# 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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