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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, q=3.8 m2/sq = 3.8\ \text{m}^2/\text{s}, n=0.012n = 0.012, S01=1/120=0.00833S_{01} = 1/120 = 0.00833, S02=1/1500=0.000667S_{02} = 1/1500 = 0.000667. For a wide channel R≈yR \approx y.

Normal and critical depths

yn=(nqS0)3/5,yc=(q2g)1/3=(3.829.81)1/3=1.138 my_n = \left(\frac{nq}{\sqrt{S_0}}\right)^{3/5},\qquad y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3.8^2}{9.81}\right)^{1/3} = 1.138\ \text{m} yn1=(0.012×3.80.00833)0.6=0.659 m (<yc:steep)y_{n1} = \left(\frac{0.012 \times 3.8}{\sqrt{0.00833}}\right)^{0.6} = 0.659\ \text{m}\ (< y_c: \text{steep}) yn2=(0.012×3.80.000667)0.6=1.407 m (>yc:mild)y_{n2} = \left(\frac{0.012 \times 3.8}{\sqrt{0.000667}}\right)^{0.6} = 1.407\ \text{m}\ (> y_c: \text{mild})

Sketch

The upstream slope is steep (supercritical, yn1=0.659y_{n1} = 0.659 m) and the downstream slope is mild (subcritical, yn2=1.407y_{n2} = 1.407 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 yn1y_{n1}:

Fr1=3.8/0.6599.81×0.659=2.27,y2=yn12(1+8Fr12−1)=1.809 mFr_1 = \frac{3.8/0.659}{\sqrt{9.81 \times 0.659}} = 2.27,\qquad y_2 = \frac{y_{n1}}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = 1.809\ \text{m}

Since y2=1.809 m>yn2=1.407 my_2 = 1.809\ \text{m} > y_{n2} = 1.407\ \text{m}, 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 yn1=0.659y_{n1} = 0.659 m upwards until its conjugate depth equals yn2y_{n2}:

yj: yj2(1+8q2gyj3−1)=1.407 ⇒ yj=0.905 my_j:\ \frac{y_j}{2}\left(\sqrt{1 + 8\frac{q^2}{g y_j^3}} - 1\right) = 1.407\ \Rightarrow\ y_j = 0.905\ \text{m}

The jump takes the depth from 0.905 m to 1.407 m.

Length of the M3 profile (direct step, 2 steps, mild slope S0=1/1500S_0 = 1/1500)

E=y+V22g,Sf=n2q2y10/3,Δx=E2−E1S0−SˉfE = y + \frac{V^2}{2g},\quad S_f = \frac{n^2q^2}{y^{10/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f}

Steps: y=0.659→0.782→0.905y = 0.659 \to 0.782 \to 0.905 m.

yy (m)VV (m/s)EE (m)SfS_fSˉf\bar S_fS0−SˉfS_0 - \bar S_fΔx\Delta x (m)
0.65945.7632.35220.008333
0.78234.8581.98490.0047140.006524-0.00585762.7
0.90524.1981.80340.0028980.003806-0.00313957.8
L=62.7+57.8=120.5 mL = 62.7 + 57.8 = 120.5\ \text{m}

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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