FLUID MECHANICS , EXPERIMENT # 3

To find the theoretical and actual center of pressure on partially submerged plane area / surface.

APPARATUS:

Hydrostatic pressure Apparatus, weights

PROCEDURE:

1. First level the hydrostatic pressure apparatus with the help of leveling screws looking to the spirit level.
2. Make the beam and quadrant balance by the adjusting the counter weight so that pointer concedes the balancing mark.
3. Add the weight in the weight pan, will un balance the beam.
4. Add water to the tank containing quadrant, till the beam again comes to the balance position.
5. Note the loads/ weights added, and level of water “r” which exerts hydrostatic pressure on the plane area.
6. Measure the width of the plane area “B”, and the moment arm of the weight pan about the pivot of the beam.
7. Repeat the experiment by adding more weights and following the same procedure.

Theoretical Center of Pressure = H_p= 2/3r + q       (q = 20- r)

 Actual Center of Pressure = (P x W)/F_r

P= Moment arm

F_r= resultant Force

F_r= ½ γ_w r² B

γ_w = 9.81 KN/m³

DIAGRAMMATICALLY REPRESENTATION   :

OBSERVATION & CALCULATION:

Distance From the pivot = P= 27.2 cm

Width of the plane area of quadrant = B= 7.5 cm

Unit weight of water = γ_w = 9.81KN/m³=62.4lb/ft³

g = 9.81 m/s² = 32.2 ft/s²

S.no	Mass (g)	Weight(N)	Water Level “r” (cm)	Distance of water level from the pivot “q”	Resultant Force (N) Fr= ½ γw r2 B	Actual Position Hp=(P x W)/Fr	Theoretical Position = Hp= 2/3r + q
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Posted in: Fluid Mechanics l