112/04/21 林再興教授編 1

Chapter 2 Fluid at Rest – Pressure and it Effects

(Chapter 2 Fluid Statics)

Fluid is either at rest or moving

-- no relative motion between adjacent particles

-- no shearing stresses in the fluid

→ surface force will be due to the pressure

In the chapter, the principal concern is to investigate

(1) pressure and its variation throughout a fluid,

(2) the effect of pressure on submerged surface.

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§2.1 Presure at a point

At rest, on any plane

Gas P2 Shear stress = 0

A or Normal stress

  Liquid P1 P3 =P1=P2=P3=P4=P

=Pressure

P4 ( positive for

compression )

θ

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normal force

  Fluid pressure : (1) = P ( fluid pressure )

(2) the surface of contact

In an area P =

At any point P =

How the pressure at a point do varies with the orientation of the plane passing through the point ?

A

F

A

FF

dA

dF

A

Flim

0A

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Equation of motion

( or equilibrium equation)

In y-direction,

Σ Fy = 0

Py δx δz – ( Ps δx δs)sin θ = 0

Py δx δz – Psδxδz = 0

Py– Ps = 0

or Py = Ps

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In z-direction

ΣFz = 0

Pz δx δy – ( Ps δx δs)cosθ –ρg δx δy δz(1/2) = 0

Since δ scos θ = δ y

Pz δx δy – Ps δx δy –1/2 ρ g δx δy δz = 0

Pz = Ps + 1/2 ρ gδz

For δx δy δz → 0 Pz = Ps

=>Py = Ps = Pz

 

Pascal's law

• Since the angle θ was arbitrarily chosen , we can obtain

• Py = Px = Pz

Py = Px = Pz= PS ~ Pascal's law

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§2-2 Basic Equation for Pressure Field

pressure net force

pressure gradient net force

Let P = the pressure at the center of the element. (x,y,z)

pressure force in y direction

on a fluid element

zyxy

pF

zxy

y

ppzx

y

y

ppF

y

y

)2

()2

(

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In the same manner

In x direction

In z direction

The total net pressure force on the element

zyxz

pF

zyxx

pF

z

x

gradientpressurePwherePV

ff

dxdydzPzyxkz

pj

y

pi

x

p

kFjFiFF yyxpress

)1.2(─)(

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Equilibrium of a fluid

Force on a fluid element

─Surface force ─ acting on the sides of the element

i,e, pressure gradient and viscous stress ( not

included in this chapter; will be considered in chapter 6)

─Body force ─ acing on the entire mass of the

element, i.e, gravitational potential

and electromagnetic potential

( neglected in this chapter )

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Surface forces

(1)   pressure gradient

(2) viscous stress

 

Body force (gravitational potential)

Pd

Fdf pressp

vz

v

y

v

x

v

d

Fdf vsvs

2

222

2

)(

kggd

gd

d

Fdf grav

grav

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Force balance

-----------------(2.2)

Eq(2.2) is the general equation of motion for a fluid in which there is no

shearing stress.

aff

afffam

f

gravp

gravvsp

akgP

0f vs

avkgP 2

The following equation is the general equation of motion for a fluid in which there is shearing stress

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§2.3 Pressure Variation in a Fluid at Rest

From Eq.(2.2) Such as ---------(2.2)akgP

0a

0k)gz

P(j

y

Pi

x

P or

0kgkz

Pj

y

Pi

x

P

0kgP

For fluid at rest

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2

1

2

1

0

0

Z

Z

P

PgdzdPorgdzdP

gdz

dPor

gdz

dP

gz

Pand

(2.3)

…(2.4)

0;0

y

P

x

PPressure does not change in a horizontal plane ( or x-y plane)

z

p2─z2

p1─z1

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§2.3.1 Incompressible Fluid

Incompressible fluid, ρ=const. & assuming g = Const.

)7.2(

)(

]:[)(

21

1221

1212

2

1

2

1

ghPPor

zzgPP

zNotezzgPPdzgdPZ

Z

P

P

z

p2─z2

p1─z1

h= where h is called pressure head

Hydrostatic pressure distribution

─ the pressure varies linearly with depth

For H2O h = 23.1 ft for ΔP = 10 psia

For Hg h = 518mm for ΔP = 10 psia

g

PP 21

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  P = P0 + ρgh

where P0: pressure at free surface;

h: Depth below the free surface)

same h same P

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Example 2.1

Given: As figure on the right

SG(gasoline)=0.68

h1 = 17 ft (Gasoline)

h2 = 3 ft (water)

Find: pressure at point (1) & (2)

in units of lb/ft2, lb/in2, and as a

pressure head in feet of water

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Solution:

(a) P1 = P0 + ρgh1

(b) P1 = P0 + ρH2OghH2O

hH2O =

(c) P2 = P1 + ρH2Ogh2

= 722.13(lbf/ft2) + 1.94(slug/ft3) 32.2(ft/s2) 3(ft)

= 722.13 + 187.4 = 909.53

psigin

ftft

lbfft

lbf

fts

ftft

slugpsfg

0.5144

112.72213.722

172.3294.168.00

2

2

22

23

ft56.112.3294.1

13.722

g

PP

OH

01

2

psia31.6in144

ft1

ft

lbf2

2

3

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(d) P2 = P0 + ρH2OghH2O

hH2O =

Transmission of fluid pressure

since p1 = p2

ft

P

OH

56.14

2.3294.1

53.909

2

2

121

212

11

22

2

2

1

1

1

)(

FFA

AAA

FA

AF

A

F

A

F

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§2.3.2 Compressible Fluid

 --- Compressible fluid

air, oxygen(O2), hydrogen(N2)

 

(ρg)gas = f(p,T)

 

Eq(2.4) dp = -ρgdz

(ρg)air = 0.076 lbf/ft3 at p=14.7psia & T=60℉

(ρg)H2O = 62.4 lbf/ft3 at p=14.7 psia &T=60℉

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If dz is small, dp=-(ρg)air dz → 0

 If dz is large

Eq(2.4) dp =-ρgdz

Eq. of state for ideal gas p = ρRT

 

dzRT

pgdp

)10.2......(

,.

)9.2......(1

ln).(

0

12

2

1

2

1

2

1

)(

12

0

1

2

RT

ZZg

z

z

Z

Z

P

p

epp

TconstTIf

dzTR

g

p

pconstgAssume

dzRT

g

p

dp

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Example2.2

The Empire State Building in New York city, one of the tallest building in the world

Given: h = 1250 ft

(1) p2(at top) / p1(at bottom) = ? if air is compressible fluid

at T = 59 ℉

(2) p2 / p1 = ? if air is assumed to be incompressible fluid

(ρg)air = 0.076 lbf/ft3 at 14.7psia

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Solution:

(1) From Eq (2.10)

(2) If air is incompressible fluid

dp = -ρgdz p2 –p1 = -ρg(z2 – z1)

956.0

)R46059)(R.slug/lb-ft1716(

)ft1250)(s/ft2.32(exp

]RT

)zz(gexp[

p

p

oo

2

0

12

1

2

955.01447.14

12500765.01

P

)zz(g1

p

p

1

12

1

2

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§2.4 Standard Atmosphere

From Eq.(2.9)

 

where Ta = temp. at z = 0 (sea level) ;β= lapse rate

where the parameters in Eq(2.10) are shown in Table

2.1 (P.50)

R=286.9 J/kg.k or 1716 ft-lb/slug.0R

zTTT

dz

R

gpdp

a

Z

Z

p

p

2

1

2

1

/

Rg

a

a T

z1pp --- (2.10)

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Fig 2.6

P.51 (fig. 2.6)

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§2.5 Measurement of Pressure

pressure measurement

─absolute pressure (with respective to a zero pressure

reference)

─gage pressure (with respective to local atm pressure)

Absolute pressure ─ PAbs > 0 psia ;

Patm ~ 14.7 psia (atmosphere pressure)

Gage pressure ─ Pgage = 0 psig (atmosphere pressure)

P =Pgage+14.7 psia [=] psia Pgage> 0 P > Patm

Pgage< 0 P < Patm

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20.83 psia

13.54 psia

20.83 psia

8.33 psia

7.29 psig

- 5.21 psig

5.21 psi vacuum

13.54 psi vacuum

0 psia

-13.54 psig

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Mercury barometer

Patm = ρgh + Pvapor

ρgh

because Pvapor = 0.000023 psia

at T=68 ℉

P = 14.7 psia h=760mmHg(abs.)

=29.9in Hg(abs.)

=34ft H2O(abs.)

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§2.6 Manometry

─A standard technique for measuring pressure

involves the use of liquid columns in vertical or

inclined tubes.

Pressure measuring devices based on the technique

are called manometers.

Piezometer tube

─Manometer U-tube

Inclined-tube

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§2.6.1 Piezometer tube

P = PA = ρgh + Po [=] psia

(absolute pressure)

P = PA = ρgh [=] psig

(gage pressure)

Disadvantages

1. PA > Patm

2. h1 should be reasonable

3. Fluid in container must be a

pressure at point(A) liquid

is desired

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§2.6.2 U-Tube manometer

─To overcome the difficulties note previously, another type of

manometer which is widely used consisted of a tube formed into

the shape of a U as is show in Fig.2.10.

─A major advantage of the U-tube manometer

Gage fluid can be different from the fluid in

the container in which the pressure is to

be determined

─"Jump across"

Same elevation within the same continuous mass of fluid.

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Method(A)

PA-P2 = -ρ1g(z1 - z2)

+) P2-P4 = -ρ2g(z3 - z4)

PA-P4= -ρ1g(z1-z2)-ρ2g(z3-z4)

=> PA= -ρ1gh1-ρ2g(-h2)

or PA= -ρ1gh1+ρ2gh2

=> PA+ρ1gh1 -ρ2gh2 = 0

=> PA=ρ2gh2-ρ1gh1

Method(B) 0

PA+ρ1gh1-ρ2gh2=P4 =>PA+ρ1gh1-ρ2gh2=0

=P2=P3

=P4 =>PA=ρ2gh2-ρ1gh1

(4)

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Example 2.4

Give Right figure

SG(oil) = 0.9 SG(Hg) = 13.6

h1 = 36 inch

h2 = 6 inch

h3 = 9 inch

Find ： PA= ？ Solution

PA+ρ0g(h1+h2) -ρHggh3 = pgage

pA= -ρ0gh(h1+h2) +ρHggh3

= -0.9 * 1.94 slug/ft3 * 32.2 ft/s2*(36+6)in * 1ft / 12inch

+ 13.6 * 1.94 * 32.2 * 9/12 = 440lbf / ft2*(1ft2 / 144in )= 3.06psig

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The U-tube manometer is also widely used to measure the

difference in pressure between two containers or two points

in a given system.

 

 

 

PA+ ρ1gh1- ρ2gh2-

ρ 3gh3=PB

PA-PB=-

ρ1gh1+ ρ2gh2+ ρ3gh3

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Example2.5

Given: Right figure

Find:(1)PA-PB=f(γ1, γ2,h1,h2)

(2)PA-PB=?

if γ1=9.8kN/m3, γ2=15.6kN/m3

h1=1.0m and h2=0.5m

Solution:

(1) PA- ρ1gh1- ρ2gh2+ ρ1g(h2+h1)=PB

PA-PB= ρ1gh1+ ρ2gh2- ρ1g(h2+h1)=h2(ρ2g- ρ1g)

(2) PA-PB=h2(ρ2g- ρ1g)

=0.5(15.6*103-9.8*103)=2.9*103 N/m2 =2.9kpa

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§2.6.3 Inclined-Tube Manometer

—To measure

Small pressure

changes

PA+ ρ1gh1- ρ2g(l2sinθ)–ρ3gh3=PB

PA-PB= ρ2gl2sinθ+ ρ3gh3–ρ1gh1

 

If fluid in ρ1 & ρ3 is gas, then ρ1gh1→0, ρ3gh3→0,and

PA-PB= ρ2g l2sinθ

l2 as sinθ 0 ↗ ↘

sing

PPl

2

BA2

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§2.7 Mechanical and Electronic pressure Measuring

Devices

Disadvantage of manometers

(1) not well suited for measuring very high pressure.

(2) not well suited for measuring pressure that are

changing rapidly with time.

(3) require the measurement of one or more column heights.

time consuming

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Other types of pressure measuring instruments.

Idea: pressure acts on an elastic structure, the structure will

deform,and then deformation can be related to the

magnitude of the pressure.

(1) Bourdon pressure gage (gage pressure)

(2) The aneroid barometer(Bourdon type)

— for measuring atmospheric pressure. (absolute

pressure)

(3) pressure transducer

— pressure → electrical output

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Example:

(a) Bourdon tube is

connected to a

linear Variable

differential transformer

(LVDT.)

(b) Thin,elastic diaphragm which is in contact with the fluid.

— pressure changesdiaphragm deflectselectrical

voltage

(i) strain gage

(ii)piezoelectric crystal

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§2.8 Hydrostatic Force on a Plane Surface

Assume ρ=Const (incompressible fluid) => P=Pa+ρgh

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The total hydrostatic force (FR)

(2.17)

for ρ=Const,θ=Const

The first moment of the area with

respect to the X-axis =

=> FR=P0A+ρgsinθyCGA or FR=P0A+ρghCGA ─ (2.18)

FR┴Surface F﹝ R=(Patm+ρghC)A or F﹞ R=PCGA , where PCG=Patm+ρghc

AyydA CGA

ydA ρgsinθAP

dAysinθρgAP

ρghdAdAρghPPdAF dF

0

0

00R

A

A

AAAAAP

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To find the center pressure or resultant force(xR,yR) :

Base on moment of the resultant force=moment of the distributed pressure force

To find yR

FRyR = = =

=ρg sinθ

{ρg sinθyCGA}yR=ρg sinθ

yCG A yR=

yR =

yR = where = Second moment of

the area(moment of inertia) = IX

A ydF A

dA)siny(gy A

2dAysing

A2dAy

A2dAy

A2dAy

Ay

dAy

CG

A

2

Ay

I

CG

x A2dAy

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The parallel axis theorem(to express IX)

IX=IXC+AyCG2 where IXC=the second moment of the

area with respect to an axis passing

through its centroid and parallel to the

X-axis.

yR= + yCG ------- (2.19)

Note:

(1) IXC /yCG A > 0 => yR > yCG

(2) Resultant force does not pass through the

centroid,but is always below it.

Ay

I

CG

XCG

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To find xR

=> FRxR = =

=> ρg sinθ yCGA xR = ρg sinθ

=> xR =

=> xR =

Ixyc = the product of inertia with respect to an orthogonal coordinate system passing through the centroid of the area

AxdFA xydA

Ay

I

Ay

xydA

CG

xy

CG

A

)20.2( CGCG

xyc xAy

I

A dAygx )sin(

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Example2.6

Given ： figure on right

Diameter of circular gate=4m

ρg = 9.8 KN/m3

Dshaft=10m

Determine ：

(a)the magnitude and location

of the resultant force

exerted on gate by the water

(b)the moment that would have to be applied to the shaft to

open the gate

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Solution ：(a)FR=PCGA=ρghCGA

=9.8*103(N/m3)*(10m)*(π·22)m3

=1231*103N=1231KN

=1.23MN

yR= + yCG ---- (2.19)

= + yCG = + yCG

(fig. 2.18 P.64)

= + = 0.0866+11.5 = 11.58m

Ay

I

CG

XC

Ay

dAy

CG

AC

22

CG

4

Ry4

R

2

4

2)60sin

10(

4/2

60sin

10

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yR-yCG=0.0866m below the shaft

xR= + xCG= +xCG = 0 + xCG = xCG

(b)Center of rotation at c

Σr · F = 0 or ΣM = 0

M - FR·(yR-yCG) = 0

M = FR·(yR-yCG) = 1231*103*0.0866

=1.07*105 N-M

Ay

I

c

xyc

Ay

xydA

C

AC