CEE 3310 Open Channel Flow, Nov. 6, 018 175 8.10 Review Open Channel Flow Gravity friction balance. y Uniform Flow x = 0 z = S 0L = h f y Rapidly Varied Flow x 1 y Gradually Varied Flow x 1 In general we take an energy equation approach. For uniform flow with D H defined based on wetted perimeter we find: V = C (R h S 0 ) 1/ Chézy formula where R h = D h /4 where C = ( ) 1/ 8g is the Chézy coefficient f which is tabulated or can be found from the Moody diagram using f(re Dh, ɛ/d h ) where we could find h f from the Darcy-Weisbach friction facter and f(re Dh, ɛ/d h ) Manning found C = ( 8g f and hence Manning s equation: ) 1/ α R1/6 h n V = α n R/3 h S1/ 0 and for b d, R h d V α n d/3 S 1/ 0 and Q = V A α n bd5/3 S 1/ 0 Fall Creek Example Flow rate from depth in a natural river. Note - Depth (y) solved from Manning s equation for a given flow rate defines what is known as the normal flow depth (y N ) - the flow depth that results under uniform flow conditions, when the free surface slope is equal to the bed slope is equal to the slope of the energy grade line, due to a friction gravity balance. It may be super or sub critical. Specific Energy a local energy energy relative to the bed elevation: E = y+ V g = y+ Q q = y+ gb y gy where q = Q b = V y The Specific Energy Curve - solution to the cubic: c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 176 8.11 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): q E 1 = E + h = y + gy + h Rearranging to be a polynomial in y we find: y 3 + ( h E 1 )y + q g = 0 The coefficient terms to the polynomial are all known (E 1 is the left-hand side boundary condition and hence y 1 and q and thus E 1 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. c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 177 a) Initial Flow Subcritical Upstream and Downstream (e.g., Fr 1 < 1 and Fr 3 < 1) b) Initial Flow Supercritical Upstream and Downstream (e.g., Fr 1 > 1 and Fr 3 > 1) c) Initial Flow Subcritical Upstream and Supercritical Downstream (e.g., Fr 1 < 1 and Fr 3 > 1) The specific energy plot for each flow looks like: c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 178 5 4.5 4 3.5 3 y/y c.5 1.5 1 0.5 0 0 0.5 1 1.5.5 3 3.5 4 4.5 5 E/y c c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 179 8.1 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 1 < 1 and Fr 3 < 1 b) Fr 1 > 1 and Fr 3 > 1 c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 180 c) Fr 1 < 1 and Fr 3 > 1 The specific energy plot for each flow looks like: 0.3 b=0.6m Q=0.03m b min =b/3 0.5 0. y (m) 0.15 0.1 0.05 0 0 0.05 0.1 0.15 0. 0.5 0.3 E (m) c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 181 8.13 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 1 ) 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 1 ) 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 1 > 1 and Fr 3 < 1 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! You were introduced to this flow in Lab #. c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 18 8.14 Rapidly Varied Flow 8.14.1 The Hydraulic Jump As described a transition from super- to sub-critical flow. Extremely efficient energy dissipater 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 we have: Continuity Q 1 = y 1 bv 1 = y bv = Q q 1 = q = y V = y 1 V 1 Energy conservation E 1 = E + h L y 1 + V 1 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 1, V 1, g unknowns y, V, h L Therefore we have two equations and three unknowns we need another equation conc 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 183 servation of linear momentum. Our picture: γ y 1 y 1b γ y y b = ṁ out ṁ in = ρq(v V 1 ) = ρv 1 y 1 b(v V 1 ) Rearranging we have y 1 y = V 1y 1 g (V V 1 ) Combing the above with continuity to eliminate V and solving for the ratio y y 1 we arrive at ( y y 1 ) + y V 1 = y 1 gy 1 ( y y 1 ) + y y 1 Fr 1 Solving this quadratic equation we have y = 1 ± 1 + 8Fr 1 y 1 y = 1 + 1 + 8Fr 1 y 1 since clearly y y 1 > 0 and Fr > 1. This result can be combined with the energy and continuity equations to solve for the head loss (h L ) through the jump.what does a hydraulic jump look like on a specific energy diagram? c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 184 5 4.5 4 3.5 3 y/y c.5 1.5 1 0.5 0 0 0.5 1 1.5.5 3 3.5 4 4.5 5 E/y c c 018 Edwin A. Cowen
CEE 3310 Open Channel Flow, Nov. 6, 018 185 8.14. Broad-Crested Weir Consider the following flow: This is known as the broad-crested weir which is characterized by: Sufficiently short that energy loss due to channel friction is negligible h L = 0 Bernoulli s equation. Sufficiently long horizontal section that hydrostatic flow is a reasonable approximation pressure over horizontal section is hydrostatic. Therefore our starting point is the Bernoulli equation Aha! E 1 = E + h w V 1 g + H + h w = V g + h w + y Since the flow upstream is subcritical and there is a region where dy dx 1 just upstream of the weir the flow over the horizontal section must be a control and hence y = y c and V = Vc = gy c. Therefore we have Solving for y c we have V 1 g + H = y c + y c = 3 y c y c = V 1 3g + 3 H 3 H if V 1 g H Therefore a reasonable approximation of the flow rate over a weir is Q = by c V c = by c gyc = b gy 3/ = b gy 3/ = c 018 Edwin A. Cowen ( ) 3/ b gh 3 3
CEE 3310 Open Channel Flow, Nov. 6, 018 186 Now the reality is while the above is reasonable, there are energy losses and often broadcrested weirs are used as flow discharge measurement devices. Hence experimental calibration is often preferred and a weir discharge coefficient C d is experimentally determined. I.e., Q = C d b gh 3 The literature is full of different formulations such as equations 10.57 and 10.58 in your textbook which yields a maximal C d of about 0.54. c 018 Edwin A. Cowen