07-Str-B11 · December 2015

Question 3 of 6: Three-Pipe Looped Distribution Network

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2015 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides) and any non-communicating calculator are permitted. Six questions of twenty marks each; candidates complete any five, and where a question has parts the parts carry equal weight. The solutions below work all six, so the paper can be used whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, series/parallel equivalent pipes, looped-network analysis and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, service-pressure criteria, control-valve characteristics and gutter/roadway hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and compound cross-sections; Henderson, F.M., Open Channel Flow (Macmillan) — the Saint-Venant equations, wave celerity and the kinematic/diffusion/dynamic hierarchy; Transportation Association of Canada, Geometric Design Guide for Canadian Roads — crossfall, curb reveal and roadway drainage conventions.

Notation and conventions used throughout. The paper supplies the SI Hazen-Williams form $Q = 0.278\,C\,D^{2.63}S^{0.54}$ with $Q$ in m³/s, $D$ in metres and $S = h_f/L$; it is used exactly as printed. Writing $K = 0.278\,C\,D^{2.63}$ inverts it to the head-loss form $h_f = L\,(Q/K)^{1.852}$, the exponent $n = 1.852$ being the value the paper itself quotes in its loop-correction note. Following Note 6, local losses and velocity head are neglected, so the hydraulic grade line (HGL) and the energy grade line coincide and “pressure head at a node” means $p/\gamma = \mathrm{HGL} - z$. Water is taken as $\rho = 1000$ kg/m³ with $\nu = 1.31\times10^{-6}$ m²/s.

Question 3: Three-Pipe Looped Distribution Network (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A single-loop network: the source reservoir feeds node N1 through P1 and node N2 through P2, and P3 closes the loop between N1 and N2 (Figure 3 of the exam paper). Every pipe is identical and both nodal demands are equal.

Given data — Question 3
Quantity	Symbol	Value
Pipe diameter (P1, P2, P3)	$D$	250 mm = 0.250 m
Pipe length (P1, P2, P3)	$L$	500 m
Hazen-Williams coefficient	$C$	110
Demand at N1	$q_1$	5 L/s = 0.005 m³/s
Demand at N2	$q_2$	5 L/s = 0.005 m³/s
Source reservoir water level	$H_R$	90 m

Find. The discharge carried by each of the three pipes, including its direction, and the head at N1 and N2.

Figure 3.1 — The single-loop network with the solved flows. P1 and P2 each deliver exactly their own node’s demand and the closing pipe P3 carries nothing, because the loop is geometrically and hydraulically symmetric about the reservoir.

Approach. Recognise that the loop is symmetric — identical pipes from a single source to two identical demands — so the symmetric flow distribution satisfies both continuity and energy simultaneously and is therefore the unique solution. The Hardy Cross correction is then applied from a deliberately wrong starting guess to demonstrate that the iteration converges on that same answer.

Establish the pipe resistance. Using the standard inversion, $$K = 0.278 \times 110 \times (0.250)^{2.63} = 30.58 \times 0.026096 = 0.79803\ \mathrm{m^3/s}$$ $$R = \frac{500}{0.79803^{1.852}} = \frac{500}{0.65846} = 759.33$$ so $h_f = 759.33\,Q^{1.852}$ applies identically to P1, P2 and P3.
Set up the unknowns. Let $q$ be the flow in P3, taken positive from N1 to N2. Continuity at N1 requires $Q_{P1} = 0.005 + q$, and continuity at N2 requires $Q_{P2} = 0.005 - q$. Total supply from the reservoir is $Q_{P1} + Q_{P2} = 0.010$ m³/s regardless of $q$, as it must be. One equation remains: the energy equation around the closed loop R → N1 → N2 → R, $$h_{f,P1} + h_{f,P3} - h_{f,P2} = 0$$
Exploit the symmetry. Substituting the resistance law, the loop equation reads $$R\left[(0.005+q)^{1.852} + q^{1.852} - (0.005-q)^{1.852}\right] = 0$$ Try $q = 0$. The first and third terms become identical and cancel, and the middle term vanishes, so the equation is satisfied exactly. Because the left-hand side increases monotonically with $q$ over the admissible range $-0.005 < q < 0.005$, that root is the only one: $$\boxed{Q_{P3} = 0, \qquad Q_{P1} = Q_{P2} = 5\ \mathrm{L/s}}$$ Physically, N1 and N2 are hydraulically indistinguishable — same pipe from the same reservoir, same demand — so they sit at the same head, and a pipe with no head difference across it carries no flow. P3 is a redundant link under this loading.
Confirm with one Hardy Cross correction. Suppose a candidate had guessed $q = +2$ L/s. The loop correction supplied on the exam cover page is $$\Delta q = -\frac{\sum_{\mathrm{loop}} k_i Q_i^{1.852}}{1.852\sum_{\mathrm{loop}} \left|k_i Q_i^{1.852}/Q_i\right|} = -\frac{\sum h_{f,i}}{1.852\sum |h_{f,i}/Q_i|}$$ Circulating clockwise round the loop R → N1 → N2 → R, pipes P1 and P3 are traversed with their flow and P2 against it. With $Q_{P1} = 7$ L/s, $Q_{P2} = 3$ L/s and $q = 2$ L/s, the head losses are 0.07755 m, 0.01615 m and 0.00762 m respectively, so the numerator is $0.07755 + 0.00762 - 0.01615 = 0.06902$ m and the denominator is $1.852(11.078 + 5.382 + 3.810) = 37.54$, giving $$\Delta q = -\frac{0.06902}{37.54} = -1.84\ \mathrm{L/s}$$ The correction drives $q$ from 2.00 L/s to 0.16 L/s in a single round, and successive rounds shrink it toward zero — the iteration converges on the symmetric answer already obtained in closed form.
Compute the nodal heads. With 5 L/s in each supply pipe, $$h_f = 759.33 \times (0.005)^{1.852} = 759.33 \times 5.476\times10^{-5} = 0.0416\ \mathrm{m}$$ $$\boxed{H_{N1} = H_{N2} = 90 - 0.042 = 89.96\ \mathrm{m}}$$ The loss is trivially small because 5 L/s in a 250 mm pipe is a mean velocity of only 0.102 m/s — the network is enormously oversized for this loading, which is normal for a distribution grid designed around fire flow rather than average day demand.
Express the result as pressure head. The paper gives no ground elevations for N1 and N2, so the pressure head follows only once a node elevation is adopted: $p/\gamma = 89.96 - z$. Taking the nodes at a datum of 0 m, as the figure implies by showing no elevation offset, the pressure head at each node is 89.96 m. If the nodes were instead at, say, 45 m, the pressure head would be 44.96 m at each. The hydraulically meaningful and elevation-independent statement is that both nodes lie 0.042 m below the reservoir surface, and that they are at identical head.

Check: node elevations not supplied. The question asks for “the pressure heads at the nodes” but the paper gives no ground elevation for N1 or N2, and Figure 3 shows none. The heads are therefore reported as hydraulic grade line values (89.96 m at both nodes), with the conversion $p/\gamma = \mathrm{HGL} - z$ stated explicitly. An examination candidate should state this assumption on the answer paper, as Note 1 of the cover page invites.

Final results — Question 3
Quantity	Result
Pipe resistance $R$ (each pipe)	759.3 (with $h_f = RQ^{1.852}$)
Flow in P1 (reservoir → N1)	5.0 L/s
Flow in P2 (reservoir → N2)	5.0 L/s
Flow in P3 (N1 – N2)	0 L/s
Head loss in P1 and in P2	0.042 m each
Head (HGL) at N1 and N2	89.96 m — pressure head 89.96 m − $z$
Total supply from reservoir	10 L/s

Topic. Looped-network analysis by the Hardy Cross loop-correction method, and the use of symmetry to short-circuit the iteration.

Key relations. Nodal continuity $\sum Q_{\mathrm{in}} = \sum Q_{\mathrm{out}} + q_{\mathrm{demand}}$; loop energy $\sum_{\mathrm{loop}} h_f = 0$ with signs by circulation direction; Hazen-Williams loop correction as printed on the cover page, $\Delta q = -\sum h_f / \left(1.852\sum |h_f/Q|\right)$.

Why this works. A pipe network is determinate when the number of independent continuity and loop-energy equations equals the number of unknown pipe flows. This network has three pipes, two demand nodes and one loop — two continuity equations plus one loop equation, exactly three. Hardy Cross solves that set by guessing a flow distribution that already satisfies continuity, then adding the same correction to every pipe around a loop, a step which preserves continuity by construction while driving the loop head imbalance to zero. When a network possesses a mirror symmetry in geometry, pipe properties and loading, the symmetric distribution satisfies every equation identically, so it is the answer and no iteration is needed. Recognising that saves a great deal of exam time, and it also carries a design message: a tie such as P3 does nothing under balanced demand but becomes essential the moment demand is unbalanced or one supply pipe is taken out of service.

Common pitfall. Assuming a non-zero flow in P3 because “a loop must circulate,” and then reporting whatever the first Hardy Cross round happens to produce without checking convergence. The other frequent error is forgetting that the demand at N1 is drawn at the node, so P1 must carry the demand plus whatever it passes on through P3 — not merely the demand.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 4 (Hardy Cross and linear-theory network analysis); Chin, Water-Resources Engineering, 3rd ed., §2.5 (pipe-network equations, loop and nodal formulations).

← Question 2 Next: Question 4 →

Quantity	Symbol	Value
Pipe diameter (P1, P2, P3)	\(D\)	250 mm = 0.250 m
Pipe length (P1, P2, P3)	\(L\)	500 m
Hazen-Williams coefficient	\(C\)	110
Demand at N1	\(q_1\)	5 L/s = 0.005 m³/s
Demand at N2	\(q_2\)	5 L/s = 0.005 m³/s
Source reservoir water level	\(H_R\)	90 m

Quantity	Result
Pipe resistance \(R\) (each pipe)	759.3 (with \(h_f = RQ^{1.852}\))
Flow in P1 (reservoir → N1)	5.0 L/s
Flow in P2 (reservoir → N2)	5.0 L/s
Flow in P3 (N1 – N2)	0 L/s
Head loss in P1 and in P2	0.042 m each
Head (HGL) at N1 and N2	89.96 m — pressure head 89.96 m − \(z\)
Total supply from reservoir	10 L/s