Question 5 of 7: Double-Ring Infiltrometer — Horton's Infiltration Model

Question 5: Double-Ring Infiltrometer — Horton's Infiltration Model (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.

Approach. Convert each measured volume increment to an infiltration rate using the inner ring's area (only the inner ring's reading is used — see part d), fit Horton's exponential decay f(t) = f_c + (f₀−f_c) e^−kt through the early, mid, and steady-state rates, then integrate it to get cumulative infiltration.

1. Inner ring area and rate conversions. A_ring = (π/4)(30)² = 706.9 cm². Converting each volume increment to a depth over that area and dividing by its time interval gives the rate at that time: $$f_0 \approx \frac{21.9/706.9}{1\ \text{min}}\times 60 = \boxed{1.86\ \text{cm/hr}}\ (t\approx0)$$ $$f(17.5\,\text{min}) = \frac{32/706.9}{5\ \text{min}}\times 60 = 0.543\ \text{cm/hr}$$ $$f_c = \frac{29.5/706.9}{5\ \text{min}}\times 60 = \boxed{0.501\ \text{cm/hr}}\ (t\to\infty)$$
2. b) Decay constant. Using the first-minute rate as f₀ (t ≈ 0) and the 15–20 min rate at its interval midpoint (t = 17.5 min = 0.2917 hr) in Horton's equation: $$k = \frac{1}{t}\ln\!\left(\frac{f_0-f_c}{f(t)-f_c}\right) = \frac{1}{0.2917}\ln\!\left(\frac{1.86-0.501}{0.543-0.501}\right) = \boxed{11.9\ \text{hr}^{-1}}$$ The three rates decrease monotonically (1.86 → 0.543 → 0.501 cm/hr) exactly as Horton's model predicts, which is a useful internal check on the fit.
3. c) Cumulative infiltration, 0–20 min. Integrating Horton's equation from 0 to t gives F(t) = f_ct + [(f₀−f_c)/k] (1−e^−kt). With t = 20 min = 0.3333 hr: $$F(20\,\text{min}) = (0.501)(0.3333) + \frac{1.86-0.501}{11.9}\left(1-e^{-11.9(0.3333)}\right) = 0.167 + 0.1142(0.981) = \boxed{0.279\ \text{cm}}$$ Over the inner ring's area this is a volume of $$(0.279\ \text{cm})(706.9\ \text{cm}^2) \approx \boxed{197\ \text{cm}^3}$$

d) Purpose of the outer ring. The outer ring is a hydraulic buffer: keeping it ponded at the same head as the inner ring suppresses the lateral (radial) spreading of water beneath the inner ring's edges, forcing flow under the inner ring to be essentially one-dimensional and vertical. Without it, water escaping sideways under the inner ring's walls would make the inner-ring reading over-estimate true vertical infiltration capacity.

e) Other methods for infiltration parameters. Single-ring infiltrometer (simpler, but subject to the edge-effect the double ring avoids), tension/disc infiltrometer (applies a controlled negative head to exclude macropore-dominated flow), rainfall/sprinkler simulators, the Guelph permeameter (constant-head well method, yields field-saturated K), rainfall–runoff hydrograph separation at the watershed scale, and fitting alternative models (Philip's two-term equation, Green–Ampt) to ponding-test data.