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Chapter 3: Fluid Statics

By

Dr Ali Jawarneh

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Outline

We will discuss:

• The definition of pressure.

• Pascal’s principle.

• The concepts of absolute, gage and vacuum pressures.

• Pressure variation with elevation for uniform density fluids.

3

Outline

• We will become familiar with different pressure measurement techniques and devices:

– Simple and differential manometers

– Bourdon-tube gages

– Pressure Transducers

4

Outline

• become familiar with hydrostatic forces on plane surfaces.

• Calculate the magnitude of the resultant hydrostatic force.

• Locate the line of action of the resultant hydrostatic force.

• Analyse hydrostatic forces on curved surfaces.

• Discuss the principle of buoyancy.

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3.1: Pressure, p

• For fluids at rest, only normal forces exist, which are basically known as pressure forces.

• At every point in a static fluid, a pressure defined as follows exists:

• Pressure is a scalar quantity. It is unit: N/m2

(pa), Ibf/in2 (psi), or Ibf/ft2 (psf)

• Pressure at a point in a static fluid acts with the same magnitude in all directions.

A

Fp

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Pressure is defined as a normal force exerted by a fluid per unit area. We speak of pressure only when

we deal with a gas or a liquid. The counterpart of

pressure in solids is normal stress. Since pressure is

defined as force per unit area, it has the unit of

newtons per square meter (N/m2), which is called a

pascal (Pa).

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Pressure

Considering an element of fluid in

equilibrium:

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Pressure

• The summation of forces in the x-direction:

• The summation of forces in the z-direction:

0)sin(sin)( lyplyp xn

0sincos2

1)cos(cos)( ylllyplyp zn

xn pp

zn pp

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Pressure Transmission• Pascal’s principle states that a

pressure change produced at one point in the system will be transmitted throughout the entire system.

• A consequence of the pressure in a fluid remaining constant in the horizontal direction is that the pressure applied to a confined fluid increases the pressure throughout by the same amount. This is called Pascal’s law, after Blaise Pascal(1623–1662).

• This principle is utilised in the design and development of hydraulic controls used in a wide range of applications.

Pascal’s law states

that increase in

pressure on

the surface of a

confined fluid is

transmitted

undiminished

throughout the

confining vessel or

system

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. This important

result is known as

Pascal’s law named in

honor of

Blaise Pascal

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Pressure Transmission

11

12

Absolute, Gage, and Vacuum Pressure

• Pabs=pgage+ patm

• Pressure values measured with reference

to the atmospheric pressure are referred

to as gage pressures.

• Most pressure measurement devices

measure gage pressure, such as the

Bourdon-tube gage.

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Absolute and Gage Pressure

• Gage pressures can take negative values.

• Negative gage pressures are also referred

to as vacuum pressures.

• If the atmospheric pressure is 101.3 kPa

which is measured at sea level at T=23 oC:

• 50 kPa gage = 151.3 kPa absolute.

• - 50 kPa gage = 51.3 kPa absolute.

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3.2: Pressure Variation with

Elevation

• The general equation

for pressure variation

in the fluid element

shown:

dl

dz

dl

dp

dz

dpThe pressure changes

inversely with elevation. If

one travels upward in the

fluid, the pressure

decreases

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Pressure Variation with Elevation

• Pressure variation occurs only along a

vertical path through the fluid.

• No pressure variation occurs along a

horizontal path through the fluid.

• Pressure changes inversely with elevation;

i.e.: going down increases the pressure

and going up decreases the pressure.

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Pressure Variation with Elevation

• For a fluid with uniform density, integrating the

basic equation results the following:

• This sum is called the piezometric pressure.

• Another form is:

• This sum is called the piezometric head.

constant zp

constant zp

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Pressure Variation with Elevation

• Therefore, the pressure and elevation at

one point can be related to the pressure

and elevation at another point according to

the following:

2211 zpzp

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- Pressure Variation for

Compressible Fluids

• For Ideal gas: p=RT or =p/RT

• Multiply by g: g=pg/RT, then

=pg/RT

• =fn (p, T)

• dp/dz = - = fn (p, T)

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Figure 3.5

Temperature

variation with

altitude for the

U.S. standard

atmosphere in

July (2).

Example: Find the pressure at the tank bottom.

Solution: =9790 N.m3 (Table A.5)

Another practical and fast way:

or

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22

11 z

γ

pz

γ

p

)gauge(kpa.p)(γ

p5819200 2

2

kpa.pp))((

pγhp

5819297900 22

21

1212 pγhppγhp

Example: Find the pressure at the tank bottom

Solution:

Second way:

21

kpa.p).()(.

p

zγ

pz

γ

p

oiloil

924350981080

00 22

22

11

kpa.p)(p

zγ

pz

γ

p

waterwater

5042329790

09790

39243

3

33

22

kpa.pp)().)((.

phγhγp wateroil

5042329790509810800 33

3211

Example: Find the pipe pressure if h1=1.2 m,

h2=1 m, and h3=3 m

Solution:

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21231 phγhγhγp waterHgoil

002198101981061339810901 ....p

kpa.pp pipe 157951

Example: Given F1= 200 N, Find the F2. Neglect

the weights of the pistons

Solution:

23

21 pγhp

kpa.

.A

Fp 23159

0404

200

21

11

kpa.).(p 558142298108501592302

kN..APF 1191104

1425582

222

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Manometry• Pressure is one of the most important

variables in fluid mechanics.

• Numerous instruments and devices have

been developed for its measurement.

• One of the simplest and most common

measurement techniques is based on

hydrostatics or manometry.

• The principle is basically to utilise the

change in pressure with elevation to

determine the value of the pressure.

3.3: Pressure Measurements

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a: Piezometer

• A simple manometer,

or a piezometer can

be attached to a pipe

and the height of the

liquid’s column is an

indicator of the

pressure in the pipe:

hp

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b: U-tube Manometers

• A simple manometer

becomes impractical

in the case of gases

or in the case of

high pressures.

• For such cases, a

U-tube filled with

another liquid can

be used. 4 m fp h L

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c: Differential Manometers

• A differential

manometer can also

be used to measure

the pressure

difference between

two points in a pipe,

according to:

hp fm )(

Example: Find the pipe pressure.

Solution:

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015131 HgairoilA γ.γ.γp

019810513031981090 ...pA

kpa.pA 912143

0.0

• Example: Find the pipe pressure

• Solution:

29

0102050101033 L.pB

m.L.

L.sin 20

50

10

5

3

kPapB 1

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- Bourdon-Tube Gage

• The Bourdon-tube

gage is a very

common device that

utilises the deflection

in a spring tube to

measure the

pressure.

• Bourdon gages need

to be periodically

calibrated.

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- Pressure Transducers

• Similarly, pressure transducers utilise the

deflection of a diaphragm to produce an electrical

signal that can be related to the pressure.

• Pressure transducers are widely used with

applications requiring extensive data acquisition

and processing.

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3.4:Hydrostatic Forces on Plane

Surfaces

• For a horizontal surface in a fluid, the

pressure is uniformly distributed and

the resulting force is simply equal to

the product of the pressure and the

area.

• For a vertical or an inclined surface the

pressure is not uniformly distributed

and thus some derivation is required to

determine the resulting force.

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34

35

Hydrostatic Forces on Plane Surfaces

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Hydrostatic Forces on Plane

Surfaces

• The pressure on the differential area is

equal to:

• Consequently, the differential force on the

differential area is equal to:

• And the total force on the area is equal to:

sinyp

dAydF sin

dAyFA

sin

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Hydrostatic Forces on Plane

Surfaces

• Rewriting,

• Integrating:

• In short, the magnitude of the resultant

hydrostatic force on a plane surface is the

product of the pressure at the centroid of

the surface and the area of that surface.

dAyFA

sin

ApAyF sin

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Location of Resultant Hydro. Force

• The location of the line of action of the

resultant hydrostatic force lies at a point

called the center of pressure.

• Writing the moment equation:

• Or:

dApydFyFycp

dAyFycp sin2

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Location of Resultant Hydro. Force

• Re-writing:

• Substituting with the area moment of

Inertia ( I ):

• Applying the parallel-axis theorem:

dAyFycp

2sin

ocp IFy sin

)(sin 2 AyIFycp

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Location of Resultant Hydro. Force• Substituting the value of F:

• Reducing:

• where is the 2nd moment of area (area moment of inertia): see Fig. A.1

• This equation implies that the center of pressure is always below the centroid, but comes closer to the centroid as the depth increases.

Ay

Iyycp

I

ApAyF sin

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Fig. A.1

Centroids

and

moments

of inertia

of plane

areas

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• Example: The gate shown is rectangular and has

dimensions 6 m x 4 m. What is the reaction at point

A? Neglect the weight of the gate.

• Solution:

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m.LL

cos 46433

30

44

sinyP

APF

m..Ly 4646464333

pa.sin.P 26549166046469810

N][.F 1317990462654916

Ay

Iyycp 72

12

64

12

33

hb

I

m.yor

.][.Ay

Iyy

cp

cp

9286

4640644646

72

kNR

R).(

R).(F

.M stop

557

6464031317990

646403

00

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3.5: Hydrostatic Forces on Curved

Surfaces

• The hydrostatic forces

on curved forces can

be found by

equilibrium concepts

on systems

comprised of the fluid

in contact with the

curved surface.

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Forces on Curved Surfaces• For the free body

diagram:

ApFF AChorizontal

CBvertical FWF The resultant hydrostatic force acting on the

curved solid surface is then equal and opposite

to the force acting on the curved liquid surface

(Newton’s third law).

Horizontal force component on curved

surface:

Vertical force component on curved

surface:

FR=F

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• Example: Find the vertical and horizontal

forces on the given gate

(L=1 m).

• Solution:

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tV cV

49

N)()L(hAPF 1962011298101111

N.))L(

(

)VV(VVWWW ctctct

15210914

2

41119810

2

kN..WFFvertical 5117152109196201

2222 AsinyAPF m.LL

y 5112

1

2

190sinsin 2

2 111 mA

kN..F 71514115198102

2Ay

Iyycp 121

12

11

12

33

/bh

I

m.ycp 5551

kN.FFhorizontal 715142

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3.6: Buoyancy

• Considering the

submerged body in

the figure:

• The force acting on

the lower surface of

the body is equal to

the weight of the

liquid needed to fill

the volume above this

surface.

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Buoyancy

• The force acting on

the upper surface of

the body is equal to

the weight of the

liquid above this

surface.

• Summing the forces:

B up down fluid bodyF F F V

A body completely submerged in a liquid

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A body partially submerged in a liquid

B fluid DF V

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Buoyancy

• Consequently, the buoyant force on a

submerged body is equal to the weight of

the liquid that would be needed to occupy

the volume of the body.

• Similarly, for a floating body, the buoyant

force is equal to the weight of the liquid

that would be needed to occupy the

submerged volume of the body.

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Buoyancy

• This principle of buoyancy is basically

what is commonly known as the

Archimedes’ principle: “For an object

partially or completely submerged in a

fluid, there is a net upward force equal

to the weight of the displaced fluid.”

• The buoyancy force is an upward force

that passes through the centroid of the

displaced volume.

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Hydrometry• Device [glass bulb] to measure the or S of a

liquid based on the principle of buoyancy

• Example: Find the tension in the given figure.

• Solution:

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kN.T

kN.dπ

γVγF

WFT

B

B

359

85176

3

• Example: find the specific gravity of the given

unknown fluid, the hydrometry weight is 0.015 N.

• Solution:

57

3

2232

9202

0150100

361010101

0150

m/N

.).

}{.}{(VF

N.WF

oil

oiloilB

B

93809810

9202.S

water

oil