07-Str-B11 · May 2014

Question 1 of 6: Branched Supply Network — Fire Flow and Maximum-Day Pressures

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 1: Branched Supply Network — Fire Flow and Maximum-Day 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 single-source branched (tree) network fed by reservoir R1, with every pipe identical and every node at the same ground elevation.

Given data — Question 1
Quantity	Symbol	Value
Reservoir R1 water level	$H_{R1}$	110 m
Ground elevation, all nodes	$z$	45 m
Hazen-Williams coefficient (ductile iron)	$C$	115
Internal diameter, every pipe	$D$	452 mm = 0.452 m
Length, every pipe	$L$	200 m
Maximum-day demand, each of N1–N5	$q_{\mathrm{MDD}}$	1.0 L/s
Fire flow superimposed at N5	$q_{f}$	33 L/s

Find. (a) the pressure head at N4 under maximum-day demand plus fire flow at N5, and (b) the pressure head at N5 under maximum-day demand alone.

[Figure not reproduced: Figure 1.1 — Branched network read off Figure 1 of the examination paper. Flow leaves R1 through P1 to N1, continues along P3 to the junction N3, and divides there into P5 (to N4) and P4 (to N2), from which P7 feeds N5. Because the network is a tree there are no loops to balance: continuity al. See the official exam paper.]

Approach. A branched network has exactly one path from the source to each node, so the pipe flows follow from continuity alone — no Hardy Cross iteration is needed — and the head at any node is the reservoir level less the sum of the Hazen-Williams losses along that single supply path.

Reduce the printed Hazen-Williams equation to a head-loss law. With $S = h_f/L$, the form given in Note 5 rearranges to $$h_f = L\left(\frac{Q}{K}\right)^{1.852}, \qquad K = 0.278\,C\,D^{2.63}$$ where $K$ is the pipe conveyance, numerically the discharge the pipe would carry at unit hydraulic gradient. Substituting the common pipe data, $$K = 0.278 \times 115 \times (0.452)^{2.63} = 31.97 \times 0.12389 = 3.9606\ \mathrm{m^3/s}$$ Every pipe in this network shares the same $K$ and the same $L = 200$ m, so head loss depends only on the flow each pipe happens to carry.
Accumulate the pipe flows from the leaves back to the source, case (a). Working inwards from the far end of the tree, each pipe carries its own downstream node demand plus everything beyond it. With the 33 L/s fire flow superimposed on N5’s 1.0 L/s demand: $$Q_{P7} = 1.0 + 33 = 34\ \mathrm{L/s}, \qquad Q_{P4} = Q_{P7} + 1.0 = 35\ \mathrm{L/s}$$ $$Q_{P5} = 1.0\ \mathrm{L/s}, \qquad Q_{P3} = Q_{P4} + Q_{P5} + 1.0 = 37\ \mathrm{L/s}$$ $$Q_{P1} = Q_{P3} + 1.0 = 38\ \mathrm{L/s}$$ The total leaving the reservoir, 38 L/s, equals the five nodal demands plus the fire flow, $5 \times 1.0 + 33$, which is the continuity check on the whole tree.
Evaluate the head loss in each pipe on the path R1 → N1 → N3 → N4. Only P1, P3 and P5 lie on the supply path to Node 4, but the fire flow still matters because it loads P1 and P3 upstream of the branch point at N3. $$h_{P1} = 200\left(\frac{0.038}{3.9606}\right)^{1.852} = 0.0366\ \mathrm{m}$$ $$h_{P3} = 200\left(\frac{0.037}{3.9606}\right)^{1.852} = 0.0349\ \mathrm{m}$$ $$h_{P5} = 200\left(\frac{0.001}{3.9606}\right)^{1.852} = 0.00004\ \mathrm{m}$$ The last term is negligible because P5 carries only 1 L/s in a 452 mm main, a mean velocity of about 6 mm/s.
Convert the accumulated loss to a pressure head at Node 4. The HGL at N4 is the reservoir level less the three losses, and the pressure head is the height of that HGL above the node’s ground elevation: $$H_{N4} = 110 - (0.0366 + 0.0349 + 0.00004) = 109.93\ \mathrm{m}$$ $$\frac{p_{N4}}{\gamma} = H_{N4} - z = 109.93 - 45 = \boxed{64.93\ \mathrm{m}}$$ That is about 637 kPa, comfortably above the 140 kPa residual normally required at a hydrant during a fire event in Canadian municipal practice.
Repeat the flow accumulation for case (b), maximum day only. Removing the fire flow leaves each node drawing 1.0 L/s, so $$Q_{P7} = 1.0,\quad Q_{P4} = 2.0,\quad Q_{P5} = 1.0,\quad Q_{P3} = 4.0,\quad Q_{P1} = 5.0\ \mathrm{L/s}$$ and the supply path to Node 5 is now the longest one in the tree: P1, then P3, then P4, then P7.
Sum the losses along that path and convert to pressure head at Node 5. Applying the same head-loss law pipe by pipe, $$h_{P1} = 8.56\times10^{-4},\quad h_{P3} = 5.66\times10^{-4},\quad h_{P4} = 1.57\times10^{-4},\quad h_{P7} = 4.3\times10^{-5}\ \mathrm{m}$$ $$\sum h_f = 1.63\times10^{-3}\ \mathrm{m} \quad\Rightarrow\quad H_{N5} = 110 - 0.0016 = 109.998\ \mathrm{m}$$ $$\frac{p_{N5}}{\gamma} = 109.998 - 45 = \boxed{65.00\ \mathrm{m}}$$ Under maximum-day demand the whole network is hydraulically flat: the total loss across four pipes is under 2 mm of water, so every node sits essentially at reservoir level. Mains of this size are sized for fire flow, not for domestic demand, and the pair of answers demonstrates exactly that — adding a 33 L/s hydrant draw costs only about 70 mm of head.

Question 1 — results
Part	Quantity	Result
(a)	Flow in P1 / P3 / P5 (MDD + fire)	38 / 37 / 1.0 L/s
(a)	Head loss R1 → N4	0.071 m
(a)	Pressure head at Node 4	64.93 m (≈ 637 kPa)
(b)	Flow in P1 / P3 / P4 / P7 (MDD only)	5.0 / 4.0 / 2.0 / 1.0 L/s
(b)	Head loss R1 → N5	0.0016 m
(b)	Pressure head at Node 5	65.00 m (≈ 638 kPa)

Topic: Steady-state analysis of a branched (dendritic) distribution network by continuity plus a single head-loss path.

Key relations: $K = 0.278\,C\,D^{2.63}$; $h_f = L\,(Q/K)^{1.852}$; at any node $\sum Q_{\mathrm{in}} = \sum Q_{\mathrm{out}}$; $p/\gamma = \mathrm{HGL} - z$.

Why this works: A tree network has no independent loops, so the number of unknown pipe flows equals the number of continuity equations available. Every flow is therefore fixed before any head-loss calculation is attempted, and the energy equation is applied once, marching downstream along the unique supply path, rather than iteratively as in a looped system.

Common pitfall: Loading only the pipe immediately upstream of the hydrant with the fire flow. That flow must be carried by every pipe between the reservoir and N5, which is why it degrades the pressure at N4 even though N4 sits on a different branch. The second trap is the exponent: $1/0.54 = 1.852$, not 1.85 and certainly not 2.

Source: Mays, Water Resources Engineering, 3rd ed., Ch. 4 (pipe flow, Hazen-Williams) and Ch. 12 (water distribution — branched systems, maximum-day and fire-flow loading); Chin, Water-Resources Engineering, 3rd ed., §2.3 and §7.2 (design flows and minimum service pressures).

← Paper overview Next: Question 2 →

Quantity	Symbol	Value
Reservoir R1 water level	\(H_{R1}\)	110 m
Ground elevation, all nodes	\(z\)	45 m
Hazen-Williams coefficient (ductile iron)	\(C\)	115
Internal diameter, every pipe	\(D\)	452 mm = 0.452 m
Length, every pipe	\(L\)	200 m
Maximum-day demand, each of N1–N5	\(q_{\mathrm{MDD}}\)	1.0 L/s
Fire flow superimposed at N5	\(q_{f}\)	33 L/s