16-Civ-A5 Hydraulic Engineering · May 2017

Question 3 of 6: Distribution network — HGL and water age

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

Question 3: Distribution network — HGL and water age (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. Source reservoir R1 feeds node N3 through the parallel pair P1∥P2 and node N2 through the parallel pair P3∥P4, with a tie pipe P5 between N3 and N2; every pipe is identical. In part a) the P3 valve is shut, so N2 is fed only by P4 and the tie P5.

Find. (a) HGL at N2 and N3 with P3 closed; (b) the qualitative change in water age at N2 and N3 if both P1 and P3 are closed.

Approach. With P3 shut, write nodal continuity at N3 and N2 using Hazen-Williams $h_f = kQ^{1.852}$ for each link; identical parallel pipes split their flow equally. Solve the two nodal heads, then recover pipe flows and pressure heads.

1. Per-pipe resistance. $k = 420\,[0.278(117)(0.205)^{2.63}]^{-1.852} = 1495.7$, so each link obeys $h_f = 1495.7\,Q^{1.852}$ (Q in m³/s).
2. Continuity equations (P3 closed). Let $H_3,\ H_2$ be the HGLs at N3, N2. The parallel feeders P1∥P2 to N3 each carry $Q_{P1}=\left((H_{R1}-H_3)/k\right)^{1/1.852}$; P4 feeds N2 and the tie P5 carries $Q_{P5}=\left((H_3-H_2)/k\right)^{1/1.852}$: $$2\,Q_{P1} = 0.050 + Q_{P5}\ \ (\text{N3}),\qquad Q_{P4} + Q_{P5} = 0.200\ \ (\text{N2}).$$
3. Solve the nodal heads. Simultaneous solution gives $$H_3 = \boxed{60.6\ \text{m}}, \qquad H_2 = \boxed{44.5\ \text{m}}.$$ The heavily-loaded N2 sits well below N3, so the tie P5 delivers flow from N3 toward N2.
4. Pipe flows and pressures. Back-substituting: $Q_{P1}=Q_{P2}=68.3$ L/s, $Q_{P4}=113.3$ L/s, $Q_{P5}=86.7$ L/s (N3→N2); supply $2Q_{P1}+Q_{P4}=250$ L/s balances the total demand. Pressure heads $p/\gamma = H - z$ are $60.6-11 = 49.6$ m at N3 and $44.5-11 = 33.5$ m at N2.

Part b) — water age when P1 and P3 are both closed. Water age at a node is the travel time from the source, and for a pipe carrying steady flow that is its volume divided by its discharge, $t = LA/Q$. The demands (200 and 50 L/s) do not change when valves close, so the same 250 L/s must now pass through fewer open pipes: N3 is fed by P2 alone, and N2 by P4 plus the tie P5. Re-solving the two nodal-continuity equations with P1 also shut gives $H_3 = 43.8$ m and $H_2 = 34.3$ m, with $Q_{P2} = 115.0$ L/s (was 68.3), $Q_{P4} = 135.0$ L/s (was 113.3) and $Q_{P5} = 65.0$ L/s (was 86.7). The flow per open pipe rises, so velocities rise (P2 2.07→3.48 m/s) and residence times fall. With $A = 0.0330$ m² and $L = 420$ m:

• N3: age = travel time in its single feeder P2, falling from about 3.4 min to about 2.0 min — water age decreases, strongly.
• N2: a flow-weighted mix of direct P4 water and water that has travelled P2 then P5. More of N2’s demand now comes straight down P4 (135 of 200 L/s), and the tie carries less, so the mean age falls from about 3.8 min to about 3.0 min — water age also decreases, but less than at N3, because part of N2’s supply still travels the two-pipe route R1→N3→N2.

The hydraulic reason is that the demands set the throughput, and taking pipes out of service removes storage volume from the flow path without reducing that throughput, so water turns over faster in the pipes that stay open. The age penalty sits in the closed pipes instead: the water trapped in P1 and P3 stops moving, loses its chlorine residual, and will reach customers as a slug of old water (possibly with discoloration) when the valves are reopened unless those mains are flushed first. The price of the faster turnover is hydraulic, not water quality: the larger flows raise friction losses, so pressure heads drop to about 32.8 m at N3 and 23.3 m at N2, and each node is left with less redundancy.