07-Str-B11 · May 2013

Question 2 of 6: Ten-Pipe Transfer Network Between Two Reservoirs

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

Notes on this paper

National Examinations — May 2013 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and one approved Casio or Sharp calculator are permitted. The paper prints six questions of 20 marks each and instructs the candidate to complete any five; where a question has more than one part, the parts carry equal marks. Candidates are urged to state any interpretive assumptions with their answer. All six questions are worked below, because this set is intended as a study resource.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, pipe networks, pump-system curves and quasi-steady reservoir routing; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning's equation, compound and divided channel sections, and the specific-energy and momentum treatment of channel obstructions; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American design practice for transmission mains, minimum service pressures and pump selection; Henderson, F.M., Open Channel Flow (Macmillan) — surges, hydraulic jumps and the unsteady response of a channel to a sudden blockage.

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. Local losses and velocity head are neglected, as instructed in Note 6, 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 properties are $\rho = 1000$ kg/m³ and $\nu = 1.31\times10^{-6}$ m²/s.

Question 2: Ten-Pipe Transfer Network Between Two Reservoirs (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. Ten identical mains form a branching-and-rejoining network between two fixed-level reservoirs, and the ground contours on the figure fix the elevation of every junction.

Given data — Question 2
Quantity	Symbol	Value
Water elevation, Reservoir A	$z_A$	90 m
Water elevation, Reservoir B	$z_B$	85 m
Available head across the network	$\Delta H$	5.00 m
Diameter, every pipe	$D$	300 mm = 0.300 m
Length, every pipe	$L$	250 m
Hazen-Williams coefficient, every pipe	$C$	140
Ground elevation, junctions 1, 2 and 3 (90 m contour)	$z_{1,2,3}$	90 m
Ground elevation, junctions 4 and 5 (80 m contour)	$z_{4,5}$	80 m
Ground elevation, junction 6 (75 m contour)	$z_6$	75 m

Find. (a) the total discharge delivered from A to B, and (b) the largest and smallest pressure heads occurring anywhere in the network.

[Figure not reproduced: Figure 2-S. Network topology read from the examination figure, with the computed discharge marked on each pipe. Junctions 1, 2 and 3 sit on the 90 m contour, junctions 4 and 5 on the 80 m contour and junction 6 on the 75 m contour. See the official exam paper.]

Approach. Because all ten pipes are identical they share one resistance coefficient, so the network collapses by series and parallel reduction into a single equivalent resistance between A and B; solve that for the total discharge, unwind the reduction to recover every individual pipe flow, then march the HGL from A to B to obtain the pressure head at each junction.

Express head loss as a single power law. With $K = 0.278\,C\,D^{2.63} = 0.278(140)(0.300)^{2.63} = 1.6406\ \text{m}^3\text{/s}$, inverting Hazen-Williams gives each pipe the identical resistance $$h_f = r\,Q^{n}, \qquad r = \frac{L}{K^{\,n}} = \frac{250}{1.6406^{1.852}} = 99.95, \qquad n = \frac{1}{0.54} = 1.852$$ so every pipe in the network obeys $h_f = 99.95\,Q^{1.852}$ with $Q$ in m³/s and $h_f$ in metres.
Read the topology off the figure. Three pipes leave A: P1 to junction 1, P2 to junction 2 and P3 to junction 3. P4 and P5 bring junctions 1 and 2 together at junction 4, while the twin pipes P6 and P7 run in parallel from junction 3 to junction 5. P8 and P9 then bring junctions 4 and 5 together at junction 6, from which the single pipe P10 discharges to B. The network is therefore two independent routes from A to junction 6, in parallel, followed by P10 in series.
Collapse the upper route. Path A–1–4 and path A–2–4 are identical two-pipe series legs in parallel, so each takes half the upper-route discharge $Q_U$; adding P8, which carries all of $Q_U$, $$h_{A\to 6}^{\text{upper}} = 2r\left(\tfrac{Q_U}{2}\right)^{n} + rQ_U^{\,n} = r\,Q_U^{\,n}\left(2^{\,1-n} + 1\right) = 1.5541\,r\,Q_U^{\,n}$$
Collapse the lower route. P3 carries the whole lower-route discharge $Q_L$, the twin pipes P6 and P7 split it in half, and P9 carries it whole again, giving $$h_{A\to 6}^{\text{lower}} = rQ_L^{\,n} + r\left(\tfrac{Q_L}{2}\right)^{n} + rQ_L^{\,n} = r\,Q_L^{\,n}\left(2 + 2^{-n}\right) = 2.2770\,r\,Q_L^{\,n}$$
Split the flow between the two routes. Both routes span the same two nodes, so their head losses are equal: $$1.5541\,Q_U^{\,n} = 2.2770\,Q_L^{\,n} \quad\Rightarrow\quad \frac{Q_U}{Q_L} = \left(\frac{2.2770}{1.5541}\right)^{0.54} = 1.2292$$ so the lower route takes $Q_L = Q_T/2.2292 = 0.4486\,Q_T$ and the upper route the remaining $0.5514\,Q_T$. The upper route carries more because two of its three legs are doubled up, whereas the lower route doubles only one leg.
Impose the total available head. Adding the A-to-6 loss and the P10 loss and setting the sum to the 5.00 m difference in reservoir levels, $$2.2770\,r\,(0.4486\,Q_T)^{n} + r\,Q_T^{\,n} = 5.00 \quad\Rightarrow\quad 99.95\,Q_T^{\,1.852}(1.5161) = 5.00$$ $$\boxed{Q_T = 0.1585\ \text{m}^3\text{/s} = 158.5\ \text{L/s}}$$ which answers part (a).
Unwind the reduction for the individual pipes. The route flows are $Q_U = 87.4$ L/s and $Q_L = 71.1$ L/s, so P1, P2, P4 and P5 each carry 43.7 L/s, P6 and P7 each carry 35.5 L/s, P3 and P9 each carry 71.1 L/s, P8 carries 87.4 L/s and P10 carries the full 158.5 L/s. Applying $h_f = 99.95\,Q^{1.852}$ to each gives the losses in the table below.
March the hydraulic grade line from A to B. Starting at $\mathrm{HGL}_A = 90.00$ m and subtracting each loss in turn, $$\mathrm{HGL}_1 = \mathrm{HGL}_2 = 90.00 - 0.303 = 89.697\ \text{m}, \qquad \mathrm{HGL}_3 = 90.00 - 0.747 = 89.253\ \text{m}$$ $$\mathrm{HGL}_4 = 89.697 - 0.303 = 89.393\ \text{m}, \qquad \mathrm{HGL}_5 = 89.253 - 0.207 = 89.046\ \text{m}$$ $$\mathrm{HGL}_6 = 89.393 - 1.095 = 88.298\ \text{m}$$ The lower route is a useful check: $89.046 - 0.747 = 88.299$ m at junction 6, the same value. Finally $88.298 - 3.298 = 85.00$ m at B, which recovers Reservoir B exactly and confirms the whole solution.
Subtract the ground elevations to obtain pressure heads. Using $p/\gamma = \mathrm{HGL} - z$ with the contour elevations, $$\frac{p}{\gamma}\bigg|_{6} = 88.298 - 75 = \boxed{+13.30\ \text{m}} \qquad \frac{p}{\gamma}\bigg|_{3} = 89.253 - 90 = \boxed{-0.75\ \text{m}}$$ These are the maximum and the minimum respectively, which answers part (b). Junctions 1 and 2 are also slightly sub-atmospheric at $-0.30$ m, while junctions 4 and 5 sit at $+9.39$ m and $+9.05$ m.

Final results — Question 2
Pipe or junction	Discharge (L/s)	Head loss (m)	HGL (m)	Ground elevation (m)	Pressure head (m)
Reservoir A	158.5 out	—	90.000	—	—
P1 / P2 (A to 1, A to 2)	43.7 each	0.303	—	—	—
P3 (A to 3)	71.1	0.747	—	—	—
P4 / P5 (1 to 4, 2 to 4)	43.7 each	0.303	—	—	—
P6 / P7 (3 to 5, twin)	35.5 each	0.207	—	—	—
P8 (4 to 6)	87.4	1.095	—	—	—
P9 (5 to 6)	71.1	0.747	—	—	—
P10 (6 to B)	158.5	3.298	—	—	—
Junction 1	—	—	89.697	90	−0.30
Junction 2	—	—	89.697	90	−0.30
Junction 3	—	—	89.253	90	−0.75 (minimum)
Junction 4	—	—	89.393	80	+9.39
Junction 5	—	—	89.046	80	+9.05
Junction 6	—	—	88.298	75	+13.30 (maximum)
Total flow A to B	158.5 L/s = 0.1585 m³/s

Check — the three junctions on the 90 m contour run below atmospheric pressure. Reservoir A is itself at 90 m, so any friction at all puts junctions 1, 2 and 3 beneath the hydraulic grade line datum they start from. The computed heads of $-0.30$ m and $-0.75$ m are only a fraction of a metre of vacuum and are far from the roughly $-8$ m at which cold water would flash, so the line will not cavitate. It would nevertheless draw air through any leaking joint or air valve, and no service connection could be made at those junctions. In practice a designer would lower the alignment, raise Reservoir A, or accept that these three points are transmission junctions and not supply points.

Topic. Series and parallel reduction of a pipe network under the Hazen-Williams power law, and the recovery of nodal pressure heads from a hydraulic grade line.

Key relations. $h_f = rQ^{n}$ with $r = L/K^{n}$, $K = 0.278\,C\,D^{2.63}$ and $n = 1/0.54 = 1.852$; series pipes add resistances at equal $Q$; $m$ identical pipes in parallel give $h_f = r(Q/m)^{n} = m^{-n}rQ^{n}$; two routes spanning the same node pair share head loss so $Q_1/Q_2 = (R_2/R_1)^{1/n}$; and $p/\gamma = \mathrm{HGL} - z$.

Why this works. Head loss under Hazen-Williams is a pure power of discharge with the same exponent in every pipe, so the algebra of series and parallel combination closes on itself exactly as it does for linear resistors, only with the exponent 1.852 in place of unity. That is what allows the ten-pipe network to be reduced by hand without any Hardy Cross iteration: no loop here carries an unknown circulating flow, because the network is a series and parallel tree rather than a genuine looped grid. Once the total discharge is known the reduction unwinds mechanically, and the closure of the HGL march back onto Reservoir B at exactly 85.00 m is an independent proof that both the topology and the arithmetic are right.

Common pitfall. Treating the twin pipes P6 and P7, or the two-pipe legs through junctions 1 and 2, as if halving the flow halved the head loss; the correct factor is $2^{-1.852} = 0.277$, not 0.5. The other frequent error is reporting HGL elevations as pressure heads and so missing the sub-atmospheric result entirely, which is the whole point of part (b) and of the contour information on the figure.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 5 (equivalent pipes, series and parallel systems, network analysis); Chin, Water-Resources Engineering, 3rd ed., Section 2.5.

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Quantity	Symbol	Value
Water elevation, Reservoir A	\(z_A\)	90 m
Water elevation, Reservoir B	\(z_B\)	85 m
Available head across the network	\(\Delta H\)	5.00 m
Diameter, every pipe	\(D\)	300 mm = 0.300 m
Length, every pipe	\(L\)	250 m
Hazen-Williams coefficient, every pipe	\(C\)	140
Ground elevation, junctions 1, 2 and 3 (90 m contour)	\(z_{1,2,3}\)	90 m
Ground elevation, junctions 4 and 5 (80 m contour)	\(z_{4,5}\)	80 m
Ground elevation, junction 6 (75 m contour)	\(z_6\)	75 m