lecture 02_fluid_statics

8/7/2019 Lecture 02_fluid_statics

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Announcement

Class on Thursday, September 23rd Types of fluid flow

Application of Bernoulli Equation


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PRESSURE : Static FLUID1. Pressure is the same over any

horizontal plane

2. Assumption: Density is same overany horizontal plane

3. Pressure at a point is independentof the direction of the PLANE/AXISused to define it.

gdz

dp


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Pressure at a point is independent of thedirection of the surface used to define it.

P

3


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Pressure at a point is independent of thedirection of the surface used to define it.

P

B

B

C

C

4


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PRESSURE : Definition

Pressure = Force exertedArea

Under static conditions a fluid will

exert a force normal to a solidboundary or any plane drawn through

the fluid.

p = F/A p *A = F

5


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Variation of Pressure withPosition in a Fluid

An Element of fluid

Hydrostaticforces normal tothe planes of theelement

F = p * A

6


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P

Q

(p+p) A

pA

gAL

z+ z

z

Figure 1


Assuming UP as +veZ.

Datum 7


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Resolving forces along the length of the cylinder

0cosLAgApApp

0zgp 0z

gz

p

That is Pressure decreasesas elevation increase.

Since fluid is static resultant force is equal to zeroand there is no shear forces because there is nomotion.

As the elementreduces to apoint.

8


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+ve

h = depth

Z is the axis of

elevation.Pressure decreasesas elevation increase.

+ve

z

-ve

z

+ve p

p/zFluid surface

Depth is + [+h]In the directionof negative

elevation [-Z] 9


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Pz


z+ zQ

Points P and Q are vertically apart by a distanceof Z. The pressure differencep = gZ.

10


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Same Pressure on Horizontal Plane

QP R

T S

All the points on a horizontal plane are atthe same elevation i.e.Z = 0Therefore p = 0 and all the points havethe same pressure value.

Horizontal plane through fluid11


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Variation of Pressure with Position

CD

BA

pA = pB = FB

AB

= FA

AA

If ratio AA:AB is 10then FA = 10*FB .

12


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z

Same Density on Horizontal Plane

g

z

p

zz

pp

PQ S

R

g

dz

dp

13


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PRESSURE1. Pressure at a point is independent of

the direction of the surface used todefine it.

2. Pressure same over any horizontalplane

3. Assumption: Density same over any

horizontal plane

4. gdz

dp


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Change in Density

of water withTemperature

Density of liquid water

Temp (C)Density

(kg/m3)[10][11]

+100 958.4

+80 971.8+60 983.2

+40 992.2

+30 995.6502

+25 997.0479

+22 997.7735+20 998.2071

+15 999.1026

+10 999.7026

+4 999.9720

0 999.8395

10 998.117

20 993.547

30 983.854

The values below 0 C referto supercooled water

See graph in next slide

15
http://en.wikipedia.org/wiki/Properties_of_waterhttp://en.wikipedia.org/wiki/Supercoolinghttp://en.wikipedia.org/wiki/Supercoolinghttp://en.wikipedia.org/wiki/Properties_of_waterhttp://en.wikipedia.org/wiki/Properties_of_water


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955960

965

970

975980

985

990

9951000

1005

-50 0 50 100 150

(kg/m3)

T (oC)

Density of water VS Temperature

16


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PRESSUREg

dzdp

ghph

gdzp ttanconsgzp

Qpghp

a

hzQAt , +vep

atm

pa -veh

Q- Z

Pressure at Q = atmospheric pressure +

water pressure at depth h


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Atmospheric pressure

MEASUREMENT OF PRESSURE

Gauge pressure

Absolutepressure A

Vacuum = negativegauge pressure

Absolutepressure B

Barometer reading(varies with altitudeand slightly withweather)

18


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Pressure Head

ghp

g

ph

19


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1. Piezometer

P

h

P

h = Pressure Head20

Pi t i P


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Piezometric Pressure

and Head

DATUM

gh = p

z

gzp

pressure

cpiezometri

zh

g

gzp

head

cpiezometri

Pressure, p

(h = p/g)

= total from Datum

21


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2. U-Tube Manometer

A

B

P Q

x

y

CQP pp

gxp mQ gypp wCP

gygxp wmC

22


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Pressure Difference

PQ x

p1p2

y

QP

pp

gxgyppmwQ

2 xygpp

wP

1

Mercurymanometer

x

g

pp

w

m

w

1

21

23


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Announcements


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HYDROSTATIC THRUSTS ONSUBMERGED SURFACES

Magnitude of the force

Direction of the force

Line of action

25

HYDROSTATIC THRUSTS ON


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HYDROSTATIC THRUSTS ONSUBMERGED SURFACES

26

S i i


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StatisticsConstruction period: 1931 to 1936

Construction cost: $49 M (= $676 M)

Volume of water: 35.2 km (35.2 x 109 m3)

Area: 639 km, backing up 177 km behind the dam.

Dam height: 221.4 m.

Dam length: 379.2 m.

Dam thickness:200 m at its base;

15 m thick at its crest. 27


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28


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Location and Magnitude of Resultant Force

PivotMid point

Pivot & mid point coincide when force isevenly distributed

Evenly distributed force

29


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Resultant forceEvenlydistributed

force

Pivot, center &resultant force

coincide

30


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Location and Magnitude of Resultant Force

Pivot Mid point

31


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Pivot & resultantforce coincide BUT

not mid point

Force not uniformly distributed

32

Fluid surface


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Hydrostatic force

distribution. Forceincrease with depth

Fluid surface

33


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Location and Magnitude ofResultant Force

Resultant force below mid point

34

Hydrostatic Forces


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Hydrostatic Forces

0.5 m diameter

3 m water, = 997 kg/m3

patm

2/34.290.381.9997 mkNp

kNApF 76.5

?

35


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Magnitude of Thrust on a Submerged Plane

C

h

X

Y

O

F

Free surface (atmosphere)

Edge viewof plane

View normal to

plane

O

h-

y-

36


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Magnitude of Force on a Submerged Plane


C

hF

Edge

view

of

plane

View

normal to

plane

O

h-

y-

F


h

Increasing Force with

depth h

37



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P (x, y)C

A

F

hX

Y

O

F


Edge viewof plane

View normal to

plane

O

yh-

AygAhgApF )sin()(

AA ydAsingdAsingyF

hgAyAsingF

y-

Determine Resultant force F

38


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P (x, y)C

F

hX

Y

O

F

Free surface (atmosphere)O

yh-

y-y '

AdAsingy'Fy A2

yAsingF

yA

AkyAsingdAysing'y OX

2A

2

yA

Aky

yA

yAAk'y C

22C

2

Location of

Resultant

ForceAsingyM 2

39


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hgAF

River

Sewer Gate

1.0 m

0.6 m

45

mh 212.145cos3.00.1

kNF 352.3212.13.081.9997 2

Circular Gate D = 0.6 m

40


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River

Sewer Gate

1.0 m

0.6 m

45

yAAkyy C

2

'

m

y

Ry

yA

Akyy C 727.1

4'

22

4

42 RAkC

my 714.13.045cos1

41


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End

Tutorial

42

H d t ti Th t C d S f


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Hydrostatic Thrust on Curved Surfaces

43

Submerged plate

Free surface Volume of fluid above


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Free surfaceof fluid

Volume of fluid abovecurved surface

FRFV

FH

FR

44

Free surface Volume of fluid above


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45



Free surfaceV l f fl id b


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W



46

Distribution ofhydrostatic force

F h

Free surfaceV l f fl id b


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F1 F2ahh

h/3

Ah

gF2

1 A

hgFa2

2



47

=

No influence on the curved surface


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F1 F2ah

LBLhgFF Hb 22

W

FRFV

FH

L/2F2b L

48

Force on curved surfaceresolved into Vertical and

horizontal components

[Horizontal projection]

gVWFV [Weight of fluidabove curved

surface]

H d t ti Th t C d S f


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Hydrostatic Thrust on Curved Surfacewith Fluid Below it

APRIL 2004A sluice gate AB, 1.0 m wide and having a

circular cross-section of radius 1.0 m, is

hinged at one edge B to a vertical plate asshown in the figure.

If the depth of water, h, to the bottom of thesluice gate is 2.0 m, determine the horizontaland vertical components of the resultanthydrostatic force acting on the sluice gate.

49

Hence determine the resultant


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Hence determine the resultanthydrostatic force, its line of action, andits direction from the horizontal.

If the gate is now held closed at its lowerend, A, by a horizontal force determinethe magnitude of this force.

(Note the 2nd Moment of area of arectangle of width b and depth d is

given by: )

12

3bd

50


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B

A

1.0 m

water

FH

2.0 m

Horizontal Force is given by the height ofthe submerged curved surface.

Shaded area represent equivalent


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B

A

water 2.0 m

FV

Shaded area represent equivalent[vertical] force acting on curved surface


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kN67.1411)2/11(81.9997

AhgFH

kNgVFV 46.1714/0.1)11(81.99972

FR

kNFR 81.2246.1767.14 22

96.4967.14

46.17tan

1

FV

FH53

Additional Review


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h

yA

Akyy

LBLhgAhgF

C)('

22

L/2

F L

verticalaxiswhenyh

y

PC

h 'y

Additional Review

54

Buoyancy and Stability


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Relevance: Stable designs of structures(boats, rafts, submarines, etc.) totallysubmerged or floating in fluids.

Examples: Hot air balloons that are whollyimmersed in air and submarines and

diving bells, which are wholly immersed in

water; Rafts, boats, which are floatingvessels.



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Buoyancy is the tendency of a fluid to exerta supporting force on a body placedin the fluid.

Stability refers to the ability of a body toreturn to its original position after being

tilted out of ist initial position.

Buoyancy


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BuoyancyA body in a fluid, whether floating or submerged, is

buoyed up by a force equalto the weight of the fluid displaced.

The buoyant force acts vertically upward through

the centroid (= center of buoyancyB) of thedisplaced volume (Vd) and can be definedmathematically by Archimedes principle as:

where, Fb is the buoyant force, is the density of thefluid, Vd is the displaced volume of fluid.

db gVF


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Stability of Bodies in Fluids


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Stability of Bodies in Fluids A body is in stable equilibrium if it returns to its

original position after being given a smalldisplacement.

It is in unstable equilibrium if it continues to moveaway from its original position on being given a

small displacement. It is in neutral equilibrium if it adopts a new

equilibrium position at the small displacement it wasgiven.

There are two cases to consider:(1) for bodies completely submerged in a fluid; and(2) for bodies only partially submerged, that is, for

bodies that float in a fluid.

St bilit


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StabilityBodies Completely Submerged in a Fluid

e.g. a hot air balloon two forces acting on it

- weight vertically down (constant)- buoyancy vertically up (since Vd is constant)

Stable equilibrium (G below B)

Fb=W Fb=W

Wx

Unstable equilibrium (G above B)

B

G

W

G

B

Wx

G

BW

Fb=W

x Wx

B

G

WFb=W

St bilit


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StabilityBodies Completely Submerged in a Fluid

Fora body completely submerged in a fluid,stable equilibrium is achieved oncethe centre

of gravity is below the centre of buoyancy.

Stable equilibrium (G below B) Unstable equilibrium (G above B)

B

G

W

Fb=W

G

B

W

Fb=W

x

Wx

G

BW

Fb=W

x Wx

B

G

WFb=W

Stability


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StabilityBodies Partially Submerged (Floating) in a Fluid

two forces acting on it

- weight vertically down (constant)- buoyancy vertically up (not necessarily constant)

Stable equilibrium (G below M) Unstable equilibrium (G above M)

Vd

W

Fb

Vd

W

Fb

Fb=W

BG G B (old/new)

Vd

W

Fb

BG

Vd

WFb

Fb=W

B (old/new)G

M

M

M = metacentre

Stability


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StabilityBodies Partially Submerged (Floating) in a Fluid

Vd

W

Fb

Vd

W

Fb

Fb=W

BG G B (old/new)

Vd

W

Fb

BG

Vd

W

Fb

Fb=W

B (old/new)G

M

M

M = metacentre

Fora body partially submerged (floating) in a fluid,the condition for stable equilibrium is that the

metacentre be located above the centre of gravity.

lecture 02_fluid_statics

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