Question 2 of 4: Unit-Hydrograph S-Curve Transformation and Green–Ampt Infiltration

Question 2: Unit-Hydrograph S-Curve Transformation and Green–Ampt Infiltration (33.3 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. A triangular 1-hr unit hydrograph and a 2-hour design storm (Part 2.1); Green–Ampt soil-infiltration parameters (Part 2.2):

Find. (2.1a) The 2-hr unit hydrograph ordinates by the S-hydrograph method. (2.1b) The direct-runoff hydrograph for the 3 cm / 6 cm two-hour storm. (2.2a) The infiltration-rate curve (constant, rate-limited) up to the time of ponding and that time itself. (2.2b) The qualitative shape of the infiltration curve after ponding.

Approach. Build the S-curve by superposing the 1-hr UH lagged every hour to its equilibrium value, lag it by the new 2-hr duration and rescale to get the 2-hr UH (2.1a); convolve the original 1-hr UH with the two hourly rainfall-excess pulses for the storm hydrograph (2.1b); locate the Green–Ampt ponding time as the instant the rain-limited infiltration rate equals the declining potential infiltration capacity (2.2a), then reason qualitatively about the post-ponding decline (2.2b).

1. Part 2.1(a) — S-curve from the 1-hr UH. Discretising the triangular UH at hourly ordinates, $U(0)=0$, $U(1)=6.7$, $U(2)=6.7\times\tfrac{3-2}{2}=3.35$, $U(3)=0\ \text{m}^3/\text{s}$. The S-curve is the running sum of the UH lagged every 1 hr: $$S(t)=\sum_{n=0}^{\infty}U(t-n)$$ giving $S(0)=0$, $S(1)=U(1)=6.7$, and from $t=2$ hr onward the curve reaches its equilibrium value $$S(2)=U(2)+U(1)+U(0)=3.35+6.7+0=\boxed{10.05\ \text{m}^3/\text{s}}$$ (and $S(3)=S(4)=\dots=10.05\ \text{m}^3/\text{s}$, unchanged thereafter). Lagging the S-curve by the new 2-hr duration and rescaling by $t_R/t_R'=1/2$ gives the 2-hr UH ordinates $U_2(t)=\tfrac12\left[S(t)-S(t-2)\right]$: $$\begin{aligned}U_2(1)&=\tfrac12(6.7-0)=\boxed{3.35}\\U_2(2)&=\tfrac12(10.05-0)=\boxed{5.025}\\U_2(3)&=\tfrac12(10.05-6.7)=\boxed{1.675\ \text{m}^3/\text{s}}\end{aligned}$$ with $U_2(0)=U_2(4)=0$. The 2-hr UH spans a 4 hr base and peaks at $t=2$ hr, but it is no longer a simple triangle — its rising limb kinks at $t=1$ hr (slope 3.35 then 1.675 m³/s per hr) and its recession kinks at $t=3$ hr, which is the normal outcome of an S-curve duration change. Its area still integrates to the same 1 cm of excess spread over the (unchanged) drainage area as the original 1-hr UH, confirming the transformation.

2. Part 2.1(b) — Storm hydrograph by convolution with the 1-hr UH. With rainfall-excess pulses $P_1=3$ cm (hour 1) and $P_2=6$ cm (hour 2) applied to the ORIGINAL 1-hr UH (the unit pulse length matches the 1-hr UH, not the derived 2-hr UH), linearity gives $$Q(t)=P_1\,U(t)+P_2\,U(t-1)$$ $$Q(1)=3(6.7)+6(0)=\boxed{20.1\ \text{m}^3/\text{s}}$$ $$Q(2)=3(3.35)+6(6.7)=10.05+40.2=\boxed{50.25\ \text{m}^3/\text{s}}\quad(\text{peak, at }t=2\ \text{hr})$$ $$Q(3)=3(0)+6(3.35)=\boxed{20.1\ \text{m}^3/\text{s}}$$ $$Q(4)=0$$

3. Part 2.2(a) — Green–Ampt time to ponding. Working in cm/hr, $K_s=0.78$ cm/hr, $S_f=10$ cm, $\Delta\theta=\theta_s-\theta_i=0.45-0.18=\boxed{0.27}$. Before ponding, every drop of rain infiltrates (rate-limited), so the infiltration RATE curve is simply the constant rainfall rate: $$f(t)=i=2.9\ \text{cm/hr},\qquad 0\le t\le t_p$$ Ponding occurs once the accumulated infiltration $F$ has reduced the Green–Ampt potential capacity $f_p=K_s\!\left(1+\dfrac{S_f\Delta\theta}{F}\right)$ down to the rainfall rate $i$; solving $i=K_s(1+S_f\Delta\theta/F_p)$ for the cumulative infiltration at ponding, $$F_p=\frac{K_s\,S_f\,\Delta\theta}{i-K_s}=\frac{0.78\times10\times0.27}{2.9-0.78}=\frac{2.106}{2.12}=\boxed{0.993\ \text{cm}}$$ and, since $f=i$ up to that point, the time to ponding is $$t_p=\frac{F_p}{i}=\frac{0.993}{2.9}=\boxed{0.343\ \text{hr}\approx20.6\ \text{min}}$$
4. Part 2.2(b) — Shape of the curve after ponding (qualitative). Once $t>t_p$, the soil surface can no longer transmit rain as fast as it arrives, so the infiltration rate drops BELOW the rainfall rate and follows the full Green–Ampt capacity curve $f(t)=K_s\!\left(1+\dfrac{S_f\Delta\theta}{F(t)}\right)$: it starts at $f(t_p^+)=i=2.9$ cm/hr, falls steeply at first, and decays smoothly and asymptotically toward $K_s=0.78$ cm/hr as $t\to\infty$ and $F$ keeps growing — the same concave, Horton-like monotonic decline seen in every Green–Ampt curve, never below $K_s$ and never rising again.