Cumulative Infiltration from Horton Curve
Mediuminfiltrationintegrationlosses
Under continuously ponded conditions with enough water supply, infiltration capacity is f(t) = 5 + 20e^(−t) mm/h, with t in hours. How much water infiltrates during the first hour?
- A12.36 mm
- B25.00 mm
- C17.64 mm
- D12.64 mm
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Answer: C. 17.64 mm
Depth is the integral of the rate: F = ∫₀¹(5 + 20e^(−t))dt = 5(1) + 20(1 − e^(−1)) = 5 + 20(0.6321) = 5 + 12.64 = 17.64 mm. 12.36 mm is f(1) = 5 + 20e^(−1), the end-of-hour rate (mm/h) reported as if it were a depth. 25.00 mm is f(0) held for one hour. 12.64 mm integrates only the exponential term and omits the constant 5 mm/h.
mm
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