06 infiltration

7/28/2019 06 Infiltration

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3/29/03Colorado State University

Department of Civil EngineeringCE 322

Notes on Infiltration and Soil MoistureJose D. Salas

Water in the soil and in the groundwater are generally replenished from infiltrated water.

Infiltration generally occurs naturally as precipitation falls on watersheds and river basins,although sometimes infiltration is artificially induced to recharge the groundwater storage.Infiltration is an important component of the hydrologic cycle since it plays a major role in

determining other processes such as surface runoff. In this chapter we begin by reviewing someof the basic properties of soils, the factors affecting infiltration, how infiltration is measured, and

how infiltration is estimated.

1. Definitions

As rainfall reaches the land surface it may enter into the soil through the soil surface pore

space, cracks, fissures, etc. Thus, infiltration is the passage of precipitation through the landsurface into the ground. The rate at which precipitation enters the ground is called infiltrationintensity and the total amount of water infiltrated in a given time period is called cumulative

infiltration. Also when referring to infiltration, it does not necessarily have to come fromprecipitation but it may result from ponded water from ponds, canals, flooded irrigation systems,

etc. Likewise, infiltration may occur after snow is accumulated in the ground and melts. Oncewater infiltrates and enters into the soil it may continue moving in some direction (e.g.downward). The water movement within the soil mass is called percolation. Percolation will

be studied in a subsequent chapter dealing with subsurface hydrology. In this chapter we willconcentrate on infiltration. Infiltration depends on a number of factors some of which are related

to soil characteristics. These are discussed in section 2.

2. Elementary Properties of Soils and Soil Moisture

Some of the elementary properties of soils are reviewed herein. They include properties

related to their physical geometry such as porosity, their water storage and water holdingcapacities, and dynamic related properties such as hydraulic conductivity.

(a) Porosity. It is the volume of the pore space that is available in a soil sample. Commonly it is

expressed as a fraction of the total volume of the soil sample, i.e.

V

Vv=0 (1)

where vV = volume of the voids or pore space and V = total volume of the soil sample.


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(b) Soil moisture content. It is the amount of water that is contained in a soil sample. It may beexpressed in terms of volume or weight and commonly it is given in relative quantities. For

example, Eq.(2a) gives a volumetric soil moisture content as

V

V= (2a)

where V = volume of pore space filled with water. Note that for full saturation vVV = , so that

0=s . Generally though the condition 0


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is positive and equal to ho. On the other hand, if the soil is not saturated, a point 2 that is

located h cm above point 1 has a negative pressure equal to - h. Such negative pressure iscalled soil moisture tension or capillary potential. This concept and the notation generally used

are further described below.

Fig.1 Pipe filled up with sand. Point 0 below the water surface has a positivepressure while point 2 above the surface has a negative pressure.

A typical notation for soil moisture tension is , which is assumed to be negative, i.e.

h = (3)

As h increases the soil moisture tension increases while the soil moisture content decreases. Thus the plot of orh versus decays as shown in Fig.2 (a). Also if we assume

that a soil sample is initially saturated (with no air present) and then we dry the soil, the plot of versus will be as shown in (1) in Fig.2 (b). If we saturate the soil again (allowing air to betrapped), curve (2) will apply. If the soil is dried again, curve (3) will result. The different curvesare due to the so-called hysteresis effect (because of the effect of the air trapped in the soil).

(a) (b)

Fig.2 (a) soil moisture tension versus soil moisture and (b) hysteresis effect

Referring to the sketch the pressure at point 1is:

hpp += 21

Because app =1 = atmospheric pressure,then

hpp a =2

Considering only the relative pressure we

have

hp =2

0

2

1

h

h0

h

Soil moisture content

(1) drying

(3)

Soil moisture content s

Soil moisture

tension

(2) wetting


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(c) Hydraulic conductivity. According to Darcy, Phillip, and others (refs.) the following lawapplies to the movement of water in the soil. The water velocity at a point located at an elevation

zis given by

zKqz

= )( (4)

wherez+= (5)

and qz = water velocity at point 0, =hydraulic head, )(K = hydraulic conductivity, =capillary potential, and z= elevation head. Figure 3 defines some of the variables involved inEqs. (4) and (5). Thus hydraulic conductivity is a soil property that expresses a dynamical

feature of the soil. It is an increasing function of as shown in Fig. 4. And relative hydraulic

conductivity is defined as sr KKK /= .

Fig. 3.

Fig. 4

0

datum

tensiometer

ground surface

z

Hydraulic

conductivity K

K0

Soil moisture content 0 s

Ks

K


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(e) Specific retention (field capacity) and specific yield. It is a measure of the water holdingcapacity of a soil sample. Assume that we have a soil sample with volume V and the soil is

saturated (with water). Then we let the soil drain by gravity until the drain stops. The amount ofwater remaining in the soil Vr is called specific retention or field capacity while the amount of

water drained Vd is calledspecific yield. Commonly these properties are expressed relative to the

total volume of the soil sample V. Thus one can write:

yrs += (6)

where VVrr /= = specific retention, VVyy /= = specific yield, and s = saturated soil

moisture content. Also note that sometimes s in Eq.(6) is replaced by 0 the soil porosity.

Soil moisture quantities related to field capacity are wilting point and available water. Wiltingpoint is defined as the soil moisture level below which plant roots will not be able to extract

water from the soil. Available wateris the difference between field capacity and wilting point.Refer to Tables 1 and 2 for typical values of these properties for various soil types.

Example 1. Assume that the root zone in sandy loam is 1.5 m thick. Estimate the amountof available water (available soil moisture) in cm of depth that the soil is expected to

hold.

Table 2 gives, available water = 7% (dry weight) and specific weight = 1,440 kg/m3. Weight of water in 1.5 m3 of soil (Eq.2b):

KgmKgmWW s 2.151/440,15.107.033 ===

Volume of water: 33 1512.0)/000,1/(2.151 mmKgKgV ==

Depth of water: cmmmmh 12.151512.01/1512.0 23 ===

3. Factors Affecting Infiltration

Rainfall rate

Type of soil (see Fig.)

Initial soil moisture (antecedent conditions)

Soil compaction

Vegetative cover

Temperature (see Fig.)

Season (see Fig.)

Other factors


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4. Measurement of Infiltration

Infiltrometers Rainfall simulators

5. Infiltration Capacity and Actual Infiltration

Infiltration capacity of a soil surface is the maximum infiltration rate that the soil can take for

given conditions of soil moisture and atmospheric conditions such as rainfall and evaporation.

The soil moisture content and the type of soil only limit the infiltration capacity. In other

words for a given type of soil, the infiltration capacity will vary with and as increases theinfiltration capacity will decrease. In fact as the soil moisture content reaches saturation theinfiltration capacity will reach a constant value. Assuming that a 4-hour rainstorm occurs with a

constant intensity of i in/hr and that i is greater than the initial infiltration capacity f0, then it isexpected that the infiltration capacity will decrease throughout the 4-hr period because the soil

moisture content will increase as a result of the infiltrated water into the soil. In this case the

actual infiltration is the same as the infiltration capacity. On the other hand, if i < f0, then theactual infiltration will the same as i. In this case the infiltration capacity will also decrease with

time but at a smaller rate that would occur in the first case. Furthermore, after the rain stops(after the 4-hr period) it is expected that the infiltration capacity will increase with time becausethe soil moisture content will be depleted (through time) because of evapotranspiration and

percolation losses. Such increase of the infiltration capacity with time is known as infiltrationcapacity recovery. Further details on infiltration capacity recovery are given in section 6.2.

A general equation for determining actual infiltration )(aft given that the infiltration

capacity )(cft and the rainfall rate ti are known is

],)(min[)( ttt icfaf = (7)

Some examples on estimating infiltration capacity and actual infiltration are given in section 6.1.

6. Methods for Estimating Infiltration Capacity

Several methods are available for estimating infiltration capacity. They include:

Hortons model Green-Ampt model Holtans model

Phillips modelThis section includes Hortons model only. In addition, it includes a method for estimating therecovery of infiltration capacity after rainfall stops and a dry period follows.

6.1 Hortons Model for Estimating Infiltration Capacity

Hortons model provides the decay of the infiltration capacity rate as a function of time giventhat the rainfall intensity is greater or equal (to the infiltration capacity). The model is


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tKcct efffcf+= )()( 0 (8)

where cf (in/hr), 0f (in/hr), and K(hr-1) are parameters of the model. The cumulative infiltration

capacity is obtained by integrating )(cft of Eq.(8) from time 0 to tas

+==t

tKcc

t

tt dtefffdtcfcF

0

0

0

])([)()(

)1()()/1( 0tK

cc effKtf+= (9)

The graphs in Fig.4 show the functions )(cft and )(cFt versus time tand )(cft versus )(tFt .

Fig. 5 (a) constant rainfall intensity of duration tt, infiltration capacity ft (c) decays at

full rate when rainfall is bigger or equal to the infiltration capacity, and infiltrationcapacity recovers (increases) after rainfall stops; (b) infiltration capacity decays at full

rate because it is smaller than rainfall intensity; (c) cumulative infiltration capacity; and(d) relationship between infiltration capacity and cumulative infiltration capacity.

ft(c)

f0

fc

ft(c)

0 t

t

ft(c)

f0

ft(c)

0 Ft(c)

Ft(c)

Ft(c)

Ft(c)

0 t

t

(b)

(c)(d)

ft(c)

f0

0 t tr

t

(a)

infiltrationcapacity ft(c)

rainfall

intensity i

recovery of

ft(c)


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Example 2. For the storm hyetograph shown below use Hortons model to determine theinfiltration capacity curve assuming the following parameters: f0=0.6 in/hr, fc=0.2 in/hr,

andK=0.4 hr!1. In addition, calculate the cumulative infiltration capacity.

Example 3. For the storm hyetograph shown below determine the infiltration capacity

curve by using Hortons model assuming the following parameters: f0=0.6 in/hr, fc=0.2in/hr, and K=0.4 hr!1. Note that the difference between this example and the previous

one is that the rainfall rate during the first hour is 0.3 in/hr instead of 1 in/hr.

In this case becausef0>i,the infiltration capacity decay during the first 2-hours should beless than that if i 0.6 (as it was in Example 2). Thus assuming that such infiltrationcapacity decay is bigger that the rainfall intensity for the first two hours, we will applythe infiltration mass method. For this purpose we build the relationship betweenft (c) andFt(c) using Hortons Eqs.(8) and (9), respectively. The results are shown in columns (3)

and (4) of the Table below and also in the sketch shown below. Considering the firsthour, the cumulative rainfall is 0.3 in and this amount is also infiltrated in that period.

We enter with this value (Ft (c)=0.3) in column 4 of the table below and find byinterpolation in column 3 that ft (c) = 0.525. Alternatively we use the figure as shownbelow. We follow a similar approach for the second hour and find that ft (c) = 0.453.

Thus, our initial assumption that the infiltration capacity was bigger than 0.3 during theperiod 0-2 is correct.

Since f0< rainfall rate one can apply Hortons equationdirectly to find the variation of the infiltration capacity

through time. The table below shows the calculationsand the figure shows the variation offt during the time

period 0-6. The table also shows the cumulativeinfiltration capacityFt(c). Also note that in this case theactual infiltration is equal to the infiltration capacity.

t ft (c) in/hr Ft(c) in t ft (c) in/hr Ft(c) in

0 0.6 0 4 0.281 1.598

2 0.38 0.951 6 0.236 2.109

1.0

1.4

t

0.9

0 2 4 6

it(in/hr)

f0 = 0.6ft(c)

0 2 4 6

0.9

1.4

0.3

t

i(in/hr)

f0 = 0.6

0 0.4 0.8 1.2 1.6

0.2

ft(c)

0.6

Ft(c)

0.4

0.4530.525

0.3 0.6


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We now proceed in calculating the infiltration capacity for the period 2-6 hr. Thecalculations are shown in columns 5-7 of the table. We reset the time scale so that t'=0

fort=2, then apply Hortons Eq.((8) with a new value off0=0.453 as shown in column

(6). The cumulative infiltration capacity during the time 2-6 is shown in column (7).Column (8) summarizes the final infiltration capacity for the whole time period 0-6 hr.

Also the actual infiltration rate ft (a) is shown in column (9) and the cumulative actualinfiltration is listed in column (10). The sketch above shows the variation of the final ft(c) during the period 0-6 hours. Note that the actual infiltration is the red curve while theinfiltration capacity is the blue curve. Also note that the two curves overlap in the period2-6 hrs.

t

hours

(1)

ti

(in/hr)

(2)

)(cft

in/hr

(3)

)(cFt

in

(4)

t hours

(5)

)(cft

in/hr

(6)

)(cFt

in

(7)

Final

)(cft

(8)

Actual

infiltration

rate )(aft

(9)

Cumulative

actual infiltration

)(aFt

(10)0 0.3 0.600 0.000 0.600 0.300 0.000

1 0.3 0.468 0.530 0.525 0.300 0.300

2 1.4 0.380 0.951 0 0.453 0.000 0.453 0.453 0.600

3 1.4 0.320 1.299 1 0.370 0.409 0.370 0.370 1.009

4 0.9 0.281 1.598 2 0.314 0.748 0.314 0.314 1.348

5 0.9 0.254 1.865 3 0.276 1.042 0.276 0.276 1.642

6 0 0.236 2.109 4 0.251 1.305 0.251 0.000 1.905

Example 4. For the storm hyetograph shown below determine the infiltration capacity

curve by using Hortons model assuming the following parameters: f0=0.6 in/hr, fc=0.2in/hr, and K=0.4 hr!1. Note that the difference between this example and the previous

one is that the rainfall rate during the first hour is 0.45 in/hr instead of 0.3 in/hr.

t

hrs

ft(c )

in/hr

Ft(c)

in

t

hrs

t

hrs

)(cft

in/hr0 0.600 0 1.37 0 0.450

1 0.468 0.530 2 0.63 0.394

2 0.380 0.951 4 2.63 0.287

6 4.63 0.239

f0 = 0.6

i=0.45

t

0.9

0 2 4 6

it(in/hr)

1.4ponding

time


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Firstly we construct the table that relates ft (c) andFt (c). It is shown above for 0, 1, and

2 hours. Since i0< f0 we will assume thatit < ft (c) for 20 t , i.e. during the first 2-hours the infiltration capacity decay will be bigger than the rainfall intensity (as it was in

Example 1). Thus we apply the infiltration mass method. Since the rainfall during thetime 0-2 hrs is 0.9 in, we enter with this value in the table above and find that f= 0.391.

Because this value is smaller than 0.45, our initial assumption that the infiltrationcapacity would be bigger than 0.45 during the period 0-2 is incorrect. Therefore, insteadwe find the ponding time (time in which the infiltration capacity curve will cross the

rainfall rate of 0.45). The ponding time can be obtained by entering the table above withf = 0.45 in/hr for infiltration capacity rate and by interpolation we get F=0.616 inches

(note that we could also get this value by using Hortons Eq.(8) with ft (c)=0.45 andsolving for t. It gives t=1.175 hr.. Then, Eq.(9) gives Ft (c) =0.61 for t=1.175 hr).Therefore, the ponding time will be equal to 0.616 in/(0.45 in/hr) =1.37 hr. After this

time one can apply Hortons equation again since the rainfall rate through the rest of thestorm is greater than the infiltration capacity. One can do this by reinitializing the time

scale so that time 1.37 hr becomes time zero. The rest of the computed values for times t

= 2,4,6 are also shown in the table above.

6.2 Recovery of the infiltration capacity

We have seen in section 6.1 the application of Hortons equation for estimating infiltrationcapacity given that a storm rainfall occurs. Naturally under this condition the infiltrationcapacity decays with time. However, as rainfall stops and a dry period follows, the soil begins to

dry up due to evaporation and deep percolation. Thus, we will expect that as the soil dries theinfiltration capacity will increase converging to the value f0. A simple procedure for estimating

the recovery of infiltration capacity is given by (ref.)

tK

r

tK

rrtrr

efffefffcf

=+= )()1()()( 000 (10)

in which for convenience a new time reference t' is used and ft' (c) = infiltration capacity during

recovery,fr= value of infiltration capacity at the beginning of the recovery period (at the end ofthe rain storm), Kr= recovery parameter, and f0 = the initial infiltration capacity under dry soil

conditions. Figure 6 illustrates the concept and defines some of the variables used in Eq.(10).

Fig. 6. Infiltration capacity and infiltration capacity recovery

fr

f0

t

fr

t

)(cft

Infiltration capacity recovery


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Example 5. Consider the rainfall hyetograph as shown in the figure below. Use Hortonsmodel to determine the infiltration capacity curve assuming the following parameters:

f0=0.6 in/hr,fc=0.2 in/hr,K=0.4 hr!1, andKr=0.16 hr

!1. In addition, calculate the actual

infiltration rate, the total infiltration during the time period 0-12 hrs, and the total excessrainfall during the same period.

The first storm during the period 0-6 hr. is the same as that for Example 2. Likewise theHortons model parameters are the same. Therefore, using the results shown in Example

1 the infiltration capacity at time 6 is equal to 0.236 in/hr. Since rainfall is zero duringthe time 6-10 hr. the infiltration capacity must recover from fr= 0.236 in/hr at time 6 to

some bigger values at times 8 and 10 hr. Thus, the infiltration capacity recovery iscalculated applying Eq.(10) as shown below. The bulk of the calculations are shown inthe table below. The infiltration capacity throughout the period 0-12 hr is given by the

columns (3), (6), and (8). The actual infiltration rate is: (i) equal to the infiltrationcapacity during 0-6 hr. (refer to Eq. 7), (ii) equal to zero during 6-10 hr., and ( iii) equal tothe infiltration capacity during 10-12 hr. The total infiltration during the period 0-12 hr.

is equal to 2.109 + 0.686 = 2.795 inches (refer to columns (4) and (9) in the table below).The total excess rainfallPe during the period 0-12 hr. is obtained by substracting the total

infiltration from the total precipitation, i.e. Pe =P-F= (1+1.4+0.9+1.2)2 2.795 = 6.2

in.

thr

itin/hr

ft(c )in/hr

Ft(c)in

t'hr

ft'(c )in/hr

thr

ft(c )in/hr

Ft(c)in

(1) (2) (3) (4) (5) (6) (7) (8) (9)

0 1.0 0.6 0

2 1.4 0.380 0.951

4 0.9 0.281 1.598

6 0.0 0.236 2.109 0 0.236

8 0.0 2 0.336

10 1.2 4 0.408 0 0.408 0

12 0.0 2 0.293 0.686

Notes: (5) time is reset to calculate recovery of infiltration capacity; (6) Eq.(10) is applied;

(7) time is reset again to calculate infiltration capacity during 10-12 hr. In this case Eq.(8)is used withf0=0.408.

t

t

tKrt

e

e

efffcf r

=

=

=

16.0

16.0

00

364.06.0

)236.06.0(6.0

)()(

1.0

1.4

0.9

it(in/hr)

f0 = 0.6 ft(c)

1.2

t(hours)

0 2 4 6 8 10 12


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7. Methods for Estimating Infiltration and Runoff

index (needs prior estimate of excess rainfall) Soil Conservation Service or SCS method (indirect estimates of infiltration)

7.1 Index

The index is a simple empirical method for determining a constant infiltration ratethroughout the duration of the excess rainfall. It requires a prior knowledge of the amount ofexcess rainfall (direct runoff). Figure 7 describes the concept. The estimation of the indexgenerally requires a trial and error approach unless the rainfall hyetograph is simple. Examples 6

and 7 below further illustrates the method.

t t

0 1 2 3 4 0 1 2 3 4

Fig. 7

Example 6. For the rainfall hyetograph shown below calculate the index assumingthat the excess rainfall amounts are: (a)Pe= 0.5 in, (b)Pe = 1.5 in, and (c)Pe = 2.6 in.

known excess

rainfall Pe

i(in/hr)i(in/hr)

index

initial losses

duration of excessrainfall

duration of excess

rainfall

By observing the rainfall pattern it is clear that:(a) forPe=0.5 the index must be bigger

than 1, then (2- ) in/hrx

1 hr= 0.5 in,

whichgives =1.5 in/hr(b) forPe=1.5

(2- +1- )=1.5, then =0.75in/hr(c) forPe=2.6

(0.5- +2- +1- )=2.6, then =0.3 in/hr

2 in

0.51 in

2.0

1.0

0 1 2 3

i in/hrPe=0.5 in

5.1= in/hr

infiltration


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Example 7. For the rainfall hyetograph shown below calculate the index and theduration of the excess rainfall assuming the following two cases: (a) Pe= 0.9 in, and (b)

Pe = 2.4 in.

i in/hr 2

1

0.5

1.5

t

0 1 2 3 4

(a) Let assume that the index is between0.5 and 1 in/hr. Then we can write:

(2- ) +(1- )+(1.5- )=0.9Solving the equation gives =1.2.Thus our original assumption was

incorrect. For our second trialweassume that 1<


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Problems

1. The thickness of a layer of clay loam soil is 1.25 m. Calculate the specific retention in cm ofdepth that this type of soil is expected to hold.

2. A storm hyetograph is shown in the sketch below. The 6-hr storm begins at time 2. At time0 the soil has been very dry. Calculate for the time period 0-8 hr: (i) the infiltration capacityrate, (ii) the actual infiltration rate, and (iii) the cumulative infiltration capacity. Theparameters of Hortons model are:f0=0.5 in/hr,fc=0.25, andK=0.45/hr.

3. For the storm hyetograph shown below determine for the time period 0-12 hr the following:

(a) the infiltration capacity rate, (b) the actual infiltration rate, (c) the cumulative infiltrationcapacity, (d) the cumulative actual infiltration, and (e) the excess precipitation. In addition,

assuming that the storm occurs over a 100 sq-mi basin, and the flood generated by the storm

(at the outlet of the basin) occurs during a 24 hr. period, estimate the average flood flow incfs. Assume that Hortons model applies with: f0=0.5 in/hr, fc=0.25, K=0.45/hr, and

Kr=0.15/hr.

thr

0 2 4 6 8

0.75

1.4

0.9

it(in/hr)

f0 = 0.5

t hours

0 2 4 6 8 10 12

0.25

1.4

0.9

it(in/hr)

0 = 0.5

1.0

0.7


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4. The storm hyetograph shown below produces 375 acre-ft of direct runoff in a 1,000- acrewatershed. Determine the F infiltration index.

5. The rainfall pattern on an irrigation field during a 7-hr period is shown below. Theparameters of Hortons model for infiltration capacity (including infiltration capacity

recovery) are: K = 0.5 hr-1, Kr = 0.15 hr-1, fo = 0.7 in/hr, and fc = 0.3 in/hr. Calculate the

infiltration capacity at times: (a) 2, (b) 3, (c) 5, (d) 6, and (e) 7 hrs.

6. Referring to the storm hyetograph of Problem 3 above estimate the infiltration index andthe duration of excess rainfall assuming that (a)Pe = 0.8 in, (b) Pe = 1.2 in, and (c)Pe = 2inches.

5

4

1

3

2

thours

in/hr

0 1 2 3 4 5

0 1 2 3 4 5 6 7

fo=0.7

1.2

0.50

in/hr

thrs

0.5

1.0


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7. A 2-hr storm occurs over a small watershed covered with turf. The storm hyetograph is asshown in the graph below. The infiltration capacity can be modeled by using Hortons

equation with parameters f0=0.6 in/hr,fc=0.2 in/hr, and K=0.4 hr!1. The depression storage

capacity for turf can be assumed to be inSd 25.0= . Determine the following:

(note: perhaps I should include an example within the text with a simpler version of this or movethis as example and make another for this section)

0.65

0.5

t

0 2 4

it(in/hr)

f0 = 0.6

(a) The actual infiltration during the time period 0-4 hrs.

(b) The cumulative infiltration during the time period 0-4hrs.

(c) The cumulative excess precipitation during the time

period 0-4 hrs.(d) The overland flow rate during the time period 0-4 hrs

assuming that overland flow begins only after alldepression storages has been filled.

(e) The overland flow rate during the time period 0-4 hrs

assuming that both overland flow and depression

storage occur simultaneously.


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References

Linsley et al, 1986

06 infiltration

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