implying the solution must have the form
-1
-1
1
( x + c1 )
( t + c2 )
ηcf
U ( x, t ) = κ cf
ηcf
ηcf
.
(7.15)
Co-ordinates are then normalized by defining
χ = x + c1
(7.16)
τ = t + c2
(7.17)
thus giving
1
⎛ χ ⎞ ηcf
U ( χ ,τ ) = ⎜
⎜κ τ ⎟
.
(7.18)
⎟
⎝ cf ⎠
In the normalized co-ordinates ( χ ,τ ), each particular solution represents a wave of
volume flux with its origin at the point ( χ = 0,τ = 0 ). In other words, each differential
element of mobile water, dAm, generates its own particular solution where τ = 0 occurs at
the time the water volume becomes mobile within the snow (surface melt or rainfall), and
χ = 0 occurs at the vertical height of the snowpack at the time the water volume becomes
mobile.
To find the depth of penetration of a wave as a function of time is simply a matter of
integrating saturation from the point of generation of the wave to whatever depth
necessary to allow the total volume enclosed by the integration to be equal to the original
volume of the wave. For an input volume per unit area of A (cm) over an area σwa, if the
depth of penetration of the resulting flux wave is denoted as xU, then at any point τ in
time
χ = xU
∫χ
Seφ (1 - Swi )σ wa d χ = Aσ wa
(7.19)
=0
where A is the volume/unit area runoff in cm and σwa is the area. Therefore we have
A
χ = xU
∫χ =0
Sed χ =
.
(7.20)
φ (1 - Swi )
Using Se from Equation (7.4) and U from Equation (7.18), substituting into Equation
(7.20), then solving the integral yields the equation for the depth of penetration of a wave
as a function of time
η
2
ηcf -ηcf
⎛
⎞ ⎛ ρwkw0 g ⎞
A
(
κ cf τ ) .
1-ηcf
xU = ⎜
⎜ η φ (1 - S ) ⎟ ⎜ η
(7.21)
⎟
⎟
wi ⎠ ⎝
⎠
⎝ cf
w
Next, consider the time it will take a wave of volume flux to penetrate to a given depth
(most useful in calculating the delay between influx and outflux). In order for the water
mass to balance in time, the volume of water passing by a given depth D must equal the
initial mobile water volume
τ =∞
∫τ τ U ( D,τ )dτ = A
.
(7.22)
m
=
0
Substituting from Equation (7.15) yields
-1
-1
1
τ =∞
ηcf
ηcf
ηcf
∫τ τ
κ cf D τ
dτ = Am
(7.23)
=
0
where D is the depth of penetration at time τ0.
60
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