Question 4 of 6: Rigid water-column (mass-oscillation) draining
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Worked solutions. Closed-book format, 3 hours; six questions, candidates complete any five — all six are solved here as a study resource. All questions are of equal value (20 marks).
Reference texts. Mays, Water Resources Engineering, 2nd ed. (pipe systems, equivalent pipes, valves, network analysis, rigid water-column / mass-oscillation model); Chow, Open-Channel Hydraulics (Manning uniform flow, critical depth, specific energy); Crowe, Elger & Roberson, Engineering Fluid Mechanics (unsteady flow, St. Venant equations). Exam-supplied relations: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI units), Manning $Q=\tfrac{1}{n}A R^{2/3}S^{1/2}$, and Total Dynamic Head $\mathrm{TDH}=H_S+H_f$.
Figure 4 (schematic). Two reservoirs connected by a single pipe; the 40 m head difference drives the transient as the levels equalise.
Find. (a) the initial steady-state pipe flow; (b) $Q$ and both reservoir levels after two rigid-column time steps; (c) the physical reason the 450 mm response is a smooth decay while the 750 mm response oscillates.
Approach. The rigid water-column momentum equation is $\dfrac{L}{gA_p}\dfrac{dQ}{dt}=(H_{u/s}-H_{d/s})-h_f(Q)$, coupled to the reservoir continuity $\dfrac{dH_{u/s}}{dt}=-\dfrac{Q}{A_{\text{res}}}$, $\dfrac{dH_{d/s}}{dt}=+\dfrac{Q}{A_{\text{res}}}$. Start from the steady flow (so $dQ/dt=0$) and march forward explicitly.
(a) Steady flow at $t=0$. Steady state means $dQ/dt=0$, so all 40 m of head is spent on friction: $h_f=kQ_0^{1.852}=40$, with $k=73.82$ (SI). Thus $$Q_0=\left(\frac{40}{73.82}\right)^{1/1.852}=\boxed{0.718\ \text{m}^3/\text{s}.}$$ This matches the $t=0$ value read from Figure 4a ($\approx0.70$ m³/s).
(b) Discretise. Explicit (Euler) update with $\dfrac{dQ}{dt}=\dfrac{gA_p}{L}\big[(H_{u/s}-H_{d/s})-h_f(Q)\big]$: $$Q_{n+1}=Q_n+\Delta t\,\frac{gA_p}{L}\big[\Delta H_n-kQ_n^{1.852}\big],\quad H_{u/s,n+1}=H_{u/s,n}-\frac{Q_n\Delta t}{A_{\text{res}}},\quad H_{d/s,n+1}=H_{d/s,n}+\frac{Q_n\Delta t}{A_{\text{res}}}.$$ The source does not state $\Delta t$; we adopt $\Delta t=20\ \text{s}$ (see check note).
(b) Step to $t_1=20$ s. At $t_0$, $\Delta H=40$ and $h_f=40$, so $dQ/dt=0$ and $Q$ is unchanged; the tiny reservoirs, however, move quickly: $$Q_1=0.718\ \text{m}^3/\text{s},\quad H_{u/s}=80-\tfrac{0.718\cdot20}{5}=77.13\ \text{m},\quad H_{d/s}=40+2.87=42.87\ \text{m}.$$
(b) Step to $t_2=40$ s. Now $\Delta H=34.25$ m but $h_f(0.718)=40$ m, so the column decelerates: $$\tfrac{dQ}{dt}=\tfrac{9.81\cdot0.159}{1000}(34.25-40)=-8.96\times10^{-3},\quad Q_2=0.718-20(8.96\times10^{-3})=\boxed{0.539\ \text{m}^3/\text{s},}$$ with $H_{u/s}=74.25$ m, $H_{d/s}=45.75$ m. The flow is now falling as the levels close on the common equilibrium of 60 m.
(c) Why 450 mm decays but 750 mm oscillates. The momentum equation balances an inertia term $\tfrac{L}{gA_p}\tfrac{dQ}{dt}$ against a friction term $h_f\propto Q^{1.852}/D^{4.87}$. Enlarging the pipe to 750 mm slashes friction (the $D^{4.87}$ in the denominator) while the inertia coefficient $L/gA_p$ also falls but far less. The friction that used to over-damp the system is now too small to dissipate the column’s kinetic energy in one swing, so the mass of water overshoots the equilibrium, reverses (negative flow), and executes a slowly-decaying mass oscillation (Figure 4b). The 450 mm case is friction-dominated (over-damped): the flow simply decays to zero without reversing (Figure 4a). In short, the diameter sets the damping ratio friction/inertia — large pipe → under-damped oscillation, small pipe → over-damped decay.
Check: the exam does not give a numerical time step. A value of $\Delta t=20\ \text{s}$ is adopted to demonstrate the rigid-column march; the method (not the specific $\Delta t$) is what the question tests. A smaller $\Delta t$ tracks Figure 4a more closely (explicit Euler with 20 s slightly over-shoots the decay).
