Question 2 of 6: Ten-pipe system between two reservoirs
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 16-Civ-A5 Hydraulic Engineering — December 2017. 3 hours, closed book (one 8.5×11 aid sheet permitted). Six questions of equal value; candidates answer any five. All six are solved here as a study resource. Unless stated, local losses and velocity head are neglected; water ρ = 1000 kg/m³, ν = 1.31×10−6 m²/s.
Reference texts: D. A. Chin, Water-Resources Engineering (3rd ed.), Ch. 2 (flow in closed conduits) & Ch. 7 (water-distribution systems); A. Osman Akan, Open Channel Hydraulics; M. Hanif Chaudhry, Open-Channel Flow (2nd ed.), Ch. 3 (hydraulic jump); Chadwick, Morfett & Borthwick, Hydraulics in Civil and Environmental Engineering.
Question 2: Ten-pipe system between two reservoirs (20 marks)
Given. Ten identical pipes join reservoir A to reservoir B. Figure 2 (as printed on the exam page) shows they form two independent A→B routes: an upper route of three pipes in series (A–J2–J3–B), and a lower route in which a single pipe runs from A to junction J1, from which three parallel two-pipe branches (via J4, J5, J6) run on to B (3 + 1 + 3×2 = 10 pipes).
Find. (a) the total discharge A→B; (b) the maximum and minimum pressure head in the system (at the junction nodes).
Figure 2. Ten-pipe system as drawn on the exam: upper route A–J2–J3–B (three pipes in series); lower route A–J1 (one pipe), then three parallel two-pipe branches J1–J4/J5/J6–B. Not to scale.
the printed drawing instead shows the two-route layout above (one feeder pipe from A to a hub junction that fans into three branches, plus a separate three-pipe upper route). The solution follows the printed drawing.
Approach. Write every pipe loss as $h_f=k\,Q^{1.852}$. The two routes share only the reservoirs, so each spans the full 15 m on its own: reduce each route to an equivalent resistance, solve its flow, and add the two. Then trace the HGL along each route and subtract the 25 m pipe elevation.
Upper route (3 pipes in series) carries $Q_u$. Series losses add, $3k\,Q_u^{1.852}=15$:
$$Q_u=\left(\dfrac{15}{3(75.76)}\right)^{0.54}=0.230\ \text{m}^3/\text{s},$$
and each upper pipe drops $15/3 = 5.0$ m.
Lower route carries $Q_l$. The feeder A–J1 carries all of $Q_l$; the three identical branches split it equally ($Q_l/3$ each), and each branch is two pipes in series:
$$k\,Q_l^{1.852}\left[1+2\left(\tfrac13\right)^{1.852}\right]=15,\qquad 1+2\left(\tfrac13\right)^{1.852}=1.2615,$$
$$Q_l=\left(\dfrac{15}{75.76(1.2615)}\right)^{0.54}=0.368\ \text{m}^3/\text{s}.$$
HGL at the junctions. Feeder loss $k\,Q_l^{1.852}=11.89$ m; each branch pipe $k\,(Q_l/3)^{1.852}=1.55$ m. Hence
$$\text{HGL}_{J2}=80-5.0=75.0,\quad \text{HGL}_{J3}=70.0,\quad \text{HGL}_{J1}=80-11.89=68.11,\quad \text{HGL}_{J4,J5,J6}=68.11-1.55=66.56\ \text{m},$$
and both routes close on 65.0 m at B (upper: $70.0-5.0$; lower: $66.56-1.55$).
Maximum and minimum pressure head (part b). Pressure head = HGL − 25 m:
$$p/\gamma\big|_{\max} = 75.0-25 = \boxed{50.0\ \text{m (at J2)}},\qquad p/\gamma\big|_{\min} = 66.56-25 = \boxed{41.6\ \text{m (at J4, J5, J6)}}.$$
Along the pipes themselves the pressure head is bounded by the reservoir ends, 55 m where the pipes leave A and 40 m where they enter B.
