Question 4 of 6: Quasi-steady simulation of two draining tanks
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 98-Civ-A5 Hydraulic Engineering, May 2013 · 3 hours, closed book (one aid sheet). Six questions; any five constitute a complete paper — all six are solved here as a study resource. All parts of a question are of equal value.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, pumps, network analysis); Chow, Open-Channel Hydraulics (Manning flow, compound sections); Crowe, Elger & Roberson, Engineering Fluid Mechanics (energy/continuity). Exam-supplied relations used throughout: Hazen–Williams $Q=0.278\,C\,D^{2.63}\,S^{0.54}$ with $S=h_f/L$ (SI), Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, and total dynamic head $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected, and water has $\rho=1000\ \text{kg/m}^3$.
Check (Q4 data consistency): the pipe/valve data in Question 4 are internally inconsistent — the stated “initial valve flow = 400 L/s” corresponds to a node head of only 9.5 m, but the two supply pipes driven by the 96 m and 89 m tank levels deliver far more than that at 9.5 m. Continuity at the node fixes a network-consistent initial discharge of ≈693 L/s at a node head of ≈28.4 m. The simulation below is run from that physically consistent state, with the discrepancy noted (per Note 1, candidates may state assumptions).
Question 4: Quasi-steady simulation of two draining tanks (20 marks)
Given. Two cylindrical tanks ($D_t=5$ m, area $A_t=\pi(2.5)^2=19.63$ m²), initial levels 96 m and 89 m; valve $Q_v=C_v\sqrt{H}$ with $C_v=0.13$; supply pipes $C=110$, $D=0.250$ m, $L=300$ m; $\Delta t=10$ s.
Find. The node pressure head $H$ and the two pipe flows over the first three time steps.
Figure 2 (Q4). Two elevated tanks feed a demand node whose valve discharges to atmosphere; both supply pipes are 300 m long.
Approach. At each instant treat the flow as steady (quasi-steady): solve node continuity $Q_1+Q_2=Q_v(H)$ for the node head $H$, where each pipe obeys Hazen–Williams with driving head $(z_i-H)$ and the valve obeys $Q_v=C_v\sqrt H$. Then drop each tank level by $Q_i\,\Delta t/A_t$ and repeat.
Pipe conductance. $0.278(110)(0.250)^{2.63}=0.799$, so each pipe $Q_i=0.799\big((z_i-H)/300\big)^{0.54}$.
Node continuity (each step). $$0.799\Big(\tfrac{z_1-H}{300}\Big)^{0.54}+0.799\Big(\tfrac{z_2-H}{300}\Big)^{0.54}=0.13\sqrt{H}.$$ Solve for $H$ given the current tank levels $z_1,z_2$.
Initial step ($t=0$: $z_1=96,\ z_2=89$). Iterating gives $$\boxed{H_0\approx 28.4\ \text{m}},\quad Q_1\approx 357\ \text{L/s},\ Q_2\approx 336\ \text{L/s},\ Q_v\approx 693\ \text{L/s}.$$ (See the check note above: this network-consistent value replaces the stated but inconsistent 400 L/s.)
Update tank levels. $z_i\leftarrow z_i-\dfrac{Q_i\,\Delta t}{A_t}$: Tank 1 drops $0.357(10)/19.63=0.182$ m, Tank 2 drops $0.336(10)/19.63=0.171$ m per step.
March forward. Repeating the solve after each update gives the table below. Because the tanks are large and the driving heads (>60 m) far exceed the node head, the flows and node head decline only gradually.
Report the node pressure head. With the node at datum, the pressure head equals $H$: it starts at 28.4 m and eases to 28.3 m by the end of the third step — the system is effectively quasi-static over 30 s.
