BFC 32002 Hydrology
Chapter 3. Evaporation, Transpiration & Infiltration
Prepared by:
Siti Nazahiyah Rahmat
Wan Afnizan Wan Mohamed
Hartini Kasmin
Learning Outcomes
After completing this chapter, the students should be able to :
• simulating the rate of evaporated and transpired water over time in
modeling conceptual.
• defining the infiltration process and estimate the infiltration rate.
2
What is Evaporation, Transpiration and Infiltration?
Three types of water loss/outflow (also known as initial loss) in hydrologic cycle. Each of losses has its own
character and contribution; and occurs at different location
Transpiration is the process
where plants absorb water
through the roots and then
give off water vapour
through pores in their
leaves. (2)
Evapotranspiration (ET) is the sum
of
evaporation
and
plant
transpiration from the earth's
land surface to atmosphere. (3)
Evaporation is the process by which water is
transformed from the liquid phase to vapour
phase (transferred from the land and water
masses of the earth to the atmosphere) (1)
Infiltration is the
process by which
precipitation or
water soaks into
subsurface soils
and moves into
rocks
through
cracks and pore
spaces (4)
3
Infiltration
• The flow of water/precipitation into the ground through
the soil surface and the process can be easily understood
through a simple analogy.
• It replenishes soil moisture, recharge aquifer & support
stream flows duirng dry periods
• This analogy can be simplified in two important aspects,
which are:
• INFILTRATION RATE - maximum rate at which the
ground can absorb water
• FIELD CAPACITY - volume of water that ground can
hold.
Infiltration Capacity
BFC32002_Ch3/ZARINA'S
Infiltration Capacity, fc
f0 = initial infiltration capacity,
cm/hr or mm/hr
fc = final constant infiltration
capacity, cm/h or mm/h
Horton model
BFC32002_Ch3/ZARINA'S
Factors Affecting Infiltration
Three main factors:
(a) Characterictics of Soil
• TEXTURE, STRUCTURE, PERMEABILITY, UNDER DRAINAGE and
TYPE OF SOIL.
• Loose soil texture will has larger fc
• Soil with a good underneath drainage would obviously have a higher
infiltration capacity. Dry type of soil can absorb more water than one
whose has full pore.
• Land use has a significant influence on fc , for instance, a forest soil which
is rich with organic matter will have much higher value of constant
infiltration rate that the similar types of soil in an urban area where is
subjected to compaction.
Factors Affecting Infiltration
(b) Soil Surface
• At the soil surface, the impact of raindrops causes the fines in the soils to
be displaced and these in turn can clog the pore spaces in the upper layers.
This is an important factor affecting the infiltration capacity.
• Thus a surface covered by grass and other vegetation which can reduce
this process has a pronounced influence on the value of fc >> high
infiltration.
• Viessman and Lewis (2003) stated that infiltration rate for bare-soil is 2.5
mm/h - 25 mm/h.
Factors Affecting Infiltration
(c) Fluid Characteristics
• Water infiltrating into the soil will have many impurities, both in solution and
suspension.
• The turbidity of water, especially the clay and colloid content is an important
factor as suspended particles block the fines pores in the soil and reduce its
infiltration capacity.
• The temperature of the water is also a factor in the sense that it affects the
viscosity of the water which in turn affects the infiltration rate (high
temperature will increase fc).
• Besides that, contamination of the water by dissolved salts also affects the soil
structure and then the infiltration rate (high contamination will increase fc).
How to measure infiltration
• Infiltration characteristics of soil can be obtained by conducting controlled
experiment on small areas.
• The experiment set-up is called an infiltrometer, which are flooding type
infiltrometer and rainfall simulator.
How to measure infiltration
(a) Flooding Type Infiltrometer
• consist a metal cylinder and open at both ends (30
cm dia & 60 cm long), planted into the ground to a
depth of 50 cm.
• water is poured to a depth of 5 cm and pointer is set to mark the water level.
• add water to keep the water level at the tip of the pointer as infiltration proceeds,
and may take 2 to 3 hours till reach uniform rate.
• experiments are continued is obtained, surface of the soils is usually protected by
a perforated disk to prevent formation of turbidity and its settling on the soil
surface.
• Disadvantage of simple ring: infiltered water spreads at the outlet from the tube,
and can’t be figured as area in which infiltration takes place.
How to measure infiltration
(a) Flooding Type Infiltrometer
• double ring is used to overcome problem of area.
• the rings are inserted in to the ground and water is maintained
on the soil surface to a common fixed level.
• the outer ring provides a water jacket to the infiltering water of the inner ring and
hence, prevents the spreading out of the water from the inner tube.
• the measurement of water volume is done in the inner ring only.
• main disadvantages of flooding type infiltrometer are:
1. The raindrop effect is not simulated.
2. The driving of the tube or rings disturbs the soil structure.
3. The results of the infiltrometer depend to some extent on their size with the larger
meters give less rates than the smaller ones and this is due to the border effect.
How to measure infiltration
(b) Rainfall Simulator
• this instrument give low values than
flooding type infiltrometers, due to the
rainfall effect and turbidity of the surface
soil
• consist a small plot of land (about 2 m × 4 m size), series of nozzles and
measures apparatus.
• the nozzles produce raindrops fall a height of 2 m and capable in producing
various intensities of rainfall.
• Using the water budget equation involves volume of rainfall, infiltration and
runoff, infiltration rate and its variation with time can be calculated.
• If the rainfall intensities is higher than the infiltration rate, infiltration capacity
values are obtained.
Infiltration Methods

Two (2) methods :-

Horton model - focused
 Green – Ampt Model
Infiltration Methods
(a) Horton Model
The
temporal
variation
in
infiltration rate is applicable when
the water is continuously ponded
above the soil column.
Figure 3.13 : Infiltration rate graph
Where:f = Infiltration rate capacity
(depth/time) at a given time (mm/hr
or cm/hr)
k = Decay constant which is representing
the rate of decrease in f
capacity
fc = Asymptotic/ constant infiltration
rate (t → ∞)
fo = Initial infiltration capacity, t = 0
t = time
Infiltration Methods
(a) Horton Model
In cases where water is not continuously ponded above the soil column, the
potential infiltration fp can be expressed in terms of the cumulative
infiltration, F by implicit relationship of total infiltration:
t
t
0
0
[
F( t ) = ∫ f ( t )dt = ∫ f c + (f o − f c ) e − kt
(
fo − fc )

− kt 
F = f c t +
1−e

k

0
(
)
t
] dt
Example 3.7
A catchment soil has Horton infiltration parameters: fo = 100 mm/h, fc = 20
mm/h and k = 2 min-1. What rainfall rate would result in ponding from
beginning of the storm? If this rainfall rate is maintained for 40 minutes,
describe the infiltration as a function of time during the storm.
Given :fo = 100 mm/h
fc = 20 mm/hr
k = 2 min-1
Find :f?
Example 3.7
Solution:
The potential infiltration rate varies between a maximum of 100 mm/h (fo) and
minimum of 20 mm/h (fc). Any storm in which the rainfall rate exceeds 100
mm/hr during the entire storm will cause ponding from the beginning of the
storm. Under these circumstance, the infiltration rate, f as a function of time is
given as equation as
From Horton’s equation :
Substitude fo, fc and k values :
( 0 < t < 40 min )
Example 3.8
An initial infiltration was recorded as 5.5 cm/hr during 10 hours of rainfall.
Given that fc and k is 0.4 cm/hr and 0.32 respectively, determine;
a) Infiltration at 5 hours.
b) Total infiltration within first 8 hours.
c) Total infiltration between 5 and 10 hours from rainfall begin.
Solution:
fo = 5.5 cm/hr, fc = 0.4 cm/hr dan k = 0.32 h-1
a) Infiltration at 5 hours.
f = fc +( fo − fc )e( −kt )
f 5 = 0.4 + (5.5 − 0.4 )e −0.32 (5) = 1.43cm / hr
Solution:
b) Total infiltration within the first 8 hours.
c) Total infiltration between 5 and 10 hours from rainfall begin.
F = ∫ (f t )dt
( f − fc )
F = [ fc t + o
( 1 − e( −kt ) ) ] 10
5
K
5 .1
5 .1

 
(1 − e −0.32 x 5 )
F = ( 0.4 ) ( 10 ) +
( 1 − e −0.32 x10 ) − (0.4() 5 ) +
0.32
0.32

 

F = 4.56cm
Infiltration Methods
(b) Green-Ampt Model
The Green – Ampt model sometimes called as the delta function
model. Today known as one of the most realistic models of
infiltration available to the engineer in designing a storm water
management systems.
What is Infiltration Index?
• The average infiltration rate.
• Average rainfall at which P volume = R volume
• If the rainfall intensity is larger than Φ index,
the difference between rainfall and infiltration in
an interval of time represents the runoff volume
as shown as in figure.
• the amount of rainfall in excess of the index is called rainfall excess.
• the Φ index thus accounts for the total abstraction and enables runoff
magnitudes to be estimated for a given rainfall hyetograph
P−R
Φ index =
te
P = Total rainfall or precipitation (cm)
R = Total runoff (cm)
te = Time of rainfall excess
Example 3.9
A storm with 10 cm rainfall produced a direct runoff of 5.8 cm. Table
below shows the time distribution of the storm, estimate the Φ index.
Time (hour)
1
2
3
4
5
6
7
8
Rainfall (cm/h) 0.4 0.9 1.5 2.3 1.8 1.6 1.0 0.5
Example 3.9
(i) Sketch first rainfall hyetograph
Rainfall hyetograph
Ranfall Intensity (cm/hr)
2.5
2.3
2
1.8
1.5
1
0.9
1
0.5
1.6
1.5
0.5
0.4
Φ (1st assumption)
( 0 < Φ < 0.4 cm/hr )
Φ
Solution:
0
1
2
3
4
5
Time (hour)
te = 8 hours
6
7
8
Example 3.9
Solution:
Total rainfall, P = 0.4 (1) + 0.9 (1) + 1.5 (1) + 2.3 (1) + 1.8 (1) + 1.6 (1)
+ 1(1) + 0.5 (1) = 10 cm
Total runoff, R = 5.8 cm
Assume te is 8 hours,
P - R 10 − 5.8
Index φ =
=
= 0.525cm/h
te
8
But this value of Φ makes the rainfall of the first hour and eight hour
ineffective as their magnitude is less than 0.525 cm/h. The value of te is
need to modified.
(ii)
Do checking (is Φ follows the 1st assumption??)
Cheking ….
Rainfall hyetograph
Ranfall Intensity (cm/hr)
2.5
2.3
2
1.8
1.5
Φ (calculated)
1
0.9
1
0.5
1.6
1.5
0.5
0.4
0
1
2
3
4
5
Time (hour)
te = 8 hours
6
7
8
Φ = 0.525 cm/hr
NO !
Φ (1st assumption)
( 0 < Φ < 0.4 cm/hr )
Then, assume te is 6 hours.
Total rainfall, P = 10 - 0.4 – 0.5 = 9.1 cm
Then,
P - R 9.1 − 5.8
Index φ =
=
= 0.55cm/h
6
te
This value of Φ is satisfactory and by calculating the rainfall excess
Time (hour)
Rainfall excess
(cm)
1
2
3
4
5
6
7
8
0
0.35
0.95
1.75
1.25
1.05
0.45
0
Total rainfall excess = 5.8 cm = total runoff
Time (hour)
.
Rainfall (cm/h)
1
2
3
4
5
6
7
8
0.4 0.9 1.5 2.3 1.8 1.6 1.0 0.5
(iii)
Modify the te value :
Rainfall hyetograph
2.3
2
1.8
1.5
Φ (2nd assumption)
( 0.5 < Φ < 0.9 cm/hr )
1
0.9
1
0.5
1.6
1.5
0.5
0.4
Φ
Ranfall Intensity (cm/hr)
2.5
0
1
2
3
4
5
Time (hour)
te = 6 hours
6
7
8
(v)
Again! Do checking (is Φ follows the 2nd assumption??)
Rainfall hyetograph
2.3
2
1.8
1.5
Φ (calculated)
1
0.9
1
0.5
1.6
1.5
( 0.5 < Φ < 0.9 cm/hr )
Φ (2nd assumption)
0.5
0.4
Φ = 0.55 cm/hr
Φ
Ranfall Intensity (cm/hr)
2.5
0
1
2
3
4
5
6
Time (hour)
te = 6 hours
7
8
Answer → YES !! GOT IT
Thus :
Φ = 0.55 cm/hr
Checking Rainfall Excess with Runoff depth given in the question :
Rainfall hyetograph
Ranfall Intensity (cm/hr)
2.5
2.3
2
1.8
1.5
1
0.9
1
0.5
1.6
1.5
Φ = 0.55 cm/hr
0.5
0.4
0
1
2
3
4
5
Time (hour)
6
7
8
Hence :
P excess = (0.9 – 0.55)(1) + (1.5 – 0.55 )(1) + (2.3 – 0.55)(1)
+ (1.8 – 0.55)(1)+ (1.6 – 0.55)(1) + (1 – 0.55)(1)
P excess = 5.8 cm = R (5.8 cm) ok!
Example 3.10
need to convert unit (m)
The rainfall intensity in the 50 hectar of catchment area is given below. If volume of
surface runoff is 30000 m3, estimate Φ index for the catchment area and sketch the
circumstances in form of hyetograph.
P = Intensity × time
Solution:
= mm
hr
× hr
Runoff, R = (3 × 104)/(0.5 × 1000 × 1000) = 60 mm
Total rainfall = (5+10+38+25+13+5)(1) = 96 mm
P - R 96 − 60
=
= 6mm/h
Then, Index φ =
te
6
But this value of Φ makes the rainfall of the first hour and six hour
ineffective as their magnitude is less than 6 mm/h.
Time
(hour)
1
2
3
4
5
6
7
Rainfall
intensity
(mm/hour)
5
10
38
25
13
5
0
(i)
Sketch first rainfall hyetograph
Rainfall hyetograph
38
35
30
25
25
20
13
15
10
10
5
Φ (1st assumption)
( 0 < Φ < 0.5 cm/hr )
5
5
0
0
1
2
3
4
5
Time (hour)
te = 6 hours
6
7
Φ
Ranfall Intensity (mm/hr)
40
(ii)
Do checking (is Φ follows the 1st assumption??)
Answer → NO !
Rainfall hyetograph
38
35
30
25
25
Φ (calculated)
Φ = 6 mm/hr
20
13
15
10
10
5
Φ (1st assumption)
( 0 < Φ < 0.5 cm/hr )
5
5
0
0
1
2
3
4
5
Time (hour)
te = 6 hours
6
7
Φ
Ranfall Intensity (mm/hr)
40
(iii)
Modify the te value :
Rainfall hyetograph
38
35
30
Φ (2nd assumption)
( 5 < Φ < 10 cm/hr )
25
25
20
13
15
10
10
5
5
Φ
Ranfall Intensity (mm/hr)
40
0
1
2
3
4
5
Time (hour)
te = 4 hours
5
0
6
7
(iv)
Calculate again the Φ index :
P−R
Φ index =
te
However, P become new :
P = 96 – (5 + 5)(1) = 86 mm
therefore :
86 − 60
Φ index =
= 6.5 mm/hr
4
(v)
Again! Do checking (is Φ follows the 2nd assumption??)
Rainfall hyetograph
38
35
Φ (calculated)
Φ = 6.5 mm/hr
30
25
25
Φ (2nd assumption)
( 5 < Φ < 10 cm/hr )
20
13
15
10
10
5
5
5
Φ
:
Ranfall Intensity (mm/hr)
40
0
1
2
3
4
5
Time (hour)
te = 4 hours
Answer → YES !! GOT IT
6
0
7