Question 1 of 6: Channel Capacity, Culvert Minor Losses and Inlet-Control Profiles

Question 1: Channel Capacity, Culvert Minor Losses and Inlet-Control Profiles (10 + 4 + 6 = 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.

Part (a) — Discharge of the trapezoidal channel (10 marks)

Given.

Find. The volume of water conveyed past any section in one day, in cubic metres per day.

Approach. The depth is stated to be constant along the whole length, so the flow is uniform and Manning's equation applies directly with the bed slope as the friction slope; compute the geometric properties of the trapezoid, then the mean velocity, then the discharge, and finally scale to a day.

1. Flow area of the trapezoid. A trapezoid of bottom width $b$ and side slope $z$ horizontal to 1 vertical carries, at depth $y$, an area $$A=(b+zy)\,y=\bigl(5+3(2)\bigr)(2)=\boxed{22.0\ \text{m}^2}$$ The two side triangles contribute $zy^2 = 12\ \text{m}^2$ and the rectangular core $by = 10\ \text{m}^2$, which is a useful arithmetic check.
2. Wetted perimeter. Each sloping side has length $y\sqrt{1+z^{2}}$, so $$P=b+2y\sqrt{1+z^{2}}=5+2(2)\sqrt{1+3^{2}}=5+4\sqrt{10}=17.649\ \text{m}$$ Only the bed and the two wetted sides count; the free surface is not a solid boundary and contributes nothing.
3. Hydraulic radius. Dividing the area by the wetted perimeter, $$R=\frac{A}{P}=\frac{22.0}{17.649}=1.2465\ \text{m}$$ Note that $R$ is appreciably smaller than the depth, which is normal for a wide trapezoid.
4. Choice of roughness coefficient. The paper gives no value of $n$, so Note 2 is invoked. For a trowelled or float-finished concrete lining, Chow's table gives $n$ between 0.011 and 0.015 with 0.013 as the usual design value; that value is adopted here and its influence is quantified in the callout below.
5. Mean velocity from Manning's equation. In SI units Manning's equation reads $$V=\frac{1}{n}\,R^{2/3}\,S_0^{1/2}=\frac{1}{0.013}\,(1.2465)^{2/3}(0.02)^{1/2}$$ Evaluating the two factors, $(1.2465)^{2/3}=1.1580$ and $(0.02)^{1/2}=0.14142$, so $$V=76.923\times1.1580\times0.14142=\boxed{12.60\ \text{m/s}}$$
6. Discharge. Continuity gives the volumetric flow rate as the product of area and mean velocity, $$Q=AV=22.0\times12.60=\boxed{277.2\ \text{m}^3/\text{s}}$$
7. Convert to a daily volume. One day contains $24\times3600=86\,400\ \text{s}$, therefore $$Q_{\text{day}}=277.2\times86\,400=2.395\times10^{7}\ \text{m}^3/\text{day}$$ that is, $\boxed{Q_{\text{day}}\approx 2.40\times10^{7}\ \text{m}^3/\text{day}}$, or roughly 24 million cubic metres per day.

Part (b) — Entrance and exit minor losses (4 marks)

Minor losses are the local energy losses caused by changes in the geometry of the flow boundary, as distinct from the distributed friction loss along the barrel. In culvert hydraulics they are conventionally expressed as a multiple of the velocity head in the barrel, $h = k\,V^{2}/2g$, and only two of them normally matter.

The entrance loss occurs where water accelerates from the relatively still approach pool into the much smaller barrel. The streamlines cannot turn a sharp corner, so they separate at the barrel lip and form a vena contracta a short distance inside; the eddying and re-expansion downstream of that contraction dissipates energy. It is written $h_e=k_e\,V^{2}/2g$, where the entrance loss coefficient depends entirely on the geometry of the end treatment. HDS-5 gives $k_e\approx0.9$ for a thin projecting pipe, 0.5 for a square-edged headwall, 0.2 for a groove-end or socket-end pipe in a headwall, and about 0.2 for a bevelled or wingwall-flared box entrance. Improving the entrance geometry is the cheapest way to increase the capacity of an inlet-controlled culvert, which is why tapered and side-tapered inlets exist.

The exit loss occurs where the jet leaving the barrel expands abruptly into the downstream channel or pool and its kinetic energy is dissipated in turbulence. It is written $h_o=k_o\bigl(V^{2}/2g - V_d^{2}/2g\bigr)$, where $V_d$ is the downstream channel velocity. When the culvert discharges into a large, comparatively still pool, $V_d\approx0$ and the coefficient reduces to $k_o=1.0$: the entire barrel velocity head is lost. When the downstream channel is of similar size and carries a similar velocity, the loss is much smaller, and it is usual (and conservative) to keep $k_o = 1.0$ unless the tailwater channel is well defined.

The practical significance is that both losses appear in the outlet-control headwater equation, $HW = TW + \bigl(1+k_e+\frac{2gn^{2}L}{R^{4/3}}\bigr)\frac{V^{2}}{2g} - S_0L$, whereas under inlet control only the entrance geometry matters and the exit loss does not affect capacity at all. Since a design must satisfy both control conditions, the headwater is computed under each and the larger governs.

Part (c) — Hydraulic profiles under inlet control (6 marks)

Inlet control means the control section — the section at which critical depth occurs and which therefore fixes the discharge for a given headwater — lies at or just inside the culvert entrance. The capacity is governed only by the headwater depth, the barrel cross-sectional area and the inlet edge geometry; the barrel roughness, the barrel length, the barrel slope and the tailwater have no influence on capacity. Inlet control is typical of steep, hydraulically short culverts. Downstream of the control the barrel flows part full and supercritical, so the profile is drawn from the inlet forward.

(i) Inlet and outlet both unsubmerged (HDS-5 Type A). The headwater is below about 1.2 times the barrel rise, so the entrance behaves as a weir. Critical depth forms at the inlet, the water surface drops through critical to the supercritical normal depth of the steep barrel, and the jet leaves the barrel freely at a depth below the crown with a low tailwater. This is the classical steep-culvert case.

(ii) Inlet unsubmerged, outlet submerged (HDS-5 Type B). The entrance is still acting as a weir with critical depth at the control, so capacity is unchanged — this is the point of the sketch. However the high tailwater forces the downstream end of the barrel to run full, and the supercritical flow leaving the inlet must return to the subcritical, pressurised condition through a hydraulic jump inside the barrel. The jump position adjusts itself so that the sequent depth matches the tailwater. Air entrained at the jump must be able to escape, which is why such culverts are vented.

(iii) Inlet submerged, outlet unsubmerged (HDS-5 Type C). The headwater now exceeds roughly 1.2 to 1.5 times the barrel rise and the entrance acts as an orifice rather than a weir; the discharge follows an orifice relation in which $Q$ varies with the square root of the effective head. Immediately inside the entrance the flow contracts to less than the full barrel section, then runs part full and supercritical for the remaining length and discharges freely to a low tailwater. A mitred or bevelled inlet in a fill slope is the usual field example.

In all three sketches the same feature identifies inlet control: the water surface passes through the control at the entrance and the barrel does not flow full over its whole length. If the barrel were full from end to end and the headwater depended on the barrel friction and the tailwater, the culvert would be in outlet control and a completely different set of profiles would apply.