TABLE OF CONTENTS
To my mother lluminada,
my ro,ife Imelda,
anil our Children Kim D eunice,
KenDainiet,
and KarlaDenise
I
CHAPTER 2
Principles of Hydrostatics
Unit Pressure ..............;
............27
Absolute and Gage Prcssures..
............29
Variations in Pressure
............ .........31
Pressure below I .rr urs of I)rtlerent LiryurJs...... .................32
Pressure Hea.l. ............
Manonr.,tcrs ........
.......... 3+
SOLVED PR(}TIT-LMS
SUPPI t:MtNTARY I'I{UBLEMS
...........69 to72
TABLE OF CONTENTS
lt
TABLE OF CONTENTS
CHAPTER 3
CHAPTER 5
Total Hydrostatic Force on Surfaces................ .......73
Fundamentals of Fluid Flow
Total Hydrostatic Force on Plane Surface
................73
Properties of Common Geometric Shapes...
........76
Total Hydrostatic Force on Curved Surface,.....
.......78
Dams
Types of Dams
Analysis of Gravity Dams........
Buoyancy.
Archimedes' Principles .................
Statical Stability of Floating Bodies
Stress on Thin-Walled Pressure Vessels
Cylindrical Tank..........
..................... 81
...................81
........... 84
............. SS
......88
........90
................,..96
......96
............:................98
.....98
Spherical Shell .........
Wood Stave Pipes......................:.....
SOLVED PROBLEMS.................
SUPPLEMENTARY PROBLEMS .................................. 1 9 6 to 200
Discharge.
Definition of Terms...
Energy and Head
Power and Efficiency...........,
Bernoulliis Energy Theorem...
Energy and Hydraulic Grade Lines
SOLVED PRO8LEMS................
iii
...........247
...........241
.................244
..................245
...............246
.....248
250 to 273
SUPPLEMENTARY PROBLEMS .................................. 27 4 to 27 6
iv
TABLE OF CONTENTS
TABLE OF CONTENTS
:
I
i
\
I
{
.t
V
I
VI
TABLE OF CONTENTS
CHAPTER 9
I
I
"""""""""' 551
I lytlroclynamics........'........'
""""""" 551
Ij,rrcc against Fixed Flat Plates
"""' 553
Irorcc a[ainst Fixed Curved Vanes
""""""""' 554
Ijorce a[ainst Moving Vanes ..'......""":"'
""""" 555
Work Done on Moving Vanes
"' 556
Force Developed on Closed Conduit
""""""557
Drag and Lifi.............
"""" 559
Terminal Velocity....
""""""""" 560
Water Hamrner...
563 to597
SOLVED PROBLEMS""""""""
SUPPLEMENTARY PROBLEMS .'.. " " " " " " " " " " " " " " " 5 97 to 598
APPENDIX
l- and
on,t (-nnrrprqin'n
Conversion Factors ""!"""";'.r' .......8g9
Properties of Fluids
Table A - 1: Viscosity and Density of Water at l atm""" """"S99
600
Table A - 2:Viscosity and Density of Air at 1 atm"""""""""'
TableA_3:PropertiesofCommonLiquidsat].atm&20.C..601
Table A - 4: I'roperties of Comrnon Gases at 1 atm & 20'C ""' 601
Table A - 5: Surface Tensiory Vapor Pressure,
""'602
ancl sound SPeed of Water
Atmosphere
Standard
""""""""""' 603
Table A - 6: Properties of
Units
""""""" 604
Table A - 7: Coriversion Factors from BG to SI
"""""""""' 605
Table A - 8: Other Conversion Factors
INDEX I - IV
FLUID MECHANICS
& HYDRAULICS
CHAPTER ONE
Properties of Fluids
Chapter I
Properties of Flui ds
FLUID MECHANICS & HYDRAULICS
Fltid Mechanlcs is a physical science dealing with the action of fluids at rest or
in motion, and with applications and devices in engineering using fluids.
Fluid mechanics can be subdivided into two m4jor areas, Jluid stntics, which
deals with fluids at rest, and fluid dynamics, concerned with fluids in motion.
The term hydrodynanrics is applied to the flow of liquids or to low-velocity gas
flows where the gas can be considered as being essentially incompressible.
Hydratilics deals with the application of fluid mechanics to engineering devices
involving liquids, usually water or oil. Hydraulics deals with such problems
as the,flow of fluids through pipes or in open channels, the design of storage
dams, pumps, and water furbines, and with other devices for the control or
use of liquids, iuch as nozzles, valves, jets, and flowmeters.
TYPES OF FTUID
Fluids are generally divided into two categories: ideal fluids and real fluids.
Ideal Jluids
. Assumed to have no viscosity (arr.l lrence, no resistance to shear;
. Incompressible
. Have uniform velocity when flowing
. No friction between moving layers oi fluid
. No eddy currents or turbulence
ReaI fluids
. Exhibit infinite viscosities
. Non-uniform velocity distribution when flowing
. ComPressible
. Experience friction and turbulence in flow
CHAPTER ONE
2
FLUID MECHANICS
& HYDRAULICS
Properties of Fluids
[{t'al fltrids arc fttrther divided into Newtonian
fluids and non-Nerutonian fhrirls.
Most l'luid problems assume real fluids with Newtonian characteristics for
<'ortvenierrce. This assumption is appropriate for water, air, gases, steam, and
.thcr simple fluids like alcohol, gasoline, acid solutions, etc. However,
slurries, pastes, gels,' suspensions may not behave according to simpre fluid
rcltrtionships.
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
3
where: p = absolute pressure of gas in Pa
R = gas constant Joule / kg-"K
For air:
R= 287 l/kg- "K
R ='1,7L6 lb-ft/siug-sP
T: absolute temperature in oKelvin
"K: "C + 273
"R="F+460
Table 1 - 1r Approximate Room-Temperature
Densities of Common Fluids
p in kg/mr
Fluid
Air (STP)
't.29
Air (21'F, a 1tm)
1..20
Alcohol
790
602
Ammonia
Pseudoplastic Fluids
Fluids
Gasoline
720
Glvcerin
Mercury
1.,260
13,600
1,000
Water
Figure 1 - 1: Types offluid
MASS DENSITY, p (RHO)
SPECTFTC VOLUME, Vs
The density of a fluid is its mass per unit of volume.
Specific volume, %, is the volume occupied by a unit mass of fluid
massof fluid, M
volume, V
Eq. 1- 1.
Units:
English
Metric
SI:
:
:
slugs/ft3
gram/cm3
Note: psrug, = ptbnf g
UNIT WEIGHT OR SPECIFIC WEIGHT, Y
Specific weight or unit weight, y, is the weight of a unit volume of a fluid.
k8lms
lior an ideal gas, its density can be found from the specific gas constant and
icleal gas law:
o=
'RT
P
Eq.1. -2
weightof fluid, W
volume,V'
v:Pg
Eq.1, - 4
1-5
CHAPTER ONE
4
FLUID MECHANICS
& HYDRAULICS
Properties of Fluids
SI
5
the upper plate will adhere to it and will move with the same velocity U while
the fluid in contact with the fixed plaie will have a zero velocity. For small
values of U and y, the velocity gradient can be assumed to be a straight line
and F varies as A, U and y as:
U rrits:
Iinglish
Metric
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
: lb/ ftt
: dyne/cm3
: N/m3 or kN/m3
^AUFU
loc
-
SPECIFIC GRAVITY
Specific gravlty, 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
Eq.1-6
P water
L)
dv
dy
(from the figure)
5A : Sheuri.tg stress, r
T oc
Pliuui.t
yAy
dv Or T =k-.dv
dv
-dv
where the constant of proportionality k is called the dynamic of
absolute viscosity denoted as p.
dv
In gases, the standard reference to calculate the specific gravity is the density
'dy
of air.
.
s=
Pgas
__.::_
t-l=--:--.--' dv/dy
Eq. "I _ 7
P air
For water at 4oC:
where:
r : shear stress in lb/ftz or Pa
p = absolute viscosity in lb sec/ft2 (poises) or Pa-sec.
y = distance between the plates in ft or m
U: velocity in ft/s or m/s
y = 62.41b / tF = 9.81 kN/mr
p = 1.94 slugs/ft3 = 1000 kg/mt
s=1.0
VISCOSIW, p (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
Eq. 1-8
KINEMATIC VISCOSITY v (NU)
Kinematic viscosity is the ratio of the dynamic viscosity of the fluid, p, to its
tnass density, p.
U
Area = A
fixed plate
where:
Lr = absolute viscosity in Pa-sec.
p = density in kg/m3
6
CHAPTER ONE
FLUID MECHANICS
& HVDRAULICS
Properties of Fluids
Table 1 - 2: Common Units of Viscosity
Svstem
English
Metric
Absolute, p
CHAPTER ONE
Properties of Fluids
7
Capillarity
Kinematic, v
lb-sec/ ftz
fA/ sec
(slus/ft-sec)
dyne-s/cmz
s.l
FLUID MECHANICS
& HYDRAULICS
cm2/s
(stoke)
(ooise)
Pa-s
mz/s
(N-s/rnz)
Note:
1 poise = 1 dyne.s/cm2 = 0.1 Pa-sec (1 dyne = 10-5 N)
1 stoke = 0.0001 m2ls
(a) Adhesion > cohesion
SURFACE TENSION o (SIGMA)
The membrane of "skin" that seems to form on the free surface of a fluid is
due to the intermolecular cohesive forces, and is known as surface tension.
Surface tension is the reason that insects are able to sit on water and a needle is
able to float on it. Surface tension also causes bubbles and droplets to take on
a spherical shape, since any other shape would have more surface area per
unit volume.
(b) Cohesion > adhesion
Capillarity (Capillary nction) is the name given to the behavior of the liquid in a
thin-bore tube. The rise or fall or a fluid in a capillary tube is caused by
surface tension and depends on the relative magnitr"rdes of the cohesion of the
liquid anct the adhesion of the liquid to the walls of the containing vessel.
Liquids rise in tubes they wet (adhesion > cohesion) and fall in tubes they do
not wet (cohesion > adhesion). Capillary is important when using tubes
smaller than about 3/8 inch (9.5 mm) in diameter.
Pressure inside a Droplet of Liquid:
'4o
P=T
'
Eq.1 - 10
For complete wetting, as with water on clean glass, the angle e is 0o. Hence
the formula becomes
where:
o = surface tension in N/m
d = diameter of the droplet in m
p= gage pressure in Pa
where:
/l = capillary rise or depression in m
y
: unit weight in N/m3
d = diameter of the tube in m
o = surface tension in Pa
I
FLUID MECHANICS
& HYDRAULICS
CHAPTER ONE
Properties of Fluids
Table I - 3: Contact Angles, 0
_ stress =
_ Ap
rLs=--------,strain LV
Angle,0
Materials
mercury-glass
140'
water-paraffin
water-silver
107"
kerosene-glass
slvcerin-elass
water-glass
ethvl alcohol-glass
26"
I
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
D^ 1-15
1 1r
Eq.
v
dn
orEs=-..'.-- dv/v
90'
Eq. 1-16
19"
00
00
PRESSURE DISTURBANCES
Pressure disturbances imposed on a fluid move in waves. The velocity or
celerity of pressure wave (also known as acoustical or sonic aelocity) is
expressed as:
COMPRESSIBILIW, B
fractional
ompressibilify (also known as the cofficient of compressibillty) is the
in a constantchange in the volume of a fluid per unit change in pressure
temperature process.
C'
tr;
tr
p
c=.1 " = l-
I
lFp
Eq."I_17
AV
-\/Y 1
B=
'LPEB =j-
Eq.1-13
dv /v
Eq.1, - 1.4
orP:-----;- qp
^- a
PROPERTY CHANGES IN IDEAL GAS
For any ideal gas experiencing any process, the equation of state is given by:
PrVt - PzVz
Tt
where:
AV = change in volume
V = original volume
Ap = change in Pressure
dV /V = change in volume (usually in percent)
Tz
Eq.1-18
When temperature is held constant, Eq. 1 - l8 reduces to (Botlle's l.arc)
ptVt=pzVz
Eq.1-19
When temperature is held constant (isothermal condition), Eq. 1 - 18 reduces
lo (Clurle's Lato)
BULK MODULUS OF ELASTICITY' E8
of the
The bulk moclulus ot t,lasticity r,f the lluicl expresses the compressibility
fluid. It is the ratio ol the change in urrit pressure to the corresponding
volume change Per unit ot vt'lume'
vt : v,
T1 Tz
Eo.L -20
q{
lli
li
CHAPTER ONE
to
FLUID MECHANICS
& HYDRAULICS
Properties of Fluids
For Adiabatic or Isentropic Conditions (no heat exchanged)
pt Vtk = pz Vzk
i
or Pt\u = u = Constant
\vz )
8q.1, - 21
Fluid
kPa,20"C
Eq.1, - 22
mercury
turoentine
water
ethvl alcohol
0.000173
0.0534
k-^L
and
lPzlk
r, - l.p,l
tl
Table 1 - 4r Typical Vapor Pressures
Pt
t2
CHAPTER ONE
Properties of Fluids
FTUID MECHANICS
& HYDRAULICS
ether
butane
Freon-12
Eq.1-23
where:
2.34
5.86
58.9
278
584
DroDane
855
ammonla
888
pr = initial absolute pressure of gas
pz= final absolute pressure of gas
Vi = initial volume of gas
Vz = final volume of gas
Tr = initial absolute temperature of gas in .K (.K =.C + 273)
Tz = final absolute ternperafure of gas in .K
lc = ratio of the specific heat at constant pressure to the specific heat at
constant volume. Also known as adiabatic exponent.
VAPOR PRESSURE
Molecular activity in a liquid will allow some of the molecules to escape the
liquid surface. Molecules of the vapor also condense back into the liquid. The
vaporizalion and condensation at constant temperature are equilibrium
processes. The equilibrium pressure exerted by these free molecules is known
as the papor pressure or saturation pressure.
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 aolatile 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 liquicl's tendency to vaporize.
Solved Problems
Problem 1- 1
A reservoir of glycerin has a mass of '1,200 kg and a volume of 0.952 cu. m.
liincl its (a) weight, W, (b\ untt weight, y, (c) mass density, p, and (d) specific
gravity (s).
Solution
(a) Weight *-_** t
(1,200)(9.81)
Weight, W=77,772 N or 11.772 kN
(1,) Unit weight, , =
+
_ 77.772
0.952
Unit weight, y:72.366 kN/m3
(,') Density, p= M
v
DensitY, P =
12oo
0.952
Density, p = 1,,260.5 kglm3
t2
CHAPTER ONE
FLUID MECHANICS
& HYDRAUTICS
Properties of Fluids
(d) Specific gravity, s =
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
r3
Solution
Pglv
(a) W=*g=22(9.75)
Pwater
.
1.260.5
Spectrlc gravrty, s = -:'
1,000
W= 214.5 N
(b) Since the mass of an object is absolute, its mass will still be 22kg
Specific gravity, s ='L.26
Problem 1- 2
Probtem 1- 5
The specific gravity of certain oil is 0.82. Calculate its (a) specific weight in
lblft3 and kN/m3, and (b) mass density in slugs/ft3 and kglm'.
What is the weight of a 45-kg boulder if it is brought to a place where the
Solution
Solution
(a) Specific weight, y : ywater x s
acceleration due to gravity is 395 m/s per minute?
W= Mg
m/s r lmin
Specific weight, y = 62.4 x 0.82 = 51.168 tb/ft3
Specific weight, y = 9.81 x 0.82 = 8.044 kN/mr
'c=3g5 min
g = 6.583 m/s2
(b) Density, p = p*ot.,.x s
w= 4s(6.s83)
Density, p = 1.94x 0.82 = 1.59 slugsftl
Density, p = 1000 x 0.82 = 820 kglm3
W=296.25N
Problem l - 5
If the specific volume of a certain gas is 0.7848 m3/kg, what is its specific
Problem 1- 3
A liter of water weighs about 9.75 N. Compute its mass in kilograms.
weight?
Solution
Mur, =
Mass =
Solution
W
1
Vr= -
s
p
4.71
'_-'i"
9.8L
,
Mass = 0.994 kg
Problem 1 - 4
If an object has a mass of 22 kg at sea level, (a) what will be its weight at a
point where the acceleration due to gravity g = 9.75 m/ sz? (b) What will be its
mass at that point?
60sec
'
11
v, 0.7848
p = 1.2742k9/mu
specific weight,,
]=l.irtn *n.r.,
Specific weight y = 12.5 N/m3
l4
CHAPTER ONE
FLUID MECHANICS
a nvonAuucs
Froperties of Fluid5
& HVDRAULICS
l5
$olution
Problem 1- 7
l,Vhat is the specific weight of air at 480 kPa absolute and 21'C?
1 :
Solution
Density, p = I
I
{
- 13.7
g;81,
T=q"8
= 1.397 kglxrf
p
where R=287J/kg:K
= 4=
'RT
480xL03 ,
p
t)
l
,
5.689kg
Densitv; o = '
:
287(21.+273)
p=
CHAPTER ONE
Properties of Fluids
Fr.urD MEcHnrrircs
'J,.397
'
=
RT
(205+L0'l'.325)x 103
Note: P.tm = 101.325 kPa
-RAr.r?3)
Gas constant, R = 778.87 1441. "K
5.689 x 9.8L
y = 55.81 N/rn3
,
1. 10
io kept at:a pressure of 200 iiPa absolute and a temperatirre of 30bC in
Jiter container. What is the mass of air?
Problem 1- 8
a
Find the mass density of helium at a temperature of 4 "C and a pressure of 184
kPa gage, if atmospheric pressure is 10'1'92 kPa. (R = 2A79 J /kg :, 'K)
o=
.RTP
Solution
n
Densitv.
_-_r. 6
r - -I-
RT
p=p ag"+p*^
184+ 101.92
?=285.92kPa
9
T = 4+ 273= 277"K
,1..'
.
Densrtv.
- J o=
"
285.92x103
rp*2.3kg/m3
,'ltdass= pxV'
'', ,=2.3xffi
It,hfass = 1.L5 kg
2,079(277)
Density, p = 0. 4g65kg6.f'
Problem 1- 9
200 x 103
287(30 + 273)
-
and 205 kPa gage, the specific weight of a certain gas,was 13.7 N/mr.
.)t.32"C
I )etermine the gas constant of this gas.
1t
t-11
tank 80 cm in diameter and 90 cm high is filled with a liquid
and the liquid weighed 420 kg. ,The weight of thb empty tank is 4(
io the unit weight of:the liquid in kN/m3
CHAPTER ONE
t6
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
l7
Solution
Solution
M
o: v
Es=
dP
dv
/v
dp=pz-pt
420- 40
= 840 kg/m:
(0.8)'(0.e0)
f
h=o
y=pg
dp=p,
= 840(9.81) = 8240.4 N/m3
y = 8.24 kN/ms
dV = Vz- Vt
dV=-0.6%V=-0.006V
F -
"
Problem L - Lz
A lead cube has a total mass of 80 kg. what is the length of its side? sp. gr. of
lead =-11.3.
Pz
0.006v
/v
-.^
pz= 0.0132GPa
pz=13.2MPa
Solution
Let L be the length of side of the cube:
M= pV
33 = (1000 x 11.3) Ls
L= 0.192 m = 19.2 cm
Problem 1- 15
Water in a hydraulic press, initially at 137 kPa absolute, is subjected to a
pressure of 176,280 kPa absolute. Using Ee = 2.5 GPa, determine the
percentage decrease in the volume of water.
Solution
Problem 1- 13
r-
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 (Er) of the liquid is:
2.5x10e=-
Solution
dV
= -0.0465
v
-- dP
dv/v
dP
- dv/v
2-1
dV
(0.ee5-7)/1,
v
(116,280-137)x103
dv /v
= 4.65"ludecrease
Ea = 200 MPa
Problem ! - 14
\Alhat pressure is required to reduce the volume of water by 0.g percent? Bulk
modulus of elasticity of water, Ee = 2.2GPa.
Prcblem 1- 16
lf 9 m3 of an ideal gas at24 oC and 150 kPaa is compressed to 2 ma, (n) what is
the resulting pressure assuming isothermal conditions. (b) What would have
heen the pressure and temperature if the process is isentropic. Use k = 1.3
CHAPTER ONE
IB
ll
I
Properties of Fluids
l9
Problem 1- 18
Two large plane surfaces are 25 mm aPart and the space between them is filled
with a liquid of viscosity p = 0.958 Pa-s' Assuming the velocity gradient to be
a straight line, what force is required to pull a very thin plate of 0.37 m2 area al
a constant speed of 0.3 m/s if the plate is 8.4 mm from one of the surfaces?
For isothermal condition:
pr Vt : pz. Vz
tts0(9) = pz (2)
Pz = 675 kPa abs
Solution
For isentropic process:
p'rVtk = pzVzk
(b)
Properties of Fluids
& HYDRAULICS
Solution
(n\
CHAPTER ONE
FLUID MECHANICS
FLUID MECHANICS
& HYDRAULICS
r_ Fr+Fz
r-
150(9)'.3 = p2 (2)r.t
' u/v
/A
' u/v
T
pz = 7,O60 kPa abs
/
Tr= lPz
T1
lp'
T.
I
.J
:tt
25r
F
\(k-l)/k
( 1.060 \(r'3-r)/ l
F= VUA
j
v
24+273 [ tso J
11
Tz = 466.4'K or 193.4'C
-
0.958(0.3)(0.37)
00166
- o.4 iri
0.958(0.3)(0.3n
F -+-12.o6N -,
rz0.00g4
Probfem I - L7
F=6.4+12.66
If the viscosity of water at 70 "C is 0.00402 poise and its specific gravity is 0.978
determine its absolute viscosity in Pa - s and its kinematic viscosity in m2/s
F = 19.05 N
and in stokes.
Problem I - 19
Solution
A cylinder of 725 mm radius rotates concentrically inside a fixed cylinder o
130 mm radius. Both cylinders are 300 mm long. Determine the viscosity o
the liquid which fills the space betweep the cylinders if a torque of 0.88 N-m i
rerluired to maintain an angular velocity of 2n radians/sec Assume th,
Absolute viscosity:
u = 0.00402 poise x
0.1 Pa-s
lpoise
vekrrrity gradient to be a straight line
I = 0.000402 Pa - s
Kinematic viscosity:
v=:=
u
p
0.000402
(1000 x 0.978)
v = 4.11 x1V7 mzfs
v = 4.11 x 1o-7 n'z/ s*
l stoke
0.0001m2/s
v = 4.11 x 1(}3 stoke
i
I
CHAPTER ONE
20
FLUID MECHANICS
& HYDRAULICS
Properties of Fluids
Solutlon
Nt = o.oo5
i*-\.*
U--ra
u =-olles
F =tA
fixed cylinder
V = 0'005 m
Torque: tA (0.125)
lF.=tA=ug-Al
v
ojmL
(0.3)
U = 5.614 m/s
pr = 5.614 m/s
0.88 = r [2n(0.12s)(0.3)l (0.125)
t = 29.88 Pa
L=0.3m
liquid
29.88
Problem 1- 21
Estimate the height to which water will rise in a capillary tube of dianreter 3
mm. Use a = 0.0728 N/m and y = 9810 N/m3 for water.
0.005
p = 0.19 Pa-s
F =Wsin0
F'= 176.58 sin 15'
t76.5|sin 15o = 0.0si4
Torque = F(0.125)
t,.= o.zsslo.oo5
Ia t.
Wsin0-F =0
m
rotating
rylinder
U = 0.1,25(2n)
U = 0.785 m/s
Properties of Fluids
[tr, = 0]
fixed
u= u/v
'
CHAPTER ONE
FLUID MECHANICS
& HYDRAULICS
0.13 m
Solution
Problem 1- 20
Note: 0 = 90o for water in clean tube
An 18-kg slab slides down a 15' inclined plane on a 3-mm-thick film of oil
with viscosity Lr = 0.0814 Pa-sec. If the contact area is 0.3 m2, find the terminal
velocity of the slab. Neglect air resistance.
Solution
W = 18(9.81) = 176.58 N
s
Capillaryrlse,h= I
yd
CaPillary rise,l =
4(0'0728)
e810(0.003)
Capillary rise, /r = 0.0099 m = 9.9 mrn
Problem t - 22
llrtimate the capillary depression for mercury in a glass capillary tube 2 mm in
dlameter. Use o = 0.514 N/m and 0 = 140".
y = 0.003 m
lolutlon
plane
0=
Terminal velocity is attained when the sum of all forces in the direction of
mction is zero.
4ocos0
4(0'514Xcos140')
yd - (9810x13.6X0.002)
(the negative sign indicates capillary depression)
Capillary rise, I = -0.0059 m
Capillary rise, h =
Capillary depression, lr = 5.9 mm
CHAPTER ONE
22
Properties of Fluids
FLUID MECHANICS
& HYDRAULICS
Problem 1- 23
What is the value of the surface tension of a small drop of watgr 0,9 Tall
diameter which is iri contact with air if the pressure within the droplet is 561
CHAPTER ONE
FLUID MECHANICS
or"p-"l,;;
d nYoRnuucs
"?il;'
23
Problem L - 26
Pa?
A sonar transmitter operates at 2 impulseS per second. lf the device is held tb
pidwal'
tf'ta t"ifu.. of fresh water'(Er = 2'04 x10e P;) and the echo is received
between impulses, how deep is the water?
Solution
Solution
The velociry of the pressure wave (sound wave) is:
4a
P= -.:
lr
d
ILB
'- {;
4o
56"1=
0.0003
o = 0.042 N/m
=7,428m/ s
Problem L - 24
An atomizer forms water droplets 45 pm in diarneter. Determine the excess
pressure within these droplets using o = 0.0712 N/m.
covered is 2h, then;
Solution
p=
p=
the echo is received
way between impulses, tlien
total time of kavel of sound,
1/z(0.5) = % sec and the total
4o
2]t=ct
d
Eh ='1,,428(1/4
4(0'0712)
= G,329 pa
'li = L78,5 m
45 x-10-o
t -27
Problem t - 25
bistiiled water stands in a glass tube of 9 mm dfmeter at a height of 24 mm
What is the true static height? Use o = 0.0742 N/m.
pressure will 80 "C water boil?
pressure of water at 80oC = 47.4l<Pa)
Solution
will boii if the atmospheric pressure'equals the vapor piessure
, 4ocos0
n='yd
where 0 : 0" for water in glass tube
tI= 4(0'0742) = o.oo336 m = 3.35 mm
9810(0.009)
True static height = 24 - 3.36
True static,height = 20.54 mm
water at 80 "C will boil at 47.AkPa
2+
CHAPTER ONE
FLUID MECHANICS
6. HYDRAULICS
Properties of Fluids
upplementary Problems
FLUID MECHANICS
& HYDRAULICS
CHAPTER ONE
Properties of Fluids
zaE
J
Problem 1 - 33
Problem 1- 28
what would be the weight of 1 3-kg mass or a planet where the acceleration
due to gravity is 10 m/s2?
(a) If 12 m3 of nitrogen at 30oC and 125 kPa abs is permitted to expand
isothermally to 30 m3, what is the resulting pressure? (b) What would the
pressule and iemperature have been if the process had been isentropic?
Ans: (a) .50 kPa abs
(b) 34.7 kPa abs; -63'C
Ans:30 N
Problem 1- 29
Problem 1 - 34
A vertical cylindrical tank with a diameter of 12 m and a depth of 4 m is filled
with water to the top with water at 20"C. If the water is heated to 50.C, how
much water will spill over? Unit weight of water at 20'C and 50oc is 919
A square block weighing 1.1 kN and 250 mm on an edge slides down an
incline on a film of oil 6.0 pm thick. Assuming a linear velocity profile in the
oil and neglecting air resistance, what is tl're terminal velocity of the block?
T'he viscosity of oil is 7 mPa-s. Angle of inclination is 20o.
Ans:4.7 m3
Ans'.5.1.6 m/ s
kN/m3 and 9.69 kN/m3, respectively.
Problem 1- 3O
Problem I - 35
A rigid steel container is partially filled with a liquid at 15 atm. The volume of
the liquid is 1.23200 L. At a pressure of 30 atm, the volume of the liquid is
1.23100 L. Find the average bulk modulus of elasticity of the liquid over the
given range of pressure if the temperature after compression is allowed to
return to its initial value. What is the coefficient of compressibility?
Ans" Ea = 1872 GPa; 0 = 0.534 GPa-'
Problem t - 31
Calculate the density of water vapor at 350 kPa abs and 20'c if its gas constant
is 0.462 ppu-6r/kg-"K
llenzene at 20'C has a viscosity of 0.000651 Pa-s. What shear slress is required
to deform this fluid at a strain rate of 4900 s-1?
Atts: r -- 3.19 Pa
Problem 1 - 36
A shaft 70 mm in diameter is being pushecl at a speecl of 400 mm/s through a
lrt'irring sleeve 70.2 mm in diameter and 250 mm long. The clearance, assumed
ltlliform, is filled with oil at 20.C with v = 0.005 m2/s and sp. gr. = 0.9. Find
llrc force exerted by the oil in the shaft.
Atzs: 987 N
Ans:2.59 kg/ m,
Problem t - 37
Problem 1- 32
Air is kept at a pressure of 200 kPa and a temperature of 30oC in a 500-L
contajner. What is the mass of the air?
'l'w<r clean parallel glass plates,
separated by a distance d ='J..5 mm, are ctipped
ln n bath of water. How far does the water rise due to capillary action, if o =
(),()730 N/m?
Ans: 1.15 kg
Ans:9.94mn
26
CHAPTER ONE
Properties of Fluids
FLUID MECHANICS
& HVDRAULICS
Problem 1- 38
Iiincl the arrgle the surface tension film leaves the glass for a vertical tube
irnmersed in water if the diameter is 0.25 inch and the capillary rise is 0.08
inc-'h. Use o = 0.005 lb/ft.
Arts:64.3"
Problem 1 - 39
What force is required to lift a thin wire ring 6 cm in diameter from a water
surface at 20'C? (o' of water at 20"C = 0.0728 N/m). Neglect the weight of the
ring.
Ans:0.0274N
FLUID MECHANICS
CHAPTER T\vO
Principles of Hydrostatics Z I
& HYDRAULICS
Chapter 2
Principles of Hydrostatics
UNIT PRESSURE OR PRESSURE, p
Pressure is the force per unit area exerted.by a liquid or gas on a body or
surface, with the force acting at right angles to the surface uniformly in all
directions.
Force. F
P=
.--- A.
' Area,
Eq.2-1
ln the English system, pressure is usually measured in pounds per square inch
(psi); in international usage, in kilograms per square ce.timeters (k[/cm2), or
in atmospheres; and in the international metric system (sI), in Newtons per
Bquare meter (Pascal). The unit atmosphere (atm) is defined as a pressure of
1.03323 kg/cm2 (74.696 lb/inz), which, in terms of the conventional mercury
batometer, corresponds to 760 mm (29.921in) of mercury. The unit kilopascal
(kPa) is defined as a pressure of 0.0102 kg/crft (0.1a5 lb/sq in).
PASCAL'S LAW
Ittrst:nl's lazo, cleveloped by French mathemati cian Blaisc pnscal,states that the
l'lrcssure on a fluid is equal in all directions and in all parts of the container. In
Figure 2 -'1,, as liquid flows into the large container at the bottom, pressure
pusl'res the liquid equally up into the tubes above the container. The liquid
tlgcs to the same level in all of the tubes, regardless of the shape or angle of the
ttrbc.
2B
CHAPTER T\YO
FLUID MECHANICS
6. HYDRAULICS
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAUTICS
CHAPTER T\vO
Principles of Hydrostatics
29
ABSOLUTE AND GAGE PRESSURES
;. ;ifi*iiM;s.il
Gage Pressure (Relative pressure)
Gage pressures are presswes above or below the atrnosphere and can be
measured by pressure gauges or manometers. For srnall pressure differences, ua
fube manometer is used. It consists of a u-shaped tube
o.ru
end
connected
to
-ith
the container and the other open to the atnosphere. Filled with a liquid, such
as
Figure 2 - 1: Illustration of Pascal's Law
The laws of fluid mechanics are observable in many everyda-v situations. For
example, the pressure exerted by water at the bottom of a prnd 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.
water, oil, or mercury, the difference in the liquid surface levels in the two
manometer legs indicates the pressure difference from local ahnospheric
conditions. For higher pressue differences, a Bourdon gauge, named after the
French inventor Eugdne Bourdon, is used. This consists of i hollow metal
tube
with an oval cross sectiory bent in the shape of a hook. one end of the tube is
closed, the other open and connected to the measurernent region.
Atmospheric Pressure & Vacuum
Atttrospluic Pressute is the pressure at any one point on the earth's surface from the
weight of the air above it. A anarum is a space that has all matter removed from it.
It is impossible to create a perfect vacuum in the laboratory; no matter how
aclvanced a vacuum system is, some molecules are always present in the vacuum
ett'a, Even remote regions of outer space have a small amount of gas. A vacuurn
r:an rrlso be described as a region of space where the pressure ii less than the
nrrrnral atmospheric pressure of 760 mm (29.9 n) of mercury.
l-Jtrtlt.r Normal conditions at sea level:
It"r" = 2166lb/ft2
= 1.4.7 psi
= 29.9 inches of mercury (hg)
= 760 mm Hg
= 101.325 kPa
r, =Pi=P.
Figure 2 - 2: Illustration of Pascal's Law
lbrolute Pressure
Alrrrrtrrlr. pressure is the pressure above absolute zero (unumnt)
r Ab5(llute zero is attained if all air is removed.
I Aurhtte pressure can never be negative. It is the lowest possible pressure attainable.
r lll€ nrtallest gage pressure is equai to the negative of the ambient atmospheric pressure.
30
CHAPTER T\vO
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Principles of Hydrostatics r
'
VARHTIOI{S IN PRESSUR,E
Consider any two points (1 & 2), whose difference in elevation is Il, to lie in the
ends of an eiemeniary prism having a cross-sectional area 4 and a length of L.
Since this prism is at rest, all forces acting upon it must be in equilibrium'
sB.6
Standard
GHAPTERTWO 2 t
FLUID MECHANICS
& HYDRAULICS
atmosphere =
Free liquid surface
Current atmosphere = 100 abs
-40 gage
pressurcS
Absolute zero = -101.325 gage
or -100 gage
All pressure units in kPa
Figure 2 - 3: Relationship betvveen absolute and gage pressures
=Lsine
Note: Unless otherwise specified in this book, the term pressure signifies gage pressure.
MERCURY BAROMETER
A mercury barometer is an accurate and relatively
simple way to measure changes in atmospheric
pressure. At sea level, the weight of the atmosphere
forces mercury 760 mm (29.9 in) up a calibrated
glass tube. Higher eleVations yield lower readings
because the atmosphere is less dense there, and the
thinner air exefts less pressure on the mercury,
A
Flgure 2 - 4: Forces acting on elementary prism
Atmospheric
p le tsu re
w
Note: Free Uquid Surface refers to liquid surface subject to zero gage press{Jre or with
atnospheric pressure only.
Wlth reference to Figure 2 - 4:
w-f v
1r'11= y @Ll
ANEROID BAROMETER
In an aneroid barometer, a
partially evacuated metal drum
expands or contracts in response
to changes in air pressure, A
series of levers and springs
translates the up and down
movement of the drum top into
the circular motion of the
pointers along the aneroid
barometer's face.
[lF, - 0]
Fz-Fr=Wsin0
P24-Pta=Y(aL)sin0
p2-p| =1Lsin0
butLsin0=h
=yh
.2-3
Threfore; the difference in pressure behueen any huo points in a lnmogeneous fuid
4l nst ia equnl to thc product of the unit weight of the fluid (^il to tlu aerticnl distnnn
(h) llrhueen the points.
32
CHAPTER T\vO
FLUID MECHANICS
Princlples of Flydrostatics
& HY'DRAULICS
Also:
CHAPTER T\vO
FTUID MECHANICS
-Principles of Hydrostatics 55
& HYDRAULICS
Consider the tank shown to be filled with liquids of different
clensities and
with air at the top under a gage pressure of pa, the pressure at the
bottom of
the tank is:
This means that arty change in pressure nt point 1. tuould cnu.se an equal chantge at
point 2. Thereforq a pressufe applied at any point in a Liquid at rest is
trqnsmitted equally and undiminished to every other point in the liquid.
Let us assume that point o in Figure 2 - 4 lie on the free liquid surface, then
the gage pressure pt is zero and Eq. 2 - 4 becomes:
=]i
h,! p'= y1h1 + yl lb + ^bhs +
'PRessunr nrno
'Pressure head is the height "rt" of a corumn of homogeneous liquid of unit
o -- --weight y that will produce an intensity of pressure p.
= zoh
This means that the pressure at any point "rt" beruo a
free liquid surface is equnl to
tlrc product of the unit weight of the
fluid (y) antt h.
consider that points o and o in Figure 2 - 4keon the same elevation, such
that h = 0; then Eq. 2 -
Convert Pressure head (height) of liquid A to liquid B
becomes:
This means that the prcssure along the same horizontal plane in a homogeneous
fluid
at rest are equal.
Pressure below Layerc of Different Liquids
h2
hr
Fuotom
convert pressure head.(height) of any riquid to water,
iust murtipry its
Cht by its specific aravrty
34
CHAPTER TWO
Principles of Hydrostatics.
CHAPTER TU(/O
FLUID MECHANICS
& HVDRAULICS
FTUID MECHANICS
& HYDRAULICS
2q
Principles of Hydrostatics J J
Steps in Solving. Manometer Problems:
MANOMETER
t. Decide or, ,f," fluid in feet or meter, of which the heads are to be
A ntunometer is a tube, usually bent in a form .of a U, corttaining a liquid of
expressed, (water is most advisable).
known specific gravity, the surface of which moves proportionally to changes
of pressure. 'lt is used to measure pressure
2. Starting from an end point, number in order, the interface of different
fluids:
3. Identify points of equal pressure (taking into account that for a
homogeneous fluid at rest, the pibssure along the same horizontal plane
are equal). Label these points with the same number.
Types of Manometer
Open Type - has an atmospheric surface in one leg and is capable ot
measuring gage preisu res.
Differential Typ"
without an atmospheiic surface and capable of
measuring only differences of pressure.
Piezometer - The sirnplest form of open manometer. lt is a tube'tapped hto a
wall of a container or conduit"for the purpose of measuring pressure. The
fluid in the container or qonduit rises in this tube to form a free surface
Limitations of Piezometer:
. Large pressures in the lighter liquids require long tubes
. Gas pressrtres can not be measured because gas can not form a free
i;
4. Proieed from level to level, adding (if going down) or subtracting (il
going up) pressure ,heads as the elevation decreases or increases,
respectively with due regard for the specific gravity of the fluids.
i
PrOblems
2-'
a depth of liquid of L m causes a pressure of 7 kPa, what is the specifi<
ty cif the liquid?
suiface
Pressure, p=lh
7 = (s.81. x s) (1)
s:0.714 ) Specific Gravity
2-.2
(a) Open manometer
(b) Differential rnanometer
le the pressure 12.5 m below the ocean? Use sp. Cr-. = 1.03 for salt water.
'
,P*Yh
*
p
P=1:26.3kPa
(c) Piezometer
:
(9.81. x1.03)(12.5)
CHAPTER T\vO
36
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
GHAPTER
T\vO
2a
Principles of Hydrostatics J t
Problem 2 - 3
Problem 2 - 5
lf the pressure 23 meter below a liquid is 338.445 kPa, deterrnine its unit
weight y, mass densilv p , and specific gravity s
If 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 2 m below the surface?
Sotution
Solution
(n) Unit weight, y
p=vh
338.445 = y (23)
y = 14.715 kN/m3
) Mass density, p
p=l
I
x'l'03
P= 4,7'15
9s1
1
p:pturtacelfll
Psurtace=115-101.325
Note: patm.= 101.325 kPa
Psurrace = 73.675 kPa gage
p =13.675 + (9.81x0.7s)(2)
P = 28.39 kPa
1-6
the absolute pressure in kPa at a deplh of 10 m belovy the free surface of
of sp. gr. 0.75 1f the'barornetric reading is 752 mmHg
p = 1,500 kg/m3
(c) Specific gravrty, s
Prtria
' /* l,)i,,'
pnbs=pntn+pgagr
=
= (9.81 x 13.6)(0.752)
P"t. = 100.329 kPa
P water
.5-
1,500
1,000
,s=1.5
Pnu,
= 100.329+ (9.81 x 0.75X10)
puts = 173.9 kPa
Problem 2 - 4
7.-7
tf the pressure at a point in the ocean is 50 kPa, what is lhe pressure 27 meters
below this point?
?q*" gage 6 m above the bottom of the tank containing a liquid rea{s 90
{Another gage height 4 rri reads 103 kPa. Deterrriine the speeific'weight of
Solution
lhe difference in pressure belween any two point€ in a
liquid is pz - pt = t h
" =?L..'|nr
1 x 1 .03) (27)
p2 = 332.82kPa
Pz-n=Yh.
103-90=y(2)
y.6.5 kN/6r
FLUID MECHANICS
& HYDRAULICS
CHAPTER T\vO
3B
Principles of HYdrostatics
CHAPTER T\vO
FLUID MECHANICS
39
Principles of Hydrostatics
& HYDRAULICS
Solution
Problem 2 - 8
Arr open tank contains 5.8 m of water covered with 3.2 m of kerosene (f = 3
kN/m3). Ilirrd the pressure at the interface and at the bottom of the tank.
Since the density of the mud varies with depth, the pressure
should be solved by integration
dp=ydh
Solution
dp = (L0 + 0.5 h)dh
(u\ Pressure at the interface
p5
aa
Pa = fxht^
lao = |rc*o.snran
J'
J'
00
= (8X3.2)
pa = 25.6kPa
Kerosene
(b) Pressure at the bottom
Pa =2 Ylt
= Y'' h'' + Ys Iq
= e.81(s.8) + 8(3.2)
pn = 82.498 kPa
15
v = 8 kN/m3
p = 10h + 0.25h2 |
lo
Water
= 9.81 kN/m3
= [10(s) + [./$($)z] -
[
P = 56.25kPa
Problem 2 - 9
lf atmospheric pressure is 95.7 kPa ancl the gage attached to the tank reads 188
mmHg rru.,r.t*, find the absolute pressure within the tank
Solution
Problem 2 - 11
ln the figure shown, if the atmospheric
pressure is 101.03 kPa and the absolute
pressure at the bottom of the tank is
231.3 kPa, what is the specific gravity
uf olive oil?
I .5m
'i#'flno;tr**
Water
2.5 m
+ PgrgL
Ptus -- Ptnt
Pgty = f mercu'Y fut""u'Y
', oiuoJ-t
= (9.81 x 13.6)(0'188)
= 25.08 kPa vacuum
Pg"g,' = -25 08 kPa
"'"'
Mercury, s = 13.6
pots=95.7 + (-25.08)
P^u, = 70'62 kPa abs
Problem 2 - 10
The weight density of a mud is given by y = 10 + 0'5k, where y is in kN/m3
h is in mlters. Determine the pressure, in kPa, at a depth of 5 m
.
and
.
2.9 m
'ri
0.4 m
tolutlon
(iagepressureatthebottomof thetank, p=23'l .3 -101 .03
t inge' pressure at the bottom of the tank, p = 130.27 kPa
l;r - )-Y[]
P = T, hr, + Yo ho + ln, htu + lot l].,t
130.27 = (9.81 x 13.5)(0.4) + (9.81 . s)(2.e) + e.81(2.5) + (9.81 x 0.89X1.5)
.r - 1.38
CHAPTER TWO
40
FLUID MECHANICS
& HYDRAULICS
Prlnclples of Hydrostatics
.FLUID MECHANICS
CHAPTER TN/O
Principles of Hydrostaticl
& HYDRAULICS
4l
Problem 2 - 12
Probtem 2- 14
lf air had a constant specific weight of 72.2 N/mr and were incompressible,
'Compute the barometric pressure in kPa at an altitude of 1,200-_m if the
.,pressure at sea level is 101.3 kPa. Assume isothermal conditions a 21oC. Use
R = 287loule /kg-"K.
what would be the height of the atmosphere if the atmospheric pressure (sea
level) is 102 kPa?
Solutlon
Height of atmosphe re, h = L
v
_ 102 x 103
n
.RT
72.2
Height of atmosphere, h = = 8,350.55 m
=U
287(21 + 273)
Problem 2 i 13 (CE Board May 1994)
p = 0.00001185 p
Assuming specific weight of air to be constant at 12 N/m3, 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)(e.s-t) dtr
adn = 0.00011,63 dlt
p
p
7200
Solution
=.0.0001163
,rrr!fr
jp
lnp I
p,
0
11200
. =-0.0001163/r'lI o
J 101.3x10'
lnp - ln (101.3 x 103) = - 0.0001163(1200 - 0)
lnp = 11.tt'
p = e77.386
p = 8$080 Pa
pmt-p.op=I,h
- (y,, hr)rop = (l h)""
(9,810 x 13.6'X0.654) - (9,s10 x 13.6)Q.a4 ='t2h
h = 1,934.53 m
(Yn, h,,)uotto^
CHAPTER T!/O
4Z
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS^
CHAPTER T\YO
Principles of Hydrostatics
6. HYDRAUTICS
43
Problem 2 - 15
Problem 2 - t8 (CE November 199g)
Convert 760 rnrn of mercury to (n) oil of sp. gr. 0.82 and (b) water.
Piston A has a cross-section of 7,200 sq. cm while
that of piston B is 950 sq. cm.
with the latter higher than piston A by 1.7s m. If the intervening
passages are
filled with oil whose specific gravity is 0.g, what is the differer.,iu
i^ pressure
between A and B.
Solution
s--
(rr) /rn,; = It^.rrur--!!!!!S oil
13.6
= 0.76 OU
hnir= 12.605 m of oil
Solution
pa-ps=ypho
= (9,810 x 0.8)(1.25)
Pa-Pa=13,734Pa
(b) /r*,u1", = h,,,"r.ury smercury
= 0.76(13.6)
ft*ater = 10.34 m of water
1200 cm2
Problem 2 - 16 (CE Board May 1994)
950 cm2
A barometer reads 760 mmHg and a pressure gage attached to a tank reads
850 cm of oil (sp. gr. 0.80). What is the absolute pressure in the tank in kPa?
Solution
pabs=parn+pgage
= (e.81 x"13.6)(0.76) + (e.81 x 0.8)(8.5)
Pnr', = 168.1 kPa abs
Problem 2 - L9
ln the figure shown,
300 mm O
llctermine the weight W
lhnt can be carried bv the
1,5 kN force acting on tt
"
phton.
Problem 2 - 17
A hydraulic press is used to raise an 80-kN cargo 'truck. If oil of sp. gr. 0.82
acts on the piston under a pressure of 10 MPa, what diameter of piston is
required?
lolutlon
Solution
Since the pressure under the piston is uniform:
Force=pressurexArea
80,000 = (10 x 10!) LD2
D=0.1 m=100mm
Since points 1 and 2lie on the
rrrnre elevatiorr, pr = ?z
1.5
-_-_____= =
[ (().03)z
W* 150 kN
w
iQ.3)'
300 mm Z
CHAPTER T\x/O
44
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Problem 2 - 20
A clrunr 700 mm in diameter and filled with water has a vertical pipe, 20 mm
in cliamctc.r, attached to the top. How many Newtons of water must be
pourcd into the pipe to exert a force of 6500 N on the top of the drum?
CHAPTER T\vO
FLUID MECHANTCS
Principles of Hydrostatics
& HYDRAULICS
'fA JE
Solution
Solution
Force on the top:
f=pxArea
6500=px f,(7002-202)
p = 0.016904 MPa
p :1.6,904 Pa
,s=0.78
lp: v hl
1.6,904 = 9810 h
@
h=1.723n
Weight=yxVolume
= 9810 x ! (0.02)2(1.723)
Weight = 5.31N
tpz - p'' = y,, ttl
trF
lt1-
n
-
0.00323
P=309.6F (kPa)
w44
Area on top
0)= _
A
700 mm A
0.323
Pz = 136.22 kPa
Problem 2 - 2L
136.22 - 309.6 F = (9.81 x 0 7ti)(4.6)
The figure shown shows a setup with a vessel containing a plunger and a
cylinder. What force F is required to balance the weight of the cylinder if the
weight of the plunger is negligible?
Cylinder
W=44kN
F=
A = 0.323 m2
l'= 0.326 kN = 325 N
lhs hydraulic press shown rs filled with oil with sp. gr. 0.82. Neglecting the
gFlght of the two pistons, wlrat force F on the handle is required to support
tlte lllkN weight2
.4.6 m
J
Oil, 5 = 6.73
Oil, s = 0.78
CHAPTER T\vO
46
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
47
fLUID MECHANICS
6. HYDRAULICS
since the gage reads "FULL'trren the reading is equivarent to 30 cfl of gasoJine
$olution
Sinct' points I and 2 lie on the same
t'lt'vation, then;
Reading (pressure heaci) when the tank contair.r
l)t = Itz
water : (y * 2# ) cm of gasoline
f,
=F,
A1 A2
fher.r; y+2'fu =30
10
F2
tQoTrz
i (0.025)2
t1 = 27.06 cm
F, =.1 .11 kN
Problem 2-24 (cE B@
[IMo=0]
F'or the tarrk shown in the Figure, llr = 3m and /ir
F(0.42s) = Fr(0.02s)
F(0.425) = 1.11(0.025)
F = 0.0654 kN
<tf 14
r=65.4N
FBD of the lever arm
Problem 2 - 23
The fuel gage for a gasoline (sp. gr. = 0.68) tank in a car reads proportional to
its bottom gage. If the tank is 30 cm cleep an accidentally contaminated with 2
cm of water, how many centimeters of gasoline does the tank actually contain
when the gage erroneously reads "FULL"?
Solution
Solution
Summing-up pressure head
from 1 to 3 in meters of water
D"
tL
+hr(0.84)-t= 41
Yy
0+0.84h2_g_r=o
hz = 1|1.9 m
r
2.rL
T_
I
30 cm
I
J
"Full"
= 4 m Deterrine
the value
CHAPTER I\vO
48
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
Problem 2 - 25 (CE Board May 1992)
ln the figure shown, what is the static pressure in kpa in the air chamber?
FLUID MECHANICS
& HYDRAULICS
e.i,,.ipr.,l?ffi*rY,.: 49
Problem 2 - 26
For the manometer showry
determine the pressure at the
center of the pipe,
t--
= 13.55
L
m
Solution
The pressure in the air space
equals the pressure on the surface
of orl, pt.
9um-up pressure head from
1 to 3 in meters of water:
D.
Yt
+1(13.55) + 1.5(0.8)= Iq
Pr=0
fy
0+14.75= la
Pz = fu, hu
f
= e.81(2)
pz = 19.62kPa
pz - pt
Solution
tl
t yu lt,,
= u.zsm of water
v
19.62- pt= (9.81' 0.S0X4)
pt = -1'1,.77 kPa
Another solution.
Sum-up pressure head fronr l to 3 in meters of water
. Pt *2-4(0.80)= ta
Yy
o+z-3.2= Pt
9.81
p. = _1.1.77 kpa
Pt= M.75(9.81)
pt = 144.7 kPa
, s = 13.55
50
CHAPTER TUTO
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Problem 2 - 27 (CE Board November 2001)
CHAPTER T\I/O
FLUID MECHANICS
& HYDRAULICS
principles of Hydrostatics
Problem 2 - 2A (CE May 1993)
L)c.tcnrrirre the value of y in the manometer shown in the Figure.
ln 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 cm. Determine the new
-T1m
.,
T-
deflection of mercury when the
funnel is filled with water to B
I
3m
I
+-
1m
J
Solution
Solution
-T-
Summing-up pressure head from
A to B in meters of water:
14+ e(0.8) + 1.s - y(13.6) = ru-
yy
5 * g.9-13.6r,=!-E" y
9.8't
where pa = 0
1m
l'l,
I
3m
tI
1m
.'
11= 0.324m
Figure (b): Level at B
Figure (a):'Level at A
Solve for y in Figure (a):
sum-up pressure head from A to 2 rn nleters ot water
o^ *
Y"y r-
0.15(r3.5s) =
0 + y - 2.03 = 0
7=2.03m
la
5l
5Z
CHAPTER T\VO
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
In Figure (b):
When the funnel is filled with water to B, point 1 will move d.own to 1,
with the same value as point 2 moving up-to Z'
o.s * y +
Y-y
CHAPTER TWO
Principles of Hydrostatics
53
Sum-up pressure head from Z to m jn ineters of water:
. P'*rt0.3.6\-y=P*
y
Y "'
13.5y - x =
Sum-up pressure head from B to 2':
P-P +
FLUID MECHANICS
& HYDRAULICS
70
9.81
Eq (1)
In Figure (b):
*- (r+ 0.t5 + x)(Ig.ss) = 2
Sum-up pressure head from 2, to m,in meters of water:
0 + 0.80 + 2.03+ x -27.'tx - 2.03 = 0
26.1 x = 0.80
0.031 m = 3.L cm
2 +Q.2sino+ y+0.2)(1g.6)-(r+0.2) - Pnt'
Ty
0 + 2.7Zsin 0 + 12.6y + 2.72 _ x_ 0.2
= ffi
x:
13.6y - x = 8.183 - 2.72 sin 0
New reading, R = 15 + ?t = 1,5 + 2(3.1)
New reading,R=X.2cm
Eq.(2)
[13.6y-x=L3.5y-x]
8.183-2.72sine=#
Problem 2- 29
The pressure at point rz in the figure
shown was increased. from 70 kPa to
105 kPa, This causes the top level of
mercury to move 20 mm in the sloping
tube. What is the inclination, 0?
sin 0 = A3852
0 = 22.660
2-30
closed cylindrical tank contains 2 m of water, 3 m of oil (s
= 0.g2) and the air
ove oil has a pressure of 30 kpa. If an open mercury manorneter at the
of the tank has 1 m of water, determine the deflection of mercury.
Solution
Sum-up pressure head from
1 to 4 in meters of water:
n
!!tL
a3(0.82) +z+L-y(13.0y= ll
Yy
# *2.+o+3-13.6y=0
y = 0.626m.
Figure (a)
In Figure (a):
Figure (b)
CHAPTER T\vO
54
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Problem 2 - 31
FLUID MECHANICS
CHAPTER T\x/O
& HYDRAULICS
Frinciples of Hydrostatics
In Figure (&):
llrt' Lj-tubc shown is 10 mm in diameter
Summing-up pressure head from 1 to 3 in mm of water:
,rrrtl contains mercury. If 12 ml of water is
lL +752.8_R(13.6): u
Yy
pourccl into the right-hand leg, what are
lhc ultirlate heights in the two legs?
R = 11.24 mm
--1
In Eq. (2):
I
E
E
11.24+2x=240
r:1'14.38 mm
N
I
I
*
L ,ro * --'l
Solution
Ulhimate heights in each leg:
Right-handleg,hn=h+x
= 1.52.8 + 't14.38
Right-han d leg, hn = 267 ;1,8 rnm
5olvrng for /r. (sar ftgttrr'h1
Volume of water = + (#F ir = l2 cml
Note: 1ml =1cm3
h = 15.28 crn = 152.8 mm
Lefrhand teg' ht =
fr.rX .
114.38
Left-hand Leg, ht = 1%.62 rnm
Since the quantity of mercury before and after water is poured
remain the same, then;
Problem 2 - 32
120(3)=R+-r+120+r
Fnr a gage reading of -17.7kpa,
R+2x=240 )Eq.(1)
dE'termine the (a) elevations of
the liquids in the open
plezometer columns E, F, and
G and (b) the deflection of the
tnercury in the U-tube
manometer neglecting the
welght of air.
I
-l
E
E
E
E
N
N
1
I
I
I
J
J
t_
-t
120 mm
Figure (b)
-J
55
CHAPTER I\vO
Princaples of Hydrostat,cr
56
FLUID MECHANICS
& HYDRAULICS
Solution
FLUID MECHANICS
& HYDRAULICS
CHAPTER T\VO
Principtes of Hydrostatics
FA
, I
Column G
Sum-up pressure head from 1 to g in meters of water;
1-
lL + se.z)+ a(1) - tr{1.6) = !-Evy
# *2.1+4-7.6fu=s
E!. _15
E!.
_r)_m_
Jm
ht= 2.'U^
El _12-*_
Surface elevation = 8 + /rs
Surface elevation = 8 + 212 = 7O.72 m
E_l
1
4m
Deflection of mercury
EI
El._8 ln _
Sum-up pressure head from 1 to 5 in meters of water;
Pr + zp.4 + 4 + 4 - h4(1s.6) : !-9v
1
4m
#
*'t0.1.-13.6tu
ha= 0.6'L4m
t ---v
El.4m
Problem 2 - 33
Column E
Sum-up pressure head from 1 to c in metes of water;
P' *trr(0.7)= P,
yy
#
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
water. Calculate the pressure at the centerline of the pipe.
Solution
*ftr(07)=o
/rr = 2.5 m
Surface elevation = 15 - /rr
Surface elevation = 15 - 2.5 = 1Z5 m
Column F
Sum-up pressure head from I tp /in meters of water;
Dt
t-
yy
Pr
+3(07)-hr(1\=''
i#
*7'1 -h'=o
ltz = g'3t'
^ = 12 + lu
Surface elevation
' Surface elevation = 12 + 0.357 = 1'2.3ST m
8um-up pressure head from 1
9ln meters of water;
b- + 0.45- 0.15(13.6) = l
v
P, *0.45-2.04=o
9.81
;r1 = L5.6 kPa
CHAPTER T\VO
58
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
FLUID MECHANICS
CHAPTER T\vO
& HYDRAUTICS
Principles of Hydrostatics
Problem 2 - 34
Solution
lior llre configuration shown, calculate the
wt.ililrt of the piston if the pressure gage
(a) Gage liquid = mercury, /e = 0.1 m
Sum-up pressure head from
1 to 4 in meters of water;
rt,irrling is 70 kPa.
U
Pr + +
x h - tx(13.6) - x - 1.5 =
v
Pt
v
Pt
v
v
- ' a - 1.5-0.1 +0.1(13.6)
f),
v'
- Pq =2.76mof water
v
Solution
Surn-up pressure head from
A to B in meters of water;
Ps = 70 kPa
o -t (0.86) : r-t
B
P
(1,)
Sum-up pressure head from 1 to 4 in meters of water;
vv
tl
P"t + x +
Pe _o.se = 7o
9.81
9.81
FA=pAxArea
Weight = Fn
h - tr('1.5g\ - x - 1.5 =
Yy
Pr -P+ =1.5+0.59/r
yy
:rt,t = 78.44kPa
= PA x Area
7v.aa x
Gage liquid = carbon tetrachloride
reading, h = ?
i e)2
Weight = 61.61kN
Problem 2 - 35
Two vessels are connected to a differential manorneter using mercury, tht'
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 tetrachloriclc
(.s = 1.59) were.used instead of mercury, what would be the manomett't'
re.rding for the same pressure difference?
lL
where Pt - P+ = 2.76 m) from (a)
yv
II
2.76=1.5+0.591t
Ir= 2.136m
ln thr. figure shown, determine
the height /r of water and the
lrge reading at A when the
lbnolutc pressure at B is 290
lPr,
TI
h
f
I
700 mm
J,c
Air, p = 175 lPu u6t
o
oo
Water
59
CHAPTER T\YO
60
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Solutlon
CHAPTER T\vO
FLUID MECHANICS
& HYDRAULICS
Sum-up pressure (gage) head from L to 4 in meters of water;
Sum-up absolute pressure head
from B to 2 in rneters of water;
Air, p = 175 ;,Pu u5t
f-- €
lL -o.z1s.e)-h= lz
Yy
fh
ffi -s.sz-t'= #
h = 2.2O3 m
700 mm
Jc
wt
.tL + xe.s)+ 1.3(0.e) -1.3(12.6)= lL
yy
4o *0.9x-16.51=o
watei
9.81.
tl
M;r*ry/
r = 13.81m
o
o
Then,x+!/=28.42m
B
Sum-up absolute pressure head from B to A in meters of water;
0o
!!L - o.z(tZ.o) + o.z = !-t
Yy
Pa
ffi -s.sz+ s.7 = 9.81
Problem 2 - 38
,For the manometer sefup shown,
determine the difference in pressure
between A and B.
pa = 203.5 kPa abs
Problem 2 - 37
In the figure shown, the atmospheric
pressure is 101 kPa, the gage
reading at A is 40 kPa, and the vapor
pressure of alcohol is 72 kpa
absolute. Computer+y.
Solution
x+0.68=y+1..7
x-y=1.02m )Eq.(1)
Sum-up pressure head from A to B
in meters of water;
?
-.- 0.68(0.85) + y = a_
v
? + =x-v+0.s78 )
Solution
Sum-up absolute pressure head from
l" to 2 in meters of water;
lL _ yp.t1= ?L
Y.r
40 +
101_u.9a= _12
9.81
y = 74.67 m
"
9.81
Eq.(2)
Substitute x - y = L.02 in Eq. (1) to Eq. (2):
Pn -Ps =1..02+o.sr|
yv
Pa-Pn =1.SSS
9.81
Pa - Ps = 15.58 kPa
, t
Principles of Hydrostatics O I
CHAPTER T\VO
62
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatacs
Problem 2 - 39
CHAPTER T\)(/O
FLUID MECHANICS
Principles of Hydrostatics
6. HYDRAULICS
Problem 2 - 40
A differential manometer rs
In thg figure shown, the
attached to a pipe as shown
Calcu late the pressure
.lifference between points ,4
and ll
deflection of mercury is initially
250 mm. If the pressure at A is
increased by 40 kPa, while
maintaining the pressure at B
constant, what will be the new
mercury deflection?
Solution
2
t00 mm
l
T-
Solution
+ffi-rr
1
Sum-up pressure head from A to B in rneters of water
n"
-r(Tn\ 01(136) +01(09) +y(09) = {g
''^
VV
Pa
U = 01(13 6) () l(0 e)
vy
Pa - Pa
= t.L/
trt
9.81
PA - PB = 12.46 kPa
Figure (a)
Figure (b)
In Figure d, sum-up pressure head from A to B in meters of water;
!t
yv-s.6-0.25(13.6)+0.25+ 2.1=
Pa
Pn
v
v
= 1.65 m of water
!j-
63
64
CHAPTER T\VO
FLUID MECFIANICS
& HYDRAULICS
Principles of Hydrostatics
ln Figure b, po = pt + 40
Sum-up pressure head from A lo B tn meters ot water
FLUID MECHANICS
& HYDRAULICS
CHAPTER T\x/O
Principles of Hydrostatics
,, r
(,)
Solution
Dd
ft- (0.6- r) (0.25 + 2xlt36 + (2.35 * ,) = Pt
YV
o" t40
250 mm
Vv
Pa * 4o -r 65 - 25.2:r = Pa
v .9.81
Pa U =zs.zt-2.423
VV
But PA
U = t.ts
t65=25.2t-2.423
r=0162m=162mm
New mercury deflection = 250 + 2r = 250 * 2(162\
New mercury deflection = 574 mm
Sum-up pressure head from A to B in meters of water;
!-4- + 0.2(0.8s)- 0.09(13.6)- 0.31(0.82) + 0.25 - 0.1(0.0012) = Ps
P
Problem 2 - 4t
e - Pn = Losi23m of water
fl
pA - pB = 9.81 (1.0523) = 10.32 kPa
ln the figure shown, determrne the difference in pressure between points A
and B
2 - 42 (CE Board)
Assuming normal barometric pressure, how deep in the ocean is the point
here an air bubble, upon reaching the surface, has six Limes its volume than
had at the bottom?
Applying Boyle's Law
(4ssuming isothermal condition)
lp Vt = pzVzl
pt = 101..3 + 9.81(1.03)lr
pt = 101.3 + 10.104 h
l,
v'l -- 1,
v
pz=101.3 + 0 = 101.3
Vz- 6V
(101.3 + 10.104/r)14 = 101.3 (6 t/)
10.104 h=10t.3(6) - 101.3
h = 50.13 m
th
66
CHAPTER T\vO
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
CHAPTER TWO
FLUID MECHANICS
& HYDRAULICS
Principles of Hydrostatics
Problem 2 - 43
Since the pressure in air insicle the tube is uniform,
A vertical tube, 3 m long, with one end closed is inserted vertically, with the
open end down, into a tank of water to such a depth that an open manometer
connected to the upper end of the tube reads 150 mm of mercury. Neglecting
vapor pressure and assuming normal conditions, how far is the lower end of
the tube below the water surface in the tank?
then pc - p6 = 20.0124 kPa
Solution
r a
Ot
Pr=T*h
20.0124 = 9.81.h; h = 2.04 m
Then; x=h+ y =2.04+ 0.495
x = 2.535 m
Area = A
Probfem 2 - 44
0.15 m
?: Y;
Mercury
x,.
Applying Boyle's Law:
prVt= pzVz
A bottle consisting of a cylinder 15 cm in diameter and 25 cm high, has a neck
which is 5 cm diameter and 25 cm long. The bottle is inserted vertically in
water, with the open end down, such that the neck is completely filled with
water. Find the depth to which the open end is submerged. Assume normal
barometric pressure and neglect vapor pressure.
,:':,,:,
Solution
Applying Boyle's Law
15cmO
ptVt = pzVz
Before the bottle was inserted:
Volume of air:
Before the hrbe was inserted;
Absolute pressure of air inside, pt = 101.3
Volume of air inside, Vt = 3A
When the tube was inserted;
Absolute pressure of air inside, pz -- \01,.3 + 9.81 (13.6X0.15)
Absolute pressure of air inside, pz -- 721.31 kPa
Volume of air inside the tube, Vz= (3 - y)A
lptVr= pzVzl
101.3 (3 A) = 121.3t t (3 - y) a I
3-Y=2'sos
A=0495m
From the mahometer shown;
pt = y,,, h,,,
= (9.81 x 13.6)(0.15)
pn = 20.0l24kPa
V, = i(1s), (2s) +
f (5)r(25)
Vt = 4,908.V4 cm\
Absolute pressure in air:
7 ='10'l'325
When the bottle is inserted:
Volume of air:
vz= (15), (25)
i
Vz= 4,417.9 cml
Pressure in air:
pz='10"1.325 + 9.8-l h
lp Vr = pzVzl
101.325(4,908,7a) = $01.32s + 9.81 h)(4,417 9\
101.325 + 9.8t h = 112.58
ft = 1.15 cm
x=ltr25=26.15cm
CHAPTER T\vO
68
Principles of Hydrostatics
FLUID MECHANICS
& HYDRAULICS
Problem 2 - 45
A bicycle tire is inflated at sea level, where the atmosfiheric pressure is 101.3
kPaa and the temperature is 27 "C, to 445 kPa. Assuming the tire does not
expand, what is the gage pressure within the tire on the top of a mountain
where the altitude is 6,000 m, atmospheric pressure is 47.22 kPaa, and the
temperature is 5'C.
FLUID MECHANICS
& HYDRAULICS
e,i,,.ipru,1?fi;it*#,.: 69
Supplementary Problems
Problem 2 - 46
A weather report indicates the barometric pressure is 28.54 inches of mercury'
What is the atmospheric pressure in pounds per square inch?
Ans: t+.02 ps;
Solution
Ptvt = Pzvz
T1
T2
At sea level:
Problem 2 - 47
The tube showrr is fillect with oil. Determine the pressure heads at B and C irr
meters of water.
Ans:
Absolute pressure of afu, pt ='107.3 + 445
Absolute pressure, p1= = 546.3kPaa
Volume of air, Vt = V
Absolute temperature of air. Tr = 21 + 273 = 294 "K
Pn
= -2.38ln
v
Pc
= -0.51m
v
On the top of the mountain:
Absolute pressure of air, pz= 47.22 + p
Since the tire did not expand, volume of ait, Vz= V
Absolute temperature of air ; Tz = 5 + 273 = 278 "K
tT=hr
, PtVt - PzVz ,
s46.3(v)
'294 - $7.22+P)V
278
47'22+ P = 5L6'57
P = 469'35kPa
s = 0.85
2- 48
For thb tank shown in the figure, compute the pressure at points B, C, D, and E
kPa. Neglect the unit weight of air
Arts pu = 4.9; pc = po = 4.9; Pt = 21.64
70
CHAPTER T\VO
FLUID MECHANICS
& HYDRAULICS
Principtes of Hydrostatics
Problem 2 - 49
FLUID MECHANICS
& HYDRAUTICS
CHAPTER T\VO
Principles of Hydrostatics
2-52
A'gtass U-tube open to the ahnosphere at both ends is shoWn. lf the U-tube
contains oil and water, determine the specific gravity of the oil
Ans: 0.86
cylindrical tank contains water at a height of 55 mm, as shown' Inside is a
open cylindrical tank containing cleaning fluid (s.g. = 0.8) at a height /re pressure pe = 13.4 kPa gage and pc =
-13.42
kPa gage. Assume the cleaning
id is prevented from moving to the top of the tank. Use unit'weight of
= 9.79 kN/m3. (a) Deterrnine the pressure pe in kPa, (b) the value of h in
and (c) the value of y in millimeters.
Ans: (a) 12.88; (b) 1,0.2; (c) 101
Problem 2 - 50
A glass 12 cm tall fillecl with water is inverted. The bottom is open. What is
the pressure at the closed end? Barometric pressure is 101.325lclL
rOO.rU OO",
Problem 2 - 51
ln Figure'13, in which fluid will a pressure of 700 kPa first be
Po = 90 kPa
".nt?,ilt rr*rr"
2-53
tial manometer shown is.measuring the differen'ce in pressure two
pipes. The indicating liquid is mercury (specific gravity + 13.6),ln is 675
Itrn is 225 mm, and h,,,2 is 300 mrn. What is the pressure differential
ethyl alcohol
p = 773.3 kglm3
60m
oil
p = 899.6 k9/m3
10m
water
p = 979 kglmr
5m
glycerin
p = 1236 kglm3
5m
the two pipes.
Ans:89.32kPa
72
CHAPTER T\vO
Prlnclples of Hydrostatics
FLUID MECHANICS
FLUID MECHANICS
& HYDRAULICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
73
Problem 2 - 54
Chapter 3
A force of 460 N is exerted on lever AB as shown. The end B is connected to a
piston which fits into a cylinder having a diameter of 50 mm. what force Fo
dcts on the larger piston, if the volume between c and D is filled with water?
Total Hydrostatic Force
on Surfaces
Ans: 15.83 kN
TOTAL HYDROSTATIC FORCE ON PLANE SURFACES
lf tlre pressure over a plane area is uniform, as in the case of a horizontar
;Nurface submerged in.a liquid or a plane surface i"rla. u-g*';;;;^;;
total hydrostatic force (or total pressure) is
liven by:
.
Problem 2 - 55
An open tube open tube rs attached to a tank as shown. lf water rises to a
height of 800 mm in the tube, what are the pressures pa ancl pB of the air above
water? Neglect capillary effects in the tube.
where p is the uniform pressure and A is the area.
In the case of an incrined. or-vertical plane submerged in a
liquid, the total
Pressure can be found by the followingior*rtu,
Ans: pe = 3.92kPa; pz = 4.9Q kPa
of gravity, cg
of pressure, cp
74
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Consider the plane surface shown inclined at an angle 0 with the horizontal.
-[o
get thc' total force F, consider a differential element of area dA. Since this
element is horizontal the pressure is uniform over this area, then;
CHAPTERTHREE qF
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
LOCATION OF F (yp):
In Figure 3 - l, taking moment of force about s, (the intersection of the
prolongation of the plane area and the liquid surface),
ry,=
dL=pdA
where p = yh
f
Jyar
where dF=yysin0dA
P=YYsin0
F=ysin0 Ay
dF = .r y sinO dA
y sin 0 AT yp
[n,=rsi,,e Jvan
I
-
v
!v0 "i^e
ysin0Ayyr=ysin0
From calculus, ly dn = Ay
aa)
I
JV'aa
J
P=ysin0 Al
F=y(tsin0)A
From catcutu
", Ir2 dA = Is
(moment of inertia about S)
AYYp=ls
From the figure, y sin 0 = ir
Then,
P=yhA
Eq.3-2
By {ansfer formula of moment of inertia:
Is=\+ n!2
Since y lr is the unit pressure at the centroid of the plane area, pcs, the formula
may also be expressed as:
F-
A
lfo--
Eo.3-3
Eq. 3 - 2 is convenient to use if the plane is submerged in a single liquid and
without gage pressure at the surface of the liquid. However, if the plane is
submerged under layers of different iiquids or if the gage pressure at tht'
liquid surface is not zero, Eq.3 - 3 is easier to apply See Problem 3 - 15
L
I"+AY2
6
AY
Snce yo = 7 + e, fromFigure 3 -'L, then
I
Table 3 - 1 in Page 76 for the properties of commbn plane sections.
76
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
TABLE 3 - 1: Properties of Common plane sections
Triangle
CHAPTER THREE
FLUID MECHANICS
Total Hydrostatic Force on Surfaces
& HYDRAULICS
Half ellipse
Quarter ellipse
Rectangle
X
Area = h nnb
v.=
k-ttz-$-6124
l-- r *-l
r = i+b
a
Area = bd
.
'
y,=tt/3
Area=r/zblt
r',_bd3
- --;-
blt3
!8r bhl
72
36
. =- bd3 I
db3
I
='st
72 '&i 1,2
Iy
r ab3
:
6'
db'
=
3n r'
- ntb3
I=-l=-
Area = n2 = y4 jr D2
lq,
-
nbn.t
,lr. = 0.055nb3 Iru= 0.055bn3
8
Sector of a circle
Parabolic seqment
Quarter circle
r
-
3n
"16"'16
t bn3
-;
f (20) = 12 0
2 rsin9
"30
4r
-4
h
Area = 7/z
tvr
4b
1_=-tt=-
let=0."11 nb|
I ='sY
3
.f
Circle
4n
", -3t
l,=
Area = ln ntb
4b
oro nDa
464
Area = l/t z 12; x, = yr
l
Ir = -f,' (0 - % sin 2e)
.1fr
,14
,t - lu -
Irj
'76
/r,=1r=0.055/
Semicircle
t4
n
Area = ?ar,
2,
x"=
'58-il
Spandrel
)J
' ltbt
'15"7
I_=
(e+hsn20)
3
1t"=
3,
-D
1..= 'bh3
Segment of arc
Ellipse
Y=kxn
l*o---*l
+t'
Area=%nrr; y,= ;-
Area= rab
lr-l{r-
tr
'8
-. d
ls, = 0.11
4
Length of arc = r(20) = 2tfi
'i
oa3
.tn
'g,
Area =
- -Ti
n ba1
'w- --i-
'
'
1
,1
+7
b/r
1 1t.= ,r+1 tr
tr+2o: "'
4rt+2
-
,1. =
rsn9
----*
0
\ /hen 0 = 90' (semicircle)
2r
E
.1
I I
16
t O
CHAPTERTHREE
FLUID MECHANICS
& HYDRAULICS
Totat Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
TOTAL HYDROSTATIC FORCE ON CURVED SURFACES
CASE
CHAPTER THREE aO
Total Hydrostatic Force on Surfaces t
A
orFs=yiA
Fv=IV
Eq.3-8
tan}=Fy/Fy
Eq.3-70-
Fu = p,c
I : FLUID IS ABOVE THE CURVED SURFACE.
Eq.3 -7
Eq.3-9
where:
A = vertical projection of submerged curve (plane area)
pcg = pressure at the cenhoid of A
Note: The procedure used in solving Fn is the same are that presented in Page 73.
Vertical projection
of the curved surface
CASE II: FLUID IS BELOW THE CURVED SURFACE
Fx
q
Curved surfacb
1,"
t
FLUID BELOW AND ABOVE THE CURVED SURFACE
CHAPTER THREE
80
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
CHAPTER THREE
& HYDRAUTICS
Total Hydrostatic Force on Surfaces
BI
DAMS
l)ams are strucfures that block the flow of a river, stream, or other waterway.
some dams divert the flow of river water into a pipeline, canal, or channel.
others raise the level of inland waterways to make them navigable by ships
and barges. Many dams harness the energy of falling water to ge'nerate electric
power. Dams also hold water for drinking a'd crop irrigatiJn, and provide
flood control.
' up strranl
t'
lates 0p en
Eoat
Ocwnttre am
qates
5tr enn
|latiJ
cl!!e
PURPOSE OF A DAM
Dams are built for the following purposes:
1. Irrigation and drinking water
2. Power supply (hydroelectric)
:
oprn
3. Navigation
4. Flood control
5. Multi purposes
sJuire!
llgure 3 - 3: Boat passing through canar Lock.
Canar rocks are a series of gates designed
to allow a boat or shio to pass from one rever
of water to another. Here, afier a boat has
rntered the rock and ail gates ur. t".ula,-il.rJoo*nrtr"am
through them' when the water tever is equat-o;
the stuices; the do;"ro..,
il:tt#il:gJhrou,eh
sruices open and
flows
;;.,. side of the downstream water
gate, water
sr* opens, and tnu ooui.ontinues on at
PorterhDuse
To transmiiSion line!
TYPrS oF DAMs
t
cereralot,
Drait tube
PenrtocL
Eedrock
Figure 3 - 2: Section of a dam used for hydroelectric
the force of
grarziry ro resist water pressure_
?^r:r::, _(::.,1]1r,":"tI
"
b;
;."
y"'".
:11'.i: Ji":,1:li-1"*'l:
o.ynwald ro do this, gru,rity d^;;;;;;#;;;;:r:
l:rll'lg
so
hea'uy tr-rat the water in a reser"voir
"
cannot push the darn
downstream or tip it over. 'Ihey afe
much,hi.Lu. at the base than the
top-a shape that reflecrs the distribution of the
forces oiin" -ui",
against the dam. As water becomes
deeper, it exerts more horizontal
pressure on the dam. Gravity <lams
are relativery thi.
the surface
t",::.voir,.where the warer p.";;;;;-;;
^ear
il.i.
:j"*:
;;-;;;; ;;in:;?:;;
;il;
enables the dam to 44withstand
the more i;;;;-;;,*
the bottom of the reservoir
;il:
pressure at
82
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAUTICS
|LUID MECHANICS
& HYDRAULICS
4.
'"^:J5:#'j.:: 83
Total Hydrostatic Forc€
A btttttess dan consists of a wall, or face, suPPorted by several
buttresses on the downstream side. The vast majority of buttress
,dams are macle of concrete that is reilforcecl with steel. Buttresses are
typically spaced across the dam site every 5 to 30 m (20 to 100 ft),
depending upon the size and clesign of the dam. Buttress dams are
sometimes called hollow dams beciuse the buttl:esses do not form a
solid wall stretcl-ring across a river valley.
Figure 3 - 4: Gravity dam
Nn entbankment dam is a gravity dam formed out of loose rock,
earth, or a combination of these materials. The upstream and
downstream slopes of ernbankment 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
Arch danrs are concrete or lnasonry structures that curve upstream into
a reservoir, stretching from one wall of a river czrnyon to the other. This
design, based on tl're same principles as the architectural arch and vault,
transfers some water pressure onto the walls of thg 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
Figure 3 - 5: Buttress dam
carry parfof the pressure exerted by the reservoir
Figure 3 - 5: Arch dam
Figure 3 - 7: Multiple arch darn
84
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
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 sride horizontaily
on its
foundation and overfurn it about its downstream edge or toe.
These
tendencies are resisted by friction on the base of the
dari and gravitational
forces which causes a moment opposite to the overturning
moment. The dam
rnay also be prevented from sliding by keying its base
Upstream Sjde
FLUID MECHANICS
& HYDRAULICS
A. Vertical forces
1. Weight of the dam
W't:l,Vt Wz=f,Vz; Wz=l,Vz
2. Weight of water in the upstream side (if any)
Ws = yVs
3. Weight or permanent structures on the dam
4. Hydrostatic Uplift
Ut = f Vu't
Uz=fV,z
Downstream Side
(Tailwater)
Headwater
I
t'
I
I
CHAPTER THREE
Totat Hydrostatic Force on Surfaces
B. Horizontal Force
1. Total Hydrostatic Force acting at the vertical projection
of the submerged portion of the dam,
p=yE,l
2. Wind pressure
3. Wave Action
4. Floating Bodies
5. Earthquake Load
-t
I
I
I
I
I
I
I
I
I
I
I
IIl. Solve for the Reaction
A. Vertical Reaction, R,,
P =sE
Lr t,
t\y -
Rv = W, + Wz + Wz + Wq - L)t - l)z
Uplift Pressure
B. Horizontal Reaction, R.
Diagram
R* = IFr,
R'=P
IV. Moment about the Toe
Figure 3 - 8: Tvpicar section of a gravity dam showing ,n"Torlo,"
A. Righting Moment, RM (rotntibn toroards tlrc ttpstrcnm
forces acting
Steps of Solution
with r.eference to Figure 3 - 8,'for purposes of illustration, an assumptiori
was
made in the shape of the uplift pressure diagram.
L Consider 1 unit (1 m) length of dam (perpendicu.lar to the sketch)
ll Determine all the forces acting:
RM = Wt xt + Wz xz + Wt xj + Wa x1
side)
B. Overturning Moment, OM (rolatton totoards the downstreant
sid.e)
OM=PU+Lftzt+l-Izzz
V Location of R, ( t )
85
CHAPTER THREE
86
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
where:
y = unit weight of water = 9.81 kN/m3 (or 1000 kg/m.)
y, = unit weight of concrete
y, = 2.4y (usually taken as 23.5 kN/m)
FLUID MECHANICS
& HY,DRAULICS
a=-:
,B
CHAPTER THREE
Total Hydrostatic Force on Surfaces
87,
6R,e
R,r
-82
,= +(t-#),where e<8/6
Factors of Safety
Eq.3-14
Factor of safety against sliding, FS_s:
uR..
Y >'l
FS.-R,
=
Eq.3 -1.2
.Note: Use (+) to get the stress at point where R, is nearest. In the diagram
thown above, use (+) to get qr and (-) to get qH. A negative stress indicates
pressive stress and a positive stress indicates tensile stress
soil cannot carry any tensile stress, the result of Eq. 3 - 14 is invalid if the
is positive. This will happen if e > 8/6. Should this happen, Eq. 3 - 15
Factor of safety against overturning, Fso:
be used.
,to=
Eq.3-13
Hr,
where:
p = coefficient of friction between the base of the dam and the foundation
Foundation Pressure
r Middle Third I
For e 38/6
Bl3
lBt3i
Blt
e>B/6
i =a/3
a=3V
From combined axial and bending
R, = 1/z(a)(q,)(1)
stress formula:
R, = 1/z(37-)q,
n=_!r!!
'AI
P=R,
A=B(L)=B
M= R,,e
1
=
Heel
2R,v
qe= 4
QH
'" 3i
1(B)3
12
c =.82
O=-r+
'
Ru $,, e)(B /2)
B ' 83/12
1m
cg{
oR,
Eq.3-15
88
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
BUOYANCY
FLUID MEcHANtcs
CHA1flE\THREE 89
where:
y_ = unit weight of the
fluid
vp = volumJ displacecl. Volume of the bocly belor,v fhe Iiquid surface
ARCHIMEDES' PRINCIPLE
A principle discovered by the Greek scientist Archimedes that states that "any
h<tdy immersed in a fluid is acted upon by an upward
weight of the displaced fluid".
force (buoynnt force) equal to the
This principle, also known as the law of hydrostatics, applies to both floating
and submerged bodies, and to all fluids.
Consider the body shown in Figure 3 - 9 immersed in a fluid of unit weight y.
The horizontal components of the force acting on the body are all in
equilibrium, since the vertical projection of the body in opposite sides is the
same. The upper face of the body is subject to a vertical downward force
which is equal to the weight of the fluid above it, and the lower face is subject
to an upward force equal to the weight of real or: imaginary liquid above it.
solae problenrs in hrcynncy, tderirfu tlrc
ilibium:
I,Fs = I
forces n,.lirtg ortr! +ypl!/
conditions of stntic
lFy=6
LM= 0
homogeneous solid body of volume V "floa$ng,,6
u homogeneous
y, = sP 8r olbody- V = luoay U
sp. gr. of
liquid
fluid at
Eq.3 -17
Yrquirr
The net upward force acting on the body is the buoyant force.
octional area such as
the body of height.H has a constant horizontal cross-9e
cylinders, blocks, etc.:
Vo=Volz-Volr
,=
Figure 3 - 9: Forces acting on a submerged body
BF = Fvz- Fvt
= yffolz) - y(Voh)
sP'gr'olbod.y_
H= lboay ,
sp.gr.of liquid --
Eq.3-18
yhquid
the body is of uniform vertical cross-sectional area A, rhe area submerged A.
BF=y(Vol2-Volr)
:' ,
BF=yVo
,
gq.3-te
a - sP.gr.ofbody -n^
,-^s---n==
sp. gr. of liquid
Tbody
l uquid
A
Eq.3 - 19
90
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FTUID MECHANICS
& HYDRAULICS
FTUID MECHANICS
CHAPTER THREE
& HYDRAULICS
Total Hydrostatic Force on Surfaces
STATICAL STABILIW OF FLOATING BODIES
A floating body is acted upon by two equal opposing forces. These are, the
body's weight w (acting at its center of gravity) and its buoyant force BF
(acting at the center of buoyancy that is located at the ce$pr of gravity of the
displaced liquid)
when these forces are collinear as shown in Figure 3 - 10 (a), it floats rr.r 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(r). The body will not overturn if this couple makes the body rotate towards
its original position as shown in Figure 3 - 10 fb), ancl will overturn if the
situation is as shown in Figure 3 - 10 (c).
Bo.{
Y
rhe point of intersection between the axis of the body and the line of action of
the buoyant force is called the metacenter. The distance from the metacenter
(1w)_to the center of gravity (G) of the body is called the metacentric height
(MG). It can be seen that a body is stable if 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) tf M
coiniides with G, the body is said to be iust stnble
Figure 3 - 10 (c): Unstable position
Figure 3 - 1O: Forces on a floating body
MOMENT AND OVERTURNING MOMENT
Wedge of
Immersion
OF A FLOATING BODY:
W = weight of the body
Wedge of
emersion
BF = buoyant force (always equal to W for a floating bocly)
G = center of gravity of the body
Bo = center of buoyancy in the upright position
,
Figure 3 - 10 (a): Upright position
Figure 3 - 10 (b): Stable pbsition
(centroid of the displaced liquid)
!o' = center of buoyancy in the tilted position
YD = volume dispkiced
M = metacenter, lhe point of intersection between the line of action
of the buoyant force and the axis of the body
c = center of gravity of the wedges (imrnersion and emersion)
s = horizontal distance between the cg,s of the wedges
u = volume of the wedge of immersion
,0 = angle of tilting
MBo = distance from M to Bo
GBo = distance from G to Bo
MG = metacentric height, distance from M to G
V I
i!
92
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Metacentricheight, MG: MB"+ GB"
Eq.3-2'l
FLUID MECHANICS
CHAPTER THREE
E HYDRAUTICS
fotal Hydrostatic Force on Surfaces
Moment due to shifting of BF = moment
93
due to shifting of wedge
BF(z)=F(s)
BF=yVo
Use (-) rf G is above Bo
Use (+1 if G is below Bo
F=lz)
z=MBosin0
Notet M ts alwavs above 8"
fVoMBosin0=yas
VALUE OF MB"
l-he stability of the body depencis on the amount of the rrghtrng momenl
which in turn ls 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 weclge o' to shift to a new position o The moment due to the shifting of
the buoyant force BF(z) is must equal to moment due to wedge shift F(s)
vAtuE oF MBo
small values of 0, (0 . 0 or 0 = 0):
Volume of
Pitchtng
e
Wedge, volume = v
Figure 3 - 1l: Rectangular body
Wirterline Section
a body in the shape of a rectangular parallelepiped
length L as
inFigure3-11;
Volume of wedge,'u = Vz(B 2)t@
/
/2) tan OJt
Volume of wedge, o =
+LB2tan e
For small values of 0,
s. I B
CHAPTER THREE
94
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
MB,=
CHAPTER THREE
Total Hydrostatic Force on Surfaces
95
US
V, sin 0
MBn=
MB" =
FLUID MECHANICS
& HYDRAULICS
Vp sin 0
But for small values of 0, sin 0 * tan 0
#rs'
'vn
But $ tar t the moment of inertiaof the waterline section, 1
MB"=
Eq,,3-23
*
Note: This formula can be applied to any section.
since the metacentric height MG is dependent with MB", 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.
Centroid of wedge
MB.= ot
Vp sin0
Vo =
BDL
a = Vz(B / z)t(B / 2) tan }lL
where L is the length perpendicular to the figure
a= f LBztan0
MOMENT
Centroid of hiangle, i
The righting or overturning moment on a floating body is:
From geome try,i =
R.Azf or OM = W x = W (MG
sin9
3 *24
xl + x2 + x3
J
- : 9* (B/2)sec0+(B/Z)cos}
3
* = B ( 1 *.ore)= 4fr+tos2e)
6\cos0
s 2
-=Y=_tl
n(
s(t+.or2e)
6|
coso )
s=_t_l t+cor2 e)
3[ coso )
/ dl
cosO )
CHAPTER THREE
96
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
MB,=
FTUID MECHAN'CS
& HYDRAUIICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
(BDL) sin 0
LB'3 sin0 1 + cos2 e
Projection of
curved
: 24 cosO cosO
^/R
internal pressure and this is to be
sted by T which is the total skess of
BDL sin e
MB-= B' l+cos2o
24D
MB,,--
pipe wall.
e
".r.2
B' l,-a-.r)
24D l..or2 e
Applying equilibrium condition;
'J
[>Fs = o]
F=27
,U"= !-(sec20 + 1;
24D'
p= pA=pDS
butbec2e=1+tan2e
.
,r" = !-[ "(1 + tan2 e) + U
24D
pDs=2x[Sr(sxf)]
B' (2 +
Bt * tu"t e )
MB, =
12(2\D tan2e) = .t2D l?
\2 2 |
MB^=
B' [, * tut" e)
12Dt
2 )
T = SrA*arr
I=Sr(sxf)
Eq'3-25
determine the longitudinal stress, let us
the cylinder across its length as shown.
lEFs = ol
F=T
F=pA
F=p f,Dz
STRESS ON THIN.WALLED PRESSURE VESSELS
THIN-WAILED CYLINDRICAL TAN K
A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile
forces, which resist bursting, developed across longifudinal and transverse
sections.
T = S1 A*";1
,
Anut= fiDt
T = StrDt
p +Dr=SrrDt
lst
Sr
*-fit-t'
\ s.
97
CHAPTER THREE
98
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
FIUID MECHANrcS
! HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
SPHERICAL SHELL
lf a spherical tank of diameter D ancl thickness I contains gas under a pressure
of p, the stress at the wall can be expressed as:
3-1
vertical rectangular plane of height d and base b is submerged,in a liquid
its top edge at the liquid surface. Determine the total force F acting on
side and its location from the liquid surface.
walsrress, s=+
4t
Eq.z-29
p=yhA
i=a1z
A=bd
F = y(d/z)(bd)
F =lz"y b iP
SPACING OF HOOPS OF A WOOD STAVE PIPE
I.
Ai
' v =i =a1z
*ua3
LZ
(bd)(d / 2)
e=d/6
Yr= i *,
Pressure diagram
(triangular prism)
yP= d/2+ d/6
Vp= 2iU3
spacing,
f ?t#
where:
Si : allowable tensile stress of the hoop
A1, = cross-s€ctional area of the hoop
p = internal pressure in the pipe
D = diameter of the pipe
Eq.3- 3o
Using the pressure diagram:
F = Volume of pressure diagram
F =1/z(Yd)(d)(b) =1/2yb dz
The location of F is at the centroid of the pressure diagram.
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 I/3 of its
height.
I
oo ftril;l.H[.'[
Force on surraces
FLUID MECHANICS
& HYDRAUTICS
Problem 3 - 2
A vertical triangular surface of height d and horizontal base width b is
FLUID MECHANICS
CHAPTER THREE
6. HYDTIAULICS
Total Hydrostatic Force on Surfaces
tot
Solution
r=via
sul'rmet'ged in a liquid with its vertex at the liquid surface. Determine the total
forc* f acting on one side a'd its locafion from the liquid surface.
F = y(r)(rc rz)
F=nyF
Solution
t.
F=yh A
= +a
6
i
AV
1 _-.4
p= !!__
A = l/zbd l
(nr')(r\
yp=r+e
yr= r + rf4
F=y*3dx1/zbd
F -- | ybdz
=y14
yn = 5r/4
a=
I
5
Using the pressure diagram for this case
Ai
n= i = za/s
('--
showrl its volume
is quiet complicated, with the
b"
computed by in'tegration. Hence, pressure
"u.,
m is easy to use only if the area is rectangular,
!ua3
with one side horizontal.
-1b
Ghd)(zd / 3)
e = d/12
3-4
Pressure diagram
(pyramid)
ry,=l-r
vr= 1d+d/12=3d/4
Using the pressure diagram.
f = Volume of pressure diagram
vertical rectangurar gate 1.5 m wide and 3
Fini u," t.iJ ;;:;," u. ti., g
TJ'",T:
:,..,n
:i"l::."q:
.,,,,ruce. .
one side of
the igate
-y;r:u, from
and its location
the bottom
r=yia
i =t.S+2=3.5m
F= +Ar,","^ height
f'= {tt'"yd\et\= tryua,
f rs located at the centroid of the diagram, which is % of the altitude
F = e.81(3.5)[(1.bX3)]
F = 154.51 kN
from tlre base
e-
Problem 3 - 3
A vertical circular gate or radius r is submerged in a liquid with itd top edgerl
flushed on the liquid sru'face. Determine the magnitude and. Iocation of tlrt,
total force acting on one side of the gate
m high is submerged ih water
t"6
AV
e=
(1.5 x 3)(3.s)
Y=1.5-e
y=1.5-0.214
y =1.286m
= 0.2L4 m
1.5 m
t
02 ;i*i;;H:'i::l: Force on surraces
FLUID MECHANICS
& HYDRAULICS
FTUID MECHANICS
r=yiA
F = (5Y !2Y ,. e) 1r.sy
F = 1,5.75y
F = 1s.7s(e.81)
F = 154.51 kN
Totat Hydrostatic Force
Solution
Using the pressure diagram:
F = Volume of pressure diagram
l\-/
CHAPTER THREE
& HYDRAULICS
h =2+ *fsl
i =zm= g
''
: [e.81(0.82)] (3)t%(1.5X3)
F = 54.3 kN
l"
J
*(i.sxs)3
Ay t+(r.sxs)l(a)
e = 0.1,67 m
yr=i +e
y,= 3JL57 m from the oil surface
Pressure diagram
(trapezoidbl prism)
Location of F:
3 - 6 (CE Board May 1994)
h=2y(3)=6y
Az=lz(3y)(3): a.5y
A = At + Az= 1,0.5y
[Ay = zay]
1,0.5y y = 6y(1.5) + a.5y(1)
y =7'286m
(much complicated to get than using the formula)
Fo.ce on upper half:
Fs=y,i A
Fo = (y,, x 0.8)(d/ +)[b(d / z))
Fo=0.1y,,bd3
Problem 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. of 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.
Force on lower half:
Fw= pcgzx A
Pcg2=fo14* Yruh,,,
P,e2 = (f ,u x 0.8)(d/2) + y",(d/ 4)
P,sz = 0.65 Y, d
F6, = (0.65 y,,, d)[b(d/ 2)]
Fw = 0.325 Tu b d2
Ratio =
Ratio =
Fw
Fo
o'225v,b72
0.1,y,ubd2
= s.zs
#";;.;
I 03
CHAPTER
l04 Total
Hydrostatic Force on Surfaces
THREE
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
CHAPTER THREE
q HYDRAUL'CS
Total Hydrostatic Force on Surfaces
r05
Problem 3 - 7 (CE Board May 1994)
Problem 3 - 8 (CE Board May 1992)
A vertical circular gate in a tunn€l 8 m in diameter has oil (sp. g. 0.8) on one
side and air on the other side. If oil is 12 m above the invert and the air
pressure is 40 kPa, where will a single support be located (above the invert of
the tunnel) to hold the gate in position?
2 m in diameter and m deep with axis verticar
contains 6 m deep of o'(sp gr.= 0.g) The
air above theiiquid surface has a
pressure of 0.8 kg/cm2. Determine the total
norrirar force in tg
on the
wall at its location from the bottom of the tank.
".u"g
Solution
Solution
tt
A closed cylindricar tank
g
^^
I I
oir;s=o'8
I'i
12m
tlI .i -,- *
Air;P=46gPu
I
* -'*qt -
F.i,
-.-.-.-.-8m@
4-y
4m
Fe1 =yq1; hA
F"rr = (9.81 x 0.80)(8)
Ft=Pu*A
* f (8),
Pair = 0.8 kg/ cmz = 8,000 kg/ m,
Fr = 8,000(2n x 2) = 32,000n kg
6 +'1, =7 m
F.l = 3,156 kN
e=
I-
h=
AV
Fz= prcA
E
e=
,l"l
+(8)4
b4\_,
p., = (1000 x 0.8)(3) + 8,000
p,e= 10,400kg/m,
=0.5m
f (8)'(8)
z=4-e=3.5m
Fz ='1,0,400(2r x 6) ='124,800n kg
Solve for e:
Fair=pauA,=40t t(8),
F"i' = 2,011 kN
F2=y.i A
724,800n = (1000 x 0.8) /i (2n
The support must be located at point O where the moment due to Fuo
and Foir is zero. Since Fou ) Fai,, O must be below F"l.
1.s6e(3.s -y)=4-V
5.493-1'.569Y=4-Y
V =2'52m
" 61
-_ I, _ $12n11013
AV
[>Mo = o]
Foir(z-!)=F^t(4-y)
(3,1s6X3.5 - y) = 2,011.(4 - y)
h = v =13m
(2n x 6)(13)
b = 0.23077 m
!
yz=3-e=217m
F = Fr + Fz = 155,8002r kg
) Total normal force
u
I
t
06 ;i.:il;:.T*:l: Force on surraces
Fy=Fryt+Fzy)
(156,800n) y = (32,000n) (n + 924,800n) (2.77)
) Location of F from the bottom
3t = 3.63 m
Using the pressure diagram:
FLUID MECHANICS
& HYDRAULICS
.
FLUID MECHANICS
& HYDRAUTICS
CHAPTER THREE
t07
Total Hydrostatic Force on Surfaces
Solution
[XMr,i"e"=0]
Fz=a}e)
r=yhA=9.8th (1)(1.5)
F =M.7tI5h
Ir
(--- ---:= Wnefe l/
Ay
=i
e= #(r.s)(r)3
1
(1.5 x 1)/r
12h
6
1
z=0.5+e=0.5+
800(6) = 4800 8000
Pressure Diagram
12h
1
\
Pl = 8000(8)(2n) = 128,000n kg
P2= 1/2,(4,800)(6)(2n) = 28,800n kg
\
-/
't4.7'L5h
lo.s+_]:l=+o
1.2h )
0.5i + 0.08333 :2.718
i =S.ZZm= l, +0.5= 5.77m ) critical water depth
P = Pr + P2= 156,800nkg ) Total normal force
IP
y: Pt yt + Pzyzf
(156,800n) y = (128,000n)( ) + (28,800n)(2)
1/ = 3.63
m ) Location of P from the bottom
3-10
vertical circular gate is submerged in a liquid
so that its top edge is flushed
th the liquid surface. Find thJ rafio of the total
force acting on the lower
to that acting on the upper half.
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
1.5 m
Rauo =
I
0.5756r
F1
Ratio =
("\
yht A,t
"34a
At=Az
i
Ratio =
I
h1
; .. 1..424r =2.42s
r(ano=
LS7S6,
x-.
x
x=4r/
I
oB ;Hi;;:#[.'i: Force on surraces
FLUID MECHANICS
& HYDRAULICS
Problem 3 - 11
FLUID MECHANICS
& HYDRAULICS
I otal Hydrostatic
Solution
A 30 m long dam retains 9 m of
L=30m
p=yi A
water as shown in the figure. Find
the total resultant force acting on the
dam and the location of the clnter of
pressure from the bottom.
i =z.s+21s
h = 4.767 m
A = %(1)(2.61)
A = 1.305 m2
| = (?s_t!" 0.ss)(4 ftn (t .305)
L = M,277N
F =44.277kN
SoluUon
p=yhA
F = e.81 (4.s) [(30)(10.3e2)]
F = L3,753 kN
L=30m
e=
3-13
inclined, circular
with water on one
is shown in the
I^
"
Ay
Determine the
(30- x 10.392)(4.5
resultant force
/ sin 60')
2m
I
on the gate.
e =1.732m
o
!=lz(10.392)-1.j32
y = 3.464ln
o
o
o
y= +(10.3e2)=3.464m
Problem 3 - 12
The isosceles hiangle gate shown
.yEA
weighs 1500 N. What is the total
hydrostatic forceacting on one side
of fhe gate in kiloNewton?
h =2+ 0.5sin60.
h = 2.433
in the figure is hinged at A and
.9.81,(2.433)t
18.746 kN
Oi 1s,=rg.8a),;
$),
CHAPTER THREE
Force on Surfaces
t09
CHAPTER THREE
I to
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Problem 3 - 14
'llrr.
1',ate in tlre figure shown is 1.5 m wide, hinged at point A, and rests
ag,ainst a smooth wall at B. Compute (a) the total force on the gate due to
scawater, (b) the reaction at B, and (c) the reaction at hinge A. Neglect tl-re
wcight of the gate.
ts
A,
(b)
CHAPTER THREE
fotal Hydrostatic Force on Surfaces
#(1.5X3.6)3
(1.5x3.6)(7.21)
e=0.15m
x=1.8-0.15
x=1.65m
[>Ma - 0]
F(r)-Rr(2)=0
218.25(1.65) = 2 RB
Ra = 180 kN
Seawater
s = 1.03
l
[:Fs = 0]
Rar,+Fsin0-Rs=Q
o
I
I
I
J5m
rrr
1
I
o
9
tgp"
Rar, = L80 - 278.25 sin 33'69'
)
.
Rar, = 58.94 kN
[I F,, = 0]
Rau-Pcos0:0
Ra, = 218.25 cos 33.69"
Ra" = 181.6 kN
t*Ro,
(1S1.6)2 + (58.94)2
Solution
d2=32+22
d=3.6rn
tan0 = 2/3
0 = 33.69"
1l =-
i4fiiqr:i?:kle:
Determine the magnitude
lnd location of the total
hydrostatic force acting on
the2mx4mgateshown
ln the figure.
m
1.5 m
Water
i
" sin0
*4
l/ =
sin 33.69'
J = 7.21 rn
-
r=yi n
F: (e.81 x 1.03)(a)[(1.sX3.6)]
F = 218.25 kN
m
W3%/
ltl
\
Ia tI G
a
CHAPTERTHREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Solution
FTUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
I t3
Problem 3 - 16 (CE Novembe1t997)
Determine the magnitude of
m
1.5 m
Oil, s = 0.80
Wdter
m
the force on the inclined gate
1,5 m by 0.5 m shown in the
001. The tank of
lrrater is completely closed
fnd the pressure gage at the
of the tank reads
90,000 N/mz. Use 9,300
/cu. m. for water.
Figure 001
F = p,tA
Pcg= IYh + P
p,s= (9.8LxL 26)(3) + (e 81)(1.5) + (9.81x0.80)(1) + 32
P,g:9-l'645 kPa
f=91.645(2x4)
F = 733.16 kN
F = prrA
Solving for e:
Solve for i
and y :
r=yi a
733;1.6 = (9.81x1.26)h Qx a)
Pz-p,s=ylt
90000 - p,g= 9,800(2.G5)
p"g= 64030Pa
F = 64030 (0.5 x 1.5)
F = 48,022.5 N
i =7.414m
v =n /sin60"=7.4'14f sin60"
7 = 8'561m
Ie
#Q)$)u
Av
(2 x a)(8.561)
e = 0.156 m
z=2-e=1.844m
Therefore, F is located 1.844 from the bottom of the gate.
-L7
9l,T.tT*" in the figure is hinged at A and rests on a smooth floor at B.
g,,or 0.s2,;;;;o
,t"1t:r;:#"'-T":*fl1*lt
?t_f"i.:"r
re the hinge A. rhe air auove
"i"ish, "r, s
thelii1fJ;f*";;;r#;;;r#'"'ii'uo"
If the gate weighs 5 krrr, determin" .h"';;;;;"rir."
,1T":l*l:
.
to open it.
CHAPTER THREE
I I+
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDTTAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
I t5
Prcblem 3 - 18 (CE Board)
n pinp 20 mm in diameter are used for supporting flashboards at the crest
masdnry dams. Tests show that the yield point of iron to be 310 Mpa
fiber st'ess). Neglecting the dynamic effect of water on flashboards
cl assuming static conditions, what is the proper spacing, S, of the iron
pins,
that the flashboards 600 mm high will yield when wateiflows 150 mm deep
the top of the flashboards.
-:-
acl
F
Floor'
""ffi
B
Solution
P = p,rA
h = 0.45
Pts= Pau + t}Jtu
p,s : 7 + 9.81(0.S2)(2.56)
p,r = 27.59 kPa
P = 27.ss t(3X3)l
P = 248.34 kN
1.5 sin 45'
= 1.06 m
p=yhA
248.34= (9.81x0.82)[ (a. S)
l, =l+g.t't
h
sin 45'
V
i =485m
I
E_
At
3.43
sin 45"
rment capacity of one iron pin (20 mn A\
lFa= Mc/Il
# (3X3)''
(3 3)(4.85)
"
r = 0.155 m
r='l .5+a
r = 1.655 nr
I'MA = 0l
P(r) + W(1.06)- f (2.12) = 0
2.72F = 248.34(1 655) + 511 s61
r = 195.37 kN
310 =
&Qq4
M=243,473.43 N-mm
M= 0.24347kN-m
Itrtent caused by F (considering S m width of flashboard):
Mra'Fxy
P=yhA whereA=0.6S
F = 9.81(0.45)[0.6 S]
I I.
I IO
CHAPTERTHREE
Total Hydrostatic Force on surfaces
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
tl7
Problem 3 - 20
| -- 2.6495
v=03-,'
,At 20 'C, gage A in the figure reads 290 kPa absolute. The tank is 2 m wide
' perpendicular to the figure. Assume atmospheric pressure to be 1 bar. Sp. gr.
' of mercury = 13.6. Determine the total pressure acting on side CD.
,,=',
- #(sxo'6)j
Av (0 s)(0 4s)
6
r-
e=0067m.
y=03 -0067 =0233m
1m
Mt=f ,y=M
.l,
2.6495"0.233=0.24347
1
h
S=0394m=394mm
J
Problem 3 - 19
70 cm
l-he semi-circular gate shown
rn Figure 28 ts hinged at B
Gage A
Determtne the iorce F required
to hold the gate rn Position
for h:
Solution
4ft
L
i =f = lo-l6elt
i =,t =B3o2fl
pt=2yhtptop
290 = (e.81 x 13.6)(0.70) + (9.81)h + 17s
h=2.2m
force on side CD: (Note: 1 bar = 100 kPa)
1:=yh A
P = 62.4(8.302)lt/znl4l 2l
I'] = 13,01 9 89 lbs
r 175 - 100
r 75 kPa
r 9.81(2.9)
,=L
nt
n \.
:28.449kPa
lx=01098'I
ir = 0 1098(4)'
/' = 28 11 fto
=-
28.11
\ "(4\' (8 302)
'Pr(3.e)(2)
.75(3.e)(2\
4ft
l_
,'= itzsz t,
h -- 1 698 - O 1347 = 1 5633 fl
[IMr, = 0l
P(b) = r(4)
13,019.89(i 5633) = t(4)
t = 5088.5|bs
r 585 kN
l1m
l
t
2.2
l_
0.7 m
tr/zpz(2.9)(2)
tVz(28.aa\(2.e)(2)
r E2.5 kN
Fr+Fz
667.5 kN
lI
IE
ldt
E r"i
qr
NI
2m
CHAPTER THREE
r 18
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Problem 3'21
I'he funnel shown in the figure is
full of water. The volume of the
CHAPTER THREE
TLUID MECHANIES
Total Hydrosiatic Force on Surfaces
HYDRAULICS
ll9
F=yhA
F=9.81(2.6X1.6x1.2)
F = 48.97 kN
upper part is 90 liters and the
lower part is 74 liters. What is the
force tending to Push the Plug
I"
e= --L
AY
t'lu t?
1.2(1..6)3
'12
g=
'
(1.6x1.2)(2.6)
1.6 m
a = 0.082 m
7=0.8-e
z = 0.718 m
= 460 cm2
totlrl.lll
rn" plug area in contact with water is horizontal, the pressure all over
tht'
it is uniform. The shape of the container does not affect the pressure on
Plug
Force=p^A
Force = e,810(3XiS )
Force = L353.78 N
Problem 3'22
ln the figure shown, the gate AB
rotates about an axis through B
The gate width is 1.2 meters. A
T=Fxz
T*48.97x0.718
T = 35.16 kN-m
3 - 23 (CE Board)
box, 1.5 m on each edge, has its base horizontal and is half-filled
water. The remainder of the box is filled with air under a gage pressure
kPa. One of he vertical sides is hinged at the top and is free to swing
To what depth can the top of this box be submerged in an open body
water without allowing any wate-r to enter?
torque T is applied to the shaft
through B. Determine the torque T
-T
to keep the gate closed
I
1.5'm
0.25
9.81(0.75) 82 kPa
1.5mx1.5m
= 7.36 kPa
,l
. -A
I ZU
CHAPTER THREE
rotal Hydrostatic Force on surfaces
[I Mh'^g" = 0l
Fr (r)- Fr (0.75)- Fz(1.2S; = s
FLUID MECHANICS
& HYDRAULICS
) Eq. (1)
Ft = Pn,A
F, = 82[(1.5)(1.s)l = 184.5 kN
FLUID MECHAN'CS
& HYDRAULICS
rotar Hydrostatic
Ftoblem 3 - 24
the magnitude and location of
force exerted by water on one
of the vertical annular disk
n.
Fz
4m
= Vz(7 .361 (0. 75) (1 .5)
-
F: = 4.14 kN
l_
Pr=Yh n
Fr = e.81 ii ttr.sxr.s)l
Ft = 22.07i
t = 075 + c
l'5(1'5'l
t"o_
I
1.2
AV
P=:
t(15X15)1l?
0.'t875
-
,=
- Av
'r - f(1.s)4 -f(1)4
d(Gf11)r(4)
--:'r=0.75+ 0.1875
h
e = 0.203 m
tn Equation (1):
22.07i (OZS+
'Ir
p=yi.A
F = e.81,$)[n(1.5)'- n(l)z]
F = 154.1 kN
Location of F:
h
0'187s
) - 184.s(0.75) - 4 14(1.25) = 0
Ap = 4 + 0.203 = 4.203
m below the w.s.
16.55t + 4.138 - 138.375 - 5.'t75 = 0
16.55 t =:.3.s412
i =s.szm
h= h -0.75 :
h=7.67m
CHAPTER THRFF
rorclPJf:#fi: I Zl
,gate in the figure shown
5 kN for each meter
to the paper. Its center of
is 0.5 m from the left face
0,6 m above the lower face.
h for the gate just to come
the vertical position.
=1.5m
| ,1
I zz
CHAPTER THREE
FTUID MECHANICS
6. HYDRAULICS
Total Hydrostatic Force on Surfaces
Solution
,Solution
Considering 1 m length
Fr = 4.905 h2 kN
.
p=vy
0.6 I
el
dA* 2x dy
fw=srr,t
h
By squared property of parabola:
*'=2'
cg
[tMo = o]
F{h / 3) + w (0.6) - Fz(1.5 / 2) = 0
4.e05t* (h/3) + s(0.6) - 14.71sh (0.75) = 0
1..6351f -17.04h + 3 = 0
CHAPTER THREE
Total Hydrostatic Force on Surfaces
dF=pdA
F'r =Vz (9.81/,XhX1)
F, :9.S1h(1.sX1)
Fz = 14.715h kN
FLUID MECHANICS
& HYDRAULICS
vq
ty
2Jin
x2=
9.81h
9.81 h.
x-
Solve h by trial and error
h= 0.2748m
eJi /3 ) dvj
dF = yy 12
dF = 2.31yy3/2 dy
F3
Problem 3 - 26
In Problem 3 - 25, find h when the force against the ,'stop,, is a maximum.
p, =221y [r',,n,
00
Solution
F = 2.31y
[:Mo = 0]
F{h/ 3) + w (0.6) + p(1.5)- pz(1.5 / 2) = 0
4.e05t* (h/3) + s(0.6) + p(1.s)
-14.775h (0.7s) = 0
P = L.09h3 -7.358h + 3
'liV,!l'| = 2.31(9.81)t lzsrz - osrzl
L"
l
Jo
F = 141.3 kN
Location:
3
# = 3.27 rp- 7.3s8 = o
F yp=
IP= 2.25
/ =1.5m
[,rn,
0
3
1.41.3
Problem 3 - 27
y,=
_0
Determine the force due to water
[rbtt rttzav)
.1
acting on one side and its location on
the parabolic gate shown using
integration.
yr= 0."1604 lvu/'a,
J0
l*--,---- o
*"
y,=0.1504lQtnrr,,
I
)'o
y, = 0.1604 (217) 137/2 - 07/2 |
yo= 2.14 m below the w.s.
t23
t
24 ;liil;:.:H:,? ,o,..
on surfaces
FLUID MECHANICS
6( HYDRAULICS
FLUID MECHANICS
CHAPTER THREE
& HYDRAUI.ICS
Total Hydrostatic Force on Surfaces
125
Problem 3 - 28
Problem 3 - 29 (CE Board)
ln the figure shown, find the
width b of the concrete dam
necessary to prevent the dam
I d."t", tt triangular in cross-section with the upstream face vertical. water is
dam is 8 * hth and 6 m wide at the base and
$f,:O ln1th the top. -Themeter.
'weighs 2.4 tons per cubic
The coefHcient of friction between the base
gmd the foundation is.0s. Determine (a) the maximum
and minimum unit
Pressure on the foundation, and the. (b) iactors of safety against overturning
lnd against sliding.
from sliding. The specific gravity
of concrete is 2.4 and the
coefficient of friction between the
base of the dam and the
founilation is 0.4. Use 1.5 as the
factor of safety against sliding. Is
sp.s' of conc, s6on. = Js9!l
the dam also safe from
fw
overturning ?
sp.st of conc, ,.on. = 33.IJ!oo = 2.4
Solution
1000
Consider 1 m length of dam
der L m length of dam
W,--1, V,
w, = yQ.a)l(bx6)(1)l
w=Yrv
= (yx2.4) [+fultrlAlJ
W,=1'4.44v
W = 57.6 y
r=yiA
F: y(2.25)t(a.5)(1)l
p-yiA
F = 10.125y
. y(4)(8 x 1)
- 3Zy
R'=f:10.125Y
Rv=W,
Rv=
-14.4
l.e.-
-W=57.6.1
frR,
- W(4)
-
- 57,6y(4)
- 230.4y
- P(8/3)
0.4(14.aby\
'
w.s.
-P=321
W
-R,
I-sc=
where y = unit wt. of water
10.125y
b=2.637 m
- 32y(8/3)
.85.33y
r.'oM
: 444
.,RM-oM
Rv
Dc _ wc(b /2)
I J^
'
-
FS" =
, tr30.4y - 85.33y =2.519m<B/2
F(i.s)
M.aQ.637\y(2.637 /2)
-
10.12sy(1.s)
.:3.3>1(Safe)
57.6y
t/2- i
l'2,5t9=0.481 m<B16
B/6 ='l m
B = 6m
THREE
t26 CHAPTER
Total Hydrostatic Force oh Surfaces
Rr(-
6e\
B\
B/
rl= _ jl
,
Ll = -
l+-1
57.6(e.87)
-
F-S,,
=
CHAPTER THREE
Total Hydrostatic Force on Surfaces
o1o.+s11 I
[. ------:-I
t:1t -- - 139.47 kPa
r7H=-48.88kPa
ir,lt,,
0.8(57.6y)
R.
?2y
" =
FLUID MECHANICS
& HYDRAULICS
Figure:
I r r
6L6J
Using (+).
Using (-)
FLUID MECHANICS
& HYDRAULICS
) soil pressure at the toe
) lsoil .pressure at the heel
F5,, = 1.44
RM
FS'= ona =
230.4t
85.33y
l'-5. = 2.?
Problem 3 - 30 (CE Board May 1992)
A gravity dam of traprezoidal cross-section with one face ,vertical and
horizontal base is 22 m high and has a thickness of 4 m at the top. Water
upstream stands 2 m below the crest of the dam The specific gravity o(
masonry is 2.4
A
Neglecting hydrostatic uplift:
1
Fincl the base width B of the ctam so that the resultant force will cut
the extreme edge of the middle third near the toe.
Compute the factors of safety against slicling and overturning
Use pt = 0.5.
Consiclering uplift pressure to vary uniformly from full hydrostatrt
pressure at the heel to zero at the toe:
1 Find the base width B of the clam so that the resultant force will act al
the extremity of the rniddle third near the toe.
Compute the maximum and.minimum compressive stresses achng
against the base of the dam
Neglecting hydrostatic uplift
L Consider 1 m length of dam
ll. Forces
r, = (, x 2.4)I(4)(22)(1)l
Wt = 211.2y
W.r = t,
Wz= (y x 2.4)Iv, (B-4)(20)(1)l
Wz= 24By - 96y
F=yhA=y(ro)t(20)(r)l
, F = 200to
lll. Reaction
&=If,=P
' R, =20Q
. Ry=IFv=Wr+Wz
:21"1.2Y+248Y-96y
Rr=248y +115.2y
127
THREE
t28 CHAPTER
Total Hydrostatic Force on Surfaces
TTUID MECHANICS
I HYDRAULICS
rotar Hydrost"ti. rc#fJ,E:J,?*:: I Zg
,4m
B= -
M.8t (M.q2 - 4$)(-1,4ss.ry
2(8)
B=r't..'r,7sm
Factors of Safety:
-
Factor ofsafety against slicling
FS.=uR.y
R,
(0.s)[24(1 1.1 75)1 + t15.Zy
.
)
200y
F',S. = 0.958s
Factor of
Uplift
pressure
diagram
lafety against overturnlng
RM
f 5. = -__
OM
_ t6(71.175)2 y + s3.2(tt.t75)y _ r66.4y
'
lV Monrent about the toe
RM = Wt(B - 2) * W.,t+ (B - 4)l
= 211.2y(B - 2) + (248y - e6y) [ 4 tS s))
= 277.28y - 422.4y + 168) - 1288y + /56y
RM = 1682y + 83.28y - 166.4y
oM = F(20/3)
= 200y(20/3\
t)M = 133333v
V
l.ocatron t-rf R
R.,i=RM_OM
-
1333.33v
FS" = 2.A7
Considering hydrostatic uplift:
Uplift force, U = th (Z0y)(B)(1) = t0Bzl
Rr=Wt+Wz-l)
= 24BY + 715.2y - 10By
Rr=148y+115.2y
RM=w@-2)+wzl4ts-sil
RM = 768) + 83.28y - 166.4y
.OM=FQT/s)+U(28/3)
= 200y(20/3) + 10By (28/3)
OM=6.6782y+1333.33y
Since the resultant force will pass through the extreme edge ol
the middle thirds near the toe, i = B/3 Then.
l24BY + 11s.2y)(B/3) = l6Bzy + 83.28y - 166 4y 1333.33r
j
8B2y + 33.49r ='l6B?t + 832h 14997?y
882+44.88-149973 =0
Rrf=RM-OM
(1.48y + "t't'.zy)(B
/ 3) = 1 6B2y
4.6682 + 44.88 - 1499.73
8=
=0
166.4y - (6 678) + 1333.33y)
- 44.8t l(M.q2 _ 49.66)(_14ss.73\
2(4.66\
B = 1.3.766 m
I 30
;i.:ili:.:H:[ Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
MECHANICS
HVDRAULICS
rotarHydrostuti.FTXPJ"EIJ.?*:: I 3 I
lioundation stress:
v =B/3
r = 13.766/3 = 4.59 m
r=B/2-i =2.2943nt
F =.y,"i A
* 9.81(3)(6 x 11
F
,,=
, Ir(rr99l
B\
B]
Rv = t4$s.206x9.81 ) + I 15.2(9.81 )
R,r = 3020.73 kN
3,020.731
'' - ' 1!766 " -
l''
W=Y'v''
= 23.s[2(8)(1)l
Wl=376kN
11766 )
= 23.5[vr(2)(B)(7)]
-
li
wl
i
= 188 kN
qr=-438.87kPa
r7H
x- {(o)=zm
Wz=y,Vt
O1Z.ZS+Z1I
2m
- 176.58 kN
: 0 kPa
t4-Yz(Z)=i
;
'(2/3)(2\ = 1.333 m
Problem 3 - 31 (CE Board May 2002)
;r=rzo.sstN
Ihe section of a concrete gravity
- W-r + Wz = 376 + lgta
* 564 kN
dam shown in the figure. The
depth of water at the.upstream side
is 6 m. Neglect hydrostatic uplift
and use unit weight of concrete
equal to 23.5 kN/mt. Coefficient of
friction between the base of the
dam and the foundation is 0.6.
Determine the following: (a) factor
of safety against sliding, (b) the
factor of safety against
bverturning, and (c) the
overturning moment acting against
the dam in kN-m
=
trIi,
R,
0.6(50qt
=ffi'l.elb
'Wtrt+Wzxz
5
= 376(3) + 188(1.333)
= 1378.604 kN-m
=F^y
= 1.76.58(2)
= 353.16
kN-m ) overturnlng moment
RM
OM
" 1378.604
35316 = 3.eo4
--*t
4m
132
CHAPTER THREE
FLUID MECHANICS
6. HYDRAULICS
Total Hydrostatic Force on Surfaces
Problem 3 - 32 (CE Board November 2OO1)
I|| n:*:lT:"T::*y
varies rino,,h,
r.^*
linearly from
as shown in the figure. Assume
maximum hydrostatic
uplift from the heel to zero at
re at the heel to zero al
the location of the drain,
ine the (a) location of
resultant force, (b) factor
safety against sliding if
:fficient of friction is 0.75,
factor of safety against
rning, (d) the stress at
heel and at the toe, and (e)
unit horizontal shearing
at the base
the toe Determine the total
reaction per unit length at the
base of the dam Use sp. gr of
concrete = 2 4
Solution
Consicler 1-foot length of dam
B,=P=y/rA
. 30' ,|. t0' 1
I
= 52.4(30)(60 , l)
R, = 112,320 tbs
Ru= LV1+Wz+Wt-U
I
I
10
i
\
i
I
I
I
I
60'
I
:;l1lr. i
Wz = t, Vt = (62,4 x Z.Q vzQal(40\(7\
l0
Wz
I
I
" Wz=l,Vt
w.= gz.4 " 2.ay 99'i91oy1r1
W'z = 44,928lbs
t'
,,60
I
)
wt = 224,640]bs
I
60'
4
Wq = 56,160lbs
U = y^, Vu = (62.4) Yz(60)(70)(1) = 131,040lbs
R,,= 224,640 + 44,928 +71,8848 + 56160 - 131,040
R,,=266,5728Ibs
R=
6,
I
Wt=y,Vt
Wq = tu, Vq = (62.4\ vz(30)(60)
duT
!/ (1 12,320\' , (266,5728\' = 289,269 tbs per foot
t33
l fown rhe specific weight of water is
is 23.54kN/#';5r:;;;"i,1;:,;rffi;
9.81 kN/mr and that of: concrete 1s
hydrostatic uplift to vary
urriformly from full hydrostatic
Wq = 71,884.8|bs
CHAPTER IHREE
Total Hydrostatic Force on Surfaces
Problem 3 - 33 (CE Board May 1986)
I'lre sectiorr of a gravity dam is
wt -- (62.4 ^ 2 4) ]q'je (bo)(t
FIUID MECHANICS
& HYDRAUTICS
-'
,1,
,l, Wg
7m
fl.52m
i -I '
I 5+
CHAPTER THREE
Total Hydrostatic Force on Surfaces
I
Consider 1 m length of dam
I
Forces
I
FLUID MECHANICS
& HYDRAULICS
FIUID MECHANICS
O HVDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
(c) Fs" = RM - 683,900'"12
oM
377,758
FS,, = 1.81
Wt=y, Vt
Wt = 23.54 [vz(5.2)(52)(1)] - 3,183 tN
w2= /t.$4 t(7X52)(1) = 8,56e kN
wt = 23.541v,(26)(52)(1\l = 1s,e]3 kN
Wa = e.81[rz(5xs0x1)] = 1,226.3 kN
u = vz(490.5)(%.2)(1\ = s,6e0 kN
r = yi A = e.8i (2s)t5o(1)l = 12,263 kN
(d) Foundalion pressure:
e=B/2-i
e = 38.2/2 - 13.2 = 5.9 m < B
,t=
, Yh tlel
B\
,.t
B/
- - 23,201.3f , * ots.gl
38.2 L
i,
lll Reaction
Rv=Wt+Wz+Wt+Wq-L)
= 3,183 + 8,569 + 15,913 + 7,226.3 - s,690
R'l = 23,201.3 kN
3s.2 l
qr = -1;1.70.2't kpa
Stress at the heel, (use ,,-,,)
qn =.-44.52kPa
lV Moment
RM = W t(34.73) +'vt1r129.5) + Ws(17 .33) + W4(36.53)
= 3,183 (34.7 3) + 8,569 (29.5) +'.r5,91 3 ('17 .33\ + 1,226.3(36.53)
RA4 = 683,900.12 kN-m
(e) Unit horizontal shearing stress, S.
S.= R'
-12,263
Aho"" 3g 2(1) = 321 kPa
oM=F(50/3)+U(30.47)
= "12,263(50/3) + 5,690(30.47)
OM = 377,758 kN-m
V Location of R,
11,,f=RM-OM
23,201.3 T = 583,900jt2 - 377,758
T =13.2m
lnl The resultant force is 13.2 nr from the toe
uR
v
R-,
L\
=
0.75Qg.207.3\
\
/
12,263
-]
Stress at the toe, (use ,,+,,;.
R, = F= 12263 kN
(h\ FS. =
/6
-1An
3-34
cubmerged curve AB is one
of a circle of radius 2 m
ls Iocated on the lower
of a tank as shown. The
of the tank perpendicular
sketch is 4 m. Find the
and location of the
and vertical
ts of the total force
on AB
I55
I A,
I 5()
CHAPTER THREE
Total Hydrostatic Force on surfaces
fLUID MECHANICS
HYDRAUTICS
Solutlon
way of solving t r
tn=yi A
FH = e.81(s)[(4)(2)]
Fp = 392.4 kN
t1
=l+p
unit pressure is always normal to the
and a normal to the circle passes
gh its. center, then the total force F
also pass through the center of the
O, hence the moment about O clue to
t"
6
Or due to Fa and Fy is zero.
Av
4e)3 / D
t(4x2)l(5)
e = 0.067 m
v=1+0.067
y = 1.067 m
4o=01
Fyi-Fsy=Q
437.13i =392.4(7.06n
7 = 0.9578 m
Note: This is true to all cylindrical or spherical sudaces.
Therefore; Fp is acting 1.067 m below B
Fy = Wcrglrtasslt
Fv = yVMsD,
V as(-D = 4(A)
A=At+Az
A'r=(4X2)=8m2
A2 = r7n T(2)2 = 3.14 m2
A=8+3.1,4
A ='l'1.'14 m2
V aa(D = 4("1"1 .14\ = 44 56 mr
Fv = 9.81(44.56)
3 - 35 (CE Board)
crest . gate shown
of a cylindrical
of which AB is the
supported by a
tural frame hinged at
The length of the gate
10 m. Compute the
and location of
horizontal and vertical
nents of the total
ure on AB.
Fr = 437.13 kN
Locahon if Fv
Ax =Arrt+Azx.t
rr=1m
41 4(21
t7
-
3n
-
3n
12 = 0.849 m
11.14 i =8(1) +3.14(0.849)
i = 0.957 m
Therefore; Fv is acting 0.957 to the right of .4
E
(o
q
€O
il
o
io
c
o
L=10m
CHAPTERTHREE
fI J2o(,
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
FIUID MECHANICS
& HYDRAULICS
lolution
l'11=yh A
I: r r = 9.81(4.33) [10(8.66)]
Considering 1 meter length:
tir = 3679 kN
y=
Fs=yh A
Fu=9.81(3)(6x1)
FH = L76.58 KN
\ p.oo1= 2.887 m
Therefore; Fs is acting 2.887 m above O
Fv =
Fv=^lV,
As=Asetor-Atnangre.
I Vasc
Vasg=Vn1sg-V16s
vasc= *(8.65) x t0 -y2(70)2 [00"6]
Vasg = 125.9 m3
Fv -- 9.81.('].25.90)
Fv = 1235 kN
CHAPTER THREE
"
o,:
-vz(elzsin 6oo
" ^tu]]{9')
360.
to
A'=J./$12
'
fu= 9.81(3.26x1)
Fv = 31.98 KN
F= Jir, *pr,
Moment about O due to Fa and fy = Q
Fv (x) = Fu (y)
7235 x = 3679(2.877)
.t 8.57 m
F = 179.45 KN
:
3-37
Therefore; Fv is acting 8.57 m to from O
Problem 3 - 36 (CE May 1999)
Calculate the magnitude of the
the magnitude of the
Itant pressure on a 1-ft-wide strip
I semicircular taintor gate shown
Flgure-12.
Q
resultant force per meter length due
to water acting on the radial tainter
gate shown in Figure 021.
Ftr = p,r A
Pr, = (62.4x 2.5)(5 x 1) = 736 16t
Fv - f VaBc
Fv - 62.4 x
.
-A
rotarHydrostaticForce;"1"##; 139
If (s)r(1)] = 1225tbs
fr
(Fr)2 +(Fy)z
fr
(780)2 + (7225\2 = 1452lbs
I 40
;if,';;:,X[ff Force on surraces
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Problem 3'38
Problem 3 - 39 (CE Board November 1993)
l)etermine the magmtude 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
diameter Qylinder, 1.20 m long is
acted upon by water on the left and
oil having sp. gr. of 0.80 on the right.
Determine the components of the
reaction at B if the cylinder iveighs
s'hown
19.62 kN.
6olution
fm=yh A
Fm =9.81(1..2)(1.2 x 1.2)
Fnr = 16.95 kN
Solution
'
Fvr= TVt
Fw, = 9.81.V/z n (0.6)2(1,.2)l
Fvr = 5.557 kN
Fuz=yh A= (9.81x 0.8)(0.6)(1.2 x 1.2)
Fsz = 6.78 kN
Fvz= yVz= (9.81 x 0.8)11/zn (0.6)2(1.2)l
Fvz = 5.32 kN
Fu=yh A
r+=01
Fm - Fuz- RsH = 0
FH = e.81(3)[(1X2)]
Fs = 58.85 kN
RsH=16.95-6j8
RsH = 10.17kN
l,,Trl,i:'-t1
-^'ffi^
[tFv = 0]
Rsv+Fy1 +Fvz-W=0
Rsv=19.62-6.657 -5.32
Rrv = 7.64 kN
Fv=fVol
Fv
= e.8U4(2) (1) + 0.7sln(2),(1)l
Fv = 170.94 kN
l4l
. ,^
I+Z
CHAPTERTHREE
rotal Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
FLUTD MECHANICS
& HYDRAULICS
Problem 3 - 40
P' = (9.81 x 0.82)(0.00628)
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
Fz = 0.0505 kN
rD- -F ,-7- f1
sp. gr. of 0.82. Determine the total vertical force acting on the plug.
F"=0.11,4-0.0505
Solution
r--
o+m
F
CHAPTER THREE
Total Hydrostatic Force on Surfaces
143
N
F" = 0.0635 kN - 63.5 N downward
--11
3-4L
2 m diameter horizontal cylinder 2 m long plugs a 1m by 2m rectangular
at the bottom of a tank. With what force is the cylinder pressed against
bottom of the tank due to the 4-m depth of water?
ll
ution
hr=2x(1cos30')
h=1.732ll:.
hz=4-hr
hz=2.268m
F.t = yVr
Vt=Atx2
Area, Ar = Area of rectangle DEFG - A4
Area of segment, o,=
Fr = TVt
n = (e.81 x 0.82)[n(0.1)'z(0.45)]
Fr = 0.114 kN
Fz-- YVz
V2 = Vrrrrtu- - Vcyltnd..
% = '(0u15) ft0.21' * (0.2)(0.1) + (0.1)'] - n(0.1)'z(o.r5j
Vz = 0.00628 m3
^2;#0",
Area of segm ent, Aq= 0.09059 m2
Area, A1= 1'(2.268) - 0.09059
Area,Al =2:1,774m2
Vr= 2.1774(2) = 4.355 mg
Fr = 9.81(4.355)
Fr - 42.72 kN
Fz= Fs=yVz
Vz= Azx 2
-y,(I)(l)sin 60o
I 44 fi*i,];H:*:i:ForceonSurraces
Area of segment, A2 -
r(1)2
FLUID MECHANICS
& HYDRAULICS
!l?0") - %e)g)sin
120o
'-r
360"
Area of segment, A2= 0.614nf
Vz= 0.674(2) = 1..228 m3
Fz= h= 9.81(1..228)
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
F,1=yh A
F u = e.81 (6.12)t9.24) (11
Fr = 254.56 kN
Fv = Y %roa"a
Fz = Fs = 12.05 kN
V"nua"a = (Asemtcircle 1- A1r.rp"ro14) x 1
%nua"a
Net force = Fr - Fz - Fl
Net force = 42.72 - 12.05 - 12..05
Net force = 18.62 kN
: Ltn(s)t * uy 1+..i+lg4
%hu,i"a = 40'1 m3
Fv = 9.81(40.7)
Fv: 393.38 kN
Problem 3 - 42
In
the figure shown,
determine the horizontal and
Problem 3 - 43
vertical components of thq
guarter circle 2.5 m
wide. Find the force F
gate shown is a
total force acting on the
cylinder per m of ib length.
sufficient to prevent
n about hinge B.
the weight of
gate.
Solution
m
c ''!
all
fp=yE,A
Fn=9.S1(1)(2.5x2)
Fu = 49.05 kN
Fv = f Vasc
F v = 9.81[(2 x 2) - 0.25n(2)z](2..s)
Fv = 21.05 kN
2.5m
t45
146
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
Solve for z and x.
Since the surface is circular, LMo -- 0 due to Fri and Fr
Total Hydrostatic Force on Surfaces
Forces due to oil:
Fuo = preo A
Fv (z) = Fu (2/3)
Fno = (9.81 x 0.80)(7 1.273) *
Vrn(3)2
Fso : 635.4 kN
2-1.05(z) = 4e.05(2/3)
z=1.55m
x=2-z=2-1.55
Fvo : Tn V,,
r=0.45m
V, = Volume of imaginary oil above
the surface
Z, = Volume of halfiylincler _ Volume of
[tMa = o]
Fu (2/3) + Fv (x) - F(2) = 0
% spher.e
V" = Vzn(3)z(7) _ 1/<
2F = ae.05(2/ 3) + 21.05(0.45)
n(z)z
u!
V" = 70.686 m3
Fvo = (9.81 x 0.80)(70.636)
Fvo = 554.74 kN
F = 21.09 kN
Problem 3 - 44
he cylindrical tank showr-r has a
CHAPTER THREE
6( HYDRAUTICS
Forces due to water:
flt=p,e*A
hemispherical end
cap.
Compute the horizontal and
jw= \(!,st x 0 8)(7) + s.8r.(1..273)l xrzn(3)z
\Fr*v = 953.19 kN
vertical components of the total
force due to oil and water acting
Fvr.ri = Weight of real and
on the hemisphere
imaginary oil above the surface
+ weight of real water above
Fvw= (9.87 x 0.8)x Vzn(3)r(7) + 9.81 x
the surface
%! ne)3
Fvw = 1,054.07
Solution
Total horizontal force, FH = FH() + FHry
lotal horizontal force, Fu = 635.4+ SSS.f
S
Total horizontal force, Fs = 1,5gg.59
kN
,
'l'otal vertical force,
Fv = Fvw -
lvo
Total vertical force, Fv = 1,054.01 _ 554.74
Total'vertical force, Fv = 499.27 kN
.
Another wray to solve for the total ,r".ti"ufi..",
Fy = weight of fluid within
the hemisphere
Fv=ToVnty*V.
fy = (e.81x0.4)tix n (3)u)l + e.81[ix
f
ln (3),)]
I:v = 499.27 kN
Fu,
147
I
+8 ;tril;il:fiff
Force on surraces
FLUID MECHANICS
& HYDRAULICS
Problem 3 - 45
Pressurizecl water fills the tank shown in the figure Compute the net
hydrostatic force acting on the hemispherical surface.
FLUID MECHANICS
6. HYDRAULICS
CHAPTER THREE
fotal Hydrostatic Force on Surfaces
149
Problem 3 - 46
Determine the force required to
open the quarter-cyri.der gate shown
The
weight of the gate is so [N u.ti"g
i;-_ ioi# ,r*n, o, o
Hemispherical
suface
-T
I
l=
l")
l-
Solution
Convert 100 kPa to its
equivalent pressure
_t
head, fi.,,
h"u=
p
t the gate has circular surface,
to-tal water pressure passes
gh point O which is alio the
v
100
,
nP,' =
' 9.81
of the hinge therefore the
-
due to water pressure
h.q = 10 194 m
the hinge is zero
h=10.794-5
-01
h -- 5."194 m
F(Z.S;=s0(1.2)+Fr(g)
F-24kN
F = Weight of imaginary water.above
the hemispherrcal surface
F = yu, Vu,
V,,, = Volume of cylincler + Volume of hemisphere
V,,, = n(2)2(5.'194\
+L
"
nispherical dome shown is filled with
oil G = 0.9) and is attached to the
by eight diametrically opposed bolts.
What f
in each bolt is required
a ,h"?.-"l.-",?in"
dome weigh" 50 kN?ot""
t n(2)3
V^ = 82.025 m3
/. = 9.81(82.025)
F = 804.7 kN
3-47
,
I 50
;ifi;;I,.:H:[ Force on surraces
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
fotal Hydrostatic Force on Surfaces
l5r
Solution
[Erv=o]
-20 kPa
F + Foir - F.,r ="0
F=Foir-Fo,,
Foil = Y Vorl at ove rhe curre
F"ir = (9.81x0.S)[r(0.805)z(5)
+ n(0.s0s)r(3)l
F"ir = 63.91 kN
Fou = Pu* A
F"i' = 20 I+ Q.61), J = 40.72 kN
I F.,r
i,t
il
r.or mJ
F=63.9't-40.72
Solution
'
F = 23.19 kN
Fy = Y Vi.ug,na11, orl above rhe rtome
Fy = (e.81x0. 9)ln(2)2(8) - I n1zf1
Fr.
3-49
: 739.66 kN
8F6n11 +
W=Fv
r_
739.66 - 50
Th,|ll -
300 mm diameter steel pipe 12 mm thick carries water
water. Determine the stress in the steel.
under a head of 50 m
8
Fr.rr = 86.2 kN
l5r='
'
I
2t'
- 9.81(50)(300)
' " / =6131.25kpa
)/=
Problem 3 - 48
Determine the force F
required 'to hold the cone
DD
2(12)
,
s,= 6'13 MPa
shown. Neglect the weight ot
the cone
3-50
the required thickness of a 450 mm diameter steel
pressure of 5500 kPa if the allowable working
[5,= PD t
2!'
124 x.tooo
_ p(qso)
2t
t = 9.98 mm say 10 mm
pipe to carry a
stress of steer is 124 Mpa.
I F-t
I )Z
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
6. HYDRAULICS
Problem 3 - 51
l)etermine the stress at the walls of a 200 mrn diameter pipe, l0 mm thick.
uncler a pressure of .150 m of water and submerged to a aepir, or 20 m in salr
FTUID MECHANICS
CHAPTER THREE
& HYDRAULICS
Total Hydrostatic Force on Surfaces
r53
Pipe diametet, p =6 m = 6000 mm
Maximum pressure the tank (at bottom), p
p = 9.81(0.8)(7) = 54.936 kPa
water
= ^lnith
. - 2(110x103)(300)
Solution
s4.936(6000)
PD,
,,.=
rv'
2r
S = 200.23 run say 200 mm
I
p=prnsrrte-/oursrrle
p=e.81(1s0)-e.81(1.03)(20)
"
3-54
p = 7269.4 kPa = 1.269 Mpa
, , ="12.69MPa
\,= 1.2o9(200')
2(10)
thin-walled hallo.ry sphere 3.5 m in diameter holds
helium gas at 1700 kpa.
the minimum wall thickness of the sphere if
its allowable
stress is
MPa.
Problem 3 - 52
A 100-mm-lD steel pipe has a 6 mm wall thickness. For an allowable tensile
stress of 80 MPa, what maximum pressure can the pipe withstand?
Solution
Wall srress. S, = !2
4T
1,700(3.5 x 1000)
60,000 _
t=24.79mm
49r
rs,=
'
4t
l/'
80=
p(100)
3-55
2(6)
p = 9.6 MPa = 9,600 kPa
vertical cylindrical tank is 2 meters in diameter
and 3 meters high.
Its sides
rI
*."i":-.?j two steel hoops, one at the top,and the other
i*tl:^ryrtg.o:
the bottom' If the tank is filled with water t;;
d"p;h';f ii i.,,*d"*r*""
t tensile stress in
each hoop.
Problem 3 - 53
A wooden storage vat is 6 m in diameter hnd is filled with 7 m of oil, s = 0.g
Ihe 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?
Solution
Spacing ot hoops, 5 =
25,Ar,
pD,j
Allowable tensile stress of hobps, Sr = 110 Mpa
Cross-sectional area of hoops. At, = 50(6) = 300 mm2
2Tz
154
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
CHAPTERTHREE . FF
I 55
Total Hydrostatic Force on Surfaces
3-57
lI M6u = 0l
2T2(3) = F(2.3)
'Ir = 0.3833F ) Eq. (1)
cylindrical container 8 m high and 3 m in diameter is rein-forced with two
1 meter from each end. when it is filled with water, what is the tension
F=yiA
each hoop due to water?
F = e.s1(# tttzltz rtl
: 43.26 kN
ln Eq. (1)
T2 = 0.3833(43.26)
l"z = 16.58
kN (tension in the bottom hoop)
-T-----T-
tt
H
I rt
[tFH=o]
ZT2+27,=P
ZTt=P'",
8m 6m
7Tt = 43.26 - 2(16.58)
Ir = 5.05 kN (tension in the top hoop)
Problem 3 - 56
A vertical cylindrical tank, open at the top, is filled with a liquid. lts sides are
held in position by means of two steel hoops, one at the top and the other at
the bottom. Determine the raho of the stress in the upper hoop,to that in the
lower hoop
6m 8m
t.
t'"-l
I
p=yhA
F = e.81(8/2X8(3)l
F = 941.76 kN
[IMtop noop = 0]
272(6): F (13/3)
Solution
Raho=Tr/ Tz
[IM1.,o = 0l
2T2(h) -- F(zh/3)
Tz= F/3
[EMuouuu' = 0l
Tz = 73F /36
T2= 13(94L.76)/36
T-
Tz = 340.08 kN
h
[EMro,.o. r,oop = 0]
2T{6) = F(5/3)
l_
Tr = 5F / 36 = 5(e47.76) / 36
Tr = 130.8 kN
zrlh) = F(h/3)
Tt = Fl6
Raho=
3 - 58 (CE Board November t9B2)
F/u =o.u
F
drical tank with its axis vertical is 1 rneter in diameter and 6 m high. It
together by two steel hoops, one at the top and the other at the bottom.
r liquids A, B, and C having sp. gr. of 1,.0, 2.0, and 3.0, respectively fill this
-on
each having a depth of t.iO m.
the surface of A there is adospheric
Find the tensile stress in each hoop if each has a cross-sectional area
/3
2rz
mm2.
THREE
t56 CHAPTER
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Solution
CHAI'TER THREE
Total Hydrostatic Force on Surfaces
Tz= 3.6(9870) = 35316 N
2T, pr=0
Liquid A
+
1.6 m
FLUID MECHANICS
& HYDRAULICS
1.2 m
lrFn = ol
271 + 272= p', + Fz+ Ft+ F+ + Fs
2T1= 9.7r, + 1.44y +'t.44y + 4.32y + 2.L6y _
2(3.6y)
Tt = 7.44y
Tt ='1.44(9870)
Tr='1.4126.4N
Liquid B
s=2.0
iiquid C
s=3.0
'f^
2- 35316
A2 7250
Stress, Sz = 28.25 Mpa ) stress in bottom
hoop
Stress, g, =
1.2 m
stress,s,=A-74,126.4
A1 1.,250
Stress, Sr =L1.3 Mpa
Pt=0
) stress in top hoop
p2= pt + yAh^
Pz=0+ (y'1)(1 2\=1.2y
3-s9
pt=p?+y|lt!<
pt = 1 2y + (i'2)(1 2\ -- 3.6y
p4=p\+y(ltl
pq = 3.6y + (yx3)(1 2\ = 7 2y
f1 = t/2(p2l(1 2\("1)
Ft = t/z(12y1{d, 2)(11= 0.72t
F1 = pr( I 2)(1 )
Fz=12y(12)(1)=1.aay
F\ = t/2(p, pz)(12)(1)
1.86 m2
Fr= V:(3rrv t2y\(1 2\(l)= l44Y
0.93 m2
f t = pt("| .2t(1 t
F, = 3 6y(1 2'10) = a32t
Fs = Ltz(h p1)(1 2)(1)
Fq = t/z(7 2y . 3 6y'10 2)q't1 = 21Ur,,
[lM.p=0]
t
,
3.6(2T2)= F'(0.8) + Fz(].8) * F{2)+ Fa(3) + Fs(3.2)
7.272= 072y(0.8\ + 1 44y(1.8\;1 44y(2) + a32v(3) + 2.16ip.2)
Tz=36y
€ After lowering
t57
t58
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
In Figure @
2,831,
Volume of water = (1.86 - 0.93)(b + h) + 0.93b = x000
't.86b+0.93h=2.831,
CHAPTER THREE
fotal Hydrostatic Force on Surfaces
- 7.522- 0.5h ) Eq. (r)
t59
Problem 3 - 60 (CE Board November Lg77)
having specific graviry of 0.92 is floating on salt water
1l^a:l"lg
of sp. gr.
1.03. If the volume
of ice above the water surface is 1000 cu. m., what
total volume of the ice?
2h+h=3.044
b
FLUID MECHANICS
& HYDRAULICS
is the
Solution
Let V = total volume of ice
lP Vt = Pz vz)
p1= 101.325 kPa (atmospheric pressure)
Vo = volurne displaced
Vr = 0.93(3.05) = 2.8365 m3
pz= 101.325 + 9.81,h
Vz= 0.93(3'05 - b\
Vo=V-7000
7V." = yr." V = (9.87x0.92)(1,)
Wir":9.0252 V
101.32s (2.836s) = (101.32s + 9.81h) [0.93(3.0s - b)]
309.04 = 309.04 - 1,01,.325b + 29.92h - 9.81bh
29.92h - 1,0"L325b - 9.81bh = 0
BF=10.1043(y_1000)
29.92h - 101.32s(1.522 - 0.5h) - 9.8L(L.s22 - 0.5rt)/1 = 0
29.9Ax - 154.22 + 50.651t - L4.931r + 4.9051* = 0
4.905h2 + 65.65h - 154.22 = 0
. -6s.65t@
h=
" 2(4.e05)
=2.039
b = 0.5027 m
2r = pD(l)
+Dr:1.86 m2
D=1.539m=1,539mm
2T = 0.024937 (t,53s) (1)
T=19.2N
Wi,"= BF
9.0252V = 10.1043(V _ 1000)
1.0791= 10704.3
V :9,364 cu. m.
For hornogeneous solid L-,,cly floating on a homoge.eous
The maximum tensile stress occurs at the bottom of the tank.
Tension, T:
[IFy = 0]
Solution:
H=b+h=2.542m
p=yH=e.81.(2.s42)
p = A.937 kPa= 0.024937 MPa
BF = yr"u*o1". Vp
BF = (e.8111.0.3)(y _ 1000)
yo =
t""
s
liclr-iicl
t
vr,uay = '', ',., y,,,,.,,.
liquirl
I liqutd
then; V-1000 =ffiv
T
0.106796V = 1000
V:9,364cu.m.
3 - 61 (.t to"rd
:k o{ wood 0.60 m x 0.60 m x /l meters in dimension
was thrown into the
and floats with 0.18 m projecting above the water
surfac". fn" ,"^. Lfo.t
thrown into a container of a liquid having a specific gr.avity
of 0.90 and it
above the surface.-Determine i.,e following,
1Tl|.lli
(a)
the value":projecting
of ir,
. , A
I OU
CHAPTERTHREE
FLUID MECHANICS
& HYDRAULICS
rotal Hydrostatic Force on surfaces
Solution
'Ystone --
.In Water:
S*ood
Draft =
7,
Sn ater
18
h-0;Ig- S*ood
FLUID MECHANICS
460
= 28,204N/ms
h - 0.18
3 - 63 (CE Board May 1993)
body having a sp. gr. of 0.7 floats on a liquid of sp. gr. 0.g. The volume
of
l body above the liquid surface is what percent of iis total volume?
ln another liquid
iir'-.
'iri;:::r
S*ood
l..rf:r
1,
0.14
Suquia
S*nod
,,YD _ sbo.iy ,,l/bodv
-
;,
0.9
S*oo4 /r = 0.9h - 0.126
- 0.14
sliquid
Vo= #Vuoay
= 0.B75Vb..ru
) Eq. (2)
In other liquid (S = 0.9)
[S.ooa /r = S*ooa /r]
h-0.18 =0.91t-0.126
ft=0.54m
) height of the block
Substitute /r to Eq. (1):
S*'"a(0.54) = 0.54 - 0.18
) Specific gravity of wood
S-ooa : 0.667
Weight of block = |*oocl Vbtock
Weight of block = (9.S1 x 0.667)l(0.6 x 0.6)(0.54)l
Weight of block = 1.272 kN
Since the volurne of the body displaced (below the liquid surface)
is 0.875
87.5% of its total volume, then the volume of the body above
the liquid
.is 72.501' of its total volume.
3.- 64 (CE November 1997)
block of wood 0.20 m thick is floating in sea water. The specific
gravif, of
is 0.65 while that of seawater is 1.03 Find the minimum area
of a block
will support a man weighing 80 kg.
Wvlr'r = B0 kg
Problem 3 - 62
A stone weighs 460 N in air. When submerged in water, it weighs 300 N,
{
Solution
Weight of stone = 450 N
Weight of stone in water = 300 N
Buoyantforce, BF = 460 - 300 - 160 N
,l
Wwooo
Find the volume and specific gravity of the stone.
IBF = y*",", %,on"]
160 = 9810 (%"*)
Vrtun" = 0.0163 cu. m.
Ib I
%ro."
0.0153
L
It _ 0."14_
Total Hydrostatic Force on Surfaces
Wrro."
Ysrone = -
7,
S*,oah=h-0.18)Eq.(f)
Draft =
CHAPTER THREE
& HYDRAULICS
[>Fv: 0]
BF= W'"n *W-ood
.
fsu,Vwood = ?Vnr"n + ]*ood V.ood
(L000 x 1.03) V.oo6 = 80 + (1000 x 0.65) V*oo<,1
V*ood = 0.2105 m3 - Area x 0.2
Area = 1.05 square meter
CHAPTER THREE
162
FTUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
Problem 3 - 65 (CE November 1997)
A cube of wood (s.g. = 0.60) has 9-in sides. Compute the magnitude and
direction of the force required to hold the wood completely submerged in
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Problem 3 - 67
A unifonn block of steel (s = 7.g5)
will float at a mercury_water
k
interface as shown in Figure 2/.
What is the ratio of the distances
t'a" and a'b"
for this cohdition?
water.
Solution
t63
Water
Weight of wood = (62.4 x 0.60) gI = 15.795 lbs
Buoyant force when completely submerged in water:
BF=62.4(+I :26.325tbs
Required force = 23.325 - 15.795
Required force = 10.53 lbs downward
Problem 3 - 66 (CE Nov 2000)
The block shown in Figure 04
weighs 35,000 lbs. Find the
L2' x 12'
A be the horizontal cross-sectional
of the block
value of ft.
B\+ BPr=17y
lt,Von,*f,VD,u=f, V
Figure O4
9.81.(A x a) + (9.81 x 73.56)(A x D) (e.81
=
x 7.85)[A(a + b)]
a + '1.3.56b 7.85;a + 7.85b
:
5.71.b=6.85a
Solution
From the figure shown:
a/b = 0.$a
lIFv = 0l
BFr+BFz=35,000
BF1 = ysil Vp
3P, = (62.4x0.8) (12x12x3)
Bh= 21,5rU.nnrbt
21,,565.44 + BF: = 35,000
BFz= 13,ntn.U'tO"
BFz= h,Vo
73,434.56 = 62.410.2)(1,2) hl
h = 1.495 ft
3 - 68 (CE May 199E
5.kg steel plaie is attached to one end of a 0.1
m x 0.3 m x 1.20 m wooden
t, what is the length of the pole above
water? Use s.g.
;;;l of 0.50.
lect buoyant force on steel
"f
CHAPTER THREE
164
Total Hydrostatic Force on Surfaces
MECHANICS
,FTUID
& HYDRAULICS
CHAPTER THREE
l6s
Total Hydrostatic Force on Surfaces
Problem 3 - 70
Solution
Neglecting the buoyant force on steel:
wooden buoy (s.g. =:::? is
I se-a water (s.g. 1.025).
i1
=
BF*ooa = LVsteet * |V*ood
1000(0.1x0'3xV)=S+
:ffi51*ff;::ff"1j.
1000(0.5)[0.1 x 0.3 x 1.2]
y=0'77m
mm by 50 mm by 3 m long is made to float
.toy Tu", N of steel tr g li.slJ should be
make the buoy noat with exaJny 450;m
exposed
_50
Solution
h=1.2_!/
h=1.2-0.77
ll
FTUID MECHANICS
& HYqRAULICS
iEFv = 0
BF"t*r + BF*ooa -W*oo6- tr/Vrt""l = 0
/r = 0.43 m
0.05 m
sffi
0.45 m
BFrt*r = 7r* %r"ur
BF,t*r = (9810 x 1.025)%t*r
BF,tu"r = 10,055.25 %t""r N
1,,
Problem 3 - 69
If a 5-kg steel plate is attacl'red to one end of a 0.1 mx0.3mxl,.20mwooden
pole, what is the length of the pole above water? Use sp. gr. of wood of 0.50
and that of steel 7.85.
BF*ooa = ls* Vp
ll--o = (9810 x 1.025)[(0,05)r(2.5s)]
.
2.55 m
BF*ooa = 64.1 N
W*ood = |wood Vwood
W*ood= (9810 x 0.62)[0.05)r(3)]
W.ood = 45.62 N
Solution
W"t*f = Tsteel Vsteel
LV*ood = (1000 x 0.5)(0.1 x 0.3 x 1.2)
TVsteer = (9810 x 7.85) V,t""r
W*ooa = 18 Kg
LVsteer = 77008.5 V,ua
10055.25 V"t""r * 64.1 - 45.62 - 7700g.5
%t*r = 0
66953.25 %tu"r = 18.48
l/:t""r = 0.000276 m3
Wrro"r = 5 kg
BFw- 1000(0.1x 0.3 x d)
BFw=30d
BF. = 1000 Vs
Ws = (1000 x 7.85) Vr = 5
1.2 m
lV,tu"r = 9810(7.85) (0. 0027 6)
Wrt.r = 21.255 N
V5 = Q.[QQ$]/ 6t
BFs= 0.637 Kg
I/V*ood
3 -71
t WrL*t = BF5 + gP*
ce of lead (sp' gr' 11'3) is tied to a 130
18+5=0.637+30d
d=0.745m
x=1,.2-d
r = 0.455 m
cc of cork whose specific gravity is
They float just submerged in water. what
t*'
is the weight of the lead?
THREE
t66 CHAPTER
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
t67
Solution
Solution
lIFy = ol
Wc+Wt=BFc+ BFt
Wc=kVc
w6= (1x 0.25)(130)
Wc = 32.5 granls
BFc = Yu, Vc
Brc = (1X130)
BF6 = 130 grm
Wr= IrVr
Wr = (1 x 11.3) Vr
Wt= 11'3 Vr
BFr= Yu' Vr
BFr = (1) Vt=Yr
32.5 + 11.3 Vt.= L30 + Vt,
Vt= 9.47 cc
Wt= 11'.3(9.47)
Wr=106.97 grams
1
(a) Lead is fastened outside the cylinder
(e) Lead is fastened outside
BFc= y* Vo
Probfem 3 - 72 (CE November 1993)
A hallow cylinder 1 m in diarneter and 2 m high weighs 3825 N. (a) How
many kN of lead weighing 110 kN/mr must be fastened to the outside bottont
of the cylinder to make it float with 1.5 m submerged in water? (b) How many
kN of lead if it is placed inside the cylinder?
Brc = e.81[f (1)r(1.5)]
BFc = 11'5U U*
BFt= Y, Vt
BFt = 9.8'l'Vr
Wr=ytVt = 110Vr
[XFv=0]
BFs+ gPr=Wc+Wt
11.55 + 9.8'J.Vt= 3.825 + 1,70Vt
Vt= A.0772xn3
Wr = 110(0.07721= g.4gr*
$) Lead is inside the cylinder
[EFv = o]
Wt+V\/r= gP,
Wt+ 3.825 = 11.55
Wr = 7.735 kN
(b) Lead is placed inside the cylinder
l6g ;ifil;H:lff Force on surfaces
FLUID MECHANICS
& HYDRAULICS
Problem 3 - 73
FLUID MECHANICS
& HYDIIAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
t69
Solution
A stone cube 280 rlm on each side ancl weighing 425 N is lowered into a tank
containing a layer of water 1..50 m thick over a layer of mercury. Determine
the position of the block when it has reached equilibrium.
Solution
W= 425 N
BFu = !uVou
BF,ra = (9,810 x 13.6)[0.28'?(x)]
BFm:19,45n.U' *
BFw = yw Votv
BFyy= 9,810[(0.28)z(0.28 - r)]
BFw=769.1(0.28-.r)
2:= !9?n
[IFy = 0]
BFpl+ BP*:ryn
10,459.81 7 + 769.1(0.28 - x) = 425
9690.71,x 209.652
r = 0.0216 m
:
r = 21.6 mm
Therefore; the block will float with 21.6 mm below the mercury surface.
Problem 3 - 74
A cube 2.2 feet on an
. edge has its lower half of
s.g. = 1.6 and upper half
of s.g. = 0.7. lt rests in a
two-layer fluid, with
lower s.g. = L.4 and
upper s.g. =
0.8.
Determine the height ft
of the top of the cube
above the interface. See
Figure 33.
x 7,\[(2.2)z(2.2 - n)] + (62.4 x
BF = [62.4 x 2.22] (3.08 _ Lah
Y,= !r?
0.8)[(2.2)z(h)]
i g.SpS /r\-
n x 1. 6)l(2.2)z (1.1)l + (62.4 x 0.4[e.z),
W=162.4x2.221(2.53)
(1.1)]
IBF =w1
\6?.^a "
2.211 (3.08 - 1
y * O.s/r) = [62.4 x z.z?](2.ss)
3.08 -1.4h + 0.8/z = 1..76 + O.Zi
h = o.917 ft
3 - 7s (CE May 1e97)
100-mm diameter solid cylinder is
mersed in a Iiquid (y = g.17S
95 mm high and weighing 3.7F
N is
IN/*) contained in a tall meiui.yti.,au.
ling a diameter of 125 mm. Before i*-"rri*,
what level will the solid cylinder floati----'--'
the liquid _u,
ii-^L a""p.
THREE
t70 CHAPTER
Total Hydrostatic Force on Surfaces
Solution
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Iotal Hydrostatic Force on Surfaces
l7t
Probfem 3 - 76
L25mmA
'A wooden beam of sp. gr. 0.64 is 150 nm
by 150 mm ancl is hinged at A, as
;chown in the Figure. At what angle 0 will the beam float i. water?
Solution
x--
LIt
T
E
I
1m
C
y1
N
l25mmb
- (a) Before immersion
(b) After immersion
Solve for the draft D in figure (b):
TBF=W
ItVo=W
.
(8,175) Vo = 3.75
Vo = 0.0004587 m3 = 458/716, mrn3
+(100)'?xD=458,7't6
Draft,D=58.4mm
When the solid cylinder is immersed, the liquid in the tall cylinder riser
due to volume of liquid displaced. Therefore, the volume of liquid
displaced equals the total volume of real and imaginary liquid above the
original level
Vaboue orig. level = VD
ft (1?5)2(x) = 458,716
r = 37.38 mm
Weight of beam, LV= yr,"u^ Vt".,^
Weightof beam, W= (9,870 x 0.64)[(0.15)z(5)]
Weight of beam, W = 706.32kN
Buoyant force, BF = f *ater Vp
Buoyantforce, BF = 9,g10[(0.15), /]
Buoyant force, BF = 220.725x
lE tvto = g1
BF(5 - 0.5x) cos 0 = W(2.5 cos 0)
2_20.725 x(5 - 0.5r) = 706.32(Z.s)
5x-0.5x2=8
0.512-5r+8=0
x=
2(0.s)
From Figure b:
/$+a=P+y
!=75+37.38-58.4
V = 53'98 mm
Therefore; the solid cylinder will float with its bottom 53.976 mm above
bottom of the hallow cylinder.
x=2m
sin0=
=rru
JJ-x
Q = 19147"
s.
li
CHAPTER THREE
t72
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Probfem 3 - 77 (CE Board May 2003)
F'rom the figure below, it is shown that the gate is 1.0 m wide and is hinged at
the bottorn of the gate. Compute the following:
, (a) the hydrostatic force in kN acting on the gate,
(b) the location of the center of pressure of the gate from the hinge,
(c) the minimum volume of concrete (unit weight = 23.6 kN/m3) needed to
keep the gate in closed position.
FTUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
173
Problem 3 - 78 (CE Board November 1993)
A boat going from salt water (sp gr. = 1.03) to fresh water
(sp. gr. = 1.0) sinks
7,82 cm and after burning Z2,ZZO-kg of .out rises
up by 15.24 cm. Find the
original displacement of the boat in sea water in kN.
Dolution
D+
Jlii: r)'.',i ii, "
llll r, rilil :irl::l :,
I;:
Figure (a)
Figure (b)
Solution
F=yhA=9.s1(1)(2x1)
F = 19.62 kN
v= +Q)=0.667r..
.dS
f-sF6+r
;r
V
IT
D+0.0762-0.1524
= D - 0.0762
[IM,a = 0]
F*y=Tx2.5
19.62(0.66n = 2.57
T = 5.232 kN
Figure (c)
From the FBD of the
concrete block:
lxFy = 0l
have to assume that the boat have a constant cross-sectional
area A below
Water surface and use |water = 1000 kg/ma
T+BF=W
BF = y,uVron, = 9.81 V.on
W = Yron, V.on. = 23.6 Vror*
5.232 + g.g1 y.o,,. = 23.6 V,o,,"
V.on. = 0.3796 m3
Figure (a):
Bh=W
lryVo=W
W= (1000 x 1.03)[A(D)]
W=1030AD
) Eq. (1)
I I +
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
-
For any floating body; Buoyant force
1V'1
lr*VD=W
w= 1000[,4(D+0.0762)]
W=1000A(D+0.0762)
CHAPTER THREE
Total Hydrostatic Force on Surfaces
t75
6olution
In Figure (b):
BF2
FTUID MECHANICS
& HYDRAULICS
) Eq. (2)
= Weight
Solving for displacement in sea water:
|sea water Vot = W
(&) Vr, = 24,A00 x 2,240
ln Figure (c):
Bh= W -72730
i000[A(D - 0.0762)l = w - 72730
Vor = 840,000 ftr
) Eq. (3)
Solving for displacement in fresh water:
"L
|freshwater Voz=W
From Eq. (1) and Eq. (2):
Iw=w
(62.2)(V D2) = 24,000 x 2240
Voz= 864,308.68 ft3
1030AD = 1000A(D + 0.0762)
1030D=1.000D+76.2
D = 2.54 m (draft in sea water)
From Eq. (1)
w=1,o3oA(2.s4)
w = 2516.2A
From Eq. (3)
7000A(2.54 - 0.07 62) = 261,6.2A - 72730
2463.8A = 26L6.2A - 72730
A = 477.23m2
Therefore:
w=261.6.2@n.n)
Figure (a)
Let /l be the difference in trre drafts in fresrr
& seawater:
Voz- Vu = Area x /r
fr_ 864,308.68_940,000
32,000
h = 0.76 ft
Draft in fresh water, D = 34 + 0.76 = 34.7G ft
3-80 lcrao@
rr an arbitrary shaped body with a submerged
volurne Vs and a
p', submerged in a fluid of density p1.
What is the net vertical force on
body due to hydrostatic forces?
W = 1,248,529 kg (9.81/1000)
W= 12248 kN
Fnet= Tf Vs
yf=ilxg
Problem 3 - 79
A ship having a displacement of 24,000 tons and a draft of 34 feet in
enters a harbor of fresh water. If the horizontal section of the ship at
waterline is 32,000 sq. ft, what depth of fresh water is required to float
ship? Assume that marine ton is 2,240 lb and that sea water and fresh w
weight 64 pcf and 62.2pcf, respectively.
Fn* = plg Ve
3-81
)herical balloon, 9 m in diameter is fined with herium
gas pressurized to
u.,a unchor"d by a rope to the ground.
H::,,L^,1Tli'_1*.':
the dead weieht?fof?2"c:
Frting
the bailoon, derermini ,r.." iJ*r", iri,fr.
,oo"
R-212mloK for helium gas and ,",, = tilJ N/-..
.q,
II O
CHAPTERTHREE
Total Hydrostatic Force on surfaces
FLUID MECHANICS
& HYDRAULICS
Solution
'
FIUI!i MECHANIC5
& HyDRAUUCS
Wu,=45.52
nerrutrr
P
RT
272(273 + 2o)
Yhelrun, = 1'787 N/mr
N (fronr Figure3_ l)
[IMo = 0J
Vbulonn = \n(9/2)3
Vboloon = 381.7 m3
[r!,i;:* ?,;z{ltit'),3;'r
12.57 Lz
[IFv = 0]
=u
58.43
L*2,33m
BF-W-T=0
h/-
t77
BF = Y,n Vn
BF = 9810(1.025)[(0.05),Ll
BF = 25,138L
"177 x 103
BF = y"1, V5o1lnon
BF ='t1..76(381.f = a43.u
CHAPTERTHREE
-rotat Hyctrostatic Force
"" surracll
sin 0 * 2/l
sin 0 * 2/2.93
0 *59o
*
yh.riunr Vboiloon
W=1..787 (381..7) - 682.1 N
4488.8-682."t=T
T = 3806.7 N
Problem 3 - 82
The buoy in Figure 3 - t has 80 N of steel weight attached. The buoy has
lodged against a rock 2 m deep. Compute the angle 0 with the horizontal at
which the buoy will lean, assuming the rock exerts no mornent on the buoy.
Solution
t.
and 200 mm hisrr and weighs
If:ir:t';*il;:ffiij:r mnr in diameter
r.6
to-push..'h"
o a body of tiquid h":".:.:-Tquired
:r:1,
*l
:on-u
d;;:;nward)
;ffi :ffilffi,111i j 3,,1?,,::":",::i
The reqrrirecl downward vertical
F=BF-W
BF * yriqud %uuo
BF - (e,810 x 0.S)
BF*4.11 N
force is:
[(r/3)( 0.1/212(0.21!
F*4,11-1.6
F-2.51N
This force F. L5l N becomes
tt no matter how deep
cone is submerged
further
f;"T:i:*,jT;
t7 8
;i*il;:#[:,? ,o,." on surraces
Problem 3 - 84
'Io what depth
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
CHAPTERTHREE
_
Total Hydrostatic Force
j";;:;;
""
,& HYDRAUTICS
Solution
a 2-m-diameter log, 4 nr long and of sp gr 0.425 sink in
F=40N
fresh water.
Solution
For a homogeneous solid body floating on a
homogeneous liquid:
yo = j!!9L vu.o,
sliquid
ast=ffiar
As = 0.425nr2 (shaded area)
Let 7 = volume of woocl
ln water:
From geornetery:
As = Ar".,or - At ,ongl*
0.425nra = lz ra Q" - lz 12 sin 0
0, - sin e -- 2.67
Solve 0 by trial and error:
Try 0 = 170'
170" (n / 1"80") - sin170" = 2.7 6
IXFY = 91
Bh-W-F=o
9810V_W =40
Y
(*2.67)
es10 ) Eq' (1)
ln glycerin:
lxFv = 0l
BFz-W-F=0
Try Q --'1,66"
1,66'(n/'180") - sin166" =2.655
= 49+w
(+2.67)
Try 0 = L66.M"
1.66.M" (n / 1,80') - sin166'44" = 2.67
It=r-A
ft=1-(1)cos(0/2)
h = 1 - (1) cos (166.44' /2)
ft = 0.882 m
(9,810x 1.3)V_W=t00
(e,810
xr.sr14*wl -w=loo
'L lsro J
52+1361_W=100
W= 160 N
From Eq. (1):
, _ 40+160
9810
V = 0.0204 ml
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 weighl
and sp. gr. of the wood.
Unitweight, r=
+=#
Unit weight, f = Zg43 N/pr
SP.gt.,r- Iwoo'l -7843
Iwater 9910
. Sp. gr.,s = 0.8
t79
I
CHAPTER THREE
tB0
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Problem 3 - 86
FLUID MECHANICS
& HYDRAULICS
I otat Hydrostatic
Since the volume of oil
A rectangular tar-rk of interrral width of
5 m, as shown, contains oil of sp. gr. =
0.8 and water. (a) Find the depth of oil,
h. (b) If a 1000-N block of wood is
floated in the oil, wllat is the rise in free
surface of the water in contact with air?
/orr linirrar,l = voir
CHAPTER THREE
Force on Surfaces
tBt
remain unchangecl;
lrnalt
5)J5](1 2s) = (0 s)(s)(/,) _
0.1274
!9
/r = 1.301 m
-
As shown in Figure b'
if theoil-water interface crrops
rree surface of water
by a distance of y, the
o;;I;,';;;the cross-sectionar
right compartnrent is twice
1i' rrge
area of the
thar ;riffi;;ilrnparrment.
Sum^-up pressure head
frorn oil surface to wat
'- vvat€rsurfaceinmof water:
0+.1.301(0.8) + (3_ y)-4_V/;:;
1.0408- 1,-3ry/2=0
3v/? = o.o4o8
y/2 = 0.0136 m or 13.6 mur
Solution
1000 N
o
{vtz
I
o
oit
s=0.8
J
I
4m
lit,l
I
irilj
i.ll
I
riiil
Figure (a)
open cylindrical tank 350
mnr in diameter and 1.8
lcallyinro a body
m high is inserted
d: wn and
"i;;";;'-I:::1T"'"r
a ncr floats
,ocr< : of
n""1,' with
*i,r, a 1300
('p g.. = 2.4j;;;p-";;il:iir'i.-n
rsoo
ll1.
,concrete ;. ;Jtf'!XiT,,':"^]li
"
or the
cyrindei, io what
ir .il,"-t
a"pin *iri;J:;:X"":T1"
Figure (b)
.oJ
(a) Depth of oil: (Refer to Figure a)
Sum-up pressure head from oil surface O to water surface € in m of water
a+/r(0.8) +g-4= Pz
fv
0+0.8ft-1=0
/r = 1.25 m
(b) Rise of the water surface: (Refer to Figure b)
mm.
3-87
I
4m
,0;5 rnrr,:{;,ri; ,i :;., 1m,
of water will rise 13.6
-T-
I
I
Therefore; the free surface
+BF.vr -W=0
BF.on. = I*ater Vcon.
V.on.
=
) Eq. (r)
W"on"
I.onc
t,TCOnC -_
1300
e870(2.4)
%on. = 0.0552 m3
BF =W
lotVo=W
BF.un. = 541.2 N
(9810 x 0.8) Vo = 1000
V6 = Q.l)/Q sP
tFn,r = l*arc, Vn
tF.on. = -9310(0.0552)
EF.yr = 9Si0lf (0.eS;27,;
c1t =943.83
lt
Tff::;%T
182
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
l;ronr liq. (1)
541.7+943.83h-1300=0
FLUID MECHANICS
& HYDRAUI-ICS
CHAPTERTHREE
_
Totat Hydrostatic
Force on Surfaces
,Solutlon
/r = 0.803 m
Wr = 205 kg
Applying Boyles Law (taking p^nn = ^101.325 kPa)
Before irrsertion:
Absolute pressure in au, pt = 101.325 kPa
Voiume of air inside the cylinder, V, = i (0.35)'z(0.18)
6me
Wr
- = 205 k9
0.8.4
__{
1.8m
I
Volume of air insicle the cyiinder, % = 0.0173 m3
0.6m O
-------r.2.1 m
0.96
Y
1.Bm
After insertion:
Absolute pressure in air, p2= 101.325 + yh
^ Absolute pressure in air, p2= 10L.325 + 9'81(0.803)
Absolute pressure in a.-r, 1tz = 109.2 kPa
Volume of air inside the cylinder, V, = i (0'35)2 r
l
*
',
D
H
l'',
-rl
Volume of air inside the cylinder, Vz = 0.0962x
[pr Vt = pz Vz]
101.325 (0.173) = L09.2(0.09 62t)
x ='1..67 m
x-h+U= 1.8
y =1.8-1.67+0.803
t/ = 0.933 m
Therefore, the open end is submerged 0.933 m below the water surface.
Problem 3 - 88 (CE Board)
A cylindlical buoy 600 mrn in diameter and 1.8 m high weighs 205 kg. It is
moored in salt water to a 12 m length of chain weighing 12 kg per m of its
length. At high tide, the height of buoy protruding above water surface is
0.84. What could be the length of protrusion of the buoy if the tide dropped
2.'I m? Density of steel is 7,790 kg/ mt. Use density of water : 1000 kg/m..
Figure a: HiSh Tide
Figure b: Low Tide
,ht of-chain = 12kg/m
rty ot steel = 7,790 kg/rfi
of steel (chain) =1Z1ZZOO
of steel (chain) = 0.00154
m3 per nreter length
Figure a:
lxrv = 0l
B\+Vpr-Wt-Wz=0
BFt = y,* Vo
= (1000 x 1.03)[* (0.6), (0.e6)]
Bh = 279.58kg
BFz = 7,* %noin
= (1000 x 1.03)[0.0015a(r)]
BFz= 1.586L
Wz=72L
279.58 + 1.586L _ 205 _ 12L
L=7.'1,6m
Depth of wa ter, [! = L + 0.96
Depth of water, iLI = g.12 m
l*,
=0
t83
I
84 fifil;:#[:[ ro,.." on surfaces
FLUID MECHANICS
& HYDRAULICS
Irr liigrrre h:
Depth of water, H' = H - 2.1,
Depth of water, H' = 6.02 nr
Draft,D=H'-L'
FLUID MECHANICS
BF' t = 1753.1,8 - 29L.23L'
BF'2 = (1000 x 1.03)[0.00154(L'))
BF''= 1'56Ut'
W'z=12L'
1753.1,8 - 291,.23L' + 1.586L', - 205 - 12L', = 0
L'= 5.13 rn
r85
Weight of ball:
W = yutt Vaou
S4r
= (9870 x O.aZ)
W= 5s.2S N
[IFv = 0]
BF',r: 291.rU (6.02 - L',)
Force on Surfaces
Solution
Draft,D=6.02-L'
BF'1+gP'r-Wt-W'r=g
3P" = (1000 x 1.03))[f (0.6)'zD]
CHAPTERTHREE
_
Total Hydrostatic
& HYDRAUTICS
!n(0.15)3
Buoyant Force:
BF = y*otu, Vb"rr
BF = (9810)
+r(0.1s)3
BF = 138.69 N
Depth of pool:
Work done by W = Work done
W(4.3+h)=bpfu\
by BF
58.25(4.3+h)=138.69h
h= 3.llm
D=6.02-5.13=0.89
y=1'8-D
y=1.8-0.89
y = 0.9'1, m (Iength of protrusion)
3-90
hydrometer weighs 0'0214N
illHT:: rT;#*"'n
and has a stem at the upper
auuf"'
end which is 2.7g
'", t- o""t in ol^(sp. s,. = ;.;;, that in
Problem 3 - 89 (CE Board)
A wooden spherical ball with specific gravity of 0.42 and a diameter of 30()
mm is dropped frorn a height of 4.3 m above the surface of water in a pool of
. unknown depth. The ball barely touched the bottom of the pool before it
began to float. Determine the depth of the pool.
In alcohol:
BF =W
(9810 x 0.821)V s,
= 0.0214
Vo" = 2.657 x 10-6 m3
Von= 2,657 mms
oir:
BF =W
(9810 x 0.78)VD,= 0.0214
Voo = 2.797 x 10-6 m3
Voo= 2,797 mmt
VDn Voo- Vo,
O - 2,797 - 2,657 = |40 mm3
Yo. tQ.7e1zh=740
229 mm
Alcohol, s = 0.82I
Oil, s = 0.78
I UO
CHAPTER THREE
rotal Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Problem 3 - 91
A plastic cube of side L and sP gr. 0.82 is placed vertically in water. ls the
cube stable?
m
'
Solution
Ihe body is stable if M is above C.
Drarr,D=optDraft, D = 0.82L
lvtD^:
I
+ (L)(L)3
VD
(L x L)(0.821)
-
LxL
i i r-+-6c
liT-l F".
t
I
I
rit
lll
GB"= L/2- D/2
GB,, = 0.091
Since MBo > GB", M is above C.
The body is stable.
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Problem 3 - 93
Note: The body is stable when M is
lbove G and unstable if M is below
10cmx10cm
G. Mth'smaller value of H, the
tnetacenter M will become higher
than G making it much stable.
When H increases, M will move
closer to G making it less
. Hence, the maximunr
for stable equitbrium is
Mcoincides with G, or MBn=
Seawater, s = 1,03
Waterline Section
Problem 3 - 92
A solid wood cylinder of specific gravity 0.6 is 600 mm in diameter and L2(x)
mm high. If placed vertically in oil (sp. gr. = 0'85), would it be stable?
the figure:
GB"= H/2- D/2
Draft,D=*#H=0.62.tH
GB"-0.5H -0,621,H/2
GBo=g.1tt,
Solution
sP gr' wood
Draft, D -
,
wB"=
Draft, D = ffi trzoo) : 847 mnr
MB"=
+ (300)4
MB,,= O' ;'
*r
MBo= #(roxro)3
(10x10)D
I
v,
E
,E
a
o
N
n(300)'(8a7)
MB"-- 26.56 mn
GB" = 699 -1/z(847)
GB" = 176.5
Since MB, < GBo, the metacenter is below G
Therefore, the bqdY is unstable.
t87
A block of wood (sp. gr: = 0,64)-is_ in the shape of a rectangular parallelPpiped
a 10-cm square base. If the block fltats in salt w"ater *i*, it"'rqriur"
|avin-g
base horizontal, what is its maximum height for stable equilibrium in the
upright position?
folution
t-.l
l
,lil
MB,,:0.'1,02L
FLUID MECHANICS
& |{YDRAUUCS
MBn=
100 _ 73.419
12(0.621,H\ H.
oR: MB.= B' [, * tu"2 e)
12D( , )whereo=oo
lo' (1 + o) 73.41.9
MB,, =
1,2(0.621H)',
=
H
CHAPTER THREE
r88
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
rotar Hydrostut'. F."?1PJ"E:JH:: I Bg
Initial metacentric height, MG = MBo _ GBo
Initial metacentric heighg MG = 77 .49 _ tt/.SS
Initial metacenkic height, MG = -39.96 mm
lMB"= GB"l
73'419
= 0.189H
H
H = 8.43 cm
Problem 3.94
A wood cone, 700 rnm diarneter and 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.
3-9s
Solution
V-ood =
{ n(350)'z(1000)
V*ood = 1,28,281,,700 m3
---T--
ttYD - --0.60 tt
Ywood
Vo = 0.6 V*ooa
E
E
o
o
o
Vo = 0.6 (L28,281.,700)
Vo = 76,969,020 mm3
!--1, * tan2 el
ut"= 12DL
2 )
where 0 = 0"
MB"= e)2 f, * tunt o"l
L2(2.4)l z
By similar solids:
_l
MBo=2.8125m
GBo = 2., - 1.2 = 1.5 m
vruo,l _1rooo13
v, - l. D,/
osv-* -l D
Initial metacenkic height:
v u,uu,l _ r 1000 \'1
Initial metacenkic height, MG = MB"_ GB"
Initial metacentric height, MG = 2.g725 _ 1-.5
J
il metacenkic height, MG = l.31^2i m
D = 843.4 mm
350
843.4 1000
r = 295.2 mm
Waterline Section
e=+3
-'14.93
MBn=
I
+Q95.2\4
76,969,020
G =-=//.+vmm
Fronr the Figure:
GB"=759-3D/4
GB"=750 -3(U3.4)/4
GB" = 117 'nU
*
Since MB,, < GB", M is below C and the cone is UNSTABLE.
=#l*o+--l
=2.9'l.m
t
90 ;l|3|I;:'.H[:,1,o,.." on surraces
FLUID MECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
B=10m
gince MG > MBn, the moment is righting moment.
MBo=
Righting moment, RM = W (MC sin 0)
tul'el
I-lr*
12DL
;l
Bf = yVp
w= (9.8.1, x 1.03)[9(1sX24)l= 3,273.8 kN
MBo=
Rrghting n"roment, RM = 3,273.8[(1.41) sin 74.93"1
Righting nloment, RM = 1,189.3 kN-m
where o = oo
(1 + o) = 5.45 m
"i*,
Metacentric height, MG = 5.45 3.2gs
Metacenkic heighf MG = Z.21Sm (the barge
A barge floating rn fresh water has the iornr of a parallelepiped having
dimensions 10 m x 30 m by 3 m lt weighs 4,500 kN when loaded with center
of gravity along its vertical axis 4 m from the bottom. Find the metacentric
height about its longest and shortest centerline, and determine whether or rrol
B=30m
B'| *8.2ol
MB"=
"'"":nDl'*-T-l
where0 = 0'
the barge is stable
*u"=
Sotution
Metacenkic height, MG = 49.02 3.2gs
Metacenkic height, MG = 4S.7gSm (the barge
W = 4,500 kN
.l
I-I
Solve for the draft, D
IBF =vq
YVo=W
9.81 [10 x 30 * Dl = 4,500
D=1.53m
GB,,= 4 - D/2
GB,,= 4 -1.53/2
GB"= 3.235 m
is stable in rolling)
Along kansverse direction (pitching):
Problem 3 - 95
3m
19l
Along longitudinal axis (ro[ing):
Metacentric height, MC = MB" - GB"
Metacentric height, MC -- 2.91 - 1.5 = 1.41 m
1111=
FLUID MECHANICS
& HYDRAULICS
#ru)(l+
o) = 4e.o2m
is stable in pitching)
4 m above the bottom
fr
Rollinq
3 - 97 (Ce eugust rSZS;
crane barge, 20 m long, g meters wide, and
2 meters high
roaded at its center
*"lqt*g 20 short to;H;; on fresh water
with a draft of
l|r::::l
meters l:]fi
and has its center of giavity located
along its ,er'.ai
t L'50 meters above its bottoml Compute the horizontar
"_i,uut
"
distance
out", to
t(J
side from the centerlin"
barge thiough which the crane could swing
?f,g:
q" lenter of the deck, and tip the
:r":a
with the 2O-meter edge just touching the water
surface?
i::il*l-ilj:1ti{Td ,i*
t92
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
CHAPTER THREE
q HYDRAULICS
Total Hydrostatic Force on Surfaces
t93
z=dsin0
Solution
I short torr = 2000 lb
z = 0.552sin 11.31"
= 900 kg
GB"=1.4-d=0.848m
Wp= (20 x 900) 9.81
Wn = 176.58 kN
Road roller
Wp = 20 short tons
= 0.10g m
MB"= B' lr* tan2el
12DL 2 J
MB^=
,vLDo-
82 |-., . ,u!t 11.31'l
45 Lt*___r_J =a.saam
MG = 4.533 - 0.848.= 3.685 m
r=MGsin0
r = 3.685 sin 11.31" = 0.723 m
BF = yVe
lrMc = ol
(BQr=Wn(L+z)
Br=9.81[8 x7.2x70]
BF ='1,883.52 kN = Wr
].,883.52(0.723) = (12d.58)(L + 0.108)
Weight of barge, Ws = BF -Wx
Weight of barge,Ws= 1,883.52 - 176.58
Weight of barge, Wn = 1,705.94 kN
L =7,604 m
) Horizontal distance from the center of the deck
3-98
wooden barge of rectangurar cross-section is g m- wid",
Tilted position
4 m high, and 16 m long.
Xjii*:Tg: ::Ta1if
I i..:p l"1d
*."ieh: of.rf
25 kN (included t";;ilidilii'.rffi;
=
or r,ioo kN:;:r"dt"; ib ow.
.*1".fl9";
" side, it will cause
, ot z.J m to {one
Wsl
iennn
bur"lu i*,*6;t
*"igh,
3*l*::,f*:*^
how far above the !::.T,:l:,".,k*r)
""i,ii"
waterline is the centerof
;6;r;: tLJ;
te
tut' 0 = 9$s
0 = '11.31'
polve for the new position of C in the tilted Position:
Wr(0.5) = ws(d)
1,883.52(0.5) ='t,706.e4(d)
d = 0.552 m
"
the barge to go down +SO mm in ttre
'p of imrnersion and also rise 450 mm in th" .oltopona ing wedge
oy
ffij
THREE
t94 CHAPTER
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
FLUIDMECHANICS
& HYDRAULICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
Problem 3 - 99
Solve for the draft, D:
BF =W
(9.81 x 1.03)[8 x 16 x D] = 1,500
The waterline section of a 1,500-kN
l;1,n*ffithe'center
D=1.16m
barge is as shown. Its center of gravity
or u,ovu,,.v
E;;;ffi;'il; ;ilil"" height
ln the Tilted position:
wB"=
tan0=f
^r
*t
f = frrctangle * /triangl" t lsernirircle
e:6.42
'l
tu1'0
::-lr
12Dl* 2 l
MB^= 8' lr-tur'6.42'I =n.uu.n
MB"=
t95
1= #(12)(8)3+
+(6,X4)3 x2+ #(4)4
I = 676.53ma
I
" 't2(1.16)l
2
)
[IMs" = 0]
1,425(b)+75(q)=BF(c)
, =MBosin0
:
c 4.63 sin 6.42"
c = 0.5'18 rn
a = 3.42 sin 6.42o + 2.5 cos 6.42o
a = 2.867 m
b = (lt + 0.58) sin 6.42"
1,4251(h + 0.58) sin 6.42"1 + 75(2.867) = 1,s00(0.518)
lt -- 2.947 m ) distance of G from the w.s.
IBF = vvl
9.81' Vo = 1,500
Vo ='1,52.9 m3
MBo=
ffi
=4.4?sm
IMG= MBo-GB"l
MG = 4.425 -1.5
MG = L9E m ) initial metacentric
height
is
l96
CHAPTER THREE
Total Hydrostatic Force on Surfaces
FLUID MECHANICS
& HYDRAULICS
Supplementary Problems
FTUID MECHANICS
& HYDRAUTICS
CHAPTER THREE
Total Hydrostatic Force on Surfaces
197
3-104
Problem 3 - 100
A vertical rectangular gate 2 m wide. and 1.2 rn high has water on one side
with surface 3 m above its top. Determine the magnitude of the total
Water in a tank is pressurized to g0 cmHg.
Determine the total force per meter
width on panel AB.
Ans: 482 kN
hydrostatic force acting on the gate and its distarrce from the water surface
Ans: [ = E4.b kN. yy, = 3.63 nr
Problem 3 - 1O1
A vertical serni-circular area of radius r is submerged in a liquid with its
diameter in the liquid surface How far is the center of pressure from tht'
liquid surface?
Ans: 0.589r
3-10s
Problem 3 - 102
An open vat holding oil (s = 0.80) is 8 m long and 4 m deep and has .r
trapezoidal cross-section 3 m wide at the bottom and 5 m wide at the top
Determine the following: (a) the weight of oil, (b) the force on the bottom of tht'
vat, and (c) the force on the trapezoidal end panel.
Ans (a) 1002 kN; (b) 752 kN
Ans: Fu =7491b
(c) 230 kN
Fv = 2,7341bs
)
{
Problem 3 - 103
Freshly poured concrete approxirnates 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
'oL
the lower bolts.
Oil, s = 0.75
Ans:19,170lbs
3. 106
tute the hydrostatic force and its location
on semi-cyrindrical indentation
Consider only 1 mete. reng*,
ii:,H.
oi.yri'a".';d;;;;.;;;to the
Ans: Fs = 109.5 kN @ 1.349 m below D
Fv = 20.5 kN @ 0.531 m to the left
of B
198 fHili:#*:,? ro.." on surfaces
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
rotarHydros,ur,.rTf;J't:J#fi: 199
3-109
Problem 3'LO7
The 1-m diameter solid cylinder shown is 8 rn long perpendicular to the figure
and rests in static equilibrium against a frictionless wall at O. Determine thc'
unit weight of the cylinder'
Ans: 10.5 kN/nrl
Ans: s. = 10.5'
Water
Problem3-108
i
The section of a concrete dam is shown in the figure. Concrete weights 23'!
kN per cubic meter an<l water weighs 9,790 N per cubic meter. Coefficient of
friction between the dam and foundation is 0.55. Determine the factors of
safety against sliding and against overturning, and also the soilpressure at thc
heel and toe. Assume hydrostatic uplift varies unifornrly fro5n full hydrostatlc
head at the heel of the dam to zer6 at the toe. Consider L m length of dam.
Ans FSs= 2.20; FSs = 1.66
qL'*r = 85.2 kPa; q6= 300'2 kPt
-AA
ZUU
CHAPTER THREE
FLUID MECHANICS
& HYDRAULICS
Total Hydrostatic Force on surfaces
Re, a ti ve E
Problem 3 - 11O
s u ifi
20t
f"Tff Tu:;iT
Chapter 4
Two spheres, each 1.3 m in diameter, weigh 5 kN and 13 kN, respectively.
They are connected with a short rope and placed in water. What is the tension
in the rope and what portion of the lighter sphere prokudes from the water?
Ans: T = 1".74 kN; 40.1%
Relative Equilibrium
of Liquids
Problem 3 - 111
A block weighing 1,25 pcf is 1 ft square and 9 inches deep floats on a stratified
liquid composed of a 7-in layer of water above a layer of mercury. (o)
er certain conditions,
Determine the position of the bottom of the block. (b) If a downward vertical
force of 260 lb is applied to the center of mass of this block, what is the new
position of the bottom of the block?
Ans: (a) 0.8 " below mercury
(b) 4.67" below mercury
the
particles of a fluid
mass rnay have no
o' betwee' each other ,"r';;';:""::.:,tjtto
relahve
.14y ue
r'" tn
rr<r is
$ moving with a constanr
* motion.
,'",i,iol ',i'"',i'",,
If a nrass of
onrrun, speed
.,
.,,.-;;,.:;
(uniform velocity), ,f.r"-.".ai,ons
"""JlilT,::T',.1Y*"t
are
.h;;i;;'
;;
u,li'il,n"
;il,"*;
ooo,
1urscteJ
reguired, and this wilr lTfJilHTil; #;:'"".
be dir.";;";';;ilffi'ililotation), special rreahnent
e s.ame as in fluid ,ruri,
Problem 3 - 112
Would a wooden cylinder (rp. g.. : 0.61) 660 mm in diameter and 1.3 m long
be stable if placed vertically in oil (sp. gr. : 0.85)?
TR ANSLATToN (
Ans: Not stable
ivrovrNc VESSET)
Motion
Problem 3 - 113
A rectangular scow 7 ftby 18 ft by 32 ft long loaded with garbage has a drnft
of 5 feet in water. Its center of gravity is 2 ft above the waterline. Is the scow
stable? What is the initial metacentric height?
'
Ans: The scow is stabla
,
MG=0.90ft
rrDr(ler a lnass of fluid
*n,m*l;:.:m:T$jF.t+;*::H"i:,1.,.1nff Tj;ilT:di;
,
"
i ,' t;;
"
* i'^'j':#.,l.ili lffi"'l1;T" i;ffi i::i:i*
,j.$
h i c ri i s
w=M9
Problem 3 - 1.t4
A cube of dimension L and sp. gr. 0.82 floats horizontally in water. Is the
stable?
Ans:
:o
lt
>
REF = Ma
ZUZ
CHAPTER FOUR
Relative Equilibrium of Liguids
FLUID MECHANICS
& HYDRAULICS
tan0:
Rerative
Vertical Motion
F'ronr the force polygon showtr
REF
tan0=
FLUID MECHANICS
6( HYDRAUTICS
Consider a mass of fluid
w
-Ma
;.':",fi :'Tili; J :* :,*"
Mg
ro,,u,=tuti,"n.oli"
-
,a
tanO=:
o
Eq.4-1
accelerated uo
;flJ;lt, :L'm n*, ;x1, #
"''il;;;"
I'q'id
.b;;; il';;;ri;
"i
ii:,:T"-,:'.t:t
the i,rerria
fr:
lEF,, = ol
F=Ma+yV
,
Therefore; the surface and all planes of equal hydrostatic Pressure must be
inclined at this angle 0 with the horizontal
Esu',fffftjrfi:ir| ZO 3
M=py= I y
I
o
0 'ffi"#ffi
o
o-
F= LVa+yV
o
ff:
\
Volume, V = Ah
lnclined Motion
F=PA
Consider a nrass of fluid being accelerated upwards at arr inclinahon o v
the horizontal so that al = a cos o and n,, = n sin cr.,
l1A= oY (AI)a+y14111
6
P=y141+ a/g)
I,=,^
REF' = 1Y uu
W=Ms
3s
REFs = [t4 3n
Use 1+; for upward motion
and (-) for downward
Note: a is positive.for acceleration
From the force polygon shown
"
tan0=
tarr e =
Mou
M g+ May
aH
8+av
Use (+; sign for upward motion and (-) sign for downward motion
motion
and negative for deceleration.
204 ;5#J5[;"""trium or Liquids
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAUTICS
CHAPTER FOUR
Relative Equilibrium of Liquids
From calculus, slope = dy
dx
205
=tan0
:dv =tane= _,2,
dxg
. ur2
clY=
o
6
-r4^
lntegrate both sides:
Hgurel-t1ay
Figure4-1(b)
For cylindrical container of radius r revolved about ils vertical
axis, the height
of paraboloid is:
W=M9
W=M9
CF
e
=
(W/g) o2 x
where ol is the angular speed in radians per second.
'
Figurea-1(c)
NOTE: 1 rpm = n/30 radlsec
Figure 4 - 1: Paraboloid of revolution
4 - 1 (b),the relationship between any two points
in the parabora can
figule
grven by (squared property of parabola):
From the force polygon
tant-CF
- w
-
;= v
Eq'4-7
tano= \w/g).'zx
w
of P"araboloid of Revolution
tu^e=
tlt
g
Eq.4-4
Where tan 0 is the slope of the paraboloid any point x from the hxis of rotation
Volume=l/znrah
Eq.4_g
206
CHAPTER FOUR
Relative Equilibrium of Liquids
':""':'::::;::'"':?:::,::ers
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
CHAPTER FOUR
207
Relative Equilibrium of. Liquids
For closed cylindrical containers more than half-full of liquid, rotated
more than harr-ur or riquid, rotated about
its vertical axis (h > H/2):
about its vertical axis (/r > H/2):
€
€
I
Y/2<D
yl2=D
YIz<D
Yl2=D
(liquid surface just
touching the top rim)
Liquid surface just
touching the top rim
(No liquid spilled)
(No liquid spilled)
I
Y/2>D
Y=H
Yl2>9
Vortex at the bottom
(Some liquid spilled)
(with imaginary
paraboloid above)
-'----------t---_______-'1
ll
rl
;''l
Y = H2l2D
(vortex just touching the
bottom)
(Some liquid spilled)
For closecl vessels, there ca.n
be any liquid spilled, so the initial
of liquid (before rotation) is
s equal to the final volume of the
y>H
Vortex (imaginary) below
the bottom
(Some liquid spllled)
id (after rotation) or tht initinlaolunr
air inside is equnl to the fnnl aolume of air
The oolume of nir relation is more
ient to use in solving this type ol
vr= Q/H\(y - K1; K= Hz/2D
"-l-
y > H2l2D
(Vortex below
the bottom)
20
g Fl,#ff [;"?'Tium
FLUID MECHANICS
& HYDRAULICS
or Liquids
FLUID MECHANICS
& HYDRAULICS
For closed cylindrical containers completely filled with liquid:
U-tube revolved about its own axis:
Note: the pressure head at any point in
T-ir::-:-:l
I t\
:l
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
rivj
Yl\. ' i',
"
Without pre'ssure at top
For pipes and tubes:
t
I
i
I
I
i
i
;
<:
F-!
i
I
I
I
I
i
!
I
!
I
I
I
'i
I
I
l* *, --'l ' t'
rl-/
Without initial pressure inside
With initial pressure inside
CHAPTER FOUR
Relative Equilibrium of Liquids
209
2lo
CHAPTER FOUR
FLUID MECHANICS
& HYDRAULICS
Relative Eguilibrium of Liquids
lS"lu"d Pr"bl.-t
FLUID MECHANICS
& HYDRAULICS
CHAPTER FOUR
Relative Equilibrium of Liquids
2tt
(c) When a=6m/sz
tan0!,a6=931
Problem 4 - 1
0 = 31.45'
An open rectangular tarrk nrouuted on a truck is 5 m long, 2 m wide and 2.5 m
high is filled with water to a depth of 2 m (a) What maximum horizontal
acceleration can be tmposed or.r the tank without spilling any water and (b)
deterrnine the accelerating force on the liquid mass? (c) If the acceleration ts
increased to 6 m,/s2, lrow nruch water is spilled out?
x = 2.5 cot31.45o
x=4.0875<5m
V un = 1/z(4. 082 S) (z.s) (2)
Vut= 10.22rrlf
Vorisinar = (2X2X5)
Solution
Voriginat = 20 m3
(al
,
%piiled = Vortginar - Vleft
Vspilled = 20-1'0.22
Vspiled = 9.78 m3
Problem 4 - 2
Ihe figure shows the water level under nlaxtmunr a when no water ts
spilled out
ta60= 9; ={) 2
tano= ! =oz
compl'tely-filled with gasoline (sp. gr. = 0.g2) and accererated horizontaly
at 3
rn/s2. Find the total force acting aithe rear'wall and at the front wall of
the
tank. Find also the accelerating force on the fluid mass.
Solution
I
tan0= 3
a = 0.2(9.81)
't
A closed horizontal cyrindrical tank 1.5 m in diameter and 4 m long is
o
= 1,.962 m/s)
6
a=3m/s2
y_ 3
4 9.81
(1, )
Accelerating Force, F = Mn
Mass, M = p(Voiun-re of liqurd)
y =1..223 m
h =t.zSg+0j5
Mass,M=1000[5"2"21
Mass,M-20,000k9
i : t.gzz rn
Acceierating Force, F = 20,000 x 1.962
Accelerating Force, F = 39,240 N
F**=yi A
F*u, = (e.81 x 0.82)(1,.973)l.f
1r.sy1
F*u. = 28.05 kN
or;
F = F."o, rvall - Fhonr *atl
F = e.sl(
!
ltz.stzll e 81(+ )tl.s(2)l
F = 39.24 kN
'
i
Fmnt= yii A
Fho.t = (9.81 x 0.82)(0.75) t.f
Fr.."t =10.66 kN
1r.S1r1
2t2
CHAPTER FOUR
FLUID MECHANICS
& HYDRAULIC5
Relative Equilibrium ot Lrqurds
FLUID MECHANICS
& HYDRAULICS
CHAPTER FOUR
Relative Equitibrium of Liguids
213
Also:
Accelerahng Force
[=Ma
Tair (original) = Vair (finat)
Mass, M = p(Volurne)
4(0.2) (2) = (1. / 2) x z (2)
xz = 1,.6
| = {(1000 ' 0 ri2)l+ (t s)r(4)ll (3)
I = 17,3q() N
t = 17.39 kN
+ Eq. (2)
Substitute z and xz to Eq. (1)
aQ.6/ x) - 1.6 = 4.1x
) multiply by r
6.4 - 7.6x = 4.'t xz
or
I)
rL_D.
-rredr-rlroilt
4.Lx2+1..6x-6.4=0
F=28.05-10.66
f = 17.36 kN
:ljj
X- --__-Jtt.u
y2 - +1+.ry1-0.+y
2(4.1)
= 1.0695 m
z = 1.6/1.0695 = 1.496 m
Problem 4 - 3
A closed i'ectangular tarrk 4 nr [ong, 2 m wide, and 2 m high is filled with
water to a depth of 1.8 nr lf the allowable force at the rear wall of the tank is
200 kN, how fast can it lre accelerated horizontally?
6.ylg =!
a
t
=Z
8)c
7.496
= 1-o6ed
a =13.72 m/s2 (horizontal acceleration)
Solution
1-
4-4
f-l
open tank 1_82 m square weighs 3,425
h
N and contains 0.91m of water. It is
N p","ua to a pair of sides. what is
*j:::::?:tT::1,.*.:.of-ro,+00
force
acting in the side with the ,*uffuriJ"ptnZ
P=yha
Solvefor aand,y:
F=Ma=1O400
p=yn a
M= M*^t", f Munk
200=e.81 i1Z1Zy
Y = 1,goot(t .82)(1.82)(0.s1)l + 3,42s/s.81
M=3;363.42kg
lr =51rn
10,400=3,363.42x a
y= h I =4.1 nr
a=3.092m/sz
By similar triangles
4-x
4.1
,
.r
z
4z-xz=4.7t
) Eq. (i)
I
tan$= a = !
I
0.91,
1.82mx1,82m
2l+
CHAPTER FOUR
FLUID MECHANICS
& HYDRAULICS
Relative Equilibrium of Liquids
FLUID MECHANICS
CHAPTEII FOUR
& HYDRAULICS
3.092 _ y
Relative Equilibrium of Liquids
2t5
AasE = 5.11 m2
9.81 0.91
VaM =5.11(1.5) = 7.665 m3
y=0.29m<0.91m(OK)
h=0.91-y
Ir= 0.62m
Vrcn=12.4635 -7.665
V:".n= 4.7985 ms
P = 9,810 (0.62/2) 10.62 x 1'.821
P = 3,432 N
%pined=8.6985 -4198s
%pilled = 3.9 m3
Problem 4 - 5
An open trapezoidal tank having a bottom width of 3 m is 2 m high, 1.5 m
wide, and has its sides inclined 60o with the horizontal. It is filled with water
to a depth of 1..5 m. If the tank is accelerated horizontally along its length at
4.5 mf s2, how much water is spilled out?.
4-6(CEBoard)
A vesser 3 m in diametercontaining
2.4 m ofwater is being raised. (a)
Find the
pressure at the bottom of the vesser
in kpa when the
(b) find the pressure at the bottom
""I;iry;;;.1-,stant, and
"rirti'"*r"r when it i, u."h"rui^ g 0.6 m/ sz
upwards.
Solution
3 + 2(2 cot 60') = 5,399
t
3 + 2(1.5 cot 600) = 4,732 m
a 4.5
tan0=t =9.81
Til
0 = 24.64o
.il 1.5
EI
For vertical motion:
p=yh(1xa/g)
It=2.4m
(a) When the velocity is constant, 4
= 0, then
p=vh
p = e.81,e.a)
Vspilled=Vorig.-%eft
v*'r= lEZ (1.5)x 1.5
Vo,ig = 8.6985 m3
p = 23.544 kpa (pressure at the bottom)
(b) When a = 0.6m/s2,(use',
,
\,, forupward motion)
p=e.81(2.4X1+(1**f)
V*n= Vasco - VaeE
p = 24.984kpa
vAsco = 3rry92 (2x1.s)
Vasco = 12.4635
rfi
Varc: Aase(1.5)
Aale = 1/z (AB)(AE)sinl
cr=180o-60"-24.64'
or = 95.35o
AE = sin 60o
AE = sin50' = 4.618 m
A as r = 1/z(5.309) (4.618) sin 24. 64'
4-7
containing oil is accelerated on a plane
incrined 15. with the
at 7.2 m/ sz. Determine the inclination
i3ontal
of the oil surface when the
vessel
-
is (a) upwards, and (b) downwards,
CHAPTER FOUR
216
Relative Equilibrium of Liquids
FLUID MECHANICS
& HYDRAULICS
Solution
FTUID MECHANICS
& HYDRAULICS
CHAPTER FOUR
Relative Equilibrium of Liquids
(c) Downward motion with a positive acceleration (use,,_,,with
tan0=
AU
p = (e.81x 0.s)(3) (t -
Stav
2t7
a = +g m/s2)
r*-)
p =4.34kPa
All=OCOSd.
(d) Downward motion with a negative accereration (use "-" with
as = L.2 cos 15o
as = 1.159 nf sz
p = (e.81. x 0.8)(3) U p =42.74kPa
av= asincx
av = 1'.2 sin 15o
ay = 0.31, mf s2
4-9
(a) When the motion is upwards:
1.759
tan0=
a = -grn/s2)
Dt)
9.81+ (+0.31)
0 = 6.533'
A cylindrical water tank used in lifting water to the top of a tower is
1.5 m
high- If t{re pressure at the bottom of th*e tank is must not
16 Kpa, what
"x."ed
,ntaximum vertical acceleration can be imposed in the cylinder
when it is filled
tvith water.
(b) When the motion is downwards:
tanH=
1.1.59
P=yh(1+a/S)
e.81- (+0.31)
16 = 9.81(1.5)(1+ a/9.81)
a = 0.857 m/sz
0 = 6.955"
4-to
Problem 4 - 8
An open tank containing oil (rp. g.. = 0.8) is accelerated vertically at 8 m/s2.
Deterrnine the pressure 3 m below the surface if the motion is (a) upward with
a positive acceleration, (b) upward with a negative acceleratiorL (c) downward
with a positive acceleration, and (d) downward with a negative acceleration.
Solution
/
\
The pressure at a depth ft is given by,.p = *ltxZ)
(a) Upward motion with a positive acceleration (use "+" with a = +8 m/s2)
p = (e.81. x 0.s)(3) (t.
#r)
P = 42'74kPa
(b) Upward motion with a negative acceleration (use "t" with a = -8 m/s2 )
p = (s.81x 0.s)(3) (t .
P = 4'34kPa
rt)
open rylindrical vessel having a height equal to ib diameter is
half-filled with
er and revolved about its own vertical uiis *ith a constant angular
speed of
rpm. Find its minimum diameter so that there can be no riquid Jpil"a.
ttrat there's no liquid spilled, base of the
must just coincide with the
rim of the cylinder. Since the cylinder
initially half-fuIl, the height of the
aboloid is therefore equal to the height of
cylinder.
,lt= _,2r2
2g
h=H=D
or = 1.20 rpm x n/f,Q
a = 4n rad/sec
-l
ll'
2 t 8 [:fi,J5:;.?',Y#ium or Liquic,s
FLUID MECHANICS
& HYDRAULICS
CHAPTER FOUR
Relative Equilibrium of Liquids
2t9
%pilled = Vairltinag - Vuirlinli"tl
(4n)2 \D / 2))
,.
l,=
FLUID MECHANICS
& HYDRAULICS
2(e.81)
l) = 0.497 rn or 497 mm
%piued = 1/ztr(0.6)2(1,63)
- n(0.6)2(0.7)
Vsplled = 0.13 m3 x 1000
lit/ml'
%pired = 130 liters
-
Problem 4 - Ll
An open cylindncal tank 1.6 nr tu dtameter and 2 m hrgh rs iull ol water
When rotated about its vertical axis at 30 rpnr wlrat would be the slope of tlrt
water surface at the rim of the tank?
Solution
2
u)I
the water stand when brought to rest?
(d). what angular speed ro (in rpm) will just zero the depth of water
I
2n rad
I mrtr
rev
60 ser
at the
center of the tank?
(e) If or = 100 rpm, how much area at the bottom of the tank is uncovered?
to = n rad/sec
Solution
(n)2 {0.8)
a212
981
2g
'
(a) ro=3radlsec
Problem 4 - L2 (CE Board November 1978)
/, =
An open cylindrical vessel 1.2 m in dian-reter and 2.1 m hrgh rs 2/3 full
water Compute the anrount of water ilr liters that will be spilled out is
vessel is rotated about its vertical axis at a constant angular speed of90 rpm
(?):
(l)'z
2(e.81)
22
h = 0.46
ir=1m
->
t\ t'
The maximum <o so that there is no liquid spilled is such that
h/2=lmor/r=2m
.
)o
-6
ur = 90 rpnr
4
h/2=0.23<7m
.'. no liquid is spilled out
F
Solution
,
.
(a) If ro = 3 rad/ sec., is there any liquid spilled?
(b) what maximum value of ar (in rpm) can be imposed without spilring
(r) If o = 8 radf s, how much water is spilled out and to what depth wilr
SloPe = _-
Slone =
of 3 m. It is rotated about its own vertical axis with a c-onstant angular speed ol
any liquid?
Slope = tar.r 0
rev
trr = -]0
min
Problem 4 - 13
An open cylindrical tank, 2 m in diameter and 4 m high contains water to a clepth
' n/3()
^2r2
o = 3n rad/s
2g
,'(1)'
, 1sn.1210.0;2
n=
"-
/r = 1.63 m
a = 6.26 rad/sec x 30
2(9.81)
h/2 : 0 815 > 0 7 nr (sorne hqurd sprlled)
-
2(e.81)
to = 59.78 rpm
=0.6m
CHAPTER FOUR
--A
ZZV
Relative Equilibrium of Lrqurds
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
{c} r,r=8racl/ser
Ir='
(8)2(1)2
2(e.81)
CHAPTER FOUR
& HYDRAULICS
Relative Equilibriurn of Liquids
Area,l=nyz
ro = 100 rpm
y=5.59_4
€
=J26ur
.y = 1.58 m
h/2=1.63nr>lnr
By the squared property
of parabola;
some liquid spilled but the vortex of the
paraboloid is inside tlre tank si4ce /r < 4nr
x?
-12
!/h
x, = *
ysprllerl = Tarr (trral) - Vo,,
t,u,l,"l,
Varr (hnal) = Vparatrlor<l
Varr (rrnar) = L/z n(1)2 (3 2ol
(1.s8)
x2 = 0.283
V.r (finat) = 5.I 21 nr '
Area, A = n(0.253)
l/arr{rrrrrrat) = tr(1)2(1)
Area,A=0.889m2
Varr (rnrtrat) = 3.142 nt'
V;r,,re,r = 5.121 .a 142
Vspiloa = '/,,.979 m3
Another solution
When the tank rs brought to rest, the
water level will rest at /r/2 fronr top
a:?
E5,
v=h/2-1
v = 1.63 '1
r/=063rrr
= n r) !
4piled = 7r (1)2(0.63)
V
'r'11n'1
'
4pillecr = 7,979 m3
(d) The vortex touches the bottonr when lr = 4nr
" ,2 0\2
+=
2(e.8't)
r,r=g.g6* #
rrl = 84.6 rpm
lel When o = 100 rpnr
t,r = 100(n/30) = 3.33n radlset
,.: ot2r2 _ (3.33n)2(1)2
2g
2(e 81 0)
h=5.58>4nr
the vortex of the paraboloid rs already below the tank (imaginary)
,
a2r2
2g
Solve for ft (by squared property)
Q)2 (0.s)2
h h-4
h-4=0.25h
0.75h = 4
/r = 5.33 m
5.33 = g-11'
2(e.81)
o = 10.23 rad/sec x !!
a = 97.65 rpm
22t
CHAPTER FOUR
--ZZZ
(bl
FLUID MECHANICS
& HYDRAULICS
Relative Equilibrium of Liquids
FLUID MECHANICS
& HYDRAULICS
Vten = Vc),linae, - Vf.urtuu, of paraboloid
CHAPTER FOUR
Relative Equilibrium of Liquids
223
hz= pz/y
Vron = 7r (1)' (4) - [1/z n (1)2 (5.33) - 1/ztr (0.5)z (5.33 - 4)]
lu= 74/0.0008
Vv* = 4.776 m3
hz = 17500 cm = 175 nr
h=hz-2.75-p1/ru
h=175-2.75-62.5
Problem 4 - 15 (CE Board November 1993)
h ='109.75
A 1.90 m diameter closed cylinder, 215 m high is completely filled with oil
tog.zs =
having sp. gr. of 0.8 under a pressure of 5 kg/cm2 at the top. (a) What angular
speed can be imposed on the cylinder so that the maximum pressure at tl're
bottom of the taflk is 14 kg/cm2? (b) Compute the pressure force exerted by oil
on the side of the tank in kg.
Solution
ft
o = 466.44 rpm
[,
p=yn A
i
tr=0.95
I
2(e.81)
co = 48.84 rad/sec x 30
(bl
lmaginary L.S.
a2 (o'gs)2
= hz - 2.75/2 = 173.62s m
F = 800 (17 3. 62s)
[zn (O.es) (2.7 s\]
F = 2.28 x 106 kg
l6v
I
I
!
I
F
4 - 16 (CE MayigsD
'f
hz = Pzly
E
b
N
tri
o{"" cylindrical tank having a radius of 300
lTmm and a height of
m is
full of water. How fast should i", U"
,.i"i"i about irs own vertical axis1.2
o/"
so
that
of its volume will be spilled out?
,
Oil (s! = 0.8)
a2r2
2g
:e 75o/o of the total volume is
spilled out,
paraboloid will be formed a part
outside
vessel (i.e. with its vortex beiow
the tank)
2n(0.95) = 5.969 m
Unit weight of oil, y = 1000(0.8)
Unit weight of oil, y = 800 kg/m3 = 0.0008 kg/cm3
22
(al l, = '='
z8
Solve for lr :
Pt/r = 5/0.0008 = 6250 cm
fi/y = 62.5 rn
%piled = V
= 0.75[nrz(1.2)]
^"
V^i, = 0.9nr2
t
V^ir= Tbrgparabolotd - Vsnrallparabolord
0.9na = yz nr2 h _ 1/2 r.x2 y
7.8f=r.zh_x2y >Si.(1)
o=7
I
224
CHAPTER FOUR
Relative Equilibrium of Liquids
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
Rerative
lly squared properiy of parabola:
*2 ,2
vhh
a ='1.2.528 rad/sec x
,2
co = 119.64 rpm
ln Eq. (1)
1.8 P = ,'h -
Ec,i,if"ryf;:?t;?il: ZZs
rea 60 sec
2rrnd lmin
'J,
-x_
t2
;lt y(y) ) multiply both sides by h/ta
1.81t=lP-t/2
4-18
An open vessel,500 mm.in diameter
butY=1,-1.,
1.8h=h2_(h_t.z\z
and fillecl with water, is rotated
about is
100 r'm rrom the axis
angte of 40o with the
fi':.TlTi;:il'-i
makes
;,ll':y,*:ilt::$;;;..
hoiizo'tat. C;;ilil:;".1':T;.;"il,1i:
illl
1..81r = l? - (hz - 2.4h + 1.44)
0.6h = 1.44
h=2.4m
The slope of the paraboloici
tuno='2,
I
Finally:
2.4 =
,210.3;2
2(e.81)
at any clistance ,,r,, fronl
Where 0 = 40. and;r = 0.1 m
a :22.87 rad/sec x 30T
tan4oo=
trl = 218.4 rpm
a = 9.07 rad/sec x !Q = g6.64rpm
Probfem 4 - 17
$fo.rl
4-19
Ar"r open cylindrical tank 1 m in diameter and 3 m high is full of water. At
what speed (in rpm) must it be rotated to discharge 1/3 of its content.
Solution
open cylindricar tank 1.2 m in
diameter and 1.g m deep is
filred with water
rotated
vs qvvqr
about rrb
its uwn
own axrs
axis at 50
rewnl,,r,^-. per
6O.revolutions
--_, ^^_ rninut,e.
How
and wllat is the pressure
at the center of its bottom?
".r"Jfiq.,ia
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
Yz n (0.5)2 y =
t (0.s), (3)
"n
y =2m
,= #r
., _ .2 10.5;2
2(e.81)
the axis is given by:
.r=o5m
ro = 60 rpm
,
a2r2
I
2g
st-
r€:
co = 60 x (n/30) = 2n
rad./ sec
r ='1 .2/2 = 0.6 m
r,= Qt-9.t=0.724m
%pillea = Tparatroloid
%pilied = lz n2 It
V,spirhcr = Vz rc(0.6)z (0.724
%pilled = 0.409 m3
)
r=0.6m
226
CHAPTER FOUR
FLUID MECHANICS
& HYDRAULICS
Relative Eguilibrium of Liquids
FLUID MECHANICS
CHAPTER FOUR
& HYDRAULICS
Relative Equilibrium of Liquids
h/2=3.67m>1m
[)rcssure at the center:
(part of the paraboloid is above the vessel)
P=vy
l/ = 1.6 - ft = 1.8' 0.724 = 7.076 r^t'r
p = 9.8"1(1.075) = 10.555 kPa
Verify tl're position of the vortex (See page 207)
H2 _ $)'
2D
Problem 4 - 20
A closed cylindrical vessel, 2 m in diameter and.4 m high is filled with water
to a depth of 3m and rotated about its own vertical axis at a constant angular
speed, ol . The air inside tl-re vessel is under a pressure of 120 kPa.
(o) If io = 12 radf sec, what is the pressure at the center and circumference al
the bottom of the tank?
(b) What angular speed tu will just zero the depth of water at the center?
(c) lf a = 20 rad f sec, how much area at the bottom is uncovered?
Solution
a = L2 radls
=
->
I
T--fI r'
|ml
| ,'.
ll
I
)o
-d
,., -_
2(e.81)
Y nir (final) --
r,
Y arr (irlitidl)
lznxzy = nrz111
x2y=2P )Eq.(1)
By squared property of parabola:
r-)1 r'
ylt
^12
It '
>Eq.(2)
Substitute 12 in Eq. (2) to Eq. (1)
t
( r' !t)v=2rz
u
!/2 = 2lt: 2(7.34)
y=3.83m<4rn
Pressure at the center, (at O)
pt=ylh+pait
ht=4-lt
lq=4-3.83=0.17rn
fi = 121,.66 kPa (pressure at the center)
p2=ylh+ p^n
lh= h't + Ir
Iu= g.t + 7.34 = 7.51 m
/s
(t2)2 (1)2 :7
Il
the vortex is inside the vessel
Pressure at circumference, (at O)
(al Refer to Figure (h)
22
. oJr
lt = _
n',2
= 8nr>7.3{m
2D
p=9.87(0.77)+tzO
I
Figure (a)
rt: -- "12 rad
xtx
i
2(1)
pz = 9.81,(7 .51) + 120
z,t
pz= 193.67 kPa (pressure at the ci_r.curnference)
227
CHAPTER FOUR
ZZA
Relative Equilibrium of Liquicls
FLUID MECHANICS
& HYDRAULICS
(b)
' [,et us first derive the general value of ft
FLUID MECHANICS
& HYDRAULICS
,
By squared property of parabola
xr22
Hh
2
r,='
) Eq. (b-z)
H
It
,2r2
2g
rt_ Qo).(\2
'Varr (final) = Varr (rntttall.
x2H=2rzD )Eq.(b-1)
2(e.81)
,
h=20.4m
(Lr\H=zt2D
lh
)
Simplify:
,=
fteight of the paraboloid when it touches the bottom)
#
E
tf
In Figure (d):
V"i. liniai"ly = Vai.
ct
N
1finu4
lt
ni(7) = rTrn*tz yt - lznx2z y2
2r2 = xt2 !/t - xz2 yz >nq.
1c_r;
-c
By squared property of parabola:
*t2
Substitute.r2 to Eq. (b-1)
CHAPTERFOUR
(c) or = 20 rad/ sec
when the vortex of the paraboloid
reaches the bottorn of the vessel
L/znx2H = niD
-
Relative Equitibrium of Liquiai ZZc)
xr2
12
atVzh
I
^1- - --h" V7
) Eq. (c-2)
t2-^12
-n
) Eq. (c-3)
llt
Figure (d)
Substitute rr2 and *22 to Eq. (c-7)
2
2t= + v,(yt-
{ yr 0lz) multiply both
2
side by h/r2
2h=y-r2-y2z
h.
11
=
(4)2
2(1)
-8m
,2r2
2g
-
Butyr=4+ y
n=@+yz)2-y2z
2h=15+8yz+!22-y22
8yz= 2(20.4) - 16; yz= 3.1 m
6 = '2(t)2
2(e.81)
Eq. (c-3)
a = 12.528 rad/sec , 30r
or = 11.9.5 rpm
^12
X2'=
-hr-
*n=
U,
#(3.r; =e.15,
Area = nxz2= n(0.1521
Area = 0.48 m2 (area uncovered
at the bottom)
FLUID MECHANICS
& HYDRAULICS
23o ;i,:i'J5[;:,,Y*um or Liquids
FLUIDMECHANICS
& HYDRAULICS
CHAPTER FOUR
Relative Eguilibrium of Liquids
Problem 4'21
A closed vertical cylindrical vessel, 1.5 m in cliameter and 3.6 m high is 3/a
full of brine (s = 1.3) and is revolved about its vertical axis with a constant
angular speed. The vessel is made up of steel 9 mm thick with an allowablt'
Determine the position of the vortex:
tensile stress of 85 MPa and has a small opening at the center of the top cover
(a) If the angular speed is 2L0 rpm, what is maximum the stress in the walls?
(b) To what maxirnum angular speed can the vessel be revolved?
Since /r = 13.86 > 7.2, the vortex is below the
H2 rc.G\z = 7.2rn
2D 2(0.e)
vessel, See Figure (b)
V"ir 1iniu"l1 = Vair 1nnat1
nr2(0.9) = Yznxf lt - l/znx2z y2
1.8 rz = xr2 yt - xz2
Solution
(a) a=210rpn"rxnl30
2t
) Eq. (1)
By squared property of parabola:
a=Trradfs
PD
S, =
yz
x'r2
xr2
^=_=Ur !/z
Note: The maximum pressure ls at the circumference at the bottom
I r=0.75m
P = lt,in" h'r
G'
il
r=0.75m
,2
h
^12
xt'=
T lt
) Eq. (2)
^12-; y2
x2'=
) Eq. (3)
Substitute rr2 and rz2 to Eq. (1)
'
-.2
,2
't.8r?=
r rr(rr)- |V,0)
1..8h=yf_y2z
But yr = 3.6 + y2
1.97=(3.6+yz)z-y22
1.8h= 12.96 + 7.2y2+ yz2 _ y22
7 .2 yz = '1,.8h - 12.96
111
Figure (b)
Figure (a)
) multiply both sid.es by h/p
) Eq. (a)
7.2y2 = 1.8(13.8d) - 12.96
lz= 1.665 m
lu: 1g.8G - 7.66s
h=12.195rr.
p = e.81, (1,.3) ('I2.1es)
p = 155.52 kPa
Solve for hr:
h=h-Vz
, ,2r2
2g
/r = 13.86 m
(7n)7 (0.75)2
2(9.8"1)
-
s,=
(L55.52)(1500)
i
,\
/
2(e)
St= 12,960kPa
&=12.96 MPa (maximumwall skess)
231
CHAPTERFOUR
-t-5 Z
Z
Relative Equilibrium of Liguids
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAUTICS
CHAPTER FOUR
Relative Equilibrium of Liquids
233
(lt) Iror nraximum value of trl, Sr = 85 MPa
82x103=
e.81(1.3)h] (1500)
85 x 103
2(e)
pz:455.5 kPa
h1 = 79.98 m
Pz = Ylt2
y2=h-79.98
455.5 = 9.81(L.6)h2
ln= 29.02n..
In Eq. (a)
7.2(h - 79.98) = 7.8h - 12.96
h=lu-pt/v-2.7
5.4h = 562.896; h ="104.24 rn
h: 29.02 - 2.7 - 15.61
22
trr =
9-I-
h = 10.77 m
1
,/o
-6
104.24
-t'= ,rtz (0.g\2
=10'77
2PJ'L,
= '2(o'zs)'
to = 16.1 rad/sec x 39
2(e.81)
o = 60.3 rad/sec " 30 / n
o = 576 rpm
co = 1.53.8
4- 23
Problem 4 - 22
A 1.8 m diameter closed cylinder, 2.7 m high is completely filled with glycerirr
lraving sp. gr. of L.6 under a pressure of 245 kPa at the top. The steel plakrr
which form the cylinder are 5 rnm thick with an ultirnate tensile stress of 82
MPa. How fast can it be rotated about its vertical axis to the point of bursting?
Solution
water pump
is rotated at
rpm. If the casing is full of water, what pressgre is developed
by
tion?
p
- = il
v
, _a2r2
tt-
)o
-6
.29
'Solve for /r
y
1..5 m diameter impeller of a closed centrifugal
Pressure head,
, a2r2
tl = _
Pt-
rpm (maximum allowable angular speed)
245
ro = 50n rad/sec
e.81(1.6)
rr = 15.61 m
h-
r=0.75m
, o - 1500xn/30
hz = Dzlt
,, =
(sol)1(o,zs)2
2(e.81)
=707.4rn
Y
The maximum tensile stress occurs
at point @:
pD
From Sr =
2t
*n =707.4mof water
r 5,940 kPa
234
CHAPTER FOUR
FLUID MECHANICS
& HYDRAULICS
Relative Equilibrium of Liquids
Problem 4 - 24 (CE Board)
FLUID MECHANICS
CHAPTER FOUR
& HYDRAULICS
235
Relative Equilibrium of Liquids
Solution
A conical vessel with sides inclined 30' with its vertical axis is revolved about
another axis L m from its own and parallel. How many revolutions per
minute rnust it rnake in order that water poured into it will be entirely
t
Pz=vlh
!
I
Solvirrg for /r2:
lu=yz-yt
)1
,lI2= o-x^'
discharged by the rotative effect?
Solution
--32g
i 0= 27.5 rad/sec
I
I
-
I
I
. Q7.sf
t,,=
The water in the vessel will entirely be
discharged at a speed when the
paraboloid is tangent to the cone at the
vertex, hence, tfie inclination, 0, of the
paraboloid at r = 1 m is 60o or its slope is
tan 60o.
VOf. [(2.s), - (0.s),]
I
!
i
I
I
I
I
lu= 23L.27 m
,+ 0.5 -+l
iol
i 30.
zZm
::-r-=1:_: --:: tr_
_____t
i
3Oo
Yz
I
p = (e.81, x 0.s22)(237.27)
i*-- r.,."
p = 7,,865kpa
1
From the formula:
-12
f2n$=
o
-1d
2
(r)
tan 60o = *
9.81
'.
a=4.12rad/secx 39
--
rr ,/i
i
f-------7-i
i/:/
t/
--i-"
i
4-26
glass u-tube whose verticar stems
are 300 mm apart is fited with
mercury
verrical ;,-"*"- ii is rotatea about
a vertical
:*T,1,:ti,rri::
ugh the midpoint i,p
of the horizo.,,ur ,".uor,];il;i#il:";:::.: axis
ill;
luce a pressure of absolute
zero in ;i";;;y
at the axis?
ro = 39.36 revolutions per minute
a2 *2
Problem 4 - 25 (CE November 1992)
A 75 mm diameter pipe,2 m long is just filled with oil (sp. gr. = 0'822) and
then capped, and placed on a horizontal position. It is rotated at27.5 rad/str€
about a vertical axis 0.5 m from one end (outside the pipe)' What is tht
pressure in kPa at the far end of the pipe?
"29
ll=1t,,,+9.15
r=0.15m
btnce the pressure at the center
is absolute zero, then the gage
T
-760 nrmHg, therefore lu,'
.L
f
0.15
pressure at the center is _pot. or
Y=0.75+0.75
y=0.91 m
g.91 = <o2(0.t5)2
2(e.81)
a = 28.17 radlsec x 30
<o = 269
rpm
= [.7U
^
h,
I
CHAPTER FOUR
236
FLUID MECHANICS
& HYDRAULICS
Relative Equilibrium of Liquids
Problem 4 - 27
FLUID MECHANICS
& HYDRAULICS
RerativeEq,i,,ff,;1::1;fl:T 237
$olution
A glass U-tube whose vertical stems are 600 mm apart is filled with mercury
to a depth of 200 mm in the vertical stems. It is rotated about a vertical axis
lhrough its horizontal base 400 rnm from one $tem. How fast should it be
rotatecl so that the difference in the mercury levels in the sterns is 200 rnm?
d-
\>u,
:
I
Solution
T
I
xz = 0.4m
_______+k_xr
I
ht
I
Initial mercury
level
0.2 m
tce there is no liquicl spilled
out, its initial volume is equal
to its final
lu're or as show^ in the hgure,
th"'.;;;;'arL nurgn, of water in the
vertical
ms before and after rotatio]r
,"rrt t
.""'
" "lu'uf
In the figure shown:
1/z- lr = 0.2
a2 *22
2g
4r+4r=5(0.5)
,2*2
-
where
Yt + yz = 1..25
1n
"
- ,'*"r2 = 0.2
.
2g
,2
[(0.q)z- (0.2)r]= 0.2
4r.81)
a = 5.72 rad/sec x 30T
ro = 54.6 rpm
) Eq (1)
By squared property of parabola:
Q)2 (o.s)z
Uz
!12=
4Yr
Ut
) Eq. (2)
Substitute y: to Eq. (1)
y1+ 4y1= 7.25
y = 0.25
Problem 4 - 28
A glass tubing consist of 5 vertical sterns which are 500 mm apart connected to
a 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 the
middle stem to just zero the depth of water in that stem?
(1)2)I"l/i = ------:-
"29
g.2u = ^2 (o.s)2
2(e.81)
a = 4.43 rad/sec x 30
n
or = 42.3 rpm
238
CHAPTER FOUR
Relative Equilibrium of Liguids
FLUID MECHANICS
& HYDRAULICS
Problem 4'29
FLUID MECHANICS
CHAPTER FOUR
& HYDRAULICS
Mirrimize /r', differe'riate Eq. (1) with respect to
A 75 rnm diameter pipe, 2 m long is filled with water and capped at both ends
:dlr = 2.5n8* - tan 60o = 0
vertical axis through its lower end with a constant angular speed of 5 rad/sec
(a) Compute the pressure at the upper end of the pipe and (b) determrne the
minimum pressure and its location in the pipe
a = .r sec60"
n = 0.68 sec60o = I .36 nr
It is held on a plane inclined 60" with the horizontal and rotated about a
Solution
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 rn, and
therefore the vortex of the
paraboloid will be 1.73 m above
the lower end.
, ,2r',
)o
,
,t=
-i- arrd equate
to zero:
r = 0.68 n'r
In Eq. (1):
ll = 1.27aQ.68), + t.Zg - (0.68) tan60o
h' = 1.747 nt
p.,u = 9.81(1.141)
p,ri,,=11.196 kPa Iocated 1.36 m frorn the lower.errcr
(arong the
pipe)
4-30
cyli'drical bucket 150 mm in diameter and 200 mm
high contains 150 mm of
rter: A boy swings the bucket on a vertical plane so that
the bottom of the
:ket describes a circle of radius 1 m. How fast
should it be rotated so that
water will be spilled?
(s)2(1)',
+
2(9.81)
O.7mA
h = 1..274 m
*-
(a) Pressure at the uPPer end
Pus,y*, = Ih
/v
,r\
pupper = 9.81(1.274)
''' ":'- .."i.
/
Minimum pressure
|
*
flt'
I
pn r, =
s
c
1r:
.
| \"\. -\'
.i i\..
puppe, = 12.497 kPa
(b)
239
Relative Equilibrium of Liquids
il
-- .....
I
\
Solve for ft'
,\
h'--Y+2
z='1.73-rtan60o
,-'
,2
y
1.2 '!.2
lt L.274
a = 1'274 xz
h' ='1.274 x2 + (1.73 - x tan60") ) Eq. (1)
\
'-i
\
:
l::i i
t...
rri':-]i
\tf/--:!t
\ ---l
Figure (b)
Figure (a)
240
CHAPTER FOUR
Relative Equilibrium of Liquids
FLUID MECHANICS
& HYDRAULICS
l he critical position for the liquid to fall is at the highest point
From Figure (b).
CF =W
CF=Mn,.
FLUID MEcHANIcs
CHAPTER FIVE
& HYDRAUTICS
Fundamentals of Fluid Flow
241
Chapter 5
Fundamentals of
Fluid Flow
w
CF=-rrr2r
I
W
-f, ul) r=W
The previous chapters
t.r2r=g
(02 (0.925) = e.8l
ot = 3.26 rad/sec
only with fluids at rest in
which the only
tigruficant property used-dears
is the weight or *re Ruia. Thrs chapter
--iit aua *itt
fluids in motion which is based on ihe fo'owingprincipres: (a) the principle of
eonservation
" J9
of mass, (bl
.
feof-energy
rnergies), and (c) the principle
r'r = 31.13 rpm
principle (the' kinetic ani potential
-oril.,i rr.,
Problem 4 - 3l
A cubical tank is filled with 2 nr of oil having sp. gr of 0.8. Find the force
acting on one side of the tank when the acceleration is 5 m/s2 (a) vertically
upward, and (b) vertically downward
oR FLOW RATE, Q
harge or flow rate is the amount of fluid
passing through
as a max ftur,.,i,
.:,{H:; *:::.::l:"'"q
kN/sec),
and aolume floto rate or
Solution
(al
Volume flow rate, e = Aa --
F = p,rA
F = fy lurQ + (g)) A
Mass flow rate, M= p
F = (e.81 x 0.8)(1)(1 + 5 / e.81)[(2)'?ti
E+ 5- 1
Eq.S _2
where:
F = p,rA
F = [7lr.r(1 - a/9] a
F = (9.81
e
.--
Weightflow rate,W=y e
F = 47.392 kN
tbt
a section per
1u*"ug1,",i,,,,J,t 7o,, rate
floto iaten(ex. m3/s, litls)
" 0 8)(1)(1 - 5/e.87\(2x2)
F = 15.392 kN
2m
Q = discharge in m3/s or fF/ s
A = cross-sectional area of flow in m2 or
ft2
z, = mean velocity of flow
in m/s of ft/ s
'p = mass density in kg/mr or
slugs/fts
y = weight density in N/mr
or lb/fF
OF TERMS
Flow.
1a1 be steady or unsteady; uniform or n on-untform ; continuou s ;
' or turbulent;
one-dimensional, ftio_dimensional
or urotatronal
or three-dimensional; and
Z+Z
CHAPTER FIVE
Fundamentats of Fluid Flow
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
& HYDRAULICS
CHAPTER FIVE
Fundamentals of Fluid Flow
243
Steady Flow
Turbulent Flow
This occurs when the discharge Q passing a given cross-section is constant with
time. If the flow Q at the crosi-section varies with time, the flow is wtsteady.
The flow is said to be turbulent when the path
of individual particles are
irregular and continuously cross each other. Turbulent
flow normallv occurs
when the Reynolds number exceed 2,100, (although
the most .o-*on
situation is when it exceeds 4000).
Uniform Flow
Laminar flow in circular pipes can be maintained
up to values of & as high as
50'000' However, in such cases- this rype of flo* i,
i"r,"r"rrirf ,r*tabre, and
the least disturbance wilr transform ii instantly
into turburent flow. on
This occurs if, with steady flow for a given length, or reaclr, of a stream, the
average velocity of flow is the same at every cross-section. This usually occurs
when"an incompressible fluid flows through a stream with uniform crosssection. In stream where the cross:sections and velocity changes, the flow is
the
other hand, it is practicaily impossible for t'rbul""in.*
*
pi;;
persist at values of & much below 2100, because
"'r""lgn,
any turbulence
that is set up
will be damped out by viscous friction.
said to be non-uniform.
Continuous Flow
One-Dimensional Flow
This occurs when at any time, the discharge Q at every section of the stream
This occurs when in an incompressible fluid, the direction
and magnitude of
he velocity at all points are identical
the same (principle of conseruation of mass).
Flow
occurs when the fluid particles move in planes
or parallel planes and the
mline patterns are identical in each planei
Continuity Equation
F
Q=A{ut=Azuz=Atat=constant Eq.s-a
.l
F
are imaginary curves drawn through a fluid
to indicate the direction of
r in various sections of the flow of the fluid
system.
or incompr essible f luids:
tubes
or co mp re ssible fluids:
PrArar - PzAzaz= Py'r:roI = constant
arhA;rr\= y2A:llr;z* lsAs:oe = constant
I
represents elementary portions of a flowing
Eq' 5 - 5
Itreamlines which confine the flbw.
fluid bounded by a group
n+ !: 6
Nets
Laminar Flow
The flow is said to be laminar when the path of individual fllrid particles tlo
not cross or intersect. The flow is always laminar when the Reynolds nurrtlre!
R" is less than (approximately) 2,100.
are drawn to indicate flow patters in case
of two-dimensional flow, or
three-dimensional flow
CHAPTERFIVE
-,,
Z++
FLUID MECHANICS
6. HYDRAULICS
Fundamentals of Fluid Flow
ENERGY AND HEAD
'l'he energy possessed by a flowing fluid consists of the kinetic and the potential
energy. Potential energy may in turn be subdivided into energy due to
position or eleuation above a given datum, and energy due to pres;sure in the
fluid. The amount of energy per pound or Newton of fluid is called thelrcad.
FLUID MECHANICS
& HYDRAULICS
a fluid which,has a-small
.>
pt"rrrrruenergylseqivalent-a,"-**'
Pressure Energy = Wl-
=*
+w29
Eq.5-8
z = position of the fluid above
p = fluid pressure
r) = mean velocity of
_
_;p
Eq.5_13
(+) or below (_) the datum
plane
flow
Flow Energy, E
,2 _(e/A)2 = e,
29
29
2gA2
zs(iD ,Y
W
where:
For circular pipe of diameter D flowing full:
)o
-=
Eq. 5 _ 12
v
Eq.5 -7
I
q_
Hence the
Pressure head = PressureEnergy
_
K.E. =1/z M az = yrATp
245
opening at the top
,H:Yi.l:::'."1" "r,,1" top, ihe n"ia-pr".u.aily will
w'r not tlow.
flow. In
rn Chapter
ch"-r.* 2,
the equivalent head (pr"rr.r;1;;;;
i;;:;""uv
tvt o pr€ssure of p it p/y
The ability of the fluid mass to do work by virtue of its velocity:
u')
Fundarnentals of Fluid Flow
Pressure Energy (potential
Energy)
consider a crosed tank filled with
Kinetic Energy
Kineticorvelocityhead"=
CHAPTER FIVE
]
total energy or head in a fluid
flow is the sum of the krnehc
arrd the
rtial energies. It can be summarized
as:
Total Energy = Ki""fi
o'
2s
=
=
TotalHead, ,=
8Q''
Eq"s-e
WF
*.
f,.z
Eq.5_15
The energy possessed by the fluid by virtue of its position or elevation with
respect to a datum plane.
AND EFFICIENCY
lver is the rate at which work
is done. For a fluid of unit
weight r (N/m.ti
I moving at a rate of e (m3ls) with
a totJ eiergy of E (m), the power
in Ne floule/sec) or watts is:
Elevation Energy =Wz = Mgz
Eq.5-10
Power = ey E
Eq.5-16
Eq.5-11
Efficiency, ,., = OttPttt x 700%
lnput
Eq.5 -77
Elevation Energy (Potential Energy)
Elevation head =
Elevation-Energy
w
-,
Note; 1 Horsepower (hp) = 246 Watts
1 Horsepower (hp) = 550
ft_lblsec
1 Waft = 1 N_m/s = 1
Joule/sec
Z+O
CHAPTER FIVE
FLUID MECHANICS
& HYDRAULICS
Fundamentals of Ftuid Flow
BERNOULLI'S ENERGY TH EOREM
l'hc Bernoulli's energy theorem results from the application of the pnnuples ol
r'ottsen,ntion of energrl. This equation may be surnmarized as follows:
FLUID MECHANICS
& HYDRAULICS
CHAPTER FIVE
Fundamentals of Fluid Flow
247
Energy Equation with Head Lost:
Considering head lost, the values
that we can attain are called
nctual oqlttes.
With reference to Figure 5 _ 4:
Bernoulli's Principle, in physics, the concept
that as the speed of a moving fluid (liquid or
gas) increases, the pressure within that fluid
decreases. Originally formulated in 1738 by
Swiss mathematician and physicist Daniel
Bernoulli, it states that the total energy in a
steadily flowing fluid system rs a constant
along the flow path. An increase in the fluid's
speed must therefore be matched by a
decrease in its pressure.
f'*
|
.5-18
Eat sechon I * Eadctecl - Elost or extlacted = Eat section 2
nerr.cgdgg",ecl
-r-
Pll
\
-.
Energy .Equation without Head Lost:
If the f-luid experiences no head lost in movrng from section 1 to section 2 tht'rr
.l
the total energy at section must be equal to the total energy at section 2
Neglecting head lost ir-r fluid flow, the values that we get are called ideal ot
theoretical ualues. With reference to Figure 5 - 3:
Et: Ez
2g
Pt
+a=
2
0o
2g
v
-f
v12l2g
+22
Eq.5 - 20
vz'}l29
I
I
+
I
PzJt
Zt
I
Figure 5 - 3
Ll
Z2
_D1tg
_t
Equation with pump:
I
Qrlt
22
Datum
Energy grade Line, EGL
F*
Z1
Eq. 5 - 1()
a
tr'12
5-4
+
.tr "::o-
basicany to increase the head. (usualry
to raise water from
tlq"l.oo,u, (p,^*,) of the pump is erectricara
1jr1_t*::T::::1,Tn"
and its output power (p.u,p,,) is the
flo* #;gy
248 Fl#"tiii,',",for Fruid Frow
FLUID MECHANIC5
& HYDRAULICS
Et+ HA-HL|2=E,
Eq.5-23
FLUID MECHANICS
CHAPTER FIVE
Fundamentals of Fluid Flow
249
Characteristics of HGL
'. HGL slopes downward in the direction of flow but it may
rise or falr due
to changes in velocity
. P1 + 21+ HA= ?' * P2 + zz+ Hh-t Eq.5-24
+29v2gr
Output Power of Pump = Q y IU
'
6. HYDRAUTICS
or pressure.
cross_section, HGLisparailel
to the EGL
: X::;11l:::ttry
;;;;d;;il;""..i#1p,".,o,"h"ud.
i:n:1':ly,*,n::.,1""if
anv two points is urro uq,rui-io;:T"1Tiil
.5-25
?:"ff::: ;:::
i:ffi*
points.
Energy Equation with Turbine or Motor:
Iurbines or motors exh'act flow energy to do rnechantcal work which in turrr
converted into electrical energy for furbines
Grade Line (EGL)
:lffti]l:ff"fi :i:.#.;:J,txi;:"..:.l:fix,"Hffi
+ !_ +2.
;lffJ,,""
v
cteristics of EGL
a
a
iT?{'ilini;i"i:,:::iff;1,';
The drop of the EGL retween
those points.
direction or now, and it w*r
'n"
any two points is the head
I
For uniform pipe cross_section,
onrv
lost between
EGL is parallel to the HGL.
h - HE - HLr.z= Ez
Eq.S - 26
,r^2
!r' * Pt * r,= -if
* Dt * zz.+ HLn+ HE
ES! * always above the \tsL
HGLby
vy an amount equal to the
velocity head,
u2/2g.
Eq.S - 27
NeglecHng head loss, EGL
2g
;
v
is horizontal
ut Power of Turbine = Qy HE
HLz
T--
ENERGY AND HYDRAULIC GRADE LIN-ES
I
I
Hydraulic Grade Line (HGL)
p/y
Also known as pressure gradient, hydrauhc grade lme rs the graphrcal
v
i,r__
-Frcg
I
I
lrr
Pa/yrrl
HLv
i-i-r---->--r\
;l;
r\
!il
ril
HLs
HE
i-*1
HLn
vn2/29
gepresentation of the total potential energy of flow. It is the line that conrru,tl
the water levels in successive piezometer tubes placed at intervals along tlu
pipe. lts distance fronr the datum plane is 4 + z
vqzl29
_lq=
Valve
e
i
Illustration showing the
behavior of energy and
IURBINE
I
h
hydraulic grade lines.
CHAPTER FIVE
Z5U
Fundamentals of Fluicl Flow
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
CHAPTER FIVE
& HYDRAULICS
Fundamentals of Fluid Flow
25t
Solution
W= yQ=yAa
Solved Problems
Problem 5 - 1
Water flows through a 75 mm diameter pipe at a velocity of 3 m/sec. Find (a)
the volume flow rate in m3/sec and lit/sec, (b) the mass flow rate in kg/sec,
and (c) the weight flow rate in N/sec.
,=
,RTP
110 x
103
=12'39N/mr
' 1= 2%(go+273)
20 = 72.3e10.1 6 (0.32)lu
o = 31.53
(average velocity)
Solution
m/s
(a) Q=Au
Q= Aa
= -i (0.075),(3)
= 0.013
(0.16X0.32X31.s3)
1*674
=
m3/sec (volume flux or discharge)
a
m3/s x '1000 lit/mj
Q = 13lit/sec
5-
(b) M=pQ
4.110-.1y" diameter plunger is being pushed at 60 mm/sec
= 1000(0.013)
q ::.
11ltis :|,l11lq
"f out
being forced
at a 30-mm
M = 13 kglsec (mass flow rate)
o.f 0
(c) w=vQ
into a tank filled
rf tne fluia is incompressibte, how many N/s of
diameter hole?
= e810(0.013)
W = 1,27 N/sec (weight flow rate)
the fluid is incompressible:
Qt= Qz
Problem 5 - 2 (CE November 1995)
Qr = AtVt
What is the rate of flow of water passing through a pipe with a diameter of 20
mrn and speed of 0.5 m/sec?
=
t (0.1)'z(0.06)
Qr = 0.00047 rfi/ s
Qz = 0.00047 m3 / s
Solution
w=yQ
Flow rate, Q= At,
Q=
= (9810 x 0:82)(0.00042)
f (0.02)'z(o's)
W= 3.78 N/s
Q = L.57 ,, 1g_r 6:/sec
Problem 5 - 3
Air at 30'C and 110 kPa flows at 20 N/s through a rectangular duct tlrill
measure 160 mm x 320 mm. Compute the average velocity and volume fltrx
Use gas constant R = 29,3 m/'K
3
5the velocity of flow in a zS-mm diameter fire hose is 0.5
m/s, what is the
locity in a 25 mm diameter iet issuing from a nozzre attached
at the end of
pipe. Compute also the power available in the jet.
CHAPTERFIVE
Fundamentars of Fruid Frow
-FZ5Z
FLUID MECHANICS
& HYDRAULICS
Solution
lly continuity equation:
FLUID MECHANICS
& HYDRAULICS
CHAPTER FIVE
Fundamentats of Fluid Flow
253
Solution
PE=Wz
W=y * Volume
Qnnr" = Ql",
An at, = Ai at
. t (0.075)2 (0 s) = f, (0.025)2 z',
oi= 4.5 rc/s
(velocity of the iet)
=9.81,x f,(s)r(19)
w- 1,925.2kN
PE='1.,926.2x7
PE = 13183.32 kN-m
Power, P = QV E
'
Q= Au
Q = + (0.025)'z (4.5) = 0.002209 rn3/ s
,r2
L \2
E=:2s 2(e.81)
Power, P = 0.002209(9,810)(1.0321)
Power, P = 22.37 walts (power available in the yet)
Problem 5 - 6
A turbine is rated at 600 hp when the flow of water through it is 0.61 m3/s
Assuming an efficiency of 87%, what is the head acting on the turbine?
Solution
Given: Power output = 600 hp
EfficiencY, 11 : g7o1'
Power input =
'
600
- 589.555 hof
0.87
Power input = 514,483 watts
Power input = Qy HE
514,483 = 0. 61 (9,81 0)HE
HE = 85.97 rn
Problem 5 - 7
A standpipe 5 m in diameter and 10 m high is filled with water. ialculate tht'
potential energy of the water if the.elevation datum is'taken 2 m below the
base of the standpipe.
Determine the kinetic energy flux of
discharging through a So-mmtia*",*
0.02 m3/s of oil (rp. gr. = 0.85)
;";;i".
Solution
Kinetic energy flux = Kinetic Energy per
second = power
Power,P=QyE
A * 0.02 m3/s
^a2
2g
,= I - o'02
A t(0.05)2
a = 1.0..1.96 m/ s
(10.186)2
^
ts=-:____-,
=5.2ggm
2(e.81)
P * 0.02(9810 x 0.8s)(5.288)
P = 882 watts
Iecting air resistance, d"::rTiT to what height
'-o- a vertical jet of water could
if projected with a velocity of 20 m/ s?
25 4 Fl#.ti:i,'"",f or Fruid Frow
FLUID MECHANICS
& HYDRAULICS
FLUIE MECHANICS
& HYDRAULICS
Problem 5 -
Solution
d,
r-l
KE_PE
L/
trr'"
,,1 =yy71
,,2
l*
)o
-6
','l
.20\2
Ir= ' '
$OlUtion
;t
'l
8
It='
l
$l
$n
1/2M7'1 =Wh
i
450
4s0 mm O
a
= 60 kPa
Qr= Q, = 0.15 m3ls
0.15
"t
I
I
= :-::--:; =8.49m/s
f, (0.1s)z
a'z=
=20.4m
2(e.81)
oil of specific gravity 0.g77
changes in size from 150 mm at
;arrying
section 1 and 450 mm at sectiori2.
section 1 is 3.5 m berow section 2 and
the
pressures are g0 kpa and 60_tpa,respectivery.
If the dischargJl,lso [t/sec,
determine the head lost and the direction;ifir*.
:: .l-llffi-
Neglecting ait resistance:
0.15
TQ'45f
=o'e43m/s
'
Taking O as datum:
Problem 5 - 10
Water is flowing in an open channel at a depth of 2 mand a velocity of 3 n"r/s
It flows down a chute into another channel where the depth is 1 m and tht'
velocity is 10 m/s. Neglecting friction, determine the difference in.elevation ol
the channel floors
r-f
Solution
ilrl
I
v22l2g
lt
to .irl,
I
----i --li
--t+
- iA J1'n
I
Neglecting friclion
(head lost):
E't = Et
-2
(Jt
I
-.2
U.L
rAr--
)o
-d
L1
)o
-d
32
aL--I
2(e.81)
I
z = 3.64 rn
L
lo2
L_
2(e.81)
11
255
lt
A pip"
As the jet rises, its krnetic erlergy is transformed into potential energy
CHAPTER FIVE
Fundarnentals of Fluid Flow
o
Pr=gokPa
90
E,=++Pl
+zr= g.492 +
29v
2(9.s1) (e.81x 0.8??) +0
Er = 14.L35 m
o"2
Do
---:- + J-:E2=
+ z-
z8y
_ 0.9432
T-r,fe
60
2(9.81) (9.81x0.877) ' "'"
Ez='1.0.62m
Since Er > E2, the flow is from L to 2
Head Lost, HL = Et - Ez = 14.135 -'t0.62
Head Lost, HL 3.515 m
=
CHAPTERFIVE
^F '
Z5O
FLUID MECHANICS
& HYDRAULICS
Fundamentats of Ftuid Ftow
Problem 5 - L2
()il flows from a tank through 150 nr
of 150 mm diameter pipe and then
discharges into air as shown in ti-re
I.igure. If the head loss frorn point 1
to point 2 is 600 mm, determine the
pressure needed at point 1 to cause
FT-UID MECHANICS
& HYDRAULICS
El. 30 m
CHAPTER FIVF
Fundamentals of Fluid Flow
M = p2A2a2
oz[(0.3)(0.3)] ( 2) = 0.7s75
Pz = 0.875
(mass density at
kglmt
N
section 2)
E
E
o
5-
El.20
I
17 lit/ sec of oil to flow
Iter flows at the rate of 7.5 m/ s
through Z5_mm diameter pipe (pipe 1) and
diameter and 65lmm diamet", pipes
at the rate of 3
nxt=.'"i'dl",op
and 3'5 m/s, -l*
/sl":jlTX*,U^O
respectivery u,
E
o
h
,.o*^l'tth;;;';;
Solution
Assume thu
Energy equation between O and @
Et - HL1-r= 2,
zgyZgy
zt - HLt t:
('22
of the
cqlculate dtt/dt and the
*,ff.:rv"",:
*:.:ygl
.l,s9rmm_aia1".",:.r,".,..
of air flow
through
the vent.
Q:0.017 m3/5
r,r2 * Pt +
flo_=;;;;,"..*rl"Slr"
* P, , Z2
' ; +0+ 30
0+ "Dt +20-0.6 = 8(0.017\2
;'
n'g(0.15)'
Y
Pt
10.65 m of oil
v2=3m/s
v
P1 = 10 65(9.81
' 0.84) = 87.76kPa
Q:
Cas is flowing tluough a square conduit whose sechon gradually changes frorrr
150 mm (section 1) to 300 rnm (section 2). At section 1, the velocity of flow is 7
m/s and the density of gas is 1kg/mi wlrile 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
p) = I kg/m'
dM
e
''
300 mm
Q'n = Qou,
(0.075)2(7.5) =
t
f (0.0s),(3) * f (0.065)r(3.5) + ! (0.6), dh/ dt
Considering the air above the tank:
[Qn = Q"i.]
E
o
f, (0.05)zu = 4L (0.6)2 dh/ dt
M=pQ
M = prA'ttr,
= 1[(0.15X0.1s)](7)
M : 0.1575 kg/sec (mass flow rate)
vr = 3.5 m/s
Assuming the flow to be incompressibie:
dlt/ dt = 0.0553 m/s
150 mm
257
f, (0.05)2 ut = * (0.6)r(0.0553)
a+ = 7.963
m/s (velocity of air flow)
258 Fffi#H'Y"torFruid Frow
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
Fundamentals of Fluid Flow
Sqlution
Problem 5 - 15
A liquid having sp. gr. of 2.0 is flowing in a 50 mm diameter pipe. The total
head at a given point was found to be 17.5 Joule per Newton. The elevation ol
the pipe above the datum is 3 m and the pressure in the pipe is 65.6 kPa
compute the velocity of flow and the horsepower in the stream at that point.
Q, = Qz = 0.03 m3ls
u2 ge2
29 n'gDn
ur2= g(0.03)2
,s F(rrtxrrf = o'0465 m
Solution
Totalenersv,E=
u' * L *,
E=17.5m
a
uz2
29v
E ='17.5 Joule/N x (1 N-m/Joule)
-,
65'6
*3
v5=L*
29 e.81.(z)
)
!- = n.so
^
2g
o = 14,79
CHAPTER FIVE
& HYDRAULICS
m/s (velocity of flow)
Power, P=QyE
= tf (0.05)'z(1 a.7e)l x (e810 x 2) x't7 '5
= 9970.92watts x (Ihp/7a6 watts)
Power, P =73.37 hp
Energy Equation between A and B:
Ea-HLz.l+HA-HLLr= Es
lto2
D^
-;-T*r^_HLet+HA
May 1994, May 2004
The pump shown draws water from reservoir A at elevation 10 m and lifts it
reservoir B at elevation 60 m. The loss of head from A to L is twb times
velocity head in the 200 mm diameter pipe and the loss of head from 2 to ll
ten times the velocity head in the 150 mm diameter pipe. Determine the r
hcirsepower of the pump and the pressure heads at 'I and 2 in meters when
discharge is 0.03 m3/sec.
-HLz-"=
_.2
us
29v*
Ps +ZB
0+0+10_2(0.0465)+ HA-70(0.147)=0+O+OO
HA = 51.563 m
Power output = ey HA = 0.03(9,S10)(51.563)
=15,175 watts x (1hp/7a6watts)
Power output = 2}.B4horsepower'
irated power of the pump)
Pressure heads at 1 and 2:
, Energy Equation between A and 1:
Ea- HL+t= Er
or' * p, +21
29v
0+0+ L0-2(0.0465)=0.046S * Pt +0
uo2
5-1
8(0.03)2
,s = 7p rx""f = o'147 m
PA +za-HLA-1
2g
v
=
v
o.
-:- = 9.86 m of water
Energy Equation between of 2 and, B:
Ez-HLz-e=Ee
0r2
Do
E. i
0.'1.4V + Pz
v
Pz
v
_,2
*"-HL2-B= uB
29v
Pa
+0-1,4(0.747)=0+0+60
= 61.;323 m of water
+zB
259
r-
?,60 Fl##:i#,:or F,uid F,ow
FLUID MECHANICS
& HYDRAULICS
Problem 5 - L7 (CE November 1986)
A pipeline with a pump leads to a nozzle as shown. Find the flow rate when
pump develops an 80 ft (24.4 m) head. Assume head lost in the 6-inch (152
mm) pipe to be five times its velocity head while the head lost in the 4-inch
(102 mm) pipe to be twelve time its velocity head. (a) Compute the flow rate,
(b) sketch the energy grade line and hydraulic grade line, and (c) find the
Pressure head at the suction side
Er 80,(24.4
m)
3"jet
(76.2 mm)
o
6" (r52 mm)
-
HL2 = )'lge.lj pz
uB2
Be2
,s - }'rc.*F = 24so'8 Q')
0 * 0 + 2t.3 - 723.96
Q = 0.0415 m3/s
e, + 24.4 _ g"160.12e2 = 2450.8 e2 + 0 + 24.4
) Discharge
(b) Energy and Hydraulic grade lines:
a7t
8(0.0415)2
uB, _
8(0.0415)2
,s = ,(rs1x01st. =0.266m
uz2 _
8(0.0415)2
-;irflxolorf
=131 m
,s
N
El 70' (21.3 m)
CHAPTEfi
Fundamentar, orrr.rialFiovui Zb I
72,384.89 Q2 = 2I.g
B
E
E
FLUID MECHANICS
& HYDRAUL'CS
El 30'(15.2 m)
\
,s - }e.Blra;;,r =4'22m
HLt = 773.96 Q2 = 1.33 m
Solution
HLz= 9,1,60.13 Q2 = 15.78 m
(a) Discharge
Qt= Qz= Qa = Q'
Energy Equation between A and B:
Ea - HLt + HA - HLz= Ea
t
* pA +zA-HL1 +HA-HL2= ,,u' Ps +ZB
'!2gr
29t
HA = ?4.4 rn
HL, = 5 "t2 =5 =sQt
29
=-
nr gDrn
8Q2
n21v.sr;1o.rsz;a
HL1= //]')$ Qz
HL,=n+=-#
="12
8Q2
n21e.ar;1o.roz;a
El 30'(1S.2
El. 19.97 m
El. 19.704 m
m)
CHAPTER FIVE
ZOZ
FTUID MECHANICS
& HYDRAULICS
Fundamentals of Fluid Flow
FLUID MECHANICS
Energy Equation between A and S
Et- HLr: Es
29
=
v
u129
263
datum)
Et-HE=Ez
ttA')1 +PA +ze_HL.t=,.t- * Ps +zs
as'
Fundamentals of Fluid Flow
Energy Equation between L and 2
(neglecting head lost and taking point 2 as
(c) Pressure head at S
2gr2g
CHAPTER FIVE
& HYDRAUTICS
u.2
Pr
2g
v
-+
= 0.266 m
+21 -HE=or'*
2g
14
8(0.5)2
Pz
I z^
Y
8(o.s)2 *
7(rEnorf * n^s, +2's- nr= 7ffi+
o+o+ 27.3-1,.g3=0.266+ lL +ts.2
HE = 3.647 m
4
# *o
v
Ps
Power, p=eyHE
= 4.504 m
= 0.5(9810)(3.647) = 1Z88S.S wans x (irhp/746 watts)
v
Power, P = 23.98 horsepower
Or from the figure shown above, the pressure head at S is the vertical distanctr
from the pipe to the HGL.
lL =D.7oa-:rs.z
May
'f
v
lt
'
= 4.gg4^
20-h-n suction p*mp operating at 70% efhciency
draws water from a suction
line whose diameter is 200 mm-and discharges into
air through a line whose
diameter is 150 mm. The verocity in the 150 ti* trnu
is 3.6 m/s. If the pressure
at point A in the suction pipe is 34 kpa below the
v
aknosphere, *nu* a is 1.g m
below B on the 150 mm line, determine the maximum,erevation
Problem 5 - 18 (CE November 1980)
Water enters a motor through a 600-mm-diameter pipe under a pressure of 14
kPa. It leaves through a 90O-mm-diameter exhaust pipe with a pressure of 4
kPa. A vertical distance of 2.5 m separates the centers of the two pipes at thr'
sections where the pressures are measured. If 500 liters of water pass the
motor each second, compute the power supplied to the motor'
which water can be raised assuming a head loss of
az=3.6m/s=o6
-.2
uC
2g
= 0.65 m
Qz= 0.0636m3/s
Q: Qr: Qr= 0.0636 m3ls
14 kPa
UA=01=
0.0636
1p.42
oa = 0t = 2.025 rn/ s
u^2
f
zg
Qr: Q, = 0.5 m3/s
900mmz O
* d;;;d;;;..
above B to
*Q
Qr= no(0.15)'?(3.6)
Solution
3
=0.21 m
Poweroup,s = Qy HA
^, ^
ZO+
CHAPTER FIVE
FLUID MECHANICS
& HYDRAULICS
Fundamentars of Fruid Frow
20 x746 = 0.0536(9810)HA
FLUID MECHANICS
& HYDRAULICS
CHAPTER FIVE
Fundamentals of Fluid Flow
265
Energy equation betweenA and M
HA = 23.91, m
Ea - HLn-u - HLu - HLua
-HL,v = Erv
PA +za-z-2--j.o-o1.aN
4^
+.
29 y
Energy equation bqtween A and C (datum at A):
Ea+HA-HL=Ec
o!-,) *pA +zA+HA-Ht='l +pc +zc
2gy2gy
+ 0 + 23.91, -z = 0.6G+ 0 + (1.8 + h)
0.21 + !!
-
"-E
='au2
,g
*pN +zn
v
* Pn *za-1s=rr.gn,!' + !!- *7*
429y'ozgy
8Q2
;,di*r. # +o-15=104e#ffiorrr +o+16
9.81
h= 15.19 m
sso
Q = 0.0'1.067q5 m3ls
Problem 5 - 20 (CE November 1978)
A fire pump delivers water through a 300-mm-diameter main to a hydrant to
which is connected a cotton rubber-lined fire hose tr 00 mm in diameter
terminating to a 2S-mm-diameter nozzle. The nozzle is 2.5 m above the
hydrant and 16 m above the pump. Assuming frictional losses of 3 m from the
pump to the hydrant, 2 m in the hydrant, 10 m from the hydrant to the base of
the nozzle, and the loss in the nozzle of 4% of the velocity head in the jef to
what vertical height can the jet be thrown if the gage pressure right after thc
uN-
O
o.o1o674s
AN f (o.o2s)2
-
aN = 2'1,.74m/ s
,,
_ ,N2
29
21.742
2(e.81)
Ir=24.\02m
pump is 550 kPa?
Solution
25 mmo
$e pipe shown in the Figure z1 = uz = 'J,.2 m/s. Determine the total head
between l and 2.
150 mm @ fire
2,5 m
Qa=Qe=Q,u=Q
300 mm A maln pipe
4.3 m
J----.-D.atvm
^, ,
ZOO
CHAPTER FIVE
FLUID MECHANICS
& HYDRAULICS
Fundamentals of Fluid Flow
Solution
,Under the
Energy equation between L and2:
a
t
Z2
)1
)o
-6
P"t +21 -HL:P?
=u2)o
-d
+zz
vv
Et. 3.21 m
the Figure. If the
,,2
*P, *..-HL=u2 +!z +
"2gv29
v
280 +
267
conditions shown in
Er-HL=Ez
9.81
CHAPTER FIVE
Fundamentals of Fluid Flow
Oil of sp. gr. 0.84 is
flowing in a pipe
ut=o2=1.2m/s
Sinceor =o.,
FLUID MECHANICS
& HYDRAULICS
4.g - HL= ?oo + q.os
9.81.
HL = 3.375 m
total head loss from
point 1 to point 2 is
900 mm, find the
,pressure at point 2.
6olution
Qt= Qr= 0.056 m3ls
225mma
El. 1.2 m
Energy equation between O and
Et - HLtz= Ez
I
Problem 5 - 22 (CE November 2O00)
A nozzle inclined at an angle of 60" with the horizontal issues a 50-mur
diameter water jet at the rate of 10 m/s. Neglecting air resistance, what is tht'
area of the jet at the highest point of the projectile?
Pl
or'
+.
'- 29 * Pz *z^
zg y +,l-HLtz=
v
8(0.0s6)2 44s
q(o.os6)2.
+ p
tr'g(0.1.5)a 9.81x o.gZ "'" _o.eo=n2
g(0.225)a 9.81x
0.84
D
Solution
Solving for the velocity of the jet at the summit (highe-st point, A)
av=0
?, = tlo- cos 0
t, = 10 cos 60" = 5 m/s
, *ao1/
=5m/s
7)=
Since the flow is continuous:
lQo = Q^l
Aoao = Aa u
f (0.05)'z (10) = AA (5)
Aa = Q.QQJ)I'/ pz
Problem 5 - 23
,"-_ *,
= 55.52 m of oil
p = 457.53kpa
550-mm diameter siphon discharges
o' (sp. gr. = 0.g2) from a reservoir (elev.
hdjl;; r,om ae,e,"*"i, rp",,., rl
,*l*:L"::;(:tA
f Lin:
summit (point 2, elev.22m)
is 1.5 m and fro; th;;;;;;"";:"ffi;i;:
is 2'4 m' Determine the flow rate in
the plpe in lit/sec and
.
the absolute
, ri"u" to:ioio..
CHAPTERFIVE
-, ^
ZOU
FLUID MECHANICS
& HYDRAULICS
Fundamentals of Fluid Flow
Solution
FLUID MECHANICS
CHAPTER FIVE
& HYDRAULICS
Fundamentals of Fluid Flow
Problem 5 - 25
El.22 m
Determine the velocitv and
discharge through the tiO mm
Et.30 m
diameter pipe shown (a) assuming
h;ad loss and (b) considering i
1o
head lost of 200 mm.
Qz= Qz= Q
HLr-z = 1.5 m
El. 28 m
El. 27.5 m
HL2. = 2.4
Solution
(a) Assuming no head loss:
Energy equation between
Energy equation between 1 and 3:
Et - HLp- HLzt = Ez
,)
p7
p3
!-)o * v + 21 - HLt-z- HLz-t - lt)o + v + z3
-6dl
o+o+ 20-7.5-2.4=
t-Ql-., +o+15
O and G) neglecting head lost:
Er=Ez
.)
)
at- r _Pr +r.= az_Pz +r^
29v'
z8y
_
-2
'
n'g(0.05)*
0+0+ 30= 02
2g
Q = 0.00912 nr'/s
Q = 9.12 IiVsec
+0+24.9
_-2
"2g =5.r.^
Energy equation between 1 and 2:
oz=10 m/s
Et - HLtz= Ez
,r,2 -:--.1,) t
+
+.-r
-L
2gv2gv -HLt.-
Q= Azoz= i (0.15)r(10)
,r^2
L +'' Dt +-.'-
-' ' +
0+0+20-1.5= 8f',0.00912)2
"Lto
n2g(0.05)a 9.81x 0.82
p2= -37 kPa
Absolute pressure at O : 1,01,.3 + (-3i)
Absolute pressure at € = 64.3 kPa
Q=0.1n m3ls = 777es
+22
(b) Considering head loss of 0.2 m:
Et-HL=Ez
. Pt +zt-HL= a22 + p, *,^
tzgy29f'
0+0+ g0-0.2= T^ +o+24'9
a.2
-L- = 4.9 m
"
-d
vz= 9.805 m/ s
Q = Az a, = t(0.15)r(9.805)
Q = 0.173 yp/s = 173IJs
269
ZI U
CHAPTER FIVE
Fundamentals of Flurd Flow
FLUID MECHANICS
& HYDRAULICS
Problem 5 - 26
Water flows freely tronr the reservotr shown through a 50-mnr diameter prpe
at the rate of 6.31 tit/sec lf the head lost in the system is 11.58 Joule/N,
determine the elevation of the water surface in the reservoir if the discharge
errd is at elevation 4 nr
FLUID MECHANICS
CHAPTER FIVE
& HYDRAULICS
27t
Fundarnentals of Fluid Flow
Problem S - 27
Neglecting head loss, determine
the manometer reading in the
system shown when the velocity
of water flowing in the 75_mm
25mm@
cuameter pipe is 0.6 m/ s.
E
o
rr
.i
I
I
75mma
I
*I
750
at= 0.6m/s
[Qr= Q21
25mma
I
P.075),(0.6) = f, (0.02s)2a2
uz= 5.4m/s
Solution
Q = 6.31 L/s = 0.00631 rnr/s
HL = 1-|'.58 N-m/N = 11.58 nr
Energy equation between O and @
Er-HL=Ez
ut-)) * Pt * :,.H1 "2
=
2gv2g
()+0*zr - ll58
zr = 16.11 m
l)t
* 4,zt
8(0.00631)2
Energy equation between 0 and
€:
r
L1, -F
- E2
_-2
a1
71n"
-:-T-tzt h
= - + Pz +22
zgl2g
v
0.62 * pt - 5.42
(r31) ? *o=
l! = S.SOS m of water
frt
E
o
75mme
+o+2.4
Sumriring-uppressure head from O
yy
3.868+0.75-1,g6h=0
ft=0.3395m=339.5mm
I
-
Jr*,'
7S0 mm
I
v
P' *o,Ts-hft35\= la
oi
75mmA
r-g(0 05)"
) Elevation of w s rn tlre tank
!f,
to g in meters of water:
CHAPTER FIVE'
Zf Z
FLUID MECHANICS
& HYDRAULICS
Fundamentals of Ftuid Ftow
FIUID MECHANICS
& HYDRAULICS
CHAPTER FIVE
Fundamentals of Ftuid Ftow Z l3
Problem 5 - 28
A lrorizontal prpe gradually reduces from 300 rnrn diameter sectiorr to 100 mnr
clianreter section. The pressure at the 300 mnr section is 100 kPa and at the 100
mm section is 70 kPa lf the flow rate is 15 liters/sec of water, compute the
head lost between the two sections
Solution
Energy equation between 1 & 3
(Neglectinghead loss & datum
along point 3)
Er=Eg
{.Pl
29
300 mm
Q = 0.015 mr/s
+zt='t-+P3-ar^
T ' zg y'"
o+o+rc=ry:+o+o
2g
u3=14m/s
or = 1oo kPa
,
Q= Q'= t pi.22s)2(14)
er = ez = 0.015 *i/s,
Q = 0.557 m3ls
Energy equatlon between O and €)
Pressure at the throat:
Er-[-IL=Ez
t
l'1 ,Pt
2gy28y
8(0.015)2
+ l oo +0-H/ = 8(0.01s)
-n'g(0.3)'
Energy equation between O and O:
,,n2
D.
u = --3:+ !-:- + '1
-1 "L
q 81
HI = 2.872 m
2
n2g(0 1)a
_.2
+22='t2 +P3 +,.
{-P2
zgy2gy"
+ 70
+()
c.81
-
Problem 5 - 29
A divergrng tube discharges water from a reservoir at a depth of l0 nr belon
the water surface. The diameter of the lube gradually increases from 150 nrrrt
at the throat to 225 r'r'rrn at the outlet Neglecting friction, determine: (a) the
maximunr possible rate of discharge through this tube, and (b) the
corresponding pressure at the throat
s!o.ss7)2
n2
*p2 *r=++o+o
g(0.15)a y
pz= -398.75kPa
225mm b
-, I +,
Z
S
CHAPTER FIVE
Fundamentats of Fluid Ftow
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
CHAPTER FIVE
& HYDRAULICS
Fundamentals of Fluid Flow
Problem 5 - 33
If the water level in Problem 5 - 32 varies
upplementary Problems
Problem 5 - 30
Air is moving through a square 0.50-m by 0.50-m duct at 180 m3/nrin. What is
change dh/ dt.
and. az = 1.0 m/s, find the
Problem S - 34
yi$r"i:::l:jo; rl_o;yff::,,};
Problem 5 - 31
The piston of a hypodernlc apparatus shown in Figure 5 - 8 is beurli
withdrawn at 6 mm/sec; air leaks around the piston at 20 n-rm3/sec. What ir
the average speed of blood flow in the needle?
Anl 498 rnm/st.r
rate of
Ans: -9 mrn/s
the mean velocity of the air?
Arts: 12 m/s
27s
:ylindricar
arrangement shown in Figure
and at
tection B. Assume radial flow at B.
Ans: at = 7.47 mf s; oz = 2.46 cm/
80mmU
Figure 5 - 8
Figure 5 - 10
Problem 5 - 32
The water tank in Figure 5 - 9 rs being filled through section 1 at 6 m/s arul
through section 3 at 15 L/s. If water level h is constant, determine the t,xil
yelocity o2.
15 L/s
Ans:7.97
5-35
a jet is inclined upward. 30o from
the horizontal, what must be its
velocity to
tch over a 3-m walt at a horizont"t
diri;;;;; rs *,
^ugi"J"g
ii"l.",
Ans:L6.93 m/s
r
t-]
5-36
chng air resistance, determine the
height a vertical jet of water will
rise if
ted with velocity of 21, m/ g?
Ans:22.5 m
Figure 5 - 9
l.-O.S m a-t
5
;]3"*iJi?lf;x'"il',)f*,I1"-:r**-il;"f;;#J;$ll,tril::;
*t;::lt :::"dy flow,.d_etermine the ulrurug" lr.ro.trty at section A
s
ZI 6
CHAPTER FIVE
Fundamentats of Ftuid Ftow
FLUID MECHANICS
& HYDRAULICS
Problem 5 - 37
High velocity water flows up an inclined plane, as shown in Figure 5 - 1.1
Wlrat are the two possible deptlr of flow at section 2? Neglect al losses.
Ans:0.775m&2.74nr
FLUID MECHANICS
CHAPTER SIX
& HYbRAULICS
Fluid Flow Measurement
277
Chapter 6
llgid Ftow Measurement
There are numerous number of devices
used to measure the flow of fluids.
any of these devices, the.,Bernoulti,, Errergy
Figure 5 - 11
9.806 m/s
Tlteorem is greatly utilized
In
and
knowredge of the characterisrici'and
coefficients of each device is
,111L:""1
lmportant. In the absence of reriabre varues
and coefficients, a device shourd
be calibrated for the expected opuruti^g
"or,iirio*.
DEVICE COEFFICIENTS
of Discharge, C or Ca
coefficient of discharge is the ratio
of the actual discharge through
j: n"^'_1"_1^ 11,tq"oiuuc ail, J ;;s;;;,.h wo uld occ ur witho u t lo the
Xmay
es
be expressed as:
s s
corCa= _e"ggd.:lgg9_ =
a
Th"*"tt"rldtr.h"tg" = A,
actual discharge may be accomplished
Eq'6-1
by series of observation, usuaty
by
fluii passinj,nro"gll
the device for a known
-bu
::rt"1*",1r^.1^amountof
' The theoreticar value can
'm neglecting losses.
u.;*;iisL"d using the Bernoulli,s
of Velocity, C"
coefficient of velocity is the ratio of the
actual mean velocity to the ideal
or
retical velocity which would occur
withoui u.,y torrur.
Actualvelocity a
(_ =_=__:_
fneor"ti*t"J*it11 =
o,
Eq.6-2
A-A
CHAPTERSIX
Z l6
FTUID MECHANICS
& HYDRAULICS
Fluid Flow Measurement
FLUID MECHANICS
Coefficient of Contraction, C.
Fluid Flow Measurernent
279
rable 6 - l: Discharse c"fl:::::
v:Tf4 Sharp_Edsed Circutar orifice
Discharging inro lqr
Air at rS.O.C
The coefficient of contraction is the ratio of the actual area of the contracted
section of the stream or jet to the area of the opening through which the fluid
flows.
a
_ Areaof streamorjet _
-'' -.
- A
Aluuof op""itrg
CHAPTER SIX
6. HYDRAULICS
titi.,rl
Eq. 6- 3
Relationship between the Three Coefficients
Actualdischarge, Q= C"Q'
) Eq. (1)
Also
Q = Actual atea, A x Actual velocity, c'
Q= C,A x C,,u1
Q = C, C,, Aut
but Aor = Qr
) Eq. (2)
Q: C,C,QI
From Equations (1) and (2)
C- C"x C.,
Eq.6 - 4
tosT
head lost through Venturi
as:
The coefficient of discharge varies with Reynolds Nurnber. It is not constarrl
for a given device. Table 6 - 1 gives the coefficients for verhcal sharp edged
orifice.
meters, orifices, tubes, and
Flgure 6 - 2
The-ideal energy equation between
E
Ll-E2-F
/ -aa
Q-1'0
':::::'7,,,,t
c; -'0.86
Co = 0.86
Square shoulder
-.::.:jt .! ti ,
a! 1.0
c,- 0.qt
cd= 0,91
at2 , Pt
+21
;:4'8 - -f
Ath = Azaz
7t"2
2g
_ +22
T,Pz
Thick plate
Figure 6 - 1: Orifice coefficients
7,t= 12 a, und. a12 =
A7
29
1 and 2 is:
v
(!.\' 'r'
{.A' j
29
nozzles may be
FLUID MECHANICS
& HYDRAULICS
2BO FllffL:l',trasurement
(4t\'t'z? +Pr +zt=oz2 *Pz +zz
[Arl 29 y
2s
FLUID MECHANICS
& HYDRAULICS
nr= Y'
y
2g
Fruid
f't*J'1
Frow,fiH:I"'*:fi zB l
+ *[, t*]']
* (# 'Jl't*l'l'
=
[' [f)'] + (+.,,). (?.,,)
If the orifice or nozzle takes off directly
from
where A1 is very much
o" then the velocity or lpfrou.ha tank
is negligible and Eq. 6 _ 5
ff111; ;|1"
Considering head lost bdtween 1 and2:
))
u7'
+ I'1 + z. _ HL= ,r-, * P2 + z,
29vzgv
,r= (J:-tl') 4
rt
lc,t
This equation sirnplifies to:
-Eq.6-6
-
Note: zr = acfual velocity
(,rt)".,'o,
1- (Az / Ar)'
ffi
Since ?u.su"1 = C0 z/theorerical = z,
u= Co
1- (Az / A,,)2
Squaring both sides and arranging terms:
,
*l' [#)'] +=(t.,,) (?.',)
From (b)
'cling to shape, orifice
*uI circular, square, or rectangular ln
cross_
n' The circular sharo-crested?"orifice
i, mo'ri*raely
used because of the
icity of its design anj construction.
figure berow shows a general case
of fluid flow through an orifice.
jLT:' normat 1".:. ;;;ill
I'J:,T
of the stream ::-'_':.
to
the ptane
Let pa
a,,", p". uvery and aa be the
J,h";;*;ffiJ#:i:#::.T;
r two points 1 and 2 such ihat ul
= ,^ ur,J
= o and writing the
quation between thes,
equation
rhocoe r.^,^
two points
-^:-r^ neglecting
-, r
losses
FLUID MECHANICS
& HYDRAULICS
282 F'Tf,tJ':ltil.asurement
Qo,
Q
o'
FLUID MECHANICS
flouin
t"ii";";Xli""::j}"Lt"i"tl$:1"
meters,or feet of the flowing
"iL"
-
Conditions
a
and 2 neglechng head lost
Et=Et
u"r2 * Pt +zt= u22 + rz
2gy29y
A
I
I
+z?
uA2 *Pa+ylt *o=!:*P,
I
*0
29y28Y
H=h(lta/g)
uA2 * P^ *r,- t'2 * PB
29y28y
'2 =lr* P^ - PB *
29yyZg
uo2
rl"-t-[+ +)]
Theoretical velocity, ,, = ,pgn
Actual velocity, a= e, rFgH
Theoretical disclrarge, Qt = n
,pn
Actual discharge, Q = CA ,pn
H=ir+
aA2
2g
* Pa - Ps
vv
Eq.6 7
Eq.
6-i
Eq.6 I
Eq.6-l0
H=h+p/1,
283
fluid. It
Ro- enersv upst uu* ress the
Chamber B
Chamber A
.l
CHAPTER S'X
Fluid Flow Measurement
where H is the total head,prod-ucing
Values of H for Various
Energy equation between
_
& HYDRAULICS
H=hz+hr(ylyz)+plyz
284 Filf,'J:t',tr asuremenr
FLUID MECHANICS
& HYDRAULICS
FTUID MECHANICS
& HYDRAULICS
CHAPTER SIX
Fluid Flow Measurement
285
Orifices under Low Heads
when the head on a verticar orifice
is sma' in comparison with
the height or
an appreciable dlrfe'-enle u.t-u"'"
tr-'" J,.t.'u.,", using rhe
ffi;t*:"tlXis
Submerged Orrflce
(Neglecting v.)
H=hr-h:=h
sider the rectangular secf,on of length
L ancr height D as shown in the
re with both the surface and the
luu;e.i io atmospheric pressure.
iet
The
reticar discharge through an elementary'roip
of length L and height dft is
dQ, = (L dh)
dQ, =
Contraction of the Jet
The figure shown represents a cross-section of fluid ilow through a vertical
sharp-edged orifice from a reservoir to the almosphere. The fluid flowing ir
coming from all direction upstream fronr the orifice and as they leave tht,
orifice, they cannot make an abrupt change in their direction and they move irr
curvilinear paths, thus causil-lg tl-re jet to contract for a short distance beyond
the orifice. The pherromerron is referred to as the contrnction of tlrc 1et. Thr
section on the jet where the contraction ceases is called the oatn contraclt
which is approximately located at one half of the orifice dianreter (D/2) front
the upstreanr face
Jrsh
J2S L tr/z dh
Q, = J2s
t
Q, = ,l2g
t
!,,t
0,
I o ,1ht
L;rt ]^.
Q, = t JzS t thz3/2 - h,t3/2 )
Q=CQ,
METER
meter is an instrumer,
the drscharge through pipes
"::d,: measuring
Fi;;;;"_
ii,"r,i.r, is connected io the
l'j""?,::::lT:.s;u!"
1, Q:" ii a cylindrical
ln pipe
at rhe inlet at
section ,. .jil":
{,an{3nding
oat' and a diverging section
cp whiih is colnected again to the main ;l:
pipe
the outlet D The ancre of divergence
ir t"pt sma' to reduce
tne by turbulence as tte velocity
is reduced
the head rost
CHAPTERSIX
^A,
26O
FLUID MECHANIC5
& HYDRAULICS
Fluid Ftow Measurement
T_I
CHAPTER SIX
Fluid Flow Measurement
The theoretical or ideal discharge ,,er,,
vl/2s
f-
FLUID MECHANICS
& HYDRAULICS
:"::,2:,H:',:;z;^;',f;;;f
-f-
canbe found oric€ o;
":,1:ilr,::F,,,y,ryj;;;:il
287
,rm
",
,"ff^:::J,ll,
,,ur,," uu
I
I
Note: If we neglect the heart lnqi in
I
+\
9tlv
ir ;;:;;;;;Z;i17i"iiill ^,,. ^^^_^..
theoretical
tost, we set the actuat ualues
(actuat
NozztE
A nozzle is a converging tube
purpose of increasing the
Piezomet€r ring
Z2
!:['illi'llr"l,ijji.l;li:::,fiJ: '.:::,f"IffiT"j;
""t;;;;.ffi;ilat discharqe).
instated at trre end of a pipe
velocity of tn" i.rrirrg;"t
or hose for the
0r<0t
t
Figure 6 - 3: Venturi meter
Consider two points in the system, 0 at the base of the in_tet and O at the
throat, and writing the energy equation between these two points neglecting
head lost:
ur2 * P't *
'2g
,y
clischarge through a nozzlecan
be calculated using the
equation
,'r' * !_2
0o
+ 7,
2., =
2g
Y
zs ,s -|.t*",l (1"*,.\
\ v -)
u22 _r1',=(pr_.\
Ihe left side of the equation is the krnetic energy which shows an increase in value
where:
H= total head at base of nozzle
A, = area at the nozzle tip
whjle the left side of the equation is tlre potential energy which shows a decrease ln
value. Therefore, neglectilrg head lost, the increase in kinefic mergy is equal to tl*
dea'ease in potentinl alerry. This statement is known as the Ventui Princtple.
followi'g table gives the mean
,tgh a nozzle having a base
values of coefficients for
diameter.i 40;
The difference in pressure between the illet and the throat is commonly measu
by means of a differential manometer connecting the inlet and throat.
lf the elevations and the difference in pressure betweeu O and O are known, I
discharge can be solved
head lost through a nozzleis
given by Eq. 6 _ 5.
water crischargr'g
and C. = 1.6.
FLUID MECHANICS
& HYDRAULICS
2BB FilffL:l',fi.asurement
PITOT TUBE
Named after the French physr.cist and engineer Henri Pitot, Pitot tube is a bent
(L-shaped or U-shaped) tube with both ends open and is used to measure the'
velocity of fluid flow or velocity of air flow as used in airplane speedometer
FLUID MECHANICS
& HYDRAULICS
CHAPTER SIX
Fluid Flow Measurement
This equahon shows that the velocity
head at point .l is transformed
into
pressure head at point 2
.
(b)
When the tube is placed in a moving stream with open end oriented into tht'
direction of flow, the liquid enters the opening at point 2 until the surface in
the tube rises a distance of /r above the stream surface. An equilibriunr
condition is then established, and the quantity of liquid in the tube remaitrs
unchanged as the flow remains steady. Point 2 at the face of the tube facing
tlre stream is called the stngmtion point
(c)
I
P/:v
l
h2
I
hr
J_
Figure 6 - 5
v:=0
Figure 6 - 4
Consider a particle at point 1 to moving with a velocity of u As the partich'
approaches point 2, its velocity is gradually retarded to 0 at point 2. Writing
tlre energy equation between 1 and 2 neglecting friction:
Et=Ez
ut2 *L*r(=
2gv'
u1
u
"2'
Pz
2g
Y
-t. ,
=o; Pl -n1
-
Pz
*?1
=hz
Y
u2 *1r",=lr,
2g
u'1=2s(7)
Figure 6 - 6: pitot tube rn a ptpe
o = ,fzgn
Eq.6-16
289
/d\
29o F,TfJ,:l'#"asurement
FLUID MECHANICS
& HYDRAUTICS
FLUID MECHANICS
CHAPTER SIX
& HYDRAUL'CS
Fluid Flow Measurement
29t
GATES
A gate is an opening lr1 a dam or other hydraulic structure to control the
Actual o = C,, rl 2S(d, _ d2) + a.,2
passage of water. It has the same hydraulic properties as the orifice. In using
gates, calibration test are advisable if accurate tneasurements are to be
obtained since its coefficient of discharge varies widely
Eq.6-18
Actual Q = CA rl zS@., _ d,r)+ u.,2
Eq.6 - 19
Coefficient of contractio n, C, = lZ
Eq.6-20
The following illustrations show the two different flow conditions through the
sluice gate
where:
C = C, C,, (varies from 0.61 to 0.91)
A=bv
ll = width of the flunre
TUBES
Figure 6 - 7 (b): Submerged Flow
Figure 6 - 7 (a): Free Flow
. Figure 6 - 7: Flow through a gate
ln Figure 6 - 7 (a), wrihing the errergy equation betweetr 7 and 2 neglecting
e.trance and has a
the flow started suddenry
-id., i,igi i";r"Ji;
aiu*"tll u,., hown in Fis;;" ; ;' "Fis",:
jm ::when
f -Xll
;*:."ii
.,111.:1i
8 (a):jshows
a condition
j"":Til_ i:.^,:I:l_,T_ 1"5 ol.'h" pllu
-Figurerhis condruon is-,,ery rnuch
:::r:
e same
as that of a sharp-crested orifice.
6 - g (b) shows a condition
the rube. Th? dischargl ,nro,rgh *,]s
|;:|"^Y:"1hes.rhe..wails.of
t one-third greater than that of the standu.d ,hu.p'-"Jg"J3rtr,." t,rUe i,
,h"
ity of flow is lesser
i",
head lost:
h= Et
,.2
rar
6tandard Short Tube
A standard short rube is.the one with a square-cornered
01
29v2g
t Ll -
u"2
-
Do
I !-2 a7,
Y
where Pt = dt und, P2 = d,
C. = 1.0
vy
u2 *dr*0
ul2 *d,+o= _.2
2g
C=C"=0.82
-4-\--lLl--"Dr
( .t _t:-_=-:
2g
L\-_.2_!__-
ur' -!,f- =dt_dt
29 29
ttzz-ut2==29@t-dz)
uzz: 2g (dt - dz1 + o'tz
Figure6-8(a)
- d2\+ a12
Figure6-8(b)
Figure 6 - 8: Standard Short tubes
-A^ Z
Ztf
CHAPTERSIX
FLUID MECHANICS
& HYDRAULICS
Fluid Flow Measuremenr
FLUID MECHANICS
& HYDRAUL'CS
F,uidF,ow,fH:15*:# z9J
Converging Tubes
Re-entrant Tubes
Clorricral converging tubes has the form of a
These are tube having
l:rustunr of a right circular cone with the
larger end adjacent to the tarrk or reservoir
as slrown in Figure 6 - 9
C,,pgH
Q = C A,FgH
u=
Eq.6 - 21
Eq.6 -22
their ends projecting
inside a reservoir or
tank.
::#{;##'!0.,:;';Jn: f a special case or.a re-enkan*ube,
T,T :;:fi:H', :1,:" il::i
# n?.1,H,*?
:: j,n" fi
**
Tubes
Table 6 - 2: Coefflcients for Conical Converging Tubes
]ill""o
Angle of Convergence, o
Coefficient
cu
-:1,u*pl:
0o
5o
100
15"
200
25"
300
400
500
0.829
0.911
0.947
0.965
0.971
0.973
0-976
0.981
0.984
a
1.000
0.999
0.992
0.972
0.952
0.935
0.918
O.BBB
0.859
c
0.829
0.910
0.939
0.938
0.924
0.911
0.896
0.871
0.845
Diverging Tubes
A diverging tube has the form of a frustum of a right circular cone with thl
of submerged tube is a
The discharge through
conyeying water through
a
^.:l::1,
submerged tube is
given by tl:;
cre C is the coefficienf
of discharge, A is the
area of the opening,
difference in elevarion
and H is
of the tiquilJ;;A;:
smaller end adjacent to the reservoir or tank.
i1
--.------F0 =T';;*-/
mr-) e
;
'-
Figure 6 - lO: Submerged
Tube (Culvert)
29
4 FiltrJ,:l',ilasurement
FLUID MECHANICS
& HYDRAULICS
FTUID MECHANICS
UNSTEADY FLOW
'l'he flow through orifice, weirs, or tubes is said to be steady or-rly if
the total
Asdh
dt=
0--4";
head producing flow, H, is constant. The amount of fluid being discharged for
a time I can therefore be computed using the formula
VoI=Qf
CHAPTER
!
FtuidFtowt"rrrr.-J#
295
& HYDRAUL'CS
-
Eq.6-24
where Q is the discharge, which is constant or steady. ln some conditions,
however, the head over an orifice, tube or weir may vary as the fluid flows oul
and thus causing the flow to be unsteady
When there is no inflow (ei,,
t=
= 0), the formula becomes:
ft" A,dr,
J
r, - Qou,
nging the limits to change the
Note: If A, is variable, it must be
Figure 6 - 11
tanks with constant cross_
Consider the tank shown in the figure to be supplied with a fluid (inflow) and
simultaneously discharging through an outlet (either an orifice, tube, weir or
pipe). Obviously, if Qin > Qout, the head will rise and if Qout > Qin, the head will
fall. Suppose we are required to compute the time to lower the level from /rr to
ftz (assuming Qou, > Qin), the amount of fluid which is lost in the tank will be
ional area and the outflow is
rugh an orifice or tube (with
inflow), the time for the head
change from Hr to H2 is:
r= lH' e,au
Ju, CA",l2fi
dV = (Qt^ - Q""r) r//
at=
dV
Qin -Q.u,
where dV is the differenhal volume Iost over a differential timedt. If the hcnd
over the outlet is /2, then the level will drop dh, thus dV : A" dh, where A" is thr
surface area in the reservoir at any instant and may be cbnstant or varialrl€,
then
expressed in terms of fi.
e outflow is through and orifices
ugh any other openings, use the
'=#
Ii:"",
n'
t=
Al,tH
lru* l"'
cA"J2g L-" _ln,
sign of the integrand:
or tube, eout = CA
On.
If the flow is
corresponding formula for discharge.
- ,
ZtlO
CHAPTER SIX
FLUID MECHANICS
& HYDRAULICS
Fluid Flow Measurement
,= -34=(,f+
L/.o.,l Zg
-W)
sq.6-27
tf liquid flows through a submerged orifice or tube connecting two tanks as
shown, the time for the head to change from Hr to Hz is: (See the derivation of
these formulas in PROBLEM 6 - 36)
FLUID MECHANICS
CHAPTER SIX
& HYDRAUTICS
Fluid Flow Measurement
297
WEIR
weirs are overflow shucfures which
are buirt across an open
channer for the
Purpose of measurinS or coltr$ing th" flo*.
of
liquids.
Weirs
commonly used to measure the
have been
florar"of wate
measure ihe now of other riquids.
discussed on this chapter u.u
gur,"rut, i.".
"pffi."frf" to any type of liquid.
il;#l|J'jfffiLi",rr;:iffii;i
Classification of Weirs
ti;z:i::z::i*i";.il"?;fi?L?:#:'K:,T^:::1ryn,*apezoidat,circurar,
th o
J, ;XT ;ilt".L:;
of the crest, weirs mai be
" "p ", " ra J', "; J
commonly used shapes are
the
ihe rec
rectangutur,
tan gu ar, tr
diurr*otur
i an gu la r r-r,
"i1:
an;d rho ,---^-^,,-To,tt
I
r,'
slnrp_crested or broaa_tesled-
flow over a weir may either be
free or subnterged. If the water surface
,h:;;:st, the now is fiee, but the
il$::il surface
fff.His higher
il:fjj,l"::I,i.'""
than the.r"ri n" n"_'i:r;;;r;;I
ir
of Terms
Tank 1
Nappe - the overflowing stream
Tank 1
in a weir.
""' "friiir|;"*:rflt"#top surface of a weir wirh which the flowing
1.
t= --.-..--.-.:
CA",,!2g
-AstA,-z g-1/2 dH
fi' A4 A,2
+
Eq.6-2g
Contracted ueir - weirs having
contracted
two ends.
Suppressed zoeir or
where A51 and Arz is the water surface areas in the tanks at any time, and H it
the difference in water surfaces in the two tanks at any time. If A.r and/or A,r
will vary, it must be expressed in terms of H
[f Asr and A"2 dre constant, i.e. the two'tanks have uniform cross-sectional arcn,
the formula becomes:
sides sharp_edged,
so that the nappe is
in width or having end contr"actio;;,
;i;J.re
full-tuidth rceir _ weirs havit
its length L being equal to
the width'.r,r-,""
width '.,f il,. .n"""a-;",i:"rtii
rLa--^r ^^ ,, . 19
the
nappe suffers no end
contractions.
Drop-dorun cun)e - the downward
weir.
curvature of the riquid surface
Head' H - the distance between
Art A"z
As + Ar2
dmtrq-ftr;)
Eq.6 -2e
before the
the liquid surface and the
crest of the weir,
measured be.fore the drop_dow.
.rrrr"-.
f=
end or
\
29a Flitffiltf.asurement
FLUID MECHANICS
& HYDRAULICS
FLUID MECHANICS
6. HYDRAULICS
Fruid Frow,;H:l"t#:#
299
Ion,= JEt [t.h,)ian
e,=,fg rEA.r"trf
Q,= t JU rt(H+11,12y2-(o+11s2121
Actual Q =Ce,
't
,'v
t/i\r
--_<--
a=+cJ4LL@+h,ym
l\.\.
'r\-
___*____---,' .._l],
I
a common practice
Figure 6 - 12: Pathlines of flow over a rectangular sharp-crested weir
to comhino 2 t;^- _
a singre coefficient
c, caned
in*
r""",",11H111;,ffI;::
1i,,i1gt*
;idering velocity"of
.
Q = C,"L l(u + tr,,)3 i2 *
RECTANGULAR WEIR
€'-"
lh
jan
I
I
-
through a recta.gular
upfrou.h d;;;"..arge
tr-- J
small, the verocity or
ff#5:#{J,ji,i,Y,',0,ff"Tv
approach
Lay be neglected.
Eq.6-37
o*"-;
The discharge formula
becomes
Q= CruLgt
P
rifuations where the
Figure 6 - 13: Section A-A of Figure 6 - 12.
Consider a differential area of length L and height dh to be located ft nrt
below the liquid surface. By orifice.theory, the theoretical velocity thrt
this area is:
,fu:1, *i,,g E; ol;r::;ilt:
the verociry or
;' ;:Tffi$,considerine
woulcl lead to successive
" - ur
(since
velocityof aonroa.h
kials to solve
Q.Jsince the velocity
r.
:^
h' ^.ld
La
"q"ui"r,;;to""tr"Ttoach
dQ,=Latr,[zg1tr+nj
is a function
c'ruti'.o(#)'J
,, = JTgIt
where the total head producing flow H' = h t llo, where ho is the velocity
of approach and is equal to tto2/28. The discharge through tlre elemt'r
strip is then,
dQt= dAq
weir
C, =
=
{ 9-uO*
d=H+p
"f Q). H;;;".;*,
Eq.6-33
g*!1
22g
of water upstream
Eq.6 -34
300
CHAPTER SIX
FLUID MECHANICS
& HYDRAULICS
Fluid Flow Measurement
Standard Weir
'l'he following specifications must be applied
to a standard rectangular weirs
without end conkactions:
1. The upstream face of tl're weir plate should be vertical and smooth.
2. The crest edge shall be level, shall have a square upstream corner,
and shall be narrow that the water will not touch it agairi after
FTUID MECHANICS
& HYDRAUTICS
CHAPTER SIX
tstuid Flow Measurement
BAZINFORMULA
For rectangular weirs
of length from
rnm to 600 mm.
0.5 m to 2.0 m under
30r
heads from 50
passing the upsh'eam corner.
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
Rectangular Weirs
effective length of L
ofa contracted weir
weir, and the water surface is free of waves or surges.
is given by:
L=L'-0.1NH
where l, = measure length
Standard Weir Factor (C*) Formulas
of crest
N = number or""ia-.or,t
Numerous equations have been developed for finding the discharge
coefficient C"' to be used in Eq. 6 - 31 and Eq. 6 - 32. some of these are givcn
H = measured head
Jluon
--- -. \r
or z;
1t v,
below.
FRANCIS FORMULA
Based upon experiments on rectangular weirs frorn'1..07 m (3.5 ft.) to 5.1g ur
(17 ft.) long under heads from 180 mm to 490 mm.
C,,=7.841+0.26(H / 421 (S.I. Unirs)
Eq.6-35
F-L
For H/P < 0.4, the following value of C,, may be used.
One-end Contraction
(N=1)
S.I. Unit, C,, =
1.84
Two-end Contraction
(N=2)
Eq. 6 - 36
Ij.uLAR WEIR (V_NOTCH)
the nappe of a recrangular
.j::^Tf:,
weir nas
h a tendency to adhere
'nsheam face.
REHBOCK AND CHOW FORMULA
A
.,.,"""r1:"::: -*"t
,ccurate,esurt'o.-ui.
English Unit, C, = 3.22 + O.EOY.
'Eq.6-38
S.l. Unit, Co, = 1.8 + ().22!-
Eq.6-39
p.
to
"#q?;::T;:.1;n*n*;##,:
y of measurement
is required- rrr" r*t"-']"gre
0 0f a v-notch weir
Detween 10o to 90o
but rarely larger.
is