07-Str-B11 · May 2014

Question 2 of 6: Nine Identical Pipes Linking Two Reservoirs

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

Notes on this paper

National Examinations — May 2014 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and any non-communicating 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 submit a clear statement of any interpretive assumptions with the answer paper. 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, branched and looped network analysis, transmission mains and valve characteristics; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, fire-flow criteria, minimum service pressures and gutter hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and the momentum treatment of the hydraulic jump; Henderson, F.M., Open Channel Flow (Macmillan) — specific force, conjugate depths and jump classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the control-volume force balance that links wall shear stress to the Darcy friction factor.

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. 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: Nine Identical Pipes Linking 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. Nine identical pipes arranged as shown in Figure 2 of the paper, delivering water from Res. A to Res. B through five intermediate junctions, with no demand drawn off at any junction.

Given data — Question 2
Quantity	Symbol	Value
Upstream reservoir water level	$H_A$	120 m
Downstream reservoir water level	$H_B$	70 m
Elevation of all pipework and junctions	$z$	17 m
Diameter of every pipe	$D$	400 mm = 0.400 m
Length of every pipe	$L$	350 m
Hazen-Williams coefficient	$C$	145
Number of pipes (counted off Figure 2)	—	9

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

[Figure not reproduced: Figure 2.1 — Pipe layout traced from Figure 2 of the examination paper: Res. A feeds J1, J2, J3 and J5 directly; J1 and J2 discharge directly to Res. B; J3 and J5 both feed J4, which discharges to Res. B through a single pipe. Red labels are the hydraulic grade line elevations computed in the . See the official exam paper.]

Approach. Because every pipe is identical, each obeys $h_f = r\,Q^{1.852}$ with the same resistance $r$; the network then collapses by inspection into three parallel branches between the reservoirs, two of which are simple two-pipe series paths and the third of which is a small parallel pair in series with a common outlet pipe.

Establish the resistance of one pipe. Using the same inversion of the printed Hazen-Williams equation as in Question 1, $$K = 0.278 \times 145 \times (0.400)^{2.63} = 40.31 \times 0.089837 = 3.6210\ \mathrm{m^3/s}$$ $$h_f = r\,Q^{1.852}, \qquad r = \frac{L}{K^{1.852}} = \frac{350}{3.6210^{1.852}} = \frac{350}{10.838} = 32.29\ \mathrm{s^{1.852}/m^{4.556}}$$ Every one of the nine pipes carries this same $r$, so the network can be handled entirely in terms of $r$ and the branch flows.
Read the topology off Figure 2 and identify the branches. Tracing the drawing, Res. A is connected to four junctions (J1, J2, J3, J5); J1 and J2 each run straight on to Res. B; J3 and J5 each run to J4; and a single pipe carries the combined flow from J4 to Res. B. That is nine pipes and three independent routes between the reservoirs:
- Branch I: A → J1 → B, two pipes in series;
- Branch II: A → J2 → B, two pipes in series;
- Branch III: (A → J3 → J4 in parallel with A → J5 → J4) in series with J4 → B.
All three branches span the same two reservoirs, so all three lose the same total head.
Fix the driving head and solve the two simple branches. Since local losses and velocity head are neglected, the whole reservoir difference is spent on pipe friction: $$\Delta H = H_A - H_B = 120 - 70 = 50\ \mathrm{m}$$ For a branch of two identical pipes in series carrying $Q_{\mathrm{I}}$, the two losses simply add: $$2\,r\,Q_{\mathrm{I}}^{1.852} = 50 \quad\Rightarrow\quad Q_{\mathrm{I}} = \left(\frac{50}{2 \times 32.29}\right)^{1/1.852} = (0.7743)^{0.540} = 0.8709\ \mathrm{m^3/s}$$ Branch II is geometrically identical, so $Q_{\mathrm{II}} = Q_{\mathrm{I}} = 0.8709$ m³/s.
Collapse the composite branch. The two legs A–J3–J4 and A–J5–J4 are identical two-pipe paths in parallel, so they must share the branch flow $Q_{\mathrm{III}}$ equally, each carrying $Q_{\mathrm{III}}/2$. Adding the loss through the shared outlet pipe J4–B, $$\Delta H = 2\,r\left(\frac{Q_{\mathrm{III}}}{2}\right)^{1.852} + r\,Q_{\mathrm{III}}^{1.852} = r\,Q_{\mathrm{III}}^{1.852}\left(2^{\,1-1.852} + 1\right)$$ $$2^{-0.852} = 0.5540 \quad\Rightarrow\quad \text{equivalent resistance} = 32.29 \times 1.5540 = 50.18$$ $$Q_{\mathrm{III}} = \left(\frac{50}{50.18}\right)^{0.540} = 0.9980\ \mathrm{m^3/s}$$ The composite branch out-performs the simple two-pipe branches because doubling up its upstream half roughly halves the loss there.
Add the branch flows to obtain the system discharge. The three branches are hydraulically in parallel between fixed reservoir levels, so their flows simply sum: $$Q_{\mathrm{total}} = Q_{\mathrm{I}} + Q_{\mathrm{II}} + Q_{\mathrm{III}} = 0.8709 + 0.8709 + 0.9980 = \boxed{2.74\ \mathrm{m^3/s}}$$ As a check on the arithmetic, the mean velocity in a two-pipe branch is $0.8709/(\pi \times 0.4^2/4) = 6.93$ m/s — high, but consistent with a 50 m head being dissipated over only 700 m of 400 mm pipe.
Compute the junction heads so the pressure extremes can be located. On Branch I the loss is shared equally between two identical pipes carrying the same flow, so J1 (and J2) sit exactly half way down: $$H_{J1} = H_{J2} = 120 - \tfrac{1}{2}(50) = 95.0\ \mathrm{m}$$ On the composite branch each upstream leg carries $Q_{\mathrm{III}}/2 = 0.4990$ m³/s, so $$h_{A \to J3} = 32.29 \times (0.4990)^{1.852} = 8.91\ \mathrm{m} \quad\Rightarrow\quad H_{J3} = H_{J5} = 120 - 8.91 = 111.1\ \mathrm{m}$$ $$H_{J4} = 120 - 2(8.91) = 102.2\ \mathrm{m}, \qquad H_{B} = 102.2 - 32.29(0.9980)^{1.852} = 102.2 - 32.17 = 70.0\ \mathrm{m}\ \checkmark$$ The closure on the downstream reservoir level confirms the branch flows.
Convert every head to a pressure head and pick out the extremes. All pipework lies at $z = 17$ m, so $p/\gamma = H - 17$ everywhere: $$\frac{p}{\gamma}\Big|_{J1,J2} = 95.0 - 17 = 78.0\ \mathrm{m}, \quad \frac{p}{\gamma}\Big|_{J3,J5} = 111.1 - 17 = 94.1\ \mathrm{m}, \quad \frac{p}{\gamma}\Big|_{J4} = 102.2 - 17 = 85.2\ \mathrm{m}$$ The pipework also touches the two reservoirs, and those connections bound the whole system: $$\frac{p}{\gamma}\Big|_{\max} = H_A - z = 120 - 17 = \boxed{103.0\ \mathrm{m}}, \qquad \frac{p}{\gamma}\Big|_{\min} = H_B - z = 70 - 17 = \boxed{53.0\ \mathrm{m}}$$ Restricting attention to the junctions themselves, the highest pressure head is $94.1$ m at J3 and J5 and the lowest is $78.0$ m at J1 and J2. Both readings are reported below because the wording “in the system” admits either.

Check: which extremes the marker wants. “Maximum and minimum pressure head in the system” is read here two ways, and both are reported. Taken over the entire pipework including the reservoir connections, the extremes are simply the two reservoir levels referred to the 17 m pipe elevation, 103.0 m and 53.0 m. Taken over the five junctions — the reading that actually requires the network solution — they are 94.1 m at J3/J5 and 78.0 m at J1/J2. State whichever interpretation is used before quoting a number.

Question 2 — results
Part	Quantity	Result
(a)	Resistance of one pipe, $r$	32.29
(a)	Flow, branch A–J1–B and A–J2–B	0.871 m³/s each
(a)	Flow, composite branch through J4	0.998 m³/s
(a)	Total system flow	2.74 m³/s
(b)	HGL at J1, J2 / J3, J5 / J4	95.0 / 111.1 / 102.2 m
(b)	Pressure head at J1, J2 / J3, J5 / J4	78.0 / 94.1 / 85.2 m
(b)	Maximum pressure head in the system	103.0 m at the Res. A connection (94.1 m if junctions only)
(b)	Minimum pressure head in the system	53.0 m at the Res. B connection (78.0 m if junctions only)

Topic: Equivalent-pipe reduction of a looped network of identical pipes between two fixed-head reservoirs.

Key relations: series pipes, $h_f = \sum r_i Q^{1.852}$; $m$ identical pipes in parallel, $h_f = r\,(Q/m)^{1.852}$; branches spanning the same two nodes share the same head loss and their flows add.

Why this works: Fixed reservoir levels impose the same 50 m head difference on every route, so no iteration is required — each branch is solved independently for the flow that dissipates exactly 50 m, and continuity at the reservoirs sums them. The symmetry of the two identical legs feeding J4 is what allows the composite branch to be reduced in closed form instead of by Hardy Cross.

Common pitfall: Treating parallel pipes as though the head losses added, or the flows were equal to the total. With $n = 1.852$, $m$ identical parallel pipes carry $m$ times the flow of one for the same head loss, so the equivalent resistance is $r\,m^{-1.852}$, not $r/m$. The second trap is forgetting that pressure head is measured from the pipe elevation $z = 17$ m and not from the reservoir datum.

Source: Mays, Water Resources Engineering, 3rd ed., Ch. 4 (series, parallel and equivalent pipes) and Ch. 5 (network analysis); Chin, Water-Resources Engineering, 3rd ed., §2.4 (pipe networks and equivalent-pipe reduction).

← Question 1 Next: Question 3 →

Quantity	Symbol	Value
Upstream reservoir water level	\(H_A\)	120 m
Downstream reservoir water level	\(H_B\)	70 m
Elevation of all pipework and junctions	\(z\)	17 m
Diameter of every pipe	\(D\)	400 mm = 0.400 m
Length of every pipe	\(L\)	350 m
Hazen-Williams coefficient	\(C\)	145
Number of pipes (counted off Figure 2)	—	9

Part	Quantity	Result
(a)	Resistance of one pipe, \(r\)	32.29
(a)	Flow, branch A–J1–B and A–J2–B	0.871 m³/s each
(a)	Flow, composite branch through J4	0.998 m³/s
(a)	Total system flow	2.74 m³/s
(b)	HGL at J1, J2 / J3, J5 / J4	95.0 / 111.1 / 102.2 m
(b)	Pressure head at J1, J2 / J3, J5 / J4	78.0 / 94.1 / 85.2 m
(b)	Maximum pressure head in the system	103.0 m at the Res. A connection (94.1 m if junctions only)
(b)	Minimum pressure head in the system	53.0 m at the Res. B connection (78.0 m if junctions only)