Question 2 of 6: Multi-path pipe network between two reservoirs — total flow and pressure extremes
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 16-Civ-A5, Hydraulic Engineering. Closed book (one aid sheet; approved calculator); 3 hours; six questions, candidates complete any five — all six are solved here as a study resource. Each question 20 marks; parts of equal value.
Governing relations supplied on the exam cover sheet: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with slope $S=\Delta h/L$ (SI: $Q$ in m³/s, $D$ in m); Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$; Darcy–Weisbach $\Delta h = 0.0826\,\tfrac{fL}{D^{5}}Q^{2}$; total dynamic head $TDH=H_s+H_f$. Unless stated, local losses and velocity head are negligible, diameters are nominal, and the fluid is water ($\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$).
Question 2: Multi-path pipe network between two reservoirs — total flow and pressure extremes (20 marks)
Given. Ten identical pipes ($D=0.400\ \text{m}$, $L=350\ \text{m}$, $C=145$, invert elevation 17 m) join Res A ($120\ \text{m}$) to Res B ($70\ \text{m}$), so the total available head is $\Delta H=120-70=50\ \text{m}$. From Figure 2 the pipes form two independent A→B routes: an upper route of three pipes in series (via $J_2,J_3$), and a lower route in which one pipe leaves A to junction $J_1$, whence three parallel two-pipe branches (via $J_4,J_5,J_6$) run on to B.
Given data
Quantity
Value
Res A / Res B water level
120 m / 70 m
Available head $\Delta H$
50 m
Pipe $D$ / $L$ / $C$
0.400 m / 350 m / 145
Pipe invert elevation
17 m
Number of pipes
10 (6 junction nodes)
Find. (a) the total discharge $Q_{tot}=Q_{upper}+Q_{lower}$; (b) the maximum and minimum nodal pressure head.
Figure 2. Ten identical pipes forming two parallel A→B paths (upper 3-in-series; lower single pipe + three parallel two-pipe branches).
Approach. With identical pipes, write every friction loss as $h_f=k\,Q^{1.852}$ where $k=L/a^{1/0.54}$ and $a=0.278\,C\,D^{2.63}$. Reduce each route to an equivalent resistance, impose $\Delta H=50\ \text{m}$ across both parallel routes to get the two route flows, and add them. Then walk the HGL from A to find the highest- and lowest-pressure nodes.
Single-pipe resistance. $a=0.278(145)(0.400)^{2.63}=3.620$, so
$$k=\frac{L}{a^{1/0.54}}=\frac{350}{3.620^{1.852}}=32.30\ \ (\text{SI, }Q\ \text{in m}^3/\text{s}).$$
Upper route (3 pipes in series) carries $Q_u$. Series losses add: $3k\,Q_u^{1.852}=\Delta H$, so
$$Q_u=\left(\frac{50}{3(32.30)}\right)^{0.54}=0.700\ \text{m}^3/\text{s},$$
and each upper pipe drops $50/3=16.67\ \text{m}$.
Lower route carries $Q_l$. The single pipe A–J1 takes the full $Q_l$; the three parallel branches beyond J1 split it equally ($Q_l/3$ each) and each branch is two pipes in series. Equating the route loss to $\Delta H$:
$$k\,Q_l^{1.852}\left[1+2\left(\tfrac{1}{3}\right)^{1.852}\right]=50,\qquad 1+2(1/3)^{1.852}=1.262,$$
$$Q_l=\left(\frac{50}{32.30(1.262)}\right)^{0.54}=1.117\ \text{m}^3/\text{s}.$$
Total flow (part a).
$$\boxed{Q_{tot}=Q_u+Q_l=0.700+1.117\approx 1.82\ \text{m}^3/\text{s}.}$$
Trace the HGL for pressures (part b). Losses: upper pipe $16.67\ \text{m}$ each; A–J1 $=k\,Q_l^{1.852}=39.64\ \text{m}$; each lower parallel pipe $=k\,(Q_l/3)^{1.852}=5.18\ \text{m}$. Hence the nodal HGLs are
$$HGL_{J2}=120-16.67=103.33,\quad HGL_{J3}=86.67,$$
$$HGL_{J1}=120-39.64=80.36,\quad HGL_{J4,5,6}=80.36-5.18=75.18\ \text{m}.$$
Both routes correctly close on $70\ \text{m}$ at B. Subtracting the 17 m invert, the pressure-head extremes are
$$\boxed{p_{max}/\gamma = 103.33-17 \approx 86.3\ \text{m (at }J_2),\qquad p_{min}/\gamma = 75.18-17 \approx 58.2\ \text{m (at }J_4,J_5,J_6).}$$
