Fluid statics

Fluid Statics: Pressure intensity and

pressure head: pressure and specific weight

relationship, absolute and gauge pressure

Dr. Mohsin Siddique

Assistant Professor

1

Fluid Mechanics

Fluid Statics

2

� Fluid Statics: It is the branch of fluid mechanics that deals with the behavior/response of fluid when they are at rest.

� Pressure, (average pressure intensity): It is the normal force exerted per unit area. It is denoted by P and is given by;

� Units

� SI: N/m2 (called Pascal)

� BG: lb/ft2 or lb/in2 (called psi)

� CGS: dyne/cm2

1 bar=105N/m2=105Pascal

A

F

area

forceP ==

Pressure vs Water depth/height

3

� Consider a strip or column of a cylindrical fluid,

� h= height or depth of strip of fluid

� γ= specific weight of fluid

� dA=cross-sectional area of strip

� dV=volume of strip

� dW=weight of strip

� Pressure at base of strip=dF/dA=dW/dA

� P= γdV/dA

� P= γdA.h/dA

P=γh

h

Pressure vs Water depth/height

4

� P=γh

� P α h

� For h=0, P=0

� For h=h, P=γh

h

Pressure distribution diagram/pressure profile

� As you know atmospheric pressure reduces, as we move to higher elevations. Is it because of h, as h reduces, P also reduces.

PASCAL’S LAW

5

� “Pressure at any point in fluid is same in all directions when the fluid is at rest”

� Consider a wedge shape element of fluid having dimension dx, dy and dz along x, y and z axis.

� dl= dimension of inclined plane making an angle α with the vertical

� Px, Py, Pz and P are pressure acting in x, y, z and perpendicular to inclined surface

� dW=weight of element

zy

x

Py(dxdz)

Px(dydz)

P(dldz)

dx

dy

dz

α

α α P(dldz)cosα

P(dldz)sinα

dW

zy

x

PASCAL’S LAW

6

Py(dxdz)

Px(dydz)

P(dldz)

α α P(dldz)cosα

P(dldz)sinα

PP

odydzPdydzP

dldyodldydldzPdydzP

odldzPdydzP

F

x

x

x

x

x

=

=−

==−

=−

=∑

)(

/cos/)(

cos)(

0

α

α

Q

dW

PP

odxdzPdxdzP

dwdldxodldxdldzPdxdzP

odldzPdWdxdzP

F

y

y

y

y

y

=

=−

≅==−

=−−

=∑

)(

0&/sin/)(

sin)(

0

α

α

Q

PASCAL’S LAW

7

� Similarly by applying the conditions in z direction, it can be proved that

� Hence,

� The above states that the pressure acting on fluid particle is same in all directions when the fluid is at rest.

PPz =

PPPP zyx ===

Absolute and Gauge Pressure

8

� Atmospheric pressure

� Gauge pressure

� Vacuum/negative pressure

� Absolute pressure

� Atmospheric pressure: Pressure exerted by atmosphere

� Gauge pressure: Pressure more than atmospheric pressure

� Vacuum/negative pressure: Pressure less than atmospheric pressure

� Absolute pressure: Pressure measure relative to absolute zero

vacatmabs

gatmabs

PPP

PPP

−=

+=

Atmospheric Pressure

9

� It is defined as weight of air per unit surface area of earth.

� It decreases with increase in elevation w.r.t. surface of earth.

� Standard atmospheric pressure at mean-sea-level is

=101.3KN/m2

=1.013bar

=14.7psi

=760mm of Hg

=33.9ft of water

=10.3m of water

Measurement of Atmospheric Pressure

10

� Barometer: It is device used to measure the atmospheric pressure at any point on the earth.

� There are two types of barometer

� (i) Liquid barometer

� It measures the pressure with help of column of liquid

� (ii) Aneroid barometer

� It measures atmospheric pressure by its action on an elastic lid of evacuated box.

Liquid Barometer

11

� It consists of a transparent tube which is open from one end only. The tube is filled with liquid and is inserted in a jar also containing same liquid. The liquid initially drops in tube due to gravity but stabilizes at certain level under the action of atmospheric forces. The atmospheric pressure is then measured as height of liquid at which it stabilizes.

� Three forces acting on fluid are

Patm(A)=Force of atmospheric Pressure

W=Weight of liquid

Pvap(A)= Force of vapour pressure

A=Cross-sectional area of tube

Weight of liquid

Force of Patm

Vacuum Pressure

Liquid/

Liquid Barometer

12

� Three forces acting on fluid are

Patm(A)=Force of atmospheric Pressure

W=Weight of liquid

Pvap(A)= Force of vapour pressure

A=Cross-sectional area of tube

� For Equilibrium

W

PatmA

PvapA

Liquid/

h

0; =−−=∑ APWAPoFy vapatm

0=−− APAhAP vapatm γ AhW γ=Q

vapatm PhP += γ

Liquid Barometer

13

� Generally, mercury is preferred liquid because its vapour pressure is minimum. Moreover, its specific gravity is very high so that size (height) of barometer required is small.

� However, for other liquid vapour pressure must be considered in estimation.

� The barometer using mercury is called mercury barometer and while using water is called water barometer.

� Size of barometer tube should be more than ½ inches (or 13mm) to avoid capillarity.

Absolute Pressure

14

� Gauge Pressure (Pg): It is the pressure measured relative to atmospheric pressure (Patm) and is always above the atmospheric pressure

� It may be defined as normal compressive force per unit area

� Vacuum Pressure (Pvac): It is the pressure measured relative to atmospheric pressure and is less than the atmospheric pressure

� It may be defined as normal tensile force per unit area

� Absolute Pressure(Pabs): It is the pressure measured from absolute zero

vacatmabs

gatmabs

PPP

PPP

−=

+=

Problem

15

� Q.3.2.2

4600m

Surface

γ=10.05kN/m3

( )2/46700

460005.10

mkN

P

hP

=

= γ

Problem

16

� Q.3.2.4

( )

psi

P

hSPhPP woil

9.57

)12/(684.6288.032 2

2

112

+=

+=+= γγ

P2=?

P1=32psi Stream

68ft

S=0.88

γ=Sγw

Problem

17

Problem

18

� Q.3.3.1

5m

Surface

γ=10 kN/m3

( ) 2/1628 mkNhPi === γγ=8 kN/m3

2m

interfacePi

Pb( ) 2

2211

/1.655)81.9(28 mkNP

hhP

b

b

=+=

+= γγ

NUMERICALS

19

� Q.3.3.3

h=4600γ=10.05 kN/m3

2/______4600)05.10( mkNP

hP

==

= γ

P=?

Problem

20

� Q.3.3.3

h=?

Surface

γ=12N/m3

( ) ( )

mh

h

hP

8442

1210003.101

=

=×

= γ

P=101.3kPa

Determine the depth of gaseous atmosphere to cause 101.3kN/m2 over surface of earth?

Problem

21

� 3.5.1

psiaP

psiaOHftP

vap

atm

09.2

47.144.33 2

=

==

Weight of liquid

Force of Patm

Vacuum Pressure

Alcohol

w

vapatm

vapatmvapatm

vapatm

vapatm

S

PPh

PP

A

APAPh

APAhAP

APWAP

oFy

γ

γγ

γ

−=

−=

−=

=−−

=−−

=∑

0

0

;

h=?

22

Measurement of Pressure

23

� The following devices are used for pressure measurement

� 1. Piezometer

� 2. Manometer

� a) Simple manometer

� B) Differential manometer

� 3. Mechanical Pressure Transducer (Bourden gauge)

� 4. Electrical Pressure Transducer

1. Piezometer

24

� It is used to measure pressure in pipes or vessels.

� In it simplest form, it consists of a transparent tube open from other ends

� The diameter of tube should > ½” to avoid capillarity action

� Piezometers may be connected to sides or bottom of pipe to avoid eddies that are produced in the top region of pipe

� Limitations:

� It must only be used for liquids

� It should not be used for high pressure

� It cannot measure vacuum (-ve) pressure

� When connected to pipes, thewater level rises in it whichgives a measure of pressure.

2. Manometer

25

� a). Simple Manometer

� Figure shows a set up of simple manometer.

� It consists of a U shaped tube, part of which is filled with manometricfluid.

� One end of tube is connected with the pipe whose pressure is required to be determined.

� Due to pressure, level of manometric fluid rises on one side while it falls on other side.

� The difference in levels is measured to estimate the pressure.

A

Fluid, γf

Manometricfluid, γm

Y

z

� Y=Manometric reading

� γf =Specific weight of fluid in pipe

� γm =Specific weight of manometric fluid

2. Manometer

26

� Manometric Fluids

� 1. Mercury

� 2. Oils

� 3. Salt solution etc

� Properties of manometricFluid

� 1. Manometric fluid should not be soluble/intermixale with fluid flowing in pipe whose pressure is required to be determined.

� 2. Lighter fluid should be used if more precision is required.

A

Fluid,

Manometricfluid,

y

z

2. Manometer

27

� Pressure measurement

A

Fluid, γf

Manometricfluid, γm

Y

z

� Sign Convention

� -ve: upward direction

� +ve: downward direction

AatmAabs

atmmfAabs

PPP

PYZP

+=

=−+

Q

γγPatm

ZYP

YZP

fmA

mfA

γγ

γγ

−=

=−+ 0

PA

� Above is a gauge pressure equation

2. Manometer

28

� b). Differential manometer

� It is used to measure difference of pressure.

� Case 1: when two vessels/pipes are at same level

ZA

ZB

Y

A BPA

PB

Fluid A, γA

Fluid B, γB

ManometricFluid , γm

AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

γγγ

γγγ

−++=−

−++=

2. Manometer

29

( )

( )

( )( )YPP

YYPP

ZZYPP

ZYZPP

if

fmBA

fmBA

BAfmBA

AfmBfBA

fAB

γγ

γγ

γγ

γγγ

γγγ

−=−

−=−

−−=−

−++=−

==

2. Manometer

30

� b). Differential Manometer

� Case 1I: when two vessels/pipes are at different level

ZA ZB

Y

A

B

PA

PB

Fluid A, γA Fluid B, γB


AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

γγγ

γγγ

−++=−

−++=

2. Manometer

31

( )BAfmBA

AfmBfBA

fAB

ZZYPP

ZYZPP

if

−−=−

−++=−

==

γγ

γγγ

γγγ

2. Manometer

32

� b). Differential manometer

� Case 1II: when manometer is inverted

ZA

ZB

Y

A

B

PAPB

Fluid A, γA

Fluid B, γB


AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

γγγ

γγγ

+−−=−

+−−=

2. Manometer

33

( )BAfmBA

AfmBfBA

fAB

ZZYPP

ZYZPP

if

−+−=−

+−−=−

==

γγ

γγγ

γγγ

Advantages and Limitation of Manometers

34

� Advantages

� Easy to fabricate

� Less expansive

� Good accuracy

� High sensitivity

� Require little maintenance

� Not affected by vibration

� Specially suitable for low pressure and low differential pressure

� Easy to change sensitivity by changing manometric fluid

� Limitations

� Usually bulky and large in size

� Being fragile, get broken easily

� Reading of manometer is get affected by temperature, altitude and gravity

� Capillary action is created due to surface action

� Meniscus has to be measured accurately for better accuracy.

3. Mechanical Pressure Transducer

35

� Transducer is a device which is used to transfer energy from one system to other

� Mechanical pressure transducer converts pressure system to displacement in mechanical measuring system

� Bourden Gauge is used to measure high pressure either positive or negative. It gives pressure directly in psi of Pascal units

Bourden Gauge

36

� The essential mechanical element in this gage is the hollow, elastic curved tube which is connected to the pressure source as shown in Fig.

� As the pressure within the tube increases the tube tends to straighten, and although the deformation is small, it can be translated into the motion of a pointer on a dial as illustrated.

Fig. Bourden gauge

Bourden Gauge

37

� Since it is the difference in pressure between the outside of the tube and the inside of the tube that causes the movement of the tube, the indicated pressure is gage pressure.

� The Bourdon gage must be calibrated so that the dial reading can directly indicate the pressure in suitable units such as psi, psf, or pascals.

� A zero reading on the gage indicates that the measured pressure is equal to the local atmospheric pressure.

� This type of gage can be used to measure a negative gage pressure (vacuum) as well as positive pressures.

3. Mechanical Pressure Transducer

38

� Elevation Correction

� Bourden gauge gives pressure at the center of dial. So to calculate pressure at point A,

� Where

� γz=elevation correction

zPP gA γ+=

z

A

Pg

4. Electrical Pressure Transducer

39

� It converts displacement of mechanical measuring system to an electrical signal.

� Its gives continuous record of pressure when converted to a strip chart recorder.

� Data can be displayed using computer data acquisition system.

Problem

40

� Q 3.5.4

� x= ??in

� New Manometric reading=4+2x=34.7in

OHftpsiaPatm 292.337.14 ==

( )

psiaP

PP

A

atmmwA

??

4125

=

=−×+ γγ

( ) ( )atmmwA PxxP =+−+×+ γγ 241252

41

Problem

42

� Q. 3.5.7

0=+=

=−−

atmwmmA

atmmmwA

PhRP

PRhP

Qγγ

γγ

56.13,/31.6268@

45.13,/20.61150@

3

3

==

==

m

o

w

m

o

w

SftlbF

SftlbF

γ

γ

43

Problem

44

� 3.5.8

45

Problem

46

� 3.5.10

� Two vessels are connect to a differentialmanometer using mercury (S=13.56), theconnecting tubes being filled with water. Thehigher pressure vessel is 5ft lower in elevationthan the other. Room temperature prevails. If themercury reading is 4.0in, what is the pressuredifference in feet of water and in psi ? (b) ifcarbon tetrachloride (S=1.59) were used insteadof mercury what would be manometric readingfor the same pressure difference.

Problem

47

Problem

48

� 3.12

Problem

49

Forces on Immersed Bodies: Forces on

submerged planes & curved surfaces and

their applications, Drag and Lift forces

Dr. Mohsin Siddique

Assistant Professor

UOS Sharjah, UAE

50Date:

Fluid Mechanics

University of Sharjah

Dept. of Civil and Env. Engg.

Forces on Immersed Bodies

51

� Hydrostatic Force: It is the resultant force of pressure exerted by liquid at rest on any side of submerged body.

� It is the summation of product of uniform pressures and elementary areas of submerged body

� It is equal to the product of submerged area and pressure at the centroid of the submerged area

pAdAppdAF === ∫∫

Forces on Plane Area

52

� Center of pressure

� The point of application of resultant force of pressure on a submerged area is called center of pressure.


53

� A=total submerged area

� F=hydrostatic force

� θ=angle of submerged plane with free surface

� hc=depth of center of area

� hp=depth of center of pressure

� yc=inclined depth to center of area

� yp=inclined depth to center of pressure

� dA=elemtry area

� dF=force of pressure (hydrostatic force) on elementry area

( ) ( )dAhdApdF γ==


54

� Lets’ choose an elementary area so that pressure over it is uniform. Such an element is horizontal strip, of width, x so . The pressure force, dF on the horizontal strip is

� Integrating

� Where, yc is the distance OX along the sloping plane to the centroid C of the area A. If hc is the vertical depth to the centriod, then we have

( ) ( )dAhdApdF γ==

xdydA =

∫∫∫∫∫ ==== ydAdAyhdApdAdF θγθγγ sinsin

θ

γ

sinyh

hp

=

=

A

ydAyc

∫=

( )AyF cθγ sin=

AhF cγ=


55

� Thus, we find the total force on any plane area submerged in a liquid by multiplying the specific weight of the liquid by the product of the area and the depth of its centriod.

� The value of F is independent of the angle of inclination of the plane so long as the depth of its centroid is unchanged.

� Since is pressure at the centroid, we can also say that total pressure force on any plane area submerged is a liquid is the product of the area and the pressure at the centroid

�

AhF cγ=

chγ

Center of Pressure

56

� In order to determine location of center of pressure, yp, from OX, let’s take the moment of elementary area around OX

� Integrating

� Where, I is the 2nd moment of submerged area about OX

� Where ycA is called static moment of area

( ) ( ) ( )dAyyhdAypdAyydFdM θγγ sin====

( ) ( )

( )IFy

dAydAyyydF

p θγ

θγθγ

sin

sinsin 2

=

== ∫∫∫

( ) ( )Ay

I

Ay

I

F

Iy

cc

p ===θγ

θγθγ

sin

sinsin( )AyF cθγ sin=Q

Center of Pressure

57

� Now, according to parallel axis theorem,

� Where, Ic is 2nd moment of area about centroidal axis.

� From this equation we again see that the location of center of pressure, P, is independent of the angle θ.

� When the plane is truly vertical, i.e., θ=90o

Ay

AyI

Ay

Iy

c

cc

c

p

2+== cc AyII

2+=Q

Ay

Iyy

c

ccp +=

Ah

Ihh

c

ccp +=

Lateral Position of Center of Pressure

58

� To find the lateral position of center of pressure P, consider the area is made up of series of elemental strips. The center of pressure for each strip is at the mid point of the strip. Since the moment of the resultant force F must be equal to the moment of distributed force system about any axis, say, the y-axis

� Where, Xp is the lateral distance from the selected y-axis to the center of pressure P of the resultant force F, and xp is the lateral distance to the center of any elemental horizontal strip of area dAon which the pressure p is acting

∫= pdAxFX pp

∫= pdAxFX pp

Forces on Curved Surface

59

� Horizontal force on curved surface

� Vertical force on curved surface

area horizontal equivalenton force chydrostati'' === FFxFx

surface above liquid of volumeofWeight ' === WFzFz

Forces on Curved Surface

60

� Resultant Force

22

zx FFF +=

= −

x

z

F

F1tanθ

Hydrostatic force formulas

61

Ay

Iyy

c

ccp +=

AhF cγ= ( )AyF cθγ sin=

62

Problem

63

� Q 3.7.6: A plane surface is circular with a diameter of 2m. If it is vertical and the top edge is 0.5m below the water surface, find the magnitude of the force on one side and the depth of center of pressure.

� Solution:

Ah

Ihh

c

ccp +=

( )AyF cθγ sin=

AhF cγ=Ay

Iyy

c

ccp +=

Free surface

0.5m

D=2m

mD

hc 5.12

5.0 =+=

( )

kNF

F

AhF c

2.46

24

5.1810.9 2

=

=

=

π

γ

( ) ( )mh

DDh

Ah

Ihh

p

p

c

ccp

667.1

4/5.1/64/5.1 24

=

×+=

+=

ππ

Problem

64

� Q 3.7.8: A rectangular plate 5ft by 4ft is at an angle of 30o with the horizontal, and the 5 ft side is horizontal. Find the magnitude of force on one side of the plate and the depth of its center of pressure when the top edge is (a) at the water surface (b) 1 ft below water surface

� (a)

4ft

5ft

30o

hp

yp

hcyc

4ft

θsinyh =

ftyho

cc 130sin2sin === θ ( )( ) lbAhF c 12484514.62 =×== γ

( ) ( ) ftbdbdAy

Iyy

c

ccp 67.22/12/2 3 =×+=+= fthp 33.1=

Problem

65

� (b)

4ft

5ft

hp

yp

hcyc=4ft

6ft

θsinyh =

ftho

c 230sin21sin21 =+=+= θ ( )( ) lbAhF c 25004524.62 =×== γ

( ) ( ) ftbdbdAy

Iy

c

cp 33.44/12/44 3 =×+=+=

ftyho

pp 167.230sin33.4sin === θ

1ft

ft

hy cc

4

sin/

=

= θ

Problem

66

Q 3.7.9

(a) For critical stability resultant of all hydrostatic for must pass through B.

Thus B will be at a/3=5.4/3=1.8m above N

Schematic pressure distribution diagram

a/3

Problem

67

(a)

Since B point is hinge, therefore, moment of all force at there must be zero

Schematic pressure distribution diagram

1m

4.4/3

RB

RN

( )

mkNRN

RNM B

/59.17

08.13/4.43/4.595;0

=

=−−=∑

For reaction at B

mkNRN

FRBRNFx

/4.7759.1795

;0

=−=

=+=∑

mkNAhF c /954.4)2

4.4(81.9 === γ

For a water depth of 4.4m

Problem

68

� 3.7.12

Problem

69

� Q 3.8.8. Cross section of tank is shown in figure., where r=2m and tank is open and contain water to a depth h=3.5m. Determine the magnitude and location of horizontal and vertical force components acting on unit width of tank wall ABC.

( ) mh

mkNF

AhF

p

x

cx

33.25.33

2

/1.605.3)2/5.3(81.9

==

==

= γhp

xp

`

70

hpxp

( )ALVOLWFz γγ ===

86.3)2(5.3

4/)4(4

)2(5.3 2

=−=

−=

π

π

A

A

2m

d=4m( )

mkN

AFz

/9.37

)86.3(81.9

=

== γ

Lets take moment about AB

( ) ( )

mx

x

p

p

123.1

849.0)1(25.386.3

=

−×= π

mr

xc 849.03

4==

π

Drag and Lift Force

71

� Lift is the component of aerodynamic force perpendicular to the relative wind.

� Drag is the component of aerodynamic force parallel to the relative wind.

� Weight is the force directed downward from the center of mass of the airplane towards the center of the earth.

� Thrust is the force produced by the engine. It is directed forward along the axis of the engine.

Drag and Lift Force

72

� The drag force acts in a direction that is opposite of the relative flow velocity.

� Affected by cross-section area (form drag)

� Affected by surface smoothness (surface drag)

� The lift force acts in a direction that is perpendicular to the relative flow.

� CD= Coefficient of drag

� CL= Coefficient of lift

� A=projected area of body normal to flow

� V= relative wind velocity

Buoyancy and Floatation

73

� When a body is immersed wholly or partially in a fluid, it is subjected to an upward force which tends to lift (buoy)it up.

� The tendency of immersed body to be lifted up in the fluid due to an upward force opposite to action of gravity is known as buoyancy.

� The force tending to lift up the body under such conditions is known as buoyant force or force of buoyancy or up-thrust.

� The magnitude of the buoyant force can be determined by Archimedes’ principle which states

� “ When a body is immersed in a fluid either wholly or partially, it is buoyed or lifted up by a force which is equal to the weight of fluid

displaced by the body”


74

� Lets consider a body submerged in water as shown in figure.

� The force of buoyancy “resultant upward force or thrust exerted by fluid on submerged body” is given

Water surface

11 hP γ=

( )212 hhP += γ

2h

1h 1F

2F

� dA=Area of cross-section of element

� γ= Specific weight of liquid

( ) ( )

( )[ ][ ]volumeF

dAhF

dAhdAhhF

FFF

B

B

B

B

γ

γ

γγ

=

=

−+=

−=

2

121

12


75

� =Weight of volume of liquid displaced by the body (Archimedes's Principle)

� Force of buoyancy can also be determined as difference of weight of a body in air and in liquid.

� Let

� Wa= weight of body in air

� Wl=weight of body in liquid

� FB=Wa-Wl

[ ]volumeFB γ=


76

� Center of Buoyancy (B): The point of application of the force of buoyancy on the body is known as the center of buoyancy.

� It is always the center of gravity of the volume of fluid displaced.

Water surface

CG or GC or B

CG or G= Center of gravity of body

C or B= Centroid of volume of liquid displaced by body

Thank you

� Questions….

� Feel free to contact:

77

Engineering

Fluid statics