Question 4 of 6: Two elevated tanks — quasi-steady simulation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 98-Civ-A5 Hydraulic Engineering, December 2013 · 3 hours, closed book (one 8.5×11 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; ‘cms’ = m³/s.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, valves, network analysis); Chow, Open-Channel Hydraulics (Manning uniform flow, gutter flow); Crowe, Elger & Roberson, Engineering Fluid Mechanics (laminar film flow, 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 $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, Darcy–Weisbach, and total dynamic head $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
The Hazen–Williams relation is used in inverted form to get the friction loss carried by a known pipe discharge $Q$ (SI, $Q$ in m³/s):
Check (Question 4 data are inconsistent — NOTE 1 assumption). The stated “initial steady-state flow = 350 L/s” implies a node head of only $H=(0.350/0.15)^2=5.4$ m; but at $H=5.4$ m the two mains, driven by the 96 m and 91 m tank levels, would deliver $\approx 1{,}110$ L/s — so 350 L/s is not a steady state. Continuity at the node fixes a network-consistent initial discharge of $\approx 890$ L/s at a node head of $\approx 35.2$ m; the simulation is run from that physically consistent state. The figure also labels Tank 2 at 89 m and $L=300$ m, which conflict with the prose values (91 m, 350 m); the prose is adopted and the discrepancy noted per NOTE 1.
Given. Two supply pipes meet at a demand node whose valve discharges to atmosphere; tanks fall as they deliver.
Given data (Question 4)
Tank 1 / Tank 2 initial level
96 m / 91 m
Tank diameter
5 m ($A=19.63$ m²)
Pipe $C$ / $D$ / $L$
100 / 300 mm / 350 m
Valve coeff. $C_v$
0.15 m$^{5/2}$/s
Valve law (to atmosphere)
$Q_v=C_v\sqrt{H}$
Time step $\Delta t$
15 s
Find. Node pressure head $H$ and pipe flows $Q_1,Q_2$ at the first three time steps.
Figure 4. Two tanks feed a common node; $H$ is the node pressure head, $Q$ the valve discharge to atmosphere.
Approach. At each step the node is treated as steady: the two inflows (Hazen–Williams, driven by tank level minus node head $H$) must equal the valve outflow $C_v\sqrt{H}$. Solve that single continuity equation for $H$, read $Q_1,Q_2$, then lower each tank by $Q_i\Delta t/A_{\text{tank}}$ and repeat.
Continuity at the node. With $z_1,z_2$ the tank levels and $H$ the node head,
$$Q_1+Q_2=C_v\sqrt{H},\qquad Q_i=0.278\,C\,D^{2.63}\!\left(\dfrac{z_i-H}{L}\right)^{0.54}.$$
This is one nonlinear equation in $H$; solve by bisection.
Step 1 ($z_1=96.00$, $z_2=91.00$). The balance closes at $H=35.2$ m, giving
$$\boxed{H_1=35.2\ \text{m},\quad Q_1=455\ \text{L/s},\quad Q_2=435\ \text{L/s},\quad Q_v=890\ \text{L/s}.}$$
Update the tanks. $\Delta z_i=Q_i\Delta t/A_{\text{tank}}$: $\Delta z_1=0.455(15)/19.63=0.348$ m, $\Delta z_2=0.435(15)/19.63=0.332$ m, so $z_1\to95.65$, $z_2\to90.67$ m.
Step 2 ($z_1=95.65$, $z_2=90.67$): $H_2=35.1$ m, $Q_1=455$ L/s, $Q_2=434$ L/s, $Q_v=888$ L/s. Update: $z_1\to95.31$, $z_2\to90.34$ m.
Over the first 45 s the tanks fall only about 1 m, so the node head and discharges drift down very slowly (890→887 L/s) — the hallmark of a quasi-steady drawdown where storage change per step is small compared with the through-flow.
