Chapter Three
Hydrologic abstraction
Definition
◼ Hydrologic abstractions are the process acting
to reduce total precipitation into effective
precipitation.
◼ Effective precipitation eventually produces
surface runoff.
◼ The difference between total and effective
precipitation is the depth abstracted by the
catchment.
Hydrologic Abstractions
◼ Interception
◼ Surface or Depression storage
◼ Evaporation and
Evapotranspiration
◼ Infiltration
Interception: First abstractive process
◼
The process by which precipitation is abstracted by
vegetation or other forms of surface cover.
L = S + kEt
L = interception loss (mm)
S = interception storage depth (0.25-1.25mm)
k = ratio of evaporating surface to horizontal projection
E = evaporation rate (mm/hr)
t = storm duration (hr)
Surface or Depression storage
◼ The process by which precipitation is abstracted by
being retained in puddles, ditches, and other natural
or artificial depressions on the land surface.
Vs = S d (1 − e − kPe )
Vs = equivalent depth of storage (mm)
Pe = excess precipitation = PT – L , (mm)
Sd = depression storage capacity (mm), (10-20mm)
K = constant
Evaporation & Evapotranspiration
◼ Evaporation is the process whereby liquid water is
converted to water vapour (vaporization) and
removed from the evaporating surface (vapour
removal).
◼ Water evaporates from a variety of surfaces, such as
lakes, rivers, pavements, soils and wet vegetation.
◼ Evaporation depends on: (1) net solar radiation, (2)
the saturation vapor pressure, (3) the vapor pressure
of the air, (4) air and water surface temperature, (5)
wind velocity, and (6) atmospheric pressure.
Cont…
◼ Solar radiation is the main source of heat
energy.
◼ Ability to transport vapor away from the
evaporation surface depends on wind velocity
and specific humidity.
◼ Vapor pressure is partial pressure exerted by
water vapor.
Transpiration
◼ Consists of the vaporization of liquid water contained in
plant tissues and the vapor removal to the atmosphere.
◼ Crops predominately lose their water through stomata.
◼ The process of evaporation from the land surface and
transpiration from vegetation
are collectively termed
Evapotranspiration.
◼ Potential evapotranspiration (PET) is the evapotranspiration
that would occur from a well vegetated surface when
moisture supply is not limiting.
Evaporation measurement
◼ By the use of an evaporation pan
◼ widely used is the U.S. Class A pan
◼ made up of unpainted galvanized iron,
◼ has a diameter of 120.7cm (4ft.) and a height
of 25.4cm (10in.) and is mounted about 15 cm
(6in.) above the ground
◼ correction factor is applied to the pan
evaporation measurement
◼ This correction factor is referred to as the pan
coefficient
Class A pan
Cont…
◼ E = Kp Epan
◼ where
E = evaporation from water body [mm/day],
Kp = pan coefficient [0.6-0.8],
Epan = pan evaporation [mm/day].
◼ network of much lesser density would be
required
Evapotranspiration measurement
◼ Lysimeters
are instruments designed to
measure actual evapotranspiration.
◼ A lysimeter is a tank of soil in which
vegetation is planted that resembles the
surrounding ground cover.
◼ The amount of evapotranspiration from the
lysimeter is measured by means of water
balance of all moisture inputs and outputs.
Method of Estimating Evaporation
1. Water Budget Method
2. Energy Budget Method
3. Mass transfer approach
Water Budget Method
◼ Reservoirs or lake evaporation is calculated as;
E = P + Q − O − I − S
E = volume evaporated from the reservoir,
P = precipitation falling directly onto the reservoir,
Q = surface runoff inflow into the reservoir,
O = outflow from the reservoir,
I = net volume infiltrated from the reservoir in to the
ground, and
∆S = change in stored volume.
Example
◼ Estimate the evaporation for a month for a
lake of 500hectar surface area. The mean
discharge from the lake is estimated to be
1m3/s. The monthly rainfall is about 10cm. A
stream flows with an average discharge of
2m3/s into the lake. The water level in the
lake dropped about 5cm in the month. The
seepage loss are negligible.
Solution
◼ Monthly inflow, Q = 2*3600*24*30 = 51.84*105m3
◼ Monthly outflow, O = 1*3600*24*30 = 25.9*105m3
◼ Monthly rainfall, P = 0.1*500*10,000 = 5*105m3
◼ Change in storage, ∆S = -0.05*500*10,000 = 2.5*105m3
◼ From water balance:
E = (5 + 51.84 – 25.9 – 0 – (2.5))*105 = 33.42*105m3
E = 33.42*105/(500*104) = 0.6684m = 66.84cm
Infiltration
◼ Infiltration is the process by which precipitation
is abstracted by seeping into the soil below the
land surface
◼ The abstracted water moves either laterally, as
interflow or moves vertically, by percolation
◼ In practice, infiltration rates are determined
either by the use of infiltrometers
◼ or by the analysis of rainfall-runoff data from
natural catchments.
Infiltration Curve
Infiltration Formulas
◼ Horton’s Equation
f = f c + ( f o − f c )e − kt
f = instantaneous infiltration rate;
fo = initial infiltration rate;
fc = final infiltration rate;
k = a constant; and
t = time in hours
❑Philp’s Equation
f = (1/ 2)St −1 / 2 + A
f = instantaneous infiltration rate;
S = Sorptivity, a function of soil suction
potential:
A = hydraulic conductivity;
t = time
Infiltration Index
◼ Difficulties in measuring some of the terms of
the above theoretical infiltration formula
◼ Infiltration indexes assume the infiltration rate
is constant throughout the storm duration
◼ In practice, the most commonly used
infiltration index is the Φ-index
◼ defined as the (constant) infiltration rate to be
subtracted from the prevailing rainfall rate in
order to obtain the runoff volume that actually
occurred.
Example
◼The following rainfall distribution was measured
during a 6-h storm.
Time (h)
0 1 2
3
4 5 6
Rainfall Intensity 0 0.5 1.5 1.2 0.3 1.0 0.5
(cm/h)
The runoff depth has been estimated as 2cm.
Calculate the Φ-index and effective rainfall.
Solution
M
rd =  (Rm −  * t )
m =1
rd = direct runoff depth (cm)
Rm = observed rainfall (cm)
M = number of rainfall that actually contribute to direct
runoff depth
Φ = constant infiltration rate (abstraction rate), (cm/hr)
t = time interval length, (hr)
Trial One
◼ If M = 1, the largest rainfall depth, Rm =
1.5cm, is selected and substituted into the
above equation, the value of Φ is -0.5cm/hr.
◼ However, this value isn't physically possible.
Trial Two
◼ If M = 2, R1= 1.5cm and R2= 1.2cm are selected and
substituted in the equation, the value of Φ is
0.35cm/hr.
◼ However, the value, Φ*t = 0.35cm, isn’t greater than
1cm and 0.5cm rainfall depth occurred in the rainfall
distribution.
Trial Three
◼ If M = 3, R1= 1.5cm, R2= 1.2cm and R3 = 1cm
◼ Φ= 0.57cm/hr
◼ The value of Φ is satisfactory because it
gives Φ*t = 0.57cm, greater than all the
rainfall outside that of the three assumed to
contribute to direct runoff.
Effective rainfall
Time
0
1
2
3
4
5
6
Intensity
0.5
1.5
1.2
0.3
1
0.5
Φ
0.57
0.57
0.57
0.57
0.57
0.57
Effective
rainfall
-
0.93
0.63
-
0.43
-