Practice report work center of pressure on a surface of plane.3éme year civil.modul hydraulic engineering

INTRODUCTION :
body out partially or wholly immersed in a fluid undergoes compressive forces equivalent to a single thrust force called. This thrust, vertical and directed upwardly, is equal in magnitude to the weight of the displaced fluid (body fluid which occupies the place)

1 - Purpose of work:

Determine the position of the center of pressure on a rectangular planar surface partially or completely immersed

2.Résume theory:

A pressing force on a flat surface with arbitrary orientation is equal to the product of the area of ​​the wall surface by the pressure experienced by the center of gravity and is directed along the normal to the inner bearing action P = pg hсg A ... (1) 
Or p = the pressure-force n, p = density of the liquid in kg / m³, g = acceleration due to gravity in m / s ²; hсg = the height of the center of gravity, 
M A = the surface area of the wall, in m². 

Called the center of pressure the crossing point of the pressure force of the liquid with the surface undergoing pressure center push on a flat surface with arbitrary orientation can determine the help of the following formula.:


hсp = hcg + Isin ² θ / Ahcg ........................................ (2)
Or hcp = height of center of pressure, in m Ig = the moment of inertia of the central area of ​​the surface of the wall with respect to the axis parallel to the horizontal edge of the water that passes through the center of gravity of the wall, in m ⁴, θ = the angle which the planar wall is inclined relative to horizontal 
It follows from equation (2) that the center of grows on a flat surface to arbitrary orientation is always below the center of gravity of a value: Isin ² θ / Ahcg 
In the case o = 90 ⁰ or the formula (2) becomes: hсp = Ig + hcg / Ahcg
3 - Principle of the experiment:

The apparatus of hydrostatic pressure ( fig 1 ) is the balance that balances the moments of the pressure forces acting on the liquid immersed framing , ugly weight placed on the weighing pan .
The dial of this device that is securely attached to the beam of the balance is the baque cart with rectangular section. Center of curvature of the framing coincides with the pivot 9 (Fig. 1 ) of the balance that is ie with the point 0 ( fig.2et3 ) . therefore the resultant of the pressure forces acting on the curved surfaces of the framing speak pass is the time point O of the forces relative to the point O is equal to zero . therefore one should take odds that the action of two forces acting on the beam of the balance :
1 ) the pressure force of the fluid acting on the end strength of the dial
2) the gravity of the weight on the weighing pan .
Figures 2et3 scheme shows the device calculates hydrostatic pressure for two following cases:
Fig2 : flat surface of the rectangular end face of the dial is partially immersed in the liquid that is to say the height of the dial ( d) is greater than the height of water ( y) > y
Fig3 : flat surface of the rectangular end face of the dial is completely immersed in the liquid (y ≥ d)
Calculated for the two cases the values ​​of the forces acting on the heights of the center of pressure . Let us write the equations of equilibrium of moments of forces acting from the point O.

Here C = center of gravity .
        C = center of pressure .
        Y = water depth
         H = height of center of gravity.
         H = height of center of pressure .
        P = pressure force on the planar surface of the end face of the dial.
a = 0.10m , b = 0.075m d = 0.10m .

Or: m / y ² = (bp (a + d)) / 2L - pb/6L - there ........................... (9) 

m / hcg = (pbd (a + d / 2)) / L + (pb.d ³ / (1/hcg)) / 6L ............................... (12)
4-experimental procedure:

1 . measure , and the dimensions L and b of the face of cadran6
2 . Place the bowl on the bench scourge 7 on the pilot bushing knife
3 . Hang the plate 3 at the end of the plague 7 .
4 . Connecting the pipe 14 to the drain valve 13 of the tank and to dive into the lever bench ; connect another pipe when arriving water and the opening arranged on the triangular top of the tank .
5 . add horizontality of the tank using the screws rigging adjustable feet and level bulle2 .
6 . Move the weight against the scourge 10jusqu'a 7soit horizontal .
7 . Close the drain valve 13and fill with water until the level is flush with the inner level 6 framing .
8 . 4 meter weights on the scale pan 3 . The first weight should be equal A10G , but the latter can be recommandés20 , 30,40,50,70,100,150,200,220 , and 230 ( for the partial immersion of the end of framing strength ) and 240,250,270,300,350,400,430,440 ( for total immersion ) .
9 . Add the little water until the scourge 7 is horizontal again, that one (we control the ugly mark on the level of the beam 5.
10 . Note the water level there on the scale of the dial 11 and the value of the weight on the weighing pan 3 .
11 . adjusting the specific water level can be performed by filling the vessel
A little too well by removing excess water gently with the help of the drain valve 13.
12 . Repeat ( operation for each increase of the weight 4 until the water level reaches the maximum value on the graduated scale 11 .
13 . Then remove weight4 one by one by selecting New water levels until all weights 4 are removed.

5-Results and calculations:

To d = 100 mm 
(Partial Immersion)
1.calculer the value of m / y ² in Tab

Filling the Tank		Emptying the tank		average
Weight in g (m)	Water height in mm (y)	Weight in g (m)	Water height in mm (y)	Y in mm	Y²	m/ Y²
00 10 20 30 50 70 100 150 200 220 230	0 20 28 34 44 52 64 81 94 99 100	00 10 20 30 50 70 100 150 200 220 230	0 20 29 35 44 52 64 81 94 99 100	0 20 28.5 34.5 44 52 64 81 94 99 100	0 400 812.25 1190,25 1936 2704 4096 6561 8836 9801 10000	0 0,0250 0.0246 0.0252 0.0258 0.0259 0.0244 0.0229 0.0226 0.0224 0.0230
The calculated Y and Y2: Y=(0+0)/2=0 Y=(20+20)/2=20 Y=(28+29)/2=28.5 Y=(34+35)/2=34.5 Y=(44+44)/2=44 Y=(52+52)/2=52 Y=(64+64)/2=64 Y=(81+81)/2=81 Y=(94+94)/2=94 Y=(99+99)/2=99 Y2=(20)2=400 Y2=(28.5)2=812.25 Y2=(34.5)2=11190,25 Y2=(44)2=1936 Y2=(52)2=2704 Y2=(64)2=4096 Y2=(81)2=6561 Y2=(94)2=8836 Y2=(99)2=9801 Y2=(100)2=10000 The calculated m / Y ²: m/ Y²=10/400=0,0250 m/ Y²=20/812.25=0.0246 m/ Y²=30/1190.25=0.0252 m/ Y²=50/1936=0.0258 m/ Y²=70/2704=0.0259 m/ Y²=100/4096=0.0244 m/ Y²=150/6561=0.0229 m/ Y²=200/8836=0.0226 m/ Y²=200/9801=0.0224 m/ Y²=230/10000=0.0230 2- courbe m/y² en fonction de y    3 - experimental comparison graph with the theoretical expression (9) on the graph The theoretical expression (9): = - y  Knowing the values​​: a = 0.10 m, b = 0.075 m, d = 0.10 m, L = 0.275 m = 1000kg /  So: =m / (y ²) y = 27.27 to 45.45 - Our experimental graph is increasing, while that of the theoretical expression is decreasing. 4 - explanation of the differences between measured and theoretical values We have a large difference between the measured and theoretical values:  For values ​​were measured in g / mm ²  For theoretical values ​​were in kg / mm ² Table 2 FILLING THE TANK CHANGING THE CUV AVERAGE M(g) Y (mm) M(g) Y (mm) M(g) Y (mm) hcg=y-d⁄2 m⁄h cg 1⁄hcg 230 240 250 270 300 350 400 430 440 450 100 104 106 110 118 130 143 149 152 160 230 240 250 270 300 350 400 430 440 450 100 104 106 110 118 130 143 149 152 160 230 240 250 270 300 350 400 430 440 450 100 104 106 110 118 130 143 149 152 160 50 54 56 60 68 80 93 99 102 110 4.60 4.44 4.46 4.50 4.41 4.37 4.30 4.34 4.31 4.09 0.0200 0.0185 0.0178 0.0167 0.0147 0,0125 0.0107 0.0101 0.0098 0.0091 The calculations of: hcg=y-d/2=100-100/2=50 hcg= y-d/2=104-100/2=54 hcg= y-d/2=106-100/2=56 hcg= y-d/2=110-100/2=60 hcg= y-d/2=118-100/2=68 hcg= y-d/2=130-100/2=80 hcg= y-d/2=143-100/2=93 hcg= y-d/2=149-100/2=99 hcg= y-d/2=152-100/2=102 hcg= y-d/2=160-100/2=110 the calculated m / hcg m/ hcg=230/50=4.60 m/ hcg=240/54=4.44 m/ hcg=250/56=4.46 m/ hcg=270/60=4.50 m/ hcg=300/68=4.41 m/ hcg=350/80=4.37 m/ hcg=400/93=4.30 m/ hcg=430/99=4.34 m/ hcg=440/102=4.31 m/ hcg=450/110=4.09 the calculated 1 / hcg: 1/ hcg=1/50=0.0200 1/ hcg=1/54=0.0185 1/ hcg=1/56=0.0178 1/ hcg=1/60=0.0167 1/ hcg=1/68=0.0147 1/ hcg=1/80=0.0125 1/ hcg=1/93=0.0107 1/ hcg=1/99=0.0101 1/ hcg=1/102=0.0098 1/ hcg=1/110=0.0091 The diagram m / hcg: - Experimental comparison graph of the theoretical expression (9) on the graph: The theoretical expression (12): m / hcg = +  Knowing the values​​: a = 0.10 m, b = 0.075 m, d = 0.10 m, L = 0.275 m = 1000kg /  So: m / hcg = (1000 * 0,075 * 0.10 * (0.10 +010 / 2)) / 0,275 + (1000 * 0,075 * (0.10) 2 / (1/hcg)) / 6 * 0,275 m/hcg= 4.1 + 0.45hcg - Our experimental graph decreases, while that of the theoretical expression is increasing. Conclusion Our TP is to determine the position of the center of pressure on a flat rectangular partially Immersed. This pressure we can the determined by filling the tank of the hydrostatic unit and by increasing the mass at As, contrary emptying the vessel and reduce the weight.