fluid statics 23 module 2 fluid statics - ancalle
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Fluid Statics 23
Fluid Mechanics lecture notes by David S. Ancalle (updated 8/3/2020)
Module 2
Fluid Statics
2.1 Approaching Pressure
2.1.1 Pressure at a Point
Pressure at a point can be defined as an infinitesimal normal compressive force divided by an infinitesimal area over
which it acts.
Pressure at a point does not vary with direction. Consider a wedge-shaped element with a uniform width π₯π¦, sides
Ξπ₯ and Ξπ§, and hypotenuse Ξπ . A force, πΉπ results from a pressure ππ on the hypotenuse, and forces πΉπ₯ and πΉπ§ act on
the other sides. If we draw a free body diagram and take forces in the x and z directions, we get:
Then, taking a sum of forces in the x-axis:
β πΉπ₯ = πππ₯ βΉ πΉπ₯ β πΉπ sin π = πππ₯
ππ₯Ξπ¦Ξπ§ β ππ Ξπ Ξπ¦ sin π =1
2πΞπ₯Ξπ¦Ξπ§ππ₯
ππ₯Ξπ¦Ξπ§ β ππ Ξπ¦Ξπ§ =1
2πΞπ₯Ξπ¦Ξπ§ππ₯
ππ₯ β ππ =1
2πππ₯ Ξπ₯
As the element shrinks to a point, Ξπ₯ β 0 and the limit becomes:
ππ₯ β ππ = 0 βΉ ππ₯ = ππ
Repeating this process for the z-axis:
β πΉπ§ = πππ§ βΉ πΉπ§ β πΉπ cos π β πΉπ = πππ§
πΉπ = ππ Ξπ Ξπ¦
πΉπ₯ = ππ₯Ξπ¦Ξπ§
πΉπ§ = ππ§Ξπ₯Ξπ¦
π₯
π§
πΉπ = πΎπ = ππΞπ₯Ξπ¦Ξπ§
2
π
Ξπ₯ = Ξπ cos π
Ξπ§
=Ξ
π si
nπ
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ππ§Ξπ₯Ξπ¦ β ππ Ξπ Ξπ¦ cos π β ππΞπ₯Ξπ¦Ξπ§
2=
1
2πΞπ₯Ξπ¦Ξπ§ππ§
ππ§Ξπ₯Ξπ¦ β ππ ΞxΞπ¦ β ππΞπ₯Ξπ¦Ξπ§
2=
1
2πΞπ₯Ξπ¦Ξπ§ππ§
ππ§ β ππ β ππΞπ§
2=
1
2πΞπ§ππ§
ππ§ β ππ =1
2π(ππ§ + π)Ξπ§
As the element shrinks to a point, Ξπ§ β 0 and the limit becomes:
ππ§ β ππ = 0 βΉ ππ§ = ππ
So, we see that, at a point, the pressure in all directions does not vary. This analysis holds true for other shapes.
2.1.2 Pressure Variation
Consider an infinitesimally small element with dimensions ππ₯, ππ¦, ππ§, that undergoes a pressure π(π₯, π¦, π§), where
π0 is the pressure at the center of the element. We can determine the pressure at the sides of the element using the
chain rule:
ππ =ππ
ππ₯ππ₯ +
ππ
ππ¦ππ¦ +
ππ
ππ§ππ§
So, if we move from the center of the element to an arbitrary side π (where π can be any side, π₯, π¦, or π§), then the
pressure at that side is:
π = π0 +ππ
ππ
ππ
2
Note that the distance moved is ππ
2 because we are moving from the center of the element to the side, so we only
move half of the entire length of the element in that direction. We can use this equation to find the pressure on all of
the sides of the element, and we can use those pressures to determine the forces acting on the element:
On the top face: π = π0 +
ππ
ππ§
ππ§
2 πΉ = β (π0 +
ππ
ππ§
ππ§
2) ππ₯ ππ¦
On the bottom face: π = π0 β
ππ
ππ§
ππ§
2 πΉ = (π0 β
ππ
ππ§
ππ§
2) ππ₯ ππ¦
On the front face: π = π0 +
ππ
ππ₯
ππ₯
2 πΉ = β (π0 +
ππ
ππ₯
ππ₯
2) ππ¦ ππ§
ππ¦ ππ₯
ππ§
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On the back face: π = π0 β
ππ
ππ₯
ππ₯
2 πΉ = (π0 β
ππ
ππ₯
ππ₯
2) ππ¦ ππ§
On the left face: π = π0 β
ππ
ππ¦
ππ¦
2 πΉ = (π0 β
ππ
ππ¦
ππ¦
2) ππ₯ ππ§
On the right face: π = π0 +
ππ
ππ¦
ππ¦
2 πΉ = β (π0 +
ππ
ππ¦
ππ¦
2) ππ₯ ππ§
Newtonβs second law can now be applied in each direction, taking π = ππ = π ππ₯ ππ¦ ππ§ and πΉπ = πππ:
β πΉπ₯ = (π0 βππ
ππ₯
ππ₯
2) ππ¦ ππ§ β (π0 +
ππ
ππ₯
ππ₯
2) ππ¦ ππ§ = π ππ₯ ππ¦ ππ§ ππ₯
βππ
ππ₯ππ₯ ππ¦ ππ§ = π ππ₯ ππ¦ ππ§ ππ₯
ππ
ππ₯= βπππ₯
β πΉπ¦ = (π0 βππ
ππ¦
ππ¦
2) ππ₯ ππ§ β (π0 +
ππ
ππ¦
ππ¦
2) ππ₯ ππ§ = π ππ₯ ππ¦ ππ§ ππ¦
ππ
ππ¦= βπππ¦
β πΉπ§ = (π0 βππ
ππ§
ππ§
2) ππ₯ ππ¦ β (π0 +
ππ
ππ§
ππ§
2) ππ₯ ππ¦ β ππ ππ₯ ππ¦ ππ§ = π ππ₯ ππ¦ ππ§ ππ§
βππ
ππ§ππ₯ ππ¦ ππ§ β ππ ππ₯ ππ¦ ππ§ = π ππ₯ ππ¦ ππ§ ππ§
ππ
ππ§ = βπππ§ β ππ = βπ(ππ§ + π)
We can now apply these expressions to find the pressure differential in any direction:
ππ =ππ
ππ₯ππ₯ +
ππ
ππ¦ππ¦ +
ππ
ππ§ππ§ = βπππ₯ππ₯ β πππ¦ππ¦ β π(ππ§ + π)ππ§
2.1.3 Pressure in Fluids at Rest
An object at rest does not undergo acceleration in any direction. Therefore, for a fluid at rest, the pressure
differential becomes:
ππ = βπππ₯ππ₯ β πππ¦ππ¦ β π(ππ§ + π)ππ§ = βππππ§
which we can express as:
ππ
ππ§= βππ = βπΎ
This tells us the following: there is no pressure variation in the x and y directions, pressure only varies in the vertical
direction; pressure decreases as we move up and increases as we move down; and pressure variation for a fluid at
rest depends on the specific weight of the fluid.
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If the specific weight of the fluid is constant (e.g. an incompressible liquid with no variations in temperature or
composition), then we can integrate the pressure differential:
β« ππ = β« βπΎππ§ = βπΎ β« ππ§
Ξπ = βπΎΞπ§
2.2 Hydrostatic Pressure
2.2.1 Pascalβs Law
Fluid statics: the study of fluids at rest.
When a system is at rest: β οΏ½βοΏ½ = 0, β οΏ½βββοΏ½ = 0
We define pressure as π = ππΉ/ππ΄. Following this definition, we can express the average pressure over an area as
πππ£π = πΉ/π΄.
In 1647-1648, Blaise Pascal established that a change in pressure in an enclosed fluid at rest is equal at all points in
the fluid. This is known as Pascalβs Law.
Consider an enclosed container full of a fluid:
If we apply a force πΉ1 on the left area π΄1, then the pressure applied on the fluid is π1 = πΉ1/π΄1. Pascalβs Law states
that the change in pressure will be equal at all points in the fluid, which means that on the right area π΄2, the pressure
will be increased by π2 = π1. We can then determine the resulting force on the right area: πΉ2 = π2π΄2.
Mathematically:
π1 =πΉ1
π΄1
= π2 =πΉ2
π΄2
β΄ πΉ2 = πΉ1
π΄2
π΄1
Notice that if π΄2 is larger than π΄1, then the force applied on the right side will be increased. This makes hydraulic
lifts possible.
πΉ1 πΉ2π΄1
π΄2
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Ex. In the figure above, πΉ1 = 1 ππ, π΄1 = 2 π2, and π΄2 = 4 π2. Determine πΉ2.
Ans:
πΉ2 = πΉ1
π΄2
π΄1
= (1) (4
2) = 2 ππ
Note: If the area is twice as large, the force will be twice as large as well.
2.2.2 Pressure in Incompressible Fluids
We define hydrostatic pressure as the pressure exerted by a fluid at rest on a point submerged in that fluid. While the
term literally means βpressure of water at rest,β we will use this term to refer to pressure from any liquid. Consider a
container with an incompressible fluid. We will analyze a very small cube of fluid:
We can determine the pressure acting on the top face of the fluid as follows:
π =πΉπ
π΄=
πΎπ
ππ₯ ππ¦=
πΎ ππ₯ ππ¦ β
ππ₯ ππ¦= πΎβ
where:
β’ πΉπ is the weight of the column of water above the cube
β’ π΄ is the area of the top face of the cube
β’ πΎ is the specific weight of the column of water above the cube
β’ π is the volume of the column of water above the cube
Notice that the pressure exerted by the fluid on the top of the cube only depends on the depth of the cube, and not its
geometry. We also did not include atmospheric pressure (which can be conceptually defined as the pressure exerted
on the water by the column of air above it).
This equation agrees with the equation for pressure in fluids at rest that we derived in the previous section Ξπ =
βπΎΞπ§, where Ξπ§ = ββ (since β is measured from top to bottom).
β
ππ¦ ππ₯
ππ§
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2.2.3 Pressure Head
For incompressible fluids, the pressure can be expressed in dimensions of length by solving for β:
π = πΎβ βΉ β =π
πΎ
This quantity is called the pressure head, and is used as a unit in pressure measurements. Conceptually, this head
represents the height of a column of liquid with specific gravity πΎ that produces a gage pressure π. Pressure is
commonly measured as the length of a column of water (π = 1) or a column of mercury (π = 13.6). The unit
conversions are as follows:
1 πππ»2π = 9.81 ππ
1 πππ»π = 133 ππ
From the conversions, it can be deduced that mercury is used to measure higher pressures than water. Pressure head
can also be measured in English units.
2.2.4 Pressure in Compressible Fluids
On compressible fluids, density and specific weight vary with pressure. Therefore, for a compressible fluid (such as
an ideal gas), we will have to express density as a function of pressure. Applying the ideal gas law to the pressure
differential yields:
ππ
ππ§= βππ = β
π
π ππ
If we take the temperature of the fluid to be constant, we can integrate the above expression to get:
β«1
πππ
π
π0
= β« βπ
π πππ§
π§
π§0
lnπ
π0
= βπ
π π(π§ β π§0)
π = π0 πβπ
π π(π§βπ§0)
where π0 is a reference pressure at an elevation π§0.
2.2.5 Pressure in the Atmosphere
We have learned that the temperature in the troposphere decreases linearly by π = π0 β πΌπ§. We can now determine
the pressure variation in the troposphere. Applying the ideal gas law to the pressure differential yields
ππ
ππ§= βππ = β
π
π ππ
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Rearranging, expressing π as a function of π§, integrating between an elevation of 0 and z, and solving for pressure:
β«1
πππ
π
πππ‘π
= β« βπ
π (π0 β πΌπ§)ππ§
π§
0
lnπ
πππ‘π
=π
πΌπ ln
π0 β πΌπ§
π0
π = πππ‘π (π0 β πΌπ§
π0
)
ππΌπ
Solving for values between 0 < π§ β€ 1000 shows that the pressure decrease is very small, and changes in static air
pressure can be neglected unless the elevation difference is relatively large.
The temperature in the stratosphere, ππ , is constant. So, we determine the pressure by integrating from π§π , the lowest
elevation in the stratosphere to an elevation π§.
π = ππ πβ
ππ ππ
(π§βπ§π )
2.3 Measuring Pressure
2.3.1 U-tube Manometers
Consider pressurized flow in a pipe. If we punch open a hole in the top wall of the pipe and insert a tube, water will
flow upward until the pressure of the water is equal to atmospheric pressure. The fluid in the tube will be static, and
so we can use hydrostatic equations to find the pressure at different points in the tube and in the pipe.
Manometers are instruments that use this principle to measure pressures. In the figure below, a cross-sectional view
of the pipe is shown, and a manometer is connected to the side of the pipe. The manometer connects the pipe to a
point with a known pressure (i.e. β2 , where the pressure is atmospheric). By measuring the pressure difference
between points β1 and β2 , we can determine the pressure at the pipe. If we want to measure gage pressures, then
π2 = πππ‘π = 0.
π = πππ‘π
β
π = πΎβ
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π2 + Ξπ = π1
0 + πΎβ = π1
π1 = πΎβ
Remember that pressure increases with depth. Therefore, since we move βdownwardβ from π2 to π1, the pressure
difference is positive. If we would have started measuring from the pipe to the tube, our equation would have been:
π1 β πΎβ = π2 = 0
π1 = πΎβ
Ex. In the U-tube manometer above, β = 2ππ‘ and the fluid is water. What is the pressure in the pipe?
Ans:
πΎ = 62.4 ππ/ππ‘3
π = πΎβ = (62.4)(2) = 124.8 ππ/ππ‘2
These types of manometers are called U-tube manometers. Manometers that use a single fluid are usually only used
to measure very small pressures. To measure larger pressures, a heavier fluid can be inserted in the manometer.
Measuring the pressure differences from β1 to β3 yields:
π1 + πΎ1β β πΎ2π» = π3 = 0
π1 = πΎ2π» β πΎ1β
Notice that the pressure at β2 and β2β is the same, since there is no difference in elevation.
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2.3.2 Differential Manometers
A differential manometer is used to measure pressure differences between two pipes or between two points in a
conduit. The pressure difference is computed in the same way as the U-tube manometer, but the end of the
manometer will not be open to the atmosphere.
2.3.3 Piezometer
A piezometer is a simple type of manometer that consists of a vertical tube that is inserted in a vessel with a fluid.
Piezometers are useful for measuring small pressures. The static pressure of a point in the liquid at a depth π is:
π = πΎ(β + π)
2.3.4 Micromanometers
Another type of manometer is the micromanometer, which is used to measure very small pressure changes.
Applying the manometer equations, we get:
π1 + πΎ1(π§1 β π§2) + πΎ2(π§2 β π§3) β πΎ3(π§4 β π§3) β πΎ2(π§5 β π§4) = π5
π1 + πΎ1(π§1 β π§2) + πΎ2(π§2 β π§3 + π§4 β π§5) β πΎ3(π§4 β π§3) = π5
Note that π5 = πππ‘π = 0, π» = π§4 β π§3, and β = π§5 β π§2.
π1 + πΎ1(π§1 β π§2) + πΎ2(π» β β) β πΎ3π» = 0
π1 = πΎ1(π§2 β π§1) + πΎ2β + (πΎ3 β πΎ2)π»
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A small change in pressure Ξπ will result in a change in the elevation π§2 of Ξπ§, as well as a change in β of 2Ξπ§, and
a change in π» of 2Ξπ§π·2/π2. Therefore, we can evaluate a pressure change by:
π1 = πΎ1(βΞπ§) + πΎ2Ξβ + (πΎ3 β πΎ2)Ξπ»
Ξπ1 = πΎ1(βΞπ§) + πΎ2(2Ξπ§) +(πΎ3 β πΎ2)2Ξπ§π·2
π2
And so, the rate of change of π» with respect to π is:
Ξπ»
Ξπ1
=2Ξπ§π·2/π2
πΎ1(βΞπ§) + πΎ2(2Ξπ§) +(πΎ3 β πΎ2)2Ξπ§π·2
π2
=2π·2/π2
βπΎ1 + 2πΎ2 + 2(πΎ3 β πΎ2)π·2
π2
2.3.5 Barometers
A barometer is an instrument used to measure atmospheric pressure. It was invented by Evangelista Torricelli.
A barometer consists of a glass tube filled with mercury. The tube is inserted in a reservoir filled with mercury and
turned upside down. The weight of the mercury in the tube causes it to move downward and creates a vacuum in the
top of the tube where π β 0. The atmospheric pressure can be measured by measuring the height of the mercury
column inside the tube, so that:
πππ‘π = πΎπ»πβ
Note: Under standard engineering conditions, the atmospheric pressure is 760 πππ»π.
2.3.6 Pressure Gages
In cases where static pressures are too high to measure with manometers, a pressure gage can be used. Different
types of pressure gages are:
Bourdon gage: uses an elastic, coiled metal (Bourdon tube) to determine pressures.
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Pressure Transducer: uses an electrical strain gage to determine deformation in its diaphragm and converts the
electrical current measure into a pressure measure.
Fused Quartz Force-Balance Bourdon tube: uses the same concept as the Bourdon gage, but determines pressures
using a magnetic field that returns the Bourdon tube to its original position.
Piezoelectric gages: devices that change their electric potential when subjected to small pressure changes.
2.4 Buoyancy
2.4.1 Determining Buoyant Force
Consider an object with specific weight πΎπ, submerged in a fluid with specific weight πΎπ:
The force from the surrounding liquid acting on top of the object can be computed by multiplying the pressure on
the top face times the area of the top face:
β
ππ¦ ππ₯
ππ§
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πΉπ‘ππ = ππ‘πππ΄π‘ππ = πΎπβ ππ₯ ππ¦
The force acting on the bottom face is:
πΉπππ‘ = ππππ‘π΄πππ‘ = πΎπ(β + ππ§)ππ₯ ππ¦
We donβt consider the forces on the sides since they will cancel out. The net force acting on the object is:
πΉπππ‘ = πΉπππ‘ β πΉπ‘ππ = πΎπ(β + ππ§)ππ₯ ππ¦ β πΎπβ ππ₯ ππ¦ = πΎπ(β + ππ§ β β)ππ₯ ππ¦ = πΎπ ππ₯ ππ¦ ππ§
Notice that the net force is then the specific weight of the fluid by the volume of the submerged object (i.e. the
displaced volume). We call this force the buoyant force, πΉπ΅ and define it as:
πΉπ΅ = πΎππ
Where π is the displaced volume. In order for an object to float, the buoyant force has to be larger than its weight,
applying static equilibrium, we see that a submerged or partially submerged object is static when its buoyant force is
equal to its weight:
β πΉ = πΉπ΅ β πΉπ = 0 βΉ πΉπ΅ = πΉπ
2.4.2 Hydrometers
A hydrometer is an instrument used to measure the specific gravity of liquids that uses the principle of buoyancy. It
consists of a stem with a constant area π΄. When placed in water, the hydrometer will submerge until:
πΉπ = πΎπ»2ππ0
where π0 is the initial submerged volume. When submerged in a different fluid with specific weight πΎ, the force
balance is:
πΉπ = πΎ(π0 β π΄Ξβ)
where Ξβ is the change in submerged portion of the stem. Combining both equations gives:
πΎπ»2ππ0 = πΎ(π0 β π΄Ξβ) βΉ S =V0
π0 β π΄Ξβ=
1
1 βπ΄Ξβπ0
2.5 Stability
Consider a submerged object:
β’ Center of gravity: the centroid of an object, πΊ
β’ Center of buoyancy: the centroid of the displaced volume, πΆπ΅
Gravitational force acts on the center of gravity, and buoyant force acts on the center of buoyancy. We have already
shown that a floating object is in vertical equilibrium if πΉπ΅ = πΉπ. However, we must also consider if an object has
rotational equilibrium, which results when no moment is formed by the two forces. We will consider three types of
rotational equilibrium:
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Stable Equilibrium
Occurs when πΊ is below πΆπ΅. If the object is slightly rotated, a coupled moment
between πΉπ and πΉπ΅ will restore the object to its original position.
Unstable Equilibrium
Occurs when πΊ is above πΆπ΅. If the object is slightly rotated, a coupled moment
between πΉπ and πΉπ΅ will continue to rotate the object.
Neutral Equilibrium
Occurs when πΊ and πΆπ΅ coincide. No coupled moment forms from a rotation.
For objects that are partially submerged, a rotation may move the center of buoyancy. This allows for stable
equilibrium even in instances where πΊ is above πΆπ΅.
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Consider a partially submerged object (like a ship) in unstable equilibrium. We
will define the point about which the object rotates as the origin, π, which lies at
the water surface on the vertical line that crosses πΆπ΅ to πΊ. We will call this line
the line of action.
When a rotation is applied to the object, the line of action rotates along with the
object, and the πΆπ΅ can shift. If we trace a vertical line from the new position of
the center of buoyancy, it will intersect the line of action at the metacenter, π.
If π is above πΊ in the line of action, then the object is in stable equilibrium, since
a counteracting moment acting between π and πΊ will return the object to its
original position.
If, however, π is below πΊ in the line of action, then the object is in unstable
equilibrium, as the resulting moment will continue the rotation and cause the
object to overturn.
While the theory for stability is relatively simple to grasp at a conceptual level, determining the stability of objects
becomes complicated when it is time to determine the location of π. In this class, we will consider the method
taught by Potter et. al.
Letβs define the metacentric height, πΊπΜ Μ Μ Μ Μ as the distance from πΊ to π. A positive πΊπΜ Μ Μ Μ Μ value indicates that the
metacenter is above the center of gravity. A negative πΊπΜ Μ Μ Μ Μ value indicates that π is below πΊ. Consider a floating
body with a uniform cross section, having rotated as shown below:
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The x-coordinate of the center of buoyancy, οΏ½Μ οΏ½ can be found by considering the volume of the original submerged
volume (π0), plus the volume from added wedge π·ππΈ, π1 minus the volume from the subtracted wedge π΄ππ΅, π2:
οΏ½Μ οΏ½ =οΏ½Μ οΏ½0π0 + οΏ½Μ οΏ½1π1 β οΏ½Μ οΏ½2π2
π
οΏ½Μ οΏ½π = οΏ½Μ οΏ½0π0 + οΏ½Μ οΏ½1π1 β οΏ½Μ οΏ½2π2
Taking the x-coordinate of the original center of buoyancy to be zero, we get οΏ½Μ οΏ½0 = 0:
οΏ½Μ οΏ½π = οΏ½Μ οΏ½1π1 β οΏ½Μ οΏ½2π2
οΏ½Μ οΏ½π = β« π₯ πππ1
β β« π₯ πππ2
Now, to find an expression for the infinitesimal volume ππ, letβs take a step back and take a look into the basics of
integrating. Suppose you want to find the area of the triangle formed by the lines π¦ = π π₯, π¦ = 0, and π₯ = π₯1. You
could easily do this by integrating π¦ = π π₯ from 0 to π₯1. Your integral would look like:
β« π π₯ ππ₯π₯1
0
We know that in a linear equation, the constant represents the slope of the line. If we define πΌ as the angle of the line
measured from the x-axis, then the slope is π =π₯π¦
Ξπ₯= tan πΌ. Giving us an integral of β« π₯ tan πΌ ππ₯
π₯1
0. Now, this
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integral only gives us the area of the triangle. If instead, we wanted to find the volume of the triangle, assuming a
constant width π, then we could express the volume as π = β« πππ1
= β« π₯ tan πΌ ππ΄π΄1
where ππ΄ = π ππ₯. This means
that an expression for a differential volume is ππ = π₯ tan πΌ ππ΄.
Now, back to our x-coordinate equation, we can insert our definition of ππ into it to yield:
οΏ½Μ οΏ½π = β« π₯2 tan πΌ ππ’π’1
β β« βπ₯2 tan πΌ ππ΄π΄2
Notice that the second term is negative because the area is located on the negative side of the y-axis (setting our
origin at π.) Simplifying:
οΏ½Μ οΏ½π = tan πΌ (β« π₯2 ππ’π’1
+ β« π₯2 ππ΄π΄2
) = tan πΌ β« π₯2 ππ’π’1+π΄2
= tan πΌ β« π₯2 ππ’π΄
Notice that π΄1 + π΄2 is equivalent to the total waterline area A. From statics we define the second moment of area
(moment of inertia) about an axis perpendicular to x as πΌ = β« π₯2 ππ΄. This means we can substitute our equation
with:
οΏ½Μ οΏ½π = πΌπ tan πΌ
where πΌπ represents an axis passing through the origin. Now, looking at our diagram, we can express οΏ½Μ οΏ½ as a function
of the distance from the original center of buoyancy to the metacenter, πΆπΜ Μ Μ Μ Μ by: οΏ½Μ οΏ½ = πΆπΜ Μ Μ Μ Μ tan πΌ, and substitute:
πΆπΜ Μ Μ Μ Μ π tan πΌ = πΌπ tan πΌ
πΆπΜ Μ Μ Μ Μ =πΌπ
π
And from the diagram, we can see that πΆπΜ Μ Μ Μ Μ = πΆπΊΜ Μ Μ Μ + πΊπΜ Μ Μ Μ Μ so, finally, we get:
πΊπΜ Μ Μ Μ Μ =πΌπ
πβ πΆπΊΜ Μ Μ Μ
Now, for a given body orientation, if πΊπΜ Μ Μ Μ Μ is positive, the body is stable. If not, the body is unstable. Note that this
equation assumes that the object is rotating about the origin. For other cases, the location of the center of buoyancy
must be determined to determine stability.
2.6 Forces on Plane Areas
2.6.1 General Formulation
Consider a submerged flat plate with an arbitrary shaped, oriented at an angle π from the horizontal. We will set an
x/y coordinate system starting at the surface of the liquid so that the positive y-axis extends downward along the
plane of the plate.
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If we consider a force ππΉ acting over a differential area ππ΄ at a depth of β = π¦ sin π, we can determine that:
ππΉ = π ππ΄ = πΎβ ππ΄ = πΎ π¦ sin π ππ΄
Remembering that we determined that the pressure at a point submerged in a liquid is π = πΎβ. The resultant force
acting on the plate is equivalent to the sum of all differential forces acting on the plate, which we can obtain by
integrating over the entire plate area π΄:
πΉ = β ππΉ = β« πΎ π¦ sin π ππ΄π΄
= πΎ sin π β« π¦ ππ΄π’
= πΎ sin π οΏ½Μ οΏ½π΄
Keeping in mind that οΏ½Μ οΏ½ =β« π¦ ππ΄π’
π΄, where οΏ½Μ οΏ½ represents the distance to the centroid of the plate. Now, the depth to the
centroid can be expressed as βΜ = οΏ½Μ οΏ½ sin π, which gives:
πΉ = πΎβΜ π΄
Since the average pressure acting over an area is πππ£π = πΉ/π΄, then we can say that πππ£π = πΎβΜ .
The center of pressure (π₯π , π¦π) is the point where the resultant force acts on. It can be determined by a balance of
moments as follows.
On the y-axis:
ππ π₯= β ππ₯
π¦ππΉ = β« π¦ ππΉπ΄
π¦ππΎ sin π οΏ½Μ οΏ½π΄ = β« π¦ (πΎ π¦ sin π ππ΄)π΄
π¦π οΏ½Μ οΏ½ π΄ = β« π¦2 ππ΄π΄
We know that the area moment of inertia πΌπ₯ = β« π¦2 ππ΄π΄
. So, we express the center of pressure as:
π¦π =πΌπ₯
οΏ½Μ οΏ½π΄
The parallel-axis theorem (from Statics) says that πΌπ₯ = πΌοΏ½Μ οΏ½ + π΄οΏ½Μ οΏ½2. Replacing into the equation above yields:
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π¦π = οΏ½Μ οΏ½ +πΌοΏ½Μ οΏ½
οΏ½Μ οΏ½π΄
Where πΌοΏ½Μ οΏ½ is the moment of inertia about the centroidal axis of the plate, and π΄ is the submerged are of the plate.
On the x-axis:
ππ π¦= β ππ¦
π₯ππΉ = β« π₯ ππΉπ΄
π₯ππΎ sin π οΏ½Μ οΏ½π΄ = β« π₯ (πΎ π¦ sin π ππ΄)π΄
π₯π οΏ½Μ οΏ½ π΄ = β« π₯π¦ ππ΄π΄
where the integral represents the product of inertia πΌπ₯π¦
π₯π =πΌπ₯π¦
οΏ½Μ οΏ½π΄
Applying the parallel-plane theorem: πΌπ₯π¦ = πΌοΏ½Μ οΏ½π¦ + π΄οΏ½Μ οΏ½οΏ½Μ οΏ½:
π₯π = οΏ½Μ οΏ½ +πΌοΏ½Μ οΏ½π¦
οΏ½Μ οΏ½π΄
2.6.2 Geometric Formulation
An alternative method to finding force and center of pressure is to treat the pressure distribution along a plate as a
geometric figure.
Pressure gradient: the distribution of pressure along a submerged area
Pressure prism: a geometric representation of the pressure gradient as a volume
Considering a force ππΉ acting over ππ΄ at a depth β, we once again determine that ππΉ = π ππ΄ = πΎβ ππ΄. We can
also represent the force geometrically as a differential volume with base ππ΄ and height π, as shown in the figure.
The resultant force can then be found by integrating the differential force elements over the volume of the pressure
prism:
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πΉ = β ππΉ = β« π ππ΄π΄
= β« πππ
= π
This means that the magnitude of the resultant force is equal to the volume of the pressure prism.
The location of the center of pressure is found by taking the moment of the resultant force about the y-axis and x-
axis. For the y-coordinate:
(ππ )π₯ = β ππ₯
π¦ππΉ = β« π¦ ππΉ
π¦π =β« π¦ ππΉ
πΉ=
β« π¦ π ππ΄π΄
β« π ππ΄π΄
=β« π¦ ππ
π
π
For the x-coordinate:
(ππ )π¦ = β ππ¦
π₯ππΉ = β« π₯ ππΉ
π₯π =β« π₯ ππΉ
πΉ=
β« π₯ π ππ΄π΄
β« π ππ΄π΄
=β« π₯ ππ
π
π
Notice, then, that the center of pressure is located at the centroid of the pressure gradient.
2.6.3 Integral Formulation
If we can express the plateβs x- and y-coordinates as a function π₯ = π(π¦) then the resultant force is found by direct
integration. Since the x-coordinate is measured from the center of the area, then ππ΄ = 2π₯ ππ¦:
πΉ = β« π ππ΄π΄
= β« π 2π₯ ππ¦π¦
= 2 β« π π(π¦)ππ¦π¦
The location of the center of pressure is found by equating to the moment of the resultant force to the moment of the
pressure distribution about the x- and y- axes. We take ππΉ to act along the centroid of ππ΄, at coordinates (οΏ½ΜοΏ½, οΏ½ΜοΏ½):
(ππ )π₯ = β ππ₯
π¦ππΉ = β« οΏ½ΜοΏ½ ππΉπ΄
π¦π =β« οΏ½ΜοΏ½ ππΉ
π΄
πΉ=
β« οΏ½ΜοΏ½ π ππ΄π΄
β« π ππ΄π΄
=β« οΏ½ΜοΏ½ π π(π¦)ππ¦
π¦
β« π π(π¦)ππ¦π¦
(ππ )π¦ = β ππ¦
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π₯ππΉ = β« οΏ½ΜοΏ½ ππΉπ΄
π₯π =β« οΏ½ΜοΏ½ ππΉ
π΄
πΉ=
β« οΏ½ΜοΏ½ π ππ΄π΄
β« π ππ΄π΄
=β« οΏ½ΜοΏ½ π π(π¦)ππ¦
π¦
β« π π(π¦)ππ¦π¦
2.6.4 Projection Method for Inclined Areas
An alternative way of determining forces on inclined areas is by dividing the resultant force into horizontal and
vertical components.
The horizontal component of the force is equal to the force from hydrostatic pressure acting on a vertical projection
of the inclined area.
πΉβ = β« π sin π ππ΄π΄
The vertical component of the force is equal to (a) the weight of the liquid of the liquid above the plate, or (b) the
buoyant force acting on the plate (if the liquid is below the plate).
πΉπ£ = β« πΎ πππ
= πΎπ
πΉπ£ = πΉπ΅ = πΎπ
On gases, where weight can be neglected, we can assume the pressure to be constant. Therefore, the horizontal and
vertical components of the resultant force can be determined by projecting the curved surface area into vertical and
horizontal planes.
The resultant force is found by performing a vector sum of the horizontal and vertical components:
πΉ = βπΉβ2 + πΉπ£
2
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The center of pressure can be determined by a weighted sum:
π₯π =πΉβπ₯πβ
+ πΉπ£π₯ππ£
πΉ
π¦π =πΉβπ¦π β
+ πΉπ£π¦ππ£
πΉ
2.6.5 Projection Method for Curved Areas
While forces on curved areas can be determined using direct integration, it is easier to determine forces using the
projection method.
2.7 Pressure on Linearly Accelerating Containers
2.7.1 Horizontal Acceleration
If a container is moving horizontally with a constant acceleration, ππ, then the fluid will be at rest relative to the
container (which we will call a reference frame).
The acceleration will result in the liquidβs surface being inclined, where the rotation occurs about the center of the
container. However, since there is no relative movement between the liquid particles, the hydrostatic equation can be
used to determine pressure. At any point in the fluid, the pressure will be computed as:
π = πΎβ
If we analyze a differential cube (like we have done to determine hydrostatic pressure and buoyancy), we can
calculate the forces acting on the left and right sides of the cube and then find the sum of forces in the horizontal
axis. Let point 1 represent the leftmost side of the cube and point 2 represent the rightmost side of the cube:
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β πΉπ₯ = πππ₯
πΉ1 β πΉ2 = πππ₯
π1ππ΄ β π2ππ΄ = π ππππ₯
(π1 β π2)ππ΄ =πΎ
π ππ₯ ππ΄ ππ₯
π1 β π2 =πΎ
πππ₯ ππ₯
Taking the pressure on each side to be π = πΎβ, we get that
πΎβ1 β πΎβ2 =πΎ
πππ₯ ππ₯
β1 β β2 =ππ₯
πππ₯
β1 β β2
ππ₯=
ππ₯
π
Notice that β1 β β2 = ππ§, the difference in elevation of the water surface above the two sides of the cube. This
means that the expression (β1 β β2)/ππ₯ can be taken as the slope of the liquid surface. If we take the angle of the
water surface to be π, we get:
β1 β β2
ππ₯=
ππ§
ππ₯= tan πΌ
So we define the slope of the water surface as:
tan π =ππ₯
π
If the fluid is enclosed, and unable to pivot about the center of the container, the acceleration results in a pressure
increase that can be resolved by picturing an imaginary liquid surface at an angle π, which pivots about point B in
the figure.
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2.7.2 Vertical Acceleration
If the reference frame is accelerating vertically, the liquid surface remains horizontal. The pressure along the z-axis
can be found by analyzing the forces acting on the vertical element in the image.
β πΉπ§ = πππ§
πππ΄ β πΉπ =πΎ
π β ππ΄ ππ§
πππ΄ β πΎβ ππ΄ =πΎ
πβ ππ΄ ππ§
π = πΎβ (1 +ππ§
π)
2.7.3 2D Acceleration
If the reference frame is accelerating in both the x- and y- directions, we can determine the slope of the liquid
surface by simplifying the pressure differential to:
ππ = βπππ₯ππ₯ β π(π + ππ§)ππ§
Integrating between two arbitrary points in the liquid gives:
π2 β π1 = βπππ₯(π₯2 β π₯1) β π(π + ππ§)(π§2 β π§1)
If the two points are selected at the liquid surface, where π = 0, we can further simplify to:
0 = βπππ₯(π₯2 β π₯1) β π(π + ππ§)(π§2 β π§1)
π§2 β π§1
π₯2 β π₯1
= βππ₯
π + ππ§
π§1 β π§2
π₯2 β π₯1
=ππ₯
π + ππ§
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where, π§1βπ§2
π₯2βπ₯1 is the slope of the liquid surface, so that:
tan π =ππ₯
π + ππ§
2.8 Pressure on Rotating Containers
If the reference frame is rotating with a constant angular velocity π, the liquid surface will take the shape of a forced
vortex. We know that the acceleration of a fluid particle at a distance π from the center is ππ = π2π acting toward
the center. Analyzing a differential ring element that surrounds the free surface point at the center, with radius π,
thickness ππ, and height πβ, we can define the pressure inside the ring as π. Then, the pressure outside the ring is
defined as:
π2 = π + (ππ
ππ) ππ
The sum of forces along a radial axis becomes:
β πΉπ = πππ
π2ππ΄ β πππ΄ =πΎ
π ππ π2π
(π + (ππ
ππ) ππ) (2ππ πβ) β π(2ππ πβ) =
πΎ
π (2ππ πβ ππ)π2π
(π + (ππ
ππ) ππ) β π =
πΎ
π ππ π2π
ππ
ππ=
πΎ
ππ2π
Integrating both sides:
π = (πΎπ2
2π) π2 + πΆ
At the center, where π = 0, π = 0, therefore πΆ = 0.
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π = πΎβ = (πΎπ2
2π) π2
β = (π2
2π) π2
This equation describes a parabola. Therefore, the liquid in the container has the shape of a paraboloid. The height
of the paraboloid above the center surface at the edges is found by solving for the entire radius of the container π :
π» = (π2
2π) π 2
The liquid inside a rotating container will be displaced π»/2 above the original surface at the edges, and π»/2 below
the original surface at the center. The volume of the paraboloid section is π = ππ 2π»/2.
As with linearly accelerating containers, the pressure at a point is calculated by π = πΎβ, and for closed-lid
containers, the pressure is estimated by picturing an imaginary liquid surface forming the paraboloid.
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