horton's infiltration model

11
7.3 Horton's Infiltration Model 181 7.3 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 = /c + (fo - /c)e-kt (7.1) where /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 deter- mining 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 min- utes, 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/ver- sus 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 rain- fall 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 dis- charge 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 f1 (in./hr) 0.50-1.00 0.10-0.50 0.01-0.10 f1 (=/h) 12.50-25.00 2.50-12.50 0.25-2.50 High (sandy soils) ~ Intermediate (loams, clay, silt) Low (clays, clay loam) Source: After ASCE Manual of Engineering Practice, No. 28.

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7.3 Horton's Infiltration Model 181

7.3 HORTON'S INFilTRATION MODEL

The infiltration process was thoroughly studied by Horton in the early 1930s [4).Anoutgrowth of his work, shown graphically in Fig. 7.1, was the following relation fordetermining infiltration capacity:

/p = /c + (fo - /c)e-kt (7.1)

where /p = the infiltration capacity (depth/time) at some time tk = 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 tendsto decrease in an exponential manner. Although simple in form, difficulties in deter-mining useful values for /0 and k restrict the use of this equation. The area under thecurve for any time interval represents the depth of water infiltrated during thatinterval. The infiltration rate is usually given in inches per hour and the time t in min-utes, although othdr time increments are used and the coefficient k is determinedaccordingly.

By observing the variation of infiltration with time and developing plots of/ver-sus tas shown in Fig. 7.5, we can estimate /0 and k. Two sets of / and t are selected fromthe curve and entered in Eq. 7.1. Two equations having two unknowns are thusobtained; 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 typicalrelation between /1 and the infiltration rate throughout a rainfall period is showngraphically in Fig. 7.6a; Fig.7.6b shows an infiltration capacity curve for normalantecedent conditions on turf. The data given in Table 7.1 are for a turf area and mustbe multiplied by a suitable cover factor for other types of cover complexes. A range ofcover factors is listed in Table 7.2.

Total volumes of infiltration and other abstractions from a given recorded rain-fall are obtainable from a discharge hydro graph (plot of the streamflow rate versustime) if one is available. Separation of the base flow (dry weather flow) from the dis-charge hydrograph results in a direct runoff hydrograph (DRH), which accounts forthe direct surface runoff, that is, rainfall less abstractions. Direct surface runoff orprecipitation excess in inches uniformly distributed over a watershed can readily becalculated by picking values of DRH discharge at equal time increments through the

TABLE 7.1 Typical f1 Values

Soil group f1 (in./hr)

0.50-1.000.10-0.500.01-0.10

f1 (=/h)

12.50-25.002.50-12.500.25-2.50

High (sandy soils)~ Intermediate (loams, clay, silt)

Low (clays, clay loam)

Source: After ASCE Manual of Engineering Practice, No. 28.

182 Chapter 7 Infiltration

6

.a::: 5'"~.g 4~u§'J:: 3

~'§ 220

.~. 1P::

------

0

3.02.82.6

, 2.4ci

Co 2.2

~ 2.08"0 1.8~~ 1.6~ 1.4

g 1.2>-'0 1.0[ 0.8'"u 0.6

0.4

0.2

O.10 20 30 40 50 60 70 80 100 120 140 160

Time (min) from beginning of infiltration capacity curve, tf

(b)

180

FIGURE7.6

(a) Typical infiltration curve. (I)) Infiltration capacity and mass curves for normalantecedent conditions of turf areas.

[After A. L. Tholin and Clint J Kiefer, "The Hydrology of Urban Runo!f,"Proc. ASCE 1.SanitaryEng. Div. 84(SA2), 56 (Mar. 1959).}

-

r-f= 0.53 + 2.47 e-O.0697tfF = 0.00883t = 0.59(1 - e-O.0697tf)f -

\ e(f>\ . '\'j c:"-l.7?>c>

'oc.?>\'..........."""\,?>\"

;\.\e\ ........-\ C. \f>.?>"o"o ........-

\ ?>\e .....-1" ....... .l'>-cC

\ ........""""'"" ./""""'"

,/...... Infiltration capacitycurve (f)

-.........

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I

j'7.3 Horton's Infiltration Model

~183

TABLE7.2 Cover Factors

Cover Cover factor

Permanent forest and grass Good (1 in.humus)Medium (i-1 in.humus)Poor «i in.humus)GoodMediumPoorGoodMediumPoor

Close-growing crops

Row crops

3.0-7.52.0-3.01.2-1.42.5-3.01.6-2.01.1-1.31.3-1.51.1-1.31.0-1.1

Source: After ASCE Manual of Engineering Practice, No. 28.

For most cases the difference between the original rainfall and the direct runoffcanbe considered as infiltrated water. Exceptions may occur in areas of excessive sub-surface drainage or tracts of intensive interception potential. The calculated value ofinfiltration can then be assumed as distributed according to an equation of the form ofEq. 7.1 or it may be uniformly spread over the storm period. Choice of the methodemployed depends on the accuracy requirements and size of the watershed.

To circumvent some of the problems associated with the use of Horton's infiltra-tion model, some adjustments can be made [6]. Consider Fig. 7.5. Note that where theinfiltration capacity curve is above the hyetograph, the actual rate of infiltration isequal to that of the rainfall intensity, adjusted for interception, evaporation, and otherlosses. Consequently, the actual infiltration is given by:

hydrograph and applying the formula [5]:

(O.03719)(k qDPe= A nd

where 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

'-f(t) = min [fp(t), i(t)]

(7.2)

..

(7.3)

where f(t) is the actual infiltration into the soil and i(t) is the rainfall intensity. Thus theinfiltration rate at any time is equal to the lesser of the infiltration capacity fp(t) or therainfall intensity. .

Commonly, the typical values of fo and fc are greater than the prevailing rainfallintensities during a storm. Thus, when Eq. 7.1 is solved for fp as a function of timealone, it shows a decrease in infiltration capacity even when rainfall intensities aremuch less than fF' Accordingly, a reduction in infiltration capacity is made regardlessof the amount of water that enters the soil.

.;,

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 infil-tration capacity curve (Fig. 7.7). The actual accumulated infiltration given by Eq. 7.5 isequated to the area under the Horton curve, Eq. 7.4, and the resulting expression issolved for tp' This equation:

F = f t + fo- fC (l - e-ktp)C p L (7.6)

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 isassumed that the actual infiltration has been equal to fF' As previously noted, this isnot usually the case, and the true cumulative infiltration must be determined. This canbe done using:

F(t) = ltf(t) dt(7.5)

cannot be solved explicitly for tp' but an iterative solution can be obtained. It should beunderstood that the time tp is less than or equal to the actual elapsed time t. Thus theavailable 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 actualamount of water infiltrated and not just a variable with time as is assumed in the origi-nal Horton equation.

In selecting a model for use in infiltration calculations, it is important to know itslimitations. In some cases a model can be adjusted to accommodate shortcomings; inother cases, if its assumptions are not realistic for the nature of the use proposed, themodel 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 thehydrologic cyc1~.In later chapters, the emphasis is on putting these componentstogether in various hydrologic modeling processes. When these models are designedfor continuous simulation, the approach is to calculate the appropriate components ofthe hydrologic equation, Eq. 1.4, continuously over time. A discussion of how infiltra-tion could be incorporated into a simulation model follows. It exemplifies the use ofHorton's equation in a storm water management model (SWMM) [6J.

First, an initial value of tp is determined. Then, considering that the value of fpdepends on the actual amount of infiltration that has occurred up to that time, a valueof the average infiltration capacity, ~, available over the next time step is calculatedusing:

- = ~ ltl=tp+At dt = F(tl) - F(tp). fp M fp M

tp

(7.7)

Equation 4.3 is then used to find the average rate of infiltration, l:

1 = {!p if ~ 2=lpi if i < fp

where i is the average rainfall intensity over the time step.Following this, infiltration is incremented using the expression:

(7.8)

F(t + At) = F(t) + AF = F(t) + 1At (7.9)

where AF = 1At 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.7.6must be solved by iteration for the new value of tpoThis can be accomplished using theNewton-Raphson procedure [6].

When the value of tp 2= 16jk, the Horton curve is approximately horizontal andfp = f C' Once this point has been reached, there is no further need for iteration sincefp 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 is0.50 in./hr. For the first 8 hours, estimate the total volume of water infiltrated in inchesover the watershed.

186 Chapter 7 Infiltration

Solution:

1. Using Horton's equation (Eq. 7.1), values of infiltration can be computed for ivarious times. The equation is:

f = Ie + (fo - le)e-kt

2. Substituting the appropriate values into the equation yields:

I = 0.50 + (2.9- 0.50)e-o.28t

3. For the times shown in Table 7.3,values ofI are computed and entered into thetable. Using a spreadsheet graphics package, the curve of Fig. 7.8 is derived.

4

.......

~ 3~~Q)

~ 2~

gco

t: 1~

.'

FIGURE7.8 00 4

Infiltration curve for Example 7.1.

8. 12

Time (hr)

16

4. To find the volume of water infiltrated during the first 8 hours, Eq. 7.1 can beintegrated 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.

20

TABLE7.3 Calculations for Example 7.1

Time Infiltration Time Infiltration

(hr) (in./hr) (hr) (in./hr)

0 2.90 5.00 1.090.10 2.83 6.00 0.950.25 2.74 7.00 0.840.50 2.59 8.00 0.761.00 2.31 9.00 0.692.00 1.87 10.00 0.653.00 1.54 15.00 0.544.00 1.28 20.00 0.51

7.10 Phi Index 203

S = [(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

0 1.0 2.0

Time (hr)

3.0 4.0 FIGURE 7.16

Using Fig. 7.14 and interpolating between curve numbers, we get Q = about5.4 inches of runoff, the same result.

7.10 PHI INDEX

Infiltration indexes generally assume that infiltration occurs at some constant or aver-age rate throughout a storm. Consequently, initial rates are underestimated and finalrates are overstated if an entire storm sequence with little antecedent moisture is con-sidered. The best application is tolarge storms on wet soils or storms where infiltrationrates may be assumed to be relatively uniform [1].

The most common index is termed the phi (<J» index for which the total volumeof the storm period loss is estimated and distributed uniformly across the storm pat-tern. 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 initialdepression storage and wetting.

4.0

';:;' 3.0,e~>-

:\::'"~Q)

.S 2.0

:§~

'",~

Representation of a <I>index.

204 Chapter 7 Infiltration

To determine the <I> index for a given storm, the. amount of observed runoff isdetermined from the hydro graph, and the difference between this quantity and thetotal gauged precipitation is then calculated. The volume of loss (including the effectsof interception, depression storage, and infiltration) is distributed uniformly across thestorm pattern as shown in Fig. 7.16.

Use of the <I> index for determining the amount of direct runoff from a givenstorm pattern is essentiallythe reverse of this procedure. Unfortunately, the <I> indexdetermined from a single storm is not generally applicable to other storms, and unlessit is correlated with basin parameters other than runoff,it is of little value. .

SUMMARY

Infiltration is a significant component of hydrologic processes. Soils have varyingcapacities 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 proper-ties also have a modifying effect.

When the rainfall intensity is less than the infiltration capacity, all of the waterreaching the ground can infiltrate. But if the rainfall intensity exceeds the infiltrationcapacity, infiltration will occur only at the infiltration capacity rate, and water inexcess of that capacity will be stored in depressions, become surface runoff, or evapo-rate. In general, the initial infiltration capacity of a dry soil is high. As rainfall contin-ues, 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 coverconditions. A number of equations have been developed to serve as models for theinfiltration 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 is0.55 in.lhr. For the first 10 hours, estimate the total volume of water infiltrated in inchesover 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. Atotal of 30 in. of water infiltrated during a 10-hr interval. Determine the value of k inHorton'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 205

30-min periodIntensity (in./hr)

14.0

22.0

36.0

45.0

a. Determine the total storm rainfall (in inches). .

b. Determine the q, index for the basinif the net stormrain is 3.0in.

7.7 The direct surface runoff volume from a 4.40-mi2 drainage basin is determinec! by plariimeterfrom the area under the hydrograph to be 10,080 cfs-hr.The hydrograph was produced by al.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-hrstorm 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 capacitymethod.

Rainfall Data for a Hypothetical Storm on April 15 of the Currnt Year(Beginning at midnight on April 14)

Rainfall Infiltration capacity Hourly deduction forintensity at beginning of hour depression storage

Hour (in./hr) (in.!hr) (in./hr)

1 0.41 0.200 0.202 0.49 0.160 0.143 0.32 0.125 0.044 0.31 0.100 0.025 0.22 0.085 0.006 0.08 0.070 0.007 0.07 0.0658 0.09 0.0579 0.08 0.052

10 0.06 0.04711 0.11 0.04412 0.12 0.04013 0.15 0.03714 0.23 0.03615 0.28 0.03516" 0.26 0.034'17 0.21 0.03318 0.09 0.03319 0.07 0.03320 0.06 0.03221 0.03 0.03222 0.02 0.03223 0.01 0.031'24 0.01 0.031

206 Chapter 7 Infiltration

7.9 Tabulated below are total rainfall intensities during each hour of a frontal storm over adrainage 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 drainagebasin. (Note that by definition the area under the hyetograph above the q, index linemust be 2.00 in.)

d. Determine the area of the drain&ge basin (acres ).if the net rain is 2.00 in. and the mea- ,.sured 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:

. '1'1

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 ~mdescnbe the factors that are consIdered when a curve number ISdetermmed.?:¥':

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? . !id

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 "x.~~and 60 percent native pasture in fair hydrologic condition. .~~. . "'.,

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;,~

<"oii

II.~'i

~I

7.15 Rework Problem 7.12 if the soil group is B and the land use is 50% straight-row cropsunder 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 ifthe direct rainfall is 20 em.

Hour 1 2 3 4Intensity (in./hr) 0.40 0.05 0.30 0.20

Hour Rainfall intensity (in./hr) I Hour. Rainfall intensity (in./hr)

1 0.41 13 0.152 0.49 14 0.233 0.22 15 0.284 0.31 16 0.265 0.22 17 0.216 0.08 18 0.097 0.07 19 0.078 0.09 .20 0.069 0.08 21 0.03

10 0.06 22 0.0211 0.11 23 0.0112 0.12 24 0.01

....