The solution to this integral gives the time required for a wave of volume flux to
penetrate to depth D
ηcf
⎛
⎞ηcf -1
1
1 -ηcf
κ cf ⎞ηcf ⎟
⎛
τ0 = ⎜
Am ⎜
.
(7.24)
⎟
⎜ ηcf
D⎠ ⎟
⎝
⎜
⎟
⎝
⎠
For modeling purposes, note that Equation (7.21) may also be written in the simpler form
1-nc
1
⎛ n -1 ⎞
nc
xU = κ cf ⎜ c ⎟ τ nc .
(7.25)
A ⎠
⎝
Having depth of penetration as a function of time, and vice versa, requires only that the
effects of preceding flux waves be incorporated in order to construct a working model of
water movement through homogeneous snow. Therefore, consider two flux waves, each
proceeding according to Equation (7.18)
1
⎛χ ⎞
ηcf
U1 ( χ1 ,τ1 ) = ⎜ 1 ⎟
(7.26)
⎝ κτ1 ⎠
and
1
⎛ χ2 ⎞
ηcf
U 2 ( χ2 ,τ 2 ) = ⎜
⎜κ τ ⎟
.
(7.27)
⎟
⎝ cf 2 ⎠
Assume that χ1 - χ2 = ε > 0 (that the wave U1 began at an earlier point in time than the
wave U2). Since there is no singularity within the region over which the U2 solution
applies, flux will be continuous and differentiable over this region, up to the singularity
caused by the U1 U2 junction. As the U2 wave flows through the snow, Equation (7.25)
must always be met, even as new water volume is absorbed into wave U2 from the
residual saturation of wave U1. Therefore, to accommodate the new water volume the
second wave has encountered while still maintaining its solution to the conservation and
flux-saturation equations, the volume flux wave U2 must travel deeper into the snowpack
than it otherwise would, thereby encountering more residual saturation in the process.
Numerically, the residual saturation encountered will take the form
1
⎡ ηw ⎤
nc
1
S1 = ⎢
(7.28)
⎥ U1
nc
⎣ ρwkw0 g ⎦
which, when combined with U1 from Equation (7.26), becomes
1
1
⎡ ηw ⎤ ⎛ χ1 ⎞
nc -1
nc
S1 = ⎢
⎜
⎟ ,
(7.29)
ρwkw0 g ⎥ ⎜ κ cf τ1 ⎟
⎣
⎦⎝
⎠
which means that if xA2 is the deepest point within the snow at which the U2 solution
applies, (the point of the singularity), then the volume of water absorbed by the U2 wave
in moving some infinitesimal distance through the snow will be given by
1
1
⎛ xA2 ⎞
⎡ ηw ⎤
nc -1
nc
∂ A2 = φ (1 - Swi ) ⎢
∂ xA2 .
⎜
⎜κ τ ⎟
(7.30)
ρwkw0 g ⎥
⎟
⎣
⎦
⎝ cf 1 ⎠
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