Geoscience Reference
In-Depth Information
sublimation needs more power than evaporation and because the surface temperature
becomes closer to the air temperature when ice grows thicker that lowers the turbulent
fl
fluxes. The averages are 56.9 mm/month and 43.3 m/month for precipitation and evap-
oration/sublimation, respectively.
Evaporation and sublimation are strongly limited by the available heat
uxes. In
practice, in cold climate conditions the mass loss per unit area is less than 10 mm liquid
water equivalent per day, while the average level is 1 mm/day. Evaporation takes place in
open water and melt ponds, while dry snow and ice surfaces lose mass by sublimation.
Occasionally atmospheric water vapour deposits as hoar crystals or condenses into liquid
water on the surface that also brings the phase change energy to the surface. Latent heat
transfer Q
e
for evaporation or sublimation, respectively, can be directly expressed as
fl
Q
e
¼
q
i
L
S
E
ð
ice
Þ
Q
e
¼
q
w
L
e
E water
ð
2
:
17
Þ
ð
Þ
where
ˁ
w
is water density. The latent heat of vaporization of liquid water depends on the
temperature,
¼
2
;
494
2
:
2
T
½
C
L
e
kJ kg
1
, the latent heat of freezing is Lf
f
=
333.5 kJ kg
−
1
, and the latent heat of sublimation is
¼ 2
;
828
0
:
39
T ½
C
L
S
kJ kg
1
ð
T
0
C
Þ
. Thus sublimation needs about 13.5 % more power than evaporation.
Q
e
30Wm
2
, we have
E
1mmday
1
. This energy
Example 2.9.
ux can be
estimated using turbulent transfer models. A common way is the bulk formula (Eq.
2.13c
).
For the relative humidity of air at 50 % and
For
fl
U
a
10 ms
1
,
the latent heat
loss is
at the temperature of 0
Q
e
100Wm
2
E ¼ 3mm day
1
°
C, but at
−
20
°
C the energy
.
20Wm
2
E ¼ 0
:
6mmday
1
would be reduced to
2.3.4 Budgets of Impurities
Closely connected to the water budget are the budgets of impurities, which are received by
atmospheric deposition and in
fl
ow from the drainage basin and lost of by out
fl
ow.
Evaporation and sublimation remove only pure water. In addition to the water
uxes,
concentrations of impurities are changed by sedimentation/resuspension and biochemical
processes.
Consider the mass of a given chemical substance in a lake, C = c
fl
S, where c is the
·
mean concentration as mass/volume. The balance is therefore
d
dt
Z
c
ðÞ
¼c
i
I
c
o
O
þ
APc
P
þ
/
dS
ð
2
:
18
Þ
where c
i
, c
o
, c
P
are the concentrations in in
fl
ow, out
fl
ow and precipitation, respectively, and
˕
is the in
fl
uence of biochemical processes, resuspension and sedimentation. The renewal
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