--- title: "Reservoir to lake channel with jump" subject: "Hydraulics" chapter: 8 key: "semester_4/hydraulics:reservoir-to-lake-channel-jump-profiles" source: https://www.bctengineer.com/q/semester_4/hydraulics/reservoir-to-lake-channel-jump-profiles site: BCT Engineering --- # Reservoir to lake channel with jump Hydraulics · Chapter 8: Non-uniform gradually varied flow (GVF) - Asked in: 2073 Magh, 3+4+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/reservoir-to-lake-channel-jump-profiles - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-8 ## Question Water is flowing from reservoir A to lake C via point B through a rectangular channel section of 4 m wide as shown in figure. The length of AB and BC are 100 m each and the corresponding elevations are shown in figure. The normal depth above point B is 0.5 m taking Manning's n = 0.025 and ignoring energy losses except in hydraulic jump. a) Determine the water surface elevation for upper reservoir. b) Is there any possibility of formation of hydraulic jump? If so find the parameters of jump and its location. c) Show all possible water surface profiles. [Figure: reservoir A (water surface 24 m, 20 m mark) feeding channel AB then BC to lake C (water surface 20.8 m, bed 19.6 m); normal depth 0.5 m above B] ## Short answer Reservoir water surface elevation is about 25.2 m. A hydraulic jump forms about 2 m downstream of B with initial depth 0.526 m, sequent depth 1.115 m, Froude number 1.82, energy loss 0.087 m, and length about 7 m. Profiles: S2 on AB, then M3, jump, and M1 on BC. ## Answer **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: | Item | Value | |---|---| | Location of jump | about **2 m downstream of B** | | Depth before jump $y_1$ | 0.526 m | | Depth after jump $y_2$ | 1.115 m | | $Fr_1$ | 1.82 (weak jump) | | Energy loss $\dfrac{(y_2 - y_1)^3}{4y_1y_2}$ | 0.087 m | | Length of jump ($\approx 6y_2$) | about 7 m | #### c) Possible profiles ```text 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. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is AB steep? A. Its normal depth 0.5 m is less than yc = 0.783 m B. Its normal depth exceeds yc C. Its slope is 0.004 D. Flow is subcritical Answer: A. yn < yc defines a steep slope with supercritical uniform flow (Fr = 1.96). (Key point: Slope AB) 2. How is the reservoir level found? A. Add yn to the bed level of A B. Use the lake level C. Flow enters at critical depth, so H = zA + 1.5 yc = 25.18 m D. Use the bed level at B Answer: C. Critical depth control at entry means specific energy is 1.5 yc, giving about 25.2 m. (Key point: Reservoir level) 3. Why does the jump not form at B? A. Sequent depth of 0.5 m (1.159 m) exceeds the M1 depth at B (1.114 m) B. BC is steep C. Sequent depth is less than the lake level D. Fr is below 1 Answer: A. The tailwater at B is too low, so supercritical flow continues as M3 until sequent depth equals the M1 depth, about 2 m downstream. (Key point: Jump on BC) 4. If the lake were higher so depth at B exceeds 1.159 m, where would the jump go? A. Further downstream B. Stay at B C. Disappear D. Up onto the steep reach AB, followed by S1 Answer: D. More tailwater pushes the jump upstream, onto AB with S1 after it. 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