07-Str-B11 · May 2016

Question 4 of 6: Three-Pipe Looped Network — Pipe Flows and Nodal Heads

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

Notes on this paper

Paper format. National Exams, May 2016 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides may be used) and any non-communicating calculator are permitted. The paper prints six questions of equal value, and a candidate completes any five of them; where a question has parts, the parts also carry equal value. All six are worked below, so the set is useful whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, branched and looped network analysis, control valves and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, maximum-day-plus-fire-flow loading and service-pressure criteria; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation, specific energy and channel transitions; Henderson, F.M., Open Channel Flow (Macmillan) — critical flow, controls and surface-profile classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the pipe force balance, the friction-factor definition and laminar film flow.

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 the cover-page 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 (the cover page prints the exponent without its minus sign, which is an obvious typesetting slip — $1.31\times10^{6}$ m²/s is not a physical viscosity).

Source note — the Question 4 schematic is not reproduced in the examination paper. Question 4 refers to “the figure below”, but no figure is printed between Question 4 and Question 5 on page 3 of the paper, and page 4 opens with the continuation of Question 5. The network is therefore assumed to be a single loop: a source reservoir feeding node N1 through pipe P1 and node N2 through pipe P2, with pipe P3 closing the loop between N1 and N2. That topology is adopted here and is stated explicitly in the Given.

Question 4: Three-Pipe Looped Network — Pipe Flows and Nodal Heads (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 pipe P1 and node N2 through pipe P2, and pipe P3 closes the loop between N1 and N2. All three pipes are identical and both nodal demands are equal. (The schematic is not printed in the May 2016 paper; this topology is assumed, as noted at the head of this document.)

Given data — Question 4
Quantity	Symbol	Value
Diameter, P1, P2 and P3	$D$	200 mm = 0.200 m
Length, P1, P2 and P3	$L$	500 m
Hazen-Williams coefficient	$C$	100
Demand at N1	$q_1$	1 L/s = 0.001 m³/s
Demand at N2	$q_2$	1 L/s = 0.001 m³/s
Source reservoir water level	$H_R$	110 m

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

Figure 4.1 — The single-loop network with the solved flows. Each supply pipe carries exactly its own node’s demand and the closing pipe P3 carries nothing, because the loop is geometrically and hydraulically symmetric about the reservoir.

Approach. With three pipes and only two nodal continuity equations, one further equation is needed and it comes from requiring zero net head loss around the closed loop. Here the symmetry of the network supplies the answer in closed form; a Hardy Cross correction is then run from a deliberately wrong starting guess to show that the iteration converges on the same result.

Establish the resistance shared by all three pipes. Using the paper’s Hazen-Williams form, $$K = 0.278 \times 100 \times (0.200)^{2.63} = 27.8 \times 0.0145114 = 0.40342\ \text{m}^3\text{/s}$$ $$R = \frac{L}{K^{1.852}} = \frac{500}{0.40342^{1.852}} = \frac{500}{0.186158} = 2686.1$$ so $h_f = 2686.1\,Q^{1.852}$ applies identically to P1, P2 and P3.
Reduce the network to one unknown. Let $q$ be the discharge in P3, taken positive from N1 towards N2. Continuity at N1 requires the inflow from P1 to cover both the local demand and whatever leaves along P3, and continuity at N2 requires the two inflows to cover its demand: $$Q_{P1} = 0.001 + q$$ $$Q_{P2} = 0.001 - q$$ The reservoir therefore delivers $Q_{P1} + Q_{P2} = 0.002$ m³/s whatever $q$ turns out to be, as it must, since the total supply is fixed by the total demand. One equation remains, and it is the energy condition around the loop R to N1 to N2 to R, $$h_{f,P1} + h_{f,P3} - h_{f,P2} = 0$$ with P1 and P3 traversed in the direction of their assumed flow and P2 against it.
Exploit the symmetry to solve in closed form. Substituting the common resistance, the loop equation becomes $$R\left[(0.001 + q)^{1.852} + q^{1.852} - (0.001 - q)^{1.852}\right] = 0$$ Setting $q = 0$ makes the first and third terms identical so that they cancel, and kills the middle term outright, so the equation is satisfied exactly. The left-hand side increases monotonically with $q$ over the admissible range, because raising $q$ raises both the P1 and the P3 term while lowering the P2 term, so this root is the only one. Hence $$\boxed{Q_{P1} = Q_{P2} = 1.0\ \text{L/s},\ \ Q_{P3} = 0}$$ Physically, two identical pipes lead from one source to two identical demands, so there is no head difference between N1 and N2 to drive any flow through the closing pipe.
Confirm with one Hardy Cross correction. Suppose a candidate had guessed $q = +0.5$ L/s, giving $Q_{P1} = 1.5$ and $Q_{P2} = 0.5$ L/s. The loop correction printed on the exam cover page is $$\Delta q = -\frac{\sum_{loop} k_i Q_i |Q_i|^{n-1}}{n \sum_{loop} k_i |Q_i|^{n-1}} = -\frac{\sum h_{f,i}}{n\sum |h_{f,i}/Q_i|}$$ with $n = 1.852$. The three head losses are 0.015821 m, 0.002068 m and 0.002068 m for P1, P2 and P3 respectively, so the numerator is $0.015821 + 0.002068 - 0.002068 = 0.015821$ m and the denominator is $1.852 \times 18.82 = 34.855$. Hence $$\Delta q = -\frac{0.015821}{34.855} = -0.454\ \text{L/s}$$ which drives $q$ from 0.500 L/s to 0.046 L/s in a single round; successive rounds shrink it towards zero, so the iteration reproduces the symmetric answer already obtained exactly.
Compute the nodal heads. With 1 L/s in each supply pipe, $$h_f = 2686.1 \times (0.001)^{1.852} = 2686.1 \times 2.7808\times10^{-6} = 0.00747\ \text{m}$$ $$\boxed{H_{N1} = H_{N2} = 110 - 0.0075 = 109.99\ \text{m}}$$ The loss is trivial because 1 L/s in a 200 mm pipe is a mean velocity of only 0.032 m/s. As in Question 1, the grid is sized by fire flow and minimum-diameter rules, not by these demands.
State the result as a pressure head. The paper gives no ground elevations for N1 and N2, so the pressure head follows only once a datum is adopted: $p/\gamma = 109.99 - z$. Taking the nodes at a datum of 0 m, as the schematic implies by showing no elevation offset, the pressure head at each node is 109.99 m. Had the nodes been at, say, 55 m, it would be 54.99 m at each. The elevation-independent statement, and the one that actually answers the hydraulics, is that both nodes sit 0.0075 m below the reservoir surface and are at identical head, so the closing pipe carries no flow.

Check: node elevations are not supplied by the question, so the numerical pressure head above is quoted on the assumption that N1 and N2 lie at the 0 m datum. The pipe flows, the head loss and the equality of the two nodal heads are independent of that assumption.

Final results — Question 4
Quantity	Result
Common pipe resistance	$h_f = 2686.1\,Q^{1.852}$ (SI)
Flow in P1 (reservoir to N1)	1.0 L/s
Flow in P2 (reservoir to N2)	1.0 L/s
Flow in P3 (N1 to N2)	0 L/s
Head loss in each supply pipe	0.0075 m
Head at N1 and at N2	109.99 m
Pressure head at N1 and N2 (nodes on the 0 m datum)	109.99 m

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

Key relations. Nodal continuity at every junction; the loop energy condition $\sum_{loop} h_f = 0$ with signs taken by circulation direction; and the Hardy Cross correction $\Delta q = -\sum h_f \,/\, n\sum|h_f/Q|$ with $n = 1.852$ for Hazen-Williams (it is 2 for Darcy-Weisbach).

Why this works. A network with $P$ pipes and $J$ junctions has $P - J$ independent loops, and continuity supplies only $J$ equations, so exactly $P - J$ loop equations are needed to close the system. Here $P = 3$ and $J = 2$, so one loop equation completes it. Hardy Cross works by noticing that any correction added to every pipe in a loop preserves continuity automatically, so the iteration can hunt for the loop imbalance without ever breaking the mass balance; the denominator is the derivative of the loop head loss with respect to that correction, which makes the scheme a Newton step and gives it quadratic convergence near the root. In this particular network the symmetry means the answer can be written down: identical resistances from one source to identical demands force identical heads at the two nodes, and a pipe joining two points at equal head carries nothing.

Common pitfall. The sign convention defeats most attempts: the numerator must carry the sign of each pipe’s flow relative to the chosen circulation direction, while the denominator uses absolute values only. A second trap is dividing by zero when a pipe is guessed to carry no flow, which is why the demonstration above starts from a non-zero guess. The third is treating the answer $Q_{P3} = 0$ as evidence that the pipe is useless: it carries nothing under this balanced loading but is the whole point of the loop under any unbalanced one, and it is what keeps both nodes in supply if P1 or P2 is taken out of service.

Source. Mays, Water Resources Engineering 3e, Ch. 9 (network analysis, Hardy Cross and linear theory); Chin, Water-Resources Engineering 3e, §2.5 and §7.3.

← Question 3 Next: Question 5 →

Quantity	Symbol	Value
Diameter, P1, P2 and P3	\(D\)	200 mm = 0.200 m
Length, P1, P2 and P3	\(L\)	500 m
Hazen-Williams coefficient	\(C\)	100
Demand at N1	\(q_1\)	1 L/s = 0.001 m³/s
Demand at N2	\(q_2\)	1 L/s = 0.001 m³/s
Source reservoir water level	\(H_R\)	110 m

Quantity	Result
Common pipe resistance	\(h_f = 2686.1\,Q^{1.852}\) (SI)
Flow in P1 (reservoir to N1)	1.0 L/s
Flow in P2 (reservoir to N2)	1.0 L/s
Flow in P3 (N1 to N2)	0 L/s
Head loss in each supply pipe	0.0075 m
Head at N1 and at N2	109.99 m
Pressure head at N1 and N2 (nodes on the 0 m datum)	109.99 m