Reservoir to lake channel with jump

Reading of the figure (assumed): bed levels A = 24 m, B = 20 m, C = 19.6 m; lake level at C = 20.8 m (depth 1.2 m). Channel width $b = 4$ m, $n = 0.025$, and $AB = BC = 100$ m. The reservoir level is unknown and is to be found.

$$S_{AB} = \frac{24 - 20}{100} = 0.04,\qquad S_{BC} = \frac{20 - 19.6}{100} = 0.004$$

a) Reservoir level

On AB the normal depth is 0.5 m ($A = 2\ \text{m}^2$, $P = 5$ m, $R = 0.4$ m):

$$Q = \frac{1}{n}AR^{2/3}S^{1/2} = \frac{1}{0.025}(2)(0.4)^{2/3}(0.04)^{1/2} = 8.686\ \text{m}^3/\text{s}$$ $$V = 4.343\ \text{m/s},\quad y_c = \left(\frac{(8.686/4)^2}{9.81}\right)^{1/3} = 0.783\ \text{m},\quad Fr = 1.96$$

AB is steep ($y_n = 0.5 < y_c = 0.783$). Flow from the reservoir enters at critical depth (the control), so

$$H = z_A + \tfrac{3}{2}y_c = 24 + 1.175 = 25.18\ \text{m}$$

(If uniform flow at 0.5 m is assumed from the entrance, the level is $24 + 0.5 + V^2/2g = 25.46$ m, a difference of 0.28 m.)

Reservoir water surface elevation $\approx 25.2\ \text{m}$.

b) Hydraulic jump on BC

BC: $S = 0.004$. Normal depth for $Q = 8.686$: $y_n = 1.085\ \text{m} > y_c = 0.783$ m, so BC is mild. Supercritical flow of 0.5 m enters it; the lake gives a depth of 1.2 m at C, which is above $y_n$, so a backwater (M1) curve falls to about 1.114 m at B.

Sequent depth of 0.5 m: $Fr = 1.96$, so $y_2 = \frac{0.5}{2}\left(\sqrt{1 + 8(1.96)^2} - 1\right) = 1.159$ m. This is greater than the M1 depth at B (1.114 m), so the jump cannot form at B; it moves downstream. The supercritical flow follows an M3 curve until its sequent depth equals the M1 depth. Marching both curves (gradually varied flow, step 0.01 m) gives:

c) Possible profiles

 A (res.)      B                  C (lake)
 ====\                              
      \ S2 (yc to 0.5)  0.5 m  M3 J   M1      1.2 m
       '-----_______________/'|-----------'''
 steep AB   uniform  | mild BC: M3, jump, M1 to lake

1. AB (steep): S2 curve from $y_c = 0.783$ m at the entrance to the uniform depth 0.5 m (reached well before B).
2. BC (mild), as computed: M3 rising from 0.5 m, a jump at about 2 m from B, then M1 rising to 1.2 m at the lake.
3. If the lake were higher (depth at B above the 1.159 m sequent depth of 0.5 m), the jump would move up onto the steep reach AB and be followed by S1, then M1 on BC.
4. If the lake were lower (depth at C below $y_n$ but above $y_c$), the profile on BC after the jump would be M2 instead of M1.