UNIT 1 UNIFORM FLOW.pptx

Uniform Flow: Basic concepts of free surface flows,
velocity and pressure distribution,
Mass, energy and momentum principle for prismatic and non-prismatic channels,
Review of Uniform flow: Standard equations,
hydraulically efficient channel sections,
compound sections,
Energy-depth relations:
Concept of specific energy, specific force,
critical flow, critical depth,
hydraulic exponents, and
Channel transitions.

1. Semester : VI th Year : 3 rd COLLEGE OF ENGINEERING ROORKEE, COER Bachelor of Technology (B .Tech. ) CIVIL ENGINEERING OPEN CHANNEL FLOW (BCET 603)

2. LEARNING OBJECTIVES Uniform Flow: Basic concepts of free surface flows, velocity and pressure distribution, Mass, energy and momentum principle for prismatic and non-prismatic channels , Review of Uniform flow: Standard equations, hydraulically efficient channel sections, compound sections, Energy-depth relations: Concept of specific energy, specific force, critical flow, critical depth, hydraulic exponents, and Channel transitions. 2

3. An open channel is a conduit in which a liquid flows with a free surface. The free surface is actually an interface between the moving liquid and an overlying fluid medium and will have constant pressure. In civil engineering applications; water is the most common liquid with air at atmospheric pressure as the overlying fluid. The prime motivating force for open channel flow is gravity. OPEN CHANNEL FLOW 3

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5. TYPES OF CHANNELS Rigid channels are those in which the boundary is not deformable in the sense that the shape, platform and roughness magnitudes are not functions of the flow parameters. Typical examples include lined canals, sewers and non-erodible unlined canals . The flow velocity and shear-stress distribution will be such that no major scour, erosion or deposition takes place in the channel and the channel geometry and roughness are essentially constant with respect to time. We have many unlined channels in alluvium —both man- made channels and natural rivers—in which the boundaries undergo deformation due to the continuous process of erosion and deposition due to the flow. The boundary of the channel is mobile in such cases and the flow carries considerable amounts of sediment through suspension and in contact with the bed. Such channels are classified as mobile-boundary channels. On the basis of the nature of the boundary open channel Rigid and Mobile Boundary Channels 5

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7. Prismatic Channel- A channel in which the cross-sectional shape and size and also the bottom slope are constant is termed as a prismatic channel. Most of the man-made (artificial) channels are prismatic channels over long stretches. he rectangle, trapezoid, triangle and circle are some of the commonly used shapes in manmade channels. Non- Prismatic Channel- All natural channels generally have varying cross-sections and consequently are non-prismatic Prismatic & Non Prismatic Channel 7

8. Steady and Unsteady Flow Uniform and Non-uniform Flow Laminar and Turbulent Flow Sub-critical, Critical and Super-critical Flow CLASSIFICATION OF FLOWS 8

9. CLASSIFICATION OF FLOWS Steady and Unsteady Flows A steady flow occurs when the flow properties, such as the depth or discharge at a section do not change with time. And if the depth or discharge changes with time the flow is termed unsteady. Flood flows in rivers and rapidly varying surges in canals are some examples of unsteady flows. Unsteady flows are considerably more difficult to analyse than steady flows. 9

10. If the flow properties, say the depth of flow, in an open channel remain constant along the length of the channel, the flow is said to be uniform. As a corollary of this, a flow in which the flow properties vary along the channel is termed as non-uniform flow or varied flow. A prismatic channel carrying a certain discharge with a constant velocity is an example of uniform flow [Fig. 1.1(a)]. In this case the depth of flow will be constant along the channel length and hence the free surface will be parallel to the bed. It is easy to see that an unsteady uniform flow is practically impossible, and hence the term uniform flow is used for steady uniform flow. Flow in a non-prismatic channel and flow with varying velocities in a prismatic channel are examples of varied flow. Varied flow can be either steady or unsteady. Uniform and Non-uniform Flows 10

11. Types of Non-uniform Flow Gradually Varied Flow (GVF) If the depth of the flow in a channel changes gradually over a length of the channel. Rapidly Varied Flow (RVF) If the depth of the flow in a channel changes abruptly over a small length of channel 11

12. Laminar and Turbulent Flow Both laminar and turbulent flow can occur in open channels depending on the Reynolds number (Re) R e = ρ VR/µ Where, ρ = density of water = 1000 kg/m 3 µ = dynamic viscosity R = Hydraulic Mean Depth = Area / Wetted Perimeter 12

13. Sub-critical, Critical and Super-critical Flow 13

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17. Velocity Distribution Velocity is always vary across channel; because of friction along the boundary The maximum velocity usually found just below the surface. the velocity vectors of the flow to have components only in the longitudinal and lateral direction but also in normal direction to the flow. In a macro-analysis, one is concerned only with the major component, viz., the longitudinal component, v x . The other two components being small are ignored. The distribution of v in a channel is dependent on the geometry of the channel. Figure show isovels (contours of equal velocity) of v for a natural and rectangular channel respectively. The influence of the channel geometry is apparent. The velocity v is zero at the solid boundaries and gradually increases with distance from the boundary 17

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19. a. Natural Channel , b. Rectangular Channel 19

20. A typical velocity profile at a section in a plane normal to the direction of flow is presented in figure. Max velocity occurs at a distance 0.0y o to 0.25y o from the free surface. Deeper and narrower channel, more deep will be the max velocity from the surface Field observations in rivers and canals have shown that the average velocity at any vertical V av occurs at a level of 0.6 y o from the free surface, where y o = depth of flow. Further, it is found that V av = (V 0.2 + V 0.8 ) /2 which V 0.2 = velocity at a depth of 0.2 y o from the free surface, and V 0.8 = velocity at a depth of 0.8 y o surface. The surface velocity V s is related to the average velocity V av as V av = k V av where, k = a coefficient with a value between 0.8 and 0.95. Depends on Channel section. 20

21. VELOCITY DISTRIBUTION – 1 D APPROACH Velocity profile : Discharge through elemental area dA having velocity v dQ = vdA So, Total discharge Q = ʃ vdA Considering Avg velocity V and area A Q= VA VA = ʃ vdA Therefore Avg velocity V = 1/A ʃ vdA dA Velocity - Depth 21

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24. PRESSURE DISTRIBUTION IN OPEN CHANNEL The intensity of pressure for a liquid at its free surface is equal to that of the surrounding atmosphere. Since the atmospheric pressure is commonly taken as a reference and of value equal to zero, the free surface of the liquid is thus a surface of zero pressure. This linear variation of pressure with depth having the constant of proportionality equal to the unit weight of the liquid is known as hydrostatic-pressure distribution. Pressure distribution in still water: W = PA – P atm W = PA -0 P = = = y. ϒ = ρ w g .y P = y. ϒ = ρ w g .y   24

25. Let us consider a channel with a very small value of the longitudinal slope θ . Let θ ~ sin θ ~ 1/1000. For such channels the vertical section is practically the same as the normal section. If a flow takes place in this channel with the water surface parallel to the bed, i.e. uniform flow, the streamlines will be straight lines and as such in a vertical direction the normal acceleration an = 0. P = y. ϒ =y and + Z = Z 1   Thus the piezometric head at any point in the channel will be equal to the water surface elevation. The hydraulic grade line will therefore lie essentially on the water surface. Channels with Small Slope 25

26. Figure shows a uniform free-surface flow in a channel with a large value of inclination θ . The flow is uniform, i.e. the water surface is parallel to the bed. An element of length Δ L and unit width is considered at the Channels with Large Slope At any point A at a depth y measured normal to the water surface, the weight of column A 1 1′ A ′ = γ Δ L y and acts vertically downwards. The pressure at AA ′ supports the normal component of the column A 1 1′ A ′. Thus pA Δ L = γ y Δ L cos θ pA = γ y cos θ pA / γ = γ cos θ 26

27. The pressure pA varies linearly with the depth y but the constant of proportionality is γ cos θ. If h = normal depth of flow, the pressure on the bed at point 0, p = γ h cos θ. If d = vertical depth to water surface measured at the point O, then h = d cos θ and the pressure head at point O, on the bed is given by p = h cos θ = d cos2 θ 27

28. For the channels given below – Calculate the discharge. 5mX3m Depth of flow = 2.30 m Velocity i ) = 2.8m/s ( 32.2 ) ii) = 3.15 m/s (36.5) Depth of flow = 4 m Velocity i ) = 2.25m/s (99) ii) = 5.12 m/s (225.8 ) 5m 1v:1.5H (b) (a) 28

29. In the case of steady-uniform flow in an open channel, the following main features must be satisfied: The water depth, water area, discharge, and the velocity distribution at all sections throughout the entire channel length must remain constant, i.e.; Q , A , y , V remain constant through the channel length. The slope of the energy gradient line (S), the water surface slope ( Sws ), and the channel bed slope (S ) are equal i.e.; S = S ws = S Flow Formulas in Open Channels (steady-uniform flow): 29

30. The Chezy Formula: (1769) Empirical formulas are used to describe the flow in open channels. The Chezy formula is probably the first formula derived for uniform flow. It may be expressed in the following form: where C = Chezy coefficient ( Chezy’s resistance factor), m 1/2 /s.   30

31. The Manning Formula: (1895) Using the analysis performed on his own experimental data and on those of others, Manning derived the following empirical relation: where n = Manning’s coefficient for the channel roughness, m-1/3/s.   31

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33. G EOMETRIC E LEMENTS OF O PEN C HANNELS A channel section is defined as the cross-section taken perpendicular to the main flow direction. Referring to Figure, the geometric elements of an open channel are defined as follows; Flow depth, y : Vertical distance from the channel bottom to the free surface. Depth of flow section, d : Flow depth measured perpendicular to the channel bottom. The relationship between d and y is d= yCos ϴ . For most manmade and natural channels Cos ϴ = 1.0, and therefore y = d. The two terms are used interchangeably. 33

34. Top width, T: Width of the channel section at free surface. Wetted perimeter, P: Length of the interface between the water and the channel boundary. Flow area, A: Cross-sectional area of the flow. Hydraulic depth, D: Flow area divided by top width, D = A/T. Hydraulic radius, R: Flow area divided by wetted perimeter, R = A/P. Bottom slope,S : Longitudinal slope of the channel bottom, S =tan ϴ = sin ϴ . 34

35. Geometric elements of channel sections 35

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37. The cross section of an open channel is a trapezoid with a bottom width of 4 m and side slopes 1:2, calculate i ) the discharge if the depth of water is 1.5 m and bed slope = 1/1600. Take Chezy constant C = 50. Ii) Measure the discharge using Manning’s formula if the channel lining is of smooth concrete. 37

38. A section of a channel is said to be most economical when the cost of construction of the channel is minimum. But the cost of construction of a channel depends on excavation and the lining. To keep the cost down or minimum, the wetted perimeter, for a given discharge, should be minimum. This condition is utilized for determining the dimensions of economical sections of different forms of channels. Most economical section is also called the best section or most efficient section as the discharge, passing through a most economical section of channel for a given cross-sectional area A, slope of the bed S and a resistance coefficient, is maximum. But the discharge Q = AV = Const. Hence the discharge Q will be maximum when the wetted perimeter P is minimum   MOST ECONOMICAL SECTION OF CHANNELS 38

39. Most Economical Rectangular Channel: Consider a rectangular section of channel as shown. Let B = width of channel, D = depth of flow. Area of flow, A = B x D ……………………………….(7.4a) Wetted perimeter, P = 2D + B……………….……(7.4b) from Eq. (7.4a) we have B = A/D P = 2D+ A/D ………………………….…….. (7.4c) For most economical cross section, P should be minimum for a given area; D =   39

40. From equations (7.5) and (7.6), it is clear that rectangular channel will be most economical when either: (a) The depth of the flow is half the width or (b) The hydraulic radius is half the depth of flow 40

41. 1. For a rectangular channel with width = 4m and bed slope = 1/1500, Calculate the max discharge for the most economical condition when Chezy’s constant C = 50. D = B/2 = 4/2 = 2m A = 2X4 = 8m2 Q = AV = 8 50 (R (1/1500)) 1/2 = 10.328 m3/s R = D/2 = 1 41

42. Consider a trapezoidal section of channel as shown. Most Economical Trapezoidal Channel 42

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45. Best side slope for most economical trapezoidal section can be shown to be when n = 1/ So far we assumed that the side slopes are constant. Let us now consider the case when the side slopes can also vary. The most economical side slopes of a most economical trapezoidal section can be obtained as follows:   45

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47. Therefore, best side slope is at 60 o to the horizontal, i.e.; of all trapezoidal sections a half hexagon is most economical. However, because of constructional difficulties, it may not be practical to adopt the most economical side slopes. 47

48. For a most economical section the discharge, for a constant cross-sectional area, slope of bed and resistance coefficient, is maximum (or P is minimum). But in the case of circular channels, the area of the flow cannot be maintained constant . Indeed, the cross-sectional area A and the wetted perimeter P both do not depend on D but they depend on the angle a . Most Economical Circular Channel Referring to the figure shown, we can determine the wetted perimeter P and the area of flow A as follows: Let D = depth of flow d = diameter of pipe r = radius of pipe 2 a = angle subtended by the free surface at the centre (in radians) 48

49. A1 a b O A2 D α M 49

50. Consequently, the cross-sectional area A and the wetted perimeter P both depend on the angle α which is the most suitable variable. Thus in case of circular channels, for most economical section, two separate conditions are obtained: 1) condition for maximum discharge, and 2) condition for maximum velocity 50

51. Condition for Maximum Discharge for Circular Section 51

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53. This means that the maximum discharge (minimum P) in a circular channel occurs when the of the pipe. The above results holds good when the Chezy formula is used. If Manning’s formula is used, results will be: α = 151 o and D = 0.94 d 53

54. Condition for Maximum Velocity for Circular Section Since the cross-sectional area A varies with α, the condition for the maximum velocity is different from the condition for the maximum discharge. 54

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56. For a most economical open channel of Trapezoidal section with side slope of 2V:3H, Evaluate the wetted perimeter if discharge flowing through it is 10m 3 /s and the velocity of flow is 1.5 m/s and bed slope is 0.014. A nearly horizontal channel has a bottom width of 3 ft , and it carries a discharge of 60 cfs at a depth of 4 ft. Determine the magnitude and direction of the hydrostatic pressure force exerted on each of the sidewalls per unit length of the channel if - (a) the channel is rectangular with vertical sidewalls (b) the channel is trapezoidal with each sidewall sloping outward at a slope 2 horizontal over 1 vertical, that is m= 2. 3. The base of the most economical trapezoidal channel section is 6m and the side slope is 1H:2V, Calculate the maximum discharge through the channel if the bed slope is 1 in 1000 and C = 50. NUMERICALS 56

57. 4. The top width of a most economical trapezoidal channel section is 8m, determine the hydraulic radius of the channel if the side slope is 1H:3V. (1.9m) 5. The top width of a most economical trapezoidal channel section is 7m and the side slope of the channel is 1H:2V, determine the depth of the channel section. (3.13m) 6. A circular channel is proposed to lay on a slope of 1 in 2000 and is required to carry 1.5cumec. What size of circular channel should be used if it has to flow half-full take n=0.015. (2.1m) 57

58. Energy Principles in Open Channel Flow Referring to the figure shown, the total energy of a flowing liquid per unit weight is given by, Total Energy E =Z +y + Where Z = height of the bottom of channel above datum, y = depth of liquid, V = mean velocity of flow. If the channel bed is taken as the datum (as shown), then the total energy per unit weight will be, E s = y + This energy is known as specific energy, Es , S pecific energy of a flowing liquid in a channel is defined as energy per unit weight of the liquid measured from the channel bed as datum. It is a very useful concept in the study of open channel flow .   58

59. which is valid for any cross section . 59

60. It is defined as the curve which shows the variation of specific energy ( Es ) with depth of flow y. Let us consider a rectangular channel in which a constant discharge is taking place. If q = discharge per unit width = = constant ( since Q and B are constants), Then, Velocity of flow, V = = = = Substituting for V into above equations, we get E s = y + = E p + E k   Specific Energy Curve (rectangular channel) 60

61. θ Energy line y Z V 2 /2g 61

62. SECTION FACTOR Z 62

63. In many computations involving a wide range of depths in a channel, such as in the GVF computations, it is convenient to express the variation of Z with y in an exponential form . The ( Z – y) relationship Z 2 = C 1 y M In this equation C1 = a coefficient and M= an exponent called the first hydraulic exponent . It is found that generally M is a slowly-varying function of the aspect ratio for most of the channel shapes. FIRST HYDRAULIC EXPONENT 63

64. The concepts of specific energy and critical energy are useful in the analysis of transition problems. Transitions in rectangular channels are presented here. The principles are equally applicable to channels of any shape and other types of transitions CHANNEL TRANSITIONS Channel with a Hump Transition with a Change in Width General Transition 64

65. a)Subcritical Flow Channel with a Hump Consider a horizontal, frictionless rectangular channel of width B carrying discharge Q at depth y1. Let the flow be subcritical. At a section 2, smooth hump of height ΔZ is built on the floor. Since there are no energy losses between sections 1 and 2, construction of a hump causes the specific energy at section to decrease by ΔZ. Thus the specific energies at sections 1 and 2 are, E 1 =y 1 +   E 2 =E 1 - Δ Z 65

66. The flow is subcritical, the water surface will drop due to a decrease in the specific energy. In Figure the water surface which was at P at section 1 will come down to point R at section 2. The depth y 2 will be given by- E 2 = y 2 +   As the value of ΔZ is increased, the depth at section 2, y2, will decrease. The minimum depth is reached when the point R coincides with C, the critical depth. At this point the hump height will be maximum, ΔZ max , y 2 = yc = critical depth, and E2 = Ec = minimum energy for the flowing discharge Q. The condition at ΔZ max is given by the relation, E 1 – Δ Z =E 2   66

67. when Δ Z > Δ z max The flow is not possible with the given conditions (given discharge). The upstream depth has to increase to cause and increase in the specific energy at section 1. If this modified depth is represented by y1`, E` 1 =   with E` 1 >E 1 and y` 1 >y 1 ) At section 2 the flow will continue at the minimum specific energy level, i.e. at the critical condition. At this condition, y2 = yc , and, 67

68. when 0 < ΔZ < ΔZmax the upstream water level remains stationary at y1 while the depth of flow at section 2 decreases with ΔZ reaching a minimum value of yc at ΔZ = ΔZmax . With further increase in the value of ΔZ, i.e. for ΔZ > ΔZmax , y1 will change to y1` while y2 will continue to remain yc . 68

69. b) Supercritical Flow If y 1 is in the supercritical flow regime, Fig. (5.13) shows that the depth of flow increases due to the reduction of specific energy. In Fig. point P` corresponds to y 1 and point R` to depth at the section 2. Up to the critical depth, y 2 increases to reach y c at ΔZ = ΔZ max . For ΔZ > ΔZ max , the depth over the hump y 2 = y c will remain constant and the upstream depth y 1 will change. It will decrease to have a higher specific energy E 1 `by increasing velocity V 1 . The variation of the depths y 1 and y 2 with ΔZ in the supercritical flow is shown in Fig. 69

70. Q. 1. A rectangular channel has a width of 2.0 m and carries a discharge of 4.80 m3/sec with a depth of 1.60 m. At a certain cross-section a small, smooth hump with a flat top and a height 0.10 m is proposed to be built. Calculate the likely change in the water surface. Neglect the energy loss. (E2 = y2 = 1.48m, ) 1. a) If the height of the hump is 0.50 m, estimate the water surface elevation on the hump and at a section upstream of the hump. 70

71. 2. Water flow in a wide channel approaches a 10 cm high hump at 1.50 m/sec velocity and a depth of 1 m. Estimate a) The water depth y2 over the hump and b) The hump height that will cause the crest flow to be critical. 71

72. Transition with a Change in Width Subcritical Flow in a Width Constriction Consider a frictionless horizontal channel of width B 1 carrying a discharge Q at a depth y 1 as in Fig. At a section 2 channel width has been constricted to B 2 by a smooth transition. Since there are no losses involved and since the bed elevations at sections 1 and 2 are the same, the specific energy at section is equal to the specific energy at section 2. It is convenient to analyze the flow in terms of the discharge intensity q = Q/B. At section 1, q 1 = Q/B 1 and at section 2, q 2 = Q/B 2 . 72

73. Since B 2 < B1, q 2 > q 1 . In the specific energy diagram drawn with the discharge intensity, point P on the curve q 1 corresponds to depth y 1 and specific energy E 1 Since at section 2, E2 = E1 and q = q2, point P will move vertically downward to point R on the curve q2 to reach the depth y2. Thus, in subcritical flow the depth is y2 < y1. If B2 is made smaller, then q2 will increase and y2 will decrease. The limit of the contracted width B2 = B2min is reached when corresponding to E1, the discharge intensity q2 = q2max, i.e. the maximum discharge intensity for a given specific energy (critical flow condition) will prevail. 73

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75.   If B2 < B2min, the discharge intensity q2 will be larger than qmax , the maximum discharge intensity consistent E1. The flow will not, therefore, be possible with the given upstream conditions. The upstream depth will have to increase to y1`. The new specific energy will be formed which will be sufficient to cause critical flow at section 2. It may be noted that the new critical depth at section 2 for a rectangular channel is, Since B 2 < B 2min , y c2 will be larger that y cm , y c2 > y cm . Thus even though critical flow prevails for all B 2 < B 2min , the depth section 2 is not constant as in the hump case but increases as y 1 ` and hence E 1 ` rises. 75

76. If the upstream depth y1 is in the supercritical flow regime, a reduction of the flow width and hence an increase in the discharge intensity cause a rise in depth y2. Point P` corresponds to y1 and point R` to y2. As the width B2 is decreased, R` moves up till it becomes critical at B2 = B2min . Any further reduction in B2 causes the upstream depth to decrease to y1` so that E1 rises to E1`. At section2, critical depth yc ` corresponding to the new specific energy E1` will prevail. The variation of y1, y2 and E with B2/B1 in supercritical flow regime is indicated in Fig. Supercritical Flow in a Width Constriction 76

77. General Transition A transition in general form may have a change of channel shape, provision of a hump or a depression, contraction or expansion of channel width, in any combination. In addition, there may be various degrees of loss of energy at various components. However, the basic dependence of the depths of flow on the channel geometry and specific energy of flow will remain the same. Many complicated transition situations can be analyzed 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. The transitions in supercritical flow are different and involve suppression of shock waves related disturbances. 77

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80. A compound channel is a channel section composed of a main deep portion and one or two flood plains that carry high-water flows. The main channel carries the dry weather flow and during wet season, the flow may spillover the banks of the main channel to the adjacent flood plains. A majority of natural rivers have compound sections. A compound section is also known as two-stage channel . The hydraulic conditions of the main channel and the flood plain differ considerably, especially in the channel geometry and in its roughness. The flood plains generally have considerably larger and varied roughness elements. COMPOUND CHANNEL SECTION 80

81. In one-dimensional analysis, Manning’s formula is applied to the compound channel by considering a common conveyance K and a common energy slope S f for the entire section to obtain the discharge as . However, to account for the different hydraulic conditions of the main and flood plain sections, the channel is considered to be divided into subsections with each subsection having its own conveyance, Ki . The sum of the conveyances will give the total channel conveyance ( Σ Ki = K ) for use in discharge computation. Various methods for defining the boundaries of the sub-sections are proposed by different researchers leading to a host of proposed methods. However, the overall method of considering the channel as a composite of sub sections is well accepted and the method is known as Divided Channel method (DCM). Currently DCM is widely used and many well-known software packages, including HEC-RAS (2006), adopt this method in dealing with compound channels   81

82. Evaluation of compound section – Vertical Interface Method Diagonal Interface Method 82

83. For the compound channel shown below (Fig. 3.32) estimate the discharge for a depth of flow of ( i ) 1.20 m, and (ii) 1.6 m, by using DCM with vertical interface procedure. A compound channel is symmetrical in cross section and has the following geometric properties. Main channel: Trapezoidal cross section, Bottom width = 15.0 m, Side slopes = 1.5 H : 1V, Bank full depth = 3.0 m, Manning’s coefficient = 0.03, Longitudinal slope = 0.0009. Flood plains: Width = 75 m, Side slope = 1.5 H : 1V, Manning’s coefficient = 0.05, Longitudinal slope = 0.0009. Compute the uniform fl ow discharge for a fl ow with total depth of 4.2 m by using DCM with ( i ) diagonal interface, and (ii) vertical interface procedures. 83

84. 3. The base of the most economical trapezoidal channel section is 6m and the side slope is 1H:2V, Calculate the maximum discharge through the channel if the bed slope is 1 in 1000 and C = 50. 84

85. 1. The top width of a most economical trapezoidal channel section is 8m, determine the hydraulic radius of the channel if the side slope is 1H:3V. The top width of a most economical trapezoidal channel section is 7m and the side slope of the channel is 1H:2V, determine the depth of the channel section. 85

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