Question 1 of 6: Branched Distribution Network — Nodal Pressure Head under Maximum Day and Fire Flow
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 1: Branched Distribution Network — Nodal Pressure Head under Maximum Day and Fire Flow (20 marks)
Given. A single reservoir supplies five demand nodes through a branched (tree) network of five identical pipes, laid out as in Figure 1 of the paper: P1 runs from R1 to N1, P3 from N1 to N3, P4 from N3 to N2, P5 from N3 to N4 and P7 from N2 to N5.
Given data — Question 1
Quantity
Symbol
Value
Reservoir water level
\(H_R\)
70 m
Elevation of every node
\(z\)
20 m
Hazen-Williams coefficient (PVC)
\(C\)
138
Internal diameter, every pipe
\(D\)
406 mm = 0.406 m
Length, every pipe
\(L\)
255 m
Maximum day demand, each of N1 to N5
\(q\)
1.5 L/s
Fire flow at N5 (case a only)
\(q_F\)
33 L/s
Find. The pressure head at N4 when every node draws its maximum day demand and N5 additionally draws the fire flow, and the pressure head at N5 when only the maximum day demands are drawn.
Figure 1.1 — The branched supply network, with the pipe discharges for the maximum-day-plus-fire-flow case of part (a). Because the network is a tree there is exactly one path from the reservoir to each node, so continuity alone fixes every flow before any head loss is computed.
Approach. A branched network is statically determinate in flow: sum the demands downstream of each pipe to get its discharge, convert each discharge to a head loss with the paper’s Hazen-Williams relation, and then walk the hydraulic grade line from the reservoir surface along the single path that reaches the node in question.
Part (a) and (b) — establish the resistance of one pipe. Every pipe in this network has the same diameter, length and roughness, so one resistance serves throughout. With \(K = 0.278\,C\,D^{2.63}\),
$$K = 0.278 \times 138 \times (0.406)^{2.63} = 38.364 \times 0.093416 = 3.5838\ \text{m}^3\text{/s}$$
and inverting the Hazen-Williams equation for head loss over a length \(L\),
$$h_f = L\left(\frac{Q}{K}\right)^{1.852} = \frac{255}{3.5838^{1.852}}\,Q^{1.852} = 23.982\,Q^{1.852}$$
with \(Q\) in m3/s and \(h_f\) in metres. This single expression is applied to each pipe in turn.
Part (a) — work back from the far end of the tree to get the pipe flows. Each pipe carries the sum of the demands lying beyond it. Starting at N5, which draws its own 1.5 L/s plus the 33 L/s fire flow,
$$Q_{P7} = 1.5 + 33 = 34.5\ \text{L/s}$$
$$Q_{P4} = Q_{P7} + 1.5 = 36.0\ \text{L/s}$$
$$Q_{P5} = 1.5\ \text{L/s}$$
$$Q_{P3} = Q_{P4} + Q_{P5} + 1.5 = 39.0\ \text{L/s}$$
$$Q_{P1} = Q_{P3} + 1.5 = 40.5\ \text{L/s}$$
The total drawn from the reservoir is \(5 \times 1.5 + 33 = 40.5\) L/s, which is exactly what P1 carries — the check that the tree has been traced correctly.
Part (a) — head loss along the path R1 to N1 to N3 to N4. Only three pipes lie on the route to N4, and each is evaluated from the resistance of Step 1:
$$h_{f,P1} = 23.982\,(0.0405)^{1.852} = 0.0632\ \text{m}$$
$$h_{f,P3} = 23.982\,(0.0390)^{1.852} = 0.0590\ \text{m}$$
$$h_{f,P5} = 23.982\,(0.0015)^{1.852} = 0.000141\ \text{m}$$
Summing along the path gives a total friction loss of 0.1223 m from the reservoir surface to N4. The contribution of P5 is three orders of magnitude smaller than that of P1 because it carries only the local demand, and head loss scales with roughly the square of discharge.
Part (a) — convert the grade line to a pressure head at N4. Neglecting velocity head, the HGL at N4 is the reservoir surface less the accumulated friction, and the pressure head is that grade line measured above the node itself:
$$\mathrm{HGL}_{N4} = 70 - 0.1223 = 69.878\ \text{m}$$
$$\boxed{\left.\frac{p}{\gamma}\right|_{N4} = 69.878 - 20 = 49.88\ \text{m}\ \ (489\ \text{kPa})}$$
That is a healthy 489 kPa, comfortably inside the 140 to 700 kPa band that Canadian distribution practice expects to hold during a fire event.
Part (b) — re-trace the tree with the fire flow removed. With every node drawing only 1.5 L/s the discharges collapse to a fifth of their earlier values on the trunk and less on the branches:
$$Q_{P7} = 1.5\ \text{L/s}$$
$$Q_{P4} = 3.0\ \text{L/s}$$
$$Q_{P3} = 6.0\ \text{L/s}$$
$$Q_{P1} = 7.5\ \text{L/s}$$
Now N5 is reached by the route R1 to N1 to N3 to N2 to N5, so four pipes contribute rather than three.
Part (b) — head losses on the route to N5, and the pressure head there. Applying the same resistance,
$$h_{f,P1} = 23.982\,(0.0075)^{1.852} = 0.002785\ \text{m}$$
$$h_{f,P3} = 23.982\,(0.0060)^{1.852} = 0.001843\ \text{m}$$
$$h_{f,P4} = 23.982\,(0.0030)^{1.852} = 0.000510\ \text{m}$$
$$h_{f,P7} = 23.982\,(0.0015)^{1.852} = 0.000141\ \text{m}$$
The four losses total only 0.005275 m, so
$$\mathrm{HGL}_{N5} = 70 - 0.005275 = 69.995\ \text{m}$$
$$\boxed{\left.\frac{p}{\gamma}\right|_{N5} = 69.995 - 20 = 49.99\ \text{m}\ \ (490\ \text{kPa})}$$
Interpret the two answers together. The difference between the two cases is 0.12 m of head out of a 50 m static lift, that is one part in four hundred. The reason is visible in the velocity: even in the worst case, P1 carries 40.5 L/s through a 406 mm pipe, so
$$V = \frac{Q}{A} = \frac{0.0405}{0.12946} = 0.313\ \text{m/s}$$
against the 1 to 2 m/s a transmission main is normally sized for. The grid is enormously oversized for these demands, which is entirely normal for a distribution network whose pipe sizes are set by the fire-flow requirement and by minimum-diameter rules rather than by average-day economics.
