lecture 2 - fluid mechanics

7/23/2019 Lecture 2 - Fluid Mechanics

http://slidepdf.com/reader/full/lecture-2-fluid-mechanics 1/61

Fluid Static

Dr. Mohammed Zakria Salih Xoshnaw



Ch Fluid Statics

• Fluid either at rest or moving in a manner that ther

is no relative motion between adjacent particles.

• No shearing stress in the fluid

• Only pressure (force that develop on the surfaces othe particles)



Outline

1. Pressure at a Point

2. Basic Equations for the Pressure Field

3. Hydrostatic Condition

4. Standard Atmosphere

5. Manometer and Pressure Measurements

6. Barometer

7. Piezometer 8. Differential manometer

9. Example Problems



Fluid Mechanics Overview

Gas Liquids Statics Dynamics

Air, He, Ar,

N2, etc.

Water, Oils,

Alcohols,

etc.

0 i F

Viscous/Inviscid

Steady/Unsteady

Compress

Incompres

0 i F

Laminar/

Turbulent

, Flows

Compressibility ViscosityVapor

Pressure

Density

PressureBuoyancy

Stability

Chapter 1: Introduction Chapter 2: Fluid StaticsFluid Dynami

Rest of Cour

Surface

Tension

Fluid Mechanics



1. Pressure at a point N/m2 (Force/Area)

am F

Y: sin s x P z x p F s y y

ya

z y x

2

Z: z z z z a

z y x s x p y x p F

2

cos

z

a z y x

2



sinz ; cos s s y

2)(:

2 :

z a p p z

ya p p y

z s z

y s y

What happen at a pt. ?0,, z y x

s z

s y

p p

p p

s z y p p p θ

is arbitrarily chosen

Pressure at a pt. in a fluid at rest, or in motion, is

independent of direction as long as there are no shearing

stresses present. (Pascal’s law)



2. Basic equation for Pressure Field

Surface & body forces acting on small fluid element

pressure weight

How does the pressure in a fluid which there are no shearing stresses vary

from pt. to pt.?



Surface forces:

z x y

y

p p z x

y

y

p p F y y

)

2()

2(:

z y x y p F y

Similarly, in z and x directions:

z y x

x

p F x

z y x

z

p F z

z y xk z

p j

y

pi

x

pk F j F i F F z y x s )(

k z

j y

i x

z y x p

)(



Newton’s second law

W F am F s

z y x

3. Pressure variation in a fluid at rest

ak p

General equation of motion for a fluid in which there

are no shearing stresses.

2Eq.(

0

0

0 0

dz

dp

y

p

x

p

k pa

z y x z y x p F

2 3 1 I ibl



2.3.1 Incompressible

hγ ) z z ( γ p pdz γdp

const g ργ

z

z

p

p

1221

1

2

2

1

Hydrostatic Distribution21

phγ p *see Fig. 2.2

21 p p

h

pressure head

Ex:

)133()62.4(

518or1.2310

32

21

m KN

ft lb

mmHg ft h p p psi

phγ p

Pressure in a homogeneous, incompressible fluid at rest: ~ reference level,

indep. of size or shape of the container.



The required equality of pressures at equal elevations

Throughout a system.

1

1

2

2 F A

A F

Transmission of fluid pressure



2.3.2 Compressible Fluid perfect gas:

RT ρ p

2

1

2

1)z(zconst., ln 21

1

2 p p

Z Z R g

T

dz

R

g

p

p

p

dp RT

gp g

dz

dp

Assume

0

12

12

210

)(

exp

conditionsisothermal ,

RT

z z g

p p

z z over T T



Troposphere:

00357.0

0065.0

0@

ft

R

ratelaposem

K

z T z T T aa

a

a )

T

z β ( p p 1

2.4 Standard Atmosphere



vapor atm phγ p

(Mercury barometer)

2.5 Measurement of Pressure

Parameter= measure atmospheric pressure

A

atom B

B A

p p

p p

A ph p

h



Manometry

1. Piezometer Tube:

2. U-Tube Manometer:

3. Inclined-tube manometer

P h p

11h p A



gasanotliquid,3.

reasonableish2.

1.

1 a

a

p p

p p

2. U-Tube Manometer:

1122

2211 0

hh phh p

A

A



)γγ( h p p

p )hh( γhγhγ p p pk ) flowtheof ratevolumethe( Q

p p p Δ , p Δ ,u

B A

B A

B A

B A

122

2112211

Small difference in gas pIf pipes A & B contain a



θ sinγ

p pl

θ sinl γ p p

B A

B A

2

2

22

Inclined – Tube manometer

2 7 Mechanical and Electronic Pressure Measuring Device



2.7 Mechanical and Electronic Pressure Measuring Device

.Bourdon pressure gage (elastic structure)

Bourdon Tube

, p curved tube

straight

deformation

dial

.A zero reading on the gage indicates that the measured

pressure

. Pressure transducer－pressure V.S. time

Bourdon tube is connected to a linear variable

differential transformer(LVDT), Fig. 2.14

coil; voltage

.Aneroid barometer－measure atmospheric pressure

(absolute pressure)

This voltage is linear function of the pressure and could



This voltage is linear function of the pressure, and could

be recorded on an oscillograph, or digitized for storage

or processing on computer.

Disadvantage-elastic sensing element

meas. pressure are static or only changing

slowly(quasistatic).

relatively mass of Bourdon tube

<diaphragm>

*strain-gage pressure transducer *

Fig. 2.15 (arterial blood pressure)

piezo-electric crystal. (Refs. 3, 4, 5 )

1Hz

Application Examples



Application Examples

Feeder Gates for Canal

Gate Valves fo

Spillway Contr

Applications (cont.)



Applications (cont.)

Spillway Drum Gates

hollow inside, use

buoyancy to control

position of the gate.

2 8 Hydrostatic Force on a Plane Surface



2.8 Hydrostatic Force on a Plane Surface

pA F R

Storage tanks, ships

Fig. 2.16 Pressure and resultants hydrostatic force

developed on the bottom of an open tank.

. For fluid at rest we know that the force must be

perpendicular to the surface, since there are no shearing

stress present.

H d t ti F I li d Pl S f



Hydrostatic Force on an Inclined Plane Surface

h y

dF

sin

Integrate over the entire surf

sin

Define centroid of the area y

1, so that

sin

In order to find equilavent sy

need to make sure that

R

C

R C C

dF PdA hdA ghdA

gy dA

F dF g ydA

y ydA A

F gAy gAh

the m

of the resultant force must eq

the moment of the distribute

x

y

x

Assume atmoshperic condition on the other

side of the surface

Free surface

Hydrostaic forces



y

FR

O

O

R

2

2

xx

C

Taking Mmoment about the x-axis: y'F

' sin sin s

Recognize that y (area mome

ITherefore, y'=

Ay

Also, from parallel axis theorem, we can

C

A

xx A

y g y A gy dA g

dA I

2 ˆ ˆxxˆ ˆxx xx C

ˆ ˆxy xy

C

C

moment of inertia about the centroid of

found in table)

II = I , therefore, y'=y

I ISimilarly, x'= x

Ay

C

C

C

Ay Ay

Ay

z

y

y

x

FR

z

y’

yc



Example



hinge

The square flood gate (2m by 2m) is hinged along its bott

shown. Determine the moment at the hinge in order to

the gate steady.

R

3

ˆ ˆ

C

First, find the resultant force:

F (1000)(9.8)(1)(2 2) 39200(Then, determine the point of action:

1 (2)(2) 1 412y'=y (1) 1 ((2 2)(1) 3 3

As expected, it falls at a depth 2/3 of the tot

C

xx

C

gh A N

I m

Ay

O

al

The holding moment (M) on the hinge O will b

4M (2 ) 0,

3

18479( . )

R M F

M N m

y

x

O

2my’

Example (cont.)



2my’

If the square gate is replaced by a circular

shaped gate as shown, recalculate the ho

moment.

x

y

2

R

4

ˆ ˆ

C 2

O

Again, find the resultant force first:

F (1000)(9.8)(1) (1) 30772

Next, the line of action:

11 54y'=y 1 1 ( )

(1) 4 4

The holding moment:

5 3M (2 ) 0

4 4

23079( . )

C

x x

C

R R

gh A

R I m

Ay R

M F M F

M N m

Example (cont.)45°



2m

y

y’

45

If the square gate is placed at an angle of 45° as

recalculate the holding moment again. Note: th

has been redefined to follow the gate for conve

R

ˆˆ

C

First, calculate the resultant force:

F (1000)(9.8)(1)(2 2 2 2) 78400(

Note: the h stays the same and is independent

of the incline angle, however, the gate area increases

1 (22 2 12y'=y

2

C

xx

C

gh A N

I Ay

3

R

2)(2 2)

2 2(2 2 2 2)2

2 4 2' 2

3 3

4 2The holding moment: M=F (2 2 ) 73916(3

y

An interesting observation



When the gas tank is low, the low fuel light will lit to warn the driver. Have you n

that the light will not always stay on for a period of time. It turns off when either

accelerate (decelerate) or climb (descend) on a sloped road. Can you explain this

phenomenon by using the principle of fluid statics.

Fuel level transducer

Accelerating (climbing) Decelerating (descending)

Hydrostatic balance can be applied to a small fluid element as shown

d



( ) , , integrate from fluid elemen

the free surface ( )

dp pA p dp A mg Agdy g

dy

p h

p gh

p

p+dp h

Free surface, p=p

Example: If a container of fluid is accelerating

with an acceleration of ax to the right as shown belothe free surface of the fluid will incline with an ang

shown.

ax

p p+dp

1

( ) ,

tan( ) , tan

x x

x

x

dp pA p dp A ma Adxa dx

dpa dy g

dpdx g a

a a

a

dx

dy

x

y

2.9 Pressure Prism

the pressure varies linearly with depth



the pressure varies linearly with depth.

Ah

bhh

e prismof pressur volume F

Ah

A P F

R

Ave R

2))((

2

1

)

2

(

No matter what the shape of the pressure prism is, the resultaforce is still equal in magnitude to the volume of the pressure

Prism, and it passes through the centroid of the volume.

First, draw the pressure prism out. dz

dp

0 p z p



Hydrostatic Force on a Curved Surface• General theory of plane surfaces does not apply to curved surfaces

• Many surfaces in dams, pumps, pipes or tanks are curved

• No simple formulas by integration similar to those for plane surfaces

• A new method must be used

Isolated Volume

Bounded by AB an AC and

BC

Then we mark a F.B.D. for the volume:

F1 and F2 is the hydrostatic force on each

planar face

FH and FV is the component of the result

force on the curved surface.

W is the weight of the fluid volume.



Hydrostatic Force on a Curved Surface

Now, balancing the forces for the Equilibrium condition:

Horizontal Force:

Vertical Force:

Resultant Force:

The location of the Resultant Force is through O by sum of Moments:

H H

V V c

x F x F

x F Wx x F

22

11Y-axis:

X-axis:



Buoyancy: Archimedes’Principle



c p e

Archimedes (287-212 BC) Story

•Buoyant force is a force that results from a floating or submerged body in a fluid.

•The force results from different pressures on the top and bottom of the object

•The pressure forces acting from below are greater than those on top

Now, treat an arbitrary submerged object as a planar surface:

Arbitrary Shape

V

Forces on the Fl

Archimedes’ Principle states that the buoyant

force has a magnitude equal to the weight of the

fluid displaced by the body and is directed

vertically upward.



Buoyancy and Flotation: Archimedes’ Principle

Balancing the Forces of the F.B.D. in the vertical Direction:

V AhhW 12

W is the weight of the shaded area

F1 and F2 are the forces on the plane surfaces

FB is the bouyant force the body exerts on the fluid

Then, substituting:

Simplifying,

The force of the fluid on the body is opposite, or vertica

upward and is known as the Buoyant Force.

The force is equal to the weight of the fluid it displaces.




Sum the Moments about the z-axis:

Find where the Buoyant Force Acts by Summing Moments:

We find that the buoyant forces acts through

the centroid of the displaced volume.

The location is known as the center of buoyancy.

VT is the total volume of the parallelpiped




We can apply the same principles to floating objects:

If the fluid acting on the upper surfaces has very small specific weight (air), the centroid

is simply that of the displaced volume, and the buoyant force is as before.

If the specific weight varies in the fluid the buoyant force does not pass through the

centroid of the displaced volume, but through the center of gravity of the displaced

volume.



Stability: Submerged Object

Stable Equilibrium: if when displaced returns to equilibrium position.

Unstable Equilibrium: if when displaced it returns to a new equilibrium position

Stable Equilibrium: Unstable Equilibrium:

C > CG, “Higher” C < CG, “Lower”



Buoyancy and Stability: Floating Object

Slightly more complicated as the location of the center buoyancy can change:



Pressure Variation, Rigid Body Motion: Linear Motion

Governing Equation with no Shear (Rigid Body Motion):

The equation in all three directions are the following:

Consider, the case of an open container of liquid with a constant acceleration:

Estimating the pressure between two closely spaced points apart some dy, dz:

Substituting the partials

Along a line of constant pressure, dp = 0: Inclined free

surface for ay≠ 0



Pressure Variation, Rigid Body Motion: Linear Motion

Now consider the case where ay = 0, and az ≠ 0:

0

x

pRecall, already:

z

a g z

p

y

p

0Then,

So, Non-Hydrostatic

Pressure will vary linearly with depth, but variation is the combination of gravity and externally

developed acceleration.

A tank of water moving upward in an elevator will have slightly greater pressure at the bottom.

If a liquid is in free-fall az = -g, and all pressure gradients are zero—surface tension is all that keep

the blob together.



Pressure Variation, Rigid Body Motion: Rotatio

Governing Equation with no Shear (Rigid Body Motion):

Write terms in cylindrical coordinates for convenience:

Pressure Gradient:

Accceleration Vector:

Motion in a Rotating Tank:

P V i ti Ri id B d M ti R t ti




The equation in all three directions are the following:

Estimating the pressure between two closely spaced points apart some dr, dz:

Substituting the partials

Along a line of constant pressure, dp = 0:

Equation of constant pressure surfaces:

The surfaces of constant pressure are parabolic

P V i ti Ri id B d M ti R t ti




Now, integrate to obtain the Pressure Variation:

Pressure varies hydrostaticly in the vertical, and increases radialy

lecture 2 - fluid mechanics

Documents