Horton infiltration – cumulative infiltration under continuous ponding For Horton’s parameters f0 = 7.62 cm/hour, fc = 1.34 cm/hour, and k = 4.182 per hour, compute the cumulative infiltration F (cm) after t = 2 hours, assuming continuous ponding over the soil surface.

Introduction:
Horton’s infiltration model describes how infiltration capacity decays exponentially from an initial value toward a constant capacity as rainfall continues. For design, we often need the cumulative infiltrated depth under continuous ponding, which integrates the capacity curve over time.

Given Data / Assumptions:

• Initial capacity f0 = 7.62 cm/h, final capacity fc = 1.34 cm/h, decay constant k = 4.182 h^-1.
• Surface is ponded (rainfall intensity ≥ infiltration capacity for the duration).
• Duration t = 2 h; soil is homogeneous during the event.

Concept / Approach:

Horton capacity: f(t) = fc + (f0 − fc) * exp(−k t). The cumulative infiltration under continual ponding is the integral from 0 to t, yielding F(t) = fc * t + (f0 − fc) * (1 − exp(−k t)) / k. Substitute values with consistent units to obtain the infiltrated depth in centimeters.

Step-by-Step Solution:

Verification / Alternative check:

Since k is large, the capacity decays quickly toward fc, so most of the 2-h total is the steady component (≈2.68 cm), plus a modest transient (≈1.5 cm), consistent with 4.18 cm.

Why Other Options Are Wrong:

2.68 cm and 1.34 cm count only one component each; 1.50 cm is only the transient term; 6.52 cm overestimates by summing incorrectly or assuming constant f0.

Common Pitfalls:

Confusing infiltration capacity with actual infiltration under non-ponded rainfall; neglecting unit consistency or forgetting to include both terms of the integral.

Final Answer:

4.18 cm