--- title: "Jump and chute block with hump control" subject: "Hydraulics" chapter: 9 key: "semester_4/hydraulics:jump-chute-block-hump-critical-5m-channel" source: https://www.bctengineer.com/q/semester_4/hydraulics/jump-chute-block-hump-critical-5m-channel site: BCT Engineering --- # Jump and chute block with hump control Hydraulics · Chapter 9: Non-uniform rapidly varied flow (RVF) - Asked in: 2079 Chaitra, 6+2 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/jump-chute-block-hump-critical-5m-channel - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-9 ## Question For a rectangular channel as shown in fig. with 5 m width the flow is 15 m³/sec. The flow over the hump ($\Delta Z$ = 0.4 m) at section 3 is critical. i) Calculate depths $y_2$ and $y_1$ at section 2 and 1 (Assume hydraulic jump occurs between 1 and 2). ii) If a single chute block of height 0.1 m and length 1 m is placed symmetrically with channel at centre ($C_D$ of block = 0.1). Find the depth $y_2$, if $y_1$ and $y_c$ remains constant as above. [Figure: depth $y_1$ at section 1, a hydraulic jump between sections 1 and 2 with depth $y_2$ at section 2, and a hump of height $\Delta Z$ = 0.4 m at section 3 where the depth is critical $y_c$] ## Short answer For q=3 m2/s, critical depth on hump is 0.972 m and E_c=1.458 m, so E_2=1.858 m gives y_2=1.70 m; the conjugate supercritical depth is y_1=0.49 m. With a 0.1 m2 block and drag 185 N, momentum gives y_2=1.696 m, about 1.70 m. ## Answer **Given:** $b = 5$ m, $Q = 15\ \text{m}^3/\text{s}$ ($q = 3\ \text{m}^2/\text{s}$), hump $\Delta z = 0.4$ m with critical flow at section 3. Losses are neglected, except in the jump. **Critical conditions on the hump:** $$ y_c = \left(\frac{3^2}{9.81}\right)^{1/3} = 0.972\ \text{m},\qquad E_c = 1.5\,y_c = 1.458\ \text{m} $$ #### i) Depths $y_2$ and $y_1$ Energy equation between section 2 (upstream of the hump, after the jump) and section 3: $$ E_2 = E_c + \Delta z = 1.458 + 0.4 = 1.858\ \text{m} $$ Section 2 is subcritical: $$ y_2 + \frac{3^2}{2g\,y_2^2} = 1.858 \;\Rightarrow\; y_2 + \frac{0.4587}{y_2^2} = 1.858 \;\Rightarrow\; y_2 = 1.699\ \text{m} $$ Section 1 is the supercritical depth conjugate to $y_2$: $$ \frac{q^2}{g} = \frac{y_1y_2(y_1 + y_2)}{2} \;\Rightarrow\; 0.9174 = \frac{1.699\,y_1(y_1 + 1.699)}{2} \;\Rightarrow\; y_1 = 0.493\ \text{m} $$ Check: $V_1 = 6.09$ m/s, $Fr_1 = 2.77$, and $\dfrac{y_1}{2}\left(\sqrt{1 + 8(2.77)^2} - 1\right) = 1.699$ m. **Answer (i): $y_2 = 1.70$ m, $y_1 = 0.49$ m.** #### ii) With a chute block Assumed: the block's frontal area is height $\times$ width $= 0.1 \times 1 = 0.1\ \text{m}^2$ (the 1 m dimension is across the channel). The drag force on the block (acting on the water in the upstream direction), using the velocity at section 1: $$ F_D = C_D\,\rho A\frac{V_1^2}{2} = 0.1 \times 1000 \times 0.1 \times \frac{6.086^2}{2} = 185\ \text{N} $$ Momentum equation between sections 1 and 2: $$ \frac{\gamma b y_1^2}{2} + \rho Q V_1 - F_D = \frac{\gamma b y_2^2}{2} + \rho Q V_2 $$ For the full width: $\gamma b y_1^2/2 = 5960$ N and $\rho QV_1 = 91\,288$ N, so the left side is $5960 + 91\,288 - 185 = 97\,063$ N. The right side is $24\,525\,y_2^2 + 45\,000/y_2$. Solving: $$ 24\,525\,y_2^2 + \frac{45\,000}{y_2} = 97\,063 \;\Rightarrow\; y_2 = 1.696\ \text{m} $$ **Answer (ii): $y_2 \approx 1.70$ m (a reduction of about 3 mm, since the block force is very small compared with the momentum flux of the jump).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is the depth y2 upstream of the hump found? A. E2 = Ec + dz with critical flow on the hump, then solving for the subcritical depth B. y2 = yc C. By Belanger equation from y1 only D. By Manning's equation Answer: A. Critical flow on the hump means E2 = 1.5 yc + 0.4 = 1.858 m; the subcritical root gives 1.70 m. (Key point: Energy at section 2) 2. How is y1 obtained once y2 is known? A. From the energy equation between 1 and 2 B. As the conjugate depth of y2 using q^2/g = y1 y2 (y1+y2)/2 C. y1 = y2/2 D. As the critical depth Answer: B. The jump links y1 and y2 by momentum, giving y1 = 0.49 m (supercritical). (Key point: Finding y1) 3. Why does the single chute block change y2 by only about 3 mm? A. The block raises the energy B. The block is submerged C. The drag force (about 185 N) is very small compared with the momentum flux of the jump D. The jump is removed Answer: C. F_D = Cd rho A V1^2/2 is tiny compared with terms of about 97,000 N. 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