Question 5 of 5: Discharge and velocity in a part-full sewer

Question 5: Discharge and velocity in a part-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. A circular gravity sewer running at uniform (normal) depth, with the geometry, roughness and depth of flow listed below; the hydraulic-elements chart on page 3 of the examination paper is supplied for the partial-flow ratios.

Find. The discharge q and the mean velocity v in the sewer at the stated depth of flow.

Approach. Uniform flow means the friction slope equals the invert slope, so Manning's equation gives the full-bore condition directly; the supplied hydraulic-elements chart then converts the full-bore discharge and velocity to the part-full values at d/D = 0.30. The chart read is confirmed against the closed-form circular-segment geometry.

1. Establish the slope. The flow is uniform, so the water surface, the energy grade line and the invert are all parallel and the friction slope is the invert slope: $$S=\frac{\Delta z}{L}=\frac{1.0}{200}=0.005\ \text{m/m}\ (1\ \text{in}\ 200)$$
2. Full-bore geometry. Running just full, the wetted area is the whole circle and the wetted perimeter the whole circumference, so the hydraulic radius of any circular pipe flowing full is one quarter of its diameter: $$A=\frac{\pi D^2}{4}=\frac{\pi(0.500)^2}{4}=0.1963\ \text{m}^2, \qquad R=\frac{A}{P}=\frac{D}{4}=0.125\ \text{m}$$
3. Full-bore velocity and discharge from Manning's equation. Substituting into V = (1/n)R^2/3S^1/2, $$V=\frac{1}{0.013}(0.125)^{2/3}(0.005)^{1/2} =\frac{1}{0.013}(0.2500)(0.07071)=1.360\ \text{m/s}$$ and multiplying by the full area, $$Q=AV=0.1963\times 1.360=\boxed{Q=0.267\ \text{m}^3/\text{s}=267\ \text{L/s}}$$ These are the reference values the chart ratios are applied to; they are not the answer.
4. Read the hydraulic-elements chart at d/D = 0.30. Entering the supplied curves on the vertical axis at a depth ratio of 0.30 and reading across to each curve gives $$\frac{q}{Q}\approx 0.20\qquad\text{and}\qquad \frac{v}{V}\approx 0.78$$ The velocity ratio is much closer to unity than the discharge ratio because velocity depends only on the hydraulic radius, which changes slowly with depth, whereas discharge also carries the full penalty of the much smaller flow area.
Hydraulic-elements chart for a circular pipe with constant n. Entering at d/D = 0.30 gives q/Q = 0.20 and v/V = 0.78.
5. Confirm the chart read against the exact geometry. A chart read to two decimal places deserves a check, and the circular segment is elementary. The angle subtended at the centre by the wetted perimeter is $$\theta=2\arccos\!\left(1-\frac{2d}{D}\right)=2\arccos(0.400)=2.3186\ \text{rad}\ (132.84^{\circ})$$ from which $$A_p=\frac{D^2}{8}(\theta-\sin\theta)=\frac{0.500^2}{8}(2.3186-0.7333)=0.04954\ \text{m}^2$$ $$P_p=\frac{\theta D}{2}=\frac{2.3186\times 0.500}{2}=0.5796\ \text{m}, \qquad R_p=\frac{A_p}{P_p}=0.08547\ \text{m}$$ Applying Manning's equation directly at this depth, $$v=\frac{1}{0.013}(0.08547)^{2/3}(0.005)^{1/2}=1.055\ \text{m/s}, \qquad q=A_p v=0.04954\times 1.055=0.05229\ \text{m}^3/\text{s}$$ so the exact ratios are q/Q = 0.196 and v/V = 0.776, reproducing the chart read to within the width of a pencil line. That agreement also identifies which family of curves the paper supplied — the constant-n family, not the variable-n family (see the callout below).
6. State the answers. Taking the exact values, at 30 per cent depth the sewer carries $$\boxed{q=52.3\ \text{L/s}\quad\text{at}\quad v=1.06\ \text{m/s}}$$ Reading the chart to the nearest half-division instead gives 53 L/s at 1.06 m/s, which is the same answer to the precision the method supports.
7. Check the result against the design criteria the numbers now make available. A sewer is not designed on discharge alone. The boundary shear (tractive) stress at this depth is $$\tau=\rho g R_p S=1000\times 9.81\times 0.08547\times 0.005=4.19\ \text{Pa}$$ comfortably above the 1.5 Pa usually required for self-cleansing and into the 3–4 Pa band that will move grit, so the pipe will not silt at this flow. The velocity of 1.06 m/s likewise exceeds the conventional 0.6 m/s minimum and is well below the 3 m/s at which abrasion becomes a concern. The Froude number, taken on the hydraulic depth D_h = A_p/T with the top width T = D sin(θ/2) = 0.458 m, is $$Fr=\frac{v}{\sqrt{gD_h}}=\frac{1.055}{\sqrt{9.81\times 0.1081}}=1.02$$ i.e. essentially critical, which is worth flagging: flow near Fr = 1 is unstable, prone to standing waves and to a hydraulic jump at any downstream control, and the designer should either flatten the grade slightly or ensure the manholes downstream are detailed to take the surging.