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- 1. Chapter 4 Open Channel Flows4.1. IntroductionWhen the surface of flow is open to atmosphere, in other terms when there is onlyatmospheric pressure on the surface, the flow is named as open channel flow. Thegoverning force for the open channel flow is the gravitational force component along thechannel slope. Water flow in rivers and streams are obvious examples of open channelflow in natural channels. Other occurrences of open channel flow are flow in irrigationcanals, sewer systems that flow partially full, storm drains, and street gutters.4.2. Classification of Open Channel FlowsA channel in which the cross-sectional shape and size and also the bottom slope areconstant is termed as a prismatic channel. Most of the man made (artificial) channels areprismatic channels over long stretches. The rectangle, trapezoid, triangle and circle aresome of the commonly used shapes in made channels. All natural channels generallyhave varying cross-sections and consequently are non-prismatic. a) Steady and Unsteady Open Channel Flow: If the flow depth or discharge at a cross-section of an open channel flow is not changing with time, then the flow is steady flow, otherwise it is called as unsteady flow. Flood flows in rivers and rapidly varying surges in canals are some examples of unsteady flows. Unsteady flows are considerably more difficult to analyze than steady flows. b) Uniform and Non-Uniform Open Channel Flow: If the flow depth along the channel is not changing at every cross-section for a taken time, then the flow is uniform flow. If the flow depth changes at every cross-section along the flow direction for a taken time, then it is non-uniform flow. A prismatic channel carrying a certain discharge with a constant velocity is an example of uniform flow. c) Uniform Steady Flow: The flow depth does not change with time at every cross section and at the same time is constant along the flow direction. The depth of flow will be constant along the channel length and hence the free surface will be parallel to the bed. (Figure 4.1). 1 Prof. Dr. Atıl BULU
- 2. y1=y0 y S0 y2=y0 x Figure 4.1Mathematical definition of the Uniform Steady Flow is, y1 = y 2 = y 0 ∂y ∂y (4.1) = 0, =0 ∂t ∂x y0 = Normal depth d) Non-Uniform Steady Flows: The water depth changes along the channel cross- sections but does not change with time at each every cross section with time. A typical example of this kind of flow is the backwater water surface profile at the upstream of a dam. y1 y2 y x Figure 4.2. Non-Uniform steady FlowMathematical definition of the non-uniform steady flow is; ∂y y1 ≠ y 2 → ≠0 Non-Uniform Flow ∂x (4.2) ∂y1 ∂y = 0, 2 = 0 Steady Flow ∂t ∂t 2 Prof. Dr. Atıl BULU
- 3. If the flow depth varies along the channel, these kinds of flows are called as varied flows.If the depth variation is abrupt then the flow is called abrupt varied flow (flow under asluice gate), or if the depth variation is gradual it is called gradually varied flow (flow atthe upstream of a dam). Varied flows can be steady or unsteady.Flood flows and waves are unsteady varied flows since the water depths vary at everycross-section and also at each cross-section it changes with time.If the water depth in a flow varies at every cross-section along the channel but does notvary with time at each cross-section, it is steady varied flow.4.3. Types of FlowThe flow types are determined by relative magnitudes of the governing forces of themotion which are inertia, viscosity, and gravity forces. a) Viscosity Force Effect: Viscosity effect in a fluid flow is examined by Reynolds number. As it was given in Chapter 1, Reynolds number was the ratio of the inertia force to the viscosity force. InertiaForce Re = Vis cos ityForce For the pressured pipe flows, VD Re = < 2000 → Laminar flow υ VD Re = > 2500 → Turbulent Flow υ Since D=4R, Reynolds number can be derived in open channel flows as, VD V 4R VR Re = = = 2000 → Re = < 500 → Laminar Flow (4.3) υ υ υ VD V 4R VR Re = = = 2500 → Re = > 625 → Turbulent Flow (4.4) υ υ υ 500 < Re < 625 → Transition zone 3 Prof. Dr. Atıl BULU
- 4. R is the Hydraulic Radius of the open channel flow cross-section which can be takenas the flow depth y for wide channels.Moody Charts can be used to find out the f friction coefficient by taking D=4R.Universal head loss equation for open channel flows can be derived as, f V2 f V2 hL = × ×L = × ×L D 2g 4R 2 g hL fV 2 S= = L 8 gR 8 gRS f = V2Since Friction Velocity is, τ0 u∗ = = gRS ρ 2 8u∗ f = (4.5) V2b) Gravity Force Effect:The ratio of inertia force to gravity force is Froude Number. InertiaForce Fr = GravityForce V (4.6) Fr = A g L L A=Area y B Figure 4.3. 4 Prof. Dr. Atıl BULU
- 5. Where , A = Wetted area L= Free surface widthFor rectangular channels, A = By L=B (4.7) V V Fr = = By gy g B Fr = 1 → Critical Flow Fr < 1 → Sub Critical (4.8) Fr > 1 → Super Critical Flow4.4. Energy Line Slope for Uniform Open Channel Flows α hL V2/2g Energy Line V2/2g y L Free Water Surface=Hydraulic Grade Line y α z1 Δx z2 Horizontal Datum Figure 4.4 5 Prof. Dr. Atıl BULU
- 6. The head (energy) loss between cross-sections 1 and 2, V12 p1 p 2 V 22 z1 + + = z2 + + + hL γ 2g γ 2g p1 p2 = =y γ γ V1 = V 2 = V h L = z1 − z 2The head loss for unit length of channel length is energy line (hydraulic) slope, hL z1 − z 2 S ener = = = Sinα L LSince in open channel flows the channel slope is generally a small value, α < 5 0 − 10 0 Sinα ≅ Tanα hL Tanα = = S 0 → (channel bottom slope) Δx S ener = S 0 (4.9)Conclusion: Hydraulic grade line coincides with water surface slope in every kind ofopen channel flows. Since the velocity will remain constant in every cross section atuniform flows, energy line slope, hydraulic grade line slope (water surface slope) andchannel bottom slope are equal to each other and will be parallel as well. S = S 0 = S ener (4.10)Where S is the water surface slope.4.5. Pressure DistributionThe intensity of pressure for a liquid at its free surface is equal to that of the surroundingatmosphere. Since the atmospheric pressure is commonly taken as a reference and ofequal to zero, the free surface of the liquid is thus a surface of zero pressure. The pressuredistribution in an open channel flow is governed by the acceleration of gravity g andother accelerations and is given by the Euler’s equation as below: 6 Prof. Dr. Atıl BULU
- 7. In any arbitrary direction s, ∂ ( p + γz ) − = ρa s (4.11) ∂sand in the direction normal to s direction, i.e. in the n direction, ∂ − ( p + γz ) = ρan (4.12) ∂nin which p = pressure, as = acceleration component in the s direction, an = acceleration inthe n direction and z = geometric elevation measured above a datum.Consider the s direction along the streamline and the n direction normal to it. Thedirection of the normal towards the centre of curvature is considered as positive. We areinterested in studying the pressure distribution in the n-direction. The normal accelerationof any streamline at a cross-section is given by, v2 an = (4.13) rwhere v = velocity of flow along the streamline of radius of curvature r.4.5.1. Hydrostatic Pressure DistributionThe normal acceleration an will be zero, 1. if v = 0, i.e. when there is no motion, or 2. if r →∞, i.e. when the streamlines are straight lines.Consider the case of no motion, i.e. the still water case, Fig.( 4.5). From Equ. (4.12),since an = 0, taking n in the z direction and integrating, p + z = constant = C (4.14) γ Figure. 4.4. Pressure distribution in still water 7 Prof. Dr. Atıl BULU
- 8. At the free surface [point 1 in Fig (4.5)] p1/ γ = 0 and z = z1, giving C = z1. At any pointA at a depth y below the free surface, pA = z1 − z A = y γ (4.15) p A = γyThis linear variation of pressure with depth with the constant of proportionality equal tothe specific weight of the liquid is known as hydrostatic pressure distribution.4.5.2. Channels with Small SlopeLet us consider a channel with a very small value of the longitudinal slope θ. Let θ ≈ sinθ≈1/1000. For such channels the vertical section is practically the same as the normalsection. If a flow takes place in this channel with the water surface parallel to the bed, thestreamlines will be straight lines and as such in a vertical direction [section 0 – 1 in Fig.(4.6)] the normal acceleration an = 0. The pressure distribution at the section 0 – 1 will behydrostatic. At any point A at a depth y below the water surface, p p = y and + z = z1 = Elevation of water surface γ γ Figure 4. 6. Pressure distribution in a channel with small slopeThus the piezometric head at any point in the channel will be equal to the water surfaceelevation. The hydraulic grade line will therefore lie (coincide) on the water surface.4.5.3. Channels with Large SlopeFig. (4.7) shows a uniform free surface flow in a channel with a large value of inclinationθ. The flow is uniform, i.e. the water surface is parallel to the bed. An element of lengthΔL is considered at the cross-section 0 – 1. 8 Prof. Dr. Atıl BULU
- 9. Figure 4.7. Pressure distribution in a channel with large slopeAt any point A at a depth y measured normal to the water surface, the weight of columnA11’A’ = γΔLy and acts vertically downwards. The pressure at AA’ supports the normalcomponent of the column A11’A’. Thus, p A ΔL = γyΔL cosθ p A = γy cosθ (4.16) pA = γ cosθ γThe pressure pA varies linearly with the depth y but the constant of proportionalityis γ cosθ . If h = normal depth of flow, the pressure on the bed at point O, p0 = γh cosθ .If d = vertical depth to water surface measured at point O, then h = d cosθ and thepressure head at point O, on the bed is given by, p0 = h cosθ = d cos 2 θ (4.17) γThe piezometric head at any point A, p A = z + y cosθ = z O + h cosθ . Thus for channelswith large values of the slope, the conventionally defined hydraulic gradient line does notlie on the water surface.Channels of large slopes are encountered rather rarely in practice except, typically, inspillways and chutes. On the other hand, most of the canals, streams and rivers withwhich a hydraulic engineer is commonly associated will have slopes (sin θ) smaller than1/100. For such cases cos θ ≈ 1.0. As such, the term cosθ in the expression for thepressure will be omitted assuming that the pressure distribution is hydrostatic. 9 Prof. Dr. Atıl BULU
- 10. 4.5.4. Pressure Distribution in Curvilinear Flows.Figure (4.8) shows a curvilinear flow in a vertical plane on an upward convex surface.For simplicity consider a section 01A2 in which the r direction and Z direction coincide.Replacing the n direction in Equ. (4.12) by (-r) direction, ∂ ⎛p ⎞ a ⎜ + z⎟ = n ⎜γ ⎟ g (4.18) ∂r ⎝ ⎠ Figure 4.8. Convex curvilinear flowLet us assume a simple case in which an = constant. Then the integration of Equ. (4.18)yields, p an +z= r +C (4.19) γ gin which C = constant. With the boundary conditions that at point 2 which lies on thesurface, r = r2 and p/γ = 0 and z = z2. p an C= +z−r γ g a C = 0 + z 2 − n r2 g p a a + z = n r + z 2 − n r2 γ g g p an = (z 2 − z ) − (r2 − r ) (4.20) γ gLet z 2 − z = y = depth of flow the free surface of any point A in the section 01A2. Thenfor point A, 10 Prof. Dr. Atıl BULU
- 11. (r2 − r ) = y = (z 2 − z ) p an = y− y (4.21) γ gEqu. (4.21) shows that the pressure is less than the pressure obtained by the hydrostaticdistribution. Fig. (4.8).For any normal direction OBC in Fig. (4.9), at point C, ( p γ )C = 0 , rC = r2 , and for anypoint at a radial distance r from the origin O, p an = (zC − z ) − (r2 − r ) γ g z C − z = (r2 − r )cosθ p an = (r2 − r )cosθ − (r2 − r ) (4.22) γ gIf the curvature is convex downwards, (i.e. r direction is opposite to z direction)following the argument above, for constant an the pressure at any point A at a depth ybelow the free surface in a vertical section O1A2 [ Fig. (4.9)] can be shown to be, p an = y+ y (4.23) γ gThe pressure distribution in vertical section is as shown in Fig. (4.9) Figure 4.8. Concave curvilinear flowThus it is seen that for a curvilinear flow in a vertical plane, an additional pressure will beimposed on the hydrostatic pressure distribution. The extra pressure will be additive if thecurvature is convex downwards and subtractive if it is convex upwards. 11 Prof. Dr. Atıl BULU
- 12. 4.5.5. Normal AccelerationIn the previous section on curvilinear flows, the normal acceleration an was assumed tobe constant. However, it is known that at any point in a curvilinear flow, an = v2 / r ,where v = velocity and r = radius of curvature of the streamline at that point.In general, one can write v = f (r) and the pressure distribution can then be expressed by, ⎛p ⎞ v2 ⎜ + z ⎟ = ∫ dr + const ⎜γ ⎟ (4.24) ⎝ ⎠ grThis expression can be evaluated if v = f (r) is known. For simple analysis, the followingfunctional forms are used in appropriate circumstances; 1. v = constant = V = mean velocity of flow 2. v = c / r, (free vortex) 3. v = cr, (forced vortex) 4. an = constant = V2/R, where R = radius of curvature at mid depth.Example 4.1: A spillway bucket has a radius of curvature R as shown in the Figure. a) Obtain an expression for the pressure distribution at a radial section of inclination θ to the vertical. Assume the velocity at any radial section to be uniform and the depth of flow h to be constant, b) What is the effective piezometric head for the above pressure distribution?Solution: a) Consider the section 012. Velocity = V = constant across 12. Depth of flow = h. From Equ. (4.24), since the curvature is convex downwards, ⎛p ⎞ v2 ⎜ + z ⎟ = ∫ dr + const ⎜γ ⎟ ⎝ ⎠ gr p V2 +z= Lnr + C (A) γ g 12 Prof. Dr. Atıl BULU
- 13. At point 1, p/γ = 0, z = z1, r = R –h V2 C = z1 − Ln(R − h ) gAt any point A, at radial distance r from 0, p V2 ⎛ r ⎞ = ( z1 − z ) + Ln⎜ ⎟ (B) γ g ⎝ R−h⎠ z1 − z = (r − R + h )cosθ p V2 ⎛ r ⎞ = (r − R + h )cosθ + Ln⎜ ⎟ (C) γ g ⎝ R−h⎠Equ. (C) represents the pressure distribution at any point (r, θ). At point 2, r = R, p = p2. b) Effective piezometric head, hep:From Equ. (B), the piezometric head hp at A is, ⎛p ⎞ V2 ⎛ r ⎞h p = ⎜ + z ⎟ = z1 + ⎜γ ⎟ Ln⎜ ⎟ ⎝ ⎠A g ⎝ R−h⎠Noting that z1 = z 2 + h cosθ and expressing hp in the form of Equ. (4.17),h p = z 2 + h cosθ + ΔhWhere, V2 rΔh = Ln g R−hThe effective piezometric head hep from Equ. (4.17), 1 ⎡V 2 ⎛ r ⎞⎤ Rhep = z 2 + h cosθ + h cosθ ∫ ⎢ g Ln⎜ R − h ⎟⎥ dr R−h ⎣ ⎝ ⎠⎦ V2 ⎡ ⎛ R ⎞⎤hep = z 2 + h cosθ + ⎢− h + RLn⎜ R − h ⎟⎥ gh cosθ ⎣ ⎝ ⎠⎦ 13 Prof. Dr. Atıl BULU
- 14. ⎡ h ⎛ 1 ⎞⎤ V 2 ⎢− + Ln⎜ ⎜ 1 − h R ⎟⎥ ⎟ ⎣ R ⎝ ⎠⎦hep = z 2 + h cosθ + (D) ⎛h⎞ gR⎜ ⎟ cosθ ⎝R⎠It may be noted that when R → ∞ and h/R → 0, hep → z 2 + h cosθ4.6. Uniform Open Channel Flow Velocity Equations Figure 4.10.The fundamental equation for uniform flow may be derived by applying the Impuls-Momentum equation to the control volume ABCD, F = MaThe external forces acting on the control volume are, 1) The forces of static pressure, F1 and F2 acting on the ends of the body, r s F1 = F2 → Uniform flow, flow depths are constant 2) The weight, W, which has a component WSinθ in the direction of motion, r Wx = WSinθ = γALSinθ 3) Change of momentum, r w M 1 = M 2 = ρQV → V1 = V 2 = V 14 Prof. Dr. Atıl BULU
- 15. 4) The force of resistance exerted by the bottom and sides of the channel cross- section, w T = τ 0 PLWhere P is wetted perimeter of the cross-section. r v v w w w M 1 + F1 + Wx − M 2 − F2 − T = 0 γALSinθ = τ 0 PL A τ0 = γ Sinθ P A = R, Sinθ = S 0 P τ 0 = γRS 0 (4.25)Using Equ. (4.5) and friction velocity equation, τ 0 = ρU ∗2 fV 2 U ∗2 = 8 fV 2 ρ = γRS 0 8 8g V2 = × RS 0 f 8g V= × RS 0 (4.26) fIf we define, 8g C= f (4.27) V = C RS 0This cross-sectional mean velocity equation for open channel flows is known as Chezyequation. The dimension of Chezy coefficient C, [V ] = [C ][R]1 2 [S 0 ]1 2 [LT −1 ] = [C ][L1 2 ][F 0 L0T 0 ] [C ] = [L1 2T −1 ] 15 Prof. Dr. Atıl BULU
- 16. C Chezy coefficient has a dimension and there it is not a constant value. When using theChezy equation to calculate the mean velocity, one should be careful since it takesdifferent values for different unit systems.The simplest relation and the most widely used equation for the mean velocity calculationis the Manning equation which has been derived by Robert Manning (1890) by analyzingthe experimental data obtained from his own experiments and from those of others. Hisequation is, 1 12 V = R 2 3S0 (4.28) nWhere n is the Manning’s roughness coefficient.Equating Equs. (4.27) and (4.28), 1 23 12 CR1 2 S 0 2 = 1 R S0 n R1 6 C= (4.29) nEqu. (4.28) was derived from metric data; hence, the unit of length is meter. Although nis often supposed to be a characteristic of channel roughness, it is convenient to considern to be a dimensionless. Then, the values of n are the same in any measurementsystem.Some typical values of n are given in Table. (4.1) Table 4.1. Typical values of the Manning’s roughness coefficient n 16 Prof. Dr. Atıl BULU
- 17. 4.6.1. Determination of Manning’s Roughness CoefficientIn applying the Manning equation, the greatest difficulty lies in the determination of theroughness coefficient, n; there is no exact method of selecting the n value. Selecting avalue of n actually means to estimate the resistance to flow in a given channel, which isreally a matter of intangibles. (Chow, 1959) .To experienced engineers, this means theexercise of engineering judgment and experience; for a new engineer, it can be no morethan a guess and different individuals will obtain different results. 4.6.2. Factors Affecting Manning’s Roughness CoefficientIt is not uncommon for engineers to think of a channel as having a single value of n forall occasions. Actually, the value of n is highly variable and depends on a number offactors. The factors that exert the greatest influence upon the roughness coefficient inboth artificial and natural channels are described below. a) Surface Roughness: The surface roughness is represented by the size and shape of the grains of the material forming the wetted perimeter. This usually considered the only factor in selecting the roughness coefficient, but it is usually just one of the several factors. Generally, fine grains result in a relatively low value of n and coarse grains in a high value of n. b) Vegetation: Vegetation may be regarded as a kind of surface roughness, but it also reduces the capacity of the channel. This effect depends mainly on height, density, and type of vegetation. c) Channel Irregularity: Channel irregularity comprises irregularities in wetted perimeter and variations in cross-section, size, and shape along the channel length. d) Channel Alignment: Smooth curvature with large radius will give a relatively low value of n, whereas sharp curvature with severe meandering will increase n. e) Silting and Scouring: Generally speaking, silting may change a very irregular channel into a comparatively uniform one and decrease n, whereas scouring may do the reverse and increase n. f) Obstruction: The presence of logjams, bridge piers, and the like tends to increase n. g) Size and Shape of the Channel: There is no definite evidence about the size and shape of the channel as an important factor affecting the value of n. h) Stage and Discharge: The n value in most streams decreases with increase in stage and discharge. i) Seasonal Change: Owing to the seasonal growth of aquatic plants, the value of n may change from one season to another season. 17 Prof. Dr. Atıl BULU
- 18. 4.6.3. Cowan MethodTaking into account primary factors affecting the roughness coefficient, Cowan (1956)developed a method for estimating the value of n. The value of n may be computed by, n = (n0 + n1 + n2 + n3 + n4 ) × m (4.30)Where n0 is a basic value for straight, uniform, smooth channel in the natural materialsinvolved, n1 is a value added to n0 to correct for the effect of surface irregularities, n2 is avalue for variations in shape and size of the channel cross-section, n3 is a value ofobstructions, n4 is a value for vegetation and flow conditions, and m is a correction factorfor meandering of channel. These coefficients are given in Table (4.2) depending on thechannel characteristics. (French, 1994).Example 4.2: A trapezoidal channel with width B=4 m, side slope m=2, bed slopeS0=0.0004, and water depth y=1 m is taken. This artificial channel has been excavated insoil. Velocity and discharge values will be estimated with the vegetation cover change inthe channel. For newly excavated channel, n0 value is taken as 0.02 from the Table (4.2).Only n0 and n4 values will be taken in using Equ. (4.30), the effects of the other factorswill be neglected. n = n0 + n4 (A) Table 4.2. Values for the Computation of the Roughness Coefficient 18 Prof. Dr. Atıl BULU
- 19. The calculated velocity and discharge values corresponding to the n values found by Equ.(A) have been given in Table. As can be seen from the Table, as the vegetation coversincreases in the channel perimeter so the velocity and discharge decreases drastically.The discharge for the high vegetation cover is found to be 6 times less than the lowvegetation cover. Vegetation cover also corresponds to the maintenance of the channelcross-section along the channel. High vegetation cover correspond low maintenance ofthe channel. It is obvious that the discharge decrease would be higher if we had taken thevariations in other factors. Table . Velocity and Discharge Variation with n n Velocity (m/sec) Discharge (m3/sec) 0.020 0.79 4.77 0.030 0.53 3.17 0.045 0.35 2.12 .070 0.23 1.36 0.120 0.13 0.79The calculated velocity and discharge values corresponding to the n values found by Equ.(A) have been given in Table. As can be seen from the Table, as the vegetation coversincreases in the channel perimeter so the velocity and discharge decreases drastically.The discharge for the high vegetation cover is found to be 6 times less than the lowvegetation cover. Vegetation cover also corresponds to the maintenance of the channelcross-section along the channel. High vegetation cover correspond low maintenance ofthe channel. It is obvious that the discharge decrease would be higher if we had taken thevariations in other factors. 4.6.4. Empirical Formulae for nMany empirical formulae have been presented for estimating manning’s coefficient n innatural streams. These relate n to the bed-particle size. (Subramanya, 1997). The mostpopular one under this type is the Strickler formula, d 506 1n= (4.31) 21.1Where d50 is in meters and represents the particle size in which 50 per cent of the bedmaterial is finer. For mixtures of bed materials with considerable coarse-grained sizes, d 906 1n= (4.32) 26Where d90 = size in meters in which 90 per cent of the particles are finer than d90. Thisequation is reported to be useful in predicting n in mountain streams paved with coarsegravel and cobbles. 19 Prof. Dr. Atıl BULU
- 20. 4.7. Equivalent RoughnessIn some channels different parts of the channel perimeter may have differentroughnesses. Canals in which only the sides are lined, laboratory flumes with glass wallsand rough beds, rivers with sand bed in deepwater portion and flood plains covered withvegetation, are some typical examples. For such channels it is necessary to determine anequivalent roughness coefficient that can be applied to the entire cross-sectionalperimeter in using the Manning’s formula. This equivalent roughness, also called thecomposite roughness, represents a weighted average value for the roughness coefficient,n. Figure 4.5. Multi-roughness type perimeterConsider a channel having its perimeter composed of N types rough nesses. P1, P2,…., PNare the lengths of these N parts and n1, n2,…….., nN are the respective roughnesscoefficients (Fig. 4.5). Let each part Pi be associated with a partial area Ai such that, N∑Ai −1 i = A1 + A2 + ........ + Ai + ....... + AN = A = Total areaIt is assumed that the mean velocity in each partial area is the mean velocity V for theentire area of flow, V1 = V2 = ...... = Vi = ......... = V N = VBy the Manning’s equation, V1n1 V2 n2 V ni V n Vn S0 2 = 1 23 = 2 3 = ....... = i 2 3 = ........ = N 2 3N = 2 3 (4.33) R1 R2 Ri RN RWhere n = Equivalent roughness.From Equ. (4.33), 23 ⎛ Ai ⎞ ni Pi 2 3 ⎜ ⎟ = ⎝ A⎠ nP 2 3 (4.34) n3 2 P Ai = A i3 2 i n P 20 Prof. Dr. Atıl BULU
- 21. ∑ (n 32 P ) ∑A i = A= A i 32 n P i n= (∑ n 32 i Pi ) 23 (4.35) 23 PThis equation gives a means of estimating the equivalent roughness of a channel havingmultiple roughness types in its perimeters.Example 4.3: An earthen trapezoidal channel (n = 0.025) has a bottom width of 5.0 m,side slopes of 1.5 horizontal: 1 vertical and a uniform flow depth of 1.10 m. In aneconomic study to remedy excessive seepage from the canal two proposals, a) to line thesides only and, b) to line the bed only are considered. If the lining is of smooth concrete(n = 0.012), calculate the equivalent roughness in the above two cases. 1.1m 1 1.5 5mSolution:Case a): Lining on the sides only,For the bed → n1 = 0.025 and P1 = 5.0 m.For the sides → n2 = 0.012 and P2 = 2 × 1.10 × 1 + 1.5 2 = 3.97m P = P1 + P2 = 5.0 + 3.97 = 8.97 mEquivalent roughness, by Equ. (4.36), n= [5.0 × 0.025 1.5 + 3.97 × 0.0121.5 ] 23 = 0.020 8.97 2 3Case b): Lining on the bottom only, P1 = 5.0 m → n1 = 0.012 P2 = 3.97 m → n2 = 0.025 →P = 8.97 m n= (5.0 × 0.012 1.5 + 3.97 × 0.0251.5 ) 23 = 0.018 8.97 2 3 21 Prof. Dr. Atıl BULU
- 22. 4.8. Hydraulic RadiusHydraulic radius plays a prominent role in the equations of open-channel flow andtherefore, the variation of hydraulic radius with depth and width of the channel becomesan important consideration. This is mainly a problem of section geometry.Consider first the variation of hydraulic radius with depth in a rectangular channel ofwidth B. (Fig. 4.11.a). Figure 4.11. A = By, P = B + 2 y A By R= = P B + 2y B R= B +2 y y=0→R=0 For B (4.36) y → ∞, R = 2Therefore the variation of R with y is as shown in Fig (4.11.a). From this comes a usefulengineering approximation: for narrow deep cross-sections R≈ B/2. Since any 22 Prof. Dr. Atıl BULU
- 23. (nonrectangular) section when deep and narrow approaches a rectangle, when a channelis deep and narrow, the hydraulic radius may be taken to be half of mean width forpractical applications.Consider the variation of hydraulic radius with width in a rectangular channel of with aconstant water depth y. (Fig. 4.11.b). By R= B + 2y y R= 2 1+ yB B=0→R=0 (4.37) B → ∞, R → yFrom this it may be concluded that for wide shallow rectangular cross-sections R≈y ; forrectangular sections the approximation is also valid if the section is wide and shallow,here the hydraulic radius approaches the mean depth.4.9. Uniform Flow DepthThe equations that are used in uniform flow calculations are, a) Continuity equation, Q = VA b) Manning velocity equation, 1 23 12 V= R S0 n 1 Q = AV = A R 2 3 S 0 2 1 n The water depth in a channel for a given discharge Q and n= Manning coefficient, S0= Channel slope, B= Channel width, is called as Uniform Water Depth.The basic variables in uniform flow problems can be the discharge Q, velocity of flow V,normal depth y0, roughness coefficient n, channel slope S0 and the geometric elements(e.g. B and side slope m for a trapezoidal channel). There can be many other derivedvariables accompanied by corresponding relationships. From among the above, thefollowing five types of basic problems are recognized. 23 Prof. Dr. Atıl BULU
- 24. Problem Type Given Required 1 y0, n, S0, Geometric elements Q and V 2 Q, y0, n, Geometric elements S0 3 Q, y0, S0, Geometric elements n 4 Q, n, S0, Geometric elements y0 5 Q, y0, n, S0, Geometry Geometric elementsProblems of the types 1, 2 and 3 normally have explicit solutions and hence do notrepresent any difficulty in their calculations. Problems of the types 4 and 5 usually do nothave explicit solutions an as such may involve trial-and-error solution procedures.Example 4.4: Calculate the uniform water depth of an open channel flow to conveyQ=10 m3/sec discharge with manning coefficient n=0.014, channel slope S0=0.0004, andchannel width B=4 m. a) Rectangular cross-section y0=? B=4m A = By 0 = 4 y 0 P = B + 2 y0 A 4 y0 R= = P 4 + 2 y0 1 Q = AV = A R 2 3 S 0 2 1 n 23 1 ⎛ 4 y0 ⎞ 10 = 4 × y 0 × ×⎜ ⎟ × 0.00041 2 0.014 ⎜ 4 + 2 y 0 ⎟ ⎝ ⎠ 23 ⎛ y0 ⎞ 10 × 0.014 ⎜ y0 × ⎜ ⎟ ⎟ = = 0.694 ⎝ 4 + 2 y0 ⎠ 4 5 3 × 0.02 y0 3 5 X= (4 + 2 y 0 )2 3 24 Prof. Dr. Atıl BULU
- 25. y 0 = 2m → X = 0.794 ≠ 0.694 y 0 = 1.90m → X = 0.741 ≠ 0.694 y 0 = 1.80m → X = 0.689 ≠ 0.694 y 0 = 1.81m → X = 0.694 ≅ 0.694The uniform water depth for this rectangular channel is y0 = 1.81 m.There is no implicit solution for calculation of water depths. Trial and error must be usedin calculations. b) Trapezoidal Cross-Section B+4y0 51/2y0 y0 y0 1 2y0 2 B=4m B + B + 4 y0 A= y 0 = (B + 2 y 0 ) y 0 = (4 + 2 y 0 ) y 0 2 P = B + 2 × 5 × y 0 = 4 + 4.472 y 0 A (4 + 2 y 0 ) y 0 R= = P 4 + 4.472 y 0Trial and error method will be used to find the uniform water depth. 1 Q = AV = A R 2 3 S o 2 1 n 23 ⎛ 4 y0 + 2 y0 ⎞ 10 = (4 y 0 + 2 y )× 2 1 2 ×⎜ ⎟ × 0.00041 2 0.014 ⎜ 4 + 4.472 y0 ⎟ o ⎝ ⎠ 10 × 0.014 =7 0.02 X= (4 y + 2 y0 ) 0 2 53 (4 + 4.472 y0 )2 3 25 Prof. Dr. Atıl BULU
- 26. y0 = 1m → X = 4.95 ≠ 7 y0 = 1.10m → X = 5.70 ≠ 7 y0 = 1.20m → X = 6.73 ≠ 7 y 0 = 1.22m → X = 6.94 ≠ 7 y 0 = 1.23m → X = 7.05 ≅ 7The uniform water depth for this trapezoidal channel is y0 =1.23 m.Example 4.5: A triangular channel with an apex angle of 750 carries a flow of 1.20m3/sec at a depth of 0.80 m. If the bed slope is S0 = 0.009, find the roughness coefficientn of the channel. y0 750Solution: y0 = Normal depth = 0.80 mReferring to Figure, 1 ⎛ 75 ⎞Area A = × 0.80 × 2 × 0.80 × tan⎜ ⎟ = 0.491m 2 2 ⎝ 2⎠Wetted perimeter P = 2 × 0.80 × sec 37.5 0 = 2.02m A 0.491 R= = = 0.243m P 2.02 AR 2 3 S 0 2 0.491× 0.243 2 3 × 0.009 0.5 1 n= = = 0.0151 Q 1.204.10. Best Hydraulic Cross-Section a) The best hydraulic (the most efficient) cross-section for a given Q, n, and S0 is the one with a minimum excavation and minimum lining cross-section. A = Amin and P = Pmin. The minimum cross-sectional area and the minimum lining area will reduce construction expenses and therefore that cross-section is economically the most efficient one. Q Q V= → = V max A Amin 26 Prof. Dr. Atıl BULU
- 27. 1 12 23 V= S0 R = C × R 2 3 n V = Vmax → R = Rmax A R= P R = R max → P = Pminb) The best hydraulic cross-section for a given A, n, and S0 is the cross-section that conveys maximum discharge. 1 Q = A R 2 3S0 2 1 n Q = C′× R2 3 C ′ = const Q = Q max → R = R max A R= P R = R max → P = PminThe cross-section with the minimum wetted perimeter is the best hydraulic cross-section within the cross-sections with the same area since lining and maintenanceexpenses will reduce substantially.4.10.1. Rectangular Cross-Sections y B A = By = Constant A B= y A P = B + 2y = + 2y y 27 Prof. Dr. Atıl BULU
- 28. Since P=Pmin for the best hydraulic cross-section, taking derivative of P with respect to y, dA ×y−A dP dy = +2=0 dy y2 A = 2 → A = 2 y 2 = By (4.38) y2 B = 2y The best rectangular hydraulic cross-section for a constant area is the one with B = 2y. The hydraulic radius of this cross-section is, A 2y 2 y R= = = (4.39) P 4y 2 For all best hydraulic cross-sections, the hydraulic radius should always be R = y/2 regardless of their shapes.Example 4.6: Calculate the best hydraulic rectangular cross-section to convey Q=10m3/sec discharge with n= 0.02 and S0= 0.0009 canal characteristics.Solution: For the best rectangular hydraulic cross-section, A = 2y2 y R= 2 1 Q = A R 2 3S0 2 1 n 23 1 ⎛ y⎞ 10 = 2 y ×2 ×⎜ ⎟ × 0.00091 2 0.02 ⎝ 2 ⎠ 10 × 0.02 × 2 2 3 y8 3 = = 5.29 2 × 0.03 y = 5.29 3 8 = 1.87 m B = 2 × y = 2 × 1.87 = 3.74m y = 1.87mImplicit solutions are possible to calculate the water depths for the best hydraulic cross-sections. 28 Prof. Dr. Atıl BULU
- 29. 4.10.2. Trapezoidal Cross-Sections L my B my 1 y y 1+ m2 m B Figure 4.12. B + B + 2my A= × y = (B + my ) y y P = B + 2 y 1+ m2 A B = − my y A P = − my + 2 y 1 + m 2 yAs can be seen from equation, wetted perimeter is a function of side slope m and waterdepth y of the cross-section. a) For a given side slope m, what will be the water depth y for best hydraulic trapezoidal cross-section? For a given area A, P = Pmin , dP dA = 0, =0 dy dy dA y−A dP dy = 2 − m + 2 1 + m2 = 0 dy y A 2 = −m + 2 1 + m 2 y 29 Prof. Dr. Atıl BULU
- 30. ( A = 2 1+ m 2 − m y 2 ) (4.40) ( ) P = 2 1 + m 2 − m y − my + 2 y 1 + m 2 P = 2 y 1 + m 2 − my − my + 2 y 1 + m 2 ( P = 2 y 2 1+ m 2 − m ) (4.41)The hydraulic radius R, channel bottom width B, and free surface width L may befound as, ( A 2 1+ m2 − m y 2 ) R= = P 2 y 2 1+ m2 − m ( ) y R= 2 A ( B = − my = 2 1 + m 2 − m y − my y ) ( B = 2 y 1+ m2 − m ) (4.42) L = B + 2my ( L = 2 y 1 + m 2 − m + 2my ) L = 2 y 1+ m 2 (4.43)b) For a given water depth y, what will be the side slope m for best hydraulic trapezoidal cross-section? P = Pmin dP dA = 0, =0 dm dm A P= − my + 2 y 1 + m 2 y dA y−0 dP dm 2m = 2 − y + 2y =0 dm y 2 1 + m2 2my 2m y= → =1 1+ m 2 1 + m2 4m 2 = 1 + m 2 → 3m 2 = 1 30 Prof. Dr. Atıl BULU
- 31. 1 m= 3 (4.44) 1 Tanα = = 3 → α = 60 0 m ( ) A = 2 1+ m 2 − m y 2 ⎛ 1 1 ⎞ 2 A = ⎜ 2 1+ − ⎜ ⎟y ⎝ 3 3⎟ ⎠ ⎛ 2 4 −1 ⎞ 2 3 2 A=⎜ ⎜ ⎟y = ⎟ y ⎝ 3 ⎠ 3 A = 3y 2 (4.45) ( P = 2 y 2 1+ m 2 − m ) ⎛ 1 1 ⎞ P = 2 y⎜ 2 1 + − ⎜ ⎟ ⎝ 3 3⎟ ⎠ ⎛ 4 −1 ⎞ 6 y P = 2 y⎜ ⎜ ⎟= ⎟ ⎝ 3 ⎠ 3 P = 2y 3 (4.46) A 3y 2 1 3 −1 B = − my = − y= y y y 3 3 2 3 P B= y= (4.47) 3 3 A 3y 2 y R= = = P 2y 3 2The channel bottom width is equal one third of the wetted perimeter and therefore sidesand channel width B are equal to each other at the best trapezoidal hydraulic cross-section. Since α = 600, the cross-section is half of the hexagon. 31 Prof. Dr. Atıl BULU
- 32. Example 4.7: Design the trapezoidal channel as best hydraulic cross-section with Q= 10m3/sec, n= 0.014, S0= 0.0004, and m= 3/2. 1.5y B 1.5y y 2 3 B Solution: B + B + 3y A= × y = ( B + 1 .5 y ) y 2 9 P = B + 2 y 1 + = B + 2 y 3.25 = B + 3.61 y 4 A B= − 1 .5 y y A A P = − 1.5 y + 3.61 y = + 2.11 y y y dA y−A dP dy = + 2.11 = 0 dy y2 A = 2.11y 2 B = 2.11 y − 1.5 y = 0.61 y P = 0.61 y + 3.61 y = 4.22 y A 2.11 y 2 y R= = = P 4.22 y 2 32 Prof. Dr. Atıl BULU
- 33. 1 Q = A R 2 3S0 2 1 n 23 1 ⎛ y⎞ 10 = 2.11 y × 2 ×⎜ ⎟ × 0.00041 2 0.014 ⎝ 2 ⎠ 10 × 0.014 × 2 2 3 y8 3 = = 5.266 2.11× 0.02 y = 5.266 3 8 = 1.86m A = 2.11y 2 = 2.11× 1.86 2 = 7.30m 2 B = 0.61y = 0.61× 1.86 = 1.13m L = B + 3 y = 1.13 + 3 ×1.86 = 6.71m P = 4.22 y = 4.22 ×1.86 = 7.85m A 7.30 y 1.86 R= = = 0.93m → R = = = 0.93m P 7.85 2 2Example 4.8: A slightly rough brick-lined trapezoidal channel (n = 0.017) carrying adischarge of Q = 25 m3/sec is to have a longitudinal slope of S0 = 0.0004. Analyze theproportions of,a) An efficient trapezoidal channel section having a side of 1.5 horizontal: 1 vertical,b) the most efficient-channel section of trapezoidal shape.Solution:Case a): m = 1.5For an efficient trapezoidal channel section, by Equ. (4.40), ( ) A = 2 1+ m2 − m y 2 A = (2 1 + 1.5 − 1.5)y 2 2 = 2.106 y 2 y R= , Q = 25 m 3 sec 2 23 1 ⎛ y⎞ 25 = × 2.106 y × ⎜ ⎟ × 0.00041 2 2 0.017 ⎝2⎠ y = 2.83m 33 Prof. Dr. Atıl BULU
- 34. ( B = 2 y 1+ m2 − m ) ( ) B = 2 × 2.83 × 1 + 1.5 2 − 1.5 = 1.72mCase b): For the most-efficient trapezoidal channel section, 1 m= = 0.577 3 A = 3 y 2 = 1.732 y 2 y R= 2 23 1 ⎛ y⎞ 25 = × 1.732 y 2 × ⎜ ⎟ × 0.0004 0.5 0.017 ⎝2⎠ y = 3.05m 2 B= × 3.05 = 3.52m 34.10.3. Half a Circular Conduit r y=r Figure 4.13.Wetted area A, and wetted perimeter of the half a circular conduit, πr 2 A= 2 P = πr (4.48) πr 2 A r y R= = 2 = = P πr 2 2Half a circular conduit itself is a best hydraulic cross-section. 34 Prof. Dr. Atıl BULU
- 35. Example 4.9: a) What are the best dimensions y and B for a rectangular brick channel designed to carry 5 m3/sec of water in uniform flow with S0 = 0.001, and n = 0.015? b) Compare results with a half-hexagon and semi circle.Solution: a) For best rectangular cross-sections, the dimensions are, y A = 2y2 , R= 2 1 23 12 Q = A R S0 n 23 1 ⎛ y⎞ 5 = 2y × 2 ×⎜ ⎟ × 0.0010.5 0.015 ⎝ 2 ⎠ 5 × 0.015 × 2 2 3 y8 3 = = 1.882 2 × 0.0010.5 y = 1.27mThe proper area and width of the best rectangular cross-section are, A = 2 y 2 = 2 × 1.27 2 = 3.23m 2 A 3.23 B = = 2y = = 2.54m y 1.27 y=1.27 B=2.54m b) It is constructive to see what discharge a half-hexagon and semi circle would carry for the same area of A = 3.23 m2. L=B+2my y 1 600 m B 35 Prof. Dr. Atıl BULU
- 36. 1 m = cot α = cot 60 0 = = 0.577 3 A= (B + B + 2my ) y = (B + my ) y 2 (A) A = (B + 0.577 y ) y P = B + 2 y 1+ m2 (B) P = B + 2 y 1 + 0.577 2 = B + 2.31yUsing Equs. (A) and (B), AB= − 0.577 y y AP = − 0.577 y + 2.31 y y 3.23P= + 1.733 y ydP 3.23 = − 2 + 1.733 = 0dy y 3.23y2 = → y = 1.37 m 1.733 3.23B= − 0.577 × 1.37 = 1.57m 1.37P = 1.57 + 2.31× 1.37 = 4.73m A 3.23 y 1.37R= = = 0.68m → R = = ≅ 0.68m P 4.73 2 2The discharge conveyed in this half a hexagon cross-section is, 1Q = AV = 3.23 × × 0.68 2 3 × 0.0010.5 = 5.27 m 3 sec 0.0155.27 − 5.00 = 0.054 5.00Half of a hexagon cross-section with the same area of the rectangular cross-section willconvey 5.4 per cent more discharge. 36 Prof. Dr. Atıl BULU
- 37. For a semicircle cross-section, πD 2A = 3.23 = 8 8 × 3.23D2 = → D = 2.87 m 3.14 πD 3.14 × 2.87P= = = 4.51m 2 2 A 3.23 DR= = = 0.716m = P 4.51 4 1Q = 3.23 × × 0.716 2 3 × 0.0010.5 = 5.45 m 3 sec 0.0155.45 − 5.00 = 0.09 5.005.45 − 5.27 = 0.03 5.27Semicircle carries more discharge comparing with the rectangular and trapezoidal cross-sections by 9 and 3 per cent respectively. Table 4.3. Proportions of some most efficient sections Channel Shape A P B R L Rectangle (half square) 2y2 4y 2y y 2y 2 Trapezoidal 3y 2 3y 2 y 4 (half regular hexagon) y y 3 2 3 Circular (semicircle) π y y2 πy - 2y 2 2 Triangle y2 2 3y - y 2y (vertex angle = 900) 2 2 (A = Area, P = Wetted perimeter, B = Base width, R = Hydraulic radius, L = Water surface width) 37 Prof. Dr. Atıl BULU
- 38. 4.10.4. Circular Conduit with a Free Surface O r θ B y Figure 4.14.A circular pipe with radius r is conveying water with depth y. The angle of the free watersurface with the center of the circle is θ. Wetted area and wetted perimeter of the flow is, 360 0 = 2πradian → A = πr 2 → P = 2πrFor θ radian, the area and the perimeter of the sector are, πr 2 r 2θ A =θ = (4.49) 2π 2 2πr PAB = θ = rθ (4.50) 2πWetted area of the flow with the flow depth y is, r 2θ A= − AOAB (4.51) 2 θ/2 r r y x 38 Prof. Dr. Atıl BULU
- 39. x θ θ Sin = → x = rSin 2 r 2 θ y θ Cos = → y = rCos 2 r 2 θ θ 2rSin × rCos 2 = r × 2 Sin θ Cos θ 2 AOAB = 2 2 2 2 2 r 2 2θ r 2 AOAB = × Sin = Sinθ 2 2 2Substituting AAOB into the Equ. (4.51) give the wetted area of the flow, r 2θ r 2 A= − Sinθ (4.52) 2 2 a) What will be the θ angle for the maximum discharge? 23 1 ⎛ A⎞ 1 1 A5 3 Q = A ⎜ ⎟ S0 2 = S0 2 2 3 1 n⎝ P⎠ n P 1 1 C ′ = S 0 2 = cons n A5 3 Q = C′ 2 3 P Derivative of the discharge equation will be taken with respect to θ. 5 2 3 dA 2 3 2 −1 3 dP 5 3 A P − P A dQ 3 dθ 3 dθ = ×C′ = 0 dθ P4 3 5 dA 2 dP A 5 3 × A2 3 × × P2 3 = × × 3 dθ 3 dθ P 1 3 2 1 5 2 dA + dP − 5× × P3 3 = 2× × A3 3 dθ dθ dA dP 5 P=2 A (4.53) dθ dθ 39 Prof. Dr. Atıl BULU
- 40. Derivatives of the wetted area and the wetted perimeter are, r2 A= (θ − Sinθ ) 2 dA r 2 = (1 − Cosθ ) dθ 2 P = rθ dP =r dθ Substituting the derivates to the Equ. (4.53), r2 2 5 (1 − Cosθ )rθ = 2r r (θ − Sinθ ) 2 2 5θ (1 − Cosθ ) = 2(θ − Sinθ ) 5θ − 5θCosθ − 2θ + 2Sinθ = 0 3θ − 5θCosθ + 2 Sinθ = 0 (4.54)Solution of the Equ. (4.54) give the θ angle as, θ = 302 0 30 ′ = 5.28radianThe wetted area, wetted perimeter, and hydraulic radius of the flow are, r2 A= (θ − Sinθ ) 2 A= r2 2 ( 5.28 − Sin302.5 0 = r2 2 )(5.28 + 0.84) A = 3.06r 2 P = rθ = 5.28r A 3.06r 2 R= = = 0.58r P 5.28r 40 Prof. Dr. Atıl BULU
- 41. The water depth in the circular conduit is, θ ⎛ 302.5 0 ⎞ y = r − rCos 2 ⎜ 2 ⎟ = r (1 + 0.88) = r − rCos⎜ ⎟ ⎝ ⎠ y = 1.88r = 0.94 D 1 Q = A R 2 3S0 2 1 n 1 Qmax = 3.06r 2 × × (0.58r ) S 0 2 23 1 n 1 1 Qmax = 2.13r 8 3 S 0 2 n b) What will be the θ angle for the maximum velocity? 1 23 12 V= R S0 n A Vmax → R = → max P dA × P − dP × A dR = =0 P2 dA dP = A P 2 r (1 − cosθ ) r 2 = r2 (θ − sin θ ) rθ 2 (1 − cosθ )θ = θ − sin θ θ − θ cosθ = θ − sin θ sin θ θ= = tan θ cosθ θ = 257 0 30′ = 2.49radian θ ⎛ 257.5 0 ⎞ ⎜ y = r − r cos = r ⎜1 − cos ⎟ = 1.626r = 0.813D ⎟ 2 ⎝ 2 ⎠ 41 Prof. Dr. Atıl BULU
- 42. Figure 4.15.4.11. Compound SectionsSome channel sections may be formed as a combination of elementary sections.Typically natural channels, such as rivers, have flood plains which are wide and shallowcompared to the main channel. Fig. (4.16) represents a simplified section of a stream withflood banks.Consider the compound section to be divided into subsections by arbitrary lines. Thesecan be extensions of the deep channel boundaries as in Fig. (4.16). Assuming thelongitudinal slope to be same for all subsections, it is easy to see that the subsections willhave different mean velocities depending upon the depth and roughness of theboundaries. Generally, overbanks have larger size roughness than the deeper mainchannel. If the mean velocities Vi in the various subsections are known then the totaldischarge is ∑ViAi. Figure 4.16. Compound sectionIf the depth of flow is confined to the deep channel only (y < h), calculation of dischargeby using Manning’s equation is very simple. However, when the flow spills over theflood plain (y > h), the problem of discharge calculation is complicated as the calculationmay give a smaller hydraulic radius for the whole stream section and hence the dischargemay be underestimated. The following method of discharge estimation can be used. Inthis method, while calculating the wetted perimeter for the sub-areas, the imaginarydivisions (FJ and CK in the Figure) are considered as boundaries for the deeper portiononly and neglected completely in the calculation relating to the shallower portion. 42 Prof. Dr. Atıl BULU
- 43. 1. The discharge is calculated as the sum of the partial discharges in the sub-areas; for e.g. units 1, 2 and 3 in Fig. (4.16) Q p = ∑ Qi = ∑Vi Ai (4.55) 2. The discharge is also calculated by considering the whole section as one unit, (ABCDEFGH area in Fig.4.16), say Qw. 3. The larger of the above discharges, Qp and Qw, is adopted as the discharge at the depth y.Example 4.10: For the compound channel shown in the Figure, determine the dischargefor a depth of flow 1.20 m. n = 0.02, S0 = 0.0002.Solution:a): y = 1.20 mPartial area discharge; Sub-area 1 and 3: A1 = 7.0 × 0.3 = 2.1m 2 P1 = 0.3 + 7.0 = 7.3m A1 2.1 R1 = = = 0.288m P1 7.3 1 Q p1 = × 2.1 × 0.288 2 3 × 0.0002 0.5 0.02 Q p1 = 0.648 m 3 sec = Q p 3 43 Prof. Dr. Atıl BULU
- 44. Sub-area 2: A2 = 3.0 × 1.2 = 3.6m 2 P2 = 3.0 + 0.9 + 0.9 = 4.8m A2 3.6 R2 = = = 0.75m P2 4.8 1 Qp2 = × 3.6 × 0.75 2 3 × 0.0002 0.5 = 2.10 m 3 sec 0.02Total discharge, Q p = Q p1 + Q p 2 + Q p 3 Q p = 0.648 + 2.10 + 0.6.48 = 3.396 m 3 secb): By the total-section method: A = 2 .1 + 2 .1 + 3 .6 = 7 .8 m 2 P = 0.3 + 7.0 + 0.9 + 3.0 + 0.9 + 7.0 + 0.3 = 19.4m A 7 .8 R= = = 0.402 P 19.4 1 Qw = × 7.8 × 0.402 2 3 × 0.0002 0.5 = 3.00 m 3 sec 0.02Since Qp > Qw, the discharge in the channel is, Q = Q p = 3.396 m 3 sec4.10 Design of Irrigation ChannelsFor a uniform flow in a canal, 1 Q= AR 2 3 S 0 .5 0 nWhere A and R are in general, functions of the geometric elements of the canal. If thecanal is of trapezoidal cross-section, Q = f (n, y 0 , S 0 , B, m ) (4.56)Equ. (4.56) has six variables out of which one is a dependent variable and the rest five areindependent ones. Similarly, for other channel shapes, the number of variables dependsupon the channel geometry. In a channel design problem, the independent variables areknown either explicitly or implicitly, or as inequalities, mostly in terms of empiricalrelationships. The canal-design practice given below is meant only for rigid-boundarychannels, i.e. for lined an unlined non-erodible channels. 44 Prof. Dr. Atıl BULU
- 45. 4.12.1. Canal SectionNormally a trapezoidal section is adopted. Rectangular cross-sections are also used inspecial situations, such as in rock cuts, steep chutes and in cross-drainage works.The side slope, expressed as m horizontal: 1 vertical, depends on the type pf canal, i.e.lined or unlined, nature and type of soil through which the canal is laid. The slopes aredesigned to withstand seepage forces under critical conditions, such as; 1. A canal running full with banks saturated due to rainfall, 2. The sudden drawdown of canal supply.Usually the slopes are steeper in cutting than in filling. For lined canals, the slopesroughly correspond to the angle of repose of the natural soil and the values of m rangefrom 1.0 to 1.5 and rarely up to 2.0. The slopes recommended for unlined canals incutting are given in Table (4.4). Table 4.3. Side slopes for unlined canals in cutting Type of soil m Very light loose sand to average sandy soil 1.5 – 2.0 Sandy loam, black cotton soil 1.0 – 1.5 Sandy to gravel soil 1.0 – 2.0 Murom, hard soil 0.75 – 1.5 Rock 0.25 – 0.5 4.12.2. Longitudinal SlopeThe longitudinal slope is fixed on the basis of topography to command as much area aspossible with the limiting velocities acting as constraints. Usually the slopes are of theorder of 0.0001. For lined canals a velocity of about 2 m/sec is usually recommended. 4.12.3. Roughness coefficient nProcedures for selecting n are discussed in Section (4.6.1). Values of n can be taken fromTable (4.2). 4.12.4. Permissible VelocitiesSince the cost for a given length of canal depends upon its size, if the available slopepermits, it is economical to use highest safe velocities. High velocities may cause scourand erosion of the boundaries. As such, in unlined channels the maximum permissiblevelocities refer to the velocities that can be safely allowed in the channel without causingscour or erosion of the channel material. 45 Prof. Dr. Atıl BULU
- 46. In lined canals, where the material of lining can withstand very high velocities, themaximum permissible velocity is determined by the stability and durability of the liningand also on the erosive action of any abrasive material that may be carried in the stream.The permissible maximum velocities normally adopted for a few soil types and liningmaterials are given in Table (4.5). Table 4.4. Permissible Maximum velocities Nature of boundary Permissible maximum velocity (m/sec) Sandy soil 0.30 – 0.60 Black cotton soil 0.60 – 0.90 Hard soil 0.90 – 1.10 Firm clay and loam 0.90 – 1.15 Gravel 1.20 Disintegrated rock 1.50 Hard rock 4.00 Brick masonry with cement pointing 2.50 Brick masonry with cement plaster 4.00 Concrete 6.00 Steel lining 10.00In addition to the maximum velocities, a minimum velocity in the channel is also animportant constraint in the canal design. Too low velocity would cause deposition ofsuspended material, like silt, which cannot only impair the carrying capacity but alsoincrease the maintenance costs. Also, in unlined canals, too low a velocity may encourageweed growth. The minimum velocity in irrigation channels is of the order of 0.30 m/sec. 4.12.5. Free BoardFree board for lined canals is the vertical distance between the full supply level to the topof lining (Fig. 4.17). For unlined canals, it is the vertical distance from the full supplylevel to the top of the bank. Figure 4.10. Typical cross-section of a lined canal 46 Prof. Dr. Atıl BULU
- 47. This distance should be sufficient to prevent overtopping of the canal lining or banks dueto waves. The amount of free board provided depends on the canal size, location, velocityand depth of flow. Table (4.5) gives free board heights with respect to the maximumdischarge of the canal. Table 4.6. Discharge Free board (m) (m3/sec) Unlined Lined Q < 10.0 0.50 0.60 Q ≥ 10.0 0.75 0.75 4.10.6. Width to Depth RatioThe relationship between width and depth varies widely depending upon the designpractice. If the hydraulically most-efficient channel cross-section is adopted, 1 2y B m= → B = 0 = 1.155 y0 → = 1.155 3 3 y0If any other value of m is use, the corresponding value of B/y0 for the efficient sectionwould be from Equ. (4.18), B y0 ( = 2 2 1+ m2 − m ) (4.57)In large channels it is necessary to limit the depth to avoid dangers of bank failure.Usually depths higher than about 4.0 m are applied only when it is absolutely necessary.For selection of width and depth, the usual procedure is to adopt a recommended valueExample 4.11: A trapezoidal channel is to carry a discharge of 40 m3/sec. The maximumslope that can be used is 0.0004. The soil is hard. Design the channel as, a) a lined canalwith concrete lining, b) an unlined non-erodible channel.Solution: a) Lined canalChoose side slope of 1 : 1, i.e., m = 1.0 (from Table 4.4)n for concrete, n = 0.013 (from Table 4.2) 47 Prof. Dr. Atıl BULU
- 48. From Equ. (4.57), B y0 ( = 2 2 1+ m2 − m ) ( − 1) B = 2 × 2 1 + 12 y0 B = 3.67 y0Since, A B A = (B + my0 ) y0 → = + m = 3.67 + 1 = 4.67 y0 y0 y0For the most-efficient hydraulic cross-sections, R = = 0.5 y 0 2 1 Q= AR 2 3 S 0 .5 0 n 1 × 4.67 y0 × (0.5 y 0 ) × 0.0004 0.5 23 40 = 0.013 y0 = 8.838 → y0 = 3.70m 53 A = 4.67 y 0 = 4.67 × 3.70 = 17.28m 2 Q 40 V= = = 2.31 m sec A 17.28 B = 3.67 y 0 = 3.67 × 3.70 = 13.58mThis velocity value is greater than the minimum velocity of 0.30 m/sec, and further is lessthan the maximum permissible velocity of 6.0 m/sec for concrete. Hence the selection ofB and y0 are all right. The recommended geometric parameters of the canal are therefore: B = 13.58 m, m = 1.0, S0 = 0.0004Adopt a free board of 0.75 m. The normal depth for n = 0.013 will be 3.70 m. b) Lined canalFrom Table (4.), a side slope of 1 : 1 is chosen. From Table (4.2), take n for hard soilsurface as n = 0.020. 48 Prof. Dr. Atıl BULU
- 49. From Equ. (4.57), B y0 ( = 2 2 1+ m2 − m ) ( − 1) B = 2 × 2 1 + 12 y0 B = 3.67 y0Since, A B A = (B + my0 ) y 0 → = + m = 3.67 + 1 = 4.67 y0 y0 y0For the most-efficient hydraulic cross-sections, R = = 0.5 y 0 2 1 × 4.67 y 0 × (0.5 y0 ) × 0.0004 0.5 23 40 = 0.020 y0 3 5 = 13.60 → y 0 = 4.79mSince y0 = 4.79 m > 4.0 m, y0 = 4.0 m is chosen, B = 3.67 → B = 3.67 × 4.0 = 14.68m y0 A = 4.67 × 4 = 18.68m 2 Q 40 V= = = 2.14 m sec A 18.68But this velocity is larger than the permissible velocity of 0.90 – 1.10 m/sec for hard soil(Table 4.5). In this case, therefore the maximum permissible velocity will control thechannel dimensions.Adopt V = 1.10 m/sec, 40 A= = 36.36m 2 1.10 y 0 = 4 .0 49 Prof. Dr. Atıl BULU
- 50. A = (B + my 0 ) y 0 A 36.36 B= − my 0 = − 1× 4.0 = 5.09m y0 4 .0 P = B + 2 m 2 + 1y0 P = 5.09 + 2 1 + 1 × 4.0 = 16.40m 23 1 ⎛ 36.36 ⎞ Q= × 36.36 × ⎜ ⎟ × 0.0004 0.5 0.020 ⎝ 16.40 ⎠ Q = 61.82 m 3 sec〉 40 m 3 secThe slope of the channel will be smoother, A 2 3 0.5 Q= R S0 n 2 ⎛ Qn ⎞ S0 = ⎜ 2 3 ⎟ ⎝ R A⎠ 2 ⎡ ⎤ ⎢ ⎥ 40 × 0.02 S0 = ⎢ ⎥ ⎢ ⎛ 36.36 ⎞ 2 3 ⎥ ⎢⎜ ⎟ × 36.36 ⎥ ⎣ ⎝ 16.40 ⎠ ⎦ S 0 = 0.000167Hence the recommended parameters pf the canal are B = 5.09 m, m = 1, and S0 =0.000167. Adopt a free board of 0.75 m. The normal depth for n = 0.02 will be y0 = 4.0m. 50 Prof. Dr. Atıl BULU

Full NameComment goes here.abush16we really 10qArus Edo, Instructor at haromaya university institute of technology 8 months agoKrishan Sameera, Army Officer at Sri Lanka Army 8 months agoyawowusu-asante1 year agoroseate921 year ago08351810192 years ago