Q = α n AR2/3 h S1/2 0. bd 2d + b d if b d. 0 = ft1/3 /s

Transcription

1 CEE 330 Open Channel Flow, Nov., 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 ) 8.9 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? R h = A P = Q = α n AR/3 h S/ 0 bd d + b d if b d therefore Q = α n bd5/3 S / 0 =.486 ft/3 /s (50 ft)(.39 ft) 5/ 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

2 7 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. Therefore: P = 0 + ( d 5 ) + d. Q = α n A 5/3 P /3S/ 0 =.486 ft/3 /s [( ) ] 0 + d 5 5/3. d ] (d ) / = 69 ft 3 /s 5 + d [ 0 + 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.

3 CEE 330 Open Channel Flow, Nov., used the function fzero in Matlab which is a nonlinear root finder and found the zeroes to the expression: residual = Q α n A 5/3 P /3S/ 0 with an initial guess of d=3 ft. The result is d =.8 ft. 8.0 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 V g + y + S 0 L = V g + y + S f L hence V g + y = V g + y + (S f S 0 )L 8.0. Specific Energy Define E = y + V g which is known as the specific energy.

4 74 The specific energy is seen to be the height of the EGL above the bed. Let q = Q/b = V y 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 = V y = 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 = Hence at the inflection point in the specific energy curve we have Fr =, y = y c, E = E min = 3 y C

5 CEE 330 Open Channel Flow, Nov., 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. 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): Rearranging to be a polynomial in y E = E + h = y + q gy we find: + h 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). 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.

6 76 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 > ) The specific energy plot for each flow looks like:

7 CEE 330 Open Channel Flow, Nov., y/y c E/y c 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:

8 78 a) Fr < and Fr 3 < b) Fr > and Fr 3 > c) Fr < and Fr 3 > The specific energy plot for each flow looks like:

9 CEE 330 Open Channel Flow, Nov., y/y c E/y c