Horton Infiltration — Fitting the Decay Constant and Total 4-Hour Infiltration: 22-Agric-A2 Soil Physics and Mechanics, December 2015, Q5

Question 5: Horton Infiltration — Fitting the Decay Constant and Total 4-Hour Infiltration (20 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.

Quantity	Value
Initial infiltration rate, f₀	90 mm/hr
Final (constant) infiltration rate, f_c	10 mm/hr
Observation window	t = 30 to 35 min after test start
Water infiltrated in that window	1.3 mm

Find. (a) the shape of f(t) under Horton's model; (b) cumulative infiltration over the first 4 hours.

Approach. Horton's equation, $f(t) = f_c + (f_0-f_c)e^{-kt}$, has one unknown, the decay constant k, which is fixed by the single observed increment of cumulative infiltration; integrate f(t) to get the cumulative form $F(t) = f_c t + \frac{f_0-f_c}{k}\left(1-e^{-kt}\right)$, solve for k from the 30–35 min data, then evaluate F(4 h).

Horton infiltration-rate curve fitted to the double-ring data, decaying from f₀ = 90 mm/hr toward the asymptote f_c = 10 mm/hr.

Set up the cumulative-infiltration equation for the observed window. With $t_1 = 0.5\ \text{h}$, $t_2 = 0.5833\ \text{h}$, the increment $\Delta F = F(t_2)-F(t_1)$ must equal the observed 1.3 mm: $$\Delta F = f_c(t_2-t_1) + \frac{f_0-f_c}{k}\left(e^{-kt_1}-e^{-kt_2}\right) = 1.3$$ $$10(0.0833) + \frac{80}{k}\left(e^{-0.5k}-e^{-0.5833k}\right) = 1.3 \ \Rightarrow\ \frac{80}{k}\left(e^{-0.5k}-e^{-0.5833k}\right) = 0.4667$$
Solve for k numerically. This transcendental equation has no closed-form root; solving it numerically gives $$k = \boxed{4.92\ \text{hr}^{-1}}$$
b) Cumulative infiltration over 4 hours. With $e^{-4k}=e^{-19.7}\approx 0$, $$F(4) = f_c(4) + \frac{f_0-f_c}{k}\left(1-e^{-4k}\right) = 10(4) + \frac{80}{4.92}(1-0) = 40 + 16.26 = \boxed{56.3\ \text{mm}}$$

Quantity	Value
Fitted decay constant, k	4.92 hr⁻¹
Cumulative infiltration over 4 h, F(4)	56.3 mm

Concept & Theory

Topic: Horton's empirical infiltration-capacity model and its cumulative (integrated) form.

Key relations: $f(t) = f_c + (f_0-f_c)e^{-kt}$; $F(t) = f_ct + \dfrac{f_0-f_c}{k}\left(1-e^{-kt}\right)$.

Why this works: Infiltration capacity is highest on a dry soil (large matric-suction gradient pulling water down) and decays toward a constant rate once the surface layer saturates and gravity drainage dominates; Horton's model captures that with a single exponential decay between the two measured extremes, and only the rate constant k needs fitting from field data.

Common pitfall: Confusing the infiltration rate f(t) with the cumulative depth F(t) — the single 1.3 mm data point is a cumulative increment over a finite window, not an instantaneous rate, so it must be matched against ΔF, not against f(32.5 min); and forgetting the final e^−4k term is negligible only because k turns out to be large (a slow-draining soil with small k would keep contributing at t = 4 h and could not be dropped).

Source: G.O. Schwab et al., Soil and Water Conservation Engineering, 5th ed., Ch. 4 (infiltration, Horton's equation).

Question 5 of 7: Horton Infiltration — Fitting the Decay Constant and Total 4-Hour Infiltration

Question 5: Horton Infiltration — Fitting the Decay Constant and Total 4-Hour Infiltration (20 marks)