Chapter Four.pptx

WOLAITASODOUNIVERSITY
DEPARTMENT OF HYDRAULIC AND WATER RESOURCES ENGINEERING
FLUID MECHANICS (HENG-2111)
CHAPTER FOUR
HYDROSTATIC FORCES ON SURFACES
2ND year HWRE (2023 G.C.)
By: ManamnoB.(MSc.)
manamnobeza@yahoo.com

 TOPIC OUTLINE
Introduction
Hydro Static Force on a Plane Surface
Hydro Static Force on a Curved Surface
Translation and Rotation of a Fluid masses

INTRODUCTION
 When a surface is submerged in a fluid, forces develop on
the surface due to the fluid.
 The determination of these forces is important in the design
of storage tanks, ships, dams, and other hydraulic
structures.
 For fluid at rest we know that the force must be
perpendicular to the surface since there are no shearing
stresses present.
 We also know that the pressure will vary linearly with

Intro Cont.….
 Total Pressure and Center of Pressure
 Total Pressure: It is defined as the force exerted by
static fluid on a surface either plane or curved when
the fluid comes in contact with the surface
 This force is always at right angle (normal ) to the
surface
 Center of pressure: It is defined as the point of
application of the total pressure on the surface

Hydrostatic Forces On Surfaces
1. Forces on plane surface
 The distributed forces resulting from the action of fluid on a finite area can be
conveniently replaced by resultant force.
 The magnitude of resultant force and its line of action (pressure center) are
determined by integration, by formula and by using the concept of the
pressure prism.
Herein, we are considered three cases
i. Horizontal surfaces
ii. Vertically immersed surfaces
iii. Inclined plane surface
2. Forces on curved surfaces

Hydrostatic Forces On Surfaces Cont.…
i. Horizontally Immersed surfaces
 A plane surface in a horizontal position in a
fluid at rest is subjected to a constant pressure.
 Consider a plane horizontal surface immersed
in a liquid.
Let A= area of immersed surface
𝑥 = Depth of horizontal surface from the liquid
𝛾= Specific weight of the liquid
The total pressure Force on the surface =
FP = 𝛾 × 𝐴 × 𝑥
Specific weight of liquid × area of
surface × depth of liquid

ii. VERTICALLY IMMERSED SURFACE
 Total Pressure: It can be determined by dividing the entire surface into a
number of small parallel strips.
 The force on small strip is then calculated and the total pressure force on the
hole area is calculated by integrating the force on small strip
 Consider a strip of thickness dh and width b at a depth of h from the free
surface of liquid as shown in the fig.
Pressure strip intensity on the strip, 𝑃 = 𝜌𝑔ℎ
Area of the strip, 𝑑𝐴 = 𝑏 × 𝑑ℎ

 Total pressure force on strip, 𝑑𝐹 = 𝑃 × 𝑑𝐴
𝑑𝐹 = 𝜌𝑔ℎ × 𝑏 × 𝑑ℎ
Total pressure force on the whole surface:
𝐹 = 𝑑𝐹 = 𝜌𝑔ℎ × 𝑏 × 𝑑ℎ = ρ𝑔 𝑏 × ℎ × 𝑑ℎ
But 𝑏 × ℎ × 𝑑ℎ = ℎ × 𝑑𝐴
= Moment of surface area about the free surface of liquid
= Area of surface × 𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝐶. 𝐺 𝑓𝑟𝑜𝑚 𝑓𝑟𝑒𝑒 𝑠𝑢𝑟𝑓𝑎𝑐𝑒
= 𝐴 × ℎ
F = 𝝆𝒈𝑨𝒉

 Center of Pressure (𝒉∗
): Is calculated using the “ principle of Moments” which states
that the moment of the resultant force about an axis is equal to the sum of moments of
the component about the same axis
 The resultant force F is acting at P, at distance ℎ∗
from free surface of the liquid as
shown in fig above
 Hence moment of the force F about free surface of the liquid = F× ℎ∗
 Moment of force dF, acting on a strip about free surface of liquid
= dF× ℎ
= 𝜌𝑔ℎ × 𝑏 × 𝑑ℎ × ℎ

 Sum of moments of all such forces about free surface of liquid
= 𝜌𝑔ℎ × 𝑏 × 𝑑ℎ × ℎ = 𝜌𝑔 𝑏 × ℎ × ℎ𝑑ℎ
= 𝜌𝑔 𝑏ℎ2
dh = 𝜌𝑔 ℎ2
dA
But ℎ2 dA = 𝑏ℎ2 dh
= Moment of Inertia of the surface about free surface of liquid
= 𝐼𝑜
∴ Sum of moments about free surface =𝜌𝑔 𝐼𝑜
Now, equating the two moment forces
F× ℎ∗= 𝜌𝑔 𝐼𝑜 , But F = 𝝆𝒈𝑨𝒉

 Therefore, 𝝆𝒈𝑨𝒉 × ℎ∗
= 𝜌𝑔 𝐼𝑜
ℎ∗
=
𝜌𝑔 𝐼𝑜
𝝆𝒈𝑨𝒉
=
𝐼𝑜
𝐴𝒉
By theorem of parallel axis, we have
𝐼𝑜 =𝐼𝐺 + A× ℎ2
Where 𝐼𝐺 = Moment of Inertia of area about an axis passing through the C.G of the area and parallel
to the free surface of liquid
Thus Substituting, 𝐼𝑜
𝒉∗
=
𝑰𝑮 + A×𝒉𝟐
𝑨𝒉
=
𝑰𝑮
𝑨𝒉
+ 𝒉

 𝒉 is the distance of C.G of the area of the vertical surface from free
surface of the liquid.
 From the above equation, we conclude that
 For a vertical plane surface, the CP does not coincide with the centroid of
the area.
 Since the pressure increases with depth, the CP lies below the centroid of
the surface area.
 The distance of center of pressure from free surface of liquid is
independent of the density of the liquid

 Pressure Prism
 The pressure prism is an approach, which is developed for determining the resultant
hydrostatic force and line of action of the force on a plane surface.
 It is a prismatic volume with its base the given surface area and with altitude at any point
of the base given by p=h. Where h is the vertical distance to the free surface

 Force acting on the elemental area dA is:
 dF = γhdA = dV
 The volume of pressure prism equals to the magnitude of the resultant force acting on one
side of the surface
 The center of pressure is given by
 This shows that the resultant force passes through the centroid of the pressure prism.
 Therefore; the pressure force is the volume of the prism in magnitude acting at the centroid
of the prism normal to the surface.

iii. Inclined plane surface submerged in liquid
 A plane surface which is inclined to the water surface may be subjected to hydrostatic
pressure.
 For a plane inclined 𝜃0from the horizontal, the intersection of the plane of area and the
free surface is taken as the x-axis.
 The y-axis is taken in the plane of the area with origin 0 at the free surface.
 Thus, the x-y plane portrays the arbitrary inclined area.
 We wish to determine the magnitude, direction and line of the action of the resultant
force acting on one side of this area due to the liquid in contact with the area.

 Consider a plane surface of arbitrary shape immersed in a liquid
in such a way that the plane of the surface makes an angle 𝜃 with
the free surface of the liquid as shown in the fig
Let A = Total area of the inclined surface
𝒉 = Depth of C.G of inclined area from free surface
𝒉∗ = Distance of center of pressure from free surface of liquid
𝜃 = Angle made by the plane of the surface with free liquid surface

 Let the plane, if produced meet the free liquid surface at O. Then O – O is the
axis perpendicular to the plane of the surface
 Let 𝑦 = distance of C.G of the inclined surface from O - O
 𝑦∗
= distance of center of pressure from O – O

 Consider a small strip of area dA at a depth h from free surface and at a
distance y from the axis O – O as shown in the fig above
Pressure intensity on the strip, 𝑃 = 𝜌𝑔ℎ
∴ 𝑃𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑓𝑜𝑟𝑐𝑒, 𝑑𝐹 = 𝑃 × 𝐴𝑟𝑒𝑎 𝑜𝑓 𝑡ℎ𝑒 𝑠𝑡𝑟𝑖𝑝 = 𝜌𝑔ℎ × 𝑑𝐴
 Total pressure force on the whole area, F = 𝑑𝐹 = 𝜌𝑔ℎ𝑑𝐴
 But from the figure
ℎ
𝑦
=
ℎ
𝑦
=
ℎ∗
𝑦∗ = sin𝜃, Thus h = ysin𝜃
∴ 𝐹 = 𝜌𝑔 × 𝑦 × 𝑠𝑖𝑛𝜃 × 𝑑𝐴 = 𝜌𝑔 𝑠𝑖𝑛𝜃 𝑦𝑑𝐴
Since 𝑦𝑑𝐴 = A𝑦
𝐹 = 𝜌𝑔 𝑠𝑖𝑛𝜃 A𝑦
𝑭 = 𝝆𝒈 𝐀𝒉 (Total pressure Force on the plane)

 Center of Pressure (𝒉∗)
 Moment of the force dF about axis O – O = dF× 𝑦
= 𝜌𝑔𝑦𝑠𝑖𝑛𝜃dA× 𝑦 = 𝜌𝑔𝑠𝑖𝑛𝜃𝑦2
dA
Sum of moments of all such forces about O – O
= 𝜌𝑔𝑠𝑖𝑛𝜃𝑦2dA = 𝜌𝑔𝑠𝑖𝑛𝜃 𝑦2dA
Since 𝑦2
dA = M.O.I of the surface about O – O =𝐼𝑜
∴ Sum of moments of all such forces about O – O = 𝜌𝑔𝑠𝑖𝑛𝜃 𝐼𝑜
 Moment of the total force F about O – O is also given by = F× 𝑦∗
Summation of moment is equal to zero
F× 𝑦∗ = 𝜌𝑔𝑠𝑖𝑛𝜃 𝐼𝑜

𝑦∗
=
𝜌𝑔𝑠𝑖𝑛𝜃 𝐼𝑜
𝐹
Since 𝑦∗
=
ℎ∗
𝑠𝑖𝑛𝜃
and F = 𝜌𝑔 Aℎ
𝐼𝑜 =𝐼𝐺 + A× 𝑦2 (Parallel axis theorem)
ℎ∗
𝑠𝑖𝑛𝜃
=
𝜌𝑔𝑠𝑖𝑛𝜃
𝜌𝑔𝐴ℎ
𝐼𝐺 + A × 𝑦2
ℎ∗=
𝑠𝑖𝑛2𝜃
𝐴ℎ
𝐼𝐺 + A × 𝑦2 Since
ℎ
𝑦
= sin𝜃 → 𝑦 =
ℎ
𝑠𝑖𝑛𝜃
ℎ∗=
𝑠𝑖𝑛2𝜃
𝐴ℎ
𝐼𝐺 + A ×
ℎ2
𝑠𝑖𝑛2𝜃
𝒉∗=
𝑰𝑮𝒔𝒊𝒏𝟐𝜽
𝑨𝒉
+ 𝒉 ( Center of of pressure from the free surface)

2. Forces on curved surfaces
 When the elemental forces pdA vary in direction, as in the case of a curved
surface, they must be added as vector quantities that is, their components in
three mutually perpendicular directions are added as scalars and then the three
components are added vectorally
 With two horizontal components at right angle and with vertical component-
all easily computed for a curved surface the resultant can be determined.
 The lines of action of the components also are readily determined.

Horizontal and Vertical component of Forces on a curved surface
 The horizontal component of pressure force on a curved surface is
equal to the pressure force exerted on a vertical projection of the
cured surface.
 The vertical plane of the projection is normal to the direction of the
component.
 The vertical component of pressure force on a curved surface is
equal to the weight of liquid vertically above the curved surface and
extending up to the free surface
 It acts through the center of gravity of the fluid mass within the
volume.

 Consider a Curved surface Ab
 dA area of the strip
 Pressure force on the strip, dF = 𝜌𝑔ℎ × 𝑑𝐴
 Total pressure force, F = 𝜌𝑔ℎ𝑑𝐴

 The direction of the force on the small strip is not in the same, but varies point to point.
 Hence, integrating the force over the total area is impossible since the plane is curved surface
 So the total force is resolved into the x and y directions
Now, resolving into the x and y direction
𝑑𝐹𝑥 = 𝑑𝐹𝑠𝑖𝑛𝜃 = 𝜌𝑔ℎ𝑑𝐴𝑠𝑖𝑛𝜃
𝑑𝐹𝑦 = 𝑑𝐹𝑐𝑜𝑠𝜃 = 𝜌𝑔ℎ𝑑𝐴𝑐𝑜𝑠𝜃
Total forces in the x and y directions are:
𝐹𝑥 = 𝑑𝐹𝑥 = 𝜌𝑔ℎ𝑑𝐴𝑠𝑖𝑛𝜃 = 𝜌𝑔 ℎ𝑑𝐴𝑠𝑖𝑛𝜃
𝐹𝑦 = 𝑑𝐹𝑦 = 𝜌𝑔ℎ𝑑𝐴𝑐𝑜𝑠𝜃 = 𝜌𝑔 ℎ𝑑𝐴𝑐𝑜𝑠𝜃

 From the figure above (b) shows the enlarge area dA i.e ∆𝐸𝐹𝐺
EF = dA
FG = dA𝑠𝑖𝑛𝜃
EG = dAcos𝜃
→ dA𝒔𝒊𝒏𝜽 = FG = Vertical projection of the area dA and hence the expression 𝜌𝑔 ℎ𝑑𝐴𝑠𝑖𝑛𝜃 represents the
total pressure force on the projected area of the curved surface on the vertical plane
→ dAcos𝜃 = EG = horizontal projection of dA and ℎ𝑑𝐴𝑐𝑜𝑠𝜃 is the volume of the liquid on the elementary
area dA.
ℎ𝑑𝐴𝑐𝑜𝑠𝜃 = the total volume on the curved surface
𝜌𝑔 ℎ𝑑𝐴𝑐𝑜𝑠𝜃 = the total weight of the liquid supported by the curved surface

 Then the total Force on the curved surface is:
F = 𝐹𝑥
2
+ 𝐹𝑦
2
 And inclination of resultant with the horizontal is 𝑡𝑎𝑛∅ =
𝐹𝑦
𝐹𝑥

Translation and Rotation of fluid masses
 A fluid may be subjected to translation or rotation at constant accelerations without relative motion
between particles. This condition is one of relative equilibrium, and the fluid is free from shear. There is
generally no motion between the fluid and the containing vessel
 Translation is the type of motion where the element retains its shape. Its side do not undergo any change
in length and the four angles do remain square
 Rotation of a fluid particle can be caused only by a torque applied by shear forces on the sides of the
particle.
 Since shear forces are absent in an ideal fluid, the flow of ideal fluids is essentially irrotational. Generally
when the flow is viscid, it also becomes rotational
 Rotational flow also known as Vortex flow

Translation and Rotation of fluid masses Cont.….
 Vortex flow: The flow of a fluid along a curved path or the flow of a rotating
mass of fluid is known as vortex flow. The vortex flow is of two types namely:
forced vortex flow and free vortex flow.
 Forced vortex flow: It is defined as that type of vortex flow, in which some
external torque is required to rotate the fluid mass. The fluid mass in this type
of flow rotates at constant angular velocity,𝜔
 The tangential velocity of any fluid particle is given by 𝑣 = 𝜔𝑟, Where r =
radius of fluid particle from the axis of rotation.

 Hence angular velocity  is given by
 𝜔 =
𝑣
𝑟
= constant

 Examples of forced vortex are:
1. A vertical cylinder containing liquid which is rotated about
its central axis with a constant angular velocity, as
shown in figure above.
2. Flow of liquid inside the impeller of a centrifugal pump.
3. Flow of water through the runner of a turbine.

 Free vortex flow : Is a type of flow when no external torque is
required to rotate the fluid mass.
 Thus the liquid incase of free vortex is rotating due to the rotation
which is imparted to the fluid previously.
 Examples of the free vortex flow are:
1. Flow of a liquid through a hole provided at the bottom of a
container.
2. Flow of liquid around a circular bend in a pipe.
3. Flow of fluid in a centrifugal pump casing.

 Equation of forced vortex flow
 Consider a small element of fluid to move in a circular path about
an axis with radius r, and angular velocity 𝜔, as shown in the
figure

 Then, the depth of the paraboloid can be determined using the equation as:
𝑦 =
𝜔2𝑟2
2
2𝑔
 Thus y varies with the square of r. Hence, the equation is an equation of
parabola.
 This means the free surface of the liquid is a paraboloid.

Basic points
 Pressure force when the plane immersed horizontally = FP = 𝛾 × 𝐴 × 𝑥
 Pressure force when the plane immersed vertically = F = 𝝆𝒈𝑨𝒉
 Center of pressure force = 𝒉∗=
𝑰𝑮 + A×𝒉𝟐
𝑨𝒉
=
𝑰𝑮
𝑨𝒉
+ 𝒉
 Pressure force when the plane immersed Inclined = 𝑭 = 𝝆𝒈 𝐀𝒉
 Center of pressure force = 𝒉∗=
𝑰𝑮𝒔𝒊𝒏𝟐𝜽
𝑨𝒉
+ 𝒉
 Pressure force on a curved surface = By resolving the total pressure force
in to x and y
 Equation of vortex's flow = 𝑦 =
𝜔2𝑟2
2
2𝑔

Example #1.
 A rectangular plate 3m*5m is immersed vertically in water such that the 3m side is parallel to
the surface. Determine the hydrostatic force and center of pressure if the top edge of the
surface is
a) Coincides with the surface
b) 2m below the water surface

Example
 A 2m long curved gate in the figure below is hinged at o.
A. Find the horizontal component of force and its line of action
B. Find the vertical component of force and its line of action
C. What force is required to open the gate ?

Example #3.
 A 375mm high open cylinder, 150mm in diameter, is filled with water and rotated about its
vertical axis at an angular speed of 33.5rad/s2. Determine
a. The depth of water in the cylinder when it is brought to rest
b. The volume of water in the cylinder if the speed is doubled.
Solution
Height of parabola
𝑦 =
𝜔2
𝑟2
2
2𝑔
=
(33.5 × 0.075)2
2 × 9.81
= 0.32𝑚
Amount of water spilled out=volume of the paraboloid
= 1/2*volume of circumscribing cylinder
= 0.5 ∗ 𝜋 ∗ 0.0752
*0.32
= 2.83 × 10−3𝑚3
original volume of water = 𝜋 ∗ 0.0752*0.32 = 6.63 × 10−3𝑚3
Remaining volume of water = (6.63 − 2.83) × 10−3
𝑚3
= 3.8 × 10−3
𝑚3

Solution #3.
 Hence depth of water at rest =
3.8×10−3𝑚3
0.0752∗0.32
= 0.215m

Chapter Four.pptx

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