Hydrostatic Pressure 3f1p37
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1.0 TITLE HYDROSTATIC PRESSURE
2.0 OBJECTIVES 2.1
To determine the center of pressure on both submerged and partially submerged a plane surface.
2.2
To compare the center of pressure between experimental result with the theoretical values.
2.3
To determine experimentally the magnitude of the force of pressure hydrostatic force.
3.0 INTRODUCTION The Hydrostatic Pressure (Model: FM 35) apparatus has been designed to introduce students to the concept of centre of pressure of an object immersed in fluid. It can be used to measure the static thrust exerted by a fluid on a submerged surface, either fully or partially, and at the same time allowing the comparison between the magnitude and direction of the force with theory. The apparatus consists of a specially constructed quadrant mounted on a balance arm. It pivots on knife edges, which also correspond to the centre of the arc of quadrant. This means that only the hydrostatic force acting on the rectangular end face will provide a moment about the knife edges (SOLTEQ, n.d.). The force exerted by the hydraulic thrust is measured by direct weighing. With no water in the tank, and no weights on the scale, the arm is horizontal. As weights are added one by one to the scales, water can be added to the tank so that the hydrostatic force balances the weights and bring the arm back to horizontal. Figure 1 is a sketch of the Hydrostatic Pressure (Model: FM 35).
1
Figure 1: Hydrostatic Pressure (Model: FM 35). The design of many engineering systems such as water dams and liquid storage tanks requires the determination of the forces acting on the surfaces using fluid statics. The complete description of the resultant hydrostatic force acting on a submerged surface requires the determination of the magnitude, the direction, and the line of action of the force (Fluid Mechanics, Cengel & Cimbala, 2010). 3.1 THEORY
Figure 2
Inclined plane submerged in a static fluid as shown in the Figure 2. The resultant force FR is acting perpendicular to the plane since no shear force is present when the fluid is at rest. FR has a line of action that es through the point (x, y), which is called the center of pressure.
2
Now take a small differential element dA at a depth of h. The differential force dF acting on dA is given by: dF
=
ρgh dA
The magnitude of the resultant force can be obtained by integrating the differential force over the whole area: FR
=
∫A ρgh dA
=
∫A ρgy sin θ dA
=
ρg sin θ ∫A y dA
The integral ∫A y dA represents the first moment of the area about the x axis, which is equal to: ∫A y dA
=
ycA
where yc is the y coordinate of the centroid of the plane surface.
From trigonometry, hc
=
yc sin θ
where hc is the vertical distance from the fluid surface to the centroid of the plane surface, then the resultant force is simplified to: Fr = ρghcA
Equation 1 :
The center of pressure, x and y can be obtained by summing moments about the 2y and x axis, respectively. First, by equating the sum of moments of all pressure forces about the x axis to the moment of the resultant force: ∫ y dF
=
FRy
∫A ρgy2 sin θ dA
=
(ρgyc sin θ dA)y
y
=
∫A y2 dA ycA
3
where ∫A y2 dA is the second moment of the area or the area moment of inertia (Ix) about the x axis. According to the parallel axis theorem, the moment of inertia can also be written as: Ix
=
Ix′ + Ayc2
where Ix′ is the second moment of area with respect to the centroidal axis, which is parallel to the x axis. Hence, the center of pressure coordinate y is given by:
Equation 2:
y = Ix′
+ yc ycA
Similarly, x can be obtained by equating the sum of moments of all pressure forces about the y axis to the moment of the resultant force.
∫ x dF
=
FRx
∫A ρgxy sin θ dA
=
(ρgyc sin θ dA)x
x
=
∫A xy dA ycA
where ∫A xy dA is the product of inertia (Ixy) of the area about the x and y axes. Once again according to the parallel axis theorem, it also can be written as:
Ixy = Ixy′ + Axcyc where Ixy′ is the product of inertia of the area with respect to the centroidal axes. Hence, the coordinate x is given by:
Equation 3:
X = Ixy′ + xc ycA
From the formulations of x and y, it is noted that the center of pressure is always lower than the centroid of the plane area. Figure 3 illustrate a hydrostatic pressure demonstration setup. 4
Figure 3: Hydrostatic Pressure Demonstration.
Referring to Figure 3, L
= Distance between suspended mass and fulcrum/pivot point.
D
= Height of end surface.
B
= Width of end surface.
H
= Total depth of quadrant.
C
= Centroid of end surface.
P
= Center of pressure of plane surface.
d
= Depth of immersion in water.
F
= Hydrostatic force exerted on quadrant.
Although the basic theory for the partially submerged and fully submerged plane is the same, it will be clearer to consider the 2 cases separately.
5
1. Partially submerged vertical plane surface
Figure 4: Partially Immersed Vertical Plane Surface. h
= Depth of centroid from water surface.
h′
= Depth of center of pressure from water surface.
h′′
= Distance between fulcrum/pivot point and the center of pressure.
Thrust on surface: For a partially submerged plane surface, from Equation 1,
F
=
ρghcA
where A
=B×d
hc
=d÷2
Thus,
Equation 4:
ρg (Bd2/2)
F =
Moment of thrust about fulcrum/pivot point: Moment, M
=
F × h′′
where h′′
= depth of line of action of thrust below fulcrum/pivot point. 6
At equilibrium condition, a balancing moment is produced by the weight (W) applied to the hanger at the end of the balance arm = W × L. For static equilibrium the two moments are equal. Hence, F × h′′
=
W×L
=
m×g×L
=
mgL
Thus, h′′
F =
2mgL ρgBd2
Equation 5
h’’
=
2mL/ρBd2
mgL/F =
Theoretically, the center of pressure from the water surface, h’ is expressed in equation below Y =
-(ρg sin Өlxy)/(PcgA)
Gives, h’ = h – Y = h – [-(Ixx sin Ө)/(hcg A)] The orientation of the surface from the horizontal = 90⁰ gives 1 for sin Ө, then
h′ =h
.
Ixx Ah
where Ix
= 2nd moment of area of immersed section about an axis in the free water surface.
So, Ix
=
Ic + Ah2
7
Thus Ixx
=
Bd3 12
+ (rectangular)
Where Ixx is the moment of inertia of the submerged surface. The center of pressure bellow the fulcrum is finally determined by, h′′
h′ + H – d
=
and by substitute Ix into h′ we will get: Equation
h’’
=
H – d/3
As the theoretical result. In other words, the distance from the pivot to the center of pressure is the depth to the bottom of the vertical plane, minus one third the depth of submerged part of the vertical plane. So the center of pressure on a partially submerged plane will always be one third of d up from the base of the plane surface.
2. Fully submerged vertical plane surface
Figure 5: Fully Immersed Vertical Plane Surface. 8
h
= Depth of centroid from water surface.
h′
= Depth of center of pressure from water surface.
h′′
= Distance between fulcrum/pivot point and the center of pressure.
Thrust on surface: For a fully submerged plane surface, from Equation 1, F
=
ρghcA
where A
=B×D
hc
=D÷2
Thus, Equation 7: F
=
ρgAh =
ρgBD(d – (D/2))
Moment of thrust about fulcrum/pivot point: Moment, M
=
F × h′′
Where
h′′
= depth of line of action of thrust below fulcrum/pivot point.
At equilibrium condition, a balancing moment is produced by the weight (W) applied to the hanger at the end of the balance arm = W × L. For static equilibrium the two moments are equal. Hence, F × h′′
=
W×L
=
m×g×L
9
But this time, h′′
=
mgL F
h’’
Equation 8
=
mL/ρBD(d-(D/2))
mgL/F =
Theoretically, the center of pressure from the water surface, h’ is expressed in equation, h’ =
h – Y
h – [(-Ixx sin Ө)/ (hcg A)]
=
The orientation of the surface from the horizontal = 90⁰ gives 1 for sinӨ, then The theoreticasult for depth of center of pressure, P, below the free surface of fluid is:
h′ =h+
Ixx Ah
h’
=
(d – (D/2)) + [(BD3 /12) X (1/BD) X (1/(d-(D/2)))]
=
[(d-(D/2))2 + (D2/12)] / (d – (D/2))
where Ix
= 2nd moment of area of immersed section about an axis in the free water surface.
Thus
Ix = BD[
D2
+ (d -
12
D
)2]
2
So, the depth of ‘P’ below the fulcrum/pivot point will be: h′′
=
h′ + H - d
10
and by substitute Ix into h′ we will get: Equation 9:
h’’
=
{[(d – (D/2))2 + (D2/12)] / ( d – (D/2))} + H – d
as the theoretical result. Apparatus The hydrostatic pressure (model : FM35) apparatus consists of a specially constructed quadrant mounted on a balance arm which pinots on knife edges and it also correspond to the center of arc of the quadrant. This means that only the hydrostatic force acting on the rectangular end face will provide a moment about the knife edge. The balance arm also has an adjustable counterbalance, and a balance pan for the weight supplied. The quadrant assembly is mounted on top of a clear reservoir tank provided with leveling screws on its feet. A spirit level attached to the base of the tank assures accuracy of the alignment. The tank is filled with water from the top and can be drained through a valve at the side. A scale is provided on the side of the quadrant to indicate the water level in the tank The torque exerted by the hydraulic thrust is measured by direct weighing. Tests may be carried out over a range of water level
3.3 Description and assembly
Figure 6: Assembly Diagram of Hydrostatic Pressure (Model: FM 35). 11
Quadrant i. Material: PVC ii. Total Depth of Quadrant, H: 200 mm iii. Height of Fulcrum above Quadrant: 100 mm iv. Height of End Surface, D: 100 mm v. Width, B: 75 mm vi. Length of Balance: 300 mm
Reservoir Tank i. Material: Clear Acrylic ii. Volume: 5.5 L o Balance Arm i. Distance between Suspended Mass and Fulcrum, L: 290 mm ii. Set of Weight: 50 g/each
Overall Dimension i. Length: 340 mm ii. Width: 130 mm iii. Height: 300 mm
4.0 Method 1. The Hydrostatic Pressure Apparatus was placed on top of a Hydraulic Bench. 2. The apparatus was installed properly as in Figure 7.
Figure 7 : Hydrostatic Pressure equipment by SOLTEQ 12
3. The apparatus was been levelled using the adjustable feet facilitated by the spirit level attached. 4. The counter-weight was been adjusted to level the balance arm to horizontal position. 5. The drain valve was been made sure is closed and water was added slowly into the tank until the surface just touched the quadrant base, thus establishing a datum level. 6. The reading of datum level was recorded at the scale. 7. A 50 g weight was placed on the balance pan and water was added slowly into the tank until the balance arm is again horizontal.
Figure 8 : Weight part 8. The mass of weight, m and depth of immersion, d in the experimental data sheet were recorded. The new depth of immersion was been subtracted with the datum level before recorded. 9. Steps 6 to 7 were repeated with increasing number of weights until the tank is full of water. 10. The water was been drained off after the experiment was completed. 11. Data was recorded in a table.
13
5.0 Result and discussion 5.1 Result There are several important properties is need to record to find the Thrust Force and second moment for this experiment. The important data of this experiment was recorded and it is show in table 5.1. and the experimental data of this experiment is show in table 5.2 Height of Quadrant,D (m)
0.075
Width of Quadrant,B (m)
0.073
Length of balance,L (m)
0.27
Quadrant to pivot, H (m)
0.205
Density (kg/m3)
1000
Gravity (m/s2)
9.81 Table 5.1: Properties Experiment.
14
Depth of Immersion Mass, Immersion, (Full/ m (kg) d (m) Partial) 0.05 0.040 Partial
0.5729
2nd Moment Experimental, h" (m) 0.2311
2nd Moment Theory, h" (m) 0.1917
Thrust F (N)
Error % 20.61
0.10
0.059
Partial
1.2464
0.2125
0.1853
14.66
0.15
0.075
Full
2.0141
0.1973
0.1800
9.59
0.20
0.091
Full
2.8735
0.1844
0.1763
4.5919
0.25
0.105
Full
3.6254
0.1827
0.1744
4.70
0.30
0.121
Full
4.4848
0.1772
0.1731
2.35
0.35
0.136
Full
5.2904
0.1752
0.1723
1.73
Table 5.2: Experimental Data.
Calculations For partially immersed, Thrust,
F = ρg (Bd2/2) F = 1000 × 9.81 ×
0.073 × 0.0402 2
= 9810 × 5.84 × 10-5 = 0.5729N
15
2nd Moment Experimental,
= mgL/F = 2mL/ρBd2
h’’
h′′ = =
2 × 0.05 × 0.27 1000 × 0.073 × 0.0402 0.027 0.1168
= 0.2311 m 2nd Moment Theory,
h’’ = H – d/3 h′′ = 0.205-
0.040 3
= 0.2 - 0.0157 = 0.1917 m For fully immersed, Thrust,
F = ρgAh = ρgBD(d – (D/2)) 16
0.075
F = 1000 × 9.81 ×.073 ×.075 × (0.075-
)
2
= 53.7097× (0.075 - 0.0375) = 2.0141 N 2nd Moment Experimental,
h’’ = mgL/F = mL/ρBD(d-(D/2))
h′′
0.15 × 0.27
=
1000×.073 ×.075 × (0.075 -
.075
)
2
=
0.0405 5.475 × 0.0375
= 0.1973 m 2nd Moment Theory,
h’’
= h’ + H – d
17
= {[(d – (D/2))2 + (D2/12)] / ( d –
h’’
(D/2))} + H – d (.075 – .075/2)2 + (.0752/12)
h′′ = =
+ 0.205 - 0.075
.075 – 0.075/2
0.1800 m Error
Theoretical actual Theoretical
20.61%= [(0.1917-0.2312)/0.1917] X 100
Moment Theory & Experimental 0.25
0.2
0.15
2nd Moment Experimental, h" (m)
0.1
2nd Moment Theory, h" (m)
0.05
0 1
2
3
4
5
6
7
Figure 9: Graph for 2nd Moment Theory & Experimental
18
5.2 Discussions The comparison had made between the second moment experimental values and second moment experimental values expressed as a percentage. The result show second moment experimental value doesn’t agree well with the second theory values. Overall, all theoretical measurements were consistently slightly higher than the actual moment. This may due to human error, as in reading the measurements accurately at the correct time, or apparatus error such as the vertical ruler being slightly off, or the pivot arm or counter weight being not completely accurate when displaying equilibrium. 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 through the center or so called the pivot. Thus, no moments are created and hence no effect 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. Consider the buoyancy forces acting normal to the surface, then the buoyancy force do 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 ing for the buoyancy effect.
6.0 Conclusion The data collected appears to follow a trend similar to the theoretically derived equation. However, the actual data appears to be shifted approximately 0.02m above the theoretical model. This may be partially the result of the friction in the hinges and the pulley, which was not taken into . This friction would increase the closing moment, requiring the opening moment to
19
be larger to open the door. To make the opening moment larger, the depth of water would have to increase also. Another contributing factor to the actual depth being higher than the theoretical is that water was constantly leaking around the seal on the door. This velocity of the water created a lower pressure around the seal, increasing the closing moment on the door, and again requiring a greater depth of water to open the door.
The theoretical model showed that the relationship between the mass of the weight and the depth of the water is a cubic one. This is a logical conclusion, since as the water depth increases, the volume of water (a cubic value) is increasing, and it is this volume which is applying the pressure over the area.
1.0 Reference Y.A. Cengel & J. M. Cimbala, 2006. Fluid mechanics: fundamental 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/centre_of_pressure.htm [Accessed 31 July 2010].
A. Nurulhadi, n.d. Hydrostatic force on plane surface. [Online] Available at: http://atapaje.blogspot.com/2008/02/hydrostatic-force-on-plane-surface.html [Accessed 31 July 2010].
20
2.0 OBJECTIVES 2.1
To determine the center of pressure on both submerged and partially submerged a plane surface.
2.2
To compare the center of pressure between experimental result with the theoretical values.
2.3
To determine experimentally the magnitude of the force of pressure hydrostatic force.
3.0 INTRODUCTION The Hydrostatic Pressure (Model: FM 35) apparatus has been designed to introduce students to the concept of centre of pressure of an object immersed in fluid. It can be used to measure the static thrust exerted by a fluid on a submerged surface, either fully or partially, and at the same time allowing the comparison between the magnitude and direction of the force with theory. The apparatus consists of a specially constructed quadrant mounted on a balance arm. It pivots on knife edges, which also correspond to the centre of the arc of quadrant. This means that only the hydrostatic force acting on the rectangular end face will provide a moment about the knife edges (SOLTEQ, n.d.). The force exerted by the hydraulic thrust is measured by direct weighing. With no water in the tank, and no weights on the scale, the arm is horizontal. As weights are added one by one to the scales, water can be added to the tank so that the hydrostatic force balances the weights and bring the arm back to horizontal. Figure 1 is a sketch of the Hydrostatic Pressure (Model: FM 35).
1
Figure 1: Hydrostatic Pressure (Model: FM 35). The design of many engineering systems such as water dams and liquid storage tanks requires the determination of the forces acting on the surfaces using fluid statics. The complete description of the resultant hydrostatic force acting on a submerged surface requires the determination of the magnitude, the direction, and the line of action of the force (Fluid Mechanics, Cengel & Cimbala, 2010). 3.1 THEORY
Figure 2
Inclined plane submerged in a static fluid as shown in the Figure 2. The resultant force FR is acting perpendicular to the plane since no shear force is present when the fluid is at rest. FR has a line of action that es through the point (x, y), which is called the center of pressure.
2
Now take a small differential element dA at a depth of h. The differential force dF acting on dA is given by: dF
=
ρgh dA
The magnitude of the resultant force can be obtained by integrating the differential force over the whole area: FR
=
∫A ρgh dA
=
∫A ρgy sin θ dA
=
ρg sin θ ∫A y dA
The integral ∫A y dA represents the first moment of the area about the x axis, which is equal to: ∫A y dA
=
ycA
where yc is the y coordinate of the centroid of the plane surface.
From trigonometry, hc
=
yc sin θ
where hc is the vertical distance from the fluid surface to the centroid of the plane surface, then the resultant force is simplified to: Fr = ρghcA
Equation 1 :
The center of pressure, x and y can be obtained by summing moments about the 2y and x axis, respectively. First, by equating the sum of moments of all pressure forces about the x axis to the moment of the resultant force: ∫ y dF
=
FRy
∫A ρgy2 sin θ dA
=
(ρgyc sin θ dA)y
y
=
∫A y2 dA ycA
3
where ∫A y2 dA is the second moment of the area or the area moment of inertia (Ix) about the x axis. According to the parallel axis theorem, the moment of inertia can also be written as: Ix
=
Ix′ + Ayc2
where Ix′ is the second moment of area with respect to the centroidal axis, which is parallel to the x axis. Hence, the center of pressure coordinate y is given by:
Equation 2:
y = Ix′
+ yc ycA
Similarly, x can be obtained by equating the sum of moments of all pressure forces about the y axis to the moment of the resultant force.
∫ x dF
=
FRx
∫A ρgxy sin θ dA
=
(ρgyc sin θ dA)x
x
=
∫A xy dA ycA
where ∫A xy dA is the product of inertia (Ixy) of the area about the x and y axes. Once again according to the parallel axis theorem, it also can be written as:
Ixy = Ixy′ + Axcyc where Ixy′ is the product of inertia of the area with respect to the centroidal axes. Hence, the coordinate x is given by:
Equation 3:
X = Ixy′ + xc ycA
From the formulations of x and y, it is noted that the center of pressure is always lower than the centroid of the plane area. Figure 3 illustrate a hydrostatic pressure demonstration setup. 4
Figure 3: Hydrostatic Pressure Demonstration.
Referring to Figure 3, L
= Distance between suspended mass and fulcrum/pivot point.
D
= Height of end surface.
B
= Width of end surface.
H
= Total depth of quadrant.
C
= Centroid of end surface.
P
= Center of pressure of plane surface.
d
= Depth of immersion in water.
F
= Hydrostatic force exerted on quadrant.
Although the basic theory for the partially submerged and fully submerged plane is the same, it will be clearer to consider the 2 cases separately.
5
1. Partially submerged vertical plane surface
Figure 4: Partially Immersed Vertical Plane Surface. h
= Depth of centroid from water surface.
h′
= Depth of center of pressure from water surface.
h′′
= Distance between fulcrum/pivot point and the center of pressure.
Thrust on surface: For a partially submerged plane surface, from Equation 1,
F
=
ρghcA
where A
=B×d
hc
=d÷2
Thus,
Equation 4:
ρg (Bd2/2)
F =
Moment of thrust about fulcrum/pivot point: Moment, M
=
F × h′′
where h′′
= depth of line of action of thrust below fulcrum/pivot point. 6
At equilibrium condition, a balancing moment is produced by the weight (W) applied to the hanger at the end of the balance arm = W × L. For static equilibrium the two moments are equal. Hence, F × h′′
=
W×L
=
m×g×L
=
mgL
Thus, h′′
F =
2mgL ρgBd2
Equation 5
h’’
=
2mL/ρBd2
mgL/F =
Theoretically, the center of pressure from the water surface, h’ is expressed in equation below Y =
-(ρg sin Өlxy)/(PcgA)
Gives, h’ = h – Y = h – [-(Ixx sin Ө)/(hcg A)] The orientation of the surface from the horizontal = 90⁰ gives 1 for sin Ө, then
h′ =h
.
Ixx Ah
where Ix
= 2nd moment of area of immersed section about an axis in the free water surface.
So, Ix
=
Ic + Ah2
7
Thus Ixx
=
Bd3 12
+ (rectangular)
Where Ixx is the moment of inertia of the submerged surface. The center of pressure bellow the fulcrum is finally determined by, h′′
h′ + H – d
=
and by substitute Ix into h′ we will get: Equation
h’’
=
H – d/3
As the theoretical result. In other words, the distance from the pivot to the center of pressure is the depth to the bottom of the vertical plane, minus one third the depth of submerged part of the vertical plane. So the center of pressure on a partially submerged plane will always be one third of d up from the base of the plane surface.
2. Fully submerged vertical plane surface
Figure 5: Fully Immersed Vertical Plane Surface. 8
h
= Depth of centroid from water surface.
h′
= Depth of center of pressure from water surface.
h′′
= Distance between fulcrum/pivot point and the center of pressure.
Thrust on surface: For a fully submerged plane surface, from Equation 1, F
=
ρghcA
where A
=B×D
hc
=D÷2
Thus, Equation 7: F
=
ρgAh =
ρgBD(d – (D/2))
Moment of thrust about fulcrum/pivot point: Moment, M
=
F × h′′
Where
h′′
= depth of line of action of thrust below fulcrum/pivot point.
At equilibrium condition, a balancing moment is produced by the weight (W) applied to the hanger at the end of the balance arm = W × L. For static equilibrium the two moments are equal. Hence, F × h′′
=
W×L
=
m×g×L
9
But this time, h′′
=
mgL F
h’’
Equation 8
=
mL/ρBD(d-(D/2))
mgL/F =
Theoretically, the center of pressure from the water surface, h’ is expressed in equation, h’ =
h – Y
h – [(-Ixx sin Ө)/ (hcg A)]
=
The orientation of the surface from the horizontal = 90⁰ gives 1 for sinӨ, then The theoreticasult for depth of center of pressure, P, below the free surface of fluid is:
h′ =h+
Ixx Ah
h’
=
(d – (D/2)) + [(BD3 /12) X (1/BD) X (1/(d-(D/2)))]
=
[(d-(D/2))2 + (D2/12)] / (d – (D/2))
where Ix
= 2nd moment of area of immersed section about an axis in the free water surface.
Thus
Ix = BD[
D2
+ (d -
12
D
)2]
2
So, the depth of ‘P’ below the fulcrum/pivot point will be: h′′
=
h′ + H - d
10
and by substitute Ix into h′ we will get: Equation 9:
h’’
=
{[(d – (D/2))2 + (D2/12)] / ( d – (D/2))} + H – d
as the theoretical result. Apparatus The hydrostatic pressure (model : FM35) apparatus consists of a specially constructed quadrant mounted on a balance arm which pinots on knife edges and it also correspond to the center of arc of the quadrant. This means that only the hydrostatic force acting on the rectangular end face will provide a moment about the knife edge. The balance arm also has an adjustable counterbalance, and a balance pan for the weight supplied. The quadrant assembly is mounted on top of a clear reservoir tank provided with leveling screws on its feet. A spirit level attached to the base of the tank assures accuracy of the alignment. The tank is filled with water from the top and can be drained through a valve at the side. A scale is provided on the side of the quadrant to indicate the water level in the tank The torque exerted by the hydraulic thrust is measured by direct weighing. Tests may be carried out over a range of water level
3.3 Description and assembly
Figure 6: Assembly Diagram of Hydrostatic Pressure (Model: FM 35). 11
Quadrant i. Material: PVC ii. Total Depth of Quadrant, H: 200 mm iii. Height of Fulcrum above Quadrant: 100 mm iv. Height of End Surface, D: 100 mm v. Width, B: 75 mm vi. Length of Balance: 300 mm
Reservoir Tank i. Material: Clear Acrylic ii. Volume: 5.5 L o Balance Arm i. Distance between Suspended Mass and Fulcrum, L: 290 mm ii. Set of Weight: 50 g/each
Overall Dimension i. Length: 340 mm ii. Width: 130 mm iii. Height: 300 mm
4.0 Method 1. The Hydrostatic Pressure Apparatus was placed on top of a Hydraulic Bench. 2. The apparatus was installed properly as in Figure 7.
Figure 7 : Hydrostatic Pressure equipment by SOLTEQ 12
3. The apparatus was been levelled using the adjustable feet facilitated by the spirit level attached. 4. The counter-weight was been adjusted to level the balance arm to horizontal position. 5. The drain valve was been made sure is closed and water was added slowly into the tank until the surface just touched the quadrant base, thus establishing a datum level. 6. The reading of datum level was recorded at the scale. 7. A 50 g weight was placed on the balance pan and water was added slowly into the tank until the balance arm is again horizontal.
Figure 8 : Weight part 8. The mass of weight, m and depth of immersion, d in the experimental data sheet were recorded. The new depth of immersion was been subtracted with the datum level before recorded. 9. Steps 6 to 7 were repeated with increasing number of weights until the tank is full of water. 10. The water was been drained off after the experiment was completed. 11. Data was recorded in a table.
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5.0 Result and discussion 5.1 Result There are several important properties is need to record to find the Thrust Force and second moment for this experiment. The important data of this experiment was recorded and it is show in table 5.1. and the experimental data of this experiment is show in table 5.2 Height of Quadrant,D (m)
0.075
Width of Quadrant,B (m)
0.073
Length of balance,L (m)
0.27
Quadrant to pivot, H (m)
0.205
Density (kg/m3)
1000
Gravity (m/s2)
9.81 Table 5.1: Properties Experiment.
14
Depth of Immersion Mass, Immersion, (Full/ m (kg) d (m) Partial) 0.05 0.040 Partial
0.5729
2nd Moment Experimental, h" (m) 0.2311
2nd Moment Theory, h" (m) 0.1917
Thrust F (N)
Error % 20.61
0.10
0.059
Partial
1.2464
0.2125
0.1853
14.66
0.15
0.075
Full
2.0141
0.1973
0.1800
9.59
0.20
0.091
Full
2.8735
0.1844
0.1763
4.5919
0.25
0.105
Full
3.6254
0.1827
0.1744
4.70
0.30
0.121
Full
4.4848
0.1772
0.1731
2.35
0.35
0.136
Full
5.2904
0.1752
0.1723
1.73
Table 5.2: Experimental Data.
Calculations For partially immersed, Thrust,
F = ρg (Bd2/2) F = 1000 × 9.81 ×
0.073 × 0.0402 2
= 9810 × 5.84 × 10-5 = 0.5729N
15
2nd Moment Experimental,
= mgL/F = 2mL/ρBd2
h’’
h′′ = =
2 × 0.05 × 0.27 1000 × 0.073 × 0.0402 0.027 0.1168
= 0.2311 m 2nd Moment Theory,
h’’ = H – d/3 h′′ = 0.205-
0.040 3
= 0.2 - 0.0157 = 0.1917 m For fully immersed, Thrust,
F = ρgAh = ρgBD(d – (D/2)) 16
0.075
F = 1000 × 9.81 ×.073 ×.075 × (0.075-
)
2
= 53.7097× (0.075 - 0.0375) = 2.0141 N 2nd Moment Experimental,
h’’ = mgL/F = mL/ρBD(d-(D/2))
h′′
0.15 × 0.27
=
1000×.073 ×.075 × (0.075 -
.075
)
2
=
0.0405 5.475 × 0.0375
= 0.1973 m 2nd Moment Theory,
h’’
= h’ + H – d
17
= {[(d – (D/2))2 + (D2/12)] / ( d –
h’’
(D/2))} + H – d (.075 – .075/2)2 + (.0752/12)
h′′ = =
+ 0.205 - 0.075
.075 – 0.075/2
0.1800 m Error
Theoretical actual Theoretical
20.61%= [(0.1917-0.2312)/0.1917] X 100
Moment Theory & Experimental 0.25
0.2
0.15
2nd Moment Experimental, h" (m)
0.1
2nd Moment Theory, h" (m)
0.05
0 1
2
3
4
5
6
7
Figure 9: Graph for 2nd Moment Theory & Experimental
18
5.2 Discussions The comparison had made between the second moment experimental values and second moment experimental values expressed as a percentage. The result show second moment experimental value doesn’t agree well with the second theory values. Overall, all theoretical measurements were consistently slightly higher than the actual moment. This may due to human error, as in reading the measurements accurately at the correct time, or apparatus error such as the vertical ruler being slightly off, or the pivot arm or counter weight being not completely accurate when displaying equilibrium. 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 through the center or so called the pivot. Thus, no moments are created and hence no effect 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. Consider the buoyancy forces acting normal to the surface, then the buoyancy force do 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 ing for the buoyancy effect.
6.0 Conclusion The data collected appears to follow a trend similar to the theoretically derived equation. However, the actual data appears to be shifted approximately 0.02m above the theoretical model. This may be partially the result of the friction in the hinges and the pulley, which was not taken into . This friction would increase the closing moment, requiring the opening moment to
19
be larger to open the door. To make the opening moment larger, the depth of water would have to increase also. Another contributing factor to the actual depth being higher than the theoretical is that water was constantly leaking around the seal on the door. This velocity of the water created a lower pressure around the seal, increasing the closing moment on the door, and again requiring a greater depth of water to open the door.
The theoretical model showed that the relationship between the mass of the weight and the depth of the water is a cubic one. This is a logical conclusion, since as the water depth increases, the volume of water (a cubic value) is increasing, and it is this volume which is applying the pressure over the area.
1.0 Reference Y.A. Cengel & J. M. Cimbala, 2006. Fluid mechanics: fundamental 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/centre_of_pressure.htm [Accessed 31 July 2010].
A. Nurulhadi, n.d. Hydrostatic force on plane surface. [Online] Available at: http://atapaje.blogspot.com/2008/02/hydrostatic-force-on-plane-surface.html [Accessed 31 July 2010].
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