ENT 257 Fluid Mechanics
Laboratory Module
EXPERIMENT 3
HYDROSTATIC PRESSURE
1.0 OBJECTIVE
1.1
1.2
To determine the centre of pressure on both submerged and partially submerged
plane surface.
To compare the centre of pressure between experimental and theoretical.
2.0 EQUIPMENT/APPARATUS
Figure 1: Assembly Diagram of Hydrostatic Pressure
DESCRIPTION
The Hydrostatic Pressure apparatus consists of a specially constructed quadrant
mounted on a balance arm which pinots on knife edges and it also correspond to the
centre 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 weights supplied.
The quadrant assembly is mounted on top of a clear reservoir tank provided with levelling
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.
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Specification
Quadrant
Material: PVC
Total Depth of Quadrant, H:175mm
Height of Fulcrum above Quadrant: 100
mm
Height of End Surface, D: 75 mm
Width, B: 73 mm
Reservoir Tank
Material
:
Clear Acrylic
Volume: 5.5 L
Balance Arm
Distance between
Suspended Mass and
Fulcrum, L:
260 mm
Set of Weights:
50g/each
Overall
Dimension
Length :
340 mm
Width :
130 mm
Height :
290 mm
3.0 INTRODUCTION AND THEORY
3.1
Hydrostatic Forces on Plane Surface
Consider a plane surface of arbitrary shape and orientation submerged in a static fluid as
shown in Figure 1. If P represents the local pressure at any point on the surface and h
represents the depth of fluid above any point on the surface, from the basic physics, we
can easily show that the net hydrostatic force on a plane surface is given by:

F  PdA
A
(1)
Free Surface
p = pa
q
h
hcg
Resultant Force
F = PcgA
dA = dxdy
CG
Side
View
CP
Plan view of arbitrary
plane surface
Figure 2: Hydrostatic Pressure on Plane Surface
The hydrostatic force on one side of a plane surface submerged in a static fluid equals to
the product of the fluid pressure at the centroid of the surface times the surface area in
contact with the fluid. Thus, basic physics says that the hydrostatic force is a distributed
load equal to the integral of the local pressure force over the area. Pressure acts normal
to a surface, therefore the direction of the resultant force will always be normal to the
surface.
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Laboratory Module
In most cases, since it is the net hydrostatic force that is desired and the contribution of
atmospheric pressure Pa will act on both sides of a surface, the result of atmospheric
pressure Pa will cancel and the net force is obtained by:
F   g h cg A
(2)
F  Pcg A
(3)
Pcg is now the gauge pressure at the centroid of the area in contact with the fluid.
Therefore, to obtain the net hydrostatic force, F on a plane surface,
Determine depth of centroid, hcg for the area in contact with the fluid
Determine the gauge pressure at the centroid Pcg
Calculate F = PcgA
Table 1 shows the centroid and other geometric properties of several areas.
Table 1
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Laboratory Module
The effective point of application of force which is normally called the “Center of
Pressure, CP” of the hydrostatic force and this is not necessary the same as the centroid.
The location of the resultant force is determined by integrating the moment of the
distributed fluid load on the surface about each axis and equating this to the moment of
the resultant force about that axis. Therefore, for the moment about the y axis:

F y cp  y P dA
A
(4)
Applying a procedure similar to that used previously to determine the resultant force, we
obtain:
Ycp  
 g sinq I xx
Pcg A
0
(5)
Where, Ixx is defined as the Moment of Inertia, or

I xx  2 nd moment of the area
(6)
Therefore, the resultant force will always act at a distance ycp below the centroid of the
surface (except for the special case of sin θ = 0). Proceeding in a similar manner for the x
location, and defining I xy = product of inertia, we obtain:
X cp  
 g sinq I xy
Pcg A
(7)
Where Xcp can be either positive or negative since I xy can be either positive or negative.
For areas with a vertical plane of symmetry through the centroid, the y-axis (e.g. squares,
circles, isosceles triangles, etc.), the center of pressure is located directly below the
centroid along the plane of symmetry, Xcp = 0.
For most problems where we have a single, homogeneous fluid and the surface pressure
is at atmospheric, the fluid specific weight  cancels in the equation for Ycp and Xcp and we
have the following simplified expressions:
Ycp  
X cp  
I xx sinq
h cg A
I xy g sinq
hcg A
(8)
(9)
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ENT 257 Fluid Mechanics
3.2
Laboratory Module
Hydrostatic Pressure on a Vertical Plane Surface
Figure 2 , illustrates a hydrostatic pressure demonstration setup where the center of
pressure of an immersed and partial immersed vertical plane surface can be determined.
L
H
W
C
P
D
Figure 3: Hydrostatic Pressure Demonstration
Referring to Figure 2,
L
D
B
H
C
P
= Distance between Suspended Mass and Fulcrum
= Height of End Surface
= Width of End Surface
= Total Depth of Quadrant
= Centroid of End Surface
= Center of Pressure of Plane Surface
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Partial Immerged Vertical Plane Surface
Fulcrum
H
Quadrant
d
h h’ h”
C
P
D
F
B
Figure 4: Partial Immersed Vertical Plane Surface
d
= Depth of Immersion in Water
F
= Hydrostatic Force exerted on Quadrant
h
= Depth of Centroid from the Water Surface
h’
= Depth of Center of Pressure from the Water Surface
h”
= Distance between Fulcrum and the Center of Pressure
For a partially immersed plane surface as shown in Figure 3,
A  Bd
(10)
d
h
2
(11)
Therefore, the hydrostatic force, F (Equation 2) is simplified as:
Bd 2
F g
2
(12)
If the system as shown in Figure 3 is equilibrium with the moment of horizontal arm, then,
M
fulcrum
 Fh "WL
Fh "  WL  mg L
(13)
Substituting Equation 12 into Equation 13, we obtain,
h"
mg L 2mL

F
Bd 2
Experiment
(14)
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Laboratory Module
Theoretically, the center of pressure from the water surface, h’ is expressed in Equation 8:
 I sinq 

h '  h  Ycp  h    xx
 hcg A 


(15)
The orientation of the surface from the horizontal = 90° gives 1 for sin θ, then
h' h
I xx
hA
(16)
and
I xx 
Bd 3
12 (For a Rectangular Surface)
(17)
where I xx is the moment of inertia of the submerged surface (please refer to Table 1)
The center of pressure bellow the fulcrum is finally determined by,
h "  h ' H  d
(18)
Thus, substituting Equation 16 and 17 into Equation 18 will finally give,
h"  H 
d
3
Theory
(19)
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Fully Immerged Vertical Plane Surface
Fulcrum
H
Quadrant
d
h
D/2
h’ h”
C
P
D
F
B
Figure 5: Fully Immersed Vertical Plane Surface
Referring to Figure 4,
D
F
h
h’
h”
= Depth of Immersion in Water
= Hydrostatic Force exerted on Quadrant
= Depth of Centroid from the Water Surface
= Depth of Center of Pressure from the Water Surface
= Distance between Fulcrum and the Center of Pressure
The hydrostatic force of a fully immersed vertical plane surface, F is determined as:
D

F   g Ah   gBD d  
2

(20)
If the system as shown in Figure 4 is equilibrium with the moment of horizontal arm, then,
M
fulcrum
 Fh "WL
Fh "  WL  mg L
Thus,
h '' 
mgL

F
mL
Experiment (21)
D

BD  d  
2

Theoretically, the center of pressure from the water surface, h’ is expressed in Equation 8:
 I sinq 

h '  h  Ycp  h    xx
 hcg A 


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Laboratory Module
The orientation of the surface from the horizontal = 90° gives 1 for sinθ, then
h' h
I xx
hA


 3

D   BD
1
1 

h '  d    


D
2
12 BD 


 d   
2


2
D   D2 

 d     
2  2 


D

d  
2

(22)
The center of pressure bellow the fulcrum is finally determined by,
h "  h ' H  d
D   D2 


 d    
2   2 

''
h 
 H  d Theory (23)
D

d  
2

2
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Laboratory Module
4.0 EXPERIMENTAL PROCEDURE
No.
1.
Procedure
Place the Hydrostatic Pressure Apparatus on top of a Table.
2.
Make sure that the apparatus is installed properly as in Figure 5.
3.
Level the apparatus using the adjustable feet facilitate by the spirit level
attached.
4.
Adjust the counter-weight to level the balance arm to horizontal position.
5.
Make sure that the drain valve is closed and slowly add water into the tank until
the water surface just touches the quadrant base, thus establishing a datum
level corresponding to zero at the level scale.
6.
Place a 50 g weight on the balance pan and slowly add water into the tank until
the balance arm is again horizontal.
7.
Record the mass of weight, m and depth of immersion, d in Table 1.
8.
Repeat steps 6 and 7 with increasing number of weights until the tank is full of
water.
9.
Drain off the water after the experiment is completed.
10.
End of experiment.
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ENT 257 Fluid Mechanics
Name :
Laboratory Module
______________________________
Date : ______________
Matrix No : ______________________________
5.0 DATA & RESULTS
Observation Table:
Height of Quadrant , D
=______________________ m
Water Density
=___________________kg/m3
Width of Quadrant , B
=______________________m
Gravity
=___________________m/s2
Length of Balance, L
=______________________m
Quadrant to pivot, H
=______________________m
Table 1
Mass,
m
Depth of
Immersion ,
d
kg
m
Immersion Sketch
(refer figure 2)
Show water level and all
dimensions
Immersion
Type
Thrust, F
2nd Moment
Experimental
2nd Moment
Theory
Error
N
h”
h”
%
0.05
0.10
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Mass,
m
Depth of
Immersion ,
d
kg
m
Laboratory Module
Immersion Sketch
(refer figure 2)
Show water level and all
dimensions
Immersion
Type
Thrust,F
2nd Moment
Experimental
2nd Moment
Theory
Error
N
h”
h”
%
0.15
0.20
0.25
0.30
Show all your calculations in column.
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ENT 257 Fluid Mechanics
Name :
______________________________
Matrix No :
_____________________________
Laboratory Module
Date : ______________
7.0.
DISCUSSION / EVALUATION & QUESTION
7.1.
Briefly summarize the key results of each experiment
7.2.
Explain the significance of your findings
7.3.
Explain any unusual difficulties or problems which may have led to poor results
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Laboratory Module
7.4.
Offer suggestions for how the experimental procedure or design could be improved.
7.5.
Compare your experimental values with theoretical values given
7.6.
What will happen if Quadrant weight is increased?.
7.7.
What have you understand about Hydrostatic Pressure?. Explain in your own words.
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ENT 257 Fluid Mechanics
Name :
______________________________
Matrix No :
______________________________
Laboratory Module
Date : ______________
8.0 CONCLUSION
(Based on data and discussion, make your overall conclusion by referring to experiment objective.)
The conclusion for this lab is…
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