Question 2 of 6: Ten-pipe twin-path network between two reservoirs

Question 2: Ten-pipe twin-path network between two reservoirs (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.

Given. Ten identical pipes ($D=0.300$ m, $L=250$ m, $C=130$, all at elevation 25 m) join reservoir A (80 m) to reservoir B (70 m), so the available head is $\Delta H=80-70=10$ m. The printed figure (page 3) shows that the ten pipes are not a parallel bundle: they form two independent A→B routes. The upper route is three pipes in series (A–J2–J3–B); the lower route is a single pipe A–J1, from which three parallel two-pipe branches (via J4, J5, J6) run on to B.

Find. (a) total discharge $Q_{\text{tot}}=Q_u+Q_l$; (b) maximum and minimum pressure head.

[Figure not reproduced: Figure 2. Ten identical pipes forming two A→B routes: upper 3 pipes in series; lower single pipe A–J1 then three parallel two-pipe branches (redrawn from the exam figure). See the official exam paper.]

Approach. Write every pipe loss as $h_f=k\,Q^{1.852}$. The two routes share only the reservoirs, so each spans the full $\Delta H=10$ m on its own. Reduce each route to an equivalent resistance, solve its flow, add the two flows, then trace the HGL through the junctions to find the pressure extremes.

1. Single-pipe resistance. $a=0.278(130)(0.30)^{2.63}=1.523$, so $$k=\dfrac{L}{a^{1/0.54}}=\dfrac{250}{1.523^{1.852}}=114.7\quad(\text{SI, }Q\text{ in m}^3/\text{s}).$$
2. Upper route (3 pipes in series) carries $Q_u$. Series losses add, $3k\,Q_u^{1.852}=10$: $$Q_u=\left(\dfrac{10}{3(114.7)}\right)^{0.54}=0.148\ \text{m}^3/\text{s},$$ and each upper pipe drops $10/3=3.33$ m.
3. Lower route carries $Q_l$. Pipe A–J1 carries all of $Q_l$; the three identical branches split it equally ($Q_l/3$ each), each branch being two pipes in series: $$k\,Q_l^{1.852}\left[1+2\left(\tfrac13\right)^{1.852}\right]=10,\qquad 1+2(1/3)^{1.852}=1.261,$$ $$Q_l=\left(\dfrac{10}{114.7(1.261)}\right)^{0.54}=0.236\ \text{m}^3/\text{s}.$$
4. (a) Total flow. $$\boxed{Q_{\text{tot}}=Q_u+Q_l=0.148+0.236\approx 0.384\ \text{m}^3/\text{s}\ (384\ \text{L/s}).}$$
5. (b) Trace the HGL. Upper pipes lose 3.33 m each; A–J1 loses $k\,Q_l^{1.852}=7.93$ m; each lower branch pipe loses $k(Q_l/3)^{1.852}=1.04$ m. Nodal HGLs: $$\text{HGL}_{J2}=76.67,\quad \text{HGL}_{J3}=73.33,\quad \text{HGL}_{J1}=72.07,\quad \text{HGL}_{J4,5,6}=71.04\ \text{m},$$ and both routes close on 70 m at B (check). Subtracting the 25 m pipe elevation: $$\boxed{p_{\max}/\gamma\approx 51.7\ \text{m (at J2)},\qquad p_{\min}/\gamma\approx 46.0\ \text{m (at J4, J5, J6)}.}$$ At the pipe connections themselves the pressure head is bracketed by the reservoir levels, 55 m at A and 45 m at B (velocity head and entrance/exit losses neglected per NOTE 6). The junction values above are the ones the network layout controls.

The single feeder A–J1 carries the whole lower-route flow and uses up 7.9 of the 10 m, so the lower junctions sit lowest on the HGL. The upper route has no such bottleneck, so its first junction J2 keeps the highest pressure. The lower route still carries more water (236 vs 148 L/s) because its three branches in parallel give it the lower overall resistance.