3.hydrostatic pressure

8/9/2019 3.Hydrostatic Pressure.

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Tafila Technical

University Faculty

of Engineering

Experiment #3

Hydrostatic Pressure.

Name:

Salam F.Albaradie.

Lecturer name:

Dr. kaled alrababa

&

Eng. Zaki alqasi.

Date of submission: 25/11/2014.


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Table of Contents

Object .... p. 1

Theory ....p p. 1-6

Procedure ....p. 7

Results ... ...pp. 7-8

Conclusion ....p. 9

References ....p. 9


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Object

1. To determine the hydrostatic thrust on a plane surface partly immersed in water.2. To determine the position of the line of action of the thrust.3. To compare the position determined by experiment with the theoretical position .4. To verify the formula for calculating hydrostatic thrust.

Theory

When the quadrant is immersed in water it is possible to analyze the forces acting on thesurfaces of the quadrant as follows:

The hydrostatic force at any point on the curved surface is normal to the surface and

therefore resolves through the pivot point because this is located at the origin of the radii.Hydrostatic forces on the upper and lower curved surfaces therefore have no net effect no torque to affect the equilibrium of the assembly because all of these forces passthrough the pivot.

The forces on the sides of the quadrant are horizontal and cancel out (equal andopposite).

Once the moment caused by the weight of the quadrant is nullified by adjusting thecounterweight so that balancing arm is perfectly collinear with the level indicator a

balancing weight of known mass is added and water meticulously added until the


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Balancing arm has once again returned to its previous precisely noted position aided byThe hydrostatic force of water creating a moment which counter acts the torque of theWeight on the balancing arm and thus holds the system in equilibrium

The magnitude of the hydrostatic force can be calculated from the value of the balance

Weight and the depth of the water as follows:

mgL = Fh

Where:

m = mass of weight in kilograms.

g = gravitational acceleration (9.81m/s 2).

L = distance between fulcrum and balancing weight in meters.

F = Hydrostatic Thrust in Newton.

h is the distance between the fulcrum and the centre of pressure in meters.

By calculating the Thrust and the centre of pressure on the end face of the quadrant wecan compare the experimental results with those obtained from theoretical idealcalculations

Partially submerged quadrant (fig. B)


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Where:


H = distance from fulcrum to bottom of quadrant in meters.

D = height of quadrant in meters.

d = depth of immersion in meters.

h = depth of the centre of pressure in meters.

h = distance of the line of trust below the fulcrum in meters.

The forces present are F (the hydrostatic thrust) and mg (the balancing weight)

Hydrostatic Force - Theoretical

Hydrostatic force can be defined as: F = gAh

Where:

F = The Hydrostatic force in Newton

= density of water in kg/m 3

A = Area defined as: B*d in m 2

h = depth of the centroid defined as: d/2, in metres

Thus: F = (1/2) (gBd 2) Newton (Formula 1)

Experimental position of the line of force action below thepivot (h)

The moment described previously in the section that causes the system to reachequilibrium again along the levelling indicator can be described as:

M = Fh

A weight is placed on the balancing arm, the moment created is proportional to the lengthof the arm defined as: L

The moment is thus defined as M = WL = mgL

The two forces have put the system back into its previous state of equilibrium, thereforethe moments are equal i.e.


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Fh = mgL

Thus: h= mgl/F

But: F = (1/2) (gBd 2)

Therefore, by substitution:

h = 2mL/ Bd 2 (Formula 2)

Where:

h = distance of the line of trust below the fulcrum in meters.

m = mass of weight in kilograms.


= density of water in kg/m 3.

B = width of the end face of the quadrant in meters

d = depth of immersion in meters.

Theoretical position of the line of force action below the pivot (h)

The theoretical formula for the depth of pressure (h) is:

h = I x/Ah

Where:

h = depth of the center of pressure in meters.

Ix = the area moment of inertia of the immersed plane relative to its centroid.

A = Area defined as: B*d in m 2.

h = depth of the centroid defined as: d/2, in meters.

Thus: Ah = Bd 2/2

By substitution:


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h = 2Ix/Bd 2

Utilizing the parallel axis theorem yields the following formula:

Ix = Ic + Ah

Where:

Ix = the area moment of inertia of the immersed plane relative to its centroid.

Ic = the moment of inertia of the object about an axis passing through its centre of massin this case: Bh

A = Area defined as: B*d in m 2.

h = depth of the centroid defined as: d/2, in metres.


Ix = Bh + Ah

But:

h = d/2

Therefore

h = d 2/4

h= d 3/12

Substitute values:

Ix = Bd 3/12 + Bd(d/2) 2

= Bd 3/12 + Bd 3/4

= 4Bd 3/12 = Bd 3/3

But:

h = 2Ix/Bd 2



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h = (2Bd 3/3) Bd 2

. = 2d/3

By examining fig. B we can derive from it the following:

h = H + h d

But: h = 2d/3


h = H + (2d/3 d)

= H + (2d/3 3d/3)

= H + (-d/3)

Therefore:

h = H d/3 (Formula 3)

Hydrostatic Force Actual

By utilizing Formula 3 we can postulate a new formula by substituting another variablefor h into the sum of moments equation i.e.

Fh = mgL

But: h = H d/3


F (H d/3) = mgL

Thus:

F = (mgL)/(H d/3) (Formula 4)


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Procedures

1. The instruments used was adjusted as the tank was filled with water till it touched the bottom surface of the surface and that point was considered to be the zero of the verniercaliper.

2. Masses were added to the balance pan in increments of about 50g, and the watersurface was raised in the tank until the balance arm is horizontal again.

3. The vernier reading was taken to measure the depth of immersion, which restores the balance arm to its balanced position.

4. A series of readings with increasing values of "m" are then taken.

ResultsTable (1) Result

# M(Kg)

H(m)

d(m)

X(m)

hEXP(m)

hTO (m)

hEXP(m)

hTO (m)

F(N)

1 0.1 0.148 0.052 0.026 0.049 0.0346 0.197 0.182 0.994

2 0.2 0.126 0.074 0.037 0.068 0.049 0.194 0.175 2.02

3 0.3 0.108 0.092 0.046 0.081 0.061 0.0189 0.169 3.1

4 0.4 0.091 0.109 0.059 0.089 0.073 0.180 0.164 4.3

5 0.5 0.072 0.128 0.078 0.098 0.096 0.170 0.168 5.7

6 0.6 0.052 0.148 0.098 0.111 0.106 0.163 0.158 7.22

7 0.7 0.037 0.163 0.113 0.129 0.120 0.165 0.157 8.32

8 0.8 0.2 0.18 0.13 0.144 0.136 0.164 0.156 9.57


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GRAPH

Graph 1. depth Vs. center of gravity

Graph 2. Depth Vs. center of pressure.

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

d ( m )

X (m)

0

0.02

0.04

0.06

0.08

0.1

0.12

0.14

0.16

0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2

d ( m )

h' (m)


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Conclusion

The obtained results showed large discrepancies between the theoretical and experimentalvalues of the center of pressure, where the experimental ones were larger than thetheoretical ones.

These discrepancies might be a result of errors occurred in the experimental proceduresor apparatus.

some of the possible errors that might have caused the large discrepancies:

1. Neglecting the weights of the balance and the pan.

2. Errors in determining the depth "h", either due to errors in taking the reading from thevernier or from parallax errors in determining the touching point between the watersurface and the pin of the measuring device.

References

1. Introduction to Fluid Mechanics, 3rd Edition William S. Janna (1993)2. Fluid mechanics: fundamentals and applications Yunus A. engel, John M. Cimbala. 1st ed.3. A Manual for the Mechanics of Fluids Laboratory William S. Janna (2008)

3.hydrostatic pressure

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