--- title: "Step and width reduction to produce critical flow" subject: "Hydraulics" chapter: 7 key: "semester_4/hydraulics:critical-flow-step-width-reduction-combination" source: https://www.bctengineer.com/q/semester_4/hydraulics/critical-flow-step-width-reduction-combination site: BCT Engineering --- # Step and width reduction to produce critical flow Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow - Asked in: 2077 Chaitra, 2+2+4 marks - Video: [Open Channel Flow-20 | Channel Width Reduction | Shubham Sarathe](https://www.youtube.com/watch?v=V123w0PH0I8) — EIQ - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/critical-flow-step-width-reduction-combination - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7 ## Question A 50 m wide rectangular channel is carrying a flow of 250 m³/s at a flow depth of 5 m. To produce critical flow in this channel, determine: (i) The height of the step in the channel bottom if the width remains constant. (ii) The reduction in the channel width if the channel-bottom level remains unchanged. (iii) A combination of the width reduction and the bottom step. ## Short answer (i) Step height Δz = E1 − Emin = 3.00 m. (ii) With unchanged bed, reduce width to 12.92 m (reduction 37.08 m). (iii) Combining a 1.5 m step, reduce width to 21.91 m (reduction 28.09 m). ## Answer Critical flow at the transition requires $E_2=E_{min}=\tfrac32y_c$ with $y_c=\sqrt[3]{q_2^2/g}$ at that section. #### Given data $b=50$ m, $Q=250$ m³/s, $y_1=5$ m. $q_1=\dfrac{250}{50}=5$ m²/s, $V_1=1$ m/s. $$ E_1=5+\frac{1^2}{19.62}=5.0510\ \text{m},\qquad y_{c1}=\sqrt[3]{\frac{25}{9.81}}=1.366\ \text{m},\quad E_{min1}=2.049\ \text{m} $$ The approach flow is subcritical ($Fr_1=0.143$). #### (i) Step in the bed (width constant) $$ E_1=E_{min1}+\Delta z\;\Rightarrow\;\Delta z=5.0510-2.0489=3.002\ \text{m} $$ **Height of step $=3.00$ m.** #### (ii) Width reduction (bed unchanged) Critical flow at the contraction with $E_2=E_1$: $$ y_c=\frac23E_1=3.3673\ \text{m},\qquad q_2=\sqrt{gy_c^3}=\sqrt{9.81\times3.3673^3}=19.354\ \text{m}^2/\text{s} $$ $$ b_2=\frac{Q}{q_2}=\frac{250}{19.354}=12.92\ \text{m} $$ Reduction in width $=50-12.92=37.08$ m. #### (iii) Combination of step and width reduction Many combinations are possible. Take a step of $\Delta z=1.5$ m (assumed). Then energy at the contraction: $$ E_2=E_1-1.5=3.5510\ \text{m},\qquad y_c=\frac23E_2=2.3673\ \text{m} $$ $$ q_2=\sqrt{9.81\times2.3673^3}=11.408\ \text{m}^2/\text{s},\qquad b_2=\frac{250}{11.408}=21.91\ \text{m} $$ Reduction in width $=50-21.91=28.09$ m. | Case | Step (m) | Contracted width (m) | |---|---|---| | (i) step only | 3.00 | 50 | | (ii) width only | 0 | 12.92 | | (iii) combination (assumed step) | 1.50 | 21.91 | **Answer: (i) step $=3.00$ m; (ii) width reduced by 37.08 m to 12.92 m; (iii) e.g. step 1.5 m with width reduced by 28.09 m to 21.91 m.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For critical flow by a bed step with constant width, the step height is A. E1 + Emin B. E1 - Emin C. Emin - E1 D. yc - y1 Answer: B. Critical flow needs E2 = Emin, so Δz = E1 - Emin = 5.051 - 2.049 = 3.00 m. (Key point: Step only) 2. For critical flow by narrowing the width (no step), what is used? A. Emin2 = E1, so yc = 2E1/3 and q2 = sqrt(g yc^3) B. E2 = 0 C. yc = E1 D. q2 = q1 Answer: A. With no step, E2 = E1; yc = (2/3)E1 = 3.367 m, q2 = 19.35 m^2/s, b2 = Q/q2 = 12.92 m. (Key point: Width reduction only) 3. Narrowing the channel does what to q and Emin at the contraction? A. Decreases q and Emin B. Increases q and Emin C. Increases q but leaves Emin unchanged D. Leaves both unchanged Answer: B. q2 = Q/b2 is larger, so yc and Emin = 1.5yc increase. (Key point: Effect of narrowing) 4. Why does part (iii) have many possible answers? A. The flow is not critical B. Any step Δz is paired with a width b2 so that E1 - Δz = 1.5yc(b2); the problem fixes only one condition C. The discharge is unknown D. Energy is lost in the step Answer: B. One equation, two unknowns (step and width), so the step is assumed (here 1.5 m) and b2 follows (21.91 m). 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