Merge Fluid Mechanics

FLUID MECHANICS

CENTRE FOR PROFESSIONAL DEVELOPMENT & LIFELONG LEAR NING UNIVERSITY OF MAURITIUS

Support Materials

Fluid Mechanics Aug 2008 ii

CONTRIBUTORS

FLUID MECHANICS (Support Materials)

was prepared by

Associate Professor M. Nowbuth,

from the

Faculty of Engineering,

University of Mauritius.

August 2008

All rights reserved. No part of the work may be reproduced in any form, without the

written permission from the University of Mauritius , Réduit,

Fluid Mechanics Aug 2008 iii

TABLE OF CONTENTS

ABOUT THE COURSE

Unit 1

Properties of Fluids

Unit 2

Fluid Pressure

Unit 3

Measurement of Fluid Pressure

Unit 4

Hydrostatic Forces on Plane Surfaces

Unit 5

Hydrostatic Forces on Curved Surfaces

Unit 6

Pressure Diagrams

Unit 7

Buoyancy

Unit 8

Hydrodynamics – Fluid Dynamics (in Motion)

Unit 9

Principles Of Conservation Of Mass & Energy

Unit 10

Flow Rate Measurements – Orifices & Weirs

Unit 11

Flow Rate Measurements – Venturimeters

Fluid Mechanics Aug 2008 iv

ABOUT THE COURSE

Welcome to FLUID MECHANICS 1 (CIVE1104) is the introductory module of a series of Fluid Mechanics modules which comes in levels 2, 3 and 4, namely, CIVE2211, CIVE3104, CIVE4007.

The aims of this module are to help you:

• Appreciate the fundamental principles governing fluid at rest and in motion.

• Learn about flow measuring devices.

• Learn about forces exerted by fluids within a system.

This module introduces you to the fundamentals behind the approach of analysis fluids at rest or

in motion. In general, you tend to find this introductory module of Fluid Mechanics confusing

and vague, since you are most probably used to analysing the more concrete solid mechanics

behaviour.

This manual has thus been structured in such a way that you are gradually introduced to the

various concepts, through a set of theoretical notes, diagrams and simple examples. The

complexity of examples will grow as you proceed through the contents of the modules stepwise.

Your attention is drawn to the fact that all the units forming this module are interlinked, so

the units should be studied progressively for the first time. Once you become familiar with the

various units, you can then consult each unit on its own.

You are strongly advised to ensure that the contents of this module are well assimilated and that

you have properly understood the fundamental concepts governing the analysing of fluids. These

basic concepts will be used again and again in the more advanced level modules of Fluid

Mechanics. A series of tutorials are included in the manual, and unless you actually attempt

them on your own, you will fail to get the thorough understanding of the individual units.

Fluid Mechanics Aug 2008 v

LEARNING OBJECTIVES FOR THE COURSE By the end of the course, you will be able to do the following:

Unit 1 Identify the basic concepts, theories and equations related to the properties

of fluids.

Unit 2 Derive the equations governing variation of pressure with depth and

variation of pressure along a horizontal plane

Unit 3 Be conversant with the various apparatus used to measure the pressure

exerted by a fluid.

Unit 4 Learn how this fluid pressure is expressed in terms of forces, commonly

termed hydrostatic forces when the fluid is at rest.

Unit 5 Calculate the individual forces acting at different points along solid and

curved surfaces.

Unit 6 Use the different approaches of calculating the resultant hydrostatic force

acting on vertical plane surface only.

Unit 7 Learn about equilibrium conditions in liquid medium and governing factors

influencing stability of structures in such medium.

Unit 8 Discuss about the concepts and approaches used to analyse fluids in motion

(hydrodynamics).

Unit 9 Define and differentiate between two main principles used to analyse fluids

in motion, the principles of Continuity and the principles of Conservation

of Energy.

Unit 10 Apply the two main principles: the principles of Continuity and the

principles of Conservation of Energy.

Unit 11 Measure with a venturimeter, the flow rate within a closed conduit.

Fluid Mechanics Aug 2008 vi

HOW TO PROCEED

COURSE MATERIALS

The manual is self-contained. HOW DO I USE THE COURSE MANUAL?

Take a few minutes now to glance through the entire manual to get an idea of its structure.

Notice that the format of the different units is fairly consistent throughout the manual. For

example, each unit begins with an OVERVIEW, and LEARNING OBJECTIVES sections.

The OVERVIEW provides a brief introduction to the unit and provide perquisite skills and

knowledge you will have to possess to proceed successfully with the unit.

You should then read the LEARNING OBJECTIVES . These objectives identify the

knowledge and skills you will have acquired once you have successfully completed the study of

a particular unit. They also show the steps that will eventually lead to the successful completion

of the course. The learning objectives also provide a useful guide for review.

WHERE DO I BEGIN?

You should begin by taking a look at the TABLE OF CONTENTS in the MANUAL . The

table provides you with a framework for the entire course and outlines the organisation and

structure of the material you will be covering. The Course Schedule indicates how you should

allocate your workload and what you should be working on in each week to be ready for the

respective class. You should stick to the Course Schedule to ensure that you are working at a

steady space and that your workload does not pile up.

Fluid Mechanics Aug 2008 vii

Proposed * Course Schedule (CIVE 1104)

Session Student’s Workplan

01 Read Unit 1

02 Read Unit 2

03 Read Unit 3.

04 Read Unit 4

05 Read Unit 5.

06 Read Unit 6

07 Read Unit 7

08 Read Unit 8.

09 Read Unit 9

10

11

12 Read Unit 10

13 Read Unit 11

14

15

* Any change will be communicated by your respective tutor in class.

Revision

CLASS TEST

Revision

Fluid Mechanics Aug 2008 viii

NOTE (to b confirmed by lecturer): For this module you are required to submit/present three practical reports which will be assigned

to you by your lecturer during the course of the module.

ASSESSMENT → COURSE GRADING SCHEME:

Continuous Assessment: 30 marks

Examinations: 70 marks

→ FINAL EXAMINATIONS:

■ Scheduled and administered by the Registrar’s Office

■ A two-hour paper at the end of the Semester.

STUDY TIPS Much of your time in the course will be spent reading. Your comprehension and assessment of

what you read are likely to be best if you heed the following tips:

1. Organise your time. It is best to complete each assigned reading in one sitting. The

logical progression of thought in a chapter/unit can be lost if it is interrupted.

2. Be an active reader. Use question marks to flag difficult or confusing passages. Put

exclamation marks beside passages you find particularly important. Write short

comments in the margins as you go. For example, if you disagree with an author’s

argument or if you think of examples which counter the position presented, note your

opinions in the margin.

If you prefer to leave your book pages unmarked, you can make your notations on “post-

it-notes”.

3. Read critically. You must evaluate, as well as appreciate and understand, what you read.

Ask questions. Is the author’s argument logical? Are there alternatives to the author’s

Fluid Mechanics Aug 2008 ix

explanations or to the conclusions drawn? Does the information fit with your

experience?

4. Take notes. If you make notes on an article or chapter right after finishing it, you reap a

number of benefits. First, note-taking allows you an immediate review of what you have

just read. (You will find that this review helps you recall information). Second, it gives

you an opportunity to reassess your flagged or margin comments. Finally, it gives you a

second shot at deciphering any confusing passages.

5. Review your scribbling! Whether or not you make separate notes on your readings,

review your flags, underlining and marginalia. Study closely those passages you

considered significant or difficult.

6. Write down your ideas in a course journal. As you progress through the course, the

new information you absorb will stimulate new thoughts, questions, ideas, and insights.

These may not be directly related to the subject matter, but may be of great interest to

you. Use these ideas to focus your personal involvement in this and other courses.

7. Your ability to explain the subject matter to others is a good test of your true

comprehension of the material. Try explaining the material you are learning to others,

classmates or friends, without resorting to jargon. Even if some of them are not directly

involved with the techniques discussed in this course, many of the concepts may be of

interest to them.

8. Activities found in units will not be marked. We strongly recommend that you do not

skip any of them. They will help you prepare for the graded assignments.

Now, it’s time to get to work. Good luck and enjoy the course!

Fluid Mechanics Aug 2008 1

UNIT 1 PROPERTIES OF FLUIDS

Unit Structure

1.0 Overview

1.1 Learning Objectives

1.2 Introduction

1.3 States of Matter

1.3.1 Difference Between Solid and Fluid

1.3.2 Technical Terms commonly Used to Describe the Properties of Fluids

1.3.2.1 Mass Density or Density, ρρρρ

1.3.2.2 Specific weight, w

1.3.2.3 Relative Density ( R) or Specific gravity, s

1.3.2.4 Specific volume, v

1.3.2.5 Viscosity

1.3.2.6 Coefficient of Dynamic (µ) and Kinematic (γγγγ) Viscosity

1.3.3 Real & Ideal fluids

1.3.3.1 Surface Tension

1.3.3.2 Vapour Pressure

1.4 Bulk Modulus

1.5 Cavitation

1.6 Activities

1.7 Summary

1.8 Worked Examples

1.9 Tutorial

1.0 OVERVIEW

This first unit emphasizes the need to be familiar with technical terms related to the description

and analysis of fluids in motion and at rest, their exact definitions and their S.I. (Standard

International) units.

When analysing fluids at rest or in motion, students will need to know how the equations being

used for the analysis were derived, what are the assumptions behind the derivation of the


equations and does the equation hold true for this particular situation. The results of an analysis

are very much dependent on the validity of the equations used in the analysis, and students will

have to be well conversant with the equations used. The implication is that there will be no

more learning equations by heart and replacing numbers. Students will need to derive

equations before applying them and from there, get a feel about whether the results they obtain

have a practical and realistic meaning.

So in this unit, you will gradually be introduced to technical terms commonly used in describing

properties of fluids, the basic units of length, mass and time, the units of various fluid parameters

and the need for assumptions in the analysis of fluid mechanics. This is a very important unit,

it is simple, and it should be properly studied and appreciated, for it will come up throughout

the entire course on Fluid Mechanics, until the final level of your course, in basically all the

different units.

1.1 LEARNING OBJECTIVES

At the end of this unit, students should be able to do the following:

1. State the main difference between a solid and a fluid.

2. Differentiate between real & ideal fluid.

3. Identify situations when a real fluid has to be treated as an ideal fluid

4. Elaborate on the different ways of expressing fluid properties such as density, viscosity

and compressibility.

5. Define cavitation.

6. Explain how cavitation is taken care of in design of pipelines.

1.2 INTRODUCTION

Unit 1 introduces students to the basic concepts, theories and equations related to the properties

of fluids. The main properties differentiating behaviour of a solid and a fluid, basic properties

used to describe fluids, importance of SI units of each term within an equation and the need to

resort to assumptions during analysis of fluids behaviour will be introduced and discussed in the

course of the unit.


1.3 STATES OF MATTER

Matter can exist in 3 different states - Solid, Liquid and Gas. The state of a substance dictates to

a large extent the behaviour of that substance under static or dynamic conditions. Both liquids

and gases are considered to be under the category fluids.

1.3.1 Difference between a Solid & a Fluid

A solid has a definite shape while a fluid takes the shape of the vessel containing it. In a more

technical term, a solid offers resistance to a force while a fluid cannot resist an applied force.

Hence, in order to better distinguish between a solid and a fluid medium, we shall examine the

response of each substance to an applied shear force:

Apply of a force:

Solid: Offers resistance to the deforming force

Liquid and Gas (Fluid): Deform continuously as long as the force is applied.

Other properties of a fluid:

It flows under its own weight.

It takes the shape of any solid body with which it comes into contact.

Definition of a fluid:

A fluid (liquid or gas) is a substance which deforms continuously under the action of shearing

forces.


1.3.2 Technical Terms Commonly Used to Describe the Properties of Fluids

The properties of a fluid are characterised by its density, its viscosity and its degree of

compressibility. The fluid property density can be expressed in several ways, mass density also

commonly referred to as density, specific weight, specific gravity, relative density or specific

volume. Similarly viscosity of a substance can either be expressed as the coefficient of dynamic

viscosity or the coefficient of kinematic viscosity. The degree to which a fluid can be

compressed is expressed by the term bulk modulus of compressibility. Different symbols

are used to differentiate between different terms and each term is associated with a specific SI

unit.

(Students need to pay particular attention to the definition, symbol and unit for each of

these terms. Students would need to learn and remember these basic technical terms, to

be able to recognise precisely in which way the density of a substance has been stated in

a given question.)

1.3.2.1 Mass Density or Density, ρρρρ

The density of a substance is defined as the mass per unit volume,

ρ = mass / volume

units: kg m-3

Example: Mass density of water at 4oC is 1000 kg m-3


1.3.2.2.1.1 Specific weight, w

Another way of expressing the density of a substance, is by the term specific weight, where

specific weight is the weight per unit volume,

w = weight / volume

units: Mass x Acceleration due to gravity / volume

kg m s-2 / m-3 or N / m3

Example: Specific Weight of water at 4oC is 9.81 kN/ m3

1.3.2.3 Relative Density (R) or Specific gravity, s

The specific gravity of a substance, also known as the relative density of the substance, is the

ratio of its density or specific weight to that of water under standard conditions of temperature

and pressure. The standard pressure for water is usually taken as one atmosphere and

temperature as 4oC,

s = Density of substance / Density of water

or

Specific weight of substance / Specific weight of water

Example: Relative Density or Specific gravity of mercury = 13.6


1.3.2.4 Specific volume, v

The specific volume v, is defined as the volume occupied by a unit mass of the substance, it is

the reciprocal of the density, ρ

v = 1/ρ

Units: m3 per unit weight of substance

Example: The specific volume of 1 kg of water = 10-3 m3 per unit weight of water kg

1.3.2.5 Viscosity

When a shearing force is applied to a liquid which is initially as rest, the fluid cannot resist the

shearing forces. The fluid will flow in such a way, that the fluid in contact with the boundary,

will have the same velocity as the boundary, while successive layers of the fluid will move with

increasing velocity, away from the boundary (Figure 1.1).


Figure 1.1: Application of a shearing force to a f luid initially at rest

Open channel running completely full

A A’ B’ B

C D

Applied force, F

Stationary base of channel Velocity of fluid at surface, U

Velocity of fluid in contact with the base of the channel is 0.


To explain the behaviour of a fluid when shearing force is exerted, consider an element of a fluid

(ABCD), which is initially at rest between two solid surfaces separated by a small distance, h,

along the y direction (Figure 1.1). Now, suppose that the upper solid surface is moved in the x-

direction, by applying a force, F (Figure 1.1). The elemental fluid will undergo a change in

shape as illustrated by element A’B’CD (Figure 1.1). The bottom layer of the fluid immediately

in contact with the boundary will have the same velocity of the boundary, i.e, the velocity at y=0,

will be equal to 0 (Figure 1.1). The layer of fluid in contact with the moving upper solid surface,

will have the same velocity as the velocity of the moving surface, velocity U. Successive fluid

layers will have velocity which increases from bottom to the top surface (Figure 1.2).

Figure 1.2: Consider a small elemental fluid under the application of a shearing force

With reference to Figure 1.2; application of the force F, has caused a shearing action to be

exerted within the fluid element, whereby

Shearing stress, ττττ = F/shearing area = F/AB x thickness of elemental fluid Shear Strain = δθ = δx / δy Shear Stress α Rate of change of shear strain

ττττ α δθ / δt

ττττ α (δx/ δy) / δt = δu/ δy (where δx/ δt = δu)

A A1 B1 B

CD Stationary base of channel

APPLYING A SLIDING FORCE AT SURFACE OF LIQUID

Successive fluid layers sliding above each other, since each has a different velocity, Top layer, velocity, V Bottom layer, velocity 0

δ

δδ

Applied force, F


Shearing force per unit area, shear stress, τ = F / AB x s

The consequent, shear strain acting on the fluid element, ϕ , can be represented by the term x/y.

The shear strain will continue to increase with time and the fluid will flow. It is found

experimentally that the rate of shear strain is directly proportional to the shear stress, for a fluid:

Assuming that shear stress is proportional to rate of shear strain,

τ α δϕ / time

τ α δx / δy per unit time

where distance x per unit time, can also be expressed as the velocity, u

τ α δu/δy

Removing the symbol of proportionality results in the following equation:

τ = µ δu/δy ………………………equation A

whereby the term µµµµ is known as the coefficient of absolute or dynamic viscosity of the fluid,

Units: kg m -1 s-1

The relationship defined by equation A is known as Newton’s Law of Viscosity.

1.3.2.6 Coefficient of Dynamic (µ) and Kinematic (γγγγ) Viscosity

Viscosity is the property of a fluid which offers resistance to fluid deformation by the application

of a tangential, shearing force (Figure 1.2). For liquids, viscosity arises mainly from the cohesive

force of molecules.

The coefficient of dynamic or absolute viscosity, µ , is a function of temperature, for a specific

fluid. When the temperature of a fluid increases, the cohesive forces within fluid body


decreases. This results in a decrease in the shear stress, hence, eventually, in a decrease in the

coefficient of viscosity.

Another way of defining coefficient of viscosity, is by using the coefficient of kinematic

viscosity, which is the coefficient of dynamic viscosity divided by the density of the fluid:

γ = µ / ρ

where γ = the coefficient of kinematic viscosity,

units : m2 s-1

1.3.3 Real and Ideal Fluids

Fluids that obey Newton’s equation, (equation A), are classified under the group of Newtonian

fluids, e.g air and water. Fluids such as tar and polymers do not obey equation A, the relation

between shear stress and the rate of shear strain in these cases is non-linear (Figure 1.3).

The concept of an ideal or perfect fluid is based on theoretical considerations because all real

fluids exhibit viscous property. When µ = 0, then u/y = 0, hence shear stress, τ , vanishes.

Therefore, a real fluid, with coefficient of viscosity very small, and velocity gradient very small

can be considered as being frictionless, i.e, no shearing action takes place.

An ideal fluid is one which has zero viscosity, and in many problems arising in fluid mechanics,

we often have to resort to the assumption that the fluid we are dealing with is an ideal one,

(however, only theoretical solutions are obtained in these cases).


Figure 1.3: Real & Ideal fluids

1.3.3.1 Surface Tension

Consider a molecule p, of fluid well inside the body of the fluid (Figure 1.4).

Figure 1.4: Liquid molecules and forces of attraction

Shear Stress, ττττ

Velocity gradient, δδδδv/δδδδy

Newtonian fluid ττττ αααα δδδδv/δδδδy µµµµ is constant – LINEAR RELATIONSHIP

Ideal fluid , ττττ = 0, µµµµ is zero

Non-newtonian fluid ττττ αααα δδδδv/δδδδy µµµµ is not constant – NON LINEAR RELATIONSHIP

Molecule p is at rest, hence equal & balancing forces are acting on p

Molecule q is at the surface, and unequal forces are acting on B, a resultant downward attractive force is exerted on q


Molecule p, is attracted equally in all directions by the surrounding molecules within a small

sphere of radius, a. But molecule, q, which is on the surface of the liquid, will experience a

resultant pull inward due to the unbalanced cohesive force of attraction. The surface molecules

are being pulled inward towards the bulk of the liquid, this effect causes the liquid surface to

behave as if it were an elastic membrane under tension.

The surface tension, σ , is measured as the force acting across unit length of a line drawn in the

surface, and surface tension in a liquid has a tendency to contract to a minimum surface area for

a given volume.

1.3.3.2 Vapour Pressure

Molecules of liquids that possess sufficient kinetic energy leave the liquid surface and become

vapour. If the vapour is confined to a space, an equilibrium condition is obtained when the

amount of vaporisation is equal to the amount of condensation. The vapour pressure at this

condition is called saturation vapour pressure, which depends on the temperature and increases

with its rise. The degree of molecular activity increases with increasing temperature, and

therefore, the vapour pressure will also increase (Figure 1.5). Boiling will occur when the

vapour pressure is equal to the pressure above the liquid. By reducing the pressure, boiling can

be made to occur at temperatures well below the boiling point.


1.4 BULK MODULUS – COMPRESSIBILITY PROPERTIES OF FL UIDS

Compressibility is basically the change in volume of a substance when subjected to a

compressive force. All matter whether solids, liquids or gases is to some extent compressible.

While gases are highly compressible, solids and liquids offer much resistance to compressive

forces. Compressive forces causing major changes in volume, also result in changing the density

of a substance (density=mass/volume).

The degree of compressibility of a substance is expressed by the term Bulk Modulus, K.

Consider a substance of original volume V, subjected to a compressive force, and thereby

undergoes a change in volume, expressed by term δv. If the force which is applied to compress

the fluid is increased from P to P + δP, then the relationship between the change in pressure and

the change of volume as defined by the property Bulk Modulus is given by:

Figure 1.5: Saturated vapour conditions

1

2

3

4

Few water molecules possess enough energy & go into vapour state

More water molecules

possess enough energyto go into vapour state

Increasing heat, further increases the number of molecules in vapour state

Air space eventually becomes completely saturated & cannot take more vapour molecules & At this point an equilibrium is achieved,number of water Molecules going into vapour state is the same as the number of molecules going back into liquid state


Change in volume / Original Volume = Change in Pressure / Bulk Modulus

-δδδδV/V = - δδδδP/K

where K is the Bulk Modulus.

K = -V δδδδP/δδδδV

Bulk Modulus of a fluid can also be expressed in terms of density - Considering unit mass of a

substance V=1/ρ;

Then K = ρρρρ dp/dρρρρ

NOTE:

When the change in volume is very small then a liquid can be assumed to be incompressible

and this happens to be an assumption often made in fluid mechanics problems. As far as a gas is

concerned, the compressibility of a gas is very large, and can rarely be ignored. If the pressure

applied is very small, only in such a case, a gas can be assumed to be incompressible.

Units: same as pressure (N/m2)

Typical values: 2.05 x 109 N/m2 (water)

1.62 x 109 N/m2 (oil)

1.5 CAVITATION

Consider a pipeline running full (Figure 1.6). Gases are known to be soluble in water under given

temperatures and pressures. Under certain conditions along a pipeline, localised zones of low

pressures (at A & B) can occur. If the pressure in such areas falls below the vapour pressure at

which certain gases were initially soluble in water, at those points the gases are no longer

soluble, and are thus released. Vapour bubbles are formed at A and B, but being in vapour state

and thus light, they will move towards the highest points within the pipeline (C).


The highest points within the pipeline may be regions of high pressures, whereby the gas

molecules are once again soluble in the water. At this point, (C), the gas molecules will collapse

suddenly to go back to the liquid phase. To change from gaseous to liquid state, the gaseous

molecules will have to lose excess energy. This is achieved by striking either against each other

or by striking against the walls of a container. This phenomenon can cause serious problem, and

is known as cavitation. As this process takes place over time, usually years in the case of pipes,

at location C, the pipe gradually gets damage, until it breaks. To avoid such damage to pipelines,

air valves are normally positioned at highest points along pipelines, thus allowing the gaseous

molecules to escape.

Figure 1.6: Cavitation within a pipeline

1.6 ACTIVITIES

(a) True/False Statements

1. Solids and fluids behave in the same way under the action of stress.

2. Liquids and gases behave in the same way under the action of all types of stresses.

3. Liquids are often assumed to be ideal since the fluid property viscosity is often an

unknown parameter in analysis of fluids.

A B

CGaseous molecules going back

to liquid state

Gas molecules released from water at zones of low pressures


4. Density and specific weight are terms describing the same property of a fluid.

5. Temperature affects the property of liquids and gases in the same way.

6. Cavitation is the process of release of gases from a liquid.

7. Air valves are always needed in water pipeline owing to cavitation process.

8. Under all conditions, a liquid can be safely assumed to be incompressible.

(b) Main Questions:

1. Derive with the help of sketches, the equation relating shear stress and velocity gradient,

for a fluid in motion.

2. Show that the bulk modulus of compressibility can also be expressed in terms of the

pressure and density parameters.

3. Describe the process of cavitation with the help of well labelled sketches.

4. Differentiate with the help of curves, Newtonian and non-Newtonian fluids.

1.7 SUMMARY

This basic unit introduced the basic technical terms commonly used to describe the properties of

fluids. The terms introduced will be used in the coming units.

In this unit, we explained the:

1. Importance of getting the right meaning of various technical terms in fluid mechanics.

2. Importance of getting the right units for the various technical terms in fluid mechanics.

3. Meaning of the different ways of expressing the fluid properties, density and viscosity.

4. Derivation of shear stress and velocity gradient relationship.

5. Concept behind the process of cavitation.


6. Importance of the fluid property bulk modulus of compressibility for a liquid and for a

gas.

The next unit will be concerned with fluid pressure: what is fluid pressure, how it varies over

space and how it is measured. Once again you are strongly advised to ensure that the various

concepts illustrated in this unit are clear to you before you proceed to UNIT 2.

1.8 WORKED EXAMPLES

Example 1 – Shearing force & Viscosity properties

Example 1 – Shearing forces & Viscosity properties…1/2

• The space between two large flat and parallel walls 25 mm apart is filled with a liquid of absolute viscosity 0.7 N m-2 s. A thin flat plate 100mm x 100mm, is located mid way between the two large walls. If the thin flat plate is being towed at 75mm/s, determine the force exerted by the liquid on the thin flat plate.

25mmV=75mm/s


Example 1 – Shearing forces & Viscosity properties…2/2

• Shear stress = coefficient of dynamic viscosity x v elocity gradient

• τ = µ δV/δy = 0.7 x ((75 x 10-3 – 0) / 25/2 x 10-3} = 4.2 N/m2

• Shearing force acting on the thin plate– Force = Shearing stress x surface area of the plate– Force = 2 x 4.2 x (100 x10-3)2 = 0.084 N

• NOTE: – Shearing forces are acting on the thin plate on the top surface

as well as on the bottom surface.– The velocity distribution is assumed to be linear in this example,

maximum at the centre and zero at the contact of the walls.

25mm V=75mm/s

1.9 TUTORIAL Question 1

The space between two large flat and parallel walls 25 mm apart is filled with a liquid of

absolute viscosity 0.7 N m-2 s. Within this space a thin flat plate, 250 by 250 mm is towed at a

velocity of 150 mm s-1 at a distance of 6 mm from one wall, the plate and its movement being

parallel to the walls, determine the force exerted by the liquid on the plate.

Question 2

The velocity distribution for viscous flow between stationary plates is given as follows:

v = dP/dx (By - y2) 2 µ

If glycerine (µ = 0.62 N s m-2) is flowing and the pressure gradient dP/dx is 1.6 kN/m3, what is

the velocity and shear stress at a distance of 12 mm from the wall if the spacing B is 5 cm? What

are the shear stress and velocity at the wall?


Question 3 The velocity profile in laminar flow through a round pipe is expressed as

U = 2V (1- r2/ro2)

Where V is the average velocity, r is the radial distance from the centre line of the pipe, and ro is

the pipe radius. Draw the dimensionless shear stress profile τ/τo against r/ro. What is the value of

wall stress when fuel oil having absolute viscosity µ = 4 x 10-2 N s/m2 flows with an average

velocity of 4m/s in a pipe of diameter 150mm?

Question 4

A bush of 165mm length and 103mm internal diameter slides on a vertical column of 100mm

diameter, the clearance space being filled with oil. If a 3.5kg bush mass slides with a velocity of

1m/s, determine the coefficient of dynamic viscosity of the oil.


UNIT 2 FLUID PRESSURE

Unit Structure

2.0 Overview


2.2 Introduction

2.3 Pascal’s Law for pressure at Point

2.4 Variation of pressure vertically in a fluid under gravity

2.5 Equality of pressure at the same level in a static fluid

2.6 Basic general equation for fluid statics (at rest)

2.7 Activities

2.8 Summary

2.9 Worked Examples

2.10 Tutorial

2.0 OVERVIEW

In this unit, student will be introduced to the concept of pressure exerted by a fluid. Student will

need to understand and appreciate variation of pressure with depth and along a horizontal plane.

These concepts will form the basis for the coming units, measurement of pressure and

hydrostatics. Some of the concepts learnt in this unit will also be used in flow measurement in

later units.

Students are strongly advised to make sure that the concept illustrated in this unit is clear,

for this unit forms an introductory unit to many ot hers that will follow later on in this

course.



At the end of this Unit, students should be able to do the following:

7. Define and prove Pascal’s Law

8. Derive the equation governing variation of pressure with depth

9. Derive the equation governing variation of pressure along a horizontal plane

2.2 INTRODUCTION

Fluids exert a force on the walls on the vessel containing it. The magnitude of this force which

can also be expressed as pressure (force per unit area) varies with the depth of fluid from the top

water surface. In this unit, you will learn how fluid exerts pressure at a point and how this

pressure varies both with depth and along horizontal planes.

2.3 PASCAL’S LAW FOR PRESSURE AT A POINT

When a fluid is at rest, there are no shearing forces acting on the fluid, and consequently, the

fluid is in equilibrium. Under these conditions, the only force that such a fluid can sustain acts

normally on a surface within the fluid. The normal force per unit area (acting inwards) is termed

the fluid pressure, p.

The famous seventeenth century mathematician, B. Pascal, first established that the pressure at

any point within a stationary fluid is the same in all directions.

Consider equilibrium of a small fluid element in the form of a triangular prism ABCDE

surrounding a point in the fluid (Figure 2.1):


Figure 2.1: Pressure at a Point

If the fluid is at rest, the following pressure forces will be acting over the prism:

the pressure acting at right angles to plane ABFE, px,

the pressure acting at right angles to plane CDEF, py

the pressure acting at right angles to plane ABCD, ps

Since the fluid is at rest, the sum of all the forces in any direction must be equal to zero:

Solving for resultant forces in the x direction:

= px (area ABFE) - ps sin θ (Area ABCD)

= px (δ zδy) - ps (δy/δs) δsδz

= px δ zδy - ps δyδz………………..equation A

For equilibrium conditions, equation A, is equal to 0

Therefore, px = ps

θθ

A

B

F

E

px

C

D py

ps

θ δδδδx

δδδδs

δδδδyy


Similarly resolving for forces in the y direction

py = ps

Hence, pz = py = ps

Given that ps, is the pressure on a plane at any angle θ , the x, y and z axes have not been chosen

with any particular orientation, and the element is so small that it can be assumed to be a point,

therefore indicating that the pressure at a point is the same in all directions. This is known as

Pascal’s Law to a fluid at rest.

2.4 VARIATION OF PRESSURE VERTICALLY IN A FLUID UND ER GRAVITY

Consider a cylindrical element of fluid of constant cross-sectional area A, totally surrounded

with fluid of mass density, ρ: (Figure 2.2)

Figure 2.2: Cylindrical elemental fluid The pressure acting on the bottom of the elemental fluid is p1, and on the top of the elemental

fluid is p2

Cross sectional area, A

Elevation = Z1

Elevation = Z2

Pressure = P1

Height = h

Liquid density, ρ


Resolving for vertical forces acting on the elemental fluid:

Force due to pressure p1 = p1 * A

Force due to pressure p2 = p2 * A

Force due to weight of the fluid = Volume x Density x gravity

= A (z2 - z1) ρ g

Since fluid is at rest, and consequently under equilibrium conditions, the resulting vertical force acting on the fluid is equal to 0: p1 A - p2 A + A (z2 - z1) ρ g = 0 p1 - p2 = - (z2 - z1) ρ g Thus in any fluid under gravitational attraction pressure decreases with increase of height h.

2.5 EQUALITY OF PRESSURE AT THE SAME LEVEL IN A STA TIC FLUID

Consider a cylindrical element (Figure 2.3):

Figure 2.3: Cylindrical elemental fluid

If P and Q are two points at the same level in a fluid at rest, a horizontal prism of fluid of

constant cross-sectional are A will be in equilibrium: Forces acting on the fluid element in the

horizontal direction is P1A and P2A, similar in magnitude but in opposite direction. Since the

fluid is at rest, there is no horizontal resultant force acting on the elemental fluid.

2


Pressure = P1

Pressure = P2

Liquid density, ρ

P

h h

Q


For static equilibrium the sum of the horizontal forces must be equal to zero:

P1 A = P2 A

Hence P1 = P2

Thus, the pressure at any two points at the same level in a body of fluid at rest will be the same,

or fluid pressure is the same along a horizontal plane.

2.6 BASIC GENERAL EQUATION FOR FLUID STATICS (AT RE ST)

Consider an elemental cylindrical fluid, with pressure p acting on one end and pressure p + δp

acting on the other end of the fluid (Figure 2.4). The weight of the fluid also acts downwards.

Figure 2.4: Cylindrical inclined elemental fluid

Since the fluid is at rest, the all forces must be in equilibrium.


Pressure = P

Pressure = P + δp

Liquid density, ρ

δs

Weight of elemental fluid mg

θ

θ


Resolving horizontal forces acting on the elemental fluid:

pA - (p + δp )A - mg cos θ = 0

pA - (p + δp )A - A δs ρ g cos θ = 0

δp =- δs ρ g cos θ

In differential form : dp/ds = - ρ g cos θ

In a horizontal plane (x, y direction) When θθθθ =0 then cos θθθθ =0, hence : dp/dx = 0, implying

that pressure is constant everywhere in a horizontal plane.

In a vertical plane ( z direction ) When θθθθ =90 then cos θθθθ =1, hence : dp/dz = ρρρρ g, implying

that in a vertical plane, pressure varies with height, whereby integrating with respect to z,

gives, p = z ρρρρ g.

Replacing dscos θ by dz, then we end up with the general equation:

dp/dz = - ρρρρ g

Hence p = z ρρρρ g ……………….equation 1

NOTE: From equation 1, pressure can also be expressed as an equivalent head (z) of liquid

with density ρρρρ.


2.7 ACTIVITIES

Main Questions:

5. Define Pascal’s Law and illustrate it with the help of sketches

6. Derive Pascal’s Law

7. Show that the pressure exerted by a fluid is dependent of the depth of the liquid

8. Show with the help of sketches that the pressure of a fluid is the same along a horizontal

plane.

9. Explain why the pressure of a liquid can either be given in terms of N/m2 or in terms of a

head of liquid.

2.8 SUMMARY

Unit 2 introduced the student with the concept of fluid pressure, how it varies at a point, with

depth and along a horizontal plane. The mathematical expression relating pressure to depth of

liquid must be clear to the student before he moves on to the next unit.

The next unit will be concerned with the measurement of the pressure exerted by a fluid, using

the basic concepts illustrated in unit 2. Unless these concepts are clear, student will not be

able to appreciate contents of Unit 3.


2.9 WORKED EXAMPLES

Example 1 – Pressure & Force

Example 1 – Pressure & Force

• A mass of 50kg acts on a piston of area 100cm2. What is the intensity of pressure on the water in contact with the underside of the piston, if the piston is in equilibrium?

• Intensity of pressure = Force / Area• P = (50 x 9.81) / (100 x 10-4) = 49.05 kN/m2.

2.10 TUTORIAL

Question 1 A gas holder at sea level contains gas under a pressure head equal to 9cm of water. If the mass

densities of air and gas are assumed to be constant and equal to 1.28kg/m3 and 0.72 kg/m3

respectively, calculate the pressure head in cm of water in a distribution main 260 m above sea

level.

Question 2

A pump delivers water against a head of 15 m of water. It also raises the water from a reservoir

to the pump against a suction head equal to 250mm of mercury. Convert these heads into N/m2

and find the total head against which the pump works in N/m2 and in metres of water.

Question 3


In a hydraulic jack a force F is applied to the small piston to lift the load on the large piston. If

the diameter of the small piston is 15mm and that of the large piston is 180 mm calculate the

value of F required to lift 1000kg.

Question 4

A mercury manometer is used to measure pressure drop between two points along a

horizontal pipe through which water drops. If the manometer shows a reading of 0.8m,

what is the corresponding pressure drop? Assume the relative density of mercury is 13.6.

Question 5 Two pipes, A and B, are in the same elevation. Water is contained in A and rises to a level of

1.8m above it. Carbon tetrachloride, specific gravity 1.59 is contained in B. The inverted U-

tube is filled with compressed air at 300kN/m2 and barometer reads 760mm of mercury.

Determine (a) the pressure difference in kN/m2 between A and B if the elevation of A is 0.45m,

and (b) the absolute pressure in mm of mercury in B.


UNIT 3 MEASUREMENT OF FLUID PRESSURE

Unit Structure

3.0 Overview


3.2 Introduction

3.3 Gauge and Absolute Pressure

3.4 Piezometer

3.4.1 Advantages and disadvantages of using Piezometer

3.5 U tube Manometers

3.5.1 Pressure difference measurement using U tube manometers

3.6 Enlarged ends U tube manometers

3.7 Inverted U tube manometers

3.8 Activities

3.9 Summary

3.10 Worked Examples

3.11 Tutorial

3.0 OVERVIEW

Unit 3 deals with the various apparatus which are used to measure the pressure exerted by a

fluid. Students are introduced from the simplest pressure measuring device to the more complex

ones. The advantages and disadvantages related to each of these apparatus are highlighted, so

that the students are not only aware of the existence of the various apparatus, but also of the need

for the use of a particular apparatus. Students should however note that the concepts already

described in Unit 2 form the basis behind pressure measurement, thus, the concepts of Unit 2 will

need to be well appreciated and understood before students set out to study Unit 3.




1. Differentiate between Gauge Pressure and Absolute Pressure

2. Describe how a Barometer and a Piezometer works

3. State the advantage(s) and disadvantage(s) of using piezometers for pressure measurements

4. Describe how a vertically and inverted U tube manometer works

5. List the advantage(s) and disadvantage(s) of using U tube manometers for pressure measurements

6. Analyse how an enlarged and an inverted U tube manometer measures pressure difference

3.2 INTRODUCTION

Fluids can exert either a positive or a negative pressure (suction pressure). There are several

devices which are commonly used to measure pressure, but not all of them are able to measure

both positive and negative pressures. While some devices can measure positive pressures, they

are limited physically by the magnitude of the pressure to be measured, hence the need to

appreciate their limitations.

So, while various devices can be used to measure pressure, it is very important to appreciate and

understand the advantages and disadvantages associated with the use of a particular apparatus.

3.3 GAUGE AND ABSOLUTE PRESSURE

In practice, pressure is always measured by the determination of a pressure difference, i.e,

relative to a particular datum. (A datum being a reference point or a reference line which forms

the basis of a particular analysis, for example in the case of land surface elevation, the mean sea

level is taken as datum). If the datum is total vacuum, then the difference between the pressure

of the fluid in question and that of a vacuum is known as the ABSOLUTE PRESSURE of the

fluid.

More commonly, the pressure difference is determined between the pressure of the fluid

concerned and the pressure of the surrounding atmosphere, i.e, the atmosphere is taken as the

datum. So, when measuring the pressure of a fluid with respect to the atmosphere, this pressure


is referred to as GAUGE PRESSURE. Pressure gauges are mechanical devises which are used

to measure the pressure of either liquids or fluids. They are calibrated on gauge pressure, i.e,

taking atmosphere as datum.

Hence, ABSOLUTE PRESSURE = ATMOSPHERIC PRESSURE + GAUGE PRESSURE

Example - GAUGE AND ABSOLUTE PRESSURE Consider a vessel containing liquid, as shown in Figure 3.1, Point A is located at the top water

surface, and point B is located at the bottom of the vessel.

From what we have learnt in Unit 2, Pressure = z ρ g, where z is the vertical distance from the

top water surface to the point under consideration, ρ is the density of the liquid and g, the

acceleration due to gravity.

IN TERMS OF ABSOLUTE PRESSURE

Consider Figure 3.1:

Figure 3.1: Vessel containing liquid

Figure 3.1 illustrate a vessel containing liquid to a maximum depth, hb. Points A, B, C and D

have been positioned on the diagram with the objective of defining the relationship between

gauge pressure and absolute pressure. Now, Point A is open to the atmosphere, so pressure at

point A is simply Atmospheric pressure.

Top water surface A

D C

B

hb

hd hc


PA = Atmospheric Pressure

Similarly, Point B is located at vertical distance hb from top water surface, so the pressure at

point B is Pressure at top water surface + Pressure due to the column of liquid above point B, so

PB = Atmospheric Pressure + hb ρρρρ g

Therefore, pressures at point C and D:

PC = Atmospheric Pressure + hc ρρρρ g

PD = Atmospheric Pressure + hd ρρρρ g

NOW IN TERMS OF GAUGE PRESSURE

As mentioned in earlier, gauge pressure is measured, taking the atmosphere as datum, so

PA = 0 PB = hb ρρρρ g PC = hc ρρρρ g PD = hd ρρρρ g

NOTE: Gauge pressure refers only to the pressure recorded by the column of liquid above the

point at which the pressure is being measured.

Refer to relationship: ABSOLUTE PRESSURE = ATMOSPHERIC PRESSURE + GAUGE

PRESSURE


3.4 PIEZOMETER

A piezometer is simply a small diameter tube open at both ends. It is usually connected to

pipelines for measuring the pressure of the fluid in the pipe at different positions (refer also to

the Bernoulli’s experiment carried out during the Practical sessions of Fluid Mechanics).

Consider a pipeline running full under pressure (Figure 3.2). Imagine now that a piezometer is

connected to the top of the pipeline as shown in Figure 3.2. Liquid will rise in the piezometer

tube to a height h. Now the relationship between the height of a column of liquid and the

pressure at its base is given by the following equation:

P = h ρ g

Figure 3.2: Pressure measurement by a Piezometer tube

Hence Gauge Pressure at point A = H ρ g.

A

H

h

Pipeline running full under pressure

Piezometer


If the tube were closed at the top, the space above the liquid surface were a perfect vacuum, then

the height of the column of liquid would then correspond to the absolute pressure of the liquid,

and Absolute Pressure at A = P (gauge pressure) + Atmospheric pressure.

This principle is used in the well-known mercury barometer.

Advantages and disadvantages of using PIEZOMETER

Piezometers are simple tube and relative inexpensive, so a cheap pressure measuring device.

With reference to the equation (P= h ρ g), it can be noted that the higher the pressure the higher

will be the value of h. When a piezometer is being used for pressure measurement, it is limited

by the density of the fluid being measured, so density cannot be changed. Hence, if a piezometer

were to be used to measure very high pressure, we would need a very long tube, and practically

this is not possible. So piezometers can only be used to measure relatively small pressure, and

besides piezometers cannot be used to measure negative pressure (pressure lower than

atmospheric pressure), also commonly known as suction pressures.

3.5 U TUBE MANOMETERS

Manometers, usually U shaped (Figure 3.3), are devices in which columns of a suitable liquid

(usually mercury) are used to measure the difference in pressure between a certain point and the

atmosphere, or between two points neither of which is necessarily at atmospheric pressure. For

measuring small gauge pressures of liquids, simple piezometer tubes may be adequate, but for

larger pressures some modifications are necessary. A common type of manometer is employing

a transparent U - tube set in a vertical plane (Figure 3.3).


Figure 3.3: U tube mercury manometer

This mercury U tube manometer is connected to a pipe or other container containing liquid (A)

under pressure. The lower part of the U - tube contains a liquid B (mercury in this case)

immiscible with A and of greater density (Figure 3.4).

Figure 3.4: U tube mercury manometer connected to pipe flowing full under pressure with

liquid A

After equilibrium is achieved, the pressure is the same at any two points in a horizontal plane

when equilibrium is achieved. To solve for the value of P1, the first step is to identify a

horizontal plane, commonly known as the DATUM LINE . For ease of analysis, this line is

taken at the lowest level of mercury level in the U tube manometer. Having drawn the DATUM

line, the next step is then to identify two reference points along this line and referring to Figure

DATUM LINE

P1

QP

liquid B - mercury liquid

A y

h

mercury


3.4, these reference points are P and Q. Therefore P and Q are in the same horizontal plane and

hence, the pressure at P and Q are equal and in equilibrium.

Let the pressure in the pipe at is centre line be P1. Then provided the fluid is of constant density,

the pressure at P is P1 + ρA g y

If the other side of the U tube is open to the atmosphere, the gauge pressure at Q is given by

ρB g h

From unit 2, we have learnt that pressure at two points lying on the same horizontal plane is

similar, therefore, since P and Q are along the same horizontal line:

P1 + ρA g y = ρB g h

Hence, P1 = ρρρρB g h - ρρρρA g y


3.5.1 Pressure difference measurement -U tube MANOMETERS

Consider the U tube manometer in Figure 3.5, connected in such a way so as to measure the

pressure difference between points 1 and 2.

Figure 3.5: U tube manometer measuring pressure difference Pressure at P = P1 + ρA g (y+x)

Pressure at Q = P2 + ρhg g y + ρB g x

Therefore P1 + ρA g (y+x) = P2 + ρhg g y + ρB g x

Thus: P1 - P2 = (ρρρρB g x - ρρρρA g x - ρρρρA g y + ρρρρhg g y)

y

x

P1

Q P

P2

Liquid B

Mercury

Liquid A


3.5.2 Advantages and Disadvantages of Using U Tube Manometers

U tube manometers can be used to measure both positive and negative pressures and besides by using a manometer liquid of high density, U tube manometers can be used to measure high pressures.

3.6 ENLARGED ENDS U TUBE MANOMETERS

Consider a U tube manometer with enlarged ends (Figure 3.6). The right hand side of the U tube

manometer contains liquid A and the left hand side contains liquid B. Liquid B is lighter than

liquid A, hence it floats on liquid A.

The first step in the analysis is to define the relationship between ha and hb when this tube is not

connected to any system.

Drawing a horizontal line at the intersection of the two liquids: P1 = P2

P1 = ha ρa g , while P2 = hb ρB g

Liquid B

Liquid A

ha hb

1 2

Figure 3.6: Enlarged ends U


Hence, ha ρA g = hb ρB g ha = hb (ρρρρB / ρρρρA)……………………………………..equation 1

The second step is then to apply the pressure difference. Suppose pressure in the right hand side

is higher than pressure in the left hand side. Then the pressure will cause the liquid in the right

hand side to move downwards. This downward movement in the right limb will be accompanied

by an upward movement small limb at the point of contact of the two liquids and another upward

movement, this time in the left limb, as shown by the small arrows located next to the side of the

tubes, Figure 3.7. The volume of the liquid moving is the same in all three cases.

Figure 3.7: Enlarged ends U tube manometer – measuring pressure difference

Note: When both enlarged ends of the U tube manometer have similar diameter, then on both

sides the liquid moves by the same height, X.

The diameter of the tube in which the two liquids meet is smaller, hence the liquid moves by a

height h.

The third step in this analysis is now to draw a NEW DATUM, which is a line drawn at the new

position of the interface of the liquid, as shown in Figure 3.6. Points 3 and 4 are positioned in

the right and left limb, and since both lies on the same horizontal line, P3 = P4.

ha hb

PA PB

x

y

x

old datum

new 3 4


Pressure at Point 3 = PA + (ha – x ) ρA g…………………….equation 2 Pressure at Point A = PB + (hb – y + x) ρB g………………….equation 3 The final step would be to combine equations 1,2 and 3, and simply for the difference in

pressure:

PA - PB = x ρρρρA g - y ρρρρB g + x ρρρρB g NOTE: Students are advised to follow all the steps while solving similar problems. 3.7 INVERTED U TUBE MANOMETER

U tube manometer can also be connected in an inverted position as shown in Figure 3.8. As

described in the above sections, the first step is to draw the DATUM line, and in the case of an

inverted U tube manometer, this datum is at the highest position of the mercury level.

If we refer once again to the relationship between pressure and head of liquid, P = h ρ g, we will

find that this relationship implies that as the head of liquid above the point under consideration

increases, the pressure also increases. Thus starting from a particular point and moving upward,

then will imply a decreasing pressure.

Referring to Figure 3.8:

Along the Datum Line, at points X and Y, we know that Px = Py since both points X and Y lie on

the same horizontal plane.

Px = P1- ha ρa g- H ρhg g

Py = P2- hb ρb g

Hence, equating Px and Py, we have,

P1- ha ρa g- H ρhg g = P2- hb ρb g

P1 - P2 = ha ρρρρa g + H ρρρρhg g - hb ρρρρb g


3.8 ACTIVITIES

Main Questions:

10. Differentiate between absolute and gauge pressures.

11. Explain how a piezometer tube works.

12. List the limitations of using piezometers for pressure measurements.

13. Why is the manometer liquid usually denser than the liquid whose pressure is being

measured?

14. Explain with the help of sketches how pressure difference is measured using enlarged

ends U tube manometers.

15. Explain with the help of sketches how pressure difference is measured using inverted U

tube manometers.

P1

yx

P2

Liquid B

Liquid A

Mercury

Figure 3.8: Inverted U tube

hb

ha

H

DATUM LINE


3.9 SUMMARY

Unit 3 has introduced students to the various apparatus commonly used to measure pressure and

pressure difference. Their advantages and limitations has been highlighted, together with

detailed explanation of how the pressure is actually measured using these apparatus.

The next unit, Unit 4, will now introduce the student to the approach adopted to calculating the

force exerted by a fluid on the walls of its container, the hydrostatic force. In Units 2 and 3, you

have learnt that pressure varies with depth, so bearing this in mind, the next unit will explain

how hydrostatic forces are calculated.

3.10 WORKED EXAMPLES

Example 1 – Simple application of pressure head equation, P = h ρρρρ g

Example 1 – Pressure = h ρρρρg

• Calculate the pressure in the ocean at a depth of 2000m assuming that salt water is (a) incompressible with a constant density of 1002kg/m3, (b) compressible with a bulk modulus of 2.05 GN/m2 and a density at the surface of 1002 kg/m3.

(a) Assuming constant densityPressure = h ρ g

= (2000) x 1002 x 9.81 = 19.66 MN/m2.

(b) Considering the salt water as being compressible

incomplete part (b)

NOTE: Student should pay particular attention to the S. I. unit of each termin the equation of P=hρg, and at the end of the calculation always specify the S. I. unit associated with the answer.


Example 2 – Difference between Absolute and Gauge Pressure

Example 2 – Gauge & Absolute Pressure

• What will be (a) the gauge pressure and, (b) the absolute pressure of water at a depth of 12m below the free surface. Assume the density of water to be 1000kg/m3 and the atmospheric pressure 101 kN/m2.

Absolute Pressure = Atmospheric Pressure + Gauge Pr essure

Gauge pressure = h ρ g = 12 X 1000 x 9.81= 117.72 kN/m2

Absolute pressure = 101 + 117. 72

= 218.72 kN/m2

Example 3 – Pressure head as a function of different types of liquid (density dependency)

Example 3 – Pressure & density dependency

• Determine the pressure in N/m2 at (a) a depth of 6m below the free surface of a body of water and (b) at a depth of 9m below the free surface of a body of oil of specific gravity 0.75.

Pressure P = h ρρρρ g

Specific gravity of a substance = Density of the substance/ density of water at standard temperature and pressure (1000 kg/m3)

Pressure at point A = 6 x 1000 x 9.81= 58.86 kN/m2

Pressure at point B = 9 x (1000 x 0.75) x 9.81= 66.22 kN/m2

6mwater

9mOil

A

B


Example 4 – Pressure expressed as a depth of a given liquid

Example 4 – Expressing pressure as a depth of liquid

• What depth of oil, specific gravity 0.8, will produce a pressure of 120 kN/m2. What would be the corresponding depth of water.

Pressure = h ρρρρ g

Depth of oil corresponding to pressure of 120kN/m2:

120 x 1000 = hoil x (0.8 x 1000) x 9.81

hoil = (120 x 1000 ) / ( 0.8 x 1000 x9.81)= 15.29 m

Corresponding depth of water,

hwater = (120 x 1000) / (1000 x 9.81)= 12.23m

Example 5 – Negative gauge pressure

Example 5 – Negative gauge pressure & absolute pressure

• A U tube manometer is connected to a pipe in which a fluid is flowing under a negative gauge pressure of 50mm of mercury. What is the absolute pressure in the pipe in N/m2, if the atmospheric pressure is 1 bar?

Gauge pressure is negative and its magnitude is 50m m mercury.

Atmospheric pressure = 1 bar = 101 kN/m 2

Absolute pressure = Atmospheric pressure + Gauge pr essure

= (101x 1000 ) - (50 x 0.001 x 1000 x 13.6 x 9.81)

= 94.32 kN/m2


Example 6 – Barometric pressure expressed as head of mercury

Example 6 – Barometric pressure in terms of equivalent head of mercury…1/2

• What is the gauge pressure and absolute pressure of the air in the figure below, if the barometric pressure is 780mm of mercury and (a) the liquid is water of density 1000kg/m3 and (b) oil of specific weight 7.5 x102 N/m3.

0.5m

Air

liquid

Pressure at A = Pressure at B (along same horizontal line)

Absolute Pressure at A = atmospheric pressure

Absolute Pressure at B = Pair + (Pressure due to 0.5m of liquid)

Therefore:Absolute pressure of air = Atmospheric pressure - (Pressure due to 0.5m depth of liquid)

A B

Example 6 – Barometric pressure in terms of

equivalent head of mercury…2/2

Liquid is water:Absolute Pressure of air = Atmospheric pressure – pressure due to 0.5m depth of water

= (780 x 0.001 x 1000 x 13.6 x 9.81) – (0.5 x 1000 x9.81)= 99.16 kN/m2

Gauge pressure of air = Pressure due to 0.5m depth of water = - (0.5 x 1000 x 9.81)= -4.9 kN/m2

Liquid is oil:Absolute Pressure of air = Atmospheric pressure – pressure due to 0.5m depth of oil

= (780 x 0.001 x 1000 x 13.6 x 9.81) – (0.5 x 7.5 x 102)= 103.7 kN/m2

Gauge pressure of air = Pressure due to 0.5m depth of oil= - (0.5 x 7.5 x 102)= - 3.75 kN/m2


Example 7 – U tube manometer

E xa m p le 7 – U tu b e m an o m e te r… 1 /2

• A U tube m a no m e te r as illu s tra ted in the f igu re b e lo w is used to m e asu re the p ressu re a bov e a tm osp he ric o f w a te r in a p ipe . If the m e rcu ry , liqu id Q , is 30 cm b e lo w A in th e le ft ha nd lim b and 20c m a bov e A in th e rig h t ha nd lim b , w ha t is the ga uge p ressu re a t A . S p ec ific g rav ity o f m e rcu ry is 13 .6 .

D

A

B C

L iqu id Q (m e rcu ry)

w a te r

20 cm

3 0cm

Example 7 – U tube manometer…2/2

• Pressure at B = Pressure at C, since B and C are on the same horizontal linePB = PA + Pressure due to the height of water between A and BPC = Pressure due to the height of mercury between D and C

PB = PC

PA + (30 x 10-2 x 1000 x 9.81) = (30+20) x 10-2 x 13.6 x 1000 x 9.91PA = 63.7kN/m2

D

A

B C


Example 8 – U tube manometer to measure negative pressure

Example 8 – Negative gauge pressure (suction pressure)

• In the figure below, fluid P is water and fluid Q is mercury. If the specific weight of mercury is 13.6 times that of water and the atmospheric pressure is 101.3kN/m2, what is the absolute pressure at A when h1 is 15cm and h2is 30cm.

h1

h2

A

B C

NOTE: When the pressure in the pipeis lower than atmospheric pressure, the gauge pressure is negative and isknown as a Suction Pressure.

PB = PC

PB = PA + h1ρPg+ h2ρQgPC=0 (gauge pressure) or atmosphericpressurePA + h1ρPg+ h2ρQg =101.3 x 103

PA = (101.3 x 103) – (15x10-2)x1000x9.81-(30x10-2)x13.6 x1000x9.81

Absolute pressure, PA = 59.5 kN/m2

Fluid P

Fluid Q


Example 9 – U tube manometer with enlarged ends

Example 9 – Enlarged U tube manometer…1/5

• An oil and water manometer consists of a U-tube 4mm diameter with both limbs vertical. The right-hand limb is enlarged at its upper end to 20mm diameter. The enlarged end contains oil with its free surface in the enlarged portion and the surface of separation between water and oil is below the enlarged end. The left-hand limb contains water only, its upper end being open to the atmosphere.

• When the right hand side is connected to a cylinder of gas the surface of separation is observed to fall by 25mm, but the surface of the oil remains in the enlarged end. Calculate the gauge pressure in the cylinder . Assume that the specific gravity of the water is 1.0 and that of the oil is 0.9.

Diameter = 4mm

Diameter= 20mm

oil

water


• Step 1 : Work out the relationship between hoil and hwaterwhen there is no pressure difference between the two top ends of the U tube.

• Working in terms of Gauge pressure:P1 = hoil ρoil gP2 = hwater ρwater gP1 = P2 (points 1 and 2 lie on the same horizontal line within the

system)

hoil ρoil g = hwater ρwater gHence : hoil = hwater ρwater / ρoil = hwater/0.9….equation 1

hwaterhoil

1 2



• Step 2 : Calculating x

• When the pressure is applied, the level of oil in the larger end goes down by a value of x, while the level of oil in the smaller diameter pipe, moves down by an amount y, since the change in volume in the two cases is the same.

• Volume change in smaller diameter pipe = (25x10-3) x Π/4 x (4 x 10-3)2

• This volume equals (X) x Π/4 x (20 x 10-3)2

• Hence X = 1mm

hwaterhoil

1 2y

yx

3 4


• Step 3:Working in terms of Gauge pressure:

P3 = Pg + (y + hoil - x)ρoil gP4 = (2y + hwater ) ρwater gP3 = P4Hence: Pg + (y + hoil - x)ρoil g = (2y + hwater ) ρwater g

Pg = (2y + hwater ) ρwater g – (y + hoil - x)ρoil g …equation 2

hwaterhoil

1 2

PressurisedGas cylinder Pg

y

yx

3 4



• Step 4 : Replacing values of X, Y and relationship from equation 1 into equation 2:

X=1mm, Y=25mm and hoil=hwater/0.9

Pg = (2 x 0.025 + hwater) x 1000 x 9.81 – (0.025 + hwater/0.9) ) x 0.9 x1000 x 9.81

Pg = 278.6N/m2

hwaterhoil

1 2y

yx

3 4

Pg = (2y + hwater ) ρwater g – (y + hoil - x) ρoil g …equation 2


3.11 TUTORIAL

Question 1

(a) For a U tube manometer with enlarged ends containing two manometer liquids, derive the

formula for the difference in pressures applied to the two enlarged ends.

(b) An oil and water manometer consists of a U tube 7 mm diameter with both limbs vertical.

The right-hand limb is enlarged at its upper end to 25 mm diameter. The enlarged end

contain oil, of density 900 kg/m3, the free surface of the oil is in the enlarged portion of

the limb and oil/water interface is in the smaller diameter tube. The left hand limb

contains water only, and its upper end is opened to the atmosphere. A pressure of 350

N/m2 is applied to the right hand side of the manometer. If the surface of the oil remains

in the enlarged end and the oil/water interface remains in the smaller diameter tube,

calculate the depth by which the surface separation between the oil and the water will

move.

Question 2

A manometer consists of two tubes A and B open to the atmosphere, with vertical axes and

uniform cross-sectional areas 500 mm2 and 800 mm2 respectively, connected to a U tube C of

cross-sectional area 70 mm2 throughout. Tube A contains a liquid of relative density 0.8; tube B

contains one of relative density 0.9. The surface of separation between the two liquids is in the

vertical side of C connected to tube A. Tube B is now close and the space above the liquid is

pressurised such that the surface of separation in tube C rises by 60 mm. Determine the pressure

applied.

Question 3

A manometer consists of a U tube of diameter d, the upper part of each limb being enlarged to

diameter D, where the ration of D/d is equal to 5. The small tube contains water and on top the

water surface there is a liquid of relative density 0.95 in both limbs. The free surfaces are in the

enlarged parts of the U tube, while the surfaces of separation between the two liquids are in the

small tube. Initially the surfaces are level.


When a pressure difference is applied to the top of the U tube in the enlarged ends, this causes

the interface to move by a depth of 1 cm. Calculate the difference in pressure.

Question 4

The sensitivity of a U-tube gauge is increased by enlarging the ends, and one side is filled with

water and the other side with oil, (specific gravity, 0.95).

If the area A of each enlarged end is 50 times the area of the tube, calculate the pressure

difference corresponding to a movement of 25 mm of the surface separation between the oil and

water.

Question 5

The inclined U tube manometer as indicated in the diagram below, gives zero reading when A

and B are set at the same pressure. The cross sectional are of the reservoir is 50 times that of the

tube. For an angle of inclination θ of 30o and with a manometric fluid of specific gravity 0.8,

find the difference in pressure between A and B in N/m2, when the gauge reads 110mm.

Lθθθθ

A

B

hL

θθθθ

A

B

h


UNIT 4 HYDROSTATIC FORCES ON PLANE SURFACES

Unit Structure

4.0 Overview


4.2 Introduction

4.3 Magnitude of the Hydrostatic Force – Vertical plane

4.4 Magnitude of the Hydrostatic Force – Inclined plane

4.5 Centre of Pressure

4.6 Activities

4.7 Summary

4.8 Additional Reading Materials

4.9 Worked examples

4.10 Tutorial

4.0 OVERVIEW

In units 2 and 3, you have been introduced to the concepts of a fluid exerting pressure on the

walls of the vessel in which it is contained. You are now familiar with the relationship between

pressure (P) and an equivalent head of fluid (h), from the relationship (P = h ρ g). You have also

learnt about the various devices that can be used to calculate pressure or pressure difference.

In this unit, you will now learn how this fluid pressure is expressed in terms of forces, commonly

termed hydrostatic forces when the fluid is at rest. Since pressure exerted by a fluid is depth

dependent, this particular characteristic is taken into consideration when calculating hydrostatic

forces.



At the end of this unit, you should be able to do the following:

1. Derive the equation governing the resultant force exerted by a fluid on plane surfaces.

2. Clearly define the exact meaning of each term making up the equation for hydrostatic force

on plane surface.

3. Analyse how the worked examples were solved.

4.2 INTRODUCTION

Fluid exerts pressure on the walls of the vessel in which it is contained, and this pressure can also

be expressed in terms of a force, given that Force = Pressure x Cross Sectional Area. Now we

also know that the pressure exerted by a fluid varies with depth, increasing as the depth of liquid

above the point under consideration, increases. Thus, the hydrostatic force acting on the plane

surface also varies from point to point, being higher in magnitude, at the bottom of the vessel,

ending up with a series of individual forces.

These individual forces can then be combined to get the resultant force acting. Similarly, the

point of action of the resultant force can be obtained by taken moments of each individual force,

as being described in the next section.

4.3 MAGNITUDE OF THE RESULTANT HYDROSTATIC FORCE –VERTICAL

PLANE

Consider a vessel containing liquid, of density ρ , to a depth H, Figure 4.1, and consider the

surface ABCD. As we move down from AB to CD, we know that the pressure will be increasing

from a value of 0 to a value of Hρg.

Now imagine that surface ABCD is now divided into very small layers of area δA. Consider

now point 1, which is located at depth h1 from top water level. The pressure acting at point 1 is

given by h1ρg, and from there, the hydrostatic force acting at point 1, will be given by the


relationship, force= pressure x area. Thus force acting at point 1, will be F1, where F1= h1ρg

δA. Similarly, the forces acting at point 2, 3 and 4, can be expressed by the following equations:

h2ρg δA, h3ρg δA and h4ρg δA

top water surface

C

D

B

Ah1

h2

h3

h4

Cross sectional area of each layer, δA

H

Figure 4.1: Magnitude of Hydrostatic

Force


These individual forces can be represented as shown in Figure 4.2, as a series of parallel

forces acting perpendicularly to the surface ABCD.

Similarly, going down to the bottom of the side ABCD, there would be a series of individual

forces. When vertical surface of a plane is being analysed, these individual forces are parallel

and thus can be linearly combined (refer to section 4.8), to get the final resultant force, which

will be:

Resultant force acting on ABCD = ∑ hi ρ g δA where i will vary from 1 to the total number of

layers into which the solid surface has been divided for analysis purposes.

If the area is made extremely small, then the above equation can also be written as ∫∫∫∫ ρρρρg h δδδδA,

where ∫ h δA represents the first moment of area about the horizontal surface of the liquid and

this can also be written as A h, where A is the total surface area of the body and h, is the distance

from the centroid of the body to the horizontal surface of the liquid. This results in equation 1

(I think it is equation 1instead of 2) being written as follows:

F1=h1 ρ g δA

F2=h2 ρ g δA

F3=h3 ρ g δA

F4=h4 ρ g δA

h1

h4 h2

h3

Figure 4.2: Individual hydrostatic forces

B

C

h

Position of centroid of ABCD


FR= ρg A h ……………………..equation 1

Where ρ = density of the liquid

g = acceleration due to gravity

A = cross sectional area of the surface in contact with the liquid

h = vertical depth from top liquid level to the centroid of the solid

surface

Note: 1. The centroid of a rectangular body lies at the centre of that body.

2. You need to know how to derive this equation, and also to be clear about the exact

meaning of each of the term in the equation.

4.4 MAGNITUDE OF THE RESULTANT HYDROSTATIC FORCE – INCLINED PLANE

Consider surface XY, which is submerged in the liquid. We will consider here, the hydrostatic

force acting on the left hand side of surface XY. Four points 1, 2, 3 and 4 have been identified

on that side, and these points are located at depth h1, h2, h3 and h4, respectively from the top

water level.

Refer to the relationship between depth of liquid and pressure, P = h ρ g, where h is the vertical

depth of liquid above point under consideration. Thus applying this relationship, we have

pressure acting at point 1, being P1 = h1 ρ g. Similarly at points 2, 3 and 4, the pressures will be

h2 ρ g, h3 ρ g and h4 ρ g. Once again imagine that surface XY has been divided into small layers,

δA. Hence the corresponding forces acting at point 1, 2, 3 and 4, will be h1 ρ g δA, h2 ρ g δA, h3

ρ g δA and h4 ρ g δA. Note here, that these individual forces are also parallel to each other, and

they can thus be combined linearly as was the case for the vertical plane surface.


So referring to equation 1, the resultant hydrostatic force acting on an inclined plane

surface will be given by

FR = ρg A h …………………..equation 2

h3

X

Y

h1 h2 h4

12

34

Figure 4.3: Inclined plane surface

h

Position of centroid of XY




A = cross sectional area of the surface in contact with the liquid

h = vertical depth from top liquid level to the centroid of the solid

surface

4.5 CENTRE OF PRESSURE

The hydrostatic force, as illustrated in previous sections, is given by equation 1 or 2. This force

is, as indicated before, the summation of individual forces acting on the surface of an inclined

body.

Now each individual force acts at a particular point. The next step will be then to find out the

location of the resultant hydrostatic force. This location is also known as the point of action of

the hydrostatic force, or the Centre of Pressure.

Consider an elemental force acting on the surface, whereby δF = ρg δA h (Figure 4.4). Take

moments about point O on the surface of the liquid then δM = δF x s.

X

Y

Figure 4.4: Centre of Pressure

h

FR = h ρ g A

Position of centroid of XY

D

Position of centre of pressure

δF = h ρ g

O

s

θθθθh


Total moments of individual elemental forces acting on the surface of the inclined body:

M = ∑ δF x s = ∑ ρg δA h x s

= ∑ ρg δA (s sin θ) x s

= ∫∫∫∫ ρg δA s2 sin θ …… ……………………..equation 3

The total moments can also be given in terms of the resultant hydrostatic force, at position of

centre of pressure, D from point O, given by:

M = FR x D sin θ

M = ρg A h D sin θ ……………………………….equation 4

Hence, equating 3 and 4 gives:

ρg A h D sin θ= ∫ ρg δA s2 sin θ

∫∫∫∫ s2 δδδδA is the second moment of area about the surface of the liquid and about the axis

passing through the point O.

This can be written (Theorem of Parallel Axes ) as the second moment of area about the

centroid of a body + A y2, where y is the distance between the axis passing through the centroid

of the body and the axis passing through point O (refer to section 4.8).

Therefore,

ρg A h D / sin θ= ρg sin θ ( Icg + A (h / sin θ) 2)

D = Icg sin2 θ / A h + h

Where D = vertical distance from position of centre of pressure to the top water level

Icg = second moment of area of solid surface about its centroid


A = cross sectional area of solid surface in contact with the liquid

h = vertical distance from centroid of body to the top water level

θ = angle of inclination of body with the horizontal plane

4.6 ACTIVITIES

1. Illustrate with the help of sketches the centroid of a rectangle and a triangle.

2. A rectangular body, of width 1m and depth 2m, is inserted in a vessel containing water.

If the rectangular body is just submerged in the vessel, what is the value of h in the

equation of resultant hydrostatic force?

3. Refer to the case in part 2. If the top of the rectangular body is now 1m below the top

water level, what is the value of h in the equation of the resultant hydrostatic force?

4. Refer to the case in part 2. Now the top of the rectangular body is 0.5m above the top

water level in the vessel. What is the value of h in the equation of the resultant

hydrostatic force?

5. Derive the equation for the hydrostatic force acting on the inclined plane rectangular

surface of a dam.

6. Derive the centre of pressure for the hydrostatic force acting on a plane vertical

rectangular surface.

7. The resultant hydrostatic force acts as the centroid of the solid body on which the force is

being exerted Discuss.

8. How does your equation derived in part 5 change, when the body is now circular in

shape, such as a valve in the side of a reservoir?


4.7 SUMMARY

Unit 4 has illustrated how to calculate the hydrostatic force exerted by fluids on the surface of

solids bodies, and also the position of location of that force. You are advised to go over the

derivations of the equations and make sure that you have understood them and can easily

reproduce them. Your attention is also drawn to the exact meaning of each of the terms in the

equations derived. You will be required to derive the equations governing the resultant force and

centre of pressure, but you need not do so when solving problems all the time. You should

derive the equations when asked to work from first principles or when directly asked to do so.

Unit 5 is a continuation of the concepts described in Unit 4. Unit 5 will illustrate the governing

principles for deriving hydrostatic forces and centre of pressure when the force is being exerted

on a curved surface, as compared to a plane surface as was illustrated in Unit 4. There are some

important differences between the equations derived in Unit 4 and those to be derived in Unit 5.

So, students are strongly advised to get the concepts discussed here clear before moving on to the

next unit.

NOTE: Additional reading materials have been included in section 4.8, to help you better

appreciate the following:

1. the meaning of resultant force,

2. the position of action of a resultant force,

3. the meaning of the Theorem of Parallel Axes,

4. the application of the Theorem of Parallel Axes.

While section 4.8, is not an integral part of Unit 4, it does help you to get a better understanding

of the contents of the unit, and you are advised to go through section 4.8 at least once.


4.8 ADDITIONAL READING MATERIALS

1. The concept of resultant forces:

2. The concept of the point of action of Resultant Forces:

Solid body, A F1=20kN

F2=50kN

F3=30kN

Consider 3 forces acting on solid body A: In terms of magnitude only , the resultant force acting on A is simply the summation of the forces, taking into consideration that not all the three forces are acting in the same direction: Hence FR = (20 + 50) – 30 = 40 kN

RESULTANT FORCES

Solid body, A F1=20kN

F2=50kN F3=30kN

Consider point Z: Each of the individual forces exert a moment about point Z, the magnitude of the resultant moment of the forces about Z, will either cause the body to overturn or to stay unmoved. Hence, the resultant force, FR, should be located such that it produces the same resultant moment about Z. So taking moments about Z: F1 x h1 + F2 x h2 – F3 x h3 = FR x y, where y is the perpendicular distance form the point of action of FR and point Z. Solve for position of Z .

POINT OF ACTION OF RESULTANT FORCE

h1

h2 h3

Z

y


3. Proof of Theorem of Parallel Axes:

Consider a rectangle of width b and depth d. In this exercise, the second moment of area of the

rectangle about two axes will be derived. The first axis is indicated by line aa, which is a line

located at the bottom of the rectangle. The second axis passes through the centroid of the

rectangle, line gg.

Proof – Theorem of Parallel Axes…. (part 1)

Small element, thickness δh

a a

Second moment of area of small element about line aa: = Area x (distance)2

= b δh x h2

Hence, total second moment of area of rectangle about line aa: = ∫ b δh x h2 with limits from h= 0 to d = bd 3/3

d

b

g g

Second moment of area of small element about line gg, line gg passing through the centroid of rectangle = Area x (distance)2

= b δh x h2

Hence, total second moment of area of rectangle about line gg: = ∫ b δh x h2 with limits from h= +d/2 to -d/2 = bd 3/12 = Icg

d

b

h

h


Having derived the second moment of area of the rectangle about axes passing through aa and

gg, the next step will be to make use of the theory in the Theorem of Parallel Axes, to show that

the second moment of area passing through line aa can be obtained directly, if the second

moment of area about the centroidal axis is known. Hence this second part sets out to prove the

Theorem of Parallel Axes.

Proof – Theorem of Parallel Axes…. (part 2)

a a

Second moment of area of rectangle about line aa = bd 3/3 Second moment of area of rectangle about line gg, Icg = bd 3/12

d

b

g g

The Theorem of Parallel Axes states that: The second moment of area of a body about any axis, is the sum of the second moment of area of the body about its centroidal axis Icg and the area of the body x (perpendicular distance from the given axis to the centroidal axis)2

Ixx = Icg + (AY2) where Y is the perpendicular Distance from the centroid to the axis xx.

Hence, applying t he above theorem, to find the second moment of area of rectangle about the axis aa: Iaa = Icg + AY2

Iaa = bd3/12 + (bd) (d/2)2

Iaa = bd3/3


4.9 WORKED EXAMPLES

Example 1 – Hydrostatic force exerted on a plane surface

Example 1 – Rectangular plane surface…1/4

• A rectangular plane area, immersed in water, is 1.5m by 1.8m with the 1.5m side horizontal and the 1.8m side vertical. Determine the magnitude of the force on one side and the depth of its centre of pressure if the top edge is:

a. in the water surfaceb. 0.3m below the water surface

c. 30m below the water surface

Width = 1.5m

1.8m

(a)

1.8m

30m

(c)

1.8m

(b)

0.3m


Hydrostatic force acting on one side of rectangular surface = F = ρ g A y

Where is the density of the liquid in contact with the plane surface (kg/m

3)

g is the acceleration due to gravity (m/s2)

A is the wetted area of the rectangular surface in contact with the liquid (m

2)

y is the vertical distance from the top liquid surface to the centre of gravity of the wetted surface (m).

Hydrostatic force acting on one side of rectangular body = 1000 x 9.81 x (1.8 x 1.5 ) x 1.8/2

= 23.8 kN

Position of centre of pressure D = Icg/Ay + y = (0.729/(1.5x1.8X0.9) + 0.9 = 1.2 m

Width = 1.5m

b=1.8m

(a)

Icg of a rectangular body:Icg =bd3/12

Icg = 1.5 x (1.8)3/12= 0.729 m4



Hydrostatic force F = ρ g A y

Hydrostatic force = 1000 x 9.81 x (1.8 x 1.5 ) x 1.2 = 31.8 kN



Icg = 1.5 x (1.8)3/12= 0.729 m4

1.8m

(b)

0.3m

Since the top edge of the rectangular body is below the water surface, there is a need to calculate the new value of y, i.e the vertical distance from the top water surface to the centre of gravity of the rectangular body:y= 0.3 + (1.8/2) = 1.2 m


Hydrostatic force F = ρ g A y

Hydrostatic force = 1000 x 9.81 x (1.8 x 1.5 ) x 30.9 = 818 kN



Icg = 1.5 x (1.8)3/12= 0.729 m4

New value of y in this case:y = 30 + (1.8/2) = 30.9 m

1.8m

30m

(c)

Example 2 – Hydrostatic force and its dependency on density of the liquid exerting it


Example 2 – Hydrostatic force & density of liquid

• One end of a rectangular tank is 1.5m wide by 2m deep. The tank is completely filled with oil of specific weight 9kN/m2. Find the resultant pressure on this vertical end and the depth of the centre of pressure from the top.

2m

width = 1.5m

Side under study

A

B

Hydrostatic force F = ρ g A y = γ A y

F = 9 x 1000 x (2 x 1.5) x 2/2 = 27 kN

Centre of Pressure D = Icg/Ay + y= 1/ (2 x 1.5)x1 + 1 = 1.33 m

Icg = bd3/12= 1.5 x 23 = 1 m4

Vertical depth from top liquid surface to centre of gravity of plane vertical surface y= 2/2 = 1m

NOTE: Specific weight = Density x acceleration due to gravity

γ = ρg

Example 3 – Hydrostatic force exerted on a circular opening

Example 3 – Hydrostatic force on a circular opening..1/2

• A culvert draws off water from the base of a reservoir. The entrance to the culvert is closed by a circular gate 1.25m in diameter, which an be rotated about its horizontal diameter. Show that the turning moment on the gate is independent of the depth of water if the gate is completely immersed and find the value of this moment.

Diameter of circular gate = 1.25mHinged about the horizontal diameter

Circular opening


Example 3 – Hydrostatic force on a circular opening..2/2

Hydrostatic force F = ρ g A y = 1000 x 9.81 x (Π x 1.252/4) x y = 12043.5 y kN

Centre of pressure D = Icg/Ay + y = 0.12/(Πx1.252/4)x y + y = (0.1/y)+y

Icg for circular body = ΠΠΠΠ r4/4= 22/7 x (1.25/2)4 / 4 = 0.12m4

D y

The hydrostatic force acts at D and since the opening is hinged along its horizontal diameter, the moment of the hydrostatic force is given by the hydrostatic force multiply by the vertical distance from D to the hinge.

Taking moment about the horizontal diameter of the circular opening, M = F (D-y)

M = 12043.5 y x { (0.1/y)+y – y } = 1177.2 y x 0.1/yNote : y is eliminated in the equation for MM = (12043.5 x 0.1) = 1204 Nm, thus independent of y.

D-y

Example 4 – Turning moment exerted by hydrostatic force about a hinge

Example 4 – Hydrostatic force and turning moments..1/4

• A rectangular sluice door is hinged at the top at A and kept closed by a weight fixed to the door. The door is 120cm wide and 90cm long and the centre of gravity of the complete door and weight is at G, the combined weight being 9810N. Find the height of the water h on the inside of the door which will just cause the door to open.

600

hA

B

W

30cm

G



• Hydrostatic force acting on door AB = (1000 x 9.81 x 1.2 x 0.9) x (h-0.78/2)

= 10595(h-0.39) N

• Centre of pressure D = (1.2 x 0.93/12)/(1.2x0.9)(h-0.39) + (h-0.39) m

=0 .07/(h-0.39) + (h-0.39) m

• Under equilibrium conditions: Clockwise moment about hinge A = Anticlockwise moments about hinge A

600

hA

B

W

30cm

G0.9sin60= 0.78m


• Under equilibrium conditions: Clockwise moment about hinge A = Anticlockwise moments about hinge A

• Hydrostatic force will create a clockwise moment trying to open the gate, while the weight of the door and the gate will induce an anticlockwise moment trying to close the gate:

hW

30cm

D

Clockwise moment about A = {10595(h-0.39) } x (D-h+0.78)/sin60o

= {10595(h-0.39) } x { [0 .07/(h-0.39) + (h-0.39) ] – h + 0.78} /sin60o

= 10595(h-0.39) x { 0.07(h-0.39) + 0.39 }/sin60o

Anticlockwise moment about A = (W x 0.3) = 9810 x 0.3 = 2943 Nm

Equating these two moments and solving for h, yields h=0.88m

F

h-0.78



Mathematical solution :

Replacing (h-0.39) by X

Clockwise moments: 10595(h-0.39) x { 0.07(h-0.39) + 0.39 }/sin60o

= (10595X) x (0.07X+0.39)/sin60o

= (7416.5+10595X2) /sin60o Nm= 8564.6X2 + 4771.3X

Anticlockwise moments: 2943 Nm

Equating moments: 8564.6X2 + 4771.3X- 2943 =0

X = -4771.3 E e ( 4771/32-4x8564.5x-2943) =

h-0.39 = , therefore h = m

Example 5 – Circular opening and moment exerted by hydrostatic force

Example 5 – Circular openings & moment of hydrostatic force…1/2

• A sluice gate closes a circular opening 0.30m diameter and is hinged 1m below the surface of the water which acts on its face. If the centre of the opening lies at a depth of 1.25m find the force on the gate due to the fluid pressure. Find also the minimum force that must be applied by a clamp which lies 0.5m below the hinge, in order to keep the gate closed.

1.25m1m

clamp

0.3m

0.5m

Sluice gate

Circular opening, 03m diameter


Example 5 – Circular openings & moment of hydrostatic force…2/2

• Hydrostatic force, F = ρ g A y = 1000 X 9.81 x (π x 0.32/4) x 1.25= 867N

• Centre of pressure, D = Icg/Ay + y = (π x 0.154/4)/[(π x 0.32/4)x1.25] + 1.25= 1.255m

• Taking moments about the hinge:867 x (1.255-1) = Fclamp x 0.5

Therefore Fclamp = 442 N

1.25m1m

clamp

0.3m

0.5m

Fclamp

Hydrostatic force, F

D-11m

4.10 TUTORIAL

Question 1

A square aperture in a vertical side of a tank has one diagonal vertical and is completely covered

by a plane plate hinged along one of the upper sides of the aperture. The diagonals of the

aperture are 2m long and the tank contains water to a height of 1.5 m above the centre of the

aperture. Calculate the force exerted on the plate by the water, the moment of this force about

the hinge, and the position of the centre of pressure. How will the above values change if,

instead of water, the tank contained a liquid of relative density 1.25?

Question 2

The figure below shows a rectangular gate AB hinged at the top A and kept closed by a weight

fixed to the door. The door is 120 cm wide and 90 cm long and the combined centre of gravity

of the complete door and the weight is at G, the combined weight being 1000N. Find the height

of water h on the inside of the door that will just cause the door to open.


Question 3

(a) A closed channel full of water has a cross-section in the form of an equilateral triangle of

sides 2.5 m and lies horizontally on one of its sides. Its end is closed by a triangular vertical

gate having similar side lengths, i.e, 2.5m. The gate is supported by a bolt at each corner.

Calculate the magnitude and position of the force acting on the gate. Find also the force

acting on each bolt.

Icg = bh3/36 Question 4 (a) A vertical dock gate is 5.5 m wide and has water to a depth of 7.3 m on one side and to a

depth of 3 m on the other side. Find the resultant horizontal force acting on the dock gate

and the position of its line of action.

(b) To what position does this line tend as the depth of water on the shallow side rises to 7.3 m ?


Question 5

(a) Derive from first principles, an expression for the force acting on one side of a plane

surface, cross sectional area A submerged in a liquid of density ρ.

(b) A square shaped gate, of side length 0.3 m, closes an opening. The gate is hinged at the

top, (Figure 1), below the surface of the water which acts on its face. If the centre of the

opening lies at a depth of 1.25 m below top water level, calculate the magnitude and

position of the force acting on the gate. Find also the force that must be applied by a

clamp which lies 0.5 m below the hinge in order to keep the gate closed.

Square opening 0.3m side length


UNIT 5 HYDROSTATIC FORCES ON CURVED SURFACES

Unit Structure 5.0 Overview


5.2 Introduction

5.3 Magnitude of the Horizontal Component of the Hydrostatic Force – Curved Surface

5.4 Magnitude of the Vertical Component of the Hydrostatic Force

5.5 Resultant Hydrostatic Force Acting on Curved Surface

5.6 Activities

5.7 Summary

5.8 Worked Examples

5.9 Tutorial

5.0 OVERVIEW

In Unit 4, the concept of fluid exerting a pressure has been converted to the fluid exerting a

resultant hydrostatic force on the walls of the vessel in which it is contained. At this point, it is

advisable that you once again go back to the contents of Unit 4, to appreciate the methodology

guiding the approach to deriving the resultant hydrostatic force and the position of action of this

force, the centre of pressure.

You must probably have noted that the individual forces acting on the solid plane surface were

all parallel , and thus could be linearly combined. In the case of a curved surface, there will

once again be individual forces acting at different points along the solid surface, but this time

they will not be parallel forces. So your attention is drawn to this particular difference.



1. Derive the equation governing the resultant force exerted by a fluid on curved

surfaces.


2. Clearly define the exact meaning of each term making up the equation for hydrostatic force on plane surface.

3. Analyse how the worked examples were solved.

4. Differentiate between solution of problems between plane and curved surface.

5.2 INTRODUCTION

Referring to what has been explained in Unit 4, a fluid exerts pressure on the surface of the

vessel in which it is contained. This pressure varies with depth, and in terms of gauge pressure,

this fluid pressure will be a minimum (zero) at the top liquid surface and a maximum at the

bottom liquid level. Force is equal to pressure x cross sectional area, and since fluid pressure

varies with depth, this implies that the force exerted by the fluid also varies with depth. The

force exerted by a fluid, also known as the hydrostatic force, can be illustrated as a series of

individual forces acting on the surface of the solid.

When the surface area is plane, as shown in Unit 4, these forces can be linearly combined.

However, when the surface is curved, a slightly different approach is adopted to combine these

forces and to eventually find their point of action of the resultant force.

5.3 MAGNITUDE OF THE RESULTANT HORIZONTAL COMPONENT OF THE

HYDROSTATIC FORCE – CURVED SURFACE Consider a vessel containing liquid, of density ρ , to a depth H, Figure 5.1, and consider the

curved surface ABCD. As we move down from AB to CD, we know that the pressure will be

increasing from a value of 0 to a value of Hρg.

Now, as was the case with the plane surface in Unit 4, imagine that the curved surface ABCD is

now divided into very small layers of area δδδδA. Consider now layer 1, which is located at depth

h1 from top water level. The pressure acting at 1 is given by h1ρg, and from there, the

hydrostatic force acting at point 1, will be given by the relationship, force= pressure x area.

Thus, force acting at point 1, will be F1, where F1= h1ρg δA. Similarly, the forces acting at


top water surface

C

B

Ah1

h2 h3 h4

H

Width of AB = b

D

δδδδA

point 2, 3 and 4, can be expressed by the following equations: h2ρg δA, h3ρg δA and h4ρg δA

(same situation as described for plane surface in Unit 4).

Figure 5.1: Hydrostatic force on a Curved Surface

These individual forces can be represented as shown in Figure 5.2, as a series of individual

forces acting perpendicularly at their point of contact with the curved surface ABCD. Note

in the case of a curved surface, that these individual hydrostatic forces are NOT parallel to each

other.


Figure 5.2: Individual hydrostatic forces

Similarly, going down up to the bottom of the side ABCD, there would be a series of individual

forces acting perpendicularly on the curved surface, but these forces are not parallel to each

other, and therefore cannot be combined linearly, as was the case for the plane surface.

However each inclined individual force can be considered in a different way. An inclined force

has a vertical and a horizontal component, Figure 5.3.

F1=h1 ρ g δA

F2=h2 ρ g δA

F3=h3 ρ g δA

F4=h4 ρ g δA

h1

h4

h2

h3

B

C

H


Figure 5.3: Horizontal and Vertical Components of Individual forces

Consider a general inclined individual force Fi, making an angle θ with the horizontal. The

corresponding horizontal force is Fi cos θθθθ and the corresponding vertical force is Fi sin θθθθ, as

illustrated in Figure 5.3.

Now consider all the individual inclined forces acting on the curved surface, each will have a

vertical and a horizontal component, Figure 5.4.

C

HF

Fi cos θ θ

Fi sin θ


Figure 5.4: Horizontal and vertical components of individual forces

Now consider only all the horizontal component of these individual forces. From Figure 5.5, it

can be noted that when all the horizontal components of the individual forces are being

considered, they can be considered as if they are acting on a vertical plane surface, given by B’C

x width of curved surface (b), i.e, the projected area of curved surface ABCD.

C

BB’

H


Figure 5.5: Resultant horizontal force acts on

Projected Area = B’C x width of curved surface BC

Thus the horizontal components of the individual forces therefore act on the projected area of the

curved surface. Since here we are once again dealing with a vertical plane surface (the projected

area), we can resort back to the methodology adopted for plane vertical surface in Unit 4, i.e,

combine the individual forces linearly.

Therefore, the Resultant Horizontal component of the hydrostatic force acting on AB’CD =

∑ hi ρ g δA where i will vary from 1 to the total number of layers into which the solid surface

has been divided for analysis purposes, and which eventually yields,

FH= ρρρρg A h

C’

B

B’

H




A = cross sectional area of the projected surface in contact with the liquid

h = vertical depth from top liquid level to the centroid of the projected surface;

and this force acts at the centre of pressure related to the projected vertical surface.

5.4 MAGNITUDE OF THE RESULTANT VERTICAL COMPONENT OF THE HYDROSTATIC FORCE

Having obtained the horizontal component of the resultant force, we need now to work out the

vertical component of the vertical force.

Figure 5.6: Vertical component of resultant forces = weight of liquid above or below curved

surface

Consider Figure 5.6, the vertical component of the hydrostatic force is exerted by the weight of

the liquid. Hence the vertical component of the resultant hydrostatic force acting on surface BC,

will simply be the weight of the fluid above the curved surface or opposing the curved

surface.

C’

BB’

H


Weight of liquid = shaded area x width of curved surface x density x acceleration due to gravity

Weight of liquid = A b ρ g = Vρ g , and this weight acts at the centroid of the body of liquid,

and is also the vertical component of the resultant hydrostatic force.

Hence, FV= V ρρρρ g

Where V is the volume of liquid above or against the curved surface

ρ is the density of the liquid

g is the acceleration due to gravity

5.5 RESULTANT HYDROSTATIC FORCE ON CURVED SURFACE

From section 5.3, it was shown that the horizontal component FH acts at the centre of pressure

related to the projected vertical surface B’C. Similarly, in section 5.4, it has been shown that the

vertical component of the hydrostatic force FV acts at the centroid of the liquid above the curved

surface, and these are illustrated in Figure 5.7.


Figure 5.7: Resultant force acting on Curved Surface

Thus the resultant hydrostatic force acting on the curved surface, is reported as follows:

FR = { (FH)2 + (FV)2 } ½

Acting at an angle, θ, where θ is given by tan-1 (FV/FH).

C

BB’

H

FH

D

FV

Position of the centroid of the liquid above curved surface

FR FV

FH

θ


5.6 ACTIVITIES

1. Differentiate between the concepts behind the analysis of the resultant hydrostatic force

acting on curved surface as compared to those used in the analysis of a plane surface.

2. What do you understand by the term projected area? What is the shape of the projected

surface of a sphere?

3. Given the equation of the curved surface under study, explain, based upon the principles

of first moment of area, how you would locate the position of centroid of the liquid above

the curved surface?

4. Derive the equation governing the resultant hydrostatic force acting on a curved surface.

5.7 SUMMARY

Unit 5 was a continuation of Unit 4, whereby in this unit you have has been explained how to

work out the resultant force acting on a curved surface. The next unit will also be concerned with

the calculation of hydrostatic force on plane surface based upon a slightly different approach, the

Pressure Diagrams.

You need to learn both methodologies. However, before moving to Unit 6, you are strongly

advised to go back once more to Units 4 and 5, to make sure the concepts described there are

clear, else you might get confused while going through Unit 6.


5.8 WORKED EXAMPLES

Example 1 – Hydrostatic force on curved surface in the shape of the quadrant of a circle

Example 1 – Quadrant of circle …1/4

• A sluice gate as shown below, consists of a quadrant of a circle of radius 1.5m pivoted at its centre O. Its centre of gravity is at G as shown. When the water is level with the pivot O, calculate the magnitude and direction of the resultant force on the gate due to the water and the turning moment required to open the gate. Thewidth of the gate is 3m and it has a mass of 6000kg.

1.5m

O

G

W

1.5m radius

0.6m

0.6m

water


Example 1 – Quadrant of circle..2/4

• Horizontal component of force acting on Projected area,

FH = ρ g A yWhere is the density of the liquid exerting the hydrostatic force (kg/m3)g the acceleration due to gravity (m/s2)A is the projected area of the curved surface (m2)y is the vertical distance from the top liquid surface to the centre of

gravity of the projected area (m)

• Projected area of curved surface is a rectangular surface, width 3m, and depth 1.5m.

• y is given by 1.5/2 = 0.75m, measured from the top liquid surface.

• Hence, horizontal component of force acting on Proj ected area,

FH = ρ g A y = 1000 x 9.81 (1.5 x 3) x (1.5/2)= 33.1 kN


• Vertical component of the hydrostatic force is given by the weight of the liquid either above the curved surface or opposing the curved surface.

• In this example, the vertical component of the force will be given by the weight of the water which is opposing the presence of the curved surface.

• Weight of the water acting against the curved surface

Weight = volume of quadrant x density x g

Weight = (πr2)/4x3 x 1000 x 9.81Weight = Vertical component of the hydrostatic force

= (πr2)/4x3 x 1000 x 9.81FV = (πx 1.52)/4x3 x 1000 x 9.81 = 52 kN



• Resultant hydrostatic force acting on curved surfac e, FRFR = ? (FH

2 + FV2)

FR = ? {33.12 +522}FR = ? {33.12 +522} = 61.6 kN

Point of action of resultant hydrostatic force ,θ = tan -1 (52/33.1)θ = 52.5o

• Turning moment required to open the sluice gate:

Taking moment about the centre of the quadrant,

W x 0.6 = 6000 x 0.6 x 9.81 = 35.3 kNm

Note: Since the line of action of the resultant hydrostatic force passes through the centre of the quadrant, the hydrostatic force exerts no moment about the pivot O.

Example 2 – Hydrostatic force acting on a curved surface, the surface of a dam

Example 2 – Force acting on a Dam..1/9• The face of a dam is curved according to the relation Y=X2/2.4

where y and x are in metres. The height of the free surface above the horizontal plane through A is 15.25m. Calculate the resultant force F due to the fresh water acting on unit breadth of the dam, and determine the position of the point B at which the line of action of this force cuts the horizontal plane through A.

H =15.25m

Y

XA


Example 2 – Force acting on a Dam..2/9

• Horizontal component of force acting on Projected area,

FH = ρ g A y = 1000 x 9.81 (15.25 x 1) x (15.25/2) = 1143.7 kN per m width of dam

15.25m

Y

XA


• Vertical component of the hydrostatic force:– Weight of the water above the curved surface

= Shaded area x width of the dam x density of water x g

• Equation of dam: Y=X2/2.4, and when Y = 15.25, X = 6.05 • Shaded area = ∫ (2.4 Y)1/2 dy from y=0 to y=12.25 gives • Shaded area = {2/3 x 2.4 x Y} 3/2 from y=0 to y=15.25• Shaded area = 120.5 m2

A



• Vertical component of the hydrostatic force:– Weight of the water above the curved surface FV

= Shaded area x width of the dam x density of water x g= 120.5 x 1 x 1000 x 9.81 = 1.18 MN per m width of the dam

• Resultant hydrostatic force acting on dam:

FR = ? (FH2 + FV

2)FR = ? {1143.72 +11802} = 1643 kN per m width of

dam

A


• Distance AB = Distance AZ + ZB

• AZ = horizontal distance from point A to the line o f action of thevertical component of the force, F V

• ZB = (H-D) tan θθθθ

FR

FV

FH

Dcentre of pressure

for force FH H

(H-D) tan θθθθ

θAZ B



• Calculating distance D :

D = Icg/Ay +y= (bd3/12)/(bd)xd/2 + d/2D = (1x15.253)/{12x(1x15.25)x15.25/2} + 15.25/2D=10.17m (distance ZB)

FRFV

FHH

(H-D) tan θθθθ

θAZ B


• Calculating position of centroid of the curved surface :

Consider a small element of thickness dX, within the curve surface: First moment of area of this small element about line AA’:

? {(H-Y)dX x X} = area above curved surface x distance of centroid of curved surface from line AA’.

H

A

XB

A’

Y=X2/2.4

H-Y

Small elementthickness dX



• ? {(H-Y)dX x X} = area above curved surface x distance of centroid of curved surface from line AA’.

• ? {(H-X2/2.4)dX x X} , Integrating from X =0 to X =6.05 (see slide no. 3)

• Total area above curved surface = 120.5 (see slide no. 3)

H

A

XB

A’

Y=X2/2.4

H-Y



• ?{(H-X2/2.4)dX x X} , Integrating from X =0 to X =6.05 (see slide no. 3)

Integrating equation results in{ HX – X3/(3x2.4)} and from X=0 to

X=6.05, gives 61.51.

• Total area above curved surface = 120.5 (see slide no. 3)

• Therefore X = (120.5/61.51) == 1. 96m (distance AZ)

H

A

XB

A’

Y=X2/2.4

H-Y


Distance AB = AZ + ZB = 1.96 + 10.17 = 12.13m


5.9 TUTORIAL

Question 1 The curved surface of a dam retaining water is shaped according to the relationship y=x2/4,

where x and y are measured in metres from the origin. The origin is being defined as the point of

intersection of the base of the dam to the horizontal.

Calculate

(a) the resultant force acting on the dam, in terms of magnitude and direction.

(b) the point of action of the resultant force, measured from the origin.

Question 2 (a) The profile of the water face of a dam is given by the equation 44.75 y = x 2.5, where

the coordinates of x and y in metres are measured from an origin set at the point of

intersection of the flat floor and the curved face of the dam. The depth of the water is 4 m.

Calculate the magnitude and direction of the resultant hydrostatic force per meter

width of the dam.

(b) Calculate also the horizontal distance at which the resultant force cuts a horizontal line at

floor level.

Question 3

(a) A sluice gate consists of a quadrant of a circle of radius 1.5m pivoted at its centre at O. Its

centre of gravity is at G as shown. When the water is level with the pivot O, calculate the

magnitude and direction of the resultant force on the gate due to the water and the turning

moment required to open the gate. The width of the gate is 3 m and it has a mass of 6000 kg.


Question 4

The face of a dam is curved according to the relation y=x2/2.4, where y and x are in metres. The

height of the free surface above the horizontal plane through A is 15.25 m. Calculate the

resultant force F due to the fresh water acting on unit breadth of the dam, and determine the

position of the point B at which the line of action of this force cuts the horizontal plane through

A.



UNIT 6 PRESSURE DIAGARAMS

Unit Structure

6.0 Overview


6.2 Introduction

6.3 Pressure diagrams

6.4 Magnitude of the Vertical component of the Hydrostatic Force

6.5 Position of Centre of Pressure

6.6 Activities

6.7 Summary

6.8 Worked Examples

6.9 Tutorial

6.0 OVERVIEW

In Units 3, 4 and 5, the student was introduced to the concept of pressure exerted by fluids and

the conversion of this fluid pressure into a resultant hydrostatic force on a plane and a curved

surface. In Units 4 and 5, one particular approached to calculating resultant hydrostatic force

was presented.

In Unit 6, the student will now be introduced to a second approach of calculating the resultant

hydrostatic force acting on vertical plane surface only.




1. Understand the meaning of pressure diagrams

2. Understand why the method of pressure diagrams can be applied only to plane surface

3. How to derive the equation for calculating resultant hydrostatic force on vertical plane surface

using the pressure diagram method.

4. How to derive the equation for calculating the centre of pressure for the resultant hydrostatic

force on a vertical plane surface using the pressure diagram method.

6.2 INTRODUCTION

As discussed in Unit 4, a fluid at rest exerts pressure and this pressure can be converted to a

force, the hydrostatic force exerted on the walls of the vessel in which it is contained. One

particular method for calculating this pressure was illustrated in Unit 4.

A second, more simplified way of calculating the resultant hydrostatic force and its centre of

pressure will be illustrated in this unit.

6.3 PRESSURE DIAGRAMS

The method of Pressure Diagram is a graphical method for calculating hydrostatic forces on

solid surfaces and the centre of pressure of these forces. Pressure diagrams are convenient for

plane vertical surfaces only.

Pressure diagram is simply the graph showing the distribution of pressure from one extremity of

the plane vertical surface to the other. For example, from Figure 6.1, we need to calculate the

hydrostatic force acting on vertical plane surface AB. The two extremities of the plane vertical

surface is A and B, where A is located at the top water level and B is located at the bottom edge.

The fluid pressure at A in terms of Gauge Pressure (Unit 3), is zero, and the fluid pressure at B is

(Hρg). Thus plotting these two points yields a triangular shaped graph as shown in Figure 6.1:


Figure 6.1: Pressure Diagram – vertical plane surface AB

Similarly, consider a vertical plane surface, submerged in a body of fluid, Figure 6.2. The

extremities of this plane vertical surface is A and B, where the fluid pressure at A is (Yρg) and

the fluid pressure at B is [(H+Y)ρg], yielding a trapezoidal shaped pressure diagram.

Figure 6.2: Pressure Diagram – a submerged vertical surface AB

A

B Pressure head

depth

PA=0

PB=H ρρρρ g

H Liquid

density, H

A

B Pressure head

depth

PB=(H+Y) ρρρρ g

H

Liquid density, ρρρρ

Y

PA=Y ρρρρ g


Note: The plane surface AB should either be rectangular or square, else the width of the plane

surface will vary and so will the shape of the pressure diagram.

6.4 MAGNITUDE OF THE RESULTANT VERTICAL COMPONENT OF THE HYDROSTATIC FORCE

The magnitude of the resultant hydrostatic force on a vertical plane surface using the pressure

diagram is simply, the AREA of the pressure diagram multiply by the width of the plane

surface.

Referring to Figure 6.1, the hydrostatic force acting on vertical plane surface AB is the

area of the triangular pressure diagram multiply by the width of AB:

Resultant hydrostatic force = (AREA of Pressure diagram) x Width of surface AB

= { ½ Hρg x H } x B

= ½ H2ρρρρg B

Referring to Figure 6.2, the hydrostatic force acting on vertical plane surface AB, will

therefore be given by

Resultant force acting on AB (Figure 6.2) = Area of trapezium x Width of AB

= { ½ (Yρg + (H+Y)ρg} H x B

= ½ H B (H+2Y)ρρρρg

NOTE: Student should know about how to make use of the method of pressure diagram to

calculate the resultant hydrostatic force acting on a plane surface. The key point in this method

is to have a clear diagram and draw your pressure diagram accurately.

The rest is just calculating the area of simple diagrams, triangles or trapezium.


6.5 POSITION OF CENTRE OF PRESSURE

The position of action of the resultant hydrostatic force, i.e, the centre of pressure, is simply the

CENTROID of the pressure diagram.

Referring to Figure 6.1, the pressure diagram in this case is in the form of a triangle. The

CENTROID of a triangle is located at position 1/3 the height of the triangle. Hence in

Figure 6.1, the centroid will be H/3 from point B (Figure 6.3):

Figure 6.3: Pressure Diagram – CENTROID OF PRESSURE DIAGRAM

Similarly, the position of centre for the resultant force in Figure 6.2, will be the position of

centroid of the trapezium.

The position of centroid of a trapezium can be calculated by dividing the trapezium into 2 simple

shapes, a rectangle and a triangle, the position of centroid of which are known. Next by taking

moments about the base of the trapezium, the position of centroid of the trapezium can be

calculated.

Pressure

dept

PA=0

PB=H ρρρρ g

H

H/

Position of centroid of triangle


Figure 6.4: Pressure Diagram – centroid of Trapezium

Calculation of the centroid of a trapezium is best illustrated with a numerical example.

Pressure head

depth

x x

H

H/

H/2

Take moments about line XX to find the position of centroid of the trapezium


6.6 ACTIVITIES

5. Draw pressure diagrams acting on the plane vertical surface AB for the following cases:

6. Calculate the position of the centroid of the following diagrams in part 1.

7. Calculate the resultant and centre of pressure of the various cases illustrated in part 1.

8. A closed tank rectangular in plan with vertical sides is 1.8m deep and contains water to a

depth of 1.2m: (a) Calculate the hydrostatic force acting on one side of the tank if its

width is 3m and its centre of pressure, and (b) If the air space above the water is filled

with pressurised air, 35 kN/m2. Calculate the hydrostatic force and its centre of pressure.

4. A rectangular plane area, immersed in water is 1.5m by 1.8 m with the 1.5 m side

horizontal and the 1.8 m side vertical. Determine the magnitude of the force on one side

and the depth of its centre of pressure (measured from the base), if the top edge is 30 m

below the water surface.

(a)

1.5m

A

B

(b)

1.5m A

B

1m

(c)

0.5m

A

B

(d)

1.5m

A

B

Pressurised air

5kN/m2

Width of AB = 2.5m in all cases & fluid is vessel is water


6.7 SUMMARY

In this unit you have been introduced to a different method for calculating resultant hydrostatic

force and centre of pressure of pressure, for vertical plane surface. Your attention is drawn to the

fact that the plane vertical surface needs to be either rectangular or square for making use of the

method of pressure diagrams.

Unit 7 will be a continuation of hydrostatics, i.e, fluids at rest, but bears no direct relation to what has

been discussed in units 3 to 6.

6.8 WORKED EXAMPLES

Example 1 – Calculation of hydrostatic force with the top surface under atmospheric pressure

Example 1 – Simple application…1/4

• A tank, rectangular in plan with vertical sides, is 1.8m deep and contains water to a depth of 1.2m. If the length of one wall of the tank is 3m, calculate the resultant force on this wall and the height of the centre of pressure measured from above the base.

1.8m

A

B

Width of wall (AB) = 3m



• At position A, the gauge pressure = 0 and at position B, the gauge pressure is given by the height of the liquid exerting the pressure, hρg.

• Hydrostatic force per m width of the wall is given by the area of the pressure diagram = Area of the shaded triangle

• Area of shaded triangle = {½ x base x height } = {½ x 1.8ρρρρg x 1.8 }

1.8m

A

B Pressure

heig

ht

H=0, Gauge pressure=0

H=1.8Gauge pressure = 1.8ρρρρg


• Area of shaded triangle = {½ x base x height } = {½ x 1.8ρρρρg x 1.8 }

= 1.62 ρρρρg

• Total hydrostatic force acting on AB = shaded area x width of AB

= 1.62 ρρρρg x 3 = 47.7 kN

1.8m

A

B Pressure

heig

ht





• Position of action of the hydrostatic force, i.e, Centre of pressure, is given by the position of the centroid of the pressure diagram.

• Position of Centroid of a triangle, is located as shown in the diagram above:

– Hence measured from the base, the position of centroid, hence Centre of Pressure of the hydrostatic force acting on AB is 1.8/3 = 0.6m measured from the base.

1.8m

A

B Pressure

heig

ht



b/3

h/3

Centroid

Example 2 – Calculation of hydrostatic force with the top surface under pressure

Example 2 –Slightly more complex example…1/4

• A closed tank, rectangular in plan with vertical sides, is 2.4m deep and contains water to a depth of 1.8m. Air is pumped into the space above the water until the air pressure is 35kN/m2, if the length of one wall of the tank is 3m, calculate the resultant force on this wall and the height of the centre ofpressure measured from above the base.

1.8m

A

B

Width of wall (AB) = 3m

Pressurised Air

2.4m


Pressurised Air

Example 2 – Slightly more complex example…2/4

• At A the pressure at the water surface, is equal to the same as the pressure of the air, i.e, 35kN/m2.

• At B, the pressure is equal to the pressure exerted by the pressurised air and the pressure exerted by the 1.8m column of water.

1.8m

A

B Pressure

heig

ht

H=0, Gauge pressure= Pressure of Air = 35kN/m2

H=2.4m, Gauge pressure = 1.8ρρρρg + Pressure of Air

Pressurised Air


• Hydrostatic force acting on AB is given by the product of the shaded area and the width of the wall (3m).

– FAB = { (35x1000 x 2.4) + (½ x 1.8ρg x 1.8)} x 3 = 299.7 kN

1.8m

A

B Pressure

H=0, Gauge pressure= Pressure of Air

H=2.4mGauge pressure = 1.8ρρρρg + Pressure of Air

Pressure

heig

ht



• The position of action of the hydrostatic force acting on AB, is the position of centroid of the pressure diagram.

• The pressure diagram is therefore divided into two simple diagrams, a rectangle and a triangle, with known position of centroid, and the centroid of the pressure diagram is calculated by taking the first moment of area about the base:

{ (35x1000x2.4)x1.2+ (½x1.8ρgx1.8) x 0.6} = {(35x1000x2.4) + (½x1.8ρgx1.8)} Y

Therefore, Y = 1.11m

1.8/3=0.6m2.4/2=1.2m

Shaded Area =(35x1000x12.4)

Shaded Area = (½ x 1.8ρg x 1.8)

BASE

6.9 TUTORIAL

Question 1 (a) A gate 3 m wide and 2 m deep divides a storage tank. On one side of the gate there is petrol

of specific gravity 0.78 stored at a depth of 1.8m, while on the other side there is oil of

specific gravity 0.88 stored to a depth of 0.9m. Using the pressure diagram method,

determine the resulting hydrostatic force acting on the gate, and the position of action of this

force.

(b) 'The centre of pressure is always located below the centroid of the wetted surface', Discuss.


Question 2

A lock gate is 3m wide. Find the initial turning moment necessary to force the gate to open

when the water levels across the gate are 3m and 4m respectively above the sill, using the

pressure diagram method.

Question 3 Find the resultant pressure and the center of pressure on a vertical square plate 1.8m side, given

the center of the plate is 1.2m below the surface of the water, using the pressure diagram method.

Question 4 a. A gate 3 m wide and 2 m deep divides a storage tank. On one side of the gate there is

petrol of specific gravity 0.78 stored at a depth of 1.8m, while on the other side there is

oil of specific gravity 0.88 stored to a depth of 0.9m. Using the pressure diagram

method, determine the resulting hydrostatic force acting on the gate, and the position of

action of this force.

b. 'The centre of pressure is always located below the centroid of the wetted surface',

Discuss.


UNIT 7 BUOYANCY

Unit Structure

7.0 Overview


7.2 Introduction

7.3 Resultant Force Acting on a Completely Submerged Body

7.4 Definition –Buoyancy

7.5 Sink or Float?

7.6 Stability of Submerged Bodies

7.7 Metacentre

7.8 Summary

7.0 OVERVIEW

Unit 7 deals with the forces exerted by a fluid on a solid body in contact with the fluid. This unit

just like units 1 to 6 is concerned with fluids at rest.



1. Define what you understand by the term Buoyancy & Archimedes Principle.

2. Differentiate with the help of sketches between, stable equilibrium, unstable equilibrium

and neutral equilibrium.

3. Explain what is meant by the term metacenter.

4. To relate the position of metacentre, centre of gravity of a body to the stability of the

body in the fluid.


7.2 INTRODUCTION

Buoyancy is the term used to describe the force of resistance exerted by a fluid on a body which

is either partially or completely submerged into it. Buoyancy is a term which is very much

relevant in boat construction.

7.3 RESULTANT FORCE ACTING ON A COMPLETELY SUBMERGE D BODY

Buoyancy is a force exerted by a fluid on a solid body completely or partially immersed in

the fluid and buoyancy always pushes upwards against the pull of gravity.

How buoyancy works:

a) Buoyancy is based on Newton's Third Law of Motion, which states: "For every action, there

is an equal and opposite reaction." Here is the process:

(i) objects placed in a fluid push particles of the fluid away (displace particles)

(ii) following Newton's Third Law, the particles push back on the object

(iii) the force with which the particles push back is the buoyant force measured in newtons (N)

Referring to Unit 5, it is to be noted here that the method use for calculating forces on

curved surfaces applies to any shaped body. The approach is to work out the resultant

horizontal component and the resultant vertical component, and combine them to obtain

the resultant hydrostatic force acting on the curved surface. Consider the following

completely submerged body:


Figure 7.1: Completely submerged body

Calculating the resultant HORIZONTAL COMPONENT

To obtain the resultant horizontal force acting, we divide the body into two and then project the

curved surface, which is a circle in this case, as indicated in Figure 7.1. The result is that there is

an equal and opposite horizontal component of hydrostatic force acting on the body, hence no

resultant horizontal force will be acting on the completely submerged body.

Calculating the resultant VERTICAL COMPONENT

For the resultant vertical component, we once again divide the body into two parts from a

horizontal plane. Consider equilibrium of vertical forces on the upper surface and on the

lower surface of the body: Vertical force acting on the upper surface of the body = weight

of fluid displaced by this curved surface

Similarly, the vertical force acting on the lower surface = weight of fluid displaced by this

curved surface

Therefore the resultant vertical force acting on the curved body, also called the upthrust on the

body = The weight of the fluid displaced by the body (Archimedes’ Principle).

This vertical force will act through the centroid of the volume of fluid displaced by the body,

known as the centre of buoyancy.


cgb

cg d

cgb – centre of gravity of solid body ABCD cg d – centroid of the displaced fluid, space EFCD

A

F E

C D

B

7.4 DEFINITION - BUOYANCY Any body in a fluid, whether floating or completely submerged, is buoyed up by a force equal to

the weight of the fluid that has been displaced – Archimedes Principle.

The buoyancy force is given by the product of the volume of fluid displaced, the density of the

liquid and the acceleration due to gravity, V ρρρρ (liquid) g. The buoyancy force acts at the

centroid of the liquid which has been displaced (Figure 7.2) in the case of a rectangular body

ABCD floating in a liquid. In the case of a completely submerged spherical body, the centroid

of the liquid which will have been displaced, is the centre of the sphere.

Figure 7.2: Centroidal position of displaced liquid

NOTE: For a completely submerged body, the centroid of the displaced fluid is located at the

same position as the center of gravity of the body.


7.5 SINK OR FLOAT

If the downward force exerted by the weight of the body is higher than the vertical upwards

buoyancy force, the body will sink, else it floats. The larger the surface area of a solid body in

contact with the fluid, the greater the amount of fluid displaced, and hence the higher the vertical

upthrust force.

7.6 STABILITY OF SUBMERGED BODY

A submerged or partially immersed body can be in stable, unstable or neutral equilibrium

position. The body is in stable equilibrium if when titled it regains its original position. If body

acquires a different position after having been tilted, then it is said to be under unstable

equilibrium .

Consider the body ABCD floating in liquid (Figure 7.3): the weight W=mg acts through the

centre of gravity of the body cgb, and the upthrust R acts through the centre of buoyancy cgd.

Figure 7.3: Position of action of Weight and Upthrust acting on ABCD

When the body ABCD is titled, this is accompanied by a change in the volume of fluid displaced

by the tilted body, and consequently a change in the position of centroid of the dispersed fluid.

cgb

cgd

A

F E

C D

B

R

W=mg


The position of cgd changes, but the position of the center of gravity cgb of the body which is

independent of the fluid displaced, stays the same, as indicated by Figure 7.4.

Figure 7.4: Titled object – Resulting restoring moment (anticlockwise)

The two forces, the weight of the body ABCD acting at the center of gravity cgb, and the upthrust

R acting at the new position of centroid of the displaced fluid (A’B’CD), cgd, exert a moment as

indicated by the arrows in Figure 7.4 & Figure 7.5. When this moment is such that it makes the

body regain its original position, this is knows as the STABLE condition.

If the however the moment is such that the body comes to equilibrium in another position,

then we have UNSTABLE condition (Figure 7.6).

cgb

cgd

A

C

D

B

Figure 7.4: Tilted object – Resulting restoring moment (anticlockwise)

R

mg

A’ B’cgb

cgd

A

C

D

B

Figure 7.4: Tilted object – Resulting restoring moment (anticlockwise)

R

mg

A’ B’


Figure 7.5: Titled Object – Resulting Restoring moment (clockwise)

Figure 7.6: Titled object – Resulting Overturning moment (clockwise)

cgb

cgd

A

C

D

B

Figure 7.5: Tilted object – Resulting Restoring moment (clockwise)

R

mg

A’ B’cgb

cgd

A

C

D

B

Figure 7.5: Tilted object – Resulting Restoring moment (clockwise)

R

mg

A’ B’

cgd

cgb

Figure 7.6: Tilted object – Resulting Overturning moment (clockwise)

R

mg

cgd

cgb


R

mg


7.7 METACENTRE

The metacentre is defined as the theoretical point at which an imaginary vertical line through the

centre of buoyancy (cgd) intersects another imaginary vertical line through a new centre of

buoyancy created when the body is displaced, or tipped, in the water, however little. Basically

the centre of buoyancy of a floating body is the point about which all the body parts exactly

buoy.

A body is under stable equilibrium if its center of gravity cgb, always lies below the position of

metacentre M, as indicated by figures 7.7-7.9.

Figure 7.7: Position of Metacentre


cgb

cgd

M

Note : The center of gravity cgb of the solid body lies below the metacentre point M, hence body is under stable equilibrium


cgb

cgd

M

cgb

cgd

M

Note : The center of gravity cgb of the solid body lies below the metacentre point M, hence body is under stable equilibrium


M

Figure 7.8: Position of M etacentre

Note : The center of gravity cgb of the solid body lies below the m etacentre point M , hence body is under stable equilibrium

cgb cgd

M

Figure 7.8: Position of M etacentre

Note : The center of gravity cgb of the solid body lies below the m etacentre point M , hence body is under stable equilibrium

cgb cgd




cgd

cgb


M

Note : The center of gravity cgb of the solid body lies above the metacentre point M, hence body is under unstable equilibrium

cgd

cgb


M

Note : The center of gravity cgb of the solid body lies above the metacentre point M, hence body is under unstable equilibrium


Figure 7.6: Titled object – Resulting Overturning moment (clockwise)

cgd

cgb


R

mg

cgd

cgb


R

mg


7.8 SUMMARY

In this unit you have learnt how a fluid exerts a vertical force on a solid body. You will also

have learnt about how the stability of a body is related to the density of the fluid in which the

body is being immersed and the relative position of the metacentre and the centre of gravity of

the solid body.

Unit 8 will be the first chapter dealing with fluids in motion. Some of the concepts discussed in

units 1 to 8 will be referred to in the coming units, and students will need to make sure the

concepts behind the analysis of fluids at rest are well understood before they move on to the next

part of the module, i.e. fluids in motion.

7.9 TUTORIAL

Question 1

A rectangular pontoon 10m by 4m in plan weighs 280kN. A steel tube weighing 34kN is placed

longitudinally on the deck. When the tube is in a central position, the centre of gravity for the

combined weight lies on the vertical axis of symmetry 250mm above the water surface. Find (a)

the metacentric height, (b) the maximum distance the tube may be rolled laterally across the deck

if the angle of heel is not to exceed 5o.

Question 2

A rectangular tank 90cm long and 60cm wide is mounted on bearings so that it is free to turn on

a longitudinal axis. The tank has a mass of 68kg and its centre of gravity is 15cm above the

bottom. When the tank is slowly filled with water it hangs in stable equilibrium until the depth

of water is 45cm after which it becomes unstable. How far is the axis of the bearings above the

bottom of the tank?

Question 3

A cylindrical buoy 1.35m in diameter and 1.8m high has a mass of 770kg. Show that is will not

float with its axis vertical in sea water of density 1025kg/m3. If one end of a vertical chain is


fastened to the base find the pull required to keep the buoy vertical. The centre of gravity of the

buoy is 0.9m from its base.


UNIT 8 HYDRODYNAMICS – FLUID DYNAMICS (IN MOTION)

Unit Structure 8.0 Overview


8.2 Introduction

8.3 Types of Flow

8.4 Uniform & non-uniform Flow

8.5 Steady & unsteady flow

8.6 Reynolds’ Number

8.7 Pathlines & Streamlines

8.8 Streamlines & Streamtubes

8.9 Rotational & Irrotational Fluid

8.10 Activities

8.11 Summary

8.0 OVERVIEW

This unit illustrates the approach and the various concepts commonly used when analyzing a

fluid in motion (hydrodynamics). In hydrodynamics assumptions often have to be made while

carrying out analysis. In many cases, except when dealing with very complex fluid flows, the

motion of fluid can be analysed with high level of accuracy.

The concepts used in the analysis of hydrodynamics are very much similar to those used in the

motion of solids, with a few minor changes induced by the behaviour of the fluids under stress

conditions.



At the end of this Unit, you should be able to do the following:

1. Learnt about the concepts and approaches used to analyse fluids in motion.

2. Differentiate between laminar and turbulent fluid flow, steady and unsteady fluid

motion, and uniform and non-uniform fluid motion.

3. Differentiate between streamlines and stream tubes.

4. Illustrate with the help of sketches streamlines in a pipeline, in a river, over an

obstacle

5. Differentiate between a pathline and a streamline.

6. Differentiate between rotational and irrotational fluids.

8.2 INTRODUCTION

Before dealing the mathematical concepts behind the analysis of fluids in motion, it is first of all

important to understand the characteristics of different types of flows. The approach used in the

analysis of fluids in motion is very much dependent upon the prevailing flow conditions. Some

assumptions may not always hold true is all circumstances.


8.3 TYPES OF FLOW

Consider the flow in a river (Figure 8.1):

Figure 8.1: River flow

Points 1 to 5 have been highlighted in Figure 8.1, so as to make an important point about

the conditions affecting flows in a river.

Consider point 1 and 2, both of which lie more or less in the middle of the river. The velocity of

flow at these two points should have been the same, however most likely they will not be. The

reason being that slightly downstream of point 1 the river meanders slightly and this feature will

induce disturbance in flow pattern upstream.

Figure 8.1: River flow5

4

3

2

1

flowdirection

Figure 8.1: River flow5

4

3

2

1

flowdirection


Similarly compare points 3 and 5, both of which are located close to the river bank. The velocity

of flow at both points will be influenced by the presence of the solid boundary of the river bank.

However point 3 lies within the zone where the river meanders, so the velocity at 3 and 5 will not

be similar.

Point 4 was highlighted so as to emphasise on the influence of boundary (river bank) on the

velocity of flow, the closer the point to the boundary (in this case 3), the higher the impact of

boundary. Hence though both points 4 and 5 lie within the meandering zone, the velocity of

flow at point 4 will be higher than that at 3.

8.4 UNIFORM & NON-UNIFORM FLOW

In section 8.3, it has clearly been illustrated that the flow both along and across a river channel

varies from point to point, and such a situation is also technically termed – NON-UNIFORM

FLOW. Thus, had the velocity been similar at any point along and across the river channel, the

flow would have been termed – UNIFORM FLOW. Uniform or non-uniform flow is the

variation of velocity with distance.

A simplified illustration of uniform and non-uniform flow is given in Figure 8.2:


Figure 8.2 – Uniform and non-uniform flow 8.5 STEADY & UNSTEADY FLOW

Steady or unsteady flow is the variation of velocity with respect to time. A flow may either be

uniform or non-uniform, but if it stays the same all the time, then the flow is considered as being

steady. Figure 8.3 and Figure 8.4, are simplified illustrations of steady and unsteady flow.

Figure 8.2 - Uniform and non-uniform flow

Note : The direction and magnitude of the velocityis illustrated by the arrow at the various points under consideration.

Uniform flow – same velocity at any point

Non-uniform flow – varying velocity at different poi nts

Figure 8.2 - Uniform and non-uniform flow

Note : The direction and magnitude of the velocityis illustrated by the arrow at the various points under consideration.

Uniform flow – same velocity at any point

Non-uniform flow – varying velocity at different poi nts


Figure 8.3: Steady flow

Figure 8.4: Unsteady flow

When a fluid flow is uniform and steady it is commonly termed as being LAMINAR FLOW.

Unsteady and non-uniform flow is known as TURBULENT FLOW.

Figure 8.3 - Steady flow

Note : The direction and magnitude of the velocity is illustrated by the arrow at the various points under consideration.

Velocity pattern at time t=0 Velocity pattern at time t=t 1

Steady uniform flow


Steady non-uniform flow

Figure 8.3 - Steady flow



Steady uniform flow


Steady non-uniform flow

Figure 8.4 - Unsteady flow



Unsteady uniform flow

Velocity pattern at time t=0 Velocity pattern at tim e t=t1

Unsteady non-uniform flow

Figure 8.4 - Unsteady flow



Unsteady uniform flow

Velocity pattern at time t=0 Velocity pattern at tim e t=t1

Unsteady non-uniform flow


8.6 REYNOLDS’ NUMBER

The difference behind laminar and turbulent flows has been best illustrated by Reynolds’s

Experiment, Figure 8.5:

Figure 8.5: Reynolds’ Experiment

In this experiment a coloured dye in injected in a pipeline, diameter d, running full, with a

velocity V and liquid of density, ρ . So long as the flow in the pipeline is laminar, the dye

appears as a thin coloured thread or streamline within the pipeline. Gradually the flow in the

pipeline is varied, hence V varies, laminar flow changes first to a transition flow, as illustrated in

Figure 8.5. Finally when the flow rate is even higher, the entire flow in the pipeline becomes

turbulent, causing the entire liquid in the pipe to be coloured.

Dye

Dye

Dye

Laminar flow

Transition

Turbulent flow


d

ρ

v

DyeDye

DyeDye

DyeDye

Laminar flow

Transition

Turbulent flow


d

ρ

v


Apart from the flow rate, Reynolds’ also varied the diameter of the pipe and the type of liquid

within the pipeline to investigate how these three parameters, V, ρ and d, were influencing type

of flow within a pipe. Finally he came up with the relationship:

Re = ρρρρ v d / µµµµ where µ is the dynamic viscosity of the liquid

Laminar flow conditions will prevail (provided the entire system is undisturbed), if Re≤≤≤≤ 2000.

8.7 PATHLINES & STREAMLINES

A fluid flow (both liquid and air) may be described in two different ways, with respect to

distance and with respect to time, commonly referred to as (1) the Lagrangian approach (named

after the famous French mathematician Joseph Louis Lagrange), and (2) the Eulerian approach

(named after Leonhard Euler, a famous Swiss mathematician), respectively.

In the Lagrangian approach, one particle is chosen and is followed as it moves through space

with time, i.e, the variation of the particle flow characteristics with time. The line traced out by

that one particle is called a particle pathline or a streakline, Figure 8.6.

Figure 8.6: Pathline described by moving boat

PATHLINE – Consider the movement of a boat every hour

Time=t1

Time=t5

Time=t4

Time=t2

Time=t3

Time=t9

Time=t8

Time=t7

Time=t6

Figure 8.6 – Pathline described by moving boat

PATHLINE – Consider the movement of a boat every hour

Time=t1

Time=t5

Time=t4

Time=t2

Time=t3

Time=t9

Time=t8

Time=t7

Time=t6

Figure 8.6 – Pathline described by moving boat


A Eulerian approach is used to obtain a clearer idea of the airflow at one particular instant. One

can imagine taking the picture of the flow of, for instance, surface ocean currents at a particular

fixed time. The entire flow field is easily visualized. Just like drawing contour lines, here we

join particle having similar velocity together and we produce what is known as streamlines,

Figure 8.7. Similar to the properties of contour lines, streamline also does not cross each other.

In Figure 8.7, fluid is flowing over a solid body. At different points, the fluids will have different

velocity. Lines are used to join points of equal velocity and these produces the streamlines

shown below. No flow takes place across steamlines.

Figure 8.7: Flow over a solid body

Thus, a pathline refers to the trace of a single particle in time and space whereas a streamline

represents the line of motion of many particles at a fixed time. Refer to section 8.4, Figures 8.4

Streamline – Snapshot of flow at a point in time

solid body

Figure 8.7 – Flow over a solid body

Streamline – Snapshot of flow at a point in time

solid body

Figure 8.7 – Flow over a solid body


to 8.6, it can be seen that under steady conditions, pathlines and streamlines will always be

the same.

8.8 STREAMLINES & STREAMTUBES

As mentioned in section 8.7, streamlines are formed when fluid particles of similar velocity are

joined together. Bearing in mind that the velocity of flow is not constant all throughout,

streamlines formed in a pipeline and in a river will appear as shown in Figures 8.8 and 8.9.


Figure 8.8: Streamlines pattern in a pipeline running full

V1

V2V3

V4

V5

V6

NOTE: Consider a pipeline running full.Across the pipe, at each point the fluid has a different velocity (boundary effects).The same pattern will repeat itself at each section (aa, bb, cc).By joining points of equal velocity, we get streamlines.

a

a

b

b c

c

streamlines

Figure 8.8 – Streamlines pattern in a pipeline running full

V1

V2V3

V4

V5

V6

NOTE: Consider a pipeline running full.Across the pipe, at each point the fluid has a different velocity (boundary effects).The same pattern will repeat itself at each section (aa, bb, cc).By joining points of equal velocity, we get streamlines.

a

a

b

b c

c

streamlines

Figure 8.8 – Streamlines pattern in a pipeline running full

flow direction

Figure 8.9 – Streamlines in a river

flow direction

Figure 8.9 – Streamlines in a river


To draw a streamtube, we refer here to the flow through a pipeline running full. Consider a

small cylindrical element inside the pipeline, as indicated in Figure 8.10. From this circle, we

can draw a series of streamlines from one end of the cylindrical element to the other. This

process eventually forms a Streamtube, Figure 8.10.

Figure 8.10: Streamtube in a pipeline running full

Since a streamtube is formed by a series of streamlines, there cannot be any flow inside a

streamtube.

8.9 ROTATIONAL & IRROTATIONAL FLUID

As well as steady or unsteady, fluid flow can be rotational or irrotational. If the elements of

fluid at each point in the flow have no net angular (spin) velocity about the points, the fluid flow

is said to be irrotational. One can imagine a small paddle wheel immersed in a moving fluid. If

the wheel translates (or moves) without rotating, the motion is irrotational. If the wheel rotates in

a flow, the flow is rotational.

Streamtubes


streamtube

Consider a small circle inside a pipeline running full,based upon the concept of streamlines, we can imagine a seriesof streamlines stemming from the circle.This gives rise to a Streamtube.

Streamtubes


streamtube

Consider a small circle inside a pipeline running full,based upon the concept of streamlines, we can imagine a seriesof streamlines stemming from the circle.This gives rise to a Streamtube.


8.10 ACTIVITIES

1. Explain with the help of sketches what how boundaries such as river banks influence the

flow in a river.

2. Differentiate between steady and uniform flow.

3. Illustrate an unsteady but uniform flow.

4. Does unsteady uniform flow exist in practice?

5. Describe the Reynolds’ Experiment and define the purpose of the Reynolds’ number.

6. Differentiate between a pathline and a streamline.

7. Explain why there cannot be flow across a streamtube.

8.11 SUMMARY

This unit has introduced the student to the various terms used to describe the characteristics of

fluids in motion. The student need to be familiar with the terms and their meanings, for these

will eventually guide the assumptions that can be made during analysis of fluids in motion. The

next unit, will now introduce the student to the first two principles used to mathematically

analyse fluids in motion, the Principle of Continuity and the Conservation of Energy.


UNIT 9 PRINCIPLES OF CONSERVATION OF MASS & ENERGY

Unit Structure

9.0 Overview


9.2 Introduction

9.3 Continuity

9.4 Mass Flow Rate

9.5 Principles of Continuity

9.6 Discharge and Mean Velocity

9.7 Conservation of Energy

9.8 Bernoulli’s Equation – Pipelines

9.9 Hydraulic Grade Line

9.10 Frictional Losses

9.11 Activities

9.12 Summary

9.0 OVERVIEW

This unit introduces the student to two main principles used to analyse fluids in motion, the

principles of Continuity and the principles of conservation of Energy. The third principle is the

conservation of Momentum, which will be introduced at a later stage, in the second level of the

course. These basic governing principles will always be used in the analysis of the simple or

complex cases of fluids in motion, and hence, the need to understand and learn them.



1. Define Conservation of Mass and Conservation of Energy.

2. Differentiate between the governing equation for conservation of mass for a compressible

and an incompressible liquid.


3. Differentiate between mass flow rate and discharge or volume flow rate.

4. Appreciate the application of Continuity equation in a branched pipeline.

5. Derive Bernoulli’s equation.

6. Explain the term Energy Head.

7. Differentiate between Total Energy Head and Hydraulic Grade Line

8. Modify Bernoulli’s equation to include frictional losses.

9. Appreciate the application of Bernoulli’s equation.

9.2 INTRODUCTION

Analysis of fluids in motion is no very much different from that of solids in motion. Because we

are here dealing with a liquid there is some modifications which have to be considered during the

analysis. The concepts described below are simple and easy to understand. The principle of

conservation of mass is one concept which is used in most if not all the analysis of fluids in

motion. The principle of conservation of Energy is used a lot in the analysis of fluids in motion.

The most important step in this unit is to learn how to derive the Continuity and Bernoulli’s

equation, to understand the meaning of each and every term in these equations and to get used to

the units of each term.


9.3 CONTINUITY The concept of Continuity is best illustrated using the anology of a road junction (Figure 9.1).

Figure 9.1: Continuum analogy – cars

If 15 cars pass by location A, and no car is allowed to stop along the road, then the sum of the

total number of cars passing location B, C and D, will be 15.

Similarly, if a given volume of liquid flows through a main pipeline (A) in unit time, and this

pipe branches out, the sum of the volume of liquid flowing through the two branched pipes will

be the same as that which was flowing through the main pipe (Figure 9.2).

15

3 4

8

A

B

C

D


10m3/s

3m3/s

7m3/s

A

Figure 9.2: Continuum – flow through a branched pipeline

9.4 MASS FLOW RATE

A simple way of measuring the flow of water through a pipe is to allow the water to collect in a

bucket at the end of the pipe. The volume of water (V) collected in a given time (t) is noted.

The volume of water collected in a given period of time can be converted into mass of water by

multiplying the volume by density (V x ρ ).

Hence Mass flow rate = m = (V ρ )/ t , and its units are kg/s.


9.5 PRINCIPLES OF CONTINUITY

Consider a pipeline with varying diameter along its length (figure 9.3). The velocity of flow at

section 1, is V1, where V1, can also be represented by the distance L1 per unit time. Applying the

principle of conservation of mass, gives, mass flow per unit time at section 1 is equal to the mass

flow per unit time at section 2.

Mass flow per unit time at section 1 = (volume x density )/ time, or (cross sectional area of flow

x horizontal distance moved my fluid in unit time x density) = A1 L1 ρ1 / t . Similarly the mass

flow rate at section 2, will be given by A2 L2 ρ2 / t. Since in general liquids have very low

degree of compressibility, they can be safely assumed to be incompressible, hence ρ1 = ρ2 = ρ.

Thus, the mass flow rate per unit time, given by A1 L1 ρ1 / t, can be simplified to m = A1 V1 ρ.

For continuity of flow, then A2 V2 ρ = A2 V2 ρ, which simplifies to: A1 V1 = A2 V2 = Q, where Q

is referred to as the Volume Flow Rate (units are m3/s).

Hence, Q = AV, volume flow rate equation, also known as the Discharge.

9.6 DISCHARGE & MEAN VELOCITY

Consider the flow of water through a pipeline. The velocity of flow in any system is very much

influenced by the surface in which it is in contact. The closer the liquid to the boundary the more

significant the impact of the boundary of the liquid. Hence in a pipeline under laminar flow

conditions, the maximum velocity is located at the centre.


Vmax = V

V=0

V=0

VPipeline running

full

Velocity flow profile

Figure 9.4: Boundary effects on flow velocity in a pipeline

Thus if the flow rate (Q) through the pipeline is known (collect water in a given time period),

and the cross sectional area (A) of the pipe is known, then the mean velocity (Vmean) through the

pipe is calculated using the discharge equation, Q. Hence, Vmean = Q / A.

9.7 CONSERVATION OF ENERGY

In solid mechanics you have learnt about the meaning of Potential energy and Kinetic energy.

When a body is located at a given height above a datum, it possesses potential energy by virtue

of its position. When a body is moving with a given velocity, V, then this body will possess

kinetic energy by virtue of its movement. The same concepts apply for a fluid. Consider Figure

9.5, which shows a small element of a fluid, characterised by pressure P, velocity V, cross

sectional area, a, and mass M.


Figure 9.5 – Derivation of Bernoulli’s equation

1. Potential energy: The elemental fluid is located at an elevation Z above the datum,

hence the potential energy possessed by the elemental fluid is (mass x elevation x

acceleration due to gravity) = m g z.

2. Kinetic energy: The elemental fluid is moving with a velocity V. Hence, kinetic energy

possessed by the elemental fluid will be, ½ m V2.

3. Pressure energy: The fluid is also under pressure P. This pressure can be converted to

a force, by multiplying it by the cross sectional area of flow, which gives, P a. The

elemental fluid is moving with a velocity V. In unit time the force will have moved a

distance L, which can also be given by the (volume of shaded element/cross

sectional area of flow), or L = (mass/density)/a.

L = m/ρ a

Pressure Energy = Work done by this force = force x distance moved = ( P a) x m/ρ a

Therefore, the Pressure energy = P m/ρ

flow direction

A small fluid element from the main fluid body:

A

B

A’

B’

P, V

cross sectional area of flow, a

weight, mg Z


z1

V1

z2

V2

P1

P2

A fluid possesses energy by virtue of its position as well as by virtue of its movement.

DATUM

P1/ρρρρg + V12/2g + Z1 = P2/ρρρρg + V2

2/2g + Z2

Total Energy Contained by the small fluid element – Bernoulli’s equation:

E = P m/ρρρρ + ½ m V2 + m g z Units: Joules

Total energy contained by the small fluid element per unit mass, H:

H = P/ρρρρ g + V2/2g + Z Units: m

Owing to its units, this equation is also commonly known as the total energy Head.

9.8 BERNOULLI’S EQUATION – IN PIPELINES

The meaning of Bernoulli’s equation is best illustrated in a system, and a pipeline running full is

being considered in Figure 9.6. Assuming that the energy losses from section 1 to section 2 are

negligible, this would mean that the total energy head at section 1 is equal to the total energy

head at section 2. Usually this equation is applied at two different sections within a system, to

work out missing information. Principle of Continuity tends to be used in most of the problems

involving application of Energy principles.

Figure 9.6: Principle of conservation of Energy – Bernoulli’s Equation


9.9 HYDRAULIC GRADE LINE

Hydraulic grade line is simply the sum of the Pressure head and the elevation head, and this is

being illustrated in Figure 9.7. Figure 9.7 shows a reservoir located at a higher elevation Z1

feeding another one located below, at elevation Z2, both reservoirs being connected by a

pipeline. At any point along the pipeline, if a piezometer is connected, water will rise into the

piezometer tube, to a height equivalent to the pressure at that point (Unit 3 – section 3.4), points

A and B in Figure 9.7.

A line can then be drawn at points A and B, that join the sum of their pressure head and elevation

head at respective points, this line is known as the hydraulic grade line. The Hydraulic Grade

Line is always below the Total Energy Head line, and the difference is the velocity head.

Datum

Z1

Z2

h1

h2

Total energy head line

Pressure head + Elevation head line commonly known as the Hydraulic Grade Line

Velocity head

B

A


9.10 FRICTIONAL LOSSES

In practice as liquid flows frictional forces have to be overcome, so some energy is lost. The

Bernoulli’s equation has then to include the term for frictional losses, hf.

P1/ρρρρ g + V12/2g + Z1 = P2/ρρρρ g + V2

2/2g + Z2 + hf

In a pipeline connection, each time there is a feature which disturbs the regular flow pattern,

frictional losses are high. Such features may be in the form of sudden change in diameter, bends,

reducers, valves and connections to branched pipes.

9.11 ACTIVITIES

8. Derive Continuity equation

9. Derive Bernoulli’s equation

10. Explain under what circumstances frictional losses cannot be ignored when applying

Bernoulli’s equation.

11. How does Principle of conservation of energy different when it is being used to analyse

solid mechanics and fluid mechanics.

9.12 SUMMARY

This unit has introduced you to the basic principles which govern the analysis of fluids in

motion. The next unit will show you how these principles are being used in practice in flow

measurement devices to measure flow of liquid in both closed and open systems. Students are

strongly advised to ensure that the concepts described in this unit are clear, for these will crop up

in most of the remaining units of the subject, in all levels of the course.


UNIT 10 FLOW RATE MEASUREMENTS – ORIFICES & WEIRS

Unit Structure

10.0 Overview


10.2 Introduction

10.3 Small Orifice

10.4 Vena Contracta

10.5 Flow Measurement by an Orifice

10.6 Coefficient of Contraction and Velocity

10.7 Discharge Through a Large Orifice

10.8 Velocity of Approach (V1)

10.9 Flow Measurement Through a Rectangular Weir

10.10 Flow Measurement Through a Triangular weir

10.11 Flow Measurement Through a Trapezoidal Weir

10.12 Activities

10.13 Summary


10.15 Tutorial

10.0 OVERVIEW

In the previous unit, two of the main principles which form the basis of the analysis of fluids in

motion were introduced. In this unit, the student will learn how to apply these principles to

obtain further information about a system and also how to these principles are used in estimating

flow rates of fluids in motion.




10. Differentiate between a sharp edged orifice and a streamline office.

11. To applying Bernoulli’s equation to derive the velocity of flow through a small orifice.

12. To justify any assumption made in the derivation of the equation of flow velocity through

an orifice.

13. To understand why a larger orifice is analysed in a different approach.

14. To derive the equation governing flow of a fluid over a rectangular and a triangular weir

15. To derive the equation from first principles for flow rate estimation using an irregular

trapezoidal weir.

10.2 INTRODUCTION

One of the many important characteristics of a fluid in motion is the rate at which it flows. In

fluid mechanics practicals, the students will have learnt that to get an estimate of the flow rate of

a fluid, fluid can be collected in a bucket for a given time and the flow rate worked out.

However in a long complex pipe network this is not possible along the pipes. In such cases a

different approach has to be adopted, and the applications of both Continuity equations and

Energy equations, have proved to be useful tools for such estimates. Similarly when the flow

rate of a river is needed, here also there are different approaches are available. A floating object

can be timed over a distance and knowing the cross sectional area of the river, the flow rate of

the river can be estimated. This is however a rough approach. Weirs are devises which have

proved to be effective in such measurements. In this units, the student will be introduced to

several devices which are used to measure flow rate in different situations, and with different

accuracies.


10.3 SMALL ORIFICE

A small orifice is simply a small circular opening most of the time located on the side of a

reservoir or a container. A small orifice can also be located on the bottom of the container. The

opening can either be sharp edge or streamline edged (Figure 10.1). The sharp edged opening

gives rise to significant disturbances as compared to the streamline opening.

Figure 10.1: Small orifice

The approach in estimating the flow through a small orifice is to apply the Continuity equation

and Bernoulli’s equation, at two points where maximum information is present. This helps in the

simplifying the general equations (Figure 10.2).

Small orifice

Sharp edged opening

Streamline opening

A

B


Figure 10.2 – Flow pattern through small orifice

From Figure 10.2, it can be noted that if the pathway of a fluid particle is highlighted, it would be

straight vertical line from the surface of the liquid and curved as it reached the opening. Thus a

large flow area is suddenly constricted into a small opening. This sudden change in path gives

rise to turbulence and hence heavy losses of energy. Though an orifice is a simple device to

measure the flow rate, it does not offer much accuracy owing to the structure of the system.

Enlargement – at orifice


10.4 VENA CONTRACTA The basic principle behind the application of Bernoulli’s Equation is to identify two points at

which maximum information is known. In the case of an orifice, the first such point is at the

surface of the liquid. At the surface we know that the pressure is zero gauge pressure or

atmospheric pressure and that the velocity of flow is so small that it can safely be considered as

being negligible. (Imagine Mare Aux Vacoas reservoir with a small circular opening on its side.)

A second such point is just outside the opening of the orifice, since just outside the flow lines

becomes straight and parallel as compared to the curved flow lines at the opening of the orifice.

This point is also known as the Vena Contracta. The Vena Contracta is smaller in size than that

of the orifice and it lies just outside the orifice. At the vena contracta, the pressure of the fluid is

once again zero gauge pressure or atmospheric pressure (Figure 3).

Figure 10.3: Vena Contracta

Enlargement – at orifice

vena contracta

1

2


10.5 FLOW MEASUREMENT BY AN ORIFICE

Having identified the two reference points within the system, the next step is to apply Bernoulli’s

equation and Continuity equation (Figure 10.4).

Figure 10.4– Reference points for application of Bernoulli’s equation

As mentioned in the section above, the pressure at points 1 and 2 are both zero gauge pressure,

the velocity of flow at point 1 is zero and the elevation at point 2 is zero since in this analysis the

datum has been drawn at point 2 itself. Similarly the elevation at point 1 is the height of liquid

from the datum to the top water surface, which is equal to H.

1

2


2/2g + Z2

Where P1 = 0 gauge pressure, V1≅ 0, Z1= H P2= 0 gauge pressure, Z2= 0 being at datum

Datum

H


Figure 10.5– Governing equation for flow through an orifice

Simplifying Bernoulli’s equation , yields the governing equation for the velocity of flow through

an orifice, as being equal to √2gH (Figure 10.5). The flow rate through the orifice, will simply

be the given by Continuity equation, where Q = A V (Figure 10.6).

Note, since the reference point 2, was taken at the Vena Contracta position, then the cross

sectional area of flow that should be considered, is the area of flow at the vena contracta.

However, in practical situation, it is difficult to locate the vena contracta, let alone measure its

cross section. Thus assuming that the cross sectional flow area at the vena contracta is not very

different from the cross sectional area of flow at the orifice the cross sectional area of the orifice

is used in the continuity equation. We end up with a Theoretical flow rate through the orifice.


2/2g + Z2

0 + 0 + H = 0 + V22/2g + 0

V2= (2gH)½

1

2 Datum

H


Figure 10.6– Flow rate through a small orifice

Two assumptions have been made in the derivation of this theoretical velocity, first the velocity

of flow at point 1 is negligible and secondly the cross sectional area of flow at the vena contracta

is similar to that at the orifice. To cater for the errors likely to be induced by these assumptions,

the theoretical discharge is multiplied by a factor of safety known as the Coefficient of

Discharge, Cd.

Cd is either equal to 1, if any assumptions made is true, or 0 is they are completely wrong. In the

case of small orifice, Cd can lie between 0.5 to 0.6.

V2= (2gH)½

1

2 Datum

H

Applying Continuity Equation: Q = A V

Q = A(vena contracta) x V2

Q = A(vena contracta) x (2gH)½

Q = Aorifice x (2gH)½ x Cd

Where Cd = coefficient of discharge


10.6 COEFFICIENTS OF CONTRACTION & VELOCITY

In the case of an orifice, the coefficient of discharge can further be subdivided into two

coefficients, the coefficient of velocity and the coefficient of contraction.

Figure 10.7: Coefficients of contraction, velocity and discharge

The coefficient of velocity as illustrated in Figure 10.7, simply relates the velocity at the orifice

to the velocity of flow at the vena contracta. Similarly the coefficient of discharge relates the

cross sectional area of flow at the orifice and that at the vena contracta. The product of Cc and

Cv yields the coefficient of discharge, Cd.

Discharge through an orifice:

Q = Aorifice x (2gH)½ x Cd

Cc = coefficient of contraction = Area of orifice/a rea of vena contracta

Cv= coefficient of velocity = velocity of flow at orifice/velocity of flow at ven a contracta

Cd= coefficient of Discharge = Cc x Cv


10.7 DISCHARGE THROUGH A LARGE ORIFICE

When the opening is small then the height of liquid causing flow above the centre of the orifice

can be safely assumed to be H. However, the opening is large, this assumption may not

necessarily hold true.

Figure 10.8– Discharge through a large orifice

The flow is more turbulent a large orifice as illustrated by the curved flow lines at the opening

(Figure 10.8).

Large orifice

1

2


Figure 10.9: Discharge through a large rectangular orifice

In the case of a large orifice, then the opening is considered as being made up of a series of small

orifices, through which the fluid will flow. Consider the case of a large rectangular opening

(Figure 10.9). The opening will be assumed to be made up of a series of small openings, through

which the velocity of flow will be equal to √2gh. The discharge is obtained by multiplying this

flow velocity by the cross sectional area of flow of the small element, given by B dh. Finally to

obtain the total discharge, this equation is integrated with limits of h varying from h2 to h1.

10.8 VELOCITY OF APPROACH (V 1)

As discussed in section 10.4, the velocity of flow at the surface of the container or reservoir, V1,

is often safely assumed to be negligible and hence equal to zero in the analysis. This velocity is

commonly known as the velocity of approach. As illustrated in Figure 10.10, when the cross

sectional area of flow at the surface is much larger than the cross sectional area of the orifice,

then the velocity of approach will be much smaller than the velocity at the orifice, in which case

the velocity of approach can be safely assumed as being negligible.

1

h2

h1

Velocity of flow through an orifice: V = (2gh) ½

Since the orifice this time is large, it is not

accurate enough to assume that h varies from 0 to H2.

In this case the variable h in the velocity

equation varies from h 1 to h 2.

B

dhh

dQ= A V dQ = B dh √√√√2gh

Q = ∫∫∫∫ B √√√√2g h dh

where the limits of h varies

from h 2 to h 1.


Figure 10.10: Incoming velocity – velocity of approach

However, the cross sectional area of flow of the container may not always be much much larger

than the cross sectional area of flow of the orifice, in which case the assumption of considering

the velocity of approach as being negligible may not always hold true. In such a case the

analysis is carried out using an iterative approach (Figure 10.11).

The first step behind this method is to assume that the velocity of approach is negligible and

work out as described in section 10.5, the theoretical discharge through the orifice. Having now

got a first estimate for the discharge within the system, the next step is to work out a first

estimate for the velocity of approach, which is simply the discharge/cross sectional area of the

container. This velocity head (V12/2g) is then included when the Bernoulli’s equation is being

applied now and a new velocity of flow through the orifice V2 is now calculated and hence a new

discharge value.

1

2


2/2g + Z2

Where P1 = 0 gauge pressure, V1≅ 0, Z1= H

P2= 0 gauge pressure, Z2= 0 being at datum

Datum

H

V1 – Velocity of approach, often safe to assume being so small that it is equal to zero.

A1V1 = A2V2

A1>>>A2, hence V 1<<<V2, thus V 1 is relatively much smaller than V 2, that it is

considered as zero


This iterative process is continued until two consecutive discharge values are within close

agreement.

Figure 10.11: Considering velocity of approach in the analysis

10.9 FLOW MEASURMENT BY A RECTANGULAR WEIR

Weirs are another type of flow measuring devices which are used to measure flow rates through

open channels such as rivers, canals and culverts. A weir is simplying a small opening located

most of the time at the end of an open channel, and the opening can have different shapes,

ranging from the simplest rectangular ones to complex irregular shapes. The analysis of flow

over weirs is very similar to that used in the derivation of flow rate via a large orifice (Figure

10.12).

1

2


2/2g + Z2

Datum

H

For a large orifice , it may at times not be safe to ignore the velocity of approach . Thus, an iterative process is adopted. The first step is to assume V1 being equal to zero, and work out the discharge through the orifice. Step 2, once a rough estimate of Q is obtained, calculate V1, and incorporate it in the Bernoulli’s Equation, and work out the new discharge through the orifice. Step 2 is once again repeated until the 2 consecutive values of discharge are within agreeable limits.


Figure 10.12: Rectangular weirs

Here also a small element is considered as being a small orifice, with the velocity of flow being

given by the equation √2gh. The cross sectional area of flow of the small element is B dh. To

obtain the total discharge the equation is integrated with the depth of flow h varying from a value

of 0 to H. The final equation yields the theoretical velocity of flow.

The actual flow velocity Qactual = Q theoretical x Cd, where the coefficient of discharge Cd for

a rectangular weir lies being 0.7 to 0.8. These values are higher than those for an orifice, since

the disturbance to the flow pattern is relatively lower here.

dQ = Area of elemental fluid x Velocity of flow through elemental fluid

dQ = (bdh) (2gh)½

Q =∫ (2g) ½ b h½ dh

Q(theoretical) = 2/3 (2g) ½ b H3/2

h

H dh

B

H

B

0

H

Open channel


10.10 FLOW MEASURMENT BY A TRIANGULAR WEIR

Similarly if the weir has triangular shape, then the analysis for flow rate is given as shown in

Figure 10.13.

Figure 10.13: Triangular Weirs

The difference between the analytical process here, is that unlike the case of a rectangular weir,

the width of the opening is not constant here. However, the angle of inclination at the V shaped

is constant. In which case, the width of the small elemental fluid is worked out in terms of the

angle of inclination, θ. Finally to obtain the actual discharge, the coefficient of discharge is

used.

dQ = Area of elemental fluid x Velocity of flow through elemental fluid

dQ = [2 (H-h) tanθ dh] (2gh)½

Q =∫ 2 x (2g) ½ tan θ (H h½- h3/2) dh

Q(theoretical) = 8/15 (2g) ½ tan θ H5/2

0

H

Open channel

h

H

dh

θ

θθθθ H-h

(H-h) tan θθθθ

Very small element can be considered as being rectangular in shape


10.11 FLOW MEASURMENT BY A TRAPEZOIDAL W EIR

A trapezoidal weir is simply a weir having its opening in the shape of a trapezium. This

trapezium can either be regular or irregular. A trapezium can basically be subdivided in simple

shapes, a rectangle and two right angled triangles. The equation of discharge through a

trapezoidal weir can be obtained directly by combining the equations derived in sections 10.9

and 10.10, or for better understanding the equation can best be obtained by working through first

principles.

As usual the flow rate through a small elemental fluid is considered and the entire discharge is

obtained by integrating this equation, with depth of flow varying from 0 to H, Figure 10.14.

Figure 10.14: Irregular trapezoidal Weirs

In the case of an irregular trapezoidal weir, the angle of inclinations are different on either side, θ

and α respectively. The final theoretical equation is as illustrated in Figure 10.14.

B

Hθθθθ

αααα

h

Discharge through small elemental fluid:

dQ = area x velocity

dQ = [B + (H-h) tan θθθθ + (H-h) tan αααα] dh (2gh) ½

Integrating with respect to dh, from h=0 to h=H, yi elds

Q(theoretical) = (2g) ½ { 2/3 BH 3/2 + 4/15tanθθθθH5/2 + 4/15tanααααH5/2 }

dh


The best approach for any shaped weir is to work through first principles and derive the

governing equation.

10.12 ACTIVITIES

1. Derive the governing equation of velocity of flow through an orifice. Explain the choice of your

reference points for application of Bernoulli’s equation.

2. Derive the equation for the actual flow rate through a large orifice, stating the difference in approach

to small orifice.

3. Can an orifice be used to measure flow rate through an open channel?

4. Can an orifice be used to measure flow rate through a closed channel, such as a pipe?

5. Explain how the analysis of flow rate through weirs is related to the equation of flow through

an orifice.

6. Working from first principles derive the actual flow rate through the following weir:

10.13 SUMMARY

In this unit the student have been introduced to the application of both continuity equation and

Bernoulli’s equation in flow measuring devices such as orifices and weirs. In the next unit, the

student will be introduced to another flow measuring device the Venturimeter. The Venturimeter

is a flow measuring device specially for pipelines, i.e closed conduits, as compared to weirs,

which are used for flow measurements in open channels.

θ

B

H


10.14 WORKED EXAMPLES Example 1 – Flow measurement through an open channel using a rectangular weir

Weirs..1/1Example 1

• The discharge over a rectangular notch is to be 0.14m3/s, when the water level is 23 above the sill of the notch. Assuming coefficient of discharge of the notch is 0.6, and working from first principles calculate the width of the notch.

– Q theoretical =2/3 B √√√√ 2g H 3/2

– Q actual = Cd Q theoretical

– Q actual = 0.14 m3/s

– 0.14 = 0.6 x 2/3 x B x (√√√√ 2 x 9.81 ) x (0.23) 3/2

– Hence B = 0.72m

Example 2 – Flow measurement using a trapezoidal weir


Weirs..1/3Example 2

• A sharp edged notch is in the form of a symmetrical trapezium. The horizontal base is 10cm wide, the top is 50cm wide and the depth is 30cm. Assuming that the coefficient of discharge is to be 0.6, calculate the height of the water level above the base of the notch, if the discharge is 0.043 m3/s. Work from first principles.

10cm

50cm

θ

Step 1 : calculate angle θθθθ

Tan θ = [(0.50-0.10)/2] / 0.3 = 0.67

Step 2: Consider elemental fluid, located distance h from top of free water surface

Area of elemental fluid, dA= [ 0.10 + 2 x (H-h) tan θ] dh = [0.10 + 2 x (H-h) x 0.67] dh

dA = {0.10 + 1.33 ( H-h)} dh

H

h

Weirs..2/4Step 3: Calculate discharge through small element, dq

dq= dA x √ 2g h

dq = [0.10 + 1.33 (H-h)] dh x √ 2g h

dq = √ 2g { 0.10 h1/2 + 1.33H h1/2 – 1.33h3/2} dh

Step 4: Integrate to obtain the total discharge through the trapezoidal weir

Q = √ 2g ∫∫∫∫ 0.10 h1/2 + 1.33Hxh1/2 – 1.33h3/2dh

Q = √ 2g {2/3 *0.1 h3/2 +2/3 x 1.33 x H x h3/2– 2/5 x 1.33 x h5/2 } limits from 0 to H

Q = √ 2g {0.07 H3/2 + 0.89H5/2 –0.53H5/2}

Q theoretical = √ 2g {0.07 H3/2 + 0.36H5/2}


Weirs..3/3Step 5: Calculate the value of H

Qtheoretical= √ 2g {0.07 H3/2 + 0.36H5/2}

Q actual = Cd Q theoretical

0.043 = 0.6 x (2 x 9.18)1/2 { 0.07 H3/2 + 0.36 H5/2}

H = 22.9 cm.

10.15 TUTORIAL Question 1 a. Derive an expression governing the discharge of a liquid through an orifice, explaining

clearly the meaning of the following: coefficient of contraction, coefficient of velocity

and coefficient of discharge.

b. Oil of relative density 0.85 flows through a 50mm diameter orifice under a pressure of

100kN/m2 (Gauge pressure). The diameter of the vena contracta is 39.5 mm and the

disharge through the orifice is 18 litres per second, what is the coefficient of discharge,

coefficient of contraction and the coefficient of velocity.

Question 2

A large tank has a circular orifice 20 mm diameter in the vertical side near the bottom. The tank contains

water to a depth of 1 m above the orifice with oil of relative density 0.8 for a depth of 1 m above the


water. Acting on the upper surface of the oil is an air pressure of 20 kNm-2 gauge. The jet of water

issuing from the orifice travels a horizontal distance of 1.5 m from the orifice while falling a vertical

distance of 0.156 m. If the coefficient of contraction of the orifice is 0.65, estimate the value of the

coefficient of velocity and the actual discharge through the orifice. State any assumptions made.

Question 3 Discharge of water through an open channel is to be measured by means of an unsymmetrical

trapezoidal weir as shown below. The width of the weir (B) is 1.5m.

If the height of water above the base of the weir is 0.65m, working from first principles,

calculate the volume flow rate through the channel, taking Cd as 0.62.

Question 4 A tank of square cross section, each side measuring 0.3 m, is open at the top and is fixed in an

upright position. A 6 mm diameter circular orifice is situated in one of the vertical sides near the

bottom. Water flows into the tank at a constant rate of 280 x 10-3 m3/hr. At a particular instant,

the jet strikes the floor at a point 0.63 m from the vena contracta, measured horizontally 0.53 m

below the centre line of the orifice measured vertically. Determine whether the water surface in

the tank is rising or falling at the instant under consideration. Also find the height of the surface

above the centreline of the orifice and the rate of change of height, taking Cv as 0.97 and Cd as

0.64.


Question 5

Water flows along a channel over a rectangular weir 1.25 m wide. The head of water above the

weir is 40 cm and after passing over the weir, the water falls from a vertical height up of 3.7m

onto a turbine. The energy of the water is used to drive the turbine which develops a power of

30kW. Calculate the efficiency of the turbine.


UNIT 11 FLOW RATE MEASUREMENTS – VENTURIMETERS

Unit Structure

11.0 Overview


11.2 Introduction

11.3 Venturimeter

11.4 Horizontal Venturimeter

11.5 Coefficient of Discharge, Cd

11.6 Inclined Venturimeter

11.7 Pressure Difference Measurement – U Tube Manometer

11.8 Eddy Zones

11.9 Activities

11.10 Summary


11.12 Tutorial

11.0 OVERVIEW

In this unit the student will learn how the flow rate within a closed conduit, a pipeline, is

measured by venturimeter. Emphasis is being placed here on the application of energy and

continuity equations to work out theoretical flow rate through a pipe. The important issues to be

noted in this unit will be the approach behind the application of the equations based upon

different ways of pressure measurement.



1. Describe the main features of a venturimeter.

2. Explain why it is better to work along the stretch between the converging section and the throat, rather than the throat and the diverging section.


3. Derive the equation governing the theoretical flow rate in a venturimeter whereby the pressure difference between the reference points are being measured by a U tube manometer.

4. Derive the equation governing the theoretical flow rate in a pipeline fitted with an orifice,

and whereby the pressure difference between the reference points are being measured by

pressure gauges.

5. Differentiate between the accuracy of flow rate measurement by a venturimeter and by an

orifice.

11.2 INTRODUCTION

The flow rate within a pipeline is an important information which is required within water

distribution networks. Pipelines joined together to form networks and branch from one point to

another to cater for water needs within residential or commercial areas. In practice it is not

possible to actually collect the water in a given time and work out the flow rate. One indirect

and fairly accurate way of measuring flow rate in a pipeline is by a venturimeter. The

venturimeter is a simple device which makes use of the variation of pressure and velocity

conditions at varying cross sectional area, and provide a good basis for a good estimate of the

flow rate in pipelines. The main stress in this unit is the approach adopted to measure the flow

rate with the venturimeter. The stress is also on a good understand of the importance of the

coefficient of discharge and how it compares to other the coefficient of discharge of other flow

measuring devices.

11.3 VENTURIMETER A venturimeter is a flow measuring device used specially in the measurement of flow of closed

systems, most commonly pipelines. A venturimeter is made up of converging section, a throat

and a diverging section (Figure 11.1).


Figure 11.1: Venturimeter

Reference points for application of Bernoulli’s equation are taken at the converging section and

at the throat. The reason behind this approach is that the energy losses accompanying the

change in flow pattern within the converging section is lower than that associated in the

diverging section. Most of the time ideal fluid conditions are assumed, where the head loss

between the converging section to the throat is considered negligible.

11.4 HORIZONTAL VENTURIMETER

Consider a horizontal pipeline in which fluid is flowing under pressure (Figure 11.2). Note, that

fluid flows from a point of high pressure to a point of low pressure. Since Bernoulli’s equation

is to be applied, two reference points are first selected, and as indicated in section 11.3, these

reference points are at the converging section and at the throat (Figure 11.2).

flow direction

Converging end

diverging end

throat

Venturimeters are used to measure flow rate in closed conduits, such as pipelines.


Figure 11.2: Venturimeter in a horizontal pipeline

Pressure at points 1 and 2 are being measured by pressure gauges, as indicated by the standard

symbol in Figure 11.2. Applying Bernoulli’s equation at 1 and 2 and assuming ideal fluid

conditions, yields:

P1/ρg + V12/2g +Z1 = P2/ρg + V2

2/2g +Z2

Now, since the pipeline is horizontal, the elevation head on both sides are similar, Z1 = Z2.

P1/ρg + V12/2g = P2/ρg + V2

2/2g

Since pressure gauges have been used to measure the pressures, P1 and P2, these values may be

known. At the end you may be left with two unknowns in the equation, V1 and V2.

V12/2g - V2

2/2g = P2/ρg - P1/ρg

1

1 2

2

At section 11: conditions are as follows: P 1, V1 and Z 1

Similarly, at section 22, conditions are as follows : P2, V2 and Z 2

Most of the time in the measurement of flow using a Venturimeter, the Principles of Continuity and the Principles of conservation of mass

need to be applied:


2/2g + Z2

Q = A1V1 = A2V2

Note here: Z 1 = Z2

Pressure Gauges

flow direction


Applying Continuity equation, where Q=A1V1=A2V2, we can replace V2 in terms of V1 and

finally end up with only one unknown in the equation. Once the value of V1 is known, the next

step will be to work out the theoretical flow rate through the pipe using the Continuity equation.

Q(theoretical) = V1 x A1

11.5 COEFFICIENT OF DISCHARGE`, Cd

In the analysis described in section 11.4, a major assumption has been made, that of IDEAL fluid

conditions, meaning that energy losses are negligible, and hence do not appear in the Bernoulli’s

formula. Such an assumption may or may not hold completely true, depending upon the

prevailing flow conditions. If the venturimeter is providing minimum disturbance to the

incoming flow that such an assumption is a valid one. However the presence of the venturimeter

is upsetting the flow pattern, then this is accompanied generally by significant energy losses, in

which case the assumption of IDEAL fluid conditions, will induce errors in the flow

measurement.

Here also to cater for such error sources, a coefficient of discharge is used. So the actual velocity

will therefore be:

Qactual = Qtheoretical x Cd

The coefficient of discharge of a venturimeter is a function of the shape of the instrument, the

material in which it is constructed, its size and also the flow rate it measures. Overall the

coefficient of discharge of a venturimeter lies between 0.8 to 0.9. This may be one of the most

accurate flow measuring device for closed conduits.


11.6 INCLINED VENTURIMETER

In the case of an inclined pipeline in which a venturimeter is fitted, the application of Bernoulli’s

equation has to cater for the difference in elevation. Consider an inclined venturimeter (Figure

11.3), in which the datum has been considered much lower than the pipeline.

Figure 11.3: Venturimeter in an inclined pipeline

The pressure at the two reference points 1 and 2 is here also being measured by pressure gauges.

The elevation of the reference points from the datum is Z1 and Z2 respectively. Usually the

position of the datum is arbitrary and hence it is not possible to know the elevation heads at the

reference points. However, the difference in elevation can be worked out where (Z1 – Z2 = L

sinθ) as shown in Figure 11.4.

Then as described in section 11.4, given the pressure difference, the difference in elevation is

known, the final equation will have only two unknowns, V1 and V2. By applying Continuity

equation, one of the unknowns can be eliminated and the theoretical discharge through the

pipeline worked out.

In such a case elevation heads should be considered in the analysis:

P1/ρρρρg + V1

2/2g + Z1 = P2/ρρρρg + V22/2g + Z2

Datum

Z1

Z2

P

P


Figure 11.4: Difference in elevation – inclined venturimeter

11.7 PRESSURE DIFFERENCE MEASUREMENT – U TUBE MANOMETER

The pressure difference between the two reference points along a venturimeter can also be

measured by connecting the U tube manometer to the venturimeter (Figure 11.5).

Datum

θθθθ

(Z2-Z1) = Lsin θθθθ

Z

Z2

L


Figure 11.5: Venturimeter & Manometer

Whenever a manometer is connected to a Venturimeter the difference in elevation (Z2-Z1) is

often not an issue, for the difference in level of mercury in the manometer takes care of that as

will be discussed below (Figure 11.6).

Datum h

Z

Z

Pressure difference is being measured by a Manomete r tube


Figure 11.6: Manometer - analysis

By analysing the U tube manometer as explained in unit 3, and also as shown in Figure 11.6, we

end up with the following equation:

P1-P2= Z1ρρρρ(fluid)g - (Z2-h) ρρρρ(fluid)g - h ρρρρ(mercuy)g

Which can still be further simplified as:

P1-P2= ρρρρ(fluid)g { Z1 - Z2 } +h g {ρρρρ(fluid) - ρρρρ(mercuy)}…………(1)

The next step is now to apply Bernoulli’s equation to the two reference points 1 and 2, which

will give:

Datum h

Z1

Z2

a b

A1,V1

A2,V2

Pa=P1+Z1ρρρρ(fluid) g

Pb=P2+ (Z2-h) ρρρρ(fluid) g+ h ρρρρ(mercuy) g

Pa=Pb

P1-P2= Z1ρρρρ(fluid) g - (Z2-h) ρρρρ(fluid) g - h ρρρρ(mercuy) g


P1/ρ(fluid)g + V12/2g +Z1 = P2/ρ(fluid)g + V2

2/2g +Z2…………..(2)

Rearranging equation 2:

P1/ρ(fluid)g - P2/ρ(fluid)g = V22/2g - V1

2/2g +Z2 - Z1

Replacing P1-P2 from equation 1 derived above:

1//ρρρρ(fluid)g {ρρρρ(fluid)g { Z1 - Z2 } +h g {ρρρρ(fluid) - ρρρρ(mercuy)} = V22/2g - V1

2/2g +Z2 - Z1

And this equation simplifies to:

(Z1 - Z2) + h { 1 - ρρρρ(mercuy)/ ρρρρ(fluid) } = V22/2g - V1

2/2g +Z2 - Z1…………(3) It can be seen that the difference in elevation terms cancel out in equation 3, leaving:

h { 1 - ρρρρ(mercuy)/ ρρρρ(fluid) } = V22/2g - V1

2/2g

Finally, Continuity equation can be applied to reduce the number of unknowns and the equation

simplified accordingly to work out the theoretical discharge.

11.8 EDDY ZONES

An orifice can also be used to measure the flow rate in pipelines. When an orifice is fitted into a

pipeline, the resulting flow pattern ressembles that within a venturimeter (Figure 11.7).


Figure 11.7: Flow through an orifice fitted inside pipeline

The reference points are located upstream the orifice under no disturbed conditions and at the

vena contracta position downstream the orifice (see Unit 10). Just like in the case of an orifice,

the cross sectional area at the vena contracta is unknown, as well as the velocity of flow there.

The cross sectional area of the orifice and the velocity of flow at the orifice are considered in the

analysis.

The analysis of flow in such a case is similar to that described in section 11.4, by applying both

Bernoulli’s and Continuity equation. However in the case of an orifice the coefficient of

discharge is lower than that of a venturimeter.

flow direction

Eddy currents – turbulent zone

Much higher turbulence, on the downstream section

Flow pattern tend to behave similar to the case of a venturimeter.

Analysis is carried out in the same way as a horizontal venturimeter.

A1, V1 and P1 are the upstream conditions while A2, V2 and P2 are the conditions at the opening or orifice.

P1 P2

A1,V1

A2,V21


11.9 ACTIVITIES

1. Derive the equation governing the flow rate in an inclined venturimeter.

2. Explain why is it more accurate to apply Bernoulli’s equation at the converging section rather

than at the diverging section.

3. How does the range of the coefficient of discharge compare to that of an orifice, explain the

difference?

4. The distance between the converging edge and the throat has not been measured for a

venturimeter. The pressure difference is being measured with the use of a manometer.

Discuss whether this data will affect the calculation of the discharge through the pipeline.

5. ‘By inserting an orifice in the middle of a pipeline, it can be made to operate as

accurately as a venturimeter’, discuss.

11.10 SUMMARY

In this unit the student has once again been introduced to the application of Bernoulli’s equation

and Continuity equation for flow measurements, but this time in closed channels, such as

pipelines. The student should clearly appreciate, understand and learn the concepts behind the

application of these two equations, since these will constantly appear in most of the units to

follow in higher levels of fluid mechanics.



Example 1 – Pressure difference measurement in Venturimeter by Pressure Gauges

Example 1 – Pressure measurement by differential pressure gauge…1/7

• (a) What are the relative advantages of using a venturimeter to measure the flow compared with an orifice meter?

• (b) A horizontal venturimeter has a main diameter of 65mm and a throat diameter of 26mm . When measuring the flow of a liquid of density 989 kg/m3 the reading of a mercury differential-pressure gauge was 71mm. Working from first principles or proving any formula used, calculate the flow through the meter in m3/h. Take the coefficient of the meter as 0.97 and the specific gravity of mercury as 13.6.

65mm 26mmQ ?

Cd=0.97

ρ = 898 kg/m3



• (a) What are the relative advantages of using a venturimeter to measure the flow compared with an orifice meter?

1. When liquid flows through an orifice it is subjected to much larger frictional losses, since the flow has to converged from a large surface area to a much smaller cross sectional area of flow (diameter of the orifice). This large change in cross sectional area of flow induces a relatively larger frictional loss. Compared to the orifice, when the flow converges into the throat section, the change in cross sectional area is not as drastic. Hence, the frictional losses in a venturimeter is relatively lower, making the coefficient of discharge for the venturimeter larger than that for an orifice.


• (b) A horizontal venturimeter has a main diameter of 65mm and a throat diameter of 26mm . When measuring the flow of a liquid of density 989 kg/m3 the reading of a mercury differential-pressure gauge was 71mm. Working from first principles or proving any formul a used, calculate the flow through the meter in m3/h. Take the coefficient of the meter as 0.97 and the specific gravity of mercury as 13.6.

65mm 26mmQ ?

Cd=0.97

ρ = 898 kg/m3

NOTE: The pressure difference from the inlet to the throat section in a venturimeter can be measured by a differential pressure gauge (symbol as shown in diagram above), or by a U tube manometer.



65mm 26mmQ ?

Cd=0.97

ρ = 898 kg/m3

Applying Bernoulli’s equation:P1/ρg + V1

2/2g + Z1 = P2/ρg + V22/2g + Z2

•Z1 and Z2 cancel out since venturimeter is horizontal

P1/ρg – P2/ρg = V22/2g - V1

2/2g …..(1)

•(P1-P2)/ ρρρρg = 71 mm head of mercury (as given in the question)………….(2)



26mm65mmQ ?

Cd=0.97

ρ = 898 kg/m3

Applying Continuity equation:A1 V1 = A2 V2

ΠΠΠΠ (65 X 10-3)2/4 V1 = ΠΠΠΠ (26 X 10-3)2/4 V2V1 = (26 X 10-3)2/ (65 X 10-3)2 V2V1 = 0.16 V2………………(3)


26mm65mmQ ?

Cd=0.97

ρ = 898 kg/m3

Combining 1, 2 and 3:

P1/ρg – P2/ρg = V22/2g - V1

2/2g

(71 x 10-3) x 13.6 x 1000 x 9.81 / 898 x 9.81 = V22/2g –(0.16)2 V2

2/2g

V2 = 4.65 m/s



65mm 26mmQ ?

Cd=0.97

ρ = 898 kg/m3

Applying Continuity equation to obtain flow rate:V2 = 4.65 m/sQtheoretical = ΠΠΠΠ (26 X 10-3)2/4 V2 = 2.47 x 10-3 m3/s

Qactual= Qtheoreticalx Cd

= 8.89 x 0.97 m3/h

= 8.62 m3/h

Example 2 – Pressure difference measurement using U tube manometer

Example 2 – Pressure measurement by U tube manometer…1/6

A venturimeter having a throat 100mm in diameter is fitted in a pipeline 250mm in diameter through which oil of specific gravity 0.9 is flowing at the rate of 0.1m3/s. The inlet and throat of the meter are connected to a differential U tube manometer containing mercury of specific gravity 13.6 with oil immediately above this. Working from first principles, find the coefficient of discharge for the meter if the difference in mercury levels is 0.63m.


Example 2 – Pressure measurement by U tube manometer….2/6

entry

Convergingzone

Throat section

Diverging section

datum

a

b

Hha

hb

1 2


1. Applying Bernoulli’s equation at a and b:

Pa/ρg + Va2/2g + Za = Pb/ρg + Vb

2/2g + Zb

(Pa-Pb)/ ρg = Vb2/2g -Va

2/2g + Zb - Za


2/2g + H + hb – ha………………1

2. Applying Continuity Equation at a and b:

AaVa = AbVb

Vb = (Aa/Ab) x Va………………2

Vb can be replaced by Va to reduce one unknown in above equation



3. Pressure difference between a and b is obtained from Manometer readings:

Consider reference points 1 and 2:

P1 = Pa + ha ρg P2 = Pb + hb ρg + H ρ (Hg) g

P1 = P2

Pa + ha ρg = Pb + hb ρg + H ρ (Hg) g

Pa – Pb = hb ρg + H ρ (Hg) g - ha ρg

(Pa – Pb) / ρg = hb – ha + H ρ (Hg) / ρ………………3


2/2g + H + hb – ha………………1

Vb = (Aa/Ab) x Va………………2

(Pa – Pb) / ρg = hb – ha + H ρ (Hg) / ρ………………3

4. Replacing equations 2 and 3 in equation 1 above:

hb – ha + H ρ (Hg) / ρ = (Aa/Ab)2 x Va2/2g – Va

2/2g + H + hb – ha

Va = {( H ρ (Hg) / ρ – H) x 2g } 1/2 / { (A a/Ab} 2 –1}1/2

When Velocity of flow is known, for a known value of H,then Q can be found.




Va = {( H ρ (Hg) / ρ – H) x 2g }1/2/ { (A a/Ab} 2 –1} 1/2

Qactual= Cd Aa Va

0.1 = Cd (π/4 x 0.252) x [ (0.63 x 13.6/0.9 – 1) x 2 x 9.81]1/2 / { 0.254/0.14 –1}½

Cd = 0.97

11.12 TUTORIAL Question 1

A venturimeter is tested with its axis horizontal and the flow measured by means of a weighing

tank. The pipe diameter is 76 mm, and the throat diameter is 38 mm and the pressure difference

is measured by a U tube containing mercury, the connections being full of water. If the

difference in levels in the U tube reads 266 mm mercury while 2200 kg of water are collected in

4 minutes, what is the coefficient of discharge?

Question 2 A servo-mechanism is to make use of a venturimeter contraction in a horizontal 350 mm

diameter pipe which carries a liquid of relative density 0.95. The upper end of a vertical cylinder

100 mm diameter is connected by a pipe to the throat of the venturimeter and the lower end of

the cylinder is connected to the inlet. A piston is to be lifted when the flow rate through the

venturimeter is 0.15m3/s. The piston rod is 20 mm diameter and passes through both ends of the

cylinder. Calculate the diameter of the throat if the effective load on the piston rod is 180N.


Question 3

A venturimeter with a throat diameter of 100mm is fitted in a vertical pipeline of 200mm

diameter with oil of specific gravity 0.88 flowing upwards at a rate of 0.06m3/s. The coefficient

of the venturimeter is 0.96. Two pressure gauges calibrated in kilonewtons per square meter are

fitted at tapping points one at the throat and the other in the inlet pipe 320mm below the throat.

The difference between the two gauge pressure readings is 28kN/m2. Working from first

principles determine the difference in level in the two limbs of a mercury manometer if it is

connected to the tapping points and the connecting pipes are filled with the same oil.


UNIT 12 MOMENTUM EQUATION & ITS APPLICATIONS

Unit Structure

12.0 Overview


12.2 Introduction

12.3 Momentum Equation

12.4 Rate of change of momentum and applied force

12.5 Application of Momentum equation along a streamline

12.6 Possible forces acting on the control volume

12.7 Momentum correction factor

12.8 Momentum correction factor – circular pipe

12.9 Force exerted by a jet striking a flat plate

12.10 Force exerted by a jet striking an inclined plate

12.11 Force exerted by a jet striking a moving plate

12.12 Force exerted by a jet striking an inclined moving plate

12.13 Force due to the deflection of a jet by a curved vane

12.14 Force exerted on pipe bends

12.15 Activities

12.16 Summary

12.17 Worked examples

12.18 Tutorial Sheet

12.0 OVERVIEW

In this unit, you will learn about the force which is exerted to cause a change in the physical properties of

a liquid as it moves along a channel of varying orientation and size. The student will also get to

appreciate the application of Newton’s Laws of motion to fluid flows. The main points to note in this

chapter is the internal forces which are induced in a system as the physical properties of a fluid changes.



1. Define the momentum equation


2. Derive the equation for the force which is being exerted onto a fluid to cause a change in its

momentum.

3. Derive the general equation for the force to be exerted by a jet of liquid when it strikes a plate

which is at rest.

4. Understand and implement the appropriate approaching conditions when the jet strikes a moving

plate.

5. Calculate the resultant force acting on a vertical bend through which a liquid is flowing, where in

addition, pressure forces and gravity forces are acting.

12.2 INTRODUCTION

Newton’s Second and Third laws of motion will be applied in this unit to work out the force exerted on

fluids by the boundaries of a system when the flow conditions changes along the path, and consequently

the fluid offers a similar but opposite resistance to the boundaries. The force exerted is equated to the rate

of change in momentum of a liquid. The force exerted by a moving liquid on either a plate, a curved vane

and within a bend will be calculated from these principles. This chapter emphasises on the need to prove

firm support to locations where there are bends and curves within a pipe network, to ensure that the forces

which are induced owing to the change in momentum of the liquid, do not damage the piping system.

12.3 MOMENTUM EQUATION In solid mechanics, the momentum of a particle or object is defined as the product of the mass of the

object, m and its velocity, v:

Momentum = mv

Similarly, the particles of a moving fluid will also possess momentum. When the velocity of the fluid

particles changes, the momentum of the fluid particles also change.

The change in momentum is induced by a force, and in accordance to Newton’s Second Law of Motion, a

force required to produce a change in momentum is proportional to the rate at which the change in

momentum occurs (change in momentum with respect to time).


In the case of a moving fluid, the force may be provided by the contact between the fluid and the solid

boundary or by one part of the fluid acting on the other part. If the solid boundary exerts a force on the

moving fluid, consequently, the fluid will exert an opposite and equal force on the solid boundary,

Newton’s Third Law of motion.

12.4 RATE OF CHANGE OF MOMENTUM & APPLIED FORCE Consider a small element within the body of a fluid:

The fluid is assumed to be steady (same velocity conditions with respect to time), and there is no

storage within the control volume from section AB to section CD. The flow is non-uniform in

this case (velocity changes with respect to distance), hence velocity at section AB is not the same

as the velocity at section CD.

Applying Continuity equation at sections AB and CD, Q=AV, gives:

A1 V1 = A2 V2

Area, A1

Velocity, V1

Density, ρ1

Area, A2

Velocity, V2

Density, ρ2

A

B

D

C

Figure 12.1: Elemental fluid


Continuity equation can also be expressed in terms of mass flow rate (m), where m = ρ A V,

hence, in terms of mass flow rate equation:

ρ1 A1 V1 = ρ2 A2 V2……………………equation 1

Now, the rate of change of momentum = Change in momentum with respect to time = mv/t

mv/t can also be written as {volume x density x velocity}/time, or {area x length x density x

velocity}/time, or {area x velocity x density x velocity} = A V2 ρ

Initial momentum, Mi = A1V12 ρ1

Final momentum, Mf = A2V22 ρ2

The rate of change of momentum as the liquid passes from section AB to section CD= Force acting on the

liquid to cause the change in momentum (F) = Final Momentum – Initial Momentum

F = A2V22 ρρρρ2 - A1V1

2 ρρρρ1 ………………………..equation 2

Replacing velocity (A1V1 ρ1) by ( A2V2 ρ2) in equation 2, gives:

F = A2V22 ρ2 - A2V2 ρ2 V1 = A2V2 ρ2 (V2 – V1)

and the equation simplifies to:

F = m (V2 – V1), where m is the mass flow rate

and F is the Resultant force acting on the fluid element ABCD in the direction of motion. By Newton’s

third law of motion, the fluid will exert an equal and opposite reaction on its surroundings.


Flow direction

A

B

C

D

α

θV1

V1cosθ

V1sinθ

V2cosα

V2sinα

Figure 12.2 – Momentum change in a streamline

Sign convention, +ve X and +ve Y direction

12.5 APPLICATION OF MOMENTUM EQUATION ALONG A STREA MLINE

Consider a streamtube, or flow within a curved pipe of non uniform diameter, whereby the velocity at

section AB is V1 and the velocity of flow at section CD is V2.

NOTE: The velocity of flow V1 and V2, being inclined at angles and respectively, they have therefore

two components, one in the horizontal direction and one in the vertical direction. Hence, the force

causing the change in momentum, will also have two components, in the horizontal and in the vertical

direction.

Resultant force in the HORIZONTAL direction:

FX = Rate of change of momentum of the fluid in the x direction

= mass flow rate x change in velocity in the x direction

= m (V2cosα - V1cosθ)

Resultant force in the VERTICAL direction:


FY = Rate of change of momentum of the fluid in the y direction

= mass flow rate x change in velocity in the y direction

= m (V2sinα - V1sinθ)

Resultant force required to cause this change in momentum:

FR = √ {FX + FY}

NOTE:

FR is the force exerted on the liquid to cause a change in momentum, and in accordance to Newton’s third

law of motion, the liquid will exert an equal and opposite reaction on its surroundings.

12.6 POSSIBLE FORCES ACTING ON THE CONTROL VOLUME

Within a control volume, the possible forces which can act on the system are as follows:

1. F1 = Force exerted in the given direction on the fluid in the control volume by any solid body

within the control volume or coinciding with the boundaries of the control volume.

2. F2=Force exerted in the given direction on the fluid in the control volume by body forces such as

gravity.

3. F3=Force exerted in the given direction on the fluid in the control volume by the fluid outside the

control volume.

Thus:

FR = F1+F2+F3 = m(Vout-V in)

12.7 MOMENTUM CORRECTION FACTOR

When deriving the momentum equation, the velocity at a given cross section is assumed to be constant.

In reality, the liquid in contact with the solid boundary is under the influence of significant shearing

forces. Consequently, the velocity of the fluid in contact with the boundary is equal to zero, and the

velocity of the successive fluid layers increase gradually to the incoming velocity. For example, in a pipe,

the velocity will be zero at the point of contact with the boundary and maximum at the centre of the pipe.


Taking the velocity distribution into consideration, the rate of change of momentum for the whole flow

can be found by summing the rate of change of momentum over small fluid elements, whereby the

velocity changes are so small that they can be considered to be constant.

Consider the momentum of a small elemental fluid:

Mass flow rate = Density x cross sectional area of flow x velocity of flow = ρ δA u

Momentum per unit time passing through the small element of fluid = Mass flow rate x velocity of flow

δM = ρρρρ δA u2

Total Momentum per unit time passing the whole cross-section:

M = ∫ ρρρρ δA u2………………………..equation 3

This equation is solved for known velocity distribution.

12.8 MOMENTUM CORRECTION FACTOR – Circular Pipe

Consider flow through a circular pipe, with linear velocity distribution:

r

R

dr

y

Figure 12.3 – Flow through a circular pipe


Radius of pipe is R, and the velocity of flow at a general distance y is given by an empirical formula

derived by Prandtl’s one seventh power law:

U=Umax (y/R)1/7……………………………………..equation 4

with the maximum velocity occurring at the centre of the pipe.

Area of elemental fluid δA = 2π r δr………………………… ………….equation 5

Replacing equations 4 and 5, in equation 3:

M = ∫ M = ∫ M = ∫ M = ∫ ρρρρ δA u δA u δA u δA u2 2 2 2 = ∫ = ∫ = ∫ = ∫ ρρρρ 2π r δr {Umax (y/R)1/7}2

And changing r in terms or R-y and dr in terms of –dy, and integrating with respect to dy, yields:

Total momentum per unit time = 49/72 π ρ R2 Umax2.

12.9 FORCE EXERTED BY A JET STRIKING A VERTICAL FLA T PLATE

The velocity of the incoming jet is V in the x direction. When the jet strikes the plate, the final

velocity component of the jet in the x direction is zero.

Force acting on the fluid to produce this rate of change of momentum, F=m(Vout – Vin)


Vin= V, while Vout=0 in the x direction

Mass flow rate = ρ A V

Force acting perpendicular to the plate, is the force exerted by the jet on the plate:

Force = mass flow rate (Change in velocity) = ρ A V V= ρ A V2

Jetv

Stationary Plate

Figure 12.4 – Jet striking stationary plate


Jet

v

Inclined stationary Plate

Figure 12.5 – Jet striking an inclined stationary plate

θ

θv

vcosθvsinθ

12.10 FORCE EXERTED BY A JET STRIKING AN INCLINED P LATE

When the plate is inclined, it is required to work out the velocity component acting in the x direction. The

initial velocity in a direction perpendicular to the plate is zero, and the final velocity in the direction

perpendicular to the plate is vcosθ.

Force exerted by the jet onto the plate: F = mass flow rate x change in velocity = m (Vout – Vin)

F = ρ A V (Vcosθ - 0) = ρ A V2cosθ


Jetv

Moving Plate, velocity U

Figure 12.6 – Jet striking vertical moving plate

U

12.11 FORCE EXERTED BY A JET STRIKING A MOVING PLAT E The mass flow rate approaching the plate is given by the product of the density of the liquid, the cross

sectional area of flow and the relative velocity of flow, given that as the fluid approaches the plate with a

velocity V, the plate moves away with a velocity U.

Hence, force exerted on the moving plate: F = mass flow rate approaching the plate x change in velocity F = ρ A (V-U) (V-U) = ρ A (V-U)2


12.12 FORCE EXERTED BY A JET STRIKING AN INCLINED MOVING PLATE In the case of a jet striking an inclined moving plate, the first step is to work out the mass flow rate

approaching the plate, which is similar to the situation described in section 12.11. Therefore the mass

flow rate approaching the plate is given by

Mass flow rate, m = density x cross sectional area of flow x relative velocity

m = ρ A (V-U)

The initial velocity component normal to the plate is zero, and the final velocity component normal to the

plate is given by (V-U)cosθ.

Hence, the force exerted by the jet on the moving inclined plate:

F = ρρρρ A (V-U){ (V-U)cosθθθθ - 0 } = ρρρρ A (V-U)2cosθθθθ

12.13 FORCE DUE TO THE DEFLECTION OF A JET BY A CURVED VANE

A jet of liquid strikes a curve vane, which results in either in change in the magnitude of the velocity or

no change in the magnitude of the velocity of the outgoing liquid. There is however a change in the

direction at which the liquid leaves the curve vane as compared to the direction in which it entered it, this

change in velocity, results in a change in the momentum of the liquid.

θ

V2

V1

Figure 12.8 – Jet striking a curved vane


An assumption in this present analysis, is that the fluid enters and leaves the curve vane tangentially

without

impact, thus the force will be exerted between the liquid and the surface of the curved vane.

Force acting in the HORIZONTAL direction:

Initial velocity in the X direction = V1

Final velocity in the X direction = V2 cosθ

Force acting on the fluid to cause a change in momentum = mass flow rate x change in velocity = ρ Q {

V2 cosθ-V1)

Force, FX=ρ A1V1 ( V2 cosθ-V1)

Similarly, Force acting in the VERTICAL direction:

Initial velocity in the Y direction = 0

Final velocity in the Y direction = V2sinθ

Force acting on the fluid to cause a change in momentum = mass flow rate x change in velocity = ρ Q {V2

sinθ-0)

Force, FY=ρ A1V1 V2 sin θ

Therefore, resultant force acting on the fluid, FR:

FR = √√√√ (FX + FY)


12.14 FORCE EXERTED ON PIPE BENDS

When liquid flows through a pipe connected to a bend, this induces a change in the direction of the

flow, hence a change in the velocity components in both the horizontal and vertical directions,

consequently a force is being exerted on the bend.

Consider a pipe running under pressure:

The conditions at inlet (section 11) is as follows: Cross sectional area of flow A1, Diameter of pipe d1,

Velocity of flow V1 and Pressure P1 and the conditions at the outlet (section 22) is as follows: Cross

sectional area of flow A2, Diameter of pipe d2, Velocity of flow V2 and Pressure P2.

θ

V2

V1 θ

V2

V1

Figure 12.9 – Jet striking a curved vane

θ

V2

V2cos θ θ θ θ

V2sin θθθθ

Velocity component in the X direction = V2cos θ θ θ θ&Velocity component in the Y direction =V2sin θθθθ

Velocity V1,Has component in theX direction ONLY


Now refer to item 12.6 of this same unit, whereby it was being stressed that there are 3 possible forces

which can act on a control system, F1, F2 and F3.

In this case F1, which is the force exerted by the walls of the pipe onto the fluid, F2 is the force by

gravity, and F3 which is the force due to the pressure under which the liquid is flowing:

FR = F1+F2+F3 = m(Vout-V in)

F2 is negligible since gravity does not act on this system.

Hence F1 + F3 = m(Vout-V in)

The force exerted by the liquid on the bend will be - F1, denoted here by R1. Thus

-F1 - F3 = - m(Vout-V in)

R1 = F3 - m(Vout-V in)

P1V1d1A1

P2V2d2A2

θ

1

1

2

2

Figure 12.10 – Force exerted on pipe bends

+ve sign convention


P1V1d1A1

P2V2d2A2

θ

1

1

2

2

Figure 12.11 – Force exerted on pipe bends – velocity components

θ

V2

+(V2cos θθθθ)

-(V2sin θθθθ)

NOTE:V1 has components in the x direction ONLY

+ve sign convention

Calculating the Horizontal and Resultant components of the force due to change in momentum,

m(Vout-V in):

Mass flow rate passing through pipe = ρ Q

Calculating resultant force in the HORIZONTAL direction:

F(H) = ρρρρ Q (V2cosθθθθ-V1)

Calculating resultant force in the VERTICAL direction:

F(V) = ρ Q (-V2sinθ- 0) = -ρρρρ Q V2sinθθθθ

Where FH and FV are the forces exerted on the liquid to cause the change in momentum.

Consequently the force exerted by the liquid on the bend will be equal and opposite.


Calculating the components of the resultant force due to the pressure under which the fluid is

acting (F3):

Calculating resultant force in the HORIZONTAL direction:

F(PH) = P1A1 - (P2cosθ) A2 = P1A1 - P2cosθθθθ A2

Calculating resultant force in the VERTICAL direction:

F(PV) = 0- (-P2sinθA2) = P2sinθθθθA2

OVERALL Resultant force:

Recall equation R1 = F3 - m(Vout-V in)

OVERALL Resultant force in the HORIZONTAL direction acting ON THE BEND:

F(RH) = (P1A1 - P2cosθ A2) - ρ Q (V2cosθ-V1)

OVERALL Resultant force in the VERTICAL direction acting ON THE BEND:

F(RV) = P2sinθA2 - {- ρ Q V2sinθ} = P2sinθθθθA2 + ρρρρ Q V2sinθθθθ

Hence, Resultant force acting on the bend:

F(R) = √ F(RH)2+ F(RV)

2

P1V1d1A1

P2V2d2A2

θ

1

1

2

2

Figure 12.12 – Force exerted on pipe bends – pressure force components

θ

P2

+(P2cos θθθθ)

-(P2sin θθθθ)NOTE:P1 has components in the x direction ONLY and it is positive,being in the same positive direction as the sign convention

+ve sign convention


12.15 ACTIVITIES

1. Define Newton’s second Law of Motion and derive an equation governing the resultant force

required to cause a change in momentum of a moving liquid.

2. State the 3 forces acting of a moving liquid, that may influence the change in momentum of that

moving liquid.

3. What do you understand by the Momentum correction factor? Explain how you would account

for the momentum correction factor for a liquid flowing through a pipe.

4. Derive the general equation for a jet striking (i) a vertical flat stationery plate, and (ii) a moving

vertical flat plate.

5. Derive the general equation for a jet whose path is deviated by a bend through 240o.

12.16 SUMMARY

In this unit the student will have been introduced to the principles governing the Momentum Equation and

resultant force required to cause a change in momentum of a moving fluid. Consequently the moving

fluid exerts a reaction onto the boundary of its container, and students have been shown how such forces

are calculated and also why it is important to consider them in the design of pipe networks. The next unit

will be concerned with the application of the momentum equation to estimate the retardation forces at

boundaries.



Example 1 – Force exerted by a jet on a flat plate

Example 1 – Force exerted by a jet…1/4

• A flat plate is struck normally by a jet of water 50mm in diameter with a velocity of 18m/s. Calculate (a) the force on the plate when it is stationery, (b) the force on the plate when it moves in the same direction as the jet with a velocity of 6m/s and (c) the work done per second.

V=18m/sD=50mm


• PLATE IS STATIONERY

• Initial velocity of jet = 18m/s• Final Velocity of jet=0• Force exerted on the liquid to cause the change in Momentum of the

liquid, F = Rate of change of Momentum = Mass flow rate x change in velocity

• Hence, Force exerted on the plate = R which is equal to -F

• Force on the plate = -m(Vout-V in) = - ρ Q (0-18)

= 1000 x {π x (50x10-3)2/4} x 18 x 18= 636.2 N

V=18m/sD=50mm



• PLATE IS MOVING IN DIRECTION OF THE JET

• Initial velocity of jet = (18-6)= 12m/s• Final Velocity of jet=0• Mass Flow rate striking the plate = ρ A (V-U), where U is the velocity of

the plate

• Hence, Force exerted on the plate R• Force on the plate = -m(Vout-V in) = - ρ Q (0-12)

= 1000 x {π x (50x10-3)2/4} x (18-6) x 12= 282.7 N

V=18m/sD=50mm

V=6m/s


• PLATE IS MOVING IN DIRECTION OF THE JET

• Work done = Force x distance moved in the direction of the force per unit time

• Work done = 282.7 x 6 = 1696.2 W

V=18m/sD=50mm

V=6m/s

Example 2 – Force exerted by a jet of water striking a moving curved vane


Example 2 – Force exerted on vanes…1/4 • A jet of water delivers 85dm3/s at 36m/s onto a series of vanes moving in the same

direction as the jet at 18m/s. If stationary, the water which enters tangentially would be diverted through an angle of 135o. Friction reduces the relative velocity at exit from the vanes to 0.8 of that at entrance. Determine the magnitude of the resultant force on the vanes and the efficiency of the arrangement. Assume no shock at entry.

135o

V

0.8V

18m/s

+ve X and Y direction

Diameter of the jet:

Q = A V 85 x 10-3 = πd2/4 x 36d = 54.8 mm

Example 2 – Force exerted on vanes…2/4

• Horizontal component of the force exerted on the liquid to cause a change in momentum:– Mass flow rate striking the vane = ρρρρ A V – Initial horizontal velocity = (36-18)m/s– Final horizontal velocity = -(0.8V-U)sin45o

• Horizontal force = mass flow rate striking the vane x change in velocity component

• F(H) = ρρρρ A V { -(0.8V-U)sin45o – (V-U)}= 1000 x {πx(54.8x10-3)/4}x36 x {-0.8x36-18) sin45o – (36-18)}= -2.44kN

135o

V

0.8V

U=18m/s

45o45o

0.8V -(0.8V-U)cos45o

-(0.8V-U)sin45o



• Vertical component of the force exerted on the liquid to cause a change in momentum:– Initial vertical velocity = 0 m/s– Final vertical velocity = -(0.8V-U)cos45o

• Vertical force = mass flow rate striking the vane x change in velocity component

• F(V) = ρρρρ A V { -(0.8V-U)cos45o}= 1000 x {πx(54.8x10-3)/4}x36 x {-(0.8x36-18)cos45o}=--0.64kN

135o

V

0.8V

U=18m/s

45o45o

0.8V -(0.8V-U)cos45o

-(0.8V-U)sin45o


• Resultant force acting on the liquid:

F(R) = ?F(v)2 + F(H)2

= {2.44 + 0.64}1/2

= 2.52kN

135o

V

0.8V

U=18m/s

45o45o

0.8V -(0.8V-U)cos45o

-(0.8V-U)sin45o


12.18 TUTORIAL SHEET

Tutorial – Applications of Momentum Principles 1. A water nozzle is directed vertically downward against a flat metal plate as shown in the figure

below. Find the force exerted on the plate by the water jet, neglecting friction. 500 kNm-2 Diameter = 10 cm 0.5 m

Diameter = 4 cm

Fixed Plate (Diagram not to scale) 2. A pipe bend tapers from a diameter of 500 mm at inlet to a diameter of 250 mm at outlet and turns the

flow through an angle of 45o. The pressure at inlet is measured as 60 kN/m2 above atmospheric

pressure. If the pipe is conveying oil of density 850 kg/m3, and if the bend is in a horizontal plane,

calculate the magnitude and direction of the resultant force on the bend when the oil is flowing at a

rate of 0.45 m3/s. State any assumption made.

The outlet pipe is disconnected from the bend and the bend is turned so that I now lies in a vertical

plane and issues a jet of oil into the atmosphere. The volume of the bend is 0.25 m3, and the outlet is

1.75 m higher than the inlet. If the same pressure is maintained at the inlet of the bend, determine the

magnitude and direction of the resultant force on the bed. Determine also the new discharge and the

maximum height above the bend outlet to which the jet will rise.


3. A 40 mm diameter water jet strikes a hinged vertical plate of 800 N weight, normally at its centre

with a velocity of 15 m/s. Determine:

(a) the angle of deflection of the plate about the hinge

(b) the magnitude of the force F that must be applied at the lower edge of the plate to keep the plate

vertical.

4. A square plate of mass 25 kg of side 500 mm is free to swing about its upper horizontal edge. When

the plate is vertical, a horizontal water jet of diameter 25 mm and nozzle velocity 25 m/s strikes it

normally at its centre.

(a) What force must be applied to the lower edge of the plate to keep it vertical.

(b) What inclination to the vertical the plate will assume under the action of the jet if it is allowed to

swing freely?

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