Question 10.1: Design a siphon aqueduct for the following data:...

DRAINAGE WATERWAY

Perimeter P=4.75 Q^{1 / 2} (régime width, equation (9.9) B = 4.75 \ Q^{1/2})≈106 m. Providing 12 piers of 1.25 m thickness, we have 13 spans of 7m each. Therefore waterway provided =13 \times 7+12 \times 1.25=106 \, m (satisfactory). Assuming a maximum velocity through the siphon barrels of 2 ms ^{-1} , height of barrel =500 /(13 \times 7 \times 2)=2.747 \, m . Provide rectangular barrels, 7 m wide and 2.75 m high (shown in Fig. 10.5).

CANAL WATERWAY

Since the drainage width is large (106 m at the crossing) it is economical to flume (concrete, n=0.014) the canal. Adopt a maximum flume ratio of 0.5. Therefore the flumed width of the canal (trough) = 0.5×25=12.5 m. Providing a splay of 2:1 in contraction and a splay of 3:1 in expansion (Hinds, 1928),

length of transitions in contraction = 12.5 m,
length of transitions in expansion = 18.75 m.
The length of the trough from abutment to abutment = 106 m.

DESIGN OF FLUMED SECTION WITH TRANSITIONS

Referring to Fig. 10.3, the following results can be obtained to maintain a constant depth of flow of 2.0 m (given). The calculations are achieved from section 44 and proceed towards section 11 as tabulated below:

Note that the contraction loss =0.2\left(V_2^2-V_1^2\right) / 2 g the expansion loss = 0.3 \left(V_3^2-V_4^2\right) / 2 g ; the flume friction loss =V_{ f }^2 n^2 L_{ f } / R_{ f }^{4 / 3}  (the suffix f denotes the flume, and o the original canal \left.-V_{ f }=V_3=V_2 ; V_{ o }=V_4=V_1\right) .

DESIGN OF TRANSITIONS

For a constant depth of flow the transition may be designed such that the rate of change of velocity per metre length of transition is constant. This approach yields the bed width of the transition at a distance x from the flume section as

which, modified after experimental studies (UPIRI, 1940), gives

where L is length of the transition.
The following table shows the calculated geometries of the transition provided:

The transitions are streamlined and warped to avoid any abrupt changes in the width.
Transitions with a cylindrical inlet with an average splay of 2:1 and a linear outlet with a splay of 3:1 provided with flow deflectors (Fig. 10.3; Ranga Raju, 1993) have been found to perform better than lengthy curved expansions.
As the flow is accelerating in a contracting transition and the energy loss is minimal any gradual contraction with a smooth and continuous boundary should be satisfactory, e.g. an elliptical quadrant is an alternative to a cylindrical quadrant for inlet transitions. The bedline profile for an elliptical quadrant transition has the equation

and the length of transition given by

At any location (x) from flume end of the transition y is computed and the bed width B_x   calculated by

The side slope (m) of the transition (m = 0 for flume section and m ≥ 2 for canal side slope) and bed elevation may be varied linearly along the transition length.
The expansion experiences considerable energy loss and care must be exercised in designing a hydraulically satisfactory transition.
On the basis of theoretical and experimental investigations Vittal and Chiranjeevi (1983) proposed the following design equations for bed width and the side slopes of an expanding transition. The bed widths B_x are fixed by

where

and the transition length, L=2.35\left(B_0-B_{ f }\right)+1.65 \, m_o y_{ o }, y_{ o } being the flow depth in the canal, and m_{ o } its side slope. The side slopes (m) along the transition are given by

\frac{m}{m_{ o }}=1-\left(1-\frac{x}{L}\right)^{1 / 2} .

Using the constant specific energy condition in the transition between canal and flume the depth in the flume, y_{ f } , and depths \left(y_x\right) along the transition length can be obtained. The energy balance between adjacent sections within the transition with expansion loss as 0.3\left(V_{ i }^2-V_{i+1}^2\right) / 2 g  gives the bed elevations to be provided at successive sections so that the specific energy remains constant throughout the transition. Worked example 10.2 provides detailed design calculations for an expanding transition based on the Vittal and Chiranjeevi method.

WATER SURFACE PROFILE IN TRANSITION

The water surface in the transition may be assumed as two smooth parabolic curves (convex and concave) meeting tangentially. Referring to Fig. 10.4, the following equations give such profiles in transitions:

inlet transition,                    y=8.96 \times 10^{-4} x^2 ;
outlet transition,                  y=2.33 \times 10^{-4} x^2 .

A highway 6m wide is provided alongside the canal by dividing the flume into two compartments by a 0.3 m thick partition. The entire trough (flume section) can be designed as a monolithic concrete structure. Provide side walls and a bottom slab of about 0.4 m (to be fixed by the usual structural design methods).

SIPHON BARRELS

Thirteen barrels, each 7 m wide and 2.75 m high, are provided; assume that the effective roughness, k=0.6 mm (concrete). The length of the barrel, L=12.50+0.30+2 \times 0.40=13.60 \, m . The head loss through the barrel, h_{ f }=(1.5+\lambda L / 4 R) V^2 / 2 g . The velocity through the barrel, V=500/(13×7×2.75) = 1.998 m s ^{-1} . The hydraulic radius, R=7 \times 2.75 /\{2(7+2.75)\}= 0.987 m. Therefore the Reynolds number =4 V R / v=8 \times 10^6 and k/4R = 1.5 \times 10^{-4} . Hence, from Moody’s chart, the friction factor \lambda=0.015 , giving h_{ f }=0.316 \, m . Therefore, the upstream HFL=200.500+0.316=200.816 m AOD.

The uplift pressures on the roof of the barrel are as follows. The RL of the bottom of the trough = 200.00-0.40=199.60 m AOD. The entry loss at the barrel =0.5 V^2 / 2 g=0.102 \,m . Therefore the pressure head inside the barrel just downstream of its entry = 200.816-0.102-199.600=  1.114 \,m \simeq 11 \,kNm ^{-2} \text {. }

The most critical situation arises when the canal is empty and the siphon barrels are full. The weight of the roof slab = 0.4×2.4×9.81 =9.42 kNm ^{-2} (assuming the relative density of concrete to be 2.4). Hence the roof slab needs additional reinforcement at its top to resist the unbalanced pressure forces (uplift pressures).

The total weight of the trough (when empty) needs to be checked against the total upward force and suitable anchorages to piers provided, if necessary. Equally, the trough floor slab has to be checked when it is carrying water at FSL and the level in the drainage is low, i.e. barrels running part full.

The uplift on the floor of the barrel (assuming the barrel floor thickness to be 1 m initially) is as follows:

RL of the bottom of the barrel =199.60-2.75-1.00=195.85 m AOD;

RL of the drainage bed =198.00 m AOD.

Therefore the static uplift on the floor=198.00-195.85=2.15 m (the worst condition with the water table as the drain bed level). The seepage head (a maximum when the canal is at FSL and the drainage is empty)=202.00-198.00=4.00 m.

In spite of the three-dimensional seepage flow pattern, Bligh’s creep length may be approximated as follows. Creep flow commences from the beginning of the upstream transition (downstream of this the floor is impervious) and enters the first barrel floor; from its centre the flow follows downstream of the drain and emerges at the end of the impervious concrete floor of the barrel. Therefore the total creep length can be approximated as

inlet transition length +\frac{1}{2} barrel span +\frac{1}{2} length of barrel impervious floor.

Let us assume that the total length of the impervious floor of the barrel is 25 m, consisting of the following:

Therefore provide the upstream floor (1:3) length= 25.00-21.55=3.45 m. The total creep length=12.5+7/2+25/2=28.5 m. The creep length up to the centre of the barrel=12.5+7/2=16.0 m. Therefore the seepage head at the centre of the barrel=4(1-16.0/28.5)=1.75 m. The total uplift is then 2.15+1.75=3.90 m \simeq 38 \,kN m ^{-2} , and the weight of the floor = 1.00×2.4×9.81=23.54 kN m ^{-2} . Hence additional reinforcement has to be designed to resist the unbalanced uplift forces.

UPSTREAM AND DOWNSTREAM PROTECTION WORKS

The scour depth, R_{ s } (regime scour depth, equation (9.10) ) =0.47(500 / 1)^{1 / 3} = 3.73 m.

R_{ s }=0.475(Q / f)^{1 / 3}           (9.10)

The upstream cut-off below HFL = 1.5 R = 5.6 m. Therefore

RL of upstream cut-off wall = 200.816 – 5.60 = 195.00 (say) m AOD.

The downstream cut-off below HFL = 1.75 R_{ s } =6.53 m. Therefore

RL of downstream cut-off wall =200.50-6.53=194.00 (say) m AOD,

downstream apron length =2.5(198.00-194.00)=10 m,

upstream apron length =2.0(198.00-195.00) =6 m.

The detailed layout (longitudinal section) of the design is shown in Fig.10.5.