TITLE: HYDROSTATIC FORCE ON PLANE SURFACE

INTRODUCTION:

  • The determination of force which are exerted by liquid which are at rest on surface immersed in liquids.
  • From the study by hydrostatic, the following principles have been established :

ü There are no shear stresses present when the fluid is not in motion.

ü The pressure exerted by a fluid under hydrostatics conditions. This pressure acts perpendicular to an immersed surface.

ü Hydrostatic pressure various linearly, increasing with an increase in depth.

OBJECTIVE:

  • The objective of this experiment are:

ü To determine experimentally the magnitude of the force of pressure (hydrostatic force) and its point of action (center of pressure) a plane surface.

ü To compare the experimental result with the theoretical values

HYDROSTATIC FORCE ON PLANE SURFACE

  • A plane surface of total area, A which is immersed in a liquid at depth, h is subjected to hydrostatic pressure P.

  • If the plane surface is horizontal as shown, then anywhere on the plane surface, the pressure is given;

P= ρgh

  • Another explanation is shown in the THEORETICAL content.

THEORY

TOTAL PRESSURE

For a static fluid, the shear stress is zero and the only stress is true normal stress. Pressure which when in contact with solid surface exerts a normal force towards the surface. It is also call hydrostatic force.

F=ρghA

Where;

F=total pressure (hydrostatic force)

P=mass density of the liquid

g=gravitation acceleration

A=submerge area of the plane surface

H=vertical distance from the liquid surface to the centroid of the submerge plane surface

CENTRE OF PRESSURE

Centre of pressure is general below centroid since pressure increases with depth. Centre of pressure is determine by equating the moments of the result and distributed force about any arbitrary axis. The application point of the total pressure (hydrostatic force) on the surface.

h= Ia sin O + h

hA

Where;

h=vertical distance from the liquid surface to the centre of pressure

Ig=moment of inertial of the plane surface about its centroid G

PROCEDURE

Part A: vertical plane surface (O=0)

Counter balancing the water vessel.

  • The water vessel is set to angle of O = O by using the detent
  • The unit was counterbalance with using the rotating cylinders the stop pin is precisely in the true middle of the hole

Measurement

  • The rider then mounted and the level arm is set at L=22 mm. The level arm (the distance from the rider to the centre of rotating of the water vessel ) was recorded
  • The appended weight was placed and the value is recorded
  • Water is added into the water vessel until the unit is balanced (until the stop pin is at the centre of the hole). The level of water in the vessel was recorded
  • The water shape was determine for a different profile of pressure distribute (triangular profile or trapezoidal profile)
  • Steps 4 and 5 were repeated for at least 5 value of appended weight.

Part B: Inclined plane surface (O=O)

· Water rider and appended weight was removed during counter balancing process.

· Steps 1 and 2 of the counterbalancing process were repeated for an angle O=O.

· Steps 3 to 6 of part A are repeated for measurement.

Results from the Experiment

Part A α=0°

lever arm

Appended weight

Water Level

Center of Pressure

Hydrostatic Force

L(mm)

Fc(N)

S1(mm)

S2(mm)

S(mm)

ℓ(mm)

h*(mm)

F(N)

1

22

1

0

58

58

254.167

112.167

0.0866

2

22

2

0

82

82

176.042

58.042

0.25

3

22

3

0

104

104

165.432

69.432

0.399

4

22

4

0

124

124

161.261

85.261

0.5456

5

22

5

0

144

144

158.865

102.865

0.6924

Part B α=10°

lever arm

Appended weight

Water Level

Center of Pressure

Hydrostatic Force

L(mm)

Fc(N)

S1(mm)

S2(mm)

S (mm)

ℓ(mm)

h*(mm)

F(N)

1

22

1

2

60

58

243.688

101.024

0.09

2

22

2

2

86

84

173.609

58.01

0.2534

3

22

3

2

106

104

164.986

69.518

0.4

4

22

4

2

126

124

160.977

85.57

0.5467

5

22

5

2

146

144

158.661

103.289

0.6933

CALCULATION

Part A

A= 7.5x10 m

h= 0.028 m

F= ghA

= (1000)(9.81)(0.028)(7.5x10 )

= 2.0233 N

Ig= bd

12

= 75(100)

12

= 6.25x10 mm

h=Ig sin O + h

hA

= (6.25x10) sin O + 28

(28)(7500)

= 28 mm

= 200 – (55-0)

3cos 0

= 181.667 mm

Part B

Case: triangular profile of pressure distribution

F=L Fe

= 200 (1)

180.490

= 1.11 N

h= 2s

3

= 2 (65 – 10)

3

= 36.67

Case: trapezoidal profile of pressure distribute

h= ( - 200) cos + s

= (165.657 – 200 ) cos 20 + (1070- 10)

= 64.73 mm

= 150 + (100)

12(s - 50 )

cos X

= 150 + (100)

12 (s - 50 )

cos X

= 165.657 mm

DISCUSSION

  • From the experiment that were done;

ü The data is applied to the formula from theoretical value

ü From the data, it has a different between vertical plane surface and inclined plane surface.

CONCLUSION

  • Conclusion from the experiment we can get that the fluid static it is the study of fluids which are not in motion.
  • The are no relative motion between the fluid particles. The only stress will be normal stress which is equal to the pressure.

RECOMMENDATION

  • There was an error when doing this experiment. The error is;

ü When the reading is taken, parallax error has occurred at the left and right water level

ü When pouring the water it may be over the limit that we want.

  • So, we recommended that;

ü Change to the new equipment and apparatus.

ü Built the platform when taking the parallax error.