Ch2 Fluid Statics

• Fluid either at rest or moving in a manner that there is no relative motion between adjacent particles.

• No shearing stress in the fluid

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

2.1 Pressure at a point N/m2 (Force/Area)

amF

Y: sinsxPzxpF syy yazyx

2

Z:zzzz azyx

sxpyxpF2

cos za

zyx

2

sinz ; cos ssy

2)( :

2 :

zappz

yappy

zsz

ysy

What happen at a pt. ? 0 , , zyx

sz

sy

pp

pp

szy ppp θ is arbitrarily chosen

Pressure at a pt. in a fluid at rest, or in motion, isindependent of direction as long as there are no shearingstresses present. (Pascal’s law)

2.2 Basic equation for Pressure Field

Surface & body forces acting on small fluid elementpressure weight

How does the pressure in a fluid which there are no shearing stresses vary from pt. to pt.?

Surface forces:

zxy

y

ppzx

y

y

ppFy y )

2()

2(:

zyxy

pFy

Similarly, in z and x directions:

zyxx

pFx

zyxz

pFz

zyxkz

pj

y

pi

x

pkFjFiFF zyxs )(

kz

jy

ix

zyxp

)(

Newton’s second law

zyxzyxpWFamF s

azyx

2.3 Pressure variation in a fluid at rest

akp

General equation of motion for a fluid in which thereare no shearing stresses.

)2.4 Eq.(

0 0

dz

dpz

p

kpa

2.3.1 Incompressible

hγ)zz(γppdzγdp

constgργz

z

p

p

1221

1

2

2

1

Hydrostatic Distribution 21 phγp *see Fig. 2.2

21 pp

h

pressure head

Ex:

)133( )62.4(

518or 1.2310

32

21

mKN

ftlb

mmHgfthpppsi

0phγp

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 elevationsThroughout a system. )52 ( 1

1

22 ..FigF

A

AF

Transmission of fluid pressure

2.3.2 Compressible Fluid perfect gas: RTρp

2

1

2

1)z(z const. , ln 21

1

2pp

ZZ Rg

T

dz

R

g

p

p

p

dpRT

gpg

dz

dp

Assume

0

1212

210

)(exp

conditions isothermal ,

RT

zzgpp

zzoverTT

Troposphere:

003570

00650

0

ftR.

ratelaposeβmK.

z@TzβTT aa

βRg

a

a )T

zβ(pp 1

2.4 Standard Atmosphere

vaporatm phγp (Mercury barometer)Example 2.3 )pascal(

m

Npa

2

2.6 Manometry 1. Piezometer Tube: 2. U-Tube Manometer: 3. Inclined-tube manometer

gasa not liquid, 3. reasonable is h 2. 1 1 不大aa pppp.

1122 hγhγpA

see examples

*explain Fig. 2.11 Differential U-tube manometer

113322 hγhγhγpp BA Example 2.5

2.5 Measurement of Pressure See Fig. 2.7 Absolute & Gage pressure

Ex. 2.5

)γγ(hpp

p)hh(γhγhγp

ppk)flowtheofratevolumethe(Q

pppΔ,pΔ,uΔ

BA

BA

BA

BA

122

2112211

2.6.3 Fig. 2.12 Inclined tube manometer

θsinγ

ppl

θsinlγpp

BA

BA

2

2

22

Small difference in gas pressureIf pipes A & B contain a gas

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 couldbe recorded on an oscillograph, or digitized for storageor 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

2.8 Hydrostatic Force on a Plane Surface

pAFR

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.

. Pressure varies linearly with depth if incompressibleg

dz

dp hp for open tank, Fig. 2.16

The resultant force acts through the centroid of the area

* Exercise 1.66dARd i

torque shearing stress

lRdlR

dlRd

ldRdA

ii

i

i

2

)(

20

2

2

Assume velocity distribution in the gap is lineari

i

RR

wR

0

i

i

RR

wlR

0

32

Ri

R0

θd

τ

hdAdF

constants. are , if

sin

A AR dAyhdAF

AR ydAF sin

first moment of the area∫A cAy=ydA

AhAyF cCR sin Indep. Of

The moment of the resultant force must equal the moment of theDistributed pressure force

AARR dAyydFyF 2sin

Ay

dAyyAF

c

ARCR

2

sin

Ax dAyI inertia) of(moment area theofmoment second2

2 ; cxcxc

xR AyII

Ay

Iy

2.18 Fig. seeect ,

xycxcccA

xycR

cRccA

xcR

IIxy

Ix

yyyy

Iy

Note: Ixy-the product of inertia wrt the x& y area. Ixyc-the product of inertia wrt to an orthogonal coord. system passing through the centroid of the area.

If the submerged area is symmetrical wrt an axes passingthrough the centroid and parallel to either the x or y axes,the resultant force must lie along the line x=xc, since Ixyc= 0.

Center of pressure (Resultant force acts points)

Example 2.6 求 a. b.

),( ; RRR yxF)(momentM

a.

11.6m

0 2.20 2.19, Eq.

1023.1 2.18 Eq. 6

RR

RR

RR

yy

xx

NFF

b. mNyyFM

M

cRR

c

51001.1)(

water);(shaft 0

2.9 Pressure Prism the pressure varies linearly with depth. See Fig. 2.19

Ah

bhh

e prismof pressurvolumeF

Ah

APF

R

AveR

2))((

2

1

)2

(

No matter what the shape of the pressure prism is, the resultantforce 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

0pzp

my

mFmFyF

KNFFF

NAhh

F

NAphF

R

R

s

296.0

)2.0()3.0(

4.25

10954.0)2

(

1044.2)(

0

210

21

3122

411

Example 2.8

2.10 Hydrostatic Force on a Curved Surface. Eqs. Developed before only apply to the plane surfaces

magnitude and location of RF

. Integration: tedious process/ no simple, general formulas can be developed.

. Fig. 2.23

surface plane→F ;F 21

)gravity of C.G(center through ;xVW

fluid. on the exerts

tank that theforce of compoments The , VH FF

For equilibrium,

;2FFH WFFV 1

collinear. through pt

( )

lb281=

ft1×3×ft2

3×

ft

lb4.62=

Ahγ=F

23

c1

lb441=

ft1×ft4

3×π×

ft

lb4.62=∀gρ=∀γ=ω 2

2

3

at C.G

(Centroid; center of pressure, CP; center of gravity)

ft2=

ft3×

23

3×121

+ft2

3=

Ay

C×I+y=y

2

4

cCR

See Fig. 2.18

Example 2.9 排水管受力情形

ft27.1≈xR

ft27.1=π×3

3×4=

π3

R4

°5.32=F

Ftan=θ⇒

F

F=θtan

lb523=F+F=F∴

0=F;lb441=ω=F;lb281=F=F∴

V

H1-

V

H

2V

2HR

2VH1

Similarly

2.11 Buoyancy, Flotation, and Stability

2.11.1 阿基米德原理請看圖 2.24, 來分析其受力情形

VBF 任意形狀的物體之體積2.11.2 Stability stable equilibrium

stable neutral unstable

Light Heavy Heavy LightStable unstable

L

LH

H

Explain Fig. 2.25; 26; 27; 28