Operator`s manual

65
Manual
Annex 9: Calculation of Harbeck Evaporation for a Five-Minutes Interval
The Harbeck equation, which is an exponent of the Mass-Transfer Approach, is expressed by:
E = x va x (e
s
– e
a
)
= mass-transfer coefficient
E = evaporation [cm day
-1
]
v
a
= wind speed at 2 m [cm s
-1
]
e
s
= vapor pressure at the water surface [mb]
e
a
= vapor pressure at 2 m [mb]
HADA uses the standard value of = 1.26 x 10
-4
for the mass-transfer coefficient.
The datalogger program calculates evaporation in millimeters for a five-minutes interval. This requires the
following transformation:
1 day contains 24 x 12 = 288 intervals of 5 minutes. Hence, • needs to be divided by 288;
Wind speed is measured in m/s instead of cm/s; Hence, • needs to be multiplied by a factor 100;
Evaporation is expressed in mm (instead of cm). Hence, • needs to be multiplied by a factor 10.
This leads to:
mod = 1.26 x 10
-4
x 100 x 10 / 288 = 4.375 x 10
-4
In conclusion, Harbeck evaporation in mm for a five-minutes interval is expressed by:
E = 4.375 x 10
-4
x v
a
x (e
s
– e
a
)
The above equation is incorporated in the datalogger program. The five-minutes values are aggregated to hourly
and daily values.
Vapor pressure is calculated with the below equations:
e
s
= 6.11 EXP((17.3 * T
s)
/ (T
s
+ 237.3)
in which:
T
s
= surface water temperature [C]
e
a
= RH
a
x 6.11 EXP((17.3 * T
a
) / (T
a
+ 237.3)
in which:
RH
a
= relative humidity of the air at 2 m
T
a
= air temperature at 2 m
Annexes