Chapter 2: Pressure distribution in a fluid (유체내의 압력분포)

■ Introduction

1) Motivation: Why hydrostatics?

Many fluid problems do not involve motion. They concern the pressure distribution in a static

fluid to study the effect of the fluid on solid surfaces or on floating and submerged bodies.

2) Definition: What is hydrostatics?

Hydrostatics is the study of pressures throughout a fluid at rest and the pressure forces

on finite surfaces.

As the fluid is at rest, there are no shear stresses in it. Hence the pressure at a point on a

plane surface always acts normal to the surface, and all forces are independent of

viscosity

Therefore, it is conclusive that the pressure variation is due only to the weight of the fluid

As a result, the controlling laws are relatively simple, and analysis is based on a

straightforward application of the mechanical principles of force and moment. Also,

solutions are exact and there is no need to have recourse to experiment

3) Examples

∙ Pressure distribution in the atmosphere and the oceans

∙ Design of manometer, mechanical and electronic pressure instruments

∙ Forces on submerged flat and curved surfaces

∙ Buoyancy on a submerged body

∙ Behavior of floating bodies

■ Pressure

1) Definition: Normal stress on any plane through a fluid element at rest

As we are considering the situation in which there are no shearing stresses, the only external

forces acting on the wedge are due to the pressure and the weight.

Based on equilibrium of forces, the sum of forces should be zero in x and z direction. As 𝑝𝑛 can

be divided into x and z direction, forces in the x and z direction can be written as;

∑ 𝐹𝑥 = 0 = 𝑝𝑥𝑏∆𝑧 − 𝑝𝑛𝑏∆𝑠 ∙ sin 𝜃

∑ 𝐹𝑧 = 0 = 𝑝𝑧𝑏∆𝑥 − 𝑝𝑛𝑏∆𝑠 ∙ cos 𝜃 −1

2𝜌𝑔𝑏∆𝑥∆𝑧

Figure 1. Pressure on a fluid element at rest (White, 2012, Fluid Mechanics)

The geometry of the wedge is such that

∆𝑠 ∙ 𝑠𝑖𝑛 𝜃 = ∆𝑧, and ∆𝑠 ∙ cos 𝜃 = ∆𝑥

Substitution and rearrangement give

𝑝𝑥 = 𝑝𝑛

𝑝𝑧 = 𝑝𝑛 +1

2𝜌𝑔∆𝑧

These relations illustrate two important principles of the hydrostatic or shear-free 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.

In the limit as the fluid wedge shrinks to a point, ∆𝑧 → 0 and the we can get the following

equality.

𝑝𝑥 = 𝑝𝑧 = 𝑝𝑛 = 𝑝

This suggest that the pressure at a point in a fluid at rest, or in motion, is independent of

direction as long as there are no shearing stresses present, called Pascal’s law.

2) Pascal’s law

An important application of Pascal's law is the hydraulic press. A force F1 is applied to a small

piston of area A1. The pressure is transmitted through a liquid to a larger piston of area A2. Since

the pressure is the same on both sides, we see that P = F1/A1 = F2/A2. Therefore, the force F2 is

larger than F1 by multiplying factor A2/A1. That is, with larger area where pressure is acting,

greater force is producible. In other words, the hydraulic press is the equivalent of a lever, for a

small force on one piston can produce a large force on another.

For example, car lifting system uses a small cylinder. Thought the pressure is the same it is

exerted over a much larger area giving a multiplication of forces that lifts the car.

■ Basic equation of pressure field

Figure 2. Pressure acting on a fluid element (White, 2012, Fluid Mechanics)

Pressure causes a net force on a fluid element when it varies spatially. To see this, consider the

pressure acting on the two x faces.

𝑑𝐹𝑥 = 𝑝𝑑𝑦𝑑𝑧 − (𝑝 +𝜕𝑝

𝜕𝑥𝑑𝑥) 𝑑𝑦𝑑𝑧 = −

𝜕𝑝

𝜕𝑥𝑑𝑥𝑑𝑦𝑑𝑧

𝜕𝑝

𝜕𝑥𝑑𝑥 means variations of pressure with a change of distance by 𝑑𝑥. (예를 들어, 압력이 위치에

따라 차이가 있기 때문에 𝑑𝑥 만큼 변하면 압력도 변하게 되고, x 방향만 생각하기 때문에 x에

대한 편미분의 형태로 쓸 수 있다.)

Likewise, forces in y- and z-directions are can be expressed as following.

𝑑𝐹𝑦 = −𝜕𝑝

𝜕𝑦𝑑𝑥𝑑𝑦𝑑𝑧

𝑑𝐹𝑧 = −𝜕𝑝

𝜕𝑧𝑑𝑥𝑑𝑦𝑑𝑧

The resultant surface force acting on the element can be expressed in vector form as

𝑑𝐹𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = 𝑑𝐹𝑥𝑖̂ + 𝑑𝐹𝑦𝑗̂ + 𝑑𝐹𝑧�̂�

or

𝑑𝐹𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 = − (𝜕𝑝

𝜕𝑥𝑖̂ +

𝜕𝑝

𝜕𝑦𝑗̂ +

𝜕𝑝

𝜕𝑧�̂�) 𝑑𝑥𝑑𝑦𝑑𝑧

𝑖,̂ 𝑗̂, and �̂� are the unit vectors along the coordinate axes.

This equation considers pressure variation through all the directions.

The group of terms in parentheses represents in vector form of the pressure gradient (압력구배)

and can be written as

𝛻𝑝 =𝜕𝑝

𝜕𝑥𝑖̂ +

𝜕𝑝

𝜕𝑦𝑗̂ +

𝜕𝑝

𝜕𝑧�̂�

♠ Del (∇) operator = gradient: 어떤 함수에 대해서 모든 요소에 대해 각각의 편미분을

취하는 수학 연산자

▶ Considering gravity

The pressure gradient is a surface force that 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. For this reason, when considering forces in the z-direction, it is necessary to include the

effect of gravity or weight of the element.

The gravitational force can be expressed by multiplying density, volume, and g (gravitational

acceleration).

𝑓𝑔𝑟𝑎𝑣 = 𝜌𝑔 ← 𝑑𝐹𝑔𝑟𝑎𝑣 = 𝜌𝑑𝑥𝑑𝑦𝑑𝑧𝑔

With Newton’s second law, it is possible to express the change in total forces as the sum of

changes in forces caused by pressure and gravity.

∑ 𝑑𝐹 = 𝑑(𝑚𝑎)

𝑑𝑚 = 𝜌𝑑𝑥𝑑𝑦𝑑𝑧

∑ 𝑑𝐹 = 𝑑𝐹𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 + 𝑑𝐹𝑔𝑟𝑎𝑣 = 𝑑(𝑚𝑎) = 𝜌𝑑𝑥𝑑𝑦𝑑𝑧 ∙ 𝑎

Therefore, the equation can be written as

−∇𝑝 + 𝜌𝑔 = 𝜌𝑎

This is the general equation of motion for a fluid in which there are no shearing stresses, which

relates pressure gradient, weight of the element and motion if the element moves.

■ Hydrostatic pressure distributions

As we are studying a fluid at rest, acceleration term disappears and pressure gradient can be

expressed by

∇𝑝 = 𝜌𝑔 = 𝛾

This is hydrostatic distribution.

In our customary coordinate system z is “up”. Thus, the local gravity vector for small-scale

problem is

𝑔 = −𝑔𝑘

For this coordinate system, components directing into x and y become zero, while z component

that is equivalent to pressure gradient only has a term of specific weight.

𝜕𝑝

𝜕𝑥= 0,

𝜕𝑝

𝜕𝑦= 0,

𝜕𝑝

𝜕𝑧= −𝜌𝑔 = −𝛾

This equations show that an important issue. That is, pressure does not depend on x or y. For

example, as we move from point to point in a horizontal plane, the pressure does not change.

From the above equations, a first of differential equation of pressure changes in z direction is

consequently established with respect to the specific weight.

𝑑𝑝

𝑑𝑧= −𝛾

This is the fundamental equation for fluid at rest and can be used to determine how pressure

changes with elevation.

The equation is easily solvable with the method of separation of variables for ordinary 1st order

differential equation.

𝑝2 − 𝑝1 = − ∫ 𝛾𝑑𝑧2

1

This 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.

▶ Hydrostatic condition

The above equations (differential equation and its solution) states 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”

This statement is well illustrated in the figure. Point a, b, c and d are at equal depth in water and

therefore have identical pressures. Point A, B, and C are also at equal depths in water and have

identical pressures higher than a, b, c, and d. However, point D has a different pressure from A, B,

and C because it is not connected to them by a water pathway. In other words, pressure is

different because of different density of mercury against water ← Pressure depends on density in

the equation.

1) Hydrostatic pressure in liquids

Liquids are so nearly incompressible that we can neglect their density variation in hydrostatics.

Thus, we may assume constant density in liquid hydrostatics calculations and the result in the

following equation

𝑝2 − 𝑝1 = −𝛾(𝑧2 − 𝑧1)

If we set the height difference of (𝑧2 − 𝑧1) = ℎ, following equations are obtainable.

𝑝1 − 𝑝2 = 𝛾ℎ

𝑧1 − 𝑧2 =𝑝2

𝛾−

𝑝1

𝛾

In the equation, pressure over specific weight is called pressure head of the fluid. That is, the

height of a column of fluid of specific weight (𝛾) required to give a pressure difference. The

dimension of head is length (압력수두: 물기둥 높이에 따른 압력 또는 압력을 물의 높이로 표현)

The figure illustrates pressure distributions in oceans and atmosphere. The pressure at any depth z

can be written by the difference between atmospheric pressure and the pressure caused by depth

and specific weight of the fluid.

𝑝 = 𝑝𝑎 − 𝛾𝑧

2) Hydrostatic pressure in gases

Gases are compressible with density nearly proportional to pressure. Thus, density should be

treated as a variable. With the perfect-gas law, integration can be performed.

𝑑𝑝

𝑑𝑧= −𝜌𝑔 ← 𝑝 = 𝜌𝑅𝑇

𝑑𝑝

𝑑𝑧= −𝜌𝑔 = −

𝑝

𝑅𝑇𝑔

By separating the variables and integrating between points 1 and 2, following equations are

resulted.

∫𝑑𝑝

𝑝

2

1

= 𝑙𝑛𝑝2

𝑝1

= −𝑔

𝑅∫

𝑑𝑧

𝑇

2

1

In this case, the integral over z requires an assumption about the temperature variation. One

common approximation is the isothermal atmosphere where 𝑇 = 𝑇0.

𝑝2 = 𝑝1𝑒𝑥𝑝 [−𝑔(𝑧2 − 𝑧1)

𝑅𝑇0

]

This equation well approximates the pressures in earth, but actual earth’s temperature drops off

linearly with z up to an altitude of about 11,000 m (but, we assume that T is constant to integrate

the equation).

3) Pressure measurement

▶ Mercury barometer: measures atmospheric pressure

Generally, height of barometer is 760 mmHg.

♠ Why mercury? → Because it’s the heaviest liquid. Water barometer would be 10 m high.

In the mercury barometer, 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 at room temperatures.

As the mercury is liquid, we can apply the previous equation to measure pressure at point p2.

𝑝2 − 𝑝1 = −𝛾(𝑧2 − 𝑧1)

With given condition, it is possible to compute p2.

𝑝1 = 0 𝑎𝑡 𝑧1 = ℎ and 𝑝2 = 𝑝𝑎 𝑎𝑡 𝑧2 = 0

𝑝2 − 0 = −𝛾𝑀(0 − ℎ)

ℎ =𝑝𝑎

𝛾𝑀

That is, atmospheric pressure can be calculated with the height and specific weight of mercury. At

sea-level, 𝑝𝑎 = 101,350 𝑃𝑎 and 𝛾𝑀 = 133,100 𝑁/𝑚3, and ℎ = 0.761 𝑚 𝑜𝑟 29.96 𝑖𝑛.

▶ Manometer: Pressure measuring device based on the use of liquid columns in vertical or

inclined tubes. A static column of one or more liquids or gases can be used to measure pressure

difference between two points.

Fundamentals of manometer: pressure increases downward and decreases upward. That is, if point

2 is a distance h blow point 1 in a uniform liquid, pressure at point 2 is computable by the

following equation.

𝑝2 = 𝑝1 + 𝜌𝑔ℎ

The picture displays the evaluation of pressure changes through a column of multiple fluids. In

this case, it is possible to calculate the pressure at bottom by keeping adding on pressure

increments down through the layered fluid

① Piezometer: the simplest type of manometer which consists of a vertical tube, open at the top

and attached to the container in which the pressure is desired. If we know atmospheric pressure,

we can determine the ambient pressure based on the specific weight and the height of the fluid.

♠ There are two types of pressures;

1. Gauge pressure: The pressure relative to the local atmospheric or ambient pressure

2. Absolute pressure: The pressure against a perfect vacuum, which is equal to gauge

pressure plus atmospheric pressure.

② U-tube meter: consists of a glass tube bent into the shape of a U uses to some unknown

gauge pressure. The chamber fluid 𝜌1 is separated from the atmosphere by another, heavier fluid

𝜌2, perhaps because fluid A is corrosive, or more likely because a heavier fluid 𝜌2 will keep 𝑧2

small and the open tube can be shorter.

We firstly apply the hydrostatic formula from A down to 𝑧1. Note that we can then go down to

the bottom of the U-tube and back up on the right side to 𝑧1, and then pressure will be the same.

Thus, we can ‘jump across’ and then up to level 𝑧2.

𝑝𝐴 + 𝛾1|𝑧𝐴 − 𝑧1| − 𝛾2|𝑧1 − 𝑧2| = 𝑝2 ≈ 𝑝𝑎𝑡𝑚

The reason why we can jump across to the other side: 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)

■ Hydrostatic forces on plane surface

When a surface is submerged in a fluid, forces develop on the surface due to the fluid. The

determination of these forces is important in the design of storage tanks, ships, dams, and other

hydraulic structures.

With considering a container with a flat, horizontal bottom of area Ab and depth H, this plane

surface will experience a downward force Fb which is equal to γHAb

𝐹𝑏 = 𝛾𝐻𝐴𝑏

In those problems, if we neglect density changes in the fluid, the previous hydrostatic equation

for liquid 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 strength-of material theory. The hydrostatic problem thus reduces to

simple formulas involving the centroid and moments of inertia of the plate cross-sectional

area.

The picture shows a plane panel of arbitrary shape completely submerged in a liquid. The panel

plane makes an arbitrary angle 𝜃 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, the pressure there is

𝑝 = 𝑝𝑎 + 𝛾ℎ.

Figure 3. Hydrostatic forces on plane surface immerged on a fluid (White, 2012, Fluid Mechanics)

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 𝜉 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 + 𝛾ℎ)𝑑𝐴 = 𝑝𝑎𝐴 + 𝛾 ∫ ℎ 𝑑𝐴

Because

ℎ = 𝜉 𝑠𝑖𝑛 𝜃

and by the definition of the centroidal slant distance from the surface to the plate

𝜉𝐶𝐺 =1

𝐴∫ 𝜉𝑑𝐴

The above equation can be re-express by the following equation

𝐹 = 𝑝𝑎𝐴 + 𝛾 𝑠𝑖𝑛 𝜃 ∫ 𝜉 𝑑𝐴 = 𝑝𝑎𝐴 + 𝛾 𝑠𝑖𝑛 𝜃 𝜉𝐶𝐺𝐴

Finally, unravel this by noticing that

𝜉𝐶𝐺 sin 𝜃 = ℎ𝐶𝐺

This indicates that the depth straight down from the surface to the plate centroid.

Therefore, the force acting on the surface of plate submerged in a fluid is

𝐹 = 𝑝𝑎𝐴 + 𝛾ℎ𝐶𝐺𝐴 = (𝑝𝑎 + 𝛾ℎ𝐶𝐺)𝐴 = 𝑝𝐶𝐺𝐴

This equation suggests that 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 𝜽 at which it is slanted.

In other words, the equation prove that combined compression and bending of a beam of the

same cross section 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.

▶ Center of the pressure

However, to balance the bending-moment portion of the stress, the resultant force F acts not

through the centroid but below it toward the high-pressure side. Its line of action passes through

the center of pressure CP of the plate. To fine the coordinates (xcp, ycp), we sum momnets of the

elemental force 𝑝𝑑𝐴 about the centroid and equate to the moment of the resultant F.

To compute ycp, we equate

𝐹𝑦𝐶𝑃 = ∫ 𝑦𝑝 𝑑𝐴 = ∫ 𝑦(𝑝𝑎 + 𝛾𝜉 𝑠𝑖𝑛 𝜃)𝑑𝐴 = 𝛾 sin 𝜃 ∫ 𝑦𝜉 𝑑𝐴

The term ∫ 𝑝𝑎𝑦𝑑𝐴 vanishes by definition of centroidal axis. Introducing 𝜉 = 𝜉𝐶𝐺 − 𝑦, we obtain,

𝐹𝑦𝐶𝑃 = 𝛾 𝑠𝑖𝑛 𝜃 (𝜉𝐶𝐺 ∫ 𝑦𝑑𝐴 − ∫ 𝑦2𝑑𝐴) = −𝛾 sin 𝜃 𝐼𝑥𝑥

∫ 𝑦𝑑𝐴 = 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. Thus, ycp can be calculated by the following equation.

𝑦𝐶𝑃 = −𝛾 sin 𝜃𝐼𝑥𝑥

𝑝𝐶𝐺𝐴

The negative sign in the equation show that ycp is below the centroid at a deeper level and, unlike,

F, depends on angle 𝜃. IF we move the plate deeper, ycp approaches the centroid because every

term in the above equation remains constant except pCG, which increases.

xCP can be obtainable with the exactly the same method.

𝐹𝑥𝐶𝑃 = ∫ 𝑥𝑝 𝑑𝐴 = ∫ 𝑥(𝑝𝑎 + 𝛾(𝜉𝐶𝐺 − 𝑦) 𝑠𝑖𝑛 𝜃)𝐴 = −𝛾 sin 𝜃 ∫ 𝑥𝑦 𝑑𝐴 = −𝛾 sin 𝜃 𝐼𝑥𝑦

𝑥𝐶𝑃 = −𝛾 sin 𝜃𝐼𝑥𝑦

𝑝𝐶𝐺𝐴

If 𝐼𝑥𝑦 = 0, usually implying symmetry, xcp=0 and the center of pressure lies directly below the

centroid on the y axis.

♠ Area moments of inertia (=second moment of area): is a property of a cross-section that can

be used to predict the resistance of a beam to bending and deflection around an axis that lies in

the cross-sectional plane. The stress in, and deflection of, a beam under load depends not only on

the load but also on the geometry of the beam's cross-section: larger values of second moment

cause smaller values of stress and deflection

Figure 4. Area moments of inertia (White, 2012, Fluid Mechanics)

▶ Gauge pressure formulas

In most cases, the ambient pressure 𝑝𝑎 is neglected because it acts on both sides of the plate. In

this case 𝑝𝐶𝐺 = 𝛾ℎ𝐶𝐺 , and the center of pressure becomes independent of specific weight.

𝐹 = 𝛾ℎ𝐶𝐺𝐴, 𝑦𝐶𝑃 =𝐼𝑥𝑥 sin 𝜃

ℎ𝐶𝐺𝐴, 𝑥𝐶𝑃 = −

𝐼𝑥𝑦 sin 𝜃

ℎ𝐶𝐺𝐴

The above picture gives the area and moments of inertia of several common cross sections for

use wth these formulas.

■ Hydrostatic forces on curved surface

Previous equations were developed for plane surfaces. However, many surface of interest such as

pipes and tanks are non-planar.

In the hydrostatic forces on curved surface, the resultant pressure force on a curved surface is

most easily computed by separating it into horizontal and vertical components. 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. 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.

FH vanished because it acts in both side, while Fv can be calculated by the following equation that

sums the weights of fluids in the space of interest.

𝐹𝑉 = 𝑊1 + 𝑊2 + 𝑊𝑎𝑖𝑟

If FH and F’H are different, the magnitude of the resultant is obtained by the following equation.

𝐹𝑅 = √(𝐹𝐻)2 + (𝐹𝑉)2

♠ Pressure distribution around fish

A fish’s body is designed such that its eye is located

near the zero-pressure point so that its vision is not

distorted while it swims.

1. Fish eye is located at zero-pressure point →

Fish can swim at any speed without distorting

its vision.

2. The back of the gills is located near the

aerodynamic shoulder (where Cp is smallest), so that the suction pressure there helps the

fish to exhale

3. The heart is also located near this lowest-pressure point to increase the heart’s stroke

volume during rapid swimming

■ Buoyancy

When a body is completely submerged in a fluid, or floating so that it is only partially submerged,

the resultant fluid force acting on the body is called the buoyant force. A net upward vertical

force results because pressure increases with depth and the pressure forces acting from below are

larger than the pressure forces acting from above. In short, buoyancy is defined as a force exerted

by a fluid that opposes an objects weight.

♠ Buoyant force of water is greater than weight of iceberg

♠ Archimedes deduced the principles of buoyancy when he realized how light he was when

sitting in a bathtub

▶ Archimedes’ laws of buoyancy

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

Buoyant force is given by the Archimedes’ principle, saying the force is proportional to specific

weight and volume of the body.

𝐹𝐵 = 𝛾𝑉

The equation can be easily rearranged by Archimedes’ two laws, that is, the buoyant force is the

difference between fluid weight above 2 and above 1, which is equal to the weight of the fluid

equivalent to body volume.

𝐹𝐵 = 𝐹𝑉(2) − 𝐹𝑉(1)

Alternatively, we can sum the vertical forces on elemental vertical slices through the immersed

body.

𝐹𝐵 = ∫ (𝑝2 − 𝑝1)𝑏𝑜𝑑𝑦

𝑑𝐴𝐻 = −𝛾 ∫(𝑧2 − 𝑧1) 𝑑𝐴𝐻 = 𝛾𝑉

This is identical results and equivalent to Archimedes’ law 1. This equation assumes that the fluid

has uniform specific weight.

▶ Center of buoyancy

The buoyant force passes through the centroid of the displaced volume and the point through

which the buoyant force acts is called the center of buoyancy. In the picture, the center of

buoyancy is the point through which FB acts (normally labeled B or CB on a drawing). The point B

may or may not correspond to the actual center mass of the body’s own material, which may

have variable density.

For floating bodies, only a portion of the body is submerged, with the remainder poking up out

of the free surface. In this case, the equation is modified to apply the volume corresponding to

the submerged portion.

𝐹𝐵 = 𝛾(displaced volume) = floating body weight

This state is mathematically equivalent of Archimedes’ law 2.

♠ The picture of bottle shows the submerged or floating bodies with respect to weight of the

body and buoyancy.

▶ Stability

Stability considerations are particularly important for submerged or floating bodies since the

centers of buoyancy and gravity do not necessarily coincide.

Stable equilibrium means if when displaced, it returns to its equilibrium position. In contrast,

unstable equilibrium indicates if when displaced, it moves to a new equilibrium position.

예시) 무게중심이 부력중심 보다 아래에 있는 완전히 잠겨있는 물체에 대해서 평형위치로부터

회전 시키면 무게와 부력에 의해 형성되는 복원 짝힘이 생겨나는데, 이것이 물체를 반대로

회전시켜 원래 위치로 돌아가게 한다 – 안정하다.

무게중심이 부력중심보다 위쪽에 있는 완전히 잠긴 물체는 무게와 부력에 의하여 형성되는

힘은 물체가 뒤집혀서 새로운 평형 위치로 옮아가게 만든다 – 불안정하다.

■ Pressure distribution in rigid-body motion

In rigid-body motion, all particles are in combined translation and rotation, and there is no

relative motion between particles. With no relative motion, there are not shear strains or strain

rates, so that the viscous term vanishes, leaving a balance between pressure, gravity and particle

acceleration.

−𝛻𝑝 + 𝜌𝑔 = 𝜌𝑎

▶ Uniform linear acceleration

In uniform rigid-body acceleration the below equation applies a having the same magnitude and

direction for all particles.

예시) 유체가 고정된 축에 관하여 회전하는 탱크 안에 담겨있다면 유체는 탱크와 함께

강체처럼 단순 회전할 것이다.

The pressure gradient acts in the direction 𝑔 − 𝑎, and lines of constant pressure are perpendicular

to this direction.

𝛻𝑝 = 𝜌(𝑔 − 𝑎)

The surface of constant pressure must be perpendicular to this and are thus tilted at a down ward

angle 𝜃.

𝜃 = tan−1𝑎𝑥

𝑔 + 𝑎𝑧

The parallelogram sum of 𝑔 and – 𝑎 gives the direction of the pressure gradient or greatest rate

of increase of p.

𝑑𝑝

𝑑𝑠= 𝜌𝐺, 𝑤ℎ𝑒𝑟𝑒 𝐺 = √𝑎𝑥

2 + (𝑔 + 𝑎𝑧)2

The resultants are independent of the size or shape of the container as long as the fluid is

continuously connected throughout the container

Figure 5. Pressure distribution in uniform linear acceleration (White, 2012, Fluid Mechanics)

▶ Rigid-body rotation

Figure 6. Pressure distribution in rigid-body rotation (White, 2012, Fluid Mechanics)

Consider rotation of the fluid about the z axis without any translation with rotational velocity of Ω

which is constant for the fluid to have attained rigid-body rotation. In this case, the fluid

acceleration will be a centripetal term. With force balance, pressure gradient can be expressed as

the sum of pressure distributions along with radius (𝑖) and elevation (𝑧) is equal to the sum of

gravitational and angular accelerations.

∇𝑝 = 𝑖𝑟

𝜕𝑝

𝜕𝑟+ 𝑘

𝜕𝑝

𝜕𝑧= 𝜌(𝑔 − 𝑎) = 𝜌(−𝑔𝑘 + 𝑟Ω2𝑖𝑟)

Equating like component, it is possible to solve the problem by solving two first-order partial

differential equations.

𝜕𝑝

𝜕𝑟= −𝜌𝑎𝑟 = 𝜌Ω2𝑟,

𝜕𝑝

𝜕𝑧= −𝛾

The equation can be solved by following method. Integrate the first equation partially, holding z

constant with respect to r.

𝑝 =1

2𝜌𝑟2𝛺2 + 𝑓(𝑧)

Now differentiate this with respect to z and compare with the second equation.

𝜕𝑝

𝜕𝑧= 0 + 𝑓′(𝑧) = −𝛾

𝑓(𝑧) = −𝛾𝑧 + 𝐶

Therefore, the pressure distribution can be written as follows.

𝑝 = 𝐶 − 𝛾𝑧 +1

2𝜌𝑟2𝛺2

With the boundary condition

𝑝 = 𝑝0 𝑎𝑡 (𝑟, 𝑧) = (0,0) 𝑡ℎ𝑒𝑛 𝐶 = 𝑝0

It is possible to get the exact solution of pressure distribution in the rigid-body rotation. The

pressure is linear in z and parabolic in r.

𝑝 = 𝑝0 − 𝛾𝑧 +1

2𝜌𝑟2𝛺2

𝑧 =𝑝0 − 𝑝

𝛾+

𝑟2𝛺2

2𝑔

The equation shows that the surfaces are paraboloids of revolution, concave upward, with their

minimum points on the axis of rotation.

Figure 7. Paraboloids of revolution (White, 2012, Fluid Mechanics)

When considering a rotation about its central axis, the minimum points of pressure is the z axis,

resulting p=p0. Thus, the height of the elevation is equal to

z =𝑟2𝛺2

2𝑔