Chapter 3_Hydrological Losses Apr2024.pdf

hydrological losses, infiltration, evaporation, phi-index, initial losses, infiltration capacity curve, double ring infiltrometer

1. Hydrological Science
Chapter 3: Hydrological Losses
(5 hours)
Prof. Dr. Hari Krishna Shrestha
Contact: [email protected]
April 2024Engineering Hydrology_Chapter 3_HKS
3. Hydrological Losses (5 hrs) (17%)
1.Initial losses (Interception and Depression Storage)
2.Evaporation process: factors affecting evaporation (vapor pressure, radiation,
temperature, humidity, wind, atmospheric pressure, soluble salts, heat storage in water
bodies)
3.Evapotranspiration (AET, PET), evaporation equations (Penman’s equation)
4.Infiltration
1.Measurement of infiltration, (Flooding type and rainfall simulator)
2.Infiltration indices (-index, W-index)

2. 3.1 Initial Losses (interception and depression storage)
•Losses refer to precipitation amount not available for surface runoff. Major
loss factors are: interception, depression storage, evapotranspiration, and
infiltration. The combination of the first two factors is initial losses.
•The amount of rainfall caught by the vegetation before it reaches the ground is
interception; this amount may be evaporated, drip to the ground or flows
along the stem to the ground.
•The ratio of interception to total rainfall decreases as rainfall intensity and/or
duration increase.
•The factors affecting interception are rainfall intensity (amount versus
duration), air temperature, humidity and wind speed during rainfall duration,
and the nature of the vegetation (density, type and orientation of the leaves).
•Depression Storage: precipitation amount needed to fill the local depressions
(natural and/or artificial). Surface runoff occurs only after filling the depressions.
•Major factors of depression storage are: size of the depressions, initial water
volume in the depressions, soil type, antecedent moisture condition, and rainfall
intensity.
•If initial losses are high, the peak flow due to a rainfall event will be low.

3. 3.2 Evaporation Process
3.2.1 Meteorological factors of Evaporation
(1. Radiation, 2. Temperature, 2. Vapor Pressure, 4. Humidity, 5. Wind)
Evaporation Process: Absorption of solar radiation results
in increased heat energy in the water body, which
initiates high level of motion within water molecules
(increase in kinetic energy), resulting in escape of
water molecules from the water surface once the
kinetic energy overcomes the surface tension.
The evaporation “loss” is important during dry-season/
dry-year or in arid regions. The evaporation loss from
shallow reservoirs (large surface area compared to
depth) can be “significant” in terms of available water.
•How does each factor affect evaporation rate?
•What is the annual evaporation rate of Pokhara?
Annually, how much water evaporate from Fewa Lake?
Estimate the annual water balance of:
Fewa/Begnas/Rupa/Rara lake.
Humidity
E
Solar Radiation
Air Temperature
Wind Speed
E

4. Solar Radiation
• Sunshine: Potential  hours (N)
•N is a ƒ(latitude, day of a month)
•Actual  hours (n), a ƒ(?)
•n ≤ N?
•Role of n in weather
•As n  ET ? ? ?
•Actinometers and radiometers are
used to measure intensity of
radiant energy. The data is used in
studies of evapotranspiration,
snowmelt, Climate Change.
•Why do we care about  in WRM
projects?
Solar Radiation at DHM Mahakali Field Office, Attaria

5. 3.2.1.2 Air Temperature
Temperature: measure of object’s hotness
Directly affects the evaporation from a basin and
hence affects the catchment water balance.
Temperature of atmospheric air decreases at an
average rate of about 6˚C per 1000 m increase
in altitude within the troposphere.
Relatively constant (with altitude) in the lower
part of the stratosphere.

6. Temperature (T) 
•T is a ƒ(, U,,,, latitude/longitude)
•Role of T in water balance
•Relation between T & H
•As T  ET ? ? ?
•Why do we care about T in WRM projects?

7. Factors Affecting Temperature
A complex function of several interdependent variables
–Latitude (decrease increases temp),
–Altitude (decrease increases temp),
–ocean currents (stabilizes),
–distance from sea (increase increases temp),
–Winds (increase decreases temp),
–cloud cover (increase decreases temp), and
–aspect (land slope and its orientation, south facing slope has
higher temperature in northern hemisphere)
–GHG (increase increases temp)
–Heat trapping structures.
What effect it will have on location, design and operation of WR projects?

8. Temperature measurement
•The thermometers used to measure temperature,
must be placed where air circulation is relatively
unobstructed, and yet protected from the direct rays
of the sun and from precipitation.
•All thermometers should be placed at the same height
above the ground for the recorded temperatures to
be comparable.
•The Max-Min thermometers are used to record daily
maximum and minimum temperature.

9. Air Temperature Measurement
Daily Max.
temperature
Daily Min.
temperature
Dry bulb
temperature
Wet bulb
temperature
Weekly Normal Temp. Trend, TIA

10. Annual Mean Temperature Trend
2
y = 0.039x + 19.335
R= 0.5973
18.0
18.5
19.0
19.5
20.0
20.5
21.0
1975 1977 1979 1981 1983 1985 1987 1989 1991 1993 1995 1997 1999 2001 2003 2005
Year
Temperature (
°
C)
• All Nepal Temperature is
increasing steadily.
• 1.7°C increase between
1975 and 2005
Temperature Trend, Nepal
What effect it will have on location, design and operation of WR projects?

11. Highes t temperature record in K athmandu Airport
y = 0.0614x + 31.399
R
2
= 0.2298
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
1968 1970 1972 1974 1976 1978 1980 1982 1984 1986 1988 1990 1992 1994 1996 1998 2000 2002 2004 2006
year
Temperature (°C )

13. 3.2.1.3 Vapor Pressure and Humidity
•state of atmosphere in relation to amount of water vapor it
contains
•closely related to temperature – higher the air temperature, more
vapor the air can hold.
•saturation vapor pressure (e
w or e
s) goes up with air temp.
•as temperature goes up e
w also goes up.
•Significance of Humidity: The amount of water vapor in air
effectively controls the weather condition by controlling
evapotranspiration from land and water surfaces.
•Evaporation rate is proportional to difference between saturated
vapor pressure at water temperature (e
w) and actual vapor
pressure in air (e
a).
E
L = C (e
w – e
a),
where, E
L is lake/reservoir evaporation rate, C is Lake coefficient
How do you determine the value of C?
What effect it will have on location, design and operation of WR projects?

14. Relative
Humidity (H)
•H is a ƒ(e
w - e
a)
•e
a ≤ e
w?
•Relation betweenand e
w
•As H  ET ? ? ?
•H = (e
a/e
w) ×100%; also written as RH.
Hygrometer
measures RH

15. Relationship between air temperature
and saturation vapor pressure
Variation of temperature and
relative humidity with time
e
w
= 6.108 * (e^((17.27 * T
W
) / (237.3 + T
W
)));
wetbulb temp. T
W
in Celsius , e
w
in mb

16. Vapor Pressure
•One of the empirical equations used to calculate
vapor pressure (e) is:
•e
a = e
w – (0.000367) (5 p /9) (T – T
w) [1 + (5 T
w – 448)/14139]
where,
T and T
w are dry- and wet-bulb temperature (°C) of a
psychrometer consisting of two thermometers,
e
w is the saturation vapor pressure (mb)
corresponding to air temperature T (°C),

p is the atmospheric pressure (mb);
(1 mb = 100 Pascal (Pa); 133.322 Pa = 1 mmHg)
e
w
= 6.108 * (e^((17.27 * T
W
) / (237.3 + T
W
))); wet-bulb
temp. T
W
in Celsius, e
w
in mb

17. Atmospheric pressure (P)
The atmospheric pressure, P, is the pressure exerted by the
weight of the earth's atmosphere. Evaporation at high altitudes is
promoted due to low atmospheric pressure as expressed in the
psychrometric constant. The effect is, however, small and in the
calculation procedures, the average value for a location is
sufficient. A simplification of the ideal gas law, assuming 20°C for
a standard atmosphere, can be employed to calculate P:
??????=1013.25
293−0.0065 ??????
293
5.26
; P is atm. pressure (mb), z (in m amsl)
??????=1013251−2.25577× 10
−5
??????
5.25588
P is atm. pressure in Pascal (Pa) and z (in m amsl)
P = P
0×exp[(-g ×M ×(h -h
0)/(R ×T)]
P and P
0 have same unit; P
0 = pressure at h
0 (sea level)
g = 9.81 ms
-2
; M = molar mass of air = 0.0289644 kg/mol;
R = universal gas constant = 8.31432 (N-m/mol-K)
T = air temp at height h (in Kelvin) = °C+ 273.15
Air pressure at sea level = 1 atm = 1013.25 mb = 101325 Pa
P is ƒ(elev. & temp.)

18. Relative Humidity
It is the percentage ratio between the actual vapor pressure (e
a)
and the saturation vapor pressure (e
w) at the same temperature.
The relative humidity is not a direct measure of moisture in air.
H = 100 (e
a/e
w)
H = [(112 – 0.1 T + T
d)/(112 + 0.9 T)]
8
•The relative humidity is also defined as the percentage ratio
between the amount of water vapor actually contained per unit
volume and the amount of water vapor that it can hold at the
same temperature when saturated.
•Relation between relative humidity, air temperature and dew point
temperature:
T–Td ≈ (14.55+0.1147 T) (1 – H) + [(2.5+0.007 T)
(1– H)]
3
+ [(15.9 + 0.117 T) (1–H)]
14
where, T is in °C and H is in decimal fraction. This relation is correct
within 0.3°C.

19. Dew point
•Dew point (D): Temperature at which the space
becomes saturated when air is cooled under
constant pressure and with constant water vapor
content. It is the temperature having saturation
vapor pressure e
w = existing vapor pressure e.
D calculation from air temperature and RH
•B = (ln(RH / 100) + ((17.27 * T) / (237.3 + T))) / 17.27
•D = (237.3 * B) / (1 - B)
–Where: T = Air Temperature (Dry Bulb) in Celsius (C)
–RH = Relative Humidity in percent (%)
–B = intermediate value (no units)
–D = Dewpoint in Celsius (C)
•Depth of precipitable water: It is the amount of
water vapor in a layer of air.
https://ag.arizona.edu/azmet/dewpoint.html

20. Wind Speed (U)
•U is a ƒ(elevation, T)
•U/U
0 = (Z/Z
0)
k
•Relation between U and e
w
•As U  ET ? ? ?
•Why do we care about U in WRM projects?
anemometer
Wind is a moving air. Wind is one of the major factors that affect the climate
and evapotranspiration rate from water surface. Higher wind speed results
in higher ET rate from a water surface as the wind replaces saturated air just
above the water surface by unsaturated air.
3.2.1.4 Wind Speed

21. (V/V
0) = (Z/Z
0)
k

where V is the wind speed at height Z above the ground,
V
0 = Wind speed at anemometer level Z
0,
k = 1/7.

22. Types of
Wind roses
and its
applications
from Excel

23. Example
of wind
rose plot
in Nepal

24. 3.2.2 Effect of Soluble Salts on evaporation
Depends on salt properties and evaporation conditions
Presence of soluble salt in water:
•increases boiling point of water and reduces evaporation
•reduces rate of water molecule movement (osmotic
effects) and decreases evaporation
•salt increases surface tension and decreases evaporation
•in very high concentration results in crystallization,
forming a barrier to evaporation
•Some soluble salts are hygroscopic: absorb moisture from
air and increase evaporation rate (in dry environment)
and retain moisture and decrease evaporation rate (in
humid environment).

25. 3.2.3 Effect of heat stored in water bodies on
evaporation
Temperature stability: large water bodies can store heat without temperature
change; when air cools and water has high heat stored, the temperature does
not fluctuates along with the air, and results in sustained evaporation even in
cold air temperature. The same phenomenon occurs on a daily basis when the
daily air temperature fluctuation is high.
Seasonal variation: heat stored in large water bodies results in comparatively
higher evaporation in cold season. During cold season air temperature drops
rapidly but the temperature of the water bodies drops slowly resulting in higher
temperature, and higher evaporation.

26. 3.3 Evapotranspiration (ET)
•Potential Evapotranspiration (PET)
•Actual Evapotranspiration (AET), depends on water availability
and vegetation type, in addition to PEC factors
•AET≤ PET
•AET = PET when plenty of water is available
3.3.1 Actual Evapotranspiration

27. 3.3.2 PET from Penman’s Equation
PET = [A H
n + E
a ] / [A + ]
•PET = Daily Potential Evapotranspiration rate (mm/day)
•A = slope of the saturation vapor pressure vs. temperature curve at
the mean air temperature, mmHg/°C
•H
n = net radiation of mm evaporable water per day
•E
a = parameter including wind velocity & saturation deficit =
0.35(1+u
2/160)(e
w-e
a)
• = psychrometer constant = 0.49 mm Hg/°C
•H
n = A + B + C where,
•A = H
a (1-r) (a + b c), B =  T
a
4
(0.56 – 0.092 √e
a), C = (0.10 + 0.90 c)
•a = constant depending on latitude , a = 0.29 cos  b = 0.52
•c = n/N, n = actual bright sunshine duration, N = max. potential
sunshine duration
•r = albedo = reflection coefficient
•e
w = saturation vapor pressure, u
2 = wind speed at 2 meters above
ground
•The values of A, H
a, N, and e
w are found in standard textbooks

28. Example of Penman Equation application for PET estimation
Calculate the PET of a location with the following local data of November:
Latitude: 28°4’; Avg. Monthly temp.: 20°C, RH = 75%;
Avg. sunshine hours per day = 9 hrs; Avg. wind speed at 2 m above ground
= 85 km/day; Surface cover: plant-covered ground
Solution:
For 20°C, A = 1.05 mm/°C, e
w = 17.54 mmHg (from table)
For latitude of 28°4’ and November:
H
a = 9.506 mm water per day and N = 10.7 hours per day (from table),
r = 0.25 (from table)
Constants:  = 0.49 mm Hg/°C, b = 0.52,  = 2.01 x 10
-9
mm/day,
Calculations:
n/N = 9/10.7 = 0.84
e
a = e
w x 75% = 13.16 mmHg
a = 0.29 cos  = 0.2559, T
a = 273 + 20°C = 293K,  T
a
4
= 14.814;
H
n = H
a(1-r)(a+b n/N) -  T
a
4
(0.56 – 0.092 √e
a) (0.10+0.9 n/N) = 2.07mm
water day
-1
E
a = 0.35 (1 + u
2/160) (e
w – e
a) = 2.35 mm water per day
PET = (A H
n + E
a ) / (A + ) = 2.16 mm water per day

29. Numerical Example of Penman Equation application for PET estimation
Calculate the PET for a location with the following local data:
Latitude =28°4’ 28.0670.498 (rad)Month =NovemberSurface cover: plant-covered ground
Altitude = 230m Avg. Avg. Monthly temp = 20°C
RH = 75% Avg. sunshine hours per day = 9hrs
Avg. Wind speed at 2 m above ground = 85km/day
Solution:
For 20°C: A = 1.05mmHg/°C ew = 17.54mmHg (table) ew =17.549
For latitude of 28°4’ and November: Ha = 9.506mm water per day (from table)
r =0.25(from table) N = 10.7hours per day (from table)
Constants:  = 0.49mm Hg b = 0.52 =2.01E-09mm/day
Calculations:
n/N = 0.84 e
a
= e
w
x 75% =13.155mmHg
a = 0.29 cos  =0.2559 T
a
= 273 + 20°C = 293Kelvin
 Ta
4
= 14.8138
H
n
= H
a
(1-r)(a+b n/N) -T
a
4
(0.56 –0.092 √e
a
) (0.10+0.9 n/N) =2.070mm water per day
Ea= 0.35 (1 + u2/160) (es–ea) = 2.35mm water per day
PET = (A Hn + Ea ) / (A + ) = 2.16mm water per day
Calculate the PET of a location with the following local data of November:
Latitude: 28°4’; Avg. Monthly temp.: 20°C, RH = 75%;
Avg. sunshine hours per day = 9 hrs; Avg. wind speed at 2 m above ground = 85
km/day; Surface cover: plant-covered ground

30. Surface Range of r values
Close ground crops0.15 – 0.25
Bare lands 0.05 – 0.45
Water surface 0.05
Snow 0.45 – 0.95
Temp
(°C)
e
w
(mmHg)
A
(mm/°C)
0 4.580.30
5 6.540.45
109.210.60
1512.790.80
2017.541.05
2523.761.40
27.527.541.61
3031.821.85
3542.812.35
4055.322.95
4571.203.66
Lat.JanFebMarAprMayJunJulAugSepOctNovDec
012.112.112.112.112.112.112.112.112.112.112.112.1
1011.611.812.112.412.612.712.612.412.911.911.711.5
2011.111.512.012.613.113.313.212.812.311.711.210.9
3010.411.112.012.913.714.113.913.212.411.510.610.2
409.610.711.913.214.415.014.713.812.511.210.09.4
Lat.JanFebMarAprMayJunJulAugSepOctNovDec
014.515.015.214.713.913.413.514.214.915.014.614.3
1012.813.914.815.215.014.814.815.014.914.113.112.4
2010.812.313.915.215.715.815.715.314.412.911.210.3
308.510.512.714.816.016.516.215.313.511.39.17.9
406.08.311.013.915.916.716.314.812.29.36.75.4
Mean Monthly Solar Radiation H
a in mm evaporable H
2O day
-1
Mean Monthly Values of Possible Sunshine hours, N
28.07, 9.506

31. Variation of Potential Evapotranspiration with Elevation in Nepal Author(s): Larason Lambert & B.D.Chitrakar
Source: Mountain Research and Development, Vol. 9, No. 2 (May, 1989), pp. 145-152
Published by: International Mountain Society Stable URL: http://www.jstor.org/stable/3673477
Fewa Lake: Elevation: 742 – 784 m, area: 4.43 to 5.7km
2
; estimate annual PET,
and quarterly PET.

32. 3.3 Evaporimeters
A) Measurement of Evaporation
•Class A Pan
•ISI Standard Pan
•Colorado Sunken Pan
•USGS Floating Pan
Pan Coefficient (C
p)
Lake Evaporation = C
p × Pan evaporation
B) Empirical Evaporation Equations
a)Meyer’s Formula (1915): E
L = K
M (e
w-e
a)(1 + u
9 /16) (in mm/day)
K
M = 0.36 for large deep lake, 0.5 for small shallow lake, e
w and e
a in mmHg
u
9 = wind speed 9 m above ground (km/h), (e
w-e
a) = saturation deficit
b)Rohwer’s Formula (1931): E
L = 0.771 (1.465 – 0.000732
p
a)(0.44+0.0733 u
0) (e
w-e
a)
Air pressure p = 1013.25 [(293 -0.0065 z)/293]
5.26
; p in mb, z in m

33. Class A Evaporation Pan Colorado Sunken Pan

34. Colorado Sunken Pan
Water level
same as GL
Floating Pan

35. Example: A deep lake with a surface area of 250 ha has the following
vales of parameters during a week: temperature 20.46 °C, relative
humidity = 40%, wind velocity at 1.0 m above ground = 16 km/hr.
Estimate the average daily evaporation from the lake and volume of
water evaporated from the lake in during the week from Meyer’s formula.
E
L = K
M (e
w-e
a)(1+u
9/16)
Steps: (1) Calculate e
w from temperature, (2) Calculate e
a from e
w and H;
(3) Calculate u
9 from u1, (4) Set K
M value for deep lake, (5) use Mayer
formula for E
L. (6) Convert depth to volume.
Solution:
e
w= 0.6672 e
0.0613T
; temp. T in °C, e
win kPa;
1 mmHg= 133.322 Pascals (Pa)
e
w= 17.54 mmHg, e
w= 0.40 x 17.54 = 7.02 mmHg
u
9= windvelocity9 m aboveground
u
9/u
1= (9/1)
1/7
= 21.9 km/h; K
M= 0.36 for deeplake
Practical application: Minimum water
requirement of 50 lcd for basic
sufficiency according to WHO
standards. How many persons can get
water if this evap. can be stopped?

36. Numerical Example (lake evaporation estimation using pan evaporation): At
the beginning of a certain week, the depth of water in an evaporation pan, 1.2 meter
diameter, was 7.75 cm. During the week, the rainfall was 3.8 cm and 2.5 cm of
water was removed from the pan to keep the depth of water in it within a fixed
range. At the end of the week, the gauge indicated a depth of 8.32 cm of water in
the pan. Using a suitable evaporation pan coefficient, estimate the evaporation
during the week from the surface of a reservoir under similar atmospheric condition.

37. Numerical Example: Estimate the volume (in liter) of water evaporated from an
irrigation reservoir (surface area 1 km
2
) in a week, given the following data.
Pan type: Class A
DayInitial Water level
from surface (cm)
Final water level
from surface (cm)
Rainfall
(cm)
1 5 11 5
2 5 9 12
3 5 13 15
4 5 8 18
5 5 7 22
6 5 12 9
7 5 6 19
Elev. diff
(cm)
Pan Evap.
(cm)
E
L
(cm)
6 11 7.7
4 16 11.2
8 23 16.1
3 21 14.7
2 24 16.6
7 16 11.2
1 20 14.0
Weekly evaporation = 91.5 cm
Volume of evaporation = 91.5 cm x 1 km
2
=
In Kathmandu, on average, households' total per capita water consumption is70 litres
per capita per day (lcd) in dry season & 85 lcd in rainy season. This falls close to the min
m

water requirement of 50 lcd for basic sufficiency according to WHO specified standards.

38. Reducing evaporation
•Shade balls
•Chemical film: monomolecular
films of fatty alcohols, like
hexadecanol and octadecanol
•Mechanical cover: for small
reservoir
•Surface area reduction: when
possible
http://www.plasticsnews.com/article/20160115/NEWS/160119829/los-angeles-removing-shade-balls-from-
some-reservoirs
http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-50532005000700015

39. Fewa the second largest lake in
Nepal, lying at an altitude of
784m, covers an area of about
4.43km
2
with an average depth
of about 8.6m with maximum
water depth is 22.8m when
measured with echo-sounder
(depth measuring gauge;
PLASTIMO ECHOTEST II) on
December 5, 2009. Maximum
water capacity of the lake is
approximately 46 million m
3
.
If daily evaporation from
Fewa can be reduced by 5
mm, water volume available
for other uses = ?
The updated number of rivers and rivulets in Nepal, Energy Development
Commission (June 2016), is 11614.
http://www.renewableenergyworld.com/articles/2016/06/nepal-seeks-investors-for-
10-gw-of-electricity-by-2026.html
Begnas Lake, Kaski

40. 3.4 Infiltration
1.Soil type (gravel, sand, silt, clay: in decreasing order)
2.Permeability (hydraulic conductivity)
3.Vegetation cover (infiltration increases with vegetation cover,
as it slows surface flow in the initial phase of rainfall)
4.Ground slope (higher slope results in lower infiltration)
5.Rainfall intensity and duration (high intensity rainfall results in
less infiltration due to soil structure destruction)
6.Antecedent soil moisture content: higher soil moisture
content results in lower infiltration capacity
7.Land use / Land cover (Surface type): Black topped road,
brick/stone paved surface, dry cracked mud, tilled agricultural
field, dense forest, soft organic materials
8.Ground condition: freshly tilled soil has high infiltration
compared to compacted soil.
Factors affecting Infiltration
The process of water seeping into the ground under the action of
gravity. Higher infiltration results in lower peak flood discharge,
higher landslide susceptibility.

41. Soil type
Permeability
Vegetation
cover
Ground
slope
Rainfall
intensity/
duration
Antecedent soil
moisture
Land use /
Land cover
Ground
condition
infiltration factors

42. 3.4.1 Measurement of Infiltration
•In general, high infiltration → less runoff;
•high intensity rainfall → low infiltration;
•high AMC → high runoff
•The surface runoff resulting from a given storm is equal to
that portion of the rainfall which is not disposed of through (i)
interception and depression storage, (ii) evaporation during
the storm and (iii) infiltration. If (i) and (ii) are insignificant, or
can be assigned a reasonable values, runoff is a function only
of rainfall and infiltration. Hence, runoff can be estimated
from rainfall, if infiltration capacity curve of a particular basin
is known, and rainfall rate is continuously above the
infiltration capacity curve (ICC).
•Horton (1933) ICC equation: f
p = f
c + (f
o – f
c) e
-Kht
for 0>t>t
d
•f
p - f
c = (f
o – f
c) e
-K
ht
•ln(f
p - f
c) = ln(f
o – f
c) – K
ht (straight line in a ln-ln plot)

43. 3.4.1 Infiltrometers (flood type)
Single (Simple tube type) and Double Ring Infiltrometers are used to estimate
ICC of an area.

45. MinsSecst (secs)
Water
level
(cm)i (cm)
i
cum
measured
(cm)
1 0 0 0 17 0 0
23001800 15 2 2.0
36003600 13 2 4.0
490155415 11 2 6.0
5120457245 9 2 8.0
61801410814 7 2 10.0
72402014420 5 2 12.0
83002218022 3 2 14.0
Estimate the infiltration index (-index) from the following ring infiltration data.
0.000000
0.000100
0.000200
0.000300
0.000400
0.000500
0.000600
0.000700
0.000800
0.000900
0.001000
0.001100
0.001200
0 5000 10000 15000 20000
Infiltration Capacity Curve cm/s
Example Numerical exercise to estimate -index from infiltrometer data.-index
time (sec)
Water Level
(cm)
Net
Infiltration
Cumulative
Infiltration (cm)
Infiltration
Capacity
Curve cm/s
Segmental
Area 0.000295
0 17 0 0
1800 15 2 2 0.001111
3600 13 2 4 0.000556 1.5
5415 11 2 6 0.0003690.8393
7245 9 2 8 0.0002760.5905
10814 7 2 10 0.0001850.82272.8099
14420 5 2 12 0.0001390.58351.4966317
18022 3 2 14 0.0001110.44973.2888583
y-intercept
ratio =0.7969 Total Area4.785721.9757735
x-value at
-index =6873 1.313
-index =0.000295cm/sec 1.313
-index = 1.062cm/hr Difference in area % 0.0

46. 3.4.1 Rainfall Simulation for infiltration measurement
Under a controlled environment, a soil plot is enclosed and equipped with a
rainfall simulator, infiltrometers, meteorological factor monitoring instruments, and
outflow measuring device to measure inflow and outflow from the system. Since
the inflow (rainfall rate and volume) is known, and the outflows (discharge from
the outlet) and evapotranspiration loss from the system are also known or can be
estimated, the water balance method yields the infiltration depth and infiltration
capacity.

47. 3.4.2 Infiltration Indices ( & w)
(average rate of infiltration)
•The infiltration capacity curve is exponential in nature, however, an engineering
hydrologist is normally concerned with volume of water available, which can be
estimated by drawing a horizontal line so that the area under the curve is same.
The horizontal line represents average rate of infiltration (infiltration index)
•-index: average rate of infiltration for the duration of effective rainfall
•For flood estimation, in India:  = (I-R)/24; R = I
1.2
;  = function of soil type
•W-index: Similar to -index, except that initial losses are accounted for,
separately.
•W = (P – R – Ia)/t
e; P = total precipitation, R = total runoff,
Ia = initial losses, t
e = time duration of effective rainfall
Section 8-10 Hydrology for Engineers, 1958, Linsley,
Kohler and Paulhus, page 180

48. Three Numerical Examples of -index estimation based on rainfall and runoff data
Example 1Total Direct Runoff =2.5cmRequires only one iteration; ER duration constant after first iteration.
Time from start (hr) 12 3 456 7 8 9
Cumulative rainfall (cm)0.411.72.33.53.94.55.96
Total
RunoffDiff.
ER
Duration
Revised
-
index index (cm/hr)Incre. rainfall (cm)0.40.60.70.61.20.40.61.40.1
0.5 Runoff 0.10.20.10.7 0.10.9 2.10.4 60.4333
0.433Runoff 0.170.270.170.77 0.170.97 2.50
Example 2Total Direct Runoff =4cmRequires two iterations; ER duration change after first iteration.
Time from start (hr)12 3 456 7 89
Cumulative rainfall (cm)0.411.72.33.53.94.55.96
Total
RunoffDiff.
ER
Duration
Revised
-
index index (cm/hr)Incre. rainfall (cm)0.40.60.70.61.20.40.61.40.1
0.5Runoff 0.10.20.10.7 0.10.9 2.11.9 60.1833
0.1833Runoff 0.220.4170.520.421.020.220.421.22 4.43-0.4380.2375
0.2375Runoff 0.160.360.460.360.960.160.361.16 4 0
Example 3Catchment Area =5km
2
Total Direct Runoff =210000m
3
→ 4.2cm
Same as Example 2, but total direct runoff depth calculated from catchment area and total direct runoff given in volume.
Time from start (hr) 12 3 456 7 8 9
Cumulative rainfall (cm)0.411.72.33.53.94.55.96
Total
RunoffDiff.
ER
Duration
Revised
-
index index (cm/hr)Incre. rainfall (cm)0.40.60.70.61.20.40.61.40.1
0.5Runoff 0.10.20.10.7 0.10.9 2.12.1 6 0.15
0.15Runoff 0.250.450.550.451.050.250.451.25 4.7-0.580.2125
0.2125Runoff 0.190.390.490.390.990.190.391.19 4.20

49. 1.Discuss the relationship between and among different parameters, namely solar
radiation, air temperature, humidity, and wind speed.
2.What are the two main meteorological factors that you would consider are most
important in assessing technical feasibility of a water supply project? Justify your
answer with proper reasons.
3.Discuss the factors which affect the air temperature of a particular project location.
What impact the average and extreme air temperatures can have on the availability of
water at the project site.
4.What is the significance of temperature analysis in hydrological analysis of a water
resources development project?
5.What is the impact of the temporal variation of rainfall in the design of various
components of a water resources development project? What will be the effect if total
annual rainfall is equally divided in each of the 12 months?
6.A Class A Evaporation Pan (pan coefficient of 0.7) is placed close to the lake Fewa
(surface area 4.43 km
2
). On a particular day, the decrease in water level in the pan is
1.2 cm and rainfall on the same day is 5 mm. Estimate the volume of evaporation from
the lake on that day (in m
3
). Suggest ways to reduce evaporation loss.
7.In the current situation of low data availability, which method of reservoir evaporation
estimation do you consider to be the best? Why?
8.Discuss the role of infiltration capacity index in determining the excess runoff depth
from precipitation data.
Self-assessment: Answer these questions.

50. Sub Area
(Km
2
)
-index
(mm/hr)
Hourly rainfall (mm)
1
st
2-hours2
nd
2-hours3
rd
2-hours4
th
2-hours
CRN+5 10 26 48 22 10
CRN [phone]
CRN + 25 21 12 50 18 46
CRN+1 16 25 42 18 38
A 8-hour storm occurs over a watershed. The sub-area and average infiltration capacity of
the catchment are as follows. The CRN stands for the last two digits of your CRN. Calculate
the runoff from the catchment (in mm) and the two-hourly distribution of the effective
rainfall for the whole catchment (in mm). What will be the runoff if the -index (i) decreases
by 20% due to urbanization and (ii) increases by 10% due to afforestation?