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16-Civ-A5 Hydraulic Engineering · May 2014

Question 5 of 6: Hydraulic Jump in a Rectangular Channel

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper: National Exams — 98-Civ-A5 Hydraulic Engineering, May 2014. 3 hours, closed book (one aid sheet). Six questions of equal value (20 marks each); candidates complete any five, and all six are solved here as a study resource.

Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe networks, valves, pumps); Chow, Open-Channel Hydraulics (Manning uniform flow, hydraulic jump, gutter flow); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance, wall shear). Exam-supplied relations used throughout: Hazen–Williams $Q=0.278\,C\,D^{2.63}\,S^{0.54}$ with $S=h_f/L$ (SI), Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, Darcy–Weisbach $\Delta h=0.0826\,\tfrac{fL}{D^5}Q^2$, and $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.

Question 5: Hydraulic Jump in a Rectangular Channel (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Rectangular channel, width $b=11\ \text{m}$, discharge $Q=1.2\ \text{m}^3/\text{s}$ (unit discharge $q=Q/b=0.1091\ \text{m}^2/\text{s}$), depth on one side of the jump $y_1=0.3\ \text{m}$.

Find. (a) the momentum (specific-force) equation across the jump; (b) the downstream velocity.

upstream y1 = 0.3 mdownstream y2V1hydraulic jumprectangular channel, width b = 11 m, Q = 1.2 m^3/s
Figure. Hydraulic jump in a rectangular channel: the conjugate (sequent) depths $y_1$ and $y_2$ satisfy equal specific force.

Approach. Across a jump momentum is conserved (energy is not); equate the specific force (pressure + momentum flux) on the two sides to relate the conjugate depths, then get velocity from continuity.

  1. Momentum / specific-force equation (a). Per unit width, equating hydrostatic pressure force plus momentum flux on each side: $\dfrac{q^2}{g\,y_1}+\dfrac{y_1^{2}}{2}=\dfrac{q^2}{g\,y_2}+\dfrac{y_2^{2}}{2}$, which rearranges to the conjugate-depth relation $\dfrac{y_2}{y_1}=\tfrac12\!\left(\sqrt{1+8\,\mathrm{Fr}_1^{2}}-1\right)$.
  2. Upstream Froude number. $V_1=\dfrac{Q}{b\,y_1}=\dfrac{1.2}{11\times0.3}=0.364\ \text{m/s}$, so $\mathrm{Fr}_1=\dfrac{V_1}{\sqrt{g\,y_1}}=\dfrac{0.364}{\sqrt{9.81\times0.3}}=0.212$.
  3. Conjugate depth. $y_2=\tfrac12(0.3)\!\left(\sqrt{1+8(0.212)^2}-1\right)=0.0249\ \text{m}$. The specific force checks: $M(y_1)=M(y_2)=0.0490\ \text{m}^2$.
  4. Downstream velocity (b). By continuity $V_2=\dfrac{q}{y_2}=\dfrac{0.1091}{0.0249}=\boxed{4.38\ \text{m/s}}$ (supercritical, $\mathrm{Fr}_2=8.9$).
Question 5 results
QuantityValue
Momentum relation$q^2/(g y)+y^2/2$ conserved
Upstream Froude number Fr10.212 (subcritical)
Conjugate depth y20.0249 m
b) Downstream velocity V24.38 m/s
Check: the stated upstream depth 0.3 m is subcritical ($\mathrm{Fr}_1=0.21\lt1$; critical depth $y_c=0.107\ \text{m}$), whereas a real jump forms only when the approach flow is supercritical. The specific-force (momentum) equation is symmetric in the two conjugate depths, so it still returns a valid sequent pair: the 0.3 m depth would physically be the downstream side, with the 0.025 m supercritical depth upstream. The value reported answers the question as posed (momentum solution for the depth conjugate to 0.3 m); in practice one would re-examine the given approach depth.