INTRODUCTION

Mechanics: Deals with action of forces on bodies at rest or in motion.

State of rest and Motion: They are relative and depend on the frame of

reference. If the position with reference to frame of reference is fixed with

time, then the body is said to be in a state of rest. Otherwise, it is said to

be in a state of motion.

Scalar and heater quantities: Quantities which require only magnitude

to represent them are called scalar quantities. Quantities which acquire

magnitudes and direction to represent them are called vector quantities.

Eg: Mass, time internal, Distance traveled Scalars

Weight, Displacement, Velocity Vectors.

Displacement and Distance.Unit: m

Velocity and Speed: Rate of displacement is called velocity and Rate

and distance traveled is called Speed.

Unit: m/s

Acceleration: Rate of change of velocity is called acceleration. Negative

acceleration is called retardation.

Momentum: The capacity of a body to impart motion to other bodies is

called momentum.

The momentum of a moving body is measured by the product of mass

and velocity the moving body.

Momentum = Mass x Velocity

Unit: Kg m/s

Newton’s first law of motion: Every body continues to be in its state of

rest or uniform motion unless compelled by an external agency.

Inertia: It is the inherent property the body to retain its state of rest or

uniform motion.

Force: It is an external agency which overcomes or tends to overcome

the inertia of a body.

Newton’s second law of motion: The rate of change of momentum of a

body is directly proportional to the magnitudes of the applied force and

takes place in the direction of the applied force.

Measurement of force:

Unit: newton (N)

Mass: Measure of amount of matter contained by the body it is a scale of

quantity. Unit: Kg.

Weight: Gravitational force on the body. It is a vector quantity.

F = ma

W = mg

Unit: newton (N) g = 9.81 m/s2

Volume: Measure of space occupied by the body.

Unit: m3

1 m3 = 1000 litres

Work: Work done = Force x Displacement Linear motion.

Work done = Torques x Angular displacement Rotatory motion.

Unit: Nm or J

Energy: Capacity of doing work is called energy.

Unit: Nm or J Potential energy = mgh

Distinction between liquid and gas, fluid continuum

Liquids offer very little resistance against tensile force. Liquids offer maximum

resistance against compressive forces. Therefore, liquids are also called

incompressible fluids. Liquids undergo continuous or prolonged angular

deformation or shear strain when subjected to tangential force or shear force.

This property of the liquid is called flow of liquid. Any substance which exhibits

the property of flow is called fluid. Therefore, liquids are considered as fluids.

In case of gases intermolecular force is very small. Therefore the molecules are

free to move along any direction. Therefore gases will occupy or assume the

shape as well as the volume of the container.

Gases offer little resistance against compressive forces. Therefore gases are

called compressible fluids. When subjected to shear force gases undergo

continuous or prolonged angular deformation or shear strain. This property of

gas is called flow of gases. Any substance which exhibits the property of flow is

called fluid. Therefore gases are considered as fluids.

FLUID PROPERTIES AND CLASSIFICATION OF FLUID

Fluid mechanics deals with the behaviour of fluids, liquids or gases, at rest or in

motion. It explains the static, kinematics and dynamic aspects of fluids. Fluids

at rest are called fluid static, fluid in motion is known as fluid kinematics where

the pressure or force is called the fluids in motion.

Mass density or specific mass ()

Mass density or specific mass is the mass per unit volume of the fluid.

Unit: kg/m3 or kgm3

With the increase in temperature volume of fluid increases and hence mass

density decreases.

In case of fluids as the pressure increases volume decreases and hence mass

density increases.

Weight density or specific weight ()

Weight density or Specific weight of a fluid is the weight per unit volume.

Unit: N/m3 or Nm-3.

With increase in temperature volume increases and hence specific weight

decreases.

With increases in pressure volume decreases and hence specific weight

increases.

Relationship between mass density and weight density:

Specific gravity or relative density (S)

It is the ratio of specific weight of the fluid to the specific weight of a standard

fluid.

Unit: It is a dimensionless quantity and has no unit.

In case of liquids water at 4oC is considered as standard liquid.

Specific gravity or relative density of a fluid can also be defined as the ratio of

mass density of the fluid to mass density of the standard fluid. Mass density of

standard water is 1000 kg/m3.

Specific volume ()

( ) : It is the volume per unit mass of the fluid.

Unit: m3/kg

As the temperature increases volume increases and hence specific volume

increases. As the pressure increases volume decreases and hence specific

volume decreases.

Specific volume (): It is the volume per unit mass of the fluid

Calculate specific weight, mass density, specific volume and specific gravity of a

liquid having a volume of 4m3 and weighing 29.43 kN. Assume missing data

suitably.

Calculate specific weight, density, specific volume and specific gravity and if one

liter of Petrol weighs 6.867N.

Specific gravity of a liquid is 0.7

Find

i. Mass density

ii. specific weight.

iii. Also find the mass and weight of 10 Liters of liquid.

Vapour pressure

The process by which the molecules of the liquid go out of its surface in the

form of vapour is called Vapourisation.

There are two ways of causing Vapourisation.

By increasing the temperature of the liquid to its boiling points .

By reducing the pressure above the surface of the liquid to a value less than

Vapour pressure of the liquid.

As the pressure above the surface of the liquid is reduced, at some point, there

will be vapourisation of the liquid. If the reduction in pressure is continued

vapourisation will also continue. If the reduction in pressure is stopped,

vapourisation continues until vapours of the liquid exert certain pressure which

will just stop the vapourisation. This minimum partial pressure exerted by the

vapours of the liquid just to stop vapourisation is called Vapour Pressure of the

liquid.

If the pressure over the surface goes below the vapour pressure, then, there

will be vapourisation. But if the pressure above the surface is more than the

vapour pressure there will not be vapourisation unless there is heating.

1. In case of Hydraulic turbines sometimes pressure goes below the vapour

pressure of the liquid. This leads to vapourisation and formation of

bubbles of liquid. When bubbles are carried to high Pressure zone they

get busted leaving partial vacuum. Surrounding liquid enters this space

with very high velocity exerting large force on the part of the machinery.

This process is called cavitation. Turbines are designed such that there is

no cavitation.

2. In Carburetors and sprayers vapours of liquid are created by reducing the

pressure below vapour pressure of the liquid.

Unit of Vapour Pressure: N/m2 (pascal - Pa)

Vapour Pressure of a fluid increases with increase in temperature

Problem 1

A vertical cylinder 300mm in diameter is fitted at the top with a tight but

frictionless piston and filled with water at 700 C. The outer portion of the piston

is exposed to atmospheric Pressure of 101.3 kPa. Calculate the minimum force

applied on the piston that will cause water to boil at 700 C. Take Vapour

Pressure of water at 700 C as 32k Pa.

Viscosity

Viscosity is the property by virtue of which fluid offers resistance against the

flow or shear deformation. In other words, it is the reluctance of the fluid to

flow. Viscous force is that force of resistance offered by a layer of fluid for the

motion of another layer over it.

In case of liquids, viscosity is due to cohesive force between the molecules of

adjacent layers of liquid. In case of gases, molecular activity between adjacent

layers is the cause of viscosity.

Let us consider a liquid between the fixed plate and the movable plate at a

distance 'Y' apart , 'A' is the contact area (Wetted area) of the movable plate ,

'F' is the force required to move the plate with a velocity 'U' According to

Newton.

Velocity gradient or rate of shear strain is the difference in velocity per unit distance between any two layers.

If the velocity profile is linear then velocity gradient is given by . If the

velocity profile is non-linear then it is given by .

Unit of force (F): N

Unit of distance between the twp plates (Y): m

Unit of velocity (U): m/s

Unit of velocity gradient :

Unit of dynamic viscosity (): =

NOTE:

In CGS system unit of dynamic viscosity is and is called poise (P).

If the value of is given in poise, multiply it by 0.1 to get it in .

1 Centipoises = 10-2 Poise

Pressure has very little or no effect on the viscosity of fluids.

Effect of temperature on viscosity of fluids

1. Effect of temperature on viscosity of liquids: Viscosity of liquids is

due to cohesive force between the molecules of adjacent layers. As the

temperature increases cohesive force decreases and hence viscosity

decreases.

2. Effect of temperature on viscosity of gases: Viscosity of gases is due

to molecular activity between adjacent layers. As the temperature

increases molecular activity increases and hence viscosity increases

It is the ratio of dynamic viscosity of the fluid to its mass density.

NOTE:

Unit of kinematics Viscosity in CGS system is cm2/s and is called stoke (S)

If the value of KV is given in stoke, multiply it by 10-4 to convert it into m2/s.

Problem 1

Problem 2

A Plate at a distance 0.0254mm from a fixed plate moves at 0.61m/s and

requires a force of 1.962N/m2 area of plate. Determine dynamic viscosity of

liquid between the plates.

Problem 3

A plate having an area of 1m2 is dragged down an inclined plane at 450 to

horizontal with a velocity of 0.5m/s due to its own weight. Three is a cushion of

liquid 1mm thick between the inclined plane and the plate. If viscosity of oil is

0.1 PaS find the weight of the plate.

A=1m2

U= 0.5m/s

Y= 1x10-3m

= 0.1NS/m2

W= ?

F= W x cos 450

= W x 0.707

= 0.707W

Problem 4

A shaft of 20mm and mass 15kg slides vertically in a sleeve with a velocity of

5 m/s. The gap between the shaft and the sleeve is 0.1mm and is filled with oil.

Calculate the viscosity of oil if the length of the shaft is 500mm.

D = 20mm = 20x10-3m

M = 15 kg

W = 15 x 9.81

W = 147.15N

y = 0.1mm

y = 0.1 x 10-3mm

U = 5m/s

F = W

F = 147.15N

= ?

A = D L

A = x 20 x 10-3 x 0.5

A = 0.031 m2

Problem 5

If the equation of velocity profile over 2 plate is V= 2y2/3. in which 'V' is the velocity in m/s and 'y' is

the distance in 'm' Determine shear stress at (i) y = 0 (ii) y = 75mm. Take = 8.35P.

Problem 6

A circular disc of 0.3m dia and weight 50 N is kept on an inclined surface with a slope of 450. The

space between the disc and the surface is 2 mm and is filled with oil of dynamics viscosity . What

force will be required to pull the disk up the inclined plane with a velocity of 0.5m/s.

Problem 7

Dynamic viscosity of oil used for lubrication between a shaft and a sleeve is 6 P. The shaft is of

diameter 0.4 m and rotates at 190 rpm. Calculate the power lost in the bearing for a sleeve length of

0.09 m .Thickness of oil is 1.5 mm.

Problem 8

Two large surfaces are 2.5 cm apart. This space is filled with glycerin of absolute viscosity 0.82

NS/m2. Find what force is required to drag a plate of area 0.5m2 between the two surfaces at a speed

of 0.6m/s. (i) When the plate is equidistant from the surfaces, (ii) when the plate is at 1cm from one

of the surfaces.

Case (i)

Let F1 be the force required to overcome viscosity resistance of liquid above the plate and F2 be the

force required to overcome viscous resistance of liquid below the plate. In this case F1 = F2. Since the

liquid is same on either side or the plate is equidistant from the surfaces.

Total force required to drag the plate = F1 + F2 = 19.68 + 19.68

F = 39.36N

Case (ii)

Problem 9

Through a very narrow gap of ht a thin plate of large extent is pulled at a velocity `V'. On one side of

the plate is oil of viscosity 1 and on the other side there is oil of viscosity 2. Determine the position

of the plate for the following conditions.

i. Shear stress on the two sides of the plate is equal.

ii. The pull required, to drag the plate is minimum.

Conditions 1

Conditions 2

Surface tension ()

Surface tension is due to cohesion between the molecules of liquid and weak adhesion between the

molecules on the exposed surface of the liquid and molecules of air.

A molecule inside the surface gets attracted by equal forces from the surrounding molecules whereas

a molecule on the surface gets attracted by the molecule below it. Since there are no molecules above

it, if experiences an unbalanced vertically downward force. Due to this entire surface of the liquid

expose to air will have a tendency to move in ward and hence the surface will be under tension. The

property of the liquid surface to offer resistance against tension is called surface tension.

Liquid surface supports small loads.

Formation of spherical droplets of liquid

Formation of spherical bubbles of liquid

Formation of cylindrical jet of liquids.

Surface tension is measured as the force exerted by the film on a line of unit length on the surface of

the liquid. It can also be defined as the force required maintaining unit length of film in equilibrium.

Unit: N/m

Force due to surface tension = x length of film

NOTE:

Force experienced by a curved surface due to radial pressure is given by the product of intensity of

pressure and projected area of the curved surface.

To derive an expression for the pressure inside the droplet of a liquid

Let us consider droplet of liquid of surface tension ''. 'D' is the diameter of the droplet. Let 'p' be the

pressure inside the droplet in excess of outside pressure (p = pinside - poutside).

For the equilibrium of the part of the droplet,

As the diameter increases pressure decreases.

To derive an expression for the pressure inside the bubble of liquid

`D' is the diameter of bubble of liquid of surface tension . Let 'p' be the pressure inside the bubble

which is in excess of outside pressure. In case of bubble the liquid layer will be in contact with air both

inside and outside.

To derive an expression for the pressure inside the jet of liquid

Let us consider a jet of diameter D of liquid of surface tension and p is the intensity of pressure

inside the jet in excess of outside atmospheric pressure. For the equilibrium of the part of the jet

shown in the above figure,

Force due to Radial pressure = Force due to surface tension

p x Projected area = x Length

Effect of temperature on surface tension of liquids

In case of liquids, surface tension decreases with increase in temperature. Pressure has no or very

little effect on surface tension of liquids.

Problem 1

What is the pressure inside the droplet of water 0.05mm in diameter at 200C, if the pressure outside

the droplet is 103 kPa Take = 0.0736 N/m at 200C

Problem 2

A liquid bubble 2cm in radius has an internal pressure of 13Pa. Calculate the surface tension of liquid

film.

Capillarity

Any liquid between contact surfaces attains curved shaped surface as shown in figure. The curved

surface of the liquid is called Meniscus. If adhesion is more than cohesion then the meniscus will be

concave. If cohesion is greater than adhesion meniscus will be convex.

Capillarity is the phenomena by which liquids will rise or fall in a tube of small diameter dipped in

them. Capillarity is due to cohesion adhesion and surface tension of liquids. If adhesion is more than

cohesion then there will be capillary rise. If cohesion is greater than adhesion then will be capillary fall

or depression. The surface tensile force supports capillary rise or depression.

NOTE:

Angle of contact:

The angle between surface tensile force and the vertical is called angle of contact. If adhesion is more

than cohesion then angle of contact is obtuse.

Let us consider a small tube of diameter 'D' dipped in a liquid of specific weight . 'h' is the capillary

rise. For the equilibrium,

Vertical force due to surface tension = Weight of column of liquid ABCD.

It can be observed that the capillary rise is inversely proportional to the diameter of the tube.

NOTE:

The same equation can be used to calculate capillary depression. In such cases '' will be obtuse 'h'

works out to be -ve.

Problem 1

Capillary tube having an inside diameter 5mm is dipped in water at 200. Determine the heat of water

which will rise in tube. Take =0.0736N/m at 200 C.

Problem 2

Calculate capillary rise in a glass tube when immersed in Hg at 200c. Assumefor Hg at 200c as

0.51N/m. The diameter of the tube is 5mm. = 1300c.

Problem 3

Determine the minimum size of the glass tubing that can be used to measure water level if capillary

rise is not to exceed 2.5mm. Take = 0.0736 N/m.

Problem 4

A glass tube 0.25mm in diameter contains Hg column with air above it. If = 0.51N/m, what will be

the capillary depression? Take = - 400 or 1400.

Problem 5

If a tube is made so that one limb is 20mm in and the other 2mm in and water is poured in the

tube, what is the difference in the level of surface of liquid in the two limbs. = 0.073 N/m for

water.

Compressibility

It is the property by virtue of which there will be change in volume of fluid due to change in pressure.

Let 'v' be the original volume and 'dv' be the change in volume due to change in pressure 'dp'

, i.e., the ratio of change in volume to original volume is called volumetric strain or bulk strain.

The ratio of change in pressure to the volumetric strain produced is called Bulk modulus of elasticity of

the fluid and is denoted by 'K'

-ve sign indicates that as there is increase in pressure, there is decrease in volume. Reciprocal of Bulk

modulus of elasticity is called compressibility of the fluid.

Unit of Bulk modulus of elasticity is N/m2 or Pa. Unit of compressibility is m2/N.

Problem 1

The change in volume of certain mass of liquids is observed to be the of original volume when

pressure on it is increased by 5Mpa. Determine the Bulk modulus and compressibility of the liquid.

Problem 2

Find the pressure that must be applied to water at atmospheric pressure to reduce its volume by 1%

.Take K = 2 GPa.

FLUID PROPERTIES AND CLASSIFICATION OF FLUID Rheological classification of fluids: (Rheology Study of stress - strain behavior

Newtonian fluids

A fluid which obeys Newton's law of viscosity i.e., is called Newtonian fluid. In such fluids

shear stress varies directly as shear strain.

In this case the stress strain curve is a stress line passing through origin the slope of the line gives

dynamic viscosity of the fluid.

Example: Water, Kerosene.

Non- Newtonian fluid

A fluid which does not obey Newton’s law of viscosity is called non-Newton fluid. For such

fluids, =

Ideal plastic fluids

In this case the strain starts after certain initial stress (0) and then the stress-strain relationship will

be linear. 0 is called initial yield stress. Sometimes they are also called Bingham’s Plastics.

Example: Industrial sludge.

Thixotropic fluids

These require certain amount of yield stress to initiate shear strain. After wards stress-strain

relationship will be non – linear.

Example: Printers ink.

Ideal fluid

Any fluid for which viscosity is assumed to be zero is called Ideal fluid. For ideal fluid = 0 for all

values of

Real fluid

Any fluid which posses certain viscosity is called real fluid. It can be Newtonian or non – Newtonian,

thixotropic or ideal plastic.

FLUID PRESSURE AND ITS MEASUREMENT

Introduction

Fluid is a state of matter which exhibits the property of flow. When a certain mass of fluids is

held in static equilibrium by confining it within solid boundaries, it exerts force along direction

perpendicular to the boundary in contact. This force is called fluid pressure.

Pressure distribution

It is the variation of pressure over the boundary in contact with the fluid. There are two types of pressure distribution.

a. Uniform Pressure distribution. Uniform Pressure distribution.

b. Non-Uniform Pressure distribution

Uniform pressure distribution

If the force exerted by the fluid is same at all the points of contact boundary then the pressure

distribution is said to be uniform.

Non-Uniform pressure distribution

If the force exerted by the fluid is not same at all the points then the pressure distribution is said to be

non-uniform.

Intensity of pressure or unit pressure or pressure

Intensity of pressure at a point is defined as the force exerted over unit area considered around that

point. If the pressure distribution is uniform then intensity of pressure will be same at all the points.

Calculation of intensity of pressure

When the pressure distribution is uniform, intensity of pressure at any points is given by the ratio of

total force to the total area of the boundary in contact.

Intensity of Pressure 'p' =

When the pressure distribution is non- uniform, then intensity of pressure at a point is given by .

Unit of intensity of pressure

N/m2 or pascal (Pa).

Note: 1 MPa =1N/mm2

To study the variation of intensity of pressure in a static mass of fluid: or derive hydrostatic

law of pressure

Atmospheric pressure

Air above the surface of liquids exerts pressure on the exposed surface of the liquid and normal to the

surface.

This pressure exerted by the atmosphere is called atmospheric pressure. Atmospheric pressure at a

place depends on the elevation of the place and the temperature.

Atmospheric pressure is measured using an instrument called ‘Barometer’ and hence atmospheric

pressure is also called Barometric pressure.

Unit: kPa.

‘bar’ is also a unit of atmospheric pressure 1bar = 100 kPa.

Absolute pressure and gauge pressure

Absolute pressure at a point is the intensity of pressure at that point measured with reference to

absolute vacuum or absolute zero pressure.

Absolute pressure at a point can never be negative since there can be no pressure less than absolute

zero pressure.

If the intensity of pressure at a point is measurement with reference to atmosphere pressure, then it

is called gauge pressure at that point.

Gauge pressure at a point may be more than the atmospheric pressure or less than the atmospheric

pressure. Accordingly gauge pressure at the point may be positive or negative.

Negative gauge pressure is also called vacuum pressure.

From the above figure, It is evident that, Absolute pressure at a point = Atmospheric pressure ±

Gauge pressure.

NOTE:

If we measure absolute pressure at a Point below the free surface of the liquid, then,

p = . Y + patm

If gauge pressure at a point is required, then atmospheric pressure is taken as zero, then,

p = . Y

Pressure head

It is the depth below the free surface of liquid at which the required pressure intensity is available.

For a given pressure intensity ‘h’ will be different for different liquids since, ‘’ will be different for

different liquids.

Whenever pressure head is given, liquid or the property of liquid like specify gravity, specify

weight, mass density should be given.

Example:

i. 3m of water

ii. 10m of oil of S = 0.8.

iii. 3m of liquid of = 15 kN/m3

iv. 760mm of Mercury.

v. 10m not correct.

NOTE:

1. To convert head of 1 liquid to head of another liquid.

Pressure head in meters of water is given by the product of pressure head in meters of liquid and

specific gravity of the liquid.

Example:

10meters of oil of specific gravity 0.8 is equal to 10 x 0.8 = 8 meters of water.

Example:

Atm pressure is 760mm of Mercury.

NOTE:

Problem 1

Calculate intensity of pressure due to a column of 0.3m of

a. water

b. Mercury

c. Oil of specific gravity - 0.8.

Problem 2

Intensity of pressure required at a points is 40kPa. Find corresponding head in

a. water

b. Mercury

c. oil of specific gravity - 0.9.

Problem 3

Standard atmospheric pressure is 101.3 kPa Find the pressure head in

a. Meters of water

b. mm of mercury

c. m of oil of specific gravity 0.8.

Problem 4

An open container has water to a depth of 2m and above this an oil of S = 0.9 for a depth of 1m. Find

the intensity of pressure at the interface of two liquids and at the bottom of the tank.

Problem 5

Convert the following absolute pressure to gauge pressure

a. 120 kPa

b. 3 kPa

c. 15m of H2O

d. 800mm of Hg.

Pascal ‘s law

Statement: Intensity of pressure at a point in a static mass of fluid is same along the directions.

Proof:

Let us consider three planes around a point as shown in the above figure. It shows intensity of

pressure and force along different directions. The system of forces should be in equilibrium.

Fx=0

- px dy.dz + ps ds dz cos (900 ) = 0

ps ds sin= px dy

ps dy = px dy

ps = px

Fy=0

- ps ds.dz cos + py dx dz = 0

py dx = ps ds cos

py dx = ps dx

py = ps

px = py = pz

Intensity of pressure at a point is same along all the directions.

Measurement of pressure

Various devices used to measure fluid pressure can be classified into,

Classification of manometers

Manometers are broadly classified into

Types of simple manometers

Common types of simple manometers are

a. Piezometers

b. U-tube manometers

c. Single tube manometers

d. Inclined tube manometers

Piezometers

Piezometer consists of a glass tube inserted in the wall of the vessel or pipe at the level of point at

which the intensity of pressure is to be measured. The other end of the piezometer is exposed to air.

The height of the liquid in the piezometer gives the pressure head from which the intensity of pressure

can be calculated.

To minimize capillary rise effects the diameters of the tube is kept more than 12mm.

A U-tube manometers consists of a glass tube bent in U-Shape, one end of which is connected to

gauge point and the other end is exposed to atmosphere. U-tube consists of a liquid of specific of

gravity other than that of fluid whose pressure intensity is to be measured and is called monometric

liquid.

Manometric liquids

Manometric liquids should neither mix nor have any chemical reaction with the fluid whose

pressure intensity is to be measured.

It should not undergo any thermal variation.

Manometric liquid should have very low vapour pressure.

Manometric liquid should have pressure sensitivity depending upon the magnitude. Of

pressure to be measured and accuracy requirement.

Gauge equations are written for the system to solve for unknown quantities.

To write the gauge equation for manometers

1. Convert all given pressure to meters of water and assume unknown pressure in meters of

waters.

2. Starting from one end move towards the other keeping the following points in mind.

o Any horizontal movement inside the same liquid will not cause change in pressure.

o Vertically downward movement causes increase in pressure and upward motion cause

decrease in pressure.

o Convert all vertical columns of liquids to meters of water by multiplying them by

corresponding specify gravity.

o Take atmospheric pressure as zero (gauge pressure computation).

3. Solve for the unknown quantity and convert it into the required unit.

Problem 1

Determine the pressure at A for the U - tube manometer shown in the below figure. Also calculate the

absolute pressure at A in kPa.

Problem 2

For the arrangement shown in the below figure, determine gauge and absolute pressure at the point

M.

Problem 3

If the pressure at ‘A’ is 10 kPa (Vacuum) what is the value of ‘x’?

Problem 4

The tank in the accompanying figure consists of oil of S = 0.75. Determine the pressure gauge reading

in

Problem 5

A closed tank is 8m high. It is filled with Glycerine up to a depth of 3.5m and linseed oil to another

2.5m. The remaining space is filled with air under a pressure of 150 kPa. If a pressure gauge is fixed

at the bottom of the tank what will be its reading. Also calculate absolute pressure. Take relative

density of Glycerine and Linseed oil as 1.25 and 0.93 respectively.

Problem 6

A vertical pipe line attached with a gauge and a manometer contains oil and Mercury as shown in the

below figure. The manometer is opened to atmosphere. What is the gauge reading at ‘A’? Assume no

flow in the pipe.

Differential manometers

Differential manometers are used to measure pressure difference between any two points. Common varieties of differential manometers are:

a. Two piezometers.

b. Inverted U-tube manometer.

c. U-tube differential manometers

d. Micromanometers.

The arrangement consists of two pizometers at the two points between which the pressure difference

is required. The liquid will rise in both the piezometers. The difference in elevation of liquid levels can

be recorded and the pressure difference can be calculated.

It has all the merits and demerits of piezometer.

Inverted U-tube manometers

Inverted U-tube manometer is used to measure small difference in pressure between any two points.

It consists of an inverted U-tube connecting the two points between which the pressure difference is

required. In between there will be a lighter sensitive manometric liquid. Pressure difference between

the two points can be calculated by writing the gauge equations for the system.

Let 'hA' and 'hB' be the pr head at 'A' and 'B'in meters of water

hA – (Y1 S1) + (x SM) + (y2 S2) = hB.

hA – hB = S1 y1 – SM x – S2 y2,

pA – pB = (hA – hB)

U-tube differential manometers

A differential U-tube manometer is used to measure pressure difference between any two points. It

consists of a U-tube containing heavier manometric liquid, the two limbs of which are connected to the

gauge points between which the pressure difference is required. U-tube differential manometers can

also be used for gases. By writing the gauge equation for the system pressure difference can be

determined.

Let ‘hA’ and ‘hB’ be the pressure head of ‘A’ and ‘B’ in meters of water

hA + S1 Y1 + x SM – Y2 S2 = hB

hA – hB = Y2 S2 – Y1 S1– x SM

Problem 1

An inverted U-tube manometer is shown in the below figure. Determine the pressure difference

between A and B in N/M2.

Let hA and hB be the pressure heads at A and B in meters of water.

Problem 2

In the arrangements shown in the below figure. Determine the ho ‘h’.

Problem 3

In the below figure, given the air pressure in the left tank is 230 mm of Mercury (Vacuum). Determine

the elevation of gauge liquid in the right limb at A. If the liquid in the right tank is water.

Problem 4

Compute the pressure different between ‘M’ and ‘N’ for the system shown in the below figure.

Problem 5

Petrol of specify gravity 0.8 flows up through a vertical pipe. A and B are the two points in the pipe, B

being 0.3 m higher than A. Connection are led from A and B to a U–tube containing Mercury. If the

pressure difference between A and B is 18 kPa, find the reading of manometer.

Problem 6

A cylindrical tank contains water to a height of 50mm. Inside is a small open cylindrical tank

containing kerosene at a height specify gravity 0.8. The following pressures are known from indicated

gauges.

pB = 13.8 kPa (gauge)

pC = 13.82 kPa (gauge)

Determine the gauge pressure pA and height h. Assume that kerosene is prevented from moving to

the top of the tank.

Problem 7

Find the pressure different between A and B if d1 = 300mm, d2 = 150mm,d3 = 460mm, d4 = 200mm

and 13.6.

Problem 8

What is the pressure pA in the figure given below? Take specific gravity of oil as 0.8.

Problem 9

Find ‘d’ in the system shown in the below figure. If pA = 2.7 kPa

Problem 10

Determine the absolute pressure at 'A' for the system shown in the below figure.

Single column manometer

Single column manometer is used to measure small pressure intensities.

A single column manometer consists of a shallow reservoir having large cross sectional area when

compared to cross sectional area of U – tube connected to it. For any change in pressure, change in

the level of manometeric liquid in the reservoir is small and change in level of manometric liquid in the

U- tube is large.

To derive expression for pressure head at A

BB and CC are the levels of manometric liquid in the reservoir and U-tube before connecting the point

A to the manometer, writing gauge equation for the system we have,

+ y x S – h1 x Sm = 0

Sy = Smh1

Let the point A be connected to the manometer. B1B1 and C1 C1 are the levels of manometeric liquid.

Volume of liquid between BBB1B1 = Volume of liquid between CCC1C1

Inclined tube single column manometer

Inclined tube SCM is used to measure small intensity pressure. It consists of a large reservoir to which

an inclined U – tube is connected as shown in the above figure. For small changes in pressure the

reading ‘h2’ in the inclined tube is more than that of SCM. Knowing the inclination of the tube the

pressure intensity at the gauge point can be determined.

hA = h2 sin (Sm-S)+h2 sin Sm

If ‘ ’ is very small then hA = (h2 = Sin) Sm.

Mechanical gauges

Pressure gauges are the devices used to measure pressure at a point.

They are used to measure high intensity pressures where accuracy requirement is less.

Pressure gauges are separate for positive pressure measurement and negative pressure

measurement.

Negative pressure gauges are called Vacuum gauges.

Basic principle

Mechanical gauge consists of an elastic element which deflects under the action of applied pressure

and this deflection will move a pointer on a graduated dial leading to the measurement of pressure.

Most popular pressure gauge used is Bordon pressure gauge.

The arrangement consists of a pressure responsive element made up of phosphor bronze or special

steel having elliptical cross section. The element is curved into a circular arc, one end of the tube is

closed and free to move and the other end is connected to gauge point. The changes in pressure

cause change in section leading to the movement. The movement is transferred to a needle using

sector pinion mechanism. The needle moves over a graduated dial.

HYDROSTATICS

Vertical plane surface submerged in liquid

Consider a vertical plane surface of some arbitrary shape immersed in a liquid of mass density as

shown above.

Let

A = Total area of the surface

= Depth of Centroid of the surface from the free surface

G = Centroid of the immersed surface

C = Centre of pressure

= Depth of centre of pressure

Consider a rectangular strip of breadth b and depth dy at a depth y from the free surface.

Total Pressure:

The pressure intensity at a depth y acting on the strip is

p = gy

Total pressure force on the strip = dP = (gy)dA

The Total pressure force on the entire area is given by integrating the above expression over the

entire area

But y dA is the Moment of the entire area about the free surface of the liquid given by

y dA = A

Substituting in equation 01, we get

Where is the specific weight of water.

For water, =1000 kg/m3 and g = 10 m/s2. The force will be expressed in Newtons (N).

Centre of Pressure:

This is computed on the principle of Theorem of moments.

The moment of the pressure force about the free surface is given by

Similarly, the moment of the differential pressure force about the free surface is given by

dM = dPxy = (gy)dAxy = gy2dA

The moment of the pressure force about the free surface is given by integrating the above expression

over the entire area A

M = gy2dA=gy2dA

But y2dA is the Moment of Inertial of the entire submerged area about the free surface given by Io

Hence,

From equations 03 and 04, we get

But from parallel axis theorem of Moment of inertia, the moment of inertia of the given area about any

axis is sum of the moment of inertia of the area about its centroidal axis and product of area and

square of the distance between the two axes. Mathematically

Inclined plane surface submerged in liquid

Consider a vertical plane surface of some arbitrary shape immersed in a liquid of mass density in

such a way that the plane is making an angle with the free surface as shown above.

Let

A = Total area of the surface

= Depth of Centroid of the surface from the free surface

= Depth of Centroid of the surface from the free surface along the plane

G = Centroid of the immersed surface

C = Centre of pressure

= Depth of centre of pressure

= Depth of centre of pressure from the free surface along the plane

Consider a rectangular strip of breadth b and depth dy at a depth y from the free surface. From the

triangle, we have

Total Pressure:

The pressure intensity at a depth y acting normal to the plane on the strip is p = gy

Total pressure force on the strip = dP = (gy)dA

The Total pressure force on the entire area is given by integrating the above expression over the

entire area along the plane

P = dP = (gy)dA= g y dA

But y and dA are on different planes and hence substituting for y from equation 1, we get

But y* dA is the Moment of the entire area about the free surface of the liquid given by

y* dA= A sin = A

Substituting in equation 02, we get

Where is the specific weight of water.

Centre of Pressure:

This is computed on the principle of Theorem of moments.

The moment of the pressure force about the free surface is given by

Similarly, the moment of the differential pressure force about the free surface is given by

dM = dPxy* = (gy)dAxy* = g(y* sin) y*dA=g y*2dA

The moment of the pressure force about the free surface is given by integrating the above expression

over the entire area A

M = g y*2sindA =g siny*2dA

But 2dA is the Moment of Inertial of the entire submerged area about the free surface given by Io

Hence,

From equations 04 and 05, we get

But from parallel axis theorem of Moment of inertia, the moment of inertia of the given area about any

axis is sum of the moment of inertia of the area about its centroidal axis and product of area and

square of the distance between the two axes. Mathematically

Where is the centroidal depth, is the centre of pressure and is the angle of inclination of the

plane with the free surface.

From the above equation it is seen that is always either equal to or greater than , which means

that the in general for a plane surface immersed in a liquid, the Centre of pressure always lies below

the centroid.

Curved surface submerged in liquid

Consider a curved surface AB immersed in a liquid of mass density as shown below. Let dA be a

differential area located on the curved surface at a depth y from the free surface such that it makes

an angle with the horizontal.

The pressure intensity p acting on the differential area dA is given by

p = g y

The total differential pressure force dF acting normal to dA is given by

dF = p x Area = g y dA

As can be seen from the above figure. this dF goes on changing its inclination and is not acting

constantly at a particular angle. Hence, let us resolve dFinto horizontal and vertical directions

as dFx and dFyas shown above.

Where,

dFx = dF sin

= g y dA sin

dFy = dF cos

= g y dA cos

dFx acts on the vertically projected area of dA = dA sin = dAy and dFy acts on the horizontally

projected area of dA = dA cos = dAx.

Hence,

dFx = g y dAy and dFy = g y dAx

Integrating we get

dFx = Fx = g y dAy = g y dAy = gAy

Where Ay is the projected area of the curved surface on vertical plane equivalent to the vertical plane

surface and y is the centroidal distance of the projected area from the free surface.

Similarly, its location can be obtained by

Where Ig is the moment of inertia of the projected area about its centroidal axis.

dFy = g y dAx = g y dAx

Fy = g x volume of fluid present above the curved surface.

= Weight of fluid present above the curved surface.

This force acts vertically downwards at the centre of gravity of the volume of the fluid. The resultant

hydrostatic pressure force acting on the curved surface can be obtained by composing the two

components.

In case, the curved surface is not supporting any fluid above the same and instead supporting on the

sides as shown, the vertical component of the force acting on the curved surface Fy is acting upwards

as the upward thrust equivalent to the imaginary weight of the fluid present above the curved surface.

Problem 1

A square tank with 2 m sides and 1.5 m high contains water to a depth of 1 m and a liquid of specific

gravity 0.8 on the water to a depth of 0.5 m. Find the magnitude and location of hydrostatic pressure

on one face of tank.

Solution:

The problem can be solved by considering hydrostatic pressure distribution diagram for water as

shown above.

From hydrostatic law, the pressure intensity p at any depth y w is given by p = So g y w where is the

mass density of the liquid.

Pressure force P = p x Area .

Pw = 1000 x 10 x 2.0 x 1.5 x 1.5 = 45 kN ()

Acting at 1.5 / 3 = 0.5 m from the base.

Problem 2

A rectangular tank 10 m x 5 m and 3.25 m deep is divided by a partition wall parallel to the shorter

wall of the tank. One of the compartments contains water to a depth of 3.25 m and the other oil of

specific gravity 0.85 to a depth of 2 m. Find the resultant pressure on the partition.

Solution:

The problem can be solved by considering hydrostatic pressure distribution diagram for both water

and oil as shown above.

From hydrostatic law, the pressure intensity p at any depth yw is given by

p = So g yw

where is the mass density of the liquid.

Pressure force P = p x Area.

Pw = 1000 x 10 x 3.25 x 5 x 3.25 = 528.125 kN ()

Acting at 3.25 / 3 m from the base

Po = 0.85 x 1000 x 10 x 2.0 x 5 x 2.0 = 200 kN ()

Acting at 2 / 3 m from the base.

Net Force P = Pw – Po = 528.125 – 200.0 = 328.125 kN ()

Location:

Let P act at a distance y from the base. Taking moments of Pw, Po and P about the base, we get

P x y = Pw x yw /3 – Po x yo /3

328.125 y = 528.125 x (3.25/3) – 200 x (2/3) or y = 1.337 m.

Problem 3

Determine the total force and location of centre of pressure for a circular plate of 2 m dia immersed

vertically in water with its top edge 1.0 m below the water surface.

Problem 4

A rectangular plate 2 m x 3 m is immersed in oil of specific gravity 0.85 such that its ends are at

depths 1.5 m and 3 m respectively. Determine the total pressure acting on the plate and locate it.

Solution:

A = 2 x 3 = 6 m2

So = 0.85

Assume

= 1000 kg/m3

g = 10 m/s2

= GG1

= CC1

Sin = 1.5 / 3 = 0.5

= 30o

GG1 = G1A1 + A1G = G1A1 + AG Sin

GG1 = 1.5 + (3/2) Sin 30 = 2.25 m

We know that the total pressure force is given by

P = SogA = 0.85 x 1000 x 10 x 6 x 2.25 = 114.75 kN

Centre of Pressure:

The Centre of pressure is given by

Problem 5

A Circular plate with a concentric hole is immersed in water in such a way that its greatest and least

depth below water surface are 4 m and 1.5 m respectively. Determine the total pressure on the plate

and locate it if the diameter of the plate and hole are 3 m and 1.5 m respectively.

Solution:

Assume

= 1000 kg/m3 and g = 10 m/s2

Problem 6

A circular plate of dia 0.75 m is immersed in a liquid of relative density of 0.8 with its plane making an

angle of 30o with the horizontal. The centre of the plate is at a depth of 1.5 m below the free surface.

Calculate the total force on one side of the plate and location of centre of pressure.

Solution:

Assume = 1000 kg/m3 and g = 10 m/s2

Problem 7

A vertical gate closes a circular tunnel of 5 m diameter running full of water, the pressure at the

bottom of the gate is 0.5 MPa. Determine the hydrostatic force and the position of centre of pressure.

Solution:

Assume = 1000 kg/m3 and g = 10 m/s2

Pressure intensity at the bottom of the gate is = p = So gy

Where y is the depth of point from the free surface.

0.5 x 106 = 1000 x 10 x y

y = 50 m

Hence the free surface of water is at 50 m from the bottom of the gate

Problem 8

Find the horizontal and vertical component of force and its point of application due to water per meter

length of the gate AB having a quadrant shape of radius 2 m shown below. Find also the resultant

force in magnitude and direction.

Solution:

Assume = 1000 kg/m3 and g = 10 m/s2

R = 2 m, Width of gate = 1 m

Horizontal force Fx

Fx = Force on the projected area of the curved surface on the vertical plane

= Force on BO = P = So gA

A = 2 x 1 = 2 m2

Problem 9

A cylinder holds water in a channel as shown below. Determine the weight of 1 m length of the

cylinder.

Solution:

Radius of Cylinder = R = 2m

Length of cylinder = 1 m

Weight of Cylinder = W

Horizontal force exerted by water= Fx

Fx = Force on vertical area BOC

= Sog A = 1000 x 10 x (4 x 1) x (2/2) = 40 kN ()

The vertical force exerted by water = Fy = Weight of water enclosed in BDCOB

For equilibrium of the cylinder the weight of the cylinder must be equal to the force exerted by the

water on the cylinder. Hence, the weight of the cylinder is 31.416 kN per meter length.

Problem 10

The below figure shows the cross section of a tank full of water under pressure. The length of the tank

is 2 m. An empty cylinder lies along the length of the tank on one of its corner as shown below. Find

the resultant force acting on the curved surface of the cylinder.

Solution:

R = 1 m

L = 2 m

p = gh = 1000 x 10 x h = 20 x 103

h = 2 m

For this pressure, the free surface should be 2 m above A

Problem 11

Calculate the resultant water pressure on the tainter gate of radius 8 m and width unity as shown

below.

Solution:

Horizontal component of force Fx

Fx = Sog A

DB = OB sin 30 = 8 x 0.5 = 4.0 m

A = 4 x 1.0 = 4 m2

Problem 12

A 3.6 m x 1.5 m wide rectangular gate MN is vertical and is hinged at point 150 mm below the centre

of gravity of the gate. The total depth of water is 6 m. What horizontal force must be applied at the

bottom of the gate to keep the gate closed?

KINEMATICS OF FLUID FLOW

Introduction

Fluid kinematics refers to the features of a fluid in motion. It only deals with the motion of fluid

particles without taking into account the forces causing the motion. Considerations of velocity,

acceleration, flow rate, nature of flow and flow visualization are taken up under fluid kinematics.

A fluid motion can be analyzed by one of the two alternative approaches,

called Lagrangian and Eulerian.

In Lagrangian approach, a particle or a fluid element is identified and followed during the course of its

motion with time as demonstrated shown above.

Example:

To know the attributes of a vehicle to be purchased, you can follow the specific vehicle in the traffic

flow all along its path over a period of time.

Difficulty in tracing a fluid particle (s) makes it nearly impossible to apply the Lagrangian approach.

The alternative approach, called Eulerian approach consists of observing the fluid by setting up fixed

stations (sections) in the flow field shown above.

Motion of the fluid is specified by velocity components as functions of space and time. This is

considerably easier than the previous approach and is followed in Fluid Mechanics.

Example:

Observing the variation of flow properties in a channel like velocity, depth etc, at a section

Classification of flows

Steady and unsteady flows

Uniform and non-uniform flows

Example:

Flow observed at any instant, at the dam section during rainy season, wherein, the flow varies from

the top of the overflow section to the foot of the dam and the flow properties like depth, velocity etc.,

will change at the dam section at any instant between two sections, representing it as non-uniform

flow.

Consider a fluid flow as shown above in a channel. The flow is said to be steady at sections 1 and 2 as

the flow does not change with respect to time at the respective sections (y1=y2 and v1=v2).

The flow between sections 1 and 2 is said to be uniform as the properties does not change between

the sections at any instant (y1=y2 and v1=v2).

The flow between sections 2 and 3 is said to be non-uniform flow as the properties vary over the

length between the sections.

Non-uniform flow can be further classified as Gradually varied flow and Rapidly varied flow. As

the name itself indicates, Gradually varied flow is a non-uniform flow wherein the flow/fluid

properties vary gradually over a long length (Example: between sections 2 and 3).

Rapidly varied flow is a non-uniform flow wherein the flow/fluid properties vary rapidly within a very

short distance. (Example: between sections 4 and 5).

Combination of steady and unsteady flows and uniform and non-uniform flows can be classified

as steady-uniform flow (Sections 1 and 2),unsteady-uniform flow, steady-non-uniform

flow (Sections 2 and 3) and unsteady-non-uniform flow (Sections 4 and 5).

One, two and three dimensional flows

Flow is said to be one-dimensional if the properties vary only along one axis / direction and will be

constant with respect to other two directions of a three-dimensional axis system.

Flow is said to be two-dimensional if the properties vary only along two axes / directions and will be

constant with respect to other direction of a three-dimensional axis system.

Flow is said to be three-dimensional if the properties vary along all the axes / directions of a three-

dimensional axis system.

Laminar and turbulent flows

When the flow occurs like sheets or laminates and the fluid elements flowing in a layer does not mix

with other layers, then the flow is said to be laminar. The Reynolds number (Re) for the flow will be

less than 2000.

When the flow velocity increases, the sheet like flow gets mixed up and the fluid elements mix with

other layers thereby causing turbulence. There will be eddy currents generated and flow reversal takes

place. This flow is said to be Turbulent. The Reynolds number for the flow will be greater than 4000.

For flows with Reynolds number between 2000 to 4000 is said to be transition flow.

Compressible and incompressible flows

Flow is said to be Incompressible if the fluid density does not change (constant) along the flow

direction and is Compressible if the fluid density varies along the flow direction.

= Constant (incompressible) and Constant (compressible)

Rotational and irrotational flows

Flow is said to be Rotational if the fluid elements does not rotate about their own axis as they move

along the flow and is Rotational if the fluid elements rotate along their axis as they move along the

flow direction.

Rate of flow or discharge (Q)

Rate of flow or discharge is said to be the quantity of fluid flowing per second across a section of a

flow. Rate of flow can be expressed as mass rate of flow or volume rate of flow.

Accordingly

Mass rate of flow = Mass of fluid flowing across a section / time

Rate of flow = Volume of fluid flowing across a section / time

Types of lines

Path line

It is the path traced by a fluid particle over a period of time during its motion along the fluid flow.

Example:

Path traced by an ant coming out from its dwelling

Stream lines

It is an imaginary line such that when a tangent is drawn at any point it gives the velocity of the fluid

particle at that point and at that instant.

Example:

Path traced by the flow when an obstruction like a sphere or a stick is kept during its motion. The flow

breaks up before the obstruction and joins after it crosses it.

Streak lines

It is that imaginary line that connects all the fluid particles that has gone through a point / section

over a period of time in a fluid motion.

Stream tube

It is an imaginary tube formed by stream line on its surface such that the flow only enters the tube

from one side and leaves it on the other side only. No flow takes place across the stream tube. This

concept will help in the analysis of fluid motion.

Variation of a property along any given direction

If P is a Property at any point, then the property at any other location along x direction at a distance

dx is given by

New Property = Old Property + slope x Old Property x distance

Continuity equation

The derivation is based on the concept of Law of conservation of mass.

Statement:

The flow of fluid in a continuous flow across a section is always a constant.

Consider an enlarging section in a fluid flow of fluid density . Consider two sections 1 and 2 as shown

in the above figure. Let the sectional properties be as under.

A1 and A2= Cross-sectional area, V1 and V2= Average flow velocity and

1 and 2 = Fluid density at Sec 1 and 2 respectively

dt is the time taken for the fluid to cover a distance dx

The mass of fluid flowing across section 1-1 is given by

m1 = Density at section 1 x volume of fluid that has crossed section 1

1 x A1 x dx

Mass rate of fluid flowing across section 1 - 1 is given by

m1/dt = (Density at sec 1 x volume of fluid that has crossed sec 1)/dt

Similarly Mass rate of fluid flowing across section 2 - 2 is given by

From law of conservation of mass, mass can neither be created nor destroyed. Hence from equations

1 and 2, we get 1A1V1 = 2A2V2

If the density of the fluid is constant, then the equation reduces to A1V1 = A2V2

The above equations discharge continuity equation in one dimensional form for an steady,

incompressible fluid flow.

In three dimensional or differential form

Consider a parallelepiped A B C D E F G H in a fluid flow of density as shown below. Let the

dimensions of the parallelepiped be dx, dy and dz along x, y and z directions respectively. Let the

velocity components along x, y and z be u, v and w respectively.

Mass rate of fluid flow entering the section ABCD along x direction is given by x Area x flow

velocity

Similarly mass rate of fluid flow leaving the section EFGH along x direction is given by

Net gain in mass rate of the fluid along the x axis is given by the difference between the mass rate of

flow entering and leaving the control volume. i.e., equation 1 – equation 2

Similarly net gain in mass rate of the fluid along the y and z axes are given by

This expression is known as the general Equation of Continuity in three dimensional form or

differential form. If the fluid is incompressible then the density is constant and hence

Velocity

Velocity of a fluid along any direction can be defined as the rate of change of displacement of the fluid

along that direction.

u = dx / dt

Where dx is the distance traveled by the fluid in time dt.

Velocity of a fluid element is a vector, which is a function of space and time.

Let V be the resultant velocity of a fluid along any direction and u, v and w be the velocity

components in x, y and z directions respectively.

Mathematically the velocity components can be written as

u = f (x, y, z, t)

v = f (x, y, z, t)

w = f (x, y, z, t)

and

V = ui + vj + wk = u2 + v2 + w2

Where u = (dx/dt), v = (dy/dt) and w = (dz/dt).

Acceleration

Acceleration of a fluid element along any direction can be defined as the rate of change of velocity of

the fluid along that direction.

If ax, ay and az are the components of acceleration along x, y and z directions respectively, they can

be mathematically written as

ax = du/ dt.

But u = f (x, y, z, t) and hence by chain rule, we can write,

Hence

If A is the resultant acceleration vector, it is given by

For steady flow, the local acceleration will be zero

Problem 1

The velocity field in a fluid is given by

Vs=(3x + 2y) i + (2z + 3x2) j + (2t -3z)k

i. What are the velocity components u, v, and w ?

ii. Determine the speed at the point (1, 1, 1).

iii. Determine the speed at time t = 2 s at point (0, 0, 2).

u = (3x + 2y), v = (2z + 3x2), w = (2t - 3z)k

Solution:

The velocity components at any point (x, y, z) are

Substitute x = 1, y = 1, z = 1 in the above expression

u = (3*1 + 2*1) = 5, v = (2*1 + 3*1) = 5, w = (2t - 3)

V2 = u2 + v2 + w2

= 52 + 52 + (2t - 3)2

Problem 2

The velocity distribution in a three-dimensional flow is given by:

u = - x, v = 2y and w = (3-z). Find the equation of the stream line that passes through point (1,1,1).

Solution:

The stream line equation is given b

Problem 3

A fluid particle moves in the following flow field starting from the points (2,1,0) at t = 0. Determine

the location of the fluid particle at t = 3s

Solution:

Integrating we get

Problem 4

The following cases represent the two velocity components, determine the third component of velocity such that they satisfy the continuity equation.

i. u = x2 + y2 + z2 ; v = xy2 - yz2 + xy

ii. v = 2y2 ; w = 2xyz.

Solution:

The continuity equation for incompressible flow is given by

Problem 5

Find the convective acceleration at the middle of a pipe which converges uniformly from 0.4 m to 0.2

m diameter over a length of 2 m. The rate of flow is 20 lps. If the rate of flow changes uniformly from

20 lps to 40 lps in 30 seconds, find the total acceleration at the middle of the pipe at 15th second.

D1 = 0.4 m, D2 = 0.2 m, L = 2 m, Q = 20 lps = 0.02 m3/s.

Q1 = 0.02 m3/s and Q2 = 0.04 m3/s

Case (i)

Flow is one dimensional and hence the velocity components v = w = 0

Convective acceleration = u (u /x)

A1 = (/4) (D12) = 0.1257 m2

A2 = (/4) (D22) = 0.0314 m2

u1 = Q / A1 = 0.02 / 0.1257 = 0.159 m/s

and u2 = Q / A2 = 0.02 / 0.0314 = 0.637 m/s

As the diameter changes uniformly, the velocity will also change uniformly. The velocity u at any distance x from inlet is given by u = u1 + (u2 – u1) / (x / L) = 0.159 + 0.2388 x

(u / x) = 0.2388

Convective acceleration = u(u / x) = (0.159 + 0.2388 x) 0.2388

At A, x = 1 m and hence (Convective accln)x = 1 = 94.99 mm/s2

Case (ii)

Total acceleration = (convective + local ) acceleration at t =15 seconds

Rate of flow Qt = 15 = Q1 + (Q2 – Q1) (15/30) = 0.03 m3/s.

u1 = Q / A1 = 0.03 / 0.1257 = 0.2386 m/s and u2 = Q / A2 = 0.03 / 0.0314 = 0.9554 m/s The velocity u at any distance x from inlet is given by

u = u1 + (u2 – u1) / (x / L) = 0.2386 + 0.3584 x

(u / x) = 0.3584

Convective acceleration = u(u / x) = (0.2386 + 0.3584 x) 0.3584 At A, x = 1 m and hence

(Convective accln) x = 1 = 0.2139 m/s2

Local acceleration

Diameter at A is given by D = D1 + (D1 – D2) / (x / L) = 0.3 m and A = (/4)(D2) = 0.0707 m2

When Q1 = 0.02 m3/s, u1 = 0.02 / 0.0707 = 0.2829 m/s

When Q2 = 0.04 m3/s, u2 = 0.02 / 0.0707 = 0.5659 m/s Rate of change of velocity

= Change in velocity / time

= (0.5629-0.2829) / 30

= 9.43 x 10-3m/s2

Total acceleration = 0.2139 + 9.43 x 10 -3 = 0.2233 m/s2

Velocity potential ()

Velocity Potential is a scalar function of space and time such that its negative derivative with

respect to any direction gives the velocity component in that direction

Thus = (x, y, z, t) and flow is steady then,

u = - (/x); v = - (/y) ; w = - (/z) Continuity equation for a three dimensional fluid flow is given by

[(u /x)+(v/ y) +(w /z)] = 0

Substituting for u, v and w, we get

[(/x) (-/x)+(/y) (-/y) +(/z) (-/z)] = 0

i.e. [(2/x2) + (2/y2) + (2/z2)] = 0

The above equation is known as Laplace equation in For a 2 D flow the above equation reduces to

[(2/x2) + (2/y2)] = 0

We know that for an irrotational two dimensional fluid flow, the rotational fluid elements about z axis must be zero.

i.e. z = ½ [(v /x) - (u /y)]

Substituting for u and v, we get

Wz= ½ [(/x)(-f /y) - (/y) (-/x)]

For the flow to be irrotational, the above component must be zero

z = ½ [ (-2/x y) - (-2/ y x)] = 0

i.e. (-2/x y) = (-2/ y x)

This is true only when is a continuous function and exists.

Thus the properties of a velocity potential are:

1. If the velocity potential exists, then the flow should be irrotational.

2. If the velocity potential satisfies the Laplace Equation, then it represents a possible case of a

fluid flow.

Stream function ()

Stream Function is a scalar function of space and time such that its partial derivative with respect

to any direction gives the velocity component at right angles to that direction.

Thus = (x,y,z,t) and flow is steady then,

u = -( /y) ; v = (y /x)

Continuity equation for a two dimensional fluid flow is given by

[(u / x)+(v / y)] = 0

Substituting for u and v, we get

[( / x) (- / ) + ( /y) ( / x)] = 0

i.e. [ (-2 / x y) + (2 / x)] = 0

or ( 2 / x ) = ( 2 / y x)

This is true only when is a continuous function.

We know that for an irrotational two dimensional fluid flow, the rotational fluid elements about z axis must be zero.

i.e. z = ½ [(v / x) - (u /y)]

Substituting for u and v, we get

z = ½ [( /x) ( / x) - ( /y) (- /y)]

For the flow to be irrotational, the above component must be zero

i.e. [(2 /x2) + (2 /y2)] = 0

The above equation is known as Laplace equation in

Thus the properties of a Stream function are:

1. If the Stream function exists, then it represents a possible case of a fluid flow.

2. If the Stream function satisfies the Laplace Equation, then the flow should be irrotational.

Equi-potential lines

It is an imaginary line along which the velocity potential is a constant.

i.e. = Constant

d= 0

But = (x, y) for a two dimensional steady flow

d= ( / x) dx + ( / y) dy

Substituting the values of u and v, we get

d= – u dx – v dy 0

or u dx = – v dy

Where dy / dx is the slope of the equi-potential line. Line of constant stream function or stream line. It is an imaginary line along which the stream

function is a constant

i.e. = Constant

d = 0

But = (x,y) for a two dimensional steady flow

d = ( / x)dx + ( / y)dy

Substituting the values of u and v, we get

d = v dx – u dy 0

or v dx = u dy

Where dy/dx is the slope of the Stream line.

From equations 1 and 2, we get that the product of the slopes of equi-potential line and stream line is given by -1. Thus, the equi-potential lines and stream lines are orthogonal to each other at all the points of intersection.

Flow net

A grid obtained by drawing a series of equi-potential lines and stream lines is called a Flow net. The flow net is an important tool in analysing two dimensional flow irrotational flow problems.

Relationship between stream function () and velocity potential ()

We know that the velocity components are given by

u = - ( /x) v = -( / y)

and u = - ( /y) v = ( /x)

Thus u = - ( /x) = - ( /y) and v = -( /y) = ( /x)

Hence ( / x) = ( /y)

and ( / y) = - ( / x)

Problem 6 In a two dimensional incompressible flow the fluid velocity components are given by u = x – 4y and v = -y – 4x

Where u and v are x and y components of velocity of flow. Show that the flow satisfies the continuity equation and obtain the expression for stream function. If the flow is potential, obtain also the expression for the velocity potential. Solution:

u = x – 4y and v = -y – 4x

(u /x) = 1 and (v /y) = -1

(u /x)+ (v / y) = 1 - 1 = 0.

Hence it satisfies continuity equation and the flow is continuous and velocity potential exists.

Let be the velocity potential.

Then,

Where C is an integral constant, which is independent of x and can be a function of y. Differentiating

equation 3 w.r.t. y, we get

( / y) = 0 + 4x + (C /y)y + 4x

Hence, we get (C /y) = y

Integrating the above expression, we get C = y2/ 2

Substituting the value of C in Equation 3, we get the general expression as

= (-x2/ 2) + 4xy + y2/ 2

Stream Function

Let be the velocity potential.

Where K is an integral constant, which is independent of x and can be a function of y. Differentiating

equation 6 w.r.t. y, we get

( /y) = - x – 0 + (K /y) -x + 4 y

Hence, we get (K /y) = 4y

Integrating the above expression, we get C = 4y2/ 2 = 2y2

Substituting the value of K in Equation 6, we get the general expression as

= - yx - 2x2 + 2y2

Problem 7

The components of velocity for a two dimensional flow are given by

u = x y; v = x2 – y2 / 2 Check whether

i. they represent the possible case of flow

ii. and the flow is irrotational.

u = x y; and v = x2 – y2 / 2

(u /x) = y (v /y) = -y

(u / y) = x (v / x) = 2x

For a possible case of flow the velocity components should satisfy the equation of continuity.

i.e., [(u /x) + (v /y)] = 0

Substituting, we get y – y = 0.

Hence it is a possible case of a fluid flow.

For flow to be irrotational in a two dimensional fluid flow, the rotational component in z direction (z)

must be zero, where

z = ½ [(v /x) - (u /y)] = ½ (2x - x) 0

Hence, the flow is not irrotational.

Problem 8

Find the components of velocity along x and y for the velocity potential = a Cos xy. Also calculate

the corresponding stream function.

Hence u = ay Sin xy and v = ax Sin xy.

Problem 9

he stream function and velocity potential for a flow are given by

= 2xy and = x2 – y2

Show that the conditions for continuity and irrotational flow are satisfied

Solution:

From the properties of Stream function, the existence of stream function shows the possible case of

flow and if it satisfies Laplace equation, then the flow is irrotational.

(i) = 2xy

( /x) = 2 and ( /y) = 2x

(2 / x2) = 0 and (2 / y2) = 0

(2 /x y ) = 2 and (2 / y x) = 2

(2 / x y) = (2 /y x)

Hence the flow is Continuous.

(2 / x2) + (2 / y2) = 0

As it satisfies the Laplace equation, the flow is irrotational.

From the properties of Velocity potential, the existence of Velocity potential shows the flow is

irrotational and if it satisfies Laplace equation, then it is a possible case of flow

(ii) = x2 – y2

( / x) = 2 x and ( / y) = -2 y

(2 / x2) = 2 and (2 / y2) = -2

(2 /x y) = 0 and (2 / y x) = 0

(2 /x y) = (2 /y x)

Hence the flow is irrotational

(2 / x2) + (2 / y2) = 0

As it satisfies the Laplace equation, the flow is Continuous.

Problem 10

In a 2-D flow, the velocity components are u = 4y and v = - 4x

i. is the flow possible ?

ii. if so, determine the stream function ?

iii. What is the pattern of stream lines ?

Solution:

For a possible case of fluid flow, it has to satisfy continuity equation. i.e.

u = 4y and v = - 4x

(u /x) = 0 (v /y) = 0

Substituting in equation 1, we get 0.

Hence the flow is possible.

Stream function

We know that

Where C is an integral constant and a function of y.

Differentiating Equation 4, w.r.t. y, we get

( /y) = 0 + C(y) /y = - u = - 4y

Integrating the above expression w.r.t. y we get

C(y) = -2y2.

Substituting the above value in Equation 4, we get the general expression as

= - 2x2 – 2y2 = - 2 (x2+ y2)

The above equation is an expression of concentric circles and hence the stream function is

concentric circles.

Problem 11

A 250 mm diameter pipe carries oil of specific gravity 0.9 at a velocity of 3 m/s. At another section the

diameter is 200 mm. Find the velocity at this section and the mass rate of flow of oil.

Solution:

D1 = 0.25 m; D2 = 0.2 m; So = 0.9; V1 = 3 m/s;

= 1000 kg/m3(assumed);

V2 = ?; Mass rate of flow = ?

From discharge continuity equation for steady incompressible flow, we have

A1 = (/4)D12 = (/4)0.252Substituting in equation = 0.0499 m2

A2 = (/4)D22 = (/4)0.202 = 0.0314 m2

Substituting in equation 1, we get

Q = 0.0499 x 3 = 0.1473 m3/s

Mass rate of flow = Q = 0.1479 x 1000 = 147.9 kg/m3 (Ans)

V2 = (A1 / A2) x V1 = (D1 / D2)2 x V1 = (0.25 / 0.2)2 x 3 = 4.6875 m/s (Ans)

Problem 12

A stream function in a two dimensional flow is = 2 x y. Show that the flow is irrotational and

determine the corresponding velocity potential.

Solution:

For first part see problem 9 Given,

= 2 x y.

Integrating equation 1, w.r.t. x, we get

Where C(y) is an integral constant independent of x

Differentiating equation 3 w.r.t. y, we get

(/ y) = 0 + (C(y) / y) = - 2y

Integrating the above expression w.r.t. y, we get

C(y) = -y2

Substituting for C(y) in equation 3, we get the general expression for as

= x2 + C = x2 - y2 (Ans)

Problem 13

The velocity potential for a flow is given by the function = x2 - y2. Verify that the flow is

incompressible and determine the stream function.

Solution:

From the properties of velocity potential, we have that if satisfies Laplace equation, then the flow is

steady incompressible continuous fluid flow.

Given = x2 - y2

( / x) = 2 x ( /y) = -2y

(2 / x2) = 2 (2 / 2y) = -2

From Laplace equation, we have (2 / x2) + (2 / 2y) = 2 – 2 = 0.

Finding out the stream function for the above velocity potential is reverse procedure of problem 12

and the answer is = 2 x y.

FLUID DYNAMICS

Introduction

Fluid dynamics is that branch of fluid mechanics wherein we study the analysis of the fluid motion

along with the forces generating them.

The fluid motion is analysed by the Newton’s second law of motion, which states that the force

applied on a body along any direction is given by the rate of change of momentum along the

same direction.

The forces acting on the fluid can be classified as under

1. Fg Gravity forces

2. FP Pressure forces

3. Fv Viscous forces

4. Ft Turbulent forces

5. Fe Elastic forces

6. Fc Compressibility forces

The net force acting along x direction is given by

Equations of motion

Reynolds equation of motion: In the equation of motion (equation 01), the force due to

compressibility is neglected and only forces due to gravity, pressure, viscosity and turbulence are

considered, the resulting equation is termed as Reynolds equation of motion.

Navier-Stokes equation of motion: In the equation of motion (equation 01), the forces due to

compressibility and turbulence are neglected and only forces due to gravity, pressure and viscosity are

considered, the resulting equation is termed as Navier-Stokes equation of motion.

Euler’s equation of motion: In the equation of motion (equation 01), if the fluid is ideal, then the

forces due to compressibility, turbulence and viscosity are neglected and only forces due to gravity

and pressure are considered, the resulting equation is termed as Euler’s equation of motion.

Euler’s equation of motion

Assumptions

1. Only Gravitational and Pressure forces are considered

2. Fluid motion along a stream line is considered

Consider a stream line along direction x as shown in the below figure. Consider a cylindrical fluid element of cross-sectional area dA and length ds along the stream line direction. The forces acting on the fluid element are

1. The pressure force pdA along the flow direction s

2. The pressure force [s) ds] dA against the flow direction s

3. Weight of the fluid element =g dA ds acting vertically downwards at an angle with the

vertical.

From equation. 1,2 and 3, we get

Bernoulli’s equation from euler’s equation of motion

Assumptions and Limitations

1. The fluid is ideal. i.e. the viscosity is zero 2. The flow is steady 3. The flow is incompressible 4. The flow is irrotational or the flow is along a stream line

From Euler’s equation of motion, we have

Integrating the above expression

As the flow is incompressible, r is constant and we get

Statement of Bernoulli’s Energy Equation or Theorem

In a steady, ideal flow of an incompressible fluid , the total energy at any section of a flowing fluid is

always a constant. The total energy includes pressure energy, kinetic or velocity energy and potential

or datum energy.

It can also be stated as in a steady, ideal flow of an incompressible fluid, the total head at any section

of a flowing fluid is always a constant wherein the total head includes pressure head, velocity or

kinetic head and datum or potential head.

Bernoulli’s equation for real fluid or modified bernoulli’s equation

The Bernoulli’s equation has been derived for ideal and non-viscous fluid and hence it is frictionless.

But, in case of real fluid, the viscosity will be very much present and hence, there will be energy or

head loss between any two section along the flowing fluid. Hence, the Bernoulli’s equation can be

modified as The total energy at any section of a flowing fluid is equal to the total energy at the

previous section minus the energy loss between the two sections.

In other words, The difference of total head between any two sections of a flowing fluid along the flow

direction is given by the head loss between the two sections. Mathematically, it can be stated as

under:

Where hL is the head loss or the energy loss per unit weight between the two sections of a flowing

fluid.

Problem 1

The following are the data given of a change in diameter, effected in laying a water supply pipe line. The change in diameter is gradual from 200 mm at A to 500 mm at B. Pressure at A and B are 78.5 kN/m2 and 58.9 kn/m2 respectively with the end B being 3 m higher than A. If the glow in the pipe

line is 200 lps, find:

A. direction of flow and

B. the head lost in friction between A and B.

Solution:

Q = 0.2 m3/s; DA = 0.2 m; DB = 0.5 m; g = 10 m/s2 (assumed)

pA = 78.5 kN/m2; pB = 58.9 kN/m2; ZB – ZA = Z = 3 m; hf = ?

hL = 0.9345 m

The flow is always from higher pressure to lower pressure and hence from A to B.

Problem 2

Water flows up a conical pipe 450 mm diameter at the lower end and 250 mm diameter at 2.3

m above the lower end. If the pressure and velocity at the lower end are 63 kN/m2 (gauge) and 4.1

m/s, assuming a head loss in the pipe to be 10% of the pressure head at the lower end, calculate

the discharge through the pipe. Also calculate the pressure and velocity at the upper end

Solution:

DA = 0.45 m; DB = 0.25 m; ZB – ZA = Z = 2.3 m; VA = 4.1 m/s;

pA = 63 kN/m2 (gauge); hL = 10%(pA/,); Q = ?; pB = ?; VB = ?

= 1000 kg/m3, g = 10 m/s2 (assumed)

From modified Bernoulli’s equation applied between A and B, we have

Problem 3

A pipe 400 mm diameter carries water at a velocity of 2.5 m/s. The pressure head at pints A and B are

given as 30 m and 23 m respectively, while the datum head at A and B are 28 m and 30 m

respectively. Find the loss of head between A and B.

Solution:

Problem 4

A conical tube of length 2 m is fixed vertically with its smaller end upwards. The velocity of flow at the

smaller end is 5 m/s while at the lower end it is 2 m/s. The pressure head at the smaller end is 2.5

m/s of liquid. The loss of head in the tube is where v1 is the velocity at the

smaller end and v2 is the velocity at the lower end respectively. Determine the pressure head at the

lower end. Flow takes place in the downward direction.

Solution:

Let the smaller end be represented as 1 and lower end as 2 as shown below.

Applications of bernoulli’s energy equation

Bernoulli’s energy equation can be applied in all problems of fluid flow, wherein energy considerations are involved. Some of its applications are

A. Venturimeter.

B. Pitot tube.

Venturimeter

Venturimeter named after the great scientist, is a device used to measure the rate (velocity) of the

flowing fluid through a pipe line.

The main parts of a venturimeter are

Consider two sections at the gauge points 1 and 2. Applying the Bernoulli’s energy equation between

the sections, we get

As the datum passes through the centre line of the two sections, Z1=Z2. Rearranging the above

equation, we get

Let

Where h is the difference of pressure head between the two sections and hence we can write

Applying the discharge continuity equation between the two sections

The above equation is derived by assuming Bernoulli’s equation and in practice, the fluid will be real

and the actual discharge will be lesser than the theoretical discharge obtained by the above equation.

Where Cd is the coefficient of discharge or coefficient of the venturimeter, which is usually less than 1.

It is defined as the ratio of the actual discharge to the corresponding theoretical discharge. The

practical value of coefficient of venturimeter is nearly 1 and may vary from 0.95 to 0.98.

The value of h the differential head between the two sections can be obtained by manometer reading.

Applying Bernoulli’s theorem between sections 1 and 2, we get

Where h is the differential head between two sections

From the manometer reading, the equation can be written as

h = (y + z) So + x Sm – (x + y) So

= z So + x (Sm – So) ¸ So

Where h/So gives the head of Fluid and z = z2 – z1

Comparing Equations. 1 and 3, we get

From the above it can be seen that the general expression for h obtained from the manometer reading

is constant irrespective of the inclination of the venturimeter as it is independent of Z Further, if it is

an inverted U tube manometer, then

Problem 5

A horizontal venturimeter with inlet diameter 200 mm diameter and throat diameter 100 mm is used

to measure the flow of oil of specific gravity 0.8. The discharge of oil through venturimeter is 60 lps.

Find the reading of the oil-mercury differential manometer. Take Cd = 0.98.

Solution:

d1 = 0.2 m; d2 = 0.1 m; Q = 0.06 m3/s; Cd = 0.98; Sm = 13.6 (assumed) x =

?

Problem 6

A venturimeter is to be fit in a 200 mm diameter horizontal pipe line. The inlet pressure is 100 kPa. If

the maximum flow of oil (s = 0.85) is 200 Lps, calculate the least diameter of the throat, so that the

pressure does not fall below 250 mm mercury (vacuum). Assume that 3% of the differential head is

lost between the inlet and the throat.

Solution:

D1 = 0.2 m; p1 = 100 kPa; Q = 0.2 m3/s; s = 0.85; 2 - 0.25 m Hg;

hL = 3% h; Assume = 1000 kg/m3; g = 10 m/s2.

Pressure head at inlet = p1/ g = 100 x 103/(0.85 x 1000 x 10) = 11.765 m of liquid

Pressure head at outlet = p2/g = - 0.25 m of Hg = - 0.25 x 13.6/0.85 = - 4 m of liquid

Differential head = 11.765 – (-4) = 15.765 m of liquid

hL = 3% of 15.765 = 0.473 m of liquid.

Solving the above equation, we get =d2 = 12.13 mm. Note: Due to mistake in DATA, the result is not

compatible (may be in discharge).

Problem 7

A vertical venturimeter carries a liquid of relative density 0.8 and has inlet and throat diameters of

150 mm and 75 mm respectively. The pressure connection at the throat is 150 mm above that at the

inlet. If the actual rate of flow is 40 Lps and the Cd = 0.96, calculate the pressure difference between

the inlet and throat in kN/m2.

Problem 8

Water flows upward a vertical 300 mm x 500 mm venturimeter with a Cd = 0.98. The deflection of

manometer, filled with a liquid of S = 1.25 is 1.18 m. Determine the discharge if the distance

between the two pressure tapings is 457 mm. Work the problem from the first principles.

Solution:

d1 = 0.5 m; d2 = 0.3 m; Cd = 0.98; Sm = 1.25; x = 1.18 m;

Z2 – Z1 = Z = 0.457 m; Q = ?

Pitot tube

It is a device used for measuring the velocity of flow at any section in a pipe or a channel. It is mainly

based on the principle of stagnation pressure. Stagnation pressure is that pressure when the kinetic

energy of a fluid is converted pressure energy as the velocity of flow is brought to sudden rest.

Pitot tube consists of a L shaped glass tube, bent at right angles as shown. The lower end, which is

bent 90o is directed against flow direction as shown. The liquid rises up in the tube due to conversion

of kinetic energy into pressure energy. Due to the stagnation pressure developed at section 2, the

liquid rises in the vertical bent pipe by a height h above the surrounding surface.

Applying the Bernoulli’s equation between the two sections 1 and 2, we have,

Where Cv is the coefficient of velocity which is the ratio of the actual velocity of flow to the theoretical

velocity.

Velocity of flow in a pipe line

The manometer reading depends on the difference of pressure between the hydrostatic pressure head

shown by piezometer or tube perpendicular to the flow direction and stagnation pressure shown by

pipe placed parallel to the flow direction. The pressure head difference is computed in the same way

as that of venturimeter. If x is the manometer reading, then depending on the manometer whether it

is differential manometer or inverted U-tube manometer, the differential head h is respectively

computed as

Where Sm and So is the specific gravities of the manometric fluid and pipe fluid respectively. The

velocity is computed as Hence in general.

Problem 9

A submarine moves horizontally in a sea and has its axis 15 m below the surface of water. A pitot-

tube properly placed just in front of the sub-marine and along its axis is connected to the two limbs of

U-tube manometer containing mercury which reads 170 mm. Find the speed of the sub-marine

knowing that the specific gravity of sea water as 1.026 with respect to fresh water.

Problem 10

A pitot-tube is inserted in a pipe of 300 mm diameter. The static pressure is 100 mm of mercury

(vacuum). The stagnation pressure at the centre of pipe recorded by the pitot-tube is 10 kPa.

Calculate the rate of flow of water through the pipe, if the mean velocity of flow is 0.85 times the

central velocity. Take Cv = 0.98.

Momentum equation

It is based on the law of conservation of momentum or on momentum principle, which states that the

net force acting on a fluid mass is equal to the rate of change of momentum of flow in that direction.

If the force acting on a mass of fluid m is Fx along x direction, the net force along the direction is

given by Newton’s second law of motion as

Fx = m ax

Where ax is the acceleration produced due to the force Fx along the same direction.

The above equation is called momentum principle. The same equation can also be written Fx dt =

d(mu) as which is known as impulse momentum principle and can be stated as “The impulse of a

force acting on a fluid of mass m in a short interval of time dt along any direction is given by the rate

of change of momentum d(mu) along the same direction.

Force exerted by a flowing fluid on a pipe bend

Consider a flow occurring in a pipe bend which is changing its cross sectional area along the bend as

shown in the below figure. Let be the angle of bend and Fx and Fy be the force exerted by the fluid

on the bend along the x and y directions respectively. The force exerted by the bend on the mass of

fluid is -Fx and -Fy. The other forces acting on the mass of fluid are hydrostatic pressure forces at the

two sections 1 and 2 p1A1 along the flow direction and p2A2 against the flow direction respectively.

From the momentum equation, the net force acting on the fluid mass along x direction is given by the

rate of change of momentum in x direction.

Similarly the momentum equation in y direction gives

Problem 11

A 45os degree bend is connected in a pipe line, the diameters at the inlet and outlet of the bend being

600 mm and 300 mm respectively. Find the force exerted by water on the bend if intensity of pressure

at inlet to bend is 88.29 kPa and rate of flow of water is 600 lps.

Solution:

= 450, D1 = 0.6 m, D2 = 0.3 m

p1 = 88.29 kPa, Q = 0.6 m3/s

Assume g = 10 m/s2 and = 1000 kg/m3

Problem 12

Water flows up a reducing bend of weight 80 kN place in a vertical plane. For the bend, the inlet

diameter is 2 m, outlet diameter is 1.3 m, angle of deflection is 120o and vertical height (distance

between the inlet and the outlet) is 3 m. If the discharge is 8.5 m3/s, pressure at the inlet is 280 kPa

and the head loss is half the kinetic head at the exit, determine the force on the bend.

Solution:

W = Weight of the reducing bend acting downwards = 80 kN (), d1 = 2 m,

d2 = 1.3 m, = 120o, Z = 3 m, Q = 2.5 m3/s, p1 = 280 kPa. hL = 0.5

Assume g = 10 m/s2, = 1000 kg/m3, Fx = ? and Fy = ?

Substituting the values, we get

Problem 13

In a 45o bend, a rectangular air duct of 1.0 m2 cross-sectional area is gradually reduced to 0.5 m2.

Find the magnitude of the force required to hold the duct in position, if the velocity of flow is 20.0 m/s

at 1 m2 cross-section and the pressure at both sections is 40 kN/m2. Specific weight of air is 11.0

N/m3.

Solution:

= 450, A1 = 1.0 m2, A2 = 0.5m2, = 11.0 N/m3

p1 = p2 = 40.0 kPa, v1 = 20.0 m/s

Assume g = 10 m/s2

= /g = 1.10 kg/m3

From discharge continuity equation, we have

Q = A1V1= A2V2

20 = 0.5V2

Hence V2 = 40 m/s and

Q = 20 m3/s

Forces acting on the bend in x and y direction respectively are

Fx = Q [V1 – V2 cos ] + p1A1 – p2A2 cos = 25,675.61N

Fy = Q [ – V2 sin ] – p2A2 sin = -14,764.39 N

FLOW THROUGH PIPES

Definition of flow through pipes

A pipe is a closed conduit carrying a fluid under pressure. Fluid motion in a pipe is subjected to a

certain resistance. Such a resistance is assumed to be due to Friction. In reality this is mainly due to

the viscous property of the fluid.

Reynold’s number (Re)

It is defined as the ratio of Inertia force of a flowing fluid and the Viscous force. Re = (Inertia force /

Viscous force) = ( V D / )

Classification of pipe flow

Based on the values of Reynold’s number (Re), flow is classified as Follows:

Laminar flow or Viscous Flow

In such a flow the viscous forces are more predominent compared to inertia Forces. Stream lines are

practically parallel to each other or flow takes place In the form of telescopic tubes. This type of flow

occurs when Reynold’s number Re< 2000. In laminar flow velocity increases gradually from zero at

the boundary to Maximum at the center.

Laminar flow is regular and smooth and velocity at any point practically remains constant in

magnitude & direction. Therefore, the flow is also known as stream Line flow. There will be no

exchange of fluid particles from one layer to another. Thus there will be no momentum transmission

from one layer to another. Example: Flow of thick oil in narrow tubes, flow of Ground Water, Flow of

Blood in blood vessels.

Transition flow

In such a type of flow the stream lines get disturbed a little. This type of flow occurs when 2000< Re

< 4000.

Hydraulic grade line & energy grade line

A Line joining the peizometric heads at various points in a flow is known as Hydraulic Grade Line

(HGL)

Energy grade line (EGL)

It is a line joining the elevation of total energy of a flow measured above a datum, i.e., EGL Line lies

above HGL by an amount V2/2g.

Losses in pipe flow

Losses in pipe flow can be two types viz.,

Minor losses are classified as

i. Entry Loss

ii. Exit loss

iii. Sudden expansion loss

iv. Sudden contraction loss

v. Losses due to bends & pipe fittings.

Head loss due to friction

Consider the flow through a straight horizontal pipe of diameter D, Length L, between two sections

1 & 2 as shown. Let P1 & P2 be the pressures at these sections. To is the shear stress acting along

the pipe boundary.

Pipe is horizontal

Z1= Z2 Pipe is horizontal

V1 = V2

Pipe diameter is same throughout

Substituting equation 2 in equation 1

From Experiments, Darcy Found that

f=Darcy’s friction factor (property of the pipe materials Mass density of the liquid.

V = velocity

Or

But,

from Continuity equation

5 & 6 are known as DARCY – WEISBACH equation

Pipes in series or compound pipe

D1, D2, D3, D4 are diameters.

L1, L2, L3, L4 are lengths of a number of Pipes connected in series .

(hf)1, (hf)2, (hf)3 & (hf)4 are the head loss due to friction for each pipe.

The total head loss due to friction hf for the entire pipe system is given by

Pipes in parallel

D1, D2 and D3 are the pipe diameters. Length of each pipe is same, that is, L1=L2=L3

For pipes in parallel hf1=hf2=hf3 i.e.,

From continuity equation

From continuity equation

Equivalent pipe

In practice adopting pipes in series may not be feasible due to the fact that they may be of

unistandard size (ie. May not be comemercially available) and they experience other minor losses.

Hence, the entire system will be replaced by a single pipe of uniform diameter D, but of the same

length L = L1+ L2+ L3 such that the head loss due to friction for both the pipes, viz equivalent pipe &

the compound pipe are the same.For a compound pipe or pipes in series.

for an equivalent pipe

Equating 10 & 11 and simplifying

Problem 1

Find the diameter of a Galvanized iron pipe required to carry a flow of 40lps of water, if the loss of

head is not to exceed 5m per 1km. Length of pipe, Assume f=0.02.

Problem 2

Two tanks are connected by a 500mm diameter 2500mm long pipe. Find the rate of flow if the

difference in water levels between the tanks is 20m. Take f = 0.016. Neglect minor losses.

Problem 3

Water is supplied to a town of 0.5million inhabitants from a reservoir 25km away and the loss of head

due to friction in the pipe line is measured as 25m. Calculate the size of the supply main, if each

inhabitant uses 200 litres of water per day and 65% of the daily supply is pumped in 8 ½ hours. Take

f = 0.0195.

Solution:

Number of inhabitants = 5million = 5,00,000

Length of pipe = 25km = 25,000m.

Hf = 25m, D=?

Per capita daily demand = 200litres.

Total daily demand = 5,00,000 x 200 = 100 x 106 litres.

Daily supply = 65 / 100 x 100 x 106 = 65,000m3.

Supply rate

Problem 4

An existing pipe line 800m long consists of four sizes namely, 30cm for 175m, 25cm dia for the next

200m, 20cm dia for the next 250m and 15cm for the remaining length. Neglecting minor losses, find

the diameter of the uniform pipe of 800m. Length to replace the compound pipe.

Solution:

L = 800m

L1 = 175m D1 = 0.3m

L2 = 200m D2 = 0.25m

L3 = 250m D3 = 0.20m

L4 = 175m D4 = 0.15m

For an equivalent pipe

Problem 5

Two reservoirs are connected by four pipes laid in parallel, their respective diameters being d, 1.5d,

2.5d and 3.4d respectively. They are all of same length L & have the same friction factors f. Find the

discharge through the larger pipes, if the smallest one carries 45lps.

Solution:

D1 = d, D2 = 1.5d, D3 = 2.5d, D4=3.4d

L1= L2= L3= L4= L.

f1 = f2 = f3 = f4 = f.

Q1 = 45x10-3m3/sec, Q2 = ? Q3 = ? Q4 = ?

For pipes in parallel hf1 = hf2 = hf3 = hf4 ,i.e.,

Problem 6

Two pipe lines of same length but with different diameters 50cm and 75cm are made to carry the

same quantity of flow at the same Reynold’s number. What is the ratio of head loss due to

friction in the two pipes?

Problem 7

A 30cm diameter main is required for a town water supply. As pipes over 27.5cm diameter are

not readily available, it was decided to lay two parallel pipes of same diameter. Find the diameter

of the parallel pipes which will have the combined discharge equal to the single pipe. Adopt same

friction factor for all the pipes.

Equating

Problem 8

wo reservoirs are connected by two parallel pipes. Their diameter are 300mm & 350mm and

lengths are 3.15km and 3.5km respectively of the respective values of coefficient of friction are

0.0216 and 0.0325. What will be the discharge from the larger pipe, if the smaller one carries

285 lps?

Solution:

D1 = 300mm = 0.3m, D2 = -.350m

L1 = 3150m L2 = 3500m

F1 = 0.0216 f2 = 0.0325

Q1 = 0.285m3/sec Q2=?

For parallel pipes

Problem 9

Consider two pipes of same lengths and having same roughness coefficient, but with the

diameter of one pipe being twice the other. Determine (i) the ratio of discharges through these

pipes, if the head loss due to friction for both the pipes is the same. (ii) the ratio of the head loss

due to friction, when both the pipes carry the same discharge.

Problem 10

Two sharp ended pipes are 50mm & 105mm diameters and 200m length are connected in

parallel between two reservoirs which have a water level difference of 15m. If the coefficient of

friction for each pipes of 0.0215. Calculate the rate of flow in each pipe and also diameter of a

single pipe 200m long which would give the same discharge, if it were substituted for the

Original two pipes.

Problem 11

Two pipes with diameters 2D and D are first connected in parallel and when a discharge Q passes

the head loss is H1, when the same pipes are Connected in series for the same discharge the

loss of head is H2. Find the relationship between H1 and H2. Neglect minor losses. Both the

pipes are of same length and have the same friction factors.

Problem 12

Two reservoirs are connected by a 3km long 250mm diameter. The difference in water levels

being 10m. Calculate the discharge in lpm, if f=0.03. Also find the percentage increase in

discharge if for the last 600m a second Pipe of the same diameter is laid parallel to the first.

Solution:

Applying Bernoulli’s equation between (1) & (2) with (2) as datum and considering head loss due

to friction hf

Minor losses in pipes

Minor losses in a pipe flow can be either due to change in magnitude or direction of flow. They

can be due to one or more of the following reasons.

i. Entry loss

ii. Exit loss

iii. Sudden expansion loss

iv. Sudden contraction loss

v. Losses due to pipe bends and fittings

vi. Losses due to obstruction in pipe

Equation for head loss due to sudden enlargement or expansion of a pipe

consider the sudden expansion of flow between the two section 1.1 & 2.2 as shown.

P1 & P2 are the pressure acting at 1.1 and 2.2 , while V1 and V2 are the velocities.

From experiments, it is proved that pressure P1 acts on the area (a2 – a1) i.e. at the point of

sudden expansion.

From II Law of Newton Force = Mass x Acceleration.

Consider LHS of equation 1

Consider RHS of equation 1

Mass x acceleration = x vol x change in velocity / time

= volume / time x change in velocity

Substitution ii & iii in equation i

a2 (p1-p2) = pQ(V1 - V2)

Substitution ii & iii in equation i

a2 (p1-p2) = pQ(V1 - V2)

Substitution ii & iii in equation i

a2 (p1-p2) = pQ(V1 - V2)

Or

a2 (p1-p2) = pQ(V1 - V2)

Both sides by (sp.weight)

Applying Bernoulli’s equation between (1) and (2) with the centre line of the pipe as datum and

considering head loss due to sudden expansion hL only

In equation V hL is expressed in meters similarly, power (P) lost due to sudden expansion is

Problem 1

A 25cm diameter, 2km long horizontal pipe is connected to a water tank. The pipe discharges

freely into atmosphere on the downstream side. The head over the centre line of the pipe is

32.5m, f = 0.0185. Considering the discharge through the pipe.

Problem 2

he discharge through a pipe is 225 lps. Find the loss of head when the pipe is suddenly enlarged

from 150mm to 250mm diameter.

Problem 3

The rate of flow of water through a horizontal pipe is 350lps. The diameter of the pipe is

suddenly enlarge from 200mm to 500mm. The pressure intensity in the smaller pipe is 15N/cm2.

Determine (i) loss of head due to sudden enlargement. (ii) pressure intensity in the larger pipe

(iii) power lost due to enlargement.

Solution:

Q = 350 lps = 0.35m3/s

D1= 0.2m, D2= 0.5m, P1= 15N / cm2 hL= ?, p2= ?, P = ?

From continuity equation

Applying Bernoulli’s equation between 1.1 and 2. 2 with the central line of the pipe as datum and

considering head loss due to sudden expansion hLonly.

Problem 4

At a sudden enlargement of an horizontal pipe from 100 to 150mm, diameter, the hydraulic

grade line raises by 8mm. Calculate the discharge through the pipe system.

Applying Bernoulli’s equation between 1 & 2 with the central line of the pipe as datum and

neglecting minor losses (hL) due to sudden expansion

From continuity equation

Problem 5

Two reservoirs are connected by a pipe line which is 125mm diameter for the first 10m and

200mm in diameter for the remaining 25m. The entrance and exit are sharp and the change of

section is sudden. The water surface in the upper reservoir is 7.5m above that in the lower

reservoir. Determine the rate of flow, assuming f = 0.001 for each of the types.

Applying Bernoulli’s equation between 1 & 2 in both the reservoirs with the water in the lower

reservoir as datum and considering all losses.

FLOW MEASUREMENTS

Introduction

Flow Through Orifices

An orifice is an opening of any cross section, at the bottom or on the side walls of a container or

vessel, through which the fluid is discharged. If the geometric characteristics of the orifice plus

the properties of the fluid are known, then the orifice can be used to measure the flow rates.

Flow through an orifice

As the fluid passes through the orifice under a head H, the stream lines converge and therefore

the jet contracts. The stream lines which converge are mostly those from near the walls and they

do so because stream lines cannot make right angled bend in motion. This phenomenon occurs

just downstream of the orifice, and such a section where the area of cross section of the jet is

minimum is known as VENA CONTRACTA.

The pressure at Vena Contracta is assumed to be atmospheric and the velocity is assumed to be

the same across the section since the stream lines will be parallel and equally spaced.

Downstream of Vena contracta the jet expands and bends down. Figure shows the details of free

flow through a vertical orifice.

Applying Bernoulli's equation between (B) & (C) with the horizontal through BC as datum and

neglecting losses (hL)

Velocity V in Equation 1 is known as TORRICELLI’S VELOCITY.

Hydraulic Coefficients of an orifice

i)Coefficient of discharge (Cd): It is defined as the ratio of actual discharge (Qact) to the

theoretical discharge (Qth)

Value of Cd varies in the range of 0.61 to 0.65

ii) Coefficient of Velocity (Cv): It is defined as the ratio of actual velocity (Vact) to the theoretical

velocity (Vth).

Value of Cv varies in the range of 0.95 to 0.99

Coefficient of Contraction (Cc): It is defined as the ratio of the area of cross section of the jet

at Vena of cross section of the jet at Vena Contracta (ac) to the area of the orifice (a).

Value of Cc will be generally more than 0.62.

Relationship between the Hydraulic Coefficients of an orifice

From continuity equation

Actual discharge Qact = ac x Vact

Theoretical discharge Qth = a x Vth

Equation for energy loss through an orifice

Applying Bernoulli’s equation between the liquid surface (A) and the centre of jet and Vena

Contracta (C) and considering losses (hL).

Equation for coefficient of velocity (CV) (Trajectory method)

Consider a point P on the centre line of the jet, such that its horizontal and vertical coordinates

are x and y respectively. By definition, velocity

Since, the jet falls through a vertical distance y under the action of gravity during this time (t)

Equating equations 1 & 2

Problem 1

The head of water over the centre of an orifice 30mm diameter is 1.5m. If the coefficient of

discharge for the orifice is 0.613, Calculate the actual discharge.

Problem 2

Compensation water is to be discharge by two circular orifices under a constant head of 1.0m,

measured from the centre of the orifices. What diameter will be required to give a discharge of

20x103 m3 per day? Assume Cd for each notch as 0.615.

Problem 3

A jet of water issuing from an orifice 25mm diameter under a constant head of 1.5m falls

0.915m vertically before it strikes the ground at a distance of 2.288m measured horizontally

from the Vena Contracta. The discharge was found to be 102lpm. Determine the hydraulics

coefficients of the orifice and the head due to resistance.

Problem 4

The head of water over a 100mm diameter orifice is 5m. The water coming out of the orifice is

collected in a circular tank 2m diameter. The time taken to collect 45cm of water is measured as

30secs. Also the coordinates of the jet at a point from Vena Contract are 100cm horizontal and

5.2cm vertical. Calculate the hydraulic coefficients of the orifice.

Problem 5

The coordinates of a point on the jet issuing from a vertical orifice are 0.4m & 0.003m.

Neglecting air resistance, determine the velocity of the jet and the height of water above the

orifice in the tank.

Problem 6

A vertical orifice is fitted 0.2m above the bottom of a tank containing water to a depth of 2m. If

G=0.98. What is the vertical distance from the orifice of a point on the jet 0.6m away from the

Vena Contracta?

Problem 7

A closed tank contains water to a height of 2m above a sharp edged orifice 1.5cm diameter,

made in the bottom of the tank. If the discharge through the orifice is to be 4lps. Workout the

pressure at which air should be pumped into the tank above water. Take Cd= 0.6.

Problem 8

A closed tank contains 3m depth of water and an air space at 15kpa pressure. A 5cm diameter

orifice at the bottom of the tank discharge water to the tank B containing pressurized air at 25

kpa. If Cd = 0.61 for the orifice. Calculate the discharge of water from tank A.

Problem 9

A tank has two identical orifices in one of its vertical sides. The upper orifice is 4m below the

water surface and the lower one 6m below the water surface. If the value of Cv for each orifice is

0.98, find the point of intersection of the two jets.

from figure

Substituting equation 3 in equation 4 and simplifying

1.5y2 = y2 + 2

0.5y2 = 2

y2 = 4m

Again

Problem 10

Two orifices have been provided in the side of the tank, one near the bottom and the other near

the top. Show that the jets from these two orifices will intersect a plane through the base at the

same distance from the tank if the head on the upper orifice is equal to the height of the lower

orifice above the base. Assume Cv to be the same for both the orifices.

Solution:

To show that

Problem 1

A 4cm dia orifice in the vertical side of a tank discharges water. The water surface in the tank is

at a constant level of 2m above the centre of the orifice. If the head loss in the orifice is 0.2m

and coefficient of contraction can be assumed to be 0.63. Calculate (i) the values of coefficient of

velocity & coefficient of discharge, (ii) Discharge through the orifice and (iii) Location of the point

of impact of the jet on the horizontal plane located 0.5m below the centre of the orifice.

Problem 2

An orifice has to be placed in the side of a tank so that the jet will be at a maximum horizontal

distance at the level of its base. If the depth of the liquid int the tank is D, what is the position of

the orifice? Show that the jets from the two orifices in the side of the tank will intersect at the

level of the base if the head on the on the upper orifice is equal to the height of the orifice above

the base.

Solution:

Mouth pieces

A mouth piece is a short tube or pipe connected in extension with an orifice

Classification of mouth pieces

Depending on the position with respect to the tank: External, Internal

Depending on shape: Cylindrical, Convergent, Divergent

Nature of flow: Running Full, Running Free

External Cylindrical Mouthpiece

It is a short pipe whose length is two or three times the diameter.

H = Head over the centre of the mouth piece

VO= Velocity of the liquid at Vena Contracta (c) (c)

ac= Area of flow at Vena Contracta

V1= Velocity of liquid at outlet

a1= Area of mouth piece at outlet.

Cc= coefficient of contraction

Applying continuity equation between (c) (c) & (1) &(1)

ac Cc=a1v1

As the jet flows from (c) (c) to (1) (1) there will be loss of head due to sudden enlargement of

flow, and this value can be calculated from the relation.

Applying Bernoulli’s equation between (A) and (1) (1) with the centre line of the mouth piece as

datum and considering head loss hL due to sudden expansion.

By definition, Coefficient of velocity

CV= Actual velocity / Theoretical velocity

Hence, for an external cylindrical mouth piece Cd = (=0.853) is more than that of an orifice.

Pressure head at Vena contracta

Applying Bernoulli’s equation between (A) & (c) (c) with the centre line of the mouth piece as

datum & neglecting losses.

Negative sign indicates that the pressure at the Vena contracta is less than atmospheric pressure

or the pressure is negative.

Problem 1

Find the discharge from a 80mm diameter external mouthpiece, fitted to a side of a large vessel

if the head over the mouth piece is 6m.

Problem 2

An external cylindrical mouthpiece of 100mm diameter is discharging water under a constant

head of 8m. Determine the discharge and absolute pressure head of water at Vena contracta.

Take Cd = 0.855 and CC for Vena contracta =0.62. Take atmospheric pressure head =10.3m of

water.

Problem 3

An external cylindrical mouth piece 60mm diameter fitted in the side of a tank discharges under

a constant head of 3m, for which CV= 0.82. Determine

1. the discharge in lps

2. absolute pressure at Vena contracta

3. Maximum head for steady.

Flow assuming that separation occurs at 2.5m of water absolute. Local barometer reads 760mm

Hg.

2.Absolute Pressure head at Vena contracta

Applying Bernoulli’s equation between (A) & (c) (c) with the centre line of the mouth

piece as datum and neglecting losses hL

From Continuity equation Q = aV

3. Hmax for steady flow

Applying Bernoulli’s equation between Vena contracta and exit of the mouth piece with the

centre line of the mouth piece as datum & considering head loss hL due to sudden expansion of

flow.

Submerged orifice

A fully submerged orifice is one in which the entire outlet side or the downstream side is

completely under the liquid. It is also known as a drowned orifice.

Consider points (1) and (2) situated upstream of orifice and at the Vena contracta respectively.

H1= Height of water above the top of the orifice on the upstream side.

H2= Height of water above the bottom level of the orifice.

H = Difference in water level

b = width of orifice

Ca= Coefficient of discharge.

Height of water above the centre of orifice on upstream side

Applying Bernoulli’s equation between (1) and (2) with the horizontal passing through A & B as

datum and neglecting losses. (hL)

Large rectangular orifice

An orifice is said to be large when the head acting on it is five times the depth of the orifice.

Unlike in the case of a small orifice, the discharge cannot be calculated from the equation for the

reason that the velocity is not constant over the entire cross section of the jet. Consider an

elementary horizontal strip of depth “dh” at a depth h below the free surface of the liquid as

shown.

Area of the strip

Theoretical velocity through the strip =

Discharge through the elementary strip =

Therefore through the entire orifice is obtained by integrating the above equation between the

limits H1= and H2.

Problem 1

Find the discharge through a fully submerged orifice of width 2m if the difference of water levels

on both the sides of the orifice be 800mm. The height of water from the top and bottom of the

orifice are 2.5m and 3m respectively. Take Cd = 0.6

Problem 2

Find the discharge through a totally drowned orifice 1.5m wide and 1m deep, if the difference of

water levels on both the sides of the orifice is 2.5m, Take Cd = 0.62.

Problem 3

Find the discharge through a rectangular orifice 3m wide and 2m deep fitted to a water tank. The

water level in the tank is 4m above the top edge of the orifice. Take Cd = 0.62.

Problem 4

A rectangular orifice 1m wide and 1.5m deep is discharging water from a vessel. The top edge of

the orifice is 0.8m below the water surface in the vessel. Calculate the discharge through the

orifice if Cd= 0.6. Also calculate the percentage error if the orifice is treated as small.

Flow over notches & weirs

A notch is an opening made in the side wall of a tank such that the liquid surface in the tank is

below the upper edge of the opening. Generally notches are made of metallic plates and their

use is limited to laboratory channels.

A weir is a masonry/concrete structures built across an open channel so as to raise the water

level on the upstream side and to allow the excess water to flow over the entire length onto the

downstream side.

Classification

a. Depending on shape

i. Rectangular

ii. Triangular

iii. Trapezoidal

b. Depending on the shape of the crest

i. Sharp crested

ii. Broad crested

c. Depending on flow

i. Free

ii. Submerged

d. Depending on Ventilation

i. Fully aerated

ii. Depressed

iii. Clinging or Drowned

Definition sketch of a Notch shows in the below figure.

The above figure shows in the details of flow over a V – notch.

2 = Central angle

H = Head over the notch

Consider an elemental strip of thickness “dh” at a depth “h” below the free surface as shown.

Discharge through the strip dq = area x velocity

In the above equation x can be eliminated. In terms of & H. In other words

Cd = Coefficient of discharge

In deriving Equation 10, velocity of approach Va is neglected. If the head due to this velocity of

approach is considered, then,

Problem 1

A triangular notch discharges 0.0110m3/s under a head of 0.2m. Find the angle of the notch, if

Cd = 0.626.

Problem 2

A right angled triangular notch discharges 0.143m3/s. Find the head over the notch if Cd = 0.6.

Problem 3

150lpm of water is expected to flow down an irrigation furrow. Design the weir, if a minimum

head of 100mm is desired. Assume Cd= 0.61.

Problem 4

Calculate the top width and depth of a triangular notch capable of discharging 700lps. The weir

discharges 5.7 lps when the head over the crest is 7.5cm. Take Cd=0.62.

Flow over a rectangular notch

L = length of the notch

H = head over the notch

Consider a small strip of thickness dh at a depth h below the liquid surface

Discharge through the strip dq=areaxvelocity

Cd= Coefficient of discharge, its average value is about 0.62.

End Contraction

When the length of the weir(L) is less than the width of the channel (B), the nappe contracts at

the sides, and this is knows as end contractions shown in the below figrure.

According to Francis, the effective length of flow over the notch is given by substituting this

value in equation 12 and simplifying.

A notch without end contraction is known as a suppressed notch

Velocity of approach (Va)

The total head over the weir will be the sum of static head (H) and velocity head (ha), velocity

head ha = is due to the velocity of the liquid approaching the notch.

On similar lines, considering a strip of uniform thickness dh at a depth h below the liquid

surface.

Discharge through the strip dq=area x velocity. dQ = L x dh x

Therefore Total discharge is given by A notch without end contraction is known as a suppressed

notch

Velocity of approach (Va)

The total head over the weir will be the sum of static head (H) and velocity head (ha), velocity

head ha = is due to the velocity of the liquid approaching the notch.

On similar lines, considering a strip of uniform thickness dh at a depth h below the liquid

surface.

Discharge through the strip dq=area x velocity. dQ = L x dh x

Therefore Total discharge is given by

Empirical Formula:

Problem 5

Find the discharge over a rectangular notch of crest length 400mm. When the head of water over

the crest is 50mm. Take Cd = 0.6.

Problem 6

A rectangular weir 9m long is divided into 3 bays by two vertical post each 300mm wide. If the

head of water over the weir is 500mm, Calculate the discharge, given Cd = 0.62.

Problem 7

The discharge over a rectangular weir is 0.4m3/s when the head of water is 0.20m. What would

be the discharge if the head is increase to 0.3m?

Problem 8

A rectangular channel 6m wide carries a flow of 1.5m3/s. A rectangular sharp crested weir is to

be installed near the end of the channel to create a depth of 1m upstream of the weir. Calculate

the necessary height . Assume Cd=0.62.

Solution:

L=6m, Q=1.5m3/s, Cd=0.62.

Y=(Z+H)=1m

Velocity of approach

Va = Discharge / area of flow in the channel

But

Problem 9

A rectangular sharp crested weir is required to discharge 2.645m3/s of water under a head of

1.2m. If the coefficient of discharge is 0.6 and the velocity of approach near the weir is 1m/s.

Find the length of the weir.

Problem 10

A rectangular sharp crested weir is required to discharge 2.645m3/s of water under a head of

1.2m. If the coefficient of discharge is 0.6 and the velocity of approach near the weir is 1m/s.

Find the length of the weir.

Problem 11

Water passing over a rectangular notch flow subsequently over a right angled triangular notch.

The length of the rectangular notch is 600mm and Cd = 0.62. if the Cd value for the V-notch is

0.60, what will be its washing head, when the head on the rectangular notch is 20cm.

Types of nappe

The equations derived for the discharge over notches were under the assumption that pressure

under the nappe is atmosphere. However, when the liquid if flowing over the notch (suppressed),

it touches the walls of the channel and the air gets dissolved or entrained in water, continuation

of this process results in a negative pressure i.e. partial vaccum under the nappe. Finally the

nappe gets deflected closer to the weir wall.

The pressure on the inner side of the the nappe decides its type in the following ways

Free Nappe

In this type, the stream of water passing over the weir the springs clear of the weir.

Depressed Nappe

In this type, a partial vaccum is created between the nappe and the weir. Discharge for such a

flow situation is 8 to 10% greater than that with a free nappe.

Clinging Nappe

In this type the nappe totally adheres to the face of the weir. The discharge in this case would be

20 to 30% more than that in a fully aerated nappe.

Ventilation of weirs

The nappe emerging from a weir should be of a correct form, so that the discharge equations

derived for them are valid.

For accurate gauging of flow the nappe should spring clear or it should be free. In other words

the space between the weir and nappe should be maintained under atmospheric conditions,

particularly when the weir is suppressed.

In practice ventilation holes are made on the weir walls so that air circulates freely between the

weir and the nappe. This is known as ventilation or ventilation of weirs.

Submerged weir

A weir is said to be submerged when water level on both the upstream and downstream sides

are above the crest level of the weir as shown in the below figure

H1 and H2 are the heads over the weir on the upstream and downstream sides.

In the case of submerged weir, it is necessary to derive the discharge equation considering that

the flow over the weir is a combination of a free weir and a submerged orifice.

In other words, the flow Q1 between H1 and H2 is considered as a free weir and Q2 between

H2 and the weir crest as a submerged orifice.

For a free weir

For a submerged orifice

Cd1 = 0.58 and Cd2 = 0.80 are usually considered for the weir and the orifice.

As in the earlier cases the head due to velocity of approach ha = Va2 / 2g can also be

considered.

In such a case

In all the above equation L = length of the notch or weir.

Problem 1

A submerged weir 1m high spans the entire width of a rectangular channel 7m wide. Estimate

the discharge when the depth of water is 1.8m on the upstream side and 1.25m on the

downstream side of the weir. Assume Cd = 0.62 for the weir.

Problem 2

The upstream and downstream water surfaces are 150mm and 75m above the crest of a

drowned weir. If the length of the weir is 2.5m, find the discharge, the coefficients of discharge

for the free and drowned portions may be taken as 0.58 and 0.8 respectively. Allow for velocity

of approach.

Ogee weir

When the weir is suppressed and its height is large, the nappe emerging out may be subjected to

the problems of ventilation. Hence, in such cases the weir profile downstream is constructed

conforming to the shape of the lower side of the nappe. Such a weir is known as a spillway or

ogee weir.

The cross section of an ogee weir will be shown in the below figure. The coordinates of the

spillway profile can be worked out for the head H using the equation x185 = 2H085 y

The u/s face of the spillway is generally kept vertical.

Problem 1

Calculate the discharge over an ogee weir of 8.5m length, when the head over the crest is 2.15m

and Cd=0.61.

Broad crested weir

A weir is said to broad crested when its width (parallel to flow) b is greater than 0.5xmaximum

head acting on it.

.e., b > 0.5 x H

Let L = length of the weir

H = Head of water u/s of the weir w.r.t. the crest

h = Depth of water over the weir crest

V = Vel. Of flow over the weir

Applying Bernoulli’s equation between (1) and (2) with the crest of the weir as datum &

neglecting losses (hL).

Discharge over the weir Q = area of flow over the weir x vel. of flow over the weir.

i.e.

Actual discharge

From Equation 20, we see that Qact is a function of h for a given value of H.

Q is maximum when

Problem 1

Determine the discharge over a broad crested weir 26m long, the upstream level of water is

measured as 0.5m above the crest level. The height of the weir is 0.6m and the width of the

approach channel is 36m. Take Cd=0.9.

Problem 2

A reservoir discharges water at 60,000 m3/day over a broad crested weir, the head of length of

the weir, if Cd = 0.65.

Problem 3

A channel of 45m2 cross sectional area, discharging 50 cumecs of water is to be provided with a

broad crested weir. If the crest of the weir is 1.6m below the upstream water surface, find the

length of the weir, if Cd=0.85.