HYDRAUL-EXPT#4.docx

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HYDRAUL HYDRAULICS LABORATORY Experiment No.4 Determination of the Location of the Center of Pressure Pressure is defined as the force divided by the area of application and therefore the force that is impplied by the fluid is equal to the pressure multiplied by the area of the surface in contact with the fluid. The location of the equivalent force pressed by the fluid on a surface is called the center of pressure (c.p.). Consider a plane surface partially submerged in a liquid as shown below, F = PA; P = ω*( h/2 ); A = h*b Where: F = the equivalent force exerted by the liquid on the plane P = the pressure implied by the liquid equivalent to the unit weight of liquid multiplied by the depth of the center of gravity of the submerge plane from the liquid surface A = area of the submerged plane ω = unit weight or specific weight of the liquid EXPT.No.4 1

Transcript of HYDRAUL-EXPT#4.docx

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HYDRAULHYDRAULICS LABORATORY

Experiment No.4Determination of the Location of the Center of Pressure

Pressure is defined as the force divided by the area of application and therefore the force that is impplied by the fluid is equal to the pressure multiplied by the area of the surface in contact with the fluid. The location of the equivalent force pressed by the fluid on a surface is called the center of pressure (c.p.).

Consider a plane surface partially submerged in a liquid as shown below,

F = PA; P = ω*( h/2 ); A = h*b

Where:

F = the equivalent force exerted by the liquid on the planeP = the pressure implied by the liquid equivalent to the unit weight of liquid multiplied by the depth of the center of gravity of the submerge plane from the liquid surface

A = area of the submerged plane ω = unit weight or specific weight of the liquid

The center of pressure Yp is equivalent to two-third of the total height h of the plane submerged in the liquid because the distribution of the pressure on the plane is triangular, where the pressure at the liquid surface is zero while the pressure at the bottom of the surface is maximum. This can be proven, nonetheless, by the general formula for the location of the center of pressure.

When the plane is fully submerged in a liquid, however, the determination of the center of pressure will be different because the pressure distribution along the plane is no

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longer triangular as in the case of the previous presentation. This configuration is presented as follows:

In this case: F = PA; P = ω* Yc = ω* ( y – d/2 ); A = b* d; Yp = Yc + e

Where:

Yc = the location of the center of gravity of the submerged plane with reference to the liquid surfacee = eccentricity of the location of the center of pressure from the center of gravity = Ig/ ( Yc*A):

Where: Ig = bd3/12 for a rectangleIg = bh3/36 for a triangleIg = πr4/4 for a circle

For an inclined plane, Yp, Yc and e are measured on the inclined distance along the plane. In other words, Yc in this particular case is not equal to y-d/2 because h is measured vertically and the pressure P is computed similarly as when the plane is horizontal.

I. Experiment Objectives

The objective of this experiment is to determine the depth of the center of pressure in a rectangular plane area when:- Partially immersed in water- Fully immersed in water

II. Materials and Apparatuses Required

The apparatuses necessary to perform this experiment are the following:

- Center of Pressure Apparatus

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- Set of Weights- Metric Scale

III. Procedures

METHOD 1: Partial Immersion

1. Locate the torroid on the two dowel pins and fasten to the balance arm by the central screw.

2. Measure the dimensions a, b, d and the distance L from the knife-edge axis to the balance pan axis.

3. Position the Perspex rank on work surface and locate the balance arm on the knife-edge. Level the tank by placing it to a level table.

4. Adjust the counter balance screw until the balance arm is horizontal.5. Place a weight on the balance pan. Fill the tank with water until the balance arm

is horizontal again. Measure the water level. (h)6. Prove that Yp is equal to h/2 + e by taking moment at O where R passes through

point O and therefore neglected. Likewise, Yp is also equivalent to 2h/3

ΣMo = 0; W*L = F*YpYp = (W*L)/F from apparatus

Yp = (h/2) + e from theory

METHOD 2: Full Immersion

1. Increase the weight on the balance pan and add water to level the balance arm again. Make sure that the plane under consideration in method #1 is fully submerged.

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2. Measure the new water level, h.3. Compute the actual Yp by taking moment at O and compare it with the value of

Yp in theory.

ΣMo = 0; W*L = F*YpYp = (W*L)/F from apparatus

Yp = (h - d/2) + e from theory

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IV. Laboratory Data

Sample Liquid: WATER

UnitMETHOD

Partial Immersion Full Immersion

Length of L m

Length of a m

Length of b m

Length of d m

Height of water, h m

Weight, W N

Force, F N

Yp, Actual m

Yp, Theoretical m

% ERROR %

% ERROR = ( Yp theoretical – Yp actual/ Yp Theoretical ) * 100%

V. Observations

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________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

VI. Conclusions

________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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VII. Review Questions

I. Encircle the Letter of the best answer.

1. The center of pressure is the distance from the center of gravity of the submerged plane to the _____________.

a.) Water surfaceb.) Point of referencec.) Torroidd.) Point of moment.

2. The moment of the inertia Ig for a circle is given by the formula __________.

a.) Ig = π4 / 64b.) Ig = π4 / 32c.) Ig = π4 / 16d.) Ig = πD4 / 64

3. The moment of inertia Ig for the rectangle is given by the formula ________.

a.) Ig = bd³ / 36b.) Ig = bd³ / 12c.) Ig = bd³ / 24d.) Ig = bd³ / 18

4. The moment of inertia Ig for the triangle is given by the formula _________.

a.) Ig = bh³ / 36b.) Ig = bh³ / 12c.) Ig = bh³ / 24d.) Ig = bh³ / 18

5. What do you call the distance between the center of the gravity and the center of pressure?

a.) Force Gapb.) Eccentricityc.) Difference in pressured.) Pressure head

6. When computing for the total force in a submerged plane, in the formula of pressure P = γh, the variable h is always measured from the water surface to _____________.

a.) the center of pressureb.) the center of gravity of the submerged plane

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c.) the top of the submerged planed.) the bottom of the submerged plane

7. If a plane is partially submerged in a fluid, which of the following is not correct?

a.) the eccentricity e is equal to h/6b.) the location of the center of pressure Yp is equal to 2h/3c.) the simplified formula for pressure is P = ∞h/2d.) Center of pressure Yp is equal to h/2

8. If the rectangular plane 10 m in length and 6 m wide is submerged in water with its length vertical and its bottom is 15 m from the water surface, what is the total hydrostatic pressure applied to the plane?

a.) 147.105 kN/m²b.) 58.582 kN/m²c.) 98.07 kN/m²d.) None of the above

9. What is the equivalent hydrostatic force implied by water to the plane in question number 8?

a.) 5884.2 kNb.) 8826.3 kNc.) 3530.52 kNd.) None of the above

10. What is the eccentricity of the hydrostatic force from the center of gravity of the plane in question number 8?

a.) 0.8333 mb.) 0.3000 mc.) 0.2778 md.) None of the above

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