--- title: "Three successive slopes: profiles and length" subject: "Hydraulics" chapter: 8 key: "semester_4/hydraulics:three-reaches-profiles-and-length-slope-1" source: https://www.bctengineer.com/q/semester_4/hydraulics/three-reaches-profiles-and-length-slope-1 site: BCT Engineering --- # Three successive slopes: profiles and length Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF) - Asked in: 2079 Chaitra, 4+8 marks - Video: [Classification of water surface profiles for Horizontal and Adverse Slope](https://www.youtube.com/watch?v=2gjyzBB56Zc) — WATERCORE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/three-reaches-profiles-and-length-slope-1 - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8 ## Question For a rectangular channel shown in fig. below, with width = 7 m, the flow rate is 8 m³/sec, $S_{01}$ = 0.00032, $n_1$ = 0.011, $S_{02}$ = 0.0911, $n_2$ = 0.021, $S_{03}$ = 0.00020 where $S_0$ = bed slope and n = manning's coeff. i) Draw longitudinal water surface profiles for these three successive reaches. ii) Compute the length of profile in slope 1 only using direct step method or Bresse's method taking 3 steps only. [Figure: three successive reaches: slope 1 ($S_{01}$, $n_1$), slope 2 ($S_{02}$, $n_2$) and slope 3 ($S_{03}$); a depth of 1.2 m is marked at the downstream end of slope 3] ## Short answer Profile sequence: M2 on reach 1, S2 followed by hydraulic jump and S1 on reach 2, M1 on reach 3. The M2 drawdown profile in slope 1 is about 950 m long measured from 0.98 y_n to critical depth. ## Answer **Given:** $b = 7\ \text{m}$, $Q = 8\ \text{m}^3/\text{s}$ ($q = 1.143\ \text{m}^2/\text{s}$). Slope 1: $S_{01} = 0.00032$, $n_1 = 0.011$. Slope 2: $S_{02} = 0.0911$, $n_2 = 0.021$. Slope 3: $S_{03} = 0.0002$, with $y = 1.2$ m at its downstream end. $n_3$ is not given, so $n_3 = n_1 = 0.011$ is assumed. **Critical and normal depths** $$ y_c = \left(\frac{1.143^2}{9.81}\right)^{1/3} = 0.511\ \text{m} $$ | Reach | $y_n$ (m) | Class | |---|---|---| | 1 | 0.886 | mild ($y_n > y_c$) | | 2 | 0.224 | steep ($y_n < y_c$) | | 3 | 1.033 | mild | #### i) Longitudinal profiles - **Reach 1 (mild):** the flow upstream is uniform at 0.886 m. At the break to the steep slope the flow passes through critical depth, so the surface falls as an **M2** curve from $y_n$ to $y_c$ at the end of the reach. - **Reach 2 (steep):** the flow starts at $y_c$ and falls as an **S2** curve to the uniform supercritical depth of $0.224$ m. - **Reach 3 (mild):** the depth at the downstream end is $1.2\ \text{m} > y_{n3}$, so an **M1** backwater curve rises downstream from near $y_{n3} = 1.033$ m to 1.2 m. - **Jump:** the sequent depth of the supercritical flow (0.224 m) is $0.983$ m, which is less than the depth at the head of reach 3 ($\approx 1.033$ m). The tailwater is too deep for the jump to form on the mild reach, so it is pushed up onto the **steep reach 2**. After the jump an **S1** curve rises from about 0.98 m to about 1.03 m at the break. Because $S_{02}$ is very steep, this S1 curve is short. ```text R1 (mild) | R2 (steep) | R3 (mild) -- yn=0.886 | | 1.2 m \__ M2 |\ S2 J S1 | M1 _____--- yc '-.. - - -| '-._ yn=0.224 /'''''|__---' ``` #### ii) Length of the M2 profile in reach 1 The M2 curve approaches $y_n$ only asymptotically, so it is taken from $0.98\,y_{n1} = 0.868\ \text{m}$ to $y_c = 0.511$ m in 3 equal steps (the control is at the downstream end). $$ \Delta x = \frac{E_2 - E_1}{S_0 - \bar S_f},\qquad S_f = \frac{n^2V^2}{R^{4/3}} $$ | $y$ (m) | $R$ (m) | $V$ (m/s) | $E$ (m) | $S_f$ | $\bar S_f$ | $\Delta x$ (m) | |---|---|---|---|---|---|---| | 0.8680 | 0.695 | 1.317 | 0.9563 | 0.00034 | | | | 0.7488 | 0.617 | 1.526 | 0.8676 | 0.00054 | 0.00044 | 748.6 | | 0.6297 | 0.534 | 1.815 | 0.7976 | 0.00092 | 0.00073 | 171.2 | | 0.5106 | 0.446 | 2.238 | 0.7659 | 0.00178 | 0.00135 | 30.7 | $$ L = 748.6 + 171.2 + 30.7 \approx 950\ \text{m} $$ **Answer:** Profile sequence: M2 (reach 1), S2, hydraulic jump and S1 (reach 2), M1 (reach 3). The M2 profile in slope 1 is about **950 m** long (measured from $0.98\,y_n$ to $y_c$, in the flow direction). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which classification of the three reaches is correct (yc = 0.511 m)? A. Mild, mild, steep B. Steep, mild, steep C. Mild, steep, mild D. All mild Answer: C. yn1 = 0.886 and yn3 = 1.033 exceed yc (mild); yn2 = 0.224 is below yc (steep). (Key point: Slope classes) 2. Why does reach 1 carry an M2 curve ending at yc? A. The downstream steep slope forces critical depth at the break, so depth falls from yn to yc B. Tailwater backs water up above yn C. A hydraulic jump occurs in reach 1 D. Reach 1 is steep Answer: A. Flow passes through critical depth at the change from mild to steep, so depth falls as a drawdown (M2) curve. (Key point: Profile in reach 1) 3. Why is the jump forced up to the toe of reach 2? A. Because the S2 curve is subcritical B. Because sequent depth of 0.224 m (0.983 m) is below the tailwater depth (at least 1.033 m) C. Because sequent depth exceeds tailwater D. Because reach 3 is steep Answer: B. Tailwater is deeper than the required sequent depth, so the jump moves upstream to the toe of the steep reach. (Key point: Jump location) 4. Why is the M2 length computed from 0.98 yn instead of yn? A. To avoid critical depth B. Because yn is unknown C. The curve meets yn only asymptotically, so an end very near yn is used D. To make the steps equal Answer: C. At yn the distance is infinite; stopping at 0.98 yn gives a finite length (about 950 m). 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