EVAPORATION
EVAPORATION
EVAPORATION
WIND SPEED
• Wind aids in removing the evaporated water vapour from the zone of
evaporation and consequently creates greater scope for evaporation.
WATER DEPTH/ HEAT STORAGE IN
WATER BODIES
•Deep water bodies have more heat storage than shallow ones.
• A deep lake may store radiation energy received in summer and
release in winter causing less evaporation in summer and more
evaporation in winter compared to a shallow lake exposed to a similar
situation.
TURBIDITY
• Also affects the rate of evaporation by affecting the heat
transfer within the depth of water body.
ATMOSPHERIC PRESSURE
The wind velocity can be assumed to follow the 1/7 power law
Uh = C h 1/7
Where:
Uh = wind velocity at a height h above the ground and
•C = constant.
This equation can be used to determine the velocity at any desired level if Uh is
known.
Example 1 : (a) A reservoir with a surface area of 250hectares had the
following average values of parameters during a week : water
temperature = 20 deg. C, relative humidity = 40% wind velocity at 1.0 m
above ground = 16km/h. Estimate the average daily evaporation from
the lake and volume of water Evaporated from the lake during that one
week.
Solution :
ew = 17.54 mm of Hg
ea = 0.40 x 17.54 = 7.02 mm of Hg
u9 = wind velocity at a height of 9.0 m above ground
u1 = 16 km/h u9 = ?
uh = C (h) 1/7
uh = C (1) 1/7 = 16 km/h
u9/u1 = C ((9) 1/7) / C ((1) 1/7)
u9 = u1 (9) 1/7
= 16 (9) 1/7
= 21.9 km/h
By Meyer’s formula
Average daily evaporation from the lake
EL = KM (ew – ea) (1 + u9/16)
EL = 0.36 (17.54 – 7.02) (1 + 21.9/16)
= 8.97 mm/day
Evaporated volume in 7 days
EL = KM (ew – ea) (1 + u9/16)
= 7 x 8.97/1000 x 250 x10000
= 157,000 m3
• (b) An ISi Standard evaporation at the 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 occuracy if Meyer’s method
relative to the pan evaporation measurements, (ii) estimate the
volume of water evaporated from the lake in that week .
Solution:
Lake evaporation = Cp x Pan evaporation
i. Daily Evaporation as per Pan evaporimeter
= .80 x (72.00 / 7) = 8.23 mm
Mayer’s formula ( error) : 8.97 – 8.23 = 0.74 mm
Thus, Meyer’s method overestimates the evaporation relative to the
Pan.
• Percentage over estimation by Meyer’s formula
= (0.74/8.23) x 100 = 9%
ii. Considering the Pan measurements as the basis, volume of water
evaporated from the lake in 7 days
= 7 x (8.23/1000) x 250 x 10000
= 144,025 cubic meter
ANALYTICAL METHODS OF
EVAPORATION ESTIMATION
The analytical methods for the determination of lake evaporation can
be broadly classified into three categories as:
1.Water-budget method
2.Energy-balance method
3.Mass- transfer method
WATER-BUDGET METHOD
• The simplest of the three analytical method.
• Least reliable.
• Involves writing the hydrological continuity equation for the lake and
determining the evaporation from a knowledge or estimation of
other variables
Continuity Equation:
P + Vis +Vig + = Vos + Vog + EL + Δ S + TL
Where,
P = daily evaporation
Vis = daily surface inflow into the lake
Vig = daily groundwater inflow
Vos = daily surface outflow from the lake
Vog = daily surface outflow
` EL = daily lake evaporation
Δ S = increase in lake storage in a day
TL = daily transpiration loss
WATER BUDGET METHOD
The water balance equation for a basin states that all the water
that enter in a system in specified period of time must be consumed,
stored or go out as surface or subsurface flow.
Inflow = Outflow + Change in storage
I = O + ΔS
Where,
I= Inflow in system
O= Outflow in system
ΔS= Change in storage in a given time period
For a specified area and for specified period of time , the water balance
equation can be written as,
P = Q + E+ ΔS
Where,
P= Precipitation
Q = Runoff
E= Evapotranspiration
ΔS = Change in storage
EXAMPLE 1:
Water at a constant rate of 370 cumec was observed to be entering into Tarbela Reservoir
in a certain season. If outflow from the reservoir including infiltration and evaporation losses is 280
cumec, find out the change in storage of reservoir for 10 such days.
Solution:
Given: I= 370 cumec
O = 280 cumec
ΔS = ?
According to hydrologic equation,
I - O = ΔS / Δt
370 – 280 = ΔS / Δt =90 cumec
Δt =10 x 24 = 240 hours = 240 x 60 x 60 = 864x10³ sec
Total change in storage = ΔS = (ΔS / Δt) x Δt= 90 x 864x10³
= 777.6x10^4 m3
EXAMPLE 2:
Flow of River Chenab at Marala Barrage varied linearly from 34 cumec (m3/sec) to 283
cumec in 10- hours during a flood. The flow variation at Khanki Barrage, downstream of Marala was
observed to be from 28 to 255 cumec during the above mentioned time. Assuming no lateral flow
in or out of the reach, find out the rate of change of storage of the river reach between Marala and
Khanki. What is total change in storage of the reach in this period?
Given: I1 = 34 cumec I2 = 283 cumec
O1 = 28 cumec O2 = 255 cumec
Solution:
I = ( I1 + I2 ) / 2 = ( 34 + 283 ) / 2 = 158.5 cumec
O = ( O1 + O2 ) / 2 = (28 + 255 ) = 141.5 cumec
∆S / ∆t = ?
According to hydrologic equation( I1 + I2 ) /2 - ( O1 + O2 ) / 2 = ∆S / ∆t= ∆S / ∆t = 17 cumec
∆t = 10 hours = 10 x 60 x 60 = 3.6 x 104 sec
Total change in storage = ∆S = (∆S / ∆t) x ∆t= 17 x 36,000
= 6.12 x 10^5 m3
EXAMPLE 3:
A catchment of area 100 hectares received a rainfall of 10 cm in 2 hrs. At
the outlet of the catchment, the stream draining the catchment was dry before the
storm and experience a run off lasting for 8 hrs. with an average discharge of 2m^3
/sec. The stream was again dry after the runoff event. What is the loss of the
water?
Solution :
Inflow Volume = 100 x 10000 x 0.1 = 100000 m^3
Outflow Volume = ½ x 8 x 60x 60 = 14400 m^3
∆S= Inflow-Outflow
= 100000 – 14400
= 85600 m ^3
ENERGY-BUDGET METHOD
• Application of the law of conservation energy.
• The energy available for evaporation is determined by considering the
incoming energy, outgoing energy and energy stored in the water
over a known time interval.
ENERGY-BUDGET METHOD
Hn = Ha + He + Hg + Hs + Hi
Where,
Hn = net heat energy received by the water surface
= Hc (1- r) - Hb
In which, Hc (1 – r) = 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
= ρ L EL
where ρ = density of water ,
L = latent heat of evaporation and EL = evaporation in mm
MASS -TRANSFER METHOD
• When wind flows on the surface, a boundary is formed.
• This method is based on turbulent mass transfer in the boundary
layer to calculate the mass of water vapor transferred from surface to
the surrounding atmosphere.
Where,
E = Evaporation in mm/h
z1 & z2 = Arbitrary levels above water surface
e1 & e2 = Vapor pressure at z1 & z2 in km/h
v1 & v2 = wind velocity at in km/h
T = Average temperature in C between z1 & z2 .
• The mass-transfer approach is based on Dalton’s aerodynamic law
giving the relationship between evaporation and vapour pressure as:
E = k (ew – ea) (4.14)
Where E = direct evaporation
k = a coefficient and depending on the wind velocity, atmospheric
pressure and other factors
ew= saturation vapour pressure corresponding to the water-surface
temperature
ea = the vapour pressure of the air.
Mean daily temperature and relative humidity may be used in
determining mean vapour pressure ea and mean saturation deficit
(ew – ea). Equation 4.14 was originally proposed by Harbeck and
Meyers (1970)
RESERVOIR EVAPORATION
AND
METHOD FOR ITS REDUCTION
The water volume lost due to evaporation from a reservoir in a
month is calculated as:
VE= A Epm Cp
Where,
VE = volume of lost water lost in evaporation in a month
A = average reservoir area during the month
Epm = pan evaporation loss in meters in a month
= EL in mm /day x No. of days in the month x 10^ -3
Cp = relevant pan coefficient
METHODS TO REDUCE
EVAPORATION LOSSES
i. REDUCTION OF SURFACE AREA
- The volume of water lost by evaporation is directly proportional
to the surface area of the water body, therefore, the reduction of
surface are wherever feasible reduces evaporation losses.
- Measures like having deep reservoirs in a place of wider ones
and elimination of shallow areas can be considered under this category.
ii. MECHANICAL COVERS
- Permanent roofs over the reservoir, temporary roofs and floating
roofs such as rafts and light- weight floating particles can be adopted
wherever feasible.
- These measures are limited to very small water bodies such as
ponds.
iii. CHEMICAL FILMS
- This method consists of applying a thin chemical film on the
water surface to reduce evaporation.
- The only feasible method available for reduction of evaporation
of reservoirs up to moderate size.
Certain chemicals such as cetyl alcohol ( most suitable chemical
to use as an evaporation inhibitor) and stearyl alcohol form
monomolecular layers on a water surface. These layers act as
evaporation inhibitors by preventing the water molecules past them.
EVAPORATION REDUCTION
From free surfaces Evaporation losses from a fully exposed water
surface are essentially a function of the velocity and saturation deficit
of the air blowing over the water surface, and the water temperature.
Evaporation losses are held to a minimum by:
(a) Exposing the least possible water-surface area. This in turn means
that streams and reservoirs should be kept deep instead of wide;
(b) Covering the water surface
(c) Controlling aquatic growth
(d) Creating afforestation around reservoirs that would act as
windbreakers. However, this method has been found to be useful
under limited conditions for small ponds
(e) Storing water underground instead of creating a surface reservoir.
To accomplish this there are physical and legal problems in preserving
the water so stored from adverse withdrawal).
(f) Making increased use of underground water
(g) Integrated operation of reservoirs
(h) Treatment with chemical water evaporation retardents (WER).