- 2082 Kartik · 2+2+4 marks
Slope change 1/120 to 1/1500, jump and profile length
Short answer
Upstream slope is steep (y_n1=0.659 m < y_c=1.138 m), downstream is mild (y_n2=1.407 m). The jump occurs on the mild slope, from 0.905 m to 1.407 m. The M3 profile length from the break to the jump, computed by direct step (two steps), is 120.5 m.
A wide rectangular channel has longitudinal slopes changes from 1 in 120 to 1 in 1500 and carry flow of 3.8 m³/s per m width. Sketch the water surface profile for this arrangement. Will hydraulic jump occur? If yes, where does it occur: on upstream or downstream slope? Calculate also length of the profile using direct step method with 2 steps only. Take Manning's coefficient n = 0.012.
Given: wide rectangular channel, , , , . For a wide channel .
Normal and critical depths
Sketch
The upstream slope is steep (supercritical, m) and the downstream slope is mild (subcritical, m). Supercritical flow must change to subcritical through a hydraulic jump, with an M3 curve before it (see the jump location below).
STEEP | MILD
---- CDL 1.138 - - | - - - - - - - - - - - -
| J _____ yn2=1.407
______ yn1=0.659 __|__M3_____/'''|
Does a jump occur, and where?
Sequent depth of :
Since , the downstream depth is too low to force a jump at the break. A jump does occur, on the downstream (mild) slope. Flow follows an M3 curve from m upwards until its conjugate depth equals :
The jump takes the depth from 0.905 m to 1.407 m.
Length of the M3 profile (direct step, 2 steps, mild slope )
Steps: m.
| (m) | (m/s) | (m) | (m) | |||
|---|---|---|---|---|---|---|
| 0.6594 | 5.763 | 2.3522 | 0.008333 | |||
| 0.7823 | 4.858 | 1.9849 | 0.004714 | 0.006524 | -0.005857 | 62.7 |
| 0.9052 | 4.198 | 1.8034 | 0.002898 | 0.003806 | -0.003139 | 57.8 |
Answer: A jump occurs on the downstream (mild) slope, from 0.905 m to 1.407 m. The M3 profile from the break to the jump is about 120 m long.
✓ Answer checked by Claude Opus 5.5 (high effort)
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