07-Str-B11 · May 2016

Question 5 of 6: Laminar Free-Surface Flow — Velocity as a Function of Depth

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

Notes on this paper

Paper format. National Exams, May 2016 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides may be used) and any non-communicating calculator are permitted. The paper prints six questions of equal value, and a candidate completes any five of them; where a question has parts, the parts also carry equal value. All six are worked below, so the set is useful whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, branched and looped network analysis, control valves and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, maximum-day-plus-fire-flow loading and service-pressure criteria; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation, specific energy and channel transitions; Henderson, F.M., Open Channel Flow (Macmillan) — critical flow, controls and surface-profile classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the pipe force balance, the friction-factor definition and laminar film flow.

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 the cover-page 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 (the cover page prints the exponent without its minus sign, which is an obvious typesetting slip — $1.31\times10^{6}$ m²/s is not a physical viscosity).

Source note — the Question 4 schematic is not reproduced in the examination paper. Question 4 refers to “the figure below”, but no figure is printed between Question 4 and Question 5 on page 3 of the paper, and page 4 opens with the continuation of Question 5. The network is therefore assumed to be a single loop: a source reservoir feeding node N1 through pipe P1 and node N2 through pipe P2, with pipe P3 closing the loop between N1 and N2. That topology is adopted here and is stated explicitly in the Given.

Question 5: Laminar Free-Surface Flow — Velocity as a Function of Depth (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 wide open channel of total flow depth $d$, measured perpendicular to a bed inclined at $\theta$ to the horizontal, carrying a steady, uniform, laminar flow of a Newtonian fluid of density $\rho$ and dynamic viscosity $\mu$. The coordinate $y$ is measured from the bed upwards, $s$ runs down-slope, and the momentum balance on an element is $W\sin\theta - \tau\,\Delta s = 0$.

Find. A closed-form expression for the down-slope velocity $u$ as a function of the distance $y$ from the bed, together with the maximum and depth-averaged velocities that follow from it.

[Figure not reproduced: Figure 5.1 — The elemental volume used in the momentum balance, redrawn from Figure 5 of the paper. The element is taken to have unit width, length $\Delta s$ down-slope, and to extend from the level $y$ all the way up to the free surface, so that the only shear acting on it is that at its. See the official exam paper.]

Approach. Apply the supplied momentum balance not to an arbitrary slice but to the column of fluid lying above the level $y$. Because the free surface is stress-free, the entire down-slope weight component of that column must be carried by the shear at the level $y$, which gives the shear-stress distribution directly. Substituting Newton’s law of viscosity then leaves a first-order ordinary differential equation for $u$, which is integrated subject to the no-slip condition at the bed.

Choose the element so that only one shear force appears. Take a control volume of unit width, of length $\Delta s$ measured down-slope, extending from the level $y$ up to the free surface. Its volume is $(d - y)\,\Delta s$ per unit width and its weight is $$W = \rho g\,(d - y)\,\Delta s$$ The flow is uniform, so the pressure distributions on the two end faces are identical and cancel; the free surface carries no shear because the air above it is dynamically negligible. The only remaining forces are the down-slope component of the weight and the shear $\tau$ acting over the lower face of area $\Delta s$, which is exactly the balance the question supplies.
Obtain the shear-stress distribution through the depth. Substituting the weight into $W\sin\theta - \tau\,\Delta s = 0$ and cancelling $\Delta s$, $$\rho g\,(d - y)\,\Delta s\,\sin\theta = \tau\,\Delta s$$ $$\boxed{\tau(y) = \rho g\,(d - y)\sin\theta}$$ The shear therefore varies linearly across the depth, from a maximum $\tau_0 = \rho g d \sin\theta$ at the bed to exactly zero at the free surface. For the small slopes of real channels $\sin\theta \approx \tan\theta = S_0$, so $\tau_0 = \rho g d S_0$ — the same result as the pipe force balance of Question 2, with the hydraulic radius of a wide channel being the depth itself.
Substitute Newton’s law of viscosity and integrate. Writing $\tau = \mu\,du/dy$ and separating, $$\mu\,\frac{du}{dy} = \rho g \sin\theta\,(d - y)$$ $$u(y) = \frac{\rho g \sin\theta}{\mu}\int_0^{y}(d - \eta)\,d\eta + C = \frac{\rho g \sin\theta}{\mu}\left(dy - \frac{y^{2}}{2}\right) + C$$ The no-slip condition $u = 0$ at $y = 0$ gives $C = 0$, so the required closed form is $$\boxed{u(y) = \frac{\rho g \sin\theta}{\mu}\left(d\,y - \frac{y^{2}}{2}\right) = \frac{g\sin\theta}{\nu}\left(d\,y - \frac{y^{2}}{2}\right)}$$ The profile is a parabola with its vertex at the free surface, which is the signature of laminar sheet flow.
Extract the maximum, mean and unit discharge. The maximum velocity occurs where $du/dy = 0$, that is at $y = d$, the free surface: $$u_{max} = \frac{\rho g \sin\theta\,d^{2}}{2\mu}$$ The depth-averaged velocity follows by integrating the profile across the depth, $$\bar u = \frac{1}{d}\int_0^{d} u\,dy = \frac{\rho g \sin\theta}{\mu d}\left[\frac{d y^{2}}{2} - \frac{y^{3}}{6}\right]_0^{d} = \frac{\rho g \sin\theta\,d^{2}}{3\mu} = \frac{2}{3}u_{max}$$ and the discharge per unit width is $q = \bar u\,d = \rho g \sin\theta\, d^{3}/(3\mu)$. The ratio $\bar u/u_{max} = 2/3$ is the free-surface counterpart of the value $1/2$ that the same argument gives for laminar pipe flow.

Figure 5.2 — The parabolic velocity profile that the integration produces: zero at the bed by no-slip, maximum at the free surface where the shear vanishes, and a depth-mean of exactly two-thirds of the surface value.

Check the result on a realistic film. Take a 2 mm sheet on a slope of 0.001 with the cover page’s kinematic viscosity of $1.31\times10^{-6}$ m²/s. Then $$u_{max} = \frac{9.81 \times 0.001 \times (0.002)^{2}}{2 \times 1.31\times10^{-6}} = 0.01498\ \text{m/s}$$ so $\bar u = 0.00998$ m/s and $q = 1.997\times10^{-5}$ m²/s. The film Reynolds number $Re = 4q/\nu = 61$ is well below the transition value of about 500, so the laminar assumption is self-consistent — a check worth writing down, because the derivation is silently invalid the moment the sheet thickens.

Final results — Question 5
Quantity	Result
Shear-stress distribution	$\tau(y) = \rho g (d - y)\sin\theta$
Bed shear stress	$\tau_0 = \rho g d \sin\theta \approx \rho g d S_0$
Velocity profile	$\boldsymbol{u(y) = \dfrac{\rho g \sin\theta}{\mu}\left(dy - \dfrac{y^{2}}{2}\right)}$
Maximum velocity (at the free surface)	$u_{max} = \rho g \sin\theta\,d^{2}/(2\mu)$
Depth-averaged velocity	$\bar u = \rho g \sin\theta\, d^{2}/(3\mu) = \tfrac{2}{3}u_{max}$
Discharge per unit width	$q = \rho g \sin\theta\, d^{3}/(3\mu)$
Illustrative check: 2 mm film on a 0.001 slope	$u_{max} = 0.0150$ m/s, $\bar u = 0.0100$ m/s, $Re = 61$

Topic. Laminar uniform flow down an inclined plane, sometimes called Nusselt film flow: the linear shear distribution, the parabolic velocity profile and the two-thirds mean.

Key relations. The element momentum balance $W\sin\theta = \tau\,\Delta s$; the resulting $\tau(y) = \rho g (d-y)\sin\theta$; Newton’s law $\tau = \mu\,du/dy$; the profile $u(y) = (\rho g \sin\theta/\mu)(dy - y^2/2)$; and the film Reynolds number $Re = 4q/\nu$ for the validity check.

Why this works. Uniform flow is the key simplification: the depth does not change along the channel, so the pressure force on the upstream face of any element exactly balances that on the downstream face and the momentum equation reduces to a balance between gravity and shear alone. Choosing the element to run from the level of interest all the way to the free surface then removes the second shear force, because the free surface is a stress-free boundary; the shear at level $y$ has to support the entire weight of the fluid above it, which is what makes the distribution linear. From there the problem is purely mathematical: Newton’s law converts a stress distribution into a velocity gradient, and one integration with a single no-slip boundary condition delivers the profile. The same three moves — uniform-flow simplification, stress-free surface, no-slip bed — generate the laminar solution for a pipe, an annulus or a journal bearing.

Common pitfall. The commonest error is to take the element as a thin slab of thickness $dy$ rather than as the column above $y$; that is not wrong, but it leaves two shear forces and forces a second integration, and students routinely lose a boundary condition doing it. The next is to apply the no-slip condition at the free surface instead of at the bed, which inverts the parabola and puts the maximum velocity in the wrong place. The third is quoting $\bar u = u_{max}/2$ from the pipe case; for a free-surface sheet it is $2/3$, because only half the parabola lies within the flow.

Source. White, Fluid Mechanics 7e, §4.10 and §6.2 (fully developed laminar flow between and along surfaces); Chow, Open-Channel Hydraulics, §2 and §5 for the uniform-flow assumption and the bed-shear relation; Henderson, Open Channel Flow, Ch. 4.

← Question 4 Next: Question 6 →

Quantity	Result
Shear-stress distribution	\(\tau(y) = \rho g (d - y)\sin\theta\)
Bed shear stress	\(\tau_0 = \rho g d \sin\theta \approx \rho g d S_0\)
Velocity profile	\(\boldsymbol{u(y) = \dfrac{\rho g \sin\theta}{\mu}\left(dy - \dfrac{y^{2}}{2}\right)}\)
Maximum velocity (at the free surface)	\(u_{max} = \rho g \sin\theta\,d^{2}/(2\mu)\)
Depth-averaged velocity	\(\bar u = \rho g \sin\theta\, d^{2}/(3\mu) = \tfrac{2}{3}u_{max}\)
Discharge per unit width	\(q = \rho g \sin\theta\, d^{3}/(3\mu)\)
Illustrative check: 2 mm film on a 0.001 slope	\(u_{max} = 0.0150\) m/s, \(\bar u = 0.0100\) m/s, \(Re = 61\)