9. Hydrostatik I ( )

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9. Hydrostatik I (1.2–1.5)

•

Vätsketryck, tryck-densitet-höjd

•

Tryck mot plana ytor

•

Tryck mot plana ytor

Övningstal: H10 och H12

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HYDROSTATICS

• Hydrostatics: Study of fluids (water) at rest

• No motion ⇒no shear stress ⇒viscosity non-significant

• Only existing stress for a fluid at rest is normal (compression) stress, i.e. pressure

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Characteristics of pressure

1. Measurement unit [Pa]=[N/m2]

2. Pressure is transmitted normal to solid boundaries or arbitrary sections

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3. Pressure is transmitted undiminished to all other points in a fluid at rest 4. Pressure has the same magnitude in all directions at a point in a fluid at

rest (scalar quantity)

arbitrary = godtycklig skalar => ingen vektor

Relation between pressure and depth in an

incompressible liquid

Assuming constant density and no horizontal pressure variation, the liquid column in the fig below can be used to determine the pressure as a function of depth.

Vertical forces acting on column (V, A, and y are volume, area, and height

of column, respectively):

• Upward pressure force: P·A (EQ 1.2) • Weight (downward): wV = wAy

V ti l f b l

yy A

A

• Vertical force balance:

P·A = wAy⇒P = wy = ρgy (EQ 1.8) • Pressure often quoted as heads,

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ABSOLUTE AND RELATIVE PRESSURE

• Pressures are measured and

quoted in two different systems, one relative (gauge) and one one relative (gauge) and one absolute.

• The relation between them is: P_abs= P_atm+ P_gauge (EQ 1.9) • Negative gauge pressures are

often termed vacuum pressures

≈ 100 kPa

• Often only relative pressures are of interest

gauge = tryckgivare

EXAMPLE ON GAUGE AND ABSOLUTE

PRESSURES

A pressure gauge registers a vacuum of 310 mm of mercury

Solution:

Pt h i = 100 kPa

when the atmospheric pressure is 100 kPa, absolute. Calculate the corresponding absolute pressure. Patmospheric 100 kPa (Pgauge/wHg) = -310 mm Hg ⇒ Pgauge= -0.31wHg wHg= 133.0 kN/m2 (e.g., page 4; ρg) VVR145 Vatten VVR145 Vatten

Pabsolute= Patmospheric+ Pgauge⇒

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Fig. 1.5 Typical examples of situations where

hydrostatic force may have to be calculated

FORCE ON SUBMERGED PLANE SURFACES

mini summary

Example of applications:

- Design of dams, ships, gates, and tanks.

Characteristics of pressure in a fluid at rest:

• Constant pressure on plane horizontal surface Linear pressure variation with depth for constant

• Linear pressure variation with depth for constant density liquid

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• Pressure prism: “volume of pressure” on the plane

• Pressure prism: volume of pressure on the plane surface F = P A = ρgh A (EQ 1.11)

• Resultant force is equal to the volume of the pressure prism and acts through its centroid

γ= w (tunghet) prism = prisma centroid = geometriskt centrum (‘yt-tyngdpunkt’)

RELEVANT EQUATIONS – FORCES ON PLANE

SUBMERGED SURFACES

• Resultant force:

F= whGA = ρghGA (EQ 1.11)

• Point of action of resultant force: LP= IG/(A LG) + LG (EQ 1.13)

• A: area of plane surface; • hG: vertical distance liquid

surface - area center; • LP: distance O - pressure t A LP LG L LL VVR145 Vatten VVR145 Vatten center;

• L_G: distance O – area center; • I_G: second moment of area

about area center axis; • L = h/sinθ

LLp

P=ρgh

IG= second moment of the area: yttröghetsmoment Point of action = angreppspunkt

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Page 10 Page 10

Median line gives lateral position for center of pressure for regular plane areas

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H10

A

t

g l

g t 1 8

l

g

d 1 2

A rectangular gate 1.8 m long and 1.2

m high lies in a vertical plane with its

centre 2.1 m below a water surface.

Calculate magnitude, direction and

location of the total force on the gate.

H12

This rectangular gate will open

automatically when the depth of water, d,

becomes large enough. What is the

minimum depth that will cause the gate to

open?

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UPPGIFT 1 (6 poäng)

Vilken kraft P behövs för att hålla kvar den 5 m breda (in i pappret) rektangulära luckan i sin position enligt figuren nedan? Luckans längd är L = 4 m och vattendjupet till vänster om luckan (till vänster om leden) är 2 m. Antag att leden är f ikti f i h tt i h l ft å hö id l k

friktionsfri och att vi har luft på höger sida om luckan. Försumma luckans egentyngd.

LÖSNING – steg 1: Rita ut relevanta krafter som verkar på luckan.

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10. Hydrostatik II (1.6–1.7)

•

Tryck mot buktiga ytor

•

Flytkraft / Archimedes princip

•

Övningstal: H15 och H18

Fig. 1.27 Pressure on a sphere

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FORCES ON CURVED SUBMERGED SURFACES

(1) Resolve the force into two components, one vertical and one horizontal

Pressure intensity on a curved surface. F passes through the center of curvature.

curvature = krökning

(2) The horizontal force is obtained by projecting the curved surface onto a vertical plane. The horizontal force is equal to the force on this projected area: FH= ρghG,projAproj

P G

Projection of the curved surface onto a vertical plane

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(3) The vertical force is equal to the weight of the volume of liquid above the curved surface

F F = == =ρρ··gg··VV F F_V_V= = = = ρρ··gg··VV Kom ihåg: Kom ihåg: V

V = = volymvolym ovanförovanför

The vertical force component, F_V, caused by the weight of liquid above

the surface

(4) The resultant force is given by:

and the direction of the resultant force by:

2 2 H F V F F = + H F V F = φ tan Eq. 1.15 Eq. 1.15 Eq. 1.16 Eq. 1.16

The direction of the resultant force, F, which must also pass through C

(5) Remember that there is an equal

( )

and opposite force acting on the other side of the surface.

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ARCHIMEDES PRINCIPLE – BUOYANCY

FORCE

Law of buoyancy (Archimedes’ principle):

“Th th t (b f ) b d

• “The upthrust (buoyancy force) on a body immersed in a fluid is equal to the weight of the fluid displaced” Law of flotation: VVR145 Vatten VVR145 Vatten VVR145 Vatten VVR145 Vatten

• “A floating body displaces its own weight of the liquid in which it floats”

Proof of Archimedes principle

Vertical forces acting cylinder surface: “Downwards” P1: p1A = ρ·g·yA = w ·yA “Upwards” P2: p2A = ρ·g·(y+L)A= w·(y+L)A F_B ••

“Net pressure force (upthrust)”, FB:

F_B= w(y+L)A - wyA= wLA =

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H15: The quarter cylinder AB is 3 m long.

Calculate magnitude, direction, and location

of the resultant force of the water on AB.

Z Z X X VVR145 Vatten VVR145 Vatten VVR145 Vatten VVR145 Vatten C C

H18: The weightless sphere of diameter d is

in equilibrium in the position shown.

Calculate d as a function of w

₁

, h

₁

, w

₂

, and h

₂

.

Sfärs volym = Sfärs volym = ππdd33_/6_/6 w w₁ w w₂ VVR145 Vatten VVR145 Vatten VVR145 Vatten VVR145 Vatten Area = Area = ππdd22_/4_/4 equilibrium = jämvikt

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11. Hydrostatik III (1.8, 1.9, 2.1-2.7)

•

Hydrostatiska jämviktsekvationen

•

Tryckmätning, manometri

•

Övningstal: H1, H3-4 och H8

Hydrostatiska Jämviktsekvationen

(samband mellan tryck, densitet och vertikalt avstånd)

The general relation for pressure in a static fluid is:

g

dz

dp

ρ

γ

=

−

=

w

z => dp dp = = -- w w · · dzdz

Implication: pressure varies only with depth and is

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For a fluid with constant density:

or

pp₁₁–– pp₂₂= = γγ((zz22–– zz11))= = γγ ·· hh

(p z)

Implications:

• pressure varies linearly with depth γ 2 1 p p h − = •(p2, z2) •(p1, z1) h z VVR145 Vatten VVR145 Vatten

pressure varies linearly with depth

• pressure may be expressed as head of fluid of weight density w

• pressure are often quoted as head in mm Hg or m H₂O

• z Const , for all points in a fluid at rest p z p = + = + 2 2 1 1 γ γ

H1:

The weight density (w =

ρ

⋅

g) of water in the

b

l

l t d f

th

i i

l

ocean may be calculated from the empirical

relation w = w

₀

+ K

⋅

(h)

1/2

, in which h = the depth

(m) below the ocean surface. Derive an

expression for the pressure at any point h and

calculate weight density and pressure at a

depth of 3220 m assuming w

₀

= 10 kN/m

3

, K =

depth of 3220 m assuming w

₀

10 kN/m , K

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Sample problem:

An open tank contains water 1.4 m deep

d b

2

thi k l

f il ( d 0 855)

covered by a 2 m thick layer of oil (r.d.=0.855).

What is the pressure head at the bottom of the

tank, in term of a water column?

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VVR145 Vatten (r.d. = relative density)

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MEASUREMENT OF PRESSURE

MANOMETRY

Pressure is constant over horizontal planes within continuous columns of the same fluid

Conversion of manometer readings to Pressure (γ= w)

(a) p1= p2 p1= px+ γl p2= patm+ γ1h ⇒p_x= p_atm+ γ₁h -γl (absolute) (b) p4= p5 p4= px+ γ1l1 p₅= p_y+ γ₂l₂+ γ₃h ⇒p_x- p_y= γ₂l₂+ γ₃h -γ₁l₁

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H3: With the manometer reading as

shown, calculate p

_x

.

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VVR145 Vatten (r.d. = relative density)

H4: Calculate p

_x

- p

_y

for this inverted U-tube

manometer.

ρ

= r.d.

⋅ ρ

_water

).

L

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H8* The sketch shows a sectional view through a

submarine. Calculate the depth of submergence,

y. Assume that the weight density of sea water is

10.0 kN/m

3

.

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