Two elevated tanks — quasi-steady drawdown: 16-Civ-A5 Hydraulic Engineering, December 2017, Q4

Question 4: Two elevated tanks — quasi-steady drawdown (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. Two constant-diameter tanks drain through identical pipes to a common node, whose valve discharges to atmosphere.

Given data
Tank diameter (both) / plan area	5 m / 19.635 m²
Initial levels: Tank 1 / Tank 2	96 m / 89 m
Valve coefficient $C_v$ (to atmosphere)	0.15 m^5/2/s
Pipe C / D / L (both)	100 / 250 mm / 300 m
Time step $\Delta t$ / number of steps	15 s / 3

Find. The node pressure head $H$ and the two pipe flows $Q_1,Q_2$ over the first three time steps.

Figure 4. Two elevated tanks feeding a common demand node H; the node valve discharges to atmosphere. Node elevation is taken as the datum, so H is the node pressure head.

Approach. Each step is a steady network solve: find the node head $H$ that satisfies continuity $Q_1+Q_2 = Q_{\text{valve}}$, then advance the tank levels by the volume drawn, $\Delta z_i = Q_i\,\Delta t / A_{\text{tank}}$, and repeat.

Check (inconsistent given, per NOTE 1): the stated “initial valve flow 350 L/s” implies a node head $H=(0.350/0.15)^2 = 5.44$ m; but at $H=5.44$ m the 96 m and 89 m tanks would drive roughly 1.1 m³/s through the two pipes — far more than 0.35 m³/s, so continuity cannot hold. The simulation below therefore starts from the continuity-consistent initial state ($H_0 = 20.13$ m, $Q_v = 673$ L/s) and the stated 350 L/s is treated as an inconsistent datum.

Nodal equations. Taking the node elevation as datum, each pipe carries $Q_i = 0.278\,C\,D^{2.63}\left[(z_i-H)/L\right]^{0.54}$ and the valve passes $Q_v = C_v\sqrt{H}$. The unknown $H$ closes continuity: $$Q_1(z_1,H)+Q_2(z_2,H) = C_v\sqrt{H}.$$
Initial state ($t=0$). Solving with $z_1=96$, $z_2=89$ m: $$\boxed{H_0 = 20.13\ \text{m}},\qquad Q_1 = 345.3\ \text{L/s},\quad Q_2 = 327.7\ \text{L/s}\ \ (Q_v = 673.0\ \text{L/s}).$$
Advance the tank levels. Over $\Delta t = 15$ s, $\Delta z_i = Q_i\Delta t / A_{\text{tank}}$ with $A_{\text{tank}} = \tfrac{\pi}{4}(5)^2 = 19.635\ \text{m}^2$; e.g. Tank 1 drops $0.3453\cdot15/19.635 = 0.264$ m to 95.74 m, Tank 2 drops 0.250 m to 88.75 m.
Repeat for the next steps. Re-solving continuity at each new pair of levels gives the small monotonic decline tabulated below — the node head barely moves because the large tanks lose only ~0.25 m per step.

Quasi-steady simulation — Question 4
Time	Tank 1 (m)	Tank 2 (m)	Node head $H$ (m)	$Q_1$ (L/s)	$Q_2$ (L/s)	$Q_v$ (L/s)
0 s	96.00	89.00	20.13	345.3	327.7	673.0
15 s	95.74	88.75	20.07	344.8	327.2	672.0
30 s	95.47	88.50	20.01	344.3	326.7	671.1
45 s	95.21	88.25	19.95	343.8	326.3	670.1

Topic: Quasi-steady (rigid-column) drawdown of reservoirs through a pipe junction.

Key relations: node continuity $\sum Q_i = C_v\sqrt{H}$; Hazen-Williams pipe flow; tank mass balance $A\,\mathrm{d}z/\mathrm{d}t = -Q$ discretised as $\Delta z = Q\Delta t/A$.

Why this works: “Quasi-steady” means the flow adjusts instantly to the current heads (fluid inertia neglected), so every instant is an ordinary steady network problem; the only time dependence enters through the slowly falling tank levels. Because the tank plan area is large, one time step removes very little storage and the heads change by centimetres, which is why an explicit march with a 15 s step is stable and accurate here.

Common pitfall: Accepting the stated initial valve flow at face value instead of checking it against continuity; or forgetting that the valve discharges to atmosphere, so the head driving it is the node pressure head $H$, not a reservoir difference.

Source: Chin, Water-Resources Engineering, §2.5 (unsteady/quasi-steady pipe flow); Streeter & Wylie, Fluid Mechanics.

Question 4 of 6: Two elevated tanks — quasi-steady drawdown

Question 4: Two elevated tanks — quasi-steady drawdown (20 marks)