--- title: "Baffle blocks for jump, tailwater 4 m" subject: "Hydraulics" chapter: 9 key: "semester_4/hydraulics:jump-baffle-blocks-tailwater-4m-150-m3s" source: https://www.bctengineer.com/q/semester_4/hydraulics/jump-baffle-blocks-tailwater-4m-150-m3s site: BCT Engineering --- # Baffle blocks for jump, tailwater 4 m Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF) - Asked in: 2078 Chaitra, 4+1+1+2 marks - Video: [Momentum equation applied to baffles blocks in hydraulic jump stilling basin](https://www.youtube.com/watch?v=RaOBdIDkEuo) — Hubert Chanson - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/jump-baffle-blocks-tailwater-4m-150-m3s - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9 ## Question A hydraulic jump is to be formed in a rectangular channel of 10 m wide. The discharge is 150 m³/s, the tail water depth in the channel is 4 m, which is 1 m less than the depth required for the formation of the jump without any baffle blocks. Therefore, baffle blocks should be placed for the formation of the jump. The force on the baffle blocks can be estimated as $F_b = 2\gamma A E_1$, where A is the frontal area of the baffle blocks, $E_1$ is the specific energy of the supercritical flow and $\gamma$ is the specific weight of the water. (i) Find the total frontal area of the baffle blocks and energy loss during jump. (ii) Show on the specific force curve the initial and the sequent depths of the jump without baffle blocks. (iii) Show on the same specific force curve the initial and the sequent depths of the jump with the baffle blocks and the force on the baffle blocks. (iv) If height of baffle block is increased by 10%, what will be the percentage increase or decrease of energy loss? [Figure: jump with Q = 150 m³/s, supercritical depth $y_1$, tailwater $y_2$ = 4 m, water level required for jump formation without baffle blocks 1 m above the tailwater, baffle block force $F_b$] ## Short answer Initial depth 1.427 m, tailwater 4 m. Frontal area of baffle blocks A = 2.38 m², energy loss 2.34 m. If block height increases 10%, frontal area rises to 2.61 m² and energy loss increases by about 16%. ## Answer **Given:** $b = 10$ m, $Q = 150\ \text{m}^3/\text{s}$ ($q = 15\ \text{m}^2/\text{s}$), tailwater $y_2 = 4$ m. Without blocks the jump needs a sequent depth of $4 + 1 = 5$ m. #### i) Initial depth, energy loss and area of blocks **Initial depth $y_1$** (from the sequent depth of 5 m without blocks): $$ y_1 = \frac{y_2'}{2}\left(\sqrt{1 + 8Fr_2'^2} - 1\right)\ \text{or}\ \frac{q^2}{g} = \frac{y_1y_2'(y_1 + y_2')}{2} $$ Solving $\dfrac{225}{9.81} = 22.94 = \dfrac{y_1(5)(y_1 + 5)}{2}$ gives $y_1 = 1.427$ m. $$ V_1 = \frac{15}{1.427} = 10.51\ \text{m/s},\quad E_1 = 1.427 + \frac{10.51^2}{2g} = 7.056\ \text{m} $$ **Downstream (tailwater):** $V_2 = 15/4 = 3.75$ m/s, $E_2 = 4 + \dfrac{3.75^2}{2g} = 4.717$ m. **Energy loss with blocks:** $$ \Delta E = E_1 - E_2 = 7.056 - 4.717 = 2.34\ \text{m} $$ **Force on blocks** (momentum equation, horizontal bed, no friction): $$ F_b = \frac{\gamma b}{2}(y_1^2 - y_2^2) + \rho Q(V_1 - V_2) $$ $$ F_b = 99\,935 + 1\,576\,315 - 784\,800 - 562\,500 = 328\,950\ \text{N} \approx 329\ \text{kN} $$ **Frontal area** from $F_b = 2\gamma A E_1$: $$ A = \frac{328\,950}{2 \times 9810 \times 7.056} = 2.376\ \text{m}^2 $$ **Answer (i): $A \approx 2.38\ \text{m}^2$, $\Delta E \approx 2.34$ m.** #### ii) Without blocks On the specific force curve, $F = \dfrac{q^2}{gy} + \dfrac{y^2}{2}$ (per unit width, in $\text{m}^2$), the initial depth $y_1 = 1.427$ m and the sequent depth $y_2' = 5$ m have the same specific force: $$ F(1.427) = \frac{225}{9.81 \times 1.427} + \frac{1.427^2}{2} = 16.07 + 1.02 = 17.09\ \text{m}^2 $$ $$ F(5) = \frac{225}{9.81 \times 5} + \frac{5^2}{2} = 4.59 + 12.5 = 17.09\ \text{m}^2 $$ Both points lie on the vertical line $F = 17.09$. #### iii) With blocks The tailwater depth 4 m has a specific force $F_2 = 15^2/(9.81 \times 4) + 4^2/2 = 5.73 + 8 = 13.73\ \text{m}^2$, which is less than $F_1 = 17.09\ \text{m}^2$. The difference is supplied by the blocks: $$ \frac{F_b}{\gamma b} = F_1 - F_2 = 17.09 - 13.73 = 3.36\ \text{m}^2\ \Rightarrow\ F_b = 9810 \times 10 \times 3.36 \approx 330\ \text{kN}\ (\text{agrees}) $$ ```text y ^ | / 5 |- - - - - - - - - - - - - - * y2' = 5 m (no blocks) | .' 4 |- - - - - - - - - - * y2 = 4 m (with blocks) | .' yc |- - - - - - -* minimum F | '. | '-. 1.43 |- - - - - - - - - - - - - - * y1 | \ +--------------------+-------+---> F 13.73 17.09 |<----->| Fb/(gamma b) = 3.36 m^2 ``` - Without blocks: $y_1 = 1.427$ m and $y_2' = 5$ m are at $F = 17.09$. - With blocks: $y_1$ is at $F_1 = 17.09$ and the tailwater $y_2 = 4$ m is at $F_2 = 13.73$; the gap equals $F_b/(\gamma b) = 3.36\ \text{m}^2$. #### iv) Effect of 10% taller blocks Frontal area increases by 10% (width unchanged), $A' = 1.1 \times 2.376 = 2.614\ \text{m}^2$. With the tailwater fixed at 4 m, the momentum equation and $F_b = 2\gamma A'E_1$ are solved together for the new initial depth: $$ y_1' = 1.375\ \text{m},\quad E_1' = 7.438\ \text{m},\quad F_b' = 381\ \text{kN} $$ $$ \Delta E' = 7.438 - 4.717 = 2.72\ \text{m} $$ $$ \frac{2.72 - 2.34}{2.34} \times 100 = +16.3\% $$ **Answer (iv): the energy loss increases by about 16%.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the initial depth y1 found? A. From the 4 m tailwater with the Belanger equation B. From the 5 m sequent depth needed without blocks, using q^2/g = y1 y2 (y1+y2)/2 C. From critical depth D. From Manning's equation Answer: B. Without blocks the jump needs 4 + 1 = 5 m, and y1 is the conjugate of that depth (1.427 m). (Key point: Initial depth y1) 2. How does the specific force curve show the force on the blocks? A. Fb/(gamma b) equals the difference F1 - F2 (17.09 - 13.73 m^2) B. Fb equals the sum of F1 and F2 C. Fb is read from the minimum of the curve D. Fb equals the difference of the depths Answer: A. The tailwater has less specific force than the supercritical flow; the blocks supply the difference. (Key point: Force from specific force) 3. How is the total frontal area of the blocks obtained? A. A = Fb/(gamma E1) B. A = Fb/(2 gamma E1) from Fb = 2 gamma A E1 C. A = Fb x E1 D. A = Fb/(2 gamma E2) Answer: B. The given relation Fb = 2 gamma A E1 is rearranged, giving about 2.38 m^2. (Key point: Frontal area) 4. What happens to energy loss when the block height is increased by 10%? A. It decreases by about 16% B. It stays the same C. It increases by about 16% D. It doubles Answer: C. More block force gives smaller y1 and larger E1, so the loss rises from 2.34 m to 2.72 m. 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