Uniform flows and gradually varied flows
5.3 EXERCISES
What are the basic assumptions for GVF calculations?
What is the definition of the normal depth?
What is the definition of a steep slope? Does the notion of steep slope depend only on the bed slope? Discuss your answer. A discharge of 100 w?/s flows in a 12 m wide concrete channel. The channel slope is 5 m/km. Calculate: (a) normal depth, (b) critical depth and (c) indicate the slope type: steep or mild.
A smooth-concrete channel carries a water flow. The channel cross-section is rectangular with a 2 m bottom width. The longitudinal bed slope is 1.3 m/km (i.e. SQ = 0.0013). The uniform equilibrium flow depth equals 3.1m.
(a) Compute the following flow properties: flow velocity, flow rate, average bed shear stress and critical flow depth.
(b) For the uniform equilibrium flow, where would be the optimum location of a control?
Assume ks= I mm for the finished concrete lining.
An artificial canal carries a discharge of 25 m^/s. The channel cross-section is trapezoidal and symmetrical, with a 0.5 m bottom width and 1V:2.5H sidewall slopes. The longitudinal bed slope is 8.5 m/km. The channel bottom and sidewall consist of a mixture of fine sands (ks = 0.3 mm), (a) What is the normal depth at uniform equilibrium? (b) At uniform equilib- rium what is the average boundary shear stress? (c) At uniform equilibrium what is the shear velocity?
A gauging station is set at a bridge crossing the waterway. The observed flow depth, at the gauging station, is 2.2 m. (c) Compute the flow velocity at the gauging station, (d) Calculate the Darcy friction factor (at the gauging station), (e) What is the boundary shear stress (at the gauging station)? (f) How would you describe the flow at the gauging station? (g) At the gauging station, from where is the flow controlled? Why?
5.3 Exercises 109
Solution: (a) d^ = 1.23 m, (b) r^ = 51 Pa, (c) F* = 0.226 m/s, ( d ) / =0.0115, (e) r^ = 5.2 Pa, (f) turbulent, transition between smooth-turbulent to rough-turbulent (Section 4.4.1) and (g) downstream because Q/ IgA^/B = 0.56 (subcritical flow) (see Section 3.3.4).
The Cotter dam (Canberra ACT) is a 27 m high un-gated weir. The spillway is 60 m wide and a 50°
slope. At the toe of the spillway, the slope changes from 50° to 0.05°. The weir is followed by a 20 km long channel with a 60 m width. The channel slope is 0.05°. Assuming that the spillway and the channel are made of smooth concrete, compute the free-surface profiles above the spillway and in the channel for a discharge of 500m^/s.
Solution: In the reservoir upstream of the weir, the flow motion is tranquil. The flow above the spillway is expected to be rapid (to be checked). Hence a transition from slow to rapid flow motion occurs at the spillway crest: i.e. critical flow conditions. The channel downstream of the weir has a long section of constant slope, constant roughness, constant cross-sectional shape and constant shape. Uniform flow conditions will be obtained at the end of the channel. Note: if the uniform flow conditions are subcritical, a transition from supercritical to subcritical flow is expected between the end of the spillway and end of the channel.
Critical and uniform flow calculations
There is only one discharge and channel shape, and it yields to only one set of critical flow con- ditions dc = 1.92 m, and V^ = 4.34 m/s.
Assuming a roughness height k^ = 1 mm, the uniform flow conditions for the spillway and the downstream channel are:
Jo(m) To (m/s) Fr^
Spillway 0.28 29.7 17.9 Channel 2.27 3.67 0.78
Hydraulic controls
The preliminary calculations indicate that (1) the spillway slope is steep and (2) the channel slope is mild. Hence critical flow conditions occur at two locations (A) at the spillway crest, transition from subcritical to supercritical flow and (B) in the channel, transition from supercritical to sub- critical flow (i.e. hydraulic jump).
Backwater calculations
The easiest location to start the backwater calculations are the spillway crest. The flow is critical and hence we know the flow depth and flow velocity. By applying the energy equation, we can deduce the free-surface location at any position along the spillway.
In the channel downstream of the spillway, the first calculations predict the occurrence of a hydraulic jump. In such a case, we can assume that the flow depth downstream of the hydraulic jump equals the uniform flow depth. We can deduce then the flow depth immediately upstream of the jump (d = 1.61m, Fr = 1.30) by applying the momentum and continuity equations across the hydraulic jump.
The location of the hydraulic jump is deduced from the backwater calculations. Starting at the bottom of the spillway, the flow is decelerated along the channel. When the flow depth equal the sequent depth (i.e. 1.61 m), the hydraulic jump takes place.
Considering a 20 m high un-gated weir with a stepped chute spillway, the spillway is 20 m wide and the slope is 45°. At the toe of the spillway, the slope changes from 45° to 0.1°. The spillway is followed by a dissipation basin (10 m long and 20 m wide) consisting of concrete baffle blocks and then by a 800 m long concrete channel with a 25 m width. The channel slope is 0.1°.
Assume that the friction factor of the stepped spillway is 1.0, and the friction factor of the baffle block lining i s / = 2.5. Compute the free-surface profiles above the spillway and in the channel for a discharge of 160m^/s. Is the dissipation basin operating properly?
Notes
1. A dissipation basin or stilling basin is a short length of paved channel placed at the foot of a spill- way or any other source of supercritical flow. The aim of the designer is to make a hydraulic jump within the basin so that the flow is converted to subcritical before it reaches the exposed riverbed or channel downstream (Henderson, 1966).
2. Baffle blocks or chute blocks consist of reinforced concrete blocks installed to increase the resist- ance to the flow and to dissipate part of the kinetic energy of the flow.