Question 6 of 7

Problem 6 (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, series pipes A–B–C–D (8 marks)

Given.

$Q=100\ \text{L/s}=0.1\ \text{m}^3/\text{s}$ throughout (series, no branching); $p_A/\gamma=50$ m; $z_A=10$, $z_B=20$, $z_C=30$, $z_D=40$ m.

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

Approach. Continuity gives the same $Q$ (hence a different $V$) in every pipe; apply the energy equation node-to-node along the series path, accumulating elevation change and Darcy–Weisbach friction loss in each pipe.

1. Velocities from continuity. $$V_1=\frac{Q}{A_1}=\frac{0.1}{\frac{\pi}{4}(0.25)^2}=2.04\ \text{m/s}, \quad V_2=\frac{0.1}{\frac{\pi}{4}(0.35)^2}=1.04\ \text{m/s}, \quad V_3=\frac{0.1}{\frac{\pi}{4}(0.40)^2}=0.80\ \text{m/s}$$
2. Friction loss, each pipe (Darcy–Weisbach). $$h_{f1}=f_1\frac{L_1}{D_1}\frac{V_1^2}{2g}=0.02\times\frac{1500}{0.25}\times\frac{2.04^2}{19.62}=\boxed{25.4\ \text{m}}$$ $$h_{f2}=0.03\times\frac{1200}{0.35}\times\frac{1.04^2}{19.62}=5.7\ \text{m}, \qquad h_{f3}=0.04\times\frac{2000}{0.40}\times\frac{0.80^2}{19.62}=6.5\ \text{m}$$
3. Pressure head at B (same pipe, same $D$, from A to B — velocity head cancels): $$\frac{p_B}{\gamma} = \frac{p_A}{\gamma}+z_A-z_B-h_{f1} = 50+10-20-25.4 = \boxed{14.6\ \text{m}}$$
4. Continue to C, then D. $$\frac{p_C}{\gamma} = \frac{p_B}{\gamma}+z_B-z_C-h_{f2} = 14.6+20-30-5.7 = -1.1\ \text{m}$$ $$\frac{p_D}{\gamma} = \frac{p_C}{\gamma}+z_C-z_D-h_{f3} = -1.1+30-40-6.5 = \boxed{-17.5\ \text{m}}$$

(ii) Hardy Cross method and the two conservation principles (6 marks)

The Hardy Cross method solves a looped pipe network by iterative correction of an initially assumed (but continuity-satisfying) set of pipe flows. In each closed loop of the network, a flow correction $\Delta Q = -\dfrac{\sum K Q|Q|^{n-1}}{n\sum K|Q|^{n-1}}$ (with $K$ the pipe resistance coefficient and $n=1.85$ for Hazen-Williams or $n=2$ for Darcy-Weisbach) is computed from the loop's current head-loss imbalance and applied to every pipe in that loop; the process repeats, loop by loop, until all loop head-loss imbalances are acceptably small.

The method enforces the network's two governing conservation principles simultaneously: (1) continuity (mass conservation) at every junction — the initial flow assignment is chosen so inflow equals outflow at each node, and because $\Delta Q$ is applied identically to every pipe of a loop, every correction step preserves this nodal balance automatically; and (2) energy conservation around every closed loop — in a true solution the sum of head losses (signed by assumed flow direction) around any closed loop must be zero, since starting and ending at the same node/elevation the net head change must vanish; the iterative $\Delta Q$ correction is derived specifically to drive this loop head-loss sum toward zero without disturbing the nodal continuity already satisfied.

(iii) Sanitary sewer diameter, partial flow at 85% full (6 marks)

Given. $Q_d=5\ \text{m}^3/\text{s}$ at $y/D=0.85$; $S_0=3\%$; corrugated steel pipe, $n=0.02$; velocity must satisfy $0.6

Find. Required pipe diameter $D$.

Approach. Use the circular partial-flow geometric relations (central angle $\theta$ subtended by the water surface) with Manning's equation, solved for the diameter that delivers $Q_d$ at the specified depth ratio.

1. Central angle at $y/D=0.85$. From $y/D=\tfrac12(1-\cos(\theta/2))$: $$\theta = 2\cos^{-1}(1-2\times0.85) = 2\cos^{-1}(-0.70) = 4.692\ \text{rad}\ (268.9^\circ)$$
2. Partial-flow area and hydraulic radius, in terms of $D$. $$A=\frac{D^2}{8}(\theta-\sin\theta), \qquad R=\frac{D}{4}\cdot\frac{\theta-\sin\theta}{\theta}$$
3. Solve Manning's equation for $D$. $Q_d=\tfrac1n AR^{2/3}S_0^{1/2}$, with $A\propto D^2$ and $R\propto D$, reduces to $Q_d\propto D^{8/3}$; solving numerically for $Q_d=5\ \text{m}^3/\text{s}$: $$D = \boxed{1.25\ \text{m}\ (1246\ \text{mm})}$$
4. Velocity check at this diameter. $A=\frac{(1.246)^2}{8}(\theta-\sin\theta)=1.104\ \text{m}^2$, $$V = \frac{Q_d}{A} = \frac{5}{1.104} = \boxed{4.53\ \text{m/s}}$$ This satisfies $0.6
5. Select a commercial size. 1246 mm is not a standard manufactured corrugated-steel-pipe diameter; specifying the next commercial size up, $D=1350\ \text{mm}$, provides the required capacity at $y/D\le0.85$, and since velocity only decreases as $D$ increases at fixed $Q_d$, the resulting velocity remains comfortably within the 0.6–6 m/s window.