Open Channel Flow Part 2 (cont)

1. Open Channel FlowPart 2 (cont) Non-uniform flow ERT 349 SOIL AND WATER ENGINEERING Siti Kamariah Md Sa’at PPK Bioproses, UniMAP

2. Non-uniform flow • i ≠ Sw ≠ So • dy/dx ≠0 • Changes in velocity or depth of unprismatic channel • Causes of depth changes along channel: • Changes of shape of channel • Changes of discharge along the channel • Avaibility of control structures • Changes of base slope

3. Non-Uniform flow • Two types: • Rapidly varied flow (RVF) • Gradually varied flow (GVF)

4. Open Channel Flow: Energy Relations velocity head energy grade line hydraulic grade line water surface Bottom slope (So) not necessarily equal to surface slope (Sf)

5. Non-Uniform flow i v2/2g H y Sw z So Datum

6. Specific Energy • H = Total Energy • P/ρg = Pressure Head • z = Measured from datum • α = correction factor (assume as 1) • v2/2g = kinetics head Bernoulli Equation

7. Specific Energy • Total Energy = Height + flow depth + velocity energy • If z =0, • E = Specific Energy (Unit: m) = water energy at one section measured as distance from base slope to energy line

8. Specific Energy • From v = Q/A

9. Specific Energy • For rectangular channel (prismatic and straight channel), • Where • q= Q/B = flowrate/width • A = By

10. Specific Energy • From equation on previous slide, E, y and q relationship are: • Relationship of E and y when q constant • Relationship of q and y when E constant

11. Relationship of E and y when q constant

12. E-y curve when q constant Specific Energy Diagram E = y q constant y1 Sub-Critical Flow yc Critical Flow C y2 SuperCritical Flow Emin E

13. E-y curve when q constant • At Point C, specific energy, E is minimum (Emin) and flow depth is critical, yc (critical depth) • For E points, there are two value of flow depth: • y1= subcritical flow (Fr < 1.0) • y2=supercritical flow (Fr>1.0) • y1 and y2 called alternate depth y>yc or v < vc subcritical flow Y<yc or v > vc supercritical flow

14. Relationship of q and y when E constant

15. Q-y curve when E constant y E constant y1 yc Critical y2 q q qmax

16. Q-y curve when E constant • Flow maximum when q=qmax when critical depth at yc • For certain q value, there are 2 y which is: • y1>yc = subcritical depth • y2<yc = supercritical depth

17. Specific Energy: Sluice Gate sluice gate q = 5.5 m2/s y2 = 0.45 m V2 = 12.2 m/s EGL 1 E2 = 8 m 2 Given downstream depth and discharge, find upstream depth. alternate y1 and y2 are ___________ depths (same specific energy) Why not use momentum conservation to find y1?

18. Specific Energy: Raise the Sluice Gate sluice gate EGL 2 1 as sluice gate is raised y1 approaches y2 and E is minimized: Maximum discharge for given energy.

19. Specific Energy: Step Up Short, smooth step with rise Dy in channel Given upstream depth and discharge find y2 Dy Increase step height?

20. Critical Flow • Emin dE/dy = 0, y=yc 2. Q max q=Qmax dq/dy = 0 3. Fr =1.0 Critical flow = unstable condition due to changes to E and changes of flow depth

21. Critical Flow • Characteristics • Unstable surface • Series of standing waves • Occurrence • Broad crested weir (and other weirs) • Channel Controls (rapid changes in cross-section) • Over falls • Changes in channel slope,So from mild to steep Difficult to measure depth

22. Broad-crested Weir E H yc Broad-crested weir P Hard to measure yc E measured from top of weir Cd corrects for using H rather than E.

23. Critical Flow in any shape channel yc Arbitrary cross-section Find critical depth, yc & A=f(y) T=surface width dy dA A y P Hydraulic Depth

24. Changes in channel slope,So So = Sc Critical Slope (C) So < Sc Mild Slope (M) So > Sc Steep Slope (S)

25. Fr=1 (i) & (ii) Critical Slope, Sc (i) (ii)

26. Example 1: • Water flowing with normal flow on rectangular channel width 5.0m and Manning, n = 0.017 with 150m3/s and depth 6.0m. Calculate • critical depth and determine the types of flow • critical slope • Critical velocity Ans: yc=4.51 (subcritical) Sc=6.78 x 10-3 Vc=6.65 m/s

27. Example 2 • Trapezoidal Channel, Q=28 m3/s with Manning, n=0.022, B=3.0 and side slope 1:2 • Calculate: • Critical depth • Critical slope • Critical velocity Ans: yc=1.5m (try and error method) Sc= 0.00523 Vc=3.11 m/s

28. Broad crested weir: Control Structure Total Energy Line Water Surface E1 Eo y1 yo h Base slope downstream upstream Eo = E1+h

29. Minimum dam height • hmin= minimum dam/weir height cause the critical depth, yc • hmin= hc • There are 3 case will happen: • 1. h<hmin = emergent dam • 2. h=hmin = rare case • 3. h>hmin = control dam

31. dy/dx=+ve Q Gradually Varied Flow • Can be divided to two: • Backwater • Drawdown dy/dx=-ve Q

32. Example 3: • Water flowing with normal flow 1.28m in a rectangular channel with 3.0m and Q=4.25m3/s. If the broad crested weir height 0.6m constructed across the channel, are the water depth upper the weir same with critical depth? Ans: ye=yc=0.589m

33. Example 4: • Water flowing with normal depth 2.5m in rectangular channel 2.8m width with Q=10.4m3/s. If the broad crested weir with 1.2m height constructed across this channel, determine water depth above the weir and illustrate the water profile. • Ans: 1. Calculate Eo= yo+q2/2gyo2=2.62 2. Calculate yc = 1.13 m 3. Calculate Emin=1.70 m 4. Calculate hmin=0.92m h=1.2>hmin, ye=yc=1.13m, yo=2.5m>yc

34. Hydraulic Jump • Used for energy dissipation • Occurs when flow transitions from supercritical to subcritical • base of spillway • We would like to know depth of water downstream from jump as well as the location of the jump • Which equation, Energy or Momentum?

35. Examples of hydraulic jump: • At downstream of sluice gate • At upstream weir/dam • At spillway

36. Types of hydraulic jump • Undular Fr 1.0-1.7 • Weak Fr 1.7-2.5 • Oscillating Fr 2.5-4.5 • Steady Fr 4.5-9.0 • Strong Fr > 9.0