Methods of Estimating Evaporation

Methods of Estimating Evaporation:
1. Mass Transfer Methods (Myers Formula)
2. Energy Budget Methods
3. Combined Methods (Penman Equation)
4. Water Budget Method (Pan Evaporation Method)

1. Mass transfer method
Evaporation driven by
–Vapor pressure gradient
–Wind speed
E = f(u) (es-ea)
= (a + bu) ( es-ea)

es: saturation vapor pressure at temperature T of the water surface
ea: vapor pressure at some fixed level above the water surface
u : wind speed at some level above surface
a,b : empirical constants

Some formulas use a zero value for the constant “a”in the formula due to the small local air movements with velocities insufficient to remove excess vapor from a above a pan surface. Harbeck and Meyers (1970) present the following equation.
E = bu2 * ( es-e2)
Where:
E=Evaporation (cm/day)
b=0.012 for Lake Hefner, 0.018 for Lake Mead
es=vapor pressure at water surface (mb)
e2=vapor pressure 2 m above water surface (mb)
u2=wind speed 2 m above water surface (m/s)

2. Energy budget method









The overall energy budget for a lake can be written as
QN = Qe + Qh - Qv + Q?
QN: net radiation absorbed by water body [cal/cm2-day]
(solar radiation –reflection –radiation from lake)
Qe: evaporation energy
Qh: sensible heat transfer (conduction and convection to the atmosphere)
Qv: advected energy of inflow and outflow
Q?: change in stored energy in the water body the overall energy budget

Sensible heat transfer difficult to measure
Qh ? R x Qe

R = ? (T_(s-) T_a)/(e_(s-) e_a )

Ta: air temperature [°C]
Ts: water surface temperature [°C]
ea: vapor pressure of the air [mb]
es: saturation vapor pressure at water surface temp. [mb]
?: psychrometric constant = 0.66 (P/1000), P: atmospheric pressure in mb
Daily evaporation depth: E = ? Q_e/??L?_e cm/day

Energy balance QN = Qe +Qh- Qv + Q?
Qh = R * Qe
QN = Qe *(1+R) - Qv + Q?
QN = E??L?_e (1+R) - Qv + Q?
E = (Q_N + Q_v- ?Q ?_?)/(??L?_e (1+R)) cm/day
with
Q in [cal/cm2-day]
Le in [cal/g]: the latent heat of vaporization
? in [g/cm3]: mass density of evaporated water
•Most accurate method

3. Combined method (Penman, 1948)
Combined mass transfer and energy budget :
E ??L?_e= ?/( ?+?) QN + ?/( ?+?) Ea
Le = 597.3 - 0.57 (T-0 C0)
?: slope of es vs t curve or
? = ?de?_s/( d_T ) = (?2.7489*10?^8*4278.6)/?(T+242.79)?^2 e^(((-4278.6)/(T+242.79)))
QN = is net absorbed radiation
Ea: drying power from equation
? = 0.66 * ( P/1000 )
Ea= ??L?_e (a+bu) (esa-ea) cal/cm2-day
where
a,b : empirical constants
esa: saturation vapor pressure at air temp.
ea: actual vapor pressure

Example :
Assume Meyer s formula applies to a lake:
E = 0.0106 (1+0.1 u)(es-ea) [cm/day]
u in mi/h, e in mb
Given:
Ta= 32.2°C
u = 32 km/h = 20 mi/h
RH = 30%
QN= 400 cal/cm2-day
Estimate daily evaporation using Penman s formula.
Solution
?= (2.7489x?10?^8 x4278.6)/( ?(T+242.79)?^2 ) exp{-(4278.6)/((T+242.79) )}
with T = 32.2oc , ? =2.72 mb/ oc
Actual and saturation vapor pressure:
esa (T = 32.2oc) = 48.1mb
ea= RH x esa = 0.3 x 48.1 =14.4 mb
Latent heat of evaporation at air temperature:
Le = 597.3 – 0.57 x 32.2 = 579 cal/g
Ea = ? Le 0.0106 (1+0.1u) (esa-ea)
= 1 x 579 x 0.0106 (1+0.1x20) (48.1-14.4)
= 1590 cal/cm2-day
Penman s equation:
E ??L?_e= ?/( ?+?) QN + ?/( ?+?) Ea
= 2.72/( 2.72+0.66) 400 + 0.66/( 2.72+0.66) 1590
= 632 cal/cm2-day
or
E = 632/( 1 x 579) = 1.1 cm/day

4: Water budget method
Applicable to lake evaporation
? storage = input – output
? S = (I+P)-(O+E+GW)
Or
E = -?S+I+P-O-GW
I : inflow [cm]
P : precipitation [cm]
O: outflow [cm]
E : Evaporation [cm]
GW: Groundwater seepage [cm]