Abstraction from Precipitation

 All abstractions from precipitation are those due to evaporation, transpiration,

infiltration, surface detention and storage.
 Evaporation from water bodies and soil masses together with transpiration from
vegetation is termed as evapotranspiration.

Evaporation Process

 Evaporation is the process in which a liquid

change to the gaseous state at the free
surface, below the boiling point through the
transfer of heat energy.
 The rate of evaporation is dependent on
i. The vapor pressures at the water
surface and air above
ii. Air and water temperatures
iii. Wind speed
iv. Atmospheric pressure
v. Quality of water
vi. Size of the water body

Vapor Pressure

 Rate of evaporation is proportional to the difference between the saturation vapor

pressure (SVP) at the water temperature, (ew ) and the actual vapor pressure in the
air (ea)
John Dalton’s law of Evaporation
𝑬 𝑳 = 𝑪(𝒆𝒘 − 𝒆𝒂 )
Where: EL = rate of evaporation (mm/day)
C= constant
ew = in mmHg
ea = in mmHg
 Evaporation continues till e w = ea. if ew > ea condensation takes place.

Temperature

 The rate of evaporation increases with an increase in the water temperature

Wind

 Wind helps to remove the evaporated water vapor from the zone of evaporation,
thereby creating greater scope for evaporation.

Atmospheric Pressure

 Other factors remaining the same, a decrease in atmospheric pressure (as in high
altitudes areas) increases the evaporation rate.

Soluble Salts

 When a solute is dissolved in water, the vapor pressure of the solution is less than
that of pure water and hence it causes reduction in the rate of evaporation. Under
identical conditions evaporation from sea water is about 2-3% less than that from
fresh water.

Heat Storage in water bodies

 Deep water bodies have more heat storage capacity than shallow water bodies.
The effect of heat storage is to change the seasonal evaporation rates and the
annual evaporation remains more or less unaltered.

Evaporimeters

 Estimation of evaporation is of importance in many hydrologic problem associated

with planning and operation of reservoirs and irrigation systems.
 The amount of water evaporated from a water surface is estimated by the following
methods
i. Using evaporimeter data
ii. Empirical evaporation equations
iii. Analytical methods.
Types of Evaporimeters

Evaporimeters are water-containing pans which are exposed to the atmosphere

and the loss of water by evaporation measured in them at regular intervals.
Meteorological data, such as humidity, wind movement, air and water temperatures and
precipitation are also noted along with evaporation measurement.

1. Class A Evaporation Pan

 A pan of diameter
1210mm and depth 255mm
 Depth of water is
maintained between 18 and 20cm
 The pan is made of
unpainted GI sheet
 The pan is placed on a
wooden platform of height 15cm
above ground level to allow free
air circulation below the pan
 Evaporation is measured
by measuring the depth of water in
a stilling well with a hook gauge.
2. ISI Standard Pan

 A pan of diameter 1220mm and

depth 255mm
 The pan is made of copper sheet
0.9mm thick, tinned inside and
painted white outside
 The pan is placed on a square
wooden platform of width
1225mm and height 100mm
above ground level to allow free
air circulation below the pan
 A fixed-point gauge indicates the
level of water
3. Colorado Sunken Pan
 920mm square pan made
of unpainted GI sheet,
460mm deep, and buried
into the ground within
100mm of the top
 Main advantage of this
pan – its aerodynamic and
radiation characteristics
are similar to that of a lake
 Disadvantages – difficult
to detect leaks, expensive
to install, extra care is
needed to keep the
surrounding area free
from tall grass, dust, etc.
 4. US Geological Survey Floating Pan
 A square pan of 900mm
sides and 450mm deep.
 Supported by drum floats
in the middle of a raft of
size 4.25m x 4.87m, it is
set afloat in a lake with a
view to simulate the
characteristics of a large
body of water
 Water level in the pan is
maintained at the same
level as that in the lake,
leaving a rim of 75mm.
 Diagonal baffles are
provided in the pan to
reduce surging in the pan
due to wave action
 Disadvantages – High
cost of installation and
maintenance, difficulty in
making measurements.

Pan Coefficient C p

 Pan Coefficient – Evaporation Pan are not exactly models of large reservoirs and
the following drawbacks:
i. They differ in the heat – storing capacity and heat transfer from the sides
and bottom.
ii. The height of the rim in an evaporation pan affects the wind action over the
surface.
 iii. The heat – transfer characteristics of the pan material is different from that
of the reservoir.
 Thus, a coefficient in introduced as
𝐿𝑎𝑘𝑒 𝐸𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑖𝑜𝑛 = 𝐶𝑝 𝑥 𝑝𝑎𝑛 𝑒𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑖𝑜𝑛
Where: Cp = pan coefficient
S. No. Types of pan Average Value Range
1. Class A Land Pan 0.700 0.60 – 0.80
2. ISI Pan (modified Class A) 0.875 0.65 – 1.10
3. Colorado Sunken Pan 0.805 0.75 – 0.86
4. USGS Floating Pan 0.760 0.70 – 0.82

Evaporation Station

 Arid zones – one station every 30,000 km2

 Humid temperate climates – one station for every 50,000 km2, and
 Cold regions – one station for every 100,000 km2

Empirical Evaporation Equations

Most of the available empirical equations for estimating lake evaporation are a
Dalton type equation of the general form:

𝐸𝐿 = 𝐾𝑓(𝑢)( 𝒆𝒘 − 𝒆𝒂 )

Where: EL = lake evaporation mm/day

ew = saturated vapor pressure at the water surface temperature in

mmHg

ea = actual vapor pressure of over-lying air at a specified height in

mmHg

f(u) = wind speed correction function

K = coefficient
Meyer’s Formula

𝒖𝟗
𝐸𝐿 = 𝐾𝑀 ( 𝒆𝒘 − 𝒆𝒂 )(𝟏 + )
𝟏𝟔

Where: EL = lake evaporation

ew = saturated vapor pressure at the water surface temperature in

mmHg

ea = actual vapor pressure of over-lying air at a specified height in

mmHg

u9 = monthly mean wind velocity in km/h at about 9m above ground

Km = 0.36 – large deep water and 0.50 – small, shallow waters

Rowher’s Formula

𝐸𝐿 = 0.771(1.465 − 0.000732𝑝𝑎 )(0.44 + 0.0733𝑢0 )(𝒆𝒘 − 𝒆𝒂 )

Where: EL = lake evaporation

ew = saturated vapor pressure at the water surface temperature in

mmHg

ea = actual vapor pressure of over-lying air at a specified height in

mmHg

pa = mean barometric reading in mmHg

u0 = mean wind velocity in km/h at ground level, which can be taken

to the velocity at 0.6m height above ground

Analytical Methods of Evaporation

1. Water – budget method

2. Energy – balance method
3. Mass – transfer method
Water – budget method

Water – Budget Method – simplest but the least reliable. If the unit of time is kept
very large, estimates of evaporation will be more accurate.

𝑃 + 𝑉𝑖𝑠 + 𝑉𝑖𝑔 = 𝑉𝑜𝑠 + 𝑉𝑜𝑔 + 𝐸𝐿 + ∆𝑆 + 𝑇𝐿

Where: P = daily precipitation

Vis = daily surface inflow into the lake

Vig = daily groundwater inflow

Vos = daily surface outflow from the lake

Vog = daily seepage outflow

EL = daily lake evaporation

∆S = increase in lake storage in a day

TL = daily transpiration loss

Energy – budget method

Energy – Budget Method – is application of law of conservation of energy. The

energy available for evaporation is determined by considering the incoming energy,
outgoing energy and energy stored in the water body over a known time interval.
 𝐻𝑛 = 𝐻𝑎 + 𝐻𝑒 + 𝐻𝑔 + 𝐻𝑠 + 𝐻𝑖

Where: Hn = net heat energy received by the water surface

= Hc (1 – r) – Hb

Hc (1 – r) – Hb = incoming solar radiation into a surface of reflection

coefficient (albedo) r

Hb = back radiation (long wave) from water body

Ha = sensible heat transfer from water surface to air

He = heat energy used up in evaporation

= 𝜌𝐿𝐸𝐿 where 𝜌 = density of water, L = latent heat of

evaporation and E L = evaporation in mm

Hg = heat flux into ground

Hs = heat stored in water body

Hi = neat heat conducted out of the system by water flow

Ha can be estimated as:

𝐻𝑎 𝑇𝑤 − 𝑇𝑎
𝛽= = 6.1𝑥10−4 𝑥𝑃𝑎
𝜌𝐿𝐸𝐿 𝑒 𝑤 − 𝑒𝑎

Where: Pa = atmospheric pressure in mmHg.

Tw = temperature of water surface in Celsius

Ta = temperature of air in Celcius

EL can be evaluated as:

𝐻𝑛 − 𝐻𝑔 − 𝐻𝑠 − 𝐻𝑖
𝐸𝐿 =
𝜌𝐿(1 + 𝛽)
Reservoir evaporation and methods for its reduction

The water volume lost due to evaporation from a reservoir in a month is calculated
as:

𝑉𝐸 = 𝐴𝐸𝑝𝑚 𝐶𝑝

Where: VE = volume of water lost in evaporation in a month (m3)

A = average reservoir area during the month (m2)

Epm = pan evaporation loss in meters in a month (m)

= EL in mm/day x No. of days in the month x10-3

Cp = relevant pan coefficient

Methods to reduce evaporation losses

1. Reduction of surface area – as the area increases the rate if evaporation also
increases
2. Mechanical covers – permanent roods over the reservoir, temporary roods and
floating roof such as rafts and light – weight floating particles
3. Chemical films – application of cetyl alcohol (hexadecanol) and stearyl alcohol
(octadecanol)

Sample Problems

1. A class A pan was set up adjacent to a lake. The depth of water in the pan at the
beginning of a certain week was 195mm. In that week there was a rainfall of 45
mm and 15 mm of water was removed from the pan to keep the water level within
the specified depth range. If the depth of the water in the pan at the end of the
week was 190mm calculate the pan evaporation. Using a suitable pan coefficient
estimate the lake evaporation in that week.
Solution:

Given: Class A Pan

Initial depth = 195 mm

P = 45 mm

Water removed = 15 mm

Final depth = 190 mm

Find: lake evaporation

Pan Evaporation = 195 mm + 45 mm – 15 mm -190 mm

= 35 mm

𝐿𝑎𝑘𝑒 𝐸𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑖𝑜𝑛 = 𝐶𝑝 𝑥 𝑝𝑎𝑛 𝑒𝑣𝑎𝑝𝑜𝑟𝑎𝑡𝑖𝑜𝑛

= 0.700 x 35 mm

= 24.5 mm

2. A canal is 80 km long and has average surface width of 15 m. If the evaporation

measured in a class A pan is 0.5 cm/day, what is the volume of water evaporated
in a month?

Solution:

Volume = (80 x 1000)m x 15m x0.5/100m

= 6000 cu.m

EL = Cp x pan evaporation
EL = 0.70 x 6000cu.m/day x 30 days
EL = 126,000 cu.m
 3. A reservoir with a surface area of 250 hectares had the following average values
of climate parameters during a week: Water Temperature = 20 o C, Relative
Humidity = 40%, Wind Velocity at 1.0 m above ground surface = 16 km/h. Estimate
the average daily evaporation from the lake by using Meyer’s Formula.

Given: A = 250 ha = 2.5x106 m2

Water Temp = 20 o C

Relative humidity = 40 % = 0.4

Km = 0.36

Wind velocity at 1 m above the ground surface = 16km/h

𝒖𝟗
𝐸𝐿 = 𝐾𝑀 ( 𝒆𝒘 − 𝒆𝒂 )(𝟏 + )
𝟏𝟔

ew = 17.54 mmHg

ea = 17.54 mmHg x .40

= 7.016 mmHg

Uh = Ch1/7

1
𝑢 9 𝐶ℎ7
=
𝑢1 𝐶ℎ17

1
𝑢9 97
= 1
16𝑘𝑚/ℎ 17

U9 = 16km/h (9)1/7

= 21.90 km/h

𝟐𝟏. 𝟗𝟎
𝐸𝐿 = 0.36( 𝟏𝟕. 𝟓𝟒 − 𝟕. 𝟎𝟏𝟔)(𝟏 + )
𝟏𝟔

= 8.97 mm/day
 4. An ISI Standard evaporation pan at the site indicated a pan coefficient of 0.80 on
the basis of calibration against controlled water budgeting method. If this pan
indicated an evaporation of 72 mm in the week under question, i.) estimate the
accuracy if Meyer ‘s Method relative to the pan evaporation measurements. ii.)
Also, estimate the volume of water evaporated in the lake in that week.

Given: ISI standard evaporation Pan

C p = 0.80

Pan evaporation = 72mm/week

Meyer’s formula = 8.97 mm/day

Pan evaporation = 72/7

= 10.28571429 mm/day x 0.8

= 8.23 mm/day

8.23 – 8.97 = -0.74mm

. 74
𝑥100 = 8.99 𝑠𝑎𝑦 𝟗%
8.23

8.23 𝑚𝑚
Volume = 7 days x 𝑚 x 2.5x106 m2
1000𝑚𝑚

= 144,025 m3