Hidraulik-hydrostatic Pressure Apparatus 5f6zo

HYDROSTATIC PRESSURE APPARATUS

1) Water vessel 2) Detent 3) Slinder 4) Stop pin 5) Water level scale 6) Rider 7) Weights 8) Handles

Theory On The Centre Pressure The hydrostatic pressure of liquids is the “gravitational pressure” Phyd. It rises due to the intrinsic weight as the depth t increase, and is calculated from :

Phyd = p.g.t

p = Density of water g = Acceleration due to gravity (g = 9.81 m/s²) t = Distance from liquid surface

To calculated forces acting on masonry dams or ships, hull, for example, from the hydrostatic pressure, two steps are

 Reduce the pressure load on an active surfac down to a resultant force Fp, which is

applied at a point of application of force, the “centre of pressure”, ventrical to the active surface.  Determine the position of the centre of pressure by determining a planer centre of

force on the actice surface.

It is first demonstrated how the centre of pressure can be determined. The resultant force Fp is then calculated.

Determining the Centre of Pressure A linear pressure profile is 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 the active surface, but always slightly below it, at the so-called centre of pressure D! To determine the distance e of the centre of pressure from the plannar 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 planarcentre of force of this area A. A can be broken down into partial areas A1 and A2. The respective planar centres of force are identified by black dots.

A balance of moments between the areas is then established around the point O1 in arder to find the common planar centre of force (dynamic effect in direction Fp) :

/

e is the distance of the centre of pressure from the planner centre of 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 Fp, of which the line of application leads through the centre of pressure D. The size of this resultant force corresponds to the hydrostatic pressure at the planar centre of force C of the active surface.

Pc = p.g.tc

(10)

Pc = Hydrostatic pressure at the planar centre of force of the active surface. Tc = Vertical distance of the planar centre of force from the surface of the liquid.

In visual , the pressure at the planar centre of force corresponds to precisely the mean value between the highest and lowest pressure, because of the linear pressure distribution. If the wall is tilted by an angle

Pc = p.g.cosα.yc

(11)

The resultant force Fp can now be calculated : Fp = Pc.Aactive

Important! To calculated the resultant force the planar centre of force of the active surface is applied, but the line of application of the resultant force Fp runs through the centre of pressure (see section 3.1)

Mode Of Functioning Of The HM 150.05 Unit. The units water vessel is designed as a ring segment with constant cross-section. The force due to weight G of the water always produces the same moment of momentum referred to the centre of motion O as the resultant force Fp, of the active surface running through the centre of pressure D, Consequently, this apparatus can be used to determine the force of pressure Fp and the centre of pressure.

To illustrate the point, image the ring segment completely filled. The force due to weight Gg applied top the centre of volume of the water can be broken down into two component.

A radially applied component Gr running precisely through the centre of motion and A tangential component G1 with a lever arm r acting on the centre of motion O.

The radial component Gr exerts no momentum on centre O, because its lever arm is zero. Now regardless of the water level.

Fp.Yp = Gfr

(13)

That is to say the force due to weight G of the water volume always exerts the same moment of momentum as the force Fp at the centre of pressure D.

The derivation of (13) leads via determination of the centre of force of a ring segment and its volume.

Determine The Centre Of Pressure At the water level s, below the 100mm mark, the height of the active surface changes with the water level. If the water level is above that mark, the height of the active surface is always 100mm. Meaning : s = Water level e = distance of centre of pressure D from planar centre of force C of the active surface. Id = Distance to centre of motion of the unit : for a water level s < 100mm : (pressure has a triangle profile)

e = 1/6.s

(1)

ID = 200 mm – 1/3.s

(2)

for the water level s > 100 mm : (pressure has a trapezoidal profile)

e = 1.(100mm)² 12s – 50mm

ID = 10mm = e

(3)

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 : Aact = superficial content of active surface b = 75 mm – width of liquid vessel pc = Hidrostatic pressure at planar centre of force

Fp – resultant force for hydrostat. Pressure on active surface :

For s < 100mm : (triangular profile)

Pc = p.g.s and Aact = s.b

(5)

2 For s > 100mm : (trapezoidal profile)

Pc = p.g. (s/2-50mm) and Aact = 100mm.b

(6)

The resultant force is produced as Fp = pc.Aact

(7)

Balance Of Moment Calculated variables : FG = appended weight I = lever arm appended weight referred to centre of motion O checkh the theory, a balance of moment aroung the of motion O can be establish and checked £Mº = 0 ; FG.I = Fp. Id

PROSEDURE Centre Of Pressure With Water Vessel Tilted Performing the experiment

Set an angle α and counterbalance the water vessel as described under 4.1.1.1 Enter the characteristic values in the prepare worksheet : lowest water level St and highest water level Sh of the active surface perform the measurement as described under 4.1.1.2

Evaluating The Experiment

The different between evaluation of the tilted vessel and that of the vertical lies in the translation of the water levels onto the tilted active surface : A factor cos α must be taken into here.

Determining The Centre Of Pressure

When the water vessel is at a tilt too, a triangular pressure profile is produced when the water level is below Sh ; above that level a trapezoidal profile is produced. Measured values : s – water level reading α – tilt angle of vessel

Meaning : St – water level of lowest point of vessel Sh – water level of active surface at rim e – position h – height of active surface ID – distance between centre of pressure/centre of motion.

For a water level s < Sh a triangular profile as follows applies. h= S-St/ cos α e = 200mm-1/3.h For water level S>Sh a trapezoidal profile as follows applies E = 1/12. (100mm)²/(S-St cos α -50mm) Iᴅ =150mm + e

Determining The Resultant Force Meaning : A act – superficial content of active surface b – 75 mm – width of liquid vessel pc = hydrostatic pressure at palanar centre of force of active surface

For S < Sh with h from section 4.2.2.1 Pc = ρ.g.(S-St/2 and Aact = hb For S-Sh the trapezoidal profile as follow applies: Pc = p.g. (s-st-50mm.cosα) Aact = 100mm.b The resultant force is produced as Fp = Pc.Aact

Balance Of Moment The resultant can be checked with the balance of moments

(15)

Centre Of Pressure With 90º Positioning If The Water Vessel The angle α =90° represents a special case. The resultant pressure profile has form of atriangle, because the hydrostatic pressure is equal at every point on the active surface.

For this reason, the centre of pressure C lies precisely at the planar centre of force D of the active surface.

e =0 and has the lever arm Iᴅ = 150mm The resultant force is produce as Fp = ρ.g.(S-St).(100mm.b) The result can be checked with the balance of moments

PROPERTIES O F FLUIDS & HYDROSTATICS BENCH WORKSHEET FOR CENTRE OF PRESSURE

Angle α (º) 0° Lever arm I (mm)

Lowest water level St (mmHOW)

Appended weight FG (N)

0 Water Level reading s (mm)

Calculated lever arm ID(mm)

100 Resultant force Fp (N)

1 2 3 5 7.5

61 90 116 158 220

180 170 163 158 155

1.369 2.980 4.856 7.946 12.508

250 250 250 250 250 Angle α (º) 50 Lever arm I (mm) 250 250 250 250 250

Appended weight FG (N) 1 2 3 5 7.5

Angle α (º) 90 Lever arm I (mm) 250 250 250

Highest water level Sh (mmHOW)

Appended weight FG (N) 1 2 3

Lowest water level St (mmHOW) 73 Water Level reading s (mm) 66 129 153 199 255

Lowest water level St (mmHOW) 200 Water Level reading s (mm) 220 240 264

Highest water level Sh (mmHOW) Calculated lever arm ID(mm) 203 171 161 156 154

136 Resultant force Fp (N) 0.026 1.792 3.521 6.906 11.026

Highest water level Sh (mmHOW) Calculated lever arm ID(mm) 150 150 150

200 Resultant force Fp (N) 1.472 2.943 4.709

Calculation centre of pressure Example :

0˚ (S <100mm) I d = 0.2m- 1/3(S) = 0.2 – 1/3 ( 0.061 ) = 0.180m @ 180mm

(S >100mm) I d = 0.15m + e

where: e = (1/12) x (100mm)²/(S-50mm) e = (1/12) x (0.1m)²/ (0.116m-0.05m) e = 0.013m

I d = 0.15m + 0.013 =0.163m @ 163mm

Fp = Pc x Aact (S < 100mm) Pc = ρg x S/2

where: Aact = S .b = 0.061 x 0.075= 4.575x 10ˉ³m

= (10³ x 9.81) x (0.061/2) = 299.21 Fp = (299.91) x (4.575x10ˉ³m) = 1.369 N

Fp = Pc x Aact (S > 100mm) Pc = ρg (S – 50mm)

where: Aact = 100mm x b = 0.1m x 0.075m = 7.5 x 10ˉ³m

= (10³ x 9.81)x (0.116- 0.05m) = 647.46

Fp = (647.46) x (7.5 x 10ˉ³) = 4.856 N

50° Centre of pressure (S < Sh) Id = 200mm – 1/3(h)

where: h = (S –St)/(cos α)

= 0.2m – 1/3(-0.01)

= (0.066-0.073)/(cos 50°)

= 0.203m @ 203mm

= -0.01

(S > Sh) Id = 150mm + e

where: e = (1/12) x ((100mm)²/((S- St)/ (cos α )-50mm))

= 0.15m + 0.011

= (1/12) x ((0.1m)²/((0.153-0.073)/(cos 50°)-0.05))

= 0.161m @ 161mm

= 0.011

Force due plate (S < Sh) Pc = ρg ((S-St)/2) = (10³x 9.81)x((0.066-0.073)/2)

Aact = h.b h = (S-St)/(cos α) = (0.066-0.073)/(cos 50°)

= -34.335

= -0.01 h.b= -0.01x 0.075 = -7.5x10ˉ⁴

Fp = Pc x Aact = -34.335 x -7.5x 10ˉ⁴ = 0.026 N

(S> Sh) Pc = ρg (S-St- 50mm. cos α)

Aact = 100mm.b

= 10³x9.81(0.153m-0.073m-0.05m cos 50)

= 0.1m x 0.075m

= 469.51

= 7.5 x10ˉ³

Fp = Pc x Aact = 469.51 x7.5 x 10ˉ³ = 3.521 N

90˚ Centre of pressure F p= p.g (S-St) x (0.1mxb)

, b= 0.075 m

= 1000x9.81 (0.220m- 0.2m ) (0.1m x 0.075m ) = 1.472 N

DISCUSSION AND ANALYSIS

There are some problem that cause during the experiment is conduct:

I.

The differences between the true answer with our calculation (erratum) is huge.

II.

Erratum is caerratum use when up loading the load without addition.

III.

The accessories and the apparatus seem old, not up to date and not functioning very well.

IV.

Error due to human factor such as careless.

CONCLUSION AND SUGGESTION

The experiment is done and succeeds but not in perfect match with the real data due to erratum.

We can minimize the new one.

a. Change the apparatus with the new one. b. Be more careful during conduct the experiment to avoid careless.

REFERENCES

Hydraulic 2 politechnic module

Pn. Siti Balqis binti Abdul Kadir (Hydraulic lecturer).

APPARATUS