Question 2 of 6: Parallel-branch network between two reservoirs
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 16-Civ-A5 Hydraulic Engineering — May 2018 · 3 hours · closed book (one aid sheet) · six questions, complete any five (all solved here) · each question 20 marks, equal-value parts.
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 neglected, diameters are nominal, and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Question 2: Parallel-branch network between two reservoirs (20 marks)
Given. Reservoirs $H_A=95\ \text{m}$, $H_B=70\ \text{m}$; ten identical pipes $D=0.25\ \text{m}$, $L=200\ \text{m}$, $C=130$. From the contours, junctions read: J1 = J2 = J3 = 90 m (upstream tier), J4 = J5 = 80 m (middle tier), J6 = 75 m.
Find. (a) total system discharge $Q_{tot}$; (b) the maximum and minimum junction pressure heads; (c) which of the two branches carries the larger flow and why.
Figure 2. Two parallel branches from A to junction J6, then the common pipe P10 to B. Upper branch: two full two-pipe routes (P1+P4) ∥ (P2+P5) then P8. Lower branch: P3, then the parallel pair P6 ∥ P7, then P9.
Approach. Because the pipes are identical, reduce each branch by series–parallel resistance in the form $h_f=k\,Q^{1.852}$ (identical pipes in parallel split the flow equally; identical pipes in series add their losses). The two branches share the same end heads ($H_A$ and $H_{J6}$), and the common pipe P10 carries $Q_{tot}$ to B; two equations close the system.
Per-pipe resistance. With $a=0.278(130)(0.25)^{2.63}=0.6224$, one pipe obeys $h_f=k\,Q^{1.852}$ where $k=L/a^{1/0.54}=200/0.6224^{1.852}=222.9$ (SI, $Q$ in m³/s).
Reduce each branch. Upper: two identical two-pipe routes in parallel each carry $Q_u/2$, then P8 carries $Q_u$:
$$\Delta H_{A\to J6}^{\text{up}}=k\Big[2\big(\tfrac{Q_u}{2}\big)^{1.852}+Q_u^{1.852}\Big]=1.554\,k\,Q_u^{1.852}.$$
Lower: P3 (full $Q_\ell$), the pair P6∥P7 (each $Q_\ell/2$), then P9 (full $Q_\ell$):
$$\Delta H_{A\to J6}^{\text{low}}=k\Big[Q_\ell^{1.852}+\big(\tfrac{Q_\ell}{2}\big)^{1.852}+Q_\ell^{1.852}\Big]=2.277\,k\,Q_\ell^{1.852}.$$
Match branch heads. Both branches drop from $H_A$ to $H_{J6}$, so $1.554\,Q_u^{1.852}=2.277\,Q_\ell^{1.852}$, giving $Q_u/Q_\ell=(2.277/1.554)^{1/1.852}=1.229$.
Close on the full head with P10. The overall drop is $H_A-H_B=25\ \text{m}=\Delta H^{\text{up}}_{A\to J6}+k\,Q_{tot}^{1.852}$ with $Q_{tot}=Q_u+Q_\ell$. Solving simultaneously,
$$\boxed{Q_u\approx135.2\ \text{L/s},\quad Q_\ell\approx110.0\ \text{L/s},\quad Q_{tot}\approx245\ \text{L/s}.}$$
Junction HGLs and pressure heads (part b). Working the losses inward from A gives $HGL$ at each junction; pressure head $=HGL-z$:
Junction hydraulics
Node
HGL (m)
Elev z (m)
$p/\gamma$ (m)
J1, J2
93.48
90
3.48
J3
91.26
90
1.26
J4
91.97
80
11.97
J5
90.23
80
10.23
J6
86.49
75
11.49
$$\boxed{p_{\max}/\gamma\approx 12.0\ \text{m at J4};\qquad p_{\min}/\gamma\approx 1.3\ \text{m at J3}.}$$
The maximum occurs at the low-elevation node still holding a high HGL (J4), the minimum at the near-source high-ground node (J3) whose ground almost meets its HGL.
Compare branches (part c). $Q_u\approx135\ \text{L/s}\gt Q_\ell\approx110\ \text{L/s}$ — the upper branch conveys more.
The reason is purely resistance: the upper branch offers two complete two-pipe routes in parallel (equivalent coefficient $1.554\,k$), whereas the lower branch is a single pipe P3 in series with just one parallel pair and a single pipe P9 (equivalent coefficient $2.277\,k$). Lower resistance draws the larger share of flow.
