Measurement or Estimation of Evaporation
Evaporation / Evapotranspiration (ET)

Evaporation and transpiration jointly called evapotranspiration
(ET) is one of the components of hydrologic cycle. Loss of water in
the form of vapour from soil, snow, lake, streams, reservoir, seas
and depressions to the atmosphere due to energy of sun is called
evaporation. Transpiration is the process by which water leaves
the living plant body and enters the atmosphere as vapour.

The process involves collection of water in the body of the plant
and finally evaporation of it to the atmosphere from stomata of the
leaves. This loss of water by these two processes to the
atmosphere is most commonly called evapotranspiration (ET)
Measurement or Estimation of Evaporation
Factors Affecting Evaporation
There are various factors like hydrological, meteorological and
physical “that affect the rate of evaporation” from earth’s surface.
They are discussed as follows;

Radiation
Radiation is the most important factor of evaporation. Solar
radiation supplies continuous energy, which is essential for
evaporation. “Evaporation is directly proportional to radiation”.
Solar energy near the equator is more, therefore, evaporation is
much more.

Vapour pressure
“Evaporation rate varies directly with difference of vapour
pressure between air and water”. If E is the rate of evaporation
(mm/day) and, ew and ea are the vapour pressure in water and in
air, then
E = C (ew – ea) ………………………. (1)
Here C is constant, equation (1) is called “Dalton’s law”.
Measurement or Estimation of Evaporation
Factors Affecting Evaporation

Temperature
“Increase in air temperature increases evaporation” when other
factors remaining same yet the high correlation coefficient
between the two does not exit. In cold dry season although
temperature is less, rate of evaporation is more because some of
heat energy absorbed at lower depth in hot water is released in
cold season.

Wind velocity
“The increase in wind velocity increases evaporation”. Wind
removes the evaporated water vapour and thereby creates space
for new evaporated water vapour. When there is no wind above the
water body where the evaporated water vapour is in still condition,
further evaporation ceases to take place. If wind velocity over the
water body is high, it does not increases correspondingly the
evaporation. “There is always a critical velocity of wind beyond
which evaporation does not increase”.
Measurement or Estimation of Evaporation
Factors Affecting Evaporation

Atmospheric Pressure
If atmospheric pressure is more, according to “Dalton’s law”, ea is
more, hence less evaporation. “Thus, decrease in atmospheric
pressure can increase evaporation”. At higher altitude,
atmospheric pressure is low; hence evaporation should have been
more. But this is not necessary because temperature at higher
altitude is low which reduces evaporation.

Area of Water Surface
“Evaporation is directly proportional to the area exposed”. Hence,
if area is more, evaporation is more.

Quality of water
It also affects the rate of evaporation. “If water contains dissolved
salts, it reduces the saturated vapour pressure es” and by Dalton’s
law, E decreases. Also turbidity of water has some indirect effects.
Measurement or Estimation of Evaporation
Factors Affecting Evaporation

Nature of Evaporating surface
Evaporating surface is classified into three main surfaces; land
surface, water surface and snow surface. Temperature remaining
the same, evaporation from saturated soil surface is same to that
of adjacent water surface. “Evaporation decreases when the soil
surface is dry”. Thus, evaporation is dependent on the availability
of water on the surface. Further, it is again dependent on presence
or absence of vegetal cover. “Evaporation rate decreases in the
following order; bare ground, grass and croplands, light forests
and dense forest”.

Salinity of Water
It actually falls under quality of water. Yet if the soil is saline,
evaporation decreases.
Measurement or Estimation of Evaporation
Factors Affecting Evaporation

Depth of Water in the Water Body
If the depth is more, it increases evaporation in winter season. On
the other hand, low depth increases evaporation in summer as all
water gets warmed up by solar radiation.

Humidity
“If humidity is more, water holding capacity of air is less, so less
evaporation”. If water content is less in air, more evaporation will
take place.
Measurement or Estimation of Evaporation
Measurement of Evaporation
i) Empirical formulas

Fitzgerald’s Equation
Fitzgerald gave the following equation
E = (0.4 + 0.124V) (es – ea) ………………………. (1)
Where E is evaporation in mm/day, es is saturated vapour at the
temperature of water surface in (mm), ea is actual vapour pressure
of air in (mm) and V is average wind speed at the surface (km/hr).

Meyer’s Equation
E = C (1 + V/16) (es – ea) …………………………. (2)
The values of E, V, es and ea are already defined with their
magnitude and dimensions.
Measurement or Estimation of Evaporation
Measurement or Estimation of Evaporation

Rohwer’s Equation
Rohwer developed the following equation;
E=0.771(1.456 – 0.000732Pa)(0.44 + 0.0733V)(es – ea)…(3)

Horton’s Equation
E = 0.4 (2 – e-0.124V)(es – ea) …………………….(4)
the values of E, es, ea and V have the same significance as
explained

Lake Mead’s Equation
E = 0.0331V(es – ea)[1 – 0.03(Ta – Tw)] ….……….… (5)
Here Ta and Tw are average temperature in 0C of air and water
surface respectively.
Measurement or Estimation of Evaporation
ii. Water Budget Method or Storage Equation
Evaporation E from a reservoir or water body can be determined by
the following water budget or storage equation
E = P + I – O + Ou + ΔS …………………… (1)
Where P is the total precipitation, I is total inflow, O is the total
outflow, Ou is total underground inflow or outflow “which is
positive for inflow and negative for outflow” and ΔS is change in
storage (+ve for an increase in storage and –ve for decrease in
storage).
“All the parameters are converted into same units”, preferably in
terms of depth of water area, for some convenient time interval.
Measurement or Estimation of Evaporation
iii. Energy Budget Method

This method is based on the application “of the law of the
conservation of Energy in the form of heat”. The energy budget
equation from enclosed figure may be written as;
Qi – Qr – Ql – Qc – Qe = Qs – Qa
Qe = Qi – Qr – Ql – Qc – Qs + Qa
Where Qi is short solar radiation, Qr is reflected part of solar
every, Ql is long wave atmosphere radiation, Qc is conduction
energy in air, Qe is heat energy used in evaporation, Qs is
increase is stored energy and Qa is advective energy.

To calculate Qc, Bowen ration (r)
r = Qc / Qe
Since
Qe = pLE ………………….. (5.9)
Where ‘p’ is the density of water, ‘L’ is the latent heat of
evaporation and ‘E’ is evaporation in mm.
Measurement or Estimation of Evaporation
Thus,

r = Qc/Qe = Qc/pLE
Bowen ratio can again be expressed as
r = Qc/Qe = [Qi – Qr – Ql – Qe – Qs + Qa]
Qe
or
r = (Qi – Qr – Ql – Qs + Qa) – 1
Qe
or
(1 + r) = (Qi – Qr – Ql – Qs + Qa)
Qe
or
(1 + r) = [Qi – Qr – Ql – Qs + Qa]
pLE
Therefore E = [Qi – Qr – Ql – Qs + Qa]
pL(1 + r)
i.e. Qe = pLE
Measurement or Estimation of Evaporation

When parameters are measured for short period, Qs and Qa
(negligibly small) may be neglected. Then above Equation
becomes;
E=
Qi – Q r – Ql
pL(1 + r)
……………….. (a)
Again
E = (Qi – Qr – Ql) + (Qa – Qs) ……… (b)
pL(1 + r)
or
E=
Where
Qin
pL (1 + r)
……………… (c)
Qin = Net solar radiation and
r = 6.1 x 10-4 Pa [(Tw – Ta)
ew – ea
Measurement or Estimation of Evaporation
Heat Energy Budget (Picture)
Measurement or Estimation of Evaporation
iv. Pan Measurement Methods
A) US weather Bureau “Class A Pan (Surface Pan)”
The US weather Bureau Class a Pan or surface pan evaporimeter
along with its shown in the enclosed figure. It is made of galvanized
iron sheet. It is painted white.
The pan is kept in wooden platform as shown in the figure so that air
can circulate freely. Evaporation is measured by a “hook gauge in a
stilling well”. water level is measured daily. Water is added every day
upto the fixed level. “A pan coefficient of 0.7 is taken and thus pan
coefficient is”;
Pan coefficient = 0.7 = Lake evaporation / Pan evaporation
Measurement or Estimation of Evaporation
Advantages

It gives stable pan coefficient (0.6 to 0.8). The average value 0.7 is
normally adopted without appreciable error.

It is easy for observation i.e. measurement.

It has relative freedom of dirt and trash.

Cost of installation is reasonable low.
Disadvantages

The pan gives higher rate of evaporation than that of large free
water surface.

Effects of wind and radiation are more which overestimate the
expiration rate.
Measurement or Estimation of Evaporation
Pan A (Picture)
Measurement or Estimation of Evaporation
B) ISI Standard Pan

In the enclosed figure, “ISI standard pan has been sown”. It is
also known as modified class A pan. Dimensions of the pan are
shown in the figure.

It is placed in the vicinity of the lake or reservoir to determine
the evaporation of the lake. “It is covered with wire mesh of
galvanized iron to protect the water in the pan from birds”. Pan
is made of copper sheet of 0.9 mm thickness. The pan has a
stilling well with a point gauge and thermometer.

Amount of water lost can be measured by the point gauge.
Water is added to bring it back to the original level. Readings
are measured normally twice a day. “The annual pan coefficient
is 0.7”.
Measurement or Estimation of Evaporation
Advantages

As the pan is placed over wooden support, air circulation
around the pan is alright.
 Allowance for rainfall can be made if required.
 Evaporation rates obtained are quite compatible to the rate of
large water body.
Disadvantages
 If screened at the top, interference of wind takes place and
there by evaporation rate is reduced about 1.144 times as
reported from experiment.
 If left unscreened, birds-bathing in the pan and drinking water
from the pan may affect the evaporation.
Measurement or Estimation of Evaporation
Modified Pan A (Picture)
Measurement or Estimation of Evaporation
C) Colorado Sunken Pan:

The Colorado Sunken Pan with its usual dimensions is shown
in the enclosed figure. It is buried into the ground. Water level
in the pan is at ground level.
Advantages

The advantage of this pan is that radiation and aerodynamics
effects are similar to lake.
Disadvantage



It is difficult to detect leak, if any.
Extra precaution is to be taken for the effect of surrounding
tress in the area.
Installation is expensive
Measurement or Estimation of Evaporation
Sunken Pan (Picture)
Measurement or Estimation of Evaporation
D) US Geological survey floating Pan

To simulate the characteristics of large body of water, this
floating pan is adopted. “It is square pan of 90 cm side and 45
cm depth”. It is supported by drum floats in the middle of the
raft of size 4.25m x 4.87m.
 The water level in the pan is kept at the same level of the lake
with rim of 7.5 cm. To prevent wave action of the lake, diagonal
baffles are provided in the pan so that surging in the pan is
reduced.
Advantages

The pan floats in take and environment for evaporation is same
as that of the lake.
 Evaporation rates obtained is very close to lake evaporation.
Disadvantage



High cost of installation is there.
Maintenance cost is also more.
Measurement of evaporation is the main disadvantage.
Measurement or Estimation of Evaporation
Floating Pan A (Picture)
Estimation of Evapotranspiration
Pitche Evaporimeters or Atmometer

Atmometer is another device to measure evaporation. Atmometers
is provided with some special surfaces which is kept wet and from
which water loss by evaporation is recorded.

Pitche Atmometer or evaporimeter is a graduated glass tube of 1.5
cm in diameter and 30 cm long and one open end. It is filled with
water and its open end is covered with a dry paper, held in position
by metal clip.

The tube is held in inverted position so that water wets the paper
from which evaporation takes places. Water loss in the tube is
recorded as a measure of evaporation. Atmometer is shown in
(figure enclosed).
Estimation of Evapotranspiration
Picthe Evaporimeters or Atmometer
Advantages
i.
It is easy to handle as the size is small and reading are
obtained directly from graduated scale.
ii. For quick and rough estimate, it is better.
Disadvantages
i.
ii.
iii.
iv.
v.
vi.
Atmometer does not provide good measurement.
The readings obtained are more erratic them other standard
pan.
Because of its small size, rates of evaporation are in excess of
the rates of other evaporimeters.
Values are usually higher than those obtained by U.S. Weather
Bureau Class ‘A’ pan.
Readings tend to overestimate due to wind effects and
underestimate due to radiation effects.
Evaporating surface is often subject to contamination by dust
and other foreign.
Picthe Evaporimeters or Atmometer
26
Estimation of Evapotranspiration
Methods to Reduce Reservoir Evaporation

Evaporation from water body is a continuous process.
Multipurpose projects are always with a big reservoir with bigger
surface area, so loss due to evaporation is more. “Various
methods to reduce reservoir evaporation are discussed as
follows”;
Reduction of Surface Area of Reservoir
o
Evaporation is directly proportional to water surface area of the
reservoir. In selecting the site for reservoir, due consideration
should be given for deep reservoir with less surface area.
Reservoirs in deep gorges are preferable.
Wind Breakers
o
Some has advocated for wind breakers as wind velocity over the
surface increases the evaporation. If tall trees like Causerina are
allowed to grow on windward side of the reservoir, it may act as
natural wind breakers. Besides it may help in cooling the water
surface obstructing solar radiation at least for a part of the day.
Wind breakers are effective only in small reservoir.
Estimation of Evapotranspiration

Consumptive use (Cu) may be defined as the amount of water used
in evapotranspiration from an area under vegetation plus the water
used by the plants in their process for building up the plant
tissues.

The amount of water required in building up the metabolic process
is quite insignificant compared to evapotranspiration and hence,
evapotranspiration (ET) is practically equal to consumptive use
(Cu).

There
are
several
methods
of
estimating
(ET) or (Cu). The following are some of the methods;
this
Estimation of Evapotranspiration
i) Empirical Equations

Blaney-criddle Method

Thornthwaite Equation

Penman Equation

Christiansen Equation
ii) Field Measurement Methods

Lysimeters

Field Plots

Soil-moisture Depletion Studies

Water balance Method

Evaporation index Method
Estimation of Evapotranspiration
Empirical Equations

Blaney-criddle Method
Cu = ET = ∑k p (4.61t + 81.3)
100
Where ‘C’ is in cm, ‘t’ is mean monthly temperature in C0, ‘k’
monthly consumptive use coefficient determined from
experiment, ‘p’ is monthly percentage of hours sunshine.

Penman Equation
E = (ΔQin/Le) + (0.00061PEa) …………. (a)
Δ + 0.00061P
Δ is the slope of the saturated vapor pressure curve at air
temperature in (mm). ‘0C’,
From this calculated value of ‘E’, the potential
Evapotranspiration (PET) is calculated as under;
PET = KE
Where E is given in (a) above and the value of ‘k’ depends
types of crops.
Estimation of Evapotranspiration
Field Measurement Method

Water Balanced Method
Evapotranspiration (E) = [(Surface flow + Sub Surface Inflow +
Imported Water)] – [(Surface outflow
+ Sub Surface outflow + Domestic,
municipal and industrial use +
Exported water + Increase in surface
storage + Increase in ground water
storage)]

Evaporation Index Method
a high degree correlation between consumptive use (ET) and
evaporation (E) exist. Therefore, (ET) is express in terms of ‘E’
as under;
ET = KE
Where K is the coefficient giving the ratio of (ETIE) and was
found to vary with the stage of growth of the crop.
Calculation of Evaporation
Water Budget Method
32
Estimation of Evapotranspiration
Question-I
A class-A pan was setup adjacent to a lake. The depth of water in
the pan at beginning was 195 mm. in that week, a rainfall of 45 mm
came and 15 mm of water was removed from pan to keep the water
level in the specified depth. If the depth of water at end of the week
was 190 mm, calculate pan evaporation. Using suitable pan
coefficient, estimate lake evaporation in that week.
Solution
o
Initial depth of water in the pan = 195 mm
o
After the rainfall, total water in the pan = (195+45-15) mm = 225mm
o
At the end of the week depth water in the pan is 190 mm
o
Evaporation from the pan in the week = (225 – 190) mm = 35 mm
o
Assume suitable pan coefficient is 0.7
o
Evaporation from the lake in the week = (0.7 x 35) mm = 24.5 mm
Estimation of Evapotranspiration
Question-II
A reservoir has average surface area of 20 km2. In the month of June,
mean rate of inflow is 10 m3/sec. mean outflow is 15m3/sec, rainfall is
10 cm and change of storage is 16 million m3. assuming surface losses
to be 1.8cm, Estimate the Evaporation.
Solution
As per water budget equation;
Inflow + Precipitation = Outflow + Surface outflow + Change in storage + Evaporation
Now,
inflow = 10 m3/sec = 10 x 24 x 30 x 60 x 60 = 1.296 m
20 x 106
Precipitation = 10 cm = 0.1 m
Outflow = 15 m3/ sec = 10 x 24 x 30 x 60 x 60 = 1.944 m
20 x 106
Estimation of Evapotranspiration
Surface outflow = 1.8 cm = 0.018 m,
Storage = 16 x 106 = 0.8 = -0.8 m
20 x 106
Storage =ve, since outflow > inflow
Putting the value in the water budget equation
1.296 + 01 = 1.944 + 0.018 – 0.8 + E
E = 1.296 + 0.1 + 0.8 – 1.944 – 0.018 = 0.234 m = 23.4 cm
The evaporation in June is 23.4 cm
Question-III

The catchment area of an irrigation tank is 70 km2. The constant water spread
during October 2006 was 2 km2. During that month, the uniform precipitation over
the catchment was recorded to be 100 mm. 50% of the precipitation reaches the
tank. The irrigation canal discharges at a uniform rate of 1.00 m3/ sec in the month
of October.

Assuming seepage losses to be 50% of the evaporation losses, find out the daily rate
of evaporation for October 2006.
Solution:
Total Inflow
Outflow from canal
Loss of water
= 70 x 106 x 100 x 0.5 = 3.50 x 106m3
1000
= 1 x 3600 x 24 x 31 = 2.68 x 106m3
= 3.50 x 106 – 2.68 x 106 = 0.82 x 106 m3
= Seepage Loss + evaporation loss
Since seepage loss is 50% that of evaporation loss,
Loss of water
= 1.5 x evaporation loss
Therefore, evaporation loss = 0.82 x 106 = 0.55 x 106 m3
1.5
Rate of evaporation
= 0.55 x 106 x 1000 = 8.8 mm/day/m2
2 x 106 x 31
Question-IV

The catchment area of a reservoir is 10.0 km2. A uniform precipitation of 0.5 cm/h
for 2.0 h was observed on 7th of July. 50% of the runoff reached the reservoir. A
canal carrying a discharge of 1.25m3/sec is taken from the reservoir. The rate of
evaporation observed was 0.7 mm/m2/h. The seepage loss was observed to be 50%
of the evaporation loss. Find the change in the reservoir level on 4th July from 8:00
a.m. to 6:00 p.m. if the water spread of the reservoir was 0.476 km2.
Solution:
Inflow into the reservoir
Canal Outflow
= 10.0 x 106 x 0.5 x 2.0 x 0.5 = 5.0 x 104m3
100
= 1.25 x 10 x 3600 = 4.5 x 104m3
Evaporation and seepage losses = 0.476 x 106 x 10 x 1 x 0.7 x1.5
1000
= 0.5 x 104m3
Total outflow from the reservoir = 4.5 x 104m3 + 0.5 x 104
= 5.0 x 104 x 104m3 = inflow
Hence, there will not be any change in the reservoir water level.
Calculation of Evaporation
by
Emperical Formula
38
VARIOUS TERMS
40
41
42
43
Water Requirement of Crops and Soil Water Relationship
B) Empirical Methods
i) Blaney – Criddle Method
o
Blaney and Criddle derived an equation of consumptive use
(CU) of water. They initially derived the equation in FPS unit.
When that initial empirical equation is converted to MKS unit, it
is written as;
CU = ∑ kp (4.61t + 81.3)/100
o
Here CU is in cm/month, ‘t’ is mean monthly temperature in 0C,
‘k’ is monthly consumptive use coefficient determined from
experimental data, ‘p’ is monthly percentage of hours bright
sunshine, ∑refers the summation of all the months of the
season. Thus if ‘t’, ‘p’, ‘k’ are given CU can be estimated.
Water Requirement of Crops and Soil Water Relationship
Example
In a crop filed in Sahiwal in the month of November the percentage of
sunshine hours is 7.2 and mean temperature is 180C. If the
consumptive use coefficient of crop is assessed to be 0.7 for that
month, find the consumptive use of the crop in mm/day by BlaneyCridle method.
Solution
Blaney-Cridle equation is
CU = ∑ kp (4.61t + 81.3)/100
When CU in cm/month
Here
k = 0.7, p = 7.2 percent and t = 180C
CU = 0.7 x 7.2 (4.61 x 18 + 81.3)
100
= 8.264592 cm/month
= 0.2754864 cm/day