Question 10.4: Examine the stage–discharge relationship for the culvert in ......

Examine the stage–discharge relationship for the culvert in Worked example 10.3 if the bedslope is 1 in 100.

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Rising stages are as follows.

1. For the open channel, preliminary calculations now indicate that the slope is steep and hence the entrance is the control, with the critical depth at the entry. The energy equation at the inlet gives

$H=1.5 V^2 / 2 g+y=1.75 \,y_{ c } \text {. }$ (v)

Table 1

2. For the orifice (equation (iii)),

$Q=C_{ d }(1.2 \times 0.6)[2 g(H-D / 2)]^{1 / 2}$ . (iii)

Table 2

3. For pipe flow (equation (iv) changes):

$Q=2.08(H-0.57)^{1 / 2}$ (iv)

Q=2.08(H-0.3)^{1 / 2} \text { for } S_0=1 / 100 .

Table 3

Table 1

$y_c$ (m)	H (m)	Type	$Q\left(m^3 s^{-1}\right)$
0.2	0.35	Free	0.336
0.4	0.70	Free	0.951
0.6	1.50 ( >1.2D)	Submerged	–
0.411	0.72 ( =1.2D)	Just free	0.990

Table 2

H (m)	Type	$Q\left(m^3 s^{-1}\right)$	$\left.y_0( m ) \text { (equation }(i) \text { with } S_0=1 / 100\right)^*$
0.72	Orifice	1.29	0.336
1.00	Orifice	1.66	0.44
2.00	Orifice?	2.58	0.61(>D)
1.95	Orifice	2.54	0.60
${ }^* Q=9.23 y_0\left[1.2 y_0 /\left(1.2+2 y_0\right)\right]^{2 / 3}$

Table 3

H (m)	Type	$Q\left(m^3 s^{-1}\right)$	$y_{0}$ (m) (equation (i))
1.95	Pipe flow	2.67
2.00	Pipe flow	2.71
3.00	Pipe flow	3.42
Falling stages
3.00	Pipe flow	3.42
2.00	Pipe flow	2.71
1.95	Pipe flow	2.67
1.74	← Pipe flow	← 2.50	← 0.60
1.74 →	Orifice →	2.37
1.00	Orifice	1.66
0.72	Orifice	1.29
0.72	Just free	0.99
0.70	Free	0.951
0.35	Free	0.336