Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations — May 2014 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and any non-communicating calculator are permitted. The paper prints six questions of 20 marks each and instructs the candidate to complete any five; where a question has more than one part, the parts carry equal marks. Candidates are urged to submit a clear statement of any interpretive assumptions with the answer paper. All six questions are worked below, because this set is intended as a study resource.
Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, branched and looped network analysis, transmission mains and valve characteristics; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, fire-flow criteria, minimum service pressures and gutter hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and the momentum treatment of the hydraulic jump; Henderson, F.M., Open Channel Flow (Macmillan) — specific force, conjugate depths and jump classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the control-volume force balance that links wall shear stress to the Darcy friction factor.
Notation and conventions used throughout. The paper supplies the SI Hazen-Williams form \(Q = 0.278\,C\,D^{2.63}S^{0.54}\) with \(Q\) in m3/s, \(D\) in metres and \(S = h_f/L\); it is used exactly as printed. Writing \(K = 0.278\,C\,D^{2.63}\) inverts it to the head-loss form \(h_f = L\,(Q/K)^{1.852}\), the exponent \(n = 1.852\) being the value the paper itself quotes in its loop-correction note. Local losses and velocity head are neglected, as instructed in Note 6, so the hydraulic grade line (HGL) and the energy grade line coincide and “pressure head at a node” means \(p/\gamma = \mathrm{HGL} - z\). Water properties are \(\rho = 1000\) kg/m3 and \(\nu = 1.31\times10^{-6}\) m2/s.
Question 5: Momentum Across a Hydraulic Jump (20 marks)
Given. A wide rectangular channel carrying a small discharge, with a stated depth on one side of a hydraulic jump.
Given data — Question 5
Quantity
Symbol
Value
Channel width (rectangular)
\(b\)
11 m
Discharge
\(Q\)
1.2 m3/s
Depth stated upstream of the jump
\(y_1\)
0.3 m
Gravitational acceleration
\(g\)
9.81 m/s2
Unit discharge, \(Q/b\)
\(q\)
0.1091 m2/s
Find. (a) the momentum (specific-force) equation across the jump, and (b) the velocity on the downstream side.
Figure 5.1 — General hydraulic jump. Between the two control sections the roller dissipates energy, so the specific-energy equation cannot be used; the specific force, which counts only pressure and momentum flux, is the same on both sides and is what links the conjugate depths.
Approach. Apply the momentum equation to a control volume enclosing the jump — the hydrostatic end thrusts plus the momentum fluxes must balance — then evaluate the resulting conjugate-depth relation for the stated data, checking the upstream Froude number first because a jump only exists when the approach flow is supercritical.
State the momentum balance across the jump. Take a short, horizontal, prismatic control volume spanning the jump, neglecting boundary friction over its short length. Equating hydrostatic thrust plus momentum flux on the two faces:
$$\frac{\gamma b y_1^{2}}{2} + \rho Q V_1 = \frac{\gamma b y_2^{2}}{2} + \rho Q V_2$$
Dividing by \(\rho g b\) and substituting \(V = q/y\) with \(q = Q/b\) gives the specific-force (momentum function) form, which is the equation part (a) asks for:
$$\boxed{\dfrac{q^{2}}{g\,y_1} + \dfrac{y_1^{2}}{2} = \dfrac{q^{2}}{g\,y_2} + \dfrac{y_2^{2}}{2}}$$
Rearranged, this yields the familiar conjugate-depth relation \(\dfrac{y_2}{y_1} = \dfrac{1}{2}\left(\sqrt{1 + 8\,\mathrm{Fr}_1^{2}} - 1\right)\).
Reduce the given discharge to a unit discharge and approach velocity.
$$q = \frac{Q}{b} = \frac{1.2}{11} = 0.1091\ \mathrm{m^2/s}, \qquad V_1 = \frac{q}{y_1} = \frac{0.1091}{0.3} = 0.364\ \mathrm{m/s}$$
Test the approach flow with the Froude number before applying the relation. A jump requires supercritical approach flow, so this check is not optional:
$$\mathrm{Fr}_1 = \frac{V_1}{\sqrt{g y_1}} = \frac{0.364}{\sqrt{9.81 \times 0.3}} = \frac{0.364}{1.716} = 0.212$$
$$y_c = \left(\frac{q^{2}}{g}\right)^{1/3} = \left(\frac{0.1091^{2}}{9.81}\right)^{1/3} = 0.107\ \mathrm{m}$$
Since \(\mathrm{Fr}_1 = 0.21 < 1\) and \(y_1 = 0.30\ \mathrm{m} > y_c = 0.107\ \mathrm{m}\), the depth quoted as “upstream” is subcritical. As printed, the data therefore describe a state on the upper branch of the specific-force curve. This is flagged below and both readings are carried through.
Apply the momentum equation to the stated depth. The specific-force equation always has a second root, the conjugate of \(y_1\), and taking the data exactly as printed:
$$y_2 = \frac{y_1}{2}\left(\sqrt{1 + 8(0.212)^{2}} - 1\right) = 0.15\left(\sqrt{1.3594} - 1\right) = 0.15 \times 0.1660 = \boxed{0.0249\ \mathrm{m}}$$
$$V_2 = \frac{q}{y_2} = \frac{0.1091}{0.0249} = \boxed{4.38\ \mathrm{m/s}}$$
The pair is verified by evaluating the specific force on both sides: \(M_1 = M_2 = 0.04904\ \mathrm{m^2}\), so the momentum equation is satisfied exactly.
Interpret the conjugate pair physically. A real jump always runs from the supercritical member of the pair to the subcritical one, because only that direction dissipates energy. The pair here is \(\{0.0249\ \mathrm{m},\ 0.300\ \mathrm{m}\}\), so the physically admissible jump has \(y_1 = 0.0249\) m with \(V_1 = 4.38\) m/s upstream and \(y_2 = 0.300\) m with
$$V_2 = \frac{q}{0.300} = 0.364\ \mathrm{m/s}$$
on the downstream side. For \(y_1 = 0.30\) m to be genuinely supercritical the channel would have to carry
$$Q > b\sqrt{g\,y_1^{3}} = 11\sqrt{9.81 \times 0.3^{3}} = 5.66\ \mathrm{m^3/s}$$
which is nearly five times the stated 1.2 m3/s. Both velocities are reported in the results table so the answer stands whichever reading the marker intends.
Check: the printed data do not support a jump as described. With \(b = 11\) m and \(Q = 1.2\) m3/s the unit discharge is only 0.109 m2/s, giving a critical depth of 0.107 m; the stated upstream depth of 0.30 m is therefore subcritical (\(\mathrm{Fr}_1 = 0.21\)) and no hydraulic jump can form from it. The momentum equation of part (a) is unaffected and is the substance of the question. For part (b) the direct application of that equation to \(y_1 = 0.30\) m returns the conjugate depth 0.0249 m and \(V_2 = 4.38\) m/s, while the physically consistent reading — 0.30 m being the sequent (downstream) depth — gives \(V_2 = 0.364\) m/s. State the Froude check and the interpretation adopted before quoting either number; a candidate who performs that check has demonstrated the understanding the question is testing.
