07-Str-B11 · December 2015

Question 1 of 6: Ten-Pipe Reservoir-to-Reservoir Network — Total Flow and Nodal Pressures

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 1: Ten-Pipe Reservoir-to-Reservoir Network — Total Flow and Nodal Pressures (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 pure conveyance network — no demands are drawn off at any junction, so every drop of water that leaves reservoir A arrives at reservoir B. All ten pipes are identical, which is the feature that makes the whole system collapse onto a single equivalent pipe.

Given data — Question 1
Quantity	Symbol	Value
Reservoir A water elevation	$H_A$	95 m
Reservoir B water elevation	$H_B$	70 m
Pipe diameter (all ten)	$D$	250 mm = 0.250 m
Pipe length (all ten)	$L$	200 m
Hazen-Williams coefficient	$C$	130
Ground elevation, junctions of P1/P2/P3 with P4/P5/P6-P7	$z$	90 m (contour)
Ground elevation, junctions feeding P8 and P9	$z$	80 m (contour)
Ground elevation, junction upstream of P10	$z$	75 m (contour)

Find. (a) the total discharge A → B; (b) the largest and smallest pressure head anywhere in the system, with the node at which each occurs; (c) which of the two named branches carries more flow, and the hydraulic reason.

[Figure not reproduced: Figure 1.1 — The network redrawn from the exam figure. Reservoir A feeds three pipes to junctions J1, J2 and J3 on the 90 m contour; P4 and P5 rejoin at J4 and the parallel pair P6/P7 rejoins at J5, both on the 80 m contour; P8 and P9 meet at J6 on the 75 m contour, and P10 alone delivers the . See the official exam paper.]

Approach. Because every pipe is identical, replace each series run and each parallel pair by an equivalent resistance in the form $h_f = R_{\mathrm{eq}}Q^{1.852}$, collapse the network down to one pipe between the two reservoirs, recover the total flow from the 25 m of available head, and then walk the head back out through the network to obtain the HGL and hence the pressure head at every junction.

Invert the printed Hazen-Williams equation into a resistance law. With $S = h_f/L$, the form given in Note 5 rearranges to $$h_f = L\left(\frac{Q}{K}\right)^{1.852} = R\,Q^{1.852}, \qquad K = 0.278\,C\,D^{2.63}, \qquad R = \frac{L}{K^{1.852}}$$ where $K$ is the pipe conveyance — numerically the discharge the pipe would carry at unit hydraulic gradient. For the common pipe data, $$K = 0.278 \times 130 \times (0.250)^{2.63} = 36.14 \times 0.026096 = 0.9431\ \mathrm{m^3/s}$$ $$R = \frac{200}{0.9431^{1.852}} = \frac{200}{0.89722} = 222.91\ \mathrm{m\,s^{1.852}/m^{5.556}}$$ Every one of the ten pipes has this same $R$, so the only thing that distinguishes them is the flow each happens to carry.
Collapse the upper branch. P1 and P4 form a two-pipe series arm from A to J4, as do P2 and P5; series resistances simply add, so each arm has $2R$. The two arms are identical and in parallel between the same pair of points, so each takes half the branch flow: $$h = 2R\left(\frac{Q}{2}\right)^{1.852} \;\Longrightarrow\; R_{A\text{-}J4} = \frac{2R}{2^{1.852}} = \frac{2R}{3.6115} = 0.5538\,R$$ Adding P8 in series to reach J6 gives the whole upper branch $$R_{\mathrm{upper}} = 0.5538\,R + R = 1.5540\,R$$
Collapse the lower branch. P6 and P7 are two identical pipes in parallel between J3 and J5, so by the same argument $R_{6\|7} = R/2^{1.852} = 0.2770\,R$. This sits in series between P3 and P9: $$R_{\mathrm{lower}} = R + 0.2770\,R + R = 2.2770\,R$$ The lower branch is the stiffer of the two, and it is worth noticing why: it has only one parallel element where the upper branch effectively has two.
Combine the two branches and add the delivery main. The branches share the same head loss between A and J6, so $R_{\mathrm{upper}}Q_u^{1.852} = R_{\mathrm{lower}}Q_\ell^{1.852}$, which fixes the flow split $$\frac{Q_u}{Q_\ell} = \left(\frac{R_{\mathrm{lower}}}{R_{\mathrm{upper}}}\right)^{1/1.852} = \left(\frac{2.2770}{1.5540}\right)^{0.53996} = 1.2291$$ Substituting $Q_u + Q_\ell = Q$ into either branch gives the parallel equivalent, and P10 then adds in series: $$R_{A\text{-}J6} = \frac{R_{\mathrm{lower}}}{(1+1.2291)^{1.852}} = \frac{2.2770\,R}{4.4139} = 0.5159\,R, \qquad R_{\mathrm{tot}} = 0.5159\,R + R = 1.5160\,R$$ The ten-pipe network therefore behaves exactly like a single 200 m pipe of resistance $1.516 \times 222.91 = 337.9$.
Solve for the total flow. Two reservoirs with fixed surfaces impose the entire elevation difference on the system, so $h_f = 95 - 70 = 25$ m and $$Q = \left(\frac{h_f}{R_{\mathrm{tot}}}\right)^{1/1.852} = \left(\frac{25}{337.9}\right)^{0.53996}$$ $$\boxed{Q_{\mathrm{total}} = 0.245\ \mathrm{m^3/s} = 245\ \mathrm{L/s}}$$ This is the answer to part (a). The mean velocity in P10 is $V = Q/A = 0.245/0.04909 = 4.99$ m/s, which is high for a distribution main but perfectly admissible for a gravity transfer between reservoirs.
Distribute the flow through the network. Using the split ratio from Step 4, $Q_\ell = Q/(1+1.2291) = 0.1100$ m³/s and $Q_u = 0.1352$ m³/s. Within the upper branch the two identical arms halve $Q_u$, and within the lower branch the parallel pair halves $Q_\ell$:
Pipe flows
Pipe Flow (L/s) $h_f$ (m)

P1, P2, P4, P5 67.6 each 1.517 each

P3, P9 110.0 each 3.737 each

P6, P7 55.0 each 1.035 each

P8 135.2 5.475

P10 245.1 16.491

As a check on the whole arrangement, the two independent routes to J6 must lose the same head: $1.517 + 1.517 + 5.475 = 8.509$ m along P1-P4-P8, and $3.737 + 1.035 + 3.737 = 8.509$ m along P3-P6-P9. They agree, and adding the 16.491 m lost in P10 recovers the full 25 m.

Pipe flows
Pipe	Flow (L/s)	\(h_f\) (m)
P1, P2, P4, P5	67.6 each	1.517 each
P3, P9	110.0 each	3.737 each
P6, P7	55.0 each	1.035 each
P8	135.2	5.475
P10	245.1	16.491

Track the HGL and convert to pressure head. Starting at the reservoir surface and subtracting each pipe loss in turn gives the HGL at every junction; the pressure head is then $p/\gamma = \mathrm{HGL} - z$, with $z$ read from the contour on which each junction sits.

Nodal hydraulic grade line and pressure head
Junction	Ground elevation $z$ (m)	HGL (m)	Pressure head (m)
J1 (P1-P4)	90	93.483	3.48
J2 (P2-P5)	90	93.483	3.48
J3 (P3-P6-P7)	90	91.263	1.26
J4 (P4-P5-P8)	80	91.966	11.97
J5 (P6-P7-P9)	80	90.228	10.23
J6 (P8-P9-P10)	75	86.491	11.49

The extremes follow directly: $$\boxed{p_{\max}/\gamma = 11.97\ \mathrm{m\ at\ J4}, \qquad p_{\min}/\gamma = 1.26\ \mathrm{m\ at\ J3}}$$ which answers part (b). Both extremes are governed as much by topography as by hydraulics: J4 has a modest HGL but sits 10 m lower than the first tier of junctions, while J3 combines the largest first-pipe loss in the system with the highest ground elevation.

Answer part (c) — which branch carries more. From Step 6, branch P1-P2-P4-P5-P8 conveys 135.2 L/s and branch P3-P6-P7-P9 conveys 110.0 L/s, so $$\boxed{\text{the upper branch P1-P2-P4-P5-P8 carries the higher flow, }135\ \mathrm{L/s\ against\ }110\ \mathrm{L/s}}$$ The reason is structural rather than numerical. Both branches contain five pipe-lengths of material, but they are arranged differently. In the upper branch, four of the five pipes are arranged as two parallel two-pipe paths, leaving only P8 to carry the branch flow alone; in the lower branch, only two of the five pipes (P6 and P7) are in parallel, and both P3 and P9 must pass the entire branch flow by themselves. Since head loss grows as $Q^{1.852}$, splitting a flow between two identical pipes cuts the loss by a factor $2^{1.852} = 3.61$ rather than merely by two, so parallelism is disproportionately valuable. The result is $R_{\mathrm{upper}} = 1.554R$ against $R_{\mathrm{lower}} = 2.277R$ — the upper branch is 32 % less resistant, and with the same head driving both, it takes the larger share.

Final results — Question 1
Quantity	Result
(a) Total flow A → B	0.245 m³/s (245 L/s)
(b) Maximum pressure head	11.97 m at J4 (junction of P4, P5 and P8, $z$ = 80 m)
(b) Minimum pressure head	1.26 m at J3 (junction of P3, P6 and P7, $z$ = 90 m)
(c) Higher-flow branch	P1-P2-P4-P5-P8, at 135 L/s versus 110 L/s
(c) Reason	Lower equivalent resistance (1.554$R$ vs 2.277$R$) — four of its five pipes are paired in parallel, and the $Q^{1.852}$ law rewards splitting flow far more than proportionally

Topic. Series/parallel equivalent-pipe reduction of a looped conveyance network under the Hazen-Williams law.

Key relations. Hazen-Williams (SI, as printed): $Q = 0.278\,C\,D^{2.63}S^{0.54}$, inverted to $h_f = R\,Q^{n}$ with $R = L/K^{n}$, $K = 0.278\,C\,D^{2.63}$, $n = 1.852$. Series: $R_{\mathrm{eq}} = \sum R_i$. Parallel of $m$ identical pipes: $R_{\mathrm{eq}} = R/m^{n}$. Two dissimilar parallel branches: $Q_1/Q_2 = (R_2/R_1)^{1/n}$. Pressure head: $p/\gamma = \mathrm{HGL} - z$.

Why this works. A network with no intermediate demands is a pure resistance circuit: continuity forces the same flow through every element of a series run, and the common end-points of a parallel group force the same head loss across each member. Those two constraints are exactly the ones that let electrical resistances be combined, and the only complication here is that the head-loss law is non-linear, so “adding conductances” becomes the $m^{n}$ factor for identical pipes and the $(R_2/R_1)^{1/n}$ split ratio for unequal ones. Reducing the network to a single resistance turns a six-unknown nodal problem into one line of arithmetic, and because the reduction is exact, expanding back outwards recovers every individual pipe flow without iteration. Hardy Cross would give the same answer, but it would take several rounds to do it.

Common pitfall. Treating the two named branches as though five pipes in one equals five pipes in the other, and concluding they carry equal flow — the arrangement, not the pipe count, sets the resistance. A second frequent error is reporting HGL values as pressure heads: at J4 the HGL is 91.97 m but the pressure head is 11.97 m, and a candidate who forgets the 80 m contour reports a figure eight times too large.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 4 (pipe flow, equivalent pipes, series and parallel systems); Chin, Water-Resources Engineering, 3rd ed., §2.3 and §7.2 (Hazen-Williams coefficients, network analysis and nodal pressures).

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Quantity	Symbol	Value
Reservoir A water elevation	\(H_A\)	95 m
Reservoir B water elevation	\(H_B\)	70 m
Pipe diameter (all ten)	\(D\)	250 mm = 0.250 m
Pipe length (all ten)	\(L\)	200 m
Hazen-Williams coefficient	\(C\)	130
Ground elevation, junctions of P1/P2/P3 with P4/P5/P6-P7	\(z\)	90 m (contour)
Ground elevation, junctions feeding P8 and P9	\(z\)	80 m (contour)
Ground elevation, junction upstream of P10	\(z\)	75 m (contour)