Chapter 3(Immersed Surfaces) Corrected by Reviewer

FLUID MECHANICS AND MACHINERY 47 HYDROSTATIC FORCES

Chapter-3HYDROSTATIC FORCES ON IMMERSED SURFACES

INTRODUCTION:When a solid object like a plane or curved surface is immersed in a static fluid (fluid at rest), the surface experiences an upward pushing pressure exerted, by the fluid. As the surface penetrate into interior depths of the fluid, the upward pushing pressure increases due to increase in pressure head, as seen in hydrostatic law.

As per the Pascal’s law, the pressure exerted by the fluid is uniform in all the directions, through out the surface immersed in the fluid, and hence we can treat all the pressure intensities acting on the immersed surface, as a single pressure acting at a single point (named as Center of Pressure ) on the surface and in addition to this center of pressure, the surface do posses Center of Gravity where the whole weight of the surface concentrates. Hence in this section we will be focusing on this two center points namely Center of Pressure and Center of Gravity for carrying out derivations on pressures exerted on Horizontal plane surface, Vertical plane surface, inclined plane surface and Curved surface.PRESSURE EXERTED ON HORIZONTAL PLANE SURFACE:When a plane surface immersed in a liquid horizontally the Center of pressure where the total pressure acts and Center of gravity where the center of gravity of the body concentrated, coincides with each other.

Consider a horizontal plane surface having area “A” immersed in a liquid having specific weight , as shown in figure.3.1.

Let the distance of center of pressure and center of gravity on the plane surface from

the liquid surface be and respectively and as

the two center point coincides, their distances

from the liquid surface also coincide. i.e. .

The plane surface immersed in liquid displaces some volume of the liquid, which is equal to the weight of the plane surface and so the total pressure that acts on the plane surface will be equal to the weight of the liquid displaced above the immersed plane surface. I.e. Total pressure is equal to weight of the liquid above the immersed surface.

Eq:(3.0) gives the Total Pressure acting on a object immersed horizontally, in a liquid.PRESSURE EXERTED ON VERTICAL PLANE SURFACE:Consider a Vertical plane surface having area “A” and width “b” immersed in a liquid having specific weight . As the plane surface is immersed vertically, the Center of pressure and Center of gravity may or may not coincide with each other (it depends on the shape of the plane surface), as shown in figure.3.2.

Let the distance of center of pressure and center of gravity on the plane surface from the

liquid surface be and respectively.

Total Pressure:Considering a strip of the immersed object of dimension , the total pressure acting on

the strip due to pressure intensity of the liquid on the strip is


Total pressure acting on the plane surface will be

But is the moment of inertia of the immersed plane surface about liquid surface level

(first moment of inertia).

i.e.

Substituting Eq:(3.1a) in Eq:(3.1), we haveTotal pressure acting on the plane surface immersed vertically as

Total Moment:Moment of the strip due to total pressure

is . So the total moment on

the object immersed will be .

Substituting Eq:(3.1a), Eq:(3.2) in Eq:(3.3), we have

If converted in terms of moment of inertia of gravity about the surface level then according to parallel axis theorem from engineering mechanics, we have

Substituting Eq:(3.3a) in Eq:(3.3), we have

Example.3.1: An open tank of 5m x 3m x 6m is completely filled with oil (shown in figure.P3.1) of specific gravity 0.8. Determine the total pressure and center of pressures on all the faces including the base.Solution: Given data: Base EFGH is horizontal and has area

and center of gravity

And hence total pressure

Faces AEHD and BFGC are vertical and has area of

and .


And center of pressure


Faces AEFB and DHGC are vertical and has area of and .


And center of pressure

Example.3.2: A plane surface in the shape of a trapezium is immersed in a liquid having specific gravity of 0.8 such that the smaller side is at the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the trapezium plane surface and also the center of pressure. Solution: Given data:

The area of the trapezium

Center of gravity from the bigger side as the

base is

Hence the Center of gravity from the smaller

side is

Or else if the trapezium is divided into a rectangle and two triangles, then Center of gravity in terms of area will be


Center of pressure

Hence the center of gravity is 1.715m and center of pressure is 2.125m.Example.3.3: A plane surface in the shape of a trapezium is immersed in a liquid having specific gravity of 0.8 such that the smaller side is at 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the trapezium plane surface and also the center of pressure. Solution: Given data:

The area of the trapezium (refer: Table.3.1)


Center of gravity from the bigger side as the base is

Hence the Center of gravity from the free surface of the liquid is



Center of pressure

Hence the center of gravity is 3.715m and center of pressure is 3.778m.Example.3.4: A plane surface in the shape of a trapezium having a circular hole in it is immersed in a liquid having specific gravity of 0.8 such that the smaller side is at 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the trapezium plane surface with circular hole in it and also the center of pressure. Solution: Given data:


The area of the circle

Total area of the plane surface is

Center of gravity of trapezium from the bigger side as the base is

Center of gravity of circle from the bottom circumference of circle is

Hence the Center of gravity of the plane surface from the free surface of the liquid is




Center of pressure

Hence the center of gravity is 3.715m and center of pressure is 3.918m.Example.3.5: A plane surface in the shape of a rectangle having a square hole in it is immersed in a liquid having specific gravity of 0.8 such that the one of its longer side is at 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the rectangular plane surface with square hole in it and also the center of

pressure. Solution: Given data:

The area of the rectangle

The area of the square


Center of gravity of rectangle from the bigger side as the

base is

Center of gravity of square from the bigger side as the

base is


Or else Center of gravity in terms of area can be


Center of pressure

Hence the center of gravity is 4.0m and center of pressure is 5.594m.Example.3.6: A plane surface in the a triangle is immersed in a liquid having specific gravity of 1.1 such that the base of the triangle is vertical and is at 2m below the surface level of the liquid


free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the triangular plane surface and also the center of pressure. Solution: Given data:

The area of the triangle

Center of gravity of the triangle lie on the line passing through the mid point of the height “h” and

so



Center of pressure

Hence the center of gravity is 4.5m and center of pressure is 4.808m.Example.3.7: A plane surface in the shape of a rectangle having immersed in a liquid having specific gravity of 0.8 such that the one of its longer side is at the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the rectangular plane surface and also the center of pressure. Solution: Given data: The area of the rectangle

Center of gravity of rectangle from the bigger side as the base is and from the free surface of the liquid is


Center of pressure

Hence the center of gravity is 3.0m and center of pressure is 4.0m.Example.3.8: A plane surface in shape of a triangle is immersed in a liquid having specific gravity of 0.8 such that the base of the triangle horizontal and is at the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the triangular plane surface and also the center of pressure. Solution: Given data:



Centers of gravity of the triangle lie on the vertical line passing through the mid point of the base

“b” and so


Center of pressure

Example.3.9: A plane surface in shape of a circle is immersed vertically in a liquid having specific gravity of 1.1 such that its circumference touches the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the circular plane surface and also the center of pressure. Solution: Given data:

The area of the circle

Centers of gravity of the circle lie on the vertical diameter line, so


Center of pressure

Hence the center of gravity is 2.5m and center of pressure is 3.125m.Example.3.10: A plane surface in shape of a semi-circle is immersed vertically in a liquid having specific gravity of 1.1 such that its diameter touches the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the semi-circular plane surface and also the center of pressure. Solution: Given data:

The area of the semi-circle

Centers of gravity of the circle lie on the vertical diameter line, so


Center of pressure

Hence the center of gravity is 2.123m and center of pressure is 5.0m.


Example.3.11: A plane surface in the shape of a rectangle having a circular hole in it and is immersed in a liquid having specific gravity of 0.8 such that the one of its longer side is at 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the rectangular plane surface with circular hole and also the center of pressure. Solution:

Given data:


The area of the circle hole


Center of gravity of rectangle from the bigger side as the base is

Center of gravity of circle is


Or else Center of gravity in terms of area can be


Center of pressure

Hence the center of gravity is 5m and center of pressure is 5.68m.Example.3.12: A plane surface in the a triangle is immersed in a liquid having specific gravity of 0.98 such that the base of the triangle is horizontal and is at 3m below the surface level of the liquid free surface and is of dimensions as shown in the figure 3.12. Determine the total pressure acting on the triangular plane surface and also the center of pressure. Solution: Given data:



Center of gravity of the triangle lie on the line passing through the mid point of the base “b” and

so



Center of pressure

Hence the center of gravity is 4.67m and center of pressure is 4.729m.

PRESSURE EXERTED ON INCLINED PLANE SURFACE:Consider an Inclined plane surface having area “A” and width “b” immersed in a liquid having specific weight . As the plane surface is immersed inclined, the Center of pressure and Center of gravity may or may not coincide with each other (it depends on the shape of the plane surface), as shown in figure.3.3.

Let the distance of center of pressure and center of gravity on the plane surface from the

liquid surface be and respectively.

Total Pressure:Considering a strip of the immersed object of dimension , the total pressure acting on the strip due to pressure intensity of the liquid on the strip is and hence

Total pressure acting on the plane surface will

be

But is the moment of inertia of the immersed plane surface about liquid surface level

(first moment of inertia). i.e.

Substituting Eq:(3.5a) in Eq:(3.5), we have Total pressure acting on the plane surface

immersed vertically as

Total Moment:Moment of the strip due to total pressure is . So the total moment on the object

immersed will be .

Substituting Eq:(3.5a), Eq:(3.6) in Eq:(3.7), we have

If converted in terms of moment of inertia of gravity about the surface level then according to parallel axis theorem from engineering mechanics, we have


Substituting Equation.(3.8a) in Equation.(3.8), we have

Example.3.13: A plane surface in rectangular shape is immersed inclined at angle of 300 to the free surface of a liquid having specific gravity of 0.8 such that its tip is 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the rectangular plane surface and also the center of pressure. Solution: Given data:


Centers of gravity of the rectangle


Center of pressure

Hence the Center of gravity is 3.135m and Center of pressure is 3.74m.Example.3.14: A plane surface in trapezium shape is immersed inclined at angle of 600 to the free surface of a liquid having specific gravity of 0.8 such that its tip is 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure P3.14. Determine the total pressure acting on the trapezium plane surface and also the center of pressure. Solution:Given data:


Centers of gravity of the trapezium from the bigger side as the base

Hence total center of gravity is



Center of pressure

Hence the Center of gravity is 4.165m and Center of pressure is 4.29m.Example.3.15: A plane surface in circular shape is immersed inclined at angle of 600 to the free surface of a liquid having specific gravity of 1.2 such that its tip is 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the circular plane surface and also the center of pressure. Solution: Given data:


Centers of gravity of the circle

Total pressure

Moment of inertia of circle about C.G is

Center of pressure of plane surface is

Hence the Center of gravity is 4.165m and Center of pressure is 4.415m.Example.3.16: A plane surface in trapezium shape having a circular hole is immersed inclined at angle of 600 to the free surface of a liquid having specific gravity of 0.8 such that its tip is 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure P3.16. Determine the total pressure acting on the trapezium plane surface with circular hole in it and also the center of pressure. Solution: Given data:



Area of the circle hole

Total Area of the plane surface

Centers of gravity of the trapezium


Hence total center of gravity is


Moment of inertia of trapezium about C.G is


Total moment of inertia


Hence the center of gravity is 4.165m and Center of pressure is 4.18m.Example.3.17: A plane surface in rectangular shape having a circular hole in it is immersed inclined at angle of 300 to the free surface of a liquid having specific gravity of 0.8 such that its tip is 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the rectangular plane surface with circular hole in it and also the center of pressure. Solution:Given data:


Area of the circle hole

Total area of the plane surface

Centers of gravity of the rectangle from the smaller side as the base


Hence Total center of gravity is


Moment of inertia of rectangle about C.G is





Hence the center of gravity is 3.135m and center of pressure is 3.195m.Example.3.18: A plane washer surface is immersed inclined at angle of 600 to the free surface of a liquid having specific gravity of 1.2 such that its tip is 2m below the surface level of the liquid free surface and is of dimensions as shown in the figure. Determine the total pressure acting on the circular washer plane surface and also the center of pressure. Solution: Given data:

The area of the bigger circle

The area of the smaller circle


Centers of gravity of the bigger circle

Centers of gravity of the bigger circle

Total center of gravity will be


Moment of inertia of bigger circle about C.G is

Moment of inertia of smaller circle about C.G is



Hence the center of gravity is 4.165m and center of pressure is 4.449m.Example.3.19: A cover plate of 0.8m x 0.8m in dimension is hinged to a container at one of its end; to covers the oil contained in the container and oil has specific gravity of 0.85. The internal pressure due to the oil presence in the container is read as 20kPa, by the pressure gauge connected to the container. A force of “F” is applied to the cover plate at the other end so as to keep the cover plate in tight, as shown in the figure. Determine the force “F” and the reaction “R” at the hinge. Solution: Given data:


Centers of gravity of the cover plate

As there is a pressure measured due to presence of oil in the container, we need to calculate the pressure head created due to the oil presence, and hence the pressure head will be

So the total center of gravity will be the sum of center of gravity and the pressure head, which gives total center of gravity as

Hence total pressure


Hence the center of gravity is 2.399m and center of pressure is 3.28m.Taking moment about the hinge, we have two forces acting, like force applied “F” and the pressure acting at the center of pressure due to the presence of oil in the container, and hence the moments about the hinge by these two forces must be equal

Hence

Hence the Force to be applied on the cover is 101266N.And Reaction force “R” will be “P-F”, i.e. PRESSURE EXERTED ON CURVED SURFAC E :Any curved surface immersed in a liquid have pressure intensities acting on its surface along the normal direction to the surface and the force acting on the curved surface is called as Total

pressure and it can be resolved into components along the X-axis and Y-axis.Consider a curved surface AB as shown in figure.3.4a. If OA

and OB are treated as Rectangular faces, as shown, on which the horizontal component of pressure intensity and vertical


component of pressure intensity acts, Then the total pressure “P” will be

acting on the curved surface AB and the Magnitude of total pressure will be

And the Total pressure inclination is

Note: If the liquid acts from below the curved surface then the vertical pressure component will be acting upward direction.Example.3.20: A dam has the cross-sectional profile composed of a Vertical face with a circular curved section at the base as shown in Figure.P3.20. The dam is of 2m long. Calculate the resultant force and its direction of application on the curved surface of the dam design.Solution: Given data:

Horizontal Pressure Force:Horizontal pressure force on the curved surface of Dam will be acting on the vertical face “OAJF” and is equal to

Where

So Horizontal pressure force will be Center of pressure force is

Hence the horizontal pressure force will be 627.8kN acting at a point of from the base point “A”.Vertical Pressure Force:Vertical force on the Dam will be equal to the weight of the water acting on the Dams curved surface area and hence

Magnitude of the Resultant force

Direction of the Resultant force


Example.3.21: A concrete dam of width 10m has the cross-sectional profile as shown in Figure.P3.21a. Calculate the magnitude, direction and position of action of the resultant force exerted by the water on the dam.Solution: Given data: Area of the dam wall= 10m x 12m=120m2

Vertical height of the water level (h) =12m Width of dam (b) =10m.As per the figure.3.21a, we have sloping wall length,

and .

Vertical force on the Dam will be equal to the weight of the water acting on the Dams slope area, and hence

Horizontal force on the Dam will be equal to the projection of water weight acting on the vertical area, and hence



is the distance along the face to the centroid = m/2 =

6.925m

Example.3.22: A dam has a spill way gate at angle of 300, as shown in figure. The gate is hinged at one of its ends such that the end is at the level of the water surface level. The dam is of 2m in width and the gate is of 6cm long. Calculate the resultant force and its directions on the gate surface of the dam design.Solution: Given data: Height of the spill way from the dam surface will be

Horizontal Pressure Force:Horizontal pressure force on the curved surface of spill way will be acting on the vertical face “AC” and is equal

to

Where

So Horizontal pressure force will be Hence the horizontal pressure force will be 8.829Pa acting at a point of from the surface level of the Dam.


Vertical Pressure Force:Vertical force on the Dam will be equal to the weight of the water acting on the Dams curved surface area and hence

Since the sector “ABD” represents a 1/12th of a circle and so area of the sector will be

, i.e.



1. A depth of center of pressure in a rectangular lamina of height h with one side in the liquid surface is at

a) h b) h\3 c) 2h\3 d) h\2 e) 3\4 h.

2. The horizontal component of force on a curved surface is equal to the a) product of pressure at its centroid and area b) Weight of liquid retained by the curved surfacec) Force on a vertical projection of the curved surfaced) Weight of liquid vertically above the curved surfacee) None of the above

3. The vertical component of pressure force on a submerged curved surface is equal to a) Weight of liquid vertically above the curved surface and extending upon the free

surfaceb) The force on a vertical projection of the curved surfacec) The product of pressure at centroid and surface aread) Horizontal componente) None of the above

4. When will centre of pressure and centre of gravity of an immersed plane surface coincide? a) When surface immersed inclined.

b) The force on a vertical projection of the curved surfacec) When surface immersed vertically.d) When surface immersed horizontally.e) None of the above


Theory:

1. Write a note on immersion of plane surfaces.2. Derive equation for pressure exerted on a horizontal plane surface.3. Derive equation for pressure exerted on a vertical plane surface.4. Derive equation for pressure exerted on an inclined plane surface.5. Derive equation for pressure exerted on a curved plane surface.

Problems: 1. A vertical isosceles triangle gate with its vertex up has a base width of 2m and a height of

1.5m. If the vertex of the gate is 1m below the free surface of water, find the total pressure force and the position of the center of the pressure on one side of the plane.

2. A cubical tank has sides of 1.5m. It contains water in the lower 0.6m depth. The upper remaining part is filled with oil of relative density 0.9. Calculate for one vertical side of the tank a.) The pressure force b.) Position of the center of pressure.

3. A vertical gate of width “W” controls the water level in the intake reservoir of a power plant. And the reservoir has a head level “H”. What is the force of the water on the gate? What is the net hydrostatic force on the gate?

4. An Inclined gate of 300 to the horizontal and has length of 3m, width of 1.5m, controls the water level in the irrigation channel. And the channel has a head level 1m. What is the force of the water on the gate? What is the net hydrostatic force on the gate?

5. A rectangular gate of length 1.5m and width 0.5m is fitted at the opening of the bottom of a cylindrical vessel containing kerosene. The gate is hinged on one of the larger edge corners and is fixed tight on the other larger side. The vessel contains kerosene to a depth of 5m. Take the mass of the gate as 50kg.

6. The caisson for closing the entrance to a dry dock is of trapezoidal form 15m wide at the top and 12.2m wide at the bottom and 8m deep. If the water on the outside is just level with the top and dock is empty, find the total pressure on it and the depth of the ceneter of pressure.

QUICK REFERENCE OF FORMULAE:

Property Formulae Units Remarks, If anyPressure exerted on horizontal immersed plane surface (P)

or

is distance of Center of

gravity of surface from free liquid surface.

Pressure exerted on vertical immersed plane surface (P)

; is distance of Center of

pressure on surface from free liquid surface.

IO is Moment of inertia of surface about free liquid surface.

IG is Moment of inertia of surface gravity about free liquid surface.

Pressure exerted on inclined immersed plane surface (P)

;

Pressure exerted on curved immersed surface (P)

,

is the pressure inclination.

are the horizontal

and vertical pressure components .

Table 3.1: Center of gravity, Area and Moment of inertia of different types of surfaces.

Shape Center of Gravity Area Moment of Inertia

Chapter 3(Immersed Surfaces) Corrected by Reviewer

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Transcript of Chapter 3(Immersed Surfaces) Corrected by Reviewer