--- title: "Slope change 1/120 to 1/1500, jump and profile length" subject: "Hydraulics" chapter: 8 key: "semester_4/hydraulics:wide-channel-slope-break-jump-two-steps" source: https://www.bctengineer.com/q/semester_4/hydraulics/wide-channel-slope-break-jump-two-steps site: BCT Engineering --- # Slope change 1/120 to 1/1500, jump and profile length Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF) - Asked in: 2082 Kartik, 2+2+4 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/wide-channel-slope-break-jump-two-steps - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8 ## Question 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. ## 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. ## Answer **Given:** wide rectangular channel, $q = 3.8\ \text{m}^2/\text{s}$, $n = 0.012$, $S_{01} = 1/120 = 0.00833$, $S_{02} = 1/1500 = 0.000667$. For a wide channel $R \approx y$. **Normal and critical depths** $$ y_n = \left(\frac{nq}{\sqrt{S_0}}\right)^{3/5},\qquad y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{3.8^2}{9.81}\right)^{1/3} = 1.138\ \text{m} $$ $$ y_{n1} = \left(\frac{0.012 \times 3.8}{\sqrt{0.00833}}\right)^{0.6} = 0.659\ \text{m}\ (< y_c: \text{steep}) $$ $$ y_{n2} = \left(\frac{0.012 \times 3.8}{\sqrt{0.000667}}\right)^{0.6} = 1.407\ \text{m}\ (> y_c: \text{mild}) $$ #### Sketch The upstream slope is steep (supercritical, $y_{n1} = 0.659$ m) and the downstream slope is mild (subcritical, $y_{n2} = 1.407$ m). Supercritical flow must change to subcritical through a hydraulic jump, with an **M3** curve before it (see the jump location below). ```text STEEP | MILD ---- CDL 1.138 - - | - - - - - - - - - - - - | J _____ yn2=1.407 ______ yn1=0.659 __|__M3_____/'''| ``` #### Does a jump occur, and where? Sequent depth of $y_{n1}$: $$ Fr_1 = \frac{3.8/0.659}{\sqrt{9.81 \times 0.659}} = 2.27,\qquad y_2 = \frac{y_{n1}}{2}\left(\sqrt{1 + 8Fr_1^2} - 1\right) = 1.809\ \text{m} $$ Since $y_2 = 1.809\ \text{m} > y_{n2} = 1.407\ \text{m}$, 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 $y_{n1} = 0.659$ m upwards until its conjugate depth equals $y_{n2}$: $$ y_j:\ \frac{y_j}{2}\left(\sqrt{1 + 8\frac{q^2}{g y_j^3}} - 1\right) = 1.407\ \Rightarrow\ y_j = 0.905\ \text{m} $$ The jump takes the depth from 0.905 m to 1.407 m. #### Length of the M3 profile (direct step, 2 steps, mild slope $S_0 = 1/1500$) $$ E = y + \frac{V^2}{2g},\quad S_f = \frac{n^2q^2}{y^{10/3}},\quad \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f} $$ Steps: $y = 0.659 \to 0.782 \to 0.905$ m. | $y$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $S_0 - \bar S_f$ | $\Delta x$ (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 | $$ L = 62.7 + 57.8 = 120.5\ \text{m} $$ **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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is the upstream slope (1 in 120) classed as steep? A. Because its normal depth 0.659 m is less than critical depth 1.138 m B. Because its normal depth is greater than critical depth C. Because the slope value is larger than 1 in 1500 only D. Because the flow on it is subcritical Answer: A. A slope is steep when yn < yc, giving supercritical uniform flow. Here 0.659 m < 1.138 m. (Key point: Classifying slopes) 2. The sequent depth of yn1 (1.809 m) is greater than yn2 (1.407 m). What does this mean? A. The jump is swept downstream, so it forms in the mild reach after an M3 curve B. The jump forms on the steep slope upstream of the break C. The jump forms exactly at the break D. No jump occurs and flow stays supercritical Answer: A. Tailwater (1.407 m) is too low to force the jump at the break, so supercritical flow continues as an M3 curve until its conjugate depth equals yn2. (Key point: Where the jump forms) 3. How is the depth yj just before the jump found? A. Take yj as half of yn2 B. Use yj = yc C. Solve for the depth whose sequent depth equals yn2 (1.407 m), giving 0.905 m D. Use the sequent depth of yn2 Answer: C. The M3 curve rises until the conjugate of its depth equals the downstream normal depth, which gives yj = 0.905 m. (Key point: Jump location depth) 4. In the direct step formula, how is the step length obtained? A. dx = (E2 - E1) x (S0 - Sf mean) B. dx = (E2 - E1)/(S0 - mean Sf) C. dx = (y2 - y1)/S0 D. dx = (E2 + E1)/(S0 + mean Sf) Answer: B. Direct step uses specific-energy change divided by the difference between bed slope and mean friction slope. 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