--- title: "Trapezoidal to rectangular transition with bed lowering" subject: "Hydraulics" chapter: 7 key: "semester_4/hydraulics:trapezoid-to-rectangular-transition-bed-lowering" source: https://www.bctengineer.com/q/semester_4/hydraulics/trapezoid-to-rectangular-transition-bed-lowering site: BCT Engineering --- # Trapezoidal to rectangular transition with bed lowering Hydraulics · Chapter 7: Energy and Momentum Principles in Open channel flow - Asked in: 2072 Asoj, 6+6 marks - Video: [Channel Transition](https://www.youtube.com/watch?v=3WfJezMsrz8) — sami ullah - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/hydraulics/trapezoid-to-rectangular-transition-bed-lowering - Whole chapter: https://www.bctengineer.com/md/semester_4/hydraulics/chapter-7 ## Question A trapezoidal channel of base width 6 m and side slope of 2 horizontal to 1 vertical carries a flow of 60 cumecs at a depth of 2.5 m. There is a smooth transition to a rectangular section 6 m wide accompanied by a gradual lowering of the channel bed by 0.6 m. (i) Find the depth of water in the rectangular section and the change in water surface level. (ii) In case the drop in water surface level is to be restricted to 0.3 m, what is the amount by which the bed must be lowered? Assume no losses. ## Short answer (i) With 0.6 m bed lowering, y2 = 2.57 m and the water surface falls 0.528 m. (ii) To limit the surface drop to 0.3 m, the bed must be lowered by 0.86 m, resulting in y2 = 3.065 m. ## Answer #### Given data Trapezoidal section: $b=6$ m, side slope 2H:1V ($z=2$), $y_1=2.5$ m, $Q=60$ m³/s. $$ A_1=(b+zy)y=(6+2\times2.5)(2.5)=27.50\ \text{m}^2,\quad V_1=\frac{60}{27.50}=2.1818\ \text{m/s} $$ $$ E_1=y_1+\frac{V_1^2}{2g}=2.5+\frac{2.1818^2}{19.62}=2.7426\ \text{m} $$ Rectangular section: $b_2=6$ m, so $q_2=\dfrac{60}{6}=10$ m²/s, $y_{c2}=\sqrt[3]{\dfrac{100}{9.81}}=2.168\ \text{m}$. #### (i) Bed lowered by 0.6 m Energy above the new bed: $E_2=E_1+0.6=3.3426$ m. $$ y_2+\frac{q_2^2}{2gy_2^2}=y_2+\frac{5.0968}{y_2^2}=3.3426 $$ Take the subcritical root, since the upstream flow is subcritical ($T_1=6+4\times2.5=16$ m, $D_1=A_1/T_1=1.719$ m, $Fr_1=2.1818/\sqrt{9.81\times1.719}=0.53$): $$ y_2=2.5724\ \text{m} $$ Water surface level in the rectangular section (datum: bed of the trapezoidal channel) $=y_2-0.6=1.9724$ m, compared with 2.500 m upstream. Change in water surface level $=1.9724-2.5=-0.5276$ m, i.e. a **fall of 0.528 m**. #### (ii) Water surface to fall by only 0.30 m Let the bed be lowered by $x$. Then the downstream water level is $2.5-0.3=2.2$ m, so $y_2=2.2+x$. Energy balance, $E_1+x=E_2$, gives $$ (2.2+x)+\frac{5.0968}{(2.2+x)^2}=E_1+x\;\Rightarrow\;2.2+\frac{5.0968}{(2.2+x)^2}=2.7426 $$ $$ (2.2+x)^2=\frac{5.0968}{2.7426-2.2}=9.3929\;\Rightarrow\;2.2+x=3.0648 $$ $$ x=0.8648\ \text{m},\quad y_2=3.065\ \text{m} $$ **Answer: (i) depth in the rectangular section $=2.57$ m; the water surface falls by 0.53 m; (ii) the bed must be lowered by $\approx0.86$ m (depth $3.06$ m).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Upstream specific energy in the trapezoidal section uses A. A = (b + zy)y = 27.5 m^2, V = 2.18 m/s, E1 = 2.743 m B. A = b y = 15 m^2 C. A = z y^2 D. A = (b + 2y)y Answer: A. For a trapezoid A = (b + zy)y, so V = 60/27.5 and E1 = 2.7426 m. (Key point: Upstream energy) 2. With the bed lowered 0.6 m, E2 is A. E1 - 0.6 B. E1 C. E1 + 0.6 = 3.343 m D. 2.5 m Answer: C. Energy above the lower bed increases by 0.6 m. (Key point: Energy downstream with lowered bed) 3. The rectangular depth is 2.572 m. The water surface level compared with upstream 2.5 m has A. risen by 0.07 m B. fallen by 0.528 m (1.972 m vs 2.5 m) C. not changed D. fallen by 0.6 m Answer: B. Level = y2 - 0.6 = 1.972 m, so the change is -0.528 m. (Key point: Change in water level) 4. To limit the fall to 0.30 m, the lowering x is found by A. x = 0.30 B. Taking y2 = 2.2 + x in E1 = y2 - x + q^2/(2g y2^2), giving x of about 0.86 m C. x = 0.6 - 0.3 D. Using critical depth Answer: B. Solving (2.2 + x)^2 = 5.0968/(2.7426 - 2.2) gives y2 = 3.065 m, x = 0.865 m. 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