Two elevated tanks — quasi-steady simulation: 16-Civ-A5 Hydraulic Engineering, December 2015, Q4

Question 4: Two elevated tanks — quasi-steady simulation (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 5 m-diameter tanks ($A_{\text{tank}}=\pi(2.5)^2=19.63\ \text{m}^2$) at initial levels $z_1=96$ m, $z_2=89$ m feed a common demand node whose valve discharges to atmosphere as $Q_v=C_v\sqrt{H}$, $C_v=0.10\ \text{m}^{5/2}/\text{s}$. Both pipes: $C=110$, $D=250$ mm, $L=300$ m. Time step $\Delta t=10$ s.

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

Figure 4. Two elevated tanks feed a demand node; the valve discharges to atmosphere. At each step the node head is found from continuity, then the tanks are drawn down.

Approach. At each instant the node head $H$ satisfies continuity $Q_1(z_1{-}H)+Q_2(z_2{-}H)=C_v\sqrt{H}$, with each pipe flow from Hazen–Williams on its own driving head. Solve $H$, evaluate $Q_1,Q_2$, then lower each tank by $\Delta z_i=Q_i\,\Delta t/A_{\text{tank}}$ and repeat (explicit quasi-steady stepping).

Check (inconsistent datum): The stated "initial valve flow 300 L/s" implies $H=(0.30/0.10)^2=9.0$ m, but with the tanks at 96 m and 89 m the two 300 m / 250 mm pipes would deliver about 800 L/s to a 9 m node — inflow could not equal the 300 L/s outflow, so the given 300 L/s is not a valid steady state. Per Note 1, the simulation is started from the continuity-consistent initial state ($H_0\approx39.3$ m, $Q_v\approx627$ L/s) and the discrepancy is flagged.

Pipe law. $Q_i = 0.278\,C\,D^{2.63}\left(\dfrac{z_i-H}{L}\right)^{0.54}=0.799\left(\dfrac{z_i-H}{300}\right)^{0.54}$ (m³/s).
Initial node head ($t=0$). Solving $0.799\left(\tfrac{96-H}{300}\right)^{0.54}+0.799\left(\tfrac{89-H}{300}\right)^{0.54}=0.10\sqrt{H}$ gives $$\boxed{H_0 = 39.30\text{ m},\quad Q_1=0.325,\ Q_2=0.302\ \text{m}^3/\text{s},\quad Q_v=0.627\ \text{m}^3/\text{s}.}$$
Draw down the tanks. $\Delta z_i = Q_i\,\Delta t/A_{\text{tank}} = Q_i(10)/19.63$; Tank 1 falls $\approx0.165$ m, Tank 2 $\approx0.154$ m per step.
Step to $t=10,20,30$ s. Repeating the solve on the updated levels gives the table below; the heads and flows drift down only slightly because the tanks are large relative to a 10 s step.

Quasi-steady simulation — Question 4
$t$ (s)	$z_1$ (m)	$z_2$ (m)	Node head $H$ (m)	$Q_1$ (L/s)	$Q_2$ (L/s)	$Q_v$ (L/s)
0	96.000	89.000	39.30	324.6	302.3	626.9
10	95.835	88.846	39.23	324.3	302.0	626.3
20	95.669	88.692	39.16	324.0	301.7	625.7
30	95.505	88.539	39.08	323.7	301.5	625.2

Topic. Quasi-steady (rigid-column, negligible-inertia) simulation of a two-source tank system draining through a valve.

Key relations. Nodal continuity $\sum Q_i = C_v\sqrt{H}$; Hazen–Williams pipe law; tank mass balance $\dfrac{dz_i}{dt}=-\dfrac{Q_i}{A_{\text{tank}}}$, explicit as $z_i^{new}=z_i-Q_i\Delta t/A_{\text{tank}}$.

Why this works. "Quasi-steady" means the pipes reach their steady friction state instantly at each instant, so the only dynamics are the slowly falling tank levels. Each time step is a small steady-network solve followed by a bookkeeping update of storage — accurate here because $A_{\text{tank}}$ is large and $\Delta t$ small.

Common pitfall. Starting the simulation from the stated (inconsistent) 300 L/s / 9 m state instead of the continuity-consistent one; or forgetting that the node head, not the tank level, drives the valve discharge.

Source. Mays, Water Resources Engineering, Ch. 3–4 (quasi-steady network analysis, tank storage).

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

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