Unformatted text preview: Roosevelt Dam in Arizona. Hydrostatic pressure, due to the weight of a standing fluid, can cause
enormous forces and moments on largescale structures such as a dam. Hydrostatic fluid analysis is the subject of the present chapter. (Courtesy of Dr. E.R. Degginger/ColorPic Inc.)

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Chapter 2
Pressure Distribution
in a Fluid
Motivation. Many fluid problems do not involve motion. They concern the pressure
distribution in a static fluid and its effect on solid surfaces and on floating and submerged bodies.
When the fluid velocity is zero, denoted as the hydrostatic condition, the pressure
variation is due only to the weight of the fluid. Assuming a known fluid in a given
gravity field, the pressure may easily be calculated by integration. Important applications in this chapter are (1) pressure distribution in the atmosphere and the oceans, (2)
the design of manometer pressure instruments, (3) forces on submerged flat and curved
surfaces, (4) buoyancy on a submerged body, and (5) the behavior of floating bodies.
The last two result in Archimedes’ principles.
If the fluid is moving in rigidbody motion, such as a tank of liquid which has been
spinning for a long time, the pressure also can be easily calculated, because the fluid
is free of shear stress. We apply this idea here to simple rigidbody accelerations in
Sec. 2.9. Pressure measurement instruments are discussed in Sec. 2.10. As a matter of
fact, pressure also can be easily analyzed in arbitrary (nonrigidbody) motions V(x, y,
z, t), but we defer that subject to Chap. 4.
2.1 Pressure and Pressure
Gradient
In Fig. 1.1 we saw that a fluid at rest cannot support shear stress and thus Mohr’s circle reduces to a point. In other words, the normal stress on any plane through a fluid
element at rest is equal to a unique value called the fluid pressure p, taken positive for
compression by common convention. This is such an important concept that we shall
review it with another approach.
Figure 2.1 shows a small wedge of fluid at rest of size x by z by s and depth b
into the paper. There is no shear by definition, but we postulate that the pressures px, pz ,
and pn may be different on each face. The weight of the element also may be important.
Summation of forces must equal zero (no acceleration) in both the x and z directions.
Fx
Fz
0
0
pzb
pxb z
x
pnb s sin
pnb s cos
1
2
(2.1)
b
x
z

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60
Chapter 2 Pressure Distribution in a Fluid
z (up)
pn
∆s
θ
Element weight:
d W = ρ g ( 1 b ∆ x ∆ z)
2
∆z
px
∆x
θ
x
O
Fig. 2.1 Equilibrium of a small
wedge of fluid at rest.
Width b into paper
pz
but the geometry of the wedge is such that
s sin
z
s cos
x
(2.2)
Substitution into Eq. (2.1) and rearrangement give
px
pn
pz
1
2
pn
z
(2.3)
These relations illustrate two important principles of the hydrostatic, or shearfree, condition: (1) There is no pressure change in the horizontal direction, and (2) there is a
vertical change in pressure proportional to the density, gravity, and depth change. We
shall exploit these results to the fullest, starting in Sec. 2.3.
In the limit as the fluid wedge shrinks to a “point,’’ z → 0 and Eqs. (2.3) become
px
pz
pn
p
(2.4)
Since is arbitrary, we conclude that the pressure p at a point in a static fluid is independent of orientation.
What about the pressure at a point in a moving fluid? If there are strain rates in a
moving fluid, there will be viscous stresses, both shear and normal in general (Sec.
4.3). In that case (Chap. 4) the pressure is defined as the average of the three normal
stresses ii on the element
1
3
p
(
xx
yy
zz)
(2.5)
The minus sign occurs because a compression stress is considered to be negative
whereas p is positive. Equation (2.5) is subtle and rarely needed since the great majority of viscous flows have negligible viscous normal stresses (Chap. 4).
Pressure Force on a Fluid
Element
Pressure (or any other stress, for that matter) causes no net force on a fluid element
unless it varies spatially.1 To see this, consider the pressure acting on the two x faces
in Fig. 2.2. Let the pressure vary arbitrarily
p
p(x, y, z, t)
(2.6)
1

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An interesting application for a large element is in Fig. 3.7.

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2.2 Equilibrium of a Fluid Element
61
y
dz
p dy dz
(p+
dy
∂p
d x) dy dz
∂x
x
Fig. 2.2 Net x force on an element
due to pressure variation.
dx
z
The net force in the x direction on the element in Fig. 2.2 is given by
dFx
p dy dz
p
dx dy dz
x
p
p
dx dy dz
x
(2.7)
In like manner the net force dFy involves
p/ y, and the net force dFz concerns
p/ z. The total netforce vector on the element due to pressure is
dFpress
p
x
i
p
y
j
k
p
dx dy dz
z
(2.8)
We recognize the term in parentheses as the negative vector gradient of p. Denoting f
as the net force per unit element volume, we rewrite Eq. (2.8) as
∇p
fpress
(2.9)
Thus it is not the pressure but the pressure gradient causing a net force which must be
balanced by gravity or acceleration or some other effect in the fluid.
2.2 Equilibrium of a Fluid
Element
The pressure gradient is a surface force which acts on the sides of the element. There
may also be a body force, due to electromagnetic or gravitational potentials, acting on
the entire mass of the element. Here we consider only the gravity force, or weight of
the element
dFgrav
or
g dx dy dz
fgrav
(2.10)
g
In general, there may also be a surface force due to the gradient, if any, of the viscous stresses. For completeness, we write this term here without derivation and consider it more thoroughly in Chap. 4. For an incompressible fluid with constant viscosity, the net viscous force is
2
fVS
V
x2
2
2
V
y2
V
z2
∇2V
(2.11)

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where VS stands for viscous stresses and is the coefficient of viscosity from Chap.
1. Note that the term g in Eq. (2.10) denotes the acceleration of gravity, a vector act

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Chapter 2 Pressure Distribution in a Fluid
ing toward the center of the earth. On earth the average magnitude of g is 32.174 ft/s2
9.807 m/s2.
The total vector resultant of these three forces — pressure, gravity, and viscous
stress — must either keep the element in equilibrium or cause it to move with acceleration a. From Newton’s law, Eq. (1.2), we have
a
f
fpress
fgrav
∇p
fvisc
∇2V
g
(2.12)
This is one form of the differential momentum equation for a fluid element, and it is
studied further in Chap. 4. Vector addition is implied by Eq. (2.12): The acceleration
reflects the local balance of forces and is not necessarily parallel to the localvelocity
vector, which reflects the direction of motion at that instant.
This chapter is concerned with cases where the velocity and acceleration are known,
leaving one to solve for the pressure variation in the fluid. Later chapters will take up
the more general problem where pressure, velocity, and acceleration are all unknown.
Rewrite Eq. (2.12) as
∇p
(g
a)
∇2V
B(x, y, z, t)
(2.13)
where B is a short notation for the vector sum on the righthand side. If V and a
dV/dt are known functions of space and time and the density and viscosity are known,
we can solve Eq. (2.13) for p(x, y, z, t) by direct integration. By components, Eq. (2.13)
is equivalent to three simultaneous firstorder differential equations
p
x
Bx(x, y, z, t)
p
y
By(x, y, z, t)
p
z
Bz(x, y, z, t)
(2.14)
Since the righthand sides are known functions, they can be integrated systematically
to obtain the distribution p(x, y, z, t) except for an unknown function of time, which
remains because we have no relation for p/ t. This extra function is found from a condition of known time variation p0(t) at some point (x0, y0, z0). If the flow is steady (independent of time), the unknown function is a constant and is found from knowledge
of a single known pressure p0 at a point (x0, y0, z0). If this sounds complicated, it is
not; we shall illustrate with many examples. Finding the pressure distribution from a
known velocity distribution is one of the easiest problems in fluid mechanics, which
is why we put it in Chap. 2.
Examining Eq. (2.13), we can single out at least four special cases:
1. Flow at rest or at constant velocity: The acceleration and viscous terms vanish
identically, and p depends only upon gravity and density. This is the hydrostatic
condition. See Sec. 2.3.
2. Rigidbody translation and rotation: The viscous term vanishes identically,
and p depends only upon the term (g a). See Sec. 2.9.
3. Irrotational motion (
V 0): The viscous term vanishes identically, and
an exact integral called Bernoulli’s equation can be found for the pressure distribution. See Sec. 4.9.
4. Arbitrary viscous motion: Nothing helpful happens, no general rules apply, but
still the integration is quite straightforward. See Sec. 6.4.
Let us consider cases 1 and 2 here.

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2.3 Hydrostatic Pressure Distributions
63
p (Pascals)
High pressure:
p = 120,000 Pa abs = 30,000 Pa gage
120,000
30,000
Local atmosphere:
p = 90,000 Pa abs = 0 Pa gage = 0 Pa vacuum
90,000
40,000
Vacuum pressure:
p = 50,000 Pa abs = 40,000 Pa vacuum
50,000
50,000
Fig. 2.3 Illustration of absolute,
gage, and vacuum pressure readings.
Gage Pressure and Vacuum
Pressure: Relative Terms
Absolute zero reference:
p = 0 Pa abs = 90,000 Pa vacuum
0
(Tension)
Before embarking on examples, we should note that engineers are apt to specify pressures as (1) the absolute or total magnitude or (2) the value relative to the local ambient atmosphere. The second case occurs because many pressure instruments are of
differential type and record, not an absolute magnitude, but the difference between the
fluid pressure and the atmosphere. The measured pressure may be either higher or lower
than the local atmosphere, and each case is given a name:
1. p
2. p
pa
pa
Gage pressure:
Vacuum pressure:
p(gage)
p(vacuum)
p pa
pa p
This is a convenient shorthand, and one later adds (or subtracts) atmospheric pressure
to determine the absolute fluid pressure.
A typical situation is shown in Fig. 2.3. The local atmosphere is at, say, 90,000 Pa,
which might reflect a storm condition in a sealevel location or normal conditions at
an altitude of 1000 m. Thus, on this day, pa 90,000 Pa absolute 0 Pa gage 0 Pa
vacuum. Suppose gage 1 in a laboratory reads p1 120,000 Pa absolute. This value
may be reported as a gage pressure, p1 120,000 90,000 30,000 Pa gage. (One
must also record the atmospheric pressure in the laboratory, since pa changes gradually.) Suppose gage 2 reads p2 50,000 Pa absolute. Locally, this is a vacuum pressure and might be reported as p2 90,000 50,000 40,000 Pa vacuum. Occasionally, in the Problems section, we will specify gage or vacuum pressure to keep you
alert to this common engineering practice.
2.3 Hydrostatic Pressure
Distributions
0 and ∇2V
If the fluid is at rest or at constant velocity, a
the pressure distribution reduces to
∇p
0. Equation (2.13) for
g
(2.15)

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This is a hydrostatic distribution and is correct for all fluids at rest, regardless of their
viscosity, because the viscous term vanishes identically.
Recall from vector analysis that the vector ∇p expresses the magnitude and direction of the maximum spatial rate of increase of the scalar property p. As a result, ∇p

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64
Chapter 2 Pressure Distribution in a Fluid
is perpendicular everywhere to surfaces of constant p. Thus Eq. (2.15) states that a fluid
in hydrostatic equilibrium will align its constantpressure surfaces everywhere normal
to the localgravity vector. The maximum pressure increase will be in the direction of
gravity, i.e., “down.’’ If the fluid is a liquid, its free surface, being at atmospheric pressure, will be normal to local gravity, or “horizontal.’’ You probably knew all this before, but Eq. (2.15) is the proof of it.
In our customary coordinate system z is “up.’’ Thus the localgravity vector for smallscale problems is
g
gk
(2.16)
where g is the magnitude of local gravity, for example, 9.807 m/s2. For these coordinates Eq. (2.15) has the components
p
x
p
y
0
p
z
0
g
(2.17)
the first two of which tell us that p is independent of x and y. Hence p/ z can be replaced by the total derivative dp/dz, and the hydrostatic condition reduces to
dp
dz
2
or
p2
p1
dz
(2.18)
1
Equation (2.18) is the solution to the hydrostatic problem. The integration requires an
assumption about the density and gravity distribution. Gases and liquids are usually
treated differently.
We state the following conclusions about a hydrostatic condition:
Pressure in a continuously distributed uniform static fluid varies only with vertical
distance and is independent of the shape of the container. The pressure is the same
at all points on a given horizontal plane in the fluid. The pressure increases with
depth in the fluid.
An illustration of this is shown in Fig. 2.4. The free surface of the container is atmospheric and forms a horizontal plane. Points a, b, c, and d are at equal depth in a horizonAtmospheric pressure:
Free surface

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Fig. 2.4 Hydrostaticpressure distribution. Points a, b, c, and d are at
equal depths in water and therefore
have identical pressures. Points A,
B, and C are also at equal depths in
water and have identical pressures
higher than a, b, c, and d. Point D
has a different pressure from A, B,
and C because it is not connected
to them by a water path.
Water
Depth 1
a
b
c
d
Mercury
Depth 2

A
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C
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2.3 Hydrostatic Pressure Distributions
65
tal plane and are interconnected by the same fluid, water; therefore all points have the
same pressure. The same is true of points A, B, and C on the bottom, which all have the
same higher pressure than at a, b, c, and d. However, point D, although at the same depth
as A, B, and C, has a different pressure because it lies beneath a different fluid, mercury.
Effect of Variable Gravity
For a spherical planet of uniform density, the acceleration of gravity varies inversely
as the square of the radius from its center
g
g0
r0
r
2
(2.19)
where r0 is the planet radius and g0 is the surface value of g. For earth, r0 3960
statute mi 6400 km. In typical engineering problems the deviation from r0 extends
from the deepest ocean, about 11 km, to the atmospheric height of supersonic transport
operation, about 20 km. This gives a maximum variation in g of (6400/6420)2, or 0.6
percent. We therefore neglect the variation of g in most problems.
Hydrostatic Pressure in Liquids
Liquids are so nearly incompressible that we can neglect their density variation in hydrostatics. In Example 1.7 we saw that water density increases only 4.6 percent at the
deepest part of the ocean. Its effect on hydrostatics would be about half of this, or 2.3
percent. Thus we assume constant density in liquid hydrostatic calculations, for which
Eq. (2.18) integrates to
Liquids:
p2
or
p1
z1
z2
(z2
p2
z1)
(2.20)
p1
We use the first form in most problems. The quantity is called the specific weight of
the fluid, with dimensions of weight per unit volume; some values are tabulated in
Table 2.1. The quantity p/ is a length called the pressure head of the fluid.
For lakes and oceans, the coordinate system is usually chosen as in Fig. 2.5, with
z 0 at the free surface, where p equals the surface atmospheric pressure pa. When
Table 2.1 Specific Weight of Some
Common Fluids
Specific weight
at 68°F 20°C
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lbf/ft3
N/m3
Air (at 1 atm)
Ethyl alcohol
SAE 30 oil
Water
Seawater
Glycerin
Carbon tetrachloride
Mercury
v
Fluid
000.0752
049.2
055.5
062.4
064.0
078.7
099.1
846
000,011.8
007,733
008,720
009,790
010,050
012,360
015,570
133,100
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66
Chapter 2 Pressure Distribution in a Fluid
Z
p ≈ pa – b γ air
+b
Air
Free surface: Z = 0, p = pa
0
Water
g
Fig. 2.5 Hydrostaticpressure distribution in oceans and atmospheres.
p ≈ pa + hγ water
–h
we introduce the reference value ( p1, z1)
(negative) depth z,
Lakes and oceans:
p
( pa, 0), Eq. (2.20) becomes, for p at any
pa
z
(2.21)
where is the average specific weight of the lake or ocean. As we shall see, Eq. (2.21)
holds in the atmosphere also with an accuracy of 2 percent for heights z up to 1000 m.
EXAMPLE 2.1
Newfound Lake, a freshwater lake near Bristol, New Hampshire, has a maximum depth of 60
m, and the mean atmospheric pressure is 91 kPa. Estimate the absolute pressure in kPa at this
maximum depth.
Solution
From Table 2.1, take
9790 N/m3. With pa
the pressure at this depth will be
91 kN/m2
p
91 kPa
(9790 N/m3)( 60 m)
587 kN/m2
By omitting pa we could state the result as p
60 m, Eq. (2.21) predicts that
1 kN
1000 N
678 kPa
Ans.
587 kPa (gage).
The simplest practical application of the hydrostatic formula (2.20) is the barometer
(Fig. 2.6), which measures atmospheric pressure. A tube is filled with mercury and inverted while submerged in a reservoir. This causes a near vacuum in the closed upper
end because mercury has an extremely small vapor pressure at room temperatures (0.16
Pa at 20°C). Since atmospheric pressure forces a mercury column to rise a distance h
into the tube, the upper mercury surface is at zero pressure.

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The Mercury Barometer
91 kPa and z

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2.3 Hydrostatic Pressure Distributions
67
p1 ≈ 0
( Mercury has a very
low vapor pressure.)
z1 = h
p2 ≈ pa
( The mercury is in
contact with the
atmosphere.)
p
h= γa
M
z
pa
z2 = 0
ρ
M
Mercury
(b)
(a)
Fig. 2.6 A barometer measures local absolute atmospheric pressure: (a) the height of a mercury column is proportional to patm; (b) a modern portable barometer, with digital readout, uses the resonating silicon element of
Fig. 2.28c. (Courtesy of Paul Lupke, Druck Inc.)
From Fig. 2.6, Eq. (2.20) applies with p1
pa
or
0
0 at z1
M(0
h and p2
0:
h)
pa
h
pa at z2
(2.22)
M
At sealevel standard, with pa 101,350 Pa and M 133,100 N/m3 from Table 2.1,
the barometric height is h 101,350/133,100 0.761 m or 761 mm. In the United
States the weather service reports this as an atmospheric “pressure’’ of 29.96 inHg
(inches of mercury). Mercury is used because it is the heaviest common liquid. A water barometer would be 34 ft high.
Hydrostatic Pressure in Gases
Gases are compressible, with density nearly proportional to pressure. Thus density must
be considered as a variable in Eq. (2.18) if the integration carries over large pressure
changes. It is sufficiently accurate to introduce the perfectgas law p
RT in Eq.
(2.18)

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p
g
RT
g
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Chapter 2 Pressure Distribution in a Fluid
Separate the variables and integrate between points 1 and 2:
2
dp
p
1
ln
p2
p1
g
R
2
1
dz
T
(2.23)
The integral over z requires an assumption about the temperature variation T(z). One
common approximation is the isothermal atmosphere, where T T0:
p2
g(z2 z1)
RT0
p1 exp
(2.24)
The quantity in brackets is dimensionless. (Think that over; it must be dimensionless,
right?) Equation (2.24) is a fair approximation for earth, but actually the earth’s mean
atmospheric temperature drops off nearly linearly with z up to an altitude of about
36,000 ft (11,000 m):
T
T0
Bz
(2.25)
Here T0 is sealevel temperature (absolute) and B is the lapse rate, both of which vary
somewhat from day to day. By international agreement [1] the following standard values are assumed to apply from 0 to 36,000 ft:
T0
518.69°R
B
288.16 K
0.003566°R/ft
15°C
0.00650 K/m
(2.26)
This lower portion of the atmosphere is called the troposphere. Introducing Eq. (2.25)
into (2.23) and integrating, we obtain the more accurate relation
p
pa 1
Bz
T0
g/(RB)
where
g
RB
5.26 (air)
(2.27)
in the troposphere, with z 0 at sea level. The exponent g/(RB) is dimensionless (again
it must be) and has the standard value of 5.26 for air, with R 287 m2/(s2 K).
The U.S. standard atmosphere [1] is sketched in Fig. 2.7. The pressure is seen to be
nearly zero at z 30 km. For tabulated properties see Table A.6.
EXAMPLE 2.2
If sealevel pressure is 101,350 Pa, compute the standard pressure at an altitude of 5000 m, using (a) the exact formula and (b) an isothermal assumption at a standard sealevel temperature
of 15°C. Is the isothermal approximation adequate?
Solution
Part (a)
Use absolute temperature in the exact formula, Eq. (2.27):
p
pa 1
(0.00650 K/m)(5000 m)
288.16 K
101,350(0.52388)
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5.26
(101,350 Pa)(0.8872)5.26
54,000 Pa
This is the standardpressure result given at z
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Ans. (a)
5000 m in Table A.6.
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2.3 Hydrostatic Pressure Distributions
50
40
40
Altitude z, km
60
50
30
20
20.1 km
–56.5°C
Altitude z, km
60
Part (b)
30
Eq. (2.24)
Eq. (2.27)
10
Eq. (2.26)
Troposphere
0
1.20 kPa
20
11.0 km
10
Fig. 2.7 Temperature and pressure
distribution in the U.S. standard atmosphere. (From Ref. 1.)
69
– 60
– 40
101.33 kPa
15°C
– 20
Temperature, °C
0
+20
40
80
Pressure, kPa
0
120
If the atmosphere were isothermal at 288.16 K, Eq. (2.24) would apply:
pa exp
p
gz
RT
(9.807 m/s2)(5000 m)
[287 m2/(s2 K)](288.16 K)
(101,350 Pa) exp
(101,350 Pa) exp(
0.5929)
60,100 Pa
Ans. (b)
This is 11 percent higher than the exact result. The isothermal formula is inaccurate in the troposphere.
Is the Linear Formula Adequate
for Gases?
The linear approximation from Eq. (2.20) or (2.21), p
z, is satisfactory for liquids, which are nearly incompressible. It may be used even over great depths in the
ocean. For gases, which are highly compressible, it is valid only over moderate changes
in altitude.
The error involved in using the linear approximation (2.21) can be evaluated by expanding the exact formula (2.27) into a series
1
Bz
T0
n
1
n
Bz
T0
n(n 1) Bz
2!
T0
2
(2.28)
where n g/(RB). Introducing these first three terms of the series into Eq. (2.27) and
rearranging, we obtain

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pa
az
1
1 Bz
T0
n
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(2.29)
70
Chapter 2 Pressure Distribution in a Fluid
Thus the error in using the linear formula (2.21) is small if the second term in parentheses in (2.29) is small compared with unity. This is true if
z
(n
2T0
1)B
20,800 m
(2.30)
We thus expect errors of less than 5 percent if z or z is less than 1000 m.
2.4 Application to Manometry
From the hydrostatic formula (2.20), a change in elevation z2 z1 of a liquid is equivalent to a change in pressure (p2 p1)/ . Thus a static column of one or more liquids
or gases can be used to measure pressure differences between two points. Such a device is called a manometer. If multiple fluids are used, we must change the density in
the formula as we move from one fluid to another. Figure 2.8 illustrates the use of the
formula with a column of multiple fluids. The pressure change through each fluid is
calculated separately. If we wish to know the total change p5 p1, we add the successive changes p2 p1, p3 p2, p4 p3, and p5 p4. The intermediate values of p
cancel, and we have, for the example of Fig. 2.8,
p5
p1
0(z2
z1)
w(z3
z2)
G(z4
z3)
M(z5
z4)
(2.31)
No additional simplification is possible on the righthand side because of the different densities. Notice that we have placed the fluids in order from the lightest
on top to the heaviest at bottom. This is the only stable configuration. If we attempt
to layer them in any other manner, the fluids will overturn and seek the stable
arrangement.
A Memory Device: Up Versus
Down
The basic hydrostatic relation, Eq. (2.20), is mathematically correct but vexing to engineers, because it combines two negative signs to have the pressure increase downward. When calculating hydrostatic pressure changes, engineers work instinctively by
simply having the pressure increase downward and decrease upward. Thus they use the
following mnemonic, or memory, device, first suggested to the writer by Professor John
z = z1
z2
z
z3
z4

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Fig. 2.8 Evaluating pressure
changes through a column of multiple fluids.
z5

Known pressure p1
Oil, ρo
p2 – p1 = – ρog(z 2 – z1)
Water, ρw
p3 – p2 = – ρw g(z 3 – z 2)
Glycerin, ρG
Mercury, ρM
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p4 – p3 = – ρG g(z 4 – z 3)
p5 – p4 = – ρM g(z 5 – z 4)
Sum = p5 – p1
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2.4 Application to Manometry
71
Open, pa
zA, pA
A
Jump across
z1, p1
Fig. 2.9 Simple open manometer
for measuring pA relative to atmospheric pressure.
z 2 , p2 ≈ pa
ρ1
p = p1 at z = z1 in fluid 2
ρ2
Foss of Michigan State University:
pdown
pup
z
(2.32)
Thus, without worrying too much about which point is “z1” and which is “z2”, the formula simply increases or decreases the pressure according to whether one is moving
down or up. For example, Eq. (2.31) could be rewritten in the following “multiple increase” mode:
p5
p1
0z1
z2
wz2
z3
Gz3
z4
Mz4
z5
That is, keep adding on pressure increments as you move down through the layered
fluid. A different application is a manometer, which involves both “up” and “down”
calculations.
Figure 2.9 shows a simple open manometer for measuring pA in a closed chamber
relative to atmospheric pressure pa, in other words, measuring the gage pressure. The
chamber fluid 1 is combined with a second fluid 2, perhaps for two reasons: (1) to
protect the environment from a corrosive chamber fluid or (2) because a heavier fluid
2 will keep z2 small and the open tube can be shorter. One can, of course, apply the
basic hydrostatic formula (2.20). Or, more simply, one can begin at A, apply Eq. (2.32)
“down” to z1, jump across fluid 2 (see Fig. 2.9) to the same pressure p1, and then use
Eq. (2.32) “up” to level z2:
pA
1zA
z1
2z1
z2
p2
patm
(2.33)
The physical reason that we can “jump across” at section 1 in that a continuous length
of the same fluid connects these two equal elevations. The hydrostatic relation (2.20)
requires this equality as a form of Pascal’s law:
Any two points at the same elevation in a continuous mass of the same static fluid
will be at the same pressure.
This idea of jumping across to equal pressures facilitates multiplefluid problems.
EXAMPLE 2.3

v
v
The classic use of a manometer is when two Utube legs are of equal length, as in Fig. E2.3,
and the measurement involves a pressure difference across two horizontal points. The typical ap

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72
Chapter 2 Pressure Distribution in a Fluid
Flow device
(a)
(b)
L
h
1
2
E2.3
plication is to measure pressure change across a flow device, as shown. Derive a formula for the
pressure difference pa pb in terms of the system parameters in Fig. E2.3.
Solution
Using our “updown” concept as in Eq. (2.32), start at (a), evaluate pressure changes around the
Utube, and end up at (b):
pa
1gL
or
pa
1gh
pb
2gh
(
2
1gL
pb
Ans.
1)gh
The measurement only includes h, the manometer reading. Terms involving L drop out. Note the
appearance of the difference in densities between manometer fluid and working fluid. It is a common student error to fail to subtract out the working fluid density 1 — a serious error if both
fluids are liquids and less disastrous numerically if fluid 1 is a gas. Academically, of course,
such an error is always considered serious by fluid mechanics instructors.
Although Ex. 2.3, because of its popularity in engineering experiments, is sometimes considered to be the “manometer formula,” it is best not to memorize it but
rather to adapt Eq. (2.20) or (2.32) to each new multiplefluid hydrostatics problem.
For example, Fig. 2.10 illustrates a multiplefluid manometer problem for finding the
ρ3
z 2, p2
zA, pA
Jump across
z 2, p2
ρ1
A
B
z1, p1

v
v
Fig. 2.10 A complicated multiplefluid manometer to relate pA to pB.
This system is not especially practical but makes a good homework
or examination problem.
Jump across
z1, p1
z 3, p3
Jump across
z 3, p3
ρ2
ρ4

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zB, pB
2.4 Application to Manometry
73
difference in pressure between two chambers A and B. We repeatedly apply Eq. (2.20),
jumping across at equal pressures when we come to a continuous mass of the same
fluid. Thus, in Fig. 2.10, we compute four pressure differences while making three jumps:
pA
pB
(pA
p1)
1(zA
(p1
p2)
z1)
2(z1
(p2
p3)
z2)
(p3
3(z2
pB)
z3)
4(z3
zB)
(2.34)
The intermediate pressures p1,2,3 cancel. It looks complicated, but really it is merely
sequential. One starts at A, goes down to 1, jumps across, goes up to 2, jumps across,
goes down to 3, jumps across, and finally goes up to B.
EXAMPLE 2.4
Pressure gage B is to measure the pressure at point A in a water flow. If the pressure at B is 87
kPa, estimate the pressure at A, in kPa. Assume all fluids are at 20°C. See Fig. E2.4.
SAE 30 oil
Gage B
6 cm
Mercury
A
5 cm
Water
flow
11 cm
4 cm
E2.4
Solution
First list the specific weights from Table 2.1 or Table A.3:
water
9790 N/m3
mercury
133,100 N/m3
oil
8720 N/m3
Now proceed from A to B, calculating the pressure change in each fluid and adding:
pA
or
W(
pA
z)W
M(
3
(9790 N/m )(
pA
489.5 Pa
z)M
O(
0.05 m)
9317 Pa
z)O
pB
(133,100 N/m3)(0.07 m)
523.2 Pa
pB
(8720 N/m3)(0.06 m)
87,000 Pa
2
where we replace N/m by its short name, Pa. The value zM 0.07 m is the net elevation
change in the mercury (11 cm 4 cm). Solving for the pressure at point A, we obtain
pA
96,351 Pa
96.4 kPa

v
v
The intermediate sixfigure result of 96,351 Pa is utterly fatuous, since the measurements
cannot be made that accurately.

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Ans.
74
Chapter 2 Pressure Distribution in a Fluid
In making these manometer calculations we have neglected the capillaryheight
changes due to surface tension, which were discussed in Example 1.9. These effects
cancel if there is a fluid interface, or meniscus, on both sides of the Utube, as in Fig.
2.9. Otherwise, as in the righthand Utube of Fig. 2.10, a capillary correction can be
made or the effect can be made negligible by using largebore ( 1 cm) tubes.
2.5 Hydrostatic Forces on
Plane Surfaces
A common problem in the design of structures which interact with fluids is the computation of the hydrostatic force on a plane surface. If we neglect density changes in
the fluid, Eq. (2.20) applies and the pressure on any submerged surface varies linearly
with depth. For a plane surface, the linear stress distribution is exactly analogous to
combined bending and compression of a beam in strengthofmaterials theory. The hydrostatic problem thus reduces to simple formulas involving the centroid and moments
of inertia of the plate crosssectional area.
Figure 2.11 shows a plane panel of arbitrary shape completely submerged in a liquid. The panel plane makes an arbitrary angle with the horizontal free surface, so
that the depth varies over the panel surface. If h is the depth to any element area dA
of the plate, from Eq. (2.20) the pressure there is p pa
h.
To derive formulas involving the plate shape, establish an xy coordinate system in
the plane of the plate with the origin at its centroid, plus a dummy coordinate down
from the surface in the plane of the plate. Then the total hydrostatic force on one side
of the plate is given by
F
p dA
(pa
h) dA
paA
h dA
The remaining integral is evaluated by noticing from Fig. 2.11 that h
p = pa
Free surface
θ
h (x, y)
hCG
Resultant
force:
F = pCG A
ξ=
h
sin θ
y
Side view
CG
x

v
v
Fig. 2.11 Hydrostatic force and
center of pressure on an arbitrary
plane surface of area A inclined at
an angle below the free surface.
dA = dx dy
CP
Plan view of arbitrary plane surface

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(2.35)
sin
and,
2.5 Hydrostatic Forces on Plane Surfaces
75
by definition, the centroidal slant distance from the surface to the plate is
1
A
CG
Therefore, since
dA
(2.36)
is constant along the plate, Eq. (2.35) becomes
F
paA
sin
Finally, unravel this by noticing that
the surface to the plate centroid. Thus
F
pa A
dA
CG
hCG A
paA
sin
CGA
(2.37)
sin
hCG, the depth straight down from
( pa
hCG)A
pCG A
(2.38)
The force on one side of any plane submerged surface in a uniform fluid equals the
pressure at the plate centroid times the plate area, independent of the shape of the plate
or the angle at which it is slanted.
Equation (2.38) can be visualized physically in Fig. 2.12 as the resultant of a linear stress distribution over the plate area. This simulates combined compression and
bending of a beam of the same cross section. It follows that the “bending’’ portion of
the stress causes no force if its “neutral axis’’ passes through the plate centroid of area.
Thus the remaining “compression’’ part must equal the centroid stress times the plate
area. This is the result of Eq. (2.38).
However, to balance the bendingmoment portion of the stress, the resultant force
F does not act through the centroid but below it toward the highpressure side. Its line
of action passes through the center of pressure CP of the plate, as sketched in Fig. 2.11.
To find the coordinates (xCP, yCP), we sum moments of the elemental force p dA about
the centroid and equate to the moment of the resultant F. To compute yCP, we equate
FyCP
The term
yp dA
y( pa
sin ) dA
sin
y dA
pay dA vanishes by definition of centroidal axes. Introducing
Pressure distribution
pav = pCG
p (x, y)

v
v
Fig. 2.12 The hydrostaticpressure
force on a plane surface is equal,
regardless of its shape, to the resultant of the threedimensional linear
pressure distribution on that surface
F pCGA.

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Centroid of the plane surface

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Arbitrary
plane surface
of area A
Study Guide
(2.39)
CG
y,
76
Chapter 2 Pressure Distribution in a Fluid
we obtain
sin
FyCP
CG
y dA
y2 dA
sin
Ixx
(2.40)
where again y dA 0 and Ixx is the area moment of inertia of the plate area about its
centroidal x axis, computed in the plane of the plate. Substituting for F gives the result
Ixx
yCP
sin
(2.41)
pCGA
The negative sign in Eq. (2.41) shows that yCP is below the centroid at a deeper level
and, unlike F, depends upon angle . If we move the plate deeper, yCP approaches the
centroid because every term in Eq. (2.41) remains constant except pCG, which increases.
The determination of xCP is exactly similar:
xp dA
FxCP
sin
x[pa
xy dA
(
CG
y) sin ] dA
sin
Ixy
(2.42)
where Ixy is the product of inertia of the plate, again computed in the plane of the
plate. Substituting for F gives
Ixy
xCP
sin
(2.43)
pCGA
For positive Ixy, xCP is negative because the dominant pressure force acts in the third,
or lower left, quadrant of the panel. If Ixy 0, usually implying symmetry, xCP 0
and the center of pressure lies directly below the centroid on the y axis.
L
2
y
A = bL
x
Ixx =
L
2
A = π R2
y
bL3
x
12
R
Ix y = 0
R
Ixx =
π R4
4
Ix y = 0
b
2
b
2
(a)
(b)
s
y
Ixx =
x
L
3

v
v
Fig. 2.13 Centroidal moments of
inertia for various cross sections:
(a) rectangle, (b) circle, (c) triangle, and (d) semicircle.
b
2
b
2
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bL3
36
Ixx = 0.10976R 4
y
Ix y = 0
b (b – 2 s) L 2
Ix y =
72
x
R
(c )

2
A = πR
2
A = bL
2
2L
3
R
4R
3π
(d )

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Study Guide
2.5 Hydrostatic Forces on Plane Surfaces
77
In most cases the ambient pressure pa is neglected because it acts on both sides of the
plate; e.g., the other side of the plate is inside a ship or on the dry side of a gate or dam.
In this case pCG
hCG, and the center of pressure becomes independent of specific weight
GagePressure Formulas
F
hCGA
Ixx sin
hCGA
yCP
xCP
Ixy sin
hCGA
(2.44)
Figure 2.13 gives the area and moments of inertia of several common cross sections
for use with these formulas.
EXAMPLE 2.5
The gate in Fig. E2.5a is 5 ft wide, is hinged at point B, and rests against a smooth wall at point
A. Compute (a) the force on the gate due to seawater pressure, (b) the horizontal force P exerted
by the wall at point A, and (c) the reactions at the hinge B.
Wall
pa
Seawater:
64 lbf/ft 3
15 ft
A
pa
Gate
6 ft
θ
B
E2.5a
8 ft
Hinge
Solution
Part (a)
By geometry the gate is 10 ft long from A to B, and its centroid is halfway between, or at elevation 3 ft above point B. The depth hCG is thus 15 3 12 ft. The gate area is 5(10) 50 ft2. Neglect pa as acting on both sides of the gate. From Eq. (2.38) the hydrostatic force on the gate is
F
Part (b)
pCGA
hCGA
(64 lbf/ft3)(12 ft)(50 ft2)
38,400 lbf
First we must find the center of pressure of F. A freebody diagram of the gate is shown in Fig.
E2.5b. The gate is a rectangle, hence
Ixy
0
and
Ixx
bL3
12
(5 ft)(10 ft)3
12
417 ft4
The distance l from the CG to the CP is given by Eq. (2.44) since pa is neglected.
v
v
l


Ans. (a)
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yCP
Ixx sin
hCGA
6
(417 ft4)( 10 )
(12 ft)(50 ft2)
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0.417 ft
Study Guide
Chapter 2 Pressure Distribution in a Fluid
A
P
F
5 ft
CG
l
θ
B
Bx
CP
L = 10 ft
Bz
E2.5b
The distance from point B to force F is thus 10
terclockwise about B gives
PL sin
F(5
Part (c)
l
l)
P(6 ft)
P
or
5
4.583 ft. Summing moments coun
(38,400 lbf)(4.583 ft)
0
29,300 lbf
Ans. (b)
With F and P known, the reactions Bx and Bz are found by summing forces on the gate
Fx
0
Bx
F sin
or
Bx
Fz
0
Bz
F cos
or
Bz
P
Bx
38,400(0.6)
29,300
6300 lbf
Bz
38,400(0.8)
30,700 lbf
Ans. (c)
This example should have reviewed your knowledge of statics.
EXAMPLE 2.6
A tank of oil has a righttriangular panel near the bottom, as in Fig. E2.6. Omitting pa, find the
(a) hydrostatic force and (b) CP on the panel.
pa
Oil: ρ = 800 kg/m 3
5m
30°
11 m
4m
6m
pa
CG
CP
4m
8m
v

2m
4m
E2.6
v
78

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2.6 Hydrostatic Forces on Curved Surfaces
79
Solution
Part (a)
The triangle has properties given in Fig. 2.13c. The centroid is onethird up (4 m) and onethird
over (2 m) from the lower left corner, as shown. The area is
1
2
(6 m)(12 m)
36 m2
The moments of inertia are
bL3
36
Ixx
and
b(b
Ixy
(6 m)(12 m)3
36
2s)L2
72
The depth to the centroid is hCG
F
2.54
Part (b)
(6 m)[6 m
5
ghCGA
288 m4
4
2(6 m)](12 m)2
72
72 m4
9 m; thus the hydrostatic force from Eq. (2.44) is
(800 kg/m3)(9.807 m/s2)(9 m)(36 m2)
106 (kg m)/s2
2.54
106 N
2.54 MN
Ans. (a)
The CP position is given by Eqs. (2.44):
yCP
Ixx sin
hCGA
(288 m4)(sin 30°)
(9 m)(36 m2)
0.444 m
xCP
Ixy sin
hCGA
( 72 m4)(sin 30°)
(9 m)(36 m2)
0.111 m
Ans. (b)
The resultant force F 2.54 MN acts through this point, which is down and to the right of the
centroid, as shown in Fig. E2.6.
2.6 Hydrostatic Forces on
Curved Surfaces
The resultant pressure force on a curved surface is most easily computed by separating it into horizontal and vertical components. Consider the arbitrary curved surface
sketched in Fig. 2.14a. The incremental pressure forces, being normal to the local area
element, vary in direction along the surface and thus cannot be added numerically. We
Wair
d
Curved surface
projection onto
vertical plane
FV

v
v
Fig. 2.14 Computation of hydrostatic force on a curved surface:
(a) submerged curved surface; (b)
freebody diagram of fluid above
the curved surface.

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F1
b
W2
FH
a
FV
(a)

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(b)

F1
W1
c
FH
FH
e
Study Guide
FH
Chapter 2 Pressure Distribution in a Fluid
could sum the separate three components of these elemental pressure forces, but it turns
out that we need not perform a laborious threeway integration.
Figure 2.14b shows a freebody diagram of the column of fluid contained in the vertical projection above the curved surface. The desired forces FH and FV are exerted by
the surface on the fluid column. Other forces are shown due to fluid weight and horizontal pressure on the vertical sides of this column. The column of fluid must be in
static equilibrium. On the upper part of the column bcde, the horizontal components
F1 exactly balance and are not relevant to the discussion. On the lower, irregular portion of fluid abc adjoining the surface, summation of horizontal forces shows that the
desired force FH due to the curved surface is exactly equal to the force FH on the vertical left side of the fluid column. This leftside force can be computed by the planesurface formula, Eq. (2.38), based on a vertical projection of the area of the curved
surface. This is a general rule and simplifies the analysis:
The horizontal component of force on a curved surface equals the force on the plane
area formed by the projection of the curved surface onto a vertical plane normal to
the component.
If there are two horizontal components, both can be computed by this scheme.
Summation of vertical forces on the fluid free body then shows that
FV
W1
W2
Wair
(2.45)
We can state this in words as our second general rule:
The vertical component of pressure force on a curved surface equals in magnitude
and direction the weight of the entire column of fluid, both liquid and atmosphere,
above the curved surface.
Thus the calculation of FV involves little more than finding centers of mass of a column of fluid — perhaps a little integration if the lower portion abc has a particularly
vexing shape.
;;
;;
;;
EXAMPLE 2.7
A dam has a parabolic shape z/z0 (x/x0)2 as shown in Fig. E2.7a, with x0 10 ft and z0 24
ft. The fluid is water,
62.4 lbf/ft3, and atmospheric pressure may be omitted. Compute the
pa = 0 lbf/ft2 gage
FV
z
FH
z0
CP
x
x0
E2.7a

v
v
80

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((
x
z = z0 x
0

2
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2.6 Hydrostatic Forces on Curved Surfaces
81
forces FH and FV on the dam and the position CP where they act. The width of the dam is
50 ft.
Solution
The vertical projection of this curved surface is a rectangle 24 ft high and 50 ft wide, with its
centroid halfway down, or hCG 12 ft. The force FH is thus
FH
(62.4 lbf/ft3)(12 ft)(24 ft)(50 ft)
hCGAproj
899,000 lbf
103 lbf
899
Ans.
The line of action of FH is below the centroid by an amount
1
12
Ixx sin
hCGAproj
yCP
(50 ft)(24 ft)3(sin 90°)
(12 ft)(24 ft)(50 ft)
4 ft
Thus FH is 12 4 16 ft, or twothirds, down from the free surface or 8 ft from the bottom,
as might have been evident by inspection of the triangular pressure distribution.
The vertical component FV equals the weight of the parabolic portion of fluid above the
curved surface. The geometric properties of a parabola are shown in Fig. E2.7b. The weight of
this amount of water is
(62.4 lbf/ft3)( 2 )(10 ft)(24 ft)(50 ft)
3
( 2 x0z0b)
3
FV
499,000 lbf
499
103 lbf
Ans.
z0
Area =
3z0
5
2 x0z 0
3
FV
Parabola
0
E2.7b
x0 = 10 ft
3x 0
8
This acts downward on the surface at a distance 3x0 /8 3.75 ft over from the origin of coordinates. Note that the vertical distance 3z0 /5 in Fig. E2.7b is irrelevant.
The total resultant force acting on the dam is
F
2
(FH
2
FV)1/2
[(499)2
(899)2]1/2
1028
103 lbf
As seen in Fig. E2.7c, this force acts down and to the right at an angle of 29° tan 1 499 . The
899
force F passes through the point (x, z) (3.75 ft, 8 ft). If we move down along the 29° line until we strike the dam, we find an equivalent center of pressure on the dam at
xCP
5.43 ft
zCP
7.07 ft
Ans.

v
v
This definition of CP is rather artificial, but this is an unavoidable complication of dealing with
a curved surface.

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82
Chapter 2 Pressure Distribution in a Fluid
z
Resultant = 1028 × 103 1bf acts along z = 10.083 – 0.5555 x
3.75 ft
499
899
Parabola z = 0.24x2
29°
CG
CP
7.07 ft
8 ft
E2.7c
0
2.7 Hydrostatic Forces in
Layered Fluids
x
5.43 ft
The formulas for plane and curved surfaces in Secs. 2.5 and 2.6 are valid only for a
fluid of uniform density. If the fluid is layered with different densities, as in Fig. 2.15,
a single formula cannot solve the problem because the slope of the linear pressure distribution changes between layers. However, the formulas apply separately to each layer,
and thus the appropriate remedy is to compute and sum the separate layer forces and
moments.
Consider the slanted plane surface immersed in a twolayer fluid in Fig. 2.15. The
slope of the pressure distribution becomes steeper as we move down into the denser
z
F 1= p
CG1
Plane
surface
A1
z=0
p
a
ρ1 < ρ2
Fluid 1
p = p – ρ1gz
a
z 1, p1
F2 = p
p1 = p – ρ1gz1
a
A
CG 2 2
ρ2
Fluid 2
z 2 , p2
p = p1 – ρ2 g(z – z 1)

v
v
Fig. 2.15 Hydrostatic forces on a
surface immersed in a layered fluid
must be summed in separate pieces.
p2 = p1 – ρ 2 g(z 2 – z 1)

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2.7 Hydrostatic Forces in Layered Fluids
83
second layer. The total force on the plate does not equal the pressure at the centroid
times the plate area, but the plate portion in each layer does satisfy the formula, so that
we can sum forces to find the total:
F
Fi
pCGi Ai
(2.46)
Similarly, the centroid of the plate portion in each layer can be used to locate the center of pressure on that portion
yCPi
ig
sin i Ixxi
pCGi Ai
ig
xCPi
sin i Ixyi
pCGi Ai
(2.47)
These formulas locate the center of pressure of that particular Fi with respect to the
centroid of that particular portion of plate in the layer, not with respect to the centroid
of the entire plate. The center of pressure of the total force F
Fi can then be found
by summing moments about some convenient point such as the surface. The following example will illustrate.
EXAMPLE 2.8
A tank 20 ft deep and 7 ft wide is layered with 8 ft of oil, 6 ft of water, and 4 ft of mercury.
Compute (a) the total hydrostatic force and (b) the resultant center of pressure of the fluid on
the righthand side of the tank.
Solution
Part (a)
Divide the end panel into three parts as sketched in Fig. E2.8, and find the hydrostatic pressure
at the centroid of each part, using the relation (2.38) in steps as in Fig. E2.8:
pa = 0
Oi
z=0
7 ft
5.0
4 ft
(1)
l: 5
11 ft
lbf
/ft 3
8 ft
Wa
ter
(62
.4)
Me
rcu
ry
6 ft
(84
6)
16 ft
(2)
4 ft (3)
E2.8
v

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(55.0)(8)
62.4(3)
627 lbf/ft2
pCG3

(55.0 lbf/ft3)(4 ft)
pCG2
v
PCG1
(55.0)(8)
62.4(6)
846(2)
Textbook Table of Contents
220 lbf/ft2

2506 lbf/ft2
Study Guide
84
Chapter 2 Pressure Distribution in a Fluid
These pressures are then multiplied by the respective panel areas to find the force on each portion:
F1
(220 lbf/ft2)(8 ft)(7 ft)
pCG1A1
F2
F3
pCG2 A2
627(6)(7)
2506(4)(7)
pCG3A3
26,300 lbf
70,200 lbf
F
Part (b)
12,300 lbf
Fi
108,800 lbf
Ans. (a)
Equations (2.47) can be used to locate the CP of each force Fi, noting that
90° and sin
1 for all parts. The moments of inertia are Ixx1 (7 ft)(8 ft)3/12 298.7 ft4, Ixx2 7(6)3/12
126.0 ft4, and Ixx3 7(4)3/12 37.3 ft4. The centers of pressure are thus at
1gIxx
1
yCP1
F1
62.4(126.0)
26,300
yCP2
(55.0 lbf/ft3)(298.7 ft4)
12,300 lbf
0.30 ft
1.33 ft
846(37.3)
70,200
yCP3
0.45 ft
This locates zCP1
4 1.33
5.33 ft, zCP2
11 0.30
11.30 ft, and zCP3
16 0.45
16.45 ft. Summing moments about the surface then gives
FizCPi
or
12,300( 5.33)
or
26,300( 11.30)
zCP
1,518,000
108,800
FzCP
70,200( 16.45)
108,800zCP
13.95 ft
Ans. (b)
The center of pressure of the total resultant force on the right side of the tank lies 13.95 ft below the surface.
2.8 Buoyancy and Stability
The same principles used to compute hydrostatic forces on surfaces can be applied to
the net pressure force on a completely submerged or floating body. The results are the
two laws of buoyancy discovered by Archimedes in the third century B.C.:
1. A body immersed in a fluid experiences a vertical buoyant force equal to the
weight of the fluid it displaces.
2. A floating body displaces its own weight in the fluid in which it floats.
These two laws are easily derived by referring to Fig. 2.16. In Fig. 2.16a, the body
lies between an upper curved surface 1 and a lower curved surface 2. From Eq. (2.45)
for vertical force, the body experiences a net upward force
FB
FV (2)
FV (1)
(fluid weight above 2)
(fluid weight above 1)
weight of fluid equivalent to body volume
(2.48)
Alternatively, from Fig. 2.16b, we can sum the vertical forces on elemental vertical
slices through the immersed body:

v
v
FB

body
(p2
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p1) dAH
(z2
z1) dAH
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( )(body volume) (2.49)

Study Guide
2.8 Buoyancy and Stability
FV (1)
Horizontal
elemental
area d AH
p1
Surface
1
85
z1 – z 2
Fig. 2.16 Two different approaches
to the buoyant force on an arbitrary
immersed body: (a) forces on upper and lower curved surfaces; (b)
summation of elemental verticalpressure forces.
Surface
2
p2
FV (2)
(a)
(b)
These are identical results and equivalent to law 1 above.
Equation (2.49) assumes that the fluid has uniform specific weight. The line of action of the buoyant force passes through the center of volume of the displaced body;
i.e., its center of mass is computed as if it had uniform density. This point through
which FB acts is called the center of buoyancy, commonly labeled B or CB on a drawing. Of course, the point B may or may not correspond to the actual center of mass of
the body’s own material, which may have variable density.
Equation (2.49) can be generalized to a layered fluid (LF) by summing the weights
of each layer of density i displaced by the immersed body:
(FB)LF
ig(displaced
volume)i
(2.50)
Each displaced layer would have its own center of volume, and one would have to sum
moments of the incremental buoyant forces to find the center of buoyancy of the immersed body.
Since liquids are relatively heavy, we are conscious of their buoyant forces, but gases
also exert buoyancy on any body immersed in them. For example, human beings have
an average specific weight of about 60 lbf/ft3. We may record the weight of a person
as 180 lbf and thus estimate the person’s total volume as 3.0 ft3. However, in so doing
we are neglecting the buoyant force of the air surrounding the person. At standard conditions, the specific weight of air is 0.0763 lbf/ft3; hence the buoyant force is approximately 0.23 lbf. If measured in vacuo, the person would weigh about 0.23 lbf more.
For balloons and blimps the buoyant force of air, instead of being negligible, is the
controlling factor in the design. Also, many flow phenomena, e.g., natural convection
of heat and vertical mixing in the ocean, are strongly dependent upon seemingly small
buoyant forces.
Floating bodies are a special case; only a portion of the body is submerged, with
the remainder poking up out of the free surface. This is illustrated in Fig. 2.17, where
the shaded portion is the displaced volume. Equation (2.49) is modified to apply to this
smaller volume

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FB

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( )(displaced volume)
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floatingbody weight

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(2.51)
86
Chapter 2 Pressure Distribution in a Fluid
Neglect the displaced air up here.
CG
W
FB
B
Fig. 2.17 Static equilibrium of a
floating body.
(Displaced volume) × ( γ of fluid) = body weight
Not only does the buoyant force equal the body weight, but also they are collinear
since there can be no net moments for static equilibrium. Equation (2.51) is the mathematical equivalent of Archimedes’ law 2, previously stated.
EXAMPLE 2.9
A block of concrete weighs 100 lbf in air and “weighs’’ only 60 lbf when immersed in fresh water (62.4 lbf/ft3). What is the average specific weight of the block?
Solution
A freebody diagram of the submerged block (see Fig. E2.9) shows a balance between the apparent weight, the buoyant force, and the actual weight
60 lbf
Fz
or
FB
FB
40 lbf
0
60
FB
100
3
(62.4 lbf/ft )(block volume, ft3)
Solving gives the volume of the block as 40/62.4
the block is
0.641 ft3. Therefore the specific weight of
W = 100 lbf
E2.9
block
100 lbf
0.641 ft3
156 lbf/ft3
Ans.
Occasionally, a body will have exactly the right weight and volume for its ratio to
equal the specific weight of the fluid. If so, the body will be neutrally buoyant and will
remain at rest at any point where it is immersed in the fluid. Small neutrally buoyant
particles are sometimes used in flow visualization, and a neutrally buoyant body called
a Swallow float [2] is used to track oceanographic currents. A submarine can achieve
positive, neutral, or negative buoyancy by pumping water in or out of its ballast tanks.
Stability

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A floating body as in Fig. 2.17 may not approve of the position in which it is floating.
If so, it will overturn at the first opportunity and is said to be statically unstable, like
a pencil balanced upon its point. The least disturbance will cause it to seek another
equilibrium position which is stable. Engineers must design to avoid floating instabil

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2.8 Buoyancy and Stability
Small
∆θ disturbance
angle
Line of
symmetry
87
Small
disturbance
angle
∆θ
M
G
G
G
W
FB
W
Fig. 2.18 Calculation of the metacenter M of the floating body
shown in (a). Tilt the body a small
angle . Either (b) B moves far
out (point M above G denotes stability); or (c) B moves slightly
(point M below G denotes instability).
W
FB
B'
B'
B
Either
(a)
M
FB
Restoring moment
(b)
or
Overturning moment
(c)
ity. The only way to tell for sure whether a floating position is stable is to “disturb’’
the body a slight amount mathematically and see whether it develops a restoring moment which will return it to its original position. If so, it is stable; if not, unstable. Such
calculations for arbitrary floating bodies have been honed to a fine art by naval architects [3], but we can at least outline the basic principle of the staticstability calculation. Figure 2.18 illustrates the computation for the usual case of a symmetric floating
body. The steps are as follows:
1. The basic floating position is calculated from Eq. (2.51). The body’s center of
mass G and center of buoyancy B are computed.
2. The body is tilted a small angle , and a new waterline is established for the
body to float at this angle. The new position B of the center of buoyancy is calculated. A vertical line drawn upward from B intersects the line of symmetry at
a point M, called the metacenter, which is independent of
for small angles.
3. If point M is above G, that is, if the metacentric height MG is positive, a restoring moment is present and the original position is stable. If M is below G (negative MG, the body is unstable and will overturn if disturbed. Stability increases
with increasing MG.
Thus the metacentric height is a property of the cross section for the given weight, and
its value gives an indication of the stability of the body. For a body of varying cross
section and draft, such as a ship, the computation of the metacenter can be very involved.

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Stability Related to Waterline
Area

Naval architects [3] have developed the general stability concepts from Fig. 2.18 into
a simple computation involving the area moment of inertia of the waterline area about
the axis of tilt. The derivation assumes that the body has a smooth shape variation (no
discontinuities) near the waterline and is derived from Fig. 2.19.
The yaxis of the body is assumed to be a line of symmetry. Tilting the body a small
angle then submerges small wedge Obd and uncovers an equal wedge cOa, as shown.
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88
Chapter 2 Pressure Distribution in a Fluid
y
Original
waterline
area
G
Variablewidth
L(x) into paper
dA = x tan
M
dx
c
a
b
O
BG
Fig. 2.19 A floating body tilted
through a small angle . The movement x of the center of buoyancy B
is related to the waterline area moment of inertia.
G
x
d
B
x
e
Tilted floating body
The new position B of the center of buoyancy is calculated as the centroid of the submerged portion aObde of the body:
x υabOde
x dυ
x dυ
cOdea
Obd
0
x dυ
0
cOa
x L (x tan
dx)
Obd
x (L dA)
Obd
xL ( x tan
x (L dA)
cOa
dx)
x2 dAwaterline
tan
cOa
IO tan
waterline
where IO is the area moment of inertia of the waterline footprint of the body about its
tilt axis O. The first integral vanishes because of the symmetry of the original submerged portion cOdea. The remaining two “wedge” integrals combine into IO when
we notice that L dx equals an element of waterline area. Thus we determine the desired distance from M to B:
x
MB
IO
υsubmerged
MG
GB
or
MG
IO
υsub
GB
(2.52)
The engineer would determine the distance from G to B from the basic shape and
design of the floating body and then make the calculation of IO and the submerged
volume υsub. If the metacentric height MG is positive, the body is stable for small
disturbances. Note that if GB is negative, that is, B is above G, the body is always
stable.
EXAMPLE 2.10

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A barge has a uniform rectangular cross section of width 2L and vertical draft of height H, as
in Fig. E2.10. Determine (a) the metacentric height for a small tilt angle and (b) the range of
ratio L/H for which the barge is statically stable if G is exactly at the waterline as shown.

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2.9 Pressure Distribution in RigidBody Motion
89
G
O
G
L
E2.10
H
B
L
Solution
If the barge has length b into the paper, the waterline area, relative to tilt axis O, has a base b
and a height 2L; therefore, IO b(2L)3/12. Meanwhile, υsub 2LbH. Equation (2.52) predicts
8bL3/12
H
H
IO
L2
MG
GB
Ans. (a)
2LbH
2
2
υsub
3H
The barge can thus be stable only if
L2
3H2/2
or
2L
2.45H
Ans. (b)
The wider the barge relative to its draft, the more stable it is. Lowering G would help also.
Even an expert will have difficulty determining the floating stability of a buoyant
body of irregular shape. Such bodies may have two or more stable positions. For example, a ship may float the way we like it, so that we can sit upon the deck, or it may
float upside down (capsized). An interesting mathematical approach to floating stability is given in Ref. 11. The author of this reference points out that even simple shapes,
e.g., a cube of uniform density, may have a great many stable floating orientations, not
necessarily symmetric. Homogeneous circular cylinders can float with the axis of symmetry tilted from the vertical.
Floating instability occurs in nature. Living fish generally swim with their plane
of symmetry vertical. After death, this position is unstable and they float with their
flat sides up. Giant icebergs may overturn after becoming unstable when their shapes
change due to underwater melting. Iceberg overturning is a dramatic, rarely seen
event.
Figure 2.20 shows a typical North Atlantic iceberg formed by calving from a Greenland glacier which protruded into the ocean. The exposed surface is rough, indicating
that it has undergone further calving. Icebergs are frozen fresh, bubbly, glacial water
of average density 900 kg/m3. Thus, when an iceberg is floating in seawater, whose
average density is 1025 kg/m3, approximately 900/1025, or seveneighths, of its volume lies below the water.

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2.9 Pressure Distribution in
RigidBody Motion

In rigidbody motion, all particles are in combined translation and rotation, and there
is no relative motion between particles. With no relative motion, there are no strains
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90
Chapter 2 Pressure Distribution in a Fluid
Fig. 2.20 A North Atlantic iceberg
formed by calving from a Greenland glacier. These, and their even
larger Antarctic sisters, are the
largest floating bodies in the world.
Note the evidence of further calving fractures on the front surface.
(Courtesy of Soren Thalund, Green/
land tourism a/s Iiulissat, Greenland.)
or strain rates, so that the viscous term ∇2V in Eq. (2.13) vanishes, leaving a balance
between pressure, gravity, and particle acceleration
∇p
(g
a)
(2.53)
The pressure gradient acts in the direction g a, and lines of constant pressure (including the free surface, if any) are perpendicular to this direction. The general case
of combined translation and rotation of a rigid body is discussed in Chap. 3, Fig. 3.12.
If the center of rotation is at point O and the translational velocity is V0 at this point,
the velocity of an arbitrary point P on the body is given by2
V
V0
r0
where
is the angularvelocity vector and r0 is the position of point P. Differentiating, we obtain the most general form of the acceleration of a rigid body:
a
dV0
dt
(
r0)
d
dt
r0
(2.54)
Looking at the righthand side, we see that the first term is the translational acceleration; the second term is the centripetal acceleration, whose direction is from point

v
v
2

For a more detailed derivation of rigidbody motion, see Ref. 4, Sec. 2.7.
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2.9 Pressure Distribution in RigidBody Motion
91
P perpendicular toward the axis of rotation; and the third term is the linear acceleration due to changes in the angular velocity. It is rare for all three of these terms to apply to any one fluid flow. In fact, fluids can rarely move in rigidbody motion unless
restrained by confining walls for a long time. For example, suppose a tank of water
is in a car which starts a constant acceleration. The water in the tank would begin to
slosh about, and that sloshing would damp out very slowly until finally the particles
of water would be in approximately rigidbody acceleration. This would take so long
that the car would have reached hypersonic speeds. Nevertheless, we can at least discuss the pressure distribution in a tank of rigidly accelerating water. The following is
an example where the water in the tank will reach uniform acceleration rapidly.
EXAMPLE 2.11
A tank of water 1 m deep is in free fall under gravity with negligible drag. Compute the pressure at the bottom of the tank if pa 101 kPa.
Solution
Being unsupported in this condition, the water particles tend to fall downward as a rigid hunk
of fluid. In free fall with no drag, the downward acceleration is a g. Thus Eq. (2.53) for this
situation gives ∇p
(g g) 0. The pressure in the water is thus constant everywhere and
equal to the atmospheric pressure 101 kPa. In other words, the walls are doing no service in sustaining the pressure distribution which would normally exist.
Uniform Linear Acceleration
In this general case of uniform rigidbody acceleration, Eq. (2.53) applies, a having
the same magnitude and direction for all particles. With reference to Fig. 2.21, the parallelogram sum of g and a gives the direction of the pressure gradient or greatest rate
of increase of p. The surfaces of constant pressure must be perpendicular to this and
are thus tilted at a downward angle such that
ax
1
tan
g
(2.55)
az
z
ax
a
az
x
θ = tan –1
θ
–a
ax
g + az
Fluid
at rest
g
p µg – a
∆

v
v
Fig. 2.21 Tilting of constantpressure surfaces in a tank of
liquid in rigidbody acceleration.

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az
ax
S
p2
p = p1
p3
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Study Guide
Chapter 2 Pressure Distribution in a Fluid
One of these tilted lines is the free surface, which is found by the requirement that the
fluid retain its volume unless it spills out. The rate of increase of pressure in the direction g a is greater than in ordinary hydrostatics and is given by
dp
ds
G
where G
[a2
x
(g
az)2]1/2
(2.56)
These results are independent of the size or shape of the container as long as the
fluid is continuously connected throughout the container.
EXAMPLE 2.12
A drag racer rests her coffee mug on a horizontal tray while she accelerates at 7 m/s2. The mug
is 10 cm deep and 6 cm in diameter and contains coffee 7 cm deep at rest. (a) Assuming rigidbody acceleration of the coffee, determine whether it will spill out of the mug. (b) Calculate the
gage pressure in the corner at point A if the density of coffee is 1010 kg/m3.
Solution
Part (a)
The free surface tilts at the angle
az 0 and standard gravity,
given by Eq. (2.55) regardless of the shape of the mug. With
7.0
ax
tan 1
35.5°
9.81
g
If the mug is symmetric about its central axis, the volume of coffee is conserved if the tilted surface intersects the original rest surface exactly at the centerline, as shown in Fig. E2.12.
tan
1
3 cm
∆z
θ
7 cm
ax = 7 m/s2
A
3 cm
E2.12
Thus the deflection at the left side of the mug is
z
(3 cm)(tan )
2.14 cm
Ans. (a)
This is less than the 3cm clearance available, so the coffee will not spill unless it was sloshed
during the startup of acceleration.
Part (b)
When at rest, the gage pressure at point A is given by Eq. (2.20):
pA

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92

g(zsurf
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zA)

(1010 kg/m3)(9.81 m/s2)(0.07 m)
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694 N/m2

694 Pa
Study Guide
2.9 Pressure Distribution in RigidBody Motion
During acceleration, Eq. (2.56) applies, with G [(7.0)2 (9.81)2]1/2
tance ∆s down the normal from the tilted surface to point A is
s
(7.0
2.14)(cos )
93
12.05 m/s2. The dis
7.44 cm
Thus the pressure at point A becomes
pA
Gs
1010(12.05)(0.0744)
906 Pa
Ans. (b)
which is an increase of 31 percent over the pressure when at rest.
RigidBody Rotation
As a second special case, consider rotation of the fluid about the z axis without any
translation, as sketched in Fig. 2.22. We assume that the container has been rotating
long enough at constant for the fluid to have attained rigidbody rotation. The fluid
acceleration will then be the centripetal term in Eq. (2.54). In the coordinates of Fig.
2.22, the angularvelocity and position vectors are given by
k
r0
irr
(2.57)
Then the acceleration is given by
(
r0)
r
2
ir
(2.58)
as marked in the figure, and Eq. (2.53) for the force balance becomes
∇p
p
r
ir
k
p
z
(g
a)
( gk
r
2
ir)
(2.59)
Equating like components, we find the pressure field by solving two firstorder partial
differential equations
p
r
r
p
z
2
(2.60)
This is our first specific example of the generalized threedimensional problem described by Eqs. (2.14) for more than one independent variable. The righthand sides of
z, k
r, ir
p = pa
Ω
a = –r Ω 2 ir
–a
Stillwater
level

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Fig. 2.22 Development of paraboloid constantpressure surfaces in a
fluid in rigidbody rotation. The
dashed line along the direction of
maximum pressure increase is an
exponential curve.

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p = p1
Axis of
rotation
g
g–a
p2
p3

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94
Chapter 2 Pressure Distribution in a Fluid
(2.60) are known functions of r and z. One can proceed as follows: Integrate the first
equation “partially,’’ i.e., holding z constant, with respect to r. The result is
1
2
p
r2
2
f(z)
(2.61)
†
where the “constant’’ of integration is actually a function f(z). Now differentiate this
with respect to z and compare with the second relation of (2.60):
p
z
or
0
f (z)
f(z)
z
C
(2.62a)
where C is a constant. Thus Eq. (2.61) now becomes
p
const
1
2
z
r2
2
(2.62b)
This is the pressure distribution in the fluid. The value of C is found by specifying the
pressure at one point. If p p0 at (r, z) (0, 0), then C p0. The final desired distribution is
p
p0
z
1
2
r2
2
(2.63)
The pressure is linear in z and parabolic in r. If we wish to plot a constantpressure
surface, say, p p1, Eq. (2.63) becomes
z
p0
p1
r2 2
2g
a
br2
(2.64)
Thus the surfaces are paraboloids of revolution, concave upward, with their minimum
point on the axis of rotation. Some examples are sketched in Fig. 2.22.
As in the previous example of linear acceleration, the position of the free surface is
found by conserving the volume of fluid. For a noncircular container with the axis of
rotation offcenter, as in Fig. 2.22, a lot of laborious mensuration is required, and a
single problem will take you all weekend. However, the calculation is easy for a cylinder rotating about its central axis, as in Fig. 2.23. Since the volume of a paraboloid is
Still water
level
h
2
Volume =
π
2
R 2h
h
2
22
h= Ω R
2g
Ω
Fig. 2.23 Determining the freesurface position for rotation of a
cylinder of fluid about its central
axis.
R
R

v
v
†
This is because f(z) vanishes when differentiated with respect to r. If you don’t see this, you should
review your calculus.

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2.9 Pressure Distribution in RigidBody Motion
95
onehalf the base area times its height, the stillwater level is exactly halfway between
the high and low points of the free surface. The center of the fluid drops an amount
22
h/2
R /(4g), and the edges rise an equal amount.
EXAMPLE 2.13
The coffee cup in Example 2.12 is removed from the drag racer, placed on a turntable, and rotated about its central axis until a rigidbody mode occurs. Find (a) the angular velocity which
will cause the coffee to just reach the lip of the cup and (b) the gage pressure at point A for this
condition.
Solution
Part (a)
The cup contains 7 cm of coffee. The remaining distance of 3 cm up to the lip must equal the
distance h/2 in Fig. 2.23. Thus
2
22
(0.03 m)2
R
h
0.03 m
4(9.81 m/s2)
2
4g
Solving, we obtain
2
Part (b)
z
1308
or
36.2 rad/s
345 r/min
To compute the pressure, it is convenient to put the origin of coordinates r and z at the bottom
of the freesurface depression, as shown in Fig. E2.13. The gage pressure here is p0 0, and
point A is at (r, z) (3 cm, 4 cm). Equation (2.63) can then be evaluated
pA
0
(1010 kg/m3)(9.81 m/s2)( 0.04 m)
1
2 (1010
3 cm
396 N/m2
kg/m3)(0.03 m)2(1308 rad2/s2)
594 N/m2
990 Pa
This is about 43 percent greater than the stillwater pressure pA
0
Ans. (a)
Ans. (b)
694 Pa.
r
7 cm
Here, as in the linearacceleration case, it should be emphasized that the paraboloid
pressure distribution (2.63) sets up in any fluid under rigidbody rotation, regardless
of the shape or size of the container. The container may even be closed and filled with
fluid. It is only necessary that the fluid be continuously interconnected throughout the
container. The following example will illustrate a peculiar case in which one can visualize an imaginary free surface extending outside the walls of the container.
Ω
A
3 cm
3 cm
E2.13
EXAMPLE 2.14

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A Utube with a radius of 10 in and containing mercury to a height of 30 in is rotated about its
center at 180 r/min until a rigidbody mode is achieved. The diameter of the tubing is negligible. Atmospheric pressure is 2116 lbf/ft2. Find the pressure at point A in the rotating condition.
See Fig. E2.14.

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96
Chapter 2 Pressure Distribution in a Fluid
z
Solution
10 in
B
Convert the angular velocity to radians per second:
r
0
(180 r/min)
30 in
2 rad/r
60 s/min
18.85 rad/s
From Table 2.1 we find for mercury that
846 lbf/ft3 and hence
846/32.2 26.3 slugs/ft3.
At this high rotation rate, the free surface will slant upward at a fierce angle [about 84°; check
this from Eq. (2.64)], but the tubing is so thin that the free surface will remain at approximately
the same 30in height, point B. Placing our origin of coordinates at this height, we can calculate the constant C in Eq. (2.62b) from the condition pB 2116 lbf/ft2 at (r, z) (10 in, 0):
Ω
A
pB
2116 lbf/ft2
or
Imaginary
free surface
C
C
1
2
0
2116
(26.3 slugs/ft3)( 10 ft)2(18.85 rad/s)2
12
1129 lbf/ft2
3245
We then obtain pA by evaluating Eq. (2.63) at (r, z)
(0,
30 in):
E2.14
pA
1129
(846 lbf/ft3)(
30
12
ft)
1129
2115
986 lbf/ft2
Ans.
This is less than atmospheric pressure, and we can see why if we follow the freesurface paraboloid down from point B along the dashed line in the figure. It will cross the horizontal portion of the Utube (where p will be atmospheric) and fall below point A. From Fig. 2.23 the actual drop from point B will be
22
(18.85)2( 10 )2
R
12
h
3.83 ft 46 in
2(32.2)
2g
Thus pA is about 16 inHg below atmospheric pressure, or about 16 (846) 1128 lbf/ft2 below
12
pa 2116 lbf/ft2, which checks with the answer above. When the tube is at rest,
pA
2116
846(
30
12
4231 lbf/ft2
)
Hence rotation has reduced the pressure at point A by 77 percent. Further rotation can reduce
pA to nearzero pressure, and cavitation can occur.
An interesting byproduct of this analysis for rigidbody rotation is that the lines
everywhere parallel to the pressure gradient form a family of curved surfaces, as
sketched in Fig. 2.22. They are everywhere orthogonal to the constantpressure surfaces, and hence their slope is the negative inverse of the slope computed from Eq.
(2.64):
dz
dr GL
1
(dz/dr)p
1
const
r
2
/g
where GL stands for gradient line
dz
dr

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g
r
(2.65)
2
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2.10 Pressure Measurement
97
Separating the variables and integrating, we find the equation of the pressuregradient
surfaces
2
r
z
C1 exp
g
(2.66)
Notice that this result and Eq. (2.64) are independent of the density of the fluid. In the
absence of friction and Coriolis effects, Eq. (2.66) defines the lines along which the apparent net gravitational field would act on a particle. Depending upon its density, a small
particle or bubble would tend to rise or fall in the fluid along these exponential lines,
as demonstrated experimentally in Ref. 5. Also, buoyant streamers would align themselves with these exponential lines, thus avoiding any stress other than pure tension. Figure 2.24 shows the configuration of such streamers before and during rotation.
2.10 Pressure Measurement
Pressure is a derived property. It is the force per unit area as related to fluid molecular bombardment of a surface. Thus most pressure instruments only infer the pressure
by calibration with a primary device such as a deadweight piston tester. There are many
such instruments, both for a static fluid and a moving stream. The instrumentation texts
in Refs. 7 to 10, 12, and 13 list over 20 designs for pressure measurement instruments.
These instruments may be grouped into four categories:
1. Gravitybased: barometer, manometer, deadweight piston.
2. Elastic deformation: bourdon tube (metal and quartz), diaphragm, bellows,
straingage, optical beam displacement.
3. Gas behavior: gas compression (McLeod gage), thermal conductance (Pirani gage),
molecular impact (Knudsen gage), ionization, thermal conductivity, air piston.
4. Electric output: resistance (Bridgman wire gage), diffused strain gage, capacitative, piezoelectric, magnetic inductance, magnetic reluctance, linear variable differential transformer (LVDT), resonant frequency.

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The gasbehavior gages are mostly specialpurpose instruments used for certain scientific experiments. The deadweight tester is the instrument used most often for calibrations; for example, it is used by the U.S. National Institute for Standards and Technology (NIST). The barometer is described in Fig. 2.6.
The manometer, analyzed in Sec. 2.4, is a simple and inexpensive hydrostaticprinciple device with no moving parts except the liquid column itself. Manometer measurements must not disturb the flow. The best way to do this is to take the measurement through a static hole in the wall of the flow, as illustrated for the two instruments
in Fig. 2.25. The hole should be normal to the wall, and burrs should be avoided. If
the hole is small enough (typically 1mm diameter), there will be no flow into the measuring tube once the pressure has adjusted to a steady value. Thus the flow is almost
undisturbed. An oscillating flow pressure, however, can cause a large error due to possible dynamic response of the tubing. Other devices of smaller dimensions are used for
dynamicpressure measurements. Note that the manometers in Fig. 2.25 are arranged
to measure the absolute pressures p1 and p2. If the pressure difference p1 p2 is de

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98
Chapter 2 Pressure Distribution in a Fluid

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Fig. 2.24 Experimental demonstration with buoyant streamers of the
fluid force field in rigidbody rotation: (top) fluid at rest (streamers
hang vertically upward); (bottom)
rigidbody rotation (streamers are
aligned with the direction of maximum pressure gradient). (From Ref.
5, courtesy of R. Ian Fletcher.)

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Study Guide
2.10 Pressure Measurement
99
Flow
Flow
p1
p2
Fig. 2.25 Two types of accurate
manometers for precise measurements: (a) tilted tube with eyepiece; (b) micrometer pointer with
ammeter detector.
(a)
(b)
sired, a significant error is incurred by subtracting two independent measurements, and
it would be far better to connect both ends of one instrument to the two static holes p1
and p2 so that one manometer reads the difference directly. In category 2, elasticdeformation instruments, a popular, inexpensive, and reliable device is the bourdon
tube, sketched in Fig. 2.26. When pressurized internally, a curved tube with flattened
cross section will deflect outward. The deflection can be measured by a linkage attached to a calibrated dial pointer, as shown. Or the deflection can be used to drive
electricoutput sensors, such as a variable transformer. Similarly, a membrane or diaphragm will deflect under pressure and can either be sensed directly or used to drive
another sensor.
A
Section AA
Bourdon
tube
A
Pointer for
dial gage
Flattened tube deflects
outward under pressure
Linkage

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Fig. 2.26 Schematic of a bourdontube device for mechanical measurement of high pressures.

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High pressure

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Study Guide
100
Chapter 2 Pressure Distribution in a Fluid
Fig. 2.27 The fusedquartz, forcebalanced bourdon tube is the most
accurate pressure sensor used in
commercial applications today.
(Courtesy of Ruska Instrument
Corporation, Houston, TX.)
An interesting variation of Fig. 2.26 is the fusedquartz, forcedbalanced bourdon
tube, shown in Fig. 2.27, whose deflection is sensed optically and returned to a zero
reference state by a magnetic element whose output is proportional to the fluid pressure. The fusedquartz, forcedbalanced bourdon tube is reported to be one of the most
accurate pressure sensors ever devised, with uncertainty of the order of 0.003 percent.
The last category, electricoutput sensors, is extremely important in engineering
because the data can be stored on computers and freely manipulated, plotted, and analyzed. Three examples are shown in Fig. 2.28, the first being the capacitive sensor
in Fig. 2.28a. The differential pressure deflects the silicon diaphragm and changes
the capacitancce of the liquid in the cavity. Note that the cavity has spherical end
caps to prevent overpressure damage. In the second type, Fig. 2.28b, strain gages and
other sensors are diffused or etched onto a chip which is stressed by the applied pressure. Finally, in Fig. 2.28c, a micromachined silicon sensor is arranged to deform
under pressure such that its natural vibration frequency is proportional to the pressure. An oscillator excites the element’s resonant frequency and converts it into appropriate pressure units. For further information on pressure sensors, see Refs. 7 to
10, 12, and 13.
Summary

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This chapter has been devoted entirely to the computation of pressure distributions and
the resulting forces and moments in a static fluid or a fluid with a known velocity field.
All hydrostatic (Secs. 2.3 to 2.8) and rigidbody (Sec. 2.9) problems are solved in this
manner and are classic cases which every student should understand. In arbitrary viscous flows, both pressure and velocity are unknowns and are solved together as a system of equations in the chapters which follow.

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Cover flange
Seal diaphragm
Highpressure side
Lowpressure side
Filling liquid
Sensing diaphragm
(a)
Wire bonding
Stitch bonded
connections from
chip to body plug
Strain gages
Diffused into integrated
silicon chip
Fig. 2.28 Pressure sensors with
electric output: (a) a silicon diaphragm whose deflection changes
the cavity capacitance (Courtesy of
JohnsonYokogawa Inc.); (b) a silicon strain gage which is stressed
by applied pressure; (c) a micromachined silicon element which resonates at a frequency proportional
to applied pressure. [(b) and (c)
are courtesy of Druck, Inc., Fairfield, CT.]
Etched cavity
Micromachined
silicon sensor

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(Druck, Inc., Fairfield, Connecticut)
(b)

Temperature sensor
Onchip diode for
optimum temperature
performance
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(c)

Study Guide
101
102
Chapter 2 Pressure Distribution in a Fluid
P2.2 For the twodimensional stress field shown in Fig. P2.1
suppose that
Problems
Most of the problems herein are fairly straightforward. More
difficult or openended assignments are indicated with an asterisk, as in Prob. 2.8. Problems labeled with an EES icon (for
example, Prob. 2.62), will benefit from the use of the Engineering Equation Solver (EES), while problems labeled with a
disk icon may require the use of a computer. The standard endofchapter problems 2.1 to 2.158 (categorized in the problem
list below) are followed by word problems W2.1 to W2.8, fundamentals of engineering exam problems FE2.1 to FE2.10, comprehensive problems C2.1 to C2.4, and design projects D2.1 and
D2.2.
xx
P2.3
P2.4
Problem Distribution
Section
Topic
Problems
2.1, 2.2
2.3
2.3
2.4
2.5
2.6
2.7
2.8
2.8
2.9
2.9
2.10
Stresses; pressure gradient; gage pressure
Hydrostatic pressure; barometers
The atmosphere
Manometers; multiple fluids
Forces on plane surfaces
Forces on curved surfaces
Forces in layered fluids
Buoyancy; Archimedes’ principles
Stability of floating bodies
Uniform acceleration
Rigidbody rotation
Pressure measurements
2.1 – 2.6
2.7 – 2.23
2.24 – 2.29
2.30 – 2.47
2.48 – 2.81
2.82 – 2.100
2.101 – 2.102
2.103 – 2.126
2.127 – 2.136
2.137 – 2.151
2.152 – 2.158
None
P2.6
P2.1 For the twodimensional stress field shown in Fig. P2.1 it
is found that
P2.8
xx
3000 lbf/ft2
2000 lbf/ft2
yy
xy
P2.5
P2.7
500 lbf/ft2
Find the shear and normal stresses (in lbf/ft2) acting on
plane AA cutting through the element at a 30° angle as
shown.
σyy
σyx
=
σxy
*P2.9
A
σxx
A
30°
σxx
P2.10
σxy
=
σyx
σyy

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P2.1

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2000 lbf/ft2
yy
3000 lbf/ft2
n(AA)
2500 lbf/ft2
Compute (a) the shear stress xy and (b) the shear stress
on plane AA.
Derive Eq. (2.18) by using the differential element in Fig.
2.2 with z “up,’’ no fluid motion, and pressure varying only
in the z direction.
In a certain twodimensional fluid flow pattern the lines
of constant pressure, or isobars, are defined by the expression P0 Bz Cx2 constant, where B and C are
constants and p0 is the (constant) pressure at the origin,
(x, z) (0, 0). Find an expression x f (z) for the family
of lines which are everywhere parallel to the local pressure gradient Vp.
Atlanta, Georgia, has an average altitude of 1100 ft. On a
standard day (Table A.6), pressure gage A in a laboratory
experiment reads 93 kPa and gage B reads 105 kPa. Express these readings in gage pressure or vacuum pressure
(Pa), whichever is appropriate.
Any pressure reading can be expressed as a length or head,
h p/ g. What is standard sealevel pressure expressed in
(a) ft of ethylene glycol, (b) in Hg, (c) m of water, and (d)
mm of methanol? Assume all fluids are at 20°C.
The deepest known point in the ocean is 11,034 m in the
Mariana Trench in the Pacific. At this depth the specific
weight of seawater is approximately 10,520 N/m3. At the
surface,
10,050 N/m3. Estimate the absolute pressure
at this depth, in atm.
Dry adiabatic lapse rate (DALR) is defined as the negative value of atmospheric temperature gradient, dT/dz,
when temperature and pressure vary in an isentropic fashion. Assuming air is an ideal gas, DALR
dT/dz when
T T0( p/p0)a, where exponent a (k 1)/k, k cp /cv is
the ratio of specific heats, and T0 and p0 are the temperature and pressure at sea level, respectively. (a) Assuming
that hydrostatic conditions exist in the atmosphere, show
that the dry adiabatic lapse rate is constant and is given by
DALR g(k 1)/(kR), where R is the ideal gas constant
for air. (b) Calculate the numerical value of DALR for air
in units of °C/km.
For a liquid, integrate the hydrostatic relation, Eq. (2.18),
by assuming that the isentropic bulk modulus, B
( p/ )s, is constant—see Eq. (9.18). Find an expression
for p(z) and apply the Mariana Trench data as in Prob. 2.7,
using Bseawater from Table A.3.
A closed tank contains 1.5 m of SAE 30 oil, 1 m of water, 20 cm of mercury, and an air space on top, all at 20°C.
The absolute pressure at the bottom of the tank is 60 kPa.
What is the pressure in the air space?
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Study Guide
Problems 103
P2.11 In Fig. P2.11, pressure gage A reads 1.5 kPa (gage). The
fluids are at 20°C. Determine the elevations z, in meters,
of the liquid levels in the open piezometer tubes B and C.
Gasoline
1.5 m
Water
A
B
C
h
1m
2m
Air
1.5 m
Gasoline
1m
Glycerin
Liquid, SG = 1.60
P2.13
A
B
Air
P2.11
z= 0
4m
P2.12 In Fig. P2.12 the tank contains water and immiscible oil
at 20°C. What is h in cm if the density of the oil is 898
kg/m3?
2m
Air
4m
Water
2m
P2.14
15 lbf/in2 abs
h
6 cm
A
Air
12 cm
8 cm
2 ft
Oil
1 ft
Water
B
Oil
P2.12
1 ft

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P2.13 In Fig. P2.13 the 20°C water and gasoline surfaces are
open to the atmosphere and at the same elevation. What is
the height h of the third liquid in the right leg?
P2.14 The closed tank in Fig. P2.14 is at 20°C. If the pressure
at point A is 95 kPa absolute, what is the absolute pressure at point B in kPa? What percent error do you make
by neglecting the specific weight of the air?
P2.15 The airoilwater system in Fig. P2.15 is at 20°C. Knowing that gage A reads 15 lbf/in2 absolute and gage B reads
1.25 lbf/in2 less than gage C, compute (a) the specific
weight of the oil in lbf/ft3 and (b) the actual reading of
gage C in lbf/in2 absolute.

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Water
2 ft
C
P2.15
P2.16 A closed inverted cone, 100 cm high with diameter 60 cm
at the top, is filled with air at 20°C and 1 atm. Water at
20°C is introduced at the bottom (the vertex) to compress
the air isothermally until a gage at the top of the cone reads
30 kPa (gage). Estimate (a) the amount of water needed
(cm3) and (b) the resulting absolute pressure at the bottom
of the cone (kPa).
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Study Guide
104
Chapter 2 Pressure Distribution in a Fluid
P2.17 The system in Fig. P2.17 is at 20°C. If the pressure at point
A is 1900 lbf/ft2, determine the pressures at points B, C,
and D in lbf/ft2.
Mercury
Air
Air
3 ft
B
A
C
2 ft
10 cm
10 cm
Air
4 ft
10 cm
P2.19
5 ft
Water
D
2 ft
2000
lbf
P2.17
3in diameter
P2.18 The system in Fig. P2.18 is at 20°C. If atmospheric pressure is 101.33 kPa and the pressure at the bottom of the
tank is 242 kPa, what is the specific gravity of fluid X?
1 in
15 in
F
1in diameter
Oil
1m
SAE 30 oil
P2.20
Water
2m
Air: 180 kPa abs
Fluid X
Mercury
0.5 m
80 cm
P2.18
P2.21
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P2.19 The Utube in Fig. P2.19 has a 1cm ID and contains mercury as shown. If 20 cm3 of water is poured into the righthand leg, what will the freesurface height in each leg be
after the sloshing has died down?
P2.20 The hydraulic jack in Fig. P2.20 is filled with oil at 56
lbf/ft3. Neglecting the weight of the two pistons, what force
F on the handle is required to support the 2000lbf weight
for this design?
P2.21 At 20°C gage A reads 350 kPa absolute. What is the height
h of the water in cm? What should gage B read in kPa absolute? See Fig. P2.21.


Water
h?
3m
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Mercury
A
B
P2.22 The fuel gage for a gasoline tank in a car reads proportional to the bottom gage pressure as in Fig. P2.22. If the
tank is 30 cm deep and accidentally contains 2 cm of water plus gasoline, how many centimeters of air remain at
the top when the gage erroneously reads “full’’?
P2.23 In Fig. P2.23 both fluids are at 20°C. If surface tension effects are negligible, what is the density of the oil, in kg/m3?
P2.24 In Prob. 1.2 we made a crude integration of the density
distribution (z) in Table A.6 and estimated the mass of
the earth’s atmosphere to be m 6 E18 kg. Can this re
Textbook Table of Contents

Study Guide
Problems 105
Vent
Air
h?
Gasoline
SG = 0.68
30 cm
Water
mental observations. (b) Find an expression for the pressure at points 1 and 2 in Fig. P2.27b. Note that the glass
is now inverted, so the original top rim of the glass is at
the bottom of the picture, and the original bottom of the
glass is at the top of the picture. The weight of the card
can be neglected.
2 cm
Card
pgage
P2.22
Top of glass
Oil
8 cm
6 cm
P2.27a
Water
Bottom of glass
Original bottom of glass
10 cm
1G
P2.23

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2G
sult be used to estimate sealevel pressure on the earth?
Conversely, can the actual sealevel pressure of 101.35 kPa
be used to make a more accurate estimate of the atmosP2.27b
Card
Original top of glass
pheric mass?
P2.25 Venus has a mass of 4.90 E24 kg and a radius of 6050 km.
Its atmosphere is 96 percent CO2, but let us assume it to
(c) Estimate the theoretical maximum glass height such
be 100 percent. Its surface temperature averages 730 K,
that this experiment could still work, i.e., such that the wadecreasing to 250 K at an altitude of 70 km. The average
ter would not fall out of the glass.
surface pressure is 9.1 MPa. Estimate the atmospheric P2.28 Earth’s atmospheric conditions vary somewhat. On a cerpressure of Venus at an altitude of 5 km.
tain day the sealevel temperature is 45°F and the sealevel
P2.26 Investigate the effect of doubling the lapse rate on atmospressure is 28.9 inHg. An airplane overhead registers an
pheric pressure. Compare the standard atmosphere (Table
air temperature of 23°F and a pressure of 12 lbf/in2. EstiA.6) with a lapse rate twice as high, B2 0.0130 K/m.
mate the plane’s altitude, in feet.
Find the altitude at which the pressure deviation is (a) 1 *P2.29 Under some conditions the atmosphere is adiabatic, p
percent and (b) 5 percent. What do you conclude?
(const)( k), where k is the specific heat ratio. Show that,
P2.27 Conduct an experiment to illustrate atmospheric pressure.
for an adiabatic atmosphere, the pressure variation is
Note: Do this over a sink or you may get wet! Find a drinkgiven by
ing glass with a very smooth, uniform rim at the top. Fill
the glass nearly full with water. Place a smooth, light, flat
(k 1)gz k/(k 1)
p p0 1
plate on top of the glass such that the entire rim of the
kRT0
glass is covered. A glossy postcard works best. A small inCompare this formula for air at z 5000 m with the standex card or one flap of a greeting card will also work. See
dard atmosphere in Table A.6.
Fig. P2.27a.
(a) Hold the card against the rim of the glass and turn the P2.30 In Fig. P2.30 fluid 1 is oil (SG 0.87) and fluid 2 is glycerin at 20°C. If pa 98 kPa, determine the absolute presglass upside down. Slowly release pressure on the card.
sure at point A.
Does the water fall out of the glass? Record your experi

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Study Guide
106
Chapter 2 Pressure Distribution in a Fluid
pa
Air B
ρ1
SAE 30 oil
32 cm
A
Liquid, SG = 1.45
3 cm
5 cm
10 cm
ρ2
4 cm
A
Water
P2.30
6 cm
8 cm
3 cm
P2.31 In Fig. P2.31 all fluids are at 20°C. Determine the pressure difference (Pa) between points A and B.
P2.33
*P2.34 Sometimes manometer dimensions have a significant ef
Kerosine
Air
Benzene
B
40 cm
A
9 cm
20 cm
fect. In Fig. P2.34 containers (a) and (b) are cylindrical and
conditions are such that pa pb. Derive a formula for the
pressure difference pa pb when the oilwater interface on
the right rises a distance h h, for (a) d D and (b) d
0.15D. What is the percent change in the value of p?
14 cm
8 cm
Mercury
Water
D
D
P2.31
(b)
( a)
P2.32 For the inverted manometer of Fig. P2.32, all fluids are at
20°C. If pB pA 97 kPa, what must the height H be
in cm?
Meriam
red oil,
SG = 0.827
L
SAE 30 oil
H
Water
h
18 cm
Water
d
H
Mercury
A
P2.34
35 cm
B
P2.32

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P2.33 In Fig. P2.33 the pressure at point A is 25 lbf/in2. All fluids are at 20°C. What is the air pressure in the closed chamber B, in Pa?

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P2.35 Water flows upward in a pipe slanted at 30°, as in Fig.
P2.35. The mercury manometer reads h 12 cm. Both fluids are at 20°C. What is the pressure difference p1 p2 in
the pipe?
P2.36 In Fig. P2.36 both the tank and the tube are open to the
atmosphere. If L 2.13 m, what is the angle of tilt of
the tube?
P2.37 The inclined manometer in Fig. P2.37 contains Meriam
red manometer oil, SG 0.827. Assume that the reservoir
Textbook Table of Contents

Study Guide
Problems 107
with manometer fluid m. One side of the manometer is open
to the air, while the other is connected to new tubing which
extends to pressure measurement location 1, some height H
higher in elevation than the surface of the manometer liquid.
For consistency, let a be the density of the air in the room,
t be the density of the gas inside the tube, m be the density of the manometer liquid, and h be the height difference
between the two sides of the manometer. See Fig. P2.38.
(a) Find an expression for the gage pressure at the measurement point. Note: When calculating gage pressure, use
the local atmospheric pressure at the elevation of the measurement point. You may assume that h H; i.e., assume
the gas in the entire left side of the manometer is of density t. (b) Write an expression for the error caused by assuming that the gas inside the tubing has the same density
as that of the surrounding air. (c) How much error (in Pa)
is caused by ignoring this density difference for the following conditions: m 860 kg/m3, a 1.20 kg/m3,
1.50 kg/m3, H 1.32 m, and h 0.58 cm? (d) Can
t
you think of a simple way to avoid this error?
(2)
30
(1)
h
2m
P2.35
Oil
SG = 0.8
50 cm
L
Water
SG = 1.0
50 cm
P2.36
1
is very large. If the inclined arm is fitted with graduations
1 in apart, what should the angle be if each graduation
corresponds to 1 lbf/ft2 gage pressure for pA?
(tubing gas)
p1
pa at location 1
t
a
(air)
H
1 in
pA
θ
D=
5
16
Utube
manometer
in
h
m
P2.38
Reservoir
P2.37

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P2.38 An interesting article appeared in the AIAA Journal (vol. 30,
no. 1, January 1992, pp. 279 – 280). The authors explain that
the air inside fresh plastic tubing can be up to 25 percent
more dense than that of the surroundings, due to outgassing
or other contaminants introduced at the time of manufacture.
Most researchers, however, assume that the tubing is filled
with room air at standard air density, which can lead to significant errors when using this kind of tubing to measure
pressures. To illustrate this, consider a Utube manometer

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P2.39 An 8cmdiameter piston compresses manometer oil into
an inclined 7mmdiameter tube, as shown in Fig. P2.39.
When a weight W is added to the top of the piston, the oil
rises an additional distance of 10 cm up the tube, as shown.
How large is the weight, in N?
P2.40 A pump slowly introduces mercury into the bottom of the
closed tank in Fig. P2.40. At the instant shown, the air
pressure pB 80 kPa. The pump stops when the air pressure rises to 110 kPa. All fluids remain at 20°C. What will
be the manometer reading h at that time, in cm, if it is connected to standard sealevel ambient air patm?
Textbook Table of Contents

Study Guide
108
Chapter 2 Pressure Distribution in a Fluid
pA
D = 8 cm
Piston
pB
ρ1
W
10 cm
ρ1
h1
h1
d = 7 mm
Meriam red
oil, SG = 0.827
P2.39
h
15˚
ρ
2
P2.42
patm
8 cm
Air: pB
9 cm
Water
h
P2.44 Water flows downward in a pipe at 45°, as shown in Fig.
P2.44. The pressure drop p1 p2 is partly due to gravity
and partly due to friction. The mercury manometer reads
a 6in height difference. What is the total pressure drop
p1 p2 in lbf/in2? What is the pressure drop due to friction only between 1 and 2 in lbf/in2? Does the manometer reading correspond only to friction drop? Why?
Pump
Mercury
10 cm
Hg
2 cm
P2.40
P2.41 The system in Fig. P2.41 is at 20°C. Compute the pressure at point A in lbf/ft2 absolute.
45˚
1
5 ft
Water
Flow
2
Oil, SG = 0.85
5 in
A
pa = 14.7
Water
lbf/in2
10 in
6 in
6 in
Water
Mercury
P2.44
P2.41
Mercury

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P2.42 Very small pressure differences pA pB can be measured
accurately by the twofluid differential manometer in Fig.
P2.42. Density 2 is only slightly larger than that of the
upper fluid 1. Derive an expression for the proportionality between h and pA pB if the reservoirs are very large.
*P2.43 A mercury manometer, similar to Fig. P2.35, records h
1.2, 4.9, and 11.0 mm when the water velocities in the pipe
are V 1.0, 2.0, and 3.0 m/s, respectively. Determine if
these data can be correlated in the form p1 p2 Cf V2,
where Cf is dimensionless.

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P2.45 In Fig. P2.45, determine the gage pressure at point A in
Pa. Is it higher or lower than atmospheric?
P2.46 In Fig. P2.46 both ends of the manometer are open to the
atmosphere. Estimate the specific gravity of fluid X.
P2.47 The cylindrical tank in Fig. P2.47 is being filled with water at 20°C by a pump developing an exit pressure of 175
EES
kPa. At the instant shown, the air pressure is 110 kPa and
H 35 cm. The pump stops when it can no longer raise
the water pressure. For isothermal air compression, estimate H at that time.
P2.48 Conduct the following experiment to illustrate air pressure. Find a thin wooden ruler (approximately 1 ft in
Textbook Table of Contents

Study Guide
Problems 109
patm
50 cm
Air
Air
20˚ C
Oil,
SG = 0.85
75 cm
30 cm
45 cm
40 cm
H
Water
Pump
P2.47
15 cm
A
Newspaper
Water
P2.45
Mercury
Ruler
SAE 30 oil
Desk
9 cm
10 cm
P2.48
Water
5 cm
7 cm
6 cm
Fluid X
P2.49
4 cm
P2.50
12 cm
P2.46

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length) or a thin wooden paint stirrer. Place it on the edge
of a desk or table with a little less than half of it hang P2.51
ing over the edge lengthwise. Get two fullsize sheets of
newspaper; open them up and place them on top of the
ruler, covering only the portion of the ruler resting on the *P2.52
desk as illustrated in Fig. P2.48. (a) Estimate the total
force on top of the newspaper due to air pressure in the
room. (b) Careful! To avoid potential injury, make sure
nobody is standing directly in front of the desk. Perform

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a karate chop on the portion of the ruler sticking out over
the edge of the desk. Record your results. (c) Explain
your results.
A water tank has a circular panel in its vertical wall. The
panel has a radius of 50 cm, and its center is 2 m below
the surface. Neglecting atmospheric pressure, determine
the water force on the panel and its line of action.
A vat filled with oil (SG 0.85) is 7 m long and 3 m deep
and has a trapezoidal cross section 2 m wide at the bottom and 4 m wide at the top. Compute (a) the weight of
oil in the vat, (b) the force on the vat bottom, and (c) the
force on the trapezoidal end panel.
Gate AB in Fig. P2.51 is 1.2 m long and 0.8 m into the
paper. Neglecting atmospheric pressure, compute the force
F on the gate and its centerofpressure position X.
Suppose that the tank in Fig. P2.51 is filled with liquid X,
not oil. Gate AB is 0.8 m wide into the paper. Suppose that
liquid X causes a force F on gate AB and that the moment
of this force about point B is 26,500 N m. What is the
specific gravity of liquid X?
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Study Guide
110
Chapter 2 Pressure Distribution in a Fluid
pa
6m
Oil,
SG = 0.82
Water
pa
4m
h
8m
A
1m
X
1.2 m
A
B
4 ft
F
40°
P2.51
B
P2.55
P2.53 Panel ABC in the slanted side of a water tank is an isosceles triangle with the vertex at A and the base BC 2 m,
as in Fig. P2.53. Find the water force on the panel and its
line of action.
200 kg
h
m
B
A
30 cm
A
Water
Water
3m
P2.58
4m
P2.53
B, C
*P2.59 Gate AB has length L, width b into the paper, is hinged at
B, and has negligible weight. The liquid level h remains
at the top of the gate for any angle . Find an analytic expression for the force P, perpendicular to AB, required to
keep the gate in equilibrium in Fig. P2.59.
3m

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P2.54 If, instead of water, the tank in Fig. P2.53 is filled with liqP
uid X, the liquid force on panel ABC is found to be 115 kN.
What is the density of liquid X? The line of action is found
A
to be the same as in Prob. 2.53. Why?
P2.55 Gate AB in Fig. P2.55 is 5 ft wide into the paper, hinged
at A, and restrained by a stop at B. The water is at 20°C.
Compute (a) the force on stop B and (b) the reactions at
h
L
A if the water depth h 9.5 ft.
P2.56 In Fig. P2.55, gate AB is 5 ft wide into the paper, and stop
B will break if the water force on it equals 9200 lbf. For
Hinge
what water depth h is this condition reached?
P2.57 In Fig. P2.55, gate AB is 5 ft wide into the paper. Suppose
B
P2.59
that the fluid is liquid X, not water. Hinge A breaks when
its reaction is 7800 lbf, and the liquid depth is h 13 ft.
*P2.60 Find the net hydrostatic force per unit width on the recWhat is the specific gravity of liquid X?
tangular gate AB in Fig. P2.60 and its line of action.
P2.58 In Fig. P2.58, the cover gate AB closes a circular opening
80 cm in diameter. The gate is held closed by a 200kg *P2.61 Gate AB in Fig. P2.61 is a homogeneous mass of 180 kg,
1.2 m wide into the paper, hinged at A, and resting on a
mass as shown. Assume standard gravity at 20°C. At what
smooth bottom at B. All fluids are at 20°C. For what wawater level h will the gate be dislodged? Neglect the weight
ter depth h will the force at point B be zero?
of the gate.

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Study Guide
Problems 111
P2.63 The tank in Fig. P2.63 has a 4cmdiameter plug at the
bottom on the right. All fluids are at 20°C. The plug will
pop out if the hydrostatic force on it is 25 N. For this condition, what will be the reading h on the mercury manometer on the left side?
1.8 m
1.2 m
A
Water
2m
Water
Glycerin
50°
B
2m
H
P2.60
h
2 cm
Plug,
D = 4 cm
Mercury
Water
P2.63
Glycerin
*P2.64 Gate ABC in Fig. P2.64 has a fixed hinge line at B and is
2 m wide into the paper. The gate will open at A to release
water if the water depth is high enough. Compute the depth
h for which the gate will begin to open.
h
2m
A
1m
C
60°
B
P2.61
P2.62 Gate AB in Fig. P2.62 is 15 ft long and 8 ft wide into the
paper and is hinged at B with a stop at A. The water is at
EES
20°C. The gate is 1inthick steel, SG 7.85. Compute
the water level h for which the gate will start to fall.
Pulley
A
h
B
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P2.62


B
h
1m
Water at 20°C
*P2.65 Gate AB in Fig. P2.65 is semicircular, hinged at B, and
held by a horizontal force P at A. What force P is required
for equilibrium?
P2.66 Dam ABC in Fig. P2.66 is 30 m wide into the paper and
made of concrete (SG 2.4). Find the hydrostatic force
on surface AB and its moment about C. Assuming no seepage of water under the dam, could this force tip the dam
over? How does your argument change if there is seepage
under the dam?
Water
60˚
20 cm
P2.64
10,000 lb
15 ft
A
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Study Guide
112
Chapter 2 Pressure Distribution in a Fluid
5m
3m
Oil, SG = 0.83
1m
Water
A
A
P
Gate:
Side view
3m
Gate
2m
B
P2.65
;;
;;
50˚ B
P2.68
A
Water 20˚C
80 m
80 cm
1m
Dam
B
5m
Water,
20˚C
C
Hg, 20˚C
60 m
P2.66
A
2m
P2.69

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*P2.67 Generalize Prob. 2.66 as follows. Denote length AB as H,
length BC as L, and angle ABC as . Let the dam material have specific gravity SG. The width of the dam is b.
Assume no seepage of water under the dam. Find an analytic relation between SG and the critical angle c for
which the dam will just tip over to the right. Use your relation to compute c for the special case SG 2.4 (concrete).
P2.68 Isosceles triangle gate AB in Fig. P2.68 is hinged at A and
weighs 1500 N. What horizontal force P is required at point
B for equilibrium?
*P2.69 The water tank in Fig. P2.69 is pressurized, as shown by
the mercurymanometer reading. Determine the hydrostatic force per unit depth on gate AB.
P2.70 Calculate the force and center of pressure on one side of
the vertical triangular panel ABC in Fig. P2.70. Neglect
patm.
*P2.71 In Fig. P2.71 gate AB is 3 m wide into the paper and is
connected by a rod and pulley to a concrete sphere (SG

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B
2 ft
A
Water
6 ft
C
B
4 ft
P2.70
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Study Guide
P
Problems 113
2.40). What diameter of the sphere is just sufficient to keep *P2.74 In “soft’’ liquids (low bulk modulus ), it may be necesthe gate closed?
sary to account for liquid compressibility in hydrostatic
calculations. An approximate density relation would be
;;
Concrete
sphere, SG = 2.4
dp
d
a2 d
or
p
p0
a2(
0)
6m
where a is the speed of sound and (p0, 0) are the conditions at the liquid surface z 0. Use this approximation
to show that the density variation with depth in a soft liqgz/a2
uid is
where g is the acceleration of gravity
0e
8m
A
and z is positive upward. Then consider a vertical wall of
width b, extending from the surface (z 0) down to depth
z
h. Find an analytic expression for the hydrostatic
4m
Water
force F on this wall, and compare it with the incompress2
B
ible result F
0gh b/2. Would the center of pressure be
below the incompressible position z
2h/3?
P2.71
*P2.75 Gate AB in Fig. P2.75 is hinged at A, has width b into the
paper, and makes smooth contact at B. The gate has density s and uniform thickness t. For what gate density s,
P2.72 The Vshaped container in Fig. P2.72 is hinged at A and
expressed as a function of (h, t, , ), will the gate just beheld together by cable BC at the top. If cable spacing is
gin to lift off the bottom? Why is your answer indepen1 m into the paper, what is the cable tension?
dent of gate length L and width b?
Cable
C
B
A
1m
Water
3m
L
h
110˚
P2.72
A
t
P2.73 Gate AB is 5 ft wide into the paper and opens to let fresh
water out when the ocean tide is dropping. The hinge at A
is 2 ft above the freshwater level. At what ocean level h
will the gate first open? Neglect the gate weight.
A
Tide
range
10 ft
h
Seawater, SG = 1.025
Stop
B

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P2.73

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P2.75
B
*P2.76 Consider the angled gate ABC in Fig. P2.76, hinged at C
and of width b into the paper. Derive an analytic formula
for the horizontal force P required at the top for equilibrium, as a function of the angle .
P2.77 The circular gate ABC in Fig. P2.77 has a 1m radius and
is hinged at B. Compute the force P just sufficient to keep
the gate from opening when h 8 m. Neglect atmospheric
pressure.
P2.78 Repeat Prob. 2.77 to derive an analytic expression for P
as a function of h. Is there anything unusual about your
solution?
P2.79 Gate ABC in Fig. P2.79 is 1 m square and is hinged at B.
It will open automatically when the water level h becomes
high enough. Determine the lowest height for which the
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Study Guide
114
;;
;;
;;
Chapter 2 Pressure Distribution in a Fluid
A
P
θ
θ
Specific weight γ
h
Air
1 atm
2m
SA
E3
B
20
0o
il
h
60
Wa
ter
pa
Water
pa
h
cm
Mercury
80
C
P2.76
cm
cm
P2.80
Panel, 30 cm high, 40 cm wide
P2.81 Gate AB in Fig. P2.81 is 7 ft into the paper and weighs
3000 lbf when submerged. It is hinged at B and rests
against a smooth wall at A. Determine the water level h at
the left which will just cause the gate to open.
A
1m
B
1m
C
h
P
4 ft
A
P2.77
Water
8 ft
Water
h
B
Water
6 ft
A
60 cm
C
40 cm
P2.81
B
*P2.82 The dam in Fig. P2.82 is a quarter circle 50 m wide into
P2.79

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gate will open. Neglect atmospheric pressure. Is this result
independent of the liquid density?
P2.80 For the closed tank in Fig. P2.80, all fluids are at 20°C, and
the airspace is pressurized. It is found that the net outward
hydrostatic force on the 30by 40cm panel at the bottom of
the water layer is 8450 N. Estimate (a) the pressure in the
airspace and (b) the reading h on the mercury manometer.

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the paper. Determine the horizontal and vertical components of the hydrostatic force against the dam and the point
CP where the resultant strikes the dam.
*P2.83 Gate AB in Fig. P2.83 is a quarter circle 10 ft wide into
the paper and hinged at B. Find the force F just sufficient
to keep the gate from opening. The gate is uniform and
weighs 3000 lbf.
P2.84 Determine (a) the total hydrostatic force on the curved surface AB in Fig. P2.84 and (b) its line of action. Neglect atmospheric pressure, and let the surface have unit width.
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Study Guide
;;;
;;;
;;;
20 m
20 m
Problems 115
pa = 0
CP
Water
P2.82
Water
10 ft
P2.86
2 ft
P2.87 The bottle of champagne (SG 0.96) in Fig. P2.87 is under pressure, as shown by the mercurymanometer reading. Compute the net force on the 2inradius hemispherical end cap at the bottom of the bottle.
F
A
Water
r = 8 ft
B
P2.83
B
Water at 20° C
z
1m
4 in
z = x3
2 in
6 in
x
A
P2.84
P2.87
r = 2 in
Mercury
P2.85 Compute the horizontal and vertical components of the hydrostatic force on the quartercircle panel at the bottom of *P2.88 Gate ABC is a circular arc, sometimes called a Tainter gate,
which can be raised and lowered by pivoting about point
the water tank in Fig. P2.85.
O. See Fig. P2.88. For the position shown, determine (a)
the hydrostatic force of the water on the gate and (b) its
line of action. Does the force pass through point O?
6m
C
5m
Water
R=6m
Water
2m
B
6m
O
2m
P2.85
6m

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P2.86 Compute the horizontal and vertical components of the hydrostatic force on the hemispherical bulge at the bottom
of the tank in Fig. P2.86.

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A
P2.88
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116
Chapter 2 Pressure Distribution in a Fluid
P2.89 The tank in Fig. P2.89 contains benzene and is pressurized to 200 kPa (gage) in the air gap. Determine the vertical hydrostatic force on circulararc section AB and its
line of action.
3cm
4m
60 cm
p = 200 kPa
30 cm
Six
bolts
B
2m
Water
Benzene
at 20 C
60 cm
P2.91
P2.89
A
2m
P2.90 A 1ftdiameter hole in the bottom of the tank in Fig. P2.90
is closed by a conical 45° plug. Neglecting the weight of
the plug, compute the force F required to keep the plug in
the hole.
Water
p = 3 lbf/in 2 gage
Bolt spacing 25 cm
2m
P2.92
1 ft
Air :
z
Water
3 ft
1 ft
ρ, γ
45˚
cone
h
R
F
P2.90

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P2.91 The hemispherical dome in Fig. P2.91 weighs 30 kN and
is filled with water and attached to the floor by six equally
spaced bolts. What is the force in each bolt required to
hold down the dome?
P2.92 A 4mdiameter water tank consists of two half cylinders,
each weighing 4.5 kN/m, bolted together as shown in Fig.
P2.92. If the support of the end caps is neglected, determine the force induced in each bolt.
*P2.93 In Fig. P2.93, a onequadrant spherical shell of radius R
is submerged in liquid of specific gravity and depth
h R. Find an analytic expression for the resultant hydrostatic force, and its line of action, on the shell surface.

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R
z
R
x
P2.93
P2.94 The 4ftdiameter log (SG 0.80) in Fig. P2.94 is 8 ft
long into the paper and dams water as shown. Compute
the net vertical and horizontal reactions at point C.
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Study Guide
Problems 117
wall at A. Compute the reaction forces at points A
and B.
Log
2ft
Water
2ft
P2.94
Water
C
Seawater, 10,050 N/m3
*P2.95 The uniform body A in Fig. P2.95 has width b into the paper and is in static equilibrium when pivoted about hinge
O. What is the specific gravity of this body if (a) h 0
and (b) h R?
4m
A
2m
45°
B
A
P2.97
h
P2.98 Gate ABC in Fig. P2.98 is a quarter circle 8 ft wide into
the paper. Compute the horizontal and vertical hydrostatic
forces on the gate and the line of action of the resultant
force.
R
R
Water
A
O
P2.95
r = 4 ft
P2.96 The tank in Fig. P2.96 is 3 m wide into the paper. Neglecting atmospheric pressure, compute the hydrostatic (a)
horizontal force, (b) vertical force, and (c) resultant force
on quartercircle panel BC.
B
C
P2.98
P2.99 A 2ftdiameter sphere weighing 400 lbf closes a 1ftdiameter hole in the bottom of the tank in Fig. P2.99. Compute the force F required to dislodge the sphere from the
hole.
A
6m
Water
Water
45°
45°
4m
B
Water
4m
3 ft
1 ft
P2.96
C

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P2.97 Gate AB in Fig. P2.97 is a threeeighths circle, 3 m wide
into the paper, hinged at B, and resting against a smooth

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1 ft
P2.99
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F

Study Guide
118
Chapter 2 Pressure Distribution in a Fluid
P2.100 Pressurized water fills the tank in Fig. P2.100. Compute
the net hydrostatic force on the conical surface ABC.
2m
A
P2.106
C
4m
7m
B
P2.107
150 kPa
gage
P2.108
Water
P2.100
P2.101 A fuel truck has a tank cross section which is approximately elliptical, with a 3m horizontal major axis and a
2m vertical minor axis. The top is vented to the atmosphere. If the tank is filled half with water and half with
gasoline, what is the hydrostatic force on the flat elliptical end panel?
P2.102 In Fig. P2.80 suppose that the manometer reading is h
25 cm. What will be the net hydrostatic force on the complete end wall, which is 160 cm high and 2 m wide?
P2.103 The hydrogen bubbles in Fig. 1.13 are very small, less
than a millimeter in diameter, and rise slowly. Their drag
in still fluid is approximated by the first term of Stokes’
expression in Prob. 1.10: F 3 VD, where V is the rise
velocity. Neglecting bubble weight and setting bubble
buoyancy equal to drag, (a) derive a formula for the terminal (zero acceleration) rise velocity Vterm of the bubble
and (b) determine Vterm in m/s for water at 20°C if D
30 m.
P2.104 The can in Fig. P2.104 floats in the position shown. What
is its weight in N?
P2.109
whether his new crown was pure gold (SG 19.3).
Archimedes measured the weight of the crown in air to be
11.8 N and its weight in water to be 10.9 N. Was it pure
gold?
It is found that a 10cm cube of aluminum (SG 2.71)
will remain neutral under water (neither rise nor fall) if it
is tied by a string to a submerged 18cmdiameter sphere
of buoyant foam. What is the specific weight of the foam,
in N/m3?
Repeat Prob. 2.62, assuming that the 10,000lbf weight is
aluminum (SG 2.71) and is hanging submerged in the
water.
A piece of yellow pine wood (SG 0.65) is 5 cm square
and 2.2 m long. How many newtons of lead (SG 11.4)
should be attached to one end of the wood so that it will
float vertically with 30 cm out of the water?
A hydrometer floats at a level which is a measure of the
specific gravity of the liquid. The stem is of constant diameter D, and a weight in the bottom stabilizes the body
to float vertically, as shown in Fig. P2.109. If the position
h 0 is pure water (SG 1.0), derive a formula for h as
a function of total weight W, D, SG, and the specific weight
0 of water.
D
SG = 1.0
h
Fluid, SG > 1
W
P2.109
3 cm
8 cm
Water
D = 9 cm
P2.104

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P2.105 It is said that Archimedes discovered the buoyancy laws
when asked by King Hiero of Syracuse to determine

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P2.110 An average table tennis ball has a diameter of 3.81 cm and
a mass of 2.6 g. Estimate the (small) depth at which this
ball will float in water at 20°C and sea level standard air
if air buoyancy is (a) neglected and (b) included.
P2.111 A hotair balloon must be designed to support basket, cords,
and one person for a total weight of 1300 N. The balloon
material has a mass of 60 g/m2. Ambient air is at 25°C and
1 atm. The hot air inside the balloon is at 70°C and 1 atm.
What diameter spherical balloon will just support the total
weight? Neglect the size of the hotair inlet vent.
P2.112 The uniform 5mlong round wooden rod in Fig. P2.112
is tied to the bottom by a string. Determine (a) the tension
Textbook Table of Contents

Study Guide
Problems 119
in the string and (b) the specific gravity of the wood. Is it
possible for the given information to determine the inclination angle ? Explain.
Hinge
D = 4 cm
B
= 30
1m
8m
D = 8 cm
2 kg of lead
θ
Water at 20°C
P2.114
4m
B
String
θ
8 ft
Wood
SG = 0.6
P2.112
P2.113 A spar buoy is a buoyant rod weighted to float and protrude
vertically, as in Fig. P2.113. It can be used for measurements
or markers. Suppose that the buoy is maple wood (SG
0.6), 2 in by 2 in by 12 ft, floating in seawater (SG 1.025).
How many pounds of steel (SG 7.85) should be added to
the bottom end so that h 18 in?
Seawater
A
Rock
P2.115
P2.116 The homogeneous 12cm cube in Fig. 2.116 is balanced
by a 2kg mass on the beam scale when the cube is immersed in 20°C ethanol. What is the specific gravity of the
cube?
h
2 kg
Wsteel
P2.113

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P2.114 The uniform rod in Fig. P2.114 is hinged at point B on the
waterline and is in static equilibrium as shown when 2 kg
of lead (SG 11.4) are attached to its end. What is the
specific gravity of the rod material? What is peculiar about
the rest angle
30?
P2.115 The 2in by 2in by 12ft spar buoy from Fig. P2.113 has 5
lbm of steel attached and has gone aground on a rock, as in
Fig. P2.115. Compute the angle at which the buoy will
lean, assuming that the rock exerts no moments on the spar.

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12 cm
P2.116
P2.117 The balloon in Fig. P2.117 is filled with helium and pressurized to 135 kPa and 20°C. The balloon material has a
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Study Guide
120
Chapter 2 Pressure Distribution in a Fluid
mass of 85 g/m2. Estimate (a) the tension in the mooring
line and (b) the height in the standard atmosphere to which
the balloon will rise if the mooring line is cut.
P2.121 The uniform beam in Fig. P2.121, of size L by h by b and
with specific weight b, floats exactly on its diagonal when
a heavy uniform sphere is tied to the left corner, as shown.
Show that this can only happen (a) when b
/3 and (b)
when the sphere has size
Lhb
(SG 1)
D
1/3
D = 10 m
Width b << L
Air:
100 kPa at
20°C
P2.117
P2.118 A 14indiameter hollow sphere is made of steel (SG
7.85) with 0.16in wall thickness. How high will this
EES
sphere float in 20°C water? How much weight must be
added inside to make the sphere neutrally buoyant?
P2.119 When a 5lbf weight is placed on the end of the uniform
floating wooden beam in Fig. P2.119, the beam tilts at an
angle with its upper right corner at the surface, as shown.
Determine (a) the angle and (b) the specific gravity of
the wood. (Hint: Both the vertical forces and the moments
about the beam centroid must be balanced.)
;
L
h << L
γb
γ
Diameter D
SG > 1
P2.121
P2.122 A uniform block of steel (SG 7.85) will “float’’ at a
mercurywater interface as in Fig. P2.122. What is the
ratio of the distances a and b for this condition?
5 lbf
θ
Water
9 ft
Water
4 in × 4 in
a
Steel
block
P2.119
b
Mercury: SG = 13.56
P2.120 A uniform wooden beam (SG 0.65) is 10 cm by 10 cm
by 3 m and is hinged at A, as in Fig. P2.120. At what angle will the beam float in the 20°C water?
P2.123 In an estuary where fresh water meets and mixes with seawater, there often occurs a stratified salinity condition with
fresh water on top and salt water on the bottom, as in Fig.
P2.123. The interface is called a halocline. An idealization
of this would be constant density on each side of the halocline as shown. A 35cmdiameter sphere weighing 50 lbf
would “float’’ near such a halocline. Compute the sphere
position for the idealization in Fig. P2.123.
P2.124 A balloon weighing 3.5 lbf is 6 ft in diameter. It is filled
with hydrogen at 18 lbf/in2 absolute and 60°F and is released. At what altitude in the U.S. standard atmosphere
will this balloon be neutrally buoyant?
A
1m
θ
Water
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P2.120


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P2.122

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Study Guide
Problems 121
SG = 1.0
Halocline
SG = 1.025
35°/°°
Salinity
0
Idealization
P2.123
P2.125 Suppose that the balloon in Prob. 2.111 is constructed to
have a diameter of 14 m, is filled at sea level with hot air
at 70°C and 1 atm, and is released. If the air inside the balloon remains constant and the heater maintains it at 70°C,
at what altitude in the U.S. standard atmosphere will this
balloon be neutrally buoyant?
*P2.126 A cylindrical can of weight W, radius R, and height H is
open at one end. With its open end down, and while filled
with atmospheric air (patm, Tatm), the can is eased down
vertically into liquid, of density , which enters and compresses the air isothermally. Derive a formula for the height
h to which the liquid rises when the can is submerged with
its top (closed) end a distance d from the surface.
*P2.127 Consider the 2in by 2in by 10ft spar buoy of Prob. 2.113.
How many pounds of steel (SG 7.85) should be added
at the bottom to ensure vertical floating with a metacentric height MG of (a) zero (neutral stability) or (b) 1 ft
(reasonably stable)?
P2.128 An iceberg can be idealized as a cube of side length L, as
in Fig. P2.128. If seawater is denoted by S 1.0, then
glacier ice (which forms icebergs) has S 0.88. Determine if this “cubic’’ iceberg is stable for the position shown
in Fig. P2.128.
Fig. P2.128 suppose that the height is L and the depth into
the paper is L, but the width in the plane of the paper is
H L. Assuming S 0.88 for the iceberg, find the ratio
H/L for which it becomes neutrally stable, i.e., about to
overturn.
P2.130 Consider a wooden cylinder (SG 0.6) 1 m in diameter
and 0.8 m long. Would this cylinder be stable if placed to
float with its axis vertical in oil (SG 0.8)?
P2.131 A barge is 15 ft wide and 40 ft long and floats with a draft
of 4 ft. It is piled so high with gravel that its center of gravity is 2 ft above the waterline. Is it stable?
P2.132 A solid right circular cone has SG 0.99 and floats vertically as in Fig. P2.132. Is this a stable position for the
cone?
Water :
SG = 1.0
SG = 0.99
P2.132
P2.133 Consider a uniform right circular cone of specific gravity
S 1, floating with its vertex down in water (S 1). The
base radius is R and the cone height is H. Calculate and
plot the stability MG of this cone, in dimensionless form,
versus H/R for a range of S 1.
P2.134 When floating in water (SG 1.0), an equilateral triangular body (SG 0.9) might take one of the two positions
shown in Fig. P2.134. Which is the more stable position?
Assume large width into the paper.
Specific gravity
=S
M?
G
B
h
Water
S = 1.0
P2.134
P2.135 Consider a homogeneous right circular cylinder of length
L, radius R, and specific gravity SG, floating in water
(SG 1). Show that the body will be stable with its axis
vertical if
L
P2.128
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P2.129 The iceberg idealization in Prob. 2.128 may become unstable if its sides melt and its height exceeds its width. In


(b)
(a)
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R
L
[2SG(1

SG)]1/2
Study Guide
122
Chapter 2 Pressure Distribution in a Fluid
P2.136 Consider a homogeneous right circular cylinder of length
L, radius R, and specific gravity SG 0.5, floating in water (SG 1). Show that the body will be stable with its
axis horizontal if L/R 2.0.
P2.137 A tank of water 4 m deep receives a constant upward acceleration az. Determine (a) the gage pressure at the tank
bottom if az 5 m2/s and (b) the value of az which causes
the gage pressure at the tank bottom to be 1 atm.
P2.138 A 12floz glass, of 3in diameter, partly full of water, is
attached to the edge of an 8ftdiameter merrygoround
which is rotated at 12 r/min. How full can the glass be before water spills? (Hint: Assume that the glass is much
smaller than the radius of the merrygoround.)
P2.139 The tank of liquid in Fig. P2.139 accelerates to the right
with the fluid in rigidbody motion. (a) Compute ax in
m/s2. (b) Why doesn’t the solution to part (a) depend upon
the density of the fluid? (c) Determine the gage pressure
at point A if the fluid is glycerin at 20°C.
V
a?
15 cm
100 cm
28 cm
A
z
30°
P2.141
x
B
9 cm
Water at 20° C
A
24 cm
ax
P2.142
28 cm
15 cm
100 cm
A
A
pa = 15 lbf/in2 abs
Fig. P2.139
ax
2ft

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P2.140 Suppose that the ellipticalend fuel tank in Prob. 2.101 is
10 m long and filled completely with fuel oil (
890
kg/m3). Let the tank be pulled along a horizontal road. For
rigidbody motion, find the acceleration, and its direction,
for which (a) a constantpressure surface extends from the
top of the front end wall to the bottom of the back end and
(b) the top of the back end is at a pressure 0.5 atm lower
than the top of the front end.
P2.141 The same tank from Prob. 2.139 is now moving with constant acceleration up a 30° inclined plane, as in Fig.
P2.141. Assuming rigidbody motion, compute (a) the
value of the acceleration a, (b) whether the acceleration is
up or down, and (c) the gage pressure at point A if the fluid
is mercury at 20°C.
P2.142 The tank of water in Fig. P2.142 is 12 cm wide into the
paper. If the tank is accelerated to the right in rigidbody
motion at 6.0 m/s2, compute (a) the water depth on side
AB and (b) the waterpressure force on panel AB. Assume
no spilling.
P2.143 The tank of water in Fig. P2.143 is full and open to the atmosphere at point A. For what acceleration ax in ft/s2 will the
pressure at point B be (a) atmospheric and (b) zero absolute?

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Water
B
1ft
P2.143
1ft
2ft
P2.144 Consider a hollow cube of side length 22 cm, filled completely with water at 20°C. The top surface of the cube is
horizontal. One top corner, point A, is open through a small
hole to a pressure of 1 atm. Diagonally opposite to point
A is top corner B. Determine and discuss the various rigidbody accelerations for which the water at point B begins
to cavitate, for (a) horizontal motion and (b) vertical motion.
P2.145 A fish tank 14 in deep by 16 by 27 in is to be carried
in a car which may experience accelerations as high as
6 m/s2. What is the maximum water depth which will avoid
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Study Guide
Problems 123
spilling in rigidbody motion? What is the proper alignment of the tank with respect to the car motion?
P2.146 The tank in Fig. P2.146 is filled with water and has a vent
hole at point A. The tank is 1 m wide into the paper. Inside the tank, a 10cm balloon, filled with helium at 130
kPa, is tethered centrally by a string. If the tank accelerates to the right at 5 m/s2 in rigidbody motion, at what
angle will the balloon lean? Will it lean to the right or to
the left?
60 cm
A
1 atm
Water at 20°C
D = 10 cm
with the child, which way will the balloon tilt, forward or
backward? Explain. (b) The child is now sitting in a car
which is stopped at a red light. The heliumfilled balloon
is not in contact with any part of the car (seats, ceiling,
etc.) but is held in place by the string, which is in turn held
by the child. All the windows in the car are closed. When
the traffic light turns green, the car accelerates forward. In
a frame of reference moving with the car and child, which
way will the balloon tilt, forward or backward? Explain.
(c) Purchase or borrow a heliumfilled balloon. Conduct a
scientific experiment to see if your predictions in parts (a)
and (b) above are correct. If not, explain.
P2.149 The 6ftradius waterwheel in Fig. P2.149 is being used to
lift water with its 1ftdiameter halfcylinder blades. If the
wheel rotates at 10 r/min and rigidbody motion is assumed, what is the water surface angle at position A?
He
40 cm
20 cm
String
10 r/min
θ
P2.146
A
6 ft
P2.147 The tank of water in Fig. P2.147 accelerates uniformly by
freely rolling down a 30° incline. If the wheels are frictionless, what is the angle ? Can you explain this interesting result?
1 ft
P2.149
P2.150 A cheap accelerometer, probably worth the price, can be
made from a Utube as in Fig. P2.150. If L 18 cm and
D 5 mm, what will h be if ax 6 m/s2? Can the scale
markings on the tube be linear multiples of ax?
θ
D
h
Rest level
ax
30°
1
2
1
2
L
L
P2.147
L
P2.150

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P2.148 A child is holding a string onto which is attached a heliumfilled balloon. (a) The child is standing still and suddenly accelerates forward. In a frame of reference moving

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P2.151 The Utube in Fig. P2.151 is open at A and closed at D.
If accelerated to the right at uniform ax, what acceleration
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Study Guide
124
Chapter 2 Pressure Distribution in a Fluid
will cause the pressure at point C to be atmospheric? The
fluid is water (SG 1.0).
A
D
1 ft
1 ft
B
P2.156 Suppose that the Utube of Fig. P2.151 is rotated about
axis DC. If the fluid is water at 122°F and atmospheric
pressure is 2116 lbf/ft2 absolute, at what rotation rate will
the fluid within the tube begin to vaporize? At what point
will this occur?
P2.157 The 45° Vtube in Fig. P2.157 contains water and is open
at A and closed at C. What uniform rotation rate in r/min
about axis AB will cause the pressure to be equal at points
B and C? For this condition, at what point in leg BC will
the pressure be a minimum?
C
A
C
1 ft
P2.151
P2.152 A 16cmdiameter open cylinder 27 cm high is full of water. Compute the rigidbody rotation rate about its central
axis, in r/min, (a) for which onethird of the water will
spill out and (b) for which the bottom will be barely exposed.
P2.153 Suppose the Utube in Fig. P2.150 is not translated but
rather rotated about its right leg at 95 r/min. What will be
the level h in the left leg if L 18 cm and D 5 mm?
P2.154 A very deep 18cmdiameter can contains 12 cm of water
overlaid with 10 cm of SAE 30 oil. If the can is
rotated in rigidbody motion about its central axis at
150 r/min, what will be the shapes of the airoil and *P2.158
oilwater interfaces? What will be the maximum fluid pressure in the can in Pa (gage)?
P2.155 For what uniform rotation rate in r/min about axis C will
the Utube in Fig. P2.155 take the configuration shown?
EES
The fluid is mercury at 20°C.
A
30 cm
45˚
B
P2.157
It is desired to make a 3mdiameter parabolic telescope
mirror by rotating molten glass in rigidbody motion until the desired shape is achieved and then cooling the glass
to a solid. The focus of the mirror is to be 4 m from the
mirror, measured along the centerline. What is the proper
mirror rotation rate, in r/min, for this task?
C
B
Ω
20 cm
12 cm
10 cm

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P2.155
5 cm

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Study Guide
Fundamentals of Engineering Exam Problems
125
Word Problems
W2.1
Consider a hollow cone with a vent hole in the vertex at
the top, along with a hollow cylinder, open at the top, with
the same base area as the cone. Fill both with water to the
top. The hydrostatic paradox is that both containers have
the same force on the bottom due to the water pressure, although the cone contains 67 percent less water. Can you
explain the paradox?
W2.2 Can the temperature ever rise with altitude in the real atmosphere? Wouldn’t this cause the air pressure to increase
upward? Explain the physics of this situation.
W2.3 Consider a submerged curved surface which consists of a
twodimensional circular arc of arbitrary angle, arbitrary
depth, and arbitrary orientation. Show that the resultant hydrostatic pressure force on this surface must pass through
the center of curvature of the arc.
W2.4 Fill a glass approximately 80 percent with water, and add a
large ice cube. Mark the water level. The ice cube, having
SG 0.9, sticks up out of the water. Let the ice cube melt
with negligible evaporation from the water surface. Will the
water level be higher than, lower than, or the same as before?
W2.5
A ship, carrying a load of steel, is trapped while floating
in a small closed lock. Members of the crew want to get
out, but they can’t quite reach the top wall of the lock. A
crew member suggests throwing the steel overboard in the
lock, claiming the ship will then rise and they can climb
out. Will this plan work?
W2.6 Consider a balloon of mass m floating neutrally in the atmosphere, carrying a person/basket of mass M m. Discuss the stability of this system to disturbances.
W2.7 Consider a helium balloon on a string tied to the seat of
your stationary car. The windows are closed, so there is no
air motion within the car. The car begins to accelerate forward. Which way will the balloon lean, forward or backward? (Hint: The acceleration sets up a horizontal pressure
gradient in the air within the car.)
W2.8 Repeat your analysis of Prob. W2.7 to let the car move at
constant velocity and go around a curve. Will the balloon
lean in, toward the center of curvature, or out?
Fundamentals of Engineering Exam Problems
FE2.1 A gage attached to a pressurized nitrogen tank reads a
gage pressure of 28 in of mercury. If atmospheric pressure is 14.4 psia, what is the absolute pressure in the tank?
(a) 95 kPa, (b) 99 kPa, (c) 101 kPa, (d) 194 kPa,
(e) 203 kPa
FE2.2 On a sealevel standard day, a pressure gage, moored below the surface of the ocean (SG 1.025), reads an absolute pressure of 1.4 MPa. How deep is the instrument?
(a) 4 m, (b) 129 m, (c) 133 m, (d) 140 m, (e) 2080 m
FE2.3 In Fig. FE2.3, if the oil in region B has SG 0.8 and the
absolute pressure at point A is 1 atm, what is the absolute
pressure at point B?
(a) 5.6 kPa, (b) 10.9 kPa, (c) 106.9 kPa, (d) 112.2 kPa,
(e) 157.0 kPa
A
Oil
Water
SG = 1
5 cm
B
3 cm
8 cm
Mercury
SG = 13.56
4 cm

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FE2.4 In Fig. FE2.3, if the oil in region B has SG 0.8 and the
absolute pressure at point B is 14 psia, what is the absolute pressure at point B?
(a) 11 kPa, (b) 41 kPa, (c) 86 kPa, (d) 91 kPa, (e) 101 kPa
FE2.5 A tank of water (SG 1,.0) has a gate in its vertical wall
5 m high and 3 m wide. The top edge of the gate is 2 m
below the surface. What is the hydrostatic force on the gate?
(a) 147 kN, (b) 367 kN, (c) 490 kN, (d) 661 kN,
(e) 1028 kN
FE2.6 In Prob. FE2.5 above, how far below the surface is the
center of pressure of the hydrostatic force?
(a) 4.50 m, (b) 5.46 m, (c) 6.35 m, (d) 5.33 m, (e) 4.96 m
FE2.7 A solid 1mdiameter sphere floats at the interface between
water (SG 1.0) and mercury (SG 13.56) such that 40 percent is in the water. What is the specific gravity of the sphere?
(a) 6.02, (b) 7.28, (c) 7.78, (d) 8.54, (e) 12.56
FE2.8 A 5mdiameter balloon contains helium at 125 kPa absolute
and 15°C, moored in sealevel standard air. If the gas constant of helium is 2077 m2/(s2 K) and balloon material weight
is neglected, what is the net lifting force of the balloon?
(a) 67 N, (b) 134 N, (c) 522 N, (d) 653 N, (e) 787 N
FE2.9 A square wooden (SG 0.6) rod, 5 cm by 5 cm by 10 m
long, floats vertically in water at 20°C when 6 kg of steel
(SG 7.84) are attached to one end. How high above the
water surface does the wooden end of the rod protrude?
(a) 0.6 m, (b) 1.6 m, (c) 1.9 m, (d) 2.4 m, (e) 4.0 m
Textbook Table of Contents

Study Guide
126
Chapter 2 Pressure Distribution in a Fluid
FE2.10 A floating body will be stable when its
(a) center of gravity is above its center of buoyancy,
(b) center of buoyancy is below the waterline, (c) center
of buoyancy is above its metacenter, (d) metacenter is
above its center of buoyancy, (e) metacenter is above its
center of gravity
Comprehensive Problems
C2.1 Some manometers are constructed as in Fig. C2.1, where
one side is a large reservoir (diameter D) and the other side
is a small tube of diameter d, open to the atmosphere. In
such a case, the height of manometer liquid on the reservoir
side does not change appreciably. This has the advantage
that only one height needs to be measured rather than two.
The manometer liquid has density m while the air has density a. Ignore the effects of surface tension. When there is
no pressure difference across the manometer, the elevations
on both sides are the same, as indicated by the dashed line.
Height h is measured from the zero pressure level as shown.
(a) When a high pressure is applied to the left side, the
manometer liquid in the large reservoir goes down, while
that in the tube at the right goes up to conserve mass. Write
an exact expression for p1gage, taking into account the movement of the surface of the reservoir. Your equation should
give p1gage as a function of h, m, and the physical parameters in the problem, h, d, D, and gravity constant g.
(b) Write an approximate expression for p1gage, neglecting
the change in elevation of the surface of the reservoir liquid. (c) Suppose h 0.26 m in a certain application. If pa
101,000 Pa and the manometer liquid has a density of 820
kg/m3, estimate the ratio D/d required to keep the error
of the approximation of part (b) within 1 percent of the exact measurement of part (a). Repeat for an error within 0.1
percent.
To pressure measurement location
pa
a
(air)
D
h
p1
Zero pressure level
m
d
C2.1

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C2.2 A prankster has added oil, of specific gravity SG0, to the
left leg of the manometer in Fig. C2.2. Nevertheless, the

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Utube is still useful as a pressuremeasuring device. It is
attached to a pressurized tank as shown in the figure. (a)
Find an expression for h as a function of H and other parameters in the problem. (b) Find the special case of your
result in (a) when ptank pa. (c) Suppose H 5.0 cm, pa
is 101.2kPa, ptank is 1.82 kPa higher than pa, and SG0
0.85. Calculate h in cm, ignoring surface tension effects and
neglecting air density effects.
pa
Pressurized air tank,
with pressure ptank
Oil
H
h
Water
C2.2
C2.3 Professor F. Dynamics, riding the merrygoround with his
son, has brought along his Utube manometer. (You never
know when a manometer might come in handy.) As shown
in Fig. C2.3, the merrygoround spins at constant angular
velocity and the manometer legs are 7 cm apart. The
manometer center is 5.8 m from the axis of rotation. Determine the height difference h in two ways: (a) approximately, by assuming rigid body translation with a equal to
the average manometer acceleration; and (b) exactly, using
rigidbody rotation theory. How good is the approximation?
C2.4 A student sneaks a glass of cola onto a roller coaster ride.
The glass is cylindrical, twice as tall as it is wide, and filled
to the brim. He wants to know what percent of the cola he
should drink before the ride begins, so that none of it spills
during the big drop, in which the roller coaster achieves
0.55g acceleration at a 45° angle below the horizontal.
Make the calculation for him, neglecting sloshing and assuming that the glass is vertical at all times.
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Study Guide
7.00 cm
6.00 rpm
Water
h
R
5.80 m (to center of manometer)
Center of
rotation
C2.3
Design Projects
h, m
D2.1 It is desired to have a bottommoored, floating system
which creates a nonlinear force in the mooring line as the
water level rises. The design force F need only be accurate
in the range of seawater depths h between 6 and 8 m, as
shown in the accompanying table. Design a buoyant system which will provide this force distribution. The system
should be practical, i.e., of inexpensive materials and simple construction.
D2.2 A laboratory apparatus used in some universities is shown
in Fig. D2.2. The purpose is to measure the hydrostatic
force on the flat face of the circulararc block and compare it with the theoretical value for given depth h. The
counterweight is arranged so that the pivot arm is horizontal when the block is not submerged, whence the weight
W can be correlated with the hydrostatic force when the
submerged arm is again brought to horizontal. First show
that the apparatus concept is valid in principle; then derive
a formula for W as a function of h in terms of the system
parameters. Finally, suggest some appropriate values of Y,
L, etc., for a suitable appartus and plot theoretical W versus h for these values.
F, N
h, m
F, N
6.00
6.25
6.50
6.75
7.00
400
437
471
502
530
7.25
7.50
7.75
8.00
554
573
589
600
L
Counterweight
W
Pivot
Pivot arm
R
Side view
of block face
Fluid:
h
Y
Circular arc block
b
D2.2
References
2.
3.
4.
5.
6.
7.
U.S. Standard Atmosphere, 1976, Government Printing Office, Washington, DC, 1976.
G. Neumann and W. J. Pierson, Jr., Principles of Physical
Oceanography, PrenticeHall, Englewood Cliffs, NJ, 1966.
T. C. Gillmer and B. Johnson, Introduction to Naval Architecture, Naval Institute Press, Annapolis, MD, 1982.
D. T. Greenwood, Principles of Dynamics, 2d ed., PrenticeHall, Englewood Cliffs, NJ, 1988.
R. I. Fletcher, “The Apparent Field of Gravity in a Rotating
Fluid System,’’ Am. J. Phys., vol. 40, pp. 959 – 965, July 1972.
National Committee for Fluid Mechanics Films, Illustrated
Experiments in Fluid Mechanics, M.I.T. Press, Cambridge,
MA, 1972.
J. P. Holman, Experimental Methods for Engineers, 6th ed.,
McGrawHill, New York, 1993.

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8.
9.
10.
11.
12.
13.
R. P. Benedict, Fundamentals of Temperature, Pressure, and
Flow Measurement, 3d ed., Wiley, New York, 1984.
T. G. Beckwith and R. G. Marangoni, Mechanical Measurements, 4th ed., AddisonWesley, Reading, MA, 1990.
J. W. Dally, W. F. Riley, and K. G. McConnell, Instrumentation for Engineering Measurements, Wiley, New York, 1984.
E. N. Gilbert, “How Things Float,’’ Am. Math. Monthly, vol.
98, no. 3, pp. 201 – 216, 1991.
R. J. Figliola and D. E. Beasley, Theory and Design for Mechanical Measurements, 2d ed., Wiley, New York, 1994.
R. W. Miller, Flow Measurement Engineering Handbook, 3d
ed., McGrawHill, New York, 1996.
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