Unit -2

Hydrostatics

Structure:

2.1. Introduction

2.2. Objectives

2.3. Total pressure

2.4. Total pressure on an vertically immersed surface

2.5. Total pressure on inclined surface

2.6. Centre of pressure

2.7. Centre of pressure of a vertically immersed surface

2.8. Position of the center of pressure

2.9. Summary

2.10. Keywords

2.11. Exercise

2.1. Introduction

When a statics mass of fluid comes in contact with a surface, either plan or curved, a force is

exerted by the fluid on the surface. This force is known as total pressure. Since for a fluid at rest

no tangential force exists the total pressure acts in the direction normal to the surface. The point

of application of total pressure on the surface is known as centre of pressure.

2.2. Objectives

After this unit we are able to understand

Total pressure

Total pressure on an immersed surface

Total pressure on a horizontally

Immersed surface

Total pressure on a vertically immersed surface

2.3. Total Pressure on a Plane Surface

(a) Total pressure on a Horizontal plane surface.

Fig.1: Total pressure on a horizontal plane surface

Consider a plan surface immersed in a static mass of specific weight w, such that it is held in a

horizontal position at a depth h below the free surface of the liquid, as shown in fig. Since every

point on the surface is at the same depth below the free surface of the liquid. The pressure

intensity is constant over the entire plan surface, being equal to p= wh. Thus if A is the total area

of the surface then the total pressure on the horizontal surface is

P=pA | ‘.’ P= wh

P=wh A

The direction of this force is normal to the surface, as such it is acting towards the surface in the

vertical downward direction at the centroid of the surface.

Problems (1): A rectangular tank 4 m long 2 meters wide contains water up to a depth of 2.5

meters. Calculated pressure on the base of the tank

Given: length, l= 4m.

Width, b= 2m. Assume specific wt, =9.81 KN/m3

x =2.5 m.

W.K.T

P= WAx A= l x b = 4x 2 = 8m2

P= 9.81 x 8 x 2.5

P= 196.2 KN

P= 196.2 KN

(2) A tank 3m x 4m contains 1.3 m deep oil specific gravity 0.8 find (i) intensity of pressure at

the base of the tank, and (ii) total pressure on the base of the tank.

Given:

Site of tank= 3m x 4m

i.e. A = 12 m2

depth of oil, x= 1.3 m

specific Gravity =0.8(which is lighter than water).

Specific weight oil= G oil X γ H2O | ∴ γ H2O =9.810 N/m3

∴ =0.8X 9810

γ oil =7848 N/m3

Step (i) intensity of pressure at the base of the tank.

P=wh

= 7848X1.3

= 10202.4 N/m2

=10.202.4

103

= 10.202 KN/m2 ∴ one pascal =1N/m2

P= 10.202 KPa i.e. Pa=1 N/m2

2.4. Total pressure on a vertically immersed surface

Fig.2 Vertically immersed surface

Consider a plane vertically surface immersed in a liquid as shown in fig.2 let divide the whole

area a number of small parallel strips as shown in fig.2.

w-specific weight of liquid

A-total area of the immersed surface, and

x – depth of centre of gravity of the immersed surface from the liquid surface.

Let us consider a strip of thickness dx, width b and at a depth x from the free surface of the liquid

as show in fig.

Intensity of pressure on the strip and area of the strip.

Intensity of pressure = wx

Area of strip = b.dx

∴ pressure on the strip

P= intensity of pressure X area

= wx.bdx

Total pressure on the strip

P= ∫wx, bdx

P= ∫wx, bdx

∫x, bdx= moment of the surface area about the liquid level / surface

= Ax

P= W Ax

Problems (1) A circular door of 1m diameter closes on opening in the vertical side of a

bulkhead, which retains sea water. If the centre of the opening is at a depth of 2m from the water

level, determine the total pressure on the door. Take specific grarity of sea water as 1.03.

Solution

Given:

Door dia=1.0m.

x=2m.

Specific gravity of sea water, G seawater =1.03

Specific wt of seawater γ seawater =9810 x 1.03

= 10104.3 N/m3

W.K.T

Area of circular door A= π d2

4

= π X 12

4

= 0.7854m2

Total pressure on the door; P= WAx

= 10104.3 x 0.7854 x 2

= 15871.83 N/m2

= 15871.83

103

= 15.87 KN/m2

P = 15.87 KPa

2.5. Total pressure an on inclined immersed surface

Consider a plane include surface, immersed in a liquid as shown in fig.

Fig.3: Inclined immersed surface

Let us divide the whole area into a number of small || le strips as shown in fig3.

w-specific weight of the liquid

A- Area of the surface.

x- Depth of centre of gravity of the immersed surface from the liquid surface.

θ- Angle at which the immersed surface is inclined with the liquid surface.

Let us consider a strip of thickness, dx, width b and its, distance l from o (A point on the liquid

surface where the immersed surface will meet, if produced).

W.K.T

Intensity of pressure on the strip

= wt sin θ

Area of strip = b.dx

Pressure on the strip, P = intensity of pressure X Area

= W l sin θ. Bdx

Now total pressure on the surface

P = ∫ w l sinθ.bdx

= w sin θ ∫ l.bdx

∫ l .bdx = moment of the surface area about O.

= A x

sin θ

P = w sin θ x A x

sin θ

P = WAx

Problems (1) A rectangular plate 2m X 3m is immersed in water in such a way that its greatest

and least depth are 6m and 4m respectively from the water surface.

Solution

Given:

Site of the plate = 2m X 5m and greatest and least depth of the plate= 6m and 4m.

W.K.T

A = 2X3

= 6m2

Depth of centre of grarity

Fig.4

x =6+4

2

x =5m

Total pressure on the plate, P= WAx

= 9810 X 6X 5

= 294300 N/m2

= 294.3 KN/m2

(2) A horizontal passage 140 mm X 1400mm has its out let covered by a plane flap inclined at

600 with the horizontal and is hinged along the upper horizontal edge of the passage. If the depth

of the flowing water in 500mm in the passage, determine the thrust on the gate.

Solution

Given: width of passage =1400mm= 1.4 m

Depth of passage =1400mm=1.4m.inclination of flap=600

Depth of water = 500mm

= 0.5m

W.K.T area of flap

A= 1.4 X 0.5 coses 600

A= 0.808 m2

Depth of the centre of the wetted flap.

x= 0.52

= 0.25m.

Thrust on the gate

P = WAx

= 9810X0.808X0.25

= 1981.62 N/m2

= 1.98 KN/m2

2.6. Centre of pressure

The intensity of pressure on an immersed surface is not uniform, but increase with depth. As the

pressure is greater over the lower portion of the figures, therefore the resultant pressure, on an

immersed surface, will act at some point below the centre of gravity of the immersed surface and

towards the lower edge of the figure. The point, through which this resultant pressure. Acts, is

known as centre of pressure and is always expressed interns depth from the liquid surface.

2.7. Centre of pressure of a vertically immersed surface:

Considered a plane surface immersed vertically in a liquid as shown in fig.5.

Let us divide the whole area into a number of small 11th strips as shown in fig.5.

Fig.5: Vertically immersed surface

w- Specific weight of the liquid.

A- Area of the immersed surface

x- Depth of centre of gravity of the immersed surface from the liquid surface.

Let us consider a strip of thickness dx, width b and at a depth of x from the free surface of the

liquid as shown.

W.K.T

Intensity of pressure on strip

= wx

Area of the strip = bdx

∴ presure on the strip, p= intensity of pressure X area

= wx.bdx

Moment of this pressure about the liquid surface

= (wx.bdx)x

= wx2.bdx

Now some of moments of all such pressures about the liquid surface.

M= ∫wx2 .bdx

M= ∫wx2 .bdx ∴ ∫wx2. b d x=Io

Where

Io- moment of inertia of the surface about

the liquid level or second moment of area).

M= WIo------ (1)

W.K.T

Sum of the moments of the pressures =ph ----- (2)

Where

P- Total pressure on the surface and

h- Depth of centre of pressure from the liquid surface

Now equating (1) & (2) | P= WAx

W. Io= ph

W. Io= W A x h

W A x h= W. Io.

h=WIoWA x

h= IoA x

according to || le axis theorem

Io=IG +Ah2

IG – moment of inertia of the figure, about horizontal axis through its center of gravity and

h- distance between the liquid surface and the center of gravity of the figure (x in this case).

Now recurring the equation (iii)

h= I G+ Ax ̅� 2

A x

h= I GA x

+x

Note: the centre of pressure is always below the center of gravity of the area by a distance

equal to IGAX

.

Problems

(1) A circular gate of 2 m diameter is immersed vertically in an oil of specific gravity 0.84 as

shown in fig.

Find the oil pressure on the gate and position of the centre of pressure on the gate.

Solution

Given

Diameter, d=2 m

Specific gravity of 0,7 ,G oil =0.84

Specific weight of oil, γ oil =0.84 X 9.81

= 8.24 KN\m3

x=3.0m

presure on the gate; A =π X d2❑

4

= π X 22

4

= 3.142 m2

p=WAx

= 8.24 X3.142 X3

= 77.67 KN

2.8. Position of the center of pressure

Moment of inertia for circular plate is

IG =π d 4

64

= π d 4

64

IG =0.7854m4

And depth of centre of pressure from the surface

h=I G

A X+ x

h=0.7854

3.142 X 3+ 3

h= 3.08 m

(2) an isosceles ∆ le plate of base 3 meters and attitude 3 meters is immersed vertically in

water as shown in the pressure of the plate.

Given data

Base width; b= 3m

Attitude h=3m

Total pressure on the plate

W.K.T

Surface area of the ∆ lar plate.

A= b h2

= 3 X 3

2 =4.5m2

Depth of C.G. of the plate from the water surface.

x =3/3=1m.

Total pressure on the plate

P= WAx

= 9.81 x 4.5 x 1

= 44.1 KN

Centre of pressure

IG= b h3

36

= 3 X 33

36

= 2.25 m4

and depth of centre of pressure from the water surface

h= I G

A x +x

h= 2.25

4.5 X 1 +1

h= 1.5 m

(3) An isosceles ∆ le of base 3 minutes, and attitude 6 meters is immersed vertically in water,

with its axis of symmetry horizontal as shown in fig. if head of water on its axis is 9

meters. Calculate the total pressure on the plate. Also locate the centre of pressure both

vertically 4 laterally.

Given data

Split the triangle into 2 right angled triangles.

Base width of right angled ∆ le =1.5m.

Altitude of the plate=6m

And, x = 9.0m.

Total pressure on the plate.

A=12

b. h

= 12

X 3 X6=9.0 m2

P= WAx

= 9.81X 9X9

P= 794.6 KN

W.K.T moment of inertia of triangle ABD about AD

= b h3

12

= 6 X (15)3

12

= 1.6875 m4

Similarly, moment of inertia of triangle

ADC about AD.

= 1.6875 m4

Moment of inertia of the triangle ABC about AD.

IG = 1.6875+1.6875

= 3.375m4

Depth of centre of pressure of the plate from the water level

h=I G

A x+x

=3.3759 X 9

+9

= 9.04m.

2.9. Summary

In this we have studied

Centre of pressure

Centre of pressure of a vertically immersed surface

Position of the center of pressure

Centre of pressure

2.10. Keywords

Total pressure

Centre of pressure

Vertically immersed surface

2.11. Exercise

1. What do you mean by Total pressure?

2. Explain total pressure on an immersed surface.

3. Explain pressure on a curved surface.

References

1. J. T. Bottomley, Hydrostatics (London: William Collins, 1882). 2. S. L. Loney, Elements of Hydrostatics (Cambridge: Cambridge Univ. Press, 1956) 2nd

ed. (1904).