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 . 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. 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.

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Flow in Open Channels K meteolille.info - Ebook download as PDF File .pdf) , Text File .txt) or read book online. 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.

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At this condition. The energy Eq. Up to the critical depth. Solution Let the suffixes 1 and 2 refer to the upstream and downstream sections respectively as in Fig. Ec2 is less then E2. It will decrease to have a higher specific energy E'1. Calculate the likely change in the water surface. Energy—Depth Relationships 63 point R' to depth at the Section 2. Neglect the energy loss. The minimum specific energy at the Section 2 is greater than E2.

Solution a From Example 2. The upstream depth y1 will increase to a depth y'1. Energy—Depth Relationships 65 b Here. By use of Eq.

Calculate the minimum height of a streamlined.

At the Section 2 the channel width has been constricted to B2 by a smooth transition. In the spe- cific energy diagram Fig. Since there are no losses involved and since the bed elevations at Sections 1 and 2 are same. At Section 1. If B2 is made smaller. At this minimum width. The flow will not. Any further reduction in B2 causes the upstream depth to decrease to y'1 so that E1 rises to E'1. The variation of y1. At Section 2. As the width B2 is decreased. This onset of critical condition at Section 2 is a prerequisite to choking.

It is proposed to reduce the width of the channel at a hydraulic structure. 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.

Solution Let suffixes 1 and 2 denote sections upstream and downstream of the tran- sition respectively. The upstream depth y1 will increase to y'.

At a downstream section the width is reduced to 3. Details about subcritical flow transitions are available in Ref. Many complicated transition situa- tions can be analysed by using the principles of specific energy and critical depth. The transitions in supercritical flow.

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. N and Chiranjeevi. J of Hyd. Civil Engineering.

The upstream depth must rise to create a higher total head. Energy—Depth Relationships 73 Hence the contraction will be working under choked conditions. The upstream depth will therefore rise by 0.

General 2. March P K and Basak.. For Part c use Table 2A. Determine the specific energy and alternate depth. Use the trial and error method. If it is desired to keep the water- surface elevation unaffected by this change. The width beyond a certain section is to be changed to 3. For a critical depth of 0. Energy—Depth Relationships 75 2. Obtain an expression for the rela- tive specific energy at the critical flow.

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:. The hydraulic radius would then be the appropriate length parameter and prediction of friction factor f can be done by using Eqs 3. Studies on non-circular conduits. Eqs 3. Open Channels For purposes of flow resistance which essentially takes place in a thin layer adjacent to the wall.

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. Table 3. Uniform Flow 89 If f is to be calculated by using one of the Eqs 3.

Owing to its simplicity and acceptable degree of accuracy in a variety of practical applications. If Eq. Comparing Eq. This coefficient is essen- tially a function of the nature of boundary surface.

Uniform Flow 91 n2 Since from Eq. 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 fully developed velocity distributions are similar to the logarithmic.

Ganguillet and Kutter Formula 1 0. 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 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. For further details of the velocity distributions Ref.

It is zero at the intersection of the water surface with the boundary and also at the corners in the boundary. The turbulence of the flow and the presence of secondary cur-. Distributions of boundary shear stress by using Preston tube in rectangular. Preston tube5 is a very convenient device for the boundary shear stress measurements in a laboratory channel. Uniform Flow 95 rents in the channel also contribute to the non-uniformity of the shear stress distribution.

Lane9 obtained the shear stress distributions on the sides and bed of trapezoidal and rectangular channels by the use of membrane analogy.

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 Barnes11 and Arcemont and Schnieder14 are very useful in obtaining a first estimate of roughness coefficient in such situations. These act as type values and by comparing the channel under question with a figure and description set that resembles it most.

It should be realized that for open channel flows with hydrodynamically smooth boundaries. Some typical values of n for various normally encountered channel surfaces prepared from information gathered from vari- ous sources A comprehensive list of various types of channels. Estimation of correct n-value of natural channels is of utmost importance in practical problems associated with backwater computations.

These include: Photographs of selected typical reaches of canals. The photographs of man-made and natural channels with corre- sponding values of n given by Chow In the book.

According to Cowan. The roughness coefficient. The selection of a value for n is subjective. The Darcy—weisbach coefficient f used with the Chezy formula is also an equally effec- tive way of representing the resistance in uniform flow.

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. Uniform Flow 97 Table 3. Surface Characteristics Range of n a Lined channels with straight alignment 1 Concerete a formed.

Example 3. The channel is laid on a slope of 0. Solution Case a: Find the hydrodynamic nature of the surface if the channel is made of a very smooth concrete and b rough concrete. Uniform Flow 99 Since the boundary is in the transitional stage. Re is not known to start with and hence a trial and error method has to be adopted.

The most popular form under this type is the Strickler formula: For mixtures of bed materials with considerable coarse-grained sizes. For other types of vegetation. No satisfactory explanation is available for this phenomenon. The procedure is some- times also applied to account for other types of form losses.

Some important factors are: The vegetation on the channel perimeter acts as a flexible roughness element. An interesting feature of the roughness coefficient is observed in some large rivers. Channel irregularities and curvature.

For grass-covered channels. At low velocities and small depths vegetations. 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. Another instance of similar. These relate n to the bed- particle size. The dependence of the value of n on the surface roughness in indicated in Tables 3. All of them are based on some assumptions and are approximately effective to the same degree. 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.

PN are the lengths of these N parts and n1. A large number of formulae. Detailed information on this is available in standard treatises on sediment transport Section The range of variation of n is about 30 per cent.. Consider a channel having its perimeter composed of N types of roughness. One of the commonly used method due to Horton and Einstein is described below.

This equivalent rough- ness. The resistance to flow in alluvial channels is complex owing to the interaction of the flow. Canals in which only the sides are lined. Uniform Flow drains. This formula was independently developed by Horton in and by Einstein in For the sides: In an economic study to remedy excessive seepage from the canal two proposals.

Solution Case a Lining of the sides only Here for the bed: Uniform Flow i. No Investigator ne Concept 1 Horton This list is extracted from Ref. Uniform Flow For a given channel. For example. Figure 3. This depth is called the normal depth. This is also true for any other shape of channel provided that the top width is either constant or increases with 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. The normal depth 0. We shall denote these channels as channels of the first kind. While a majority of the channels belong to the first kind.

Circular and ovoid sewers are typical examples of this category. Channels with a closing top-width can be designated as channels of the second kind.

It may be seen that in some ranges of depth. The channels of the first kind thus have one normal depth only. Problem type Given Required 1 y0. The bed slope is 0. Geometry of the cross section The basic variables in uniform flow situations can be the discharge Q. Uniform Flow As can be seen form Fig. B and m for a trapezoidal channel. Compute the mean velocity and discharge for a depth of flow of 3.

Continuity equation 3. Geometric elements Q and V 2 Q. Types of Problems Uniform flow computation problems are relatively simple. There can be many other derived variables accompanied by corresponding relationships. Problems of the types 4 and 5 usually do not have explicit solutions and as such may involve trial-and-error solutions proce- dures. Geometry Geometric elements Problems of the types 1. From among the above. The available relations are 1.

Geometric elements n 4 Q. Geometric elements y0 5 Q. Geometric elements S0 3 Q. A typical example for each type of problem is given below. If the bed slope is 0. The normal depth is found to be 1. The bottom slope is to be 0. In these channels.

Such channels with large bed-widths as compared to their respective depths are known as wide rectangular chan- nels. This is true for many other channel shapes also. A few aids for computing normal depth in some common channel sections are given below. Considering a unit width of a wide rectangular channel.

Since practically all open channel problems involve normal depth. S0 and B in a rectangular channel. Table 3A. This table will be useful in quick solution of a variety of uniform flow prob- lems in rectangular and trapezoidal channels. Find the normal depth corresponding to discharges of i Use Table 3A. Qn Using this table. In practice. The graphical plot of Eq. The advantage of using Table 2A.

As noted earlier. Find the depth of flow when the discharge is 2. For convenience and ease of identification. For such hard surface lined canals the cross-section recommended by Indian Standards IS: Exposed hard surface lining using materials such as cement concrete.

These stan- dard lined sections have interesting geometrical properties which are beneficial in the solution of some uniform flow problems.

Solution For a standard lined trapezoidal canal section Fig. The side slopes are to be 1. The longitudinal slope of the bed is 1 in If a bed width of Problem 3. Uniform Flow Thus the maximum discharge will be 7. Thus it is seen that for a rectangular channel when the depth of flow is equal to half the bottom width. With the slope. Of all the various possible open channel sections. Hence a channel section having the minimum perimeter for a given area of flow provides the maximum value of the conveyance.

This channel section is also called the best section. The solution would correspondingly change.. Let O be centre of the water surface. OS and OT are perpendiculars drawn to the bed and sides respectively. In the above analysis. Using the above approach. Thus for rectangu- lar shape. Analyse the proportions of an efficient trapezoidal channel section having a side slope of 1.

The use of Kem in calculating parameters of uniform flow in most efficient channel sec- tions is shown in Example 3. Solution For an efficient trapezoidal section having a side slope of m. It is found that the second hydraulic exponent N is essentially constant for a channel over a wide range of depths. Solution For the most efficient trapezoidal section Qn From Table 3.

The longi- tudinal slope of the channel is to be 0. N is usually a slowly varying function of the aspect ratio of the channel.

To determine N for any channel. If N is con- stant between two points K1. In cal- culations involving gradually varied flow. The values of N in this curve have been generated based on the slope of the log K — log y relation using a computer.

For a trapezoidal section. The main channel carries the dry weather flow and during wet season. A compound section is also known as two-stage chan- nel. A majority of natural rivers have compound sections. The velocity of flow in the flood plain is lower than in the main channel due to relative smaller water depth and higher bed roughness. The main channel flow will have interaction with the flow in the flood plains leading to severe momentum exchange at the interface.

The hydraulic conditions of the main channel and the flood plain differ considerably. The flood plains generally have considerably larger and varied roughness elements.

Uniform Flow By equating the exponents of y on both sides. The following salient features are significant: Field observations have indicated that in the overbank flow situation. In one-dimensional analysis. These secondary flows have different directions at different corners and have influence in modifying the boundary shear stress.

Various prominent flow features at the junction of the main and flood bank flows are depicted in this figure. The interactions of the main channel flow and the flood plain flows are indeed very complex. This vortex set is believed to be responsible for momentum exchange between the main and shal- low water flows. Diagonal Interface Method In this method. Main channel is not normally subdivided. Currently DCM is widely used and many well-known software packages.

These include vertical interface. While there is no general agreement to choose a particular method. Various methods for defining the boundaries of the sub-sections are proposed by different researchers leading to a host of proposed methods.

Conveyance is calculated for each sub division by considering vertical interface. If the over bank portion has significant roughness discontinuities equivalent roughness as indicated Sec.

It is known the DCM over estimates the discharge to some extent and due to extreme complexity of the hydraulics of the problem. This interface is considered to be a surface of zero shear stress and as such the length of the diagonal interfaces are not included in the calculation of the wetted perimeters of the over bank and main channel flows.

This inter- face is considered as a surface of zero shear where in no transfer of momentum takes place. Vertical Interface Method In this method the flood banks are separated from the main channel by means of vertical interface. Numerical Methods Computation procedures of solving governing equations by using various turbulence models have been used by various researchers.

Solution The schematic representation of the channel is shown in Fig. Compute the uniform flow dis- charge for a flow with total depth of 4. Exchange Discharge Model EDM This model proposed by Bousmar and Zech focuses on exchange of discharges and momentum transfers through a computation procedures.

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