Hydrostatics - Katedra hydrauliky a hydrologie -...

Hydrostatics

K141 HYAE Hydrostatics 1

HYDRAULICS

HYDROSTATICS HYDRODYNAMICS

HYDROSTATICS

Liquid in equilibrium with regard

to Earth

to moving

system

(reservoir)

relative

equilibrium


PRESSURE IN LIQUID

- equilibrium: only normal stress pressure (tangential stress ij 0)

pressure force:

pressure:

S

F

S

F Pa

dS

dFp

- at motion: both normal and tangential stress

pressure (hydrodynamic) depends on velocity

S S

dS pdFF dS pdF

In gravitational field:

hydrostatic pressure: ph = gh

pressure in point M: p = pe+ ph = pe + gh

pressure force at horizontal bottom: F = pS


21

21

pghpSpghSSp

21 FgV F

hSVSpFSpF

22

11

Requirement of force equilibrium

in vertical direction:

h12 pghpp

Hydrostatic pressure in liquid in vertical tube closed by pistons

(considering gravity acceleration)


equilibrium of external forces

(for = const.):

SdppcosadsSSp

as → resulting acceleration in direction s change of pressure

on path ds

pressure force pressure force body force

dscosadp Euler hydrostatic equation

(one-dimensional form)

CHANGE OF PRESSURE


Euler hydrostatic equation

(component form)

component form:

dzadyadxadp zyx

zyx dpdpdp dp

dxadp xx

dyadp yy

dzadp zz

pppp zyx

Gravitational force field

dhds,ga

dhgdp

p

p

h

0e

gdhdp hgpp e hhydrostatic pressure p g h


SURFACE AREA

h = const. p = const. 0dhgdp dh = 0

22a11e hgphgp hgp

ppp

phgp

12

21

Perpendicular to resulting acceleration

in gravitational field horizontal plane level

3111e hghgp

3122a hghgp

differential pressure gauge

(U-tube)

piezometer

41ae hgpp


normal atmospheric pressure pa = 101324.72 Pa 105 Pa total static pressure ps

ps > pa overpressure

pp = (ps – pa) > 0

ps < pa underpressure

pva = (ps – pa) < 0

pe = pa hgpp as

1 a 1 p1 1

2 a 2 p2 2

3 a 3 va3 3

p p g h , p g h

p p g h , p g h

p p g h , p g h

OVERPRESSURE, UNDERPRESSURE

pva

ps=0

pa

pp

ps

ps


PASCAL’S LAW

Gradual pressure change p in a small closed volume of liquid

expends in all directions and passes on all points of liquid without

any change.

p = const., p = const. S

Fp

force [N]

area [m2]

22

2

1

11 p

S

F

S

Fp

1

212

S

SFF

21 FpF

Pressure head Suction head [m, m w.c.] g

ph

g

ph va

va

sfor p 05

va amaxp p 10 Pa .c.wm10h maxva


HYDRAULIC PRESS

PRESSURE CONVERTER

21 1

2

Ftheoretically F S

S

1

221

S

Spp

12

21 S

S

FF

... efficiency

(0,95 –1,0)

In practice losses


HYDROSTATIC FORCE

Hydrostatic force = force caused by hydrostatic pressure ph.

dSpdF

S S S

F pdS gzdS, for g const. : F g zdS

F – passes through centre of pressure body S

zdS

- perpendicular to loaded area

z ... vertical depth (depth bellow level)

If overpressure pp (underpressure pva) on level to enlarge

(reduce) real depth z by pressure head g

p,

g

pvap

total pressure force


HORIZONTAL BOTTOM

ShdShzdSSS

ShgF

h·S – volume of pressure body

hydrostatic

paradox


tF g z S

INCLINED PLANE SURFACE

SzzdS t

S

For prismatic areas with

horizontal border – possible also

... area of pressure diagram [m2]

= b ... volume of pressure body [m3]

gbgba2

zzgSzgF 21

t

baS,2

zzz 21

t

– volume of pressure body

tz S

tz S

tz S

T - centre of area S

C – point of application of

force F


DETERMINATION OF POINT OF APPLICATION

OF HYDROSTATIC FORCE

Moment condition to x-axis:

I0 ... .second moment of loaded area S about gravity centre axis o

S

c ydFyF

Ix ..... second moment of loaded area S about x-axis

t

t

o

t

2

toc y

yS

I

yS

ySIy

t

otc

yS

Iyy

dSsinygdSzgdSpdF

,dSysingdSzgFSS

S

2

Sc dSysingdSysingy

t

x

S

S

2

S

S

2

cyS

I

dSy

dSy

dSysing

dSysing

y

2

tox ySII


Determination of point of application of hydrostatic force F on

rectangular area with horizontal upper edge corresponding with

water level

baS

,ab12

1I

,2

ay

3

o

t

a3

2

2

aa

6

1

2

a

2

aba

ab12

1

yyS

Iy

3

t

t

oc

centre of loaded area S

on gravity centre axis o

point of application

of hydrostatic force F


pressure diagram - components

pressure diagram

- complex

RESOLUTION OF HYDROSTATIC FORCE IN COMPONENTS

z

x

Ftg

F

K141 HYAE Hydrostatics 16 n

ihz

n

2hz

n

1hz

i

2

1

Effective distribution of horizontal beams

11 z3

2x

1i2ii2 z3

2ωz

3

22ωωx

1iiiiii z3

2ω1iz

3

2iωωx

From equation of moments:


HYDROSTATIC FORCE ACTING ON CURVED AREAS

Two perpendicular horizontal (Fx, Fy), and vertical component (Fz):

xix dSzgcosdSzgdF

yiy dSzgcosdSzgdF

ziz dSzgcosdSzgdF

,SzgF xtxx ytyy SzgF

VgSzgF ztz

ztx (zty) ... vertical depth of projection Sx (Sy)

Sx (Sy) ... .projection of area S at plane YZ (XZ)

V ... ……..volume of vertical column of liquid above area S

2z

2y

2x FFFF ,

F

Fcos,

F

Fcos,

F

Fcos zyx

For prismatic areas also : 0F,bgF,bgF yzzxx


Course of hydrostatic pressure:

Hydrostatic force passes through centre of curvature of cylindrical

area S

Solving hydrostatic force in components:


FLOATING BODIES

Application of Archimedes principle:

buoyancy force

k … density of liquid [kgm-3]

W … volume of displaced liquid

(displacement) [m3], W = W (tn)

Vertical cylinder submerged in liquid

External surface forces:

ShgF,ShgF 2211

vzkk

12k

1k2k12

FWgShghhSg

ShgShgFF

WgF kvz Archimedes principle


Fvz < G body gravitates

Fvz > G body moves up

till Fvz = G

Fvz = G body in balance

V = W – body hovers

V > W – body floats

(V – body volume)

Fvz goes through centre C of displacement W

G goes through centre of floating body

Resolution of immersion: G = Fvz W tn


OVERVIEW OF MAIN TERMS AND TOPICS

overpressure, underpressure, static pressure

hydrostatic pressure

pressure head, suction head

Pascal´s law

hydrostatic force acting on plane and curved surface area

(dimension, direction, point of action)

pressure body - complex

- in components

floating bodies - Archimedes principle

Hydrostatics - Katedra hydrauliky a hydrologie -...

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