Water

Stagnant

not flowing

u = 0 m/s

In motion

flowing

u > 0 m/s

Hydrostatics Fluid dynamics

Pipes Open channel

[Gesloten leidingen] [Open water]

1

CU06997 Fluid Dynamics

Lecture 2 Hydrostatics [Hydrostatica]

1.1 Pressure (page 3-5) [Druk]

1.2 Pressure measurement (page 5-8)

1.3 Pressure forces on submerged bodies (page 8-11)

[Krachten op voorwerpen onder water]

example 1.1 (a) force on dam

1.4 Flotation (page 14, 15) [Opdrijven]

1

Newton Force F= m . g

F= Force [N]

m=Weight [Kg]

g=earths gravity = ± 10 [m/s2]

Water

m = ρ . V [kg/m3] . [m3]=[kg]

ρ fresh water =1000 [kg/m3]

ρ salt water =1025 [kg/m3]

1

Pressure

General pressure / druk

Pressure [Pa=N/m2]

Force [N]

Area on which the force acts [m2]

𝑝 =𝐹

𝐴

1

[N] yAgF

y

2

Fluid Pressure at a point

General pressure intensity

Pressure [Pa=N/m2]

Force [N]

Area on which the force acts [m2]

𝑝 =𝐹

𝐴=𝜌 ∙ 𝑔 ∙ 𝐴 ∙ 𝑦

𝐴= 𝜌 ∙ 𝑔 ∙ 𝑦

Gauge Pressure / Absolute Pressure 3

Pressure Head [Drukhoogte]

Pressure [Pa=N/m2]

fluid density [Kg/m3]

earths gravity = ± 10 [m/s2]

distance surface to point [m]

[Kg/m3]

[Kg/m3]

𝑦 =𝑝

𝜌 ∙ 𝑔

𝑦 =

3

Pressure Head in a pipe

y Transparent tube

Pressure 10.000 N/m2, fresh water,

Calculate pressure head

4

Pressure (p) 10.000 N/m2, fresh water

y

y

Transparent tube

4

Mano meter

4

Piezometric Head [Piezometrisch nivo]

Stagnant, not flowing

Piezometric Head

Surface / water level

[druklijn]

=Pressure Head[drukhoogte] [m]

=Potential Head[plaatshoogte] [m]

Horizontal reference line / datum

𝑦 =𝑝

𝜌 ∙ 𝑔

𝑧1 + 𝑦1 = 𝑧2 + 𝑦2 = 𝑧3 + 𝑦3

𝑦3 𝑦2 𝑦1

5

Pascal’s law

Pascal's law or Pascal's principle states that

“a change in the pressure of an enclosed

incompressible fluid is conveyed undiminished to

every part of the fluid and to the surfaces of its

container”

[Een druk die wordt uitgeoefend op een vloeistof die

zich in een geheel gevuld en gesloten vat bevindt,

zal zich onverminderd in alle richtingen

voortplanten]

6

Example Pascal’s law

???

𝑝 =𝐹

𝐴 = 1 N/m2

6

Example Pascal’s law

6

Atmospheric and gauge pressure

• 1 atmosphere = 100 kPa = 100.000 N/m2

• So 1 atm = 10 m freshwater = 10 mwc

• 1 bar = 1 atm = 100 kPa

• 1 mwc = 10.000 N/m2 = 10 kPa

• 10 mwc = 100 kPa = 1 bar = 1 atm

• mwc = meter water column [mwk]

6a

Gauge pressure = absolute pressure - atmospheric pressure

Relatieve druk = absolute druk – atmosferische druk

• Hand = 150 cm2 = 150/10000 =0.015 m2

• So downward force = 0.015 x 100.000 = 1.500 N

• According to Pascal, there is also a upward force

• Upward force = 0.015 x 100.000 = 1.500 N

6a

Atmospheric and gauge pressure

Archimedes’ principle

Buoyancy is the upward force that keeps things afloat. The

net upward buoyancy force is equal to the magnitude of the

weight of fluid displaced by the body. This force enables the

object to float or at least seem lighter.

7

Upward force = weight of fluid displaced by the body

Archimedes’ principle

7

7

EmptyCulvert

ground water level

cross-section

0 m NAP

+5 m NAP

+6 m NAP

5 m3 m

0,20 m

Exercise buoyancy [Opdrijving]

ρ concrete=2400 [kg/m3]

ρ soil=1600[kg/m3]

Will buoyancy [opdrijving] occur????

Exercise Pascal’s law

The tank contains fresh water. Width of

the tank is 2,44 m.

Question 1

Calculate the force on wall AB and CD

Question 2

Calculate the force on bottom BC

Question 3

Calculate the total weight of the water

and compare this water the answer of 2.

Explain the difference.