Question 5 of 7: Hardy-Cross Pipe Network, Sanitary Sewer Types and the Manning Formula

Problem 5: Hardy-Cross Pipe Network, Sanitary Sewer Types and the Manning 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): $Q_{AB}=0.450$, $Q_{AC}=0.550$, $Q_{BC}=0.050$ (B→C), $Q_{BD}=0.300$, $Q_{CD}=0.300$ — check: at A, $0.450+0.550=1.000$ ✓; at B, $0.450=0.050+0.300+0.100$ ✓; at C, $0.550+0.050=0.300+0.300$ ✓; at D, $0.300+0.300=0.600$ ✓.
2. Pipe head-loss coefficients (Hazen–Williams, SI), $h_L=KQ^{1.852}$, $K=10.67L/(C^{1.852}d^{4.8704})$. $$K_{AB}=445.6,\quad K_{BC}=245.4,\quad K_{CD}=109.8,\quad K_{AC}=140.2,\quad K_{BD}=41.2\ \ (\text{m per (m}^3/\text{s})^{1.852}).$$
3. Loop 1 (A-B-C-A) and Loop 2 (B-D-C-B) corrections. With the trial flows, the first-pass Hardy-Cross corrections are $\Delta Q_1\approx-0.095\ \text{m}^3/\text{s}$ (loop 1) and $\Delta Q_2\approx+0.046\ \text{m}^3/\text{s}$ (loop 2), applied with sign convention $+$AB/$+$BC/$-$AC for loop 1 and $+$BD/$-$CD/$-$BC for loop 2 (BC receives both loops' corrections each pass).
4. Iterate to convergence. Re-applying the two-loop correction drives both $\sum h_L\to0$ after several further passes ($|\Delta Q|<0.001$ L/s), converging with pipe BC's flow direction reversed from the B→C trial guess to C→B: $$Q_{AB}=354.4,\ Q_{AC}=645.6,\ Q_{CB}=91.4,\ Q_{BD}=345.8,\ Q_{CD}=254.2\ \text{(all L/s)}.$$ Loop check: $h_{AB}+h_{BC}-h_{AC}=65.27+2.92-62.35\approx0$; $h_{BD}-h_{CD}-h_{BC}=5.77-8.69-2.92\approx0$ (with $h_{BC}$ taken negative in the loop-1 traverse sense once BC's direction is C→B).

(ii) Gravity, Pressure and Vacuum Sanitary Sewers

Gravity sewer. Condition: adequate natural ground slope/relief along the alignment to maintain the self-cleansing velocity (typically ≥0.6 m/s) under gravity alone. Advantage: lowest operating and maintenance cost — no mechanical or electrical components in the collection pipe itself, making it the simplest and most reliable long-term option.

Pressure sewer (grinder-pump, low-pressure system). Condition: flat or undulating terrain, or shallow rock, where deep gravity trenching is impractical or uneconomical. Advantage: the pipe can follow the ground surface at shallow, near-uniform depth and cross ridges or high points a gravity line cannot, substantially reducing excavation cost in difficult terrain.

Vacuum sewer. Condition: flat terrain with a high water table (e.g., coastal or low-lying areas) where gravity trenching would require continuous dewatering. Advantage: very shallow, flat-grade installation with a fully sealed, negative-pressure pipe network, which reduces infiltration/exfiltration and environmental risk in sensitive, high-groundwater areas.

(iii) The Manning Formula

Given. $Q=\frac{1}{n}AR^{2/3}S^{1/2}$, the standard empirical resistance equation for steady, uniform (normal-depth) turbulent flow in an open channel or a partially/fully-flowing pipe.

Find. The meaning and consistent dimensions of $Q$, $A$, $n$, $R$ and $S$.

$Q$ is the discharge [m³/s]; $A$ is the flow cross-sectional area [m²]; $n$ is Manning's roughness coefficient, an empirical, dimensionless-in-practice friction parameter (its formal SI unit is s/m$^{1/3}$, but it is conventionally quoted as a bare number, e.g. 0.013 for smooth concrete); $R=A/P$ is the hydraulic radius (flow area divided by wetted perimeter $P$) [m]; and $S$ is the energy grade-line slope (equal to the bed slope for uniform flow) [m/m, dimensionless]. The formula works because, for fully turbulent flow, wall shear stress scales with $R^{1/3}$ times the mean-velocity gradient in a way Manning's empirical exponents capture without needing an explicit friction-factor iteration.