MM301 2 Fluid Statics-Updated

8/13/2019 MM301 2 Fluid Statics-Updated

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FLUID STATICS

Types of Problem

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 motiononly stress component present is the normal stress as in the stationaryfluids.

ObjectivesIn this chapter, an expression for the pressure distribution in a stationary

body 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 todetermine 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 surface

forces,

(1)

Body force can be expressed as,

(2)

s B F d F d F d

g dxdydz g d g dm F d B


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Surface ForceLet the pressure at the center O, of the element be P(x,y,z,t). Todetermine the pressure at each of the six forces of the element, we use

Taylor series expansion about the point O. The pressure at the left face ofthe differential element is

22)(

dy y p

pdy

y p

p y y y p

p p L L

2)(

dy

y

p p y y

y

p p p

R R

Similarly, at the right face,

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

dxdydz k z p

j y p

i x p

dxdydz k z p

j y p

i x p

k dxdydz z p

jdxdz dy y p

idydz dx x p

F d S

The term in parentheses is called the gradient of the pressure is simplypressure gradient and can be written gradP or P. In rectangular coordinatesystem,

k z p

j y p

i x p

P P grad

Pdxdydz dxdydz P grad F d S

(3)

dxdydz F d

P P grad S


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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, the

pressure gradient.

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

dxdydz g P grad F d F d F d BS

Or on unit volume baseFor a fluid particle , Newtons second law of motion gives

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

0 dxdydz g P grad F d or 0 g P grad

body force per unit volume at a point

pressure force per unit volume at a point

Components of this vector equation are

x-comp . .

ad adm F d 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 is

directed vertically, then and equationsbecome:

g g and g g z y x 0,0

g

dz

dpdz 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 pressureincreases 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 integrateEq.(4) to give

z0

y

z

x

free surface

g

An expression for pressure distribution can be obtained by solving thebasic 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

h z z 0 gh p p 0 is 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 ofwater covered with 1 m of oil. The density of water and oil are 1000kg/m 3 and 830 kg/m 3, respectively. Find the pressure at the interfaceand at the bottom of the tank. Also determine the pressure distributionat the tank. 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 the upper portion of the inverted U. Mercury (specific gravity 13.6) isin the bottom of the manometer bends. Determine the pressuredifference, P A-PB.


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Pressure Variation in a Variable-Density Fluid

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

g dz dp

A common case might involve an ideal gas . In such gases,density can be expressed as a function of pressure andtemperature. Pressure and density of liquids are related by thebulk compressibility modulus or modulus of elasticity .

d

E dP d

dp E vv /

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

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


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Example: The pressure, temperature and density of standardatmosphere at the sea level are 101.325 kPa, 15.2 C, and 1.225kg/m 3, respectively. Calculate the percent error introduced into theelevation of 8 km, by assuming 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 35.656kPa. The gas constant of air is 287 J/kgK.

Solutiona) 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 forceon the surface. This force is distributed over the surface; however, its often helpful in engineering calculations to replace the distributedforce by a single resultant. To completely specify the resultant forcewe must determine 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 ? R F Point of application ??,' y x


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Force acting on surface Ad

A pd F d 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 entirearea.

Thus, A

R A pd F . (1)

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

g dhdp

h is measured positive downward from the liquidfree surface.

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

gh p p gdhdph

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 thedistributed pressure force.

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


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

A R

R

A Pd r F r

F d r F r

dAk Ad

k F F j y xr

j y xr

R R

According to the coordinate system used ,

A

R k PdA j y xk F j y x )()()(

Evaluating the cross product, we obtain,

A R R dA yP j xP F y j F x )(

Considering the components of this vector equation, we obtain

surfacethetonormalisFof Direction

Fof Magnitude:

1

1

R

R A

R R

A R A R

A R A R

pdA F F NOTE

xpdA F

x xpdA F x

ypdA F

y ypdA F y

Moment of resultant force = Moment of distributed force


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

F d

z

y

w = 5 m

Solution


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ALTERNATIVE APPROACH FOR CALCULATION OF HYDROSTIC FORCE

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

A R A pd F

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

A gh A y g ydA g dA gy ghdA F cc A A A

R q q q 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 ycsin =hc.

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

A y ydA c A


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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 beobtaine by equating the moment of the resultant force to the moment of thedistributed pressure force.

A A

R R dA y g ydF y F 2sin q

A y

dA ydA y g

A gydA y g

F y

c

A

Ac A R

R 2

22 sinsin

1sin

1q

q q

x A I dA y 2 is the second moment of the area (moment of inertia ), with

respect to an axis formed by the intersection of the planecontaining the surface and the free surface ( x axis). Thus, we canwrite

A y

I y

c

x

R Using parallel axis theorem 2c xc x Ay I I

where I xc is the second moment of the area with respect to an axis passingthrough its centroid and parallel to the x axis. Thus,

cc

xc R y A y

I y

The x coordinate, x R, for the resultant force can be determined in a similar manner asfollows

cc

xyc R x A y

I x

where I xyc is the product of inertia with respect to an orthogonal coordinate systempassing through the centroid of the area . The point through which the resultant forceacts is called the center of pressure.


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Geometric properties of some common shapes


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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 a

submerged plane surface .

gh

y

gh1

gh2

z

y

h2 h1 h

x

dA

Considering the gage pressure at the free surface is zero, the infinitesimalpressure force, dF R, acting on the submerged plane surface is,

k d k ghdAk PdA F d P

where dA and gh are infinitesimal base area and imaginary height of thepressure prism, respectively. Thus, product of dA and gh represents the

infinitesimal volume dV P of the pressure prism. After integration, themagnitude of the resultant force may be obtained as,

k d k F P P R P

P is the volume of the prism.

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


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Point of application of the resultant force,

P

P

G P A P A R

G

P A P A R

Y yd ghdA y yPdA F

y

X xd ghdA x xPdA

F

x

111

and

111

where X G and Y G are the coordinates of the centroid of the pressureprism .


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

g(D+Lsin30)

D gwater

L=4m

D+Lsin q 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 is Ad

A pd F d

The resultant force is A

R A pd F

can be written as R F z y x R R R R F k F j F F

Where are components of in x, y and zdirections.

z y x R R R F F F and, R F

Ad


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x R

R

pdA F

A pd F

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 l direction is given by

l

l

Al R pdA F

where dA l is the projection of the area element on a plane perpendicular tol-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; the

complete 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 anincompressible fluid of density, . Determine the magnitudes and lines ofaction of the vertical and horizontal components of the resultant pressureforce on the curved part of the tank bottom.


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ALTERNATIVE APPROACH FOR CALCULATION OF RESULTANT FORCEACTING ON CURVED SURFACES

The resultant fluid force acting on a curved submerged surface can be determined byintegration. This is generally a rather tedious process, and no simple general formulascan be developed. As an alternative approach we will consider the equilibrium of thefluid volume enclosed by the curved surface of interest and the horizontal andvertical projections of this surface.

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

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

- Draw the free-body diagram for this volume as shown in Fig. 4 c.- The magnitude and location of forces F 1 and F 2 can be determined from the

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

the fluid.

- Fom the force balance, we can obtain F H and F V as follow:

W F F F F V H 12The resultant force of the fluid acting on the curved surface BC is equal andopposite in direction to that obtained from the free-body diagram.


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BUOYANCY

When a body is either fully or partially submerged in a fluid, a net forcecalled the buoyant force acts on the body. This force is caused by thedifference between the pressure on the upper and lower surface of body.Consider the object shown in the figure immersed in a static fluid. Wewant to calculate the net vertical force that pressure exerts on the body .

gd dAhh g dA gh pdA gh pdF d

z )()()( 121020

Thus the net vertical force on the body is

g gd dF F z z

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 was

reportedly 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 methodsthat used in the previous section .

xd

g

xdF

F

X B

B

11

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


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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.

STABILITY

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 displacedvolumeC: centroid of new displacedvolume


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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 stress on each element of fluid is that due

to pressure. Hence, as in the case of static fluid , the force acting on a fluidelement in rigid body motion is

d g p grad F d )(

or force on a fluid element of unit volume

g p grad d F d

Using Newtons second law, we can write

dma F d

a g p grad

The physical significance f each term in this equation is

particle

fluid of

onaccelerati

volumeunit per

mass

poaat

volumeunit per

forcebody

poaat

volumeunit per

force pressure

a g p grad

intint

From the above vector equation, following scalar equations can bewritten

z z

y y

x x

a g z p

a g y p

a g x p


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Example: An open tank is used to transport liquid. What should be themaximum height of the liquid in tank to be sure that it will not spill overduring 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 constantangular velocity , about its axis.

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

P = ?Expression for free surface = ?

Writing Newtons second law, we get ,

a g p grad

Scalar components in cylindrical coordinate system can be written as,

g z p

a

g g

a g z p

r z

z

z z r

0

r r p

r a

g

a g r p

r

r

r r

2

2

0

0

0

0

1

q

q q

q

q

q

p

a

g

a g p

r

Therefore, P = P(r,z)

NOTE: The same expressions can also be obtained by applying Newtonssecond law in the each direction to a suitable differential element.


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dz z p

dr r p

dpr z

P = P(r,z). Using chain rule, we can write,

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

gdz rdr dp 2

To obtain the pressure difference between a reference point (r1,z

1),

where the pressure is P 1, and arbitrary point (r,z), where the pressure isP, we must integrate

)()(

2

)( 12

12

2

1

2

111

z z g r r p p

gdz rdr dp p

p

r

r

p

p

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

1111 ,0, h z r p p atm

Then, we get,

)(2 1

22

h z g r

p patm

)(2 1

22

h z g r

p p atm

Solving for p, we get

h1 = ?


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Since the free surface is a surface of constant pressure ( p=p atm ), theequation of the free surface is given by

)ataxistheonvertexwith parabola(

surface.freetheof Equation

2

)()(

2

0

1

2

11

22

h z g

r h z h z g

r

We can solve for the height h1 in terms of the original height ho and R. Todo this, we use the fact that the volume of the fluid must remainconstant.

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

02 h R With no rotation,

R R rz

rzdr rdzdr 00 0

22 With rotation ,

g R

Rh g r r

h

rdr

g

r h

R

R

4822

2

2

422

1

0

422

1

0

22

1

Then equating these two expression for volume,

g

R Rhh R

4

422

102

g r

g R

h z 2

)(4

)( 220

Finally solving for z, we get

22

0 21

2)(

Rr

g R

h z Equation of the free surface

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MM301 2 Fluid Statics-Updated