Question 5 of 7: Hardy-Cross Pipe Network, Sanitary Sewer Self-Cleansing Velocity and the Harmon Peaking Formula

Problem 5: Hardy-Cross Pipe Network, Sanitary Sewer Self-Cleansing Velocity and the Harmon Peaking Formula (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.

(i) Hardy-Cross Solution for the Pipe Network

Given.

Node demands: inflow 1000 L/s at A; outflows 100 L/s at B, 300 L/s at C, 600 L/s at D (a self-consistent network — total demand equals total inflow).

Find. The converged flow in each of the five pipes.

Approach. Assume a trial flow distribution that satisfies continuity at every node, split the network into two independent loops sharing pipe BC, compute the Hazen–Williams head loss $h_L=KQ^{1.852}$ for each pipe, and apply the Hardy-Cross correction $\Delta Q=-\sum h_L/(1.852\sum h_L/Q)$ loop-by-loop until $\sum h_L\to0$ around both loops simultaneously.

1. Trial flow distribution satisfying continuity. Starting trial (m³/s): $Q_{AB}=0.550$, $Q_{AC}=0.450$, $Q_{BC}=0.100$, $Q_{BD}=0.350$, $Q_{CD}=0.250$ — check: at A, $0.550+0.450=1.000$ ✓; at B, $0.550=0.100+0.100+0.350$ ✓; at C, $0.450+0.100=0.300+0.250$ ✓; at D, $0.350+0.250=0.600$ ✓.
2. Pipe head-loss coefficients (Hazen–Williams, SI), $h_L=KQ^{1.852}$, $K=10.67L/(C^{1.852}d^{4.8704})$. $$K_{AB}=K_{CD}=K_{BD}=445.6,\qquad K_{AC}=K_{BC}=280.4\ \ (\text{m per (m}^3/\text{s})^{1.852}).$$
3. Loop 1 (A-B-C-A) first correction. With the trial flows, $\sum h_L=+94.3$ m around the loop and $\sum(1.852\,h_L/Q)=331$, giving $\Delta Q_1=-94.3/331=\boxed{-0.105\ \text{m}^3/\text{s}}$ (applied $+$ to AB, $+$ to BC, $-$ to AC in the loop-traverse sense).
4. Loop 2 (B-D-C-B) first correction. Similarly, $\Delta Q_2=\boxed{-0.039\ \text{m}^3/\text{s}}$ (applied $+$ to BD, $-$ to CD, $-$ to BC).
5. Iterate to convergence. Re-applying the same two-loop correction (BC receiving both loops' corrections each pass) drives both $\sum h_L\to0$ after five further passes ($|\Delta Q|<0.001$ L/s): $$Q_{AB}=437.0,\ Q_{AC}=563.0,\ Q_{BC}=36.0,\ Q_{BD}=301.0,\ Q_{CD}=299.0\ \text{(all L/s)}.$$ Loop check: $h_{AB}+h_{BC}-h_{AC}=96.18+0.59-96.78\approx0$; $h_{BD}-h_{CD}-h_{BC}=48.22-47.63-0.59\approx0$.

(ii) Manning Formula and the Minimum Cleansing Velocity

Given. $Q=(A/n)R^{2/3}S^{1/2}$; target minimum full-flow (self-cleansing) velocity $V_{\min}=0.6$ m/s.

Find. How the formula is used to set the sewer's design slope.

Because $V=Q/A=(1/n)R^{2/3}S^{1/2}$ is independent of $A$ alone (it depends only on the hydraulic radius, roughness and slope), the designer first fixes the pipe diameter from the required capacity, then solves the rearranged Manning relation for the minimum slope that delivers $V\ge0.6$ m/s at full flow: $S_{\min}=\left[\dfrac{V_{\min}n}{R^{2/3}}\right]^2$. For a full circular pipe, $R=d/4$, so this reduces to a direct check/solve for $S$ given $d$ and $n$; if the natural ground slope along the alignment is flatter than $S_{\min}$, the designer must either steepen the invert (accepting deeper excavation), reduce $d$ (increasing $V$ for the same $Q$, at the cost of reduced capacity), or accept periodic flushing/maintenance in lieu of self-cleansing velocity. The 0.6 m/s target itself is chosen because grit and organic solids in sanitary flow begin to deposit below this threshold shear velocity, leading to blockages, odour and H₂S corrosion if sustained.

(iii) Harmon Formula for Sanitary Sewer Peak-Flow Design

Given. $M=1+\dfrac{14}{4+p^{1/2}}$, where $M$ is the peaking factor and $p$ is the tributary population in thousands.

Find. How the formula is applied to obtain the design peak flow for a sanitary sewer.

The Harmon formula estimates the ratio of peak to average sanitary flow as a function of the tributary population alone, reflecting the same population-averaging behaviour as Babbitt's formula (Problem 2(ii)): a small service population has a comparatively high peaking factor (few simultaneous events dominate), while $M$ asymptotically decreases toward about 1.8–2 as $p$ grows large. To use it, the designer computes the average dry-weather sanitary flow from the tributary population and a per-capita generation rate, evaluates $M$ from the tributary population $p$ (in thousands), and multiplies, $Q_{\text{peak}}=M\times Q_{\text{avg}}$, to obtain the design peak flow the sewer (and, by extension, the downstream pump station or interceptor) must be sized to convey; an allowance for groundwater infiltration and rainfall-derived inflow (I/I) is normally added on top of $Q_{\text{peak}}$ for an older or leaky system.