Hydraulics of Structures

1. Hydraulics of Structures

2. Structures in this context are simply something placed in the channel to either measure or control flow. • Example: A principle spillway is used as part of a dam design to control the rate at which water is discharged from a reservoir. • Include both inlet and outlet control devices. • Control devices can operate as : • Open channel flow in which the flow has a free surface or • Pipe flow in which the flow is in a closed conduit under pressure.

3. Most basic principle of hydraulics of structures: • As head on a structure increases, the flow that is discharged through the structure increases. • Figure 5.1 (Haan et al., 1994) shows the head-discharge relationships for several flow control structures.

4. Weirs • At its most basic, just an obstruction placed in a channel that constricts flow as it goes over a crest. • The crest is the edge of the weir over which the water flows. • As the water level (head) over the crest increases, the flow rate increases dramatically. • Two basic types of weirs • sharp crested • broad crested

5. Sharp Crested Weirs • A sharp crested weir is defined by a thin crest over which the water springs free as it leaves the upstream face of the weir. • Flow over a weir is also called the nappe. • Sharp crested weirs are generally constructed of sheet metal or similar thin material.

6. nappe H Sharp Crested Weir

7. Sharp Crested Weirs • Can have several shapes • Triangular (or v-notch) • Rectangular • Trapezoidal • Classified by the shape of its notch. • V-notch weirs have greater control under low flow conditions. • Rectangular weirs have larger capacity but are less sensitive for flow measurement.

8. Using Bernoulli’s equation Sharp Crested Weir V22/2g V12/2g h H dh z

9. Making the assumption that the velocity head at the upstream point will be much smaller than the velocity head as the flow goes over the weir we assume v12/2g is negligible and: or H dh h Crest L

10. Adding a loss term to compensate for the deviation from ideal flow we get: When H 1/3 L, an approximate value for Cd is 0.6 to 0.62 leaving: Integrating this from h = 0 to h = H gives

11. H H L Coefficient of Discharge Rectangular Weirs A rectangular weir that spans the full width of the channel is known as a suppressed weir.

12. Hydraulic head (H) for weirs is simply the height of the water surface above the weir crest, measured at a point upstream so that the influence of the velocity head can be ignored. • L is the length of the weir. • The coefficient of discharge (C) is dependent upon units and of the weir shape. • For a suppressed weir with H/h < 0.4 (where h is the height of the weir) C= 3.33 can be used. • For 0.4 < H/h < 10, C = 3.27 + 0.4 H/h

13. L’ A rectangular weir that does not span the whole channel is called a weir with end contractions . The effective length of the weir will be less than the actual weir length due to contraction of the flow jet caused by the sidewalls. Where N is the number of contractions and L’ is the measured length of the crest.

14. H Q Triangular (v-notch ) weirs • Used to measure flow in low flow conditions.

15. For other angles Where Cd is based on the angle, Q, and head, H. • For Q = 90°, K = 2.5 (typically), tan (Q/2) = 1 therefore,

16. Note: Your handout with Figure 12.28 presents the equation for a v-notch weir as: with

17. Orifices • An orifice is simply an opening through which flow occurs. • They can be used to: • Control flow as in a drop inlet • Measure the flow through a pipe.

18. The discharge equation for orifice flow is: Where: C’ is the orifice coefficient (0.6 for sharp edges, 0.98 for rounded edges). A is the cross-sectional area of the orifice in ft2 g is the gravitational constant H is the head on the orifice

19. At low heads, orifices can act as weirs. • Calculate the discharge using the suppressed weir equation where L is equal to the circumference of the pipe. • Calculate the discharge using the orifice equation. • The lower discharge will be the actual discharge.

20. Example A 36-in, circular, vertical riser constructed from corrugated metal pipe (CMP) serves as the inlet for the principal spillway of a detention structure. Estimate the discharge if the head on the riser is 1ft. Estimate the discharge if the head is 3 ft.

21. H H’ Energy Grade Line D 0.6D L Elbow and Transition Pipes as Flow Control Devices

23. Head Loss Coefficients • Ke is the entrance head loss coefficient and is typically given a value of 1.0 for circular inlets. • Kb is the bend head loss coefficient and is typically given a value of 0.5 for circular risers connected to round conduits. • For risers with rectangular inlets, the bend head losses and entrance head losses are typically combined to a term Ke’ where values of Ke’ can be found in Table 5.3 and :

24. Head Loss Coefficients • Kc is the head loss coefficient due to friction. • Values for Kc are given in Tables 5.1 and 5.2 for circular and square pipes. • Kc is multiplied by L, the entire length of the pipe, including the riser.

25. Frequently, when the drop inlet is the same size as the remainder of the pipe, orifice flow will control and the pipe will never flow full. • If it is desirable to have the pipe flowing full, it may be necessary to increase the size of the drop inlet.

26. Example A 36-in diameter corrugated metal pipe is attached to a 36-in vertical riser. It is being used as the principal spillway for a detention structure. The pipe is 40 feet long and has one 90° bend. The top of the inlet riser is 10 ft above the bottom of the outlet. Assume a free outfall and estimate the discharge under pipe flow if the water elevation 30 ft from the inlet is 2 ft higher than the top of the riser.

27. Example A 48-in coated cast iron riser is connected to a 24-in coated cast iron barrel by one 90º bend. The spillway is 65 ft long. The top of the riser is 15 ft above the outlet. Assume a free outfall and estimate the discharge if the water elevation 25 ft upstream of the inlet is 1.8 ft. above the top of the riser.

28. Using Flow Control Structures as Spillways • A given drop inlet spillway can have a variety of discharge relationships, given the head. • At the lowest stages the riser acts as a weir. • As the level of the reservoir rises, water flowing in from all sides of the inlet interferes so that the inlet begins to act as an orifice. • As the level continues to rise, the outlet eventually begins to flow full and pipe flow prevails. • A stage-discharge curve is developed by plotting Q vs. H for each of the three relationships. The minimum flow for a given head is the actual discharge used.

29. Example 2 Given the previous example, develop a stage-discharge curve.

30. Broad Crested Weirs H W Where L is the width of the weir.

31. Broad Crested Weirs • Broad crested weirs support the flow in the longitudinal direction (direction of flow). • They are used where sharp-crested weirs may have maintenance problems. • The nappe of a broad crested weir does not spring free.

32. ROCKFILL HYDRAULIC PROFILE dh h1 have h2 dl

33. Modified Darcy-Weisbach Equation

34. Rockfill as Control Structure Model Reynolds Number Equation Friction factor

35. Friction Factor-Reynolds Number Relationship

36. h2 – have Relationships

37. Example 1 • A rockfill dam is composed of rock having an average diameter of 0.04 m, porosity equal to 0.46, standard deviation of 0.002 m, and length dl equal to 2.0 m. Water with a kinematic viscosity of 1 X 10-6 m/sec is flowing through the rock at a rate q of 5.0 cms/m width. Down stream conditions control the exit depth of the water h2 at 1.0 m. Find the upstream height h1.

38. Example 2 • If the rock fill in Example 1 is 3 m wide and as used as spillway from a sediment detention pond, determine the stage discharge relationship up to an upstream depth of 2 m using depths of 0.5, 1.0, 1.5 and 2.0 m. Assume that the downstream slope is such that the downstream depth is negligible.