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 terms, 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 (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
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 account 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
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)
I =150mm + eᴅ
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) (15) Aact = 100mm.b
The resultant force is produced as Fp = Pc.Aact
Balance Of Moment
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 I = 150mmᴅ
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) Lowest water level St (mmHOW)
Highest water level Sh (mmHOW)
0° 0 100
Lever arm I
(mm) weight FG (N)Appended reading s (mm)Water Level Calculated lever arm ID(mm) Resultant force Fp (N)
250 1 61 180 1.369
250 2 90 170 2.980
250 3 116 163 4.856
250 5 158 158 7.946
250 7.5 220 155 12.508
Angle α (º) Lowest water level St
(mmHOW) Highest water level Sh (mmHOW)
50 73 136 Lever arm I (mm) Appended weight FG (N) Water Level reading s (mm) Calculated lever arm ID(mm) Resultant force Fp (N) 250 1 66 203 0.026 250 2 129 171 1.792 250 3 153 161 3.521 250 5 199 156 6.906 250 7.5 255 154 11.026
Angle α (º) Lowest water level St
(mmHOW) Highest water level Sh (mmHOW)
90 200 200 Lever arm I (mm) Appended weight FG (N) Water Level reading s (mm) Calculated lever arm ID(mm) Resultant force Fp (N) 250 1 220 150 1.472 250 2 240 150 2.943 250 3 264 150 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 N50° 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) Aact = h.b = (10³x 9.81)x((0.066-0.073)/2) 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.
REFERENCES
Hydraulic 2 politechnic module
