Question 4 of 6: Two-tank quasi-steady draining simulation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 16-Civ-A5 Hydraulic Engineering — May 2018 · 3 hours · closed book (one aid sheet) · six questions, complete any five (all solved here) · each question 20 marks, equal-value parts.
Governing relations supplied on the exam cover sheet. Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with slope $S=\Delta h/L$ (SI, $Q$ in m³/s, $D$ in m); Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$; Darcy–Weisbach $\Delta h = 0.0826\,\tfrac{fL}{D^5}Q^2$; total dynamic head $TDH=H_s+H_f$. Unless stated, local losses and velocity head are neglected, diameters are nominal, and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Find. The node pressure head $H$ and pipe flows $Q_1,Q_2$ over the first three quasi-steady time steps.
Figure 4. Two elevated tanks feed a common demand node H; the valve discharges $Q$ to atmosphere. HGL dips to the node between the tanks.
Approach. At each instant the network is solved as steady: the two tank-to-node pipe flows must equal the valve discharge at the node head $H$ (nodal continuity). Solve that one nonlinear equation for $H$, evaluate $Q_1,Q_2,Q_v$, then lower each tank level by the volume it lost over $\Delta t$ and repeat.
Check / inconsistent given. The stated "initial valve flow 350 L/s" would need node head $H=(0.350/0.15)^2=5.4\ \text{m}$, but tanks at 96 and 89 m drive the 300 mm pipes to a continuity-consistent head $H_0\approx42.6\ \text{m}$ with $Q_v\approx979\ \text{L/s}$ — the 350 L/s value is not compatible with the stated geometry. Per cover-sheet NOTE 1, the simulation is run from the network-consistent initial state (and the discrepancy is flagged). Also, part (d) reads "10 mm" as printed; a tank diameter smaller than 5 m contradicts "increased," so it is taken as 10 m.
Nodal equations. With $a=0.278(110)(0.30)^{2.63}=1.289$, each pipe gives $Q_i=a\big((z_i-H)/L\big)^{0.54}$; the valve gives $Q_v=0.15\sqrt{H}$. Continuity: $Q_1+Q_2=Q_v$.
Solve for $H$ at $t=0$. Root-finding on $a\big(\tfrac{z_1-H}{L}\big)^{0.54}+a\big(\tfrac{z_2-H}{L}\big)^{0.54}=0.15\sqrt{H}$ gives $\boxed{H_0\approx42.6\ \text{m}}$, with $Q_1\approx508$, $Q_2\approx471$, $Q_v\approx979\ \text{L/s}$.
March the tanks. Update $z_i \leftarrow z_i-\dfrac{Q_i\,\Delta t}{A_t}$ (each step lowers a tank by $\approx Q_i(15)/19.635$), and re-solve. Three steps:
Quasi-steady simulation ($\Delta t=15$ s)
$t$ (s)
$z_1$ (m)
$z_2$ (m)
$H$ (m)
$Q_1$ (L/s)
$Q_2$ (L/s)
$Q_v$ (L/s)
0
96.000
89.000
42.56
507.8
470.7
978.5
15
95.612
88.640
42.38
506.7
469.7
976.5
30
95.225
88.282
42.20
505.7
468.7
974.4
45
94.839
87.923
42.02
504.6
467.8
972.3
The node head and flows drift down slowly as the tanks fall — the quasi-steady assumption (pipe/valve response fast relative to tank drawdown) is well satisfied.
(b) Lower $C$ (110→80). Higher friction for the same head means smaller pipe flows, so the valve is starved and the tanks empty more slowly — draining time increases (the node head settles lower, reducing $Q_v$).
(c) Larger pipes (300→500 mm). Friction drops steeply ($h_f\propto D^{-4.87}$), so more flow reaches the node, the head and $Q_v$ rise, and the tanks empty faster — draining time decreases.
(d) Larger tanks (5→10 m). Storage area quadruples ($A_t\propto D_t^2$, $\times4$) while the driving heads and pipe/valve hydraulics are essentially unchanged, so the drawdown rate $dz/dt=Q/A_t$ falls to about a quarter — draining time increases by roughly a factor of four.
