Question 4 of 6: Horton Infiltration and Effective Precipitation
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2014 — 04-Agric-B7, Principles of Hydrology (Soil Hydrology). Three-hour, open-book exam; any non-communicating calculator is permitted. Format: five questions constitute a complete paper, each of equal value; most questions require an answer involving calculations.
Find. (a) Horton infiltration rate $f(t)$ and cumulative infiltration $F(t)$ at $t=30$ min, assuming ponded conditions throughout. (b) The depth of effective precipitation for a 30-minute storm with a constant rainfall rate of 75 mm/hr. (c) A comparison of alternative infiltration-estimation methods.
Approach. Apply the standard Horton equation $f(t)=f_c+(f_0-f_c)e^{-kt}$ and its time-integral $F(t)=f_ct+\frac{f_0-f_c}{k}\left(1-e^{-kt}\right)$ for part (a). For part (b), because the given rainfall rate (75 mm/hr) is below the initial infiltration capacity, no runoff occurs until the Horton capacity curve decays down to meet the rainfall rate at a "ponding time" $t_p$; effective precipitation only accrues after that crossing.
(a) Infiltration rate and cumulative infiltration at $t=30$ min $=0.5$ hr.
$$f(0.5)=f_c+(f_0-f_c)e^{-k(0.5)}=10+140\,e^{-2.5}=10+11.49=\boxed{21.5\ \text{mm/hr}}$$
$$F(0.5)=f_c(0.5)+\frac{f_0-f_c}{k}\left(1-e^{-k(0.5)}\right)=10(0.5)+\frac{140}{5}(1-0.0821)=5+25.70=\boxed{30.7\ \text{mm}}$$
(b) Ponding time $t_p$ — when the Horton capacity curve decays to the 75 mm/hr rainfall rate.
$$75=10+140\,e^{-5t_p}\ \Rightarrow\ e^{-5t_p}=\frac{65}{140}\ \Rightarrow\ t_p=\frac{1}{5}\ln\!\left(\frac{140}{65}\right)=0.1535\ \text{hr}=9.2\ \text{min}$$
Before $t_p$, the rainfall rate is below the infiltration capacity, so every drop infiltrates and no runoff is generated; only after $t_p$ does actual infiltration follow the (now lower) Horton capacity curve, with the excess becoming effective precipitation.
(b, cont'd) Effective precipitation over the 30-minute storm. Total rainfall and total infiltration (rain-rate infiltrated before $t_p$, capacity-limited infiltration after):
$$P_{\text{total}}=75(0.5)=37.5\ \text{mm}$$
$$F_{\text{infil}}=75\,t_p+\big[F(0.5)-F(t_p)\big]=11.51+(30.70-16.53)=25.68\ \text{mm}$$
$$P_e=P_{\text{total}}-F_{\text{infil}}=37.5-25.68=\boxed{11.8\ \text{mm}}$$
Figure 4 — the Horton capacity curve decays from 150 to 10 mm/hr; it crosses the 75 mm/hr rainfall-rate line at $t_p=9.2$ min. No runoff occurs before that crossing; effective precipitation (part b) is the shaded gap between rainfall and capacity from $t_p$ to 30 min.
(c) Alternative infiltration-estimation methods. The Green–Ampt model derives infiltration from Darcy's law applied to a sharp wetting front, using physically measurable soil parameters (saturated hydraulic conductivity, porosity, wetting-front suction) rather than curve-fit constants — an advantage where those properties are known, but it idealizes the wetting front as a sharp step, which real soils rarely exhibit exactly. The SCS (NRCS) Curve Number method estimates total infiltration/abstraction indirectly from a single tabulated Curve Number based on soil group, land use and antecedent moisture; it is far simpler and more widely tabulated for ungauged watersheds, but it gives only a cumulative-storm abstraction rather than a true time-varying infiltration rate, and it was developed empirically from U.S. agricultural watersheds, so it can transfer poorly to very different climates or soils without local calibration. A simple $\phi$-index (a single constant loss rate that, subtracted from the hyetograph, reproduces the observed runoff volume) is the crudest and easiest to apply from measured rainfall–runoff pairs, but it ignores the physically real decline of infiltration capacity over the storm entirely, making it a calibration tool rather than a predictive model like Horton's.
Quantity
Result
(a) Infiltration rate at $t=30$ min, $f(0.5)$
21.5 mm/hr
(a) Cumulative infiltration at $t=30$ min, $F(0.5)$
30.7 mm
(b) Ponding time, $t_p$
9.2 min
(b) Effective precipitation, $P_e$
11.8 mm
Check: the source prints the decay-rate parameter as "5 /min." Taken literally, $k=5\ \text{min}^{-1}$ makes the Horton exponential collapse to the equilibrium rate within a fraction of a second, giving a physically impossible cumulative infiltration (over 300 mm in 30 minutes from a curve whose own peak rate is 150 mm/hr, i.e. a maximum possible 75 mm in that time). Typical Horton decay constants for this style of problem run 1–10 per hour; treating the printed unit as a misprint for $k=5\ \text{hr}^{-1}$ gives self-consistent, textbook-typical numbers (infiltration capacity falling smoothly from 150 to about 21 mm/hr over 30 minutes) and is the reading used throughout this solution.
