CE322 Basic
Hydrology
Jorge A.
Ramírez
Homework No. 4 - Solution
2. Compute the infiltration rate as a function of the
cumulative infiltration volume into a soil whose initial volumetric moisture
content is 0.15 and whose volumetric moisture content at saturation is 0.4.
Assume that the capillary suction head at the wetting front, S, is equal to
16.7 cm and that the saturated hydraulic conductivity Ks =
0.65 cm/h. Assume that the water supply at the surface is not limiting.
Using the Green-Ampt
equation, the infiltration capacity varies with cumulative infiltration
according to the following equation:
Thus, for the conditions
given,
3. Assume that the time evolution of the infiltration capacity
for a given soil is governed by Horton's equation (Note that this equation
assumes an infinite water supply at the surface, that is, it assumes saturation
conditions at the soil surface).
(1)
For this soil, the asymptotic
or final equilibrium infiltration capacity is fc = 0.8 cm/h;
and the initial infiltration capacity is fo
= 4.5 cm/h. The rate of decay of
infiltration capacity parameter is k = 3
h-1. For the precipitation hyetograph tabulated below, carry out
a complete infiltration analysis, including evaluation of cumulative
infiltration and rate of production of precipitation excess, s + v.
|
Time (min) |
Precipitation (cm/h) |
|
Time (min) |
Precipitation (cm/h) |
|
0 - 10 |
1.5 |
|
40 - 50 |
3.0 |
|
10 - 20 |
2.0 |
|
50 - 60 |
2.0 |
|
20 - 30 |
5.0 |
|
60 - 70 |
1.5 |
|
30 - 40 |
4.0 |
|
|
|
1)
Compute
accumulated precipitation volume as a function of time. The incremental volume
over each time period of 10 minutes is:
DP = i Dt
|
Time (min) |
Precipitation Intensity, i. (cm/h)) |
Cumulative Precipitation, P (cm) |
Time (min) |
Precipitation Intensity, i. (cm/h |
Cumulative Precipitation, P (cm) |
|
0 - 10 |
1.5 |
0.25 |
40 - 50 |
3.0 |
2.583333 |
|
10 - 20 |
2.0 |
0.583333 |
50 - 60 |
2.0 |
2.916667 |
|
20 - 30 |
5.0 |
1.416667 |
60 - 70 |
1.5 |
3.166667 |
|
30 - 40 |
4.0 |
2.083333 |
|
|
|
2)
Compute
infiltration capacity using Horton's equation for conditions of unlimited water
supply at the surface using equation 1 (Table 1 - Column 2).
3)
Compute
the accumulated infiltration that would occur under conditions of unlimited
water supply at the surface using the following equation 2 (Table 1 - Column
3),
(2)
4)
Compare
infiltration capacity with precipitation intensity (Figure 1). Observe that
during the first 20 minutes of the rainstorm, the infiltration capacity exceeds
the precipitation intensity. Thus, during this period, all of the precipitation
infiltrates. The actual infiltration rate is (Table 2 - Column 5),
(3)
5)
Because
the actual infiltration rate is less than the infiltration capacity during the
first 20 minutes, the actual infiltration capacity does not decay as predicted
by Horton's equation. This is because, as indicated above, Horton's equation
assumes that the supply rate exceeds the infiltration capacity from the start
of infiltration. Therefore, we must determine the true infiltration capacity at
t = 20 min. To do so, first determine
the time tp by solving the
following equation:
(4)
and then evaluate fp(tp)
as follows.
At t = 20 min the actual volume of
accumulated infiltration is:
F(t = 20 min) = [(1.5) cm/h + (2.0) cm/h] (10 min/60 min/h) = 0.58333
cm. Substituting this value for F(t) in equation 3 and solving for tp obtain: tp = 0.1552 h = 9.3134 min. Finally, the true
infiltration capacity at 20 minutes is obtained using equation 1 as fp(tp) = 3.1225 cm/h = fop. Alternatively,
using equations 1 and 2 to eliminate time and expressing cumulative
infiltration as a function of infiltration capacity obtain the following
equation,
(5)
6)
The
rainfall rate at 20 minutes i = 5
cm/h exceeds the corresponding infiltration capacity fop = 3.1225 cm/h.
Therefore, the actual infiltration rate equals the infiltration
capacity, and the decay of infiltration capacity follows Horton's equation with
an initial infiltration capacity equal to fop
and starting at time t* =
20 min (Table 1 - Column 5 and Table 2 Column 3). That is (see Figure 2 and
Figure 3),
(6)
7)
Because
the precipitation rate exceeds the infiltration capacity there is excess
precipitation available for runoff and depression storage, s + v
(Table 2 - Column 6).
(7)
Table 1
|
1 |
2 - Eq. 1 |
3 - Eq. 2 |
4 |
5 - Eq. 6 |
|
Time (min) |
Infiltration
Capacity, fp (cm/h) |
Cumulative
Infiltration, F (cm) |
Cumulative
Precipitation, P (cm) |
Actual
Infiltration Capacity (cm/h) |
|
0 |
4.5 |
0 |
0 |
|
|
10 |
3.044163 |
0.618612 |
0.25 |
|
|
20 |
2.161154 |
1.046282 |
0.583333 |
3.122538 |
|
30 |
1.625582 |
1.358139 |
1.416667 |
2.20869 |
|
40 |
1.300741 |
1.599753 |
2.083333 |
1.654414 |
|
50 |
1.103714 |
1.798762 |
2.583333 |
1.318228 |
|
60 |
0.984212 |
1.971929 |
2.916667 |
1.114321 |
|
70 |
0.91173 |
2.129423 |
3.166667 |
0.990646 |
|
80 |
0.867768 |
2.277411 |
|
0.915632 |
|
90 |
0.841103 |
2.419632 |
|
0.870135 |
|
100 |
0.82493 |
2.558357 |
|
0.842539 |
|
110 |
0.815121 |
2.69496 |
|
0.825801 |
|
120 |
0.809171 |
2.830276 |
|
0.815649 |
Table 2
|
1 |
2 - Eq. 1 |
3 - Eq. 6 |
4 |
5 - Eq. 3 |
6 - Eq. 7 |
|
Time (min) |
Infiltration
Capacity, fp (cm/h) |
Actual
Infiltration Capacity (cm/h) |
Precipitation
Intensity, i (cm/h) |
Actual
Infiltration Rate, f(t) (cm/h) |
Runoff
rate s + v (cm/h) |
|
0 |
4.5 |
4.5 |
1.5 |
1.5 |
0.0 |
|
10 |
3.044163 |
> 3.12254 |
2 |
2 |
0.0 |
|
20 |
2.161154 |
3.122538 |
5 |
3.122538 |
1.877 |
|
30 |
1.625582 |
2.20869 |
4 |
2.20869 |
1.791 |
|
40 |
1.300741 |
1.654414 |
3 |
1.654414 |
1.3455 |
|
50 |
1.103714 |
1.318228 |
2 |
1.318228 |
0.6817 |
|
60 |
0.984212 |
1.114321 |
1.5 |
1.114321 |
0.3856 |
|
70 |
0.91173 |
0.990646 |
0.0 |
0. |
0 |
|
80 |
0.867768 |
0.915632 |
|
|
|
|
90 |
0.841103 |
0.870135 |
|
|
|
|
100 |
0.82493 |
0.842539 |
|
|
|
|
110 |
0.815121 |
0.825801 |
|
|
|
|
120 |
0.809171 |
0.815649 |
|
|
|
4. Determine the f-index corresponding to the
situation of Problem 3. That is, assume that the total volume of infiltration
is equal to that computed in Problem 3, and that the precipitation hyetograph
is as given for Problem 3. Using this index, compare the resulting rate of
production of precipitation excess, s + n, to that obtained in Problem 3.
Volume of infiltration from Problem 3:
In the above equation, t*
= 20 min, and fop = 3.1225 cm/h. Substituting obtain:
F(t = 70 min) = 1.96 cm
From its definition:
f = Vol of
Infiltration/Rainfall Duration, thus,
f = (1.96 cm) * (60 min/h) /
(70 min) = 1.68 cm/h
Using this f-index, the actual infiltration is as
tabulated below:
|
Time (min) |
Rain Rate (cm/h)0 |
Trial Actual Infiltration Rate (cm/h) |
Actual Infiltration Rate (cm/h) |
Runoff rate s + v (cm/h) (cm/h) |
|
0 |
0 |
0 |
0 |
0 |
|
0 |
1.5 |
1.5 |
1.5 |
0 |
|
10 |
1.5 |
1.5 |
1.5 |
0 |
|
10 |
2 |
1.68 |
1.752 |
0.248 |
|
20 |
2 |
1.68 |
1.752 |
0.248 |
|
20 |
5 |
1.68 |
1.752 |
3.248 |
|
30 |
5 |
1.68 |
1.752 |
3.248 |
|
30 |
4 |
1.68 |
1.752 |
2.248 |
|
40 |
4 |
1.68 |
1.752 |
2.248 |
|
40 |
3 |
1.68 |
1.752 |
1.248 |
|
50 |
3 |
1.68 |
1.752 |
1.248 |
|
50 |
2 |
1.68 |
1.752 |
0.248 |
|
60 |
2 |
1.68 |
1.752 |
0.248 |
|
60 |
1.5 |
1.5 |
1.5 |
0 |
|
70 |
1.5 |
1.5 |
1.5 |
0 |
|
70 |
0 |
0 |
0 |
0 |
However, the total volume of
infiltration implied by the resulting (trial) actual infiltration rate is not
equal to the observed infiltration volume, as shown in the graph above. A
correction must be made to the initial index. The correction is equal to:
Df = (1.68 – 1.5)*20/60 =
0.072 cm/h
With this correction, the
f-index is :
f = 1.752 cm/h
The actual rate of
infiltration and the corresponding rate of production of rainfall excess
corresponding to the final f-index are tabulated above and shown in the graph
below.