Gradually Varied Flow I+II Hydromechanics VVR090
Gradually Varied Flow Depth of flow varies with longitudinal distance. Occurs upstream and downstream control sections. Governing equation: dy dx So Sf = 1 Fr (previously S f = 0 was studied)
Derivation of Governing Equation Total energy: H = z+ y+ u g Differentiating with respect to distance: ( / ) dh dz dy = + + d u g dx dx dx dx
dh dx = S f (slope of energy grade line) dz dx = S o (bottom slope) For a given flow rate: ( / ) d u g Q da dy Q T dy dy = = = Fr 3 3 dx ga dy dx ga dx dx Resulting equation: dy dx So Sf = 1 Fr
Definition of Water Surface Slope Water surface slope dy/dx is defined with respect to the channel bottom. Hydrostatic pressure distribution is assumed (streamlines should be reasonably straight and parallel).
Assumptions made when solving the gradually varied flow equation: The head loss for a specific reach is equal to the head loss in the reach for a uniform flow having the same R and u. Manning equation yields. S f = nu R 4/3 The slope of the channel is small No air entrainment Fixed velocity distribution Resistance coefficient constant in the reach under consideration
Classification of Gradually Varied Flow Profiles The following conditions prevail: If y < y N, then S f > S o If y > y N, then S f < S o If Fr > 1, then y < y c If Fr < 1, then y > y c If S f = S o, then y = y N
Water surface profiles may be classified with respect to: the channel slope the relationship between y, y N, and y c. Profile categories: M (mild) 0 < S o < S c S (steep) S o > S c > 0 C (critical) S o = S c A (adverse) S o < 0
Gradually Varied Flow Profile Classification I
Gradually Varied Flow Profile Classification II
Mild Slope (M-Profiles) 0 < S o < S c Profile types: 1: y > y N > y c => S o > S f and Fr < 1 => dy/dx > 0 : y N > y > y c => S o < S f and Fr < 1 => dy/dx < 0 3: y N > y C > y => S o < S f and Fr > 1 => dy/dx > 0
Steep Slope (S-Profiles) 0 < S c < S o Profile types: 1: y > y c > y N => S o > S f and Fr < 1 => dy/dx > 0 : y c > y > y N => S o > S f and Fr > 1 => dy/dx < 0 3: y c > y N > y => S o < S f and Fr > 1 => dy/dx > 0
Final Form of Water Surface Profile Asymptotic conditions: 1. y Æ, S f Æ 0, Fr Æ 0, and dy/dx Æ S o. y Æ y N, S f Æ S o, and dy/dx Æ 0 3. y Æ y c, Fr Æ 1, and dy/dx Æ dy dx So Sf = 1 Fr
Transition from Subcritical to Supercritical Flow
Transition from Supercritical to Subcritical Flow
Example: Flow into a Channel from a Reservoir
Flow Controls Locations in the channel where the relationship between the water depth and flow rate is known (or controllable). Controls: determine the depth in channel either upstream or downstream such points. usually feature a change from subcritical to supercritical flow occur at physical barriers, for example, sluice gates, dams, weirs, drop structures, or changes in channel slope
Strategy for Analysis of Open Channel Flow Typical approach in the analysis: 1. Start at control points. Proceed upstream or downstream depending on whether subcritical or supercritical flow occurs, respectively
Computation of Gradually Varied Flow Governing equation: dy dx So Sf = 1 Fr Solutions must begin at a control section and proceed in the direction in which the control operates. Gradually varied flow may approach uniform flow asymptotically, but from a practical point of view a reasonable definition of convergence is applied.
Uniform Channel Prismatic channel with constant slope and resistance coefficient. Apply energy equation over a small distance Dx: d dx u y+ = So S g f Express the equation in difference form: u Δ y+ = ( S ) o Sf Δx g
Over the short distance Dx assume that Manning s equation is suitable to describe the frictional losses (S f ): S f = nu R 4/3 The equation to be solved may be written: Δ x = o ( y u /g) ( 4/3 / ) Δ + S n u R mean
y i y i+1 Reach i u i u i+1 Dx i x Δ x = i ( ) ( y+ u / g y+ u /g) i+ 1 ( 4/3 S / ) o n u R i+ 1/ i All quantities known at i. Assume y i+1 and compute Dx i (u i+1 given by the continuity equation).
Example 6.1 A trapezoidal channel with b = 6.1 m, n = 0.05, z =, and S o = 0.001 carries a discharge of 8 m 3 /s. If this channel terminates in a free overfall, determine the gradually varied flow profile by the step method. y N b = 6.1 m 1
Solution: Compute normal water depth. = 1 n /3 Q AR So ( ) A= b+ zy y N P = b+ y 1+ z N ( + ) N b zyn yn R = b+ y 1 + z N y N = 1.91 m
Compute critical water depth: Fr u Q = 1 = = gd A ga / T c c c ( ) A = b+ zy y c c c T = b+ zy c y c = 1.14 m y N > y > y c Mild slope (y N > y c ) M profile
Table for step calculation: y A P R u u /g S f S fav Dx S (Dx) 1.14 9.55 11.0 0.85.93 0.438 0.0067 0.0058 3 3 1.4 10.64 11.64 0.91.63 0.353 0.0049 0.0044 9.3 1.3 1.3 11.54 1.00 0.96.43 0.300 0.0039 and so on 1 S = S + S S ( ) f, i+ 1/ f, i+ 1 f, i f nu = R 4/3 Δ x = i ( y+ u / g) ( y+ u /g) S o i+ 1 S f, i+ 1/ i
Other Solution Methods Problem with the step method is that the water depths is obtained at arbitrary locations (i.e., the water depth is not calculated at fixed x-locations). By direct integration of the governing equation this problem can be circumvented. Different approaches for direct integration: semi-analytic trial-and-error finite difference
Semi-Analytic Approach Find solution in terms of closed-form functions (integrals). Employ suitable approximations to these functions or some look-up tables. Approach OK for channels with constant properties. (for more information, see French)
Trial-and-Error Approach Well-suited for computations in non-prismatic channels. Channel properties (e.g., resistance coefficient and shape) are a function of longitudinal distance. Depth is obtained at specific x-locations. Apply energy equation between two stations located Dx apart (z is the elevation of the water surface): u Δ z+ = SfΔx h g u1 u z1+ = z + + Sf Δ x+ h g g e e h e : eddy losses
Estimate of frictional losses: 1 S = S + S ( ) f f1 f Equation is solved by trial-and-error (from to 1): 1. Assume y 1 Æ u 1 (continuity equation). Compute S f (and h e, if needed) 3. Compute y 1 from governing equation. If this value agrees with the assumed y 1, the solution has been found. Otherwise continue calculations.
Example 6.4 A trapezoidal channel with b = 0 ft, n = 0.05, z =, and S o = 0.001 carries a discharge of 1000 ft 3 /s. If this channel terminates in a free overfall and there are no eddy losses, determine the gradually varied flow profile by the trial-and-error step method. y N b = 0 ft 1
Solution Table Stn. z y A u u /g H 1 R S f S fav Dx h f H 0 103.74 3.74 103 9.71 1.46 105.0.81 0.00670 105.0 116 104.6 4.50 130 7.69 0.9 105.54 3.4 0.00347 0.00509 116 0.590 105.79 105.0 4.90 146 6.85 0.73 105.75 3.48 0.0051 0.00461 116 0.535 105.73 355 105.56 5.0 158 6.33 0.6 106.18 3.65 0.0001 0.006 39 0.540 106.7 105.93 5.3 173 5.78 0.5 106.45 3.85 0.00156 0.0004 39 0.74 106.47 745 106.34 5.60 175 5.71 0.51 106.85 3.89 0.00150 0.00153 490 1.14 107.59 106.96 6.1 01 4.98 0.385 107.34 4.1 0.00103 0.00130 490 0.97 107.4
Finite Difference Approach A range of numerical approaches are available to solve the governing equations based on finite differences. The equation is written in difference form and solved in terms of y: u Δ y+ = ( S ) o Sf Δx g Suitable for application on a computer (small length steps Dx might be needed). Can be applied for completely arbitrary channel configurations and properties.
Examples of Gradually Varied Flow Flow in channel between two reservoirs (lakes): 1. Steep slope, low downstream water level. Steep slope, high downstream water level 3. Mild slope, long channel 4. Mild slope, short channel 5. Sluice gate located in the channel
Steep Slope, Low Downstream Water Level Lake Critical section Hydraulic jump Lake Critical section at inflow to channel. Normal water depth occurs some distance downstream in the channel with Fr > 1 (y N < y cr ). A hydraulic jump develops before water is discharged to the downstream lake. Q in the channel depends on H 1 and critical section.
Steep Slope, High Downstream Water Level Lake No critical section Lake Fr < 1 in the channel, although it is steep Downstream water level is high enough to cause damming effects to the upstream lake. No critical section occurs in the inflow section. y > y cr > y N in the channel. Q depends on H 1 and H.
Mild Slope, Long Channel Lake Normal water depth Lake uniform flow non-uniform flow Mild slope and long channel implies that normal water depth occurs with y N > y cr. Normal water depth is also attained in the inflow section to the channel. Non-uniform flow develops in the downstream part of the channel before discharge to the lake. Q depends on H 1 and y N in the inflow section.
Mild Slope, Short Channel Lake Lake Non-uniform flow A short channel implies that normal water depth will not occur and y > y N > y cr. Non-uniform flow develops in the entire channel because of the downstream effects of the lake. Q depends on H 1 and H.
Sluice Gate Located in the Channel Lake Sluice gate (Q a function of y) Lake Jump Sluice gate cause damming upstream affecting inflow from lake. Discharge from sluice gate depends on upstream water surface elevation over gate opening. Supercritical flow occurs downstream the gate, followed by a hydraulic jump before the downstream lake is encountered. Q depends on H 1 and sluice gate properties.
Calculation Procedure for Some Gradually Varied Flows 1. Flow from a reservoir to a long, steeply sloping channel. Flow from a reservoir to a long, mildly sloping channel 3. Flow from a reservoir to a short, mildly sloping channel where a downstream water level affects the flow in the channel 4. Flow from a reservoir to a short, steeply sloping channel where a downstream water level affects the flow in the channel
Flow from a Reservoir to a Long, Steeply Sloping channel Lake Critical section occurs in inflow section. Employ energy equation from lake surface to inflow section. H 1 = y + cr Fr = 1 = u cr g u cr gy cr
Flow from a Reservoir to a Long, Mildly Sloping Channel Lake Lake uniform flow non-uniform flow Normal depth occurs in inflow section. Employ energy equation from lake surface to inflow section. un H1 = yn + g 1 u = R S n /3 1/ N N o
Flow from Reservoir to Short, Mildly Sloping Channel; Downstream Water Level Affects Flow in Channel Lake Lake non-uniform flow Downstream lake water level affects inflow from upstream lake. Non-uniform flow prevails. Q depends on H 1 and H. Assume Q = Q 1. Do a step calculation from downstream lake water level to inflow section. Employ energy equation from inflow section to upstrem lake water level. H 1 is regarded as unknown. Calculate for a new flow Q which gives a new upstream lake water level.
Make a plot of H 1 as a function of Q. Determine the correct Q based on the actual upstream lake water level H 1.
Flow from Reservoir to Short, Steeply Sloping Channel; Downstream Water Level Affects Flow in Channel Lake Lake Non-uniform flow Hydraulic Jump Non-uniform flow Critical section at inflow to channel. Make a step calculation from upstream lake and downstream lake. The hydraulic jump occur where the jump equation is satisfied.
Hydraulic jump is assumed to have negligible spatial extension. y y y y 1 1 1 = + ( ) 1 8Fr1 1 1 = + ( ) 1 8Fr 1