OPEN CHANNEL HYDRAULICS FOR ENGINEERS

Types of flow  A flow, in which the velocity does not change from point to point along any of the streamlines, is called a uniform flow.. In the course on Fluid Mechanics, we have assum

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1.1 Review of fluid mechanics

1.2 Structure of the course

Key words

Fluid mechanics; open channel flow; dimensional analysis; similitude; Reynolds number; hydraulic model

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1.1 REVIEW OF FLUID MECHANICS

This lecture note is written for undergraduate students who follow the training programs in the fields of Hydraulic, Construction, Transportation and Environmental Engineering It is assumed that the students have passed a basic course in Fluid Mechanics and are familiar with the basic fluid properties as well as the conservation laws of mass, momentum and energy However, it may be not unwise to review some important definitions and equations dealt with in the previous course as an aid to memory before starting

1.1.1 Fluid mechanics

Fluid mechanics, which deals with water at rest or motion, may be considered as one of the important courses of the Civil Engineering training program It is defined as the mechanics of fluids (gas or water) This course will mostly deal with the liquid water The following properties then are important:

(a) Density

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

standard temperature and pressure It is also fully called “mass density” and denoted by the Greek symbol  (rho) In the case of water, we generally neglect the variation in mass density and consider it at a temperature of 4C and at atmospheric pressure; then  = 1,000 kg/m3 for all practical purposes For other specific cases, the densities of common liquids are given in tables in most fluid mechanics books

(b) Specific weight

The specific weight of a liquid is the gravitational force per unit volume It is given

by the Greek symbol  (gamma) and sometimes briefly written as sp.wt In SI units, the

specific weight of water at a standard reference temperature of 4C and atmospheric pressure is 9.81 kN/m3

-(c) Specific gravity

Specific gravity is defined as the ratio of the specific weight of a given liquid to the

specific weight of pure water at a standard reference temperature Specific gravity, or sp gr., is presented as:

Sp.gr =

water pureof weight Specific

liquidof weight Specific

Specific gravity is dimensionless, because it is a ratio of specific weights

(d) Compressibility

The compressibility of a fluid may be defined as the variation of its volume, with the variation of pressure All fluids are compressible under the application of an external force, and when the force is removed they expand back to their original volume exhibiting the property that stress is proportional to volumetric strain In the case of water as well as other liquids, it is found that volumes are varying very little under variations of pressure,

so that compressibility can be neglected for all practical purposes Thus, water may be

considered as an incompressible liquid

(e) Surface tension

The surface tension of a liquid is its property, which enables it to resist tensile

stress in the plane of the surface It is due to the cohesion between the molecules at the

surface of a liquid Looking at the upper end of a small-diameter tube put into a cup of

water, we can easily see the water risen in the tube with an upward concave surface, as

shown in Fig 1a However, if the tube is dipped into mercury, the mercury drops down in

the tube with an upward convex surface as shown in Fig 1b If the adhesion between the

tube and the liquid molecules is greater than the cohesion between the liquid molecules, we will have an upward concave surface Otherwise, we get an upward convex surface The surface tension of water and mercury at 20 ºC is 0.0075 kg/m and 0.0520 kg/m, respectively

Fig 1.1a Capillary tube in water Fig 1.1b Capillary tube in mercury

The phenomenon of rising water in a small-diameter tube is called capillary rise

(f) Viscosity

The dynamic or absolute viscosity of a liquid is denoted by the Greek symbol  (mu) and defined physically as the ratio of the shear stress  to the velocity gradient du/dz:

 dudz



where u = velocity in x direction

Fig 1.2: Velocity distribution

Viscosity is its property which controls the rate of flow In the same tube, the flow of alcohol or water is much easier than the flow of syrup or heavy oil

-1.1.2 Hydrostatics

Hydrostatics means study of pressure as exerted by a liquid at rest Since the fluid

is at rest, there are no shear stresses in it The direction of such a pressure is always at right angles to the surface, on which it acts (Pascal’s law)

(a) The total force F on a horizontal, a vertical or

an inclined immersed surface is expressed as:

F = .A.hgc [kN] (1-2)

where  = g = specific weight of the liquid [N/m3

];

A = area of the immersed surface [m2];

hgc = depth of the gravity center of the

horizontal immersed surface from the liquid level [m] (see Fig 1.3)

(b) The pressure center of an immersed surface is the point through which the resultant

pressure force acts (see Fig 1.4):

Fig 1.4 Vertical and inclined surface

(c) The depth of pressure center of an immersed surface from the liquid level, hpc, (see Fig 1.4) reads:

hpc = gc

gc

h.A

hh.A

sin.I



 [m] (for inclined immersed surface) (1-4)

where IG = moment of inertia of the surface about the horizontal axis through its gravity

-(d) The pressure center of a composite section is found as follows:

 first, by splitting it up into convenient sections;

 then, by determining the pressures on these sections;

 then, by determining the depths of the respective pressure centers; and

The continuity principle is based on the conservation of mass as applying to the

flow of fluids with invariant, i.e constant, mass density The continuity equation of a liquid flow is a fundamental equation stating that, if an incompressible liquid is continuously flowing through a pipe or a channel (the cross-sectional area of which may or may not be constant), the quantity of liquid passing per time unit is the same at all sections

as illustrated in Fig 1.5

Now consider a liquid flowing through a tube

Let Q = flow discharge [m3/s];

V = average velocity of the liquid [ms-1];

A = area of the cross-section [m2];

and i = the number of section

We get:

Q1 = Q2 = Q3 = … (1-6) Fig.1.5 Continuity of a liquid flow

or V1A1 = V2A2 = V3A3 = … (1-7)

1.1.4 Types of flow

 A flow, in which the velocity does not change from point to point along any of the

streamlines, is called a uniform flow Otherwise, the flow is called a non-uniform flow

 A flow, in which each liquid particle has a definite path and the paths of

individual particles do not cross each other, is called a laminar flow This flow is

void of eddies But, if each particle does not have a definite path and the paths of

individual particles also cross each other, the flow is called turbulent

 A flow, in which the quantity of liquid flowing per second, Q, is constant with

respect to time, is called a steady flow But if Q is not constant, it is called an unsteady flow

 A flow, in which the volume and thus the density of the fluid changes while

flowing, is called a compressible flow But if the volume does not change while flowing, it is called an incompressible flow

- A flow, in which the fluid particles also rotate about their own axes while flowing,

is called a rotational flow But if the particles do not rotate about their own axes while flowing, it is called an irrotational flow

 A flow, whose streamlines may be represented by straight lines, is called a dimensional flow If the streamlines are represented by curves, the flow is called two-dimensional A flow, whose streamlines can be decomposed into three mutually perpendicular directions, is called three-dimensional

one-1.1.5 Bernoulli ’s equation

It states: “For a perfect incompressible liquid, flowing in a continuous stream, the total energy of a particle remains the same, while the particle moves along a streamline from one point to another” This statement is based on the assumption that there are no losses due to friction Mathematically it reads

g

pg

Vz

2





where z = elevation, i.e the height of the point in question above the datum; z

represents the potential energy;

 = pressure head, representing the pressure energy; p is the pressure at the

point in question and  is the liquid density

1.1.6 Euler's equation

Euler’s equation for steady flow of an ideal fluid along a streamline is based on Newton’s second law (Force = Mass  Acceleration) It is based on the following assumptions:

 The fluid is inviscid, homogeneous and incompressible;

 The flow is continuous, steady and along the streamline;

 The flow velocity is uniformly distributed over the section; and

 No energy or force, except gravity and pressure force, is involved in the flow

Euler's equation in a differential-equation form can be written as:

0g

dpg

dVV

1.1.7 Flow through orifices, mouthpieces and pipes

 An orifice is an opening (in a vessel) through which the liquid flows out The

discharge through an orifice depends on the energy head, the cross-sectional area of the orifice and the coefficient of discharge A pipe, the length of which is generally more than two times the diameter of the orifice, and which is fitted externally or

internally to the orifice is known as a mouthpiece When a liquid is flowing through

-a mouthpiece, the energy he -ad is declining due to w -all friction, ch -ange of cross

section or obstruction in the flow

 A pipe is a closed conduit used to carry fluid When the pipe is running full, the flow is under pressure The friction resistance of a pipe depends on the roughness

of the pipe inside Early experiments on fluid friction were conducted, among others, by Chezy: the frictional resistance varies approximately with: (a) the square

of the liquid velocity, and (b) the bed slope

Frictional resistance = Frictional resistance per

unit area at unit velocity

V = characteristic flow velocity [m/s];

D = characteristic length, e.g diameter of the pipe [m]

+ Darcy–Weisbach’s formula for head loss hf in pipes:

2 f

L V

h f

D 2g

where f = friction coefficient according to Darcy–Weisbach;

L = length of the pipe

+ Chezy's formula for flow velocity V in pipe: VC Ri [m/s] (1-12) where C = Chezy's coefficient [m½ s-1

];

R = hydraulic radius [m] defined as:

P

A perimeter wetted

area section cross

i = loss of energy head per unit length (= bed slope in uniform flow)

1.1.8 Flow through open channel

An open channel is a passage, through which the water flows due to gravity with atmospheric pressure at the free surface The flow velocity is different at different points in the cross-section of a channel due to the occurrence of a velocity distribution, but in calculations, we use the mean velocity of the flow In the course on Fluid Mechanics, we have assumed that the rate of discharge Q, the depth of flow h, the mean velocity V, the slope of the bed i and the cross-sectional area A remain constant over a given length L of the channel (see Fig 1.6)

Fig 1.6 Uniform flow in open channel Discharge through an open channel: QVAAC Ri (1-13)

-1.2 STRUCTURE OF THE COURSE

1.2.1 Objectives of the course

Open Channel Hydraulics is an advanced course required for all students who follow the field-study of water resources engineering The subject is rich in variety and of interest to practical problems The content is focused on the types of problems commonly encountered by hydraulic engineers dealing with the wide fields covered by open channel hydraulics Due to space and lecturing-time limitations, however, the lecture note does not extend into the specialist fields of mathematical natural flow networks required, for example, for river engineering computations

The course aims to present the principles dealing with water flow in open channels and to guide trainees to solve the applied problems for hydraulic-structure design and water system control The main objectives of the course are:

 To supply the basic principles of fluid mechanics for the formulation of open channel flow problems

 To combine theoretical, experimental and numerical techniques as applied to open channel flow in order to provide a synthesis that has become the hallmark of modern fluid mechanics

 To provide theoretical formulas and experimental coefficients for designing some hydraulic structures as canals, spillways, transition works and energy dissipators

1.2.2. Historical note for the course

Fluid mechanics and open channel hydraulics began at the need to control water for irrigation purposes and flood protection in Egypt, Mesopotamia, India, China and also Vietnam Ancient people had to record the river water levels and got some empirical understanding of water movements They applied basic principles on making some fluid machinery, sailing boats, irrigation canals, water supply systems etc The Egyptians used dams for water diversion and gravity flow through canals to distribute water from the Nile River, and the Mesopotamians developed canals to transfer water from the Euphrates river

to the Tigris river, but there is no recorded evidence of any understanding of the theoretical flow principles involved The Chinese are known to have devised a system of dikes for protection from flooding several thousand years ago Over the past 2,000 years, many dikes and canal systems have been built in the Red River delta in the North of Vietnam to contain the delta and drain off its flood water that has always been serious problems Vietnamese, under Ngo Quyen, have also known to apply the tidal law in Bach Dang river battles in 939 A.D, which has become famous in Vietnamese history

It was not until 250 B.C that Archimedes discovered and recorded the principles of hydrostatics and flotation In the 17th and 18th centuries, Isaac Newton, Daniel Bernoulli and Leonhard Euler formulated the greatest principles of hydrodynamics The work of Chezy on flow resistance began in 1768, originating from an engineering problem of sizing

a canal to deliver water from the Yvette River to Paris The Manning-equation for channel-flow resistance has a complex historical development, but was based on field observations Julius Weisbach extended the sharp-crested weir equation and developed the elements of the modern approach to open channel flow, including both theory and experiment William Froude, an Engish engineer, collaborated with Brunel in railway construction and in the design of the steamer “Great Eastern”, the largest ship afloat at that time He contributed to the study of friction between solids and liquids, to wave mechanics and to the interpretation of ship model tests

open-The work of Bakhmeteff, a Russian émigré to the United States, had perhaps the most important influence on the development of open channel hydraulics in the early 20thcentury Of course, the foundations of modern fluid mechanics were laid by Prandtl and his students, including Blasius and von Kàrmàn, but Bakhmeteff’s contributions dealt specifically with open channel flow In 1932, his book on the subject was published, based

-on his earlier 1912 notes developed in Russia His book c -oncentrated -on “varied flow” and introduced the notion of specific energy, still an important tool for the analysis of open-channel flow problems In Germany at this time, the contributions of Rehbock to weir flow also were proceeding, providing the basis for many further weir experiments and weir formulas

By the mid-20th century, many of the gains in knowledge in open channel flow has been consolidated and extended by Rouse (1950), Chow (1959, 1973) and Henderson (1966), in which books extensive reference can be found These books set the stage for applications

of modern numerical analysis techniques and experimental instrumentation to channel flow problems

open-1.2.3 Structure of the course

The lecture note is divided into three parts of increasing complexity

(a) Part 1 introduces to the basic principles: course introduction (Chapter 1),

uniform flow (Chapter 2) and hydraulic jump phenomena (Chapter 3) This part will take 15 teaching hours

(b) Part 2 includes non-uniform flow (Chapter 4) and design application as

Spillways (Chapter 5) and Transitions and Energy dissipators (Chapter 6) This part will take 20 teaching hours

(c) Part 3 deals with unsteady flow (Chapter 7) This chapter will take 10 teaching

hours

The course approach chart is presented in Fig 1.7 on the next page

-Fig.1.7 Course structure chart

OPEN CHANNEL HYDRAULICS FOR ENGINEERS

Chapter 1: INTRODUCTION

1.1 Review of fluid mechanics 1.2 Structure of the course

1.3 Dimensional analysis 1.4 Similarity and models

Chapter 2: UNIFORM FLOW

2.1 Introduction 2.2.Basic equations in uniform open channel flow

2.3 Most economical section 2.4 Channel with compound

cross-section 2.5 Permissible velocity against erosion and sedimentation

Chapter 3: HYDRAULIC JUMP

3.1 Introduction 3.2 Specific energy 3.3 Depth of hydraulic jump

3.4 Types of hydraulic jump 3.5 Hydraulic jump formulas in terms of

Froude-number 3.6 Submerged hydraulic jump

Chapter 7: UNSTEADY FLOW

7.1 Introduction 7.2 The equations of motion

7.3 Solutions to the unsteady-flow equations

7.4 Positive surge and negative waves; Surge formation

Chapter 6: TRANSITIONS AND ENERGY DISSIPATORS

6.1 Introduction 6.2 Expansions and Contractions

6.3 Drop structures 6.4 Stilling basins

6.5 Other types of energy dissipators

Chapter 5: SPILLWAYS

5.1 Introduction 5.2 General formula 5.3 Sharp-crested weir

5.4 The overflow spillway 5.5 Broad-crested weir

Chapter 4: NON-UNIFORM FLOW

4.1 Introduction 4.2 Gradually-varied steady flow

4.3 Types of water surface profiles

4.4 Drawing water surface profiles

-1.3 DIMENSIONAL ANALYSIS

Most hydraulic engineering problems are solved by applying a mathematical analysis In some cases they should be checked by physical experimental means The approach of such problems is considerably simplified by using mathematical techniques for dimensional analysis It is based on the assumption that the phenomenon at issue can be expressed by a dimensionally homogeneous equation, with certain variables

1.3.1 Fundamental dimensions

We know that all physical quantities are measured by comparison This comparison

is always made with respect to some arbitrarily fixed value for each independent quantity,

called dimension (e.g length, mass, time, temperature etc) Since there is no direct relationship between these dimensions, they are called fundamental dimensions or

fundamental quantities Some other quantities such as area, volume, velocity, force etc, cannot be expressed in terms of fundamental dimensions and thus may be called derived dimensions, derived quantities or secondary quantities

There are two systems for fundamental dimensions, namely FLT (i.e force, length, time) and LMT (i.e length, mass, time) The dimensional form of any quantity is independent of

the system of units (i.e metric or English) In this course, we shall use the LMT-system The following table gives the dimensions and units for the various physical quantities, which are important form the hydraulics point-of-view

Table 1.1: Dimensions in terms of LMT

No Quantity Symbol Dimensions in terms of LMT-system

All variables used in science or engineering are expressed in terms of a limited number of basic dimensions For example, we can designate the dimensions of velocity as:

LTT

LTime

cetan

Note:There are four systems of units, which are commonly used and universally adopted These are known as:

 SI Units (International System of Units or Syst ème International d'unités in

French): a unified and systematically constituted system of fundamental and derived units for international use have been recommended by the 11th General Conference of Weights and Measures (CGPM) SI is widely used in Vietnam as an

official unit system The fundamental units of LMT (length, mass and time) are meter, kilogram and second, respectively

 CGS Units: the fundamental units of LMT are centimeter, gram and second,

dimensionally homogeneous An equation is called dimensionally homogeneous, if the

fundamental dimensions have identical powers of LMT on both sides That is, the left-hand side (LHS) of the equation must have the same dimensions as the right-hand side (RHS) Moreover, every term in the equation must have the same dimensions Such an equation would essentially be independent of the system of measurement (i.e English or SI)

Note: Two dimensionally homogeneous equations can be multiplied or divided without

affecting the homogeneity But the two dimensionally homogeneous equations cannot be added or subtracted, as the resulting equation may not be dimensionally homogeneous

-1.3.3 Principles of Dimensional Homogeneity

The principle of dimensional homogeneity has a number of applications The following issues are important from the point of view of the subject

(a) Determining the dimension of a physical quantity

The dimensions of any physical quantity may be easily determined with this principle, e.g the dimension of energy:

Energy = Work = Force  Distance (1-16)

= [LMT-2]  [L] (Force, [F ]= [LMT-2])

= [ML2T-2]

Example 1.1: Determine the dimension of the following quantities in the LMT-system:

(i) Force (ii) Pressure, (iii) Power, (iv) Specific weight, and (v) Surface tension

Solution:

We know the dimension of ‘force‘ in the LMT-system:

(i) Force = Mass x Acceleration

Force [MLT ]

[ML T ]Area [L ]

[MLT ]

[ML T ][L ]

Example 1.2 Determine the dimension of the following quantities in the LMT-system

(i) Discharge, (ii) Torque and (iii) Momentum

-(b) Checking the dimensional homogeneity of an equation

The dimensional homogeneity of an equation may be easily checked with this principle; e.g let us consider Darcy - Weisbach’s formula for loss of energy head in pipes:

2 f

[L] =

]L][

LT[

][LT

x [L]

x ]1[

2

2 -1

Example 1.3 Check the dimensional homogeneity of the following common equations in the field of hydraulics: (i) QCdA 2gH and (ii) V  C Ri

Solution

(i) Given equation, Q = CdA 2gH

Substituting the dimensions on the LHS and RHS of the equation (the dimension of Cd, being a discharge coefficient, is taken as 1):

[L3T-1] = [1]  [L2] [[1]½ [LT-2 x L]½ = [L3T-1]

Since the dimensions on both sides of the equation are the same, the equation is

(ii) Given equation, V =C Ri

Substituting the dimensions on the LHS and RHS of the equation (the dimension of i, being dimensionless is taken as 1):

LT-1 = C  [L 1]1/2

= C [L]1/2 Since the dimensions on both sides of the equation are not the same, the equation is not

From the above equation, we find that:

]TL[]L[

]LT[

(c) Changing the coefficient of an equation while using an other system of units

The coefficient of an equation may be easily changed, while using the same equation in an other system of units, for example from English to MKS or vice versa

Let us consider Manning’s formula for the velocity

V = R 3i 2

n1

where n is a resistance coefficient, called Manning's constant Now substituting the dimensions on the LHS and RHS of the equation, we get:

Since the dimensions of both sides are not the same, the equation is dimensionally homogeneous From the above equation, we find that:

[LT ]

L T[L]

L1/3 = 3.2811/3 = 1.486

It is obvious, that the equation for English units will be V = R 3i 2

n

486.1

(1-20)

(d) Using the dimensional analysis methods

There are several methods that may be used to carry out the process of dimension analysis, such as the Step-by-Step method, the Exponent method, Students can find them in reference books In this course, Buckingham's -theorem will be introduced shortly in the next section

1.3.4 Buckingham ’s - theorem

Buckingham’s -theorem states, “If there are n variables in a dimensionally homogeneous equation and if these variables contain m fundamental dimensions such as (L, M, T), they may be grouped into (n-m) non-dimensional independent -terms.”

Mathematically, if a variable X1 depends on the independent variables X2, X3, X4, ., Xn the function may be written as:

f1 (1,2,3, ,n-m) = Constant

where  is a dimensionless term

Students can read for understanding details and how to apply in reference hydraulics books

Example 1.4: Flow through a closed conduit with rectangular cross-section Let us determine the wall friction as a dependent quantity

-Solution:

Let  be the average wall shear stress over the full perimeter

 depends on: b = internal breadth [m]

h = internal height [m]

k = dimension of wall roughness [m]

 = specific mass density of fluid [kgm-3

]

 = dynamic viscosity of fluid [kgm-1s-1]

V = fluid velocity, averaged over cross-section [ms-1]

  = f1(b, h, k, , , V)

[] = ML-1

T-2[b] = L

We have totally 7 quantities and 3 basic dimensions, viz M L and T

 There are 4 independent dimensionless parameters

-1.3.5 Limitations of dimensional analysis

Some problems may be met when applying dimensional analysis:

 In order to use dimensional analysis, we must first decide which variables are significant If we do not understand the problem well enough to make a good initial choice of variables, dimensional analysis seldom provides clarification

 One error might be the inclusion of variables whose influence is already accounted for For example, one might tend to include two or three length variables in a scale-model test, where only one may be sufficient

 Another serious error might be the omission of a significant variable If this is done, one of the significant dimensionless parameters will likewise be missing

How do we know whether a variable is significant for a given problem? Probably the proper answer is from experience After working in the field of fluid mechanics and open- channel hydraulics for several years, one develops a feeling for the significance of variables to certain kinds of application

Since the beginning of the twentieth century, the engineers engaged on the creation

or design of hydraulic structures (such as dams, spillways or large hydraulic machines) have developed a new and scientific method to predict the performance of their structures and machines This is done by preparing physical scale models and testing them in a laboratory; so as to form some opinion, about the working and behaviour of the proposed hydraulic structures, after their completion or actual installation The structure, of which the model is prepared, is known as prototype and the model is known as scale model or simply physical model

1.4.1 Advantages of model analysis

Though there are numerous advantages of model testing, yet the following is to be mentioned:

1 The behaviour and working details of a hydraulic structure or a machine can be easily predicted from its physical model The smooth and reliable working of a hydraulic structure or a machine can be ascertained by spending a relatively small sum of money, which is a negligible fraction of the total cost to be spent

on the prototype

2 If the hydraulic structure or machine is made directly, then in case of its failure,

it is very difficult to change its design Moreover, it is very costly Laboratory tests can result in saving human labour and material

3 With the help of model testing, a number of alternative designs can be studied Finally, the most economical, accurate and safe design may be selected

4 When the existing hydraulic structure is not functioning properly, then model testing can help us in detecting and rectifying the defects

5 Sometimes, it is difficult to design a particular portion of a complex hydraulic structure or machine In such a case, model testing is very essential in order to ascertain the safety and reliability of that particular portion of the prototype

-1.4.2 Hydraulic similarity

If we look at a photograph of a man, very carefully, we can have an idea of the proportion of various parts of his body The photograph will also give an idea of the features of each part of the man Similarly, to know the complete working and behaviour

of the prototype from its model, there should be a complete similarity between the prototype and its scale model This similarity is known as hydraulic similitude or Hydraulic Similarity Three types of hydraulic similarity are important, viz.:

Fig.1.8 Geometric similarity: (a) Prototype and (b) Model

Let Lp = length of the prototype

Bp = breadth of the prototype,

Dp = depth of the prototype, and

Lm, Bm and Dm = corresponding values for the model

Now, if geometric similarity exists between the prototype and the model, then the linear ratio of the prototype and the model (also called scale ratio) reads as:

(b)

Let V1p = velocity of liquid in prototype at point 1,

V2p = velocity of liquid in prototype at point 2,

V1m, V2m = corresponding values for the model

Now, if kinematic similarity exists between the prototype and the model, then the velocity ratio of the prototype and the model reads as:

V

VV

VVV

m 3 p

m 2 p

m 1

Let F1p and F1m = force acting in prototype and model at point 1;

F2p and F2m = force acting in prototype and model at point 2

Now, if dynamic similarity exists between the prototype and the model, then the force ratio

of the prototype and the model reads as:

Consider the flow over the spillway shown in Fig 1.9 Here corresponding masses of fluid

in the model and the prototype are acted on by corresponding forces These forces are the force of gravity Fg, the pressure force Fp, and the viscous resistance force Fv These forces add vectorially in Fig 1.9 to yield a resultant force FR, which will in turn produce an acceleration of the volume of fluid in accordance with Newton’s second law

F  M a

-Fig 1.9: Dynamic similarity (a) Prototype and (b) Model

1.4.6 Technique of hydraulic modelling

The technique of hydraulic modelling involves the following steps:

a Selection of suitable scale,

b Operation of the hydraulic model, and

c Correct prediction

a Selection of suitable scale

This depends on many factors But the following is important concerning this issue:

-b Operation of the hydraulic model

After selecting the type, the scale and the materials of the model, the next step is to construct the model accurately according to the plan High-tech instruments can be essential for precisely measuring the hydraulic quantities in the experiments Great care and patience are required for correctly interpreting the model results

c Correct prediction

After obtaining the precise measurements of the required hydraulic quantities in an experiment, the next step is to predict the correct working of the prototype We shall study the correct prediction of prototypes in the following pages

1.4.7 Developments in hydraulic model testing

The model testing is the most scientific and common feature of the design and successful working of hydraulic structures and machines Two types of facility are important and need to be dealt with: 1) the wind tunnel, and 2) the water tunnel

1 Wind tunnel

A wind tunnel is a standard equipment for aircraft design It provides a steady flow of air around the model which is suspended in the stream Though the walls of the tunnel will interfere, to some extent, with the stream of air, yet its effect is generally neglected

In a wind tunnel, the air is set in motion by means of a compressor The model under investigation is mounted in the path of the wind stream Sometimes, the compression of air

in the wind tunnel produces an appreciable rise in temperature, which must be dissipated

by a cooling device

2 Water tunnel

A water tunnel is a standard equipment for the design of turbines, pumps and ships In water tunnels, a uniform stream of water is produced and the model under investigation is mounted in the path of the water

The size of the water tunnel is, usually, expressed as the diameter of its best section The existing water tunnels range as size from 10 cm to 150 cm

1.4.8 Undistorted models

All the hydraulic models may be broadly classified into the following two types:

1 Undistorted models, and

2 Distorted models

A model, which is geometrically similar to its prototype is known as an undistorted model The prediction from an undistorted model is comparatively easy and the results obtained from the model, can be easily transferred to the prototype, if the basis condition (of geometric similarity) is satisfied A distorted model will be discussed in Section 1.4.10

-1.4.9 Comparison of an undistorted model and the prototype

We have discussed in Sections 1.4.3 through 1.4.5 the different types of hydraulic similarity between model and prototype If the model is to be overall similar to the prototype, then all the three similarities (i.e geometric, kinematic and dynamic) should exist between the model and the prototype But this is generally not possible in actual practice, as it is difficult to deal with two types of similarities simultaneously In general,

an undistorted model of a prototype is made applying geometric similarity only, and the remaining similarities are then compared on account of the scale ratio (i.e geometric ratio

of prototype and model)

Though the given scale ratio provides us a wide range of data of the prototype, yet the following is important to take into consideration:

1 Velocity of water in the prototype versus the given velocity at the corresponding point of the model

2 Discharge of the prototype versus the given discharge of the model

3 Time of emptying a prototype versus the given time of emptying the model

4 Power developed by the prototype versus the given power developed by the model

5 Speed of prototype versus the given speed of the model (e.g in r.p.m.)

Let V = velocity of water flowing over a weir, a dam, or a spillway, …

Q = discharge over of a weir, out of a notch, or over a spillway, …

N = speed, in r.p.m., of a centrifugal pump or a turbine, …

P = resistance power developed on a ship, or an air-plane, …

T = time needed for emptying a tank, a reservoir, …

p, m subscript characters denoting the prototype and the model

r = scale ratio of the prototype to the model

We apply the following formulas in Table 1.2 for an undistorted, geometrically similar model

Sometimes the model of an object is made and tested in the hydraulic laboratory with a difference in the specific weights of the liquids applied, for example: a ship test in sea water and fresh water, or an air-plane in a wind tunnel In such a case, we must multiple r with a ratio of the prototype specific weight to model specific weight: p

m





the prototype Such a model is called distorted Moreover, models of hydraulic structures,

such as rivers, harbours, reservoirs etc have very large horizontal dimensions, as compared

to the vertical ones If such a model would be completely geometrically similar, then the water depth would be so small that measurements can not be performed accurately, and the flow patterns cannot be represented properly

In order to overcome this difficulty, the models of such structures are made with different horizontal and vertical scales A model having complete geometric similarity with the prototype, but working under a different head of water, also behaves as a distorted model

In such models, the scale ratio of model to prototype is taken as the horizontal scale ratio and the ratio of the head of water in the model to the head of water in the prototype is taken

as the vertical scale ratio The prediction from a distorted model is relatively difficult, and the results of the models being distorted cannot be easily transferred to the prototype, as the condition (of geometric similarity) is not satisfied

-14.11 Advantages and disadvantages of distorted models

A distorted model has the following advantages and disadvantages:

+ Disadvantages

 There is an unfavourable psychological effect on the observer

 The behaviour of flow of a model in action differs from that of the prototype

 The magnitude and direction of the pressures is not correctly reproduced

 The velocities are not correctly reproduced, as the vertical exaggeration causes distortion of lateral velocity and kinetic energy

In spite of the above-mentioned disadvantages of a distorted model, it is sometimes preferred to use the distorted model However, by exercising utmost care, the results of the model may be transferred to the prototype

1.4.12 Comparison of a distorted model and its prototype

We have seen that the models of large hydraulic structures are made with different horizontal and vertical scales The comparison of such distorted models and their prototypes is done by starting from the fundamentals In the following, we shall discuss the comparison of a distorted model and its prototype

1 Velocity of water in the prototype versus the given velocity at the corresponding point of the model

2 Discharge of the prototype versus the given discharge of the model

3 Time of emptying a prototype versus the given time of emptying the model

4 Power developed by the prototype versus the given power of the model

5 Speed of the prototype versus the given speed of the model

Let us use the same symbols as in Section 1.4.9 and add rH and rV as horizontal and vertical scale ratio of the prototype to the model

-Table 1.3: Ratio of prototype value and model value for a distorted model

No Hydraulic quantity Symbol Ratio of

prototype to model

1 Water Velocity V

V m

2.1 Introduction

2.2 Basic equations in uniform open-channel flow

2.3 Most economical cross-section

2.4 Channel with compound cross-section

2.5 Permissible velocity against erosion and sedimentation

Summary

The chapter on uniform flow in open channels is basic knowledge required for all hydraulics students In this chapter, we shall assume the flow to be uniform, unless specified otherwise This chapter guides students how to determine the rate of discharge, the depth of flow, and the velocity The slope of the bed and the cross-sectional area remain constant over the given length of the channel under the uniform-flow conditions The same holds for the computation of the most economical cross section when designing the channel The concept of permissible velocity against erosion and sedimentation is introduced

Uniform equilibrium open-channel flows are characterized by a constant depth and a

constant mean flow velocity:

h0s

 

V0s

power (1/6) and the one-seventh power (1/7) formulas It should be noted that the velocity

-in open-channel flow is assumed constant over the entire cross-section

Fig 2.1 Velocity distribution profile in turbulent flow

Such flow conditions are represented schematically in Fig 2.2 Considering Bernoulli’s theorem of the conservation of energy, between cross-sections 1 and 2, leads to the expression:

Fig 2.2 Energy and hydraulic gradient in uniform-flow channel

If the flow is uniform, the cross sections at points 1 and 2 must be constant Consequently, the velocity and the depth of flow must also remain constant, or:

p h

g





2 2

p h g





2 1

V

2 V

2 genergy-gradient line

V

v

y h

Vmax

-2.1.2 Momentum analysis

Consider a control volume of length L in uniform flow, as shown in Fig 2.3

Fig 2.3 Force balance in uniform flow

By definition, the hydrostatic forces, Fp1 and Fp2, are equal and opposite In addition, the mean velocity is invariant in the flow direction, so that the change in momentum flux is zero Thus, the momentum equation reduces to a balance between the gravity force component in the flow direction and the resisting shear force:

in which  = g = specific weight of the fluid, A = cross-sectional area of flow, o = mean boundary shear stress, and P = wetted perimeter of the boundary on which the shear stress acts If Eq (2-6) is divided by PL, the hydraulic radius R = A/P appears as an intrinsic variable Physically, Eq (2-6) represents the ratio of flow volume to boundary surface area, or shear stress to unit weight, in the flow direction Eq (2-6) can be written as:

if we replace sin with S = tan for small values of  Furthermore, if we solve Eq (2-7) for the bed slope, which equals the slope of the energy grade line, hL/L, and express the shear stress in terms of the friction factor f for uniform pipe flow according Darcy-Weisbach:

fR8

V fRL

L

FP1

FP2 = FP1Wsin

-2.2 Basic equations in uniform open-channel flow

2.2.1 Chezy ’s formula

Consider an open channel of uniform cross-section and bed slope as shown in Fig 2.4:

Fig.2.4 Sloping bed of a channel Let L = length of the channel;

A = cross-sectional area of flow;

V = velocity of water;

P = wetted perimeter of the cross-section;

f = friction coefficient according to Darcy-Weisbach;

and i = uniform slope of the bed

It has been experimentally found, that the total frictional resistance along the length L of the channel, follows the law:

This water “falls” vertically down over a distance V.i in 1 second, so

Loss of potential energy = Weight of water  Height

C depends on the mean velocity V, the hydraulic radius R, the kinematic viscosity  and the relative roughness There is experimental evidence that the value of the resistance coefficient does vary with the shape of the channel and therefore with R and possibly also with the bed slope i, which for uniform flow will be equal to the slope of the energy-head line io, yielding a relationship for the velocity of the form:

where K, x and y are constants

Example 2.1: A rectangular channel is 4 m deep and 6 m wide Find the discharge through the channel, when it runs full Take the slope of the bed as 1:1000 and Chezy’s coefficient

Area of the rectangular channel: A = h  b = 24 m2

Perimeter of the rectangular channel: P = b + 2h = 14 m

Hydraulic radius of the flow: R = A

Discharge through the channel: Q = AC Ri = 49.62 m3s-1 Ans

Example 2.2 : Water is flowing at the rate of 8.5 m3s-1 in an earthen trapezoidal channel with a bed width 9 m, a water depth 1.2 m and side slope 2:1 Calculate the bed slope, if the value of C in Chezy’s formula be 49.5 m1/2

s-1 Solution:

Given: Discharge Q = 8,5 m3/s, Bed width b = 9 m,

Depth h = 1.2 m, Side slope m = 2,

Chezy’s coefficient C = 49.5 m1/2

s-1, Bed slope = ?

Surface width of the trapezoidal channel B = b + 2(h

B

-Area of the trapezoidal channel: A = b B

h2

Table 2.1: Values of Manning coefficient n [m-1/3s]

Example 2.3: An earthen trapezoidal channel with a 3 m wide base and side slopes 1:1 carries water with a depth of 1 m The bed slope is 1/1600 Estimate the discharge Take the value of n in Manning’s formula as 0.04 m-1/3

Surface width of the trapezoidal channel B = b + 2h = 5 m

Area of the trapezoidal channel: A = b B

h2

Example 2.4 : Water at the rate of 0.1 m3/s flows through a vitrified sewer with a diameter

of 1 m with the sewer pipe half full Find the slope of the water surface, if Manning’s n is 0.013 m-1/3s

Area of the flow: A =

-2.2.3 Discussion of factors affecting f and n

The dependence of f on the relative roughness in open channel flow is not as well known as in pipe flow, because it is difficult to assign an equivalent sand-grain roughness

to the large values of the absolute roughness height typically found in open channels The dependence of flow resistance on the cross-sectional shape occurs as a result of changes of both the channel hydraulic radius, R, and the cross-sectional distribution of velocity and shear

There is no substitute for experience in the selection of Manning’s n for natural channels Table 2.2 (at the end of this chapter) from Ven Te Chow (1959) gives an idea of the variability to be expected in Manning‘s n

2.3 MOST ECONOMICAL CROSS-SECTION

2.3.1 Concept

A typical uniform flow problem in the design of an artificial canal is the economical proportioning of the cross-section A canal, having a given Manning coefficient n and a slope i, is to carry a certain discharge Q, and the designer’s aim is to minimize the cross-sectional area Clearly, if A is to be a minimum, the velocity V is to be

a maximum The Chezy and Manning formulas indicate, therefore, that the hydraulic radius R = A/P must be a maximum It can be shown that the problem is equivalent to that

of minimizing P for a given constant value of A This concept has a practical application in estimating the cost for a canal excavation and /or lining

From economic considerations of minimizing the flow cross-sectional area for a given design discharge, a theoretically optimum cross-section will be introduced

2.3.2 Conditions for maximum discharge

The conditions for maximum discharge for the following cross-sections will be dealt with: (a) Rectangular cross-section, and (b) Trapezoidal cross-section

(a) Channel with rectangular cross-section

Consider a channel of rectangular cross-section

as shown in Fig 2.5

Let b = breadth of the channel, and

h = depth of the channel

Differentiating the above equation with respect to h and equating to zero yields:

or b = 2h i.e the breadth is two times the depth (2-26)

In this case, the hydraulic radius is:

Fig 2.6 Experimental relationship between

max

Q

Q and

bh

As can be seen in Fig 2.6, the maximum represented by this optimal configuration is a rather weak one For example, for aspect ratios, b

h, between 1 and 4, the flow rate is within 96% of the maximum flow rate obtained with the same area and by b/h = 2

Example 2.5.: Find the most economical cross-section of a rectangular channel to carry 0.3 m3/s of water, when the bed slope is 1/1000 Assume Chezy’s C = 60 m-1/3

s-1 Solution:

Given: Discharge Q = 0.3 m3/s, Bed slope i = 1/1000, Chezy coefficient C = 60 m-1/3s-1

Breadth of channel b = ? (m) and depth of the channel h = ? (m)

We know that for the most economical rectangular section:

b = 2h

Area: A = b  h = 2h  h = 2h2

and hydraulic radius: R = h/2 = 0.5 h

Using the relation: Q = AC Ri

b

h

A = bh = constant

1.00 0.95

bh

and squaring both sides yields:

0.09 m 7.2h

or h50.0125 h 50.0125 m

(a) Channel with trapezoidal cross-section

Consider a channel of trapezoidal cross-section ABCD as shown in Fig 2.7

Let b = breadth of the channel at the bottom,

h = depth of the channel, and

dh 

We know that: P = b2 n h2 2h2  b 2h n2 1 (2-30) Substituting the value of b from equation (2-28) yields:

2 2

We see that b + 2nh = B is the top width of the channel and 2

h n  is the length of the 1sloping side,

i.e the length of the sloping side is equal to half the top width

In this case, the hydraulic radius:

2) should be used for solving the problems

in the case of channels of trapezoidal cross-section

Example 2.6: A canal of trapezoidal cross-section has to be excavated through hard clay at the least cost Determine the dimensions of the channel for a discharge equal to 14 m3/s, a side slope for hard clay n = 1:1, a bed slope 1:2500 and Manning’s n = 0.02 m-1/3

Using Manning’s formula:

-Note: The semi-circular section (the semi-circle having its center in the surface) is the best

hydraulic section The best hydraulic cross-section for other shapes can be drawn as presented in Fig 2.8

Fig 2.8 Cross-sections of maximum flow rate: i.e “optimum design”

Students should try to proof the conditions for circular and triangular channels for the best hydraulic cross-section based on the below relationship:

 

   

5 2

Optimum wetted perimeter P

Optimum hydraulic radius R

D8

4Trapezoidal 2D

3

2

3D4

4

D2

4

2.3.3 Problems of uniform-flow computation

The computation of uniform flow may be performed by the use of two equations: the continuity equation and a uniform-flow formula When the Manning formula is used as the uniform-flow formula, the computation will involve the following six variables:

(1) the normal discharge Q (4) the mean velocity of flow V

(2) the normal depth h (5) the coefficient of roughness n

(3) the channel slope i (6) the geometric elements that

circular channel rectangular channel trapezoidal

R

D 2

90

R

D 2

60

When any four of the above six variables are given, the remaining two unknowns can be determined by the two equations The following represents some types of problems of uniform flow computation:

-A to compute the normal discharge:

In practical applications, this computation is required for the determination of the capacity of a given channel or for the construction of a sysnthetic rating curve of the channel

B to determine the velocity of the flow:

This computation has many applications For example, it is often required for the study of scouring and silting effects in a given channel

C to compute the normal depth:

This computation is required for the determination of the stage of flow in a given channel

D to determine the channel roughness:

This computation is used to ascertain the roughness coefficient in a given channel; the coefficient thus determined may be used in other similar channels

E to compute the channel slope:

This computation is required for adjusting the slope of a given channel

F to determine the dimensions of the channel section:

This computation is required mainly for design purposes

Table 2.3 lists the known and unknown variables involved in each of the six types of problems mentioned above

Table 2.3: Problems of uniform-flow computation Type of

 The known variables are indicated by the check mark () and the unknown required

in the problems by the question mark (?) The unknown variable(s) that can be determined from the known variables is(are) indicated by a dash (-)

 Table 2.3 does not include all types of problems By varying combinations of various known and unknown variables, more types of problems can be formed In design problems, the use of the best hydraulic section and of empirical rules is generally introduced and thus new types of problems are created

A compound channel consists of a main channel, which carries the base flow (frequently running off up to bank-full conditions), and a floodplain on one or both sides that carries over-bank flow during the time of flooding as sketched in Fig 2.9 The compound cross-section of a channel may be composed of several distinct subsections with each subsection different in roughness form the others

Fig.2.9 Over-bank flow in a compound channel The roughness of the side channels will be different (generally rougher) from that of the main channel and the method of analysis is to consider the total discharge to be the sum of the component discharges computed by the Manning equation The mean velocity for the whole channel section is equal to the total discharge divided by the total water area The classical method of computation of discharge, as presented by Chow in 1959, consisted in subdividing the composite cross-section into sub-areas with vertical interfaces in which the shear stresses are neglected The discharge for each sub-area is calculated by assuming a common friction slope i for the whole channel Thus in the channel, as shown in Fig 2.9, assuming that the bed slope is the same for the three sub-areas, it holds:

2

i i i

on the wetted perimeter of the channel

Note: It has been found by more recent experimentation that this hypothesis is incorrect

and that it leads to a considerable over-estimation of the discharge in the compound channel

Example 2.7: Water flows along a drainage canal having the properties shown in the figure below If the bottom slope i = 1/500=0.002, estimate the discharge

high water level

low water level

Solution:

-We divide the cross-section into three subsections as is indicated in the figure and write the discharge as Q = Q1 + Q2 + Q3, where for each section, it holds:

(2)*

(3)

1.8 2.8 1.8

3.6 3.6 3.6

0.500 0.778 0.500

0.020 0.015 0.020

Note that the imaginary portions of the wetted perimeter between the sections (denoted by the dashed lines in the figure) are not implemented in Pi That is, for section (2):

where neff is the effective value of n for the whole compound channel

With Q = 11.275 m3/s, as determined above, the value of neff is found to be:

As expected, the effective roughness (Manning’s n) is between the minimum (n2 = 0.015

m-1/3s) and maximum (n3 = 0.030 m-1/3s) values for the individual subsections

-2.5 PERMISSIBLE VELOCITY AGAINST EROSION AND SEDIMENTATION

The excavation and lining cost of open channels or conduits varies with their size With respect to water-resources-system economics, erosion of and sedimentation in channels are problems in hydraulic engineering Erosion and sedimentation must be predicted because they can change the bed slope, the channel width and therefore the flow conditions

So, if the available slope permits, the cost of the initial construction may be reduced by using the highest velocity However, if the velocity is becoming too high, the channel may

be damaged or destroyed by erosion This must be avoided by limiting the velocities according to the boundary materials For clear water in hard-surfaced water conductors, the limiting velocity is beyond practical requirements Velocities above 10 m/s for clear water

in concrete channels have been found to do no harm If the water carries abrasive material, damage may occur at lower velocities No definite relation has been established between the nature of abrasive materials, the material of channel bank and bed, and a permissible velocity

In unlined earthen channels, the limiting velocity involves many factors Generally, a fine soil is eroded more easily than a coarse one, but the effect of the grain size may be obscured by the presence or absence of a cementing or binding material The tendency to erode is reduced by seasoning Groundwater conditions can exert an important influence Seepage out of the channel, particularly if the water is turbid, tends to toughen the banks, whereas infiltration reduces the resistance to erosion Erosion can be reduced or avoided by designing for low velocities

If the water carries an appreciable amount of silt in suspension, too low a velocity will cause the canal to fill up until its capacity is impaired It is necessary to choose a velocity that will keep the silt in motion but that will not erode the bank or bottom of the canal The margin of permissible velocities depends on the amount and nature of the silt in the water, the nature of the bank material, the size and shape of the canal, and many other factors The silt content of most turbid water varies with the season, as does also the demand for water and the resultant velocity of the flow

The determination of non-scouring, non-silting velocities for earthen canals has attracted the attention of many investigators over a long period of time, and a considerable mass of data and formulas have been accumulated However, for preliminary purposes, and for design in many cases, use may be made of the approximate values purposed by Fortie and Scobey, in 1926, as shown in Table 2.4 Where the silt is important, it is better to make the slope a little too steep rather than a little too flat A gradient that proves to be too steep can

be controlled by checks In hard-surfaced channels, silting is easily controlled if fall for scouring velocity is available