Double-Ring Infiltrometer — Fitting Horton's Equation: 22-Agric-B7 Principles of Hydrology, December 2015, Q4

Question 4: Double-Ring Infiltrometer — Fitting Horton's Equation (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. Double-ring infiltrometer, outer ring diameter 1.0 m, inner ring diameter 0.5 m (measurements are for the inner ring only, so edge effects from the outer buffer ring are excluded):

Interval	Water added (inner ring)
0–5 min	1.0 L
45–60 min	1.8 L
120–135 min	1.1 L
every 15 min after 6 hr (steady)	0.75 L

Find. (a) The infiltration-rate-vs-time curve on the supplied grid. (b) An infiltration equation with all variables/parameters and units identified, its parameters fitted to this soil, and the predicted rate at $t=30$ min.

Approach. Convert each measured volume into an average infiltration rate over its own interval using the inner ring's cross-sectional area, identify the long-time steady rate directly, then fit Horton's exponential decay to the transient measurements by log-linear regression and evaluate the fitted curve at $t=30$ min.

Inner-ring area. $$A_i=\frac{\pi D_i^2}{4}=\frac{\pi(0.5)^2}{4}=\boxed{0.1963\ \text{m}^2}$$
Convert each interval's added volume to an average infiltration rate, $f=\dfrac{V/A_i}{\Delta t}$ (depth per unit time). Each interval's midpoint is taken as the representative time: $$f(t\approx2.5\ \text{min})=\frac{1.0\ \text{L}/0.1963\ \text{m}^2}{5/60\ \text{hr}}=61.1\ \text{mm/hr}$$ $$f(t\approx52.5\ \text{min})=\frac{1.8/0.1963}{15/60}=36.7\ \text{mm/hr}\qquad f(t\approx127.5\ \text{min})=\frac{1.1/0.1963}{15/60}=22.4\ \text{mm/hr}$$ $$f_c=\frac{0.75/0.1963}{15/60}=\boxed{15.3\ \text{mm/hr}}$$ The final value, from the constant rate measured after 6 hours, is the steady (basic) infiltration rate $f_c$.
Horton's equation and its parameters. Horton's model is $$f(t)=f_c+(f_0-f_c)\,e^{-kt}$$ where $f(t)$ [mm/hr] is the infiltration capacity at elapsed time $t$ [min]; $f_0$ [mm/hr] is the initial infiltration rate at $t=0$; $f_c$ [mm/hr] is the final, constant (basic) infiltration rate as $t\to\infty$ (already found above); and $k$ [min$^{-1}$] is an empirical decay constant matched to the time units used for $t$. Linearizing, $\ln[f(t)-f_c]=\ln(f_0-f_c)-kt$ is a straight line in $t$; least-squares fitting it through the three transient points from Step 2 gives $$\boxed{k=0.0149\ \text{min}^{-1}\qquad f_0=62.5\ \text{mm/hr}}$$
Predict the rate at $t=30$ min. $$f(30)=15.3+(62.5-15.3)\,e^{-0.0149(30)}=15.3+47.2(0.638)=\boxed{45.5\ \text{mm/hr}}$$

Infiltration curve sketched on the paper's own grid (0–500 min, 0–60 mm/hr): the fitted Horton curve through the three measured average rates, decaying toward the steady rate $f_c=15.3$ mm/hr.

Quantity	Value
Inner-ring area, $A_i$	0.1963 m²
Rate 0–5 min (avg.)	61.1 mm/hr
Rate 45–60 min (avg.)	36.7 mm/hr
Rate 120–135 min (avg.)	22.4 mm/hr
Steady rate, $f_c$	15.3 mm/hr
Fitted $f_0$	62.5 mm/hr
Fitted $k$	0.0149 min⁻¹
Predicted $f(30\ \text{min})$	45.5 mm/hr

Concept & Theory

Topic. Double-ring infiltrometer testing and Horton's exponential infiltration-capacity model.

Key relations. $f(t)=f_c+(f_0-f_c)e^{-kt}$; linearized $\ln[f(t)-f_c]=\ln(f_0-f_c)-kt$.

Why this works. A double-ring infiltrometer floods an inner measurement ring surrounded by an outer buffer ring kept at the same head, so the buffer ring forces the inner ring's infiltration to be purely vertical (one-dimensional), eliminating the lateral-spreading error a single ring would have. Infiltration capacity starts high on dry soil (steep matric-suction gradient pulling water in) and declines as the soil profile wets up, surface macropores seal, and the process becomes gravity-dominated — exactly the exponential decay Horton's equation captures, with $f_c$ the rate once the wetting front's influence has vanished.

Common pitfall. Using the outer ring's larger area instead of the inner ring's in the rate calculation (the outer ring's water is only there to control lateral spreading, not to be measured), and reading $f_0$ directly off the first data point rather than extrapolating the fitted exponential back to $t=0$.

Source. Chow, Maidment & Mays, Applied Hydrology, Ch. 4 (infiltration, Horton and Green–Ampt models).

Question 4 of 6: Double-Ring Infiltrometer — Fitting Horton's Equation

Question 4: Double-Ring Infiltrometer — Fitting Horton's Equation (20 marks)