EXPERIMENT OF
ELEMENTARY FLUID MECHANICS ELEMENTARY FLUID MECHANICS
Hydrostatic Pressure
Instructor: Seo, Il Won (35-310) TA: Jung, Young Jai (35-504) Tel: +82-2-880-7356; e-mail: [email protected]
Objectives Objectives
Chapter 1 Hydrostatic Pressure
• To determine the center position of pressure of a plane surface
Chapter 1. Hydrostatic Pressure
immersed in water
T th i t l iti ith th th ti l
• To compare the experimental position with the theoretical results
Background theory Background theory
•Fluid pressure ?
What is fluid statics…?
The pressure at some point within a fluid
• Fluid statics ?
- No relative motion between fluid elements - No shear stress
Can’t support shearing stresses
Pressure must be acting on g normal to surface
Background theory Background theory
• Hydrostatic pressure ? y p
The pressure at any given point of a static fluid is called
the hydrostatic pressure
Background theory Background theory
• How to calculate hydrostatic pressure ?
h
( : specific gravity, : depth)p = γ h γ h
Why?
•No shear stress
& Acting normal to surface Why?
• Force Equilibrium
F F
z2
F
2 2 0
Fx p dx p dx
p p dz
x x
⎡⎛ ∂ ⎞ ⎛ ∂ ⎞⎤
=⎢⎣⎜⎝ −∂ ⎟ ⎜⎠ ⎝− +∂ ⎟⎠⎥⎦ =
∑
⎡⎛ ⎞ ⎛ ⎞⎤
x1
F Fx2
2 2 0
Fz p dz p dz
p p dx dW
z z
⎡⎛ ∂ ⎞ ⎛ ∂ ⎞⎤
=⎢⎣⎜⎝ −∂ ⎟ ⎜⎠ ⎝− +∂ ⎟⎠⎥⎦ − =
∑
1 2
z z
F =F +dW
••
•
Background theory Background theory
• How to calculate hydrostatic pressure ?
h
( : specific gravity, : depth)p = γ h γ h
Eventually!
∂p P i i h i l l
: 0
Fx p x
∂ =
∑
∂ Pressure is constant in a horizontal plane in a static fluidz:
F p
z ρg γ
∂ = =
∑
∂ Increase of pressure with depth in a fluid of constant density – linearly increase constant density linearly increasep γ z Δ = Δ
Background theory Background theory
• Summary
2. Hydrostatic pressure:
Increase linearly with depth
1. Fluid at rest:
Pressure must be acting on
3. Same at all points on a horizontal plane Pressure must be acting on
normal to the surface
Example of
Example of hydrostatic hydrostatic pressure pressure
• Hydroelectric dam
F diff i h d h bi
From pressure difference with depth, water gets energy to operate turbine.
Example of
Example of hydrostatic hydrostatic pressure pressure
B t l k & h d t ti
•Basement leaks & hydrostatic pressure
Backfill becomes saturated
Hydrostatic pressure makes a crack at the basement wall at the basement wall
Water infiltrates the basement wall
Example of
Example of hydrostatic hydrostatic pressure pressure
• There are many kinds of problems….
- Which point should we choose to intake water from dam?
Where should we release water to get sufficient energy?
- Where should we release water to get sufficient energy?
- How can we construct safe dam?
- How can we protect water leak problem?
Basic solutions of these problems come from understanding hydrostatic pressures c so u o s o ese p ob e s co e o u de s d g yd os c p essu e
Hydrostatic
Hydrostatic pressure pressure on on submerged submerged plane plane
• Pressure distribution diagram
Pressure is linearly increase from water surface to bottom Pressure is linearly increase from water surface to bottom
h
1. What is force, F?
2. Where is the point
f i ?
h1
1 1
p =ρgh
? z=
X of action?
? F= Center of the area
Hydrostatic
Hydrostatic pressure pressure on on submerged submerged plane plane
•
Total force acting on surface- Pressure at p=ρgh
h
- Force acting on surface
h1 p1=ρgh1
? z=
F
∫
dA∫
X dAX
p ρg
? F=
F p dA gX dA
gXA ρ ρ
= =
=
∫ ∫
r
- First moments of areas X dA=X A
∫
Force
∫
X dA=X AForce
= Sum of pressure Center of the pressure
Hydrostatic
Hydrostatic pressure pressure on on submerged submerged plane plane
• Position of the total force
O
- Moment at O
2
M=
∫
X ρg dA=ρgIoo=Fz h11 1
p=ρgh X
O
(Second moment of area: )
oo oo
M g I I
z ρ
∴ = = =
2
X dA=Ioo
∫
? z= X
r
d
z F ρg A X A X
∴ = = =
- From parallel axis theorem
3
C F
2
oo gg
I =I +A X
3
gg 12 I =bd
(Rectangular : )
( )
2 3
/ 12
/ 2 2
gg gg
oo I A X I
I bd d
z X r
bd r d
A X A X A X
+ ⎛ ⎞
∴ = = = + = × − +⎜⎝ − ⎟⎠
When t=0, Center of pressure is located on 2/3 down from water surface to bottom of submerged plate
Observation
Observation for for Total Total Immersion Immersion
- In case of rectangular plane d
2 X = −r d
- Resultant Force
F = ρ gXA
- Center point of pressure
ρ g
from water surface
( )
3/ 12
/ 2 2
bd d
z r
bd r d
⎛ ⎞
= × − +⎜⎝ − ⎟⎠
from pivot
( ) ⎝ ⎠
( )
3/ 12
/ 2 2
c
bd d
X r q
bd d
⎛ ⎞
= +⎜⎝ − ⎟⎠+ o p vo ( Fulcrum)
( / 2) 2
c q
bd× −r d ⎜⎝ ⎟⎠
Theoretical
Theoretical Value Value for for Total Total Immersion Immersion
Moment arm
Force 1 Force 1
r
Force 2
1
1 2 1 2
2
c c
M M F L F X X F L
= → = → = F
F1=mg
( ) ( )
2
1 1
2 2
F = ×ρgr r b× × − ×ρg r−d × −r d ×b
p=ρgh
r
d
2 2 ρg 2 ρg
Observation
Observation for for Partial Partial Immersion Immersion
Moment arm
Force 1
y
- Center point of pressure
Force 2 3 y
Center point of pressure
z = 2 y y
X = + − a d z 3 y
c
3
X a + d
from water surface from pivot (Fulcrum)
Theoretical
Theoretical Value Value for for Partial Partial Immersion Immersion
Moment arm
Force 1
y Force 2 3 y
1
1 2 1 2
2
c c
M M F L F X X F L
= → = → = F
p ρg y F1=mg
by2
p=ρg y y
2 2
F =
ρ
g X A=ρ
gbyPrinciple
Principle of of Experiment Experiment
•Place a weight until the balance arm is horizontal.
•Record the water level on the quadrant and the weight on the balance pan.q g p
•Repeat this procedure until the water level reaches the top of the quadrant.
Procedure Procedure
①Place the quadrant on the two dowel pins and using the clamping screw,
fasten to the balance arm.
②Measure values of a, L, depth d and width b of the quadrant end face.
③With the Perspex tank on the bench, position the balance arm on the knife edges (pivot).
③W e e spe o e be c , pos o e b ce o e e edges (p vo ).
④Hang the balance pan from the end of the balance arm.
⑤Connect a hose from the drain cock to the sump and another from the bench feed to the
t i l t th t f th P t k
triangular aperture on the top of the Perspex tank.
⑥Level the tank using the adjustable feet and spirit level.
Procedure Procedure
⑦Move the counter balance weight until the balance arm is horizontal.
⑧Close the drain cock and admit water until the level reaches the bottom edge of the quadrant.
⑨Place a weight on the balance pan, slowly adding water into the tank
until the balance arm is horizontal.
until the balance arm is horizontal.
⑩Record the water level on the quadrant and the weight on the balance pan.
⑪Repeat the above for each increment of weight until the water surface level reaches the top
f h d d f Th h i f i h i i h d
of the quadrant end face. Then remove each increment of weight noting weights and water levels until the weights have been removed.
Results Results
a Arm
L
depth d
Width b
Mass m
Measured length
Theoretical Value
(m) L
(m)
d (m)
b (m)
m
(kg) Xcm Xct
Discussion Discussion
1 Confirm that the center of pressure is below the center of area 1. Confirm that the center of pressure is below the center of area,
and discuss why.
2. Compare the theoretical value with measured length, discuss why does the difference occurs.
3. Discuss why we do not consider the pressure at the surface of arc.
Why we just calculate the pressure of the rectangular section? y j p g
