Question 5 of 6: Detention Pond Design for Peak-Flow Attenuation

Question 5: Detention Pond Design for Peak-Flow Attenuation (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. Watershed area $A=100\ \text{ha}=1{,}000{,}000\ \text{m}^2$; inflow hydrograph triangular, rising linearly from $(0,0)$ to a peak $Q_p=2.0\ \text{m}^3/\text{s}$ at $t=1.0$ hr, then falling linearly to $0$ at $t=3.0$ hr. Target: attenuate the peak to $Q_{\text{out}}=1.0\ \text{m}^3/\text{s}$ (half of $Q_p$). Pond footprint limited to $0.40\ \text{ha}=4{,}000\ \text{m}^2$; outflow via an overflow weir.

Find. (a) Total runoff depth for the storm. (b) A qualitative sketch of the attenuated outflow hydrograph. (c) The active storage volume needed to cut the peak in half.

Approach. (a) The runoff depth is simply the total hydrograph volume (area under the triangle) divided by the watershed area. (c) For a simple triangular inflow, the active storage needed to shave the peak down to a target outflow is well approximated by the area of the inflow hydrograph lying above the target outflow line — the classic "peak-shaving" construction, since by continuity every cubic metre of inflow above the outflow line must be temporarily stored until the recession limb drops the inflow back below that line.

1. (a) Total runoff volume and depth. The triangular hydrograph's area is $\tfrac12\times\text{base}\times\text{height}$, with the base being the full 3-hour duration: $$V=\tfrac12(3\times3600\ \text{s})(2.0\ \text{m}^3/\text{s})=10{,}800\ \text{m}^3$$ $$P_e=\frac{V}{A}=\frac{10{,}800}{1{,}000{,}000}=0.0108\ \text{m}=\boxed{10.8\ \text{mm}}$$
2. (c) Storage triangle above the $Q_{\text{out}}=1.0\ \text{m}^3/\text{s}$ threshold. On the rising limb $Q=2.0t$ ($0\le t\le1$ hr), so $Q=1.0$ at $t=0.5$ hr; on the falling limb $Q=2.0-1.0(t-1)$ ($1\le t\le3$ hr), so $Q=1.0$ at $t=2.0$ hr. The portion of the inflow triangle above the threshold is itself a smaller triangle, base $=2.0-0.5=1.5$ hr and height $=2.0-1.0=1.0\ \text{m}^3/\text{s}$: $$V_s=\tfrac12(1.5\ \text{hr}\times3600)(1.0\ \text{m}^3/\text{s})=\boxed{2{,}700\ \text{m}^3}$$ Discussion: at the maximum permitted pond footprint (4,000 m$^2$), this storage needs an average active depth of only $2{,}700/4{,}000=0.675$ m — a modest, entirely achievable pond depth, confirming the 0.40 ha area limit does not by itself force the design to fail.

(b) Outflow sketch. With storage inserted upstream of the weir, continuity ($dS/dt=I-O$) means the outflow hydrograph cannot rise as fast as the inflow: the outflow starts at 0, climbs more gradually, and does not reach its peak until the inflow and outflow rates are momentarily equal on the inflow's falling limb (the instant of maximum storage, at $t=2.0$ hr in this case, where the outflow curve is drawn crossing the inflow's descending leg) — not at the inflow's own peak time of 1.0 hr. From there the outflow recedes on its own, slower schedule as the pond drains through the weir, so it returns to zero well after the inflow does (beyond $t=3.0$ hr). The net effect on the sketch: a lower, later, broader outflow bump that trades peak height for a longer recession tail, with the shaded area between the two curves up to their crossing point representing the stored volume computed in part (c).