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Unformatted text preview: .S 20803 kPa,
inlet diameter,
which cavita it a water tern ilet velocity is rsibilities such P R E s s U R E A N D Nithcomputa~ the effects of :hed__tube. Set then vary the sure to predict his chapter deals with forces applied by ﬂuids at rest or in rigid—body m is likely to motion. The ﬂuid property responsible for those forces is pressure, “ n 2—128 and which is a normal force exerted by a ﬂuid per unit area. We start this ‘ "' i " ‘ es denser than chapter with a detailed discussion of pressure, including absolute and gage ' ' V H ‘ ' made to ﬂoat pressures, the pressure at a point, the variation of pressure with depth in a 3‘ ‘ ' ‘ r  1 ! ould happen if gravitational field, the barometer, the manometer, and other pressure mea— ' ' ‘ 7‘7 ' ‘ ‘
surement devices. This is followed by a discussion of the hydrostatic forces ~ H v ‘ ' ‘ ‘ : ~
applied on submerged bodies with plane or curved surfaces. We then con ' ‘7 ' A '
sider the buoyant force applied by ﬂuids on submerged or ﬂoating bodies, ‘ ‘ ‘ ‘ ' ' ‘ ' “
and discuss the stability of such bodies. Finally, we apply Newton’s second i ‘ ' ' ‘l a '
law of motion to a body of ﬂuid in motion that acts as a rigid body and ana ‘ ' n: '
lyze the variation of pressure in ﬂuids that undergo linear acceleration and ‘ I “1. H in rotating containers. This chapter makes extensive use of force balances for bodies in static equilibrium, and it would be helpful if the relevant topics 7 ‘ from statics are ﬁrst reviewed. John Ninomiya ﬂying a cluster of 72 heliumﬁlled balloons over Temecula, California in April Of 2003. The helium balloons displace approximately
230 m3 of air, providing the necessary buoyant force.
' Don’t try this at home! Photograph by Susan Dawson. Used by permission. , , 74 . . . ..
PRESSURE AND FLUID STATICS 3—1 I PRESSURE Pressure is defined as a normal force exerted by a ﬂuid per unit area. We
speak of pressure only when we deal with a gas or a liquid. The counterpart
of pressure in solids is normal stress. Since pressure is deﬁned as force per
unit area, it has the unit of newtons per square meter (N/mz), which is Called
a pascal (Pa). That is, ‘ 1Pa = lN/m2 The pressure unit pascal is too small for most pressures encountered in
practice. Therefore, its multiples kilopascal (1 kPa = 103 Pa) and megapas
cal (1 "MPa = 106 Pa) are commonly used. Three other pressure units com
monly used in practice, especially in Europe, are bat: standard atmosphere,
and kilogram—force per square centimeter: 1 bar =105 Pa = 0.1 MPa = 100 kPa
1 atm = 101,325 Pa = 101.325 kPa = 1.01325 bars
1kgf/cm2 : 9.807 N/cmz = 9.807 x 104N/m2 = 9.807 X 104 Pa 150 pounds ‘ 300 pounds AW: 50 “’2 = 0.9807 bar , = 0.9679 atm Note the pressure units bar, atm, and kgf/cm2 are almost equivalent to each
P=3psi P:6psi other. In the English system, the pressure unit is pound—force per square
W 150M . inch (lbf/in2, or psi), and 1 atm = 14.696 psi. The pressure units kgf/cm2
P =‘7n= Afeet = 50in2 = 3 P51 and lbf/in2 are also denoted by kg/cm2 and lb/inz, respectively, and they are
commonly used in tire gages. It can be shown that l kgf/cm2 = 14.223 psi.
FIGURE 3—1 Pressure is also used on solid surfaces as synonymous to normal stress,
The normal stress (or “pressure”) which is force acting perpendicular to the surface per unit area. For exam
on the feet of achubby person is ple, a lSOpound person with a total foot imprint area of 50 in2 exerts a
much greater than on the feet of pressure of 150 lbf/50 in2 = 3.0 psi on the floor (Fig. 3—1). If the person
aslim person. stands on one foot, the pressure doubles. If the person gains excessive weight, he or she is likely to encounter foot discomfort because of the
increased pressure on the foot (the size of the bottom of the foot does not
change with weight gain). This also explains how a person can walk on
fresh snow without sinking by wearing large snowshoes, and how a person
cuts with little effort when using a sharp knife. The actual pressure at a given position is called the absolute pressure,
and it is measured relative to absolute vacuum (i.e., absolute zero pressure).
Most pressure—measuring devices, however, are calibrated to read zero in the
atmosphere (Fig. 3—2), and so they indicate the difference between the
absolute pressure and the local atmospheric pressure. This difference is
called the gage pressure. Pgage can be positive or negative, but pressures
below atmospheric pressure are sometimes called vacuum pressures and
are measured by vacuum gages that indicate the difference between the
atmospheric pressure and the absolute pressure. Absolute, gage, and vacuum pressures are related to each other by 3—2 Pgage Pubs _ Patm Some basic pressure gages. "  pvac = Pm“ _ pabs (32)
l l ‘ DresserInstruments, Dresser, Inc. Used by I _ _ ‘ _ permission. This 13 1llustrated 1n F1g. 3—3. :m't area. We
3 counterpart as force per
iich is called countered in
nd megapas
'e units com
‘ atmosphere, )4 Pa alent to each
e per square
inits kgf/cm2
and they are
= 14.223 psi.
ormal stress,
a. For exam
l in2 exerts a
If" the person
.ns excessive
cause of the
foot does not
can walk on
row a person lte pressure,
3ro pressure).
1d zero in the
between the
difference is
but pressures
ressures and
between the
, and vacuum (3—1 )
(3—2) atm Absolute Absolute
vacuum 7 I abs vacuum Like other pressure gages, the gage used to measure the air pressure in an
automobile tire reads the gage pressure. Therefore, the common reading of
32.0 psi (2.25 kgf/cmz) indicates a pressure of 32.0 psi above the atmos
pheric pressure. At a location where the atmospheric pressure is 14.3 psi,
for example, the absolute pressure in the tire is 32.0 + 14.3 = 46.3 psi. In thermodynamic relations and tables, absolute pressure is almost always
used. Throughout this text, the pressure P will denote absolute pressure
unless speciﬁed otherwise. Often the letters “a” (for absolute pressure) and “g” (for gage pressure) are added to pressure units (such as psia and psig) to
clarify what is meant. XAMPLE 3—1 ' * 'AbshlUte Pressure of a Vacuum Chamber a I _
‘acuum gage connected to archamber' reads, ata lecation .w_here;___ atmospheric pressure is 14.53 psi.‘ Determine the abSolute pressure inithe I
mber. ' ssure in thechamberis to be determined.
halysis The absOIUte pressure is easily determined from Eq. 3—‘2 to'be Pa... = Pam, — Pvac = 14.5 — 5.8 = 8.7 psi éiscussian Note that the local value of the atmospheric pressure is used
Then determining the absolute pressure. _ ,OLUTIDN The gage pressure Of air‘vacuum chamber is git/en. absolute Pressure at a Point Pressure is the compressive force per unit area, and it gives the impression
0f being a vector. However, pressure at any point in a ﬂuid is the same in all
directions (Fig. 3—4). That is, it has magnitude but not a specific direction,
and thus it is a scalar quantity. This can be demonstrated by considering a
Small wedge—shaped ﬂuid element of unit length (Ay = 1 into the page) in
€quﬂibrium, as shown in Fig. 3—5. The mean pressures at the three surfaces
are P1, P2, and P3, and the force acting on a surface is the product of mean P—> 75 CHAPTER‘ 3 FIGURE 3—3 Absolute, gage, and vacuum pressures. FIGURE 3—4 Pressure is a scalar quantity, not a
vector; the pressure at a point in a ﬂuid
is the same in all directions. :  76
PRESSURE AND FLUID STATICS pressure and the surface area. From Newton’s second law, a force balance in
the x— and zdirections gives 2 Fx = max = 0: P1 AyAz * P3 Ayl sin 6 = 0 (3—33) >N 1
BF, = maZ = 0: P2 AyAx —~P3 Ayl cos 6 ~ Epg Ax Ay Az = 0 (3—3h) where p is the density and W = mg = pg Ax Ay Az/Z is the weight of the
ﬂuid element. Noting that the wedge is a right triangle, we have Ax = l cos
6 and Az = 1 sin 6. Substituting these geometric relations and dividing Eq.
3—3a by Ay Az and Eq. 3—3b by Ax Ay gives 157 P 1 _ P 3 = 0 (3—4a) l
PZ—P3—EpgAz=0 (3—4h)
p2 Ax Ay The last term in Eq. 3—4b drops out as Az —> 0 and the wedge becomes
(Ay : 1) , , infinitesimal, and thus the ﬂuid element shrinks to a point. Then combining the results of these two relations gives FIGURE 3—5 m _ _ Forces acting on a wedgeshaped ﬂuid P1 " P2 _ P3 _ P (3‘5)
Element in equmbfium regardless of the angle 6. We can repeat the analysis for an element in the yz—plane and obtain a similar result. Thus we conclude that the pressure at a
point in a ﬂuid has the same magnitude in all directions. This result is
applicable to ﬂuids in motion as well as ﬂuids at rest since pressure is a
scalar, not a vector. Variation of Pressure with Depth It will come as no surprise to you that pressure in a ﬂuid at rest does not
change in the horizontal direction. This can be shown easily by considering
a thin horizontal layer of ﬂuid and doing a force balance in any horizontal
direction. However, this is not the case in the vertical direction in a gravity
field. Pressure in a ﬂuid increases with depth because more ﬂuid rests on
deeper layers, and the effect of this “extra weight” on a deeper layer is bal—
anced by an increase in pressure (Fig. 3—6). To obtain a relation for the variation of pressure with depth, consider a
rectangular ﬂuid element of height Az, length Ax, and unit depth (Ay = 1
into the page) in equilibrium, as shown in Fig. 3—7. Assuming the density of
the ﬂuid p to be constant, a force balance in the vertical z—direction gives 2Fz=maz=0z PleAy—PzAxAy—pgAxAyAz=O where W = mg = pg Ax Ay Az is the weight of the ﬂuid element and
AZ = Z2 — zl. Dividing by Ax Ay and rearranging gives AP = P2 —’P1 = — pg Az = —ys Az (3—6)
FIGURE 3‘6 ' . where ys = pg is the speciﬁc weight of the ﬂuid. Thus, we conclude that the
The Press/11f? Of a ﬂmd at FeSt pressure difference between two points in a constant density ﬂuid is propor
mcreases Wlth depth (as a res‘ﬂt tional to the vertical distanceAz between the points and the density p of the 0f adde‘i Weight) ﬂuid. Noting the negative Sign, pressure in a static ﬂuid increases linearly with depth. This is what a diver experiences when diving deeper in a lake. a balance in (3—3a) z = 0 (3—3b) eight of the
Ax = 1 cos
iividing Eq. (3—4a) (3—4b) ge becomes
1 combining (3—5) ament in the
7F6SSM7‘6 at a
his result is
)ressure is a 'est does not
" considering
1y horizontal
l in a gravity
luid rests on
layer is bal— 1, consider a
.pth (Ay = 1
he density of
:tion gives =0 element and (3—6) clude that the
lid is propor
nsity p of the
zases linearly
er in a lake. An easier equation to remember and apply between any two points in the
same ﬂuid under hydrostatic conditions is Pbelow : Pabove + : Pabove + ’yslAZI «3—7) where “below” refers to the point at lower elevation (deeper in the ﬂuid) and “above” refers to the point at higher elevation. If you use this equation consistently, you should avoid sign errors. For a given ﬂuid, the vertical distance AZ is sometimes used as a measure
of pressure, and it is called the pressure head. We also conclude from Eq. 3—6 that for small to moderate distances, the
variation of pressure with height is negligible for gases because of their low
density. The pressure in a tank containing a gas, for example, can be consid—
ered to be uniform since the weight of the gas is too small to make a signif—
icant difference. Also, the pressure in a room filled with air can be assumed
to be constant (Fig. 3—8). If we take the “above” point to be at the free surface of a liquid open to the
atmosphere (Fig. 3—9), where the pressure is the atmospheric pressure Pam,
then from Eq. 3~7 the pressure at a depth it below the free surface becomes P = Patm + pgh or Pgage = pgh (3—8) Liquids are essentially incompressible substances, and thus the variation
of density with depth is negligible. This is also the case for gases when the
elevation change is not very large. The variation of density of liquids or
gases with temperature can be signiﬁcant, however, and may need to be
considered when high accuracy is desired. Also, at great depths such as
those encountered in oceans, the change in the density of a liquid can be
signiﬁcant because of the compression by the tremendous amount of liquid
weight above. The gravitational acceleration g varies from 9.807 m/s2 at sea level to
9.764 m/s2 at an elevation of 14,000 m where large passenger planes cruise.
This is a change of just 0.4 percent in this extreme case. Therefore, g can be
assumed to be constant with negligible error. For ﬂuids whose density changes signiﬁcantly with elevation, a relation
for the variation of pressure with elevation can be obtained by dividing Eq.
3—6 by Az, and taking the limit as Az —9 0. This yields LP _ _ (3 9)
dz pg Note that dP is negative when dz is positive since pressure decreases in an
upward direction. When the variation of density with elevation is known, the pressure difference between any two points 1 and 2 can be determined
by integration to be 2 .
AP=P2—P1= —Jpgdz (3310)
1 . For constant density and constant gravitational acceleration, this relation
reduces to Eq. 3—6, as expected. Pressure in a ﬂuid at rest is independent of the shape or cross section of
the container. It changes with the vertical distance, but remains constant in ii; i 77 u
' CHAPTER ,3 . Z >1 FIGURE 3—7 Free—body diagram of a rectangular
ﬂuid element in equilibrium. FIGURE 3—8 In a room filled with a gas, the
variation of pressure with height is negligible. : 73 ,. ,
PRESSURE AND FLUID STATICS other directions. Therefore, the pressure is the same at all points on a hori~
zontal plane in a given ﬂuid. The Dutch mathematician Simon Stevin
(1548—1620) published in 1586 the principle illustrated in Fig. 3—10. Note
Pabm: Pam that the pressures at points A, B, C, D, E, F, and G are the same since they
V are at the same depth, and they are interconnected by the same static fluid.
‘ i i However, the pressures at points H and I are not the same since these two
points cannot be interconnected by the same ﬂuid (i.e., we cannot draw a
curve from point I to point H while remaining in the same ﬂuid at all
times), although they are at the same depth. (Can you tell at which point the
pressure is higher?) Also notice that the pressure force exerted by the ﬂuid
is always normal to the surface at the specified points.
A consequence of the pressure in a ﬂuid remaining constant in the hori
zontal direction is that the pressure applied to a conﬁned ﬂuid increases the FIGURE 3—9 pressure throughout by the same amount. This is called Pascal’s law, after
Pressure in aliquid at rest increases Blaise Pascal (16231662). Pascal also knew that the force applied by a
linearly with distance from the free ﬂuid is proportional to the surface area. He realized that two hydraulic
surface. cylinders of different areas could be connected, and the larger could be used to exert a proportionally greater force than that applied to the smaller. “Pas~
cal’s machine” has been the source of many inventions that are a part of our
daily lives such as hydraulic brakes and lifts. This is what enables us to lift
a car easily by one arm, as shown in Fig. 3—11. Noting that P1 = P2 since
both pistons are at the same level (the effect of small height differences is
negligible, especially at high pressures), the ratio of output force to input
force is determined to be 52 ﬂ_ﬂ P=P —9 — a — 3—11
1 2 A1 A2 F1 A1 ‘ ) PA—PB~PC—PD—PE=PF—PGPatm l 9871
PH¢PI HGURE3—1O Under hydrostatic conditions, the pressure is the same at all points on a horizontal plane in a given ﬂuid regardless of
geometry, provided that the points are interconnected by the same ﬂuid. :s on a hori—
mon Stevin
3—10. Note
l6 since they
static fluid.
:e these two
nnot draw a
ﬂuid at all
ich point the
by the ﬂuid , in the hori
increases the
PS law, after
applied by a
V0 hydraulic
ould be used
maller. “Pas—
a part of our
)les us to lift
1 = P2 since
1ifferences is
orce to input (3—1 1) irdless of The area ratio Az/Al is called the ideal mechanical advantage of the hydraulic
lift Using a hydraulic car jack with a piston area ratio of Az/A1 = 100, for
example, a person can lift a lOOOkg car by applying a force of just 10 kgf
(: 90.8 N). 32 — PRESSURE MEASUREMENT DEVICES
The Barometer AtmOSpheric pressure is measured by a device called a barometer; thus, the
atmospheric pressure is often referred to as the barometric pressure. The Italian Evangelista Torricelli (1608—1647) was the ﬁrst to conclu
sively prove that the atmospheric pressure can be measured by inverting a
mercury—filled tube into a mercury container that is open to the atmosphere,
as shown in Fig. 3—12. The pressure at point B is equal to the atmospheric
pressure, and the pressure at C can be taken to be zero since there is only
mercury vapor above point C and the pressure is very low relative to Patm and can be neglected to an excellent approximation. Writing a force balance
in the vertical direction gives Patm = pgh (3—12) where p is the density of mercury, g is the local gravitational acceleration,
and h is the height of the mercury column above the free surface. Note that
the length and the crosssectional area of the tube have no effect on the
height of the ﬂuid column of a barometer (Fig. 3—13). A frequently used pressure unit is the standard atmosphere, which is
deﬁned as the pressure produced by a column of mercury 760 mm in height
at 0°C (pHg = 13,595 kg/m3) under standard gravitational acceleration
(g = 9.807 m/sZ). If water instead of mercury were used to measure the
standard atmospheric pressure, a water column of about 10.3 m would be
needed. Pressure is sometimes expressed (especially by weather forecasters)
in terms of the height of the mercury column. The standard atmospheric
pressure, for example, is 760 mmHg (29.92 ian) at 0°C. The unit mmHg
is also called the torr in honor of Torricelli. Therefore, 1 atm = 760 torr
and l torr = 133.3 Pa. Atmospheric pressure Patm changes from 101.325 kPa at sea level to
89.88, 79.50, 54.05, 26.5, and 5.53 kPa at altitudes of 1000, 2000, 5000,
10,000, and 20,000 meters, respectively. The typical atmospheric pressure in
Denver (elevation = 1610 m), for example, is 83.4 kPa. Remember that the
atmospheric pressure at a location is simply the weight of the air above that
location per unit surface area. Therefore, it lchanges not only with elevation
but also with weather conditions. The decline of atmospheric pressure with elevation has far—reaching rami—
fications in daily life. For example, cooking takes longer at high altitudes
Since water boils at a lower temperature at lower atmospheric pressures.
Nose bleeding is a common experience at high altitudes since the difference
betWeen the blood pressure and the atmospheric pressure is larger in this
Case, and the delicate walls of veins in the nose are often unable to with
stand this extra stress. : ' For a given temperature, the density of air is lower at high altitudes, and
thus a given volume contains less air and less oxygen. So it is no surprise 79 CHAPTERS W: rghA FIGURE 3—11 Lifting of a large weight by
a small force by the application of Pascal’s law. [1 P atm FIGURE 3—12 The basic barometer. v . £50,,
PRESSURE AND FLUID‘STATICS FIGURE 3—1 3 The length or the cross—sectional area
of the tube has no effect on the height
of the ﬂuid column of a barometer,
provided that the tube diameter is
large enough to avoid surface tension
(capillary) effects. FIGURE 3—14 At high altitudes, a car engine
generates less power and a person
gets less oxygen because of the
lower density of air. that we tire more easily and experience breathing problems at high altitudes.
To compensate for this effect, people living at higher altitudes develop more
efﬁcient lungs. Similarly, a 2.0—L car engine will act like a 1.7—L car engine
at 1500 III altitude (unless it is turbocharged) because of the 15 percent drop
in pressure and thus 15 percent drop in the density of air (Fig. 3—14). A fan
or compressor will displace 15 percent less air at that altitude for the same
volume displacement rate. Therefore, larger cooling fans may need to be
selected for operation at high altitudes to ensure the specified mass ﬂow
rate. The lower pressure and thus lower density also affects lift and drag:
airplanes need a longer runway at high altitudes to develop the required lift,
and they climb to very high altitudes for cruising in order to reduce drag
and thus achieve better fuel efficiency. Measuring Atmospheric Pressure
with a Barometer EXAMPLE 3—2 Determine the atmospheric pressure at a location where the barometric read "
ing is 740 mm Hg and the gravitational acceleration is g = 9.805 m/sz.
' Assume the temperature of mercury to be 10°C, at which its density is d.
13,570 kg/m3. SOLUTION The barometric reading at a location in height of mercury col
umn is given. The atmospheric pressure is to be determined. r Assumptions The temperature of mercury is assumed to be 10°C. _' Properties The density of mercury is given to be 13,570 kg/m3. ‘Analysis From Eq. 3—12, the atmospheric pressure isdetermined to be Palm = pgh
, 1 N 1 kPa
= 13,570k / 3 9.805 m/ 2 0.740 r
( gmx S X In) 1kg  nvs2 1000 N/m2 
'= 98.5 kPa ' Discussion Note that density changes with temperature, and thus this effect
:should be considered in calculations. ' EXAMPLE 3—3 Intravenous infusions usually are driven by gravity by hanging the fluid bottle of
at sufficient height to counteract the blood pressure in the vein and to force the fILIid into the body (Fig. 3~15). The higher the bottle is raised, the
higher the flow rate of the fluid will be. (a) If it is observed that the fluid and the blood pressures balance each other when the bottle is 1.2 m above ‘7
the arm level, determine the gage pressure of the blood. (b) if the gage pres
sure of the fluid at the arm level needs to be 20 kPa for sufficient flow rate,
determine how high the bottle nfust be placed. Take the density of the fluid
to be 1020 kg/m3. Gravity Driven Flow from an IV Bottle SOLUTION It is given that an IV fluid and the blood pressures balance each
other when the bottle is at a certain height. The gage pressure of the blood
and elevation of the bottle required to maintain flow at the desired rate are
to be determined. a igh altitudes.
levelop more
L car engine
percent drop
3—14). A fan
for the same
r need to be
d mass flow
ift and drag:
required lift,
reduce drag , 81 _
CHAPTER3 Asgumptians 1 The IV fluid is incompressible. 2 The N bottle is open tofy’ffg
‘ théiatmosphere ‘ Properties The density of the IV fluid is given to be p = 1020 kg/m3. ' Analysis (a) Noting that the IV fluid and the blood pressures balance each
other when the bottle is 1.2 m above the arm level, the gage pressure of the blow in the arm is simply equal to the gage pressure of the N fluid at a", depth of 1.2 m. _ PgageJl—m : Pabs — Patm 2’ pgharm~bottle z (1020 kg/m3)(9.81 m/szxrzo “(Pa 2) 1 FIGURE 3—15
1000 kg ' W8 1 kN/m Schematic for Example 3—3. = 12.0 kPa 3(5) To provide a gage pressure of 20 kPa at the arm level, the height of the
surface of the W fluid in the bottle from the arm level is again determined Pgage, arm : pghaIm—bottle to be h = Pgage, arm
netric read— ' “meOt‘ﬂe pg
)805 W5? , I 20 kPa (1000 kg  m/s2)<1kN/m2)
’ density ‘3 ‘ (1020 kg/m3)(9.81 m/sz) lkN lkPa
= 2.00 m
nercury col— ‘ _‘ss,i_on Note that the height of the reservoir can be used to control flow ~ , ‘ gravity—driven flows. When there is flow, the pressure drop in the tube
{o frictional effects also should be considered. For a specified flow rate,‘
equires raising the bottle a little higher to overcome the pressure drop.
d to be 7' _ _ , , _ ,
MPLE3—4 Hydrostatic Pressure in a Solar Pond
kPa ) i with Variable Density
0 Mint ponds are small artificial lakes of a few meters deep that are used to 
solar energy. The rise of heated (and thus less dense) water to the sur
_is prevented by adding salt at the pond bottom. In a typical salt gradi— olar pond, the density of water increases in the gradient zone, as shown ,:
. 3—16, and the density can be expressed as l 77 S
p = po 1 + tan2(Z ere p0 is the density on the water surface, s is the vertical distance mea—
ed downward from the top of the gradient zone (5 = —z), and H is the 3 this effect :fluid bottle
and to force raised, the at the fluid l.2 m above
e gage pres
nt flow rate,
of the fluid Sun Increasing salinity
and density p0 = 1040 kg/m3 Surface zone
ialance each
of the blood ired rate are FIGURE 3—1 6
Schematic for Example 3—4. 82 P, kPa FIGURE 3—1 7 The variation of gage pressure with
depth in the gradient zone of the
solar pond. PRESSURE AND FLUID'SVTATICS ' ' thickness of the gradient zone. For H = 4 m, p0 = 1040 kg/m3, and a thick—
ness of O. *m for the surface zone, calculate the gage pressure at the bot
tom of the gradient zone. SOLUTION The variation of density of saline water in the gradient zone of a
solar pond with depth is given. The gage pressure at the bottom of the gradi
ent zone is to be determined. Assumptions The density in the surface zone of the pond is constant.
Properties The density of brine on the surface is given to be 1040 kg/m3.
Analysis We label the top and the bottom of the gradient zone as 1 and 2,
respectively. Noting that the density of the surface zone is constant, the gage
pressure at the bottom of the surface zone (which is the top of the gradient
zone) is 1 kN
P1 = pgh1 = (1040 kg/m3)(9.81m/s2)(0.8 m)( > = 8.16 kPa 1000 kg ' m/s2 since 1 kN/m2 = 1 kPa. Since 5 = z, the differential change in hydrostatic
pressure across a vertical distance of ds is given by dP = pg ds integrating from the top of the gradient zone (point 1 where s = O) to any
location 5 in the gradient zone (no subscript) gives PP1=Jpgds —> P=P1+mell+tan2(11)gds
0 0 4H Performing the integration gives the variation of gage pressure in the gradi—
ent zone to be P—P + 4:]{s‘ h“1<ta 773)
1 908 W 1H 4H
Then the pressure at the bottom of the gradient zone (5 = H = 4 m)
becomes ’ '
44
P2=8.16kPa+(1O40 kg/m3)(9.81m/s2) ( m) sinh—1<tan35l—><———1——kI—\I———2>
7T 44 1000kgm/s = 54.0 kPa (gage) V Discussion The variation of gage pressure in the gradient zone with depth is
plotted in Fig. 3—17. The dashed line indicates the hydrostatic pressure for
the case of constant density at 1040 kg/m3 and is given for reference. Note
that the variation of pressure with depth is not linear when density varies
with depth. That is why integration was required. The Manometer ' We notice from Eq. 3—6 that an elevation change of —Az in a fluid at rest corresponds to AP/pg, which suggests that a ﬂuid column can be used to
measure pressure differences. A device based on this principle is called a
manometer, and it is commonly used to measure small and moderate pres—
sure differences. A manometer consists of a glass or plastic U—tube contain—
ing one or more fluids such as mercury, water, alcohol, or oil. To keep the
size of the manometer to a manageable level, heavy ﬂuids such as mercury
are used if large pressure differences are anticipated. ‘ V Pi 1d a thick
at the bot t zone of a
ithe gradi— am. .0 kg/m3. s 1 and 2,
t, the gage
ie gradient 8.16 kPa hydrostatic = O) to any )g ds 1 the gradi— H = 4 m)
lkN >
lOkgm/s2 'ith depth is
pressure for
:rence. Note
ansity varies 1 ﬂuid at rest
in be used to
le is called a
ioderate pres—
tube contain—
. To keep the
:h as mercury Consider the manometer shown in Fig. 3—18 that is used to measure the
pressure in the tank. Since the gravitational effects of gases are negligible,
the pressure anywhere in the tank and at position 1 has the same value. Fur—
thermore, since pressure in a ﬂuid does not vary in the horizontal direction
within a ﬂuid, the pressure at point 2 is the same as the pressure at point 1,
P 2 z P 1‘ . . . . . . . . . . . The differential ﬂuld column of height h is in static equilibrium, and it is
open to the atmosphere. Then the pressure at point 2 is determined directly
from Eq. 3—7 to be P2 = Palm + pgh (3—13) where p is the density of the manometer ﬂuid in the tube. Note that the
cross—sectional area of the tube has no effect on the differential height h,
and thus the pressure exerted by the ﬂuid. However, the diameter of the tube
should be large enough (more than several millimeters) to ensure that the
surface tension effect and thus the capillary rise is negligible. EXAMPLE 3—5 Measuring Pressure with a Manometer :anometer is used to measure the pressure of a gas in a tank. The fluid
sedhas a specific gravity of 0.85, and the manometer column height is 55
as shown in Fig. 3—19. if the local atmospheric pressure is 96 kPa,
rmine the absolute pressure within the tank. OLUTIDN The reading of a manometer attached to a tank and the I
r heric pressure are given. The absolute pressure in the tank is to be ~
'umptians The density of the gas in the tank is much lower than the den»;
fthe manometer fluid. rties The specific gravity of the manometer fluid is given to be 0.85._
ke the standard density of water to be 1000 kg/m3. _ ‘ ' ' .
is The density of the fluid is obtained by multiplying its specific
y by the density of water, p = so (pHZO) = (0.85)(1000 kg/mS) = 850 kg/m?
hen from Eq. 3—13,
{Pam + pgh ' 1N lkPa
=96kP + 850k / 3 9.8111/2 0.55 a ( gm)( 1 S)( m) 1kg.m/Sz NOON/m2 _ = 100.6 kPa (Dismissian Note that the gage pressure in the tank is 4.6 kPa. Some manometers use a slanted or inclined tube in order to increase the res—
olution (precision) when reading the ﬂuid height. Such devices are called
inclined manometers. I Many engineering problems and some manometers involve multiple
1mmiscible ﬂuids of different densities stacked on top of each other: Such
Systems can be analyzed easily by remembering that (1) the pressure change.
across a ﬂuid column of height h is AP =‘ pgh, (2) pressure increases ’\ , 83 .
CHAPTER 3 FIGURE 3—18 The basic manometer. FIGURE 3—1 9
Schematic for Example 3—5. “.__.__Tv.. ‘ downward in a given ﬂuid and decreases upward (i.e., Pbottom > Pmp), and
(3) two points at the same elevation in a continuous ﬂuid at rest are at the
same pressure. ’ The last principle, which is a result of Pascal’s law, allows us to “jump”
from one ﬂuid column to the next in manometers without worrying about
pressure change as long as we stay in the same continuous ﬂuid and the
ﬂuid is at rest. Then the pressure at any point can be determined by starting
with a point of known pressure and adding or subtracting pgh terms as we
advance toward the point of interest. For example, the pressure at the bot—
tom of the tank in Fig. 3—20 can be determined by starting at the free sur—
face where the pressure is Patm, moving downward until we reach point 1 at
the bottom, and setting the result equal to P1. It gives 84. ‘ i
D FLUID STATICS PRESSURE AN FIGURE 3—20 , Patm + plghl + Pzghz + Paghs 2 P1
i In stacked—up ﬂu1d layers at rest, the
3; pressure change across each ﬂuid In the special case of all ﬂuids having the same density, this relation reduces if layer of density p and height h is pgh. t0 Patm + pg(h1 + hz ‘i‘ h3) = P]_ r Manometers are particularly wellsuited to measure pressure drops across
‘i a horizontal ﬂow section between two specified points due to the presence
A ﬂow section of a device such as a valve or heat exchanger or any resistance to ﬂow. This
or ﬂow device is done by connecting the two legs of the manometer to these two points, as
shown in Fig. 3—21. The working ﬂuid can be either a gas or a liquid whose
density is p1. The density of the manometer ﬂuid is p2, and the differential
ﬂuid height is h. The two ﬂuids must be immiscible, and p2 must be greater
than p1. A relation for the pressure difference P1 — P2 can be obtained by starting
at point 1 with P1, moving along the tube by adding or subtracting the pgh
terms until we reach point 2, and setting the result equal to P2: P1 + p1g(a + h)  ngh — plea = P2 (344) Note that we jumped from point A horizontally to point B and ignored the
part underneath since the pressure at both points is the same.._Simplifying, FIGURE 3—21 P1_ P2 2 (p2 _ pagh (345) Measuring the pressure drop across a . _ ' _ .
ﬂow section or a ﬂow device by a Note that the d1stance a must be included in the analys1s even though it has differential manometer. no effect on the result. Also, when the ﬂuid ﬂowing in the pipe is a gas,
then p1 << p2 and the relation in Eq. 3—15 simplifies to P1 — PZ E ngh. EXAMPLE3—6 Measuring Pressure with a Multifiuid Manometer The water in a tank is pressurized by air, and the pressure is measured by a
multifluid manometer as shown in Fig. 3—22. The tank is located on a moun—
tain at an altitude of 1400 m where the atmospheric pressure is 85.6 kPa.
Determine the air pressure in the tank if f71 = 0.1 m, h2 = 0.2 m, and h3 0.35 m. Take the densities of water, oil, and mercury to be 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively. SOLUTION The pressure in” a pressurized water tank is measured by a multi—
. fluid manometer. The air pressure in the tank is to be determined. > P top), and
est are at the us to “jump”
)rrying about
fluid and the
:d by starting
terms as We
.“6 at the bot—
the free sur
[Ch point 1 at ation reduces drops across
the presence
to flow. This
:wo points, as
liquid whose
1e differential
ust be greater ed by starting
.cting the pgh (3—14) .d ignored the
Simplifying, (3—15)
though it has
pipe is a gas,
P 2 E Pzgh Vlanometer easured by a
:l on a moun—
is 85.6 kPa.
m, and n3 ' 1000 kg/m3, w d by a multi—
ed. ipres$ure at the air—water interface. 1000 kg/m3, 850 kg/m3, and 13,600 kg/m3, respectively. Aha/ysis Starting with the pressure at point 1 at the air—water interface,
moving along the tube by adding or subtracting the pgh terms until we reach
point 2, and setting the result equal to Patm since the tube is open to the atmosphere gives A 0 Pl + pwaterghl + poiighz * pmmurygh3 = P2 = Pm Solving for P1 and substituting, E:P1:: Pann “ Pwaterghl T Poiighz + Pmmuryghg “1 Palm + g(pmercuryh3 "T pwaterhl _ poilhz) L = 85.6 kPa + (9.81 m/s2)[(13,600 kg/m3)(0.35 m)  (1000 kg/m3)(0.1 m) l. N l kPa
_ 5 k/ 3 0.2 (8 0 ng I10] 1kg . m/sz 1000 N/m2 _ [130 kPa ause‘of the risk of exposure to mercury vapor during an accident. MPLE 3—7 Analyzing a Multifluid Manometer with EES pressure in the tank using EES. Also determine what the differential Were replaced by seawater with a density of 1030 kg/m3. TION The pressure in a water tank is measured by a muItifluid ‘cury is replaced by seawater are to be determined using EES.
*is We start the EES program by double—clicking on its icon, open a mospheric pressure in Pa for unit consistency): g=9.81
Patm= 85600
h1=0.1; h2=0.2; h3=0.35 » rw=1000; roil= 850; rm: 13600
Pl+rw*g*h1+ r0i1*g*h2—1‘m*g*h3 : Patm Here P1 is the only unknown, and it is determined by EES to be Assamptiaﬂ The air pressUre in the tank is uniform (i.e., its variation with “““a_tion is negligible due to its low density), and thus we can determine the; properties The densities of water, oil, and mercury are given to be I Gu35ian Note that jumping horizontally from one tube to the next and L ng that pressure remains the same in the same fluid simplifies the 7
considerably. Also note that mercury is a toxic fluid, and mercury‘_'_fj
meters and thermometers are being replaced by ones with safer fluids L, i." der the multifluid manometer discussed in Example 3—6. Determine ‘5 height h3 would be for the same air pressure if the mercury in the last eter. The air pressure in the tank and the differential fluid height h3 “ Ie, and type the following on the blank screen that appears (we express I 85 CHAPTER‘ 3 ' Mercury FIGURE 3—22 Schematic for Example 3~3; drawing not to scale. ' i 86
PRESSURE AND FLUID STATICS FIGURE 3—23 Various types of Bourdon tubes used
to measure pressure. They work on the
same principle as party noise—makers
(bottom photo) due to the ﬂat tube
cross section. Photo by John M. Cimbala. P1 = 129647 Pa '=“ 130 kPa which is identical to the result obtained in Example 3—6.'The height of the
fluid column h3 when mercury is replaced by seawater is determined easily by
replacing “hi3:035” by “P12129647” and “rm=13600” by “rm=1030,”
and clicking on the calculator symbol. It gives [13 = 4.62 m Discussion Note that we used the screen like a paper pad and wrote down
the relevant information together with the applicable relations in an orga—
nized manner. EES did the rest. Equations can be written on separate lines
or on the same line by separating them by semicolons, and blank or com—
ment linesican be inserted for readability. EES makes it very easy to ask
“what if” (questions and to perform parametric studies, as explained in Appendix 3 on the DVD. Other Pressure Measurement. Devices Another type of commonly used mechanical pressure measurement device
is the Bourdon tube, named after the French engineer and inventor Eugene
Bourdon (1808—1884), which consists of a bent, coiled, or twisted hollow
metal tube whose end is closed and connected to a dial indicator needle
(Fig. 3—23). When the tube is open to the atmosphere, the tube is unde
flected, and the needle on the dial at this state is calibrated to read zero
(gage pressure). When the ﬂuid inside the tube is pressurized, the tube
stretches and moves the needle in proportion to the applied pressure. Electronics have made their way into every aspect of life, including pres—
sure measurement devices. Modern pressure sensors, called pressure trans
ducers, use various techniques to convert the pressure effect to an electrical
effect such as a change in voltage, resistance, or capacitance. Pressure trans
ducers are smaller and faster, and they can be more sensitive, reliable, and
precise than their mechanical counterparts. They can measure pressures
from less than a millionth of 1 atm to several thousands of atm. A wide variety of pressure transducers is available to measure gage,
absolute, and differential pressures in a wide range of applications. Gage
pressure transducers use the atmospheric pressure as a reference by venting
the back side of the pressuresensing diaphragm to the atmosphere, and they
give a zero signal output at atmospheric pressure regardless of altitude.
Absolute pressure transducers are calibrated to have a zero signal output at
full vacuum. Diﬁerential pressure transducers measure the pressure difference
between two locations directly instead of using two pressure transducers
and taking their difference. Straingage pressure transducers work by having a diaphragm deﬂect
between two chambers open to” the pressure inputs. As the diaphragm
stretches in response to a change in pressure difference across it, the strain
gage stretches and a Wheatstone bridge circuit ampliﬁes the output. A
capacitance transducer works similarly, but capacitance change is measured
instead of resistance change as the diaphragm stretches. a t» eight of the
ad easily by
rm=1030,” wrote down
in an orga
Jarate lines
nk or com—
aasy to ask
(plained in :ment device
:ntor Eugene
'isted hollow
icator needle
ube is unde—
to read zero
.ed, the tube
ssure. eluding pres—
:ssure trans
an electrical
:essure trans
reliable, and
HS pressures easure gage,
ations. Gage
:e by venting
ere, and they
9 of altitude.
;nal output at
ire difference
3 transducers iragm deﬂect
e diaphragm
it, the strain
1e output. A
3 is measured Piezoelectric transducers, also called solidstate pressure transducers,
work on the principle that an electric potential is generated in a crystalline
Substance when it is subjected to mechanical pressure. This phenomenon,
ﬁrst discovered by brothers Pierre and Jacques Curie in 1880, is called the
piezoelectric (or presselectric) effect. Piezoelectric pressure transducers Very suitable for highpressure applications, but they are generally not as
sensitiVe as diaphragm—type transducers, especially at low pressures. Another type of mechanical pressure gage called a deadweight tester
is used primarily for calibration and can measure extremely high pres
sures (Fig. 3—24). As its name implies, a deadweight tester measures pres
sure directly through application of a weight that provides a force per unit
area~the fundamental deﬁnition of pressure. It is constructed with an inter—
nal chamber ﬁlled with a ﬂuid (usually oil), along with a tightfitting piston,
cylinder, and plunger. Weights are applied to the top of the piston, which
exerts a force on the oil in the chamber. The total force F acting on the oil
at the piston—oil interface is the sum of the weight of the piston plus the
applied weights. Since the piston cross—sectional area A6 is known, the pres—
sure is calculated as P = F/AE. The only signiﬁcant source of error is that
due to static friction along the interface between the piston and cylinder, but
even this error is usually negligibly small. The reference pressure port is
connected to either an unknown pressure that is to be measured or to a pres—
sure sensor that is to be calibrated. 3—3 1‘ INTRODUCTION TO FLUID STATICS Fluid statics deals with problems associated with ﬂuids‘at rest. The ﬂuid can be either gaseous or liquid. Fluid statics is generally referred to as
hydrostatics when the ﬂuid is a liquid and as aerostatics when the ﬂuid is a
gas. In ﬂuid statics, there is no relative motion between adjacent ﬂuid lay—
ers, and thus there are no shear (tangential) stresses in the ﬂuid trying to
deform it. The only stress‘we deal with in ﬂuid statics is the normal stress,
which is the pressure, and the variation of pressure is due only to the weight
of the ﬂuid. Therefore, the topic of ﬂuid statics has signiﬁcance only in
gravity fields, and the force relations developed naturally involve the gravi
tational acceleration g, The force exerted on a surface by a ﬂuid at rest is
normal to the surface at the point of contact since there is no relative motion
between the ﬂuid and the solid surface, and thus there are no shear forces
acting parallel to the surface. Fluid statics is used to determine the forces acting on ﬂoating or sub
merged bodies and the forces developed by devices like hydraulic presses
and car jacks. The design of many engineering systems such as water dams
and liquid storage tanks requires the determination of the forces acting on
their surfaces using ﬂuid statics. The complete description of the resultant
hydrostatic force acting on a submerged surface requires the determination
0f the magnitude, the direction, and the line of action of the force. In the
following two sections, weconsider the forces acting on both plane and
Curved surfaces of submerged bodies due to pressure. have a much faster frequency response compared to diaphragm units and are \ . . 87 . ,
CHAPTER3‘ ' u Weights Oil
reserv Reference pressure port oir Adjustab e
plunger Crank FIGURE 3—24 A deadweight tester is able to measure
extremely high pressures
(up to 10,000 psi in some applications). 88 ‘ " a t ’ PRESSURE’ANDFLUID STATICS' FIGURE 3—25
. Hoover Dam. Courtesy United States Department of the Interim;
Bureau ofReclamationLower Colorado Region. FIGURE 3—26 When analyzing hydrostatic forces on
submerged surfaces, the atmospheric
pressure can be subtracted for
simplicity when it acts on both sides of the structure. 3—4 “ HYDRDSTATIC FORCES 0N
SUBMERGED PLANE SURFACES A plate (such as a gate valve in a dam, the wall of a liquid storage tank, or
the hull of a ship at rest) is subjected to ﬂuid pressure distributed over its
surface when exposed to a liquid (Fig. 3—25). On a plane surface, the hydro
static forces form a system of parallel forces, and we often need to determine
the magnitude of the force and its point of application, which is called the
center of pressure. In most cases, the other side of the plate is open to the
atmosphere (such as the dry side of a gate), and thus atmospheric pressure
acts on both sides of the plate, yielding a zero resultant. In such cases, it is
convenient to subtract atmospheric pressure and work with the gage pressure
only (Fig. 3—26). For example, Pgage = pgh at the bottom of the lake. Consider the top surface of a ﬂat plate of arbitrary shape completely sub
merged in a liquid, as shown in Fig. 3—27 together with its normal View.
The plane of this surface (normal to the page) intersects the horizontal free
surface at angle 6, and we take the line of intersection to be the x—axis (out
of the page). The absolute pressure above the liquid is P0, which is the local
atmospheric pressure Patm if the liquid is open to the atmosphere (but P0
may be different than Patm if the space above the liquid is evacuated or pres
surized). Then the absolute pressure at any point on the plate is P = P0 + pgh = P0 + pgy sin 6 (3—16) where h is the vertical distance of the point from the free surface and y is
the distance of the point from the x—axis (from point 0 in Fig. 3—27). The
resultant hydrostatic force F R acting on the surface is determined by inte
grating the force P dA acting on a differential area dA over the entire sur
face area, ‘ FR: JPdA= J(POFpgysin0)dA=P0A+pgsin6JydA (3—17)
A A A But the ﬁrst moment of area I y dA is related to the ycoordinate of the A
centroid (or center) of the surface by Ill pgh Patm + pgh (a) P considered (17) PaLm subtracted atm :orage tank, or
lbuted over its
1C6, the hydro
d to determine
h is called the
is open to the
iheric pressure
lCh cases, it is
i gage pressure
ie lake.
)mpletely sub normal View.
horizontal free
the x—aXis (out
ich is the local
;phere (but PO
:uated or pres
is (3—1 6) 1rface and y is
ig. 3—27). The
mined by inte
the entire sur J y dA (3—17)
A )rdinate of the Pressure
distribution :P avg P=P0+pgysin0 Center of pressine FIGURE 3—27 :
Hydrostatic force on an inclined plane surface completely submerged in a liquid. 1
yc = — Jy dA (3—18)
A
A
Substituting,
FR = (P0 + pgyC sin 0)A = (P0 + pghC)A = PCA = P..ng (3—19) where PC = P0 + pghC is the pressure at the centroid of the surface, which
is equivalent to the average pressure Pavg on the surface, and 110 = yC sin 0
is the vertical distance of the centroid from the free surface of the liquid
(Fig. 3—28). Thus we conclude that: The magnitude of the resultant force acting on a plane surface of a
completely submerged plate in a homogeneous (constant density) fluid is equal to the product of the pressure PC at the centroid of the surface
and the area A of the surface (Fig. 3~29). The pressure P0 is usually atmospheric pressure, which can be ignored in
most force calculations since it acts on both sides of the plate. When this is
not the case, a practical way of accounting for the contribution of PO to the
resultant force is simply to add an equivalent depth hequiv = PO/pg to hC;
that is, to assume the presence of an additional liquid layer of thickness
hequiv on top of the liquid with absolute vacuum above. Next we need to determine the line of action of the resultant force F R.
Two parallel force systems are equivalent if they have the same magnitude
and the same moment about any point. The line of action of the resultant
hydrostatic force, in general, does not pass through the centroid of the sur—
face—it lies underneath Where the pressure is higher. The point of intersec—
non of the line of action of the resultant force and the surface is the center
0f pressure. The vertical location of the line of action is determinedgby equating the moment of the resultant force to the moment of the distributed
pressure force about the xaXis: M = JdeA= Jy(P§+pgysin0)dA=P0JydA+pgsin6Jy2dA
A A A A Pressure prism Plane surface 89 .‘CHAPTER3’ i, ‘ ofsurface. A : FIGURE 3—28 The pressure at the centroid of a plane
surface is equivalent to the average pressure on the surface. 9.0 PRESSURE AND‘FLUID STATICS /Center of pressure Centroid of area FIGURE 3—29 The resultant force acting on a plane
surface is equal to the product of the
pressure at the centroid of the surface
and the surface area, and its line of
action passes through the center of
pressure. yPFR =Po'yCA + Pg Sin 0 [mo (3—20) where yP is the distance of the center of pressure from the x—axis (point 0 in Fig. 3—29) and [mo = J y2 dA isthe second moment of area (also called
A the area moment of inertia) about the x—axis. The second moments of area are widely available for common shapes in engineering handbooks, but they are usually given about the axes passing through the centroid of the area. Fortunately, the second moments of area about two parallel axes are related to each other by the parallel axis theorem, which in this case is expressed as Ixx,0 = Ixx,C + where [m C is the second moment of area about the x—axis passing through the
centroid of the area and 3/0 (the y—coordinate of the centroid) is the distance
between the two parallel axes. Substituting the F R relation from Eq. 3—19 and
the [my 0 relation from Eq. 3—21 into Eq. 3—20 and solving for yP yields I XX, C [3’0 + Po/(Pg Sin 6)lA For P0 = 0, which is usually the case when the atmospheric pressure is
ignored, it simplifies to yp = yc + (3—22a) Ixx, C
y C Knowing yP, the vertical distance of the center of pressure from the free sur
face is determined from hp = )2], sin 6. The I“, C values for some common areas are given in Fig. 3—30. For areas
that possess symmetry about the y—axis, the center of pressure lies on the y
axis directly below the centroid. The location of the center of pressure in
such cases is simply the point on the surface of the vertical plane of symme—
try at a distance hp from the free surface. Pressure acts normal to the surface, and the hydrostatic forces acting on a
ﬂat plate of any shape form a volume whose base is the plate area and
whose length is the linearly varying pressure, as shown in Fig. 3—31. This
virtual pressure prism has an interesting physical interpretation: its volume
is equal to the magnitude of the resultant hydrostatic force acting on the
plate since F R = f P dA, and the line of action of this force passes through
the centroid of this homogeneous prism. The projection of the centroid on
the plate is the pressure center Therefore, with the concept of pressure
prism, the problem of describing the resultant hydrostatic force on a plane
surface reduces to finding the volume and the two coordinates of the cen—
troid of this pressure prism. yp = yc + (3—2211) Special Case: Subme’rged Rectangular Plate Consider a completely submerged rectangular ﬂat plate of height [9 and
width a tilted at an angle 6 from the horizontal and whose top edge is hori—
zontal and is at a distance 's from the free surface along the plane of the
plate, as shown in Fig. 3—32a. The resultant hydrostatic force on the upper
surface is equal to the average pressure, which is the pressure at the mid—
point of the surface, times the surface area A. That is, 'u. x (3—20)
;is (point 0 in :1 (also called ments of area
ooks, but they
:1 of the area.
Les are related
9 expressed as (321) 1g through the
.s the distance
Eq. 3—19 and
P yields (3—22a) ‘ic pressure is (3—22h)
n the free sur —30. For areas
i lies on the y—
of pressure in
[1’16 of symme :es acting on a
)late area and
ig. 3—31. This
on: its volume
acting on the
)asses through
1e centroid on
Qt of pressure
toe on a plane
,es of the cen— te height [9 and
p edge is hori—
e plane of the
3 on the upper
re at the mid“ .ﬂjfs”..., A = ab, 1m C = ab3/12 A = 7R2, [m C = 77R4/4
(a) Rectangle ([7) Circle a/Z a/2 A = ab/2, I“ C = ab3/36 A = «122/2, 1% C = 0.109757R4
(d) Triangle (e) Semicircle
FIGURE 3—30 The centroid and the centroidal moments of inertia for some common geometries. Tilted rectangular plate: FR = PCA = [P0 + pg(s l b/Z) sin 6]ab (3—23) The force acts at a vertical distance of hp = yP sin 6 from the free surface
directly beneath the centroid of the plate Where, from Eq. 3—22a, I b ab3/12
' 2 [s + b/Z + PO/(pg sin 6)]ab
‘ b u b2
' 2 12[s + 12/2 + PO/(pg sin 0)] When the upper edge of the plate is at the free surface and thus s = 0, Eq.
3—23 reduces to yin—S (3—24) Tilted rectangular plate (3 = 0): FR = [P0 + pg(b sin 6)/2]ab For a completely submerged vertical plate (6 é 90°) Whose top edge is hori—
zontal, the hydrostatic force can be obtained by setting sin 6 = 1 (Fig. 3—3219) Vertical rectangular plate: FR = [P0 + pg(s + b/ 2)]ab
FR = (P0 + pgb/2)ab i3—27) hWhen the effect of P0 is ignored since it acts on both sides of the plate, the
Ydrostatic force on a vertical rectangular surface of height [2 whose top (3—25) (3—26) Vertical rectangular plate (5 : O): 91 A = ﬂab, In, C = 77ab3/4
(c) Ellipse A = 7mm, 1“, C = 0109757111;3
(f) Semiellipse Pressure prism FIGURE 3—31 The hydrostatic forces acting on a
plane surface form a pressure prism
whose base (left face) is the surface
and whose length is the pressure. ' , ,92
PRESSURE AND FLUID STATICS‘ ; FR = [P0 + pg(s + 22/2) sin 9142: y g FR = [P0 + pg(s + b/2)]ab
(a) Tilted plate (17) Vertical plate (0) Horizontal plate
FIGURE 3—32 Hydrostatic force acting on the top surface of a submerged rectangular plate for tilted, vertical, and horizontal cases. edge is horizontal and at the free surface is F R = pgabZ/Z acting at a dis
tance of 2b/3 from the free surface directly beneath the centroid of the plate.
‘ ‘ The pressure distribution on a submerged horizontal surface is uniform, f and its magnitude is P = P0 + pgh, Where h is the distance of the surface
3 from the free surface. Therefore, the hydrostatic force acting on a horizontal
rectangular surface is Horizontal rectangular plate: FR = (P0 + pgh)ab (3—28)
and it acts through the midpoint of the plate (Fig. 3—32c). EXAMPLE 3—8 Hydrostatic Force Acting on the Door
of a Submerged Car A heavy car plunges into a lake during an accident and lands at the bottom a
of the lake on its wheels (Fig. 3—33). The door is 1.2 m high and 1 m wide, and the top edge of the door is 8 m below the free surface of the water.
Determine the hydrostatic force on the door and the location of the pressure
center, and discuss if the driver can open the door. SOLUTION A car is submerged in water. The hydrostatic force on the door
is to be determined, and the likelihood of the driver opening the door is to
be assessed. Assumptions 1 The bottom surface of the lake is horizontal. 2 The passen—
ger cabin is well—sealed so that no water leaks inside. 3 The door can be
approximated as a vertical rectangular plate. 4 The pressure in the passenger
cabin remains at atmospheric value since there is no water leaking in, and
thus no compression of the air inside. Therefore, atmospheric pressure can—
cels out in the calculations since it acts on both sides of the door. 5 The
weight of the car is larger than the buoyant force acting on it. . l
i l
l FIGURE 3—33
l Schematic for Example 3—8. y "0 + pgh)ab l plate tal cases. acting at a dis
)id of the plate. 2;
ace is uniform, 3 of the surface on a horizontal if (3—28) at the botto :e on the do
the door is to: 2 The passenTg’
a door can be:
the passenget
aaking in, and
pressure canf
e door. 5 The" praparties We take the density of lake water to be 1000 kg/m3 throughout. if
Analysis  The average (gage) pressure on the door is the pressure value at
the‘Ce‘ntmid (midpoint) of the door and is determined to be Pavg = Pc = pghc : pg(s + M) ' 1kN
= 000k/ 3 9.811m2 8+ 1.2/2 (1 gm)( 8X 2 m) lOOOkg  m/s2 = 84.4 kN/m2
Then'the resultant hydrostatic force on the door becomes
L FR = PﬂVgA = (84.4 kN/m2)(1 m X 1.2 m) x 101.3 kN The pressure center is directly under the midpoint of the door, and its dis
tance from the surface of the lake is determined from Eq. 3—24 by setting
‘po‘; 0, yielding “ __ + + [,2 8+1.2+ 1.22
W” 2 12(s+b/2) 2 12(8+1.2/2) sign A strong person can lift 100 kg, whose weight is 981 N or about ‘
‘Also, the person can apply the force at a point farthest from they2
(_1 m farther) for maximum effect and generate a moment of 1 kN  m. '
sultant hydrostatic force acts under the midpoint of the door, and thus a
:e of 0.5 m from the hinges. This generates a moment of 50.6 kN ‘ m, I
about 50 times the moment the driver can possibly generate. There— _:‘
. impossible for the driver to open the door of the car. The driver’s _
L to let some water in (by rolling the window down a little, for ‘
nd to keep his or her head close to the ceiling. The driver should [
open the door shortly before the car is filled with water since at _ the pressures on both sides of the door are nearly the same and
the door in water is almost as easy as opening it in air. — 8.61 In 3—5 r HYDROSTATIC FORCES 0N
SUBMERGED CURVED SURFACES For a submerged curved surface, the determination of the resultant hydrosta—
tic force is more involved since it typically requires integration of the pres—
sure forces that change direction along the curved surface. The concept of
the pressure prism in this case is not much help either because of the com
plicated shapes involved. The easiest way to determine the resultant hydrostatic force F R acting on
a two—dimensional curved surface is to determine the horizontal and vertical
Components FH and F V separately. This is done by considering the free—body
dlagram of the liquid block enclosed by the curved surface and the two
plane surfaces (one horizontal and one vertical) passing through the two
ends of the curved surface, as shown in Fig. 3—34. Note that the vertical sur—
face of the liquid block considered is simply the projection of the curved
surface on a vertical plane, and the horizontal surface is the projection of
the Curved surface on a horizontal plane. The resultant force acting on the
Curved solid surface is then equal and opposite to the force acting on the
Curved liquid surface (Newton’s third law). CHAPTER 3 l . l ...
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 Spring '07
 Lear
 Statics, Fluid Mechanics

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