Cu06997 lecture 2_hydrostatics_17-2-2013
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Here are the steps to solve this problem using Pascal's law: 1) Wall AB and CD are both vertical surfaces with an area of 2.44 m x 5 m = 12.2 m^2 Pressure at the bottom is due to a water column of 5 m Pressure = Density x Gravity x Height = 1000 x 10 x 5 = 50,000 N/m^2 Force on wall AB/CD = Pressure x Area = 50,000 x 12.2 = 610,000 N 2) Bottom BC is a horizontal surface with an area of 2.44 x 5 = 12.2 m^2 Pressure due to water above it is the pressure at the bottom = 50,
![Water
Stagnant In motion
not flowing flowing
u = 0 m/s u > 0 m/s
Hydrostatics Fluid dynamics
Pipes Open channel
[Gesloten leidingen] [Open water]
1](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-1-320.jpg)
![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](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-2-320.jpg)
![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](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-3-320.jpg)
![General pressure / druk
Pressure [Pa=N/m2]
Force [N]
Area on which the force acts [m2]
Pressure
𝐹
𝑝=
𝐴
1](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-4-320.jpg)
![y
F g A y [N]
2](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-5-320.jpg)
![General pressure intensity
Pressure [Pa=N/m2]
Force [N]
Area on which the force acts [m2]
Fluid Pressure at a point
𝐹 𝜌∙ 𝑔∙ 𝐴∙ 𝑦
𝑝= = = 𝜌∙ 𝑔∙ 𝑦
𝐴 𝐴
Gauge Pressure / Absolute Pressure
3](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-6-320.jpg)
![Pressure Head 𝑝
𝑦=
[Drukhoogte] 𝜌∙ 𝑔
𝑦 = distance surface to point [m]
Pressure [Pa=N/m2]
fluid density [Kg/m3]
earths gravity = ± 10 [m/s2]
[Kg/m3]
3 [Kg/m3]](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-7-320.jpg)
![Piezometric Head [Piezometrisch nivo]
𝑧1 + 𝑦1 = 𝑧2 + 𝑦2 = 𝑧3 + 𝑦3
Stagnant, not flowing
𝑦3 Piezometric Head
𝑦2 𝑦1 Surface / water level
[druklijn]
Horizontal reference line / datum
𝑝
𝑦= =Pressure Head[drukhoogte] [m]
𝜌∙ 𝑔
5 =Potential Head[plaatshoogte] [m]](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-11-320.jpg)
![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](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-12-320.jpg)
![Atmospheric and gauge pressure
Gauge pressure = absolute pressure - atmospheric pressure
Relatieve druk = absolute druk – atmosferische druk
• 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](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-15-320.jpg)
![Exercise buoyancy [Opdrijving]
Will buoyancy [opdrijving] occur????
+6 m NAP
ground water level +5 m NAP
5m
0,20 m
Culvert
3m
Empty
0 m NAP
ρ concrete=2400 [kg/m3] cross-section
ρ soil=1600[kg/m3]](https://image.slidesharecdn.com/cu06997lecture2hydrostatics17-2-2013-130217135243-phpapp01/85/Cu06997-lecture-2_hydrostatics_17-2-2013-20-320.jpg)
Cu06997 lecture 2_hydrostatics_17-2-2013
- 1. Water Stagnant In motion not flowing flowing u = 0 m/s u > 0 m/s Hydrostatics Fluid dynamics Pipes Open channel [Gesloten leidingen] [Open water] 1
- 2. 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
- 3. 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
- 4. General pressure / druk Pressure [Pa=N/m2] Force [N] Area on which the force acts [m2] Pressure 𝐹 𝑝= 𝐴 1
- 5. y F g A y [N] 2
- 6. General pressure intensity Pressure [Pa=N/m2] Force [N] Area on which the force acts [m2] Fluid Pressure at a point 𝐹 𝜌∙ 𝑔∙ 𝐴∙ 𝑦 𝑝= = = 𝜌∙ 𝑔∙ 𝑦 𝐴 𝐴 Gauge Pressure / Absolute Pressure 3
- 7. Pressure Head 𝑝 𝑦= [Drukhoogte] 𝜌∙ 𝑔 𝑦 = distance surface to point [m] Pressure [Pa=N/m2] fluid density [Kg/m3] earths gravity = ± 10 [m/s2] [Kg/m3] 3 [Kg/m3]
- 8. Pressure Head in a pipe y Transparent tube Pressure 10.000 N/m2, fresh water, Calculate pressure head 4
- 9. Pressure (p) 10.000 N/m2, fresh water y Transparent tube y 4
- 10. Mano meter 4
- 11. Piezometric Head [Piezometrisch nivo] 𝑧1 + 𝑦1 = 𝑧2 + 𝑦2 = 𝑧3 + 𝑦3 Stagnant, not flowing 𝑦3 Piezometric Head 𝑦2 𝑦1 Surface / water level [druklijn] Horizontal reference line / datum 𝑝 𝑦= =Pressure Head[drukhoogte] [m] 𝜌∙ 𝑔 5 =Potential Head[plaatshoogte] [m]
- 12. 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
- 13. Example Pascal’s law 𝐹 𝑝= = 1 N/m2 𝐴 ??? 6
- 15. Atmospheric and gauge pressure Gauge pressure = absolute pressure - atmospheric pressure Relatieve druk = absolute druk – atmosferische druk • 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
- 16. Atmospheric and gauge pressure • 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
- 17. 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
- 18. Archimedes’ principle Upward force = weight of fluid displaced by the body 7
- 19. 7
- 20. Exercise buoyancy [Opdrijving] Will buoyancy [opdrijving] occur???? +6 m NAP ground water level +5 m NAP 5m 0,20 m Culvert 3m Empty 0 m NAP ρ concrete=2400 [kg/m3] cross-section ρ soil=1600[kg/m3]
- 21. 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.