Question 5 of 6: Detention Pond Design — Weir Sizing and Storage Routing

Question 5: Detention Pond Design — Weir Sizing and Storage Routing (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. Triangular design inflow hydrograph (from the attached figure) and broad-crested weir outlet:

Find. (a) Weir length $L$. (b) A sketch of the routed outflow hydrograph. (c) Pond plan area at $H_{max}$. (d) Routed outflow discharge at $t=0.5$ hr.

Approach. Size the weir directly from the target peak outflow at the maximum permitted head, size the pond's storage from the excess volume between the inflow and the allowable outflow, then apply the storage-indication (level-pool) routing method — combining mass continuity with the now-known stage–storage and stage–discharge relationships — to find the outflow at $t=0.5$ hr.

1. (a) Weir length from the target peak condition. The pond's design intent is that the weir just passes $Q_{allow}=1.0\ \text{m}^3/\text{s}$ when the pond is at its maximum allowed head $H_{max}=0.25$ m: $$L=\frac{Q_{allow}}{C\,H_{max}^{3/2}}=\frac{1.0}{1.70(0.25)^{1.5}}=\frac{1.0}{1.70(0.125)}=\boxed{4.71\ \text{m}}$$
2. (c) Required storage volume, then pond area. The inflow hydrograph exceeds the allowable 1.0 m³/s outflow between the times it crosses that level on its rising limb, $I(t)=2t=1\Rightarrow t=0.5$ hr, and its falling limb, $I(t)=3-t=1\Rightarrow t=2.0$ hr. The volume that must be stored is the area of the excess triangle between the inflow and the 1.0 m³/s line over that 1.5-hour window, peaking at $2-1=1$ m³/s excess at $t=1$ hr: $$V_s=\tfrac12(2.0-0.5)(3600)(2-1)=\boxed{2700\ \text{m}^3}$$ Treating the pond as a basin of essentially constant plan area up to $H_{max}$ (a standard preliminary-sizing simplification, flagged below), the required area is $$A_{pond}=\frac{V_s}{H_{max}}=\frac{2700}{0.25}=\boxed{10800\ \text{m}^2}$$
3. (d) Storage-indication routing to $t=0.5$ hr. With the weir length and pond area now fixed, both storage and outflow are known functions of head $H$: $S(H)=A_{pond}H=10800H$ and $O(H)=CLH^{3/2}=8H^{3/2}$ (note $CL=1.70\times4.71=8.0$ exactly, consistent with part a). Continuity over the step $\Delta t=0.5$ hr $=1800$ s, $$\frac{2S_2}{\Delta t}+O_2=(I_1+I_2)+\left(\frac{2S_1}{\Delta t}-O_1\right)$$ starts from a pond assumed empty at $t=0$ ($S_1=0$, $O_1=0$) with $I_1=I(0)=0$ and $I_2=I(0.5)=2(0.5)=1.0\ \text{m}^3/\text{s}$, so the right-hand side is simply $1.0$. Substituting $S_2=10800(O_2/8)^{2/3}$ and solving the resulting single equation in $O_2$ numerically: $$12\left(\frac{O_2}{8}\right)^{2/3}+O_2=1.0\ \Rightarrow\ \boxed{O_2\approx0.151\ \text{m}^3/\text{s}}$$

(b) Outflow sketch (no calculation required by the question, shown here for consistency with the routed answer above). Because the pond must first fill before it can spill, the outflow hydrograph is delayed relative to the inflow: it stays at zero until the pond starts receiving inflow above the weir crest, rises much more gently than the sharp inflow peak, and reaches its own (attenuated) peak of about 1 m³/s only around $t\approx2.1$ hr — well after the inflow's own peak at $t=1$ hr. The recession is likewise stretched out, because the pond continues draining stored water long after the inflow has stopped at $t=3$ hr, so the outflow does not return to zero until roughly $t\approx4.4$ hr.