Analytical Methods of Evaporation Estimation 
Analytical Methods of Evaporation Estimation
The analytical methods for the determination of lake evaporation can be broadly classified into three categories as
 Water budget method,
 Energybalance method, and
 Masstransfer method.
WaterBudget Method
The waterbudget method is the simplest of the three analytical methods and is also the least reliable. It involves writing the hydrological continuity equation for the lake and determining the evaporation from a knowledge or estimation of other variables. Thus, considering the daily average values for a lake, the continuity equation is written as
P + VIS + VIG = VOS + VOG + EL + ΔS +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
 ΔS = increase in lake storage in a day 7 daily transpiration loss
 TL =daily transpiration loss
All quantities are in units of volume (m') or depth (mm) over a reference area. Equation (3.6) can be written as
EL=P+(VISVOS)+(VIGVOG)TLΔS (3.7)
In this, the terms P. V V, and AS can be measured. However, it is not possible to measure V V and T, and therefore these quantities can only be estimated Transpiration losses can be considered to be insignificant in some reservoirs. If the unit of time is kept large, say weeks or months, better accuracy in the estimate of E is possible. In view of the various uncertainties in the estimated values and the possibilities of errors in measured variables, the waterbudget method cannot be expected to give very accurate results. However, controlled studies such as at Lake Hefner in USA (1952) have given fairly accurate results by this method.
EnergyBudget Method
The energybudget method is an application of the 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. Considering the water body as in below Fig. the energy balance to the evaporating surface in a period of one day is give by
Hn = Ho + He + Hg + Hs + Hi
Hn = net heat energy received by the water surface
=He (1r)Hb
fig:Energy balance in water body 
He (1r) = incoming solar radiation into a surface of reflection coefficient r
 Hb = back radiation 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 , EL=evaporation in mm
 Hg=heat flux into the ground
 Hs=heat stored in water body
 Hi=net heat conducted out of the system by water flow
All the energy terms are in calories per square mm per day. If the time periods are short, the terms Hs, and Hi, can be neglected as negligibly small. All the terms except Ha, can either be measured or evaluated indirectly. The sensible heat term, which cannot be readily measured is estimated using Bowen's ratio β given by the expression
β=Ha/(ρLEL) =6.1 x 104 x pa ((TwTa)/(ewea))
where p =atmospheric pressure in mm of mercury,
ew=saturated vapour pressure in mm of mercury
ea=actual vapour pressure of air in mm of mercury.
Temperature of water surface in ︒C and
, temperature of air in ︒C From Eqs (3.8) and (3.9) EL can be evauated as
EL = (Hn – Hg –Hs  Hi)/ (ρL(1+β))
Estimation of evaporation in a lake by the energy balance method has been found to give satisfactory results, with errors of the order of 5% when applied to periods less than a week
MassTransfer Method
This method is based on theories of turbulent mass transfer in boundary layer to calculate the mass water vapour transfer from the surface to the surrounding atmosphere However, the details of the method can be found in published literature With the use of quantities measured by sophisticated (and expensive) instrumentation, this method can give satisfactory results
Example:
Following observation were made for conducting the water budget of a reservoir over a period of one month of 30 days
Average surface area 10 km2  Rainfall = 10 cm 
Mean surface inflow rate=10m3/s  Mean surface outflow rate= 15m3/s 
Fall in reservoir level=150m  Pan evaporation =20 cm 
Assume pan evaporation coefficient = 0.70 

Estimate the average seepage discharge during that month
Solution:
Area of reservoir = 10 x 10 m
INFLOW  OUTFLOW  
ITEM  VOLUME( m3)  ITEM  VOLUME(m3) 
SURFACE INFLOW =15*3600*24*30  25,920,000  SURFACE OUTFLOW=15*3600*24*30  38,880,000 
Rainfall=10*106*10/100  1,000,000  Evaporation=0.7*10*106*20/100  1,400,000 

 Seepage volume = se  se 
Total inflow volume  26,920,000  Total outflow  40,280,000+se 
Reduction in storage =10*106*1.5  15,000,000 


Reduction in storage = Total inflow volume  Total outflow volume
(40,280,000+se )(26,920,000) = 15,000,000
se = 1,640,000 = 164 Mm3. 29687
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