We will assume straight channels with simple geometries (prismatic channels) and steady state flow (in time).

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1 56 Review Drag & Lift Laminar vs Turbulent Boundary Layer Turbulent boundary layers stay attached to bodies longer Narrower wake! Lower pressure drag! 8. Open-Channel Flow Pipe/duct flow closed, full, gas or liquid. Pressure friction balance. Open-channel flow free-surface (river, flume, partially filled pipe,...). Gravity friction balance. Open-channel flow is often turbulent with complex geometries (rivers, estuaries, streams!). Hence we will simplify the geometry (often an excellent approximation)! We will assume straight channels with simple geometries (prismatic channels) and steady state flow (in time). Free-Surface at constant pressure atmospheric. This helps analysis! But free-surface position is unknown! This hinders analysis! There are entire books written on open-channel flow (French, Henderson, Chow,...), we will only discuss the basics. For more details keep an eye open for CEE 330 (may get a 4000 number) that will cover the details of unsteady open channel/river flow, sediment transport, and fundamental coastal engineering/water wave theory.

2 CEE 330 Open Channel Flow, Nov. 6, D Flow Assumption Consider the energy equation. At surface P = P = P atm, therefore: V g +z = V g +z +h f From our pipe flow work we should immediately suspect that it is a reasonably good approximation to consider: h f f L D h V avg g where L = x x and D h = 4A/P as previously defined. Note P is the wetted perimeter hence only the distance along the wetted sides of the channel. Now, we can relate V to V from conservation of mass since Q = Q V = V A A. We define an appropriate local Reynolds number as Re Dh = UD h /ν. If Re Dh > 0 5 the flow is turbulent and most are (classic exception is sheet flow of paved surfaces).

3 Flow Classification by Variation of Depth with Distance Along Channel Uniform Flow constant depth dy dx = 0. Varied Flow non-constant depth dy dx 0. Gradually Varied Flow dy dx. Rapidly Varied Flow dy dx. A picture: 8.4 Uniform Flow In uniform flow y = y, V = V = V, therefore the energy equation becomes: z z = S 0 L = h f Flow is essentially fully developed, therefore we can apply Darcy-Weissbach relations. h f = f L D h V avg g In open channel flow it is more common to work with the hydraulic radius R h : R h = A P = D h 4

4 CEE 330 Open Channel Flow, Nov. 6, 0 59 where again (P) is the length of the wetted perimeter. Combining the above two equations we have: S 0 L = f L D h V avg g V = ( ) / 8g (R h S 0 ) / f 8.5 Chézy Formulas Chézy defined the coefficient C = ( ) / 8g f (now called the Chézy coefficient) and found that it varies by a factor of 3. Therefore V = C(R h S 0 ) / and Q = CA(R h S 0 ) / Manning did field tests and found C = ( 8g f ) / α R/6 h n where n is known as Manning s n and is a roughness coefficient and α is a dimensional constant that varies with systems of units (this is not a homogeneous equation, remember?!). For SI α =, for BG α =.486. It is left as an exercise for the student to find the units and verify the conversion. 8.6 Manning s Equation Substituting Manning s result into the Chézy formulas we have the celebrated Manning s equation: V α n R/3 h S/ 0 and Q α n AR/3 h S/ 0 n varies by a factor of 5 and is tabulated in your text and more extensively elsewhere.

5 60 Review Open Channel Flow Gravity friction balance. In general we take an energy equation approach. y Uniform Flow x = 0 z = S 0L = h f where we could find h f from the Darcy-Weisbach friction facter and f(re Dh,ǫ/D h ) Chezy Formulas: C = ( ) / 8g f V = C(R h S 0 ) / Q = CA(R h S 0 ) / Manning s equation: V = α n R/3 h S/ 0 Q = α n AR/3 h S/ Examples Fall Creek Flow S , n = from table 0. in text, assuming somewhere between clean and straight and sluggish with deep pools values, b = 50,d =.39 from USGS gage on web, 05:30 EST. What is Q? Q = α n AR/3 h S/ 0 R h = A P = bd d+b d if b d therefore Q = α n bd5/3 S / 0 =.486 ft/3 /s (50 ft)(.39 ft) 5/

6 CEE 330 Open Channel Flow, Nov. 8, 0 6 Q = 40 CFS Actual value from calibrated flow gage was 69 CFS. Just using ball park estimates we were in the right range. To be more accurate we would need a survey of the river slope (likely greater than 0.00) and the wetted perimeter, perhaps also a bit greater than our 50 estimate). We could get these more accurately from a USGS topographic map but to be truly accurate we would send a survey team out to measure directly in the field. Second example - Given Q, find d If d b straightforward the equation is explicit and we just solve for d directly by substitution into Manning s equation. However, let s not make this assumption and let s consider a trapezoidal shaped channel with the geometry shown below and Q = 69 CFS, S 0 = 0.00, and n = 0.035: ( ) 0+b A = d Using Pythagoras theorem we can write (b 0 ) P = 0+ +d and Therefore A = b = 0+d 5. ( 0+d 5 ) d.

7 6 Therefore: Q = α n A 5/3 P /3S/ 0 =.486 ft/3 /s ( P = 0+ d 5 ) +d. [ [( ) ] 0+d 5 5/3. d 0+ (d 5. ) +d ] / = 69 ft 3 /s Ouch - what to do? We can solve this iteratively by guessing a d and then fiddling until we find Q = 69 CFS. Alternatively, we can use an equation solver to find the result. I used the function fzero in Matlab which is a nonlinear root finder and found the zeroes to the expression: residual = Q α A 5/3 np /3S/ 0 with an initial guess of d=3 ft. The result is d =.8 ft. 8.8 Energy Equation - Revisited Our last work with the energy equation left us with: V g +z = V g +z +h L But we can write h L = S f L. Defining ξ = z y we have ξ = ξ +S 0 L and hence V g +y +S 0 L = V g +y +S f L V g +y = V g +y +(S f S 0 )L

8 CEE 330 Open Channel Flow, Nov. 8, Specific Energy Define E = y + V g which is known as the specific energy. The specific energy is seen to be the height of the EGL above the bed. Let q = Q/b = Vy where b is the width of the channel. We can express the specific energy as E = y + q gy We see that this curve is a cubic in y let s look at the shape of this plot: 5 4 y/y c E/y c Let s look at the minimum in E. We can find its location (which we will denote y c, the critical depth) by solving E y = 0 = q gy 3 c ( ) q /3 ( Q y c = = g b g ) /3 Now we can define the minimum in the specific energy curve (E min ) ( ) q /3 E min = E(y c ) = + q g g q4/3 g /3 = q/3 q/3 + g/3 q /3 g = 3 /3 g = 3 /3 y c Defining q = Vy = V c y c we can substitute into our definition of y c c y c yc 3 = q g = V g V c = gy c V c = gy c Recall from Lab #3 and dimensional analysis that Fr = V gy Fr c = V c gyc = gyc gyc =

9 64 Hence at the inflection point in the specific energy curve we have Fr =, y = y c, E = E min = 3 y C thus the inflection point separates supercritical flow from sub-critical flow (recall Lab #3 for the definitions of super- and sub-critical flow). The upper portion of the specific energy curve is known as the subcritical branch (deeper, slower flows at a given energy) while the lower branch is known as the supercritical branch) shallower, faster flows at a given energy). 8.9 Example - Flow Over a Sill (Bump) Consider the following constant width channel: If we take the bed to be horizontal (S 0 = 0) away from the sill and the flow to be frictionless (S f = 0) then the Bernoulli form of the specific energy equation gives us (where q is a constant of the flow since the flow width is constant): E = E + h = y + q gy + h Rearranging to be a polynomial in y we find: y 3 +( h E )y + q g = 0 The coefficient terms to the polynomial are all known (E is the left-hand side boundary condition and hence y and q and thus E are all known for a well specified problem).

10 CEE 330 Open Channel Flow, Nov. 8, 0 65 Clearly this equation has three roots. It turns out that for h not too large it has 3 real roots, one of which is negative (and hence is not physically possibly as a negative water depth does not make sense), leaving two real positive roots that are viable flow depths. Let s consider the range of possibilities for this flow. a) Initial Flow Subcritical Upstream and Downstream (e.g., Fr < and Fr 3 < ) b) Initial Flow Supercritical Upstream and Downstream (e.g., Fr > and Fr 3 > ) c) Initial Flow Subcritical Upstream and Supercritical Downstream (e.g., Fr < and Fr 3 > )

11 66 The specific energy plot for each flow looks like: a) 5 4 y/y c E/y c b) 5 4 y/y c E/y c c) 5 4 y/y c E/y c

12 CEE 330 Open Channel Flow, Nov. 8, Example - Flow though a Contraction Consider the following channel of constant bottom elevation with vertical banks: As in our sill example we take the bed to be horizontal (S 0 = 0) and the flow to be frictionless (S f = 0). However, now q is no longer constant throughout the flow as b = b(x) q = Q/b = q(x). We consider the same three cases: a) Fr < and Fr 3 < b) Fr > and Fr 3 >

13 68 c) Fr < and Fr 3 > The specific energy plot for each flow looks like: a) 5 4 y/y c E/y c b) 5 4 y/y c E/y c

14 CEE 330 Open Channel Flow, Nov. 8, 0 69 c) 5 4 y/y c E/y c

15 70 Review Specific Energy E = y + V g = y + Q gb y = y + q gy where q = Q b = Vy At Fr = y = y c, E = E min = 3 ( q y c where y c = g ) /3 8. Critical and Choked Flows Looking at the previous two examples if the flow is in either of the two states (c) we see that the flow over the sill or through the throat (narrowest part of the channel) is critical. If the sill is raised at all or the throat is narrowed further (e.g., q is raised further) the flow depth upstream (y ) must increase in order to pass the flow. This is known as a choked flow. Note that the flow depth at the sill (throat) is still y = y c. Hence the flow is controlled at the sill (throat) and both the upstream and downstream flows are controlled by the sill (throat). This is possible as the flow upstream is subcritical hence information can propagate upstream from the control point to set the water depth upstream (y ) and the flow is supercritical downstream hence information can propagate downstream from the control point to set the downstream water depth (y 3 ). In fact the above explains why a fourth possible solution to the two example problems does not exist. Note we did not admit any solutions of the form Fr > and Fr 3 <

16 CEE 330 Open Channel Flow, Nov., 0 7 which would look like: As this would require that a supercritical flow be controlled from downstream and a subcritical flow be controlled from upstream which is not possible. How do supercritical flows transition to subcritical flows downstream? The hydraulic jump! 8. Rapidly Varied Flow 8.. The Hydraulic Jump As described a transition from super- to sub-critical flow. Extremely efficient energy dissipator Jump characteristics determined primarily by Fr If we start with the conservation of energy and the continuity equation (our usual starting point so far) applied to a rectangular open-channel flow of width b with a hydraulic jump

17 7 we have: Continuity Q = y bv = y bv = Q q = q = y V = y V Energy conservation E = E +h L y + V g = y + V g +h L where we assume that the jump occurs over a short enough distance that h L only includes energy dissipated by the flow in the jump and not by shear stresses at the boundaries (the reasons will become clear in a moment). knowns y,v,g unknowns y,v,h L Therefore we have two equations and three unknowns we need another equation conservation of linear momentum. Our picture: γ y y b γ y y b = ṁ out ṁ in = ρq(v V ) = ρv y b(v V )

18 CEE 330 Open Channel Flow, Nov., 0 73 Rearranging we have y y = V y g (V V ) Combing the above with continuity to eliminate V and solving for the ratio y we arrive y at ( ) y + y V ( ) y = + y Fr y gy y Solving this quadratic equation we have y y y = ± y +8Fr y = + +8Fr y since clearly y y > 0 and Fr >. This result can be combined with the energy and continuity equations to solve for the head loss (h L ) through the jump. 5 4 What does a hydraulic jump look like on a specific energy diagram? y/y c E/y c