07-Str-B11 · December 2015

Question 5 of 6: Kinematic versus Dynamic Wave Model after Sudden Gate Closure

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

Notes on this paper

Paper format. National Exams, December 2015 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides) and any non-communicating calculator are permitted. Six questions of twenty marks each; candidates complete any five, and where a question has parts the parts carry equal weight. The solutions below work all six, so the paper can be used whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, series/parallel equivalent pipes, looped-network analysis and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, service-pressure criteria, control-valve characteristics and gutter/roadway hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and compound cross-sections; Henderson, F.M., Open Channel Flow (Macmillan) — the Saint-Venant equations, wave celerity and the kinematic/diffusion/dynamic hierarchy; Transportation Association of Canada, Geometric Design Guide for Canadian Roads — crossfall, curb reveal and roadway drainage conventions.

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. Following Note 6, local losses and velocity head are neglected, 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 is taken as $\rho = 1000$ kg/m³ with $\nu = 1.31\times10^{-6}$ m²/s.

Question 5: Kinematic versus Dynamic Wave Model after Sudden Gate Closure (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. A rectangular channel carrying an initially uniform flow, terminated by a sluice gate that is shut suddenly. Three successive water-surface profiles are recorded (Figure 5 of the exam paper).

Find. A reasoned selection between the kinematic and dynamic wave models, argued from the terms of the Saint-Venant momentum equation and from the flow classification implied by the observed profiles.

[Figure not reproduced: Figure 5.1 — The three observed profiles, redrawn from the exam figure (traces offset vertically for clarity, not to a common datum). At $t = 0$ the surface is a straight line parallel to the bed — uniform flow. After closure the surface departs sharply from that line near the gate, an. See the official exam paper.]

The three profiles tell the whole story before any equation is written. At $t = 0$ the water surface is a straight line running parallel to the bed: depth is constant along the channel, the friction slope equals the bed slope, and the flow is both steady and uniform. That is precisely the regime the kinematic wave model was constructed for. By $t = \Delta t$ the surface near the gate has curved away from that line, and by $t = 2\Delta t$ the disturbance is markedly larger and reaches a good deal further upstream. Depth now varies both with distance and with time, the surface slope near the gate is nothing like the bed slope, and a front is propagating upstream against the direction of flow. The flow has become unsteady and rapidly varied, and the question is whether a model built on the assumption that it is neither can still be used.

The full Saint-Venant momentum equation, written in terms of depth $y$, velocity $V$ and the bed and friction slopes, is

$$\underbrace{\frac{1}{g}\frac{\partial V}{\partial t}}_{\text{local acceleration}} + \underbrace{\frac{V}{g}\frac{\partial V}{\partial x}}_{\text{convective acceleration}} + \underbrace{\frac{\partial y}{\partial x}}_{\text{pressure gradient}} = \underbrace{S_0 - S_f}_{\text{gravity and friction}}$$

and it is accompanied by the continuity equation $\partial A/\partial t + \partial Q/\partial x = 0$. The kinematic wave model is obtained by discarding the first three terms altogether, leaving the algebraic statement $S_f = S_0$. Every consequence of that simplification is at odds with what Figure 5 shows. Setting $S_f = S_0$ asserts that the friction slope is fixed by the bed, so depth and discharge are locked in a single-valued relation $Q = Q(y)$ — a rating curve that cannot change with time. It asserts that the water surface must stay parallel to the bed, so no backwater profile and no drawdown can exist. Above all, discarding the pressure-gradient term $\partial y/\partial x$ removes the only mechanism by which a downstream disturbance can be felt upstream: kinematic waves travel in one direction only, downstream, at the celerity $c_k = \mathrm{d}Q/\mathrm{d}A$, and they neither attenuate nor steepen into a front. A gate closed at the downstream end of a kinematic-wave channel would produce no upstream response whatever, which is manifestly not what Figure 5 records.

The dynamic wave model retains all four terms and therefore captures each of the features that are visible. The local acceleration term $\frac{1}{g}\partial V/\partial t$ represents the inertia of the water column being brought to rest by the gate, and it is large precisely because the closure is sudden — that is the physical content of the word “suddenly” in the question. The convective term $\frac{V}{g}\partial V/\partial x$ accounts for the spatial redistribution of momentum as fluid decelerates approaching the obstruction. The pressure-gradient term $\partial y/\partial x$ is what allows the surface to depart from the bed slope and, mathematically, is what makes the governing system hyperbolic with two families of characteristics of celerity $V \pm \sqrt{gy}$. For the subcritical flow implied by a mild channel with a controlling gate, $V < \sqrt{gy}$, so $V - \sqrt{gy} < 0$ and one characteristic runs upstream. That negative characteristic is the mathematical carrier of the disturbance that Figure 5 shows marching back up the channel, and it exists in the dynamic model and in no simplification of it that drops the pressure term.

It is worth being precise about which mechanism is doing the work here, because the question raises compressibility explicitly. The upstream-moving disturbance in Figure 5 is a gravity wave on a free surface, propagating at the shallow-water celerity $\sqrt{gy}$ — of the order of 3 m/s for a metre of depth. It is not a water-hammer wave. Water hammer arises in closed conduits where the fluid’s elastic compressibility and the pipe-wall elasticity store the energy of the arrested column, and its celerity is of the order of 1,000 m/s. In an open channel the free surface can simply rise, so the water accommodates the closure by changing depth rather than by compressing; the flow is treated as incompressible and the storage is provided by the surface, not by the bulk modulus. The correct model is therefore the dynamic wave form of the Saint-Venant equations, not the elastic-column equations of pipeline transients — and if the surge is steep enough to break, it is handled as a moving hydraulic jump (a surge front) using the momentum equation across the front, still within the incompressible framework.

The intermediate members of the model hierarchy are worth dismissing explicitly, since a well-argued answer should show why nothing weaker will serve. The diffusion wave model keeps $\partial y/\partial x$ but discards both acceleration terms; it can reproduce attenuation and a limited amount of backwater, and it is the workhorse for slow flood routing in natural rivers. It fails here because the closure is sudden, which is exactly the circumstance that makes the local acceleration term comparable to the others. The steady gradually-varied-flow equation would give the eventual backwater profile behind a permanently closed gate, but it says nothing about the transient in Figure 5, in which the profile is still visibly evolving from one observation to the next. Only the full dynamic wave model represents an unsteady, non-uniform, rapidly-varied flow with significant inertia and a disturbance propagating upstream.

In summary: at $t = 0$ the flow is steady and uniform and any model would do; after closure it is unsteady, non-uniform and inertia-dominated, with information travelling upstream along the negative characteristic. The kinematic wave model cannot represent any of that, because setting $S_f = S_0$ forbids a variable water-surface slope, forbids a looped rating curve, and forbids upstream propagation entirely. The dynamic wave model is the appropriate choice, retaining local acceleration, convective acceleration and the pressure-gradient term alongside gravity and friction, with the fluid treated as incompressible and the free surface providing the storage.

Final results — Question 5
Aspect	Assessment
Flow classification at $t$ = 0	Steady, uniform — surface parallel to bed, $S_f = S_0$
Flow classification after closure	Unsteady, non-uniform, rapidly varied
Momentum terms that must be retained	$\frac{1}{g}\frac{\partial V}{\partial t}$, $\frac{V}{g}\frac{\partial V}{\partial x}$, $\frac{\partial y}{\partial x}$, $S_0 - S_f$ — all four
What the kinematic model assumes	$S_f = S_0$; single-valued $Q(y)$; downstream propagation only at $c_k = \mathrm{d}Q/\mathrm{d}A$
Why it fails here	Cannot represent a variable surface slope, a backwater/surge profile, or any upstream-travelling disturbance
Role of compressibility	Negligible — free-surface gravity wave at $\sqrt{gy}$, not an elastic water-hammer wave
Model selected	Dynamic wave (full Saint-Venant equations)

Topic. The Saint-Venant equations and the kinematic / diffusion / dynamic wave hierarchy of unsteady open-channel models.

Key relations. Continuity $\partial A/\partial t + \partial Q/\partial x = 0$; momentum $\frac{1}{g}\frac{\partial V}{\partial t} + \frac{V}{g}\frac{\partial V}{\partial x} + \frac{\partial y}{\partial x} = S_0 - S_f$. Kinematic wave: $S_f = S_0$, celerity $c_k = \mathrm{d}Q/\mathrm{d}A$. Diffusion wave: $\partial y/\partial x + S_f = S_0$. Dynamic wave: all terms retained, characteristic celerities $V \pm \sqrt{gy}$; Froude number $Fr = V/\sqrt{gy}$.

Why this works. The model hierarchy is a statement about which terms in one momentum equation are large enough to matter for a given disturbance, and the selection is made by comparing time and length scales rather than by preference. Slow, gently-varying flood waves on steep channels are dominated by the gravity-friction balance, so the kinematic form suffices and buys enormous computational economy. Abrupt disturbances at a control structure are dominated by inertia and by the surface slope they themselves generate, so the full equations are required. The decisive structural difference is the number of characteristic directions: the kinematic equation is first-order with a single downstream characteristic and therefore admits only a downstream boundary condition’s influence being carried forward, while the dynamic system is second-order hyperbolic with two characteristics and can therefore satisfy a condition imposed at the downstream end — which is exactly what a gate is.

Common pitfall. Reaching for water-hammer theory because a valve-like structure was closed suddenly; the elastic-column equations belong to closed conduits, and in an open channel the free surface relieves the pressure rise so that the correct treatment is a surge or moving hydraulic jump. The other frequent error is arguing purely by vocabulary — asserting “the flow is unsteady, so use the dynamic model” — without identifying which terms the kinematic simplification discards and what each discarded term physically represents. The question asks explicitly for the mathematical terms, and the marks are there.

Source. Henderson, Open Channel Flow, Ch. 8 and Ch. 9 (unsteady flow, the method of characteristics, surges and moving jumps); Chow, Open-Channel Hydraulics, Ch. 18–19 (gradually and rapidly varied unsteady flow); Chin, Water-Resources Engineering, 3rd ed., §5.4 (Saint-Venant equations and the kinematic, diffusion and dynamic approximations).

← Question 4 Next: Question 6 →

Aspect	Assessment
Flow classification at \(t\) = 0	Steady, uniform — surface parallel to bed, \(S_f = S_0\)
Flow classification after closure	Unsteady, non-uniform, rapidly varied
Momentum terms that must be retained	\(\frac{1}{g}\frac{\partial V}{\partial t}\), \(\frac{V}{g}\frac{\partial V}{\partial x}\), \(\frac{\partial y}{\partial x}\), \(S_0 - S_f\) — all four
What the kinematic model assumes	\(S_f = S_0\); single-valued \(Q(y)\); downstream propagation only at \(c_k = \mathrm{d}Q/\mathrm{d}A\)
Why it fails here	Cannot represent a variable surface slope, a backwater/surge profile, or any upstream-travelling disturbance
Role of compressibility	Negligible — free-surface gravity wave at \(\sqrt{gy}\), not an elastic water-hammer wave
Model selected	Dynamic wave (full Saint-Venant equations)