2- Fluid Statics

2. Fluid Statics: Pressure intensity and

pressure head: pressure and specific weight

Fluid Mechanics

University of SharjahDept. of Civil and Env. Engg.

pressure head: pressure and specific weight

relationship, absolute and gauge pressure

Dr. Mohsin Siddique

Assistant Professor

UOS Sharjah UAE

1Date: 15/06/2014

Fluid Statics

� 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;

2

� 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

� 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

h

3

� dW=weight of strip

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

� P= γdV/dA

� P= γdA.h/dA

P=γh

Pressure vs Water depth/height

� P=γh

� P α h

� For h=0, P=0

� For h=h, P=γh

h

4

For h=h, P=γh

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

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

zy

x

dy

dz

αzy

x

5

� 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

x

Py(dxdz)

Px(dydz)

P(dldz)

dx

α α P(dldz)cosα

P(dldz)sinα

dW

PASCAL’S LAW

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

αα

Q

dW

6

Py(dxdz)

PP

odxdzPdxdzP

dwdldxodldxdldzPdxdzP

odldzPdWdxdzP

F

y

y

y

y

y

=

=−

≅==−

=−−

=∑

)(

0&/sin/)(

sin)(

0

αα

Q

PASCAL’S LAW

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

� Hence,

PPz =

PPPP zyx ===

7

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

zyx

Absolute and Gauge Pressure

� Atmospheric pressure

� Gauge pressure

� Vacuum/negative pressure

� Absolute pressure

8

� 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

� 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

9

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

� 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

10

� (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

� 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

Weight of liquid

Vacuum Pressure

Liquid/

11

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

Force of Patm

Liquid Barometer

� 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

PvapA

Liquid/

h

12

PatmA0; =−−=∑ APWAPoFy vapatm

0=−− APAhAP vapatm γ AhW γ=Q

vapatm PhP += γ

Liquid Barometer

� 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.

13

� 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

� 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

14

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

� Q.3.2.2

Surface

( )460005.10P

hP

== γ

15

4600mγ=10.05kN/m3

( )2/46700

460005.10

mkN

P =

Problem

� Q.3.2.4

( )hSPhPP woil 112 +=+= γγ

16

( )psi

P

9.57

)12/(684.6288.032 22 +=

P2=?

P1=32psi Stream

68ft

S=0.88

γ=Sγw

Problem

17

Problem

� Q.3.3.1

Surface

( ) 2/1628 mkNhPi === γγ=8 kN/m3

2mPi

18

5m

γ=10 kN/m3

2m

interfacePi

Pb( ) 2

2211

/1.655)81.9(28 mkNP

hhP

b

b

=+=

+= γγ

Problem

� Q.3.3.3

( ) ( )hhP

1210003.101 =×= γ

19

h=?

Surface

γ=12N/m3

( ) ( )mh

h

8442

1210003.101

==×

P=101.3kPa

Problem

� 3.5.1

psiaP

psiaOHftP

vap

atm

09.2

47.144.33 2

===

Vacuum Pressure

Alcohol

20

Weight of liquid

Force of Patm

Alcohol

w

vapatm

vapatmvapatm

vapatm

vapatm

S

PPh

PP

A

APAPh

APAhAP

APWAP

oFy

γ

γγ

γ

−=

−=

−=

=−−

=−−

=∑

0

0

;h=?

Measurement of Pressure

� The following devices are used for pressure measurement

� 1. Piezometer

� 2. Manometer

� a) Simple manometer

B) Differential manometer

22

� B) Differential manometer

� 3. Mechanical Pressure Transducer (Bourden gauge)

� 4. Electrical Pressure Transducer

1. Piezometer

� 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

23

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

� 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

A

Fluid, γf

Y

z

24

� 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.

Manometricfluid, γm

� Y=Manometric reading

� γf =Specific weight of fluid in pipe

� γm =Specific weight of manometric fluid

2. Manometer

� Manometric Fluids

� 1. Mercury

� 2. Oils

� 3. Salt solution etc

� Properties of manometricFluid

A

Fluid,

y

z

25

Fluid

� 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.

Manometricfluid,

2. Manometer

� Pressure measurement

A

Fluid, γf

Y

z AatmAabs

atmmfAabs

PPP

PYZP

+=

=−+

Q

γγPatm

PA

26

Manometricfluid, γm

� Sign Convention

� -ve: upward direction

� +ve: downward direction

ZYP

YZP

fmA

mfA

γγγγ

−=

=−+ 0

� Above is a gauge pressure equation

2. Manometer

� b). Differential manometer

� It is used to measure difference of pressure.

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

A BPA

PB

27

ZA

ZB

Y

Fluid A, γA

Fluid B, γB

ManometricFluid , γm

AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

γγγγγγ−++=−

−++=

2. Manometer

( )( )

ZZYPP

ZYZPP

if

BAfmBA

AfmBfBA

fAB

γγγγγ

γγγ

−−=−

−++=−

==

28

( )( )( )YPP

YYPP

fmBA

fmBA

γγγγ

−=−

−=−

2. Manometer

� b). Differential Manometer

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

A PA

PB

29

ZA ZB

Y

BPB

Fluid A, γA Fluid B, γB


AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

γγγγγγ−++=−

−++=

2. Manometer

( )BAfmBA

AfmBfBA

fAB

ZZYPP

ZYZPP

if

−−=−

−++=−

==

γγγγγ

γγγ

30

2. Manometer

� b). Differential manometer

� Case 1II: when manometer is inverted

YFluid B, γB


31

ZA

ZB

Y

A

B

PA

PB

Fluid A, γA

AAmBBBA

AAmBBBA

ZYZPP

ZYZPP

γγγγγγ+−−=−

+−−=

2. Manometer

( )BAfmBA

AfmBfBA

fAB

ZZYPP

ZYZPP

if

−+−=−

+−−=−

==

γγγγγ

γγγ

32

Advantages and Limitation of Manometers

� Advantages

� Easy to fabricate

� Less expansive

� Good accuracy

� High sensitivity

� Require little maintenance

� 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

33

� Require little maintenance

� Not affected by vibration

� Specially suitable for low pressure and low differential pressure

� Easy to change sensitivity by changing manometric fluid

� Capillary action is created due to surface action

� Meniscus has to be measured accurately for better accuracy.

3. Mechanical Pressure Transducer

� 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

34

or negative. It gives pressure directly in psi of Pascal units

Bourden Gauge

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

35

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

� 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

36

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

� Elevation Correction

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

Where

zPP gA γ+=z

Pg

37

� Where

� γz=elevation correction

z

A

4. Electrical Pressure Transducer

� 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.

38

Problem

� Q 3.5.4

OHftpsiaPatm 292.337.14 ==

( )psiaP

PP atmmwA

??

4125

==−×+ γγ

39

� x= ??in

� New Manometric reading=4+2x=34.7in

psiaPA ??=

( ) ( ) atmmwA PxxP =+−+×+ γγ 241252

Problem

� Q. 3.5.7

41

0=+==−−

atmwmmA

atmmmwA

PhRP

PRhP

Qγγγγ

56.13,/31.6268@

45.13,/20.61150@3

3

==

==

mo

w

mo

w

SftlbF

SftlbF

γγ

Problem

� 3.5.8

43

Problem

� 3.5.10

� Two vessels are connect to a differential manometer using mercury (S=13.56), the connecting tubing being filled with water. The higher pressure vessel is 5ft lower in elevation than the other. Room temperature prevails. If the mercury reading is 4.0in, what is the pressure difference in feet of water and in psi ? (b) if carbon tetrachloride (S=1.59) were used instead of mercury

45

tetrachloride (S=1.59) were used instead of mercury what would be manometric reading for the same pressure difference.

Problem

46

Problem

� 3.12

47

Problem

48

Forces on Immersed Bodies: Forces on

submerged planes & curved surfaces and

Fluid Mechanics

University of SharjahDept. of Civil and Env. Engg.

submerged planes & curved surfaces and

their applications, Drag and Lift forces

Dr. Mohsin Siddique

Assistant Professor

UOS Sharjah, UAE

49Date:

Forces on Immersed Bodies

� 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

pAdAppdAF === ∫∫

50

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

Forces on Plane Area

� Center of pressure

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

51


52

� 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==


� 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

( ) ( )dAhdApdF γ==

xdydA =

∫∫∫∫∫ ==== θγθγγ

θγ

sinyh

hp

==

53

� 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

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

A

ydAyc

∫=( )AyF cθγ sin=

AhF cγ=


� 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

AhF cγ=

54

� 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

�

chγ

Center of Pressure

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

� Integrating

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

( ) ( )dAydAyyydF θγθγ sinsin 2== ∫∫∫

55

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

� Where ycA is called static moment of area

( ) ( )( )IFy

dAydAyyydF

p θγ

θγθγ

sin

sinsin

=

== ∫∫∫

( ) ( )Ay

I

Ay

I

F

Iy

ccp ===

θγθγθγ

sin

sinsin ( )AyF cθγ sin=Q

Center of Pressure

� 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,

Ay

AyI

Ay

Iy

c

cc

cp

2+== cc AyII 2+=Q

Ay

Iyy

c

ccp +=

56

� 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

Ah

Ihh

c

ccp +=

Lateral Position of Center of Pressure

� 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

∫= pdAxFX pp

57

� 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

Forces on Curved Surface

58

� 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

59

� Resultant Force

22zx FFF +=

= −

x

z

F

F1tanθ

Hydrostatic force formulas

Ay

Iyy

c

ccp +=

AhF cγ= ( )AyF cθγ sin=

60

Ayc

Problem

� 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:Free surface

0.5m

D=2m

mD

hc 5.12

5.0 =+=

AhF c

=π

γ

62

Ah

Ihh

c

ccp +=

( )AyF cθγ sin=AhF cγ=

Ay

Iyy

c

ccp +=

D=2m( )

kNF

F

2.46

24

5.1810.9 2

=

= π

( ) ( )mh

DDh

Ah

Ihh

p

p

c

ccp

667.1

4/5.1/64/5.1 24

=

×+=

+=

ππ

Problem

� 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

30o

hphc

ycθsinyh =

63

4ft

5ft

hp

yp

yc

4ft

ftyh occ 130sin2sin === θ ( )( ) lbAhF c 12484514.62 =×== γ

( ) ( ) ftbdbdAy

Iyy

c

ccp 67.22/12/2 3 =×+=+= fthp 33.1=

Problem

� (b)

4fthp

yp

hcyc=4ft

θsinyh =1ft

ft

hy cc

4

sin/

== θ

64

5ft6ft

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

( ) ( ) ftbdbdAy

Iy

c

cp 33.44/12/44 3 =×+=+=

ftyh opp 167.230sin33.4sin === θ

Problem

Q 3.7.9

(a) For critical stability resultant of

65

(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

(a)

1m

4.4/3

RB

RN

66

(a)

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

Schematic pressure distribution diagram

( )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

� 3.7.12

67

Problem

� 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.

68

( ) mh

mkNF

AhF

p

x

cx

33.25.33

2

/1.605.3)2/5.3(81.9

==

=== γ

hpxp

`

hpxp

( )ALVOLWFz γγ ===

86.3)2(5.3

4/)4(4

)2(5.3 2

=−=

−=

π

π

A

A

2m

d=4m( )AFz )86.3(81.9== γ

69

mkN /9.37=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

� 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

70

� 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

� 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)

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� 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

� 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

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� 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”


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

� The force of buoyancy “resultant upward force or thrust exerted by fluid on

Water surface

11 hP γ=

2h

1h 1F

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thrust exerted by fluid on submerged body” is given

( )212 hhP += γ

2F

� dA=Area of cross-section of element

� γ= Specific weight of liquid

( ) ( )( )[ ]

[ ]volumeF

dAhF

dAhdAhhF

FFF

B

B

B

B

γγ

γγ

==

−+=−=

2

121

12


� =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.

[ ]volumeFB γ=

74

� Let

� Wa= weight of body in air

� Wl=weight of body in liquid

� FB=Wa-Wl


�� Center of Buoyancy Center of Buoyancy (B):(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.

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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:

[email protected]

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2- Fluid Statics

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