Question 2 of 7

Problem 2 (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) Pressure head at B and D along the series pipes (8 marks)

Given.

Elevations: $Z_A=12$ m, $Z_B=15$ m, $Z_C=20$ m, $Z_D=30$ m. Flowrate $Q=200$ L/s $=0.200\ \text{m}^3/\text{s}$ throughout (series pipes, continuity). Pressure head at A, $p_A/\gamma=60$ m.

Find. Pressure head $p_B/\gamma$ and $p_D/\gamma$.

[Figure not reproduced: Fig. 2 — Three pipes in series with abrupt diameter expansions at B and C, as printed on the exam (schematic, not to scale). See the official exam paper.]

Approach. Because the pipe expands in diameter at B and C, velocity (and therefore velocity head) changes at each junction; velocity heads here are all under 0.5 m — small next to the friction losses — so, consistent with "assume fully turbulent flow in all cases" (i.e. $f$ is already given and Reynolds-independent), the standard simplification is to work with the hydraulic grade line (piezometric head only, velocity head and the small local/expansion losses at the two abrupt expansions neglected). Compute the friction loss in each pipe from Darcy–Weisbach with the given $f$, then step the HGL from A to D.

1. Pipe velocities (continuity check only — confirms the flow is fully turbulent in every pipe as stated). $$V_1=\frac{Q}{A_1}=\frac{0.200}{\pi(0.300)^2/4}=2.83\ \text{m/s}\qquad V_2=\frac{0.200}{\pi(0.400)^2/4}=1.59\ \text{m/s}\qquad V_3=\frac{0.200}{\pi(0.500)^2/4}=1.02\ \text{m/s}$$
2. Friction head loss in each pipe, Darcy–Weisbach $H_f=f\dfrac{L}{D}\dfrac{V^2}{2g}$. $$H_{f1}=(0.03)\left(\frac{1200}{0.3}\right)\frac{(2.83)^2}{2(9.81)}=\boxed{48.96\ \text{m}}\qquad H_{f2}=(0.04)\left(\frac{1500}{0.4}\right)\frac{(1.59)^2}{2(9.81)}=19.37\ \text{m}\qquad H_{f3}=(0.05)\left(\frac{2500}{0.5}\right)\frac{(1.02)^2}{2(9.81)}=13.22\ \text{m}$$
3. Step the hydraulic grade line from A to D. $\text{HGL}=p/\gamma+Z$; HGL falls by the friction loss in each pipe and pressure head is recovered by subtracting the new elevation: $$\text{HGL}_A=p_A/\gamma+Z_A=60+12=72.0\ \text{m}$$ $$\text{HGL}_B=\text{HGL}_A-H_{f1}=72.0-48.96=23.04\ \text{m}\ \Rightarrow\ p_B/\gamma=\text{HGL}_B-Z_B=23.04-15=\boxed{8.04\ \text{m}}$$ $$\text{HGL}_C=\text{HGL}_B-H_{f2}=23.04-19.37=3.67\ \text{m}\ \Rightarrow\ p_C/\gamma=3.67-20=-16.33\ \text{m}$$ $$\text{HGL}_D=\text{HGL}_C-H_{f3}=3.67-13.22=-9.55\ \text{m}\ \Rightarrow\ p_D/\gamma=-9.55-30=\boxed{-39.55\ \text{m}}$$

(ii) Hardy Cross method for pipe network design (6 marks)

The Hardy Cross method solves for the flow distribution in a looped pipe network by iterative balancing. An initial flow is assumed in every pipe such that continuity is satisfied at every node (inflow = outflow), even though the assumed flows will not yet satisfy the second network law (the algebraic sum of head losses around any closed loop must be zero). For each loop, a flow correction is computed from the loop's head-loss imbalance,

(with $K$ the pipe's resistance coefficient and $n=1.85$ for Hazen–Williams or $n=2$ for Darcy–Weisbach), applied to every pipe in that loop (added to flows taken positive clockwise, subtracted otherwise), and the process is repeated loop by loop until every $\Delta Q$ is negligibly small.

Two key assumptions designers must be aware of: (1) continuity is enforced exactly at every node from the very first trial flow onward — the method only ever corrects loop head-loss imbalance, so an initial guess that does not already balance every node's inflow/outflow will never converge to a valid solution; and (2) the head-loss/flow relationship used for $K$ (Hazen–Williams $C$ or Darcy $f$) is assumed known and constant for every pipe, when in reality $C$ degrades with pipe age/tuberculation and $f$ is Reynolds-number dependent — so the "solved" network is only as accurate as the assumed roughness values, and a network's real performance should be periodically re-calibrated against field-measured pressures/flows.

(iii) Sanitary sewer diameter for a peak flow of 6 m³/s (6 marks)

Given.

Find. Required pipe diameter $d$ in mm; confirm the velocity condition.

Approach. For a circular pipe flowing full, $R=D/4$; substituting into Manning's equation collapses it to a single closed-form power law in $D$, which is solved directly (no trial-and-error).

1. Collapse Manning's equation for full flow. $$Q=\frac{1}{n}A R^{2/3}S^{1/2}=\frac{1}{n}\left(\frac{\pi D^2}{4}\right)\left(\frac{D}{4}\right)^{2/3}S^{1/2}=\frac{0.3117}{n}D^{8/3}S^{1/2}$$
2. Solve for $D$. $$D=\left(\frac{Qn}{0.3117\sqrt{S}}\right)^{3/8}=\left(\frac{(6)(0.04)}{0.3117\sqrt{0.04}}\right)^{3/8}=\left(\frac{0.240}{0.06234}\right)^{3/8}=(3.850)^{3/8}=\boxed{1.658\ \text{m}=1658\ \text{mm}}$$
3. Check velocity and round to a commercial size. At $D=1658$ mm, $A=\pi D^2/4=2.159\ \text{m}^2$, so $V=Q/A=6/2.159=2.78\ \text{m/s}$ — inside the $0.7$–$8$ m/s window. Rounding up to the next standard concrete pipe size, $D=1700$ mm gives $A=2.270\ \text{m}^2$, $V=6/2.270=\boxed{2.64\ \text{m/s}}$ — still comfortably inside the velocity limits (self-cleansing and non-scouring).