I am a high school student who did field research this past summer. My
gathered data including dewpoint, air temperature, water temperature,
barometric pressure, incoming/ourgoing/net radiation, and wind speeds in
close proximity to a lake. I was wondering if you had any suggestions for
readings that might show me how to calculate evaporation rates for the
Determining evaporation from the lake is somewhat complicated, but
it is a good exercise!
I assume that you measured net radiation over land and not over the
water, so, that measurement cannot be used in calculations of lake
evaporation. The remaining measurements may be used in what is called the
"aerodynamic method". This method requires that you make measurements
of temperature and vapor pressure at two heights above the surface,
plus wind speed. Keep in mind that this is an approximate method and
can be in error by 25%.
When you say that you have measured the dew point, I assume that you mean
the dew point temperature, not the wet bulb temperature.
Hopefully you made your dew point temperature, air pressure, and
wind speed measurements at approximately the same height.
The equation to make the calculation of evaporation is:
E = - rho * L * k * ustar * (q2-q1) / ln(z2/z1)
where E is evaporation, rho is air density, L is the latent heat of
vaporization, k is von Karman's constant, ustar is the friction velocity,
(q2-q1) is the specific humidity difference between the air and near
the water surface, ln is the natural log, z2 is the height of the
dew point temperature measurement, and z1 is a height near the
I assume that your dew point and water temperature measurements
to degrees C if they are not in those units.
rho can be approximated as 0.0012 g/cm**3, where **3 means
exponent to the 3rd power.
L is approximated by 598 cal/g.
k is 0.38
Convert your wind speed to cm/s if it is not in those units.
Assume ustar to be one tenth of the wind speed.
Convert your air pressure to mb if it is not in those units.
Convert your dew point temperature measurement
height to cm if it is not in those units; this is z2.
Let z1 = 0.1 cm.
Determine the vapor pressure in mb (e2) for the air dew point
temperature (at z2) and the vapor pressure in mb (e1) for the
lake water temperature (at z1, which will be close to
the dew point near the lake surface because the air there
is almost saturated) using the following equation for both:
e = C0 + C1*T + C2*T*T + C3*T*T*T + C4*T*T*T*T + C5*T*T*T*T*T + C6*T*T*T*T*T*T
C0 = 6.11
C1 = 0.4437
C2 = 0.014289
C3 = 0.000265065
C4 = 0.00000303124
C5 = 0.000000020340809
C6 = 0.00000000006136821
and where the temperature T is in degrees C; also note that each
coefficient is multiplied by as many Ts as the number in the
The specific humidities (unitless), q2 (at z2) and q1 (at z1) can be
calculated from the following equation:
q = (0.622*e)/(P-0.378e)
Using the above information, we can estimate evaporation from the lake.
As an example, if we assume that the air dew point temperature is
20 degrees C, the lake temperature is 25 degrees C, the wind speed is
500 cm/s (about 11 mph), z2 is 300 cm, z1 is 0.1 cm, and air pressure (P) is
1000 mb, we calculate that:
ustar = 50 cm/s
e2 = 23.37 mb
e1 = 31.67 mb
q2 = 0.01467
q1 = 0.01994
E = 0.00906 cal/cm**2/s or 379.1 W/m**2
Meteorologists use Watts per meter squared (W/m**2) for energy fluxes,
such as evaporation. The conversion is, 1 cal/cm**2/s = 41,846.43 W/m**2.
This calculated evaporation for the lake is a very reasonable number
and is approximately twice the evapotranspiration from a land surface
during peak sunshine in the middle of the day.
For air temperatures even colder than what we have considered, like occur
in winter (for an ice free lake), evaporation can be much greater. In
the middle of a summer day, when air temperatures are often higher than
the water temperature, evaporation can be very small.
David R. Cook
Atmospheric Research Section
Environmental Research Division
Argonne National Laboratory
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Update: June 2012