# Evaporation

Evaporation
) The evaporation process
) Estimation of evaporation from water surface
Water balance method
Mass transfer method
Energy budget method
Combined method
) Potential evapotranspiration
1
Evaporation in hydrology
Water supply reservoir
– Loss of resources
Soil moisture condition
– Affect runoff conditions
Continuous watershed simulation models
– Overall water balance
Atmospheric science
Agricultural science
- generation of precipitation
- soil moisture available for
plants
2
e
es
Saturated air
Lake surface
3
Factors affecting evaporation
1. Availability of energy (latent heat)
•
•
•
Qe: Energy available for evaporation [cal/cm2-day]
E : evaporation [cm/day]
Le : latent heat of evaporation [cal/g]
Qe = MLe
[M : mass]
= E &times; (1 cm2 )ρ Le
E=
Qe
ρLe
[cm / day]
2. Saturation deficit, es-e
•
Evaporation rate proportional to es-e
E = C(es-e)
4
3. Temperature
•
Warm water will evaporate faster (less latent heat
required)
Le = 597.3 − 0.57 T
•
[cal/g], T in o C
Warm air can hold more vapor
4. Wind
•
Removes saturated air and maintains vapor pressure
Transpiration
•
Vegetation
5
Potential evapotranspiration
Potential evapotranspiration from land surface
≈
Evaporation from open water surface
Actual evapotranspiration depends on the dryness of the
soil.
6
Method 1: Water budget
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]
Limitations:
– Estimation of seepage (GW)
– Estimation of precipitation
7
Mass transfer method
Evaporation driven by
– Wind speed
E = f(u)(es − ea )
= ( a + b u)(es − ea )
es : saturation vapor pressure at temperature above surface
ea : vapor pressure at some level above surface
u : wind speed at some level above surface
a,b : empirical constants
8
Example
Using Meyer's formula
u 

E = 0.0269 1 +  (e − e ) [cm / day]
 16 
[u in km / h, e in mb]
8
s
a
8
determine lake evaporation for a month in which
–
–
–
–
average air temperature = 20&deg;C,
average water temperature = 15 &deg;C,
average wind speed at 8 m = 15 km/h
average relative humidity is 50%
9
Solution
Saturated vapor pressure above water surface
Air temperature above surface ≈ water temperature = 15 &deg;C
4278.6


es = 2.7489 &times; 108 exp −

 (T = 25) + 242.79 
= 17.0 mb
Vapor pressure at height of 8 m
es(20 &deg;C) = 23.3 mb
e = RH es = 0.5&times;23.3 = 11.7 mb
10
u
E = 0.0269 1 +  (e − e )
 16 
25 
= 0.0269 1 +
 (17.0 − 11.7 )
16 

= 0.37 cm / day
8
s
a
= 11.1 cm / month
11
Energy budget method
QN
Qe Qh
Qθ
Qv
Qe : evaporation energy
Qh : sensible heat transfer (water heats the air)
Qθ : change in stored energy
QN = Qe + Qh - Qv + Qθ
12
Energy budget method
Sensible heat transfer difficult to measure
Qh ≈ R &times; Qe
where
R=γ
Ts − Ta
es − ea
[Eq. 1.16]
Ta : air temperature [&deg;C]
Ts : water surface temperature [&deg;C]
ea : vapor pressure of the air [mb]
es : saturation vapor pressure at water surface temp. [mb]
γ : psychrometric constant = 0.66 (P/1000), P in mb
13
Energy budget method
Daily evaporation depth: E =
Energy balance
Qe
ρLe
[cm / day]
QN = Qe (1 + R) − Qv + Qθ
= EρLe (1 + R) − Qv + Qθ
or E =
QN + Qv − Qθ
ρLe (1 + R)
with
[cm / day]
Q in [cal/cm2-day]
Le in [cal/g]
ρ in [g/cm3]
14
Combined method (Penman)
Combined 'mass transfer' and 'energy budget':
γ
∆
EρL =
Q +
E
∆+γ
∆+γ
e
N
a
[units as before – Eq. 1.17]
∆ : slope of es vs t curve (at air temperature - equation 1.18)
Ea = ρLe ( a + bu)(esa − ea )
[cal / cm2 − day]
where a,b : empirical constants
esa : saturation vapor pressure at air temp.
ea : actual vapor pressure
15
Example (textbook Ex. 1.5B)
Assume Meyer's formula applies to a lake:
E = 0.0269 (1+0.1 u)(es-ea)
[cm/day]
u in mi/h, e in mb
Given:
Ta = 32.2&deg;C
u = 32 km/h = 20 mi/h
RH = 30%
QN = 400 cal/cm2-day
estimate daily evaporation using Penman's formula.
16
Solution
2.7489 &times; 108 &times; 4278.6
4278.6 

exp −
∆=

2
(T + 242.79)
 T + 242.79 
[eq. 1.18]
with T = 32.2oC, ∆ = 2.72 mb /oC
Actual and saturation vapor pressure:
esa (T = 32.2o C) = 48.1 mb
[eq. 1.6]
ea = RH &times; esa = 0.3 &times; 48.1 = 14.4 mb
Latent heat of evaporation at air temperature:
Le = 597.3 − 0.57 &times; 32.2 = 579 cal / g
[eq. 1.7]
17
Ea = ρLe 0.0269(1 + 0.1 u)(esa − ea )
= 1 &times; 579 &times; 0.0269(1 − 0.1 &times; 20)( 48.1 − 14.4)
= 1590 cal/cm2 −day
Penman's equation:
∆
γ
EρLe =
QN +
Ea
∆+γ
∆+γ
0.66
2.72
=
1590
400 +
2.72 + 0.66
2.72 + 0.66
= 632 cal/cm2 − day
Or:
E=
632
= 1.1 cm/day
1 &times; 579
18