1

Hydrostatics

• Study of Fluids at rest

• Dams

• Tanks or storage vessels

No Shear stresses in the fluid field

• Moving with no relative motion

• Uniformly accelerating tank

• Rotating cylinders

Nature of Forces

• Body Force

• Acts throughout the bulk of the body (fluid)

• Gravity force

• Centrifugal Force

• Inertial Force

• Electromagnetic force

Nature of Forces

• Surface Force

• Acts on the surface of the body (fluid)

• Pressure Force

• Viscous Force

• Surface tension Force

sxps δδ

x

y

z

zxpy δδ

yxpz δδ

xδyδ

zδ

2

zyxg

δδδρ

Pascal’s Law - I

• The pressure at a point in a shear stress free

fluid is independent of direction

2

zyxgcossxpp syxz

δδδρ+θδδ=δδ

2

zgpp,ycossAs sz

δρ+=δ=θδ

Vertical Force Balance

Horizontal Force Balance

θδδ=δδ sinsxpp szxy

sy pp,zsinsAs =δ=θδ

When we shrink the element to a point, δz tends to zero,

Hence pz = py = ps

Pascal’s Law - IIGoverning Equation for Pressure Field - I

δ x

δ y

δ zzxp δδ

yxHoTzz

pp δδ

+δ

∂

∂+

zyp δδ

zyHoTxx

pp δδ

+δ

∂

∂+

yxp δδ

zxHoTyy

pp δδ

+δ

∂

∂+

y

z

x

gzyx ρδδδ

2

Governing Equation for Pressure Field - II

Z-direction Force Balance

( ) 0gzyxyxzHoTzz

ppyxp 2 =ρδδδ−δδ

δ+δ

∂

∂+−δδ +

0zyx)z(HoTgz

p=δδδ

δ+ρ+

∂

∂− +

In the limit of shrinking the volume to a point after

dividing the LHS by the volume

0gz

p=ρ+

∂

∂

Governing Equation for Pressure Field - III

Similar Force Balance in x and y directions would yield

0x

p=

∂

∂0

y

p=

∂

∂and

Important conclusion is pressure varies only in z direction

This is because there was body force only in z direction We shall generalize in the next slide after making some observations

1. The weight acts downwards, whereas positive z is upwards, so we can write g = -gz

2. The final equations were force per unit volume = 0

3. In general, we can write the three equations as

Some Observations

0gz

pz =ρ+

∂

∂−0g

x

px =ρ+

∂

∂− 0g

y

py =ρ+

∂

∂−

4. We can combine the whole set into one vector equation as

0gp =ρ+∇−r

Interpretation of the final equation

0gp =ρ+∇−r

p∇− Is the net surface force (due to pressure) per unit volume of fluid in the positive direction (in terms of the components)

gr

ρ Is the net body force per unit volume of fluid in the positive direction

Evaluation of pressure distribution

• If we define positive z to be pointing upwards

0gp =ρ+∇−r

0)g(z

p=−ρ+

∂

∂−

or ∫∫ ρ=−z

z

p

p refref

gdzdp

)zz(gpp refref −ρ−= Pressure decreases with increase in elevation

• If gravity was the only body force involved, then we had concluded that there will be pressure variation only in vertical direction

prerf zref

p z

z

g

If ρ is constant

then

Manometry

1. In a constant density fluid, pressure at a given elevation from datum is same

2. This is exploited in measurement of pressure using manometers

The product ρg is called and is denoted by γ gρ=γ

3

Terminology in pressure measurement Measurement of atmospheric pressure

hpp vapatm γ+=

1. Standard sea level = 760 mm of Hg

2. This is about 10 m of water column

Toricilli Barometer

22ref3 hpp γ+=

112A hpp γ−=

Capillarity effects are negligible for large bore tube ie., diameters greater than 30 mm

Pref = 0

U-Tube Manometer

32 pp =

1122refA hhpp γ−γ+=∴

U-Tube Manometer

B332211A phhhp =γ−γ−γ+

B332211A phsinlhp =γ−θγ−γ+

θsin

1isionMagnificat

Inclined Tube Manometer

• Inclination up to 10o Ok

• Below this angle miniscis error becomes

large

ll

Pressure Gauges

There are other type of transducers.

You may learn about them in Instrumentation course

4

F2 = (A2/A1)F1

If A1 << A2 then F1 << F2

Hydraulic press

Mechanical Advantage = A2/A1

γ−=dz

dp∫∫∫ =ρ−=2

1

2

1

2

1

z

z

z

z

p

p

dzgRT

pdzgdp

Gases are compressible and hence density is not constant

∫−=2

1

z

z1

2

T

dz

R

g

p

pln

Isothermal condition

o

12

1

2

TR

)zz(g

p

pln

−−=

−−=

o

12

1

2

TR

)zz(gexp

p

p

Pressure distribution in compressible fluid - I

0TttanconsTIf ==

In general T = T (z)

T = 150 C (288.15 K)

p = 101.33 kPa (abs)

ρ= 1.225 kg/m3

γ= 12.014 N/m3

In Troposphere T varies Linearly

m/K0065.0

zmTT o

=β

−=

Airplanes

Temperature Variation in atmosphere

∫−=2

1

z

z1

2

T

dz

R

g

p

pln ∫

−−=

2

1

z

z o1

2

zmT

dz

R

g

p

pln

( )( )1o

2o

1

2

zmT

zmTln

m

1

R

g

p

pln

−

−

−⋅−=

mRg

1o

2o

1

2

zmT

zmTln

p

pln

−

−=

Pressure distribution in compressible fluid - II

Linear Temperature Variation

mRg

1o

2o

1

2

zmT

zmT

p

p

−

−=

• Determination of Hydrostatic forces is

important for the design of storage tanks,

pools, dams, ships and other hydraulic

structures

Applications to sumberged objects

• For fluids at rest

• The force must be perpendicular to the surface

since there are no shearing stresses present.

• The pressure will vary linearly with depth if the

fluid is incompressible.

FR = γγγγ hA

Pressure distribution on plane surface -I

5

yyC

yR

FRdF

hhc

Free Surface

Centroid, cLocation of resultant force (center of pressure, CP)

O - origin

xR

xC

dA

x

y

x

Pressure distribution on plane surface -II

∫∫∫ θγ+=γ+==A

o

A

o

A

R dAsinyApdA)hp(dApF

Pressure distribution on plane surface -III

AysinApdAysinAp co

A

o θγ+=θγ+= ∫

∫=A

c dAyA

1yNote y-coordinate of centroid

coR hAApF γ+=As θ= sinyh cc

hc is the vertical distance between free surface and centroid

Note that there will also be a force poA that will act in the opposite direction of FR from the other side of the plate. Hence in the net force, atmospheric pressure will cancel

[ ] [ ]∫∫∫ θγ=γ==AAA

RR dAsinyydAhydFyyF

xx

A

2 IsindAysin θγ=θγ= ∫

Pressure distribution on plane surface -IV

Position of the resultant force

Moment of the resultant force = Moment of the distributed force

Since atmospheric pressure would cancel it is not being

carried around

( )2

ccxx AyIsin +θγ= −

cccxx AysinyIsin θγ+θγ= −

cRcxxcccxx yFIsinAyhIsin +θγ=γ+θγ= −−

Pressure distribution on plane surface -V

R

cxxcR

F

Isinyy −θγ

+=∴

R

cxy

cRF

Isinxx

−θγ+=∴

Similarly we can find xR (Exercise)

Resultant force does not pass through the centroid but is always below it

The values of second

moment of area for some shapes. The

nomenclature is a bit different but easy to

grasp

cR hAF γ=

MAGNITUDE OF THE RESULTANT FORCE

LOCATION OF THE RESULTANT FORCE

Direction : FR

IS PERPENDICULAR TO THE SURFACE

R

cxxcR

F

Isinyy −θγ

+=∴

R

cxy

cRF

Isinxx

−θγ+=∴

Summary

6

2H FF = WFF 1V +=

Point O – summing moments about an appropriate axis

Pressure distribution on curved surface

Buoyancy force on the body = Weight of the fluid displaced by the body

Buoyancy Force

• When a body is completely submerged in a fluid, or floating so that it is only partially submerged, the resultant upward force acting on the body due to pressure on the surfaces is called buoyancy force

Archimedes Principle – Greek Scientist (287 BC – 212 B.C)

• The line of action of the buoyant force passes through the centroid of the displaced volume. The centroid is called the Center of buoyancy

( )( )blPPF 12B ×−=

∫∫ γ−==−2

1

2

1

z

z

p

p

12 dzdpPPBut

( ) ( )∫∫ γ×=

γ−×=∴

1

2

2

1

z

z

z

z

B dzbldzblF

y

P1

x

z

l

P2b

z1

z2

Weight of the fluid displaced by the body

gmdmg fluid

domain

fluid == ∫

Buoyancy Force - II

A simple case

Buoyancy Force - III

Liquid (ρρρρf)

A

B

C

D

General interpretation

• Consider an imaginary water lump ABCD

• The vertical force on the top surface ABC will be

equal to the weight of the liquid above it

• Similarly, the vertical force on the

bottom surface ADC will be equal to

the weight of the liquid above it

• Thus the net vertical force on the

lump will be equal to the weight of

liquid lump

• If a body replaces the lump, the force

field will not change

Archimedes Principle holds good

• for bodies of any general shape

• for both gases and liquids

• does not require density to be constant

Buoyancy force is important

• For naval vehicles

• lighter than air vehicles – hot air balloons

Buoyancy Force - IV

Vs

Water (ρρρρwater)

A

Vs

Other liquid

Ah

( )hAV

VSG

s

s

m=⇒

Principle of Hydrometer

swater VgW ρ=

• When immersed in water

( )hAVgSGW swater mρ=

• When immersed in an unknown SG fluid

=

sV

hA1

1

m sV

hA1±≈ Stem can be

calibrated to read SG

• Hydrometer is used to

find the specific gravity

of liquids

Original level

7

Stability of Floating Objects

• The point of action of buoyancy force is called Centre of

Buoyancy

• It is the CG of the submerged volume

• If CG of the body is below the centre of buoyancy, the

object is stable

Stability of Floating Objects-II

• The problem of determining stability is complex as the

centre of Buoyancy shifts when the object is tilted

• Short wide bodies are normally in stable equilibrium

Stability of Floating Objects-III

• Tall slender bodies are generally unstable • Even though a fluid may be in motion, if it moves as a

rigid body there will be no shearing stresses present

)ag(x

pxx −ρ=

∂

∂)ag(

y

pyy −ρ=

∂

∂)ag(

z

pzz −ρ=

∂

∂

Pressure variation in fluids with rigid

body motion

• An acceleration of a particle sets up an inertial force in

the direction opposite to the acceleration (called

d'Alembert’s force

• The general governing equation for fluid in a

gravitational and accelerational field can be stated as

0)ag(p =−ρ+−∇rr

)ag(prr

−ρ=∇or

Pressure variation in a tank subjected to

rectilinear uniform acceleration-I

,0x

p=

∂

∂⇒ a

y

pρ−=

∂

∂g

z

pρ−=

∂

∂

ay

pρ−=

∂

∂c)z(fayp ++ρ−=⇒

similarly gz

pρ−=

∂

∂c)y(fgzp ++ρ−=⇒ c)aygz(p ++ρ−=⇒

P is only a function of y and z

ax = az = gx = gy = 0, ay = a, gz = -g

xy

z

xy

z

ay

g

• The equation for a constant pressure line shall be

Cc)aygz(p =++ρ−= C)aygz( =+⇒ yg

aCz −=⇒

• Choosing the origin such that the free surface left hand

side is the origin, then C = 0 for the free surface

• Therefore, the equation of the free surface is

yg

az −=⇒

• Assuming the fluid to be incompressible, the mid-point

of free surface is unaffected

Pressure variation in a tank subjected to

rectilinear uniform acceleration-I

8

The case of rotating cylinder-I

)ag(r

prr −ρ=

∂

∂)ag(

p

r

1θθ −ρ=

θ∂

∂)ag(

z

pzz −ρ=

∂

∂

• In cylindrical coordinates the governing equation can be

stated as follows

aθ

= az = gr = gθ

= 0, ar = -ω2r, gz = -g

The case of rotating cylinder-II

rr

p 2ωρ=∂

∂0

p

r

1=

θ∂

∂g

z

pρ−=

∂

∂

rr

p 2ωρ=∂

∂c)z(f

2

rp

22

++ωρ

=⇒

similarly gz

pρ−=

∂

∂c)r(fgzp ++ρ−=⇒ c)

2

rgz(p

2

+ω

+ρ−=⇒

• The equation for a constant pressure line shall be

g2

rCz

2ω+=⇒C)

2

rgz(

2

=ω

−⇒Cc)2

rgz(p

2

=+ω

−ρ−=⇒

p is only a function of r

and z

• Choosing the origin such that the free surface centre is

the origin, then C = 0 (z=0, r=0)

The case of rotating cylinder-III

g2

rz

2ω=⇒

• Thus, free surface is a paraboloid of revolution