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Objectives To determine the position of the center of pressure on

the rectangular face of the toroid. To compare the measured value with that predicted

from the theoretical analysis. Introduction

Hydrostatics is about the pressures exerted by a fluid at rest. Fluid static deals with problems associated with fluids at rest. Any fluid is meant, not just water. The fluid can be either gaseous or liquid. Fluid static used to determine the force acting on a body-of course the body may be submerged or floating- and for calculating forces developed by devices like car jacks. The design of many engineering systems such as water dams and liquid tanks requires the determination of the magnitude, the direction, and the line of action of the force.

Theoretical background

Forces on Submerged Surfaces in Static Fluids 1. Fluid pressure on a surface

Pressure is defined as force per unit area. If a pressure p acts on a small area then the force exerted on that area will be

Since the fluid is at rest the force will act at right-angles to the surface.

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General submerged plane

Consider the plane surface shown in the figure below. The total area is made up of many elemental areas. The force on each elemental area is always normal to the surface but, in general, each force is of different magnitude as the pressure usually varies.

We can find the total or resultant force, R, on the plane by summing up all of the forces on the small elements i.e.

This resultant force will act through the center of pressure, hence we can say

If the surface is a plane the force can be represented by one single resultant force, acting at right-angles to the plane through the center of pressure.

Curved submerged surface

If the surface is curved, each elemental force will be a different magnitude and in different direction but still normal to the surface of that element. The resultant force can be found by resolving all forces into orthogonal co-ordinate directions to obtain its magnitude and direction. This will always be less

than the sum of the individual forces, .

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2. Resultant Force and Center of Pressure on a submerged plane surface in a liquid.

This plane surface is totally submerged in a liquid of density and inclined at an angle of to the horizontal. Taking pressure as zero at the surface and measuring down from the surface, the pressure on an element , submerged a distance z, is given by

and therefore the force on the element is

The resultant force can be found by summing all of these forces i.e.

(assuming and g as constant).

The term is known as the 1st Moment of Area of the plane PQ about the free surface. It is equal to i.e.

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where A is the area of the plane and is the depth (distance from the free surface) to the centroid, G. This can also be written in terms of distance from point O ( as )

Thus:

The resultant force on a plane

This resultant force acts at right angles to the plane through the center of pressure, C, at a depth D. The moment of R about any point will be equal to the sum of the moments of the forces on all the elements of the plane about the same point. We use this to find the position of the center of pressure.

It is convenient to take moments about the point where a projection of the plane passes through the surface, point O in the figure.

We can calculate the force on each elemental area:

And the moment of this force is:

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are the same for each element, so the total moment is

We know the resultant force from above , which acts through the center of pressure at C, so

Equating gives,

Thus the position of the center of pressure along the plane measure from the point O is:

It look a rather difficult formula to calculate - particularly the summation term. Fortunately this term is known as the 2nd Moment of Area , , of the plane about the axis through O and it can be easily calculated for many common shapes. So, we know:

And as we have also seen that 1st Moment of area about a line through O, Thus the position of the center of pressure along the plane measure from the point O is:

and depth to the center of pressure is

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How do you calculate the 2nd moment of area?

To calculate the 2nd moment of area of a plane about an axis through O, we use the parallel axis theorem together with values of the 2nd moment of area about an axis though the centroid of the shape obtained from tables of geometric properties.

The parallel axis theorem can be written

where is the 2nd moment of area about an axis though the centroid G of the plane.

Using this we get the following expressions for the position of the center of pressure

(In the examination the parallel axis theorem and the will be given)

The second moment of area of some common shapes.

The table blow given some examples of the 2nd moment of area about a line through the centroid of some common shapes.

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Shape Area A

2nd moment of area, , about an axis through the centroid

Rectangle

Triangle

Circle

Semicircle

Lateral position of Center of Pressure

If the shape is symmetrical the center of pressure lies on the line of symmetry. But if it is not symmetrical its position must be found by taking moments about the line OG in the same way as we took moments along the line through O, i.e.

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but we have so

4. Resultant force on a submerged curved surface

As stated above, if the surface is curved the forces on each element of the surface will not be parallel and must be combined using some methods. It is most straightforward to calculate the horizontal and vertical components and combine these to obtain the resultant force and its direction. (This can also be done for all three dimensions, but here we will only look at one vertical plane).

In the diagram below the liquid is resting on top of a curved base.

The element of fluid ABC is equilibrium (as the fluid is at rest).

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Horizontal forces

Considering the horizontal forces, none can act on CB as there are no shear forces in a static fluid so the forces would act on the faces AC and AB as shown below.

We can see that the horizontal force on AC, , must equal and be in the opposite direction to the resultant force on the curved surface.

As AC is the projection of the curved surface AB onto a vertical plane, we can generalize this to say, The resultant horizontal force of a fluid above a curved surface is:

RH = Resultant force on the projection of the curved surface onto a vertical plane.

We know that the force on a vertical plane must act horizontally (as it acts normal to the plane) and that must act through the same point. So we can say

RH acts horizontally through the center of pressure of the projection of the curved surface onto an vertical plane. Thus we can use the pressure diagram method to calculate the position and magnitude of the resultant horizontal force on a two dimensional curved surface.

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Vertical forces

The diagram below shows the vertical forces which act on the element of fluid above the curved surface.

There are no shear force on the vertical edges, so the vertical component can only be due to the weight of the fluid. So we can say

The resultant vertical force of a fluid above a curved surface is:

RV = Weight of fluid directly above the curved surface.

and it will act vertically downward through the center of gravity of the mass of fluid.

Resultant force

The overall resultant force is found by combining the vertical and horizontal components

Resultant force

And acts through O at an angle of .

The angle the resultant force makes to the horizontal is

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The position of O is the point of integration of the horizontal line of action of and the vertical line of action of .

What are the forces if the fluid is below the curved surface? This situation may occur or a curved sluice gate for example. The figure below shows a situation where there is a curved surface which is experiencing fluid pressure from below.

The calculation of the forces acting from the fluid below is very similar to when the fluid is above.

Horizontal force

From the figure below we can see the only two horizontal forces on the area of fluid, which is in equilibrium, are the horizontal reaction force which is equal and in the opposite direction to the pressure force on the vertical plane A'B. The resultant horizontal force, RH acts as shown in the diagram. Thus we can say:

The resultant horizontal force of a fluid below a curved surface is:

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Vertical force

The vertical force are acting are as shown on the figure below. If the curved surface were removed and the area it were replaced by the fluid, the whole system would be in equilibrium. Thus the force required by the curved surface to maintain equilibrium is equal to that force which the fluid above the surface would exert - i.e. the weight of the fluid.

Thus we can say:

The resultant vertical force of a fluid below a curved surface is:

Rv =Weight of the imaginary volume of fluid vertically above the curved surface.

The resultant force and direction of application are calculated in the same way as for fluids above the surface:

Resultant force

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And acts through O at an angle of .

The angle the resultant force makes to the horizontal is

Since there can be no shear stress in a static fluid medium, the force on the plane is due to pressure only and must act normal to the surface. This pressure force is found to be F =ρ*g*A*yc

Where A : is the area of the surface (m2) ρ: is the density of the liquid (kg/m3) g : is the gravitational acceleration (m/s2) yc : is the coordinate of the centroid (m) the force (F) may be taken as acting at the center of the pressure (CP). Now to determine the position of the center of pressure we take moments about (O). after some arrangements we can prove that the position of the center of pressure is given by : ycp = yc + (Ixx,c/yc*A) where (Ixx,c ) is the second moment of area (also called the moment of inertia) about the axis parallel to the x-axis and passing through the centroid (C). Apparatus The apparatus is shown in the figure below. A toroid is mounted on a balance and pivoted about the center of curvature of the toroid. Thus only the vertical face along (GC) of the toroid has any moment about the balancing point. A rider weight balances the weight of the toroid in the dry, so that the moment of the hydrostatic force on (C) is

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measured by the weight at the pan. The toroid is immersed in tank containing water and the depth of immersion is measured by a hook gauge. Procedure

figure shows a schematic diagram of the toroid and the balancing weight.

Level the tank under the toroid and adjust the weight

(W) to level the balance arm. Carefully, admit water to the tank until it just touches the bottom of the quadrant. Take the vernier reading.

Raise the water level in increments of about (10 mm) and add weight to the balance pan to level the balance arm for each depth of water. Note the mass (M) and the depth of immersion (h) for each reading.

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Data collected a=0.1 m b= 0.075 m c= 0.3 m d= 0.1 m Table (1) partial immersion h(cm) M

(g) Immersed area (m2)

Calculated ycp (cm)

Measured ycp (cm)

M/h2

1.5 5 1.125*10-3 5.88 -0.72 22.22222 3 21 2.25*10-3 2 1.7 23.333333 4.5 45.5 3.375*10-3 3 2.5 22.469136 6 80 4.5*10-3 4 3.8 22.22222 7.5 120 5.625*10-3 5 4.6 21.333333 9 171 6.75*10-3 6 5.9 21.11111111 Table (2) total immersion h(cm) M(g) Calculated

y (cm) Measured y (cm)

[h-d/2]

10 205 6.7 6.4 0.05 11 206 7.4 4.6 0.06 12 244 8.2 5.6 0.07 13 246 9.04 4.6 0.08 14 272 9.93 5.1 0.09

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Results an discussions

Partial immersion (y < d). To compare the measure and the calculated depth of pressure;

Measured:

Where

2*2

*2

***

2/****

bycML

LyybgcMg

yybgFLFcMg

p

p

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Then

hdaLmeasuredycp )(

Calculated:

hcalculatedycp 32)(

calculation For y = 1.5cm, m = 5g

L = 2(5*10-3)(0.3)/(1000)(0.075)(0.015)2 = 0.1778m Ycp (measured) = 0.1778-0.1-0.1+0.015 = -0.00722m Ycp (calculated) = (0.015) 2/3 = 0.0588m Error =87.7%

For y = 3cm, m = 21g L = 0.1867m Ycp (measured) = 0.01667m Ycp (calculated) = 0.02m Error =16.7%

For y = 4.5cm, m = 45.5g L = 0.1798m Ycp (measured) = 0.0248m Ycp (calculated) = 0.03m Error =17.3%

For y = 6cm, m = 80 g L = 0.1778m Ycp (measured) = 0.03778m Ycp (calculated) = 0.04m Error =5.6%

For y = 7.5cm, m = 120g L = 0.17067m Ycp (measured) = 0.04567m Ycp (calculated) = 0.05m

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Error =8.7% For y = 9cm, m = 171g

L = 0.1689m Ycp (measured) = 0.0589m Ycp (calculated) = 0.06m Error = 1.83%

Derivation of the equation:

where then For h= 1.5cm, m = 5g

22 /375.24

2)(

6mkg

cdab

cbh

hM

(theoretical)

222 /222.22

)015.0(005.0 mkg

yM

(experimental)

form the graph of (M/h2 ) (experimentally) we get:

cby

cdab

yM

ydaybcM

ydayybgcMg

yybgFYAF

ydaFcMg

yydaFcW

M

p

cp

p

p

62)(

)31(

2**

)31(*

2***

2/****

)31(**

)32)((**

0

2

2

0

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the slope = -0. 2248 kg/m3 the intercept = 23.295 kg/m2

Total immersion (y > d) (a) To compare the measure and the calculated depth of pressure;

Measured:

Where

Then

hdaLmeasuredycp )( Calculated:

where then

2*2

*2

***

2/****

bhcML

LhhbgcMg

hhbgFLFcMg

p

p

22

12

2

12

2

3

dhdh

dy

dhy

bdI

yAyIy

cp

cp

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Calculation For h = 10cm, m = 205g

L=0.164m Ycp(measured) = 0.064m Ycp(calculated) = 0.0667m Error=4.05%

For h = 11cm, m = 206g L=0.1362m Ycp(measured) = 0.0462m Ycp(calculated) = 0.0739m Error =37.5%

For h = 12cm, m = 244g L=0.1356m Ycp(measured) = 0.0556m Ycp(calculated) = 0.0819m Error =32.1%

For h = 13cm, m = 246g L=0.166m Ycp(measured) = 0.0464m Ycp(calculated) = 0.0904m Error =48.7%

For h= 14 cm, m=272 g L =0.111 m

Ycp(measured) = 0.05102m Ycp(calculated) = 0.09926m Error = 48.6%

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Derivation of the equation;

cbdda

cbddhM

12))(2/1()2/(

3

yAyIy

AyFFLMgc

cp

)2/()2/(12

2

dhbddh

bd

addh

ddhgbd

addh

ddhgbd

2)2/(12)2/(

2612)2/(

2

2

adhddh

dhddhgbd )2/(

2)2/(

)2/(12)2/( 2

]2/)[2/(

12

2

dadhdgbd

cgbdad

cbddhM

adcbddh

cgbd

1222

22123

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from the graph of (M and(h-d/2)) (experimentally) we get: the slope = 1.74 kg/m

the intercept = 0.1128 kg

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Conclusions and discussion: The measured values of the position of center of

pressure are closed to those calculated from physical dimensions within a small relative error. This error could be caused because of personal errors in reading and approximations in calculations, and because of systematic error due to the apparatus accuracy and sensitivity.

For (M/y2) vs. (y) Theoretically: Slope = - -pb/6c = 41.67 kg/m3 Intercept = pb *(a+d)/2c = 25 kg/m2 Experimentally: Slope = -0. 2248 kg/m3 Intercept = 23. 295 kg/m2

For (M) vs. (y-d/2) Theoretically: Slope = (pbd/c) (a+0.5d) = 3.75 kg/m Intercept = pbd3 /12c = 0.0208 kg Experimentally: Slope= 1.74 kg/m Intercept =0.1128 kg

The pressure forces on the four faces are not zero and

they are not neglected. They were canceled each other where (for each two opposite faces) the forces act on

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them are equal in magnitude but opposite in direction. So they cancel each other, and this is not a mistake.

The buoyancy force passes through the point (0) about which the moment was taken, so the moment of the buoyancy force is equal to zero, then it will be ignored.

The center of pressure does not depend on the type of the used fluid, but on the dimensions of the toroid.

yAyIy

AyIyy cpcp

where I, y , A are not depending on the type of the fluid or on any properties of the fluid specially( ) so, the location of the center of pressure will not be changed. References Lab manual. Fundamentals of Thermal –Fluid Sciences/Yunus

A.Gengle/Robert H. Turner. Properties of water and steam in SI-units, 2nd

Revised and Update Printing, Springer 1979. W. Wanger, A. Kruse, properties of water and

steam, springer-verlag, Berlin 1998. http://www.cheresources.com. http://www.google.com. http://www.yahoo.com. http://www.msn.com. http://www.cussons.com. http://www.uwsp.edu. http://www.engr.bd.psu.edu.

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