FLUID STATICS

Types of Problems

In fluids at rest, there is no relative motion between fluid particles. Hencethere is no shear stress acting on fluid elements. Fluids, which are at rest,are only able to sustain normal stresses. In fluids undergoing rigid-bodymotion, a fluid particle retains its identity and there is no relative motionbetween the particles. Hence, in fluids undergoing rigid-body motion,only stress component present is the normal stress as in the stationaryfluids.

ObjectivesIn this chapter, an expression for the pressure distribution in a stationarybody of fluid will be derived, and the pressure forces acting onsubmerged surfaces will be studied.

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THE BASIC EQUATION OF FLUID STATICS

Our primary objective is to obtain an equation that will enable us to determine the pressure field within the stationary fluid.

Consider a differential element of mass dm, with sides dx, dy, and dz. The fluid element is stationary relative to stationary coordinate system.

Two types of force may be acting on the fluid element.- body force → gravitational force- surface force → pressure force

The force acting on fluid element is sum of the body and surfaceforces,

(1)Body force can be expressed as,

(2)

sB FdFdFd

gdxdydzgdgdmFd B

For a fluid particle, Newton’s second law of motion gives

adadmFd

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Surface ForceLet the pressure at the center O, of the element be P(x,y,z,t). To determine thepressure at each of the six faces of the element, we use Taylor series expansionabout the point O. The pressure at the left face of the differential element is

22)(

dy

y

pp

dy

y

ppyy

y

ppp LL

2)(

dy

y

ppyy

y

ppp RR

Similarly, at the right face,

Pressure forces on the other faces of the element are obtained in the same way.Combining all such forces gives the net surface force acting on the element

dxdydzkz

pj

y

pi

x

p

dxdydzkz

pj

y

pi

x

p

kdxdydzz

pjdxdzdy

y

pidydzdx

x

pFd S

The term in parentheses is called the pressure gradient and can be writtenas gradP or P. In rectangular coordinate system,

k

z

pj

y

pi

x

pPPgrad

PdxdydzdxdydzPgradFd S

(3)

dxdydz

FdPPgrad S

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Force acting in x-direction is obtained as

𝑑𝐹𝑆𝑦 = 𝑃 −𝜕𝑃

𝜕𝑦

𝑑𝑦

2𝑑𝑥𝑑𝑧 − 𝑃 +

𝜕𝑃

𝜕𝑦

𝑑𝑦

2𝑑𝑥𝑑𝑧 = −

𝜕𝑃

𝜕𝑦𝑑𝑦 𝑑𝑥𝑑𝑧

4

Physically, the gradient of pressure is negative to the surface force perunit volume due to the pressure. We note that the level of pressure isnot important in evaluating the net pressure force. Instead, whatmatters is the rate at which pressure changes occur with distance, thepressure gradient.

Combining equations (2) and (3) in Eq. (1)

dxdydzgPgrad

FdFdFd BS

Or on unit volume base

For a fluid particle, Newton’s second law of motion gives

For a static fluid, the acceleration is zero. Thus,

0 dxdydzgPgradFd

0 gPgrad

body force per unit volume at a point

pressure force per unit volume at a point

Components of this vector equation are

x-comp. …….

adadmFd

a

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Above equations describe the pressure variation in each of the threecoordinate directions in a static fluid. To simplify further, it is logical tochoose a coordinate system such that the gravity vector is aligned withone of the axes. If the coordinate system is chosen such that z-axis isdirected vertically, then and equationsbecome:

ggandgg zyx 0,0

gdz

dp

dz

dp

z

p

x-comp

or

dz

dp(4) Basic equation of fluid statics

Note: The pressure does not vary in a horizontal direction. The pressure increases if we go down and decreases if we go up in the liquid.

Objective 2

Pressure Variation in Stationary Fluids

- Constant density fluids- Variable density fluids

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PRESSURE VARIATION IN A CONSTANT-DENSITY FLUID

If the density of the fluid is constant, we can easily integrate Eq. (4) to get an expression for pressure distribution.

z0

y

z

x

free surface

g

An expression for pressure distribution can be obtained by solving the basic equation of fluid statics as follows:

For liquids, it is often convenient to take the origin of the coordinatesystem at the free surface, and measure the distance as positivedownward from the free surface. With h measured positive downward,then

hzz 0

ghpp 0is called hydrostatic pressure

where po is the pressure at the free surface of the liquid.

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ABSOLUTE AND GAGE PRESSURES

Pressure values must be stated with respect to a reference level. Ifthe reference level is a vacuum, pressures are termed as absolute.Pressure levels measured with respect to atmospheric pressure aretermed gage pressure.

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Example: A tank which is exposed to the atmosphere, contains 2 m of water covered with 1 m of oil. The density of water and oil are 1000 kg/m3 and 830 kg/m3, respectively. Find the pressure at the interfaceand at the bottom of the tank. Also determine the pressure distribution at the tank wall. The atmospheric pressure is 101.325 kPa.

y

z

x

oil, o

h

water, w

ho=1 m

hw=2 m

Solution:

Find:

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Example: Water flows through pipes A and B. Oil, with specific gravity 0.8, is in theupper portion of the inverted U. Mercury (specific gravity 13.6) is in the bottom of themanometer bends. Determine the pressure difference, PA-PB.

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Pressure Variation in a Variable-Density Fluid

If the density is variable, we must relate it to the pressure /or elevation before we can integrate the equation,

gdz

dp

A common case might involve an ideal gas. In such gases, density canbe expressed as a function of pressure and temperature. Pressure anddensity of liquids are related by the bulk compressibility modulus ormodulus of elasticity.

dEdP

d

dpE vv

/

If the bulk modulus is assumed to be a constant, then the density is only a function of the pressure.

From two above expression, expression for pressure distribution in liquids is obtained as follows:

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Example: The pressure, temperature and density of standard atmosphereat the sea level are 101.325 kPa, 15.2 C, and 1.225 kg/m3, respectively.Calculate the percent error introduced into the elevation of 8 km, byassuming the atmosphere as,

a) to be incompressibleb) to be isothermalc) to be isentropicd) linearly decreasing temperature with rate of -0.0065 K/m.

The actual pressure at an elevation of 8 km is known to be 35.656 kPa. The gas constant of air is 287 J/kgK.

Solution:a) Incompressible air, =constant

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HYDROSTATIC FORCE ON SUBMERGED SURFACES

When a surface is in contact with a fluid, fluid pressure exerts a force onthe surface. This force is distributed over the surface; however, it’s oftenhelpful in engineering calculations to replace the distributed force by asingle resultant. To completely specify the resultant force, we mustdetermine its magnitude, direction and point of application.

Types of problems:

We shall consider both plane and curved submerged surfaces.

1. HYDROSTATIC FORCE ON A PLANE SUBMERGED SURFACE

Magnitude of resultant force ?RF

Point of application ??,' yx

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Force acting on surface Ad

ApdFd

Minus sign indicates that force acts against the surface

The resultant force acting on the whole surface is found by summing(integrating) the contribution of the infinitesimal forces over the entire area.

Thus, A

R ApdF

…………………………. (1)

In order to calculate the integral, both pressure, p, and the area elementdA must be expressed in terms of the same variables. The basicpressure-height relation for a static fluid can be written as

gdh

dp h is measured positive downward from the liquid

free surface.

p0 is the pressure at liquid free surface (h=0)

ghppgdhdph

h

p

p

0

00

……………………….. (2)

This expression can be substituted into Eq. (1). Then to perform integration,h and dA should be expressed in terms of x and/or y. (Ex: h = y Sinq , q =constant). Integration of Eq. 1 gives the resultant force due to the distributedpressure force.

The point of application of the resultant force must be such that the momentof the resultant force about any axis is equal to the moment of thedistributed force about the same axis.

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Let be r’ the position vector of the point of application of the resultant force FR and r be the position vector of any point on the surface A.

A

R

R

APdrFr

FdrFr

dAkAd

kFF

jyıxr

jyıxr

RR

According to the coordinate system used,

A

R kPdAjyıxkFjyıx

)()()(

Evaluating the cross product, we obtain,

A

RR dAıyPjxPıFyjFx )(

Considering the components of this vector equation, we obtain

surfacethetonormalisFofDirection

FofMagnitude:

1

1

R

R

A

RR

ARA

R

ARA

R

pdAFFNOTE

xpdAF

xxpdAFx

ypdAF

yypdAFy

Moment of resultant force = Moment of distributed force

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Example: The inclined surface shown, hinged along A, is 5 m wide.Determine the resultant force FR of the water on the inclined surface.

Fd

z

y

w = 5 m

Solution

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ALTERNATIVE APPROACH FOR CALCULATION OF HYDROSTIC FORCE(ALGEBRAIC EQUATIONS)

Now we will formulate an approach to determine the resultant hydrostatic force andcoordinates of its point of application. Consider the expressions developed before, i. e.

A

R ApdF

Considering that the free surface is open to atmosphere, the magnitude of the resultant force can be written as

AghAygydAgdAgyghdAF cc

AAA

R qqq sinsinsin

NOTE: is the first moment of the area with respect to the x-axis.

Where yc is the y coordinate of the centroid of the area A measure from the x axis, which passes through O, and ycsinq=hc.

hc is the vertical distance from the fluid surface to the centroid of the area.

AyydA c

A

Note: Origin of the coordinate system is placed at the intersection of the plane of the gate and the free surface.

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Point of Application of the Resultant Force

Expressions for the coordinates of the point of application of the resultant force can beobtained by equating the moment of the resultant force to the moment of thedistributed pressure force.

AA

RR dAygydFyF 2sinq

Ay

dAydAyg

AgydAyg

Fyy

c

A

AcAR

R

2

22 sinsin

1sin

1' q

qq

xA

IdAy 2

is the second moment of the area (moment of inertia), with respect to an axis formed by the intersection of the plane containing the surface and the free surface (x axis). Thus, we can write

Ay

Iyy

c

xR ' Using parallel axis theorem 2

cxcx AyII

where Ixc is the second moment of the area with respect to an axis passing through its centroid and parallel to the x axis. Thus,

c

c

xcR y

Ay

Iyy '

The x coordinate, xR, for the point of application of the resultant force can be determined in a similar manner as follows:

c

c

xyc

R xAy

Ixx '

where Ixyc is the product of inertia with respect to an orthogonal coordinate system passing through the centroid of the area. The point through which the resultant force acts is called the center of pressure.

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Geometric properties of some common shapes

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Example: Solve the previous example using the algebraic equations method.

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PRESSURE PRISM METHOD

The concept of the pressure prism provides another tool for determiningthe magnitude and point of application of the resultant force on asubmerged plane surface.

gh

y

gh1

gh2

z

y

h2 h1h

x

dA

Considering the gage pressure at the free surface is zero, the infinitesimal pressure force, dFR, acting on the submerged plane surface is,

kdkghdAkPdAFd P

where dA and gh are infinitesimal base area and imaginary height of thepressure prism, respectively. Thus, product of dA and gh represents theinfinitesimal volume dVP of the pressure prism. After integration, themagnitude of the resultant force may be obtained as,

kdkF PPR

P

P is the volume of the prism.

Therefore, the magnitude of the resultant force acting on a submerged plane surface is equal to the volume of the pressure prism.

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Point of application of the resultant force,

P

P

G

PAPAR

G

PAPAR

YydghdAyyPdAF

y

XxdghdAxxPdAF

x

111

and

111

where XG and YG are the coordinates of the centroid of the pressure prism.

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Example: Solve the previous example using the pressure prism method.

g(D+Lsin30)

Dgwater

L=4m

D+Lsinq

D=2m

30

Solution:

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HYDROSTATIC FORCE ON CURVED SUBMERGED SURFACES

Consider the infinitesimal curved surface element shown in figure. Thehydrostatic force on an infinitesimal element of a curved surface, ,acts normal to the surface. However, the differential pressure force oneach element of the surface acts in a different direction because of thesurface curvature

Usually, to sum a series of force vectors acting in different directions, we sum the components of the vectors relative to a convenient system.

The pressure force acting on area element isAd

ApdFd

The resultant force is A

R ApdF

can be written as RF

zyxRRRR FkFjFıF

Where are components of in x, y and z directions.

zyxRRR FFF and, RF

Ad

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xR

R

pdAF

ıApdıF

x

To evaluate the component of the force in a given direction, we take thedot product of the force with the unit vector in the given direction. Forexample, taking the dot product of each side of the above equation withunit vector i gives

In general, magnitude of the component of the resultant force in the ldirection is given by

l

l

A

lR pdAF

where dAl is the projection of the area element on a plane perpendicular to l-direction.

The line of action of each component of the resultant force is found byrecognizing that the moment of the resultant force component about agiven axis must be equal to the moment of the corresponding distributedforce component about the same axis.

Because we are dealing with a curved surface, the lines of action of thecomponents of the resultant force will not necessarily coincide; thecomplete resultant may not be expressed as a single force.

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Example: An open tank which is shown in the figure is filled with an incompressible fluidof density, . Determine the magnitudes and lines of action of the vertical and horizontalcomponents of the resultant pressure force on the curved part of the tank bottom.

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ALTERNATIVE APPROACH FOR CALCULATION OF RESULTANT FORCE ACTING ON CURVED SURFACES

The resultant fluid force acting on a curved submerged surface can be determined byintegration as in the above example. This is generally a rather tedious process, and nosimple general formulas can be developed. As an alternative approach we willconsider the equilibrium of the fluid volume enclosed by the curved surface ofinterest and the horizontal and vertical projections of this surface.

Consider the section BC shown in the figure above. This section has a unit length perpendicular to the plane of the paper.

- We first isolate a volume of fluid that is bounded by the surface of interest, in this instance section BC, and the horizontal plane surface AB and the vertical plane surface AC.

- Draw the free-body diagram for this volume as shown in Fig. c. - The magnitude and location of forces F1 and F2 can be determined from the

relationships for planar surfaces. - The weight, W, is simply weight of the fluid in the enclosed volume. - Forces FH and FV represent the components of the force that the tank exerts on

the fluid.

- From the force balance, we can obtain FH and FV as follow:

WFFFF VH 12

The resultant force of the fluid acting on the curved surface BC is equal and opposite in direction to that obtained from the free-body diagram.

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Example: Solve the previous example using the second method.

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BUOYANCY

When a body is either fully or partially submerged in a fluid, a net force called thebuoyant force acts on the body. This force is caused by the difference between thepressure on the upper and lower surface of body.

Consider the object shown in the figure immersed in a static fluid. We want tocalculate the net vertical force that pressure exerts on the body.

gddAhhgdAghpdAghpdF

d

z )()()( 121020

Thus the net vertical force on the body is

ggddFF zz

where is the volume of the object.

Thus the net vertical pressure force, or buoyancy force, equals the forceof gravity on the liquid displaced by the object. This relation wasreportedly used by Archimedes in 220 B.C., it is often called ‘ArchimedesPrinciple’.

The line of action of the buoyancy force may be found using the methods

that used in the previous section.

xdg

xdFF

XB

B

11

Note: The line of action of the buoyant force passes through thecentroid of the displaced volume. This centroid is called the center ofbuoyancy.

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Stability of Submerged and Floating Bodies

The location of the line of action of the buoyancy force and the line ofaction of the force due to gravity determines the stability.

CG

W

C

FB

CG

W

C’

FB

Barge

Stable

CG

W

C

FB

CG

W

C’

FB

Slender

Body

Unstable

overturningcouple

C: centroid of original displaced volumeC’: centroid of new displaced volume

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FLUIDS IN RIGID BODY MOTION

A fluid in rigid body motion moves without deformation as though it werea solid body. Since there is no deformation, there can be no shear stress.Consequently, the only surface force on each fluid element is that due topressure. Hence, the force acting on a fluid element in rigid body motion isthe same as in the case of static fluid, i.e.

dgpgradFd )(

or force on a fluid element of unit volume

gpgradd

Fd

Using Newton’s second law, we can write

dmaFd

agpgrad

The physical significance of each term in this equation is

particle

fluidof

onaccelerati

volumeunitper

mass

poaat

volumeunitper

forcebody

poaat

volumeunitper

forcepressure

agpgrad

intint

From the above vector equation, following scalar equations can be written

zz

yy

xx

agz

p

agy

p

agx

p

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Example: An open tank is used to transport liquid as shown in the figure.What should be the maximum height of the liquid in tank to be sure that itwill not spill over during the trip?

d=?

Solution

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FLUID ROTATING ABOUT A VERTICAL AXIS

A cylindrical container, partially filled with liquid, is rotated at a constant angularvelocity , about its axis.

After a short time, there is no relative motion; the liquid rotates with the cylinderas if the system were a rigid body. Determine the shape of the free surface.

P = ?Expression for free surface = ?

Writing Newton’s second law, we get,

agpgrad

Scalar components in cylindrical coordinate system can be written as,

gz

p

a

gg

agz

p

rz

z

zz

r

0

rr

p

ra

g

agr

p

r

r

rr

2

2

0

0

0

0

1

q

qq

q

q

q

p

a

g

agp

r

Therefore, P = P(r,z)

NOTE: The same expressions can also be obtained by applying Newton’s second law in each direction to a suitable differential fluid element.

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dzz

pdr

r

pdp

rz

P = P(r,z). Using chain rule, total change in pressure at any point canbe written as,

Substituting expressions for dp/dr and dp/dz, we get,

gdzrdrdp 2

To obtain the pressure difference between a reference point (r1,z1),where the pressure is P1, and arbitrary point (r,z), where the pressure isP, we must integrate

)()(2

)( 1

2

1

22

1

2

111

zzgrrpp

gdzrdrdp

z

z

r

r

p

p

Taking the reference point on the cylinder axis at the free surface gives

1111 ,0, hzrpp atm

Then, we get,

)(2

1

22

hzgr

pp atm

)(2

1

22

hzgr

pp atm

Solving for p, we get

h1 = ?

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Since the free surface is a surface of constant pressure (p=patm), the equation of the free surface is obtained as

)ataxistheonvertexwithparabola(

surface.freetheofEquation2

)()(

20

1

2

11

22

hz

g

rhzhzg

r

We can solve for the height h1 in terms of the original liquid height ho

and tank radius R. To do this, we use the fact that the volume of the fluidmust remain constant, i.e.

Volume of liquid with no rotation = Volume of liquid with rotation

02hRWith no rotation,

RR z

rzdrrdzdr00 0

22 With rotation,

g

RRh

g

rrh

rdrg

rh

R

R

4822

22

422

1

0

422

1

0

22

1

Then equating these two expression, we get

g

RRhhR

4

422

10

2

g

r

g

Rhz

2

)(

4

)( 22

0

Finally, substituting into above equation, we obtain the equation of the free surface as

22

02

1

2

)(

R

r

g

Rhz

Equation of the free surface

Solving for h1,

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ℎ1 = ℎ0 −𝜔𝑅 2

4𝑔