K Subramanya is a retired Professor of Civil Engineering . This third edition of Flow in Open Channels marks the silver jubilee of the book which first appeared . Flow in Open Channels-K Subrahmanya - Ebook download as PDF File .pdf), Text File .txt) or read book online. Flow in Open Channels-K Subrahmanya. Flow In Open Channels by K lockfollolatu.cf - Ebook download as PDF File . pdf), Text File .txt) or read book online.
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In this Flow in Open Channels By K Subramanya, the scope of the book is defined to provide source material in the form of a Text book that would meet all the. K. Subramanya in the area of HYDRAULIC AND WATER RESOURCES Flow in Open Channels 4th Edition, McGraw Hill Education (India) Private Limited. Flow in Open Channels book. Read reviews from world's largest community for readers.
Channels with Large Slope Figure 1. For simplicity consider a Section 01A2 in which the r direction and Z direction coincide. On the other hand. Replacing the n direction in Eq. Channels of large slopes are encountered rarely in practice except.
With the boundary condition that at point 2 which lies on the free surface.
Thus for channels with large values of the slope. For any normal direction OBC in Fig. Thus it is seen that for a curvilinear flow in a vertical plane. Then for point A.
If the curvature is convex downwards. The extra pressure will be additive if the curvature is convex downwards and subtractive if it is convex upwards. Normal Acceleration In the previous discussion on curvilinear flows. Determine the effective piezometric head for this section. For simple analysis. Assume the velocity at any radial section to be uniform and the depth of flow h to be constant. Solution a Consider the Section At point 2.
For this water surface. The radius of curvature of the free surface is given by d 2h 1 dx 2 d 2h 1. Taking the channel bed as the datum. It should be kept in mind that the area element and the velocity through this area element must be perpendicular to each other. Substituting for K. In open- channel flows. A budgeting of inflows and outflows of a reach is necessary. Thus in a varied flow. In a steady spatially-varied flow. Calculate the discharge per unit width of the gate.
If the top width of the canal at any depth y is T. As a result L of this the water surface 1 2 will start falling. Solution This is a case of unsteady flow and the continuity equation Eq.
The velocity is zero at the boundaries. According to this equation. If the effect of the curvature on the pressure distribution is neglected. Section OAB. The dif- ference of the ordinates between the energy line and free surface represents the velocity head at that section. For prismatic channels.
Discussions on the water-surface profiles are presented in chapter 4 and 5. Thus the energy equation is written as Eq. In designating the total energy by Eq. Introduction 23 In Fig. One can observe that for channels of small slope the piezometric head line essentially coincides with the free surface. In such cases the effective piezometric head hep as defined in Eq.
The slope of the energy line depends on the resistance characteris- tics of the channel and is discussed in Chapter 3. Thus the elevation of energy line on the line 1—1 represents the total energy at any point on the normal section through point 1. At the upstream section. The bed slope is a geometric parame- ter of the channel. Due to energy losses between Sections 1 and 2.
The energy line which is a plot of H vs x is a dropping line in the longitudinal x direction. In general. If the depths a small distance upstream y1 and downstream y2 are 2. Introduction 25 By energy equation. Solution Referring to Fig. F1 and F2. The various forces acting on the control volume in the longitudinal direction are as follows: By the linear-momentum equation in the longitudinal direction for a steady-flow discharge of Q.
In a steady flow the rate of change of momentum in a given direction will be equal to the net flux of momentum in that direction. This equation states that the algebraic sum of all external forces.
The momentum equation com- monly used in most of the open channel flow problems is the linear-momentum equation. The momentum equation is a particularly useful tool in analysing rapidly varied flow RVF situations where energy losses are com- plex and cannot be easily estimated.
Detailed information on the basis of the momentum equation and selection of the control volume are available in books dealing with the mechanics of fluids. It is also very helpful in estimating forces on a fluid mass.
In practical applications of the momentum equation. The frictional force on the bed between Sections 1 and 2 is neglected.
Simplifying Eq. Consider the control volume as shown by dotted lines in the figure. This is equal and opposite to the force exerted by the fluid on the gate. Section 2 is at the vena contracta of the jet where the streamlines are parallel to the bed. The force exerted on the fluid by the gate is F. Section 1 is sufficiently far away from the efflux section and hydrostatic pressure distribution can be assumed.
By the momentum equation. Obtain an expression for the drag force per unit length of the block. For details on the momentum equation in unsteady flow consult References 5 and 6. The momentum equation would then state that in an unsteady flow the algebraic sum of all external forces in a given direction on a fluid mass equals the net change of the linear-momentum flux of the fluid mass in that direction plus the time rate of increase of momentum in that direction within the control volume.
Introduction 29 Example 1. An application of the momentum equation in unsteady flows is given in Chapter Solution Consider a control volume surrounding the block as shown in Fig. A unit width of apron is considered. J W and Harleman. Equation 1. Manual No. Open Channel Hydraulics. Engineering Fluid Mechanics. An application of the spe- cific force relationship to obtain an expression for the depth at the end of a hydraulic jump is given in Section 6.
Fluid Mechanics. New York. American Soc. Sedimentation Engineering. Hydraulics of Sediment Transport. This fact can be advantageously used to solve some flow situations. Blackie and Son.
Fluid Dynamics. In a majority of applications the force F is taken as due to hydrostatic pressure distribution and hence is given by. Handbook of Hydraulics.. Introduction 31 d Breaking of a dam e Flow over a spillway f Sudden opening of a sluice gate g Spreading of irrigation water on a field h Flow in a sewer 1.
The depth of flow is 1. The velocity distribution at a radial section can be considered to be an irrota- tional vortex. Determine the pressure distribution and compare it with the hydrostatic distribution. Take the hydrostatic pressure distribution as the reference. Estimate the pressure at point 1 when the discharge intensity is 5. A certain flow produces a normal acceleration of 0. The depth of the stream is also simultaneously recorded.
Assume the velocity to be uniform in a vertical section. At a section that is 5 m upstream of the brink. The discharge per unit width of the channel is estimated as 4. Estimate the discharge in a river Fig. Estimate the discharge per unit width of the gate.
The velocity measurement at the end section where the flow was curvilinear is indicated in Fig. Estimate the dis- charge per unit width of the channel. The depths of water are h1 and h2 and V1 is the upstream approach velocity. The depths of flow of 1. Calcu- late the elevations of total energy and hydraulic grade lines at normal Sections A and B.
The depths of flow for a discharge of 7. Fill the missing data in the fol- lowing table: Assume the values of the kinetic energy correction factor at A and B as 1. Estimate the horizontal force on the spillway structure. At the normal Section B. At the normal Section A. Sketch the longitudinal sec- tion of the transition. Introduction 35 1. Estimate the energy loss in the transition.
If not. If the discharge per unit width is 1. The bridge pier is 0. If the depth of flow downstream of the bridge is 2. What additional assumptions are required? What is the force on the weir plate? Calculate the force on the sill per unit width. A 15 m wide rectangular canal. The space below the lower nappe is fully ventilated. Considering the control volume shown in the figure.
Assum- ing the flow to be horizontal at Section 1 and the pressure at the brink of Section 2 to be atmospheric throughout the depth.
It can be assumed that the water leaves the brink horizontally at a brink depth of ye. The hydraulic gradient line a coincides with the bed b essentially coincides with the free surface c is above the free surface d is below the free surface 1.
If this value is to be accurate within 2 per cent of the true value. Introduction 39 a 1. With the channel bed as the datum. The kinetic energy correction for this distribution is a greater than zero but less than unity c equal to unity b less than zero d greater than unity 1. If the channel has its width enlarged to 3. The canal is rectangular in cross section and has widths of 2.
Section A is upstream of B. From the data one can infer that a the discharge in the canal is constant in the reach AB. The depth of flow upstream of the contraction is 1. At a section the bed is 1. If the pond has a surface area of 1. Introduction 41 1. It may be noted that while the total energy in a real fluid flow always decreases in the downstream direction. Energy—Depth Relationships 2 2. To simplify the expressions it will be assumed. If the fric- tional resistance of the flow can be neglected.
It may be noticed that the intercept P'R' or P'R represents the velocity head. Energy—Depth Relationships 43 2. For a given Q1 as the specific energy is increased the difference between the two alternate depths increases. It is seen that there are two positive roots for the equation of E indicating that any seen that there are two positive roots for the equation of E indicating that any particular discharge Q1 can be passed in a given channel at two depths and still maintain the same specific energy E.
The condition of minimum specific energy is known as the critical-flow condition and the corresponding depth yc is known as the critical depth. These two possible depths having the same specific energy are know as alternate depths. In Fig. Of the two alternate depths.
CR of the specific- energy curve. The specific-energy diagram can be plotted. In the upper limb CR'. This denotes the subcritical flow region. At the lower limb. We thus get an important result that the critical flow corresponds to the minimum specific energy and at this condition the Froude number of the flow is unity. Discharge as a Variable In the above section the critical-flow condition was derived by keeping the discharge constant. Other channel properties such as the bed slope and roughness do not influence the critical-flow condition for any given Q.
Thus differentiating Eq. This region is called the supercritical flow region. Any specific energy curve of higher Q value i. The dotted line in Fig. Since by Eq. It is seen that for the ordinate PP'. Different Q curves give different intercepts. In this figure. The difference between the alternate depths decreases as the Q value increases. Consider a section PP' in this plot. Example 2. Solution From Eq. Calculate the alternate depths and corre- sponding Froude numbers.
What is the specific energy of the flow? Rectangular Section For a rectangular section. Hence by Eq. By Eq. Energy—Depth Relationships 49 Substituting these in Eq. Empirical relationships for critical depth in circular channels Sl. The non- Fig. This table is very useful in quick solution of problems related to criti- cal depth in trapezoidal channels. Rewriting the right-hand side of Eq. It is found that generally M is a slowly-varying. Energy—Depth Relationships 51 Further the specific energy at critical depth.
Note that the left-hand side of Eq. As a corollary of Eq. B The value of M for a given channel can be determined by preparing a plot of Z vs y on a log-log scale. If M is constant between two points Z1. For a trapezoidal channel. If the specific energy E is kept constant. The above value of M can also be obtained directly by using Eq. Tables 2A. For rectangular and triangular channel sections. The graphical solutions and monographs which were in use some decades back are obsolete now.
Energy—Depth Relationships 55 in problems connected with circular and trapezoidal channels respectively. With the general availability of computers. Examples 2. Determination of yc by empirical equations 1. I V carries a certain discharge. Solution a At critical depth. Solution a Rectangular Section The solution here is straightforward. Energy—Depth Relationships 57 1.
If the critical depth is 1. The minimum depth is reached when the point R coincides with C. At Section 2 Fig. In 1 yc Fig. The principles are neverthe- less equally applicable to channels of any shape and other types of transitions.
Let the flow be subcritical. At this point the hump height will be maximum. To illustrate the various aspects. At this condition. The upstr- yc eam depth has to increase to cause an increase in the specific energy at Sec- E2 Specific tion 1. The energy Eq. Calculate the likely change in the water surface. Ec2 is less then E2. Energy—Depth Relationships 63 point R' to depth at the Section 2.
Up to the critical depth. Neglect the energy loss. It will decrease to have a higher specific energy E'1. Solution Let the suffixes 1 and 2 refer to the upstream and downstream sections respectively as in Fig.
The minimum specific energy at the Section 2 is greater than E2. The upstream depth y1 will increase to a depth y'1.
Solution a From Example 2. Calculate the minimum height of a streamlined. Energy—Depth Relationships 65 b Here. By use of Eq. At Section 1. At the Section 2 the channel width has been constricted to B2 by a smooth transition.
Since there are no losses involved and since the bed elevations at Sections 1 and 2 are same. In the spe- cific energy diagram Fig. The flow will not. At this minimum width. If B2 is made smaller..
As the width B2 is decreased. At Section 2. The variation of y1. Any further reduction in B2 causes the upstream depth to decrease to y'1 so that E1 rises to E'1. Solution Let suffixes 1 and 2 denote sections upstream and downstream of the tran- sition respectively.
This onset of critical condition at Section 2 is a prerequisite to choking. Assuming the transition to be horizontal and the flow to be fric- tionless determine the water surface elevations upstream and downstream of the con- striction when the constricted width is a 2.
It is proposed to reduce the width of the channel at a hydraulic structure. The upstream depth y1 will increase to y'. Many complicated transition situa- tions can be analysed by using the principles of specific energy and critical depth. In subcritical flow transitions the emphasis is essentially to provide smooth and gradual changes in the boundary to prevent flow separation and consequent energy losses.
Details about subcritical flow transitions are available in Ref. The transitions in supercritical flow. At a downstream section the width is reduced to 3.
March P K and Basak. J of Hyd. Energy—Depth Relationships 73 Hence the contraction will be working under choked conditions. N and Chiranjeevi. Civil Engineering. The upstream depth must rise to create a higher total head.
The upstream depth will therefore rise by 0. General 2. If it is desired to keep the water- surface elevation unaffected by this change. Use the trial and error method. Determine the specific energy and alternate depth. For Part c use Table 2A. The width beyond a certain section is to be changed to 3.
Obtain an expression for the rela- tive specific energy at the critical flow. Energy—Depth Relationships 75 2.
For a critical depth of 0. If the depth of flow at a section where the flow is known to be at a critical state is 0. Estimate the discharge and the specific energy.
Show that for these two conditions to occur simultane- ously. Find the diameter of the conduit such that the flow is critical when the conduit is running quarter full. Estimate the value of yc. Critical depth is known to occur at a section in this canal. Estimate the discharge and specific energy corresponding to an observed critical depth of 1. Use Eq. At a section there is a smooth drop of 0. What is the water surface elevation downstream of the drop?
At a certain section it is proposed to build a hump. Calculate the water surface elevations at upstream of the hump and over the hump if the hump height is a 0.
Assume no loss of energy at the hump. A flat hump is to be built at a certain section. Assuming a loss of head equal to the upstream velocity head, compute the minimum height of the hump to provide critical flow.
What will happen a if the height of the hump is higher than the computed value and b if the energy loss is less than the assumed value? A contraction of the channel width is required at a certain section.
Find the greatest allowable contraction in the width for the upstream flow to be possible as specified. A contraction of width is proposed at a section in this canal. Calculate the water surface elevations in the contracted section as well as in an upstream 2. Neglect energy losses in the transition. If the depth in the contracted section is 0. At a certain section of the channel it is proposed to reduce the width to 2. At certain section the width is reduced to 1.
Will the upstream depth be affected and if so, to what extent? If the width is to be reduced to 2. Neglect the loss of energy in transition. What maximum rise in the bed level of the contracted section is possible without affecting the depth of flow upstream of the transition? If at a section there is a smooth upward step of 0. At a certain section the width is reduced to 2. The energy losses in the contraction can be neglected. At a section the channel undergoes transition to a triangular section of side slopes 2 horizon- tal: If the flow in the triangular section is to be critical without changing the upstream water surface, find the location of the vertex of the triangular section relative to the bed of the rectangular channel.
Assume zero energy loss at the transition. The spe- cific energy head in m is a 3. The critical depth in m for this flow is a 2. The Froude number of the flow is a 0. The critical depth in m is a 0. If the depth of flow is 1. The Froude number of flow is a 0. If, after building the hump, it is found that the energy losses in the transition are appreciable, the effect of this hump on the flow will be a to make the flow over the hump subcritical b to make the flow over the hump supercritical c to cause the depth of flow upstream of the hump to raise d to lower the upstream water surface 2.
If the width is expanded at a certain sec- tion, the water surface a at a downstream section will drop b at the downstream section will rise c at the upstream section will rise d at the upstream section will drop 2. If the flow is subcritical throughout, this will cause a a rise in the water surface on the rack b a drop in the water surface over the rack c a jump over the rack d a lowering of the water surface upstream of the rack.
Table 2A. At normal depth,. Uniform Flow 3 3. A flow is said to be uniform if its properties remain constant with respect to distance. As mentioned earlier, the term uniform flow in open channels is understood to mean steady uniform flow. The depth of flow remains constant at all sections in a uniform flow Fig. Considering two Sections 1 and 2, the depths.
Thus in a uniform flow, the depth of flow, area of cross-section and velocity of flow remain constant along the channel. It is obvious, therefore, that uniform flow is possible only in prismatic channels. The trace of the water surface and channel bottom slope are parallel in uniform flow Fig. As such, the slope of the energy line Sf , slope of the water surface Sw and bottom slope S0 will all be equal to each other.
By definition there is no acceleration in uniform flow. By applying the momentum equation to a control volume encompassing Sections 1 and 2, distance L apart, as shown in Fig. Since the flow is uniform,. R is a length parameter accounting for the shape of the channel. It plays a very important role in developing flow equa- tions which are common to all shapes of channels.
Equation 3. The coefficient C is known as the Chezy coefficient. Incompressible, turbulent flow over plates, in pipes and ducts have been extensively studied in the fluid mechanics discipline. From the time of Prandtl — and.
Von Karman — research by numerous eminent investigators has enabled considerable understanding of turbulent flow and associated useful practical applica- tions. The basics of velocity distribution and shear resistance in a turbulent flow are available in any good text on fluid mechanics1,2. Only relevant information necessary for our study in summed up in this section.
Pipe Flow A surface can be termed hydraulically smooth, rough or in transition depending on the relative thickness of the roughness magnitude to the thickness of the laminar sub-layer. The classification is as follows:. For pipe flow, the Darcy—Weisbach equation is. In the transition regime, both the Reynolds number and relative roughness play important roles.
The extensive experimental investigations of pipe flow have yielded the following generally accepted relations for the variation of f in various regimes of flow:.
Studies on non-circular conduits. Open Channels For purposes of flow resistance which essentially takes place in a thin layer adjacent to the wall. Eqs 3. The hydraulic radius would then be the appropriate length parameter and prediction of friction factor f can be done by using Eqs 3. Table 3. Simplified empirical forms of Eqs 3. How- ever. Due to paucity of reliable experimental or field data on channels covering a wide range of parameters. Uniform Flow 89 If f is to be calculated by using one of the Eqs 3.
This coefficient is essen- tially a function of the nature of boundary surface. Owing to its simplicity and acceptable degree of accuracy in a variety of practical applications. If Eq. Comparing Eq.
Uniform Flow 91 n2 Since from Eq. Ganguillet and Kutter Formula 1 0. The fully developed velocity distributions are similar to the logarithmic. This formula appears to be in use in Russia.
Many of these are archaic and are of historic interest only. A few selected ones are listed below: The maximum velocity um occurs essentially at the water surface. Assuming the velocity distribution of Eq. This equation is applicable to both rough and smooth boundaries alike.
For further details of the velocity distributions Ref. For completely rough turbulent flows. The most important feature of the velocity distributions in such channels is the occurrence of velocity-dip. It has been found that k is a universal constant irrespective of the roughness size5. The turbulence of the flow and the presence of secondary cur-. It is zero at the intersection of the water surface with the boundary and also at the corners in the boundary.
Isaacs and Macin- tosh8 report the use of a modified Preston tube to measure shear stress in open channels. A knowledge of the shear stress distribution in a channel is of interest not only in the understanding of the mechanics of flow but also in certain problems involving sediment transport and design of stable channels in non-cohesive material Chapter Preston tube5 is a very convenient device for the boundary shear stress measurements in a laboratory channel.
Distributions of boundary shear stress by using Preston tube in rectangular. Lane9 obtained the shear stress distributions on the sides and bed of trapezoidal and rectangular channels by the use of membrane analogy. Uniform Flow 95 rents in the channel also contribute to the non-uniformity of the shear stress distribution. These include: It should be realized that for open channel flows with hydrodynamically smooth boundaries. In the book.
Estimation of correct n-value of natural channels is of utmost importance in practical problems associated with backwater computations. Barnes11 and Arcemont and Schnieder14 are very useful in obtaining a first estimate of roughness coefficient in such situations.
Cowan15 has developed a procedure to estimate the value of roughness factor n of natural channels in a systematic way by giving weightages to various important fac- tors that affect the roughness coefficient.
A comprehensive list of various types of channels. The roughness coefficient. Some typical values of n for various normally encountered channel surfaces prepared from information gathered from vari- ous sources The selection of a value for n is subjective.
These act as type values and by comparing the channel under question with a figure and description set that resembles it most. According to Cowan. The Darcy—weisbach coefficient f used with the Chezy formula is also an equally effec- tive way of representing the resistance in uniform flow. The photographs of man-made and natural channels with corre- sponding values of n given by Chow Photographs of selected typical reaches of canals.
Uniform Flow 97 Table 3. Surface Characteristics Range of n a Lined channels with straight alignment 1 Concerete a formed.
The channel is laid on a slope of 0. Example 3. Find the hydrodynamic nature of the surface if the channel is made of a very smooth concrete and b rough concrete. Solution Case a: Re is not known to start with and hence a trial and error method has to be adopted.
Uniform Flow 99 Since the boundary is in the transitional stage. These relate n to the bed- particle size. For grass-covered channels. For mixtures of bed materials with considerable coarse-grained sizes.
The procedure is some- times also applied to account for other types of form losses. The most popular form under this type is the Strickler formula: At low velocities and small depths vegetations. Some important factors are: Another instance of similar. The chief among these are the characteristics of the surface. This equation is reported to be useful in predicting n in mountain streams paved with coarse gravel and cobbles.
The type of grass and density of coverage also influence the value of n. The dependence of the value of n on the surface roughness in indicated in Tables 3. Channel irregularities and curvature. An interesting feature of the roughness coefficient is observed in some large rivers. No satisfactory explanation is available for this phenomenon. The vegetation on the channel perimeter acts as a flexible roughness element. For other types of vegetation.
Uniform Flow drains. One of the commonly used method due to Horton and Einstein is described below. The range of variation of n is about 30 per cent.
This equivalent rough- ness. Detailed information on this is available in standard treatises on sediment transport Section For calculating subareas the dividing lines can be vertical lines or bisec- tor of angles at the break in the geometry of the roughness element. Table PN are the lengths of these N parts and n1. All of them are based on some assumptions and are approximately effective to the same degree.
The resistance to flow in alluvial channels is complex owing to the interaction of the flow. A large number of formulae. Canals in which only the sides are lined. Consider a channel having its perimeter composed of N types of roughness. This formula was independently developed by Horton in and by Einstein in In an economic study to remedy excessive seepage from the canal two proposals. No Investigator ne Concept 1 Horton This list is extracted from Ref.
Solution Case a Lining of the sides only Here for the bed: Uniform Flow i. For the sides: This depth is called the normal depth. Thus the normal depth is defined as the depth of flow at which a given discharge flows as uniform flow in a given channel.
Uniform Flow For a given channel. We shall denote these channels as channels of the first kind. Figure 3. For example. This is also true for any other shape of channel provided that the top width is either constant or increases with depth. The normal depth 0. The channels of the first kind thus have one normal depth only.
Channels with a closing top-width can be designated as channels of the second kind. While a majority of the channels belong to the first kind. It may be seen that in some ranges of depth. Circular and ovoid sewers are typical examples of this category. Compute the mean velocity and discharge for a depth of flow of 3. Types of Problems Uniform flow computation problems are relatively simple.
Problems of the types 4 and 5 usually do not have explicit solutions and as such may involve trial-and-error solutions proce- dures. Problem type Given Required 1 y0. From among the above.
Geometric elements S0 3 Q. Geometry Geometric elements Problems of the types 1. Geometry of the cross section The basic variables in uniform flow situations can be the discharge Q.
Geometric elements y0 5 Q. B and m for a trapezoidal channel. Geometric elements n 4 Q. The bed slope is 0. Continuity equation 3. There can be many other derived variables accompanied by corresponding relationships. Geometric elements Q and V 2 Q. Uniform Flow As can be seen form Fig. A typical example for each type of problem is given below. The available relations are 1. If the bed slope is 0. The bottom slope is to be 0. The normal depth is found to be 1. A few aids for computing normal depth in some common channel sections are given below.
Considering a unit width of a wide rectangular channel. Such channels with large bed-widths as compared to their respective depths are known as wide rectangular chan- nels. Since practically all open channel problems involve normal depth.
In these channels. This is true for many other channel shapes also. Table 3A. This table will be useful in quick solution of a variety of uniform flow prob- lems in rectangular and trapezoidal channels. S0 and B in a rectangular channel.
Use Table 3A. Qn Find the normal depth corresponding to discharges of i As noted earlier. The graphical plot of Eq.
In practice. Using this table. Find the depth of flow when the discharge is 2. The advantage of using Table 2A. For such hard surface lined canals the cross-section recommended by Indian Standards IS: These stan- dard lined sections have interesting geometrical properties which are beneficial in the solution of some uniform flow problems.
Exposed hard surface lining using materials such as cement concrete. For convenience and ease of identification. Solution For a standard lined trapezoidal canal section Fig. If a bed width of The longitudinal slope of the bed is 1 in The side slopes are to be 1. Problem 3. The solution would correspondingly change. Uniform Flow Thus the maximum discharge will be 7.
Hence a channel section having the minimum perimeter for a given area of flow provides the maximum value of the conveyance. Of all the various possible open channel sections.. Major deletions from the previous edition for reasons of being of marginal value include Pruning of Tables 2A. Chapters 1 and 2 contain the introduction to the basic principles and energy-depth relationships in open-channel flow.
Various aspects of critical flow, its computation and use in analysis of transitions xii Preface are dealt in detail in Chapter 2. Uniform flow resistance and computations are dealt in great detail in Chapter 3. This chapter also includes several aspects relating to compound channels. Gradually varied flow theory and computations of varied flow profiles are discussed in ample detail in chapters 4 and 5 with sufficient coverage of control points and backwater curve computations in natural channels.
Hydraulic jump phenomenon in channels of different shapes is dealt in substantial detail in Chapter 6. Chapter 7 contains thorough treatment of some important rapidly varied flow situations which include flow-measuring devices, spillways and culverts. Spatially varied flow theory with specific reference to side channel spillways, side weirs and bottom-rack devices is covered in Chapter 8.
A brief description of the transitions in supercritical flows is presented in Chapter 9. An introduction to the important flow situation of unsteady flow in open channels is provided in Chapter The last chapter provides a brief introduction to the hydraulics of mobile bed channels.
The contents of the book, which cover essentially all the important normally accepted basic areas of open-channel flow, are presented in simple, lucid style. A basic knowledge of fluid mechanics is assumed and the mathematics is kept at the minimal level.