !f
V,1,u,•t1 = n: ( I )2(0.63)
w2r2
v,,,u.•d = 1.979 m 1
(d) The vortex touches the bottom when II = .fm
2
- (tl ( 1)~
r =1m
-l--2(9.81)
(IJ
= 8.86 X J!l..
n
C1J
= 84.6 rpm
=
'lolve for II (by squared property)
,,
c~}r = (3.33n:) (1):!
2g
2(9.810)
11=5.58>4m
11
h=-2g
(1)
(e) Wh en w = 100 rpm
w = lOO(n:/30) = 3.33n: rad/ sec
2
m='
'olution
2
:. the vortex of the paraboloid is already below the tank (imaginary)
2
= (0.5)
2
"- 4
,, - 4 = 0.25/r
0.75/t = 4
/1 = 5.33 m
5.33 = (.1)2 (1)2
2(9.81)
w =. 10.23 rad/ sec x ~
w = 97.65 rpm
CHAPTER FOUR
Relative Equilibrium of Liquids
222
(b)
V,,.n = V,vilnd•·• - V''"''""' or pu•botood
V1,.,1 = rr (1)2 (4) - (1/2 rr (1 )2 (5.33) - 1hrr (0.5)2 (5.33 - 4)]
Vtpu = 4.716 m ·'
lz = ,,2- 2.75- Pt/W
lz = 175- 2.75- 62.5
lr = 109.75
A 1.90 m diameter closed cylinder, 2.75 m high is completely filled with
having sp. gr. of 0.8 under a pressure of 5 kg/ cm 2 at the top. (a) What angu
speed can be imposed on the cylinder so that the maximum pressure a t
bottom of the tank is 14 kg/cm 2? (b) Compute the pressure force exerted by
on the side of the tank in kg.
109.75 = w2(0.95)2
2(9.81)
w = 48.84 rad/ sec x
------- ----------!\,
1 r
-~-
=0.95
'E
I
I
I
I
h
1
-~-.,...---~
II
~
I
F=yh A
h = 112 -2.75/2 = 173.625 m
F = 800(173.625)[2rr(0.95)(2.75)]
F = 2.28 X 106 kg
II
c::::====:>
I
(l)
\
I
II
I
~
3
w = 466.44 rpm
Solution
Imaginary L.S.
h
I
I
'
'
Problem 4- 16 (CE May 1985)
I
"
,-"'--.----;..
ll
pJy
I
i----+----::
I
\n open cylindrical tank having a radius of 300 mm and a height of 1.2 m is
11111 of water. How fast should it be rotated about its own vertical axis so that
'%of its volume will be spilled out?
.,olution
E
"'....
,..;
2.75 m
Oil {sl= 0.8)
2n(0.95) = 5.969 m
w2r2
h=--
2g
tnce 75% of the total volume is spilled out,
I he paraboloid will be formed a part outside
lhl' vessel (i.e. with its vortex below the tank)
I
2
Vspolled =
(1)2/'2
h=-2g
Solve for h:
p1/y = 5/0.0008 = 6250 em
fll/Y = 62.5 m
V•., = 0.75[rrr2(1.2))
V.;, = 0.9rrr2
Unit weight of oil, y = 1000(0.8)
Unit weight of oil, y = 800 kgfm3 = 0.0008 kg/cmJ
(a)
223
1!2 = p2/y
1!2 = 14/0.0008
/r2 = 17500 em= 175m
Problem 4 - 15 (CE Board November 1993)
-
CHAPTER FOUR
Rclattve Equilibrium of Liquids
I I UID MECHANICS
1. HYDRAULICS
llut
V.;, = Vhog pu •• bolo•ct- Vsmau p.r•botood
0. 9rrr2 = 1/2 rrr2 II - 1h rrx 2y
1.8 r2 = r2h- x2y
~ Eq. (1)
fl)
='
CHAPTER FOUR
Relative Equilibrium of liquids
224
By squared property of parabola:
x2
r2
-
= -;
--
r2
xl = -
!I
"
In Eq. (1)
-7 Eq. (2)
1J
It .
I I UID MECHANICS
IIYDRAULICS
CHAPTER FOUR
Relative Equilibrium of liquids
225
1rev
60sec
c.o = 12.528 rad/ sec x - - x --.2rrrnrl 1nun
w = 119.64 rpm
1' 2
1.8 r2 = r2 11- -
h
!f(lJ)
-7 multiply both sides by h/ r2
1.811 = it2 - y2
buty=/1-1.2
1.811 = 112 - (it - 1.2)2
1.811 = 112- (/z2- 2.411 + 1.44)
0.6/t = 1.44
11 = 2.4 m
ltnblem 4- 18
11 open vessel, 500 mm in diameter and filled with water, is rotated about is
, 1t1cal axis at such velocity that the water surface 100 nu11 from the axis makes
II oll\gle of 40° with the horizontal. Compute the speed of rotation in rpm.
olution
I he slope of the paraboloid at any distance "x" fr.o m the axis is given by:
(1.)2
Finally:
2.4 = (.02 (0.3)2
2(9.81)
c.o = 22.87 rad/ sec x ~~
c.o = 218.4 rpm
Problem 4- 17
(1.)2
tan 40° = - (0.1)
9.81
c.o = 9.07 radjsec x ?;? = 86.64 rpm
l'toblem 4- 19
An open cylindrical tank 1 m in diameter and 3 m high is full of water.
what speed (in rpm) must it be rotated to discharge 1/3 of its content.
Solution
Let y be the height of the paraboloid.
Since the volume of the paraboloid represents the volume of water spilled, then:
Volume of paraboloid =%Full volume of cylinder
/2 1r (0.5)2 !f = X 1r (0.5)2 (3)
t
1
tane= - x
g
Where 0 = 40° and x = 0.1 m
1 n open cylindrical tank 1.2 min diameter and 1.8 m deep is filled with water
nd rotated about its own axis at 60 revolutions per minute. How much liquid
apilled and what is the pressure at the center of its bottom?
<•)
olution
=60 rpm
w2r2
ll=-2g
c.o = 60 x (rr/30) = 2n rad/sec
r = 1.2/2 = 0.6 m
h
y=2m
2
2
- C.O-X- j
[y 2g
(1.)2 (0.5)2
x=O.S n1
2=-..:..__~
2(9.81)
h = (2n)2(0.6)2 = 0.724 m
2(9.81)
Vsp1llcd = Vp.r•boloid
Vspillc'<l
1/2 nr2 /t
=
Vspn~c...t = Y2 1t(0.6)2 (0.724 )
Vsrmcd
= 0.409 m 3
l
1.8 m
I
I
I
y
-*-__________ iI _________ _
' r = 0.6 m
226
CHAPTER FOUR
Relative Equilibrium of liquids
II UID MECHANICS
I. HYDRAULICS
Pressure at the center
CHAPTER FOUR
Relative Equilibrium of Liquids
225
1 rev
60sec
w = 12.528 rad/ sec x - - x - 2 rr rnrl 1 min
P = YY
y = 1.8- h = 1.8- 0.724 = 1.076 111
p = 9.81(1.076) = 10.555 kPa
w = 119.64 rpm
1'1 oblem 4 - 18
Problem 4 - 20
A closed cylindrical vessel, 2 m m diameter and 4 m high is filled with wa
to a depth of 3 m and rotated about its own vertical axis at a constant angu
speed, w . The air inside the vessel is under a pressure of 120 kPa.
(n) If w = 12 radjsec, what is the pressure at the center and circu mference
the bottom of the tank?
(b) Wha l angular speed w will just zero the depth of water at the center?
(c) If c.>= 20 rad/sec, how much area at the botlom is uncovered?
.11 open vessel, 500 mm in diameter and filled with water, is rotated about is
, 1lical axis at such velocity that the water surface 100 mm from the axis makes
'" ,,ngle of 40° with the horizontal. Compute the speed of rotation in rpm.
'•olution
!'he slope of the paraboloid at any distance "x" from the axis is given by:
(j)2
tane= - x
g
Solution
Where 0 = 40° and x = 0.1 m
•
I
1
H
It
•<m~
3m
r = 1m
"
"
Air
p
= 120 kPa
I
I
Problem 4- 19
i
n open cylindrical tank 1.2 min diameter and 1.8 m deep is filled with water
I
I
nd rotated about its own axis at 60 revolutions per minute. How much liquid
w¥r
1 ~pilled and what is the pressure at the center of its bottom?
I
I
I
I
I
1r= lm
I
w2r2
h =--
2g
"=
(12)' (1) 7
= 7.34
2(9.81)
:~
I
00 2r2
0
Refer to Figure (b)
c1> = 12 rad/s
m =60 rpm
•.olution
!
F1gure (a)
(a)
(j)2
tan 40° = - (0.1)
9.81
c.o = 9.07 rad/ sec x ':? = 86.64 rpm
Figure (b)
6
Jz=-2g
c.o = 60 x (n/30) = 2rr rad/ sec
r = 1.2/2 = 0.6 m
h
h = (2rr)2(0.6)2 = 0.724 m
2(9.81)
Vspillrd = Vp•r•boloid
Y2 1tr2 It
Vspillod
=
Vspill<'<~ = Y2 1t(0.6)2 (0.724 )
1.8 m
:_ ..........j.........
Vspillro = 0.409 m 3
1
r
l
=0.6 m
CHAPTER FOUR
Relative Equilibrium of liquid.'l
226
CHAPTER FOUR
II UID MECHANICS
IWDRAULICS
Rclc~trve Equilibrium of Liquid.'l
h/2 = 3.67 m > 1m
Pressure at the center
= Y!!
(part of the paraboloid is above Lhe vessel)
lf = 1.8- h = 1.8- 0.724 = 1.076 111
p = 9.81 (1.076) = 10.555 k Pa
Verify the position of the vortex (See Page 207)
/1
Il 2 = (4)2
20
H2
Problem 4 - 20
A closed cylindrical vessel, 2m in diameter and 4 m high is· fiJJed with
to a depth of 3m and rotated about its own vertical axis at a constant angu
speed, w . The air inside the vessel is under a pressure of 120 kPa.
(n) If w = 12 radfsec, what is the pressure at the center and circumference
the bottom of the tank?
(b) What angular speed w will just zero the d e pth of water at the center?
(c) If w = 20 rad /sec, how much area at the bo t tom is uncovered?
Solution
20
1
It
•4m~
12 rad/s
•
tt •
::
..
~
'=>
r =1m
By squared property of parabola:
:
.lf
,,
,,
-7Eq. (2)
•'
Substitute x2 in Eq. (2) to Eq. (1)
h
I
I
I
I
I
j
,2
x2= - y
p = 120 kPa
I
I
I
I
I
,\ 2
,2
::
•'•'
•'
•'
•'
"4r
Wa~er
:. the vortex is inside the vessel
V,"' (looMI) = V,,,. (lnlll,ol)
~
I
3m
= 8 m > 7.34 m
1/21tx2y = nr2(1)
x2y = 2 r 2 -7 Eq. (1)
:··-----------------,----,r--r·
I
H
-
2(1)
/'2
(h .lf )y =2 r2
y2 = 2/r = 2(7.34)
y= 3.83m < 4m
Pressure at the center, (at 0)
]71 = y/11 + p,,;,
I
W~ter
h,
Jr,=4-y
/rl
= 4- 3.83 = 0.17 111
pi = 9.81(0.17) + 120
F1gure (a)
(al Refer to Figure (h)
col= 12 rad/s
w2r2
lr = - -
2g
2
,, = (12) (1)
2(9.81)
2
= 7.34
F1gure {b)
p1 = 121.66 kPa (pressure at the center)
Pressure at circumference, (at 6)
p2 = yh2 + p,,ir
lr2 = /r, +II
112 = 0.17 + 7.34 = 7.51 m
p2 = 9.81(7.51) + 120
p2 = 193.67 kPa (pressure at the circumference)
227
Let us first derive the general value of h
w hen the vortex of the paraboloid
reaches the bottom of the vessel
((ln"l)
= vAll'
229
{• ) co = 20 rad/ sec
oo2r2
h=--
(b)
vcur
CHAPTER FOUR
Relative Equilibrium of Liquids
t l UID MECHANICS
I. HYDRAULICS
CHAPTER FOUR
Relative Equilibrium of Liquids
228
2g
h = (20)2 (1)2
2(9.81)
{llllll.tll
12 nx2J I = nr2D
1
lr = 20.4 m
-? Eq. (h- 1 )
r2 // = 2 ,.2 0
In Figure (d):
By squared property of parabola
x2
H
= Vair (final)
Valr (Initial)
nr2(1) = 1/2nxt2 Yt - Vmx22 y2
2r2 = Xt2 Yt - xl y2 ~ Eq. (c-1)
r2
It
,.2
,,
r2 = -H
-? Eq. (11-2)
By squared property of parabola:
X 2
X 2
r2
Yl
Y2
h
1
Substitute x 2 to Eq (h-1)
(
r= t m
Fcgure (c)
'1~ H) H = 2r2D
~
'
_1_=_2_=-
.
...
'. iI :'
'..I '
Hl
D
2
~ Eq. (c-2)
Figure (d)
-7 Eq. (c-3)
Substitute x12 and x22 to Eq. (c-1}
(h eight of the paraboloid when it touches the bottom)
,2
,2
2r2: h y1(yt)- h
yz (yz) multiply both side by hjr2
2h = Yt2- yz2
2
h= ( 4 ) =8 m
2(1)
But y1 =4 + Yz
2h = (4 + yz)2 - y22
21! = 16 + 8y2 + yi - yz2
8y2 = 2(20.4) -16; yz = 3.1 m
0021'2
lt=--
2g
-
002(1)2
In Eq. (c-3)
8--2(9.81)
O) = 12.528 rad
c1> = 119.6 rpm
,2
I sec x 3u
"
Y2
·;..:.·--~-...L....-.1-
Simplify :
11 =
T
xz2 = - y2
It
12
x22 = ( ) (3.1) = 0.152
20.4
Area= nx22 = n(0.152)
Area = 0.48 m 2 (area uncovered at the bottom)
CHAPTER FOUR
Relative Equilibrium of Liquids
230
Problem 4 - 21
Determine the position of the vortex:
A closed vertical cylindrical vessel, 1.5 m in diametel\,.and 3.6 m high is
full of brine (s = 1.3) and is revolved about its vertica) axis with a rr..,c."'"'
angular speed. The vessel is made up of steel 9 mm thick with an allow
tensile stress of 85 MPa and has a small opening at the center of the top
(a) If the angular speed is 210 rpm, what is maximum the stress in the wal
(b) To what maximum angular speed can the vessel be revolved?
H2 = (3.6)2 = 7.2 m
20
2(0.9)
Since 1z = 13.86 > 7.2, the vortex is below the vessel, See Figure (b)
V•~r(tnitial)
= Vair(nn.•ll
nr2(0. 9) = 112nxt 2 Yt - 1/21tX22 yz
1.8 r 2 = Xt2 Yt- xz2 yz
--7 Eq. (1)
Solution
(a)
CHAPTER FOUR
Relative Equilibrium of Liquids
II UJD MECHANICS
1. HYDRAULICS
w = 210 rpm x n/30
m = 7rc radjs
By squaTed property of parabola:
X 2
X 2
r2
Y1
Y2
h
_1_=_2_ = -
pD
S,=21
Note: The max1mum pressure IS at the Circumference at the bottom
1
r = 0.75 m
Xt 2
--------~------- 1-----r---r-
~u>
~
I
I
1 r=0.75m
Xt
I
= -y1
1z
r2
xz2 = -
'•
::
Xt
r2
'•
h
y2
--7 Eq. (2)
--7 Eq. (3)
;:
Substitute Xt 2 and xz2 to Eq. (1)
r2
r2
1.8r2 = -y1(y1)- -yz (yz)
1z
h
1.8h = Yt2 -~22
Ittl--+---l
Air
-7 multiply both sides by It/ r2
>Om
h
1
I 2.t
B_r_i"- - - -~ne
ll~i13
m
._______ __
Figure (a)
Itt = 13.86- 1.665
! .
..,_,i:- - _ _ % - L_
Solve for h1 :
h1 = h - y2
w2r2
h =--
2g
h = 13.86 m
(7n) 2 (0.75) 2
2(9.81)
-7 Eq. (4)
7.2yz = 1.8(13.86)- 12.96
yz = 1.665 m
..
Figure (b)
But y1 = 3.6 + yz
1.8/t = (3.6 + yz}2 - yz2
1.8/z = 12.96 + 7.2yz + yz2 - yl
7.2 y2 = 1.8/z- 12.96
__%c_
/t 1 = 12.195 rn
p = 9.81(1.3)(12.195)
p = 155.52 kPa
5 - (155.52)(1500)
t-
2(9)
S, = 12,960 kPa
S, = 12.96 MPa (maximum wall stress)
231
CHAPTER FOUR
Relative Equilibrium of Liquids
232
IIIlO MECHANICS
I IYORAULICS
(b) For maximum value of w, S, = 85 MPa
CHAPTER FOUR
Relative Equilibrium of liquids
233
3
p2 (1.8x10
82 X 103 = ..:....::....:.._
__-'-)
2(5)
pz = 455.5 kPa
, = 9.81(1.3)1!1 (1500)
85 10
2(9)
h1 = 79.98 m
.'/2= 11 -79.98
X
I' yhz
I'>J.5 = 9.81(1.6)11z
In Eq. (4)
7.2(11 - 79.98) = 1.811 - 12.96
5.411 = 562.896; II = 104.24 m
00 2r2
'"
29.02 m
/1
hz- pl/y- 2.7
29.02- 2.7 -15.61
10.71 m
,,
/1
fh= - - J
2g
,, 0>2 (0.9)2 = 10.71
2(9.81)
104.24 = 0>2(0.75)2
2(9.81)
w = 60.3 radjsec x 30/n
w = 576 rpm
"'
111
16.1rad/secx ~
153.8 rpm (maximum allowable angul~r speed)
''l>blem 4 - 23
Problem 4 - 22
A 1.8 m diameter closed cylinder, 2.7 m high is completely filled with glycen n
having sp. gr. of ~ .6 under a pressure of 245 kPa at the top. The steel plate
lu 1.5 m diameter impeller of a closed centrifugal water pump is rotated at
110 rpm. If the casing is full of water, what pressure is developed by
It lion?
which form the cylinder arc 5 mm thick with an ullimate tensile stress of 82
MPa. How fast can it be rotated about its vertical axis to the point of bursting?
nlutlon
Solution
..:'•
00 2r2
ll=-2g
Solve for h
!2 =
y
r =0.9 m
~:·
'.
.
I
:
I
245
9.81(1.6)
l'rcssure head, E. = II
y
::
~(J).::
0
.
w2 r2
h=-2g
h
'
'
0
,•
--;----r-
j - •• _·
0
pJy
I
i---:-::
10;...__-;j I
f2 = 15.61 m
y
•
0> = 1500 X 1tj30
w = 507t r.a d/ sec
'' = (507t)2 (0.75)2 = 707.4 m
i
i
2(9.81)
I
i
The maximum tensile stress occurs
at point 6 :
From S, = pD
21
Gly~rin
s =j1.6
2.7 m
I
i
I
I
I
j
E._ = 707.4 m of water
9.81
I' "' 6,940 kPa
234
CHAPTER FOUR
Relative Equilibrium of liquids
I LUID MECHANICS
1. HYDRAULICS
Problem 4 - 24 (CE Board)
A conical vessel with sides inclined~Oo with its vertical axis is revolved
another axis 1 m from its own a d parallel. How many revolutions
minute must it make in order tlj at water poured into it w ill be
discharged by the rotative effect?
Solution
T he water in the vessel will entirely be
d ischarged at a speed when the
paraboloid is tangent to the cone at the
vertex, hence, the inclination, 9, of the
paraboloid at x = 1 m is 60° or its slope is
tan 60°.
"iolution
235
t
p2 = Y 1!2
I
I
/I
~"' = 27.5 rad/sec
Solving for 1!2:
h2 = !/2- !/1
w2x22 - w2x12
ft2= - - 2g
2g
2
275
It 2 = (
) [(2.5)2- (0.5)2 ]
2(9.81)
/12 = 231.27 m
/I
I
1
I
I
I
I
I
I
I
I
I
I
I
~ 0.5 ~
-
I
.-
- - : -e -
I
//
ol
1
......_ ...._
p = (9.81 X 0.822)(231.27)
p = 1,865 kPa
From the formula:
w2
tan 0 = - x
"'
I
2m
6
1
,..,a::=======~yr:-,
-
-
-
~.5-m- - - - - - >J j - ~
J'roblem 4 - 26
lm
g
w2
tan 60° = - - (1)
9.81
w = 4.12 rad /sec x :>O
1t
CHAPTER FOUR
Relative Equilibrium of Liquids
I
I
'-
.....
....__,_..- ,.., /
/
I
w = 39.36 revolutions per minute
Problem 4- 25 {CE November 1992)
A 75 mm diameter pipe, 2 m long is just filled with oil (sp. gr. = 0.822)
then capped, and placed on a horizontal position. lt is rotated at 27.5
about a vertical axis 0.5 m from one end (outside the pipe). What is
pressure in kPa at the far end of the pipe?
glass U-tube w hose vertical stems are 300 mm apart is filled with mercury
lt is rotated about a vertical axis
1hrough the m idpoint of the horizontal section. What angular speed OJ will
l'loduce a pressure of absolute zero in the mercury at the axis?
In c1 depth of 150 nm1 in the vertical stems.
.,olution
w2x2
y =-2g
!/=It,+ 0.15
x = 0.15 m
Since the pressure at the center
is absolute zero, then the gage
pressure at the center is -palm or
-760 mmHg, therefore""' = 0.76 m
y = 0.76 + 0.15
y = 0.91 m
0.91 = w2(0.15)2
2(9.81)
w = 28.17 rad/sec x 3~
w = 269rpm
r =0.15 m
,~---------
I
I
i
r = 0.15 m
-----1------------··;;
PobS = 0
\
\
/
I
\\
I
',
I
II
/
T
+
0.15 m
hm
',,,_~~/.:-.~ -·--·-- _t_
CHAPTER FOUR
Relative Equilibrium of liquids
236
CHAPTER FOUR
Relative Equilibrium of Liquids
II UID MECHANICS
I. HYDRAULICS.
Problem 4 - 27
olution
A glass U-tube whose vertical stems are 600 mm apart is filled with mere
to a depth of 200 mm in the vertical stems. It is rotated about a vertical a
through its horizontal base 400 mm from one stem. How fast should it
mtated so that the difference in the mercury levels in the stems is 200 mm?
Solution
level
level
J--- x2 =0.4m
\
--,
l
~2m
In the figure shown:
.'/2- .lfl = 0.2
(1)2X2
wherey= - 2g
(1)2X 2
(1)2X 2
2g
2g
= 0.2
00
-
2(9.81)
[(0.4)2- (0.2)2] = 0.2
w = 5.72 radjsec x 30
w = 54.6rpm
' ... ...
"
Problem 4 - 28
t\ glass tubing consist of 5 vertical stems which are 500 mm apart connected to
,, single horizontal tube. The tube is filled with water to a depth of 500 mm in
the vertical stems. How fast should it be rotated about and axis through
middle stem to just zero the depth of water in that stem?
I
I
..
I
I
~I, 0.5 m .:.
I
.. , 1-r
;
i x1 =0.5 m
I
·I
y,
0.5 m
=
1nce there is no liquid spilled out, its initial volume is equal to its final
nlume or as shown in the figure, the sum of the height of water in the vertical
l••ms before and after rotation must be equal.
2yt + 2y2 = 5(0.5)
Yt + y2 = 1.25
-7 Eq. (1)
By squared property of parabola:
(1)2 = (0.5)2
2
-
0.5 m
I
I
y~
,,
I
----.,.---
- ----- --- ---------
==-:::..;:_---H--..!..:....-L-.
l!:=======~~
__
2_ - _ _1_
''
4r
_t_ - '<~~~ ---:r----~-~-~ 1
I
\
---~----
---+~x, =0.2m-
l
·
02
I
\
Jl~ I\\
'
I
I
I
'
~)
Initial water
~w
Initial mercury
237
Y2
y2 = 4yt
Yt
-7 Eq. (2)
Substitute y2 to Eq. (1)
.'/1 + 4yl = 1.25
.'Jl = 0.25
(1)2X12
.'/1= - - 2g
0.25 = (1)2 (0.5)2
2(9.81)
w = 4.43 rad/ sec x 3~
w = 42.3rpm
238
CHAPTER FOUR
Relative Equilibrium of liquids
dh'
A 75 mm diameter pipe, 2m long is filled with water and capped at both
It is held on a plane inclined 60° with the horizontal and rotated about
vertical axis through its lower end with a constant angular speed of 5 rad/
(a) Compute the pressure at the upper end of the pipe and (/?) determine
minimum pressure and its location in the pipe.
Since there is no initial pressure
in the pipe the pressure head at
the lower end of the pipe will
remain equal to the static
pressure head of 1. 73 m, and
therefore the vortex of the
paraboloid will be 1.73 m above
the lower end.
w2r2
'•
= 2.548x - tan 60° = 0
-
dx
x = 0.68 m
11 = x sec60°
a= 0.68 sec60° = 1.36 m
In Eq. (1):
II'= 1.274(0.68)2 + 1.73- (0.68) tan60°
It'= 1.141 m
/1nnn = 9.81(1.141)
Pmm = 11.196 kPa located 1.36 m from the lower end (along the pipe)
Solution
tm
I
"' = 5 rad/s
l'toblem 4- 30
l
ylindrical bucket 150 mm in diameter and 200 mm hig h contains 150 mm of
1ler. A boy swings the bucket on a vertical plane so that the bottom of the
h=--
ll ket describes a circle of radius 1 m. How fast should it be rotated so that
' 'water will be spilled?
2g
h = (5)2 (1)2
2(9.81)
II= 1.274 m
'•olution
0.1 m 0
(a) Pressure at the upper end
/
/
/
w
/
Pupper = yh
~
I
Popper= 9.81(1.274)
I~~
p"''""' = 12.497 kPa
I
(b) Minimum pressure
I
I
I
10-- X ----..:
Pmm = yh'
~
/
+
I
I
CF
.,.
\
Solve for 11'
\
Jz'=y+z
z = 1.73- x tan60°
y
239
Minimize IJ', differentiate Eq. (1) with respect to x and equate to zero:
Problem 4 - 29
x2
CHAPTER FOUR
Relative Equilibrium of Liquids
I I UID MECHANICS
liYDRAULICS
12
I!
\
I
\
''
12
1.274
y = 1.274 x2
h' = 1.274 x2 + (1.73- x tan60°)
I
/
/
''
---.
__ """
Figure (a)
-? Eq. (1)
.......
/
/
/
Figure (b)
240
CHAPTER FOUR
CHAPTER FIVE
IIJID MECHANICS
I IYDRAULICS
Relative Equilibrium of Liquids
Fundamentals of Fluid Flow
241
The critical position for the liquid to fall is at the highest point
hapter 5
From Figure (b)
CF=W
CF =Mn,,
~
undamentals of
luid Flow
w
L F = - CJ12 r
~
vv
-(J) J r = W
g
<1)2
J., previous chapters deals only with fluids at rest in which the only
I 111ficant property used is the weight of the fluid . This chapter will deal with
itud~ in motion which is based on the following principles: (a) the principle of
lht'rvation of mass, (b) the energy principle (the kinetic and potential
111 •gies), and (c) the principle of momentum.
r=g
(1) 2 (0.925) = 9.81
w = 3.26 radl sec x
c11 =
Ju
31.13 rpm
Problem 4- 31
A cubical tank 1s filled with 2 m of oil having sp. gr. of 0.8. Find the
acting on one side of the tank when the acceleration is 5 ml s2 (a) vertically
upward, and (b) vertically downward
Ul"iCHARGE OR FLOW RATE, Q
1 •
harge or flow rate is the amount of fluid passing through a section per
This is expressed as a mass flow rate (ex. kg/sec), weight flow rnte
kN/ sec), and volume flow mte or flow rate (ex. m 3 Is, lit/ s).
11111 of time.
Solution
(a)
{b)
F = P<K A
F = [y llcg (1 + rv'g)J A
F = (9.81 X 0.8)(1)(1 + 519.81)[(2) :)j
F= 47.392 kN
F=p<~A
''
'
2m
ca:
~
..../·---···-··· ··---
F = [y h<g(1 - a/g)J A
F = (9.81 " 0.8)(1)(1 - 5/9.81 )(2x2)
F = 15.392 kN
2m
Volume flow rate, Q = Av
Eq. 5-1
Mass flow rate, M = p Q
Eq. 5-2
Weight flow rate, W= y Q
Eq. 5-3
where:
Q = discharge in m 3f s or ft3 Is
A = cross-sectional area of flow in m2 or ft2
v = mean velocity of flow in ml s of ft/ s
p =mass density in kg/m3 or slugs/ft3
y =weight density in Njm3 or lb/ft3
It FINITION OF TERMS
lhud Flow may be steady or unsteady; uniform or non-unifonu; continuous;
,..,llrlr or turbulent; one-dimensional, two-dimensional or three-dimensional; and
•ltll1011al or irrotational.
242
CHAPTER FIVE
Fundamentals of Fluid Flow
f I lJID MECHANICS
I tYDRAULICS
CHAPTER FIVE
Fundamentals of Fluid Flow
243
Steady Flow
I urbulent Flow
This occurs when the discharge Q passing a given cross-section is constant
time. If the flow Qat the cross-section varies w ith time, the flow is ullsteady.
h· flow is said to be turbulent when the path of individual particles are
11 ~ular and continuously cross each other. Turbulent flow normally occurs
lwn the Reynolds number exceed 2,100, (although the most common
1u.1tion is when it exceeds 4000).
Uniform Flow
This occurs if, with steady flow for a given length, o r reach, of a stream,
average velocity of flow is the same at every cross-section. This usually
when an incompressible fluid flows through a stream with uniform
section. In stream where the cross-sections and velocity changes, the flow
said to be non-uniform.
I 11111nar flow in circular pipes can be maintained up to values of Rr as high as
•11100. However, in such cases this type of flow is inherently unstable, and
Ill· least disturbance will transform it instanUy into turbulent flow.
On the
thl'r hand, it is practically impossible for turbulent flow in a straight pipe to
1 1 ~ist at values of Rr much below 2100, because any tu rbulence that is set up
til be damped out by viscous friction
Continuous Flow
This occurs when at any time, the discharge Qat every section of the stream
the same (principle of conservation of mass).
•no-Dimensional Flow
nu., occurs when in an incompressible fluid, the direction and magnitude of
th velocity at all points are identical
two-Dimensional Flow
Q
Q
lluo; occurs when the fluid particles move in planes or parallel planes and the
11 t'r\mline patterns are identical in each plane.
Figure 5-1
Continuity Equation
trcamlines
For incompressible fluids:
llwse are imaginary curves drawn through a fluid to indicate the direction of
1notion in various sections of the flow of the fluid system.
For compressible fluids:
treamtubes
PtAtVt = P~2v2 = p~3V3 =constant
or
llwse represents elementary portions of a flowing fluid bounded by a group
11 ,treamlines which confine the flow .
=Y~2V2 = Y~3V3 =constant
fl ow Nets
Laminar Flow·
The flow is said to be laminar when the path of individual fluid particles
not cross or intersect. The flow is always laminar when the Reynolds
R, is less than (approximately) 2,100.
llll'se are drawn to indicate flow patters in case of two-dimensional flow, or
't•n three-dimensional flow.
244
CHAPTER
CHAPTER FIVE
Fundamentals of Fluid Flow
11110 MECHANICS
IIYDRAULICS
ENERGY AND HEAD
The energy possessed by a flowing fluid consists of the kinetic and the
energy. Potential energy may in turn be subdivided into energy due
position or elevation above a given datum, and energy due to pressure in
fluid. The amount of energy per pound or Newton of fluid is called the ileac/
245
'' sure Energy (Potential Energy)
' " 1der a closed tank filled with a fluid which has a small opening at the top
••hout pressure at the top, the fluid practically will not flow . In Chapter 2.
1 ••quivalent head (pressure head) for a pressure of p is p/Y. Hence the
.ure energy is equivalent to:
Pressure Energy = WE.
Kinetic Energy
Eq. 5- 12
y
The ability of the fluid mass to do work by virtue of its velocity.
Pressure head = Pressure Energy = E.
w
w
K.E. = 1/2 M v2 = 1/2- v2
g
. or veIOClty
. hea d = K.E.
Kine t1c
- = -v
w
Eq. 5 -13
y
where.
z =position of the fluid above(+) or below (-) the datum plane
p = fluid pressure
ll = mean velocity of flow
2
2g
For circular pipe of diameter D flowing full:
v2
(QIA? - Q2
2g
2g
- 2gA 2
1•1tal Flow Energy, E
I Ill' to.tal energy or head in a fluid flow is the sum of the kinetic and the
t•nll•ntial energies. It can be summarized as:
Total Energy= Kinetic Energy+ Potential Energies Eq. 5- 14
v2
p
Total Head, E = + - +z
2g
y
Elevation Energy (Potential Energy)
The energy possessed by the fluid by virtue of its position or elevation with
respect to a datum plane.
I'OWER AND EFFICIENCY
1nwer is the rate at which work is done. For a fluid of unit weightY (Nfml)
11 d moving at a rate of Q (m3fs) with a total energy of E (m). the power inNm/ s Qoule/ sec) or watts is:
Power= Qy E
Elevation Energy = Wz = Mgz
Elevation head = Elevation Energy = z
w
Eq. 5-10
Efficiency, 11 =
Eq. 5-11
Eq. 5- 15
Output
x 100%
Input
Note: 1 Horsepower (hp) = 746 Watts
1 Horsepower (hp) = 550 ft-lb/sec
. 1 Watt= 1 N-m/s = 1 Joule/sec
, Eq.5-16
Eq. 5-17
246
CHAPTER FIVE
Fundamentals of
CHAPTER FIVE
Fundamentals of Fluid Flow
II UID MECHANICS
HYDRAULICS
Flow
I nurgy Equation with Head Lost:
BERNOUlli'S ENERGY THEOREM
The Bernoulli's energy theorem results from the application of the pnnczp/,•
··niiSPI'llafion of Pllrrgy. This equation may be summarized as follows:
•nsidering head lost, the values that we can attain are called actual values
1th reference to Figure 5-4:
Bernoulli's Princ1ple, 1n phys1cs, the concept
that as the speed of a moving fluid (liquid
gas) increases, the pressure within that
decreases. Originally formulated In 1738 by
Swiss mathematician and physicist Daniel
Bernoulli, It states that the total energy In •
steadily flowing fluid system is a constant
along the flow path. An Increase in the
speed must therefore be matched by a
decrease in Its pressure.
2
Figure 5-2
z,
-_'l ~ -------------
E1 - HL1.2 = E2
Eq. 5-21
p
V 2
p
+ - 1 + z, = - 2 - + - 2 + Z2 + llLt-2
2g
y
2g
y
Eq. 5-22
Vt2
G
r
>«liOn 2
---- -- ----- - -- - - -
rade Line, EGL
v12/2g
Datum
_HYdrauliC Grade L.
·-
- · - · - . ..!!!~·~L
. ........
pJy
Energy Equation without Head Lost:
If the fluid experiences no head lost in movmg hom section I to sechon 2
the total energy at section 1 must be equal to the total energy at section
Neglecting head lost in fluid flow, the values that we get are called ideal
lilPorpticalr,alues . With reference to Figure 5- 3·
2g
',_i
P1/Y
11uure 5-4
zl,
p
v 2
p
+ _ l + ZJ = - 2 - + _2 + Z2
y
2g
y
Energy grade Line, EGL
• . . - .. - .. - ..
l
vlf2g
I norgy Equation with Pump:
I •tmp is used basically to increase the head. (Us ually to raise water from a
I•• er to a higher elevation). The input power (Pmput) of the pump is electrical
n ·rgy and its output power (Poutput) is the flow energy.
Llt=======~nfo~~-~~
PUMP
z,
Figure 5-3
L_
0
Datum
+
-?.a~~ . - .. - ..- .. - .. - .r.
0
f)
z,
- ·· - ·· -··-i
j
v//2g
0
E1 = E2
v12
247
0
Figure 5-5
oc
::..
248
CHAPTER FIVE
Fundamentals of Fluid Flow
CHAPTER FIVE
Fundamentals of
+ HA - HLt-2 = E2
249
h racteristics of HGL
•
v 2
2
- 1- + E.!. + Zl + HA = !:L + !!1.. + Z2 + HLl-2
2g
y
2g
y
HGL slopes downward in the direction of flow but it may rise or fall due
to changes in velocity or pressure.
• For uniform pipe cross-section, HGL is parallel to the £GL.
• For horizontal pipes with uniform diameter, the drop in pressure heads
between any two points is also equal to the head lost between these
points.
=QyHA
Energy Equation with Turbine or Motor:
Turbines or motors exb·act flow energy to do mechantcal work which ,11
converted into electrical energy for turbines
11
rgy Grade Line (EGL)
lt••rgy grade line is a graphical representation of the total energy of flow (the
Its distance from the datum plane is
0
0
11111 of kinetic and potential energies).
+ E. + z.
y
11 racteristics of EGL
~
~====0C
IIO JM~====~
TURBINE
o
EGL always slope downward in the direction of flow, and it will only
rise with the presence of a pump.
The drop of the EGL between any two points is the ~ead lost between
those points.
• For uniform pipe cross-section, EGL is parallel to the HGL.
o
Figure 5-6
E, -HE- HL1-2 = E2
v
2
P1
2
P2
- 1- + - + Zt = - - + + Z2 + HLt-2 +HE
2g
y
2g
y
.
V2
EGL is always above the HGL by an amount equal to the velocity head,
v2/ 2g.
• Neglecting head loss, EGL is horizontal.
o
t Power of Turbine= Q y HE
ENERGY AND HYDRAULIC GRADE liNES
p/y
Hydraulic Grade Line (HGL)
Also known as pressure gradient, itydraulu grade /we ts the graphical
representation of the total potential energy of flow. It is the line that connects
the water levels in successive piezometer tubes placed at intervals along tht>
pipe. Its distance from the datum plane is E.+ z
y
(1)
b
e
TURBINE
9
Figure 5 - 7: Illustration showing the behavior of energy and hydraulic grade lines.
250
CHAPTER FIVE
Fundamentals of
CHAPTER FIVE
Fundamentals of Fluid Flow
Flow
olved Problems
W=yQ = yAv
Problem 5-1
y= _L
Water flows through a 75 mm diameter pipe at a velocity of 3 mjsec. Find
the volume flow rate in m 3/sec and lit/ sec, (b) the mass flow rate in kg/
and (c) the weight flow rate inN/sec.
110 X 103
12 39
Y= 29.3(30 + 273) = . N/m'
Q = Av
= f (0.075)2(3)
= 0.013 m 3/s x 1000 lit/m·l
Q = 13 lit/sec
M = pQ
= 1000(0.013)
M = 13 kg/sec (mass flow rate)
(b)
RT
20 = 12.39(0.16(0.32)]v
v = 31.53 rnjs (average velocity) .
Solution
(n)
251
Q=Av
= (0.16)(0.32)(31.53)
Q = 1.614 m 3jsec (volume flux or discharge)
l'roblem 5-4
100-mm diameter plunger is being pushed at 60 mmjsec into a tank filled
tlh oil having sp. gr. of 0.82. If the fluid is incompressible, how many N/s of
·•I is being forced out at a 30-mm diameter hole?
W=yQ
= 9810(0.013)
W = 127 N/sec (weight flow rate)
(c)
\olution
·•nce the fluid is incompressible:
Q,=Q2
Q1=A1V1
What is the rate of flow of water passing through a pipe with a diameter of
mm and speed of 0.5 m/ sec?
=
t (0.1)2(0.06)
Q 1 = 0.00047mJjs
011
s c 0.82
100 mm 12l,--
Plunger
v= 0.06 m/s
0
I
.___
Q2 = 0.00047 m3js
Solution
Flow rate, Q = Av
Q = f (0.02)2(0.5)
Q = 1.57 x 1()-4 mljsec
Air at 30oC and 110 kPa flows at 20 N/s through a rectangular duct
measure 160 mm x 320 mm. Compute the average velocity and volume
Use gas constant R = 29.3 m;oK.
W=yQ
= (9810 X 0.82}(0.00047)
Hole
J
30 mm f2l
W= 3.78 N/s
t•roblem 5 - 5
II the velocity of flow in a 75-mm diameter fire hose is 0.5 mjs, what is the
••locity in a 25 mm diameter jet issuing from a nozzle attached at the end of
the pipe. Compute also the power available in the jet.
252
CHAPTER FIVE
Fundamentals of
Flow
CHAPTER FIVE
Ill) MECHANICS
IVORAULJCS
Fundamentals of Fluid Flow
Solution
Sm0
By continuity equation·
Qho><• = Q,.,
A~~.u, = A1 v1
W=y x Volume
= 9.81 X f (5)2(10)
f (0.075)2 (0.5) = f (0.025)2 v,
v, = 4.5 m/s
W = 1,926.2 kN
(velocity of the jet)
PE = 1,926.2 X 7
PE = 13,483.32 kN-m
Power, P = Q y f.
Q=Av
Q = f (0.025) 2 (4.5) = 0.002209 mljs
v2
4.5 2
£ = - = - - =1.0321 m
2g
2(9.81)
Power, P = 0.002209(9,810){1.0321)
Power, P = 22.37 watts (power available in the jet)
253
-r-+--•
,l
<g
~
1
+
10m
_L.~~~~-·-·t_.
eublem 5-8
lo•rmine the kinetic energy flux of 0.02 rn3 /s of oil (sp. gr.
lr . harging through a 50-mm diameter nozzle.
0.85)
,Jutlon
~ inetic energy flux
= Kinetic Energy per second = Power
Power, P = Q y E
Q = 0.02 m3js
A turbine is rated at 600 hp when the flow of water through it is 0.61 m3/
Assuming an efficiency of 87%, what is the head acting on the turbine?
Solution
Given:
v2
E=2g
v=Q =
A
Power output= 600 hp
Efficiency, 11 = 87%
.
Power mput
= -600
- = 689.655 h p
0.87
Power input= 514,483 watts
Power input= Q y HE
514,483 = 0.61(9,810)H£
HE= 85.97m
Problem 5-7
A standpipe 5 min diameter and 10m high is filled with water. Calculate the
potential energy of the water if the elevation datum is taken 2 m below the
base of the standpipe
0.02
{-(0.05) 2
v = 10.186 mjs
E = (10.186)2 = 5.288 m
2(9.81)
p = 0.02(9810 X 0.85)(5.288)
P = 882watts
t••oblem 5-9
1 wglecting air resistance, determine to what height a vertical jet of water could
11 ,. if projected with a velocity of 20m/ s?
CHAPTER FIVE
Fundamentals of
254
CHAPTER FIVE
Fundamentals of Fluid Flow
Flow
Solution
Ptublem 5 - 11 r
As the jet nses, its kmetic energy is transformed into potential energy
Neglecting air resistance
1
1
KE = PE
1/2 M112 = W h
w
1/2 - v 2 = Wit
h
g
[12
h=-
2g
h=
255
2 0) 2 = 20.4 m
2(9.81)
(
p1pe carrying oil of specific gravity 0.877 changes in size from 150 mm at
I ton 1 and 450 mm at section 2. Section 1 is 3.6 m below section 2 and the
,, .. ures are 90 kPa and 60 kPa respectively. If the discharge is 150 lit/sec,
I t.•rmine the head lost and the direction of flow.
450mm 0
nlution
Q1 = Q2 = 0.15 m3 /s
n1 =
0 15
· 2
f(0.15)
112 =
0 15
·
= 0.943 m/s
2
f(0.45)
= 8.49 m/s
0
p1 = 90 kPa
Water is flowing m an open channel at a depth of 2 m and a velocity of 3
It flows down a chute into another channel where the depth is 1 m and
velocity is 10 m/s. Neglecting friction, determine the difference in elevation
the channel floors .
Solution
EGL
laking 0 as datum:
v1 2
p1
8.49 2
90
+0
El = - - + + Z] = - - - +
2g
y
2(9.81)
(9.81 X 0.877)
Et = 14.135 m
2
Ez = ~ + !!2. + Z2
.
2g
= 0.943
y
2
2(9.81)
+ _ _6_0_ _ +3.6
(9.81 X 0.877)
E2 = 10.62 m
Since E1 > EZJ the flow is from 1 to 2
Head Lost, HL = E, - Ez = 14.135- 10.62
Head Lost, HL = 3.515 m
Neglecting friction
(head lost):
£, = £2
v 2
v 2
+ 2 + z = - 2- + 1
2g
2g
1
- -
32
10 2
+1
2(9.81)
+2+z=---
2(9.81)
z = 3.64 m
CHAPTER FIVE
Fundamentals of
256
CHAPTER FIVE
Fundamentals of Fluid Flow
f 1 UID MECHANICS
I HYDRAULICS
uid Flow
Problem 5- 12
M = P2A2v2
p2[(0.3)(0.3)](2) = 0.1575
p2 = 0.875 k!Vm3
(mass density at section 2)
Oil flows from a tank through 150 m
of 150 mm diameter pipe ahd then
discharges into air as shown in the
Figure. If the head loss from point 1
to point 2 is 600 mm, determine the
pressure needed at point 1 to cause
17 1it/sec of oil to flow
0
E
E
-
Air, p • '
Sl
I
E
Oil
5
0
=0.84
~
Solution
Q = 0.017 mljs
l•r oblem 5 - 14
, 1ter flows at the rate of 7.5 m/s through 75-mm diameter pipe (pipe 1) and
1 1vcs through 50-mm diameter and 65-mm dia~eter pip~s at the rate of 3
111 js and 3.5 m/s, respectively as shown in the F1gure .. Au at the top of the
1 111 k escapes through a 50-mm-diameter vent.
Calculate dh/ dt ~d the
docity of air flow through the vent. Assume the flow to be incompressible.
Energy equation between 0 and 6 ·
£, - HL1-2 = £1
v~ =?
0 4 =50 mm
2
u/ +P1- + z,- HL1- 2 =
V2
P2
- + - + z2
-
2g
2g
y
Air
~
~1
y
2
0 + f!J_ + 20- 0.6 = B(0.017)
+ 0 + 30
2
4
Y
rr g (0.15)
h
y
Water
v1 = 7.5 m/s
= 10.65 m of oil
0 2 =50 mm
Pt = 10.65{9.81 X 0.84) = 87.76 kP~
v2 = 3 m/s
v3 = 3.5 m/S
Problem 5 - 13
Gas is flowing through a square conduit whose section gradually changes from
ISO mm (section 1) to 300 nu11 (section 2). At section 1, the velocity of flow is 7
m/s and the density of gas is 1 kgjm3 while at section 2 the velocity of flow is 2
m/ s. Calculate the mass flow rate and the density of the gas at section 2
Solution
._olution
Assuming the flow to be incompressible:
Qm = Qout
f (0.075)2(7.5) = (0.05)2 (3) + (0.065)2(3.5) +
t
t
dh/ dt = 0.0553 m/s
150 mm
"' a 1 k9fm'
257
a:=:::> M
V> • ?
: D ~·:------ -·-- -·- -·-·-·-·- -·- -·- -·-
-· j·-0:
f)
M=pQ
M = p,A,v,
= 1[(0.15)(0.15))(7)
M = 0.1575 kwsec
{mass flow rate)
300 mm
Considering the air above the tank:
[Q4 = Q.;,]
T(0.05)2V4 = T(0.6)2 dh/ dt
f (0.05)2v4 = f (0.6)2(0.0553)
v 4 = 7.963 mfs
(velocity of air flow)
t (0.6) 2dlz/ dt
258
CHAPTERFI)
Fundament:of\luid Flow
Problem 5 - 15
lllution
A liquid having sp. gr. of 2.0 is flowing in a 50 nun diameter pipe.
head at a given point was found to be 17.5 Joule per Newton. The elevation
the pipe above the datum is 3 m and the pressure in the p ipe is 65.6
Compute the velocity of flow and the horsepower in the stream at that point.
Q1 = Q2 = 0.03 m 3/s
8Q2
2g - n2gD4
v2
v12 =
Solution
2g
v2
Total energy, E = -
2g
v2
2g
9.81(2)
=11.156m
v = 14.79 m/s
v2 2 =
y
17.5=~+~+3
-
8(0.03)2
n 2(9.81)(0.2)
= 0.0465 m
4
p
+ - +z
E = 17.5 ]oule/N x (1 N-m/Joule)
E = 17.5m
2g
CHAPTER FIVE
Fundamentals of Fluid Flow
I I UID MECHANICS
I HYDRAULICS
(velocity of flow)
Power, P = Q y E
= [ (0.05)2(14.79)] X (9810 X 2) X 17.5
t
= 9970.92 watts x (1 hp/746 watts)
Power, P = 13.37 hp
2g
8(0.03)2
n 2(9.81)(0.15)
0.147 m
4
Energy Equation between A and B:
EA- HLA-1+ HA - HL1.a = Ea
v 2
PA
Vg2
PB
_A_ + + ZA- HLA-1 + HA - HL2-B = - - + + ze
2g
r
2g
y
0 + 0 + 10- 2(0.0465) + I-IA- 10(0.147) = 0 + 0 + 60
JJA = 51.563 m
Power output= Q y HA = 0.03(9,810)(51.563)
= 15,175 watts x (1 hp/746 watts)
Power output= 20.34 horsepower (rated power of the pump)
Pressure heads at 1 and 2:
Energy Equation between A and 1:
EA- HLA-1 = El
2
v/
PA + ZA- HLA-1 = V1
P1
-- + - + - + Zl
2g
y
2g
y
The pump shown draws water from reservoir A at elevation 10 m and lifts it
reservoir 8 at elevation 60 m. The loss of head from A to 1 i~ two times
velocity head in the 200 mm diameter pipe and the loss of head from 2 to B
ten times the velocity head in the 150 mm diameter pipe. Determine the
horsepower of the pump and the pressure heads at 1 and 2 in meters when
discharge is 0.03 m3 /sec.
0 + 0 + 10- 2(0.0465) = 0.0465 + El. + 0
y
El. = 9.86 m of water
y
Energy Equation between of 2 and B:
£2- HL2.s = Ea
2
2
!:L + E1.. + Z2 - Hha = .!2_ + E..!L + za
2g
r
2g
y
0.147 + E1.. + 0- 10(0.147) = 0 + 0 + 60
y
E1.. = 61.323 m of water
y
259
260
CHAPTER FIVE
Fundamentals of Fluid Flow
CHAPTER FIVE
Fundamentals of Fluid Flow
r I UID MECHANICS
'• HYDRAULICS
Problem 5- 17 (CE November 1986}
HL2 = 9160.13 Q 2
A pipeline w ith a pump leads to a n ozzle as shown. Find the flow rate
pump develops an 80 ft (24.4 m) h ead . Assume head lost in the 6-inch
mrn) pipe to be five times its velocity head while the head lost in the
(102 mrn) pipe to be twelve time its velocity head. (a) Compute the flow
(b) sketch the energy grade line and hydraulic grade line, and (c) find
pressure head at the suction side.
v/
El 80' (24.'1 m)
o===n::=B
3"jet
(76.2 mm)
2g
'i/A
Line 2
7t
g(0.0762) 4
= 2450.8 Q 2
0 + 0 + 21.3-773.96 Q2 + 24.4 - 9160.13Q2 = 2450.8 Q 2 + 0 + 24.4
12,384.89 Q2 = 21.3
~ Discharge
Q = 0.0415 ml/s
(b) En ergy and Hydraulic grade lines:
v, 2
2g
El 70' (21.3 m)
8Q2
2
261
7t
8(0.0415)2
= 0.266 m
2 (9.81)(0.152) 4
v2 2 =
8(0.0415)2
= 1.31 m
4
2
2g
rr (9.81)(0.102)
v82 =
8(0.0415) 2
= 4.22 m
4
2
2g
rr (9.81)(0.0762)
HL1 = 773.96 Q2 = 1.33 m
HL2 = 9,160.13 Q 2 = 15.78 m
Solution
(a) Discharge
El. 28.59 m
v22/2g • 1.31
Q, = Q2 = Qll = Q
Energy Equation between A and B:
£A - HL, + HA - HL2 = £11
VA2
p
V 2
p
+ ~ + ZA - HLI + HA - HL2 = - 8 - + _!I. + ZR
2g
y
2g
y
HA = 24.4 m
ll 2
8Q2
HLI =5-1- = 5__,....;..:::..._
2
2g
rr gD1 4
=5
8Q2
rr 2 (9.81)(0.152) 4
HLI = 773.96 Q2
HL2 = 12
1122
= 12
2g
= 12
8
2
Q
rr2 gD2 4
8Q2
2
4
rr (9.81)(0.102)
HL2 = -15.78m
v, /2g • 1.31
2
:::1
..,:
II
l?J
~
CC==~~B!~B~80~'~(24~.4~m~)~~
3" jet
(76.2 mm)
CHAPTER FIVE
Fundamentals of Fluid Flow
262
(c) Pressure head at S
Energy Equation between A and S
EA- HL, = Es
2
E,- HE= E2
2
2g
y
v 2
v 2
2g
2g
_i_ = - 1-
263
lmergy Equation between 1 and 2
(neglecting head lost and taking pmnt 2 as datum)
.!:.L + ~ + ZA - HL, = !!..L + h
2g
CHAPTER FIVE
Fundclmcntals of Fluid Flow
t I UID MECHANICS
f IYDRAULICS
y
+ zs
p
v,2
v 22
p2
+ - 1 + Zl - HE = - - + 2g
y
2g
y
+ Z2
2
8(0.5) 2
14
8(0.5)
4
--::----'-'-----:- + - - + 2.5- HE= 2
+ -- + 0
4
rr 2 (9.81)(0.6) 4
9.81
rc (9.81)(0.9)
9.81
= 0.266 m
0 + 0 + 21.3 -1.33 = 0.266 + 1!1_ + 15.2
HE= 3.647 m
E.5... =4.504 m
Power, P = Q y HE
= 0.5(9810)(3.647) = 17,888.5 watts x (lhp/746 watts)
Power, P = 23.98 horsepower
y
y
Or from the figure shown above, the pressure head at Sis the vertical
from the pipe to the HGL.
l'roblem 5 - 19 (CE May 1979)
1!1_ = 19.704-15.2
y
\ 20-hp suction pump operating at 70% efficiency draws water from a suction
l!.i. = 4.504 m
:ntl, whose diameter is 200 mm and discharges into air through a line whose
1J,1meter is 150 mm. The velocity in the 150 mm line is 3.6 m/s. lf the pressure
,, point A in the suction pipe is 34 kPa below the atmosphere, where A is 1.8 m
y
•low B on the 150 mm line, determine lhe maximum elevation above B to
,,·hich water can be raised assuming a head loss of 3 m due to friction. .
1 ..
5 - 18 (CE Nove
-Q
Water enters a motor through a 600-mm-diametcr pipe under a pressure of
kPa. It leaves through a 900-mm-diameter exhaust pipe with a pressure of
kPa. A vertical distance of 2.5 m separates the centers of the two pipes at
sections where the pressures are measured. lf 500 liters of water pass
motor each second, compute the power supplied to the motor.
'tolution
2g
t (0.15) 2(3.6)
Q
=500 lit/S
I
MOTOR
E
"'
N
4 kPa
900 mm0 6
=
VA= V1
= 2.025 m/s
VA2
2g
t (0.2)2
_l_
= 0.21 m
Poweroutput= Q Y HA
PUMP
Output power • 20 hp
Line 1:
200 mm 0
B~
1.8 m
-34 kPa
0.0636
VA= V1
h
,
600 mm 0
~~~=i(o)ll======~
o-
t I
Lme 2:
150 mm 0
_c_ =0.66m
Q2 = 0.0636 m 3 /s
Q = Q1 = Q2 = 0.0636 m 3 /s
14 kPa
Q
v 2
Q2 =
Solution
m/ s = vc
112 = 3.6
264
CHAPTER FIVE
Fundamentals of Fluid Flow
CHAPTER FIVE
Fundamentals of Fluid Flow
II UID MECHANICS
I. HYDRAULICS
20 X 746 = 0.0636(9810)HA
HA = 23.91 m
265
lncrgy equation between A and N:
LA - HLA·H - HLH - HLHB - HLN = EN
2
v 2
v 2
PN
~ + !!..!2_ + ZA - 3 - 2- 10 - 0.04 _N_ = _N_ + + ZN
2g
y
2g
2g
y
Energy equation between A and C (datum at A):
EA + HA - HL = Ec
v 2
p
.
v 2
p
_A_ + ~ + zt\ + !!A - HL = _c_ + ___£_ + zc
. 2g
y
2g
y
p
8Q2
+ 550 + 0-15 = 1.04
8Q2
4 + 0 + 16
n 2 (9.81)(0.3) 4
9.81
n 2 (9.81)(0.025) .
Q = 0.0106745 rn3js
Q
A fire pump delivers water through a 300-mm-diameter main to a hydrant to
which is connected a cotton rubber-lined fire hose 100 mm in diameter
terminating to a 25-mm-diameter nozzle. The nozzle is 2.5 m above
hydrant and 16 m above the pump. Assuming frictional losses of 3m from the
pump to the hydrant, 2m in the hydrant, 10m from the hydrant to the base of
the nozzle, and the loss in the nozzle of 4% of the velocity head in the jet, to
what vertical height can the jet be thrown if the gage pressure right after the
pump is 550 kPa?
2
+ !!..!2_ + ZA -15 = 1.04__!!_ + __.!::!_ + ZN
2g
y
2g
y
34
0.21 + + 0 + 23.91 - 3 = 0.66 + 0 + (1.8 +IT)
9.81
IT= 15.19 m
Pro
v
2
V'A
0.0106745
IJN
= AN = t(0.025) 2
VN
= 21.74 m/s
21.74 2
v 2
/z= ___!!___ = - - 2g
2(9.81)
/r=24.102m
Problem 5 - 21 (CE May 2001)
Solution
1or the pipe shown in the Figure v 1 = v2 = 1.2 m/ s. Determine the total head
lost between 1 and 2.
E.G.L.
-L-___.~-
550 kPa
Datum
- __H.G.L.
------
CHAPTER FIVE
Fundamentals of Fluid Flow
266
t 'II of sp. gr. 0.84 is
= V2 = 1.2 mj S
2
2
~ + h._ + z, - HL = !2_ + E2. + Z2
2g
2g
y
y
2
2
vl
Vz.
Smcevi =v,_, -2g =2g
Q =56 Lis
~
• p =?
~
6
~=====::J
El. 1.2 m
225 mm 0
\olution
h._ + Z1 - H L = E2. + Z2
y
445 kPa
Ilowing in a pipe
limier the
nnditions shown in
'Ill' Figure. If the
h 1lal head loss from
!'oint 1 to point 2 is
·100 rmn, find the
pressure at point 2.
Energy equation between 1 and 2:
E,- HL = E2
267
Fundamentals of Fluid Flow
1, HYDRAULICS
Solution
VI
CHAPTER FIVE
f lUID MECHANICS
Q, = Q2 = 0.056 m 3/s
y
280 + 4.3- HL = 200 + 9.08
9.81
9.81
HL=3.375 m
Energy equation between 0 and 6 :
EI - HLI-2 = E2
2
2
~ + El + ZI - HLl-2 = !2_ + E2. + Z2
2g
Pro
November 2000)
y
2g
2
A nozzle inclined at an angle of 60° with U1e horizontal issues a
diameter water jet at the rate of 10 m/s. Neglecting air resistance, what is
area of the jet at the highest point of the projectile?
Solution
Solving for the velocity of the jet at the summit (highest point, A)
y
8(0.056)
+
445
+ 3.21 - 0.90 =
2
4
1t g(0.15)
9.81 X 0.84
2
+ _ __:_P__ +
8(0.056)
4
2
9.81 X 0.84
7t g(0.225)
12
_ __:_P__ = 55.52 m of oil
9.81 X 0.84,.
p = 457.53 kPa
Vy = 0
v, = v., cos 8
v, = 10 cos 60° = 5 m/s
v = ~v 0 / +v0 /
v= ~5 2 + 0 2 =5 m/s
Since the flow is continuous:
[Qo = QA]
A.,v. = AA v
*
(0.05)2 (10) = AA (5)
AA = 0.003927 m2
Problem 5 - 23
Problem 5 - 24
A
\ 50-mm diameter siphon discharges oil (sp. gr. = 0.82) from a reservoir (elev.
'0 m) into open air (elev. 15 m). The head loss from the reservoir (point 1) to
the summit (point 2, elev. 22m) is 1.5 m and from the summit to the discharge
, nd is 2.4 m. Determine the flow rate in the pipe in lit/sec and the absolute
pressure at the summit assuming atmospheric pressure to be 101.3 kPa.
268
CHAPTER FIVE
CHAPTER FIVE
FLUID MECHANICS
Fundamentals of Fluid Flow
Fundamentals of Fluid Flow
& HYDRAULICS
Solution
El. 22m
Q2 = Q3 = Q
HL1-2 = 1.5 m
HL2.3 = 2.4
Problem 5 - 25
Determine the velocity and
d1scharge through the 150 nun
d1ameter pipe shown (a) assuming
no head loss and (b) considering a
head lost of 200 nun.
0
~
El. 30 m --"--====-=-;.......,--==rd~-1
El. 20m
El.28m --lo....
El. 27.5 m
Oil
5
= 0.82
Solution
(a) Assuming no head loss:
Energy equation between 1 and 3:
[ , - HLI-2 -JJL2-J = E3
Pl
•
v 2
p
+ - + Zl - HLl-2- HL2-3 = _1_ + _2_ + Z3
2g
y
2g
y
vl2
-
0 + 0 + 20- 1.5- 2.4 =
8Q2
2
n g(0.05) 4
+ 0 + 15
Q = 0.00912 m3j s
Q = 9.12litjsec
Energy equation between 1 and 2:
E, - HL1-2 = [z
2
V1
P1
v 2
p
+ - + Zl - HLl-2 = - 2- + _2_ + Z2
2g
y
2g
y
0 + 0 + 20 -1.5 = 8(0.00912)2 +
P2
+ 22
n 2g(0.05) 4
9.81 x 0.82
pz = -37 kPa
El. 24.9 m - - - -
2
-V1 + -P1 + Z1 = -V2 + -P2 + Z2
2g
y
2g
y
v 2
0 + 0 + 30 = - 2- + 0 + 24.9
2g
v 2
- 2-
=5.1 m
2g
v2 = 10 m/s
Q = A2 V2 = t (0.15)2(10)
Q = 0.177 rn3/s =1771/s
(b) Considering head loss of 0.2 m :
E1 - HL = E2
2
2
~ + El. + z, - HL = ~ + El_ + Z2
2g
Absolute pressure at 6 = 101.3 + (-37)
Absolute pressure at 6 = 64.3 kPa
150 mm 0
Energy equation between
0 and 6 neglecting head lost:
£1 = £2
2
2g
y
v 2
'(
0 + 0 + 30- 0.2 = - 2- + 0 + 24.9
2g
v 2
- 2-
2g
269
= 4.9m
v2 = 9.805 m/ s
Q = A2 V2 = t (0.15)2(9.805)
Q =0.173 rn3js = 1731/s
CHAPTER FIVE
Fundamentals of Fluid Flow
270
FLUID MECHANICS
& HYDRAULICS
Problem 5 - 26
Water flows freely from the reservoir shown through a 50-mm diameter pipt•
at the rate of 6.31 lit/ sec. If the head lost in the system is 11.58 Joule/ N.
determine the elevation of the water surface in the reservoir if the dischargt•
end is at elevation 4 m .
CHAPTER FIVE
Fundamentals of Fluid Flow
II UID MECHANICS
1. HYDRAULICS
f'roblem 5 - 27
,·glecting head loss, determine
1h·· manometer reading in the
1lem shown when the velocity
I water flowing in the 75-mm
li.1meter pipe is 0.6 m/ s.
271
25mm0
E
~
r-i
75mm 0
75mm 0
q -·- --·- -- - -
~-
750mm
l
Mercury
'Jolution
V1 = 0.6 m/s
[Q! = Qz]
t (0.075)2~.6) = t (0.025) 2v
2
Vz
Solution
Q = 6.31 L/ s = 0.00631 m1j s
HL = 11.58 N-m/N = 11.58 m
= 5.4 m/s
Energy equation between 0 and 8:
Et = Ez
75mm0
2
2
p + Zt = v2
+ _1
- + -P2 + Zz
2g
y
2g
y
v
_1_
Energy equation between 0 and 8 :
[I - HL = Ez
2
?
~ +!!..!.. + ::1- HL = 712 - + J:2 + :o
2g
y
2g
y
-
0 + 0 +:I -11.58 = 8(0.00631)2 + 0 + .J
rr 2 g(0.05) 4
Z1 = 16.11 m
-7 Elevation of w .sin the tank
~ + El +0= ~ +0+2.4
2(9.81)
y
2(9.81)
El = 3.868 m of water
y
Sum.riting-up pressure head from 0 to 0 1n meters ?f water:
El + 0.75- h(13.6) = E2.
y
y
3.868 + 0.75- 13.6h = 0
h = 0.3395 m = 339.5 mm
750mm
l
272
CHAPTER FIVE
Fundamenta ls of Fluid Flow
Problem 5 - 28
A ho rizon ta l pipe g radu ally redu ces fro m 300 mm diameter section to 100
dia meter sectio n. The press ure a t the 300 mm sect ion is 100 kPa a nd at the 1
m m sectio n ic; 70 kPa. lf the flo w ra te is 15 lite rs/ sec o f water, compute th
head lost be tween the two sectio ns.
olution
1•1) Energy equation between 1 & 3
Q-
00~52>2--
Jlo·-·-·- -------
!!:,~~;~,-~
Water
E1 =; E3
2g
p
v32
P3
+ _1 + Zl = -- + + Z3
y
2g
y
-\ -·-· ~-·-·:.=='.,..0
err==="""
v3 =14 m/s
1.!:::::=======..111
Q = 0.557 m3/s
l:,nergv eLJua tio n between 0 and 6 :
L~-1-ll
(II) Pressure at the throat:
= J,
Energy equation between 6 and 0:
2
v 32
p3
t•,-"
l'1
u,-"
11"
- - + - +.::1 -I l l.=---+-- + _,
2g
y
2g
y
70
8(0.01 5)" + 100 +O- l/L= 8~0.015)"
+ -- + 0
n 2 g(0.3) 4
9.81
n-g(O.J) 4
9.81
!:L + 12 +z2 = - - + 2g
•
Ill = 2.872 m
Problem 5 - 29
A diverging tu be discharges water from a reservoir a t a d e pth of 10 m beiO\\
the wate r s urface. I he d ia me te r o f the tube g radu a lly increases from 150 mm
a t the throat to 225 mm a t the outlet. Neglecting fric tio n, de termine: (a) thl•
maximum possible ra te of discharge thro ug h th is tube, a nd (b) thl
cor respond ing press ure a t the throat.
'
2g
y
8(0.557)2
10m
)150mm0
v 2
0 + 0 + 10 = - 3- + 0 + 0
2g
Q = Q3 = t (0.225)2(14)
p, = 100 kPa
273
0
n.
(Neglecting head loss & datum
along point 3)
v12
Solution
300 mm
CHAPTER FIVE
Fundamentals of Fluid Flow
I LUID MECHANICS
I. HYDRAULICS
y
+z3
+ 12 + 0 = 14 2 + 0 + 0
7t2 g(0.15)4
p2 = -398.75 kPa
y
2g
225 mm 0
274
CHAPTER FIVE
Fundamentals of Fluid Flow
Jementa
Problems
CHAPTER FIVE
I LUID MECHANICS
1. HYDRAULICS
Fundamentals of Fluid Flow
f•roblem 5 - 33
11 the water level in Problem 5- 32 varies and V2 = 10 m/s, find the rate of
h.mge dh/ dt.
Problem 5 - 30
Air is moving through a square 0.50-m by 0.50-m duct at 180 11131 min
the mean velocity of the air?
.
Aus: 12
Problem 5 - 31
T~e
piston of a hypodermic apparatus shown in Figure 5 _ 8 is
Withdrawn at 6 mmjsec; air leaks around the piston at 20 nun3jsec. What
the average speed of blood flow in the needle?
275
A11s: -9 mm/ s
11roblem 5 - 34
lluid having sp. gr. 0.88 enters the cylindrical arrangement shown in Figure 5
10 at section A, at 0.16 Njs. The 80-nun-diameter plates are 3 mm apart.
\ssuming steady flow, determine the average velocity at section A and at
o•ction B. Assume radial flow at B.
A11s: v, = 1.47 mjs; v2 = 2.46 cm/s
Figure 5-8
Figure 5-10
Problem 5 - 32
The water ta~k in Figure 5 - 9 is being filled through section 1 at 6 mj s and
throu.gh secllon 3 at 15 L/s. If water level /1 is constant, determine the exit
velocity v2•
15 l./s ,;i.
-0- 0
\j
Problem 5 - 35
1£ a jet is inclined upward 30° from the horizontal, what must be its velocity to
ll'ach over a 3-m wall at a horizontal distance of 18m, neglecting friction?
Ans: 16.93 m/s
Problem 5 - 36
Neglecting air resistance, determine the.height a vertical jet of water will rise if
projected with velocity of 21 m/s?
h
Figure 5-9
k-o.9 m 0---+!
Ans: 22.5 m
276
CHAPTER FIVE
Fundamentals of Fluid Flow
CHAPTER SIX
Fluid Flow Measurement
I I UID MECHANICS
I. HYDRAULICS
277
Problem 5 - 37
hapter 6
High velocity water flows up an inclined plane, as shown in Figure 5 - 11
What are the two possible depth of flow at section 2? Neglect a! losses.
A11s: 0.775 m & 2.74 m
Fluid Flow Measurement
llwre are numerous number of devices used to measure the flow of fluids. In
Figure 5-11
'"Y of these devices, the Bernoulli's EnergJJ Theorem is greatly utilized and
ultlitional knowledge of the characteristics and coefficients of each device is
1111portant. In the absence of reliable values and coefficients, a device should
calibrated for the expected operating conditions.
I···
UEVICE COEFFICIENTS
(;oefficient of Discharge, Cor Cd
I ~~~ coefficient of discharge is the ratio of the actual discharge through the
.lo•vice to the ideal or theoretical discharge which would occur without losses.
I his may be expressed as:
l
CorCrl=
Actual discharge
Q
=Theoretical discharge
Qy
Eq. 6-1
~----------~
1 ht• actual discharge may be accomplished by series of observation, usually by
""'"suring the total amount of fluid passing through the device for a known
1 • nod. The theoretical value can be accomplished using the Bernoulli's
llll'orem neglecting losses.
1 oefficient of Velocity, C,
llw coefficient of velocity is the ratio of the actual mean velocity to the ideal or
tlworetical velocity which would occur without any losses.
Actual velocity
Theoretical velocity
Cv= ------------~--
v
Vy
Eq. 6-2
CHAPTER SIX
Fluid Flow Measurement
278
Coefficient of Contraction, Cc
The coefficient of contraction is the ratio of the actual area of the contr
section of the stream or jel to the area of the opening through which the
flows.
C. = Area of stream or jet
·
Area of opening
a
-7 Eq. (1)
/\I so
Q =Actual area, a x Actua l velocity, v
Q = c. A X c. i't
Q =C. C,. 1\u,
but 1\v, = Q1
Q=
cc. Q,
Head
(m)
6.25
0.647
0.635
0.629
0.621
0.617
0.614
0.613
0.612
0.611
0.610
0.609
0.608
0.607
0.606
0.605
0.605
0.43
0.61
1.22
1.83
2.44
3.05
3.66
4.27
4.88
6.10
7.62
9.15
12.20
15.24
18.30
A
279
Table 6 - 1: Discharge Coeffidenls for Verttcal Sharp-Edged Circular Orifice
Discharging tnto Air at 15.6•C (60•F)
,. 0.24
Relationship between the Three Coefficients
Actual discharge, Q = C x Q1
CHAPTER SIX
Flurd Flow Measurement
I I UID MECHANICS
1. I iYDRAULICS
Diameter in mm
18.75
25.00
0.616
0.609
0.610
0.605
0.607
0.603
0.603
0.600
0.601
0.599
0.600
0.598
0.600
0.597
0.597
0.599
0.598
0.596
0.598
0.596
0.598
0.596
0.597
0.596
0.597
0.595
0.596
0.595
0.596
0.595
0.596
0.594
12.50
0.627
0.619
0.615
0.609
0.607
0.605
0.604
0.603
0.603
0.602
0.602
Q.601
0.600
0.600
0.599
0.599
50.00
0.603
0.601
Q.600
0.598
0.597
0.596
0.596
0.595
0.595
0.595
0.595
0.594
0.594
0.594
0.594
0.593
100
Q.601
Q.600
0.599
0.597
0.596
0.595
0.595
0.595
0.594
0.594
0.594
0.594
0.594
0.593
0.593
0.593
-7 Eq. (2)
From Equations (1) and (2)
IIEAD LOST
= Crx C,,
The coefficient of discharge varies with Reynolds Number. It is not cons
for a given device. Table 6 - 1 gives the coefficients for vertical sharp
orifice.
I he head lost through Venturi meters, orifices, tubes, and nozzles may be
,. pressed as:
Figure 6-2
A,
The ideal energy equation between 1 and 2 is:
E1 = E2
Cc .. 0.62
V1
-
c."' 0.98
~ .. 0.61
Sharp-edge
Square shoulder
2
2g
Al VJ
Thick plate
Figure 6- 1: Orifice coefficients
Rounded
P1
+-
V2
2
P2
+ Z1 = - - + - + Z2
y
2g
y
= A2V2
v,= ~v2 and v/ =(A2)2 v22
A1
2g
A1
2g
CHAPTER SIX
Fluid Flow Measurement
280
CHAPTER SIX
Fluid Flow Measurement
IIllO MECHANICS
I tYORAULICS
]~
_~2g [1-(A2)
2g [ 1-(~J
A
C,,
A
2
v2
HL=-
1
281
2
]
1
Eq. 6-5
II lhl• orifice or nozzle takes off directly from a tank where A, is very much
'' ollcr than A2, then the velocity of approach is negligible and Eq. 6 - 5
• olttces to:
Considering head los t bel ween 1 and 2:
"
2
_ 1_
2g
J1
2
y
2g
+ _1 + ;::, - HL = .!!1_ + E2 + Z2
y
1
HL= ( C/ -1
This equation simplifies to:
) v
2
28
Eq. 6-6
Note: 11 = actual velocity
•
ORIFICE
2g[('~ +z1 )-(~2 +z
\n orifice is an opening (usually circular) with a closed perimeter through
' •hich fluid floW'IS. It is used primarily to measure or to control the flow of
lluid. The upstream face of the orifice may be rounded or sharp. An orifice
'"'th prolonged side, such as a piece of pipe, having a length of two or three
tunes its diameter, is called a short tube. Longer tubes such as culverts under
. mbankmen ts are usually treated as orifice although they may also be treated
2)]
Squ;;·r~(~;w::,"~·r~:n: ::·):(~, .,, J
From (b)
e;• +z, Jt' u,)
<[ -(~: J']
1
;; [1-(;:Jl> <[1-(~:J}/lL
· ~ short pipes.
'\ccording to shape, orifice may be circular, square, or rectangular in cross•l'<'tion. The circular sharp-crested orifice is most widely used because of the
·•implicity of its design and construction.
+ HL
I he figure below shows a general case of fluid flow through an orifice. Let p...
.md ps be the air pressures in the chambers A and 8, respectively and VA be the
velocity of the stream normal to the plane of the orifice (velocity of approach).
Consider two points 1 and 2 such that v 1 = V A and v2 = v and writing the
t'nergy equation between these two points neglecting losses
'
CHAPTER SIX
Flurd Flow M easurement
CHAPTER SIX
Fluid Flow Measurem ent
282
($) Pa
h
-- ------fr''
·f},
0
Vht>re H is the totallzead producmg jlor11 m meters or feet of the flowing fluid. It
111 be noted in Eq. (1) that H is the sum of the flow energy u pstream less the
low energy d ownstream, or
:
Eq. 6-12
H = Head U pstream - Head Downsh·eam
.V.
Chamber A
Chamber B
V lues of H for Various Conditions
.1'7
p
=0
.1'7
T
Energy equa l:lon between 1 and 2 neglecl:lnr; hl'ad lo!>t
p=0
T
A
I
h
h
[ , - [2
2
u/ +P1- + z, = U2
P2
+ - + Z7
2g
2g
y
y
'
,~
2
L'========- p = 0
2
!:.L + PA + yil + 0 =1> - + PB + 0
2g
2g
y
2
H=h
v
o-'
2g
1
t..=:========-11p = 0
H = h(l ±a/g)
(1 2
!:.L + Jl (\ + It = - + Jl ii
2g
283
2g
y
= It+~
y
Pa
-
y
y
uA
.........
....
. . . ...
;:;:;:;f;:;:;:;:;:; :
:: :Prl!$Silte:~ p :::
2
::;.;::;:::::: :::::::
+ --
2g
Unit wt., y
T
:\:~~~~~M~ :t: f::
:;.:,: :: :::::: :::: :::
Unit wt., y,
h
l
Theoretical velocity, v, = J2gH
Actual velocity, v = Cr. J2gll
Theoretical d ischarge, Q, = A J2gH
Actua l discha rge, Q = CA J2gH
H=il + _A_ + ~
p
PB
y
y
v 2
2g
H = h + p/y
Unit wt., Yz
,.
--.;;:,
t..========:::J p =0
H = hz + Mrdrz) + Phz
284
CHAPTER SIX
Fluid Flow Measurement
..1L
r
h
285
tlflces under low Heads
It• n the head on a vertical orifin.• i'> o.,m,\11 in comparison with the height of
n11fice, there is an appreciable diffl'rl'nce between the discharges using the
11 Ill US analysis.
111>1dcr the rectangular section of length L and height D as shown in the
""' with both the surface and the jet subject to atmospheric pressure. The
It ''"'tical discharge through an elementary strip of length Land height is:
Submerged Onf1ce
( Neglect1ng v.)
H = h, - hJ
CHAPTER SIX
Fluid Flow Measurement
LillO MECHANICS
IIYORAULICS
dll
=h
riQ 1 = (L dh) ~2gh
.J2i L 11 dh
112
,/Q, =
Contraction of the Jet
The figure show n represent!. a cros!.-sectJOn of flUid flow througJ1 , , t
[ d
'('
(
GoVCflld
I ~-ec ge on ICC rom a reservoir to the atmosphere. The fluid flowin
c~~mg from all direction upstream from the orifice and as they leave~
011f1~~' the:' cannot make an abrupt change in their direction and they move 11
curvdm.eal paths, thus causmg the jet to contract for a short distance beyon
the onfice. The phenomenon is referred to as the confmct 1a11 of Llie ;et. T
sec~on on the Jel where the contraction ceases is called the nenn 1·ontrm 1
w h1ch IS approx1matclv located c1t o1w h<df of the orifice diameter (D/2) f
ron
the upsh·cam far(;'
Sha·
,,,
(.! 1 =
.J2i L Ill t dli
"I
l2· = .J2i L
Q, = f
rti" ~ ]''2
,,1
.J21 L ( 1 112 _ i1
12
13t 2)
Q=CQ,
Eq. 6-13
D/2
'.
\
\
\
VENTURI METER
,I
/
Vena contracta
l'turi meter is an instrument used in measuring the discharge through pipes.
II wnsist of a converging tube AB (See Figure 6- 3) which is connected to the
lllclin pipe at the inlet at A , and ending in a cylindrical section BC called the
lllroat, and a diverging section CD which is connected again to the main pipe
11 the outlet D. The angle of divergence is kept small to reduce the head lost
1.1use by turbulence as the velocity is reduced.
CHAPTER SIX
Fluid Flow Measurement
286
r-.---------,~~~E~.G.L~·--
v1'12g
',:-f
L
'
pJy
CHAPTER SIX
Fluid Flow Measurement
I LUID MECHANICS
1-. HYDRAULICS
287
llw theoretical or ideal discharge "Q, " can be found once 111 or l>2 ls known.
llw actual discharge "Q" is computed by multiplying the theoretical value by
Ill<' cotfficienl of disdurrge or 111efer colffinent "C'.
Q= Cx Q1
Eq. 6-14
h
Note: If we neglect the head lost 1n our energy equation, the values we get are known as
>oretical or ideal values (theoretical velocity and theoretical discharge). Considering head
I· .t, we get the actual values(aetual velocity and actual discharge).
p-Jy
NOZZLE
nozzle is a converging tube installed at the end of a pipe or hose for the
I'"• pose of increasing the velocity of the issuin~ jet.
Throat
Inlet " ' - -
z,
Piezometer nng
-- _!_ - - - - - - - - - - Figure 6 - 3: Venturi meter
~~ - - - - - _l
u1 < o,
Outlet
-------- -- ---- -
Consider two points in the system, 0 at the base of the inlet and
throat, and writing the energy equation between these two points n.,.,,.,.,.,r,•••
head lost:
2
11
_1_
2g
+
P1
y
u2 2 - u1 2
2g
2g
2
V2
+ z, = -+ -P2
2g
y
= ( El_ + z, ) -
y
Base
+ Z2
(.!!.2. + z2 )
y
The left side of the equation is the kinetic energy which shows an increase in v<~lu
while the left side of the equation is the potential energy which shows a decre<l'it' 1
value. :Oerefore, neglecting head lost, the increase in kinetic energJJ is equal fo 11
decrease 111 potenlwl energy. "This statement is known as the Venturi Principle.
fhe difference in pressure between the inlet and tht• thro,lt is commonly mt•,1s 1111
by means of a differential manometer connecting tlw inl..t .md throat.
If the elevations and the difference Ill prt>'iSllf (' h 1\V(I ll 0 cllld 6 tlr(' known. ,,
discharge can be solved
•
I h discharge through a nozzle can be calculated using the equation
I~
Q = CA.,fiiH
Eq. 6-15
\\here:
H = total head at base of nozzle
A.,= area at the nozzle tip
llu following table gives the mean values of coefficients for water discharging
1h1•11tgh a nozzle having a base diameter of 40 mm and C.= 1.0.
Tip Diameter In mm
c
19
22
25
29
32
35
0 IJ I)
() 'lHl
0.'180
0 976
0.971
0.959
II
h I t( (,
CHAPTER SIX
Fluid Flow Measurement
288
'I UID MECHANICS
1o HYDRAULICS
PITOTTUBE
CHAPTER SIX
Flu1d Flow Measurement
llu, eq uation shows that the velocity head at point 1 is transformed into
Named after the French physicist and engineer Henn Pi tot, Pitot tube is a
(L-shaped or U-shaped) tube with both ends open and is used to measure th
velocity of fluid flow or velocity of air flow as used in airplane speedometer
" -~ure head at point 2
(a)
When the tube ts placed tn a movmg stream with open end oriented into th
direction of flow, the liquid enters the opening at point 2 until the surface in
the tube rises a distance of h above the stream surface. An equilibrium
condition is then established, and the quantity of liquid in the tube remain
unchanged as the flow remains steady. Point 2 at the face of the tube facing
the stream is called the slngllafion point
( b)
(C)
(d)
E.G.L.
T
h = v1/2g
j- - -
n
Stagnation
point
h,
L~--- ________ __!
6
0
Figure 6- 5
v2 = 0
Figure 6-4
Consider a particle at point 1 to movmg with a velocity of v. As the particlt'
approaches point 2, its velocity is gradually retarded to 0 at point 2
the energy equation between 1 and 2 neglecting friction:
E, = £2
0 2
_1_
2g
+
P1
y
if
+/=
y
02
2g
D1fferent1al
manometer
)
o, = v; f!.J._ = hi '
-
289
+
P2
y
P2
y
+:/
= lt2
+ ,,, = 172
u2 = 2g(h 1 - hz)
Figure 6 - 6: Pitot tube 1n a p1pe
v = ~ 2glt
Eq. 6-16
H.G.L.
CHAPTER SIX
Fluid Flow Measurement
290
CHAPTER SIX
Flu1d Flow Measurement
I I lJID MECHANICS
lfYDRAULICS
GATES
Actual V = C,. j2.g(ci1 - d 2 ) + v 1
A gate IS an opemng 111 a dam or other hydraulic structure to control
passage of water. It has the same hydraulic properties as the orifice. In
gates, calibration test are advisable if accurate measurements
obtained s ince its coefficient of discharge varies widely
2
Actual Q = CA ~ 2g(d 1 - d2 )+ v 1
291
Eq. 6- 18
2
Eq.6- 19
d
Coefficient of contraction, Cr = _l.
Eq. 6- 20
y
The following illustrations show fhe two different flow conditions through
sluice gate
t
H
where:
C =C. C. (varies from 0.61 to 0.91)
A= by
h =width of the flume
d,
TUBES
tandard Short Tube
Figure 6 - 7 (a): Free Flow
Figure 6 - 7 (b): Submerged Flow
Figure 6 - 7: Flow through a gate
ln Figure 6 - 7 (a), writing the e nergy equation between 1
head lost:
E1 = £2
A standard short tube 1s the o ne with a square-cornered entra nce and ha5 a
lt·ngth of about 2.5 times its internal diameter as shown in Fi~ure?- 8. Figure
,, 8 (a) shows a condition when the flow started sud_d enly ~tth htgh head s so
that the je t may not touch the walls of the pipe. Thts conditio n IS very n:~ch
the same as tha t of a sharp-crested orifice. Figure 6 - 8 (b) shows a condition
vhen. th e jet tou~hes the walls of the tube. The discharge through this tube is
tbout one-third greater tha n that of the standard s harp-ed ged onftce but the
vpfocity of flow is lesser
2
v 2
~ + El_ + Zt = - 2 - + El_ + Z2
2g
2g
y
y
c..= 1.0
w I1ere -Pt -_ d1 and -P2 = d2
y
V]2
2g
v2 2
-
-
2g
y
'
V 2
+ dt + 0 = - 2 - + d2 + 0
2g
v1 2
- -
-
2g
c =c.= 0.82
' ' '{::::;:;;:::::::::::~-=---:.~~-:./
/
p, • ·0.82H
= dt - d2
H • total heao
V2 2 - V1 2 = = 2g (d1 - d2)
Vn •
0.82
.[2;
v2 2 = 2g (dt - d2) + v1 2
Figure 6 - 8 (a)
Theoretical v = ~ 2g(d 1 - d 2 )+ v 1 2
Figure 6 - 8 (b)
Figure 6 - 8: Standard Short tubes
292
CHAPTER SIX
Fluid Flow Measurement
CHAPTER SIX
I tiiO MECHANICS
IIYORAULICS
293
Fluid Flow Measurement
Converging Tubes
Conical converging tubes has the form of a
frustum of a right circular cone with the
larger end adjacent to the tank or reservoir
as shown in Figure 6 - q
lu II' are tube having their ends projecting inside a reservoir or tank.
I~
' ·' .i
......~"""· ·- ·-·-·-·-·~~7;~
· ~;<:-;-:..~:..-:::::...·
." .I
" .\
0 =angle of
convergence
Figure 6-9
u = C..~2gH
Eq. 6-21
Q = CA~2gH
Borda's Mou.thpiece - This is a special case of a re-entrant tube,
consisting of a thin tube projecting into a tank having a length of
about one diameter. The coefficient of contraction for this tube,
Cc = 0.5 and C.,= 1.0.
ubmerged Tubes
Table 6 - 2: CoeffiCients for Comcal Converging Tubes
Coefficient
c.,
Cc
c
11 example of submerged tube is a culvert conveying water through
111hankments. The discharge through a submerged tube is given by the
I • 111!1Ula:
•
1.000
0.829
0.910
0.992
0.939
0.972
0.938
0.952
0.924
0.935
0.911
Q= CA~2gH
Eq. 6-23
Where C is the coefficient of discharge, A is the area of the opening, and H is
•lu• difference in elevation of the liquid surfaces.
Diverging Tubes
A diverging tube has the form of a frustum of a right circular cone with the
smaller end adjacent to the reservoir or tank
~,:c c : : C': -: -_E
-':
/
/
·-·-·-·-·-
o =angle of
divergence
L
Figure 6- 10: Submerged Tube (Culvert)
294
CHAPTER SIX
Fluid Flow Measurement
I I UID MECHANICS
I fYDRAULJCS
UNSTEADY FLOW
The flow through orifice, weirs, or tubes is said to be steady only if the
head producing flow, H, is constant. The amount of fluid being discharged
a time I can therefore be computed using the formula
tit=
CHAPTER SIX
Fluid Flow Measurement
295
Asdh
Qin-Qout
Eq. 6-25
where Q is the discharge, which is constant or steady. In some
however, the head over an orifice, tube or weir may vary as the fluid flows
and thus causing the flow to be unsteady
When there is no inflow (Q;n = 0), the formula becomes:
,_
J
it2 A, dh
.
Jt 1 -QOIII
l11tl•rchanging the limits to change tl1e sign of the integrand:
Eq. 6-26
Note: If As is variable, it must be expressed in terms of h.
Figure 6- 11
II the outflow is through and orifices or tube, Qout = CA ~2gH . If the flow is
through any othl!r openings, use ilie correspondirlg formula for discharge.
l·or tanks with constant cross-
Consider the tank shown in the figure to be supplied with a fluid (inflow)
simultaneously discharging through an outlet (either an orifice, tube, weir
pipe). Obviously, if Qm > Qout, the head will rise and if Qout > Q;0 , the head
fall. Suppose we are required to compute the time to lower the level from /z1
it2 (assuming Qout > Qm), the amount of fluid which is lost in the tank will be
dV = (Qtn- Qout) d/
dt= __
dV
__
Qin-Qout
where dV is the differential volume lost over a differential time dt.
over the outlet ish, then the level will drop dh, thus dV =A, dlt, where A 5 is
surface area in the reservoir at any instant and may be constant or variabl
then
t•ctional area and the outflow is
through an orifice or tube (with
rH> inflow), ilie time for ilie head
to change from H1 to H2 is:
296
CHAPTER SIX
f LUID MECHANICS
lo HYDRAULICS
Fluid Flow Measurement
If liquid flows through a submerged orifice or tube connecting two tanks d
shown, the time for the head to change from f-/, to H2 is: (See the derivation of
these formulas in PROBLF.M 6- 36)
As •. 1'7
1-
f---.-
v""'
---------- I
v.,., = VA,,
_L -----------V 1n
90
. \I.
l
Tank 1
where A s1 and As2 is the water surface areas in the tanks at any time, and H 11
the difference in water surfaces in the two tanks at any time. If A,1 and/ or A
will vary, it must be expressed in terms of H
If A sl and A s2 are constant, i.e. the two tanks have uniform cross-sectional ared
the formula becomes:
As1 As2
A~1 + A s2
CA
0
WEIR
Weirs are overflow structures which are built across an open channel for the
1,urpose of measuring or controlling the flow of liquids. Weirs have been
·munonly used to measure the flow of water, but it is now being adopted to
••u•asuni the flow of other liquids. The formulas and principles that will be
lJ'ICussed on this chapter are general, i.e. applicable to any type of liquid.
( lassification of Weirs
\rcording to shape, weirs may be rectangular, triangular, trapezoidal, circular,
1 ,,rabolic, or of any other regular form. The most commonly used shapes are
llll' rectangular, triangular and the trapezoidal shapes. According to the form
,•I the crest, weirs may be sllarp-crested or broad-crested.
I he flow over a weir may either be free or submerged. If the water surface
.lownstream from the weir is lower than the crest, the flow is free, but if the
lownstream surface is higher than the crest, the flow is submerged.
Definition of Terms
Nappe- the overflowing stream in a weir.
II
Tank 1
t=
297
H,
H,
II
CHAPTER SIX
Fluid Flow Measurement
fig ((H; - .JH;)
g
Crest of weir - the edge or top surface of a weir with which the flowing
liquid c~mes in contact.
Contracted weir - weirs having sides sharp-edged, so that the nappe is
contracted in width or having end contractions, either one end or
two ends.
Suppressed weir or full-width weir - weirs having its length L being equal to
the width of the channel so that the nappe suffers no end
contractions.
Drop-down curve - the downward curvature of the liquid surface before the
weir.
Head, H - the distance between the liquid surface and the crest of the weir,
measured before the drop-down curve.
298
CHAPTER SIX
Fluid Flow Measurement
CHAPTER SIX
fluid Flow Measurement
E.G.L.
Water surface
fd Q1 =
v.2/2g
.----~
H
;
;
/
p
I
J2i L [~(It+ hv)i
Q, = f fii L [ (H + hv)31 (0 + lt,)312 ]
2 -
~
,.
J2i L f~~+ltv )l rllt
Q1 =
-·-
- · -·
f----,>1----
299
,.
,. ,.
;
Actual Q =CQ,
I ~--------.
j/
,.. .... ~--- ....,
./
I
;
'"---<-":._ ____ ,..I
I!~ .... _______
[
Eq. 6-30
It is a common practice to combine
"'4H
A._!
Figure 6 - 12: Path lines of flow over a rectangular sharp-crested weir
[
RECTANGULAR WEIR
.___,
~
~
w
L
Q = C,L [(H + lrv )3/2 - ltv3/2]
Eq. 6-31
~---------'
If the ratio H/ P is sufficiently small, the velocity of approach becomes very
mall and the term 1! 113/2 may be neglected. The discharge formula becomes
-:f."
*
J
Figure 6 - 13: Section A-A of Figure 6 - 12.
! J2i C into a single coefficient C, called
the weir factor. The general formula for a discharge through a rectangular weir
unsidering velocity of approach then becomes
Q = Cw LHt
Eq. 6-32
In situations where the discharge is required considering the velocity of
.tpproach, using Eq. 6 - 30 or Eq. 6 - 31 would lead to successive trials to solve
for Q (since the velocity of approach hv is a function of Q). The following
~•mplified equation may be used:
Consider a differential area of length L and height dh to be located 11
below the liquid surface. By orifice theory, the theoretical velocity
this area is:
Eq. 6-33
v,= ~2gH'
where the total he~d producing flow H' = 1r + lrv, where lrv is the velocity
of approach and IS equal to v,?j2g. The discharge through the
strip is then,
dQ, =d.A Vt
dQ, = L dlr ~r-2g_(_lr-+-lrv-)
Eq. 6-34
where
d = depth of water upstream
d=H+P
300
CHAPTER SIX
Fluid Flow Measurement
CHAPTER SIX
Fluid Flow Measurement
I I UID MECHANICS
I. HYDRAULICS
Standard Weir
301
BAZIN FORMULA
For rectangular weirs of length from 0.5 m to 2.0 m under heads from 50
nun to 600 mm.
The following specifications must be applied to a standard rectangular
without end contractions:
1. The upstream face of the weir plate should be vertical and smooth.
(H)
2. The crest edge shall be level, shall have a square upstream corner,
and shall be narrow that the water will not touch it again after
passing the upstream cmner.
3. The sides of the flume shall be vertical and smooth and shall extend
a short distance downstream past the weir crest.
4. The pressure under the nappe shall be atmospheric.
5. The approach channel shall be of uniform cross section for a sufficient
distance above the weir, or shall be provided with baffles that a
normal distribution of velocities exists in the flow approaching the
weir, and the water surface is free of waves or surges.
2
[ 1 + 0.55 d
c,,, = 0.5518 ( 3.248 + 0.02161)
H
]
Eq. 6-40
contracted Rectangular Weirs
11 ... effective length of L of a contracted weir is given by:
L = L'- O.lNH
where
Standard Weir Factor (Cw) Formulas
Numerous equations have been developed for finding the discharg
coefficient C,,. to be used in Eq. 6- 31 and Eq. 6- 32. .Some of these are given
below.
Eq.6-41
L' =measure length of crest
N = number of end contraction (1 or 2)
H = measured head
FRANCIS FORMULA
Based upon experiments on rectangular weirs from 1.07 m (3.5 ft.) to 5.18
(17ft.) long under heads from 180 mm to 490 mm.
Cw = 1.84 + 0.26 (H / df (S.I. Units)
~
l
One-end Contraction
(N = 1)
For H/ P < 0.4, the following value of Cw may be used.
Two-end Contraction
(N
= 2)
S.I. Unit, C..,= 1.84
Cw =3.33
1HIANGULAR WEIR (V-NOTCH)
REHBOCK AND CHOW FORMULA
t very low heads, the nappe of a rectangular weir has a tendency to adhere to
1
downstream face. A weir operating under such condition would give a
, y inaccurate result. For very low heads, a V-notch weir should be use~ ~s
English Unit, C,., = 3.27 + 0.40 H
p
S.I. Unit, Cw = 1.8 + 0.22 ~
, uracy of measurement is required. The vertex angle 6 of a V-notch werr 1s
1 t~<~lly between 10° to 90° but rarely larger.
302
CHAPTER SIX
Fluid Flow Measurement
CHAPTER SIX
flUid Flow Measurement
f'LUID MECHANICS
& HYDRAULICS
303
Q= ~ .fii ~H~H3t2)-~ust2 J -o}
Q= ~
.J2i (1\ HS/2)
(theoretical Q)
Q = 145 .J2i LH3/2
Actual Q = C x Q1
Q = t\ C
.J2i LH312
Eq. 6--1 2
I q. 6-42 can be used even if the side inclinations are unequal.
Figure 6 - 14: Tnangular (V-Notch) we1r
fhe discharge through tlw differential stnp IS
dQ = 11 dA
t• =
~2gl1
(neglectmg ve locity ot appr0ach)
2
tan(9/2)= L/
H
L =2H tan (9/2)
then,
Q = 1\ c.fii [2H tan(B/2)] H312
Jll - 'dll
by simi la r tnangles
X
L
Q = JL C
1-J - ,
1-1
L
.\ = - (/ 1- 11 )
H
15
L
(H -It) dh
H
c.lA = -
5 2
9
111
Q = 1.4fi5/2 (S.l. Units)
Q = 2.5fi5/2 (English)
dQ = ..!:_ j2i 111 12 (H - il) dli
II
dQ = .!_ j2i (Hil l 2 - 11 112) dh
u
Integrate
H
J2i C(l-!h /2 - , .~/2 ~li
J
9
.J2i tan-2 H 1 = C tan -2 HS/
2
Eq. 6-43
I or standard 90° weir:
dQ = J2gh .!_(II - h)dll
II
Q = .!_
H .
l •or triangular V-notch weir,
I
(I
Q = ~ J2g [H(t il 1/2 )- il1512
t
Eq. 6-44
Eq. 6-45
304
CHAPTER SIX
Fluid Flow Measurement
CHAPTER SIX
Fluid Fl ow Measurement
TRAPEZOIDAL SHARP CRESTED WEIR
The discharge from a trapezoidal weir is assumed the same as
rectangular wetr and a triangular weir in combination
305
UTTRO WEIR OR PROPORTIONAL FLOW WEIR
I• may be noted that, in a rectangular weir, discharge varies with 3/2 power of
11 .md, in a triangular weir, with 5/2 power of H. There exist a shape for
luch the discharge varies linearly with the head, the prc>portionnl flow or
1 ··ltger weir, also known as Suttro weir.
I
1-iyperbollc
*----- L ----.1
Figure 6 - 15 : Trapezoidal sharp-crested wetr
JS/2
whL•rt:'
b
u
I = - . ~ u h~tltuted tor tc111 - 111 l ~q o- 43
H
2
CIPOLLETTI WEIR
Cipolletti weirs are trapezotdal wetrs wtth s1d e slope ot I honzontal to
vertical. The add itional area at the sides adds approximately enough effectiv
width of the ~ln.'am to off<;et the <;ide contractions
Q = l/2C1tK
.fii H
Eq. 6-50
K=2xJY
-
Eq. 6-49
o;UBMERGED SttARP WEIR
I he discharge over a submerged sharp-crested we1r is affected not only by the
lll•ad on the upstream side H, but by the head downstream H2. The discharge
u1r a submerged weir is related to the free or unsubmerged discharge
t/illemonte expressed this relationship by the equation
Eq. 6-51
l:l =75.96'
a=1404c
... -
--- H1--------...... --..............
.............. ....
.....
/
----Q "= 1.859LH112 (S. I. Units)
Q = 3.37LH3!2 (English)
__ ,
,'
/
.............. ....
.... ,~ .
I
p
---
.;
/
I
I
Submerged sharp-crested weir
CHAPTER SIX
306
CHAPTER .SIX
t llJID MECHANICS
Fluid Flow Measurement
t lued Flow Measurement
IIYDRAULICS
wh.ere 11 IS the exponent of H in the equahon fo r free dischru·ge for the 5
we1r used . For rectangular weir, 11 = 3/2 and 11 = 5/2 for triangular weir
307
olved Problems
l•eoblem 6- 1
•olumetric tank 1.20 min diameter and 1.50 m htgh Wi\S filled with oil in 16
•llnutes and 32.4 seconds. What is th e average discharge?
UNSTEADY FLOW WEIR (VARIABLE HEAD)
Reservorr or tank
with constant water
surface area, A,
•Jiution
Volume
Discharge, Q = - - time
)
t
(1.2) 2 (1.5)
Disch arge, Q = ~---::-:,...,.--16 + 32.4
60
Discharge, Q = 0.1025 m3fmin = 102.5 lit/min
H
Problem 6-2
\ weigh tank receives 7.65 kg of liquid having sp. gr. of 0.86 in 14.9 seconds
What is the flow rate in liters per minute?
'Jolution
7 65
= 0.5134 kg/s
·
14.9
Mass flow rate, M = pQ
0.5134 = (1000 X 0.86)Q
Q = 5.97 x 104 m 3 /s
Q = 0.597 L/ sec = 35.82 lit/min
Mass flow rate, M =
If the flow is through a suppressed rectangular weir:
1
2
_ [
A~ dH
12
H1 Cw I H'
A
I= - • -
C11, L
H,
f
rr"' 12rtH
H,
t=
where
~[
1
CwL JH
1
2 -
JH 1
l
C,,, = weir factor
L = crest length
A,= cunstant water surface a re of reservoir or tank
H, = initial head
H~ =final head
Problem 6-3
( alculate the discharge in liters per second through a 100-mm diameter orifice
under a head of 5.5 m of water. Assume C. = 0.61 and C,, = 0.98
Solution
Q= CA J 2gH
C = Cc X C,, = 0.61 X 0.98 = 0.5978
H=5.5m
Q = 0.5978 t (0.100)2 J2(9.81)(5.5)
Q = 0.04877 m 3/ s = 48.77l/s
308
CHAPTER SIX
Fluid Flow Measurement
CHAPTER SIX
f lurd Flow Measurement
309
Problem 6-4
An orifice has a coefficient of dtscharge ot 0.62 and a coefficient of contractt
of 0 o3 Determme the coefficient of velocity for the ori fice
Solution
C =C.xL,
0.62 = 0.63 ' ( .
( ·,. = 0.984
Oil
Q = CA)2gH
s =0.82
1
0.82
H=1+25+1.5 . l.S
1.5
H = 3.487 m of glycerin
.--- ---Q = 0.65 X t (0.125)2 X )2(9.81)(3.487)
Water
Q = 0.066 m3fs
llrt,no... rn 6 - 5
A
a
B
~I 5 kPa
A1r
r
Air
r
Glycerin
s = 1.5
lm
l'roblem 6- 7
llu• discharge through a 75-mm diameter orifice at the bottom of a large tank
" measured be 1,734 liters in 1 minute. If the head over the orifice remain
. •nstant at 5.5 m, compute the coefficient of discharge.
\olution
Water
1---
~
Water
Solution
Q = lA ) 2gH
H = 1/up'ttn•,,m · Ho"'''"ln'""'
f/ =3+~ - ~
9.81
H = 6.568 m
2.5 m
125 mm 0
c =0.65
Calculate the d ischarge through the 140-mm diam eter orifice sho wn
c = 0.62
<S
SO kP
1.5 m
9.81
Q = 0.62 )( t (0.14)2 )2(9.81)(6.568)
Q = 0.108 m3fs
Problem 6 - 6
An open cylindrical tank, 2.4 m in diameter and 6 m tall has 1 m of glycerin (!l
= 1.5), 2.5 m of water, and 1.5 m of oil (Sn = 0.82). Determine the discharg•
thro ugh thP 125 mm diameter located at the botto m of the tank. Assume C
0 65
C=
_g_
QT
Since the head ,is constant, the flow is steady, thus;
Vol 1734/1000
Q = time = 1(60)
Q = 0.0289 m3js
Qr = A )2gH = f (0.075)2 )2(9.81)(5.5)
Qr= 0.04589 m3/s
0.0289
C= 0.04589
C = 0.63
t•roblem 6- 8 (CE May 1991)
, calibration test of a 12.5-mrn-diarneter circular sharp-edged orifice in a
··rlical side of a large tank showed a discharge of 590 N of water in 81
..,·onds at a constant head of 4.70 m. Measurement of the jet showed that it
'' ·IVeled 2 .35 m horizontally while dro pping 300 mrn. Compute the three
111fice coefficients.
310
311
Contraction, Cr:
Solution
C=Cc C.,
0.631 •= Cc X 0.989
C= 0.638
Theoretical values
p,
CHAPTER SIX
Fluid Flow Measurement
I IJID 'MECHANICS
IIYDRAULICS
CHAPTER SIX
Fluid Flow Measurem ent
= J2gH = J2(9.81)( 4.7)
Water
o, = 9.603 m/ s
Q , = Av, = f (0.0125)2 x (9.603)
V0
= V
toblem 6 - 9
Q, = 0.0011 78 mljs
Actual values.
Actual d ischarge·
Vol
Q = ti me (steady flow)
w 590
Vol =-= 9810
y
Vol = 0.0601 m '
Q = 0.0601
81
Q = 0.000743 mJ/s
,o mm diameter circular sharp-edged orifice at the side of a tank discharges
• = 2.35 m
1tcr ·under a head of 3 m. If the coefficient of contraction C. = 0.63 and the
h• .1d lost is 240 mm, Compute the discharge and the coefficients of velocity C..
u1d discharge C.
olution
.n~ o
F.nergy eq. between 1 and 2:
E1- HL = E2
2
-V1
2g
2
+ -P1 + Z1 - HL = -v2
2g
y
.
+ -P2 + Z2
Water
y
v2
0 + 0 + 3 - 0.24 = +0+0
2g
Actual velocity
x = 2.35 m
v2
e = oo
=2.76m
2g
v = 7.359 m/ s
If= -0.3 111
-7 actual or real velocity
Theoretical velocity:
.-----Vt = J2gii = J2(9.81)(3)
v 1 = 7.672 m/s -7 theoretical or ideal velocity
I'= 9502 m/~J
Coefficient of velocity, C,, = ..!:_
Coefficients
VI
.
- = -l1
Ve I OCity,
C..:,,
u,
Coefficient of velocity, Cv =
.
9.::>02
V e I OCity, C,. = - - - = 0.989
Or, using Eq. 6- 6:
2
9.603
Discharge, (_ =
R
Q,
Dischar e, l = 0.000743 = 0.631
g
0 001178
7 359
·
= 0.959
7.672
1
HL = ( C/ -1
HL = (
) v
c: -1)
2
28
(2.76) = 0.24
1.!:::====~ 50mm0
CHAPTER SIX
Fluid Flow Measurement
312
I'
l
-7 coeffic1ent of veloc1ty
= (., X (., = 0.63 X 0.959
C = 0.604
-7 coefficient at discharge
An orifice of 50 mm square, with C = 0.6 is
located on one side of a closed cylindrical
tank as s h own
An open mercury
manometer ind1cates a pressure head of ·
300 mm Hg in the air at the top of the tank
If the upper 4 m of the tank is o il (sp gr =
0.80) and the remamder IS water,
determine lht> dischargl:' through thl:'
o rifice
3l3
f 1u1d Flow Measurement
olution
C I 2 - I = 0 OR696
l.,. = 0.959
CHAPTER SIX
II UID MECHANICS
1.. HYDRAULICS
p., =·0.3 m Hg
Air
l1l'n the orifice IS opened, th~·
1 uge will sink a volume equal to
till' volume of water inside the
1 11 ge
Since the cross-sectional
"''a of the barge is consta nt and its
duckness is negligible, the barge
all sink to a depth equal to the
l1 •pth of water that goes in. Thus
1(w head over the orifice, being
11 bmerged, is kept constant at 1 .5
•II
011, s
=0.80
Hg
~
1he barge will smk to its top
l
o.srt=J!
Sm • 10m
- - - -- - - - . . -.. . - - -~
--
-
· ·, •
X
---r
hen x = 0.5 m
Water
Volum e= Q I
Volume= 5(10)(0.5)
Volume= 25 rn~
Solution
Q = CA~2gH
H = 4 + 4(0.8) + (-0.3)(13.6)
H = 3.12 m of water
,-----Q = 0.6 X (0.05)2 ~2(9.R1 )(3.12)
Q = 0.01173 ml/s
Problem 6 - 11 {CE May 1993}
A steel barge. rectangular Ill plan, floats w1th a draft of 1.5 m If the barge ts 1ll
m long, 5 m wide, and 2 m deep, compute the tim(• necessary to sink it to 1t~
top edge after operung a standard orifice, 180 mm ,,. diameter, m its bottom
Neglect the th1ckness of the vertical sides and assu me C = 0.60
Q = CA ~2gH
Q = 0.6 t
;o 18)2 J2(9.81)(1 .5) = 0.08283 m /s
1
25 = 0.08283 I
1 = 301 .83 sec = 5.03 min
Problem 6 - 12
l alculate the discharge through a 90-mm-dtameter
.harp e dged orifice in the figure shown Assumt> <
0 65
Air
p = 24 kPa
Oil
Solution
Q = CA ~2gH
24
H=3+---9.81 X 0.90
H = 5.718 m
Q = 0.65 X t (0.090)2 X ~2(9.81)(5.718)
Q = 0.0438 m 3 Is = 43.8 l/s
s =0.90
3m
~_L
~=====.l 90 mm v
314
CHAPTER SIX
t I UID MECHANICS
CHAPTER SIX
Fluid Flow Measurement
Flurd Flow Measurement
HYDRAULICS
Problem 6- 13 (CE May 2001)
6.09 = 0.98 J2(9.81)H
Water flows through an orifice at the vertical side of a large tank und~r
constant head of 2.4 m. How far horizontally from the vena contracta w ill
jet strikes the ground 1.5 m below the orifice?
H = 1.968 m
315
H = 1 + s = 1.968
s = 0.968
Solution
gx2
_II = X tan 9 - -....:;;--;;-2v2cos29
n
y=-1.5m
H = 2.4 m
v.,= ~2gH
v,, = .J~2(:-::-9.-=-81-:-:)(-:-2.4....:'-)
v. = 6.862 m/ s
- - - - -:E
e = oo
,,...;
'"'
2
9.81x
-1 .S = x tan 0° - ---::----::,....-2(6.862)2 cos 2 oo
t
•
X=?
-
-
-
-
-·-
-
-
-
-
-
-
x = 3.79 m
Problem 6 - 15
\ large closed cylindrical steel tank 4 m high with it1> bottom on a level
•.round contains two layers of liquid. The bottom l~yer is wa_ter 2 meters deep.
1he top layer is occupied by a liquid whose s~ec1fic grav1ty IS not known, to a
lt·pth of 1 meter. The air space at the top IS pre~s~nzed to 16_ kPa above
diameter orifice with a coeff1c1ent of veloetty
ts
1 Imosp he r e . A 50- rrun . . of 0.98
.
rtuated one meter from the bottom of the tank. The jet from the onftce ~tts the
•round 3.5 m horizontally away from the vena contracta Determme the
'pccific gravity of the liquid at the top layer
'.;olution
From the traJectory
gx2
lf = X tan 8 ·
A larg<' cylindrical ~;iteel tank 4 m high with its bottom on a level grou nd
contains two layers of liquid. The bottom layer is water 2 meters deep. The top
layer is occupied by a liquid whose specific gravity is not known to a depth of
1 meter. A 50 mm diameter orifice with a coefficient of velocity of 0.98 i
situated one meter from the bottom of the tank. The jet from the orifice hits th
ground 2.75 m horizontally away from the vena contracta. Determine th
spo::e1fic gravity of the liquid at the top layer.
·
Liquid
s =?
//=1+s
(10
= 7.75 m/s
c,,j2iH
9.81(2.75) 2
-1 = 2.75 tan 0°- ---=-''---:!-2 v 2 cos 2 oo
n
s = 0.56
2m
Vo
Water
lm
Water
1m
1m
----Jr2(9.81)H
16
9.81
lm
3.5 m
s =?
2m
H = 1 + 1 (s) + -
lm
- -
gx2
rrom y = ;\ tan e - ---7---,-2 v 0 2 cos 2 e
v.. = 6.09 m/s
1m
2
H=3.19m
H = 1 + l(s)
Air
p • 16 KPa
e
9 .81'(3.5)
-1 = 0 . --.,....;...._-;:-2 v n2 cos 2 oo
7.75 = 0.98
v,, = Cc, ~2gH
2
e = oo
llo =
Solution
2
2 v. cos
= 3.19
-.....-.
CHAPTER SIX
Fluid Flow Measurement
316
CHAPTER SIX
Fluid Flow Measurement
I I UID MECHANICS
HYDRAULICS
Problem 6- 16 (CE Board)
317
{r) Velocity of the jet as it strikes the ground
A jet is issued from the side of a tank under a constant head of 3 m. The s
of the tank· has an inclination of 1 H to 1 V. The total depth of water in the
is 6.70 m. Neglecting air resistance and assuming C,. = 1.0, determine
following:
(a) the maximum height to which the jet will rise,
(b) the point it sh·ike a horizontal plane 1.20 m below
tanJ<, and
(c) the velocity of the jet as it strike the ground.
Work-energy equation between 0 and 2
KEo + Wy2 = K£2
112- v. 2 + W h
w
= lf2 -w v11
2
2
g
g
7.672 + 4.9 = !:2_
2(9.81)
2g
112 = 12.45 m/s
Solution
l••oblem 6- 17
T
'· ll•rmine the diameter of an orifice that permits a tank of horizontal cross' lion 1.5 m2 to have its liquid surface draw down at the rate of 160 mm/s for
, \ 35-m head on the orifice Use C = 0.63
3m
6.7 m
+
3.7 m
ulution
The discharge through the orifice ts equal to the tank' s cross-sectiOnal
1
area times the draw down rate
1.2 m
Ground
Q = Atdnk X Vdrdw down
Q = 1.5 x 0.16 = 0.24 m3/s
Uo = C,, J2g [-[
v. = (1) ~~2(:-:-9..,.-81-)(~3) = 7.672 m/ s
[Q = CA.42gH I
0.24 = 0.63 X
(a) Maximum height (at point 1, Vv = 0)
From physics,
t
Q2 J2g(3.35)
D = 0.245 m = 245 mm
vl= Vol - 2gy
0 = (7.672 sin 45°)2- 2(9.81) y,
y1 = 1.5 m
-? maximum height above the orifice.
(b) Point is strike the ground (at point 2, y2 = -4.9 m)
From physics:
gx2
y = x tan 0- --=:..--,...2
2 v} cos 9
-4.9 = xz tan 45°-
(9.81)x 2 2
2(7.672) 2 cos 2 45°
0.167 X22 - Xz- 4.9 = 0
xz = 9.18 m -? horizontal distance from the orifice
I'IObJem 6- 18
5-mm-diameter orifice discharges 1.812 m 3 of liquid (sp. gr. = 1.07) m 82.2
. onds under a 2.75 m head. The velocity at the vena contracta is determined
, Pi tot static tube with a coefficient of 1.0. The manometer liquid is acetylene
t l~olbromide having a sp. gr. of 2.96 and the gage difference is 1 02
I It ll•rmine the three orifice coefficients
olution
!'he actual velocity of flow using Pitot static tube 1s given by
11 = Cp
2gRe:
_,!
318
CHAPTER SIX
Fluid Flow Measurement
where:
C1, = Pitot tube coefficient
R =gage reading
s., = sp. gr. of the gage liquid
.; = sp. gr. of the liquid flowing
t> =
Cl' 2gR( ~
I'= 1.0
CHAPTER SIX
f-lUid Flow Measurement
I LUID MECHANICS
"- HYDRAULICS
319
Coefficient of contraction
[C= c. XC,)
0.6792 = C. X 0.809
c.= 0.8396
Problem 6 - 19 {CE November 1999)
-1)
n1
2g(1.02)(~·~~ -1)
, losed cylindrical tank 5 m high contains 2.5 of water. A 100-mm circular
urfice is situated 0.5 m from its bottom. What air pressure must bt•
"'''rntained in the air space in order to discharge water at 10 hp
!.olution
t• = 5.9455 m/s
Power= Q y E
Theoretical velocity through the orifice
u, = J2gH
= J2(9.81)(2.75)
11, = 7.3454 m/s
v2
Power= (Av) y 2g
~
Coefficient of velocity, C,, = .!!..._
v,
. C,, = 5. 9455
. .
C oeffICient
o f ve IOC!ty,
7.3454
Coefficient of velocity, C,. = 0.809
10(746) = f (0.1)2 (9810)-u2(9.81)
u = 12.38 m /s
2.5 m
Air
Pressure= '
Water
u = J2gH = 12.38
H = 7.82 m
2.5 m
2m
'-----~ 0.5 m
H=2+ _!__=7.82
Theoretical discharge:
Q, =A., J2gH = f (0.075) 2 J2(9.81)(2.75)
9.81
p = 57.09 kPa
Q, = 0.03245 m 1/s
l'toblem 6- 20
Actual discharge:
Q = volume = 1.812
time
82.2
Q = 0.02204 m3fs
wncrete culvert 1.2 m in diameter and 5 m lo ng conveys flood water. Both
nds of the culvert are submerged and the difference in water level upstream
ml dO"Wnstream is 2.40 m . Calculate the discharge assuming C = 0.61
olution
Coefficient of discharge, C = _Q_
Q,
. .
0.02204
C oeffICient
o f d'tsc h arge, c = -0.03245
Coefficient of discharge, C = 0.6792
Q= CAJ2gH
Q = 0.61 X f (1.2)2 J2(9.81){2.4)
0 = 4. 734 m 3/s
CHAPTER SIX
Fluid Flow Measurement
320
I 1 1110 MECHANICS
IIYDRAULICS
Problem 6 - 21
321
Theoretical discharge, Q, · 0 033 m 1 /s = 33 L/s
- 23.41
l = -33
m 3 js
It ts desired to dtvert 5.1
water from a pool whose water su
elevation is 45 m, to an adjacent pond whose water surface elevation is 42
by means of a short concrete culvert 8 m long and with both ends suhrr•Prj:"'dl
What size of culvert is needed assuming C = 0.58?
c = 0.709
t oefficient of contraction.
Solution
C=C..xC,
El. 45 m
Q.7Q9 = C. X 0.78
I .
Q = CA ~2gH
l
.
\c,,,~
Pond
D-=_7_ __
H = 45-42 =3 m
5.1 = 0.58 X ;'} (0)2~2(9.81)(3)
0 = 1.21 m
em 6-22
A 75-mm-diameter orifice discharges 23.41 liters per second of liquid under 1
head of 2.85 m. The diameter of the jet at the vena contracta is found hy
callipering to be 66.25 mm. Calculate the three orifice coefficients
Solution
Coefficient of contraction
c. = .!!_ = {!__2
A
0
c = 66.25
2
c. = 0.909
El. 42 m
Pool
'
CHAPTER SIX
Flurd Flow Measurement
75 2
hoblem 6- 23
,tandard short tube 100 in diameter discharges water under a head of 4.95
A small hole, tapped in the side of the tube 50 mm from the entrance, is
11
•nnected with the upper end of the piezometer tube the lower end of which
~ubmerged in a pan of mercury. Neglecting vapor pressure, to what height
til the mercury rise in the tube? Also determine the absolute pressu re at the
•pper end of the piezometer tube
..,olution
Note: For standard short tube&, t1w
,••••ssure head at point near the entranrt>
1
0.82// <;pp pag~ 277
c. = 1.0
f.!!. = -0.82H
v
= -0.82{4.95) = -4.059
Qr
Actual discharge, Q = 23.41 Ljs
Theoretical discharge, Qr = A ~2gH
Theoretical discharge, QT = f (0.075)2 ~2(9.81)(2.85)
/
p, • -0.82H
v
Jl• = -4.059(9.81)
p.. = -39 82 kPa
~t·glecti ng vapor pressure
C= _9_
' ' ~:::;;::==J....,.:--
-~·.~~·::_.
p,
(. = 0.78
Coefficient ot discharge
'
c =c.= 0.82
II = .l!!!_ =
39.82
Y111
9.81 X 13.6
h = 0.298
11 = 298 mm
T
h
MerCur"'t'
322
CHAPTER SIX
CHAPTER SIX
flutd Flow Measurement
II UID MECHANICS
Fluid Flow Measurement
tfYDRAULICS
Problem 6 - 24
ulution
300 mm 0
A Borda's mouthpiece 150 mm in diameter discharges water under a head
m. Determine the discharge in m3/ s and the diameter of the jet at the
conb·acta.
Oil, s =0.85
Solution
Under ideal conditions, the coefficients of a Borda's mouthpiece are C. "'
and C.,= 1.0. See page 293.
~
180 mm
Diameter at vena contracta:
a= CcA
Td2 = 0.50 X T(150)2
«>
_j
Mercury
d = 106.1 rnrn
Energy equation between 1 and 2 neglecting head lost (theoretical)
E, = E2
Discharge:
11
Q = CA~2gH
C= Cc XC,
2
V2
P1
2
P2 + Z2
1 + - + Z1 = - - + 2g
y
2g
y
BQ, 2
( = 0.5 X 1.0 = 0.50
+
p
1
+ 0=
8Q·,2
+0
4
rr g(0.075)
___,_.:.:::..!.---;-
2
y
Q = 0.5 X t (0.15)2~~2(~9.8-:-1.,...,.)(3-:-:-)
_
Q = 0.0678 mljs
P1
2601Q72 - -
-7 Eq . (1)
y
Sum-up p ressure head from 3 to I in meters of oil
Problem 6 - 25
Oil discharges from a pipe
through a sharp-crested round
orifice as shown in the figure.
The coefficients of conb·action
and velocity are 0.62 and 0.98,
respectively.
Calculate the
discharge through the orifice
and the diameter and actual
velocity in the jet.
300 mm 0
12_ + 0 18lli.. - 0.75 = E2_
y
Oil, s =0.85
Pl
.
0.85
y
= 2.13 m of o il
y
In Eq. (1):
2601Qr = 2.13
Q1 = 0.0286 -7 Theoretical discharge (smce IlL IS neglected)
~
180 mm
Mercury
Actual discharge, Q = CQT = C C,, Q,
Actual discharge, Q = 0.62 x 0.98 x 0.0286
Actual discharge, Q = 0.0174 m 3 /s = 17.4l/s
323
324
CHAPTER SIX
ILUID MECHANICS
Fluid Flow Measurement
I. HYDRAULICS
hi flows through a pipe as shown in the figure. Determine the discharge of
ul m the pipe assuming C = 0.63.
02
0.62= ~
75 2
rl = 59.1 mm
250 mm 0
) 100 mm 0 orifice
Actual velocity:
Theoretical velocity, Vr = Qy
A
0.0286
Theoretical velocity, vr =
= 6.474 ml s
1-(0.075) 2
Actual velocity, v = C, vr
Actual velocity, v = 0.98(6.474)
Actual velocity, v = 6.344 mjs
z
---+...
350mm
Mercury
olution
Q, = Q2 = Q
Energy equation between 1 and 2 neglecting head lost (theoretical):
.\nother Solution:
E1 = E2
I he Discharge through this type of orifice is given by:
2gEl
Q= CAo
y
p
v2
p
+ - 1 + Z1 = - 2- + _1. + Z2
2g
y
2g
y
v2
1
- -
8Qy2
1t 2 g(0.250)
P1 0
8Q/
1 + y + = 7t 2 g(0.100) 4
805 Qrl = .!?..!_ - El_
y
y
where
C =coefficient of discharge
Ao = area of the orifice
7 Eq. (1)
Sum-up pressure head from 1 to 2 in meters of oil:
.!?..!_ = pressure head at 1 in meters or feet of the fluid flowing
.!?..!_ + z + 0.35 - 0.35 13·6 - z = El_
y
0.91
y
0 = diameter of orifice
Dp = diameter of pipe
.!?..!_ - h
y
y
y
= 4.881 m of oil
In Eq. (1):
8os Qr = 4.881
Qr = · 0.07787 rn31s
c = Cr X ell= 0.62 X 0.98
C= 0.6076
Q = 0.6076 X T(0.075)2
2 9 81 2 3
( · )( .1 )
1- (0.6076) 2 (75 1300) 4
Q = 0.0174 rn31s = 17.41/s
325
Problem 6 - 26
Actual diameter of the jet, d:
a
d2
Cc=- = -
A
CHAPTER SIX
FlUid Flow Measurement
Actual discharge, Q = C Qr = 0.63(0.07787)
Actual discharge, Q = 0.0491 m3Is = 49.11/s
& HYDRAULICS
The discharge through this orifice is given by:
2g P1- P2
Q = CAo ,_ _ _Y.:.....__
1-(A0 I Ap) 2
<;olution
Since the cross-sectional area
of the tank is constant, we can
use Eq. 6- 27.
2
t=
where
CHAPTER SIX
Flurd Flow Measurement
r LUID MECHANICS
CHAPTER SIX
Fluid Flow Measurement
326
C = coefficient of discharge
Ao = area of the orifice
1.5 m 0
~ 11
~---13m
~ [Ft -~]
CAO 2g
H1 = 2.5 rn
H2 = 1.0m
:r= ·: : : :~:::: : :
El = pressure head at 1 in meters or feet of the fluid flowing
y
!2. = pressure head at 2 in meters or feet of the fluid flowing
D
\1100 mm 0 orifice
y
Ap = area of pipe
Aol Ap = (Dol Dp)2
327
t=
2X
t (1.5)
2
( ,-;:;-;:
0.60 X t (0.1) 2 X ~2(9.81)
(:;-;;)
lc = o.Go
\..;2.5-..;1.0
t = 98.4 seconds
2g P1- P2
Q = CAo
Y
1 - (A2 I Al )
Q = 0. 63 X t (0.1)2
2
2(9.81)(4.881)
1-
Q
[(~)2
r
= 0.0491 m3ls
Q = 49.1 L/s
Problem 6 - 28
1\ 100-mm-diameter orifice on the side of a tank 1.83 min diameter, draws the
urface down from 2.44 m to 1.22 m above the orifice in 83.7 seconds.
l ·,,lculate the d4Jcharge coefficient?
Solution
Since the head vary, the flow is unsteady.
2
-27
A 1.5-m-diameter vertical cylindrical tank 3m high contains 2.5 m of water.
100-mm-diameter circular sharp-edged orifice is located at its
Assume C = 0.60.
(a) How long will it take to lower the water level to 1 m deep after openi
the orifice?
(b) How long will it take to empty the tank?
t=
~ [Ft -~]
CAD 2g
2
83.7 =
CX
t (1 '83
)
2
2
t (0.100) ~2(9.81)
[.J2.44 - -/1.22]
C= 0.8265
Problem 6 - 29 (CE May 1999)
1\n open cylindrical tank 4 rn in diameter and 10 rn high contams 6 rn of water
uld 4 m of oil (sp. gr. = 0.8). Find the time to empty the tarlk through a 100•nm diameter orifice at the bottom. Assume C.-= 0.9 and Cu = 0.98.
CHAPTER SIX
Fluid Flow Measurement
328
Solution
t=
CHAPTER SIX
Fluid Flow Measurement
I I UID MECHANICS
I. HYDRAULICS
t =?from H, = 10 to H2 = 10
2~ (JH; -JHz)
t = 3.162lJiO-
CA.y2g
2A 5
2t(4) 2
CAf2i
(0.9 x 0.98) t (0.1) 2 ) 2(9.81)
Oil, S =0.8
.J4]
11roblem 6 - 31
I h1• initial head on an orifice was 9 m and when the flow was terminated the
h•·.1d was measured at 4 m. Under what constant head H would the same
•ttfice discharge the same volume of water in the same interval of time?
::= = 819.1
CA.j2i
----:::::
Water
6m
\olution
Under variable head:
(.J9.2 -..[3.2)
ft = 819.1
t =
t, = 1019.3 sec
2A5
(.JH; _.JH,)
CAo.fii
1=
Time to empty the oil:
Ht = 4
H2=0
2A5
(
J9 _J4)
CAofii
2A
5
I=----''==
(F4 -../0)
t2 = 819.1
t2 = 1638.2 sec
CAo.fii
Under constant head:
Volume= Q t
Total time to empty= ft + t 2 = 2657.5 sec
Total time to empty= 44.3 min
,-:;-::u
A,(9 - 4) = CA• ...; "-o' •
Problem 6 - 30
2A,
,-::;-::
CA 0 ...;2g
5=2../H
H=6.25 m
A tank circular in cross-section is 10 m high. It takes 10 minutes to empty It
through a hole at the bottom when the tank is full of water at the start. How
long will it take to drop the upper 6 m of water.
11roblem 6- 32 (CE Board 1991)
Solution
t~
6• 4
t = 3.675 minutes
2A 5
Time to empty U1e water:
H, = 6 + 4(0.8) = 9.2
H2 = 4(0.8) = 3.2 m
329
2A
5
CAo.fii
t=K
[~ _ _JH;]
1
2
lFt- JfG j
t = 10 min from H1 = 10 m to H2 = 0 m
10 = K[M
K = 3.162
-../0]
' vertical cylindrical tank has an orifice for its outlet. When the water surface
the tank is 5 m above the orifice, the surface can be lowered 4 m in 20
.. unutes. What uniform air pressure must be applied at the surface if the same
olume of water is to be discharged in 10 minutes'?
111
330
CHAPTER SIX
Fluid Flow Measurement
I LUID MECHANICS
loa HYDRAULICS
CHAPTER SIX
Fluid Flow Measurement
331
Problem 6 - 33 (CE 1992)
Solution
t= CAO2/zg
[~ -~]
2g
2 As
let K =
CA 0
..j2i
Water
t=K[Ft -~]
..............................
-··
.. _........ --- .. -........... -..........
In Figure (a):
20 min
H1 = 5 m; Hz = 1 m
t=
t=K[~ -~]
K[vts -Ji]
20 =
K = 16.18
\ composite non-prismatic 5 m !ugh
ylmdrical tank shown has a frustum of a
nne at the bottom with upper base
.t,ameter 2.5 m, 1.25-m-diameter at the
•ottom, and 2 m high.
The bottom
ontains 100-mm-diameter sharp edged
11rifice with coefficient of discharge of 0.60
If fully filled to the top, determine the time
lo empty the tank in minutes.
.................................
')olution
Figure (a)
time= 20 min
I he cross sectional area from
h•vel1 to level 2 is constant.
2'fig [JH, -~]
,, =
CA 0 2g
In Figure (b):
t = 10 min
A,= f (2.5)2 = 4.91 m2
c A • ..j2i = 0.6 X f (0.1)2J2g
C A • ..j2i = 0.02087
Water
let c = !!.E..
y
____ .............
H1 =5+ l!..!!_ =5+c
--------
-...... .......................... -- .. ....
y
y
t=K[~ -~]
10 = 16.18 [.J5 + c - .J1 + c ]
..JS+C = 0.618 + ..ff+C
Squaring both sides:
5 + c = 0.3819 + 1.236../1+C + 1 + c
5+C =2.9273
Squaring both sides:
1 + c = 8.569
c = 7.569 m = !!.E..
y
H1 = 9 m; Hz= 4 m
2 4 91
t, = ( · ) ~.J9= 470.5 sec
0.02087
.J4]
Hz = 1 + E.!!. = 1 + c
Pn = 7.569(9.81)
P• = 74.25 kPa
Water
Figure (b)
time= 10 min
From level 2 to level 3:
f4 As dll
Jo CA 0 ~2gH
2
t =
As = 1tr2
r = 0.625 + x
X
- 0.625
- - x= ~11
4
II
4
r = 0.625 + 0 ·~25 h = O ~25 (4 + /r)
A, = 1t( 0 ·~25 (4 + Iz))l
As= 0.0767(16 + 811 + 112)
H=Jr
fz =
f 0.0767(16+8Jt+h )dil
Jo CA• ..j2iJh
4
2
0.625
Sm
9m
CHAPTER SIX
Fluid Flow Measurement
332
t2 =
i
4
0.0767(16 +811+11 2 )dh
0
=
0.02087
.fh
CHAPTER SIX
Fluid Flow Measurement
II Ill) MECHA NICS
IIYDRAULICS
= 3.675
£1
t1611-l/ 2 + 811 1 1 2 + 11 3/ 2 ~~~
0
3.675 [16(2)11 1 1 2 + 81.113/2 + 1.Jr5f2
3
5
J
II,
I
1
3.5 m; H2 = 0
35
2
" 0.144(12.25 + 7h + 11 )dh
-0
4
0
= 3.675 [32(4) 1 + Jf-(4) 3 /2 + !(4)5/2] .
1 2
480
0.62A 0 ~2(9.81)Jh
i&~.25!1 -l 12 + 711 112 + 17 3 ' 2 ~~~ = 9154.2A.
t2 = 439.04 sec
9154.2A., = [2511 112 +
Total time to empty, t = t 1 + b
rota! time to empty, t = 470.5 + 439.04
fotal time to empty, t = 909.54 sec= 15.16 minutes
333
~ ,3 /2 + 3.. 11 sn J?>.5
3
9154.2An = [ 25(3.5) l/ 2 +
l;
5
(I
(3.5)?>/
2
+
~ (3.5) 512
1-
0
A., = 0.0094486 m 2 = f 02
n = 0.1097 m = 109.7 mm
A tank in the form of a frustum of a right circular cone 1.50 min diameter an
t'hc botto~, 3-~-diameter a~ the top, and 3.5 m high, is full of water. A sharpe
edged onfice wtth C = 0.62 ts located at the bottom of the tank. What diamct
of orifice is needed lo empty lhe tank in eight minutes?
Solution
t=
fH' A .; dh
JH Qo
2
Qout = CA ~2gH
Qo = 0.62Ao ~2glz
A, = n[ (~:i (3.5 + 11))2
A.= 0.144(12.25 + 7/z + h2)
t = 8 min = 480 sec
•harp-edged orifice 100-mm in diameter, in the side of a tank having a
• •IIZOntal cross-section 2 m square, discharges water under a constant head
1"'' rate of inflow over which the head was kept constant is suddenly changed
• "ill 20 lit/ sec to 30 lit/ s. How long will it take, after this change occurs, for
Ito• head to become 2 meters? The coefficient of discharge may be considered
••llslant and equal to 0.60
·
ulution
Qo= CAo~2gH
H= 1z
A s = nR2
R = 0.75 +X
X
0.75
II
3.5
X= 0.75 lJ
3.5
R = 0.75 + 03.5
·75 h
·75 (3.5 + /1)
R = 03.5
1'1oblem 6- 35
= 0.60 X f (0.10)2~2(9.81)/t
Qout = 0.02087 h112
[
I
------~~-~~-0.6~
·olving for Jz1 :
Since the head was kept constant
when Q,.. = 20 L/ s, therefore
Qout = 20 L/s = 0.02 m3/s
Qout =0.02087 h, l /l = 0.02
Jz, = 0.918 m
-r-------------
,'
/1
I
I
-=-----!-------------- ,'
I
I
~ 0.75 m
1.5 m
,'
/
/
,"
_, _,L--------
,
CHAPTER SIX
Fluid Flow Measurement
334
CHAPTER SIX
Flurd Flow Measurement
It liD MECHANICS
IIYORAULICS
335
When the inflow was suddenly changed to 30 L/s
1
=
i
HI
H
1
h' rectangular tank shown ts divtded by a parhtl.on tnto two chambers and
with a round 150-mm-diameter sharp-edged orifice at the lower
dton of the partition. At a certain instant, the level in chamber B is 3 m
1 hl'r than it is in chamber A. How long will it take for the water surfaces in
I two chambers to be at the same level? Assume C = 0.62
A~ dh
1 ,, tded
Qm - Q,ut
Q,n = 0.03 m-1/ s
Q""' = 0 02087 11 111
A,= 2 x 2 =4 m2
H1 = h1 = 0.918 m
H1 = h1 =2 m
I_
2
1
2
4dh
0 911! 0.03 - 0.0208711
112
1
n91H (0.02087)(1.437 _ 111 /2)
2
I= 191.661
dil
l437 - Jt1!1 = -'
Jtl l2 = 1.437- .\
191 .66
1111224
1
1
- 2(1.437- x)d:.\
0.0229
n 470
t
--
{IJI
\
I 4371n 1 - .\
J() 474
= -383.32[1 437(1n 0.0229- In 0.479)- (0.0229- 0.479)]
I = 1.500 sec = 25 minutes
I
•
O,olution
t=
·I
~' :~
1 437
( ·
dx - dx)
1)022'1
= -383 32
Chamber A
r
047u
= -383.32
Chamber A
l
When h = 0.918; x = 0.479
When h = 2; x = 0.0229
=
7.5 m
3m
h = (1.437- x)2
dh = 2(1.437 - x)(-dx) = -2(1.437- x)dJ.
1
~3m-1
r
1 2
0 'liN 1.437 - II /
Let
~
N
4dh
Chamber A
Chamber A
;-
E
~3m-1--
fH, As2 dH2
JH
2
Qo
II
·I
dHz
Q = CA.~2gH
dH= dH, + dH2
(dV, = dV2J
As, dH, = As2 dH2
As2
dH, = - d H
A,,
I
l
r-
- - - -H
dH1 J.
A,1
1'
~·- · -·- · -
!
1'
--
I
~' ~~
2
dH = A 5 2 dH 2 + dH2 = ( A 5 2 +
A.,
7.5 m
1)
dH
- ·- ·-·- ·- ·- ·- ·- ·- - ·
CHAPTER SIX
Fluid Flow Measurement
336
CHAPTER SIX
Flu1d Flow Measurement
337
•lution '
v,
y
11------.........,jl+
v, = v2
: (2)2 y = t (3)2(1)
1
iH, A A s2
CAn fi8
H~ A,, + A,2
I=
sl
H - 1!2dH
'I= 2.25 m
7 Formula
lh+y+1=5
=5-1- 2.25
I I?= 1.75 m
When /\,, and A,2 are constant
I=
A,, As2
I
A, 1 + A, 2 CA., fii
i ~-1
H, 12rtH
I
A,lA,2
[ 2HI /2
1\ , 1 + A, 2 Cl\.,fi8
I=
A, 1 A., 2
A" + A,2
Tank 1
J:
Cl\.,~ ~~ - ~]
I=
H, =5m
A,, = f (2)2 = n m 2
7 rormula
A,2 = f (3)2 = 2.25n m 2
J2i "= 0.60 +(0.20)2 J2(9.81)
CA. J2i = 0.0835
H, = tmtial head = 3m
H2 = final l'lead = 0 m
CA.
A,, = cross-sectwnal area of tank 1 = 3 x 2 =6 m2
X
1\,2 =cross-sectional area of tank 2 = 7 5 " 2 = 15 ml
I = 6(15)
2
6+15 0.62>cf (015) 2 j2g
[.f\ _JO]
.
t = 305.91 seconds
Problem 6 - 37
Two vertical cylindncal tanks 1 and 2 having diameters 2 m and 3 111
respectively, are connected with a 200-mm-diameter tube at its lower portion
and having C = 0.60 When the tube is closed, the water surface in tank 1 is 5
meters above tank 2. How long will it take after opening the tube. for the
water surface in tank 2 to rise by 1 meter?
1=
n(2.25n) _2_
n + 2.25n 0.0835
[JS _J1 _75 J
1 = 47.57 seconds
,,roblem 6 - 38 (CE November 2000)
vertical rectangular water tank is divtded tnto two chambers whose
horizontal sections are 3 m2 and 5 m2, respectively. The dividing wall is
l'rovided with a 100 mm x 100 mm square hole located 0.5 m from the bottom
md whose coefficient of discharge is 0.60. Initially there is 5 m deep of water
111 the smaller chamber and 1 m deep of water in the larger chamber What 1s
thP difference in the water level in the two chamhers after 2 minutes?
CHAPTER SIX
Flutd Flow Measurement
CHAPTER SIX
Fluid Flow Measurement
338
339
Solution
From the formula:
A,t A,2
2
( IU !U)
t = Asl +As2 CA"fii :V Ht - v H2
H 1 =llt=4m
H2 = !12 =?
t = 2 min = 120 sec
A_.l
2
Ast + 1 CA n'>J"-:5
f2:g
120 = 3(5)
2
(J44 ~)
3+5 0.6(0.1x0.1)J2'i
- '~" 2
2As1
CAn Jfi
1
-
2
-
2
0 + 1 CA ,...;"-,'<;
~
[F, JH;]
...;n1
...;n2
79.35 seconds
T
llwblem 6 - 40 (CE Board}
.wimrning pool15 m long, 10m wide, and 3m deep at one end and 1.6 m
11 the other end is fitted with a drain pipe 200 mm in diameter at the lowest
1 •tl the pool. Compute the time required to drain the full content of the pool
~uming C = 0.80.
4m
3m 0
Solution
T (3)2
A,t = 2.25Jt m 2
A,2 =co
-
l fJT _ IJ-/]
\11lution
Asl =
2
1
2(2.25n)
[./4-~]
2
0.75 ~ t(0.2) ~2(9.81)
112 = 1.32 m
Problem 6In the figure shown, how
long does it take to raise
the water surface in the
tank by 2 meters? The
right side of the figure is a
large reservoir of constant
water surface e levation.
lF, FG]- A,l
,;.r-
-t_
4 m ·---
2m
200 mm 0
c =0.75
fi.me from Ievell to level 2 (constant water surface area)
t=
2Afig [JH; -JH;"]
CAn 2g
CHAPTER SIX
Fluid Flow Measurement
340
CA.
J2i = 0.80 -T (0.2()p .J2i
A , =10 x 15=150m~
= 2(150) l.J3 -JU']
0.111
t, = 1483.3 seconds
"
f
lime trom level 2 to level 3:
1,
•
A , dh
CA ,, ~2gh
/\, =10.r
15
0
"
1.4
J 2 =(_,X }f
d2 = 0.85 x I = 0.85 111
A = 10(10.71411) = 107.1411
C/\,.~2gl1 = 0.80 X
+(0.20) ~1gl1
2
0
Energy eq uatiOn betwt:!en
0 and t} neglecting losses
C/\ ., ~2gl1 = 0.11111 1 2
[H, = H2]
It ,= 1.4
~ + ~ = .2_ + () 85
2g
2g
"2= 0
•
lu1ce gate flows into a
ontal channe l as shown
llu· Figure Determine the
11 th rough the gate pe1
11'1 w tdth when If = 1 0 m
1I tl 1 = 6 m Ass ume thai
pressure distribution al
Ill II\~
1 cmd 2 to be
'"''sphene
and
neglect
11on losses 1n Lhe channel
, oefficie nt of contraction
0.85 a nd coefficient of
I h ltV c. = 0 95
.. lution
·'
·' = I 0. 71-lh
f, =
(CE May 2003)
1
1/, = 3 Ill
H2 = 1.-l m
f, =
iu
l07.1411dl1
0
341
X
CA ,,.fii = ll.111
l,
CHAPTER SIX
Fluid Flow M easurement
IIJID MECHANICS
I IYDRAULICS
1.4
1 2
1
= 965.225 Jt 1 dl1
-? Eq ( l 1
(Q, = Q2]
(1
= 965.225 [ 3._11 ~~ 2 J I..!
3
2
= 5. 15
2g
t•21 - v 12 = I en .04::1
0.1 1111 l / 2
•
2
Cons•oer 1 m lengtn o··
gate
0
= 965.225* (1.4''2 _ 0)
/2 = 1066 seconds
Total time to empty, t = t, + t 2
= 1483.3 + 1066
Total time to empty, t = 2549.3 seconds= 42.49 minutes
(6 X 1)711 = (0.85 X 1)71?
p,=0 1417V?
In Eq. (1).
o22- (0.1417v2) 2 = 101 .043
0.979937'?2 = 101 .043; u1 = 10. 154 m / ~
Actual veloci ty at 6 , v, =C. v2 = 0.95(10.154) = 9.6467 m /~
Disch arge = A2 v, = (0.85 x 1)(9.6467) = 8.2 m3js per meter
342
CHAPTER SIX
Fluid Flow Measurement
Problem 6- 42 {CE Board)
CHAPTER SIX
f IUid Flow Measurement
1 LUID MECHANICS
,., HYDRAULICS
..,olution
A horizontal 150 m.rn diameter pipe gradually reduces its section to 50
diameter, subsequently enlarging into 150 mm section. The pressure in
150 -mm pipe at a point just before entering the reducing section is 140
and in the 50 mm section at the end of the reducer, the pressure is 70 kPa
600 mm of head is lost between the points where the pressures are
compute the rate of flow of water through the pipe.
Solution
375 mm
] Mercury
1$ = 13.6
-------*-
(.2, - Q, = ()
ISO mm,
(a)
Energy equcllton between 1 and 2 neglectmg lwad ln., t
r:, =E)
2
2
-li t
o, = 140 kPa
+1'1
- + z,
2g
y
= ~ + 1'2 + '1
4
2g
y
+
Energy equation between 1 and 2.
E,- HL = £2
u 2
2
_l_ + El + z, - HL = !2_ + E2 + Z2
2g
y
2g
y
8Q 2
+ 140 + 0- 0.60 =
8Q2
+ ?0 + 0
rr 2 (9.81)(0.15) 4
9.81
rr 2 (9.81)(0.05) 4
9.81
161 . 2 Q2 =
P1- P2
11
-7 Eq. (I)
y
P1 - P2
So 1v 111g for "--"---'-Y
Su m-up pressure head from 2 to 1 in meters of water
13057Q2 = 6.5356
Q = 0.0224 m3/s = 22.4[/s
f!.1.. + y + 0.375(13.6)- 0.375- 11 = E1._
y
Pl - p2 = 4.725 m
y
Problem 6 - 43
A 150 mm diameter horizontal Venturi meter is installed in a 450-mm
diameter water main. The deflection of mercury in the differential manomettor
connected from the inlet to the throat is 375 mm. (a) Determine the discharg
neglecting head lost. (b) Compute the discharge if the head lost from the inlet
to the throat is 300 mm of water, and (c) what is the meter coefficient?
In Eq. (1,)
161.2 Q 2 = 4.725
Q = 0.1712 m3fs (theorettcal dtscharge)
(b) Energy eq. between 1 and 2 considering head lost
£ 1 - HL = E2
2
111
p,
-
2g
+ -
y
CJ2
2
+z,-HL=-- + £2. + Z2
y
2g
8Q 2
8Q2
+ P2 +ll
+P1
- . 0.30 + 0 = 2
4
4
2
y
y
rr g(0.15)
rr g(0.45)
343
CHAPTER SIX
Fluid Flow Measurement
344
161.2 Q 2 = Pl - p 2 - 0.30
y
-7 Eq. (1)
p
T't
-2..
+ 0.75 + y + 0.36(13.6) 0.36- lj = y
•
y
y
Q = 0.1657 m3fs
(actual discharge)
In Eq. (1):
153 Q,2 = 5.286 - 0.75
Q 1 = 0.1722 m3js -7 (theoretical discharge)
Meter coefficient, C
0.1657
0.1712
Qaclual
C=
f IUid Flow Measurement
P1 - P2 = 5.286 m of water
161.2 Q2 = 4.725- 0.30
(c)
CHAPTER SIX
llUID MECHANICS
1w. HYDRAULICS
Q theoretical
c = 0.968
Actual discharge, Q = C Q,
= 0.68 X 0.1722
Actual discharge, Q = 0.1171 m 3/s
Problem 6 - 44
A vertical Venturi meter, 150 mm in diameter is cotmected to a 300-mm
diameter pipe. The vertical distance from the inlet to the throat being 750 mm
If the deflection of mercury in the differential manometer connected from th
inlet to the throat is 360 nmt, determine the flow of water through the meter 1f
the meter coefficient is 0.68 Determine also the head lost from the inlet to th
throat.
2g
2
Energy equation between 1 and 2 considering head los.t:
£,- HL = £2
2
2
~ + E.:!_ + Z1 - HL = !2_ + f!2_ + Z2
2g
2g
y
y
2
8(0.1171)
+ f.:!_ + O _ HL =
8(0.1171)
+ .!!.2. + 0.75
4
2
4
y
1t 2 (9.81)(0.3)
y
7t (9.81)(0.15)
Energy Eq. between 1 and 2 neglecting head lost:
£1 = L2
v2
Q, = Q2 = 0.1171
2
Solution
_1_
I lead lost: (usc the actual discharge)
p
v2
p
y
2g
y
+ _ 1 + Zl = _2_ + _2_ + Z2
+ E.l. + 0 =
8Q/
rr (9.81)(0.3)
4
y
HL = Pl - p 2 - 2.098- 0.75
y
= 5.286- 2.098- 0.75
HL = 2.438 m
8Q/
rr 2 (9.81)(0.15) 4
+ !!2. + 0.75
y
153 Q,2 = Pl - p 2 - 0.75 -7 Eq. (1)
y
Sum-up pressure head from 2 to 1
in meters of water
Cv = C = 0.68
Q = 0.1171 m3/s
2
2
(0.152
)
]
8(0.1171) 4
1
)
[
HL= - -2- 1 1- - 2
( 0.68
0.3 2
1t (9.81)(0.15)
HL = 2.439m
345
346
CHAPTER SIX
I LUID MECHANICS
Fluid Flow Measurement
f. HYDRAULICS
Problem 6 - 45
347
Pl __
P2
= !/1- !/2
Air
Neglecting
losses,
calculate
the
discharge
through
the Venturi meter
shown.
CHAPTER SIX
Fluid Flow Measurement
y
y
Also, from the figure:
z + !/1 = 0.25 + !/2
!/I - !/2 = 0.25 - z
!.2 - 1!.2. = 0.25 - 2
y
y
In Eq. (1):
153.01 Q2 = 0.25 - 2 + 2
Q = 0.0404 m3js = 40.4 lfs
Solution
a
Problem 6 - 46
~ 37.5 rom Ventu ri meter (C = 0.957) is installed in a 75-mm-diameter
horizontal pipe carrying oil having specific gravity of 0.852. If the recorded
llow in the meter was 1.5 liters per second, what could have been U1c
deflection of water in the d ifferential m anom eter connected between the inlet
oll\d the throat?
Solution
Q, = Q2 = Q
Energy eq. between 1 and 2 neglecting head lost:
E, = £2
v2
p
v2
p
_1_ + _1 + ZJ = _2_ + __2 + Z2
2g
y
2g
y
1t
2
8Q2
+ El + z =
8Q2
2
4
g(0.3) 4
y
1t g(0.15)
153.01Q2 = El - J!2 + 2 -7 Eq. (1)
y
y
Sum-u p pressure head from 1 to 2 in meters of water:
Note: Neglecting density of air, the pressure in air at any point in the manometer
is equal)
Pt - y, +y2-_ P2
y
y
Actual discharge, Q = 1.5 L/s = 0.0015 m 3/s
Since the head lost is not known, the theoretical discharge will be used.
Q= CQr
0.0015 = 0.957 Qr
Q1 = 0.001567 m3js
Ql = Q2 = 0.001567
CHAPTER SIX
Fluid Flow Measurement
348
CHAPTER SIX
Fluid Flow Measurement
Energy equation between 1 & 2 neglecting head lost: (using Q1}
37.5 mm 0
£1 = £2
vl
•
-
2
2g
Pt
+ -
y
V2
2
P2
+ Z1 = - - + 2g
y
37.5 mm0
+ Z2
2
8(0.001567) + El + 0 = 8(0.001567) 2
+ 12 +0
rr 2 g(0.075) 4
Y
rr 2 g(0.0375) 4
y
y
ht
_1_·-·-·-· ·-+-·-·-·-·- ·- ·-·-·-·Mercury
El - E1. = 0.09618 m of oil
y
349
y
Sum-up pressure head from 2 to 1 in meters of oil:
P2+ y+ -IT- -h-y=Pt
y
0.852
y
0.1737fz = El - 12
y
y
0.17371! = 0.09618
h = 0.554 m = 554 mm
Q1 = Q2 = Q3 = 0.0085 m 3 /s
v2 2
8(0.0085) 2
I fLJ-2 = 0.05 = 0.05 2
2g
rr g(0.025) 4
I /L1. 2 = 0.764 m
8(0.0085) 2
_
v 2 _
//L2•3 - 0.22- -0.20 2
2g
rr g(0.025) 4
I fL2-3 = 3.057 m
J
Energy equation between 1 and 2:
£1- HL = £2
Problem 6 - 47
2
Oil (sp. gr. = 0.8) flows at the rate of 8.5 liters per second through a 25-mm
diameter horizontal Venturi meter, which is attached to a 37.5-mm-diameter
pipe as shown in the figure. A differential manometer containing mercury 11
attached from the base of the inlet to the throat and to the base of the outlet
Calculate the deflection of mercury in each tube if the head lost from the inil!t
to the throat is 5% of the velocity in the throat and from the throat to the outlt>t
is 20% of the velocity head in the throat. ·
2
!l_ + El + ZJ - HLJ-2 = .:2_ + 12. + Z2
2g
2g
y
y
2
2
8(0.0085)
+ !i + 0- 0.764 = 8(0.0085)
+ El.+o
4
2
2
4
y
7t g(0.0375)
y
7t g(0.025)
Ii - 12. = 13.03 m of oil
y
y
Sum-up pressure head from 2 to 1 in meters of oil:
37.5 mm 0
12.
+ 01- hl) + h1 13.6 -.y = 12.
y
0.80
y
16h1 = 12. - 12. = 13.03
y
y
l11 = 0.814 m = 814 mm
Mercury
Energy equation between 1 and 3:
£1 - HL1.2 - HLz.3 = £3
CHAPTER SIX
Fluid Flow Measurement
350
_.:....._...,.......:::::!___
HYDRAULICS
+ El + 0- 0.764 - 3.057 = ~..:..._..,,........,::.:_~ + 12. + 0
y
y
El. - p3 =3.821 m of oil
y
CHAPTER SIX
Fluid Flow Measurement
f I UID MECHANICS
351
t•roblem 6-49
tass tube with a 90° bend is open at both ends. It is inserted into a flowing
1
,,,:am of oil (sp. gr. = 0.90) so that one of the opening is directed upstream
,, 11 1 the other is directed upward. If the oil inside the tube is 50 mm higher
th.m the surface outside, determine the velocity measured by the tube.
.olution
y
v=
Sum-up pressure head from 3 to 1 in meters of oil:
.!2.
+ (y - 112) + , 2 13.6 - y = El.
y
0.80
y
.J2iFi (theoretical velocity)
h =50 mm
= ~2(9.81)(0.05)
v = 0.99 mfs
16112 = El y
y
h2 = 0.239 m = 239 mm
p3 = 3.821
!
Vz = 0
Oil, s = 0.90
6-48
A 30-cm by 15-cm Venturi meter is mounted on a vertical pipe with the
upward. One hundred twenty-six (126) liters per second of oil (sp. gr.
flows through the pipe. The throat section is 15 em above the upstream section
If C.,= 0.957, what is the difference in pressure between the inlet and the throat'
Solution
The discharge through a vertical Venturi
meter is given by the formula:
Solution
,,
Oil
Pt-Pz
0.126 = 0.957 X t (0.15)2..[2i
Pt- Pz _ 0 .15
-'-"-9.;..;;.8..:;:1=X=0.:,;8,;,0= - - = 1.682
~1- (0.5) 4
p, - pz = 22 kPa
(9.81 X 0.80)
Problem 6 - 50
\ Pitot-static tube (C = 1.0) is used to measure air speeds. With water in the
differential manometer and a gage difference of 75 mm, calculate the air speed
using p.,. = 1.16 ~gjm3.
_ .
0 15
1- (0.15 /0.30) 4
5
=0.80
h
= 15 em
l
·- ·- ~ · -· - ~oo- · - · -·- ·
~
-+1
75mm
Water
CHAPTER SIX
Fluid Flow Measurement
352
Energy equation between 1 and 2 neglecting head lost:
E, = E2
~ + E2. +0==+ !2
v
y
2
-
2g
y
P2 - -P1 m of au·
.
= -
y
y
m•rgy equation between 1 and 2
' glecting head lost:
-7 Eq. (1)
J, = [2
,
Sum-up pressure head from 1 to 2 in meters of air:
P1 + y + 0.075-1000
P2
- 0.075 - y = y
1.16
y
2
u,P1
1•2
P2 + :2
- + - + :, = + 2g
y
2g
y
2
Oil, s = 0.827
1
(I?
,,,
1 +0=---+----==-+0
0 +1y
2g
y
,
f2 - E2. = 64.58 m of air
y
olution
•nsider two points 1 and 2
-.hown in the figure. Point
,., the stagnation point,
nre u, = 0.
p1
v2
p
.
- - + + Z1 = - 2 - + _.3.. + Z2
2g
y
2g
y
v2
1
2g
CHAPTER SIX
Flurd Flow Measurement
II UID MECHANICS
I. HYDRAULICS
y
l'2-
2g
= 12 _ P'2 m of air
y
y
-7 Eq. (1)
In Eq. (1):
v2
= 64.58
',um-up pressure head from 1 to 2 in meters of air:
2g
v = 35.6 m/ s
1!..!_ + y + 0.08- 0.08 9810;~·827 - y = .1:!.3_
y
-7 theoretical velocity
Actual velocity, v = C v,
Actual velocity, v = 1(35.6) = 35.6 m/s
-
E.l - J7 2 = 54.006 m of air
y
.y
In Eq. (1 ):
l' 2
- 2-
Air (w = 12 Njm3) is flowing
through a system shown. If
oil (sp. gr. = 0.827) shows a
deflection
of 80
mm,
calculate the flow rate
neglecting head lost.
2g
=54.006
1•2 = 32.55 m/ s
Plow rate, Q =A? l'2
= f (0.05)2(32.55)
Plow rate, Q = 0.06391 mlfs = 63.911./s
80 mm
Oil, s = 0.827
y
353
CHAPTER SIX
fluid Flow Measurement
354
CHAPTER SIX
Fluid Flow Measurement
HYDRAULICS
Problem 6 - 52
355
In Eq. (l)
A Pitot tube in the pipe in which air is flowing is connected to a m"nn1mot,.
containing water as shown in the figure. If the difference in water levels in
manometer is 87.5 mm, what is the velocity of flow in the pipe, assuming
tube coefficient, Cp = 0.99?
,
{I -
1
- -
2g
= 71.44
u, = 37.44 m/ s (theoretical vl'iocity)
Actual velocity:
(I= (.' , X
1
87.5 mm
Ut
= 0.99 X 37.44
1• = 37.07
Water
y
-·- -
Problem 6- 53
I low nozzle is a device inserted into a pipe to measure the flow as shown in
, 11gure. If A ~ is the exit area, show that for incompressible flow,
Air
y = 12 N/m3
- · - · - ·- ·-
m/s
-·- 0- -
Q = (,, • [
6
Jl-(A:~ A,)' >g[ I~;' l]
I'
lwre C.t is the coefficient of discharge, which takes into account frictional
"' '' l~ and is determined experimenta lly.
Solution
low Nozzle
Energy equation between 1 (stagnation point) and 2 neglecting head lost:
£1 = £2
v2
p1
v2
p
1
- - + + Z1 = - 2 - + ___1_ + Z2
2g
y
2g
y
~
1
.
~ · -o- · - ·-·-·-·-·-·-· - · - · - ·
v/
0 + E.!. + 0 =
+ El_ + 0
y
2g
y
v/ =12 . .!2.
2g
y
y
-7 Eq. (1)
Sum-up pressure head from 1 to 2 in meters of water:
P1
.
9810
P2
- - y- 0.0875-- + 0.0875 + y = y
12
y
E.:!_ - !!.1_ =71.44 m of air
y
y
,
__
,. Mercury
1111111
CHAPTER SIX
356
CHAPTER SIX
I LUID MECHANICS
1t. HYDRAULICS
Fluid Flow Measurement
Solution
Fluid Flow Measurement
Problem 6 - 54
Energy equation between 1 and 2 neglecting head lost:
E1= Ez
v2
_ 1_
2g
v2
_1_
2g
p
v2
p
+ _1 + Zl = _2_ + _2 + z,
y
2g
y
-
\ Pitot tube being used to determine the velocity of flow of water in a closed
, nnduit indicates a difference between water levels in the Pitot tube and in the
J'U.'zometer of 60 mm. What is the velocity of flow?
.
'Jolution
p
v2
p
+ _1 +0= _2_ + _2 +0
y
2g
y
2
v = ~2gh
v = ~,-2(-9-.81-)(-0-.06-)
2
!2.__ - !l_ = El - 1'!.1.. 7 Eq. (1)
2g
2g
y
y
v = 1.085 m/s
[QJ = Qz]
A1v1 =Azvz
v1 = (A2/ A,)vz
Problem 6 - 55
In the figure shown, pressure gauge A reads 75 kPa, while pressure gauge B
lt•ads 82 kPa. Find the velocity of air assuming its unit weight to be 20 Njm3.
tJse C, = 0.92 and neglect compressibility effect.
In Eq. (1):
2
2
A
!2_ - (A z/ A1)2!2__ = El - E1_
2g
2g
y
y
B
2
[ 1- (A2/A1)2]!2__ = El _f2.
2g
y
y
Vz
357
=
1
1-(A2 I A1)
2
2
Air, 20 N/m3
g(P1-P2)
•
1
Q = Cn x Azvz
Solution
v = c,x~2gh
It = E.!L - !!A
y
y
82,000 75,000 - 350
f .
I1-- -- --mo au
20
20
Note: This formula can also be used for Horizontal Venhtri Meters.
t
v = 0.92 ~2(9.81)(350)
v = 76.24 m/s
358
CHAPTER SIX
Fluid Flow Measurement
CHAPTER SIX
Fluid Flow Measurement
I LUID MECHANICS
f-. HYDRAULICS
359
Problem 6 - 56
11roblem 6 - 57
Carbon
having specific gravity of 1 .6 ts
· f1 owmg
· t1uough a
Th
difftetrachloride
.
e . erential gage attached to the Pitot-static tube shows a
· of flow.
deflection of mercury. Assuming C, = 1 ·00, f'm d th eve1octty
A rectangular, sharp-crested weir 15 m long with end contractions suppressed
1 1.5 m high. Determine the discharge when the head is 300 nun.
')olution
Since the height of weir is large compared to the head H, the velocity head
of approach can be neglected.
Solution
Using francis Formula:
Q = 1.84 LH312
Q = 1.84 (15)(0.3)3/2
Q = 4.535 mJfs
Mercury
y
Problem 6 - 58
- - -·-·- yd..,.-·--·- -<>A
B
- -
Carbon tetrachloride
s = 1.60
<\ rectangular, sharp-crested weir with end contractions is 1.4 m long. How
lngh should it be placed in a channel to maintain an upstream depth of 2.35 m
lor a flow of 400 liters/ second?
Solution
V
= Ct X J2gl!
h = PB-PA
y
= R(Sgagefluid -Sfluid)
Snuid
0.08(13.6- 1.6)
1.6
h = 0.6 m
V = 1 X .J'-2(:-::-9.-=-81-)
(-0.6-)
=
v = 3.43 m/s
iE tH
~h
"0
p
! l
Using Francis Formula:
Q = 1.84LH312
L = 1.4- 0.2H
0.40 = 1.84(1.4- 0.2H)H3/2
Solve for H by trial and error:
Try H = 0.3
1.84[1.4- 0.2(0.3)](0.3)3/2 = 0.405 "': 0.4 (OK)
From the fieure shown above:
P=d-H
P = 2.35 - 0.3 = 2.05 m
360
CHAPTER SIX
CHAPTER SIX
II.UID MECHANICS
I HYDRAULICS
Fluid Flow Measurement
Problem 6 - 59
361
Fluid Flow Measurement
'Jolution
During a test on a 2.4-m suppressed weir 900 mm high, the head waa
maintained constant 1t 300 mm. In 38 seconds, 28,800 liters of water we
collected. What is the weir factor C,.,?
n-t
Solution
u
d = 1.8m
t-=:;;~~------::::;;~~0
R
i
J
d = 1.2m
~~E- - - L = 2.4 m ---~~
ti
Q = Cru L[(H + l!v)3/2. - lrv3/ 2]
Q
10.125
v.=- =
A
7.5(1.8)
Vn = 0.75 m/s
, , = v.
Q
0.7579
Velocity of approach, v. = - = - - A
2.4(1.2)
Velocity of approach, v. = 0.26316 m/ s
2
h,. = v. = (0.26316)
2g
2,g
lz,. = 0.00353
2
Q = C..,(2.4)[(0.3 + 0.00353)312- (0.00353)312] = 0.7579
c,,. = 1.891
2
2g
11,, = 0.0287 m
= Volume
(since the flow is steady)
time
28 800
'
=
= 757.9 L/s
38
Q = 0.7579 m3js
= 0.75
2g
Q = C.., L((ll + !I,.)312 -hv312]
Q
2
10.125 = 1.88(7.5)[(H + 0.0287)3/2- (0.0287)3/2]
H=0.777m
Height of weir, P = 1.8 - H
= 1.8-0.777
Height of W!!ir, P = 1.023 m
Problem 6 - 61
1>etermine the flow over a suppressed weir 3 m long and 1.2 m high under a
hl'ad of 900 mm. The weir factor C,,. = 1.91. Consider velocity of approach.
Solution
n-t
c:::;-
rd
0.
Problem 6 - 60
d =2.1m
A suppressed weir 7.5 m long is to discharge 10.125 m3js of water onto an
open channel. The weir factor C,,. = 1.88. To what height P may the weir be
built, if the water behind the weir must not exceed 1.80 m deep?
tl
1.
I<
L =3m
CHAPTER SIX
362
v/ = (Q /(3x2.1)f
2g
lr, = 0.001284Q2
Fluid Flow Measurement
1. HYDRAULICS
Q = C,,, L((H + h.,)V2- h.,J/ 2]
h,, =
CHAPTER SIX
II UID MECHANICS
Fluid Flow Measurement
2g
It can be seen that the discharge Q varies with hv which in turn
with Q. Using this formula directly would lead to trial-an-e
solution.
First, we solve the approximate velocity of approach by solving the
discharge using the formula:
Q = C.,. L l-P/ 2
Q = 1.91(3)(0.90)312 = 4.892 m3fs
h,. = 0.001284(4.892)2
h,. = 0.0307 m
11roblem 6- 62 (CE November 1995}
l111d the width, in meters, of the channl'l a t the back of a suppressed weir
u .111g the following data:
I lead, I I = 28.5 em
Depth of water, d = 2.485 m
Discharge, Q = 0.84 m 3 / s
• 111sider velocity of approach and use Francis formula.
'\olution
Q = 1.84 L[(H + lzv)312- llv3/2]
Solving for Land 11,, using the formula:
Q = 1.84LI-J3/2
0.84 = 1.84L(0.285)3/2
L=3m
h,, =
New Q = 1.91(3)[(0.9 + 0.0307)lf2- (0.0307)112)
New Q = 5.114 m3fs
363
v/ = [0.84/(3x 2.485)]2 = 0.000647 m
2g
2g
0.84 = 1.84L((0.285 + 0.000647)3/2- (0.000647)3/2)
L=3m
lr, = 0.001284(5.114)2
Using Eq. 6- 33:
" ·· = 0.03358
New Q = 1.91(3)[(0.9 + 0.03358)312 - (0.03358)312]
New Q = 5.133 m3fs
The discharge converges at 5.133 mJ/s
Q= Cwl!Ht
3 c_
C1= _
w_ = 3 (1.84)
2 2g
2 2g
2
c = 0.2588
Using Eq. 6- 33:
Q=
2 2g
2
l
2
0.84 .= 1.84(L)(0.285)3/2 1 + 0.278'
CwLHt[l+C 1 (~f]
c1 = ~ c,_c = ~ (1.91)
[1+C1(~r]
~:!:~
rl
L=3m
2
2g
c1 = 0.2789
Q = 1.91(3)(0.9)11{1 + 0.2789(
~:~
Q = 5.143 m 3/s approximately
rl
Problem 6 - 63
I he discharge from a 150-mm-diameter orifice under a head of 3.05 m and
'oefficient of discharge C = 0.60 flows into a rectangular channel and over a
1t>ctangular suppressed weir. The channel is 1.83 m wide and the weir has
height P = 1.50 m and len3th L = 0.31 m. Determin e the depth of water ·in the
'hannel. Use Francis formula and neglect velocity of app roach.
364
CHAPTER SIX
Fluid Flow Measurement
Solution
The discharge through the orifice equals the dtscharge through the wen
fLUID MECHANICS
I. HYDRAULICS
Q = CA .. ~2gl-l = 0.60 X f (0.15) 2 ~2(9.81)(3.05)
0.093 = 0 06
N2/S
·
Q = 0.08202 m'ls
N=3
Depth of water upstream of the weir·
d = If+ P = 0.274 + 1.50
d = 1.774 m
The flow in a rectangular charmel varies from 225 liters per second to 350 lilt
per second, and it is desired to regulate the depth by installing standard 4().
degree V-notch weir at the end. I low many weirs are needed to regulate th
variations i.n depth to 60 nm1?
Solution
For s tandard 90° V-notch weir, C.,."' 1.4
Q = 1.4H512
365
Variation in depth= H2- H1 = 60 mm
0.574 - 0.481 = 0.06
N2 / S
N2/S
For the orifice.
For the weir (neglecting v")
Q = '1.84LJ-1312
0.08202 = 1.84(0.31)H-ll 2
H = 0.274 m
CHAPTER SIX
Fluid Flow Measurement
Problem 6 - 65 {CE November 1996)
1he discharge over a trapezoidal weir is 1.315 m 3 Is. The crest length is 2 m
111d the sides are inclined at 75° 57' 49" with the horizontal. Find the head on
the weir in meters.
<;olution
The side inclination angle given is that for a Cipolletti weir
Q = 1.859 L J-1312
1.315 = 1.859(2)J-1312
H = 0.50 m
Problem 6- 66 {CE November 1995)
,1 spillway controls a reservoir 4.6 hectares in area. The permanent crest is at
o•levation 75 m. If water can be drawn from elevation 76.5 m to elevation 75.5
Use Francis
111 in 42 minutes, find the length of the spillway in meters.
formula neglecting velocity of approach.
Solution
Let N be the requi red number of weirs
Total flow, Q1 = N x Q = 1.4Nlf5/2
'
... ..._,........... ,
When the discharge is 0.225 m3 Is
0.225 = 1.4NH15/ 2
H1 =
~;~! {head when the discharge is 225 Ll s)
When the discharge is 0.35 m3 Is
0.35 = 1.4NH25/ 2
H2 =
~~~! (head when the discharge is 350 Lis)
El76.5 m
---.t-1.5 m
El 75.5 m
!o.s m 1 El 75 m
..
CHAPTER SIX
Fluid Flow Measurement
CHAPTER SIX
Fluid Flow Measurement
366
Length of weir, L = 1 m
Initial head, H1 =1m
Using Eq. 6 - 53:
t _ 2As [
- C 10 L
367
1
1 ]
~H 2 - fH;
The drop of water level after discharging 72 m3 is:
72
Drop= - - = 0.18 m
20(20)
Final head, H 2 = 1-0.18 = 0.82 m
Weir factor (Francis), Cw = 1.84
A s = 4.6 hectares = 46,000 m2
t = 42 minutes = 2520 seconds
}-1, = 1.5 m
H2 = 0.50111
C,, = 1.84
(Francis Formula)
2(400)
1 - 1r.; ]
1= -[ ~
1.84(1) .y0.82
2520 = 2(46,000) [-·_1_- _1_]
1.84L
.Jo.so .J1.so
L = 11.86 m
.y1
= 45.35 seconds
i•toblem 6- 68 (CE November 1991)
V-notch weir is located or cut at one end of a tank having a horizontal
• uare section 10 m by 10 m. If the initial head on the weir is 1.20 m and it
1
May 2002)
A rectangular suppressed weir of length 1 111 is consb·ucted or cut at the top
a tall rectangular tank having a horizontal section 20 m by 20 m. If the int
11kcs 375 seconds to discharge 100 m3 of water, what could have been the
\, ·rtex angle of the weir. Use C = 0.60.
head over the weir is 1 m, compute the time required to discharge 72 cu. 111
water. Use Francis formula.
o,olution
20m
Solution
I
When 100m3 is discharged from the tank, the water level drops by y meters.
t _ 2A5 [
- CTUL
1
1
~H2- ~H1
l
100 X Y = 100
y=lm
Thus, the flow is unsteady with initial head HI = 1.20 m and final head
H2 = 0.2m.
Water surface area at any time, A s= 20(20) =400m2;
368
CHAPTER SIX
Fluid Flow Measurement
JHz
dQ = dA ~2gll
Q"''
dA = 2x rill
A,= 10 x 10 =100m2
8
Qout = 15 C
J2i tan% HS/2 = 185 (0.60) .J2i tan% HS/2
Express x in terms of ll by squared property of parabola:
x2
t HS/2
1
1.5-11
2
X= 0.707 .J1.5 _,,
100dH
dA = 2(0.707 .J1.5 ~ll )rill
Qout = 1.417 tan
t = 375 seconds
375 =
1 20
.
1
5 2
0.20 1.417 tan~ H /
5.3137 tan% =
dQ = 2(0.707 .J1.5- h )d/r ~2gh
1.20
5 2
1
riQ = 6.263 .J1.5- h .Jh dll
H- 1 dH
(5
0.20
Q = 6.263 J!1 .5 -II .Jh rilr
1 20
5.3137tan% = [-l:.H-3 12 ] .
3
5.3137 tan% = -
369
( onsider the horizontal strip shown (trt!ated as an orifice under head II)
[ HlA dH
I=
CHAPTER SIX
Fluid Flow Measurement
I I UID MECHANICS
I. HYDRAULICS
0.20
By trigonometric substitution
Let I! = 1.5 sin2 e
f ((1.20)·3/2 - (0.20)·3/2]
tan % = 1.30726
fh = 1.2247 sine
0 = 105.17°
dll = 3 sin e cos e de
when"= 0, e = oo
when ll = 1.5, e = 90° = rt/2
6-69
Water flows through a parabolic weir that is 2m deep and 2m wide at the top
under a constant head of 1.50 m. Assuming C = 0.65, determine the discharg
through the weir.
Pro
r·/2 ~1.s
Q = 6.263 Jo
-1.5sin 2 e(1.2247 sin eX3sin e cosede)
22
Solution
2m
~1m
~
Q = 28.182
I
By Walli's formula:
•
X
Jii~ e cos 2 ede
11
4(2)
Q = 5.5336 m3/s
Q = 28.182[ ( ) X.!!..]
2
(theoretical discharge)
X
1.5 m
Actual discharge = CQ = 0.65(5.5336)
Actual discharge = 3.597 m 3/s
CHAPTER SIX
Fluid Flow Measurement
370
CHAPTER SIX
Flurd Flow Measurement
fLUID MECHANICS
I. HYDRAULICS
Problem 6 - 70
A trapezoidal weir having side slope of 1H to 2V discharges 50 m3j s under
constant head of 2 m. Find the length of the weir assuming C = 0.60.
Solution
~---- X----~
Problem 6 - 71
\ sharp-crested suppressed rectangular wetr 1 m long and a standard 90lt•gree V-notch weir are placed in the same weir box with the vertex of the Vlllltch weir 150 mm below the crest of the rectangular weir. Determine the
lll'ad on the rectangular weir when their discharges are equal Use Francis
lormula.
'iolution
Let H be the head on the rectangular weir:
For the rectangular weir: (HR = H)
QR = 1.84LHR3/ 2
=1.84(1)H3/ 2
QR =1.84HJ/ 2
Consider the horizontal strip shown (treated as an orifice under head h)
dQ = CdA ~2glt
dA = xdlr
x=L+2z
z = 1/2(2- h)
1.84HJ12 = 1.4(H + 0.15)512
1.727 1-P = (H + 0.15)s
By trial and error:
F-1 = 0.891 Ill
dA = (L + 2 -/r)dlt
dQ = C
J2g (L + 2. -/t)d/r ~2gh
Q = c J2g
L(th 1/2 + 2111/2- h 3/2 ~h
Q = C J2g [tL/t:\/2 + f!t3/2- !lt5/2
J:
t
50= 0.6 J2g [ L(2)3/2 + !(2)3/2 - !(2)5/2
J
L = 9.18 m
Using the combined rectangular and triangular weir formulas:
Q = t C J2g U{312 + 1~ C J2g tan f J-IS/2
From the figure, tan
50 =
t =t
t (0.6) J2g L(2)312 + 185 (0.6) J2g ( t )(2)5/2
L = 9.18 m
For the V-notch weir: (Hv = H + 0.15)
Qr = 1.4Hv51 2
QT = 1.4(H + 0.15)5/ 2
(QR = QTJ
X= L + 2[V2(2- h)] = L + 2 -It
371
Square both sides
372
CHAPTER SIX
Fluid Flow Measurement
lementa
CHAPTER SIX
Fluid Flow Measurement
I LUID MECHANICS
I. HYDRAULICS
373
Problem 6 - 75
What length of a Cipolletti weir should be constructed in order that the head
,f flow will be 0.96 m when the flow rate is 3.76 m 3/s?
Ans: 2.15 m
Problems
Problem 6 - 72
Water is being discharged through a 150-mm-diameter pipe directly into 1
container that has a volume of 6m3. Find the velocity of flow through the p ip
if the time required to fill the container is 3 minutes and 28.7 seconds.
Ans: 1.63 m/
l'roblem 6 - 76
is flowing upward through a Venturi meter as shown.
dascharge coefficient of 0.984, calculate the flow of oil.
t hl
Assuming
Ans: 158 L/s
Problem 6 - 73
Calculate the discharge through the submerged orifice shown in the figure.
A liS: 1091it/ Sl'
50 kPa
<S
3
r
A
B
15 kPa
~
Air
Air
T
[
-----
Water
Water
2r
~t~
mm 0
0.75
Problem 6- 74
The truncated cone shown has 8 = 60°. How long does it take to draw th~
liquid surface down from II = 5 m to II =2m?
Ans: 30.4 minute~
Glycerin
s = 1.26
I
I
I
I
I
I
80mm 0
c =0.80
I
I
\~1
I
I
I
I
I 'I
I I
II
'
Problem 6 - 77
A Venturi meter having a throat diameter of 150 mm is installed in a
horizontal 300-mm-diameter water main, as shown. The coefficient of
discharge is 0.982. Determine the difference in level of the mercury columns
nf the differential manometer attached to the Venturi meter if the discharge is
142 L/s.
Ans: lr = 255 mm
374
CHAPTER SIX
Fluid Flow Measurement
I I UID MECHANICS
1. HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
375
Chapter 7
300 mm 0
Throat
Outlet
Mercury
s = 13.6
Fluid Flow in Pipes
UEFINITIONS •
l'tpes are closed conduits through which fluids or gases flows. Conduits may
llnw full or partially full. Pipes are referred to as conduits (usually circular)
' hich flow full. Conduits flowing partially full are called open channels,
hich will be discussed in Chapter 8.
Problem 6 - 78
Determine the head on a 45° V-notch weir for a discharge of 200 L/s.
0.57.
lluid flow in pipes may be steady or unsteady. In steady flow, there are two
1' pes of flow that exist; they are called lami11ar flow and turbule11t flow.
Problem 6 - 79
I • minar Flow
For the s luice gate shown, if C,, = 0.98, what is the flow rate?
is the height of the opening?
I he flow is said to be laminar when the path of individual fluid particles do
nol cross or intersect. The flow is always laminar when the Reynolds number
JJ is less than 2,000.
rurbulent Flow
I he flow is said to be turbulent when the path of individual particles are
uaegular and continuously cross each other. Turbulent flow normally occurs
hen the Reynolds number exceed 2,000.
Width= 2m
2.20 m
l
...
1minar flow in circular pipes can be maintained up to values of R, as high as
>0,000. However, in such cases this type of flow is inherently unstable, and
the least disturbance will transform it instantly into turbulent flow. On tl1e
•>lher hand, it is practically impossible for turbulent flow in a straight pipe to
pPrsist at values of R, much below 2000, because any turbulence that is set up
will be damped out by viscous friction.
-r---rv
y
0.90 m
Critical Velocity
I he critical velocity in pipes is the velocity below which all turbulence are
ol,lmped out by the viscosity of the fluid. This is represented by a Reynolds
number of 2000.
CHAPtER SEVEN
Fluid Flow in Pipes
376
FLUID MECHANICS
& HYDRAULICS
REYNOLDS NUMBER
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
377
VELOCITY DISTRIBUTION IN PIPES
Reynolds numbe r. which is dimensionless, is the ratio of the inertia force to
viscous force
For pipes flowing full
Eq. 7-1
v = !::
p
Eq. 7-2
where.
u =mean velocity 111 m/s
() = pipe diameter in meter
v =kinematic viscosity of the fluid in m 2 /s
1-1 =absolu te or dynamic viscosity in Pa-s
laminar Flow
I he velocity distribution tor
l.munar flow, at a cross section.
lnllows a parabolic law of
'ariation with zero velocity at
the walls. In circular pipes, the
vt'locity va ries as the ordinates
l)f a paraboloid of revolution
with its average velocity equal
to one-ha lf of its maximu m
velocity
-
u --tl+-ol
X
Figure 7 - 1: Lam1nar flow velocity d1stnbut1on
I he equation tor the velocity profile tor lammar tlow IS gwen by
Eq. 7-5
For non-circular pipes, useD = 4R , then the for mula becomes;
Average velocity, v = 1hv,
Eq. 7-6
R = 4vRp = 4vR
•
J..l
the velocity at .any distan ce r from the center ot tht> ptpt' mav also be
computed using the squared property of parabola
v
R = Cross - sectional area of pipe, A
P ipe perimeter, P
U = Vr- ,\
Table 7 - 1: VIscosity and Density of Water at 1 atm
Temp, °C
0
10
20
30
40
so
60
70
80
90
100
p, kg/m3
1000
1000
998
996
992
988
983
978
972
965
958
Eq. 7-7
v, m2/s
~~, Pa-s
3
1.788 X 10'
1.307 X 10'3
1.003 X 10'3
0.799 X 10'3
0.657 X 10'3
0.548 X 10'3
0.467 X 10'3
0.405 X 10'3
0.355 X 10'3
0.316 X 10' 3
0,283 X 10'3
1.788 X 10~
1.307 X 10-li
l.QQS X 1Q~
0.802 X 1Q~
0.662 X 10-6
0.555 X 10-6
0.475 X 10~
0.414 X 10-li
0.365 X 10-6
0.327 X 10-6
0.295 X 10-6
where
ht = head lost in the p1pe
L = pipe length
r. =pipe radius
Ur = centerline or maxunum veloc1ty
1-1 = absolute viscosity of the liquid
y = unit weight of the fluid
u = velocity at distance r from pipe center
r• =average velocity
378
CHAPTER SEVEN
CHAPTER SEVEN
t LUID MECHANICS
Fluid Flow in Pipes
Turbulent Flow
where:
379
Fluid Flow in Pipes
F. HYDRAULICS
t 0 = maximum shearing stress in the pipE:'
f = friction factor
I he velocity distribution for turbulent flows varies with Reynolds numbt r
with zero velocity at the wall and increases more rapidly for a short distan1
from the walls as compared to laminar flow
n = mean velocity
~HEARING STRESS IN PIPES
r
E.G.L.
-J'-:-;---li:.GJ.__
:
- - - - - - """:
Pb
Figure 7 - 2: Turbulent flow velocity distribution
The velocity, u, at any point r in a pipe of radius ro and center velocity Vr is:
- - -
-
-
I
.::_
,_
-
P7h
:
I
_.l~f- ____ :~ _____ -~l(j. _L ~
I.
.I
L
<u
Sheanng stress
distribution
lonsider a mass of fluid of length Land radius r to move to the right as shown
m the figure. Due to head lost ht, the pressure p 2 becomes less than p1
or u = (1 + 1.33fj}v- 2.04
fJ v log.....!_g_
r r
0 -
The centerline or maximum velocity is given by:
J'he shearing stress, t" at the surface of the fluid can be found as tollows
[LF,1 =OJ
F,- F2- Fs = 0
F, = F,- f2
t, A,
= p1 A 1 - p2 A2
= p1 x 1tX2 - p2 x rrx 2
t, x 27t x L
v, = v (1 + 1.33fl)
= Pl- P2 X
t
s
Combining Eq. 7- 9 and Eq. 7- 11, and solving for u gives the following:
V
F:
= Vr- 3.75VP
2L
Eq. 7- 13
380
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
Multiply and divide th1s equation by unit weight, y
PI - P2 y
I lead losses in pipes may be classified into two; the major l1ead loss, which is
, <~used by pipe friction along straight sections of pipe of uniform diameter and
uniform roughness, and Minor head loss, which are caused by changes in
velocity or directions of flow, and are commonly expressed in terms of kinetic
··nergy.
21.
1 _-.;_p-=2 =It, (head loss), then
but -'-p-=-
y
'
381
HEAD LOSSES IN PIPE FLOW
{, = --'y'--- .\
{ =
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
yl!L X
2L
Eq. 7-14
MAJOR HEAD LOSS, h,
A. Darcy-Weisbach Formula (pipe-friction equation)
'
It is seen from this equation that the sheanng st:res::. at the center of the pipe (1
The maximum shearing stress, t,,, is til
fL v 2
= 0) is zero and varies linearly with _
,
the pipe wall (at x = r)
D 2g
'" = y II L r
Eq. 7-15
2L
onn= y!ILD =[yi:_
4L
h,=--
4
2g
Eq. 7- 16
Eq. 7-18
f = friction factor
L = length of pipe in meters or feet
D = pipe diameter in meter or feet
v = mean or average velocity of flow in m/ s of ft/ s
For non-circular pipes, use D = 4R, where R is the hydraulic radius defin ed in
Eq. 7-4
Shear Velocity or Friction Velocity, v 5
For circular pipes, the h~ad los.s may be expressed as:
Eq. 7-17
...·
Eq. 7-19
Eq. 7-20
wh~re Q is the discharge.
382
CHAPTER SEVEN
Fluid Flow in Pipes
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
Value off:
4.
For rough pipes, where o, < 0.3L: (Karman)
n
For laminar Flow:
f= 64 "' 64!1
Re
171
=
vDp
=2 log(~) + 1.14
Eq. 7-21
where
32!lLV
pg02
Eq. 7-27
L =absolute roughness, mm
r./ 0 .=relative roughness (dimensionless)
o, = nominal thickness of viscous sublayer
12811LQ
For circular pipes, hr= --'--7rrpgD4
01
For non-circular pipes, use Eq. 7- 22 with 0 = 4R.
5.
For Turbulent Flow:
1.
383
For turbulent flow in smooth and rough pipes, universal resistam
laws can be derived from:
=
11.6v
Eq. 7-28
~t 0 /p
For smooth and rough pipes, turbulent: (Colebrook equation)
J
E ID +2.51
-1---21og( -3.7
Jl
ReJl
Eq. 7-29
This equation was plotted in 1944 by Moody into what is now called
the MooqJI chart for pipe friction.
where v, is the shear velocity or friction velocity .
6.
2.
For smooth pipes, R. between 3,000 and 100,000: (Blasius)
-1
Jl
f= 0.316
R 0.25
t
3.
Haaland formula. This is an alternate formula for Eq. 7- 29. This
varies less than 2% from Eq. 7-29.
For smooth ptpes wtth R. up to about 3,000,000
Eq. 7-26
(E/0)
Re
=-1.8log[6.9
-+ - 3.7
1 11
· ]
Eq. 7-30
384
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
B.
Table 7 - 2: Values of Specific Roughness for. Common Pipe Materials
ft
Steel:
Sheet metal new
Stainless new
Commercial new
Riveted
Rusted
Iron:
Cast new
Wrought new
Galvanized new
Asphalted cast
Brass: Drawn new
Plastic: Drawn tubing
Glass
Concrete:
Smoothed
Rough
Rubber: Smoothed
Wood: Stave
mm
0.00016
0.000007
0.00015
0.01
0.007
0.05
0.002
0.046
3.0
2.0
0.00085
0.00015
0.0005
0.0004
0.000007
0.000005
0.26
0.046
0.15
0.12
0.002
0.0015
Smooth
Smooth
0.00013
0.007.
0.000033
0.0016
0.04
2.0
0.01
0.5
385
Manning For mula
rhe manning formula is one of the best-known open-channel formulas and is
commonly used in pipes. The Formula is given by:
Roughness, c
Material
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
,, = _!_R2/3sl/2 (S..
I uruts
. )
n
Eq. 7-31
Eq. 7-32
where
n = roughness coefficient
R = hydraulic radius
S = slope of the energy grade line = ilr/ I
11 t
Subc;t1tutmg .S = L and R = 0/4 to Eq. 7- 31 and solving for hr.
0
•
(I= ; , (
~ r/3r "[
r
2.5198nv
0 211
0
2
square both sides and solve for 1!1
n.<
0.05
0.04
0.07
0
0.05
O.Q2
O.Q15
0.04
-l5
0
Eq. 7-33
0.03
0.01
0.008
0.006
n.n• l
.!!
0.004
g n.1
u..
nnn>
~
j
1
For non-circular pipes, use 0 = 4R
l·or circular pipes:
p=Q=__g_
A
"
~:~ i
0.0002
0.0001
4
635n 2 {~]'
(l
~-
0
.!!. 0 2
ltr =
0. 00005
0.01
n.
0.0000
0
10' 2. 10'
10'' 2 ... 104
lOJ 2
10~
10• 2 • 10•
101 2 ... 107
Eq. 7-34
10'
Reynolds number R,
Figure 7 - 3: Moody Friction Factor Chart. Th1s chart IS 1dent1cal
to Eq. 7 29 for turbulent flow.
The value of 11 is given in Table 7- 4
386
C.
CHAPTER SEVEN
ILUID MECHANICS
Fluid Flow in Pipes
I. HYDRAULICS
387
Table 7 - 4: Values of n to be used w1th Mannmg Formula
Hazen Williams Formula
The Hazen Williams formula is widely used in waterworks industry. J
formula is applicable only to the flow of water in pipes larger than 50 n1.m
in.) and velocities less than 3 mjs. This formula was designed for flow in
pipes and open channels but is more commonly used in pipes.
English Unils:
For circular pipes flowing full, this formula becomes
D2.t>l so.s1
S.l. Units:
51l.5·1
For circular pipes flowing full, this formula becomes:
Q = 0.2785 c, D2 <>:l so.:;.!
10.67 L Q 1.as
and, hr = C t t.ss D 4.fl7
where:
CHAPTER SEVEN
Fluid Flow in Pipes
C1 = Hazen Williams coefficient
n
Nature of surface
Min
Max
0.013
0.010
Neat cement surface
0.010
0.013
Wood-stave pipe
0.014
0.010
Plank flumes, planed
0.010
0.017
Vitrified sewer pipe
O.Q15
0.011
Metal flumes, smooth
0.013
0.011
Concrete, precast
0.011
0.015
Cement mortar surfaces
0.015
0.011
Plank flumes, unplaned
Common-clay drainage tile
0.011
0.017
0.016
Concrete, monolithic
0.012
0.012
0.017
Brick with cement mortar
0.017
0.013
cast iron - new
0.017
0.030
Cement rubble surfaces
0.017
0.020
Riveted steel
O.Q25
0.021
Corrugated metal pipe
O.Q25
0.017
Canals and ditches, smooth earth
0.022
0.030
Metal flumes, corrugated
canals:
0.033
0.025
Dredged in earth, smooth
O.D25
0.035
In rock cuts, smooth
0.040
Rouqh beds and weeds on sides
0.025
O.Q35
0.045
Rock cuts, jagged and irregular
Natu~PI streams
O.Q25
0.033
Smoothest
0.045
0.060
Roughest
0.150
Very weedy
O.D75
Source: Fluid MechaniCS by Daugherty, Franz1111, & F1nnemore
•
D = pipe diameter in
R = hydraulic radius
0
S =slope of U1e EGL = hr/ L
Table 7 - 3: Recommended Value for C1 for Hazen Williams Formula
Description of Pipe
Extremely smooth and straight pioe
New smooth cast iron oioes
Average cast iron pipes
VItrified sewer oioes
Cast iron pipes some years in service
Cast iron pipes in bad condition
New riveted steel
Smooth wooden or wood stave
Value of C1
140
130
110
110
100
80
110
120
MINOR HEAD LOSS
Minor losses are caused by the changes in direction or velocity of flow . These
changes may be due to sudde11 contraction, sudde11 t'nlnrgemeHt, unlves, bends,
and any other pipe fittings. These losses can usually be neglected if the length
of the pipeline is greater than 1500 times the pipe's diameter. However, in
short pipelines, because these losses may exceed the friction losses, minor
losses must be consiqered.
388
A.
CHAPTER SEVEN
Fluid Flow in Pipes
II UID MECHANICS
CHAPTER SEVEN
Fluid Flow in Pipes
1. HYDRAULICS
Sudden Enlargement
The head loss, m, across a sudden enlargement of pipe diameter is:
1.2
1.1
L
- - LA,1mql ~ ~LA::'
r-.
V;
0.9
Ill =
- l1 ) 1.92
1
~
~
I
0.6
v
Head Loss = K(v1- v2)2/2g - t--
ll
o.s
I
1/
0. 2
lv
o.1
0
A s pecial app lication of Eq. 7 - 40 and Eq. 7 - 41 is the discharge from a P'l
into a reservoir. The water in the reservoir has no velocity, so a full veloul
head is lost.
o•
I
I
I
I
I
I
1
I~
I
~
0.3
I
I
~
OA
in m
2
2g
,II/
It
~
':5 0.7
Another equa tion fo r the head loss caused by sudden enlargements w
determined experimentally by Archer, and given as:
( ll
IIJ
0.8
o1 =velocity before enlargement, m/s
v2 = velocity afler enlargeme nt, m/s
I--""""
I v K lA-JA
.I ,•
1.0
where:
389
10• 2o• 30" qo• so• 60' 70" so- 90" 1oo• u o• 120• 130' 140' 1so· 160' 170' 1so·
Angle e between diverging sides of pipe
Figure 7 - 4: Head-loss coefficient for a pipe with diverging sides.
B.
Gradual Enlargement
Table 7 - 5: Loss coefficients for sudden contraction
The head loss, m, across a gradual conical enlargement of pipe diameter is:
02/01
K.:
The approximate values of K arc shown in Fig ure 7- 4.
C.
Sudden Contraction
l/2
where:
0.50
0.1
0.45
0.2
0.42
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
0.39
0.36
0.33
0.28
0.22
0.15
0.06
0.00
A special case of sudden contraction is the entrance loss for pipes connected to
1 reservoir. For this case, the values of K· are as follows:
The head loss, m, across a sudden contraction of a p ipe is:
" I. = K(. -2g
0.0
Eq. 7-43
K = the coefficient of sudden contraction, See Table 7- 5
v = velocity in smaller pipe
Flush connection .....................................Kr = 0.50
Projecting connection .............................K = 1.00
Rounded connection ................ .".............. Kr = 0.05
Pipe projecting into reservoir ................Kr = 0.80
Slightly rounded entrance ..................... Kr = 0.25
Sharp-cornered entrance ..................... .. .Kr = 0.50
390
C.
CHAPTER SEVEN
Fluid Flow in Pipes
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
391
Bends and Standard Fittings
The head loss that occurs in pipe fittings, such as valves and elbows, .m
bends is given by:
The approximate values of K are given in Table 7 - 6. K values vary not onl
for different sizes of fittings but with different manufacturers. For
reasons, manufacturer's data are the best source for loss coefficients.
The head loss due to pipe fittings may also be found by increasing the p•J
length using the values of L/D in Table 7- 6. For very smooth pipes, it
better to use the K values when determining the loss through fittings. '>
Problem 7-14.
Table 7 - 6: Loss factors for pipe fittings
Fitting
Globe valve fully open
Angle valve fully open
Close-return bend
T through side outlet
Short-radius elbow
Medium-radius elbow
Long-radius elbow
45• elbow
Gate valve wide open
Gate valve, half open
K
L/0
10
5
2.2
1.8
0.90
0.75
0.60
0.42
0.19
2.06
350
175
75
67
32
27
20
15
7
72
·-- --
For pipe with constant diameter, the difference between the water levels in the
piezometer tubes. If the pipe is horizontal and with uniform size, the
difference in pressure head measures the head lost between the two points.
If the pipe is very large such that the velocity head is very small, the total head
lost I IL can be taken as equal to H.
PIPE CONNECTING TWO RESERVOIRS
When one or m~re pipes connects two reservoirs as in the figure shown, the
total head lost in all the pipes is equal to the difference in elevation of the
liquid surfaces of the reservoir.
PIPE DISCHARGING FROM A RESERVOIR
The figure shown below shows the conditions of flow in a pipe of unifont
diameter discharging from a reservoir into opei\ air. The velocity head ami
the pressure head in the liquid surface of the reservoir are zero. If there w11l
be no head lost, the velocity head could have been equal to H, which is th
distance between the water surface in the tank and the exit end of the pipe ami
the velocity of flow could have been v = ~2gH , but such is not the case due I
losses.
..
HL = H = lzt minot + Jy
Eq. 7-45
CHAPTER SEVEN
Fluid Flow in Pipes
392
CHAPTER SEVEN
Fluid Flow in Pipes
I I UID MECHANICS
1, HYDRAULICS
PIPES CONNECTED IN SERIES
Q = Q, + Q2 + Q3
For pipes of different diameters connected in series as shown in the figur
below, the discharge in all pipes are all equal and the total head lost is equal In
the sum of the individual head losses.
HL = IILJ = lzt2 = " L3
I
~-T____
I
I
!
p/y
-..,I
I
I
Eq. 7-49
Eq. 7- 50
I
l
'"the pipe system shown, pipe 1 draws water from reservoir A and leads to
1111\Ction C which divides the flow to pipes 2 and 3, which join again in
1111\Ction D and flows through pipe 4. The sum of the flow in pipes 2 and 3
quais the flow in pipes 1 and 4. Since the drop in the energy grade line
I••·tween C and D is equal to the difference in the levels of piezometers a and b,
rhl'n the head lost in pipe 2 is therefore equal to the head lost in pipe 3.
.._
.._.._
I
---.
t-.._H.G.'
I
-......,;l...
p/y
.._"ir - - ..._ _ _ _ _ _ J
Ql
p/y
---.
---.Ql
I
pipe 1
pipe 2
1
I
393
I
Q,
pipe 3
Ql =Q2=Q1= Q
Eq. 7-46
HL = /tfl + hp. + ltp + hmmor
Eq. 7-47
If the pipe length in any problem is about 500 diameters, the error resulting
from neglecting minor losses will ordinarily not exceed 5%, and if the pipl'
length is 1000 diameters or more, the effect of minor losses can usually be
considered negligible. Neglecting minor losses, the head lost becomes:
llL = 1!11 + ltp. + l1p
Eq. 7-48
If, however, it is desired to include minor losses, a solution may be made first
by neglecting them and then correcting the results to correct them.
PIPES CONNECTED IN PARAllEL
o
noc::=:::> Q,
I he necessary equations for the system are.
Q-1 = Q4
-7 Eq. (1)
Q, = Q2 + Q~
-7 Eq. (2)
lzp. = lzp
-7 Eq. (3)
HLAR = il/1 + ltfl + 1tr4
-7 Eq. (4)
Note: The number of equations needed to solve the problem must be equal to
the number of pipes.
394
fLUID MECHANICS
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
EQUIVALENT PIPE
CHAPTER SEVEN
Fluid Flow in Pipes
395
El. 100
If a pipe system (0) is to be replaced with an equivalent single pipe (E),
equivalent pipe .must have the same discharge and head loss as the origin
pipe system.
E '7
~c----~---------~-----------------------------~~------~o--.
A
B Qo
Original pipe system, 0
Head loss =Ho
A
B
c=============================~__.
Equivalent single pipe, E
Head loss =He
<a
QE=Qo
HLr= HLo
Types of Reseavoir Problems
Type 1: Given the discharge in one of the ptpes, or gtven the pressure at the
junction P, and the required is the elevation one of the reservoirs or the
diameter or length of the one of the pipes, and
RESERVOIR PROBLEMS
In the figure shown below, the three pipes 1, 2, and 3 connects the thr
reservoirs A, B, and C respectively and with all pipes meeting at a common
junction D.
El. 100
.--1 Piezometer
Type 2: Given all the pipe properties and elevation of all reservoirs, find
the flow i.n ea~h pipe, which can be solved by trial and error.
In any of these types, the main objective is to locate the position (elevation) of
the energy at the junction P. This position represents the water surface of an
tmaginary reservoir at P. The difference in elevation between this surface and
the surface of another reservoir is the head lost in the pipe leading to that
reservoir (See .figure nbove).
Procedure in Solving Reseavoir Problems:
Type 1:
1. With known flow in one pipe leading to or flowing out from a reservOir
of known elevation, solve for its head lost hr.
2. Determine the elevation of the energy grade line at the junction of the
pipes (P) by adding or subtracting (depending on the direction of flow)
the head lost in the pipe from the elevation of the water surface in the
reservoir.
396
CHAPTER SEVEN
Fluid Flow in Pipes
3. If the known value is the pressure at P,
elevation of P + pr/Y.
4. Draw a line from P' to the surface· of the other reservoir. These linl'l
represent the EGL' s of each pipe. The difference i.n elevation betwel•n
P' and the surface of the reservoir is the head lost in the pipe.
CHAPTER SEVEN
Fluid Flow in Pipes
fLUID MECHANICS
E. HYDRAULICS
397
3. After determining the direction of Q 2 (say towards reservoir 8), express
all the head lost in terms the other, say in terms of lrn . Let lin = x.
El. 100
5. Solve for the discharge.
Type 2: (See Problem 7- 65)
1. Given all elevation''and pipe properties, determine the direction of flow
in each pipe. Of course, the highest reservoir always have an outflow
and the lowest always have an inflow, but the middle reservoir (B) may
have an inflow or outflow.
2. To find out the direction of flow in pipe 2, assume that Q2 = 0 such that
P' is at elevation B, then the values of lrfl and hp can be solved. (In th
figure shown, hfl =20m and hp =30m). With hfl and lrp known, solv
for Q 1and QJ. If Q 1 > Q3, then Q2 is towards B and P' is above reservoir
B. If Q, < Q3, then Q 2 is away from B and P' is below reservoir B.
SOm
hn=SO-x
L
l\-'liL--<s-JI
With all head lost hr expressed m terms of "-· all flow Q can also be
expressed in terms of x (usually in the form a./x + h ).
El. 100
Example, if Darcy-Weis bach or Mannmg formula is used, ltrvanes with Q2
fl!r= K j22]
lin = x = K, Q, 2
Q1 = K' 1 .[;
7 Eq. (1)
lr{2 = 20- X = K2 Q21
Q2 = K'2 ./20- X
7 Eq. (2)
lip. = 50- X = K3 Q32
Q, = K\ .)so -,.
7 Eq. (3)
IQ, = Q2 + Q,J
K', ,{; = K'2./20-). + K',./50-).
Simplify the equation and solve for x. We may also use trial-and-error
solution.
4. Once"- is detennmed, substitute it value to Equations (1), (2). and (3) to
solve for Q,, Q2, and Q3, respectively ·
398
CHAPTER SEVEN
Fluid Flow in Pipes
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
PIPE NETWORKS
The following conditions must be satisfied in any pipe network:
1. The algebraic sum of the pressure drops (head loss) around any clo~t
loop must be zero and,
2. The flow entering a junction must be equallo the flow leaving it.
a=-
l:KQ 2
a
22:KQa
399
Eq. 7-53
The first condition stales lhal there can be no discontinuity in pressure (lh
pressure drop through any rou te between two junctions must be the sanu)
The second condition is a statement of the law of continuity .
In applying the above equation:
'l:KQi "' algebraic sum of the head loss tn the circuit (clockwise positive,
counterclockwise negative)
'i.KQ" = absolute sum without regard to direction or flow (clockwise
positive, counterclockwise positive)
Pipe network problems are usually solved by numerical methods usin
computer since any analytical solution requires the usc of many simultaneou
equations, some of which arc nonlinear.
!he correction a is added or subtracted from the assumed flow 111 order to get
the true or corrected flow, It is added if the direction of flow is clockwise and
-.ubtracted if counterclockwise,
!he general formula m computing the correction a can be expressed as.
Hardy Cross Method
The procedure su ggested by Hardy Cross requires lhat the flow in each pip
be assumed so that Lhe principle of continuity is satisfied at each junction. A
correction to the assumed flow is computed successively for each pipe loop in
the network until the correction is reduced to an acceptable value.
Let
Q. =assumed flow
Q =true flow
a - correction lhen;
Q= Q,,+a
:LKQ II
ex=-
Eq. 7-54
a
II
L KQ/-1
Where 11 = 2 for Darcy-Weisbach and Manning formulas and 11
Hazen- Williams formula, The value of K are as follows
•
0.0826JL
Darcy, K = - - - " - -
os
1.85 for
Eq. 7-55
2
Using Darcy-Wcisbach formula:
0.0826JLQ~
,,, = _
___:._ _
os
ltr= KQZ
'l:KQ2 = 0
'l:K(Q., + a)2 = 0
'l:K Q,Z + 2'2:KaQ., + '2:Ka2Qn = 0
If a is small, lhe term containing a2 may be neglected.
Hence;
'l:KQ,Z + 2'2:KaQ,, = 0
K _ 10.6911 L
.
M annmg, olo/3
Eq. 7-56
.
K
10.67L
H azen-Willlams, = Ctl.ss D4.87
Eq. 7-57
400
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
!solved Problems
0:
=-
L.KQ 2
a
2'L.KQa
399
Eq. 7-53
Problem 7- 1
Water having kinematic viscosity v = 1.3 x 10·6 m2js flows in a 100-mm
diameter pipe at a veloci ty of 4.5 m/s. Is the flow laminar or turbulent?
Solution
R,. = -1!0 = _ 4.5(0.1)
_..:___:._
v
1.3 x 1o-n
Rr = 346,154 > 2000 (turbulent flow)
Problem 7- 2
Oil of specific gravity 0.80 flows in a 200 mm diameter pipe. Find the critic,ll
velocity. Use~~= 8.14 x 10·2 Pa-s.
In applying the above equation:
'i:.KQi = algebraic sum of the head loss in the circuit (clockwise positive.
counterclockwise negative)
l.KQ .. = absolute sum without regard to direction of tlow (clockwise
positive, counterclockwise positive)
[he correction a is added or subtracted from the assumed flow tn order to get
the h·ue or corrected flow, It is added if the direction of flow is clockwise and
'iubtracted if counterclockwise,
r he general formula in computing the correchon a: can be expressed as.
a:=-
Solution
L.KQ II
a
1
n I KQ/-
Eq. 7-54
/\t cri tica l velocity in pipes, R.. = 2000
Where 11 = 2 for Darcy-Weisbach and Manning formulas and 11
Hazen- Williams formula, The value of K are as followc;:
Rr = vDr
p
2000= 11c(0.2)(1000 x20.80)
8.14 x rol', = 1.0175 mjs
•
0.0826JL
Darcy, K = ---=-~
os
1.85 for
Eq. 7-55
2
.
K _ 10.69n L
M anmng, Dlb/3
Eq. 7-56
. .
K
10.67L
Hazen- W111tams, = c11.85 04.87
Eq. 7-57
Problem 7-3
For laminar flow conditions, what size of pipe will deliver 6 liters per second
of oil having kinematic viscosity of 6.1 x 10·6 m2fs?
Solution
For laminar flow conditions, R· ~ 2000.
vD
QD
A
Rr = - = - V
V
0.006 D
Jl.Q2
2000 = --:!.4.- - - , 6.1 X 10-1\
D = 0.626 m = 626 nm1
400
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
!solved Problems
Problem 7 - 1
Water having kinematic viscosity v = 1.3 x 10·6 m 2/s flows in a 100-mm
diameter pipe at a veloci ty of 4.5 m/s. Is the flow laminar or turbulent?
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
401
Problem 7-4
Oil having specific gravity of 0.869 and dynamic viscosity of 0.0814 Pa-s flows
through a cast iron pipe at a velocity of 1 m j s . The pipe is 50 m long and 150
mm in diameter. (n) Find the head lost due to friction, and (b) the shearing
~tress at the walls of the pipe.
Solution
Solution
R = vD =
4.5(0. 1)
v
1.3 x 1o-n
('
R,. = 346,154 > 2000 (turbulent flow)
(n)
Rr = VOp
1..1
Rr = (1)(0.15)(1000 X 0.869)
0.0814
R,. = 1,601 < 2,000(laminar)
Problem 7-2
Oil of specific gravity 0.80 flows in a 200 mm d iameter pipe. Find the critical
velocity. Use ~~ = 8.14 x 1 Q-2 Pa-s.
f = 64
Rc
64
f = 1601 = 0.04
Solution
fL v2
At critical velocity in pipes, R,. = 2000
ly= - 0 2g
Rr = vDp
0.04(50) _QL_
0.15
2(9.81)
lrr = 0.68 m
~~
2000
=
= ?1c(0.2)(J000x0.80)
8.14x l0 2
r•, = 1.0175 m/s
(b)
Problem 7-3
For lam inar flow cond itions, what s ize of pipe will d eliver 6 liters per second
of oil having kinem atic viscosi ty of 6.1 x 10·6 m2js?
ylr LO
··=-4L
= (9810 X 0.869)(0.68}(0.15)
4(50)
'to = 4.34 Pa
Solution
For laminar flow conditions, R,. ~ 2000.
vD
QD
A
Rr = - = - V
V
0.006 D
1!.02
1
2000= _...:!,.
" - 6.1 X 10- (,
D = 0.626 m = 626 mm
Problem 7-5
Determine the (a) shear stress at the walls of a 300-mm-diameter pipe when
water flowing causes a head lost of 5 m in 90-m pipe length, (b) the shear
velocity, and the (c) shear stress at 50 mm from the centerline of the pipe.
402
CHAPTER SEVEN
FLUID MECHANICS
Fluid Flow in Pipes
& HYDRAULICS
Solution
(a) Shear stress at walls
yhLD
(a)
t,, = - -
CHAPTER SEVEN
Fluid Flow in Pipes
403
l~r = 0.1273(0.1)
1.08x 10-4
R,. = 118 (laminar)
4/
= 9810(5)(0.3)
(b)
4(90)
t,, = 40.9 Pa
(b) Shear velocity
(1,=
=
J¥
~ :g~~
1•, = 0.2
(c)
(d)
m/s
(c) Shear stress 50 mm trom ptpe center
yhL
(e)
r= -~
2L
R = 0.1273(0.1)
r
1.51 X 10-S.
R,. = 843 (laminar)
Rr = 0.1273(0.1)
4.06x 10- 7
R,. = 31,361 (turbulent)
R = 0.1273(0.1)
<
1.02 x 10-6
R,. = 12483 (turbulent)
R = 0.1273(0.1)
,. 1.15 x 10- 7
R, = 110,716 (turbulent)
= 9810(5) (0.05)
2(90)
r = 13.6 Pa
(/)
Rr = 0.1273(0.1)
1.1~x10 -3
Rr = 10.8 (laminar)
Problem 7-6
A fluid flows at 0.001 m3 /s through a 100-mm-diameter pipe.
Determm
whether the flow 1s laminar or turbulent if the fluid is (a) hydrogen (v = 1.08
IQ-4 m2js), (b) air (v = 1.51 x 10-s m2js), (c) gasoline (v = 4.06 x 10-7 m2js), (d)
water (v = 1.02 x 10-n m 2 /s), (e) me rcury (v = 1.15 x 10-7 m2js), or (f) glycerin (
= 1.18 x 10-3 m2/s)
Problem 7-7
Water flow at the rate of 200 lit/sec through 120-m horizontal pipe having a
diameter of 300 mm. If the pressure difference betwee n the end points is 280
mrnl lg, determine the friction factor.
Solution
Solution
R.. = uD
- 0.0826JLQ
I y-
2
os
v
(I=
1•
Q = 0.001
2
A
f<0.1)
= 0.1273 m/s
For a horizontal pipe, the head lost between the points is equal to
the difference in pressure head, See page 376.
CHAPTER SEVEN
Fluid Flow in Pipes
404
J.lt-P2
hr = - - = 0.28 mm Hg (13.6) = 3.808 m of water
y
3.808 = 0.0826/(120)(0.2)
(0.3) 5
{= 0.0233
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
2
"175 = 0.0826(0.0366~(1 50)Q
405
2
(0.02)~
Q = 0.00111m 1 js
Q = 1.11 Litfsec.
Problem 7- 10 (CE November 1995)
Problem 7 - 8
A fluid having v = 4 x 10·5 m 2/s flows in a 750 m long pipe h aving a diamehr
of 20 mm. Determine the head lost required to maintain a velocity of 3 m / s.
I he head losl in 50 m of 12-cm-diamele r p ipe is know n to be 6 m when a
liquid o f sp. gr. 0.9 flows a t 0.06 m/ s. Find the s hear stress at the w alls of the
pipe.
Solution
Solution
R,= vD
R=
'
T
yh 1 D
-lL
=--=
v
''
3
2
(0.0 ) = 1500 < 2000 (laminar)
4 x 1o-s
'" = 31.78 Pa
f= 64 = ~
R,.
1,500
= 0.042667
hr = fL ~ 0.042667(750) ~
D 2g
0.02
2(9.81)
(9810 x 0.9)(6)(0 .12)
4(50)
Problem 7 - 11 (CE Board)
What commercial size of new cast iron pipe shall be used to carry 4,490 g pm
with a los l of head of 10.56 feet per mile? Assume f= 0.019.
·
Solution
2
ht = 733.95 m
h, =
0.0826JLQ
---=--=-05
_
Problem 7-9
~ x 3.79 lit x 1 m in
Q - -l,-l90 m in
Fluid flows th rough a 20-mm-diameter pipe, 150 m long at a Reynolds numbt•r
of 1,750. Calculate the discharge if the head lost is 175 m
Solut ion
- 0.0826JLQ
I lr-
2
os
Since R, = 1,750 < 2,000. the flow IS laminar
gal
60 sec
Q = 28-llitjsec = 0.284 m 3 j s
5280 ft
1m
L = 1 mile x mile x 3.28 ft
L = 1609.76 m
1m
ht = JQ.56 ft X 3.28 ft
hr= 3.22 m
f=~=~
R,.
f= 0.0366
1750
3.22 =
0.0826(0.019)(1609.76 )(0.284) 2
os
D = 0.576 m = 576 mm
406
CHAPTER SEVEN
Fluid Flow in Pipes
Problem 7- 12 (CE Board 1988)
There is a leak in a horizontal 300-rrun-diamcter pipeline. Upstream from th
leak two gages 600 m apart showed a difference of 140 kPa. Downstream fron
the leak two gages 600 m apart showed a difference of 126 kPa. Assuming f
0.025, how much water is being lost from the pipe.
Solution
Pt
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
Problem 7 - 13 (CE May 2003)
\\later flows from a tank through 160 ll'l'l of 4 inches didmelL' r pipL' and then
discharges into air as o;;hown in rigure :!0. l"he flow nf \\ dter in the pipt• is 12
d!>. Assume II = o.on and neglect minor IO'>Sl'S. Octermilw th e following:
(a) fhc velocity of How in the pipL' 111 tps.
{It) fhe total hL•ad lost in the pipe 111 feet.
(c) llw pre'isu n! at tlw top of the tanh. in p'>i.
P2
300 mm 0
0
~
- Q,
D
300 mm 0
- Ql
6
L, =600 m -----.j
407
0
=4"
20
1+---L2 = 600 m - . j
Leak
hf= 0.0826JLQ
2
os
.....
0.0826(0.025)(600) Q2
(0.3) 5
D ='I"
p = 7
LOO'
h, = 509.876Q2
Water
10'
Since the pipe is uniform and horizontal, the head lost between any to point
is equal to the pressure head difference.
D = 4'
40
[hf, = P1 - P2 I
•
y
140
509.876 Q, 2 = 9.81
Q, = 0.167 m3js
Solution
(! = 12 ft 1/'> = 0.~401 m'/ s
/) = 4" = o.::rn ft - 10 l. h mn1
II. =
20'
0.013
I = 160 feet = 48.78 m
Q2 = 0.159 m3js
[Q,. = Q,- Q2]
Q,, = 0.167- 0.159 = 0.008 irNs
Q,, = 8 Lit/sec
....
~
126
509.876Q22 = 9.81
D = 4"
p =?
El. 10'
100'
A
r
10'
El. 100'
D ='I"
Water
D = 4"
El. 0
B
40'
c
408
CHAPTER SEVEN
Fluid Flow in Pipes
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
(a) Velocity of flow in the pipe:
409
Problem 7 - 14
A 600-mm diameter pipe connects two reservoir whose difference in water
surface elevation is 48 m. The pipe is 3500 m long and has the following pipe
fittings: 2 globe valves, 4 short radius elbows, 2 long radius elbows, and one
gate valve half open. The values of loss factors for pipe fittings in given in
Table 7-6.
v= Q
A
12
t(0.333) 2
v = 137.785 ftfs
(b) Head lost in the pipe:
2
Using the equivalent length method, estimate the flow through the pipe in
L/s. Assumef= 0.015.
10.29(0.013} 2 (48.78)(0.3401) 2
(0.1016) 16 / 3
HL = 1942.33 m = 6370.86 feet
Solution
The total head lost in the system is equal to the difference in elevation of
the surfaces = 48 m
2
HL = 10.29n LQ
b16/l
=
HL = 0.0826JLQ
71 2
_A_
2g
p
2
y
2g
L = 3500 + 2[350(0.6)] + 4[32(0.6}] + 2[20(0.6)] + 1[72(0.6)] = 4064 m
+ _A + ZA - HL = ~ + E.f.. + Zc
y
2
137
0 + E. + 10- 6370.86 = ( ·785) + 0 + 100
y
2(32.2)
E. = 6755.65 feet of water
y
p = 421,552.8 psf
p = 2,927.45 psi
Using the English units for Manning's Formula:
v = 1.49 R2/3 51/2
n
v = 137.785 ft/s
R=D/4
R = 0.333/4 = 0.0833 ft
S = HL/ L = HL/160
1 49
137.785 = · (0.0833)213(HL/160)112
0.013
HL = 6356 feet
2
os
(c) Pressure in the tank:
Energy equation between A and C:
E"- HL = EB
HL = 0.0826(0.015)(4064)Q 2 = 48
5
(0.6)
Q = 0.861 m3fs = 8611./s
Problem 7- 15 (CE May 2002)
In the syringe of the figure shown, the drug had p = 900 kg/m3 and Jl = 0.002
Pa-s. The flow through the needle is 0.4 mL/ s. Neglect head loss in the larger
cylinder.
d2
Q
= 10 mm
:l§~o(
d, = 0.25 mm
F
B
. - - - 20 mm
-+-- ~
30 mm
(a) Determine the velocity at point B in m/ s.
(b) What the Reynolds number for the flow in the needle.
(c) Determine the steady force F required to produce the given flow.
410
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
Solution
(a) Velocity at point B:
CHAPTER SEVEN
Fluid Flow in Pipes
Solution
Colebrook Formula:
0.4x10-6
Q
llR = - =
FLUID MECHANICS
& HYDRAULICS
1
(E I D 2.51
11
= -2 log 3.7 + R,11
{-(0.00025) 2
va = 8.1487 rnfs
A
(b) Reynolds Number:
vDp
Rr= - l.l
5,ooo11
I
2.51
1
( O.D15
2.51
11 = - 2 og 3:7 + 5,oooJ7
8.1487(0.00025)(900)
0.002
Rr = 916.73 (laminar flow)
=
]
l
Solve for fby h·ial and error:
f= 0.0515
(c) Force F:
Energy Equation between A and B:
EA- HL = LB
2
l
_1_ = _210 (0.015 ~
g 3.7
ff
411
Using the Moody Diagram, f"" 0.05
2
~ + ~ +zA-HL= ~ + J!JL +zB
2g
y
0
2g
y
= 0 (negligible)
HL = htin the needle
0.09
11A
IUfbUI4:n, zone
lr•nst•on
zont
0.08
0.07
Since the flow is laminar:
f= 64/R,. = 64/916.73
f= 0.0698
0.06
0.05
HL = 0.0826(0.0698)(0.02)(0.4 X 10-6 ) 2
(0.00025)
5
tl
\
0.03
~'i~
0.02
0.015
fJ-_
•
~
0.03
~
HL = 18.89 m
8 487
)
0 + ~ + 0-18.89 = ( .1
y
2(9.81)
PA = 196,681 Pa
-5
0.04
0 05
0 14
:5 0.025
2
+0+ 0
u
~
0.02
0.01
0.008
0.006
0.004
0.002
~t'lj
...
0.0002
Force, F = PA x Area of piston
= 196,681 X (0.01)2
Force, F = 15.45 N
t
'
0.008
00001
!"'
'
0.01
0.009
4
7
10' 2
3 4 5 7
10'
10' 2
3 4 5 7
3 4 5 7
10'
10° 2 ' 10°
107 2
Reynolds number, R,
Problem 7 - 16
Determine the friction factor for flow having a Reynolds number of 5,000 and
relative roughness (E/ d) of 0.015 (transition zone) using Colebrook formula.
"'
"'"'c
"'e::>
"'
.t:
~
0.001 1il
0.0008 "'
0.0006 tr
0.0004
~-,~
0.01 5
'5
0 00005
0.00001
3 4 5 I
10°
107
CHAPTER SEVEN
Fluid Flow in Pipes
412
FLUID MECHANICS
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
Problem 7 - 17
413
or. using Eq . 7- 12
The velocities of flow in a 1-m-diameter pipe are 5 mjs on the centerline and
4.85 mjs at x = 100 mm. Determine discharge ifJ=.
Fu' = 5 - 75(0.21192)
Solution
~
3.75fp
(I= II,-
11 = 3.99 m/s
The velocity at any point is given by
Eq. 7- 8:
r
0
log--
II= Vr- 5.75
~'o-r
Pipe radius, r0 = 500 mm
Centerline velocity, v.. = 5 mj s
Velocity at r = 100 mm, u = 4.85 mjs
4.85 = 5- 5.75
fi
0a
vP"
g 5oo-1oo
lo
500
Discharge, Q = A o = f (1 )l(3. 9<:1)
Discharge. Q = 3.13 m 3/s
Problem 7 - 18
Oil or sp. gr 0.9 and dynamic v1scosity 1-l = 0.04 Pa-s flows at the rate ot nO
hters per second through 50 m of 120-mm-diameter ptpe If the head lost ~~
hm, determine (a) the mean velocity of flow , (b) the type of flow . (c) the tnct10n
tactor f, (d) the velocity at the centerline ot the pipe. (e) the shear stress at tht->
wa ll of the pipe. and (f) the velocity 50 mm from the centerline of the pipe
=0.2692
Solution
(a) Mean velocity
/1 =
Q =
0.06
t(0.12) 2
A
= 0.2692
(b) Type ot flow
jv2 = 0.5797
f= 0.5797
R,= uDp _ 5.31(0.12)[1000(0.9)]
R. = 14,337 > 2000 (turbulent flow)
From Eq. 7- 11:
v(1 + 1.33.J7)
v( 1.33J 0·!~97 )
-v (t + - -
5=
5-
0.04
1.1
v2
Vr =
'' = 5.31 m/s
1+
1.0126)
v
5 = v + 1.0126
v = 3.99 m/s
(c) Fnctton factor
0.0826JLQ
2
h,= ---=-~
os
0.0826/(50)(0.06)
t> = __
(0.12)5
2
...:.....,:___,:~__;.-
f= 0.01004
(d) Centerline veloctty
v, = v(1 + 1.33ft2)
= 5.31[1 + 1.33(0 01004) 1'21
11. = 6.02
m/s
CHAPTER SEVEN
Fluid Flow in Pipes
414
(e) Shear sb·ess at the wall of the pipe
From Eq. 7- 9:
~Po = v8
Uv28v2
fp
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
It' = o - 1 75
(I
squaring both sides:
fi
415
I
= 4.0 - 3.75(0 2582)
1• = 1.63 m /s
2
~ = ./:'_
8
p
= 0.01004(5.31) 2
1000x 0.9
8
't 0
Dtscharge. rJ = Au
= t (0.75)l(3.63)
I )tscharge. (.2 = 1·.6 m.ljs
'" = 31.85 Pa
Using Eq. 7- 16:
- yltL D - (9810 X 0.9)(6)(0.12)
'to- 4L 4(50)
Problem 7 - 20
What ~~ the hydrauil( radtu~ ot a rectangu lar illr duct 200 mm by 150 mm '
Solution
Hvdrauil< radtu~. /{=A/ Jl
200 X 350
I< =
200 )( 2 ... 150 )( 2
I< 63.6 mm
'to = 31.78 Pa
31.85
60
6 02 5 75
= ' - '
1000 x 0.9 log 60-50
u = 5.178 m/s
Problem 7- 21
Atr at 1450 kP& ab:. and 100 oc._ flow:. 111 a 20-mm-dta mcter tube what 1:, tht>
max tmum lam milr flo w rate7 UseR= 287 1/kg-°K. p = 2.17 • 10·~ Pa-s
Solution
Problem 7 - 19
For lammar flow . R. s 2000
a
The velocities in 750-mm-diameter pipe are measured as 4.6 m/s and 4.4
m/ s at r = 0 and r = 100 mm, respectively. For turbulent flow, determine tht
flow rate.
R.. = VDp
1.1
Solve for p
Solution
11 = Vr- 5.75
p = _L
~ log-ro_
. VP
Y0
RT
1450(1000)
-r
287(100 + 273)
~ tog-- 375
- _- - 4.4 = 4.6- 5.7s
vP"
fi
p = 13.54 kg/ m 1
375 100
R. = u(0.02)(13.54)
=0.2582
.
2.17 X 10-S
u = 0.1603 m/!>
2000
416
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
417
Problem 7 - 23
Q = Ao
= "i (0.02)2 X 0.1603
A liquid having a sp. gr. of 0.788 flows at 3.2 m/ s through a 100-mrn-diameter
pipeline if= 0.0158). (a) Determine the head loss per kilometer of pipe and (b)
the wall 'Shear stress.
= 0.0000503 m3 j s
Q = 0.0503 lit/sec.
Solution
(n) Head loss per kilometer (L = 1000 m)
Problem 7 - 22
fL v2
Glycerin (sp. gr. = 1.26 and J..l = 1.49 Pa-s) flows through a rectangular conduit
300 mm by 450 mm at the rate of 160 lit/sec.
(n) Is the flow laminar or turbulent?
(b) Determine the head lost per kilometer length of pipe.
/y= - D 2g
0.0158(1000) (3.2) 2
0.1
2(9.81)
ly= 82.5 m
Solution
(a) For non-circular conduits;
(b) Wall stress:
R.· = 4vRp
~ = fu2
J..l
p
Q
0.16
=1.185m/s
0.30x0.45
R = ~ = 0.30x 0.45
p
2(0.3 + 0.45)
R = 0.09m
v=- =
_ _-r....::.
0_ _
A
-7 See Problem 7- 18 (e)
8
1000 X 0.788
=
0.0158(3.2) 2
8
'to= 15.94 Pa
or:
Rr = 4(1.185)(0.09)(1000 X 1.26)
1.49
Rr = 360.75 < 2000 (laminar)
yhL D
to=~=
(9810 X 0.788) X 82.5 X 0.1
4x1000
'to= 15.94 Pa
(b) For laminar flow:
f= 64 =~
Rc
360.75
f= 0.1774
Problem 7 - 24
Oil with sp. gr. 0.95 flows at 200 lit/ sec through a 500 m of 200-mm-diameter
pipe if= 0.0225). Determine (a) the head loss and (b) the pressure drop if the
pipe slopes down at 10° in the direction of flow.
fL y2
lzr = - D 2g
Solution
D = 4R (for non-circular pipes)
D = 4(0.09) = 0.36 m
(a) Head loss
ly= fL ~ = 0.0826jLQ
D 2g
2
o
2
5
h = 0.1774(1000) (1.185)
I
0.36
2(9.81)
h = 0.0826(0.0225)(500)(0.2)
ly = 35.27 m
I
(0.2)5
2
=116.2m
CHAPTER SEVEN
FlUID MECHANICS
& HYDRAULICS
418 , Fluid Flow in Pipes
(b) Pressure drop:
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
v 2
0 + 0 + It -ly= - 2 - + 0 + 0
2g
v 2
h = - 2- + hr
2g
v2
fl
v2
ly = - 0 2g
0=4R=4~
p
p
+ _ 1 + Zl - JIL = _2_ + __l_ + Z2
2g
y
2g
y
A = ~ (0.1)L ~ (0.06) 2
v 2
and - 2 - cancels out
2g
2g
ll 2
Since 111 = ll2, - 1-
A= 0.005026 m2
P = 1t00 +nO;= n(0.1) + 1t(0.06)
P = 0.50265 m
E!_ + 86.82- 116.2 = !!.1._ + 0
y
-7 Eq. (1)
fL v2
Energy equation between 1 and 2 (datum at 2):
f.1 - ILL = f.2
_1_
419
y
0 = 4 0.005026 = 0.04 m
0.50265
fl._ - 12 =.29.38 m of oil
y
y
fll - ]72 = 29.38(9.81 X 0.95)
P1 - p2 = 273.81 kPa
10
v=Q=
TiiOO
A
0.005026
v =, 1.9896 m /s = v2
2
It = 0.0232(50) 1.9896
=5.85 m
I
0.04
2(9.81)
Problem 7 - 25
Water flows through commercial steel annulus 30 m long as shown in the
figure. Neglecting minor losses, estimate the reservoir level lr needed to
maintain a flow of 10 lit/ sec. Assume f = 0.0232.
ln Eq. (1):
2
= 1.9896 + 5.85
2(9.81)
It = 6.052 m
11
60mm '?'
Problem 7 - 26
Find the approximate flow rate at which water will flow in a conduit shaped
in the form of an equilateral triangle if the head lost is 5 m per kilometer
length. The cross-sectional area of the duct is 0.075 m 2 Ass umef= 0.0155.
100 mm 0
Solution
Energy equation between 1 and 2 (datu1iurt~~
E1-IIJ = f.2
<I
v12
2g
+El_ + -!tr = ~v~.oti~2 -~ill}(OOc?(~O fl)c,
Zl
y
g
y
c(~ 0)
Solution
fL v2
ly= - D 2g
ly = 5 m
L = 1000 m
<i:H'A'eTE~ SEVEN.!
f;lui~ Flow im Pipt'lS'
FLUIJD.MaiHR\tv<tl
&..+frtDAA\.11.10
D=4R
5
= 0.0155(1000)
0.24
.:
\
878 0 15
( · ) = 682 < 2000
R,. = 1.
0.000413
[Q =Av]
Q=
t (0.15)2(1 .878)
In the figure shown, the 50-m pipe is 60 mm in diameter. The fluid flowing
- (I IJ) ~
\ lfl.fl)
m o<:Odlf1.f)
!>
,C!n-t (,n
1
'O)t
rn coS.O
has mass density of 920 kgj m 3 and dynamic viscosity of 0.29 Pa-s.
pressure in the encLosed tank is 200 kPa gage. Determine the following:
(a) The amount and direction of flow?
(b) The velocity of flow in the pipe?
(c) The Reynolds Number of the fl ow?
n
mt.OfJ
n. I
Problem 7 - 27
/
Heavy fuel oil flows from 1\ to 8 tb rougb a ~dbb-m h6Hzontal 150-mmdiameter s teel pipe. The pressure at A is 1,050 kPa a'nd 'at B ls 135 kPa. The
kinematic viscosity is 0.000413 m 2/p ~d the specific ~h\vi'ty is 0.9t . What is
the flow rate?
1 , ')
' 0
Solution
PA- PB
For uniform h orizontal pipe, Jy = .!....!..:.--!....:::..
y
1
•
---11
( f8.P)S:
rn !:?.O.d 11
~L=50m_j
9.81 X 0.92
ly= 32vLv
gD2
. = 32(0.000413)(1000)v
112 46
9.81(0.15) 2
v = 1.878m
Solution
--a~
R,. = vD
v
-'\ msldo1q
11. ~161 101'! Jlrrruxc11qqr; JrlJ bm
r'
Jt,ltiJpJ llf lo r.tJO 'Jfll 01
~ ,.
(f,IIOil 1~ a I l 'Jri f riJgn • •
r7
1
I
/
H = 12m
•
f\. =- ,,
2!::
u
1\
m O' or
1
r7.
p
.j.CI
r
IH
p = 200 kPa
1
noi:lulo2
l
Check:
2
n
r
~ r~
('H,( I
D
H = 12 m
(r) .p'f nl
Assuming laminar flow:
larnma r. OK
Problem 7- 28 (CE November 2002}
X
[Q =Au]
Q = 0.075(1.232) = 0.092.4 n~3 / s
Q = 92.4 litjsec
ly = 1050 - 35 = 112 .46 m
421
Q = 0.0332 m3/s
v2
2(9.81)
o = 1.232 m/s
ti I()+ 0
0 ~ 0 -t
R = A/P = A/(3x)
1\ = 1/2 x 2 sin 60° = 0.075
x = 0.416 m
R = 0.075/ [3(0.416)]
R = 0.06 m
D = 4(0.06)
D = 0.24 m
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
~ L=SOm _j
'
2
=920 kg/m3
./
The
422
CHAPTER SEVEN
Fluid Flow in Pip es
FLUID MECHANICS
& HYDRAULICS
raking level 2 as the datum.
Energy£,= 12m
Energy.E2=0+ E.= ..E._= 200,000 =22.16m
Y
pg
920x9.81
Smce L2 > E,, the flow IS trom 2 to I
rry f= 0.03·
In Eq. (1):
0.00002
1 11
- 1 =-1.8 1og[6.9
- + (r./0)
-R,
3.7
'->ince our ass umption is correct, then
Discharge, Q = 0.00201 m3js x 3600 s/ hr
Discharge, Q = 7.24 mljhr (from 2 to 1)
Velocity of flow, v = 0.711 m/s ·
Reynolds Number = 135.4
]
!J
_,,,=
= 0.711mj1>
f(0.06) 2
_ vDp
0.711(0.06)(920)
, no Jd s N um ber, Rrl,ey
- -_ -~-...:....:....__:_
1-1
0.29
Reynolds Num ber, Rr = = 135.4 < 2000 (laminar flow, OK)
0.03 v2 = 0.471, o = 3.962 m/s
R = 3.962(0.3) = 59.430
C2-hr=L ,
22.16
12
lrr= 10.16 m
t 0 ( fl OW, U = 0.00201
Ve l OC 'L:Y
423
-7 Eq. (1)
fv2 = 0.471
hnergy equatiOn between 2 and I
Assuming laminar flow (R1 < 2UUUJ
flrr= 128~tLQ I 10 _16 =
128(0.29)(50)\,.!
4
npgD
n(920)(9.81)(0.06) 4
Q = 0.00201 mJfs
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
_I
!1
=-1. 8 log[~ +( 0.0002)
59.430
1 1
'
]
3.7
[= 0.0206
New f= 0.0206
In Eq. (1):
0.0206 u2 = 0.471.
R, = 4·78(0.3 ) = 71 700
0.00002
'
_1 =-} _8 10
11 = 4.78
[~+(0.0002) ' "]
!1
g 71,700
f= 0.0199
(OK)
3.7
In Eq . (1 ): 0.0199v2 = 0.471
o = 4.865 m/s
Q = A v = f (0.3)2(4.865)
Q = 0.344 m3/s
Problem 7 - 29
Oil, with p = 950 kg/m1 and v = 0.00002 m2js, flows through a 300-mm
diameter pipe that is 100 m long with a head loss of 8 m . r./D = 0.0002
Calculate the flow rate
Solution
Problem 7 - 30
Two tanks of a solven t (v = 0.0000613 m 2 / s, y = 8 kNfm3 ) are connected by 350
m of commercial steel pipe (roughness, r. = 0.000046 m). What size must the
pipe be to convey 60 L/ s, if the surface of one tan k if 5 m higher than the
other. Neglect minor losses
2
f1 v
[lrr=- - j
D 2g
= /(100) _u2_
8
0.3
2(9.81)
Solution
hr= 5 m
ltr= 0.0826JLQ
os
2
424
CHAPTER SEVEN
Fluid Flow in Pipes
5 = 0.0826/(350)(0.06)
05
FLUID MECHANICS
& HYDRAULICS
2
D = 0.461 / 15
-7 Eq. (ll
_9_0
n 02
K = oD = 24__ _ 4Q
v
v
1t0V
nD(0.0000613)
~ =
[)
-7 Eq. (3)
0
fry f= 0.03
In Eq. {1).
o = 0.461(0.03) 1' ' = o.228n
In Eq. (2)
R.. =
In Eq. (3):
..:. = 0.000046 = 0.000201
1246
= 5450
0.2286
1
1 = -1.8 log r6.9
[!
R;+ (r./0)1.1
3:7
l
f= 0.03692
~New f
3.7
0 = 0.461(0.03692) 1/ s = 0.2383
In Eq. (2):
R,. = 1246
In Eq. (3):
..:. =
= 5229
0.2383
2
,( 0.0826JLQ
h; = ----;:---:-::~
05
0.0826
/(1000)(0.55) 2
3=_
_..::...:...--=-...:....:....____;:..._
0 = 1.528f/5
R = 4Q
t
0.000046 = 0.000193
0.2383
Solve for new Jfrom Eq. 7 - 30:
_1 = _1 _810
]0.04
9.81(5)
os
In Eq. (1):
0
52
+6.13x10-5 (0.06)9·4 (~) '
Solution
1 11
g 5450
[
350 0 06 2 475
.
( · ) )
9.81(5)
What size of new cast iron pipe (e: = 0.00026) is needed to b·ansport 550 L/ s of
water for 1 km with head loss of 3m? Use v = 9.02 x 10-7m2Is.
r~ + ( 0.000201 ) ' ]
[!
= 0.66
(0.000046) 1 ·25 (
Problem 7- 31
Solve for new / from Eq. 7-30:
_1_ = -1.8 lo
475
52]0.04
0 = 0.66 e:1.25[LQ2l. + vQ9.4[_L_l·
[
gil!
glzf
0 = 0.234 m = 234 mm
0.2286
0
0 = 0.461(0.0374)1/5 = 0.24 m
Rr = 5192
r./ 0 = 0.000192
f= 0.0374
An approximate formula for 0 is given as follows:
-7 Eq. (2)
0.000046
0
425
The procedure has converged to the correct diameter of 240 mm.
1246
/{,. = _ _
4.!..,.(0_.0_.:.6) _
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
r~+(0.000193)u']
[!
g 5229
{ = 0.0374
Let's use this (
. 3.7
=
7t0V
~ Eq. (1)
4(0.55)
7t0(9.02 X 10-7 )
Rr = 776,366
0
~ = 0.00026
0
D
Try f= 0.03
D = 1.528(0.03)115 = 0.7578
Rr = 776,366/ (0.7578) = 1,024,500
e:/D = 0.00026/0.7578 = 0.000343
~ Eq. (2)
~ Eq. (3)
CHAPTER SEVEN
Fluid Flow in Pipes
426
.]___ =-1.8 log[
f1
FLUID MECHANICS
& HYDRAULICS
-
1 11
6.9
+(0.000343) ·
1,024,500 .
3.7
't:
5
f = 0.0163
0.915 (1.22)
= 9810(4.6) (0.261)
2(91.5)
't:5 = 64.36 Pa
New J= 0.016
D = 1.528(0.016)115 = 0.668
R, = 776,366/ (0.668) = 1,162,225
r./ D = 0.00026/0.668 = 0.000389
f1
Problem 7 - 33
Oil with p = 900 kgjm3 and v = 0.00001 m2js m flows at 0.2 m3js through 200nun-diameter cast iron pipe 600 m long. Determine the head loss.
6.9
+(0.000389)1.1ll
1,162,225
3.7
(OK)
Solution
From Table 7- 1, r. = 0.26 mm
r./ D = 0.26/200
r./0 = 0.0013
D = 1.528(0.0163)1/5 = 0.671 m
D=671mm
Using the approximate formula:
D~ 0.66[r.t.25(LQ2l4.75 + vQ9.4(_L)5.2]0.04
ghf
[v = Q J
A
ghf
v=
25 1000(0.55)2J4.75+ 9.02 10- (0.55)
( 9.81(3)
= 0.66 0.00026 1·
[
X
7
9.4 (
52]0.04
1000 ) ,
9.81(3)
02
= 6.37 m/s
·
t (0.2)2
fR,. = vD]
v
D = 0.683m
R. = 6·37 (0. 2) = 127 400
0.00001
C
Problem 7 - 32
Water is flowing through a 915 mm x 1220 mm rectangular conduit of length
91.5 m and a head loss of 4.6 m. What is the shear stress between the wall'f
and the pipe wall?
Solution
For non-circular pipes,
't:
5
=
yh L R
2L
R = hydraulic radius
A
R=p
427
R- 2(0.915 + 1.22) = 0.261 m
]
f = 0.016
.]___ = _1 .8 log[
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
I
From the moody diagram:
J:::: 0.0225
By Haaland Formula:
_1 =-1 .8 10 [ 6.9 +(0.0013)1.1ll
g 127,400
3.7
f1
f= 0.0226
CHAPTER SEVEN
428
FLUID MECHANICS
& HYDRAULICS
Fluid Flow in Pipes
0.09
0.05
0.04
r;
0.05
u~
0.04
\
0.02
O.D15
fl-_
0.01
0.008
0.006
~
0.03
~
0.004
a O.o25
.l:
1
[6.9 (e!D)l.lll
fJ
= -1.8\og R. + 3.7
0.03
~\
u ...., •
--
- - - Mfil
0.02
0.002
[1
~
c
~
.2:
2
fL v
[lrr=- - ]
D 2g
0.0002
3 4 5 7
3 4 5 7
3 4 5 7
10• 2 " 10•
' 105 2 ... 105
101 2" 101
3 4 5 7
0.00001
3 4 5 T
10' 2 )( 101
10'
Reynolds number, R,
D 2g
0.2
f
0.00005
'
ly= fL ~ = 0.0225(600) 0.37
h = 0.02(80)
0.0001
t:;..
1
6.9 + (0.0008\1.1
--1 ]
273,134
3.7 '
f= 0.0197
..~
&!
~.()"1:~
-I ; -!.Slog[
'5
0.001 1i
0.0008
0.0006
0.0004
"'·:--,
0.01 5
0.01
0.009
0.008
429
Using Eq. 7- 30
007 f'
0.06
Fluid Flow in Pipes
From the Moody diagram, f = 0.02
tuJbut.ol zone
tt•n S<':On
zone
0.08
CHAPTER SEVEN
FLUID MECHANICS
& HYDRAULICS
2
2(9.81)
0.15
2
1.83
2(9.81)
hr= 1.82 m
Pressure d rop for horizontal pipe, t:.p = y hr = p g hJ
= 998(9.81)(1.82)
Pressure drop for horizontal pipe, t:.p = 17,818.5 Pa
ly= 139.6 m
Problem 7 - 34
Compute the head loss and pressure drop in 80 m of horizontal 150-mmdiameter asphalted cast-iron pipe carrying water at 20°C with a mean velocity
o£1.83 m js.
Problem 7 - 35
What size of pipe is required to carry 450 liters per second of water with a
head loss of 3.4 m for 5000 m length? Assume friction factor f = 0.024.
Solution
hr =
Solution
From Table 7- 1, v = 1.005 x 10-6 Pa-s
From Table 7- 2, E = 0.12 mm
)
0.0826/Ld
Ds
0.0826(0.024)(5000)(0.45) 2
3.5 =
os
D = 0.895 m = 895 mm
R,. = vD = 1.83(0.15)
V
1.005 X 10- 6
R,. = 273,134
_:_ = 0·12 = 0.0008
D
150
Problem 7 - 36
Water flows in a 300 m.m x 400 mm rectangular conduit at the rate of 150
lit/sec. Assuming/= 0.025, find the head loss per km length
430
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
431
Problem 7 - 38
Solution
A 2.5-m-diameter pipe of length 2,500 m conveys water between two
reservoirs at the rate of 8.5 m 3 /s. What must be the difference in water-surface
elevations between the two reservoirs? Neglect minor losses and assume ( =
0.018.
fL y2
[llr=- --1
D 2g
A
D = 4R = 4 (p)
-
0.3(0.4)
Solution
D - 4 t(0.3 + o.4) = 0.343 m
For two reservoirs, the difference in elevation between the surfaces ts
equal to the total head.
_Q71
0.15
- A - (0.3)(0.4)
ll = 1.25 rri/ s
2
"' = 0.025(1000) 1.25
0.343
2(9.81)
hr= 5.8 m
I IL = ly = 0.0826fLd
os
0.0826(0.018) (2500) (8.5)
(2.5)5
2
HL = 2.75 m (difference in elevation)
Problem 7 - 37
A 20-mm-diameter commercial steel pipe, 30 m long is used to drain an oil
tank. Determine the discharge when the oil level in the tank is 3 m above tlw
exit of the pipe. Neglect minor losses and assume f = 0.12.
Problem 7 - 39
Water at 20 oc is to be pumped through 3 km of 200-mm-diameter wrought
iron pipe at the rate of 0.06 m 3/s. Compute the head loss and power requ.ired
to maintain the flow. Use v = 1.02 x 10·6 m 2/s and roughness E = 0.000046 m.
Solution
~'Jo ]~
I
m.r.o.u
Solution
fL v2
hr= - -
_j
f)
3m
D 2g
Solve forf
Rr= vD
v
Energy equation between 0 and 6 :
E, -lz1 = E2
0
- 1
2
2g
+ -P1 + z, -lzr = -V2
y
2
2g
v= Q =
A
+ -P2
0 + 0 + 3- 0.0826(0.12)(30)Q
(0.02) 5
Q = 0.000179 m~/s
Q = 0.1791./s
y
2
R,. =
E
D
0 06
= 1.91 m/s
·
t(0.2) 2
1 '91 (0•2) = 374,510
1.02 X 10-f,
0.000046 = O.OOQ23
0.2
CHAPTER SEVEN
Fluid Flow in Pipes
432
FLUID MECHANICS
& HYDRAULICS
,.,.
0.05
0.04
0.07
0.06
0.05
0.04
~
\
0.03
c
0.025
~'
0.02
0.015
~-'--
0.01
0.008
0.006
!I
~
Q
0.004
0.002
g 0.02
LL
.."·s
~
O.Q15
~
O.Q1
0.009
0.008
B
0.03
~
u
A pump draws 20 titjsec of water from reservotr A to reservmr 8 as shown
Ass uming f = 0.02 for all pipes, compute the horsepower delivered by tht>
pump and the pressures at points 1 and 2
...
tu,bu"ntz
trensl•on
0.08
'5
~c
1!--JOL.....:..:Av.....JI El. 10
.c
"'~
..
-~
0.001 ]j
0.0008 ~
0.0006
0.0004
0.0002
0.0001
200 mm - 500 mm
B
Solution
0.00005
3 • 5 7
104 2
10' 2. 101
3 • 5 7
3 • 5 7
10.
105 2 ( 10 5
104 2
0.00001
3 • 5 7
3 • 5 7
10°
10' 2 '( 101
104
l~l......-'JA~I El. 10
Reynolds number. R.
Q,
Using Eq. 7 - 30
200 mm - 500 m
(E/D)l.lll
- 1 = -1 .8 log[6.9
-+ - -
fJ
_1
ff
R..
3.7
=-1 _810 [
1 11
6.9 +(0.00023) '
g 374,510
3.7
f= 0.016
]
Q1 = Q 2 = 0.02 m3/s
0.0826(0.02)(500)(0.02) 2
ltn =
= 1.033 m
0.2 5
0.0826(0.02)(1200)(0.02) 2
hrz =
= 10.442 m
0.15 5
Energy equation between A and 8 :
0.0165(3000) 1.91 2
ilr = ----'----'0.2
2(9.81)
lzr= 46 m
Power required, P = Q y J-JL
= 0.06(9.81)(46)
Power required, P = 27.1 Kilowatts
433
Problem 7 - 40
From the moody diagram: f z 0.016
0.09
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
u/
u
2
PA
p
-- + + ZA - hn + HA - hn = - 8 - + _!L + z~
2g
y
2g
y
0 + 0 + 10- 1.033 + HA - 10.442 = 0 + 0 + 60
HA = 61.475 m
Power delivered by the pump (output power)
P = Q y HA
= 0.02(9,810)(61.475)
= 12,061 Watts x (1 hp/746 Watts)
P = 16.17 Hp
434
FLUID MECHANICS
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
2
.
Fluid Flow in Pipes
435
Solution
Pressure at 0:
Energy equation between A and 0:
EA -ltfl = Et
CHAPTER SEVEN
FLUID MECHANICS
& HYDRAULICS
Ell97 m
2
oA
V1
P1
- + -PA + ZA - hfl = - + + Z1
2g
y
2g
.y
0 + 0 + 10- 1.033 =
El SO m
0
0
2 2
+ El + 0
(0.0 )
1t2 g(0.2)4
• y
0
8
0
0
0
0
J't = 87.76 kPa
0
Q = 0.15 mJ/s
c = 120
f'ressure at 6:
Energy equation between 6 and B:
£2- hp. = Es
2
10.67LQ I ~5
Fnct10nal head lost. hr = -~---=:o:-=­
clss D4 R7
2
V2
Vs- + -PB + Zs
- + -P2 + Z2 - /rp. = 2g
y
2g
y
8 (0.02 ) 2
n 2 g(0.15) 4
l0.67( 100)(0.15) I K)
Frictional head lost, hr = --~~-~=­
(120) 1 RS (0.25) I R7
+ E1._ + 0- 10.442 = 0 + 0 + 60
Frictional head lost, 111 = 3.89 m
y
Energy equation between 0 and 6 (Datum at El. 0)
£, - HL- HE= £2
0 + 0 + 197 - 3.89 - /-IE = 0 + 0 + 50
H[ = 143.11 m
p2 = 690.4 kPa
Problem 7 - 41 {CE November 2002)
A hydroelectric power generating system is shown in the Figure. Water flow1
from an upper reservoir to a lower one passing through a turbine at the rate of
150 liters per second. The total length of pipe connecting the two reservoirs I
100 m. The pipe diameter is 250 nun the Hazen-Williams coefficient is 120
The water surface elevations of reservoirs 1 and 2 are 197 m and 50 m,
respectively. Determine the power generated by the turbine if it is 85~
l'fficient? Neglect minor losses.
Power,P,..QyH£
= 0.15(9.81)(143.11)
Power, P = 210.59 kW
(Input power)
Power generated (output power)
p = 210.59 X 85%
P = 179kW
Problem 7 - 42 {CE November 2002)
0
0
0
0
0
fhe pump shown in the Figure draws water from a reservoir and discharges it
mto a nozzle at D. The length of pipe from the reservoir to the pump is 150m
and from the pump to the nozzle is 1500 m . The pipe diameters before and
after the pump are 450 mm and 600 mm, respectively
0
0
0
l'he atmospheric pressure is 95 kPa absolute and the vapor pressure is 3 .5 kPa
Use f = 0.02 for both pipes. Zt = 4 m. The pump is to operate such that thE'
discharge will be the maximum possible
CHAPTER SEVEN
Fluid Flow in Pipes
436
FLUID MECHANICS
& HYDRAULICS
Determine the maximum rate at which water may be pumped from the
reservoir?
--.t,...---
c
tJI.===:::( P):=:========rfJ
B
437
v2
7.667- = 5.327 m
2g
D
.or=::::::DT
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
v = 3.692 m/s
Maximum discharge, Q = Av
= t (0.45)2 (3.692)
Maximum discharge, Q = 0.59 m3js
Problem 7 - 43
Assume that 57 titers per second of oil (p = 860 kg/m3) is pumped through a
300 mm diameter pipeline of cast iron. If each pump produces 685 kPa, how
far apart can they be placed? (Assume[= 0.031)
Solution
D
~l.Ul
B
Solution
Each pump must be spaced such tha't the head lost between any two
pumps is equal to th~ pressure head produced by each.
C
--.--- tJC===( P)=:===========rfJ
t
4m
Zz
A
p 685x 10 3
Pressure head, - = _:____ = 81.2 m
y
860(9.81)
hr = 0.0826 JLQ
2
pS
81.2 =
'Since the pump is above the water surface of the source tank, the pressure
at the inlet (at B) is always negative (vacuum).
As the discharge increases, the pressure at B drops. To avoid cavitation,
the absolute pressure at B must not fall below the given vapor pressure of
3.5 kPa.
Energy equation between A and B:
(using absolute pressure and datum at A)
EA- hfAB = Es
v/ PA + ZA- fL v 2 _ v/
PB
-- + - + - + ZB
2g
y
D 2g
2g
0.0826(0.031)L(0.057) 2
(0.3)
5
L = 23,718 m = 23.718 km
Problem 7 - 44
For a 300 mm diameter concrete pipe 3,600 m long, find the diameter of a 300m long equivalent pipe. Assume the friction factor f be the same for both
pipes.
Solution
For an equzvalent pipe system, the head loss and flowratr must be the samr as tltl'
original pipe system.
y
0 + ~ + 0 - 0.02{150) .:C.. = .:C.. + 3.5 + 4
9.81
0.45
2g
2g
9.81
Q.=Q,.
hro = 11,..
7 Eq. (1)
7 Eq. (2)
CHAPTER SEVEN
Fluid Flow in Pipes
438
FLUID MECHANICS
& HYDRAULICS
Using Manning's Formula for circular pipes,
D,,s
_
439
Solution
O.,P82PJ:Loof'
L0
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
2
If= 10.29n LQ
L,.
Das - D,.s
2
I
D16/3
5
= ~ = 10.29n Q
L
0 t6/3
2
3,600 - 300
300 5 - DC 5
D,. =182.5 mm
2
For pipes in series, Q, = Q2 = Q
For pipe 1:
Problem 7 - 45
fwo pipes, each 300 m long, are connected in series. The flow of watl•r
through the pipes is 150 lit/ sec with a total frictional loss of 15 m. If one pipl.'
has a diameter of 300 mm, what is the diameter of the other pipe? Negled
minor losses and assume f = 0.02 for both pipes.
10.29n 2 Q 2
D116/3
1 1
5, = ---:-:--:-:--"2 2
10.29111 Q
(0.5)16/3
7 Eq. (1)
5, = 414.87 111 2 Q2
Solution
Pipe 1
300m -300 mm
Pipe 2
300m-D=?
Q,
Q,
-
--
Q 1 = Q 2 = 0.15 m-1/s ·
HL = hfl + l1 fl
For pipe 2:
2
10.291122Q2
52 = --77-;-::--='D2 16/3
2 2
10.29(2nt) Q
==D 16/3
2
15 = 0.0826(0.02)(300)(0.15f +
0.0826(0.02)(300)(0.15) 2
0.3 5
Ds
D = 0.255 m
D =255mm
Problem 7 - 46
Two pipes 1 and 2 are in series. If the roughness coefficients 11 2 = 21! 1 and tlw
diameter D1 = 500 mm, find the diameter D2 if the slope of their energy gradl·
lines are to be the same.
7 Eq. (2)
52= 41.161112Q2
D216/3
[S, = S2]
414.87 ~=
41.16~
D2 16/3
D2 = 0.648 m
D2 = 648 mm
Problem 7 - 47
Two pipes 1 and 2 having the same length and diameter are in paralleL If the
flow in pipe 1 is 750 lit/ sec, what is the flow in pipe 2 if the friction factor f of
the second pipe is twice that of the first pipe?
CHAPTER SEVEN
Fluid Flow in Pipes
440
FLUID MECHANICS
& HYDRAULICS
Solution
11
fl
For pipes in parallel, the head losses are equal
ltji = ltfz
O.Qmftl..&/
It
~
fl
{I Q1 2 = (2 /t) Q22
(0.75)2 = 2 Q22
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
441
= 0.0826(0.02)(3000)(0.01) 2 = 1 .549 m
0.25
= 0.0826(0.02)(2200)(Q 2 ) 2 = 1495.64 Q 22
0.35
2
I1(3
Qz = 0.53 m3/s
Qz = 530 lit/sec
= 0.0826(0.02)(3200)(Q3) = 16 520 Q 2
0.2
5
'
IIJ~ = 0.0826(0.02)(2800)(0.01)
0.4
Problem 7 - 48 (CE May 2003)
A pipe network consists of pipeline 1 from A to B, then at B it is connected to
pipelines 2 and 3, where it merges again at Joint C to form a single pipeline 4
up to point D. Pipelines 1, 2 and 4 are in series connection whereas pipelines 2
and 3 are parallel to each other. If the rate of flow from A to B is 10 liters/sec
and assuming f = 0.02 for all pipes, Determine the flow in each pipe and thr
total head lost from A to D.
Pipelines
1
2
3
4
Length (m)
3,000
2,200
3,200
2,800
Diameter (mm)
200
300
200
400
5
[ltp = ltp]
1,495.64 Q22 = 16,520 Q32
Qz = 3.323 Q3
2
3
= 0.0452 m
-7 Eq. (1}
[Qz + Q3 = 0.01)
3.323 Q3 + QJ = 0.01
Q 3 = 0.00231 m 3 Is
Q3 = 2.31 1./s
Substitute Q3 to Eq. (1):
Qz = 3.323(0.00231)
= 0.007687 m3js
Qz t: 7.687 1./s
[HL = ltfl + lip + h14]
HL = 1.549.+ 1495.64(0.007687)2 + 0.0452
HL = 1.683m
Solution
<a
==::>
2
Q4
c
B
3
Q, = Q4 = 10 L/ s
Q, = Q4 = 0.01 m 3/s
[ltr = 0.0826 fLQ
os
2
I
==::>
Q3
==::>
4
D
Problem 7 - 49
A pipe system, connecting two reservoirs whose difference in water surface
elevation is 13 m, consists of 320 m of 600 mm diameter pipe (pipe 1),
branching into 640 m of 300 mm diameter pipe (pipe 2) and 640 m of 450 mm
diameter pipe (pipe 3) in parallel, which join again to a single 600 mm
diameter line 1300 m long (pipe 3). Assuming/= 0.032 for all pipes, determine
the flow rate in each pipe.
CHAPTER SEVEN
Fluid Flow in Pipes
442
FLUID MECHANICS
& HYDRAULICS
Solution
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
443
104.32Q12 = 13
I
13m
Q, = 0.353 m3/s
Q2 = 0.266(0.353) = 0.094 mJjs
Q3 = 2.756(0.094) = 0.259 m3/s
Q4 = Q1 = 0.353 mlfs
Problem 7 - 50 (CE May 2002)
Q, = Q4
Q, = Q2 + Q3
"h =/if,
HL = ltf, + 11/2 + ltfl = 13 m
/if= 0.0826[LQ
-7 Eq. (1)
-7 Eq. (2)
-7 Eq. (3)
-7 Eq. (4)
For the pipe system shown in the Figure, 11 = 0.015 for all pipes and the flow in
pipe 4 is 12 cfs. Determine the following:
(a) U1e head lost in pipe 1 in feet.
(b) the total head lost in terms of the total discharge Q, where Q is in cfs.
(c) total head lost in feet.
2
o~
(0.6)
5
A
= 10.877Q12
,(. - 0.0826(0.032)(640)Q2 2
/'.12- ----'---~_...:..:...=..=._
= 696.15Q22
.
(0.3) 5
.... =
//
'}>
2000 fl:- 24 tn.
1500 ft- 24 in.
/if, = 0.0826(0.032)(320)Q1 2
0.0826(0.032)(640)Q3 2
(0.45)5
= 91.67Q32
ltj4 = 0.0826(0.032)(1300)Q4
(0.6) 5
2
0
0
D
•
Solution
= 44.19 Q42
Jn Eq. (3):
696.15Q22 = 91.67Q32
Q3 = 2.756Q2
2000 ft - 24 in.
1500 ft- 24 in.
A
In Eq. (2):
Q, = Q2 + 2.756Q2
Q, = 3.756Q2; Q2 = 0.266Ql
In Eq. (4)
HL = lift + 11[2 +lifo= 13
10.877QJ2 + 696.15(0.266Q1)2 + 44.19(Q1)2 = 13
Q=Q,=Q4=12ft3/s
. 149
Q = A 11 = A -·- R2/3 S 112
II
English Version
444
CHAPTER SEVEN
Fluid Flow in Pipes
Q = 2:02
4
FLUID MECHANICS
& HYDRAULICS
1 49
· (D/4)2/3 (HL/L)l/2
445
Problem 7- 51
11
HL = 4.637112 LQ2
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
Water is flowi ng at the rate of 300 lit/sec from A toE as shown in the figure
Compute the flow in each pipe in lit/sec and the total head loss. Assume.{~
0.025 for all pipes.
in feet
D '6f3
(a) Head lost in pipe 1:
I 500m-300mm
2
HL, = 4.637 (0.015) (1500)(12) 2
{24/ 12) 16/ 3
B
A
HL 1 = 5.59 feet
C
600m·250mm
300m·450mm
D
600m·300mm
E
400m-450mm
(b) Total head lost in terms of Q:
Total! fead Lost, HL = HL, + HL2 + HL4
Total Flow= Q, = Q4 = Q
600m-200m
2
2
HL, = 4.637(0.015) (1500)Q
(24 I 12) 160
I-lL , = 0.0388 Q2
HL, = 4.637 (0.015) 2 (2000)Q 2
Solution
1500m-300mm
(24/12) 1613
H~ = 0.0518 Q2
600m·300mm
Q.
600m-200m
"'
Q, = Q6 = 0.3
[Q2 + Q3 = QJ
Q2 + 0.3034 Qz = Q
Q2 = 0.7672 Q
-7 Eq. (1)
-7 Eq. (2)
-7 Eq. (3)
-7 Eq . (4)
-7 Eq . (5)
-7 Eq . (6)
Q, = Q2 + Q3 + Q4
Q3 + Q4 = Qs
h/2 = lr/3 + lzfs
HL = 4.637 (0.015) (4000)(0.7672Q) 2
2
(18 I 12) 16/3
HL2 = 0.2826 Q2
2
HL = 0.0388 Q2 + 0.2826 Q2+ 0.0518 Q2
I-lL = 0.3732 Q2
(c) Total head lost:
Total head lost= 0.3732(12)2 = 53.74 feet
lt/3 = 11_{4
HL = hf, + il/2 + hfh
0.0826JLQ
2
lrf = ---7----'--
os
0.0826(0.025)(300)(0.3)
lzf, =
(0.45) 5
0.0826(0.025)(1500)Q2
lz/2 =
(0.3)5
E
D
---
4.637(0.015) 2 (5000)Q32
(12 /12) ' 6 13
c
600m-250mm
300m·450mm
[HLz = HL3}
4.637(0.015) 2 (4000)Q2 2
(18 /12) Hof3
Q3 = 0.3034 Q2
~
B
2
= 3.02 m
2
2
0.0826(0.025)(, 6__00...!.:)Q=3,__ = 1269Q32
h/3 = ----'--(0.25)5
400m-450mm
446
CHAPTER SF'IEN
Fluid Flow in Pipes
h{J = 0.0826(0.025)(600)Q/
(0.2) 5
h(~ = 0.0826(0.025)(600)Q 5
(0.3)
2
= ~ OQ,~
5
..
(0.45) 5
1
= 4.03 m
In Eq. (5)
1269Q12 = 3871Q,1
Q~ = 1 .747Q4
Pipe
Length, L (m)
Diameter, D (mm)
1
2
3
4
5
450
600
360
480
540
600
500
450
450
600
Solution
-
A
In Eq. (3)
1.747Q4 + Q4 = Q;.
Q .. = 2.747Q.J
Q,
Q,
(i)
In Eq. (4)
1275Q22 = 1269 (I 747Q,f + 5 10 (2 747Q.1f
Q2 = 2.461Q,
In Eq. (2)
Q, = 2.461Q4 + 1.747Q4 + Q4- 0.3
Q., = 0.0576 m"fs
Q2 = 2.46] (.0576) = 0.1418 m3fs
Q1 = l.747(0.0576) = 0.1006 m 3/s
Q, = 2.747(0.0576) = 0.1582 m3/s
-7 Eq. (1)
-7 Eq. (2)
-7 Eq. (3)
-7 Eq. (4)
-7 Eq. (5)
Q, =Qs
Q, = Q2 + Q3
Q3= Q4
I rp = lr13 + lr~
HL,u: = hfl + lrp + ht5 = 15
,_- 0.0826JLQ
D5
Check:
2
I
Q, = Q2 + Q1 + Q4
0 ~ = 0.1418 + 0.1006 + 0.0576 = 0.3 (OK)
"
f1
Problem 7 - 52 (CE Board)
l he total head lost from A toE in the figure shown is 15m. Find the d ischargt
in each pipe. 1\ssume f = 0.02 for all pipes.
B
A
CD
447
Pipe Data
= 3871Q/
/if~- 0.0826(0.025)(400)(0.3) 2
.
CHAPTER SEVEN
Fluid Flow in Pipes
I LUID MECHANICS
& HYDRAULICS
0
= 0.0826(0.02)( 450)Q/
(0.6) 5
It = 0.0826(0.02)(600)Q2
/2
(0.5) 5
" ~ = 0.0826(0.02)(360)Q3
t
2
2
(0.45) 5
II = 0.0826(0.02)(480)Q/
= 42.97Q4 2
f4
•
(0.45) 5
h
= 0.0826(0.02)(540)Q 5
f5
(0.6) 5
2
= 11.47Qs2
D
-
E
448
I n r.-,1
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
CHAPTER SEVEN
Fluid Flow in Pipes
& HYDRAULICS
(5)
15 = 9.56Q,2 + 31 .72Q22 + 11 .47Q,1
449
Solution
Note: The additional pipe should be laid in parallel (not in series) with
the original pipe in order to increase the capacity of the system.
Ru t Q., = Q,, f.rom Eq. (1)
15 = 21.03Q, 2 + ~ 1 .72Q2~
. -7 Eq . (6)
M
In Eq. (4):
31.72Q/ = 32.23Q.•2 + 42.97Q/
·-·-.
·-·-·-·-·-·-
~""''""
But Q_~ = Q.1
31.72Q22 ...,. 75.2Q,1
Q, = 0.6 ~9Q2
Q , = <.22 + O.b-19Q,
Q t = 1.649Q,
f, = f
-7 Eq . (7)
In [~q. (2):
-7 l~ll (8)
In Eq. (6)
15 = 2 1.03(1 .649Q2J2 + 31.72Q)
15 = 88.9Q2'
Q~ = 0.411 mYs
Q., = 0.649(0.41 J) = 0.267 mYs = Q1
Q1 = 0.678 mljs = Qs
0.0826 fLQ1 2
Head lost, H = --~__..:..:0,5
Additional pipe: Pipe <Zl
Required capacity, Q 2 = 1.5Q,
I lead lost = H
Check:
2
025
Q, = Qz + Q, = 0.4 11 + 0.267 = 0.678 (OK )
(since they are laid in parallel)
[H = H]
An exic;ting pipeline is to be reinfo rced with a new o ne w hose coefficient of
p1pe friction is 2/3 of the old o ne. If the leng th of lhe new pipe is equal to tl1.1l
of the o ld o ne and Lhe additional requi red capacity is 150% of the existing
capaci ty. How big the nev. pipe s hould be compared to the old one. Use tht
Darcy-Weisbach for mul a?
: . <D
Original pipe: Pipe Q)
Capacity, Q = Q1
0.0826JLQ 2
Head lost, H = ----=---'-05
H = O.Q826((2/3)f)L(l.5Qt)
Problem 7- 53 (CE November 1997}
jH
)Additional pipe <Z>
fl = (2/3)f
0.0826fLQ1 2
0.0826[(2 I 3)f)L(1.5Q1 ) 2
015
02 5
025 = 15
5
.
01
02
= 1.08
01
Therefore, 0 2 = 1.08 times D1
Problem 7 - 54
With velocity of 1 m/s in the 200-rnm-diameter pipe in the figure shown,
calculate the flow through the system and the head H required. Assume f =
0.02 for all pipes and neglect minor losses.
CHAPTER SEVEN
Fluid Flow in Pipes
450
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
451
[Q, =A, v,]
Q, = f (0.2)2(1) = 0.0314 m 3/s
A
hn = 0.0826(0.02)(300)(0.0314)
0.2 5
2
=
.
1 53 111
From Eq. (3):
hfl = hfl = 1.53
"
= 0.0826(0.02)(300)Q2
fl.
0.3 5
2
= 1.53 m
Q2 = 0.0866 m3fs
Pipe Data
Pipe
1
2
3
4
5
Length, L (m)
300
300
300
600
800
Diameter, D (mm)
200
300
500
300
300
From Eq. (1):
Q, + Q2 = Q3
Q3 = 0.0314 + 0.0866
Q3 = 0.118 m3fs
From Eq. (4):
' 14 = flj5
0.0826(0.02)(600)Q/ _ 0.0826(0.02)(800)Q 5 2
0.3 5
Qs = 0.86tiQ4
Solution
A
-
0.3 5
-7 Eq. (6)
From Eq. (2):
Q3 = Q4 + Qs
0.118 = Q4 + 0.866Q4
Q4 = 0.0632 m 3/s
From Eq. (2)
Qs = 0.118- 0.0632
Qs = 0.0548 m 3/s
G1ven 111 = 1 m/s
Equations.
Q, + Q2 = Q~
Q3 = Q4 + Qs
"fl hf2
=
ltfl = hj5
HL = hn + hn + hr.1= H
From Eq. (5)
H = ltf! + h13 + hf4
-7 Eq. (1)
-7 Eq. (2)
-7 Eq. (3)
-7 Eq. (4)
-7 Eq. (5)
= 1.53 + 0.0826(0.02)(300)(0.118)
0.55
H = 3.38m
2
0.0826(0.02)(600)(0.0632)
+ _
_
0.35
2
...!...____:_~....!....!..-~-
452
CHAPTER SEVEN
Fluid Flow in Pipes
Problem 7 - 55 (CE November 1983)
l
Three pipes of different lengths and diameters connected in series as shown
discharges 160 liters per second. If the roughness coefficient 11 = 0.012 and
disregarding minor losses, determine:
(a) the head loss in each pipe,
(b) the diameter of an equivalent- single pipe that could
three pipes, and
(c) draw Lhe approximate EGLand HGL.
A
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
CD
c
B
300 mm- 1500 m
450 mm- 1800 m
453
2
_ 10.29n LEQ/
fED£16/3
1
89.14 =
10.29(0.012)2 ( 4,100)(0.16 )2
16/3
DE
Dr= 0.304m = 304 mm
(c) EGL and HGL:
D
250 mm -800 m
;!;
Solution
t
Q1 = Q2 = Q1 = 0.16 m3js
l 1
(n) Head loss in each pipe:
2
_ 10.291l LQ
1lr-
(Manning's Formula)·
[)16/'l
2
lln = 10.29(0.012) (1800)(0.16)
2
hn = 4.83 m
CD
4i0 mm - 1800 m
2
vi/2g ' ,
I
I
I
I
I
A
(0.45) l l\j'l
1
'i
I
I
i
I
c
B
300 mm - 1500 m
D
250 mm
800 m
2
- 10.29(0.012) (1500)(0.16)
I l o-----~--~~~~~.
(0.3 )16n
iln = 34.98 m
2
(800)(0.16/
(0.25) 16 /l
hf3 = 49.33m
(b) Equivalent pipe:
Q, =O.l6m1 /s
l1fl' = I IL = llfl + llf2 + hp
= 4.83 +34.98 + 49.33
hrr = 89.14 m
Lr = 1,800 + 1,500 + 800 = 4,100 m
,fl
'
2
" 13 = 10.29(0.012)
'"'::
h0 = 49.33 :f
Problem 7- 56 (CE November 1999)
The installation shown in the Figure is designed for filling tank trucks with
water. The 10-inch line has an over-all length of 100 feet. The 6-inch line A is
10 feet long. The 10-inch line B is 40 feet long. The Da.rcy-Weisbach factor f
equals 0.02. Neglect minor losses. Determine the total discharge which can be
delivered by this system when all the gate valves are fully open.
·
CHAPTER SEVEN
Fluid Flow in Pipes
454
w.s.
+80'
HL
100'
10"
+50'
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
= 0.02(10)
0.8106QA·
(6/12) 32.2(6/12) 4
2
-A
HL2-A = 0.161 QA2
VA
2
_
2
0.8106QA
28- 32.2(6/12)
v
4
2
_A_== 0.403 QA2
2g
In Eq. (1):
80- 0.125 Q2- 0.161 QA 2 = 0.403 QA2 + 20
0.125 Q 2 + 0.564 QA2 = 60
Solution
~ Eq. (3)
Energy equation between <D and B:
£1 - HL1-2 - HL2-s = Es
w.s. o
+80'
v 82
0 + 0 + 80 - HL1-2 - HL2-B = - - + 0 + 20
2g
100'
10"
~ Eq. (4)
2
L
0.02(40) 0.8106Q 8
28
J-1 ' = (6/12) 32.2(6/12) 4
HL2-a = 0.644 Qs2
2
0.8106Q/
_
2g - 32.2(6/12) 4
VB
2
!:L = 0.403 Qs2
2g
In Eq. (4) :
~ Eq. (1)
Energy equation between 1 and A:
El - HLJ.2- JJL2-A = EA
v 2
0 + 0 + 80 - HLI-2- lll.2-A = _A_ + 0 + 20
2g
!C = 0.02(100)
D 2g
HL 1.2 = 0.125 Q2
~ Eq. (2)
Subtract: Eq. (3) - Eq. (5)
0.125 Q2 + 0.564·QA2 = 60
0.125 Q2 + 1.047 Q 82 = 60
0.564 QA2 - 1.047 Qs2 = 0
Note: _v_2 = _8_Q_2_ = 0.8106Q2
2g
n 2 gD 4
g0 4
I ILJ.2 = I L
80- 0.125 Q2- 0.644 Qs2 = 0.403 Qs2 + 20
0.125 Q2 + 1.047 Qs2 = 60
QA = 1.362 Qs
2
0.8106Q
(10 I 12) 32.2 (10 /12) 4
From Eq. (1):
Q = QA + Qs
=1.362 Qs + QB
Q = 2.362Qs
Qs = 0.423 Q
~ Eq. (6)
~ Eq. (5)
455
CHAPTER SEVEN
Fluid Flow in Pipes
456
FLUID MECHA NICS
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
From Eq. (5):
0.125 Q2 + 1.047(0.423 Q)2 = 60
Q = 13.85 ft:l/s
0.0826(0.02)(600)Q1 2
0.15 5
Q2 = 1.6422Q1
Problem 7 - 57
-
-
457
0.0826(0.025)(750)Q2 2
0.2 5
QJ + 1.6422QJ = 1
Q 1= 0.3785 m 3/s
A 250-nun -diameter pipe (f = 0.015), 150 m long, is connected in series with
another 200-mm-diameler pipe(/= 0.02), 200m long. Determine the diame ter
of an equivalent s ingle pipe of length 350m and f= 0.025 that could replace
the two pipes.
HLo= ltfl =
0.0826(0.02)(600)(0.3785) 2 =
0.15
5
1870 111
For the equivalent pipe:
Qc = 1 m3 /s
HLr = 1-ILo = 1870 m
Solution
SetQ= I m'/s
2
0.0826(0.015)(750)(1) = 1870
HLE = ..::..__-~-~~~-
For the original pipe system (two pipes in series):
o£5
Qo= Q1=Q2 =lm3 js
/fLo = flfl + fl12
Dr= 0.218 m = 218 mm
= 0.0826(0.015)(150)(1) 2 + 0.0826(0.02)(200)( 1) 2
0.25 5
0.2 5
/lLo = 1,222.81 111
For the equivalent pipe:
Qr=1m3/s
rILL = 1,222.81 m
L[J
IC
• f.
= 0.0826(0.025)(350)(1) 2
D£
5
= 1,222.81
Problem 7 - 59
In the figu re shown below, it is desired lo pump 3,411,000 lit/ day of water
from a stream to a pool. If the combined pump and motor efficiency is 70%,
calculate the following:
(n) total pumping head in meters,
(b) the power required by the pump, and
(c) the monthly power cost if electricity rate is P6.00 per kW-hr. Assume
that the pump operates for 24 hours and take 1 month = 30 days.
D~:: = 0.226 m = 226 mm
Pipe
Problem 7 - 58
A 150-nU11-diamcter pipe (f = 0.02), 600 m long, is in parallel with a 200-mm diameter pipe (f= 0.025), 750 m long. Determine the diameter of an equivalent
single pipe of length 750 m and J = 0.015 that could replace the two pipes.
1
2
3
Length
m)
1,525
1,525
915
Diameter
Hazen
c1
100 •
110
140
200
200
150
Solution
Set Q = 1 m 3/s
For the given pipes (two pipes in parallel)·
QJ + Q2 = 1
lin = ltr2
CD
CHAPTER SEVEN
Fluid Flow in Pipes
458
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
459
Power required by the pump (Input power):
Solution
"/1
H LAB =
+ lz(2
Q, = Q2 + Q,
lip. = lip,
-7 Eq. (1)
-7 Eq. (2)
-7 Eq. (3)
Po= Qy HA
= 0.03948(9.81)(69.8)
Po= 27.03 Kilowatts
I lazen Williams Formula:
10.67LQ1.85
Power input= Po/ Efficiency
= 27.03/ 0.70
Power input= 38.614 kilowatts
"f= c ,1.85 04.87
--::-::-::..:.::.,-::-=-
c
-<D
Power cost:
Cost= Power input, kW x Time in hours x Power rate per kW-hr
= 38.614 kW x (30 x 24 hr) x (P6.00 / kW-hr)
Cost = P166,812.48
Q,
Q, = 3,411,000 lit/ day x (1 day /24 hrs) x (1 hr/3600 sec)
Q, = 39.48 L/s = 0.03948 m 3/s
Solving for Q 2 and Q3:
From Eq. (3):
[11{2 = hj3]
Problem 7 - 60
How many liters per second of water must the pump shown supply when the
flow needed in the 915-mm-diameter pipe is 1.314 m 3/s? Assumef= 0.017 for
all pipes.
10.67(1 ,525)Q21.85 = 10.67(915)Q3J.SS
(110) 1.85 (0.2)4.1!7
(HO) 1.115 (0.15)4.87
El. 24.6 m 1~,._.._.,_......;1
D
Q3 = 0.787Q2
From Eq. (2):
[Q, = Q2 + Q3]
0.03948 = Q2 + 0.787Q2
Q2 = 0.0221 m 3/s
From Eq. (1):
- 10.67(1,525)(0.03948) 1.85
10.67(1,525)(0.0221) 1.!!5
HLAB+
(100) 1.85 (0.2)4.87
{110) 1.85 (0.2)4.87
FILAR = 26.801 m
El. 6.1 m
Energy equation between A and B:
[A -llLAB + HA = E,,
2
El. 0 m
A
0
2
11
VB
-11- +PA
- +zA-HL11s+HA=+Pll
- +zB
2g
y
2g
y
0 + 0 + 47- 26.801 + I-JA = 0 + 0 + 90
HA = 69.8 m
-7 Total pumping head
PU P
2440 m - 610 mm
460
CHAPTER SEVEN
Fluid Flow in Pipes
I. HYDRAULICS
Solution
~
CHAPTER SEVEN
Fluid Flow in Pipes
f I.UID MECHANICS
0.0826(0.017)(1829)Q4
hf4 =
0.5085
''
' '
2
= 9.28
Q4 = 0.35 m3/s
' '
' '
El. 24.6 m
D
hi,
''
-- '
----~8'
hf~
--
1\
----- ..._._
I
\
At junction B:
Inflow = O utflow
Q1 + Q4 = Q2 + Q3
Q l + 0.35 = 1.314 + 0.142
Q1 = 1.106 m3/s
Q1 = 1,106 Liters per second
El. 12.2 m
\
c
\
\
\
Problem 7 - 61
~
(J)
~ ~o..~lo(;'
:1-~~ \
-..:'1-
hf,
!he turbine s hown is located in the 350 mm-diam eter line. If the turbine
l'fficicncy is 90%, determine its p utput pow er in kilo watts.
\
\
El. 400 m
\
\
0,9
\
'JSilllll
A
Q, = 1.314 m3/s
El. 0
2440 m - 610 mm
= 0.0826(0.017)(2440){1.314)
B
2
0.915 5
Elevation B' = 6.1 + itrElevation B' = 6.1 + 9.22 = 15.32 m
f2
= 9.22 m
hfl = Elcv. B'- Elev. C = 15.32- 12.2
,{1 = ::1.12
- 0.0826(0.017)(1220)Q/
II f?,=).12
.
0.406 5
Q~ = 0.142 m3js
hr4- Elev. D- Elev. B'
= 24.6 - 15.32 = 9.28
11(4
E
El. 33~ m
Q2 = 1.314 m 1/s
"
461
Line 2:
610 m-150 mm
f = 0.024
Line 3 :
2440 m - 300 mm
f. 0.02
Q3 =0.23 m3/s
c
CHAPTER SEVEN
Fluid Flow in Pipes
462
Solution
El. 400 m
..-
hf, +HE
[HE
~-----------------------1(~
t
J
hf,
D'
..- _
. .-
A
~~~
/
a
EI.J30m
//
'
//
\.
Line 2:
610m-150mm
r ~ o.024
El. 280m
Line 3:
2440 m - 300 mm
Q1 = 0.23 m3js
llp = 0.0826(0.02)(2_440)(0.23)
0.3::>
llp = 87.75 m
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
r; om
2
Q3
c
Power input= Q, ., I IE
= 0.27S7t9.81)(5.425)
Power input= 14.83 kilnwatts
Power 0u tpu t = Pt>\\'L'r input x Efficiency
= 14.83 X 0.90
Power oulpul = 13.347 kilowatts
Problem 7 - 62
A 1,200·mm-diameter concrete pipe 1,800 m long carries 1.35 m 3/ s from
reservoir A, whose water surface is at elevation 50 m, and discharges into two
concrete pipes, each 1,350 m long and 750 nun in diameter. One of the 750-m.mdiametcr pipe discharges into reservoir B in which the water surface is at
elevation 4-l m. Determine the elevation of the water surface of reservoir C into
which the other 750- nun-diameter pipe is flowing. Assume f = 0.02 for all pipes.
0.23 m'/s
Solution
El. 50 m
Elev. D' = Elev. C +lip= 280 ·+ 87.75
Elev. 0' = 367.75 m
llrl = Elev. 0' - Elev. B 367.75- 330
h{2 =37.75111
I
_ 0.0826(0.024)(610)Q, 2
0.155
= 37.75
1(2-
Q2 = 0.0487 mlj s
At junction D:
[Inflow = Outflow]
Q, = Q2 + Q3
= 0.0487 + 0.23
Q, = 0.2787 m'/s
lin =
0.0826(0.018)(1220)(0.2787)'2
0.35 5
!In + Ill :~ l~lev. A - Elev. D'
26.825 + }[[ = 400- 367.75
HE= 5.425 m
It = 0.0826(0.02)(1800)(1.35)
fl
1.25
hfl = 2.18 m
= 26.825 m
463
2
Elev. P' = Elev. A- hfl =50- 2.18
Elev. P' = 47.82 m
hp = Elev. P'- Elev. B = 47.82-44
hp = 3.82 m
c
CHAPTER SEVEN
Fluid Flow in Pipes
464
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
,,,2 = 0 OR2n(O 02)f,·l5 35m,Q~- = 3.82
2
ltfl = 62.4 m
0.0826(0.02)(1500)Q1 2
0.75
Q2 = 0.6375 m~/s
0.6 5
Q'1 = 1.399 m 3 / s
(\t junction P:
[Inflow = OutflowJ
Q, = Q2 + Q,
1.35 = 0.6375 + Q:.
Q:. = 0.7125 m '/s
465
= 62.4
,, = 26.6 = 0.0826(0.025)(1000)Q2
2
(0.45) 5
fl.
Q2 = 0.488 m 3/s
hr..= 0.0826(0.02)(1,350)(0.7125)
0.75 5
2
Q, = Q2 + Q3
Q3 = 1.399- 0.488 = 0.911 m 3/s
= 4.77 111
Elev. C = Elev. P'- h,.
= 47.82- 4.77
Elcv. C = 43.05 m
Elevatio11 o[resen1oir c:
,(. - 0.0826(0.018)(900)(0.911)
/1)30.5 5
El. C = 870.6 - 35.54
El. C = 835.06 m
Problem 7 - 63 (CE May 2000)
Three reservoi rs 1\, B, and C arc con nected respectively w ith pipes 1, 2, and
joining at a common junction P w hose e levation is 366 m . Reservo ir A is ,11
e levation 933 m and reservoir B is at clcvalion 844 m. The properties of each
pipe arc as follows: Lt 1500 m, O, tlOO mm,f, - 0.02; L2 = 1000 m, 0 2 = 450
mm,f2 = 0.025; LJ = 900 m, D, = 500 mm,fJ = 0.018. 1\ press ure gage at junction
P reads 4950 kPa. What is the flow in pipe 3 in m~/ s and the elevation of
reservoir C:.
2
= 35.54m
Problem 7 - 64
Determine the flow in each pipe in th e figure sh own and the elevation of
reservoir C if th~ inflow to reservoir A is 515 Lit/sec.
El.?
c
Solution
El. 933 111 I)...J;l~....Jk:------.--
hf, = 62.4
,,
-----
'
''
Ph= 504.6 m
El. 90 m
D
IJ-o-"/.L........jl
F
''
'-
El. 80 m
A
P: El366m
p 4,950 kPa
=
600m • 600 mm
f = 0,025
E
300m • 450 mm
f m 0.03
466
CHAPTER SEVEN
Fluid Flow in Pipes
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
Solution
Elev. F' = Elev. E' + hp = 84.23 + 7.37
Elev. F' = 91.6 m
El.?
c
l!JS = Elev. F'- Elev. D = 91.6- 90
hJS = 1.6 m
" = 0.0826(0.03)(300)Q3
2
= 1.6
0.45 5
Qs = 0.199 m3fs -7 Flow in pipe 3
J5
El. 90 m
At junction F:
[Inflow= Outflow]
Q4 = Q3 + Qs
= 0.247 + 0.199
Q4 = 0.446 m 3/s -7 Flow in pipe 4
hf,
- 0.0826(0.03)(300)(0.446)
IIJ4 0.455
2
- 8 01
- . Ill
A
(J)
600m • 600 mm
r- o.o2s
Q,
It
f1
E
=0.515 m3/s
= 0.0826(0.025)(600)(0.515) 2 = 4 .23 m
0.6 5
Elev. [' = Elev. A + !tn = 80 + 4.23
Elev. [' = 84.23 m
Elev. C = Elev. F' + !t/4 = 91.6 + 8.01
Elev. C = 99.61 m
Problem 7 - 65
Determine the flow in each pipe in the th ree reservoirs shown.
El. 80 m
flp = Elev. B- Elev. [' = 90- 84.23
ltp = 5.77 m
A
, = 0.0826(0.03)(600)Q2 2 = 5.77
fl
0.45 5
Q2 = 0.268 mYs -7 Flow in pipe 2
At junclion £:
[Inflow= Outflow]
Q3 + Q2 = Q,
Q, = 0.515 - 0.268
Q3 = 0.247 mYs
-7 Flow in pipe 3
hp =
0.0826(0.03)(900)(0.247) 2
0.455
= 7.37 m
El.lO m
c
467
468
CHAPTER SEVEN
Fluid Fl ow in Pipes
FLUID MECHANICS
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pip es
FLUID MECHANICS
& HYDRAULICS
El. 80 m
Solu tion
0.0826(0.02)(1800)QJ 2
hn = ---'---'+---'---~
= 290.4 Q1 2
0.4 5
2
0.0826(0.025)5(2000)Q2 = 132.2 Q22
Ita=
0.5
2
0.0826(0.03)(4000)Q2
" ·1 = ---'------'"-;:------'-== 30.25 Q32
0.8 5
Direction of flow:
I he flow in each pipe is due lo gravity. The flow in pipe 1 is obviously
,nvay from reservoir A and the flow in pipe 3 is towa rds reservoir C, but
the flow in pipe 2 is either away or towards reservoir B. To determine the
direclion of Q2, assume Q2 = 0, then /1Jl = 0 and the EGL for pipe 2 is
horizontal.
T~A
30
h11 =X
El. SO m
p
\ I
-1----__s::.,_.:;~~---- -
---
hn=70-x
c
El. 80 m
T~A
hn =30m
hn = 290.4 Q, 2 = x,
Q, = 0.0587 j;
h12 = 132.2 Q22 =X- 30;
Q2 = 0.087 Jx - 30
/113 = 30.25 Q32 = 70 - X,
Q~ = 0.182J70 - .\
At junction P:
[Inflow =Outflow]
Q, + Q2 = Q3
ha
=40 m
0.0587JX +0.087Jx-:i0 =ll i82J70 - '
JX + 1.482 Jx- 30 = 3.1 J70-'
, + 2.964
... c
hjl = 290.4 Q,2 = 30;
lzp = 30.25 Q32 = 40;
Q 1 = 0.321 m 1/s
Q3 = 1.15 m 3 /s
Since Q3 > Q1, the supply from reservoir A is not enough for pipe 3.
Therefore, Q2 is away from reservoir Band P' is below reservoir B.
squMe both stdes
J; Jx- 30 + 2.196(.\ - 30) = q 61 (70- \)
2.964 J; Jx- 30 = 738.58- 12.806.\
square both ~tde~
8.785(x}(.:t- 30) = 545,500- 18,917' + 164 r 1
155.215x2- 18653x + 545,500 = 0
x = 50.287 m
Q1 = 0.0587 J50287 - 0.416 m 3/s
Q2 = o.o87 J50.287 30 = o.392 "''ts
Q 3 = 0.182J70-50 287 - 0.808 m'/~
469
470
CHAPTER SEVEN
FLUID MECHANICS
Fluid Flow in Pipes
& HYDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
471
Direction of flow:
Assume Q2 = O·
Check:
Ql + Q2 = Q3
0.416 + 0.392 = 0.808
(OK)
0.808 = 0.808
f
El. 90 m 1~'--A~....u ".
h 11 = 15m
'-
"".)? _l ____ ~~EI.
Problem 7 - 66
__._.._._J75
m
Determine the flow in each of the pipes shown in the figure. Assume f = 0.02
for all pipes.
/
/ I"
I
/
El. 90 m ~~~..........JI
h0 = 25m
A
El. 50 m
B
B
"
"
""
""
1
1
h,. = 45 ,.
c
p
El. 50 m
c
' fl = 12.75 612= 15;
El. 30m
-
Q.
D
Line 4 :
900m · tSOmm
/z/3 = 306 Q32 = 25;
Q1 = 1.085 m 3 /s
Q3 = 0.286 m}/s
ltr4 = 19,579 Q42 = 45;
Q4 = 0.048 m }/s
D
p
Line 4:
900m- lSOmm
Solution
' fl
0.0826(0.02)(600)Q1 2
2
= 12.75 Q1
0.65
=
0.0826(0.02)(300)Q2 2
2
0.25
= 1,549 Q2
0.0826(0.02)(
450)Q3 2 = 306 Q/
h/3 = __
_..:.__:..:--_...:..::::.:::_
0.3 5
0.0826(0.02)(900)Q4 2
0.155
= 19,579 Q42
Since Q 1 > (Q, + Q4 ), the flow trom ptpe I IS more than e nough to supply
pipes 3 and 4 Therefore. Q1 is towards reservmr B and P' IS abovE'
reservoir B
CHAPTER SEVEN
Fluid Flow in Pipes
472
''
FLUID MECHANICS
& HYDRAULICS
'
-.~====~-----JK~-~1-------------E-I.-75-m----hn,=--15---x-r/
/l\
/
\
/
B
\
\
h,.=60-x
A
B
c
E
E
E
E
E
E
0
0
LJ)
LJ)
LJ)
"'
"'
"'
"'
8..,.
E
0
E
<t"
-
Line 4:
Q, = 0.28../x
lr<2 = 1,549 Q22 = 15- x;
Q2 = 0.0254 J15 - X
hr.< "' :\06 1.),-: = 40 - X ;
Q3 = o.0572J40-x
1114 = 19,579 Q4 2 = 60 - x;
Q4 = 0.00715J60-x
At junction P:
Q, = Q2 + QJ + Q4
Q, - Q2 - QJ - Q4 = 0
0.28../x- 0.0254J15-x - 0.0572J40-x - 0.00715J60-x = o
Solve x by trial and error:
x = 3.055 m
Solution
Using llazen-Williams tormula
K=
10.67L
c,,ss 0 4.R7
KA8 =
18.67(600)
(l 20) I 85 (0.3 ) 4.87
K
10.67(400)
= 520 = KRc; = Kn = Kn1
(120) 1.85 (0.25)4.87
_
All-
--
0.3 m1/s
Q2 = 0.0254 J15- 3.055 = 0.0878 mJjs
Q3 = 0.0572J40-3.055 = 0.3477 m3fs
Q4 = 0.00715J60-3.055 = 0.0539 m3fs
--
A
B
~
(+
0.1
H
--
0.1 m1/s
0.1
c
(+
0.1
G
0.1 m'/s
-0.1
D
Loop III
Loop II
~
--
0.05
0.1
0.2
Loop I
Q, = 0.28 J3.055 = 0.4894 mJfs
0.1 m'ts
0 I m'/s
D
900m • lSOmm
"" ·- 12 75 <.h~ = .r,
E
F
0.1 m'/s
0.1 m'/s
<1"
600m- 300mm
600m · JOOmm
G
H
E
0
0
8<t"
0
600m · 300mm
El. 30m
E
E
0
E
c
D
0
LJ)
El. 50 m
600m · 300mm
600m • 300mm
600m · 300mm
\
Check:
0.4894 = 0.0878 + 0.3477 + 0.0539
0.4894 = 0.4894
(OK)
Fmd
0.3 m1/s
\
p
473
Problem 7 - 67
fhe pipe network s hown m the hgure represents c1 sprav nnst> system
the fl ow in each pipe Assume(,= 120 for all pipes
...,......,._..~~~ _ . : t _
\
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
~
(+
0.05
F
~
0.05
--
0.1 m'ts
0.05
0.1 rfl'/S
..
474
CHA PTER SEVEN
Fl uid Flow in Pipes
FLUID MECHANICS
& H YDRAULICS
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
\orrection to be applied
QIIA = 0.2 · 0.01 19 = 0. l88J
l.KQ"
<X = a
n L KQ/- 1
Ost = 0.1 + 0.014 = 0.114
Qcr = 0.05 + 0.014 - 0 = 0 064
w here ll = 1.85
Tabulated Value!>
Pipe
K
Q.
J<Qa0.8S
KQ11 L85
AB
321
520
321
520
321
321
520
321
321
520
0.20
0.20
0.10
0.10
0.10
0.10
0.05
0.05
0.05
0.05
81.73
132.40
45.34
73.45
45.34
45.34
40.75
25.16
25.16
40.75
16.346
26.479
4.534
7.345
4.534
4.534
2.038
1.258
1.258
2.038
AH
HG
BG
BC
GF
CF
FE
CD
DE
QFc= 0.1-0.014 = 0.086
Qcv = 0.05 + 0 = 0.05
Qnf = 0.05 + 0 = 0.05
Qrr = 0.05 + 0 = 0.05
Second Cycle (ustng the above corrected flow)
Pipe
K
Q.
KQ11085
KQ.L85
AB
AH
321
520
321
520
321
321
520
321
321
520
0.2119
0.1881
0.0881
0.0979
0.1140
0.0860
0.0640
0.0500
0.0500
0.0500
85.845
125.671
40.713
72.139
50.684
39.887
50.264
25.155
25.155
40.750
18.191
23.639
3.587
7.062
5.778
3.430
3.217
1.258
1.258
2.038
HG
BG
BC
Gf
CF
FE
CD
DE
Correction:
a=.
r_ KQ I.ss
a
1.85 L KQa o.ss
Loop 1: AB, BG, GH, HA:
a=-
16.364 + 7.345- 4.534- 2o.479
1.85(81.73 + 73.45 + 45.34 + 132.4)
Correction
=0.0119
Loop 11: BC, CF, FG, GB:
a = ___
4._53_4_+_2_.0_3_8_-_4_.5_3_4_-_7_.3_45_ = _
0 014
1.85(45.35 + 40.75 + 45.34 + 73.45)
Loop Ill: CD, DE, EF, FC:
1.258 + 2.038- 1.258- 2.038
-- 0
a=------------1.85(25.16 + 40.75 + 25.16 + 40 75)
Corrected flow :
QAB = 0.2 + 0.0119 = 0.2119
Qec = 0.1 + 0.0119- 0.014 = 0.0979
QcH = 0.1 - 0.0119 = 0.0881
Loop 1: AB, BG, GH, HA.
18.191 + 7.062..,. 3.587-23.639
a=---------------------~~
=0 ()0329
1 .85(85.845 + 72.139 + 40.713 + 125.671)
Loop II : BC. CF, FG, GB
5.778 + 3.217-3.43 -7.062
1 85(50.684 + 50.264 + 19.887 ... 72 119)
a=-----~~~-----~~~~~
= 0 003R
l,oop Il l · CO, DE, EF, Fe
0 =. ____
1_.2_
58_ +
__2_.0_38__
-_3_.2_
17__
-_1_.2_58-----:- = 0.00451
1 85(25.155 + 40.75 + 25.155 +50 264)
475
476
CHAPTER SEVEN
Fluid Flow in Pipes
Corrected flow:
QAa = 0.2119 + 0.00329 = 0.21519 m3fs
Qac = 0.0979 + 0.00329- 0.0038 = 0.09739 m3fs
QcH = 0.0881 - 0.00329 = 0.08481 rn3fs
QHA = 0.1881 - 0.00329 = 0.18481 rnlfs
Qac = 0.114 + 0.0038 = 0.1178 mlfs
QcF = 0.064 + 0.0038 - 0.00451 = 0.06329 mlfs
Qrc = 0.086 - 0.0038 = 0.0822 mlfs
Qco = 0.05 + 0.00451 = 0.05451 mlfs
QoE = 0.05 + 0.00451 = 0.05451 mlfs
Qer= 0.05 - 0.00451 = 0.04549 m3fs
0.3 m3/s
-
-
0.21519
A
~ 0.18481
H
-
0.1178
B
~ 0.09739
0.08481
0. I mJ/s
CHAPTER SEVEN
Fluid Flow in Pipes
I I UID MECHANICS
1. HYDRAULICS
G
-
!supplementary Problem_s _ _ _ _ _ _ _~
Problem 7 - 68
11tl having sp. gr. of 0.9 and viscos1ty ' = 0.0002 m 2 /s flow.., upward th roug h
11111lclined pipe as shown Dete rmine the flow ralP
AllY 7.64 L/ ~
-
0.05451
c
~ 0.06329
0.0822
0. 1 m3/s
477
F
-
0.04549
0.1 m3/s
p, = 350 kPa
0
~ 0.05451
Problem 7 - 69
Gasoline at 20°C (sp. gr. 0.719, ~l = 0.000292 Pa-s) flow~> al the rate of 2 L/~
throug h a pipe having an inside dia m e ter of 60 mm Dct0rmine the Reynolds
number
E
•
A11 ~
104,400
3
0.1 m /s
Problem 7 - 70
lfl40 L/s of oil flows through the system shown. dC'te rm11w tlw total head lost
hetween points R and C
Au~ · 11 H m
.s:J-
El. 35 m
A
El. 20m
r-1-
El. 5 m
I
B
300 mm C"
\SO mm 0
c
478
CHAPTER SEVEN
Fluid Flow in Pipes
FLUID MECHANICS
& HYDRAULICS
CHAPTER SEVEN
FLUID MECHANICS
& HYDRAULICS
Problem 7 - 71
Problem 7 - 75
What is the hydraulic radius of a rectangular air duct 300 mm by 500 mm?
Ans: 93.75 mm
IS
Fluid Flow in Pipes
479
A tube 15m long 1s to connect two IM)Sl' tank~ whose difterence 1n water level
2 m The tube is to carry 5 L/ s of flow What minimum size of tube '"
necessary to assure laminar flow condition?
Aus: 55 mm
Problem 7 - 72
Air at 1500 kPa absolute flows in a 25-mm-diameter tube. What is the
maximum laminar flow rate. Use density of air = 14.04 kg/m3 and J..l =
0.0000217 Pa-s.
Ans: 0.061 L/s
Problem 7 - 73
A water district which serves a suburban community needs a new water
transmission main which will supply the area from an existing reservoir. The
line will be 5.3 miles long.
l lw reservoir elevation is +280 feet, and the town elevation at point of
d1 ~charge is +100 feet. No pumping will be provided, and the water will be
d, l,vered to a non-pressurized storage tank in town.
Problem 7 - 76
fhe wa ter system tn a suburban area co~sisls ot an old 200-mm p1peline 760 ~1
long which conveys water from a pump to a reservoir whose water surface IS
107 m higher than the pump. Water is pumped at the rate of 0.07 m3/s
Determine the horsepower saved by replacing the old pipe with a new 250mm pipe Assume the value off as equal to 0.033 and 0.022, respectivelv. for
the old and new pipes Neglect losses of head except friction head
A11.' 22.82 hp
Problem 7 - 77
Paint issues from the ta nk tn shown a t Q 45 tt 1/ h
conditions and neglect entrance and exit losses
l l11• proposal is to build a single line of new smooth clean cast iron pipe.
D.trey"!' is assumed to be 0.021. The present demand is 1 MGD.
ta) Determine the minimum diameter pipe that could be used.
Ans: 8.7"
(b) Determine the commercial size of pipe that would fulfill the need of the
system?
Ans: 10"
1\ssume lammar flo"'
1
I
1'
L = 6 ft, d = 1h 1n
(c) Determine the value of the Manning's 11 which would apply to this
particular installation.
Ans: 0.0103
(11)
What 1s the veloCity of flo w 1n the p1 pt•
AilS'
Problem 7 - 74
(b) Determine the head lost 1n the ptpe
The flow rate of water through a cast iron pipe is 5000 GPM. The diameter of
the pipe is 330 mm, and the coefficient of friction is f = 0.0173. What is the
pressure drop over a 30-m length of pipe?
Ans: 10.72 kPa
(c)
Determine the kmematic viscos1 ty ot the pamt m ft~ Is
9.167 ft/ '>
A ns 7 695 teet
Ans 0.0002444
480
CHAPTER SEVEN
FLUID MECHANICS
& HYDRAULICS
Fluid Flow in Pipes
CHAPTER EIGHT
Open Channel
FLUID MECHANICS
& HYDRAULICS
481
Problem 7 - 78
A capillary tube of inside diameter 6 mm connects tan A and open container B,
as shown in the Figure. The liquid in A, B, and capillary CD is water having a
specific weight of 9780 N/m3 and viscosity of 0.0008 kg/(m-s). The pressure
PA = 34.5 kPa. Assume laminar flow.
(a) Determine the head loss in the pipe in terms of the discharge Q.
Ans: 11,058Q
(b) Determine the discharge in L/ s.
Ans: 0.00787
(c) Determine the Reynolds Number.
-r
1.4 m
Chapter 8
Open Channel
Two types of conduit are used to convey water, the open channel and the
pressure conduit (pipe) which was discussed in Chapter 7. An open chrumel is
one in which the stream is not completely enclosed by solid boundaries and
therefore has a free surface subjected only to atmospheric pressure. The flow
in such a channel is caused not by some external head, but rather by the
gravity component along the slope of the channel. In an open channel flow,
the hydraulic grade line is coincident with the stream surface since the
pressure at the surface is atmospheric. The flow in open channels may either
be uniform or non-uniform.
Container B
Air
T
lm
TankA
dl
~.~L~~~~~====~~~~~
Channel bed,
Slope= S.
,._-------------L ----------------~
Figure 8- 1: Open Channel Flow
SPECIFIC.ENERGY
The specific energy (H) is defined as the energy per unit weight relative to the
bottom of the channel. It is given by:
v2
H=- +d
2g
Eq. 8-1
482
CHAPTER EIGHT
CHAPTER EIGHT
FLUID MECHANICS
& HYDRAULICS
Open Channel
Open Channel
483
Kutter and Ganguillet Formula
CHEZV FORMULA
In Figure 8 - 1, the head lost between any two points in the channel is:
ILL=
1
0.00155
-+23+
5
sL
C=
where S is the slope of the energy grade line and L is the length or run. The
head loss balances the loss in height of the channel.
rr n (
1+ JR 23+
1.811
.65
--+41.
C=
From Darcy-Weisbach relation:
fL v2
hL= - - -
nn (
+
1 + JR 41.65+
(S.l. Units)
Eq. 8-5
(English Units)
Eq. 8-6
0.00155) '
s
0.00281
s
0.00281)'
s
0 2g
where D =4R
fL v2
Manning Formula
f1L=-4R 2g
ltL
-
L
f
=-
v2 =
v=(
88
f
ltr
where - ·
v2
-
8g R
I
L
C = .:!.R1 16 ,
=S
Eq. 8-7
(S.l. Units)
II
C= 1.486 Rl/6, (English Units)
RS
Eq. 8-8
11
8J r/2 (RS)l/2
v = .:!.R 2 os 1 12 , (S.I. Units)
Eq. 8-9
Q =A.:!.R 2 13 s1 12 ,
Eq. 8-10
11
Eq. 8-3
Q=AC.fRS
Eq. 8 -4
Bazin Formula
87
C= - - ,
These equations are called the Chezy Jonmtlas, first developed by the French
engineer Antoine Chezy in 1769. The quantity Cis called the Cltezy Coefficient,
varies from about 30 mlf2 js for small rough channels to 90 mlf2/s for large
smooth channels.
A great deal of hydraulic researchers correlated C with roughness, shape, and
slope of various open channels. Among them were Ganguillet and Kutter in
1869, Manning in 1889, Bazin in 1897, and Powell in 1950.
(S.I. Units)
n
For a given channel shape and bottom roughness, the quantity (8gjf)l/2 is
constant and can be denoted by C. The equation becomes,
v = c.JRS
.
1
m
(S.I. Units)
Eq. 8-11
+JR
c = _ _8_7- - , (English Units)
m
0.552+ JR
Eq. 8-12
484
CHAPTER EIGHT
Open Channel
FLUID MECHANICS
& HYDRAULICS
485
Table 8 - 2: Typical Bazin Coefficient
Powell Equation (S.I.)
C = -42
where:
CHAPTER EIGHT
Open Channel
FLUID MECHANICS
& HYDRAULICS
log(~+~)
R~
R
Eq. 8-13
n = roughness coefficient, See Table 8 - 1
m = Bazin coefficient, See Table 8- 2
R = hydraulic radius
E = roughness in meter
~ = Reynolds number
S = slope of energy grade line
Nature of surface
m
Smooth cement
Planed wood
Brickwork
Rough planks
Rubble masonry
Smooth earth
Ordinary earth
Rough channels
0.06
0.06
0.16
0.16
0.46
0.85
1.30
1.75
UNIFORM FLOW (S
Table 8 - 1: Values of n to be used with Manning Formula
Nature of surface
Neat cement surface
Wood-stave pipe
Plank flumes, planed
Vitrified sewer pipe
Metal flumes, smooth
Concrete, precast
Cement mortar surfaces
Plank flumes, unplaned
Common-clay drainaqe tile
Concrete, monolithic
Brick with cement mortar
Cast iron - new
Cement rubble surfaces
Riveted steel
Corruqated metal pipe
Canals and ditches, smooth earth
Metal flumes, corrugated
Canals:
Dredaed in earth, smooth
In rock cuts, smooth
Rough beds and weeds on sides
Rock cuts, jagged and irregular
Natural streams
Smoothest
Roughest
Very weedy
n
Min
0.010
0.010
0.010
0.010
0.011
0.011
0.011
0.011
0.011
0.012
0.012
0.013
0.017
0.017
0.021
0.017
0.022
Max
0.013
0.013
0.014
0.017
0.015
0.013
0.015
0.015
0.017
0.016
0.017
0.017
0.030
0.020
0.025
0.025
0.030
0.025
0.025
0.025
0.035
0.033
0.035
0.040
0.045
0.025
0.045
0.033
0.060
0.150
O.o?S
=S
0)
The simplest of all open channel problem is the uniform tlow condition. For
the flow to be uniform, the velocity, depth of flow, and cross-sectional area of
flow at any point of the stream must be constant (i.e. V1 = V21 d1 = d2, A1 = A2).
For this condition, the stream surface is parallel to the channel bed and the
energy grade line is parallel to the stream surface, and therefore the slope of
the energy grade line S is equal to the slope of the channel bed Se.
-r----2
~
-
-
E.G.l
-
__:_ ~lope , S
h, = S L
-----~ -
129
vl/29
BOUNDARY SHEAR STRESS (t0 )
The average boundary shear sh·ess, '"' acting over the wetted surface of the
channel is given by:
to= yRS
Eq. 8-14