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FLUID STATICS



Outline of Chapter 3

The Basic Equation of Fluid Statics

Types of Pressures

Pressure Variation in a Static Fluid

Example Problem

Hydrostatic Force on Submerged Surfaces

Example Problems



Hydrostatic forces

Energy conversion

Bernoulli equation

Turbine

Hydrostatic uplift



What does Static Fluid mean?

Statics means that the fluid is not moving, i.e., its

velocity =0; and its acceleration = 0.

Fluid velocity = 0 means that it does not flow.

If a static fluid does not flow, how much shear

stress the fluid is exposed to?

The word “statics” is derived from Greek word“statikos”= motionless



What does Static Fluid mean?

In this case, fluid can be exposed to only

normal forces and behaves as

“a rigid body” – no deformation



The Basic Equations

of Fluid Statics

Newton’s 2nd law:

Divide both sides d∀ by gives:



The Basic Equations of Fluid Statics

• Surface Force

x

y

z



The Basic Equations of Fluid Statics

• Body Force



• Surface Force



• Total Force



• Newton’s Second Law



Fluid statics Chee 223 2.22

Pascal’s Laws

Pascals’ laws:

– Pressure acts uniformly in all directions on a small

volume (point) of a fluid – In a fluid confined by solid boundaries, pressure

acts perpendicular to the boundary – it is a normal

force.



Direction of fluid pressure on boundaries

Furnace duct Pipe or tube

Heat exchanger

Dam

Pressure is a Normal Force

(acts perpendicular to

surfaces)

It is also called a Surface

Force



Absolute and Gauge Pressure

• Absolute pressure: The pressure of a fluid is expressed relative

to that of vacuum (=0)

• Gauge pressure: Pressure expressed as the difference

between the pressure of the fluid and that of the surroundingatmosphere.

Usual pressure guages record guage pressure. To calculate

absolute pressure:

gaugeatmabs P P P



Units for Pressure

Unit Definition orRelationship

1 pascal (Pa) 1 kg m-1 s-2

1 bar 1 x 105 Pa

1 atmosphere (atm) 101,325 Pa

1 torr 1 / 760 atm

760 mm Hg 1 atm

14.696 pounds per

sq. in. (psi)

1 atm



Pressure Variation in a Static Fluid

Reference

level and

pressure

Location and

pressure of

interest



Measurement of Pressure

Manometers are devices in which one or more

columns of a liquid are used to determine the

pressure difference between two points.

– U-tube manometer

– Inclined-tube manometer



Measurement of Pressure Differences

mambb

mmba

gR Z g P P

R Z g P P

)(

)(

3

2

Apply the basic equation of static fluids to

both legs of manometer, realizing that

P2=P3.

)( bamba gR P P



Inclined Manometer

•To measure small pressure differences need to magnify Rm some way.

sin)(1 baba gR P P




In order to fully determine the force on a surface

submerged in a liquid, we must determine the

following:

1. The magnitude of the force;

2. The direction of the force; and

3. The line of action of the force.



1. Direction of the Force on a

Plane Submerged Surface Since fluid is not moving (static), there is no

shear, i.e., only normal forces might exist.

Since this force is caused by pressure of fluid,

it will always be normal to the surface.

This determines the direction of the force.



3- Line of Action of the Force on a

Plane Submerged Surface




• Plane Submerged Surface

We can find F R , and y ́ and x ́ ,

by integrating, or …





– Algebraic Equations – Total Pressure Force





– Algebraic Equations – Net Pressure Force



Fluid statics Chee 223 2.43

Centroid Location for Common Shapes




• Curved Submerged Surface




• Curved Submerged Surface

– Horizontal Force = Equivalent Vertical Plane Force

– Vertical Force = Weight of Fluid Directly Above(+ Free Surface Pressure Force)



Buoyancy



Buoyancy• A body immersed in a fluid experiences a vertical buoyant

force equal to the weight of the fluid it displaces• A floating body displaces its own weight in the fluid in

which it floatsFree liquid surface

The upper surface of

the body is subjected

to a smaller force than

the lower surface

A net force is acting

upwards

F1

F2

h1

h2

H



Buoyancy

The net force due to pressure in the vertical direction is:

FB = F2- F1 = (Pbottom - Ptop) (DxDy)

The pressure difference is:

Pbottom – Ptop = g (h2-h1) = g H

Combining:

FB = g H (DxDy)

Thus the buoyant force is:

FB = g V



Compressible fluid

• Gases are compressible i.e. their density varies with

temperature and pressure = P M /RT

– For small elevation changes (as in engineering

applications, tanks, pipes etc) we can neglect the

effect of elevation on pressure – In the general case start from:

o

o

RT

z z M g P P

T for

)(exp

:constT

1212

g

dz

dP

Compressible



CompressibleLinear Temperature Gradient

)( 00 z z T T

z

z

p

p z z T

dz R M g

pdp

00)( 00

R

M g

T

z z T p z p

0

000

)()(



Atmospheric Equations

• Assume linear

R

M g

T

z z T p z p

0

000

)()(

• Assume constant

0

0 )(

0)( RT

z z M g

e p z p

Temperature variation with altitude

for the U.S. standard atmosphere



Compressible Isentropic

v

pC C

P constant

P

1

1 y

P

P

T

T 1

11

D

D

1

12

1

1

121111

RT z gM T T

RT z gM P P



Vocabulary List

1. Static fluid

2. Manometer.

3. Hydrostatic pressure.

4. Gauge pressure.

5. Vacuum.

6. Hydrostatic force on a submerged surface.



The nd

Terima kasih

Dam Design



Dam Design

Design concern: (Hydrostatic Uplift) Hydrostatic pressure

above the heel (upstream edge) of the dam may causeseepage with resultant uplift beneath the dam base (depends

largely on the supporting material of the dam). This reduces the

dams stability to sliding and overturning by effectively reducing

the weight of the dam structure. (Question: What prevents the

dam from sliding?)

Determine the minimum compressive stresses in the base of a

concrete gravity dam as given below. It is important that this

value should be greater than zero because (1) concrete has

poor tensile strength. Damage might occur near the heel of thedam. (2) The lifting of the dam structure will accelerate the

seeping rate of the water underneath the dam and further

increase hydrostatic uplift and generate more instability.



30 m40 m

20 m

concrete=2.5 water

First, calculate the weight of the dam (per unit width):

W=Vg=(2.5)(1000)(20)(40)(1)(9.8)=19.6106 (N)

The static pressure at a depth of y: P(y)=wgy

y

Free surface

30 22 6

w w

0

The total resultant force acting on the dam by the water pressure is:

R= P(y)dy= (1000)(9.8)(1/ 2)(30) 4.4 10 ( )2

hh

gydy g N



3

30 3 w2

w w w0

The resultant force, R, is acting at a depth h below the free surface so that

23

Rh= P(y)ydy= ( ) , 20( )3 3

Assume the load distribution under the dam is linear (it mig

hh

g h h

gy ydy g y dy g h m R

max minmin

ht not be linear if the soil

distribution is not uniform)

Therefore, the stress distribution can be written as

(x)=20

x

R

W

20 m x dam,x

20

y , max min

0

In order to reach equilibrium, both the sum of forces and

the sum of moments have to balance to zero

F 0, R=F (frictional force and the air drag force)

F 0, ( ) 10( )

1.96 1

dam yW F x dx

6

max min0 ( ) N min

max

Free surface

Example (cont.)



20

O

0

20 20

6 6 2max minmax

0 0

6

max min

The sum of moments has to be zero also: Taking moment w.r.t. the heel of the dam

M 0, (10) (10) ( ) 0

(10)(4.4 10 19.6 10 )20

240 10 133.3 66.7

Solve:

R W x xdx

xdx x dx

6 6

max min1.64 10 ( ), 0.32 10 ( )

The minimum compressive stress is significantly lower than the maximum stress

The hydrostatic lift under the dam (as a result of the buoyancy induced by water seeping

un

N N

lift

der the dam structure) can induce as high as one half of the maximum

hydrostatic head at the heel of the dam and gradually decrease to zero at the other end.1

That is ( ) (0.5)(1000)(9.8)(302

w gh 6

6

) 0.147 10 ( )

Therefore, the effective compressive stress will only be 0.173(=0.32-0.147) 10 ( ).

N

N

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