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Infiltration depth from Horton's equation, 60-min storm

Short answer

Depth is obtained by integrating the infiltration capacity curve fp = 3 + e^{-2t} over time. First 30 minutes: 1.82 cm; second 30 minutes: 1.62 cm.

The infiltration capacity in a basin is represented by Horton's equation as fp=3.0+e−2tf_p = 3.0 + e^{-2t}, where fpf_p is in cm/hr and 't' in hours. Assuming the infiltration to take place at capacity rates in a 60 minutes storm, estimate the depth of infiltration in (i) the first 30 minutes and (ii) the second 30 minutes of the storm.

Infiltration occurs at capacity, so the depth of infiltration in any period is the integral of fpf_p.

F=∫t1t2(3.0+e−2t) dt=3.0 (t2−t1)+12(e−2t1−e−2t2)F = \int_{t_1}^{t_2} (3.0 + e^{-2t})\,dt = 3.0\,(t_2 - t_1) + \frac{1}{2}\left(e^{-2t_1} - e^{-2t_2}\right)

(i) First 30 minutes (t=0t = 0 to 0.50.5 h)

F1=3.0×0.5+0.5 (1−e−1)=1.5+0.5×0.6321=1.816 cmF_1 = 3.0\times0.5 + 0.5\,(1 - e^{-1}) = 1.5 + 0.5\times0.6321 = 1.816\ \text{cm}

(ii) Second 30 minutes (t=0.5t = 0.5 to 1.01.0 h)

F2=3.0×0.5+0.5 (e−1−e−2)=1.5+0.5×(0.3679−0.1353)=1.616 cmF_2 = 3.0\times0.5 + 0.5\,(e^{-1} - e^{-2}) = 1.5 + 0.5\times(0.3679 - 0.1353) = 1.616\ \text{cm}

Answer: Infiltration in the first 30 min = 1.816 cm; in the second 30 min = 1.616 cm (total in 60 min = 3.432 cm).

✓ Answer checked by Claude Sonnet 5.5

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