Question 1 of 6: Rigid water-column transient between two reservoirs

Question 1: Rigid water-column transient between two reservoirs (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 reservoirs connected by a single pipe that starts flowing under a 40 m head difference and then equalises as the reservoirs drain/fill.

Find. (a) the steady discharge at $t=0$; (b) $Q$, $z_1$, $z_2$ over the first two time steps of the rigid-column model; (c) a physical explanation of why the 450 mm response is over-damped while the 750 mm response oscillates.

Approach. Part (a) is a steady balance — with no acceleration the full head difference is spent on friction, so Hazen–Williams gives $Q$. Part (b) advances the rigid-column momentum equation together with reservoir continuity by explicit (Euler) time-stepping. Part (c) reads the two response plots through the damping ratio, which the pipe diameter controls.

1. Steady discharge at $t=0$ (part a). At steady state $dQ/dt=0$, so the whole head difference is friction head: $\Delta h = z_1-z_2 = 40\ \text{m}$ over $L=1000\ \text{m}$, i.e. $S = 40/1000 = 0.04$. Hazen–Williams: $$Q_0 = 0.278\,C\,D^{2.63}\,S^{0.54} = 0.278(120)(0.45)^{2.63}(0.04)^{0.54}.$$ Evaluating $(0.45)^{2.63}=0.1225$ and $(0.04)^{0.54}=0.1759$ gives $$\boxed{Q_0 = 0.718\ \text{m}^3/\text{s}}$$ which agrees with the $t=0$ intercept ($\approx 0.72\ \text{m}^3/\text{s}$) of Figure Q4–A.
2. Rigid-column governing equations (part b). Treating the pipe water as one incompressible, non-deformable slug, Newton’s second law on the column and mass continuity at each reservoir give $$\frac{L}{g\,A_p}\frac{dQ}{dt} = (z_1-z_2) - h_f(Q),\qquad \frac{dz_1}{dt}=-\frac{Q}{A_{res}},\quad \frac{dz_2}{dt}=+\frac{Q}{A_{res}},$$ with $h_f(Q)$ the Hazen–Williams friction head at the current $Q$. Here $\dfrac{gA_p}{L}=\dfrac{9.81(0.15904)}{1000}=1.560\times10^{-3}$.
3. Choose a time step and advance (Euler). The exam does not state $\Delta t$; adopt $\Delta t = 20\ \text{s}$ (small relative to the $\sim$250 s natural period of this mass oscillation — see the concept note). Update rule: $Q_{k+1}=Q_k+\Delta t\,\tfrac{gA_p}{L}[(z_1-z_2)_k-h_f(Q_k)]$, then $z_{1,k+1}=z_{1,k}-\Delta t\,Q_k/A_{res}$ and $z_{2,k+1}=z_{2,k}+\Delta t\,Q_k/A_{res}$.
   At $t=0$: $Q=0.718$, $z_1-z_2=40=h_f$, so $dQ/dt=0$ — the flow is momentarily unchanged while the levels begin to close: $$z_1(20)=80-20\tfrac{0.718}{5}=77.13\ \text{m},\quad z_2(20)=40+20\tfrac{0.718}{5}=42.87\ \text{m},\quad Q(20)=0.718\ \text{m}^3/\text{s}.$$ Second step: now $z_1-z_2=34.25\ \text{m}$, which is below $h_f(0.718)=40\ \text{m}$, so the column decelerates, $dQ/dt=1.560\times10^{-3}(34.25-40)=-8.97\times10^{-3}$: $$\boxed{Q(40)=0.539\ \text{m}^3/\text{s},\quad z_1(40)=74.25\ \text{m},\quad z_2(40)=45.75\ \text{m}.}$$
4. Interpret the two plots (part c). The rigid-column equation is a damped oscillator: inertia $\tfrac{L}{gA_p}\tfrac{dQ}{dt}$ against the restoring head $(z_1-z_2)$ and the resistive (damping) term $h_f\propto D^{-4.87}$. The initial linear momentum of the slug is $\mathcal{M}=\rho L A_p V_0=\rho L\,Q_0$. Enlarging $D$ from 450 to 750 mm raises $Q_0$ from 0.72 to 2.75 m³/s, so $\mathcal{M}$ grows nearly four-fold, while friction (the damping, $\propto D^{-4.87}$) collapses to a small fraction. The 450 mm column is friction-dominated: it decays almost monotonically to zero with a single small undershoot (Figure Q4–A). The 750 mm column is inertia-dominated (lightly damped): its large stored momentum overshoots the equalised level and it oscillates about $Q=0$ with slowly-decaying amplitude (Figure Q4–B). Diameter sets the damping ratio.