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Fluid Exp. 6 Hydrostatic Pressure

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DEPARTMENT OF CIVIL ENGINEERINGCOLLEGE OF ENGINEERING

CEWB 121MECHANICS OF FLUIDS LABORATORY

EXPERIMENT 6: HB 012 HYDROSTATIC PRESSURE

Section : 02Group Member: Muhammad Taufik bin Mohd Said (CE093969) Muhamad Nur Adli Bin Zakaria (CE093697) Muhamad Faiz Bin Hamim (CE093831)Lecturer : Puan Hidayah Binti Basri

Date of Laboratory Session: 17/07/2014Date of Report Submission: 24/07/2014

Table of ContentList of the Report Headings1.0 Preamble21.1 Objective21.2 General Description of the Experiment and Apparatus21.3 Theory21.4 Theory Calculation32.0 Problem Statement43.0 Procedure44.0 Result and Analysis54.1 Observation54.2 Data64.3 Discussion95.0 Conclusion106.0 Appendix117.0 References13

EXPERIMENT NO 6 : HB 012 HYDROSTATIC PRESSURE1.0 PREAMBLE1.1 Introduction :Hydrostatic pressure on the circular side of the quadrant exerts no turning moment on the fulcrum. The same is hydrostatic pressure on the radial side of the quadrant. The only pressure exerting turning moment on the fulcrum is that pressure acting on the 100mm X 75mm surface which is maintained at vertical.1.2 Apparatus :This equipment is used to determine the center of pressure of a plane surface submerged in water and compare the theoretical and experimental results.

1.3 General Description of the experiment :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 = 280mm2. 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 = 0.200m1.4 Theory :Hydrostatic pressure of a liquid is proportional to its depthPgh................................ (1)WhenP = 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 = g (h2 h1)Since h = h2 h1Thus P2 P1 = gh orP2 = P1 + ghIf (1) is at the surface of the liquid; P1 is at the atmospheric pressureThus P2 = Patm + gh ----------------------- (2)Patm = 0 gauge pressureThus P2 = gh

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

F = gA ------------------------ (3)F = Hydrostatic force Newtonh = Depth of the centroid mA = Submerged surface area m2P and g are the same as in (1)

Pressure on small area dA at a depth of h. P = gh Force on area dA = dF = PdA = gh dA But h = X sin Thus dF = gX sin dA --------------------- (4)Integrating (4) F = = g sin = = XdA

However, X = Distace from center 0 to centroid (CG)Thus F = g sin A X ---------------------- (5) h = X sin Therefore F = g h ADetermination of Center of Pressure, CP Theoretical Method.dM = X d F = gh X dA But h = X sinThus dM = g sin X2 dAOr M = = g X2 sin dA = g sin2 dA ------------------------ (6)

Since2 dA = I0 = 2nd moment of area about 0 = LB3Thus M = g sin I0 ------------------------- (7) = Moment of hydrostatic force about 0 = F XP

Thus M = F XP = g I0 sinOr XP = -------------------------- (8)From Parallel Axis Theorem I0 = ICG + A X2Thus XP = Or XP = X + -------------------------- (9)

XP = Distance from 0 to center of pressure (Cp) mX = Distance from 0 to centroid of surface A mICG = Second moment of area A about the centroid m4A = Submerged surface m2

Determination of Center of Pressure, Experimental MethodFor 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.

When there is no water in the tank, W is the counter weight to the quadrant, the beam and weight anger. When there is water in the tank. 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)

1.5 Theory Calculation :When water level is above the quadrant lower scale.Centroid = Center of 75 mm X 100 mm quadrant surface. From equation (3),

F = g h A = Density of water = 1000 kg/m3g = acceleration due to gravity = 9.81 m/sec2h = Depth of centoid in mA = Submerged surface area = (75)(100)(10-6) m2

Thus F = ( 9.81 ) (h1 + 50) (10-3 m) (7500) (10-6 m2) =9.81 (h1 + 50) (75) (10-4) ; Newton (From = sin= ; ( = 90)Due to area of this experiment is rectangular ;(i) Horizontal width = 75 mm and (ii) Height = 100 mm

Thus ICG = mm4 = m

From (11), experimental XP = Y - R1 + h1 = - R1 + h1 = - 100 + h1 = - 100 + h1 XP = - 100 + h1

M and h1 are from the experiment = 1000 L = 280 mm A = (75) (100) mm2 = (75) (100) (10-6) m2

2.0 Problem Statement :Water in the tank puts a hydrostatic force on the block which causes a clockwise moment produce by the weight placed at the balance team. This experiment required the determination pressure of a plane surface submerged in water and compare the theoretical and experimental results.

3.0 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.

4.0 Result :Table 4.1No.m(kg)h1(mm) = (m)

h2(m)A()(mm)(mm)

10.6100150.06.25 ()2007500147156

20.576126.06.25 ()1767500122133

30.451101.06.25 ()151750096.2109

40.32676.06.25 ()126750070.787.0

50.21060.06.25 ()110750032.273.9

60.16.956.96.25 ()101750028.771.5

) (9.81

(h1 + 50) =

Since, thus

4.2 Percentage error :

1. 2. 3. 4. 20.4%5. 56.4%6. 60%Average percentage error = 27.09 %

5.0 Discussion :The comparison had made between the XP , distance from 0 to centre of pressure (Cp) experimental and theoretical values. The result show that values does not agree well with the second theory values. As the load decreasing, the and are decreasing respectively. Overall, all theoretical measurements were consistently higher than the experimental values. This may due to our first error, human error as in reading, the measurements was not accurate at correct time or apparatus error such as pivot arm or counter weight being not completely accurate when displaying equilibrium. Percentage error are between 5.77 % to 60% and average percentage error is 27.09% . This can be considered high and back to the theory we are figured out the factors.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.Another contributing factor to the actual depth being higher than the theoretical is that water was constantly leaking around the seal of the door. This velocity of the water created a lower pressure around the seal, increasing the closing moment on the door, again requiring a greater depth of water to open the door.

6.0 ConclusionAs a conclusion, the theoretical model showed the relationship between the mass of the weight and the depth of the water. This is a logical conclusion , since the water depth increases, the volume of water (a cubic value) is increasing, and this volume which is applying the pressure over the area, and hence affected ,the distance between 0 to centre of pressure (Cp). . Meanwhile, the average percentage error is 27.09% and the factors of errors have been discussed in our discussion.

7.0 ReferencesOpen Ended Lab Manual for Mechanics of Fluids Laboratory.Department of Civil Engineering, Universiti Tenaga Nasional

R. L. Daugherty and J. B. Franzini,Fluid Mechanics, 6th ed. (New York: McGraw-Hill, 1965).

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

J.F. Claydon. 2010. Centre Of Pressure [online] (updated 6 May 2010); Available at http://www.jfccivilengineer.com/centreofpressure.htm

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