Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 98-Civ-A5 Hydraulic Engineering, December 2015. Full worked solutions. Closed-book format, 3 hours; six questions, candidates complete any five — all six are solved here as a study resource. All questions are of equal value (20 marks); where a question has parts, each part is of equal value.
Reference texts. Mays, Water Resources Engineering, 2nd ed. (pipe systems, equivalent pipes, valves, quasi-steady network simulation); Chow, Open-Channel Hydraulics (Manning uniform flow, gutter/triangular sections); Crowe, Elger & Roberson, Engineering Fluid Mechanics (unsteady open-channel flow, St. Venant equations, kinematic vs. dynamic waves). Exam-supplied relations: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI units), Manning $Q=\tfrac{1}{n}A R^{2/3}S^{1/2}$, valve law $Q=\tau E_s\sqrt{H_{u/s}-H_{d/s}}$, and Total Dynamic Head $\mathrm{TDH}=H_S+H_f$. Unless stated, local losses and velocity head are neglected, $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Convention. Inverting Hazen–Williams gives a single-pipe resistance $h_f = k\,Q^{1.852}$ with $k=L\big/(0.278\,C\,D^{2.63})^{1.852}$ (SI, $Q$ in m³/s). Identical pipes in parallel between two nodes share the flow equally; pipes in series add their losses at a common flow. This collapses these "N identical pipe" networks by equivalent-pipe reduction, so no Hardy–Cross iteration is needed.
Given. Ten identical pipes ($D=250\text{ mm}=0.25\text{ m}$, $L=200\text{ m}$, $C=130$) between reservoir A (water surface 95 m) and reservoir B (water surface 70 m); available head $\Delta H = 95-70 = 25\text{ m}$. Node ground elevations read from the contours: the first-tier junctions J1, J2, J4 lie on the 90 m contour, the mid junctions J3, J5 on the 80 m contour, and the right junction J6 on the 75 m contour.
Find. (a) total system flow; (b) maximum and minimum pressure head at the junctions; (c) which of the two A→J6 branches carries more flow, and why.
Figure 1. Two parallel A→J6 branches. Upper branch: routes (P1+P4) and (P2+P5) in parallel to J3, then P8. Lower branch: P3, the parallel pair (P6∥P7), then P9. P10 carries the full flow to B. Dashed lines are ground contours (90/80/75 m).
Approach. Reduce to two equivalent branches in parallel between A and J6, impose equal head loss on the two branches and add the P10 loss to close the 25 m budget; then trace the HGL to each node and subtract its ground elevation for the pressure heads.
Reduce each branch. Upper branch carries $Q_u$: two identical series-routes (each two pipes) split $Q_u$ equally, then P8 takes the full $Q_u$, so $H_u = 2k\left(\tfrac{Q_u}{2}\right)^{1.852}+k\,Q_u^{1.852}$. Lower branch carries $Q_\ell$: P3 and P9 each pass the full $Q_\ell$ and the pair P6∥P7 splits it, so $H_\ell = 2k\,Q_\ell^{1.852}+k\left(\tfrac{Q_\ell}{2}\right)^{1.852}$.
Match the branches and close the head budget. Both branches span A→J6, so $H_u=H_\ell$; P10 then carries $Q_u+Q_\ell$: $$H_u = H_\ell,\qquad H_u + k\,(Q_u+Q_\ell)^{1.852}=25\text{ m}.$$ Solving this pair, $$\boxed{Q_u = 0.135\ \text{m}^3/\text{s}=135.2\ \text{L/s},\quad Q_\ell = 0.110\ \text{m}^3/\text{s}=110.0\ \text{L/s}.}$$
Total flow. $$\boxed{Q_{\text{tot}} = Q_u+Q_\ell = 0.245\ \text{m}^3/\text{s}\approx 245\ \text{L/s}.}$$ Check: the common branch loss is $H_{AJ6}=8.51\text{ m}$ and $h_{f,\text{P10}}=k\,Q_{\text{tot}}^{1.852}=16.49\text{ m}$, summing to $25.0\text{ m}$.
Hydraulic grade line. Starting from $\text{HGL}_A=95$ and subtracting each pipe loss (each upper feeder passes $Q_u/2=67.6$ L/s, each lower feeder $Q_\ell=110$ L/s, the pair splits to $55.0$ L/s): $\text{HGL}_{J1}=\text{HGL}_{J2}=93.48$, $\text{HGL}_{J3}=91.97$, $\text{HGL}_{J4}=91.26$, $\text{HGL}_{J5}=90.23$, $\text{HGL}_{J6}=86.49\text{ m}$.
Pressure heads. $p/\gamma = \text{HGL}-z_{\text{ground}}$: $$J1,J2:\ 3.48;\quad J3:\ 11.97;\quad J4:\ 1.26;\quad J5:\ 10.23;\quad J6:\ 11.49\ \text{m}.$$ Hence $$\boxed{p_{\max}/\gamma = 11.97\text{ m at J3},\qquad p_{\min}/\gamma = 1.26\text{ m at J4}.}$$ J3 combines a high HGL with a low (80 m) ground elevation; J4 sits at 90 m ground with only a small friction drop from A, so it is the tightest point in the system.
(c) Branch comparison. The upper branch offers two parallel two-pipe routes from A to J3 (equivalent full-flow resistance $2^{-0.852}=0.555$ of one pipe there), so its total coefficient is $1.555\,k$. The lower branch has single pipes P3 and P9 in series with only the short pair P6∥P7, giving $2.277\,k$ — more resistant. Equal head loss across a larger resistance means less flow, so $$\boxed{\text{the upper branch P1-P2-P4-P5-P8 carries the higher flow }(135\text{ vs }110\text{ L/s}).}$$
