EXAMPLE 6 : HYDRAULIC JUMP

Transcription

1 EXAMPLE 6 : HYDRAULIC JUMP A rectangular horizontal channel m. wide, carries a flow of 4 m 3 /s. The depth water on the downstream side of the hdraulic jump is 1m. a)what is the depth upstream? b)what is the loss of head? Q F 1 F 1A1 + A + ga 1 Q ga 4 1 ( 1 ) + 1 * ( * 1 ) + g 1 g 4 ( * 1 ) m V 1 4/(*0.311) 6.43 m/s V 4/(*1).0 m/s h loss m

2 Example on Critical Flow : 1 A flow of 8 m 3 /sec occurs in an earth-lined trapezoidal channel having base width 3.0m, side slopes 1V:H and n 0.0. Calculate the critical depth and critical slope. 3m 1 At critical flow Froude number is equal to unit F r Q T ga 3 1 Q 8 m3/s B 3 n0.0 slope 1V:H A 3 + ; T 3+4 ; P 3+ (5) 1/ R A/P (3 + ) / (3+ (5) 1/ ) From Manning s Q Solving for S c A n R / 3 S 1/ (8 ) 9.81 (3 (3 + + ) 4 ) solving b trial and error c 1.495m 3

3 Example on Critical Flow : A channel of rectangular section 3.5m wide, with n and S leads from a lake whose surface level is 6m above the channel bed at the lake outlet. Find the discharge in the channel. The Q max would result from critical flow at the outlet!!! V At reservoir V small thus E + 6m g No head loss between sections. For a rectangular channel at critical flow Velocit head m or velocit V c 6.64 m/s Q max m 3 /sec Ec c + c or c 4 m

4 Example on Critical Flow : 3 During a major flood events water flows over the top of a roadwa. Determine the head on the broad-crested weir for a discharge of 300 m3/s if the overflow section of the roadwa is horizontal and 150m. long. For a rectangular channel at critical flow 1 Q T ga c 3 c or q 300/150 m 3 /sec per meter c q g 1 / 3 1 / 3 c g m

5 Flow in Open Channels DESIGN OF OPEN CHANNELS FOR UNIFORM FLOW

6 Channel design for Uniform Flow The basic problem is the economical proportioning of the cross section. Channel with given n and S 0 for a known Q the objective is to minimize the area. Q A n R / 3 S 1/ o Qn S o 5 / 3 A AR / 3 P / 3 K o If A is minimum V maximum from continuit R maximum from Manning s P minimum (RA/P)

7 Hdraulic Efficienc of Cross-sections Conveance of the channel section Qn S o 5 / 3 A AR / 3 P / 3 K o It can be shown that the ideal section would be Semicircle

8 Best Hdraulic Section Qn S o 5 / 3 A AR / 3 P / 3 K o A channel section having the least wetted perimeter for a given area has the maximum conveance; such a section is known as the best hdraulic section

9 Channel design for Uniform Flow However, other economical concerns Total volume of excavation Cost of lining Construction techniques Scour in erodible bed Sedimentation for low V Short channels and variable S 0 ma require changes

10 The best hdraulic section for a rectangular channel Area A b* Perimeter P b + Perimeter must be minimum for given area P A/ + dp/d -A/ + 0 A b / b b

11 Example 7 What are the most efficient dimensions (the best hdraulic section) for a concrete (n0.01) rectangular channel to carr 3.5 m 3 /s at So0.0006? Given: n0.01 Q3.5 m 3 /s S o Find b and. b Q A n R / 3 S 1/ o Q A n A P / 3 1/ ( ) So Q b n ( b b + ) / 3 1/ S o best section b Q * n * + ( ) / 3 1/ / 3 1/ S ( ) o Q S o n Q 1/ 3 ( ) 8 / 3 1/ S ( ) 3 o n 8 / m and b.46m

12 Best Hdraulic Sections Section Most efficient Trapezoidal Rectangular Triangular Circular Base < depth Width x depth No specific relationship Semicircle if open Circle if closed

13 Precautions Steep slopes cause high velocities which ma create erosion in erodible (unlined) channels Ver mild slopes ma result in low velocities which will cause silting in channels. (Sedimentation) The proper channel cross-section must have adequate hdraulic capacit for a minimum cost of construction and maintenance.

14 Tpical Cross Sections The cross-sections of unlined channels are recommended as trapezoidal in shape with side slopes depending mainl on the kind of foundation material (considering construction techniques and equipment, and stabilit of side inclination, the United States Bureau of Reclamation (USBR) and the Turkish State Hdraulic Works (DSİ) suggest standard 1.5H:1V side slopes for trapezoidal channels)

15 Recommended side slopes Material Side Slope (H:V) Rock Nearl vertical Muck and peat soils ¼: 1 Stiff cla or earth with concrete lining ½:1 to 1:1 Earth with stone lining or earth for large channels 1:1 Firm cla or earth for small ditches 1.5:1 Loose sand earth :1 Sand loam or porous cla 3:1

16 Recommended side slopes Kızılkaa Dam

17 Fig. 1 Recommended side slopes

18 DESIGN OF NONERODIBLE CHANNELS For nonerodible channels the designer simpl computes the dimensions of the channel b a uniform-flow formula and then final dimensions on the basis of hdraulic efficienc, practicabilit, and econom. Minimum Permissible velocit In the design of lined channels the minimum permissible velocit is considered to avoid deposition if water carries silt or debris V min 0.75 m/s (non-silting velocit)

19 The determination of section dimensions for nonerodible channels, includes the following steps: All necessar information, i.e. the design discharge, the Manning s n and the bed slope are determined. Compute the section factor, Z, from the Manning equation If the expressions for A and R for the selected shape are substituted in the equation, one obtains 3 unknowns (b,, z) for trapezoidal sections, and unknowns (b,) for rectangular sections. Z Q ( ) b+ z P b+ 1+ z A AR Q, n, S o /3 1 n AR / 3 S 1 / o [( b + z) ] 5 / 3 [ ] / 3 b + 1+ z Various combinations of b, and z can be found to satisf the above section factor Z. The final dimensions are decided on the basis of hdraulic efficienc, practicabilit and econom.

20 Methods and Procedures 1. Assume side slope z. get the value of b from the experience curve, 3. Solve for. Z AR /3 [( b + z) ] 5 / 3 [ ] / 3 b + 1+ z Experience Curves showing bottom width and water depth of lined channels

21 Best Hdraulic Section: 1) substitute A and R for best hdraulic section in Eq(*), )Solve for Ex. Trapezoidal sections : z 1 3 A 3 b R T

22 Checks 1) In the proximit of critical depth, flow becomes unstable with excessive wave action, hence it is recommended that: for subcritical flows: > 1.1 c (or Fr < 0.86) for supercritical flows: < 0.9 c (or Fr > 1.13) ) ) Check the minimum permissible velocit if the water carries silt. (V min > 0.75 m/s)

23 Freeboard The freeboard, f, is determined b an empirical equation f 0. (1+) where, f is the freeboard (m) is the water depth (m) or b the curves given in Figure 1 for irrigation canals for the USBR and DSI practices.

24 Finalization 1) Modif the dimensions for practicabilit ) Add a proper freeboard to the depth of the channel section. Recommended freeboard for canals is given in figure. 3) Draw channel cross section and show dimensions and given parameter.

25 Open Channel Design Example 1a A trapezoidal channel carring 11.5 m 3 /s clear water is built with concrete (nonerodible) channel having a slope of and n Proportion the section dimensions. SOLUTION : Q 11.5 m 3 /s S n0.05 A / 3 1/ Q R So n [( b + z) ] [ ] b + 1+ z / 3 Assume b 6m and z, Solve for 1.04 m (b trial an error) b 5 / 3 z Z n * Q S AR /3 [( b + z) ] 5 / 3 [ ] / 3 b + 1 z o +

26 Q 11.5 m 3 /s S n0.05 b 6m and z, 1.04 m For given Q from DSİ s curve Height of lining above water surface 0.33m Height of bank above water surface 0.63m Check stabilit : At critical flow c 0.69m Q T F r ga 3 1 For n Fr 0.48 subcritical and within limits Velocit V Q/A 1.37 m /s OK for sedimentation. F r F r V gd V g( A/ T )

27 Open Channel Design Example 1b A trapezoidal channel carring 11.5 m 3 /s clear water is built with concrete (nonerodible) channel having a slope of and n Proportion the section dimensions. Use experience curve and z1.5 SOLUTION : Q 11.5 m 3 /s S n0.05 z 1 [( b + z) ] b 5 / 3 [ ] b + 1+ z / Take b.5m, Solve for 1.56m (b trial an error)

28 Q 11.5 m 3 /s S n0.05 b.5m and z 1.5, 1.56m Check stabilit : At critical flow c 0.69m For n Fr 0.47 subcritical and within limits Velocit V Q/A 1.5 m /s OK for sedimentation.

29 Z Open Channel Design Example 1c A trapezoidal channel carring 11.5 m 3 /s clear water is built with concrete (nonerodible) channel having a slope of and n Proportion the section dimensions. Use best hdraulic section approach! SOLUTION : Q 11.5 m 3 /s S n0.05 n * Q S o Solve for.03 m (b trial an error) b AR /3 z z T A Best Hdraulic Section for Trapezoidal Channel b R 3 3

30 Q 11.5 m 3 /s S n0.05 b.34m.03m Check stabilit : At critical flow c 0.69m For n Fr 0.38 subcritical and within limits Velocit V Q/A 1.49 m /s OK for sedimentation.