CE 356 Elements of Hydraulic Engineering

CE 356 Elements of Hydraulic Engineering

CE 356 Elements of Hydraulic Engineering Specific Energy (Alternate Depths) Hydraulic Jump (Sequent Depths) Energy in Open Channels hL v2/2g h y WS EGL HGL

v Channel Bottom 2 z 1 Datum V2 Q2 Head y z y z E z 2g 2 gA2 02/03/20

Specifi c Energ 2 Specific Energy, E E = energy (head) measured with respect to the channel bottom E = V2/2g + y = q2/(2gy2) + y Multiply through by y2 and arrange to find y3 E y2 + q2/2g = 0 What kind of equation is this? How many roots? Significance of roots? 02/03/20 3 Specific Energy Diagram

10.0 Rectangular channel: E=y 8.0 Q = 600 ft3/s. Depth (ft) B = 20 ft Fr = 1 6.0 4.0

Fr = 2 Fr = V/(gy)0.5 2.0 0.0 0.0 2.0 4.0 6.0 8.0 10.0 Specific Energy (ft) 02/03/20

4 Example: Sluice Gate q = 30 ft2/s y = 5.6 ft y = 1.8 ft 02/03/20 5 figure 1 Hydraulic jump Note that there is head loss in an hydraulic jump 02/03/20 6

jump Hydraulic jump on Rattan Creek, TX. July 2, 2002 strong jump: Fr > 9.0, rough wavy surface downstream steady jump: 4.5

8 picture 02/03/20 9 figure 2 02/03/20 10 jump cv Now lets consider an hydraulic jump We know that or

02/03/20 Q v1A1 v1 Q A1 v 2A2 ; v2 Q A2 11 momentum balance Our momentum balance

1 1 gA1y1 gA 2 y 2 v x1 Q v x 2 Q 2 2 can be written as rearranging and dividing by g: 1 Q2 Q2 g A1y1 A 2 y 2 2 A1 A2 Q2 A1y1 Q2 A 2 y 2

gA1 2 gA 2 2 Using A=By for a rectangular channel Q 2 By12 Q2 By 2 2 gB1 2 gBy 2 2 Lets go back to our hydraulic jump 02/03/20

12 solvable set Q2 By12 Q2 By 2 2 gB 2 gBy 2 2 If we have an open channel where we know Q, B, and y1, we can solve for y2 02/03/20

13 rectangular Starting from Q2 By12 Q2 By 2 2 gB 2 gBy 2 2 Recalling the definition of the flow rate per unit width we can obtain

Solving for y2 Q q B y12 y 2 2 q 2 1 1 2 2 g y 2 y1 y1 8q 2 y2 1 1 3 2 gy1 y1 solution is similar (see text) 02/03/20

14 limits Now, what are the limitations of this equation? y1 8q 2 y2 1 1 3 2 gy1 shear forces are neglected shear causes smaller head loss than turbulence in the jump requires jump occurs over a short distance rectangular channel

effect of gravity downslope is neglected again requires jump to occur over a short distance neglect non-hydrostatic pressure uniform density fluid 02/03/20 15 principle and what is the fundamental principle for this equation? y1 8q 2 y2 1 1 3 2 gy1

conservation of momentum 02/03/20 16 EM graphs 02/03/20 17 head loss The more general head loss formula is V12 h L E1 E 2 y1

2g V2 2 y 2 2g Valid for any cross section 02/03/20 18

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