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# Chapter 2 Fluid Statics

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(1)

## Chapter 2

Fluid Statics

Buoyancy and Floatation

(2)

Chapter 2 Fluid Statics

Contents

2.1 Pressure-Density-Height Relationship 2.2 Absolute and Gage Pressure

2.3 Manometry

2.4 Forces on Submerged Plane Surfaces 2.5 Forces on Submerged Curved Surfaces 2.6 Buoyancy and Floatation

2.7 Fluid Masses Subjected to Acceleration Objectives

- Study hydrostatic pressure relationship

- Study forces on surfaces by hydrostatic pressure - Study buoyant forces

(3)

## 2.1 Pressure-Density-Height Relationship

Fluid statics

~ study of fluid problems in which there is no relative motion between fluid elements

→ no shear stress

→ only normal pressure forces are present

(4)

① Pressure must be transmitted to solid boundaries normal to those boundaries.

② The pressure at a point in a fluid at rest, or in motion, has the same magnitude in all directions as long as there are no shearing stresses present.

→ Pressure is a scalar quantity.

 Pascal’s law

## 2.1 Pressure-Density-Height Relationship

Weight . W =γVol

(5)

0

∑ =F

1 3

sin 0

(a)

F

x

p dz p ds θ

∑ = − =

2

/ 2

3

cos 0

F

z

p dx ρ gdxdz p ds θ

∑ = − − =

[Pf]

Apply Newton's law for static equilibrium

(b)

sin dz = ds θ

cos dx = ds θ

1 3 1 3

( ) : a p ds sin θ p ds sin θ 0 p p

∴ − = → =

Substitute following relations into Eq. (a) & (b)

2.1 Pressure-Density-Height Relationship

(6)

2 3

( ) : cos cos cos 0

2

b p ds θ ρ − g dz ds θ − p ds θ =

2 3

1

p p 2 ρ gdz

∴ = +

### 0

dz → p

2

≈ p

3

1 2 3

p p p

∴ = =

We take the limit as then at a point

2.1 Pressure-Density-Height Relationship

(7)

• Consider static equilibrium of forces on a typical differential element of fluid

h At center;

p, γ

## 2.1 Pressure-Density-Height Relationship

(8)

2.1 Pressure-Density-Height Relationship

- Newton's first law

### 0

∑ = F

F = external force:

pressure, shear, gravity

Assume unit thickness in y direction; =1

x

0 :

∑ F = ∑ F

x

= p dz

A

− p dz

c

= 0

(2.1)

z

0 :

∑ F = ∑ p dx

B

− p dx

D

− dW = 0

(2.2)

where

p = f x y z ( , , )

(9)

2.1 Pressure-Density-Height Relationship

A

2 p p p dx

x

≈ − ∂

c

2

p p p dx

x

≈ + ∂

2

Taylor series : ( ) ( ) ( ) ( ) 2!

f x + ∆ ≈ x f x + f x ′ ∆ + x f ′′ x ∆ x + ⋅⋅⋅

D

2

p p p dz

z

≈ + ∂

(1) (2)

= variation of pressure with direction

p x

dW = ρ gdxdz = γ dxdz

(3)

Substituting (1) and (3) into (2.1) yields

2 2 0

x

p dx p dx p

## dF p dz p dz dxdz

x x x

∂ ∂ ∂

   

=   − ∂   −   + ∂   = − ∂ =

p 0

→ ∂x

∂ = (A)

Based on Taylor series, we have

B

2

p p p dz

z

≈ + ∂

(10)

2.1 Pressure-Density-Height Relationship

Substituting (2) and (3) into (2.2) yields

2 2 0

z

p dz p dz p

dF p dx p dx dxdz dzdx dxdz

z z γ z γ

∂ ∂ ∂

   

=  − ∂  − + ∂  − = − ∂ − =

p dp

z dz γ ρg

= = − = −

(partial derivative → total derivative because of (A))

p 0 x

=

(

∴ =p fn z only( )

)

p 0 x

∂ = (1) ∂

~ no variation of pressure with horizontal distance

~ the pressure is constant in a horizontal plane in a static fluid

(11)

2.1 Pressure-Density-Height Relationship

(12)

2.1 Pressure-Density-Height Relationship

(13)

(14)

## 2.1 Pressure-Density-Height Relationship

dp

dz = −γ

dz dp

→ − = γ

2 2

1 1

z p

z p

dz dp

− =

γ

∫ ∫

z

smaller.

Integrate over depth

2 1

1 2

2 1

( ) p p

p p

dp dp

z z

γ γ

− = −

=

(2.3)

(2.4)

h h

(15)

2.1 Pressure-Density-Height Relationship

For fluid of constant density (incompressible fluid; const.)γ

1 2

2 1

p p

z z h

γ

− = = −

1 2

(

2 1

)

p p γ z z γ h

∴ − = − =

1 2

p = p + γ h

(2.5)

p h γ

=

~ increase of pressure with depth in a fluid of constant density

→ linear increase

~ expressed as a head h of fluid of specific weight γ

~ heads in millimeters of mercury, meters of water - 압력수두

(16)

## 2.1 Pressure-Density-Height Relationship

( )

fn z or p γ =

no contact

[Cf] For compressible fluid, [Re] External forces

1) body force (질량력)

- forces acting on the fluid element without contact

- gravity force, centrifugal force, electro-magnetic force, Corioli's force (due to Earth’s rotation)

2) surface force (표면력)

- forces transmitted from the surrounding fluid and acting against sides of the fluid element

- pressure, shear force

(17)

## 2.1 Pressure-Density-Height Relationship

• Manometer (액주계): piezometer, U-manometer, differential manometer

h

= height of a column of any fluid (m of H2O)

h

2

2 3

(kN/m )

0.102 (kN/m ) 9.81 kN/m

p p

= = ×

γw

• For a static fluid

1 2

1 2

p p

z z

γ + = γ + =

const. (2.6)
(18)

2.1 Pressure-Density-Height Relationship

(19)

2.1 Pressure-Density-Height Relationship

γ

p

• For a fluid of variable density (compressible fluid)

~ need to know a relationship between and

~ oceanography, meteorology

[IP 2.1] The liquid oxygen (LOX) tank of space shuttle booster is filled to a depth of 10 m with LOX at -196°C. The absolute pressure in the vapor above the liquid surface is 101.3 kPa. Calculate absolute pressure at the inlet valve.

(20)

## 2.1 Pressure-Density-Height Relationship

[Sol]

2 a LOX

p = p + γ h

From App. 2 (Table A2.1)

ρ of LOX at -196°C = 1,206 kg/m3

p2 = 101.3 kPa+ (1,206 kg/m3) (9.81 m/s2) (10 m)

= 101.3 kPa+ 118,308 kg·m/s2/m2

= 101.3 kPa+ 118,308 kPa

= 219.6 kPa absolute (절대압력)

pa

(21)

2.2 Absolute and Gage Pressure

1) absolute pressure = atmospheric pressure + gage pressure for p > patm atmospheric pressure - vacuum for p < patm

atm

0 p =

2) relative (gage) pressure (상대압력) →

Bourdon pressure gage ~ measure gage pressure ⇒ open U-tube manometer

Aneroid pressure gage ~ measure absolute pressure ⇒ mercury barometer

- gage pressure is normally substituted by "pressure"

abs atm gage

p = p ± p

(22)

## 2.2 Absolute and Gage Pressure

(23)

2.2 Absolute and Gage Pressure

• Pressure measurement

1) Mercury barometer (기압계)

~ invented by Torricelli (1643)

→ measure absolute pressure/local atmospheric pressure

~ filling tube with air-free mercury

~ inverting it with its open end beneath the mercury surface in the receptacle

(24)

## 2.2 Absolute and Gage Pressure

2) Bourdon pressure gage

~ measure gage pressure

~ when pressure is admitted into the tube, its cross- section tends to become circular, causing the tube to straighten.

pressure transducer (압력변환기)

(25)

2.2 Absolute and Gage Pressure

(26)

2.2 Absolute and Gage Pressure

3) Diaphragm pressure gage

~ with pressure transducer (압력변환기)

(27)

## 2.2 Absolute and Gage Pressure

[Sol] absolute pressure = 100 kPa −310 mm Hg

101.3 kPa 100 kPa 310

760

 

= −  

 

= 58.7 kPa

[Re] App. 1

760 mmHg = 101.3 kPa = 1,013 mb

→ 1 mmHg = 101,300 / 760 = 133.3 Pa 1 bar = 100 kPa = 103 mb

760 mmHg = 760×10-3 m×13.6×9,800 N/m3 = 101.3 kN/m2

= 101,300 N/m2 / 9,800 N/m3 = 10.3 m of H2O

[IP 2.4] A Bourdon gage registers a vacuum of 310 mm of mercury when the atmospheric pressure is 100 kPa absolute. Find the absolute pressure.

(28)

2.3 Manometry

4) Manometer

~ more precise than Bourdon gage (mechanical gage)

~ The pressures are equal over horizontal planes within continuous columns of the same fluid.

p 0 x

∂ =

1

p

A

= p

1 atm 1 1

p = p + γ h

1

1 1

p

A

= p = + γ h

Patm 0

a. Piezometer

(29)

## 2.3 Manometry

(30)

2.3 Manometry

b. U-tube manometer

p 0 x

∂ =

1 2

p p

→ =

1 x

p = p + γ l

2 atm 1

p = p + γ h

1 2

;

x

### 0

1

p = p p + γ l = + γ h

1

p

x

γ h γ l

∴ = −

Patm 0

(31)

2.3 Manometry

c. Differential manometer (시차액주계)

~ measure difference between two unknown pressures

4 5

p = p

4 x 1 1

p = p + γ l p

5

= p

y

+ γ

2 2

l + γ

3

h

1 1 2 2 3

x y

p + γ l = p + γ l + γ h

2 2 3 1 1

x y

p p l l h l

∴ − = γ + − γ

1 2 w

γ =γ =γ

x

y

If and and are horizontal

3 ( 2 1)

x y w

pphll

3h w( h) ( 3 w)h

γ γ γ γ

= + − = −

w w

p p

γ h

γ γ

−  

= − 

 

-h

(32)

2.3 Manometry

[Re] Differential manometer

1 2

p = p

1 A A

p = p − γ z

2 B B M

p = p − γ z + γ y

A A B B M

p γ z p γ z γ y

∴ − = − +

( )

A B A B M

p − p = γ z − z + γ y

( )

M M

y y y

γ γ γ γ

= − + = −

A B M

1

p p

γ y

γ γ

 

− =  − 

 

γ γ =

w A B ( . .M 1) w

p p

s g y

γ

− = −

If

(33)

2.3 Manometry

(34)

2.3 Manometry

d. Inclined gages

~ measure the comparatively small pressure in low-velocity gas flows

x sin

p =

γ

h =

γ

l

θ

l h

(35)

## 2.3 Manometry

e. Measure vacuum

1 2

p = p

1 A A M

p = p + γ z + γ y

2 atm

p = p

A A M atm

p + γ z + γ y = p

A atm A M

p = p − γ z − γ y

A atm

p < p vacuum

(36)

2.3 Manometry

. . 13.55

s g =

y

. . 1 s g <

For measuring large pressure difference,

→ use heavy measuring liquid, such as mercury → makes small For a small pressure difference,

→ use a light fluid such as oil, or even air

• Practical considerations for manometry

① Temperature effects on densities of manometer liquids should be appreciated.

② Errors due to capillarity may frequently be canceled by selecting manometer tubes of uniform sizes.

(37)

## 2.3 Manometry

p

x

[IP 2.5] The vertical pipeline shown contains oil of specific gravity 0.90.

Find

l r

p = p

(0.90 9.8 10 ) 3

3

l x

p = p + × × ×

(13.57 9.8 10 ) 0.375

3

r atm

p = p + × × ×

x

23.4

∴ p =

kPa (kN/m2) [Sol]

f s g. . w

γ = ×γ

atm 0

p =

(38)

2.4 Forces on Submerged Plane Surfaces

• Calculation of magnitude, direction, and location of the total forces on surfaces submerged in a liquid is essential.

→ design of dams, bulkheads, gates, tanks, ships

p

z γ

∂ = −

p γ h

∴ =

• Pressure variation for non-horizontal planes

→ The pressure varies linearly with depth.

(2.7)

(39)

## 2.4 Forces on Submerged Plane Surfaces

Yubari dam, Japan

Hoover dam

(40)

2.4 Forces on Submerged Plane Surfaces

Spillway

(41)

2.4 Forces on Submerged Plane Surfaces

Tainter gate

(42)

2.4 Forces on Submerged Plane Surfaces

Movable weir with sluice gate

Fishway Fixed weir

with no gate

(43)

2.4 Forces on Submerged Plane Surfaces

연직평면

(44)

## 2.4 Forces on Submerged Plane Surfaces

1) 수평평면

2) 연직평면

3) 경사평면

F = pA =

γ

h A2

A A

F =

dF =

γhdA

A A sin

F =

γhdA =

γl αdA
(45)

2.4 Forces on Submerged Plane Surfaces

• Pressure on the inclined plane

• Centroid of area A (도심)

~ at a depth hc

~ at a distance lc from the line of intersection 0-0

(i) Magnitude of total force

First, consider differential force dF

(2.10) (2.9) (2.8)

## dF = pdA = γ hdA

sin h = l

α

sin

dF

γ

l

α

dA

→ =

(46)

2.4 Forces on Submerged Plane Surfaces

Center of

resultant force (작용점)

(47)

2.4 Forces on Submerged Plane Surfaces

in which = perpendicular distance from 0-0 to the centroid of area (도심) Then, integrate dF over area A

A sin A

F =

dF =γ α

ldA (2.11)

A

ldA

A l

c

= ⋅

= 1st moment of the area A about the line 0-0 (단면 1차모멘트)

lc

c

sin F γ Al α

∴ =

Substitute

h

c

= l

c

sin α

F =

γ

h Ac

(pressure at centroid) × (area of plane)

(2.12)

(48)

## 2.4 Forces on Submerged Plane Surfaces

(ii) Location of total force

Consider moment of force about the line 0-0

2 sin

dM = dF l⋅ =

γ

l dA

α sin

2

A A

M = ∫ dM = γ α ∫ l dA

sin dF =γl αdA

A 2

l dA = I0 0

where = second moment of the area A, about the line 0-0 sin 0 0

M

γ α

I

∴ =

By the way,

M = ⋅ F l

p

unknown

(total force × moment arm)

(a)

(b)

(49)

2.4 Forces on Submerged Plane Surfaces

Combine (a) and (b)

0 0

sin Fl

p

= γ I

α

(2.14) (c)

c

sin

F = γ l α A

Substitute into (c)

sin

0 0

sin

c p

l Al I

γ α = γ

α

2

0 0 c c c

p c

c c c

I I l A I

l l

l A l A l A

+

∴ = = = +

→ Center of pressure is always below the centroid by c

c

I l A

c

p c

c

l l I

− = l A

lc lplc

→ as (depth of centroid) increases decreases

(50)

## 2.4 Forces on Submerged Plane Surfaces

2

0 0 c c

I

= + I l A

• Second moment transfer equation

Ic = 2nd moment of the area A about an axis through the centroid, parallel to 0-0

→ Appendix 3

(51)

2.4 Forces on Submerged Plane Surfaces

(52)

2.4 Forces on Submerged Plane Surfaces

1) Rectangle

If

,

A = bh ,

c

2

y = h

3

c

12 I = bh

( )

c c

2

h a h y a h

∴ = + − = +

( )

c

2

F = γ h A = γ    a + h    bh

c

p c

c

h h I

= + h A

0;

a =

c

2 h = h

3

12 2

2 2 6 3

2

p

bh

h h h

h h

h bh

= + = + =

(53)

2.4 Forces on Submerged Plane Surfaces

2) Semicircle

4

4

128 ,

c

3

d r

I π y

= = π

2

c c

I = + I y A

2

c c

I I y A

∴ = −

4 2 2

4

128 3 8

d r d

π π

π

 

 

= −    

   

4 4

1 0.10976

128 π 18 d r

π

 

=  −  =

 

yc

(54)

2.4 Forces on Submerged Plane Surfaces

4

4

256 ,

c

3

d r

I π y

= = π

2

c c

I = + I y A

4 2 2

4

256 3 16

d r d

π π

π

 

 

= −    

    1

4

256 π 36 d π

 

=   −   0.05488r

4

=

(55)

## 2.4 Forces on Submerged Plane Surfaces

(iii) Lateral location of the center of pressure for asymmetric submerged area

(56)

## 2.4 Forces on Submerged Plane Surfaces

dA a. For regular plane

(i) divide whole area into a series of elemental horizontal strips of area (ii) center of pressure for each strip would be at the midpoint of the strip

(the strip is a rectangle in the limit)

(iii) apply moment theorem about a vertical axis 0-0

c

sin

dF = γ h dA = γ l α dA

c c

sin

dM = x dF = γ x l α dA

(a) Integrate (a)

c

sin

M = ∫

A

dM = ∫ x l γ α dA

(b)

F

(57)

2.4 Forces on Submerged Plane Surfaces

By the way, M = x Fp (c)

Equate (b) and (c)

p c

sin

x F = ∫ x l γ α dA

1 sin

p c

x x ldA

= F γ α

(2.15)

b. For irregular forms

~ divide into simple areas

~ use methods of statics [Re] Moment theorem

→ The moment of the resultant force is equal to the sum of the moments of the individual forces.

(58)

## 2.4 Forces on Submerged Plane Surfaces

[IP 2.9] A vertical gate: quarter circle [Sol]

(i) Magnitude

4 4

) (1.8) 0.764;

3 3

y r

π π

= = =

0.3 0.764 1.064 m

h

c

= + =

9,800(1.064) (1.8)

2

26.53 kN

4

F = γ h A =  π  =

 

 

xp

(59)

2.4 Forces on Submerged Plane Surfaces

(ii) Vertical location of resultant force

4

2

0.05488(1.8)

0.213m (1.064) (1.8)

4

1.064 0.213 1.277 m

c

p

I l A

l

π

 

= =

   

     

→ = + =

dA

(iii) Lateral location of the center of pressure Divide quadrant into horizontal strips

Take a moment of the force on about y-axis

(60)

## 2.4 Forces on Submerged Plane Surfaces

2 2 2

(1.8) x + y =

dM =

γ

hdA⋅ 9800( 0.3)( )

2 y xdy  x

= +   

2 2 2

9800 9800

( 0.3) ( 0.3)(1.8 )

2 y x dy 2 y y dy

= + = + −

1.8 2 2

0

9800 ( 0.3)(1.8 ) 18,575 N m

M 2 y y dy

∴ =

+ − = ⋅

(moment arm)

By the way,

M = F

x

p

18,575 / 26.53 10

3

0.7 x

p

m

= ×

=

right to the y - axis xp
(61)

2.5 Forces on Submerged Curved Surfaces

• Resultant pressure forces on curved surfaces are more difficult to deal with because the incremental pressure forces vary continually in direction.

→ Direct integration

Method of basic mechanics

(62)

2.5 Forces on Submerged Curved Surfaces

1) Direct integration (직접적분법)

- Represent the curved shape functionally and integrate to find horizontal and vertical components of the resulting force

i) Horizontal component

H H

F = ∫ dF = γ ∫ hb dz

where b = the width of the surface;

dz = the vertical projection of the surface element dL

(63)

## 2.5 Forces on Submerged Curved Surfaces

·

where zp = the vertical distance from the moment center to FH

location of FH: take moments of dF about convenient point, e.g., point C

p H H

z F = ∫ z dF = ∫ z hb dz γ

ⅱ) Vertical component

V V

F = ∫ dF = γ ∫ hb dx

where dx = the horizontal projection of the surface element dL

· location of FV: take moments of dF about convenient point, e.g., point C

p V V

x F =

x dF =

x hb dxγ

where xp= the horizontal distance from the moment center to FV

(64)

## 2.5 Forces on Submerged Curved Surfaces

2) Method of basic mechanics

- Use the basic mechanics concept of a free body diagram (자유물체도) and the

equilibrium of a fluid mass

- Choose a convenient volume of fluid in a way that one of the fluid element boundaries

coincide with the curved surface under consideration

- Isolate the fluid mass and show all the forces acting on the mass of free body to keep it in equilibrium

(65)

## 2.5 Forces on Submerged Curved Surfaces

• Static equilibrium of free body ABC

' 0

x BC H

F F F

∑ = − = ∴FH' = FBC =

γ

h Ac BC

' 0

z V ABC AC

F F W F

∑ = − − = ∴FV' = FAC +WABC

AC c AC AC ACDE

F = γh A = γHA =W

WABC = ABC

'

FV

∴ = ABDE

weight of free body weight of

• Location

From the inability of the free body of fluid to support shear stress,

H'

F FBC

V'

F WABC FAC

→ must be co-linear with

→ must be co-linear with the resultant of and .

D E

(66)

## 2.5 Forces on Submerged Curved Surfaces

330,000 tone 330,000 10 kg

3

W = = ×

10 103

γ = × AB

[IP 2.10] p. 59 Oil tanker

Calculate magnitude, direction, and location of resultant force/meter exerted by seawater ( N/m3) on the curved surface (quarter cylinder) at the corner.

(67)

2.5 Forces on Submerged Curved Surfaces

[Sol] Consider a free body ABC

(68)

2.5 Forces on Submerged Curved Surfaces

(i) Horizontal Comp.

m below line OA m above line BC

' 4

1.5

10 22.5 (1.5 1) 348.8

H AC c

2

F = F = γ h A = ×    +    × × =

1 (1.5)

3

23.25 12 23.25 0.0081 23.258 23.25 1.5

c

p c

c

l l I

l A

×

= + = + = + =

× 23.258 22.5 0.758 z

p

∴ = − =

24 23.258 0.742

= − =

kN/m

m

(69)

## 2.5 Forces on Submerged Curved Surfaces

(ii) Vertical Comp.

'

### 0

z BC V ABC

F F F W

∑ = − − =

'

.

V BC ABC c

F F W h A Vol

∴ = − = γ − γ

4 4

1

2

10 24 (1.5 1) 10 1.5 1.5 (1.5) 1 355.2 /

4 π kN m

 

= × × × −  × −  × =

 

• To find the location of Fv’, we should first find center of gravity of ABC using statics

Take a moment of area about line OB

4(1.5) 1

2

1.5

(1.5) 0.483 2.25

3 4 π x

c

2

π × + × = ×

x

c = 1.1646 m
(70)

2.5 Forces on Submerged Curved Surfaces

[Cf] From App. 3, for segment of square

2 2 1.5

1.165

3 4 3 4

c

x r m

π π

= = =

− −

'

FV

O Now, find location of force

Take a moment of force about point

m right of

'

0.75 1.1646

V p BC ABC

F × x = F × − W ×

355.2 × x

p

= 360 0.75 4.83 1.1646 × − × 0.744

x

p

∴ =

OB
(71)

## 2.5 Forces on Submerged Curved Surfaces

F

[Summary]

i) Magnitude of Resultant force

kN/m

2 2

(348.8) (355.2) 497.8

F = + =

θ

ii) Direction

1 1

355.2

tan tan 45.5

348.8

V H

F

θ =

  F   =

    =

 

 

iii) Location

Force acting through a point 0.742 m above BC (0.758 m below OA) and 0.744 m right of OB.

(72)

2.5 Forces on Submerged Curved Surfaces

F act through point O.

1 1

0.744

tan 44.47

0.758 α =

    =

 

1 2

348.8

tan 44.47

355.2 α =

      =

1 2

α α =

• Pressure acting on the cylindrical or spherical surface - The pressure forces are all normal to the surface.

- For a circular arc, all the lines of action would pass through the center of the arc.

→ Hence, the resultant would also pass through the center.

(73)

## 2.5 Forces on Submerged Curved Surfaces

• Tainter gate (radial gate) for dam spillway

All hydrostatic pressures are radial, passing through the trunnion bearing.

→ only pin friction should be overcome to open the gate

pin friction (radial gate) < roller friction (lift gate) Trunnion pin

(74)

2.6 Buoyancy and Floatation

• Archimedes' principle

I. A body immersed in a fluid is buoyed up by a force equal to the weight of fluid displaced.

### → 유체에 잠긴 물체는 물체의 부피에 해당하는 유체의 중량만큼 의 부력을 받는다.

II. A floating body displaces its own weight of the liquid in which it floats.

→ 유체에 떠 있는 물체는 자신의 중량에 해당하는 유체의 부피를 배제한다.

→ Calculation of draft of surface vessels, lift of airships and balloons

(75)

## 2.6 Buoyancy and Floatation

(i) Immersed body

Isolate a free body 1234 of fluid with vertical sides tangent to the body

3

4

(76)

2.6 Buoyancy and Floatation

F1’ = vertical force exerted by the lower surface (ADC) on the surrounding fluid F2’ = vertical force exerted by the upper surface (ABC) on the surrounding fluid

FB = buoyancy of fluid; act vertically upward.

For upper portion of free body

1 2 B

F ′ = F ′ + F

'

2 2 2

### 0

F

z

F W P A

∑ = − − =

(a)

(b) For lower portion

'

1 1 1

F

z

F W P A

∑ = − − + =

(77)

## 2.6 Buoyancy and Floatation

Combine (a) and (b)

' '

1 2

(

1 2

) (

1 2

)

F

B

= F − F = P − P A − W + W

1 2

( P − P A ) = γ hA =

1 2

W + W =

1 2 1 2

( P P A ) ( W W )

∴ − − +

γh

weight of free body 1234 weight of dashed portion of fluid

= weight of a volume of fluid equal to that of the body ABCD

B fluid

∴ F = γ

(volume of submerged object) (2.16)
(78)

2.6 Buoyancy and Floatation

(ii) Floating body For floating object

B f

F = γ

(volume displaced, ABCD)

F

B

= γ

f

V

ABCD

ABCDE s ABCDE

W = γ V

s ABCDE

W = γ V

γs

where = specific weight of body From static equilibrium:

F

B

= W

ABCDE

f

V

ABCD s

V

ABCDE

γ = γ

s

ABCD ABCDE

f

V γ V

∴ =

γ

E

(79)

## 2.6 Buoyancy and Floatation

[Ex] Iceberg in the sea Ice s.g.= 0.9

Sea water s.g.= 1.03 0.9(9800)

1.03(9800) 0.97

sub total total

V = V = V

(80)

2.6 Buoyancy and Floatation

• Stability of submerged or floating bodies G1 < M

G2 > M

1, 2

G G

M = metacenter (경심)

→ stable, righting moment

→ unstable, overturning moment

= center of gravity

(81)

2.7 Fluid Masses Subjected to Acceleration

• Fluid masses can be subjected to various types of acceleration without the occurrence of relative motion between fluid particles or between fluid particles and boundaries.

→ laws of fluid statics modified to allow for the effects of acceleration

(82)

## 2.7 Fluid Masses Subjected to Acceleration

• A whole tank containing fluid system is accelerated.

• Newton's 2nd law of motion (Sec. 2.1)

F Ma

∑ =

(2.17)
(83)

2.7 Fluid Masses Subjected to Acceleration

2 2

x

p dx p dx p

F p dz p dz dxdz

x x x

∂ ∂ ∂

     

∑ =  − ∂  − + ∂  = − ∂  (2.18a) First, consider force

z

F p dxdz

z

 ∂ 

∑ = − ∂ − γ (2.18b)

Then, consider acceleration

mass

: p x

x dxdz dxdz a

x g

 

∂ γ

−  =

 ∂   

   

: p z

z dxdz dxdz a

z g

 

∂ γ

− − γ =

 ∂   

   

(84)

2.7 Fluid Masses Subjected to Acceleration

x

p a

x g

∂ = − γ

(

z

)

p g a

z g

∂ γ

= − +

(2.19)

(2.20)

→ pressure variation through an accelerated mass of fluid

p 0 x

∂ =

∂ p

z γ

∂ = −

[Cf] For fluid at rest,

(85)

## 2.7 Fluid Masses Subjected to Acceleration

(86)

2.7 Fluid Masses Subjected to Acceleration

• Chain rule for the total differential for

dp

(App. 5)

p p

dp dx dz

x z

∂ ∂

= +

∂ ∂

(a)

Combine (2.19) , (2.20), and (a)

( )

x z

dp a dx g a dz

g g

γ γ

= − − +

(2.21)

0 dp =

• Line of constant pressure,

( ) 0

x z

a dx g a dz

g g

γ γ

− − + =

x x

dz a

dx g a

 

∴ = −  +  

(2.22)

→ slope of a line of constant pressure

(87)

2.7 Fluid Masses Subjected to Acceleration

(88)

2.7 Fluid Masses Subjected to Acceleration

1) No horizontal acceleration:

a

x

= 0

p 0

x

∂ =

dp g a

z

dz g

 + 

∴ = −γ 

 

a

z

= − g

dp 0

dz =

dh

2) Constant linear acceleration Divide (2.21) by

x z

dp a dx g a dz

dh g dh g dh

 + 

= −γ  + 

 

(a)
(89)

## 2.7 Fluid Masses Subjected to Acceleration

Use similar triangles

'

dx a

x

dh = g

'

dz az g

dh g

= +

' 2 2 1/2

( )

x z

g =   a + a + g  

(b.2) (b.1)

Substitute (b) into (a) dp g'

dh = −γ g

→ pressure variation along h is linear.

(90)

2.7 Fluid Masses Subjected to Acceleration

[IP 2.13] p. 70

An open tank of water is accelerated vertically upward at 4.5 m/s2. Calculate the pressure at a depth of 1.5 m.

[Sol]

3

9.81 4.5

3

( 9,800 N/m ) 14,300 N/m

9.81 dp g a

z

dz g

 +   + 

= −γ     = −     = − 14,300

dp = − dz

integrate

1.5

0p

dp =

0

− 14,300 dz

∫ ∫

1.5 2

14,300[ ]

0

14,300( 1.5 0) 21, 450 N/m 21.45 kPa

p = − z

= − − = =

9800(1.5) 14.7 kPa p = γ = h =

z

0;

If a =

(91)

Homework Assignment # 2

Homework Assignment # 2 Due: 1 week from today 1. (Prob. 2.4)

An open vessel contains carbon tetrachloride to a depth of 2m and water on the carbon tetrachloride to a depth of 1.5m. What is the pressure at the bottom of the vessel?

2. (Prob. 2.6)

How many milimetres of carbon tetrachloride (CCl4) are equivalent to a pressure of 40 kPa? How many metres of alcohol?

(92)

3. (Prob. 2.11)

The specific weight of water in the ocean may be calculated from the empirical relation (in which h is the depth below the ocean surface). Derive an expression for the pressure at any point h and calculate specific weight and pressure at a depth of 3,220 m assuming, , h in metres, and

o K h

γ γ= +

10 / 3

o kN m

γ

=

= 7.08 N/ m7/ 2

K

## Homework Assignment # 2

(93)

4. (Prob. 2.25)

A Bourdon pressure gage attached to a closed tank of air reads 141.1 kPa with the barometer at 775 mm of mercury. If barometric pressure drops to 741.2 mm of mercury, what will the gage read?

5. (Prob. 2.31)

Calculate the pressure px in Fig 2.7a if l = 760 mm, h = 500 mm ; liquid γ is water, and γ1 mercury.

Homework Assignment # 2

(94)

6. (Prob. 2.39)

## Homework Assignment # 2

(95)

7. (Prob. 2.52)

A triangular area of 2 m base and 1.5 m altitude has its base horizontal and lies in a 45o base plane with its apex below the base and 2.75 m

below a water surface. Calculate magnitude, direction and location of the resultant force on this area.

8. (Prob. 2.59)

If the specific weight of a liquid varies linearly with depth h according to the equation

γ = γo + Kh, derive expressions for resultant force per unit width on the rectangular gate and the moment of this force about O.

## Homework Assignment # 2

(96)

9. (Prob. 2.63)

Calculate magnitude and location of the resultant force of water on this annular gate.

Homework Assignment # 2

(97)

10. (Prob. 2.76)

The flashboards on a spillway crest are 1.2 m high and supported on steel posts spaced 0.6 m on centers. The posts are designed to fail under a bending moment of 5,500 N∙m. What depth over the flashboards will cause the posts to fail? Assume hydrostatic pressure distribution.

## Homework Assignment # 2

(98)

12. (Prob. 2.98)

The cylinder is 2.4 m long and is pivoted at O. Calculate the moment (about O) required to hold it in position.

Homework Assignment # 2

(99)

13. (Prob. 2.129)

Calculate the total forces on the ends and bottom of this container while at rest and when being accelerated vertically upward at 3 m/s2. The

container is 2 m wide.

Homework Assignment # 2

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