Question 5 of 5: Discharge and Velocity in a Partly Full Sewer

Question 5: Discharge and Velocity in a Partly Full Sewer (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.

Find. The discharge and mean velocity when the sewer runs full, and the discharge and mean velocity at a flow depth of 200 mm, using the supplied partial-flow curves.

Approach. Uniform flow is assumed, so the friction slope equals the invert slope. Compute the full-bore area and hydraulic radius, apply Manning's equation to obtain V and Q, enter the supplied chart at d/D = 0.667 to read q/Q and v/V, and multiply. The chart read is then confirmed independently from the exact circular-segment geometry.

1. Slope from the invert drop. For uniform flow in a gravity sewer the energy grade line, the hydraulic grade line and the invert are all parallel, so the friction slope is simply $$S = \frac{\Delta z}{L} = \frac{0.25}{100} = 0.0025\ \text{m/m}$$
2. Full-bore geometry. A pipe flowing just full has $$A = \frac{\pi D^2}{4} = \frac{\pi(0.300)^2}{4} = 0.0707\ \text{m}^2, \qquad R = \frac{D}{4} = 0.0750\ \text{m}$$ The hydraulic radius of a full circular pipe is D/4 because the area πD²/4 divided by the perimeter πD leaves D/4 — a result worth quoting, since it is the single most useful shortcut on this type of question.
3. Full-bore velocity and discharge by Manning. Substituting into $$V = \frac{1}{n}R^{2/3}S^{1/2} = \frac{1}{0.013}(0.0750)^{2/3}(0.0025)^{1/2} = \frac{1}{0.013}(0.1779)(0.0500)$$ gives V = 0.684 m/s, and hence $$\boxed{V_{\text{full}} = 0.684\ \text{m/s}, \qquad Q_{\text{full}} = AV = 0.0707 \times 0.684 = 0.0484\ \text{m}^3/\text{s} = 48.4\ \text{L/s}}$$
4. Relative depth and the chart read. The flow depth is $$\frac{d}{D} = \frac{200}{300} = 0.667$$ Entering the supplied partial-flow curves at d/D = 0.667 and reading across to each curve gives $$\frac{q}{Q} \approx 0.78, \qquad \frac{v}{V} \approx 1.11$$ Note carefully that the velocity ratio exceeds unity. This is not an error in reading the chart; above about half depth a partly full pipe flows faster than the same pipe running full, because the wetted perimeter contributed by the pipe crown is removed while most of the area remains.
   Figure 5.2 — The supplied partial-flow curves, entered at d/D = 0.667. The discharge curve gives q/Q = 0.78 and the velocity curve v/V = 1.11; the velocity curve peaks at 1.14 near d/D = 0.81 and the discharge curve at 1.08 near d/D = 0.94.
5. Partial discharge and velocity. Multiplying the ratios by the full-bore values, $$q = 0.78 \times 48.4 = 37.7\ \text{L/s}, \qquad v = 1.11 \times 0.684 = 0.76\ \text{m/s}$$ $$\boxed{q \approx 37.8\ \text{L/s}, \qquad v \approx 0.76\ \text{m/s at }d = 200\ \text{mm}}$$
6. Independent check from the exact segment geometry. The chart should never be the only evidence for a boxed answer. The wetted arc subtends $$\theta = 2\cos^{-1}\!\left(1-\frac{2d}{D}\right) = 2\cos^{-1}(-0.3333) = 3.8213\ \text{rad} = 218.94^{\circ}$$ from which the circular-segment properties follow directly: $$A_p = \frac{D^2}{8}(\theta - \sin\theta) = \frac{0.300^2}{8}(3.8213+0.6285) = 0.0501\ \text{m}^2$$ $$P_p = \frac{\theta D}{2} = 0.573\ \text{m}, \qquad R_p = \frac{A_p}{P_p} = 0.0873\ \text{m}$$ Applying Manning to the partial section, $v = (1/0.013)(0.0873)^{2/3}(0.0025)^{1/2} = 0.757$ m/s and $q = A_pv = 0.0379$ m³/s = 37.9 L/s, giving exact ratios of q/Q = 0.784 and v/V = 1.107. These reproduce the chart read to within the width of a pencil line, which confirms both the answer and the fact that the supplied curves belong to the constant-n family. Note that $R_p = 0.0873$ m is larger than the full-bore $R = 0.0750$ m — the arithmetic reason the part-full velocity exceeds the full-bore velocity.
7. Design checks the numbers now make available. A 25-mark question that asks for two numbers is asking for the engineering that follows from them. The boundary shear at this depth is $$\tau = \rho g R_p S = 1000 \times 9.81 \times 0.0873 \times 0.0025 = 2.14\ \text{Pa}$$ which clears the 1.5 Pa self-cleansing threshold, so the sewer will not silt at this flow, though it falls short of the 3–4 Pa needed to scour an established deposit. The Froude number, taken on the hydraulic depth $D_h = A_p/T$ with top width $T = D\sin(\theta/2) = 0.283$ m, is $$Fr = \frac{v}{\sqrt{gD_h}} = \frac{0.757}{\sqrt{9.81 \times 0.177}} = 0.57$$ so the flow is comfortably subcritical and free of the standing waves and depth instability that plague sewers running near Fr = 1. Finally, the velocity of 0.76 m/s exceeds the 0.6 m/s minimum required by Canadian municipal standards, so the pipe satisfies its design velocity at this depth.