Runoff and infiltration volume, Horton f=15+50e^-0.9t

Given: $f_p = 15 + 50e^{-0.9t}$ mm/h; rainfall intensity $i = 45$ mm/h for 10 h; area $= 12$ km² $= 12\times10^6$ m².

(iii) Time when runoff starts

Initially $f_p(0) = 65 > 45$, so all rain infiltrates. Runoff starts when $f_p$ falls to $i$:

$$15 + 50e^{-0.9t_p} = 45 \Rightarrow e^{-0.9t_p} = 0.6 \Rightarrow t_p = \frac{\ln(1/0.6)}{0.9} = 0.568\ \text{h} \;(\approx 34\ \text{min})$$

(ii) Total infiltration in 10 h

Up to $t_p$ infiltration equals rainfall; after $t_p$ it occurs at capacity:

$$F = i\,t_p + \int_{t_p}^{10}(15 + 50e^{-0.9t})dt = 45\times0.568 + 15(10 - 0.568) + \frac{50}{0.9}\left(e^{-0.9t_p} - e^{-9}\right)$$ $$F = 25.54 + 141.49 + 33.33 = 200.35\ \text{mm}$$ $$V_{inf} = 200.35\times10^{-3}\times12\times10^6 = 2,404,248\ \text{m}^3 \approx 2.40\times10^6\ \text{m}^3$$

(i) Total runoff volume

Rainfall $= 45\times10 = 450$ mm.

$$R = 450 - 200.35 = 249.65\ \text{mm}, \qquad V_R = 2,995,752\ \text{m}^3 \approx 3.00\times10^6\ \text{m}^3$$

(iv) Infiltration curve

 mm/h
 65 |*
    | *
 45 |--+--------------------- rainfall i = 45
    |   *  <- runoff starts, t = 0.57 h
    |     *
 15 |         *_______________ fc = 15
    +-----------------------------> t (h)
      0.57                    10
   (area under fp curve after tp = infiltration;
    area above fp, below i line = runoff)

Answer: (i) Runoff volume = 3.00 × 10⁶ m³ (249.6 mm); (ii) infiltration volume = 2.40 × 10⁶ m³ (200.4 mm); (iii) runoff starts 0.57 h (about 34 min) after the start of rain.