Contents preface (VII) introduction 1—37

Table 11.5 Designed parameters of expanding transition (Ex. 11.1)
Sec-		Vittal and Chiranjeevi method			Hinds’ method
tion	x	B (m)	m	z	h (m)	∆h (m)	h (m)	z	v (m/s)	m	B (m)
1	2	3	4	5	6	7	8	9	10	11	12
f-f	0.000	15.000	0.000	0.000	6.006	0.000	6.006	0.000	3.963	0.000	14.999
1	4.000	15.034	0.100	0.025	6.088	0.013	6.019	0.020	3.793	0.100	15.037
2	8.000	15.140	0.206	0.050	6.173	0.053	6.059	0.040	3.614	0.200	15.090
3	12.000	15.325	0.318	0.074	6.255	0.119	6.125	0.061	3.427	0.300	15.172
4	16.000	15.601	0.438	0.096	6.327	0.211	6.217	0.081	3.228	0.400	15.302
5	20.000	15.980	0.569	0.117	6.408	0.330	6.336	0.101	3.017	0.500	15.510
6	24.000	16.482	0.712	0.135	6.463	0.239	6.461	0.121	2.789	0.600	15.935
7	28.000	17.138	0.874	0.152	6.517	0.140	6.560	0.141	2.541	0.800	16.168
8	32.000	18.002	1.063	0.168	6.572	0.067	6.633	0.162	2.266	1.000	17.115
9	36.000	19.196	1.302	0.183	6.613	0.021	6.679	0.182	1.953	1.200	19.350
c-c	41.000	23.000	2.000	0.203	6.700	0.000	6.700	0.207	1.472	2.000	22.795

Steps for the design of the expanding transition using the Hinds’ method would be as follows:

Consider the same length of the transition as obtained earlier i.e., 41 m. One could, alternatively, select a suitable length on the basis of minimum splay consideration. Consider also the same condition of constant specific energy so that the velocity and the depth at the flume end are the same as obtained earlier.
Using Eq. (11.14) at the mid-section (i.e., = 41/2 = 20.5 m from the transition end) of the transition where,

	∆h = (6.7 – 6.006)/2 = 0.347 m
∴	C = ∆h/x^2 = 0.347/(20.5)^2	= 8.26 × 10^–4
	1
Thus, Eq. (11.14) becomes	∆h = 8.26 × 10^–4 x^2

For section 3-3, x = 12 m (from the flume end)
∆h = 0.119 m
h = 6.006 + 0.119 = 6.125 m
Similarly, for section 8-8, x = 9 m (from the canal end)
∆h = 0.067 m
h = 6.700 + 0.067 = 6.633 m
The values of ∆h and h for other sections of the chosen sub-reaches of the transition are computed similarly and the values have been tabulated in cols. 7 and 8 of Table 11.5.
Head loss in transition, h_{L, f – c} is given by Eq. (11.10),
h_{L,f – c} = 0.3 (v_f² – v_c²)/2g

= 0.3 [(3.963)² – (1.464)²]/(2 × 9.81) = 0.207 m

For constant specific energy, the change in bed elevation between the two ends of the transition would be equal to the head loss between those ends. Therefore,
∆z_{f – c} = 0.207 m

402 IRRIGATION AND WATER RESOURCES ENGINEERING
This drop (transition being an expanding one) of 0.207 m is assumed to be linear and the values are listed in col. 9 of Table 11.5. Writing Eq. (11.10) between flume and any section,
h_{L,f – i} = 0.3 (v_f² – v_i²)/2g




v_i² = (15.705 − 65.4 × h_{L, f − i}
Since h_{L, f – i} is equal to ∆z_{f – i} , one can compute v_i. The computed values are listed in col. 10 of Table 11.5. The transition width, B at any section can now be computed, using Eq. (11.7), if the side slopes at various sections of the transition are known. These side slopes are to be assumed arbitrarily so that the resulting profile of the transition is smooth and also feasible. For the chosen values of side slope as listed in col. 11 of Table 11.5, the computed values of the transition widths at various sections are as shown in col. 12 of Table 11.5. One needs to adjust the width suitably or try another trial with other set of the values of the side slopes.
Design of Contracting Transition:
The bed-line profile of the contracting transition can be obtained from Eq. (11.3). Adopting a value of m₁ equal to 4, Eq. (11.3) becomes

	F	x I ^2	F	yI	2
	G		J	+ G		J	= 1
	H	16K	H	4K
Length of transition	= 0.5 m1(Bc – Bf)
	= 0.5 × 4(23 – 25) = 16 m

Dividing the contracting transition reach into eight sub-reaches, one can determine the values of y and, hence, the bed width (= B_c – 2y) for known values of x measured from the flume end of the transition.

The side slopes and bed elevation may be varied linearly. The computed values of bed width and side slopes are shown in Table 11.6.
Table 11.6 Values of bed width and side slopes of contracting transition (Example 11.1)

x, (m)	0	2	4	6	8	10	12	14	16
y, (m)	4.0	3.97	3.87	3.71	3.46	3.12	2.65	1.94	0.0
B = Bc – 2y, (m)	15.00	15.06	15.25	15.58	16.08	16.76	17.70	19.12	23.00
m	0	0.25	0.5	0.75	1.0	1.25	1.5	1.75	2.00

The trough can be divided into three compartments each of width 5.0 m, separated by two intermediate walls 0.30 m thick. The two side walls may be kept 0.60 m thick and 7.3 m high so that a freeboard of 0.60 m is available over 6.7 m depth of flow. For further illustration, let the thickness of the bottom slab be 0.60 m. These dimensions yield the overall width of the trough (i.e., the length of siphon barrel) equal to 15 + (0.3 × 2) + (0.6 × 2) = 16.8 m.