UNIVERSITI TEKNOLOGI MARAKAMPUS SAMARAHAN 2 FACULTY OF CIVIL ENGINEERING DIPLOMA IN CIVIL ENGINEERING (EC110) BASIC HYDRAULICS (ECW 321) EC1104D NAME METRIC NO. NUR AMYRA HIDAYAH BINTI AMIRUL ZUBIR BIN SHIBLI 20132233456 2013651848 MOHAMMAD SYAFIQ AKMAL BIN ABDULLAH 2013431936 NORHAZERAH BINTI YUSSOP NUR HAFIZAN BINTI ASMAIL AMIRUL SIRAJ MUNIR BIN JAMALARIFFIN @ ZAINAL LECTURER’S NAME: MDM MAUREEN NEGING DATE OF SUBMISSION: 00 JANUARY 2015 TITTLE Experiment on centre of pressure 2013251252 2013617608 2013624964 OBJECTIVE To determine the hydrostatic thrust acting on a plane surface immersed in water and the position of centre of pressure. THEORETICAL BACKGROUND The effect of hydrostatic pressure is major significant in many area of engineering, such as shipbuilding, the construction of dykes,weirs and locks, and in sanitary and building services engineering. With the Hydrostatic Pressure Apparatus the following key topic can be investigated in experiment: Pressure distribution in a liquid taking into account gravity “Lateral force” of the hydrostatic pressure Centre of pressure of the lateral force The hydrostatic pressure of liquid is the “gravitational pressure” intrinsic weight as the depth t increase, and is calculated from: Phyd = ρ.g.t (6.1) ρ−¿ Density of water g−¿ Acceleration due to gravity (g= 9.81 m/ s 2 t−¿ Distance from liquid surface ρhyd . It rises due to the To calculate force acting on masonry dam or ships’ hulls, for example, hydrostatic pressure, two step are required: Reduce the pressure load on an active surface down to a resultant force F p , which is applied at appoint of application of force, the “centre of pressure” , vertical to the active surface. Determine the position of this centre of pressure by determine a planar centre of force on the active surface. It is demonstrated how the centre of pressure can be determined. The resultant force Fp is then calculate. Determine the Centre of Pressure A linear pressure profile is acting on acting on the active surface shown, because The hydrostatic pressure rises proportional to the depth t. The resultant force Fp is therefore not applied at the centre of force C of active surface, but always slightly below it, at the so-called centre of pressure D. To determine the distance e of pressure from the planar centre of force, the following model demonstration is used: Imagine an area A in front of the active surface, formed by the height h and the pressure profile of the hydrostatic pressure p1− p2 . This area is in the form of a trapezium. The centre of pressure D lies on the extension of the planar centre of force of this area A. A can be broken down into partial areas A 1 and A 2 . The respective planar centres of force are identified by black dots. A balance of moment between the areas is then established around the point O1 in order to find common planar centre of force (dynamic effect in direction F p ): O h M (¿¿1) = 0:A. ( + e ¿ = 2 ∑¿ h 2h A1 . + A2 . 2 3 (6.2) Where A 1= p1. h (6.3) p2−¿ p .h 2 A2=¿ 1 (6.4) A1+ A2 A= (6.5) The result is: p2+ ¿ p p2−¿ p ¿ 1 h.¿ 6 1 1 e= (6.6) With the hydrostatic pressure h P2=ρgcosα .( y c + ) 2 (6.7) h P1=ρgcosα .( y c + ) 2 (6.8) The result is: e= 1 h2 . 12 y c (6.9) e is the distance of the centre of pressure from the force of the active surface which we are looking for. Determining the Resultant Force The hydrostatic pressure acting on the active surface can be represented as resultant force F p of which the line of application leads through the centre of pressure D. The size of this resultant force correspond to the hydrostatic pressure at the planar centre of force C of the active surface: pc =ρ . g . t (6.10) pc =¿ Hydrostatic pressure at the planar centre of force of the active Surface t c =¿ Vertical distance of the planar centre of force from the surface of the liquid In visual terms, the pressure at the planar centre of force corresponds to precisely the mean value between the highest and lowest pressure, because the linear pressure distribution. If the wall titled by an angle α: pc =ρ . g . cosα . y c The resultant force Fp (6.11) can now be calculated: F p =p c . Aactive (6.12) CENTRE OF PRESSURE WITH VERTICAL POSITIONING OF THE WATER VESSEL 3.1 Apparatus i. Hydraulic Bench ii. The Hydrostatic Pressure Apparatus iii. A set of weights 3.2 PROCEDURE 3.2.1 Counterbalancing the Water Vessel 1. The water vessel was set to an angle of x=0O using the detent. 2. The rider was mounted, the lever arm on the scale was set at any position. 3. The unit was counterbalanced with the rotation slider. The stop pin was ensure to be precisely in the middle of the hole. 3.2.2. Performing the Measurement 1. The water is top up until the unit was balanced. 2. The water level, s was read off and record it in the prepared worksheet. 3. The appended weights was increased in increments of 0.5-1N and repeat the measurement. 3.2.3 Evaluating the experiment Measured value: S-water level reading I-Lever arm of the force due to weight FG-Force due to weight of the appended weight 3.2.3 Determine the centre of pressure At a water level s, below the 100 mm mark, the height of the active surface changes with the water level.The height of the active surface is always 100 mm if the water level is above that mark. Meaning: s - Water level e - Distance of centre of pressure D from planar centre of force C of the active surface lD - Distance to centre of motion of the unit: For a water level s < 100 mm: (Pressure has a triangular profile) e= 1 . 6 s (6.13) 1 3 .S ID = 200mm - (6.14) For a water level s > 100 mm: (Pressure has a trapezoidal profile) e= 1 12 2 . (100 mm) s−50 mm (6.15) ID = 150mm + e 3.2.4 Determining the Resultant Force The resultant force corresponds to the hydrostatic pressure at the planar centre of force C of the active surface. Thus, the height of water level, s must again be differentiated: Meaning: Aac t - Superficial content of active surface b (width of liquid vessel) = 75mm Pc - Hydrostatic pressure at planar centre of force Fp- Resultant force for hydrostatic pressure on active surface For s < 100mm: (Triangular profile) s Pc = ρ. g. 2 and Aact = s. b (6.17) For s > 100mm: (Trapezoidal profile) Pc = ρ. g(s-50mm) and Aact=100mm. b (6.18) The resultant force is produced as FP = Pc . Aact 3.2.5 Balance of Moment (6.19) Calculated variables: FG - appended weight I - Lever arm of appended weight referred to centre of motion O. To check the theory, a balance of moments around the centre of motion O can be established and checked: ΣM (O) =0: FG⋅I= FP⋅ID CENTRE OF PRESSURE WITH WATER VESSEL TILTED 3.1 APPARATUS i. ii. iii. Hydraulic Bench The Hydrostatic Pressure Apparatus A set of weights 3.2 PROCEDURE 3.2.1 Counterbalancing the Water Vessel 1. an angle α and counterbalancing the unit with a rotating slider, the top pin must be precisely in the middle of the hole for this is set. 2. The characteristic values in the prepared worksheet of the lowest water s 1 and highest water level s1 of the active surface is recorded. 1.2.1 Performing the Measurement 3. Top up with water until the unit is balanced (stop pin at centre of hole) 4. The water level is read off and entered it in the prepared worksheet. 5. The appended weights is increased in increments 0.5N-1.0N and the measurement is repeated. 1.2.2 Evaluating the experiment The different between evaluation of the tilted vessel and that of the vertical vessel lied in the translation of the water levels onto the tilled active surface. A factor cos α must be taken into account here. 3.2.3 Determine the centre of procedure When the water vessel is at a tilt, too, a triangular pressure profile is produced when the water level is below s2 above that level a trapezoidal profile is produced. Measured values: S - Water level reading α - Tilt angle of vessel Meaning: SL- Water level at lowest point of vessel SH- Water level at active surface at rim e- Position of centre of surface h- Height of active surface ID- Distance between centres of pressure/centre of motion For a water level s < sh a triangular profile as follows applies: s−st h = cosα (6.21) 1 e = 6 h( 6.22) 1 Ip = 200mm - 3 h(6.23) For a water level s> sh a trapezoidal profile as follows applies: 100 mm ¿2 ¿ ¿ e= (6.24) 1 .¿ 12 Ip = 150mm + e (6.25) 3.2.4 Determining the Resultant Force Meaning: Aact – Superficial content of active surface b (width of liquid vessel) = 75mm Pc- Hydrostatic pressure at planar centre of force of active surface Forc s <sh with h from 2.3: Pc = ρg s−st and Aact= h .b 2 (6.26) For s < sh the trapezoidal profile as follow as applies: Pc = S−St−50 mm . cosα ) ρ. g .¿ (6.27) Aac t = 100mm .b (6.28) The result force is produced as: Fp = Pc Aact (6.29) 3.2.5 Balance of moments The result can be checked with the balance of moments, as described in section 1.5 DATA Worksheet for Centre of Pressure Angle, ( 0) = 00 Lowest water level, st (mmWC) = 0 mm Highest water level, sh (mmWC) = 90 mm Lever arm, ƪ (mm) Appended weight, FG (N) Water level reading, S (mm) Calculated lever arm, ƪD (mm) Resultant force, Fp (N) 150 0.832 24 192.00 0.212 150 0.834 40 186.67 0.589 150 0.842 54 182.00 1.072 150 0.847 64 178.67 1.507 150 0.852 74 175.33 2.015 150 0.856 82 172.67 2.474 150 0.861 90 170.00 2.980 Table 1 Worksheet for data acquisition Angle, ( 0) = 100 Lowest water level, st (mmWC) = 60 mm Highest water level, sh (mmWC) = 136mm Lever arm, ƪ (mm) Appended weight, FG (N) Water level reading, S (mm) Calculated lever arm, ƪD(mm) Resultant force, Fp (N) 260 0.832 60 180.37 1.26 260 0.834 76 174.95 2.05 260 0.842 88 170.89 2.76 260 0.847 101 166.49 3.66 260 0.852 112 162.77 4.52 260 0.856 124 158.70 5.57 260 0.861 136 154.64 6.71 Table 2 DISCUSSION 1. Centre of Pressure With Vertical Positioning of the Water Vessel Based on the Table 1, the angle and the lever arm is fixed, which is 0° and 150mm. The original weight of the hanging hook is 84.8 g. We used the formula, F = ma to convert the unit of gram (g) into Newton (N). After that, the original weight of the hanging hook is added with the appended weights in increments of 5-3 N. Since the water level reading of our group is less than 100mm, the formula used: To Determine the Center of Pressure is: For water level s < 100mm e = 1/6 .s ID = 200mm – 1/3 .s To Determine the Resultant Force is: Pc = ρ.g.s/2 and Aact = s.b The resultant force is produced as: Fp = Pc .Aact To check the theory, a balance of moments around the center of motion o can be established and checked: ∑ M (0) = 0: FG.I = FP.ID However, our answer for FG.I ≠ FP.ID 2. Centre of Pressure with Water Vessel Tilted Based on Table 2, the angle and the lever arm is fixed, which is 10° and 260mm. Since the water level reading of our group’s is less than 136mm,Sh the formula used is : To determine the center of Pressure For a water level s < sh s−s t h = cos α e= 1 6 h ID = 200mm - 1 3 h To determine the resultant force s−s t Pc = ρg and Aact = h. b 2 The result force is produced as: Fp = PcAact The calculation is shown in the worksheet for calculation Some precautions should be taken during the experiments: 1. Safety must be consider when set up all the apparatus like Hydraulic Bench, because it’s heavy. 2. Make sure the stop pin is at center of the hole when filling the water into the water vessel and be careful when pouring the water into it because it may be over the limit that we want. 3. In order to get the accurate reading, avoid any movement at the table when taking the reading of water level scale 4. Make sure eyes is perpendicular to the water level scale to get an exact reading and to avoid the parallax error. 5. Make sure the water vessel is clean enough and dry before doing the experiment. CONCLUSION Therefore, we can conclude that our experiment is fail because we cannot prove that ∑ M (0) = 0: FG.I = FP.ID.It is due to the bubble that is not perfectly centered, it was definitely at the center but its position was more northward. Therefore, some precautions should be taken by the group to make sure that the experiment is success. REFERENCE 1. Mdm. Maureen Neging, Laboratory Manual of Hydraulics and Water Quality, Faculty of Civil Engineering, Uitm Samarahan 2. 2. John.F.Douglas, Janusz M.Gasiorek, John A.Swaffield, and Lynne B.Jack , Fluid Mechanics, Pearson, 5th Edition, 2006. 3. Resultant force, Retrieved from www.physicalclassroom.com/class/vectors/Lesson -1/Resultant on 9th January 2015.