chapter 3 fluid statistics

8/11/2019 Chapter 3 Fluid Statistics

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Chapter 3. Hydrostatics

1. Basic equation of fluid statics2. Pressure variation in a static fluid

3. Measurement of pressure4. Hydraulic force on a submerged surface5. Hydrostatic force on a curved surface

6. Buoyancy and stability7. Pressure variation in fluid of rigid-body motion

Linear motion and rigid-body acceleration

Scopes

1. No relative motion between fluid particles (du/dy=0)

2. No shear stress ( = du/dy = 0)3. Normal stress (pressure) exists

Fluid Statics


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Apply Newtons 2nd Law,

: sin2

: cos2 2

y y y s y

z z z s z

x y zF ma p x z p x s a

x y z x y zF ma p x y p x s a

= =

= =

cos ; sin y s z s = =Since

2

( )2

y s y

z s z

y p p a

z p p a

=

= +

When y 0 and z 0 p s = p y = p z

Pressure at a point is independent of direction ( Pascals Law)

a scalar property of fluid

(3.1)


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Forces acting on a differential fluid element

1. Body force; 2. Surface forces

X: dxdydz x pF x =

dydxdz y p

F y =

mgdydxdz y p

F z =

Y:

Z :dxdz

dy y p

p )2

(

dxdzdy

y

p p )

2(

+

x y

z

O

dy

dz

dx

p

dxdydz

z p

p )2

(+

dxdydz z p p )

2(

mg

Total force:

2. Pressure variation in a static fluid


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Applying Newtons second law:

x pF dxdydz ma x= =

y y p

F dydxdz ma

y

= =

z z p

F dydxdz mg ma z

= =

0=== z y x aaa

dxdydzm = xa x p = /

ya y p = / )( / zag z p +=

Restrictions:1. Static fluid

2. Gravity is the only body force3. The z axis is vertical and upward

Since p depends only on z, dp gdz = =

(3.4)

(3.1)

(3.3)

(3.2)

Since

0 / = x p

0 / = y p == g z p /

(3.2a)

(3.4a)

(3.3a)

Since


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For incompressible fluids, integrate Eq. (3.4)

(3.5)

(3.6)

.

.

dp dz p z Const

p z Const p

or z Const

= = +

+ =+ =

piezometric head

At two points 1 and 2 in liquid:

1 2

1 2 2 1 1 2 or ( )

p p z z p p z z

+ = + =

Pressure head (m)

= p

2.1 Pressure change Integral form:


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Example : taking the origin of the coordinate at free surface:

At z1=0, p 1=0 Const=0

2 2 / 0 p h p h g h = = =

Pressure in a liquid increases with depth.Higher elevation, lower pressure

O


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2.2 Absolute pressure and gage pressure:

Vacuum

Pressure

P absAtmospheric pressure p atm101.3kPa at standardsea level conditions

P gage

P abs = P gage + P atm (3.7)

Pressure values must be stated with respect to a reference


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2.3 Pressure transmission throughout a stationary fluid

The pressure is a constant value on a line if:

The line is horizontal

The stationary fluid is uniform andThe stationary fluid is continuous.

P 1

P = ?

Constant pressure line

P = P 1


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Example 3.1 : Hydraulic jack

The following is a Hydraulic jack. Explain why a small force F 1 canbe applied to support a large load F 2.

Since p 1 = p 2, F1 /A1 = F 2 /A2 F1 = (A 1 /A 2)F2

As A 1


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Example 3.2: Pressure on divers

Find absolute, gage pressure and pressure heads on diver.Density of water =1025 kg/m 3.

h = 30 m

Solution:


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3. Measurement of pressure

Pressure measurement devices:

Mercury barometer: for atmospheric pressure Manometers: for gage pressure using liquid columns

in vertical or inclined tubes

Mechanical and electronic pressure measuringdevices

In engineering, normally use gauge pressure. Forabsolute pressure, it must be indicated


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3.1 Mercury Barometer to find p atm

016.0 = pa pvapor

vapor atm pgh p +=

For mercury at 20 oC.

= 13600 kg/m 3 .

C atm B PPP ==

vapor C pgh p +=

C

gh p atm = (3.8)


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Example 3.3 Find absolute pressure at 40 m depth in a lake.

Temp = 10 o

C , Barometric pressure = 598 mm Hg (abs)Solution:


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3.2 Piezometer tube to find pA or p1

(abs)111 h p p atm +=

Or (gage)111 h p = (3.9)

p atm


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3.3. U-Tube manometer to find pA

32 p p =

111112 h ph p p A +=+=

223 h p p atm +=

(abs)1122 hh p p atm A +=

Constantpressure

(gage)1122 hh p A = (3.10)

p atm1 p p =


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Example 3.4: Determine pressure gauge pA

if h1 = 0.9m, h 2 = 0.15m, h 3 = 0.23m, SG of oil is 0.9.

Solution:

pA Will be given during the lecture


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3.4. Differential U-tube manometer to find P A P B

(???)

P A = P 1

P 1 + 1h 1 = P 2

P 2 = P 3

P 3 - 2 h 2 = P 4

P 4 - 3 h 3 = P B

Adding up above equations:

P A + 1h 1 - 2 h 2 - 3 h 3 = P B

Differential Pressure:

P A P B = 2 h 2 + 3 h 3 - 1h 1 (3.11)


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(5). Inclined-Tube Manometer to find P A P B

P A + 1h1 - 2l2sin - 3h3 = P B

Or P A - P B = 2l2sin + 3h3 - 1h1 (3.12)


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4. Hydrostatic force on submerged surfaces

Quantities of interestsThe magnitude of the force (N)The direction of the forceLocation of action of force

Basic equation to usep = p 0 + h or p = h with = g

Method of analysisSimple integration


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4.1 Hydrostatic force on a flat surface

Magnitude: F = p cA, p c is pressure at centroid

Direction : Perpendicular to surface

Location of action or Centre of pressure :

needs to be find


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yc

Flat surface


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pdAdF =

==== A A A A R ydAdAgyghdA pdAF sinsin

A ydA y C A =Where y C is the y coordinate of the centroid of the area A.

Where p C is the gage pressure at the centroid of the area A.

(3.13)

(3.14)

The first moment of the area

A p Ah AyF ccc R === sin

Total force = the pressure at the centroid total area

(Force magnitude)


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Location of the resultant force pressure centre

Moment of the resultant force about the x axis = Moment due tothe distributed force about the same axis

dA ydA ypF y A A R

sin' 2 ==

x A I dA y = 2

(3.15)

(3.16) A y

dA y

A y

dA y

F

dA y y

c

A

c

A

R

A ===222

sin

sinsin'

the second moment of thearea about the x axis

Use the parallel axis theorem2c xc x Ay I I +=

(3.17)

I xc : second moment of the area with respect to an axis passingthrough its centroid and parallel to the x axis.


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(3.18)' xc C C

I y y Ay= +

Substitute Eq. (3.17) into Eq. (3.16), we have

The resultant does not pass through the centroid as I xc /y c A > 0 .

Location of x can be determined following a similar procedure:

A y

I x

A y

I x

c

xycC

c

xy +=='

where: C C xyc A xy

y Ax I dA xy I +==

(3.19)

is the product of inertia with respect to x and y.

Direction of the resultant force: perpendicular to the plane.

(Force acting location)


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The second moment of some common shapes:


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Summary of force on plane surfaces

Magnitude:

Direction: normal to the surface

Location of action :

A pF c R =

2

' (3.18) x xc A cc c c

y dA I I y y

y A y A y A= = = +


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Example A: Vertical wall, width b, height hForce on the plate:

: pressure at the centroid

A : plate area.

/ 2c p h =

y, xc R C C

I y y y

= +

2( / 2)( ) / 2 R c c

F p A y A h bh bh = = = =

hpc

yc

RF

R y

press ure prism

31With and12

xc I bh A bh= =

3 /12 / 2

/ 2

= / 2 / 6 2 / 3

Rbh

y hbh h

h h h

= +

+ =

Examples of force on plane surface

Pressure center:

These give you the forcemagnitude and acting location


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Example C: Submerged vertical wall, width b, height h

y

hpc

yc

FR

yR

p1

p2

y1

y22 1( ) / 2c c p y y y = = +

Force on the plate:

, xc R cC

I y y

y= +

R cF p A=

31With and12 xc

I bL A bL= =

Pressure at the centroid:

Area: A = bh

Centroid: 2 1( ) / 2c y y y= +

Location of action (method 1):

2 1( ) / 2 RF bh y y = +Force on the plate:

(Please find y R by yourself ) y R =


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Decompose pressure prism into two:

Method 2 to find Centre of pressure y R

Part 1: Rectangular, p 1

Part 2: Triangular, p2

p1

Pressure centers:

Part 1: y R1=y1+ h/2,

Part 2: y R2=y1+ 2h/3

FRyR=F 1yR1 + F 2 yR2

Where F 1 = y1hbF2= h2b/2

yR

can be obtained

Using moment balance:

y

h FR

yR

p1

p1

y1

y2

p2-p1

F1

F2

h/2

2h/3


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Expl D: Submerged by y 1 inclined wall, width b, length L

1[ ( sin ) / 2]c c p y y L = = +

Force on the plate:

'' ,

xc R c

c

I y y Ay

= +

R cF p A=

31With and12 xc

I bL A bL= =

Pressure at the centroid:

Area: A = bL

Centroid: 1 ( sin ) / 2c y y L = +

Pressure center (method 1):

Force on the plate:

(Please find y R by yourself)

pc

pc

p2

FRyc

yc'

yR

y1

1[ ( sin ) / 2] R cF p A bL y L = = +

Method 2 for y R: Similar to method 2 as for example C

y R =


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Example 3.5: Find resultant force on inclined gate

Gate is hinged along edge A, width b = 5 m.

yc

Solution:


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Example 3.6A pressurized tank contains oil with SG=0.9 and has a square plate

(0.6m 0.6m) bolted to its side. The pressure gage reads 50kPa.Find magnitude and location of the force on the plate .


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Solution:

2 d f


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4.2 Force on a curved surface

Decompose force into F V (vertical) and F H (horizontal),

F H : Force on Projected Vertical Plane Surface,

F V : Weight of liquid directly above curved surface (Real or

imaginary). Real: downward ; imaginary: upward .F V : vertically passes through the center of gravity of the liquidvolume directly above the curved surface.

( ) 22 )( V H R F F F +=Resultant force: (3.20)

=

Pressure body

C l l i f F


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Calculation of F V

F V = Weight of Pressure bodyPressure body consists of the following boundaries:

Curved surfaceVertical lines from the ends of curved surfacesFree surface or its extensionReal pressure body: there is liquid inside FV Imagined pressure body: There is no liquid inside FV

FV:

Real P. B.

Imagined P. B.

surface

surface


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Buoyancy: the net vertical force on an object immersed in a liquid,

or floating on liquid surface Buoyancy is produced by the hydrostatic pressure

Magnitude: weight of displaced liquid: FB = ,Buoyancy on a floating object equals to weight of the object

Direction: Always upward

Line of action: passing through the centroid of the displacedvolume

5.1 Buoyancy

5. Buoyancy and stability

Example 3 8


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FB

W

T

Example 3.8An unknown object weighs 400N inair & 300N in water.Find specific weight & specific gravity of the object.

Solution:

5 2 R i l bili


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5.2 Rotational stability

(c) Unstable(b) neutral(a) stable

G-Centre of gravity

C-Centre of buoyancy (centroid of displaced volume)

(1) F (C) h bili d d i h i h

GM


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(1) For case (C), the stability depends on metacentric height

Metacentre M - point of intersection of buoyancy forces

before and after rotation

Stable if is +ve , or M is above G,Unstable - If is ve or M is below G,

Neutral if = 0 or M and G are at the same location

(a) Before rotating (b) After rotating

G-Centre of gravityC-Centre of buoyancy (centroid of displaced volume)

GM

(2) How to find the magnitude of :


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(2) How to find the magnitude of :

Neglecting the detailed derivations, can be calculated using:

Io is the second moment of the intersection area betweenthe water surface and the floating body (waterline).O is axis of rotating.For some frequently used geometries, I 0 has been given.

is the dis placed volume of water.

Example 3 9 A 0 25 m diameter cylinder is 0 25m long and


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Example 3.9 A 0.25-m-diameter cylinder is 0.25m long andcomposed of material with specific weight 8000 N/m 3. Will it float in waterwith the ends horizontal?

Solution: Assume cylinder is floatingwith axis vertical.

Waterline: circle with diameter 0.25m


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6. Fluids in Rigid-Body Motion

Fluid moves as a rigid body

No relative motion within the fluid, shear stress = 0

Can be treated as fluid statics as no shearing stress

Apply to fluids in linear acceleration & rotating motion

6 1 Linear Motion


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6.1 Linear MotionLiquid in an open container is moving along a straight linewith constant acceleration a , a y and a z

Along any constant pressure surface, dp = 0

z

y

ag

a

dy

dz

+=

z

y

ag

a

dy

dz

+== tan(3.21) (3.22)

From Eqs. (3.2) Eq. (3.5),

Pressure at any location: zag ya p z y )( +=

zy

Free surface equation:

Example 3.10


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A rectangular fuel tank is accelerated with constant

acceleration a y. (a) Derive an expression of pressure (in Pa)at the transducer for a fuel with SG=0.65. (b) The maximuma y that can occur before the fuel level drop to the transducer?

z

y

Solution:

6 2 Rigid Body Rotation


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6.2 Rigid-Body Rotation

Container of liquid rotating about z-axis at angular velocity

Use cylindrical polar co-ordinates, (r, , z) system(3.23)

As r is perpendicular to z,


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Along any constant pressure surface, dp=0

p p ,

dr r dz 2 = const gr

z +=2

22

(3.24)

(3.25)

Thus, constant pressure surfaces are parabolic surfaces. Free surface equation for a rotating container

= dzrdr dp 2 const zr p += 2

22

Thus, pressure is increasing with r 2 and decreasing with z.

To obtain pressure within the fluid, integrate Eq. (3.24),

(3.26)

The origin can be chosen either at the bottom of the container or at the surface.

Example 3.11


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pDetermine the relationship between change in fluid level, H-h o,

and angular velocity . Initial height: H.

Solution:

(a) Rotating container (b) cylindrical shell

chapter 3 fluid statistics

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