Question 2 of 6: Looped Pipe Network Between Two Reservoirs
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 98-Civ-A5 Hydraulic Engineering, May 2014. 3 hours, closed book (one aid sheet). Six questions of equal value (20 marks each); candidates complete any five, and all six are solved here as a study resource.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe networks, valves, pumps); Chow, Open-Channel Hydraulics (Manning uniform flow, hydraulic jump, gutter flow); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance, wall shear). Exam-supplied relations used throughout: Hazen–Williams $Q=0.278\,C\,D^{2.63}\,S^{0.54}$ with $S=h_f/L$ (SI), Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, Darcy–Weisbach $\Delta h=0.0826\,\tfrac{fL}{D^5}Q^2$, and $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Question 2: Looped Pipe Network Between Two Reservoirs (20 marks)
Given. Res. A level 120 m, Res. B level 70 m (driving head 50 m); ten identical pipes ($C=145$, $D=0.400\ \text{m}$, $L=350\ \text{m}$) arranged as in Figure 2, all at elevation 17 m; five interior junction nodes (labelled T, U, M, BL, BM below).
Find. (a) total conveyed flow A→B; (b) maximum and minimum nodal pressure head.
Figure 2. Looped conveyance network. Four pipes leave Res. A and three enter Res. B; the five interior nodes form loops, so the flow split is not fixed by continuity alone.
Approach. Unlike Question 1 this network contains loops, so the five interior nodal heads are unknowns fixed by writing continuity at each interior node with Hazen–Williams flows, and solving the coupled nonlinear system (Newton–Raphson).
Pipe flow law. For any pipe joining heads $H_i,H_j$, $Q_{ij}=0.278\,C\,D^{2.63}\left(\dfrac{|H_i-H_j|}{L}\right)^{0.54}\operatorname{sign}(H_i-H_j)$, identical for all ten pipes.
Node continuity. At each interior node the signed inflows sum to zero (no demands): $\sum_j Q_{ij}=0$. This is five equations in the five unknown heads $H_T,H_U,H_M,H_{BL},H_{BM}$, with $H_A=120$ and $H_B=70$ fixed.
Solve. Newton–Raphson converges to $H_T=95.00$, $H_U=102.04$, $H_M=104.84$, $H_{BL}=109.49$, $H_{BM}=98.98\ \text{m}$ (residual $<10^{-12}$). Node T is a simple pass-through, so its two identical pipes carry equal flow and its head is exactly the mid-value $\tfrac12(120+70)=95\ \text{m}$—a useful check.
Total flow (a). Sum the four pipes leaving A: $871+729+665+545=\boxed{2809\ \text{L/s}\approx2.81\ \text{m}^3/\text{s}}$ (equal to the three pipes entering B, confirming global continuity).
Pressure heads (b). All nodes sit at elevation 17 m, so $p/\gamma=H-17$: $p_{BL}=92.49$, $p_M=87.84$, $p_U=85.03$, $p_{BM}=81.98$, $p_T=78.00\ \text{m}$.
Extremes. Maximum $\boxed{p_{\max}=92.49\ \text{m}}$ at the lightly loaded node nearest A (BL); minimum $\boxed{p_{\min}=78.00\ \text{m}}$ at node T, which carries the largest share of the flow.
