Question 3 of 5: Peak discharge in an urban stormwater system

Question 3: Peak discharge in an urban stormwater system (25 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. Three subcatchments draining to three inlets; Pipe 1 (L=500 ft, r=1 ft, s=0.005 ft/ft, n=0.011) connects the Inlet 1/2 junction to Inlet 3; Pipe 2 carries the total onward. 50-yr Baltimore IDF (Figure 1).

Find. Peak runoff at each inlet and the peak flow plus velocity in Pipes 1 and 2.

Approach. Apply the rational method $Q=CiA$ at each inlet using its own $t_c$; for the pipes accumulate the contributing $\sum CA$, take the intensity at the controlling (longest) time of concentration, then use Manning at the assumed 80%-perimeter section for velocity.

1. Inlet 1 (Part 1). With $t_c=5$ min, the 50-yr curve of Figure 1 gives $i\approx9.9$ in/hr (the curve leaves the left-hand axis just below the 10 in/hr gridline). In US customary units $Q\,[\text{cfs}]=C\,i\,[\text{in/hr}]\,A\,[\text{acre}]$ (the unit factor is 1.008 ≈ 1). $Q_1 = 0.5\times9.9\times1.5 = \boxed{7.43\ \text{cfs}}$.
2. Inlet 2 (Part 2). $t_c=7$ min $\Rightarrow i\approx8.9$ in/hr. $Q_2 = 0.7\times8.9\times1.5 = \boxed{9.35\ \text{cfs}}$.
3. Inlet 3 (Part 3). $t_c=15$ min $\Rightarrow i\approx6.9$ in/hr (the 50-yr curve sits clearly above the 6 in/hr gridline at 15 min). $Q_3 = 0.6\times6.9\times2.5 = \boxed{10.35\ \text{cfs}}$.
4. Pipe 1 flow (Part 4). Pipe 1 carries the combined runoff of subcatchments 1 and 2. The controlling time of concentration is the longer of the two, $t_c=7$ min, so $i=8.9$ in/hr. With $\sum CA = 0.5(1.5)+0.7(1.5)=1.80$, the peak flow is $Q_{p1}=i\sum CA = 8.9\times1.80 = \boxed{16.0\ \text{cfs}}$.
5. Pipe 1 velocity (Part 4). With $r=1$ ft, the full circumference is $2\pi r=6.28$ ft; the wetted perimeter is $P=0.8(6.28)=5.03$ ft, i.e. a wetted central angle $\theta=0.8(2\pi)=5.03$ rad. Flow area $A=\pi r^2-\tfrac12 r^2(\alpha-\sin\alpha)=3.14-0.15=2.99\ \text{ft}^2$ (dry cap angle $\alpha=2\pi-\theta=1.257$ rad), so $R=A/P=2.99/5.03=0.595$ ft. Manning (US): $V=\dfrac{1.486}{n}R^{2/3}s^{1/2}=\dfrac{1.486}{0.011}(0.595)^{2/3}(0.005)^{1/2}=\boxed{6.75\ \text{ft/s}}$. The corresponding section capacity $Q=VA=6.75\times2.99=20.2$ cfs exceeds the 16.0 cfs inflow, so Pipe 1 is adequate; travel time $t=L/V=500/6.75=74\ \text{s}=1.23$ min.
6. Pipe 2 (Part 5). Pipe 2 carries all three subcatchments, $\sum CA = 1.80+0.6(2.5)=3.30$. The controlling $t_c$ is the longest path to Inlet 3: local subcatchment 3 gives 15 min, whereas the routed path is $7+1.23\approx8.2$ min — so 15 min governs and $i=6.9$ in/hr. Peak flow $Q_{p2}=6.9\times3.30 = \boxed{22.8\ \text{cfs}}$. Taking Pipe 2 as the same section and slope, $V_2\approx\boxed{6.75\ \text{ft/s}}$ and its capacity is only $\approx20.2$ cfs — less than the 22.8 cfs peak, so the 1-ft-radius pipe on $s=0.005$ is under-sized at this point and must be enlarged (or laid steeper) in design.