2. Properties of Open Channels  Free water surface Position of water surface can change in space and time  Many different types River, stream or creek; canal, flume, or ditch; culverts  Many different cross-sectional shapes

3. Definition of Geometric Elements R = hydraulic radius = A/P D = hydraulic depth = A/T y = depth of flow d = depth of flow section T = top width P = wetted perimeter A = flow area Gupta, Fig. 10.1

4. Circular Sections  For partially full circular sections (with diameter d o ), the geometric elements are a function of depth (y)  Values have been tabulated in non- dimensional form using d o as a scaling parameter (Gupta Table 10.1)  Note that maximum flow occurs at a depth of 0.94d o

5. Geometric Elements for Circular Pipes Gupta, Table 10.1

6. Flow Classification  Uniform (normal) flow: Depth is constant at every section along length of channel  Nonuniform (varied) flow: Depth changes along channel Rapidly-varied flow: Depth changes suddenly Gradually-varied flow: Depth changes gradually

7. State of Flow  Flow in open channels is affected by viscous and gravitational effects  Viscous effects described by Reynolds number, Re = VR/  Gravitational effects described by Froude number, F = V/(gD) 1/2

8. Viscous Effects  For Re < 500, viscous forces dominate and flow is laminar  For Re > 2000, viscous forces are weak and flow is turbulent  For Re between 500 and 2000, there is a transition between laminar and turbulent flow

9. Gravitational Effects  Critical flow is the point where velocity is equal to the speed of a wave in the water  For F = 1, flow is critical  For F < 1, flow is subcritical Wave can move upstream  For F > 1, flow is supercritical Wave cannot move upstream

10. Equations of Motion  There are three general principles used in solving problems of flow in open channels: Continuity (conservation of mass) Energy Momentum  For problems involving steady uniform flow, continuity and energy principles are sufficient

11. Conservation of Mass  Since water is essentially incompressible, conservation of mass (continuity) reduces to the following: discharge in = discharge out  Stated in terms of velocity and area: Q = V 1 A 1 = V 2 A 2 (10.5)

12. Control Volume for Open Channels Gupta, Fig. 10.4

13. Conservation of Energy  Conservation of energy applied to control volume results in the following: whereZ 1,Z 2 are elevations of the bed, y 1, y 2 are depths of flow, V 1, V 2 are velocities,  1,  2 are kinetic energy corrections, and h f is the frictional loss.

14. Energy Coefficient  The term associated with each velocity head (  ) is the energy coefficient  This term is needed because we are using the average velocity over the depth to compute the total kinetic energy  Integrating the cubed incremental velocities is not equal to the cube of the integrated incremental velocities

15. Critical Flow  Specific energy is defined as the energy in a channel section measured w.r.t. channel bed  If slope of channel is relatively flat, we can let Z 1 = Z 2 = 0  For energy coefficient (  ) = 1, specific energy (E) is thus

16. Specific Energy Gupta, Fig. 10.7

17. Critical Depth  The point of minimum specific energy corresponds to the critical depth  For higher values of specific energy there are two values that give the same specific energy  If Q is fixed, for the smaller depth there will be a higher velocity (supercritical) and for the larger depth there will be a lower velocity (subcritical)

18. Computing Critical Flow  From necessary condition for minimum specific energy, can define section factor for critical flow (Z c ): whereZ c is the critical section factor, A is the area of flow, D is the hydraulic mean depth, and Q is the discharge

19. Computing Critical Flow  If y c is known, can compute Z c for specific geometry. Then critical discharge is given as

20. Computing Critical Depth  If Q is known, can compute Z c as  For simple geometries, Z c can be written as a function of y c (see Example 1)  For complex geometries, use tabulated values (see Example 2)

21. Example 1  Determine the specific energy of water in a rectangular channel 25 ft wide having a flow of 500 cfs at a velocity of 5 ft/s.  What is the critical depth of water in the channel?  What is the critical velocity?

22. Example 2  Determine the specific energy of water in the trapezoidal channel shown having a flow of 30 m 3 /s.  What is the critical depth of water in the channel?  What is the critical velocity? 4m 4 1 y

23. Uniform Flow  Equations are developed for steady- state conditions Depth, discharge, area, velocity all constant along channel length  Rarely occurs in natural channels (even for constant geometry) since it implies a perfect balance of all forces  Two general equations in use: Chezy and Manning formulas

24. Chezy Equation  Balances force due to weight of water in direction of flow with opposing shear force  Note: V is mean velocity, R is hydraulic radius (area/wetted perimeter), S is the slope of energy gradeline, and C is the Chezy coefficient  C is a function of the roughness of the channel bottom

25. Manning Equation  The Manning equation is an empirical relationship similar to Chezy equation:  Note: V is mean velocity (ft/s), R is hydraulic radius (ft), S is the slope of the energy gradeline (ft/ft), and n is the Manning roughness coefficient

26. Manning Equation  The Manning equation for metric units is given as:  Note: V is mean velocity (m/s), R is hydraulic radius (m), S is the slope of the energy gradeline (m/m), and n is the Manning roughness coefficient

27. Manning ’ s Roughness (n)  Roughness coefficient (n) is a function of: Channel material Surface irregularities Variation in shape Vegetation Flow conditions Channel obstructions Degree of meandering

28. Manning’s n Gupta, Table 10.6

29. Manning ’ s Equation  Using continuity equation (Q=VA), Manning’s equation for English units can be written as  And for metric units

30. Conveyance  For uniform flow, A, R and n are constant thus  The term K is conveyance, given as

31. Normal Section Factor  Based on Manning’s equation, define a normal section factor for the portion dependent on geometry, namely  For simple geometries, Z n can be written as a function of y n  For complex geometries, use tabulated values (e.g., Gupta Table 10.1)

32. Computing Normal Flow Three cases to consider: Ê Find Q for known values of y n and S Ë Find S for known values of Q and y n Ì Find y n for known values of Q and S  From Manning’s equation, see that

33. ÊComputing Normal Flow  For known values of y n and S, can use Manning’s equation directly  Based on relationship between y n and Z n, determine the value of Z n  Compute discharge from Manning’s equation, namely

34. Example 3  Calculate the discharge through a 3-ft diameter circular clean earth channel running half full. The bed slope is 1 in 4500. Manning’s n = 0.018.

35. ËComputing Normal Slope  For known values of y n and Q, can use Manning’s equation directly  Based on relationship between y n and Z n, determine the value of Z n  Rearrange Manning’s equation, and compute slope as

36. Ì Computing Normal Depth  For known values of Q and S, can compute Z n as  Normal depth (y n ) is then determined from the relationship between y n and Z n

37. Example 4  A trapezoidal channel flowing at 30 m 3 /s has a bottom slope of 0.1% and n = 0.025.  Determine the normal depth.  Determine the critical slope.  What is the state of flow in the channel? 4m 4 1 y

38. Alterations to the HGL  As with pipes, there are numerous structures and channel alterations that affect the hydraulic grade line (HGL).  These alterations result in minor losses.  We will consider 5 cases: Sudden transitions Constrictions and obstructions Junctions Trash Racks Gates

39. Sudden Transitions  Changes in cross-sectional geometry over a short distance results in rapidly varied flow  This can produce a sharp drop (contraction) or rise (expansion) of the HGL  Change in geometry can be in either the horizontal or vertical plane of the channel

40. Horizontal Contractions  Writing the momentum and continuity equations between sections 1 and 3 and neglecting frictional losses and the momentum correction factor: where Source: Chow (1959) Figure 17-3

41. Horizontal contractions/expansions  The equation can be plotted against y 3 /y 1 using b 3 /b 1 as a parameter Source: Chow (1959) Figure 17-2

42. Horizontal contractions/expansions  For a given expansion (b 3 /b 1 > 1) or contraction (b 3 /b 1 < 1), the curves can be used to solve for y 3 given Q and y 1 : Compute F 1 2 from Q and y 1 Use curves to determine y 3 /y 1  To determine y 1 given Q and y 3 : Assume a value of y 1 Compute {F 1 2 } Use curves to determine F 1 2 using y 3 /y 1 and b 3 /b 1 Iterate on value of y 1 until {F 1 2 } = F 1 2

43.  The drop in the EGL for a contraction or expansion can be calculated as Horizontal contractions/expansions

44. Sudden Transitions  Based on the resulting values of F 1 2 and F 3 2, the transitions fall into four possible cases: Region 1: F 1 2 <1, F 3 2 <1 ○ Flow is supercritical throughout the transition Region 2: F 1 2 1 ○ Flow through the transition passes from supercritical to subcritical (hydraulic jump) Region 3: F 1 2 >1, F 3 2 >1 ○ Flow is subercritical throughout the transition Region 4: F 1 2 >1, F 3 2 <1 ○ Flow through the transition passes from subcritical to supercritical

45. Constrictions and Obstructions  For drastic constrictions of the channel area (b 3 /b 1 < 0.8), the calculations and resulting water-surface profiles are complicated  Likewise, calculating the effect of obstructions (e.g., piers) involves sophisticated calculations  These problems are generally solved using step-backwater computational software (e.g., U.S. Army Corps of Engineers HEC- RAS software), and are beyond the scope of this class

46. Junctions  Understanding flow through channel junctions involves numerous variables, such as number of channels, angles of intersection, cross-sectional shape and slope of the channels, directions of flow, etc.  Theoretical solutions have only been determined for simple cases.  More complex situations are usually solved using numerical techniques applied to a branching flow network

47. Special case: Combining Flow  For the special case of a rectangular channel flowing into an equal-sized rectangular channel, applying the momentum equation results in 1 3 2 yuyu ydyd

48. Special case: Combining Flow  If we know the flow in the two joining channels (Q 1 and Q 2 ), assuming normal flow upstream of the junction results in values for y 1 and y 2.  From continuity, Q 3 = Q 1 +Q 2  Setting y u =y 1, we can solve for y d using an iterative solution: Assume a value for y d Calculate LHS from known quantities Calculate RHS for assumed y d Iterate until LHS = RHS

49. Trash Racks  Head loss through trash racks can be computed using the standard minor loss equation  The minor loss coefficient (K t ) is a function of the bar thickness (s), distance between bars (b), bar shape and the angle of the rack from the horizontal plane (  )

50. Trash Racks  The shape factor (  ) is tabulated based on the shape of the bar  More sophisticated equations for the minor loss coefficient are available for racks that aren’t perpendicular to the flow in the channel Form of rack bar Shape factor (  ) Square nose and tail2.42 Square nose, semicircular tail1.83 Semicircular nose and tail1.67 Round1.79 Airfoil0.76

51. Gates  Gates can be used to control the flow in a channel or at the entrance from a reservoir to a channel  The sluice and tainter gates are common underflow structures  The design typically involves estimating the discharge rating of the structure for various gate opening heights and upstream water- surface elevations

52. Underflow Gates  For both styles of gates, the energy equation can be solved to give the following rating: Source: Chow (1959) Figure 17-37 length upstream depth discharge coefficient gate opening

53. Vertical sluice gates  The discharge coefficient is a function of the upstream depth, sluice opening, and downstream depth Source: Chow (1959) Figure 17-38 submerged conditions y 3 /h > 2

54. Tainter gates  The discharge coefficient for a tainter gate is more complicated, and depends on the radius of the gate (r) and the location of the pivot point (a) Source: Chow (1959) Figure 17-39 submerged conditions

55. Channel Design  Previously, we have assumed that channel dimensions and slope are known  In design situations, usually know discharge and need to determine geometry and slope  In an open channel, usually have allowable velocity range Minimum of 2-3 ft/s to move grit Maximum velocity to prevent scouring

56. Channel Design  For circular pipes, can determine minimum slope needed to achieve a minimum velocity of 2 ft/s Gupta, Table 10.7

57. Channel Design  From basic principles, we know that: For a given slope and roughness, Q increases with increase of section factor (Z n =AR 2/3 ) For a given area, Z n is maximum for a minimum wetted perimeter (P)  Based on these principles, can develop guidelines for efficient hydraulic sections for given channel shapes

58. Efficient Hydraulic Sections

59. Design Procedure  For a specified Q, select S and estimate Manning’s n. Compute section factor from Q, n and S  Select a shape, and express section factor in terms of y; solve for y  Try different values of S, n and shape and compute cost for each  Verify that velocity exceeds 2-3 ft/s  Add freeboard if open section