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APPLIED FLUID MECHANICS LAB MANUAL 1-4: CE 3305/ CE3305 (Complete updated 2025-26) - University of Texas, Arlington.

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APPLIED FLUID MECHANICS LAB MANUAL 1-4: CE 3305/ CE3305 (Complete updated 2025-26) - University of Texas, Arlington.

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EXPERIMENT #1: HYDROSTATIC PRESSURE




1. INTRODUCTION

Hydrostatic forces are the resultant force caused by the pressure loading of a liquid acting on
submerged surfaces. Calculation of the hydrostatic force and the location of the center of pressure are
fundamental subjects in fluid mechanics. The center of pressure is a point on the immersed surface at
which the resultant hydrostatic pressure force acts.

2. PRACTICAL APPLICATION

The location and magnitude of water pressure force acting on water-control structures, such as dams,
levees, and gates, are very important to their structural design. Hydrostatic force and its line of action
is also required for the design of many parts of hydraulic equipment.

3. OBJECTIVE

The objectives of this experiment are twofold:

« To determine the hydrostatic force due to water acting on a partially or fully submerged
surface;
To determine, both experimentally and theoretically, the center of pressure.

4. METHOD

In this experiment, the hydrostatic force and center of pressure acting on a vertical surface will be
determined by increasing the water depth in the apparatus water tank and by reaching an equilibrium
condition between the moments acting on the balance arm of the test apparatus. The forces which
create these moments are the weight applied to the balance arm and the hydrostatic force on the
vertical surface.

5. EQUIPMENT

Equipment required to carry out this experiment is the following:

« Armfield F1-12 Hydrostatic Pressure Apparatus,
« Ajug,and
« Calipers or rulers, for measuring the actual dimensions of the quadrant.

EXPERIMENT #1: HYDROSTATIC PRESSURE 3

,6. EQUIPMENT DESCRIPTION

The equipment is comprised of a rectangular transparent water tank, a fabricated quadrant, a balance
arm, an adjustable counter-balance weight, and a water-level measuring device (Figure 1.1).

The water tank has a drain valve at one end and three adjustable screwed-in feet on its base for
leveling the apparatus. The quadrant is mounted on a balance arm that pivots on knife edges. The
knife edges coincide with the center of the arc of the quadrant; therefore, the only hydrostatic force
acting on the vertical surface of the quadrant creates moment about the pivot point. This moment
can be counterbalanced by adding weight to the weight hanger, which is located at the left end of the
balance arm, at a fixed distance from the pivot. Since the line of actions of hydrostatic forces applied
on the curved surfaces passes through the pivot point, the forces have no effect on the moment. The
hydrostatic force and its line of action (center of pressure) can be determined for different water
depths, with the quadrant’s vertical face either partially or fully submerged.

A level indicator attached to the side of the tank shows when the balance arm is horizontal. Water is
admitted to the top of the tank by a flexible tube and may be drained through a cock in the side of the
tank. The water level is indicated on a scale on the side of the quadrant [1].


Level indicator . . .
Balance arm Clamping screw Knife edge pivot
»
- o AN
e
T
v‘l!




Counterbalance

. —Scale


1.‘7‘|\(




Weight hanger
‘s : ¢ Drain valve
Epitlevel HYDROSTATIC - ‘ o
. =
PRESSURE / S .
= / w g




Adjustable feet Quadrant 2l



Figure 1.1: Armfield F1-12 Hydrostatic Pressure Apparatus


7. THEORY

In this experiment, when the quadrant is immersed by adding water to the tank, the hydrostatic force
applied to the vertical surface of the quadrant can be determined by considering the following [1]:

4 APPLIED FLUID MECHANICS LAB MANUAL

, « The hydrostatic force at any point on the curved surfaces is normal to the surface and resolves
through the pivot point because it is located at the origin of the radii. Hydrostatic forces on
the upper and lower curved surfaces, therefore, have no net effect — no torque to affect the
equilibrium of the assembly because the forces pass through the pivot.
« The forces on the sides of the quadrant are horizontal and cancel each other out (equal and
opposite).
« The hydrostatic force on the vertical submerged face is counteracted by the balance weight.
The resultant hydrostatic force on the face can, therefore, be calculated from the value of the
balance weight and the depth of the water.
« The system is in equilibrium if the moments generated about the pivot points by the
hydrostatic force and added weight (=mg) are equal, i.e.:

mgXxX L=F xy (1)

where:

m : mass on the weight hanger,

L : length of the balance arm (Figure 1.2)

F : Hydrostatic force, and

y : distance between the pivot and the center of pressure (Figure 1.2).

Then, calculated hydrostatic force and center of pressure on the vertical face of the quadrant can be
compared with the experimental results.

7.1 HYDROSTATIC FORCE

The magnitude of the resultant hydrostatic force (F) applied to an immersed surface is given by:

F=PA=pgyA (2)
where:

P. : pressure at centroid of the immersed surface,

A: area of the immersed surface,

9¢: centroid of the immersed surface measured from the water surface,

p : density of fluid, and

g : acceleration due to gravity.

The hydrostatic force acting on the vertical face of the quadrant can be calculated as:

« Partially immersed vertical plane (Figure 1.2a):
EXPERIMENT #1: HYDROSTATIC PRESSURE 5

, 1
F = §prd2 (3a)

« Fully immersed vertical plane (Figure 1.2b):

FEquD(d—§> (3b)
where:

B : width of the quadrant face,

d : depth of water from the base of the quadrant, and

D : height of the quadrant face.

7.2 THEORETICAL DETERMINATION OF CENTER OF PRESSURE

The center of pressure is calculated as:

I
yp — Ayc (4)



I, is the 2°d moment of area of immersed body about an axis in the free surface. By use of the parallel
axes theorem:

I, = I, + Ay.? (5)
where 1, is the depth of the centroid of the immersed surface, and I, is the 2"4 moment of area of
immersed body about the centroidal axis. I, is calculated as:

« Partially immersed vertical plane:

Bd3 d\? Bd3


« Fully immersed vertical plane:


I, =BD 117—; + (d— 2)1 (65)
The depth of the center of pressure below the pivot point is given by:

y=yp+H—d (7)

in which H is the vertical distance between the pivot and the base of the quadrant.

Substitution of Equation (6a and 6b) and into (4) and then into (7) yields the theoretical results, as
follows:

6 APPLIED FLUID MECHANICS LAB MANUAL

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Subido en
18 de noviembre de 2025
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