Question 5 of 5: Partial flow in a 300 mm sewer

Question 5: Partial flow in a 300 mm 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 $q$ and the mean velocity $v$ at $d/D = 0.30$.

Approach. Uniform flow in a sewer means the friction slope equals the invert slope, so Manning's equation gives the full-bore condition first; the supplied hydraulic-elements curves then convert full-bore values to partial-flow values by the ratios $q/Q$ and $v/V$ read at $d/D=0.30$. The closed-form circular geometry is used alongside as an independent check on the chart reading.

1. Slope from the invert drop. The pipe runs partly full at a constant depth, so the flow is uniform and the water surface is parallel to the invert: $$S=\frac{\Delta z}{L}=\frac{1.0}{200}=\boxed{0.005\ \text{(1 in 200)}}$$
2. Full-bore geometry. For a circular pipe running just full the wetted perimeter is the whole circumference, so $$A=\frac{\pi D^2}{4}=\frac{\pi(0.300)^2}{4}=0.07069\ \text{m}^2,\qquad R=\frac{A}{P}=\frac{\pi D^2/4}{\pi D}=\frac{D}{4}=0.0750\ \text{m}$$
3. Full-bore velocity and discharge from Manning. $$V=\frac{1}{n}R^{2/3}S^{1/2} =\frac{1}{0.013}(0.0750)^{2/3}(0.005)^{1/2} =76.92\times 0.17785\times 0.07071=0.967\ \text{m/s}$$ $$Q=AV=0.07069\times 0.967=0.0684\ \text{m}^3/\text{s} =\boxed{68.4\ \text{L/s running full}}$$
4. Read the supplied partial-flow curves at $d/D=0.30$. Entering the chart on the depth axis at 0.30 and crossing to the two curves, $$\frac{q}{Q}\approx 0.20,\qquad \frac{v}{V}\approx 0.78$$ so directly $$q\approx 0.20\times 68.4=13.7\ \text{L/s},\qquad v\approx 0.78\times 0.967=0.75\ \text{m/s}$$

[Figure not reproduced: Left: the 300 mm sewer at d/D = 0.30. Right: the supplied partial-flow curves, redrawn from the closed-form geometry for constant Manning n, with the reading at d/D = 0.30 marked. See the official exam paper.]

5. Confirm the reading with the closed-form geometry. A chart read to two figures deserves a check. With $\theta$ the angle subtended at the centre by the wetted perimeter, $$\theta=2\arccos\!\left(1-\frac{2d}{D}\right)=2\arccos(0.4)=2.3186\ \text{rad}\ (132.8^{\circ})$$ $$A_p=\frac{D^2}{8}(\theta-\sin\theta) =\frac{0.09}{8}(2.3186-0.7333)=0.01783\ \text{m}^2$$ $$P_p=\frac{\theta D}{2}=0.3478\ \text{m},\qquad R_p=\frac{A_p}{P_p}=\frac{0.01783}{0.3478}=0.05128\ \text{m}$$
6. Manning applied to the partial section. $$v=\frac{1}{n}R_p^{2/3}S^{1/2} =76.92\times(0.05128)^{2/3}\times 0.07071=\boxed{0.751\ \text{m/s}}$$ $$q=A_p\,v=0.01783\times 0.751=0.01339\ \text{m}^3/\text{s} =\boxed{13.4\ \text{L/s}}$$ The implied ratios, $q/Q=0.196$ and $v/V=0.776$, reproduce the chart reading of 0.20 and 0.78 to within one part in fifty. That agreement also identifies which family of curves the paper supplies: these are the constant-n curves. The variable-$n$ family, which allows roughness to rise as the depth falls, would give about 0.17 and 0.68 here, i.e. 11.6 L/s, so the distinction is worth about 14 per cent and should be stated.
7. Check that the sewer is self-cleansing. The design test at low flow is the boundary shear the flow exerts on a deposit: $$\tau=\rho g R_p S=1000\times 9.81\times 0.05128\times 0.005 =\boxed{2.52\ \text{Pa}}$$ This clears the usual 1.5 Pa criterion for keeping organic solids moving, but it is below the 3 to 4 Pa needed to re-mobilise a consolidated grit deposit, so the pipe will stay clean if it is kept clean and will not recover on its own if it is allowed to silt. The velocity, 0.75 m/s, likewise clears the customary 0.6 m/s minimum.
8. Check the flow regime. Using the hydraulic depth, $$T=D\sin\!\frac{\theta}{2}=0.300\times 0.9165=0.275\ \text{m},\qquad D_h=\frac{A_p}{T}=\frac{0.01783}{0.275}=0.0649\ \text{m}$$ $$\mathrm{Fr}=\frac{v}{\sqrt{g D_h}}=\frac{0.751}{\sqrt{9.81\times 0.0649}} =0.94$$ The flow is subcritical but only just. A pipe operating this close to critical is prone to standing waves and to depth instability at junctions and bends, so the drop across any manhole on this reach should be checked rather than assumed.
9. Name the two errors this problem is built to catch. First, using the full-bore hydraulic radius with the partial area gives $$q_{\text{wrong}}=A_p V=0.01783\times 0.967=17.3\ \text{L/s}$$ which is 29 per cent high; $R$ must be recomputed for the partial section because a shallow flow has a disproportionately long wetted perimeter. Second, interchanging the two curves gives $0.78\times 68.4=53$ L/s, four times the correct answer. A useful guard against both is the half-depth identity, where $q/Q=0.500$ and $v/V=1.000$ exactly: any chart read must approach those values as $d/D\rightarrow 0.5$. It is also worth noting the peaks of these curves, $q/Q=1.076$ at $d/D=0.938$ and $v/V=1.140$ at $d/D=0.813$, which explain why a sewer running just below the crown carries more than the same sewer running full.