ERT 205Fluid Mechanics Engineering

Experiment 1:Analysis of Hydrostatic Pressure on Submerged Surfaces

Name:Ang Kien Hau

Members of Group A2:Huzaifah binti Ab RahmanIntan Liana binti Nordin

Ker Lee Shiang

Instructors’ Name:En. Razif Omar

Cik Khalilah Mohamad Yusuff

Date of Experiment:27th July 2010

Date of Submission:3rd August 2010

Table of Contents

1 Objectives 1

2 Introduction 1

3 Theory 2

4 Equipments and Descriptions 9

5 Procedures 10

6 Results and Calculations 11

7 Discussion 13

8 Conclusion 13

9 Appendix 14

10 References 15 ii

Objectives

The main objective of this experiment is to determine the centre of pressure on both submerged and partially submerged plane surface. Besides that, it also aims to compare the centre of pressure between experimental and theoretical.

Introduction

The Hydrostatic Pressure (Model: FM 35) apparatus has been designed to introduce students to the concept of centre of pressure of an object immersed in fluid. It can be used to measure the static thrust exerted by a fluid on a submerged surface, either fully or partially, and at the same time allowing the comparison between the magnitude and direction of the force with theory. The apparatus consists of a specially constructed quadrant mounted on a balance arm. It pivots on knife edges, which also correspond to the centre of the arc of quadrant. This means that only the hydrostatic force acting on the rectangular end face will provide a moment about the knife edges (SOLTEQ, n.d.).

The force exerted by the hydraulic thrust is measured by direct weighing. With no water in the tank, and no weights on the scale, the arm is horizontal. As weights are added one by one to the scales, water can be added to the tank so that the hydrostatic force balances the weights and bring the arm back to horizontal. Figure 1 is a sketch of the Hydrostatic Pressure (Model: FM 35).

Figure 1: Hydrostatic Pressure (Model: FM 35).

The design of many engineering systems such as water dams and liquid storage tanks requires the determination of the forces acting on the surfaces using fluid statics. The complete description of the resultant hydrostatic force acting on a submerged surface requires the determination of the magnitude, the direction, and the line of action

of the force (Cengel & Cimbala, 2006).

Theory

Figure 2According to Chean and Gramoll (n.d.), consider an inclined plane submerged in a

static fluid as shown in the Figure 2. The resultant force FR is acting perpendicular to the plane since no shear force is present when the fluid is at rest. F R has a line of action that passes through the point (xcp, ycp), which is called the center of pressure.

Now take a small differential element dA at a depth of h. The differential force dF acting on dA is given by:

dF = ρgh dAThe magnitude of the resultant force can be obtained by integrating the differential force over the whole area:

FR = ∫A ρgh dA= ∫A ρgy sin θ dA= ρg sin θ ∫A y dA

The integral ∫A y dA represents the first moment of the area about the x axis, which is equal to:

∫A y dA = ycAwhere yc is the y coordinate of the centroid of the plane surface.

From trigonometry,

hc = yc sin θwhere hc is the vertical distance from the fluid surface to the centroid of the plane surface, then the resultant force is simplified to:

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FR = ρghcA

The center of pressure, xcp and ycp can be obtained by summing moments about the y and x axis, respectively. First, by equating the sum of moments of all pressure forces about the x axis to the moment of the resultant force:

∫ y dF = FRycp

∫A ρgy2 sin θ dA = (ρgyc sin θ dA)ycp

ycp = ∫A y2 dAycA

where ∫A y2 dA is the second moment of the area or the area moment of inertia (I x) about the x axis. According to the parallel axis theorem, the moment of inertia can also be written as:

Ix = Ix′ + Ayc2

where Ix′ is the second moment of area with respect to the centroidal axis, which is parallel to the x axis. Hence, the center of pressure coordinate ycp is given by:

ycp = Ix′ + ycycA

Similarly, xcp can be obtained by equating the sum of moments of all pressure forces about the y axis to the moment of the resultant force.

∫ x dF = FRxcp

∫A ρgxy sin θ dA = (ρgyc sin θ dA)xcp

xcp = ∫A xy dAycA

where ∫A xy dA is the product of inertia (Ixy) of the area about the x and y axes. Once again according to the parallel axis theorem, it also can be written as:

Ixy = Ixy′ + Axcycwhere Ixy′ is the product of inertia of the area with respect to the centroidal axes. Hence, the coordinate xcp is given by:

Ixy′

2

Equation 1:

Equation 2:

xcp = + xcycA

From the formulations of xcp and ycp, it is noted that the center of pressure is always lower than the centroid of the plane area.

Figure 3 illustrate a hydrostatic pressure demonstration setup.

Figure 3: Hydrostatic Pressure Demonstration.

Referring to Figure 3,L = Distance between suspended mass and fulcrum/pivot point.D = Height of end surface.B = Width of end surface.H = Total depth of quadrant.C = Centroid of end surface.P = Center of pressure of plane surface.d = Depth of immersion in water.F = Hydrostatic force exerted on quadrant.

According to Claydon (2010), although the basic theory for the partially submerged and fully submerged plane is the same, it will be clearer to consider the 2 cases separately.1. Partially submerged vertical plane surface

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Equation 3:

Figure 4: Partially Immersed Vertical Plane Surface.h = Depth of centroid from water surface.h′ = Depth of center of pressure from water surface.h′′ = Distance between fulcrum/pivot point and the center of pressure.

Thrust on surface:For a partially submerged plane surface, from Equation 1,

F = ρghcAwhereA = B × dhc = d ÷ 2

Thus,

F = ρg Bd2

2

Moment of thrust about fulcrum/pivot point:

Moment, M = F × h′′whereh′′ = depth of line of action of thrust below fulcrum/pivot point.

At equilibrium condition, a balancing moment is produced by the weight (W) applied to the hanger at the end of the balance arm = W × L. For static equilibrium the two moments are equal. Hence,

F × h′′ = W × L= m × g × L

Thus,mgL

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Equation 4:

h′′ = F

= 2mgLρgBd2

h′′ = 2mLρBd2

The theoretical result for depth of center of pressure, P, below the free surface of fluid is:

h′ = Ix

AhwhereIx = 2nd moment of area of immersed section about an axis in the free water surface.

So, by using the parallel axis theorem,

Ix = Ic + Ah2

Thus

Ix = Bd3

+ Bd ( d )212 2

= Bd3

+ Bd3

12 4

= Bd3

+ 3Bd3

12 12

= 4Bd3

12

= Bd3

3

The depth of ‘P’ below the fulcrum/pivot point will be:

h′′ = h′ + H - dand by substitute Ix into h′ we will get:

h′′

= H - d3

as the theoretical result.In other words, the distance from the pivot to the center of pressure is the depth to the bottom of the vertical plane, minus one third the depth of submerged part of the vertical plane. So the center of pressure on a partially submerged plane will always be one third of d up from the base of the plane surface.

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Equation 5:

Equation 6:

2. Fully submerged vertical plane surface

Figure 5: Fully Immersed Vertical Plane Surface.h = Depth of centroid from water surface.h′ = Depth of center of pressure from water surface.h′′ = Distance between fulcrum/pivot point and the center of pressure.

Thrust on surface:For a fully submerged plane surface, from Equation 1,

F = ρghcAwhereA = B × Dhc = D ÷ 2

Thus,

F = ρgBD(d - D )2

Moment of thrust about fulcrum/pivot point:

Moment, M = F × h′′whereh′′ = depth of line of action of thrust below fulcrum/pivot point.

At equilibrium condition, a balancing moment is produced by the weight (W) applied to the hanger at the end of the balance arm = W × L. For static equilibrium the two moments are equal. Hence,

F × h′′ = W × L= m × g × L

But this time,

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Equation 7:

h′′ = mgLF

h′′ =mL

ρBD(d - D )2

The theoretical result for depth of center of pressure, P, below the free surface of fluid is:

h′ = Ix

AhwhereIx = 2nd moment of area of immersed section about an axis in the free water surface.

Thus, by using the parallel axis theorem,

Ix = Ic + Ah2

Thus

Ix = BD[ D2

+ (d - D )2]12 2

So, the depth of ‘P’ below the fulcrum/pivot point will be:

h′′ = h′ + H - dand by substitute Ix into h′ we will get:

h′′ =(d - D )2 + ( D2

)+ H - d2 12

(d - D )2as the theoretical result.

Equation 8:

Equation 9:

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Equipments and Descriptions

Figure 6: Assembly Diagram of Hydrostatic Pressure (Model: FM 35).

Quadranto Material: PVCo Total Depth of Quadrant, H: 200 mm

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o Height of Fulcrum above Quadrant: 100 mmo Height of End Surface, D: 100 mmo Width, B: 75 mmo Length of Balance: 300 mm

Reservoir Tanko Material: Clear Acrylico Volume: 5.5 L

Balance Armo Distance between Suspended Mass and Fulcrum, L: 290 mmo Set of Weight: 50 g/each

Overall Dimensiono Length: 340 mmo Width: 130 mmo Height: 300 mm

Procedures

1. The Hydrostatic Pressure Apparatus was placed on top of a Hydraulic Bench.

2. The apparatus was installed properly as in Figure 6.

3. The apparatus was been levelled using the adjustable feet facilitated by the spirit level attached.

4. The counter-weight was been adjusted to level the balance arm to horizontal position.

5. The drain valve was been made sure is closed and water was added slowly into the tank until the surface just touched the quadrant base, thus establishing a datum level.

6. The reading of datum level was recorded at the scale.

7. A 50 g weight was placed on the balance pan and water was added slowly into the tank until the balance arm is again horizontal.

8. The mass of weight, m and depth of immersion, d in the experimental data sheet were recorded. The new depth of immersion was been subtracted with the datum level before recorded.

9. Steps 7 to 8 were repeated with increasing number of weights until the tank is full of water.

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10. The water was been drained off after the experiment was completed.

11. Data was recorded in a table.

ResultsHeight of quadrant, D = 0.100 m Density, ρ = 1000 kg/m3

Width of quadrant, B = 0.075 m Gravity, g = 9.81 m/s2

Quadrant to pivot, H = 0.200 mDistance between suspended mass and fulcrum, L = 0.290 m

Mass, m (kg)

Depth of Immersion,

d (m)

Immersion (Full / Partial)

Thrust, F (N)

2nd Moment Experimental,

h′′ (m)

2nd Moment Theory, h′′

(m)

Error (%)

0.05 0.047 Partial 0.813 0.175 0.184 -5.040.10 0.067 Partial 1.651 0.172 0.178 -3.040.15 0.084 Partial 2.596 0.164 0.172 -4.420.20 0.092 Partial 3.114 0.183 0.169 7.910.25 0.111 Full 4.488 0.158 0.164 -3.170.30 0.124 Full 5.445 0.157 0.161 -2.790.35 0.137 Full 6.401 0.156 0.160 -2.520.40 0.150 Full 7.358 0.155 0.158 -2.320.45 0.163 Full 8.314 0.154 0.157 -2.16

Calculations

For partially immersed,Thrust,

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F = ρg Bd2

2

F = 1000 × 9.81 × 0.075 × 0.0472

2

= 9810 × 8.284 × 10-5

= 0.813 N2nd Moment Experimental,

h′′ = 2mLρBd2

h′′ = 2 × 0.05 × 0.291000 × 0.075 × 0.0472

= 0.0290.1657

= 0.175 m

2nd Moment Theory,

h′′

= H - d3

h′′ = 0.2 - 0.0473

= 0.2 - 0.0157 = 0.184 m

For fully immersed,Thrust,

F = ρgBD(d - D )2

F = 1000 × 9.81 × 0.075 × 0.1 × (0.111 - 0.1 )2

= 73.575 × (0.111 - 0.05)

Equation 4:

Equation 5:

Equation 6:

Equation 7:

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= 4.488 N2nd Moment Experimental,

h′′ =mL

ρBD(d - D )2

h′′ = 0.25 × 0.29

1000× 0.075 × 0.1 × (0.111 - 0.1 )2

= 0.07257.5 × 0.061

= 0.158 m2nd Moment Theory,

h′′ =(d - D )2 + ( D2

)+ H - d2 12

(d - D )2

h′′ = (0.111 – 0.1/2)2 + (0.12/12) + 0.2 - 0.1110.111 – 0.1/2

= 0.164 mDiscussion

From the results, all the theoretical and actual moments calculated are nearly similar, where all the errors are within 5%, except for one, which reach about 8%. Besides that, all the measurements show negative errors, meaning that the theoretical values are higher than the experimental one, except for the measurement at 0.20 kg. Thus, we can believe that that particular measurement was totally wrong and may ignore it. Overall, all theoretical measurements were consistently slightly higher than the actual moment. This may due to human error, as in reading the measurements accurately at the correct time, or apparatus error such as the vertical ruler being slightly off, or the pivot arm/counter weight being not completely accurate when displaying equilibrium.

In this experiment, only the forces on the plane surface were considered. However, the hydrostatic forces on the curved surface of the quarter-circle block do happen, but they do not affect the measurement. This is because no moment is created by forces acting on the curved surface of the quarter circle block. The line of action of the forces on the curved surface are perpendicular to the surface, and due to the circular shape, all lines of action that acted on the curved surface will pass through the center or so called

Equation 8:

Equation 9:

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the pivot. Thus, no moments are created and hence no effect on the results.

Buoyancy force is defined as the net pressure force acting on a submerged body, and thus in this experiment it should not being neglected in the analysis of the experimental data. Consider the buoyancy forces acting normal to the surface of the arc, then the buoyancy force do not appear because the normal forces on the curved surface do not contribute a moment about the pivot of the device. This result is due to the design of the apparatus. In other words, the circular arc shape was been chosen because it allows the measurement of hydrostatic pressure forces without accounting for the buoyancy effect.

Conclusion

The theoretical centre of pressure is located lower than the experimental one at the vertical plane surface. The centre of pressure is also getting higher at the vertical plane surface as the depth of immersion increasing. For partial immersion, the centre of pressure is definitely situated lower than the fully immersed one.

References

A. Nurulhadi, n.d. Hydrostatic force on plane surface. [Online]Available at: http://atapaje.blogspot.com/2008/02/hydrostatic-force-on-plane-surface.html[Accessed 31 July 2010].

B.R. Munson, D.F. Young, and T.H. Okiishi, 2002. Fundamentals of Fluid Mechanics. 4th Ed. New York: Wiley and Sons.

C.K. Chean & K. Gramoll, n.d. Fluids. [e-book] s.l.: eCourses.Available at: http://www.ecourses.ou.edu/cgi-bin/ebook.cgi?doc=&topic=fl&chap_sec=02.3&page=theory[Accessed 31 July 2010].

F. Stern, 2005. Fluid mechanics. [pdf]Available at: http://www.engineering.uiowa.edu/~fluids/lecture_notes/Chapter_3_Sec2/Chap_3_Sec2.pdf

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[Accessed 31 July 2010].

J.F. Claydon, 2010. Centre of pressure. [Online] (Updated 6 May 2010)Available at: http://www.jfccivilengineer.com/centre_of_pressure.htm[Accessed 31 July 2010].

L. Stonecypher, 2010. Hydrostatic force on a submerged surface. [Online]Available at:http://www.brighthub.com/engineering/mechanical/articles/68389.aspx[Accessed 31 July 2010].

SOLTEQ, n.d. Hydrostatic pressure: model FM 35. [pdf]Available at: http://www.solution.com.my/pdf/FM35%28A4%29.pdf[Accessed 31 July 2010].

Y.A. Cengel & J. M. Cimbala, 2006. Fluid mechanics: fundamental and applications. 1st Ed. Singapore: McGraw-Hill.

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