Partial Flow in a 600 mm Sewer — Discharge and Velocity: 16-Civ-B5 Water Supply and Wastewater Treatment, December 2017, Q5

Question 5: Partial Flow in a 600 mm Sewer — Discharge and Velocity (25 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.

Symbol	Quantity	Value
D	Internal pipe diameter	600 mm = 0.600 m
Δz	Drop in invert elevation	0.500 m
L	Length over which the invert drops	100 m
S	Slope of the energy grade line (uniform flow)	0.500 / 100 = 0.005
d	Depth of flow	200 mm = 0.200 m
n	Manning roughness coefficient	0.013

Find. The discharge q and the mean velocity v carried by the sewer at that depth, using the partial-flow (hydraulic-elements) curves supplied with the paper.

Figure 5.1 — Cross-section of the 600 mm sewer flowing 200 mm deep. The wetted arc subtends θ = 2.462 rad (141.1°) at the centre, giving A = 0.08250 m², P = 0.7386 m and R = 0.1117 m.

Approach. Compute the full-flow discharge and velocity from Manning's equation, read the discharge and velocity ratios off the supplied partial-flow curves at the depth ratio d/D, multiply, and then confirm the chart read by evaluating the circular-section geometry exactly.

Confirm the slope from the invert drop.
The sewer flows by gravity, and for uniform flow the water surface, the energy line and the invert are all parallel, so the friction slope equals the pipe slope:
$$S = \frac{\Delta z}{L} = \frac{0.500}{100} = 0.005 \quad (0.5\ \%,\ \text{or}\ 1\ \mathrm{in}\ 200)$$
This is a steep sanitary sewer — comfortably above the 0.1 % minimum grade normally specified for a 600 mm pipe — which is the first hint that the velocity check later in the solution will pass easily.
Full-flow geometry.
The reference condition for the chart is the pipe flowing just full, for which the geometry is elementary:
$$A_{\text{full}} = \frac{\pi D^{2}}{4} = \frac{\pi (0.600)^{2}}{4} = 0.2827\ \mathrm{m^2}, \qquad R_{\text{full}} = \frac{D}{4} = 0.150\ \mathrm{m}$$
The hydraulic radius of a full circular pipe is D/4 because the area πD²/4 divided by the perimeter πD reduces to it directly.
Full-flow velocity and discharge from Manning's equation.
Manning's equation in SI units gives the mean velocity of uniform flow in terms of the hydraulic radius and the friction slope:
$$V = \frac{1}{n} R^{2/3} S^{1/2} = \frac{1}{0.013}\,(0.150)^{2/3}(0.005)^{1/2} = 76.92 \times 0.2823 \times 0.07071$$
and the full-flow discharge follows from continuity, Q = VA:
$$\boxed{\ V_{\text{full}} = 1.536\ \mathrm{m/s}, \qquad Q_{\text{full}} = 1.536 \times 0.2827 = 0.4341\ \mathrm{m^3/s} = 434\ \mathrm{L/s}\ }$$
Enter the partial-flow curves at the depth ratio.
The chart supplied with the paper plots d/D against the ratios q/Q and v/V. Here
$$\frac{d}{D} = \frac{200}{600} = 0.333$$
Reading horizontally at d/D = 0.333 to each curve and dropping to the horizontal axis gives
$$\frac{q}{Q} \approx 0.24, \qquad \frac{v}{V} \approx 0.82$$
Note how differently the two behave: at one third of the depth the pipe carries only a quarter of its capacity, yet the water is already moving at more than four fifths of its full-bore speed. That is because the hydraulic radius — which sets the velocity — recovers much faster with depth than the flow area does.
Convert the ratios into the answers.
Multiplying the ratios by the full-flow values,
$$q = 0.24 \times 0.4341 = 0.104\ \mathrm{m^3/s}, \qquad v = 0.82 \times 1.536 = 1.26\ \mathrm{m/s}$$ $$\boxed{\ q \approx 0.104\ \mathrm{m^3/s} = 104\ \mathrm{L/s} \approx 9000\ \mathrm{m^3/d}, \qquad v \approx 1.26\ \mathrm{m/s}\ }$$
Confirm the chart read from the exact circular-section geometry.
A chart read is worth checking, and the geometry is closed-form. With the wetted arc subtending an angle θ at the centre,
$$\theta = 2\arccos\!\left(1 - \frac{2d}{D}\right) = 2\arccos(0.3333) = 2.462\ \mathrm{rad} = 141.1^\circ$$ $$A = \frac{D^{2}}{8}(\theta - \sin\theta) = \frac{0.36}{8}\,(2.462 - 0.6285) = 0.08250\ \mathrm{m^2}$$ $$P = \frac{\theta D}{2} = \frac{2.462 \times 0.600}{2} = 0.7386\ \mathrm{m} \quad\Longrightarrow\quad R = \frac{A}{P} = 0.1117\ \mathrm{m}$$
Applying Manning at this hydraulic radius and multiplying by the partial area,
$$v = \frac{1}{0.013}(0.1117)^{2/3}(0.005)^{1/2} = 1.262\ \mathrm{m/s}, \qquad q = 1.262 \times 0.08250 = 0.1041\ \mathrm{m^3/s}$$
which corresponds to q/Q = 0.240 and v/V = 0.822 — agreeing with the chart read to within the width of a pencil line, and confirming that the supplied curves are the constant-n (Manning) family.
Design checks that the numbers now make available.
Having the depth-averaged velocity, the sewer's adequacy can be judged rather than merely reported. The boundary shear stress is
$$\tau_0 = \rho g R S = 1000 \times 9.81 \times 0.1117 \times 0.005 = 5.48\ \mathrm{Pa}$$
which is well above the 1.5 Pa normally required for self-cleansing on 1 mm grit and above the 3–4 Pa needed to scour a consolidated deposit, so this reach is self-cleansing at the stated flow. The velocity of 1.26 m/s likewise sits between the 0.6–0.75 m/s minimum and the 3 m/s abrasion limit. The Froude number, using the hydraulic depth D_h = A/T with top width T = D sin(θ/2) = 0.566 m, is
$$Fr = \frac{v}{\sqrt{g D_h}} = \frac{1.262}{\sqrt{9.81 \times 0.1458}} = 1.05$$
which is essentially critical — a point worth flagging, since flow near Fr = 1 is prone to standing waves and depth instability. Finally, the chart itself shows that maximum discharge occurs not at full bore but at d/D = 0.938, where q/Q = 1.076; a sewer running just below the crown carries about 8 % more than one running full, because the wetted perimeter closes faster than the area near the crown.

Figure 5.2 — The supplied partial-flow (hydraulic-elements) curves for a circular pipe at constant n, reproduced with this problem's read-off marked: entering at d/D = 0.333 gives q/Q = 0.24 on the discharge curve and v/V = 0.82 on the velocity curve. The discharge curve peaks at q/Q = 1.076 near d/D = 0.94 and the velocity curve at v/V = 1.14 near d/D = 0.81.

Quantity	Symbol	Value
Slope of sewer	S	0.005 (1 in 200)
Full-flow area / hydraulic radius	A_full, R_full	0.2827 m², 0.150 m
Full-flow velocity	V	1.54 m/s
Full-flow discharge	Q	0.434 m³/s = 434 L/s
Depth ratio	d/D	0.333
Ratios read from the curves	q/Q, v/V	0.24, 0.82 (exact geometry: 0.240, 0.822)
Discharge at 200 mm depth	q	0.104 m³/s = 104 L/s ≈ 9000 m³/d
Velocity at 200 mm depth	v	1.26 m/s
Boundary shear stress	τ₀	5.48 Pa — self-cleansing (> 1.5 Pa) and scouring (> 4 Pa)
Froude number	Fr	1.05 — essentially critical flow

Check: the solution assumes steady uniform flow, so that the friction slope equals the invert slope, and a constant Manning n = 0.013 independent of depth, which is the family of curves the paper supplies. Some texts print the variable-n chart, on which n rises by up to 30 % at shallow depths; on that chart the same d/D would give q/Q ≈ 0.21 and v/V ≈ 0.71, i.e. q ≈ 91 L/s and v ≈ 1.09 m/s. The exact-geometry cross-check in Step 6 confirms the supplied chart is the constant-n family, so the boxed values stand. The computed Froude number of 1.05 places the flow at the critical point; in design this reach would be reviewed for standing waves, and a small change in slope or roughness would move it clear.

Topic. Uniform open-channel flow in a partially full circular conduit — Manning's equation, the hydraulic-elements (partial-flow) curves, and the sewer design checks they support.

Key relations. Manning $v = \frac{1}{n}R^{2/3}S^{1/2}$, $q = vA$; circular geometry $\theta = 2\arccos(1-2d/D)$, $A = \frac{D^2}{8}(\theta - \sin\theta)$, $P = \theta D/2$, $R = A/P$, $T = D\sin(\theta/2)$; tractive force $\tau_0 = \rho g R S$; Froude number $Fr = v/\sqrt{gA/T}$.

Why this works. Manning's equation is valid for any prismatic channel section, so a circular pipe running part full is treated exactly like any other open channel — the only work is getting A and R right for the geometry. The hydraulic-elements curves exist because, for a fixed n and S, the ratios q/Q and v/V depend on d/D alone and on nothing else: everything dimensional cancels. That is what makes one chart serve every diameter, slope and roughness, and it is why the chart read and the closed-form geometry agree to three figures. The counter-intuitive peaks — maximum velocity at d/D ≈ 0.81 and maximum discharge at d/D ≈ 0.94 — come from the wetted perimeter growing faster than the area as the surface approaches the crown, so R and hence v fall while the pipe is still filling.

Common pitfall. Applying R = D/4 to the partial-flow condition, which uses the full-bore hydraulic radius with a partial area and overstates the discharge by about 25 % here. The second is multiplying Q_full by the velocity ratio (or V_full by the discharge ratio) — the two curves cross-hatch on the same chart and are easy to swap, and doing so here would give 356 L/s instead of 104 L/s. A third is forgetting that the invert drop over a stated length is the slope only for uniform flow; in a backwater or a surcharged reach it is not.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 5 (open-channel flow; hydraulic elements of circular sections); Metcalf & Eddy, 5th ed., Ch. 3 (wastewater collection and pumping, sewer hydraulics and self-cleansing velocity); Davis & Cornwell, 5th ed., Ch. 6.

Question 5 of 5: Partial Flow in a 600 mm Sewer — Discharge and Velocity

Question 5: Partial Flow in a 600 mm Sewer — Discharge and Velocity (25 marks)