1

Done by: ahmed Majid al.Amri

Higher College of Technology

Engineering Department

Civil and architecture Section

2

UNIT -1

FLUID PROPERTIES

3

FLUID PROPERTIES

Fluid Mechanics: the branch of science dealing with the fluids and their behaviours

Properties of Fluid

1. Density (Mass Density): the ratio of the material mass to its volume

V

m

The density of water is 1000 kg / m3

2. Specific Weight (Weight Density)

w = ρ g

Specific weight of water = 9810 N / m3

3. Specific Volume

.

1v

4. Specific Gravity: the ratio of the specific gravity of the fluid to the specific gravity of

water (unit less)

Specific gravity = specific weight of liquid = specific mass of liquid

Specific weight of pure water Specific mass of pure water

5. Viscosity: the property in which we measure the flow resistance

dy

du.

Where, μ = constant of proportionality and is called as coefficient of dynamic viscosity, is

the shear stress.

4

Unit of dynamic viscosity: (N. s / m2)

1 poise = 10

1N.s / m

2

Kinetic viscosity: The ratio between the dynamic viscosity and density of fluid

ν = μ

ρ

Unit of kinetic viscosity: (m2/s)

1 stoke = 1 * 10-4

Newton’s Law of viscosity

Shear stress in a fluid is directly proportional to the shear strain

Types of Fluids:

1. Ideal Fluid: The fluid which has no viscosity ( water )

2. Real Fluid: The fluid which has some viscosity.

3. Newtonian Fluid: The fluid which follows Newton’s law of viscosity (Water,

Kerosene, air).

4. Non Newtonian Fluid: The fluid which does not follow Newton’s law of viscosity

(Polymer solutions, blood).

Newtonian Fluid

Non - Newtonian Fluid

Velocity gradient dy

du

Shear stress

5

Exercise 1.1:

A commonly used equation for determining the volume rate of flow, Q, of a liquid through an

orifice located in the side of a tank is Q = 0.61 A √(2gh),where A is the area of the orifice, g

is the acceleration of gravity, and h is the height of the liquid above the orifice. Investigate the

dimensional homogeneity of this formula.

Solution:

Q= A√ (2gh)

Q= L2√ (T-2L2)

Q= L2.T-1.L

Q= L3 T-1

Exercise 1.2:

Determine the mass density, specific volume and specific weight of a liquid whose specific

gravity is 0.85.

Solution:

Specific gravity = specific mass of liquid

Specific mass of pure water

0.085= specific mass of liquid

1000

1) Mass density ( specific mass )= 0.085 *1000= 850 kg/m3

2) Specific mass= density * g = 850 * 9.81= 8338.5 N/m3

3) Specific Volume =

1v = 1\ ( 850 )= 1.17*10

-3 m

3/kg

6

Exercise 1.3

Determine the viscosity of a liquid having kinematic viscosity 6 stokes and specific gravity

1.9.

Solution:

Specific gravity = specific mass of liquid

Specific mass of pure water

Specific mass of a liquid= specific gravity * Specific mass of water

= 1000 * 1.9 = 1900 kg/m3

ν = μ

ρ

µ= v * p = 6*10-4 * 1900

= 1.1388 Ns/m2

Exercise 1.4

Determine the specific gravity of a fluid having dynamic viscosity 0.05 poise and kinematic

viscosity 0.035 stokes.

Solution:

µ= 0.05 poise= 1\ (0.05) Ns/m2= 5*10

-3 Ns/m

2

p= 0.035 stokes= 0.035*10-4

m2\s

ν = μ

ρ

p =

=

10*0.035

3-10*54-

= 1428.57 kg/m3

Specific gravity = specific mass of liquid = 1428.57

Specific mass of pure water 1000

= 1.428

7

Exercise 1.5:

Two horizontal plates are placed 1.25 cm apart, the space between them being filled with oil

of viscosity 14 poises. Calculate the shear stress in oil if upper plate is moved with a velocity

of 2.5 m/sec.

Solution:

dy

du.

du (velocity) = 2.5 m/sec

dy (thickness) = 1.25 cm= 1.25*10-2

m

µ= 14 poises= (14/10) Ns/m2 = 1.4 Ns/m

2

2- 10*1.25

5.2*4.1

2/280 mN

Exercise 1.6

A liquid has a specific gravity of 1.9 and kinematic viscosity of 9 stokes. What is its dynamic

viscosity?

Solution:

Density (p) = specific gravity * 1000 = 1.9 *1000= 1900 Kg/m3

ν = 9 stokes = 9 * 10-4

m2\s

ν = μ

ρ

µ= v * p = 9*10-4 * 1900

= 1.71 Ns/m2

8

Exercise 1.7

The space between two parallel plates 5 mm apart is filled with crude oil of specific gravity

0.90. A force of 2 N is required to drag the upper plate at a constant velocity of 0.8 m/s. The

lower plate is stationary. The area of the upper plate is 0.09 m2. Determine the dynamic

viscosity and Kinematic viscosity of the oil.

Solution:

dy (thickness) = 5mm = 5*10-3

, specific gravity= 0.9

F= 2N, du (velocity) =0.8

A=0.09 m2, µ=? ν=???

dy

du.

= 3-10*5

8.0.

09.0

2

Exercise 1.8:

A large movable plate is located between two large fixed plates as shown in figure below.

Two Newtonian fluids having the viscosities indicated are contained between the plates.

Determine the magnitude and direction of the shearing stresses that act on the fixed walls

when the moving plate has a velocity of 4 m/s as shown. Assume that the velocity distribution

between the plates is linear

Solution:

dy

du.1

31106

402.0

= 13.33 N/m

2

dy

du.2

32103

401.0

= 13.33 N/m

2

dy

du

A

F.

23

/1388.009.08.0

)105(2

.

.mNs

duA

dyF

9

Exercise 1.9

A plate has an area of 1 m2. It slides down an inclined plane, having angle of inclination 45

0

to the horizontal, with a velocity of 0.5 m/s. The thickness of oil film between the plane and

the plate is 1mm. Find the viscosity of the fluid if the weight of the plate is 70.72 N.

Solution:

A= 1 m2

Θ= 45

du= 0.5 m/s

dy= 1*10-3

mm

F= 70.72

dy

du

A

F.

dy

du

A

F.

sin

101

5.0.

1

45sin72.703-

2

3

/1.0

5.01

)101(45sin72.70

.

.sin

mNs

duA

dyF

10

Unit 1

PROPERTIES OF FLUIDS

Problem Sheet:

1. Calculate the specific weight, specific mass, specific volume and specific gravity of a

liquid having a volume of 6m3 and weight of 44 KN.

[Ans: 7.333kN/m3, 747.5 kg/m

3, 0.00134 m

3/kg, 0.747]

Solution:

Specific weight = weight = 44 = 7.333kN/m3

Volume 6

Specific mass = specific weight = 7.333 * 10 3

= 747.5 kg/m3

g 9.81

Specific volume = 1 = 1 = 0.00134 m3

p 747.5

Specific gravity = specific mass of liquid = 747.5 = 0.747

Specific mass of water 1000

2. A plate 0.05 mm distant from a fixed plate moves at 1.2 m/s and requires a force of 2.2 N

/ m2 to maintain this speed. Find the viscosity of the fluid between the plates.

[Ans: 9.16 x 10-4

poise]

Solution:

dy

du.

dy

du

A

F.

poisemNs

duA

dyF

425

3

1016.9/1016.9

2.11

)1005.0(2.2

.

.

11

3. A flat thin plate is dragged at a constant velocity of 4 m/s on the top of a 5 mm deep

liquid layer of viscosity 20 centipoises. If the area of the plate is 1 m2, find the drag force.

Assume variation of velocity in the liquid to be linear.

[Ans: 16N]

Solution:

= 16 N

4. A plate having an area of 0.6 m2 is sliding down the inclined plane at 30

0 to the horizontal

with a velocity of 0.36 m/s. There is a cushion of fluid 1.8 mm thick between the plane

and the plate. Find the viscosity of the fluid if the weight of the plate is 280 N.

[Ans: 11.66 poise]

Solution:

5. A flat plate weighing 0.45 KN has a surface area of 0.1 m2. It slides down an inclined

plane at 300 to horizontal, at a constant speed of 3 m/s. If the inclined plate is lubricated

with an oil of viscosity 0.1 N.s / m2, find the thickness of oil film.

[Ans: 0.133mm]

Solution:

dy

du

A

F.

dy

duAF

..

3

3

105

41.1020

F

dy

du

A

F.

sin

dy

du

A

F.

poisemNsduA

dyF667.11/1667.1

36.06.0

)108.1(30sin280

.

.sin 23

dy

du

A

F.

mmm

F

duAdy

133.01033.130sin1045.0

1.031.0

..

4

3

12

6. Two large fixed parallel planes are 240 mm apart. The space between the surfaces is filled

with oil of viscosity 0.81 N s / m2. A flat thin plate 0.5 m

2 area moves through the oil at a

velocity of 0.6 m/s. Calculate the drag force,

(i) When the plate is equidistant from both planes and

(ii) When the thin plate is at a distance of 80 mm from one of the plane

surfaces.

[Ans: (i) 4.05N (ii) 4.56N]

Solution:

(a) In middle: ( 120 mm First plane, 120 mm second plane )

F= F1+F2

F= F1+F2= 2.025+2.025= 4.05 N

(b) (80 mm First plane, 160 mm second plane)

F= F1+F2

F= F1+F2= 3.0375+1.51875= 4.55625 N

dy

du

A

F.

NF 025.210120

6.05.081.0

31

NF 025.210120

6.05.081.0

32

dy

du

A

F.

NF 0375.31080

6.05.081.0

31

NF 51875.110160

6.05.081.0

32

13

7. The space between two square flat parallel plates is filled with oil. Each side of the plate

is 720 mm. The thickness of oil film is 15 mm. The upper plate, which moves at 3 m/s,

requires a force of 120 N to maintain the speed. Determine.

a. The dynamic viscosity of oil;

b. The kinematic viscosity of oil, if the specific gravity of oil is 0.95

[Ans (i) 1.16 N.s / m2 (ii) 0.00122 m

2/s]

Solution:

(a)

(b)

sm /1021.1

100095.0

157.1

23

dy

du

A

F.

25

33

3

/157.1

)1072010720(3

)1015(120

.

.

mNs

duA

dyF

14

8. The velocity distribution for flow over a plate is given by u = 2y – y2, where u is the

velocity in m/s at a distance y meters above the plate. Determine the velocity gradient and

shear stress at the boundary and 0.15 m from it. Take viscosity of fluid as 0.9 N.s / m2.

[Ans: Vel. Gradient 2/s and 1.7 / s, Shear Stress 1.8N/m2 and 1.53 N/m

2]

Solution:

u = 2y – y2

9. The velocity distribution over a plate is given by2

2

1

2

3yyu , where u = velocity in

m/s and y = distance from the plate boundary. If the viscosity of the fluid is 8 poise, find

the shear stress at the plate boundary and at y = 0.15 m from the plate.

[Ans: 1.20 N/m2, 1.08 N/m

2]

Solution:

2

2

1

2

3yyu

ydy

du22

sdy

du

boundry

/2)0(22)(

sdy

du

At

/7.1)15.0(22)15.0(

dy

du.

2

)( /8.129.0 mNboundry

2

)15.0( /53.17.19.0 mNAt

ydy

du

2

3

dy

du.

2

)( /2.15.18.0 mNboundry

2

)15.0( /08.135.18.0 mNAt

sdy

du

boundry

/2

30

2

3

)(

sdy

du

At

/35.115.02

3

)15.0(

15

UNIT - 2

Pressure Measurement

16

PRESSURE MEASUREMENT

Atmospheric Pressure (Parametric): it is happening when the atmospheric air comes in

contact with all surfaces and it is equal to 10.33 m of water, 101.3 KN, or 760mm of mercury.

Gauge Pressure: The pressure measured by using the atmospheric pressure as datum.

Absolute Pressure: The pressure measured using the absolute zero as datum

Vacuum Pressure: The pressure below the atmospheric pressure.

The relationship between the absolute pressure and vacuum pressure are as follows:

Absolute pressure= (Atmospheric pressure + Gauge pressure)

Vacuum Pressure= (Atmospheric pressure –Absolute pressure)

17

Manometers: Devices used for measuring the pressure at a point in a fluid by balancing the

column of fluid by the same or another column of a fluid. These are classified as follows.

1. Simple manometers.

2. Differential manometers.

3.

1. Simple Manometers: A simple manometer is one which consists of a glass tube whose

one end is connected to a point where pressure is to be measured and the other end remains

open to atmosphere.

Common types of simple manometers are discussed below.

(a) Piezometer: A device used to

measure the pressure in a

particular point by using a

glass tube which has two sides,

one is opened to the to the

atmosphere and the other side

is placed where the pressure is

going to be measured.

(b) U - tube manometer: A glass

tube in the form of “ U” shape

, one side is opened to the

atmosphere and one is placed

where the pressure is going to be

measured.

Common

level

18

4. Differential Manometers : A device used to determine the difference in pressure

between two different points by using a manometric fluid

1. U - tube differential manometer: (high pressure)

2. Inverted U – tube differential manometer: (low pressure)

Equations used:

( a ) U-Tube/bizometer ( equalling left side with right side )

(PA) + ρ1.g.h1= ρ2g.h2

Heavy

Liquid

Light Liquid

U-tube Differential Manometer Inverted U-tube Differential

Manometer

19

( b ) Differntial manometer ( equalling left side with right side )

(PA) + ρ1.g.h1= (PB) + ρ2g.h2 + ρ3g.h3

( c ) Inverted Differntial manometer ( equalling left side

with right side )

(PA) - ρ1.g.h1= (PB) - ρ2g.h2 - ρ3g.h3

20

How to write down the equation for any shape:

( equalling left side with right side )

21

Exercises 2.1:

Calculate the pressure due to a column of 0.3 of (a) water (b) an oil of specific gravity 0.8(c)

mercury of specific gravity 13.6.

Solution:

(a)

P= w * h = 9810 * 0.3= 2943 N/m2

(b)

P= w*h = 9810 * 0.8 * 0.3= 2354.4 N/m2

(c)

P= w*h = 9810 * 13.6 * 0.3= 40024.8 N/m2

Exercise 2.2:

The pressure intensity at a point in a fluid is given 3.924 N/cm2.Find the corresponding

height of fluid when the fluid is (a) water (b) oil of specific gravity 0.9.

Solution:

P= 3.924 N/cm2= 3.924*10

4 N/m

2

(a)

P = w * h

(b)

P = w * h

mw

Ph

water

49810

10*3.924 4

mw

Ph

oil

44.498108.0

10*3.924 4

22

Exercise 2.3:

What are the gauge pressure and absolute pressure at a point 3 m below the free surface of a

liquid having a density of 1.53 x 103 kg/m3, if the atmospheric pressure is equivalent to 750

mm of mercury? The specific gravity of mercury is 13.6 and density of water =1000 kg/m3.

Solution:

Gauge pressure = specific weight * g

= w.g

= 1.53*103 * 9.81 * 3

= 45027.9 N/m2

Abslloute pressure= gauge pressure + atmospheric pressure

= 45027.9 + (13.6 * 0.75 * 9810)

= 45027.9 + 100062

= 145089.9 N/m2

23

Exercise 2.4

A U tube manometer is used to measure the pressure of oil of specific gravity 0.85 flowing in

a pipe line. Its left end is connected to the pipe and right limb is open to the atmosphere. The

centre of the pipe is 100 mm below the level of mercury (specific gravity 13.6) in the right

limb. If the difference of mercury level in the two limbs is 160 mm, determine the absolute

pressure of the oil in the pipe.

Solution:

Exercise 2.5:

AU tube manometer containing mercury was used to find the negative pressure in the pipe,

contain water. The right limb was open to atmosphere. Find the vacuum pressure in the pipe,

if the difference of mercury level in the two limbs was 100 mm and height of water in the left

limb from the centre of the pipe was found to be 40 mm below.

Solution:

h1

100 m

m 160 mm

X X

2211 )( hghgPh

16081.910006.13))100160(81.9100085.0( hP

16081.910006.13))100160(81.9100085.0( hP

2/84625.20 mNPh

0)( 2211 hghgP

)10081.910006.13()4081.91000( hP

26 /10734.13 mNPh

24

Exercise 2.6

A differential manometer is connected at the two points A and B in a pipe containing an oil of

specific gravity 0.9. The manometer is filled with mercury (Sp. Gravity 13.6). The difference

in mercury levels in the two limbs is 150 mm. Find the difference in pressure at the two

points.

Solution:

332211 )()( hgPhghgP BA

)81.910009.0()15.081.910006.13())15.0(81.910009.0( hPhP BA

BA PhhP )81.910009.0()15.081.910006.13())15.0(81.910009.0(

)15.081.910006.13()15.081.910009.0( BA PP

2/05.18688 mNPP BA

25

Exercise 2.7

Figure shows a U tube differential manometer connecting two pressure pipes at A and B. The

pipe A contains a liquid of specific gravity 1.6 under a pressure of 110 KN/m2. The pipe B

contains oil of specific gravity 0.8 under a pressure of 200 KN/m2. Find the difference ‘h

‘between the levels of Mercury

Solution:

A

B

)()()( 332211 hgPhghgP BA

))1(81.910008.0(10200)81.910006.13()6.381.910006.1(10110 33 hh

)81.910006.13()81.910008.0()181.910008.0(10200)6.381.910006.1(10110 33 hh

h1255684.41342

m

h

329.0

125568

4.41342

26

Exercise 2.8

Figure shows an inverted differential manometer having an oil of specific gravity 0.80

connected to two different pipes carrying water under pressure. Determine the pressure in the

pipe B. The pressure in pipe A is 2.0 m of water.

Solution:

A

B

300 m

m

100 m

m150 m

m

)()()( 332211 hghgPhgP BA

)1.081.91000()15.081.910008.0()3.081.91000(29810 BP

BP )1.081.91000()15.081.910008.0()3.081.91000(29810

2/2.18835 mNPB

27

Unit 2

MEASUREMENT OF PRESSURE

Problem Sheet:

1. Find the depth of a point below water surface in sea where the pressure intensity is 100.55

KN/m2. Specific gravity of sea water is 1.025.

[Ans: 10.m]

Solution:

P = w * h

2. Convert a pressure head of 100 m of water to

(i) Kerosene of specific gravity 0.81 and

(ii) Carbon tetra chloride of specific gravity 1.6

[Ans: (i) 123.4 (ii) 62.5 m]

Solution:

(i)

(ii)

mw

Ph

seawater

109810025.1

10*100.55 3

1009810981081.0 h

mh 5.626.1

100

100981098106.1 h

mh 456.12381.0

100

28

x x

A

B

h1.5

m2.5

m

3. As shown in fig 2P.1, pipe A contains carbon tetrachloride of specific gravity 1.594,

under a pressure of 103 kN/m2 and pipe B contains oil of specific

Fig 2P.1

Gravity 0.8. If the pressure in the pipe B is 171.6 KN/m2, and the manometric fluid is

mercury, find the difference h between the levels of mercury.

[Ans: 142 mm]

Solution:

)()()( 332211 hgPhghgP BA

))5.1(81.910008.0(106.171)81.910006.13()481.91000594.1(10103 33 hh

)81.910006.13()98108.0(106.171)5.198108.0()49810594.1(10103 33 hh

h12556844.17823

m

h

145.0

125568

44.17823

29

4. The pressure of water in a pipe line was measured by means of a simple manometer

containing mercury. The reading of the manometer is shown in fig 2P.2. Determine the

pressure in pipe. [Ans : 105.75 KPa]

x x

B

100 m

m700 m

m

x x

0.5

m

1.0

m

Fig 2P.2 Fig 2P.3

Solution:

5. A U tube containing mercury is used to measure the pressure of an oil of specific gravity

0.8 as shown in fig 2P.3. Calculate the pressure head of the oil.

[Ans: 14 m]

Solution:

P = w * h

6. A simple U – tube manometer

2211 )( hghgPB

)( 1122 hghgPB

)1.09810(8.081.910006.13 BP

)1.09810(8.081.910006.13 BP

2/75.105 mKNPB

2211 )( hghgPA

)( 1122 hghgPA 2/137340)198106.13(5.098108.0 mNPA

)(14)(5.178.09810

137340

6.1

100watermoilmh

30

containing mercury is connected to a pipe in which an oil of specific gravity 0.8 is

flowing. The pressure in pipe is vacuum. The right end of the manometer is open to

atmosphere. Find the vacuum pressure in the pipe, if the difference of mercury level in

the two limbs is 200 mm and the height of oil in the left limb from the centre of the pipe

is 150 mm below.

[Ans: -27.86 KPa]

0)( 2211 hghgP

2211 )( hghgPA

2/4.27860 mNPA

)2.081.910006.13()15.098108.0( AP

31

7. Calculate the pressure difference between two points A and B shown in fig 2P.4

[Ans : 13.83 kN/m2]

B

A

x xx x

A

B

y0.6

m

1.5

m

1m

0.7

m0.8

m

Oil

(S= 0.85)

Water

Water Water

Water

Air

Fig 2P.4Fig 2P.5

Solution:

8. Find the difference in pressure between points A and B in fig 2P.5. Neglect the weight of

the air. [Ans : 17.66 KN / m2]

Solution:

332211 )()( hgPhghgP BA

))9.0(9810()6.081.9100085.0()9810( yPyP BA

)9.09810()9810()6.081.9100085.0()9810( yPyP BA

)9.09810()6.081.9100085.0( BA PP

2/1.13832 mNPP BA

2211 )( hgPhgP AB

5.29810)7.09810( AB PP

)7.09810(5.29810 AB PP

2/17658 mNPP AB

32

UNIT -3

HYDROSTATIC FORCES ON

SURFACES

33

Definitions to be memorized:

1) Hydrostatics: Studying the pressure exerted by the fluid at rest

2) Pressure force: The force exerted by the fluid on the surface when the fluid comes in

contact with the surface and that fore is always normal to the surface

3) Centre of pressure: The point of application of the total pressure acting on the

surface

Rules to be used while solving the Questions:

(a) Horizontal\Vertical:

1)

2)

3)

(b) Inclined:

1)

2)

3)

depthedgeupperx 2

1)(

xA

Ixh

g

.

xAwF ..

sin2

1)( depthedgeupperx

xA

Ixh

g

.

sin. 2

Diameter

depthleastdepthgreatestcircle

)()()(sin

34

Exercise 3.1

A rectangular plane surface is 2 m wide and 3 m deep. It lies in a vertical plane in water.

Determine the total pressure and position of centre of pressure on the plane surface when its

upper edge is horizontal and (a) coincide with water surface (b) 2.5 m below the free water

surface.

Solution:

(a) (b)

Exercise 3.2

An isosceles triangular plate of base 3 m and altitude 3 m is immersed vertically in an oil of

specific gravity 0.80. The base of the plate coincides with free surface of oil. Determine (a)

Total pressure o the plate and (b) centre of pressure.

Solution:

For Triangle (instead of using half the depth, we will use the centre of gravity)

Exercise 3.3:

depthedgeupperx 2

1)(

5.132

10 x

xAwF ..

NF 882905.1329810

xA

Ixh

g

.

mh 25.16

12

)3(2

5.1

3

depthedgeupperx 2

1)(

432

15.2 x

xAwF ..

NF 2354404329810

xA

Ixh

g

.

mh 1875.446

12

)3(2

4

3

depthedgeupperx 2

1)(

1)(3

10 basehx

xAwF ..

NF 353161332

198108.0

xA

Ixh

g

.

mh 5.115.4

36

)3(3

1

3

35

A circular plate of diameter 1.2 m is placed in water in such a way that the centre of the plate

is 2.5 m below the free surface of water. Determine (a) Total pressure on the plate and (b)

Position of the centre of pressure.

Solution:

Exercise 3.4

A rectangular plane surface 2 m wide and 3 m deep lies in a water in such a way that it s

plane makes an angle of 300 with the free surface of water. Determine the total pressure and

position of centre of pressure when the upper edge is 1.5 m below the free water surface.

Solution:

Exercise 3.5

xAwF ..

NF 1324355.2)6.0(9810 2

xA

Ixh

g

.

depthedgeupperx 2

1)(

)(5.2 tergiventocenx

mh 528.25.2)6.0(

64

)2.1(

5.22

4

sin2

1)( depthedgeupperx

xA

Ixh

g

.

sin. 2

mx 25.230sin32

15.1

xAwF ..

NF 13243525.269810

mh 33.25.13

125.125.2

36

A circular plate 3.0 m diameter is submerged in water in such a way that it’s greatest and

least depths below the free surface are 4 m and 1.5 m respectively. Determine

(a) The total pressure on one face of the plate, and

(b) The position of the centre of pressure

Solution:

(a)

Unit-3

sin2

1)( depthedgeupperx

mx 75.2833.032

15.1

Diameter

depthleastdepthgreatestcircle

)()()(sin

833.02

5.14)(sin

circle

xA

Ixh

g

.

sin. 2

mh 953.2437.19

693889.097.375.2

xAwF ..

NF 71.190692)5.1(981075.2 2

37

Hydro static forces on surfaces

Problem sheet

1. A rectangular plate 2m × 4m is vertically immersed in water in such a way that 2 meter

side is parallel to the water surface and 2.50 meters below it. Find the total pressure on the

rectangular plate and centre of pressure

(Answer: 353.16 KN, 4.80m)

2. An isosceles triangular plate of base 5m and altitude 5m is immersed vertically in an oil of

specific gravity 0.8m. The base of plate is 1m below the free water Surface Determine

i) The total pressure

ii) The centre of pressure (Answer: 261.93 KN, 3.19m)

3. A circular plate of 1m diameter is immersed in water in such a way that its plane makes an

angle of 30˚ with the horizontal and its top edge is 1.25 m below the water surface. Find the

total pressure on the plate and the point where it acts.

(Ans: 11.56KN, 1.51m)

4. A circular plate 3 meters in diameter is submerged in water in such a way that the

greatest and least depths of the surface below water surfaces are 2m and 1m respectively.

Calculate the total pressure and the position of centre of pressure.

(Ans: 104 KN, 1.54m)

.

5. A rectangular plane surface 1m wide and 3m deep lies in water in such a way that its plane

makes an angle of 30˚ with the free water surface. Determine the total pressure and position

of centre of pressure when the upper edge is 2m below the free surface.

(Ans: 80.93 KN, 2.82 m)

6. A triangular gate which has a base of 1.50 m and an altitude of 2m lies in a vertical plane.

The vertex of the gate is 1m below the oil surface which has specific gravity of 0.80. Find the

force exerted by the oil on the gate and the position of the centre of pressure.

(Ans: 27.43 KN, 2.42m)

7. A square disc of 1m side is immersed vertically in water as shown in figure below.

38

If the highest corner of the disc is at a depth of 1.50m below the water surface, find the total

pressure and the depth of the centre of pressure.

(Ans: 21.65KN, 2.245m)

1.50M

OG G

X

A

B

C

D