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9. Hydrostatik I (1.2–1.5)
• Vätsketryck, tryck-densitet-höjd
• 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
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)
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arbitrary = godtycklig
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):
y
A
•
Upward pressure force: P·A (EQ 1.2)
•
Weight (downward): wV = wAy
•
V ti l fforce balance:
Vertical
b l
P·A = wAy ⇒ P = wy = ρgy
•
(EQ 1.8)
Pressure often quoted as heads,
h = P/w (in m H2O or mm Hg)
heads = tryckhöjd
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ABSOLUTE AND RELATIVE PRESSURE
• Pressures are measured and
quoted in two different systems,
one relative (gauge) and one
absolute.
≈ 100 kPa
• The relation between them is:
Pabs = Patm + Pgauge
(EQ 1.9)
• Negative gauge pressures are
often termed vacuum pressures
• Often only relative pressures are
of interest
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gauge = tryckgivare
EXAMPLE ON GAUGE AND ABSOLUTE
PRESSURES
A pressure gauge registers a
vacuum of 310 mm of mercury
when the atmospheric pressure
is 100 kPa, absolute. Calculate
the corresponding absolute
pressure.
Solution:
Patmospheric
t
h i = 100 kPa
(Pgauge/wHg) = -310 mm Hg ⇒
Pgauge = -0.31w Hg
wHg = 133.0 kN/m2 (e.g., page 4; ρg)
Pabsolute = Patmospheric + Pgauge ⇒
Pabsolute = 100 - 0.31⋅133.0 = 58.8 kPa
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Fig. 1.5 Typical examples of situations where
hydrostatic force may have to be calculated
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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
density liquid
• Pressure acts perpendicular to the surface
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• Pressure prism: “volume
volume of pressure”
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 =
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prisma Vatten centroid = geometriskt centrum (‘yt-tyngdpunkt’)
RELEVANT EQUATIONS – FORCES ON PLANE
SUBMERGED SURFACES
•
Resultant force:
F= whGA = ρghGA
•
LP = IG/(A LG) + LG
A
LP LG L
Lp
(EQ 1.11)
Point of action of resultant force:
(EQ 1.13)
•
A:
area of plane surface;
•
h G:
vertical distance liquid
surface - area center;
•
LP:
distance O - pressure
center;
t
P=ρgh •
L G:
distance O – area center;
•
I G:
second moment of area
about area center axis;
•
L = h/sinθ
Compare with Figure A1.1, page 554 !
IG = second moment of the area: yttröghetsmoment
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Point of action = angreppspunkt
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Page 10
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Median line gives lateral position for center of
pressure for regular plane areas
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H10
A rectangular
t g l gate
g t 1.8
1 8 m long
l g and
d 1.2
12
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.
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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
friktionsfri
f i och
h att
tt vii har
h luft
l ft på
å höger
hö
sida
id om luckan.
l k
Försumma luckans egentyngd.
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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
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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.
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(2)
curvature = krökning
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 = ρg hG,proj Aproj
G
P
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
FV = = ρ·
ρ·g
g·V
Kom ihåg:
V = volym ovanför
The vertical force component, FV, caused by the weight of liquid above
the surface
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(4) The resultant force is given by:
F =
F2 + F2
V
H
and the direction of the resultant force by:
tan φ =
Eq. 1.15
F
V
F
H
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
“The upthrust
th
t (buoyancy
(b
force)
f
) on a body
b d
immersed in a fluid is equal to the weight of the
fluid displaced”
Law of flotation:
• “A floating body displaces its own weight of the
liquid in which it floats”
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Proof of Archimedes principle
Vertical forces acting cylinder
surface:
“Downwards” P1:
p1A = ρ·g·yA = w ·yA
“Upwards” P2:
FB
•
p2A = ρ·g·(y+L)A= w·(y+L)A
“Net pressure force (upthrust)”, FB:
FB = w(y+L)A - wyA= wLA =
= wV = ρ·g·V
Eq. 1.14
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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
X
C
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H18: The weightless sphere of diameter d is
in equilibrium in the position shown.
Calculate d as a function of w1, h1, w2, and h2.
w1
w2
Sfärs volym = πd3/6
Area = πd2/4
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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
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Hydrostatiska Jämviktsekvationen
(samband mellan tryck, densitet och vertikalt avstånd)
The general relation for pressure in a static fluid is:
dp
dz
= −γw = − ρ g =>
z
dp = - w · dz
Implication: pressure varies only with depth and is
constant in a horizontal plane
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OBS: z pekar uppåt
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For a fluid with constant density:
p1 – p2 = γ (z2 – z1) = γ · h
or
h=
• (p2, z2)
p −p
1
2
h
γ
• (p1, z1)
z
Implications:
•
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
•
p
1
p
2
+z =
+ z = Const , for all points in a fluid at rest
1
2
γ
γ
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H1:
The weight density (w = ρ⋅g) of water in the
ocean may be
b calculated
l l t d from
f
the
th empirical
ii l
1/2
relation w = w0+ K⋅(h) , 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 w0 = 10 kN/m3, K =
7.08 N/m7/2.
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Sample problem:
An open tank contains water 1.4 m deep
covered
d by
b a 2 m thick
thi k layer
l
off oil
il (r.d.=0.855).
( d 0 855)
What is the pressure head at the bottom of the
tank, in term of a water column?
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(r.d. = relative density)
Mini summary:
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MEASUREMENT OF PRESSURE
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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
(b) p4 = p5
p1 = px + γ l
p4 = px + γ1l1
p2 = patm + γ1h
⇒ px = patm + γ1h - γ l
p5 = py + γ2l2 + γ3h
(absolute)
⇒ px - py = γ2l2 + γ3h - γ1l1
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H3: With the manometer reading as
shown, calculate px.
L
R
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(r.d. = relative density)
H4: Calculate px - py for this inverted U-tube
manometer. ρ = r.d. ⋅ ρwater).
L
R
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(r.d. = relative density)
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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/m3.
R
L
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