Geoscience Reference
In-Depth Information
The total evapotranspiration
E 1 consists of:
Transpiration (
T ): the part of the total water vapour lux that enters the atmosphere
from the soil through the vegetation (stomata and cuticula).
Evaporation of intercepted water (
E int ): evaporation of water that has been intercepted
by plants.
Soil evaporation (
E soil ): evaporation of water from the soil (the soil may either be sat-
urated or partly dry).
Potential evapotranspiration
E pot is the theoretical evapotranspiration that would occur if
a given vegetation, completely covering the soil, is exposed to prevailing meteorological
conditions (without itself affecting the meteorological conditions). The term 'potential
evaporation' sometimes leads to confusion when no reference is made to a speciic type of
vegetation. In that case referring to ' the potential evapo(transpi-)ration' becomes useless.
Reference evapotranspiration
E ref : is the theoretical evapotranspiration that would
occur if a well-deined, theoretical vegetation, completely covering the soil is exposed
to prevailing meteorological conditions (without affecting the meteorological condi-
tions).
Furthermore, in this context a short discussion of units is needed. Whereas meteo-
rologists consider evapotranspiration as an energy term, for practical applications the
depth of the water layer that evaporates is important (usually in mm d -1 ). To arrive
from energy lux densities (in W m -2 ) to luxes in mm d -1 , the following steps are
needed:
A lux density of 1.0 W m
-2 that continues for 1.0 day (86 400 s) amounts to a total
energy lux of 86 400 J m -2 d -1 .
With this amount of energy one can evaporate 86 400/
L v kilograms of water per square
metre in one day. Taking L v = 2.45·10 6 J kg -1 (strictly speaking, L v is temperature depen-
dent) the mass of water evaporated by this amount of energy (1.0 W m -2 during one day)
is approximately 3.53 · 10 -2 kg m -2 d -1 .
This mass of water corresponds to 3.53 · 10
-2 kg/ ρ w cubic meters of water ( ρ w is the
density of water: 1000 kg m -3 ). This is 3.53 ·10 -5 m 3 m -2 d -1 .
This volume per square meter of surface per day is equivalent to a layer of water of
3.53 ·10 -5 m d -1 , or 3.53 ·10 -2 mm d -1 .
We could also start at the other end and pose the question what daily mean lux
density is needed to evaporate 1 mm of water in one day. The answer is (1 mm d -1 )/
(3.53 ·10 -2 mm d -1 /W m -2 ), which equals 28.4 W m -2 .
Question 8.1: The latent heat of vaporization is temperature dependent (see Appendix
B ). Which temperature (air temperature, surface temperature, or another temperature)
should be used to compute the value of L v that is needed to convert the evaporation in
terms of a mass-lux into evaporation in terms of an energy lux. Explain your answer.
1 Note that the literature on crop water use evapotranspiration is often denoted by ET , rather than E .
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