UNIVERSITI TENAGA NASIONAL

COLLEGE OF ENGINEERING

DEPARTMENT OF CIVIL ENGINEERING

CEWB121 MECHANICS OF FLUID LABORATORY

EXP. TITLE : HB 012 HYDROSTATIC PRESSURE

EXP. NO : 6

STUDENT NAME : NUR FAREHA BINTI ABDUL GHAFAR

STUDENT ID : CE096508

SECTION : 01

GROUP : 02

GROUP MEMBERS: 1. AZRUL AFFAN BIN MUHAMAD RASHIDI

CE096502

2. HARIGARAN A/L KANDASAMY

CE096504

3. MOHAMMAD OMAR HAMID WAGIEALLA

CE097089

INSTRUCTOR : PROF. IR. DR. MARLINDA BINTI ABDUL MALEK

Performed Date Due Date Submitted Date06 JULY 2015 13 JULY 2015 13 JULY 2015

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TABLE OF CONTENT

TITLE PAGE

Objective 3

Theory 3-5

Anticipated Result 6

Apparatus 7

Procedure 8

Data, Observations and Results 8 - 9

Discussions 10

Conclusions 10

Critique 10

References 10

Appendix 11

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OBJECTIVE

The objective of the experiment is to determine the center of pressure on both submerged and partially submerged plane surface. The purpose of this experiment also to to compare the center of pressure between experimental result with the theoretical values.

THEORY

We consider a submerged surface in a stationary fluid. When a fluid is stationary, it has only normal stress, which is called pressure, but it has no shear stress. Hence, any submerged surface in a stationary fluid would experience hydrostatic force. Another characteristic of stationary fluid is that its free surface is always perpendicular to the direction of gravitational acceleration. The fluid pressure acts normal to the surface of an object and is positive in the direction into the surface. Integration of the pressure over a submerged surface yields the total hydrostatic pressure force acting on that surface. Similarly the resultant moment about a suitable specified point can be obtained by integrating the moments from the pressure over the body surface. Through total moment of momentum balance, the rotating part of the equipment is balanced with the load W on the scale.

Hydrostatic pressure of a liquid is proportional to its depth

P gh................................ (1)

When

P = pressure Newton/

density Kg/

G = acceleration due to gravity 9.81m/

h = depth of liquid m

If P1 is pressure at depth h1 and P2 is pressure at depth h2P2 – P1 = ρg (h2 – h1)Since h = h2 – h1

Thus P2 – P1 = ρgh orP2 = P1 + ρgh

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If (1) is at the surface of the liquid; P1 is at the atmospheric pressureThus P2 = Patm + ρgh ----------------------- (2)Patm = 0 gauge pressureThus P2 = ρgh

Hydrostatic force on the submerged surface is equal to pressure at the centroid times area of the submerged surface

F = ρgA ------------------------ (3)F = Hydrostatic force Newton

H͞h = Depth of the centroid mA = Submerged surface area m2

P and g are the same as in (1)

Pressure on small area dA at a depth of h.

P = ρgh Force on area dA = dF = PdA = ρ gh dA

But h = X sin

Thus dF = ρ gX sin dA --------------------- (4)

Integrating (4)

F = = ρ g sin

= = HXdA

However, HX = Distace from center 0 to centroid (CG)

Thus F = ρ g sin A HX ---------------------- (5)

Hh = HX sin

Therefore F = ρ g Hh A

Determination of Center of Pressure, CP Theoretical Method.

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dM = X d F = ρ gh X dA

But h = X sin

Thus dM = ρ g sin X2 dA

Or M = = g X2 sin dA

= g sin 2 dA ------------------------ (6)

Since 2 dA = I0

= 2nd moment of area about 0

= LB3

Thus M = ρ g sin I0 ------------------------- (7)

= Moment of hydrostatic force about 0 = F XP

Thus M = F XP = ρ g I0 sin

Or XP = -------------------------- (8)

From Parallel Axis Theorem

I0 = ICG + A HX2

Thus XP =

Or XP = HX + -------------------------- (9)

XP = Distance from 0 to center of pressure (Cp) mH͞X = Distance from 0 to centroid of surface A m

ICG = Second moment of area A about the centroid m4

A = Submerged surface m2

Determination of Center of Pressure, Experimental Method

For HB 012 Hydrostatic Pressure, the submerged surface is always vertical or = 90.

This surface 75 mm wide and 100 mm high. The quadrant inner radius is 100 mm and outer radius is 200 mm . Fulcrum is at the center of the quadrant.

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When there is no water in the tank, W is the counter weight to the quadrant, the beam and weight anger. A weight m is required to balance the hydrostatic force. M is at a distance 280 mm from 0.

FY = mgL

Y = --------------------------- (10)

At the same time Y = XP + (R1 – h1)Thus XP = Y-R1 + h1 ---------------------------- (11)

ANTICIPATED RESULT

The expected result for Xp in this experiment are shown below:

By using the formula;

1. 600g +( (75)(100) 3 mm4 / (227mm x 7500 m2) ) = 230 12

2. 500g +( (75)(100) 3 mm4 / (204mm x 7500 m2) ) = 208 12

3. 400g +( (75)(100) 3 mm4 / (181mm x 7500 m2) ) = 185 12

4. 300g +( (75)(100) 3 mm4 / (158mm x 7500 m2) ) = 163 12

5. 200g +( (75)(100) 3 mm4 / (128mm x 7500 m2) ) = 134 12

6. 100g +( (75)(100) 3 mm4 / (106mm x 7500 m2) ) = 113 12

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APPARATUS

1. A quadrant with the following dimensions.

Inner radius = 100mmOuter radius = 200mmWidth = 7mmCenter of the quadrant is the same as fulcrum of the quadrant. Fulcrum to weight hanger distance = 280mm

2. W is a counter weight to the quadrant and weight hanger when there is no hydrostatic pressure. The position of W is adjustable horizontally.

3. A clear acrylic tank with fulcrum support for the quadrant. When the tank is filled with water, hydrostatic pressure will turn the quadrant counter clockwise which requires a balancing weight with the mass m.

Pre-determined Dimensions: Width of Quadrant B = 0.075m Height of Quadrant D = 0.100m Length of Balance L = 0.275m Quadrant to Pivot H = 0.200m

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PROCEDURE

1. Weight W is adjusted to balance the quadrant, beam and weight hanger when

there is no water in the tank.

2. The tank is filled so that the quadrant is nearly completely submerged. The beam

at the weight hanger end is now tilted upward.

3. Weight (w) of 600g is added until the beam is about to but not tilted downward.

4. The water is drained slowly from the tank and the valve is closed immediately

when the beam is horizontal. Then the water level and the weight m is recorded.

5. 100g is taken off, the beam will again tilted upward and step No. 4 is repeated.

6. Step No. 5 is repeated until all the weights are removed.

7. The % error is between Xpexp and Xptheo is got.

DATA, OBSERVATION AND RESULTS

No. m(g)

h1(mm)

= (mm)

h2(mm)

A

( ) (mm) (mm)

1 600 177 227 6.25 ( ) 277 7500 77.0 230

2 500 154 204 6.25 ( ) 254 7500 54.0 208

3 400 131 181 6.25 ( ) 231 7500 31.00 185

4 300 108 158 6.25 ( ) 208 7500 8.0 163

5 200 78 128 6.25 ( ) 178 7500 -22.0 134

6 100 56 106 6.25 ( ) 156 7500 -44.0 113

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Calculation for the result ;

) (9.81

So, force can be calculated using the formula:

(h1 + 50) =

1.

2.

3.

4. %

5. 116.41%

6. %

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Average percentage error = 101.54%

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DISCUSSION

Based on the results, we can see the comparison between the XP (distance from 0 to centre of pressure (Cp) ) of experimental and theoretical values. What can be observed is that the theoretical value is much higher than the experimental value. This may due to error that occurred during conducting the experiment. Human error might have occured when reading the height of water level. The percentage error are very high that is from 66 -138%. 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, all lines of action that acted on the curved surface will pass through the centre or so called the pivot. Thus, no moments are created and hence no effects 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. By considering the surface buoyancy forces acting normal to the surface, then the buoyancy force does 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

As a conclusion, the initial objective is met because we are able to conduct the experiment well. We are also able to calculate the force, ICG values and Xp values by using the provided formula in the lab manual. The result varies because there are some errors occurred during conducting the experiment as stated above. Other than that there is no problem during conducting the experiment.

CRITIQUE

We successfully conduct our experiment by referring the lab manual. The problem is just that the lab manual has some typos when explaining the formula derivation.

REFERENCES

1. Mechanics of Fluid Laboratory CEWB121, Lab Manual, Experiment 6:

HB012 Hydrostatic Pressure

2. Hydrostatic Pressure Experiment - https://www.youtube.com/watch?v=Gi4qBOjVAXk

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