# Horton's Infiltration Model

advertisement ```7.3
7.3
Horton's Infiltration
Model
181
HORTON'S INFilTRATION MODEL
The infiltration process was thoroughly studied by Horton in the early 1930s [4). An
outgrowth of his work, shown graphically in Fig. 7.1, was the following relation for
determining infiltration capacity:
/p
where
= /c
(7.1)
+ (fo - /c)e-kt
/p = the infiltration capacity (depth/time) at some time t
k = a constant representing the rate of decrease in / capacity
/ c = a final or equilibrium capacity
/0 = the initial infiltration capacity
It indicates that if the rainfall supply exceeds the infiltration capacity, infiltration tends
to decrease in an exponential manner. Although simple in form, difficulties in determining useful values for /0 and k restrict the use of this equation. The area under the
curve for any time interval represents the depth of water infiltrated during that
interval. The infiltration rate is usually given in inches per hour and the time t in minutes, although othdr time increments are used and the coefficient k is determined
accordingly.
By observing the variation of infiltration with time and developing plots of/versus tas shown in Fig. 7.5, we can estimate /0 and k. Two sets of / and t are selected from
the curve and entered in Eq. 7.1. Two equations having two unknowns are thus
obtained; they can be solved by successive approximations for /0 and k. .
Typical infiltration rates at the end of 1 hr (fl) are shown in Table 7.1. A typical
relation between /1 and the infiltration rate throughout a rainfall period is shown
graphically in Fig. 7.6a; Fig.7.6b shows an infiltration capacity curve for normal
antecedent conditions on turf. The data given in Table 7.1 are for a turf area and must
be multiplied by a suitable cover factor for other types of cover complexes. A range of
cover factors is listed in Table 7.2.
Total volumes of infiltration and other abstractions from a given recorded rainfall are obtainable from a discharge hydro graph (plot of the streamflow rate versus
time) if one is available. Separation of the base flow (dry weather flow) from the discharge hydrograph results in a direct runoff hydrograph (DRH), which accounts for
the direct surface runoff, that is, rainfall less abstractions. Direct surface runoff or
precipitation excess in inches uniformly distributed over a watershed can readily be
calculated by picking values of DRH discharge at equal time increments through the
TABLE 7.1
Typical f1 Values
Soil group
High (sandy soils)
~
Intermediate
(loams, clay, silt)
Low (clays, clay loam)
f1 (in./hr)
f1 (=/h)
0.50-1.00
0.10-0.50
0.01-0.10
12.50-25.00
2.50-12.50
0.25-2.50
Source: After ASCE Manual of Engineering Practice, No. 28.
182
Infiltration
Chapter 7
6
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2.8
2.6
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r-
f= 0.53 + 2.47 e-O.0697tf
F = 0.00883t f = 0.59(1 - e-O.0697tf)
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Infiltration capacitycurve (f)
-
/'
10 20 30 40 50 60 70 80
100
120
140
160
180
Time (min) from beginning of infiltration capacity curve, tf
(b)
FIGURE7.6
(a) Typical infiltration curve. (I)) Infiltration capacity and mass curves for normal
antecedent conditions of turf areas.
[After A. L. Tholin and Clint J Kiefer, "The Hydrology of Urban Runo!f,"Proc. ASCE 1.Sanitary
Eng. Div. 84(SA2), 56 (Mar. 1959).}
j'
7.3 Horton's Infiltration Model
183
~
TABLE7.2
Cover Factors
Permanent forest and grass
Close-growing crops
Row crops
Cover
Cover factor
Good (1 in.humus)
Medium (i-1 in.humus)
Poor «i in.humus)
Good
Medium
Poor
Good
Medium
Poor
3.0-7.5
2.0-3.0
1.2-1.4
2.5-3.0
1.6-2.0
1.1-1.3
1.3-1.5
1.1-1.3
1.0-1.1
Source: After ASCE Manual of Engineering Practice, No. 28.
hydrograph and applying the formula :
(O.03719)(k qD
Pe=
A nd
where
(7.2)
Pe = precipitation excess (in.)
qi = DRH ordinates at equal time intervals (cfs)
A ;", drainage area (mi2)
nd =. number of time intervals in a 24-hr period
For most cases the difference between the original rainfall and the direct runoff
canbe considered as infiltrated water. Exceptions may occur in areas of excessive subsurface drainage or tracts of intensive interception potential. The calculated value of
infiltration can then be assumed as distributed according to an equation of the form of
Eq. 7.1 or it may be uniformly spread over the storm period. Choice of the method
employed depends on the accuracy requirements and size of the watershed.
To circumvent some of the problems associated with the use of Horton's infiltration model, some adjustments can be made . Consider Fig. 7.5. Note that where the
infiltration capacity curve is above the hyetograph, the actual rate of infiltration is
equal to that of the rainfall intensity, adjusted for interception, evaporation, and other
losses. Consequently, the actual infiltration is given by:
'(7.3)
f(t) = min [fp(t), i(t)]
..
where f(t) is the actual infiltration into the soil and i(t) is the rainfall intensity. Thus the
infiltration rate at any time is equal to the lesser of the infiltration capacity fp(t) or the
rainfall intensity.
.
Commonly, the typical values of fo and fc are greater than the prevailing rainfall
intensities during a storm. Thus, when Eq. 7.1 is solved for fp as a function of time
alone, it shows a decrease in infiltration capacity even when rainfall intensities are
much less than fF' Accordingly, a reduction in infiltration capacity is made regardless
of the amount of water that enters the soil.
.;,
184
Chapter 7
Infiltration
To adjust for this deficiency, the integrated form of Horton's equation may be ~
used:
.
:
F(tp) = ltPfp dt = fctp + fa
fc (1 - e-ktp)
(7.4)
where F is the cumulative infiltration at time tp' as shown in Fig. 7.7. In the figure, it is
assumed that the actual infiltration has been equal to fF' As previously noted, this is
not usually the case, and the true cumulative infiltration must be determined. This can
be done using:
F(t)
=
ltf(t)
(7.5)
dt
where f(t) is determined using Eq. 7.3.
Equations 7.4 and 7.5 may be used jointly to calculate the time tp, that is, the
equivalent time for the actual infiltrated volume to equal the volume under the infiltration capacity curve (Fig. 7.7). The actual accumulated infiltration given by Eq. 7.5 is
equated to the area under the Horton curve, Eq. 7.4, and the resulting expression is
solved for tp' This equation:
F = f C tp + fo-
L
fC ( l - e-ktp)
(7.6)
cannot be solved explicitly for tp' but an iterative solution can be obtained. It should be
understood that the time tp is less than or equal to the actual elapsed time t. Thus the
available infiltration capacity as shown in Fig. 7.7 is equal to or exceeds that given by
fa
~
f(tp)
fp
FIGURE7.7
Cumulative infiltration.
°0
tp tp1 t1
Equivalent time
7.3 Horton's Infiltration Model
185
Eq. 7.1. By makip.g the adjustments described, fp becomes a function of the actual
amount of water infiltrated and not just a variable with time as is assumed in the original Horton equation.
In selecting a model for use in infiltration calculations, it is important to know its
limitations. In some cases a model can be adjusted to accommodate shortcomings; in
other cases, if its assumptions are not realistic for the nature of the use proposed, the
model should be discarded in favor of another that better fits the situation.
The first eight chapters of this book deal with the principal components of the
hydrologic cyc1~. In later chapters, the emphasis is on putting these components
together in various hydrologic modeling processes. When these models are designed
for continuous simulation, the approach is to calculate the appropriate components of
the hydrologic equation, Eq. 1.4, continuously over time. A discussion of how infiltration could be incorporated into a simulation model follows. It exemplifies the use of
Horton's equation in a storm water management model (SWMM) [6J.
First, an initial value of t p is determined. Then, considering that the value of fp
depends on the actual amount of infiltration that has occurred up to that time, a value
of the average infiltration capacity, ~, available over the next time step is calculated
using:
-
. fp
=
~
M
l
tl=tp+At
tp
dt
fp
=
F(tl) - F(tp)
M
(7.7)
Equation 4.3 is then used to find the average rate of infiltration, l:
1 = { !p
i
if ~ 2=lp
if i < fp
(7.8)
where i is the average rainfall intensity over the time step.
Following this, infiltration is incremented using the expression:
F(t + At) = F(t) + AF = F(t) +
1 At
(7.9)
1
where AF = At is the added cumulative infiltration (Fig. 7.7).
The next step is to find a new value of tpo This is done using Eq. 7.6. If
SF = ~M,tpl = tp + M.Butifthenewtplislessthantp
+ M(seeFig.7.7),Eq.7.6
must be solved by iteration for the new value of tpoThis can be accomplished using the
Newton-Raphson procedure .
When the value of tp 2= 16jk, the Horton curve is approximately horizontal and
fp = f C' Once this point has been reached, there is no further need for iteration since
fp is constant and equal to f c and no longer dependent on F.
Example 7.1
Given an initial infiltration capacity fa of 2.9 in./hr and a time constant k of 0.28 hr-\
derive an infiltration capacity versus time curve if the ultimate infiltration capacity is
0.50 in./hr. For the first 8 hours, estimate the total volume of water infiltrated in inches
over the watershed.
186
Infiltration
Chapter 7
Solution:
1.
Using Horton's equation (Eq. 7.1), values of infiltration can be computed for
various times. The equation is:
= Ie + (fo - le)e-kt
f
2.
Substituting the appropriate values into the equation yields:
I =
3.
+ (2.9- 0.50)e-o.28t
0.50
For the times shown in Table 7.3,values of I are computed and entered into the
table. Using a spreadsheet graphics package, the curve of Fig. 7.8 is derived.
TABLE7.3
Calculations for Example 7.1
Time
(hr)
Infiltration
(in./hr)
Time
(hr)
Infiltration
(in./hr)
0
0.10
0.25
0.50
1.00
2.00
3.00
4.00
2.90
2.83
2.74
2.59
2.31
1.87
1.54
1.28
5.00
6.00
7.00
8.00
9.00
10.00
15.00
20.00
1.09
0.95
0.84
0.76
0.69
0.65
0.54
0.51
4
.'
.......
~~ 3
~
Q)
~~ 2
gco
t: 1
~
FIGURE7.8
Infiltration
4.
00
curve for Example
7.1.
4
8.
12
Time (hr)
16
20
To find the volume of water infiltrated during the first 8 hours, Eq. 7.1 can be
integrated over the range of 0-8:
V = f[0.50 + (2.9 - 0.50)e-o.28t]dt
V
= [0.5t + (2.40 - O.28)e-o.281S
V = 11.84in.
The volume over the watershed is thus 11.84 in.
i
7.10
S
Phi Index
203
= [(1000)/(78)] - 10 = 2.82
2. Equation 7.27is then used to estimate Q:
Q
= [8 -
(0.2 X 2.82)]2/[8 + (0,8 X 2.82)] = 5.40 inches of runoff
Using Fig. 7.14 and interpolating between curve numbers, we get Q = about
5.4 inches of runoff, the same result.
7.10
PHI INDEX
Infiltration indexes generally assume that infiltration occurs at some constant or average rate throughout a storm. Consequently, initial rates are underestimated and final
rates are overstated if an entire storm sequence with little antecedent moisture is considered. The best application is tolarge storms on wet soils or storms where infiltration
rates may be assumed to be relatively uniform .
The most common index is termed the phi (<J» index for which the total volume
of the storm period loss is estimated and distributed uniformly across the storm pattern. Then the volume of precipitation above the index line is equivalent to the runoff
(Fig. 7.16). A variation is the W index, which excludes surface storage and retention.
Initial abstractions are often deducted from the early storm period to exclude initial
depression storage and wetting.
4.0
';:;' 3.0
,e
~>:\::
'"
~
Q)
.S 2.0
:§
~
'",
~
0
1.0
2.0
Time (hr)
3.0
4.0
FIGURE 7.16
Representation
of a <I>index.
204
Chapter 7
Infiltration
To determine the
<I>
index for a given storm, the. amount of observed runoff is
determined from the hydro graph, and the difference between this quantity and the
total gauged precipitation is then calculated. The volume of loss (including the effects
of interception, depression storage, and infiltration) is distributed uniformly across the
storm pattern as shown in Fig. 7.16.
Use of the <I> index for determining the amount of direct runoff from a given
storm pattern is essentiallythe reverse of this procedure. Unfortunately, the <I> index
determined from a single storm is not generally applicable to other storms, and unless
it is correlated with basin parameters other than runoff,it is of little value.
.
SUMMARY
Infiltration is a significant component of hydrologic processes. Soils have varying
capacities to infiltrate water. Influencing factors include soil type, degree of saturation,
and nature of ground cover. Activities that change the soil surface or alter its properties also have a modifying effect.
When the rainfall intensity is less than the infiltration capacity, all of the water
reaching the ground can infiltrate. But if the rainfall intensity exceeds the infiltration
capacity, infiltration will occur only at the infiltration capacity rate, and water in
excess of that capacity will be stored in depressions, become surface runoff, or evaporate. In general, the initial infiltration capacity of a dry soil is high. As rainfall continues, and as the soil becomes saturated, it diminishes to a relatively constant rate
(ultimate capacity).
Infiltration rates have been determined for a variety of soils and ground cover
conditions. A number of equations have been developed to serve as models for the
infiltration process. They are exemplified by Eqs. 7.1, 7.11, and 7.16.
PROBLEMS
7.1 Given an initial infiltration capacity fa of 3.0 in.lhr and a time constant k of 0.29hr-1,
derive an infiltration capacity versus time curve if the ultimate infiltration capacity is
0.55 in.lhr. For the first 10 hours, estimate the total volume of water infiltrated in inches
over the watershed.
7.2 Gross rain intensities during each hour of a 5-hr storm ove~ a 1000-acre basin were 5,4,1,
3, and 2 in.lhr, respectively. The direct surface runoff from the basin was 375 acre-ft.
.
Determine the basin <j> index.
7.3 The infiltration rate for excess rain on a small area was observed to be 4.5 in.lhr at the
beginning of rain, and it decreased exponentially toward an equilibrium of 0.5 in.lhr. A
total of 30 in. of water infiltrated during a 10-hr interval. Determine the value of k in
Horton'sequationf
= fe + (fa - fc)e-kt.
7.4 Rework Problem 7.2 if the storm occurred over a basin of 2.5 km2; the rainfall intensities
were 12, 10, 3, 8, and 5 cm/hr; and the direct surface runoff was 46~,000 m3.
7.5 Rework Problem 7.3 if the initial infiltration capacity was 10 cm/hr, the ultimate capacity
was 1.2 cm/hr, and a total of 33 em of water infiltrated during the 10-hr interval.
7.6 Precipitation falls on a 500-acre drainage basin according to the following schedule:
Problems
30-min period
Intensity (in./hr)
1
4.0
2
2.0
3
6.0
205
4
5.0
.
a. Determine the total storm rainfall (in inches).
b. Determine the q, index for the basinif the net storm rain is 3.0in.
7.7 The direct surface runoff volume from a 4.40-mi2 drainage basin is determinec! by plariimeter
from the area under the hydrograph to be 10,080 cfs-hr.The hydrograph was produced by a
l.71-in./hr rainstorm with a duration of 5 hr. Determine (a) the net rain and (b) the q, index.
7.8 The following table lists the storm rainfall data and infiltration capacity data for a 24-hr
storm begifming at midnight on April 14 ofthe current year.
a. Plot the rainfali hyetograph and the f capacity curve on rectangular coordinate paper.
b. Determine the total storm precipitation in inches.
c. By counting squares or by planimeter, d,etermine the net storm rain by the f capacity
method.
Rainfall Data for a Hypothetical Storm on April 15 of the Currnt Year
(Beginning at midnight on April 14)
Hour
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16"
17
18
19
20
21
22
23
24
Rainfall
intensity
(in./hr)
Infiltration capacity
at beginning of hour
(in.!hr)
Hourly deduction for
depression storage
(in./hr)
0.41
0.49
0.32
0.31
0.22
0.08
0.07
0.09
0.08
0.06
0.11
0.12
0.15
0.23
0.28
0.26
0.21
0.09
0.07
0.06
0.03
0.02
0.01
0.01
0.200
0.160
0.125
0.100
0.085
0.070
0.065
0.057
0.052
0.047
0.044
0.040
0.037
0.036
0.035
0.034'
0.033
0.033
0.033
0.032
0.032
0.032
0.031'
0.031
0.20
0.14
0.04
0.02
0.00
0.00
206
Chapter 7
Infiltration
7.9 Tabulated below are total rainfall intensities during each hour of a frontal storm over a
drainage basin.
a. Plot the rainfall hyetograph (intensity versus time).
b. Determine the total storm precipitation amount in inches.
c. If the net storm rain is 2.00 in., determine the exact q, index (in./hr) for the drainage
basin. (Note that by definition the area under the hyetograph above the q, index line
must be 2.00 in.)
d. Determine the area of the drain&ge basin (acres ).if the net rain is 2.00 in. and the measured volume of direct surface runoff is 2,015 cfs~hr.
e. Using the q, index calculated in part c, determine the volume of direct surface runoff
(acre-ft) that would result from the following:
Hour
1
Intensity (in./hr) 0.40
2
0.05
3
0.30
Hour
Rainfall intensity (in./hr)
I Hour.
Rainfall intensity (in./hr)
1
2
3
4
5
6
7
8
9
10
11
12
0.41
0.49
0.22
0.31
0.22
0.08
0.07
0.09
0.08
0.06
0.11
0.12
13
14
15
16
17
18
19
.20
21
22
23
24
0.15
0.23
0.28
0.26
0.21
0.09
0.07
0.06
0.03
0.02
0.01
0.01
,.
4
0.20
.
7.10 The SCS curve number method of estimating rainfall excess (net rain) is based on the
assun:ption that the curve numb~r for a watershed depends ~n sever~ factors. Name or
descnbe the factors that are consIdered when a curve number ISdetermmed.?:¥':
'1'1
.~
~m
~
7.U A 7-mi2 drainage basin has a cOmposite curve number CN of 50. According to the SCS,
exactly how much rain must fall before the direct runoff commences?
.
7.12 Determine a composite SCS runoff curve number for a 600-acre basin that is totally within
. soil group C.The land use is 40 percent contoured row crops in poor hydrologic condition
and 60 percent native
. pasture in . fair hydrologic condition.
~;.~
"x.~~
.~~
"'.,
7.13 Which SCS-c1assified soil would have the highest infiltration rate: A, B, C, or D?'~~
7.14 Rework Problem 7.11 if the drainage basin is 12 km2 and the CNs are 55 and 80. Give your
answers in centimeters.c;,~
7.15 Rework Problem 7.12 if the soil group is B and the land use is 50% straight-row crops
under poor conditions and 50% meadow under good conditions.
7.16 For the composite curve number found in Problem 7.15, estimate the amount of runoff if
the direct rainfall is 20 em.
!id
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