Infiltration

Infiltration
Infiltration: The movement of water through the soil surface in to the soil.
Infiltration Capacity (fp) (mm/hr ) (Potential Infiltration Rate) : The maximum rate at which soil can absorb water through its surface.
Infiltration Rate, f(t):Rate of water entering the soil surface. If there is no limit on the water supply for infiltration, f(t) = fp. Otherwise,
0 £ f(t) < fp.
fo = initial infiltration rate
fc = ultimate infiltration rate
f(t) - fc = excess infiltration rate



Cumulative Infiltration, F(t) : Depth of infiltration from the beginning of rainfall to any time, t.
F(t) = Area under the infiltration curve
Experimental Methods

Double Ring Infiltrometer
Purpose:
–measure infiltration capacity
Principle:
–fill rings
–measure rate of drop in water level of inner ring over time
Infiltration capacity curve
Definition
-maximum infiltration rate [cm/hr or in/hr]
-time dependent to some extent

Infiltration process
Infiltration rates depend on both surface and subsurface conditions:
Surface conditions: Availability of water
Subsurface conditions: Ability of water to infiltrate
Rainfall intensity = I
Actual Infiltration rate = f
Infiltration capacity = fp

Infiltration capacity and infiltration rate
Case 1: Rainfall intensity exceeds infiltration capacity
–Water will pond
–Actual infiltration rate = infiltration capacity
f = fp
–Surface runoff
Case 2: Infiltration capacity exceeds rainfall intensity
–All rain infiltrates no ponding
–Actual infiltration rate = rainfall intensity
f = i
–No surface runoff
Factors Affecting Infiltration
physical properties of the soil
soil moisture
rainfall intensity
land use
temperature
water quality

Infiltration Equations
1- Horton’s Equation










f(t) =fc+(fo-fc) e-Ct (Horton equ.)
Where:
Infiltration rate [in/hr or cm/hr]:
f(t) =Rate of disappearance of water.
fo= initial infiltration rate at t = 0
fc= ultimate (final) infiltration rate (constant value)
k = exponential (time) decay constant [hr-1]


Horton’s Equation
Advantages:
– Simple
– Can be applied graphically
Disadvantages:
– Parameters hard to estimate
– Only valid for i > f


Cumulative infiltration depth [in or cm]:







F(t) = ?_0^t?fdt = ?_0^t?f_c +( f_o - f_c ) e^(-Kt) dt
= f_c t + (( f_o - f_c )(1-e^(-Kt)))/K (Cumulative infiltration depth) [in or cm]
Example:
A watershed has the following Horton parameters:
fo= 1.5 in/hr
fc= 0.2 in/hr
k = 0.35 hr -1
a) Determine infiltration capacity at t=10 min, 30 min, 6 hrs.
b) Total depth of infiltration during a 6-hr period, assuming rainfall intensity exceeds infiltration capacity.
t
(hr) f
(in /hr )
1/6 1.43
0.5 1.29
6 0.36
Solution:
Infiltration capacity:
f(t) = 0.2 + 1.3 e^(-0.35 t) ( in / hr )

Cumulative depth:
F (t) = f_c t + (( f_o - f_c )(1-e^(-Kt)))/K
= 0.2 t + 3.71 (1-e^(-0.35 t))
F (6) = 4.46 inches

Example: H.W.
the parameters for Horton s equation are fc= 1.0 cm/hr, fo=5.0 cm/hr and k=2 hr-1 . Determine the infiltration rate and cumulative infiltration after 0, 0.5, 1.0, 1.5, 2.0 hours if the rainfall rate is 6 cm/hr. Plot as a function of time. What would be the infiltration rate if the rainfall rate were 0.6 in/hr?
Infiltration indices
1-The ?-Index:
? Index is the average rainfall intensity above which the volume of rainfall equals the volume of runoff units of [in/hr] or [cm/hr]
The area above the dashed line
represents measured runoff over the
catchment area
The area below the dashed line
is the measured rainfall that did not
appear as runoff but represents all
the losses including interception,
evaporation and infiltration



To determine the ? Index for a given storm, the amount of bserved runoff is determined and the difference between this quantity and the total gauged rainfall is then calculated
The volume of loss is then distributed uniformly across the storm pattern
It should be kept in mind that ? Index varies as the storm intensity varies with time and thus ? Index is of limited value and that many determinations should be made and averaged before the index is used
Example: The rainfall intensities during each 30 min of a 150-min storm over a 500-acre basin are 5.5, 3, 1, 3.5, and 2 in/hr, respectively , The direct runoff from the basin is 105 acre-ft , Determine ? Index for the basin
Solution:





Find the total rainfall as follows:
30/60 × (5.5 + 3 + 1 + 3.5 + 2) = 7.5 in or 0.625 ft
Rainfall volume = 500 × 0.625 = 312.5 acre-ft
Runoff volume = 105 acre-ft
Volume under ? Index = 312.5 – 105 = 207.5 acre-ft
Infiltration depth (losses depth) = 207.5/500 = 0.415 ft or 5 in
? Index = 5 × (1/150) × (60) = 1.98 in/hr

Example:
You have two storm events of 75 mm of a total duration of 6 hours as shown in the figures
Both produced a total runoff equivalent to 33 mm
Find out the ? Index for the two storm events






solution:












Example:
Compute the depth of runoff and the infiltration considering the rainfall event summarized in the table
Assume a ? Index value of 0.6





solution :
Compute the intensity for each duration
If the ? Index is higher than the rainfall intensity then the infiltration equals the rainfall
If the ? Index is less than the rainfall intensity then the infiltration equals the ? Index
Net rainfall intensity = the rainfall intensity – ? Index







Example:
Use the rainfall data below to determine the ?-index for a watershed that is 0.875 mi2, where the runoff volume is 228.7 ac-ft.

Solution:




Depth of runoff
A = 0.875 sq mi × 640 acres/sq mi = 560 acres
Q = volume/area = (228.7 ac-ft x 12 in/ft)/(560 acr) = 4.9 inches
Area above ?-index must equal 4.9 inches.
2(1.4 – ?) + 3 (2.3 – ? ) + 2(1.1 – ? ) + 3(0.7 – ? ) + 2(0.3 – ? ) = 4.9
? = 0.8

2. The w-Index:

W= (P-R)/t
P = total precipitation (cm)
R = total runoff (cm)
t = duration of rainfall (hour)
W= defined average rate of infiltration (cm)













Example1: Find the Index of a certain catchment has the change of the rainfall intensity which given below. If the runoff=33mm, P=75mm



Example2: A tabulated below are data for a number of storms happened on a river. Compute the W-index for all storms, what would be the av. Error and av. Percentage error in estimated runoff .If the av. W-index was used to compute the runoff?
Error=Rcomp.-Rob. Rcomp.=P-W*t W-index
(cm/hr.) Rob.
(cm) P
(cm) Duration,t
(hr.) Storm no.
-0.12 1.2 0.125 1.32 2.82 12 1
-1.02 0 0.041 1.02 2.98 48 2
-1.15 1.31 0.087 2.46 4.55 24 3
-2.92 4.5 0.094 7.42 14.22 72 4
0.01 0.44 0.136 0.43 2.87 18 5
0.19 0.67 0.143 0.48 3.91 24 6
1.31 3.24 0.171 1.93 8.1 36 7
1.12 2.79 0.228 1.67 4.41 12 8
-0.72 1.26 0.111 1.98 5.31 30 9
1.4 4.55 0.213 3.15 6.98 18 10
?=-1.9 ?=19.96 ?=1.35 ?=21.86
Error=-0.19 (W) ave = 1.35/10 = 0.135
For check:
Av. Error= (?Rcomp.- ?Rob.)/10
= (19.96-21.86)/10
=-0.19 ok.
-------------------------------------------------------------------------------------Example3: A rain storm with intensity=10cm and direct runoff=5.8cm If the distribution of the storm as given in table below. Find the
?-Index for the storm
8 7 6 5 4 3 2 1 t(hr.)
0.5 1 1.6 1.4 2.3 1.5 0.1 0.4 Excess rainfall/hr.

Example4:The equation of the f(t) curve for a certain catchment is given by:
f(t) =1.2+4.2 e-0.33t
1-Compute the runoff volume for the following rain storm:
2- Compute ?-Index & W- index and choose the best from them and explain cause that
6 5 4 3 2 1 t(hr.)
3 6.5 12.2 12.8 8.5 6 Rainfall(mm)



Ic t(hr.)
5.4 0
4.2 1
3.4 2
2.8 3
2.3 4
2 5
1.8 6
Solution:

P=(6+8.5+12.8+12.2+6.5+3)
= 49 mm
Depth of the infiltration =?y dt
=(1.2+4.2 e-0.33t)dt
= 1.2t-(4.2/0.33)e-0.33t]
] *e-0.33*0 4.2/0.33) )- =[1.2*6-(4.2/0.33)*e-0.33*6]-[1.2*0
= 18.19mm
R=P-I
=49-18.19
=30.81mm
W-index=(P-R)/t
=18.19/6
=3.032 mm/hr.
To Find The ?-Index:
R=30.81 mm
30.81=(12.8-12.2)*1+(12.2-8.5)*2+(8.5-6.5)*3+(6.5-6)*4+(6- ?)
30.81=16+6*5-5* ?
?=3.038 mm/hr.
We choose the w-index because it gives larger surface runoff.
Example:
A catchment soil has the following Horton infiltration Parameters:
f0 = 100 mm/min, fc = 20 mm/min, and k = 2 min-1 the required.
1- Plot the infiltration capacity curve with time for this catchment
2- Plot the potential cumulative infiltration for this catchment





Solution:
f(t) =fc+(fo-fc) e-Kt (Horton equ.)
F (t) = f_c t + (( f_o - f_c )(1-e^(-Kt)))/K (cumulative infiltration rate)





Example:
?? Find out the sensitivity of the infiltration capacity curve to different decay coefficients (k) assuming that f0 = 2.9 in/h and fc = 0.5 in/h
?? Assume k values = 0.15, 0.30, and 0.45 hour -1


Solution:
f(t) =fc+(fo-fc) e-Kt (Horton equ.)
t (hr) fp(in/hr) fp(in/hr) fp(in/hr)
K= 0.15 K= 0.30 K= 0.45
0 2.9 2.9 2.9
1 2.6 2.3 2
2 2.3 1.8 1.5
3 2 1.5 1.1
4 1.8 1.2 0.9
5 1.6 1 0.8
6 1.5 0.9 0.7
7 1.3 0.8 0.6
8 1.2 0.7 0.6
9 1.1 0.7 0.5
10 1 0.6 0.5
11 1 0.6 0.5
12 0.9 0.6 0.5
13 0.8 0.5 0.5
14 0.8 0.5 0.5
15 0.8 0.5 0.5
16 0.7 0.5 0.5
17 0.7 0.5 0.5
18 0.7 0.5 0.5
19 0.6 0.5 0.5
20 0.6 0.5 0.5
21 0.6 0.5 0.5
22 0.6 0.5 0.5
23 0.6 0.5 0.5
24 0.6 0.5 0.5
25 0.6 0.5 0.5
26 0.5 0.5 0.5