MECHANICS OF FLUIDS 1
CHAPTER 3 NOTES
SEPTEMBER 9, 2020
NJ RAMANAMANE
UNISA
Open Rubric
STUDY UNIT 3 – HYDROSTATIC FORCES
At the end of this study unit, you should must be able to:
•
Calculate hydrostatic force exerted by a body on fluid
•
Determining the resultant force and centre of pressure on flat
surfaces
•
Determining the resultant force and centre of pressure on curved
surfaces
•
Present the action of pressure graphically, read from page 94 to page
100 of Fluid Mechanics in SI Units (R.C Hibbeler).
1. FORCES EXERTED BY THE FLUIDS ON A SURFACE
I will start by making a real-life example so that we will all understand what
this topic is all about. A figure below shows a body submerged in a fluid, when
a surface is submerged in a fluid, forces develop on the surface due to the
fluid as shown by the arrows. This force is commonly termed hydrostatic
force.
Figure 1: Forces exerted on the surface
Why do we need to know this information, the determination of these forces
is important in the design of storage tanks, ships, dams, and other hydraulic
structures. Note that for fluids at rest we know that the force must be
perpendicular to the surface since there are no shearing stresses present.
Since we know that a liquid does not have a fixed shape this force is not
limited to a vertical direction but includes in the horizontal direction. If the
plane area is purely horizontal, the pressure exerted is simply the pressure
calculated in study unit 2 and the force is the pressure multiply by the area.
Therefore, the hydrostatic force equation is given by:
Fhydrostatic = Pressure × Area
= ρghA
Units: kN
This study requires you to study from page 94 to page 100 of a
prescribed book (R.C Hibbeler Fluid Mechanics in SI Units) and internet.
2. HYDROSTATIC FORCE ON VARIOUS SURFACES
Hydrostatics force simply means the resulting force due to the pressure of a
liquid on a submerged surface. However, hydrostatic forces can be divided in
the various groups as mentioned below.
Hydrostatic forces groups are:
•
On vertical flat surface
•
On an inclined flat surface
•
On a curved surface
•
On a horizontal flat surface
HYDROSTATIC FORCE ON VERTICAL SURFACES
However, most submerged surfaces are not horizontal and are either at an
incline or vertical. In such situations, the determination of the hydrostatic
force on the surface is not so simple. The idea is still the same, pressure under
water multiply by the area of the surface. However, the challenge is in finding
the “h”.
Let us assume the situation of a vertical plate.
Figure 1: Hydrostatic force on vertical surface
For a vertical plate shown figure above, the pressure on the plate is not
uniform but increase with depth. Because the relationship is a linear one you
get a triangular pressure profile. Hence, the total pressure can be calculated
with “h” halfway down the plate.
Figure 2: Centre of pressure
From figure above it is important to understand that the pressure is not at
the same point as the height “H”, but rather acts on the bottom of the plate
as shown by the red arrow. This point is called centre of pressure, and acts
at 1/3 from the base of the pressure triangle.
HYDROSTATIC FORCE ON INCLINED FLAT SURFACES
For incline surfaces, the pressure profile, and the location of the centre of
pressure is the same.
Figure 4: Hydrostatic force on inclined flat surface
1.1.1 HYDROSTATIC FORCE ON CURVED SURFACES
Figure 5: Hydrostatic force on curved surface
When dealing with curve surfaces, there is a horizontal and vertical
component. The resultant hydrostatic force on the curve surface is a vector
addition of both components.
Thus, we can see that the curvature have a direct impact on the resultant
hydrostatic force.
Let us solve examples based on the learned principles of the
hydrostatic forces, also look for the added examples on chapter 3
presentation on Myunisa.
1. Example of the hydrostatic force calculation
The sea wall in Figure (a) below is in the form of a semi parabola. Determine
the resultant force acting on 1m of its length. Where does this force act on the
wall? Take 𝜌 = 1050
kg⁄
m3.
Figure 6: Sea wall
Solution
Horizontal force components
We will name the bottom part point A,
Therefore, 𝑊𝐴 = (ρw gh)(1𝑚) = (1050)(9.81)(8)(1) = 82.40 kN⁄m
1
Thus, Fx = 2 (8)(82.40) = 329.62 k⁄N
Position where component is acting,
ȳ=
2
(8) = 5.33m
3
Vertical force components
1
Fy = (ρw g)Atriangle (1) = (1050)(9.81) [ (2)(8)] (1) = 54.94 k⁄N
3
X=
3
(2) = 1.5m
4
Therefore, Resultant force is:
FR = √(329.62)2 + (54.94)2 = 334 k⁄N
Activity 1
The tank is filled with water and kerosene to the depths shown. Determine the
total resultant force the liquids exert on side AB of the tank. The tank has a
width of 2m. Take ρw = 1000 kg⁄m3 , ρk = 814 kg⁄m3
Figure 7: Tank filled with water and keosene
Solution to this problem will be posted on myunisa.
Look for more examples Fluid Mechanics in SI Units (R.C.
Hibbeler) and also on the internet.