sg
sg dAs,m
g
s
dp = dp + d γ
g
(7)
dVm
and
ls
ls dAs,m
l
s
dp = dp + d γ
l .
(8)
dVm
By subtracting eq 4 from eq 6 and eq 6 from eq 5, then subtracting the latter differ-
ence from the former, the following relation is obtained after rearrangement:
Sm,H2O Sm,H2O
S*s
*g
*l
*s
m,H2O Sm,H2O
dT
*l
Vm,H2O Vm,H2O
Vm,H O V *g
*s
*s
m,H2O
2
*g
sg
Vm,H
sg dAs,m
2O
=
d γ
g
(9)
*g
*s
dVm
Vm,H 2O
Vm,H
2O
ls
*l
Vm,H2O
ls dAs,m
d γ
l .
*l
*s
dVm
Vm,H2O Vm,H2O
dAs,g
s
*g
m
*l
sg
>> Vm,H2O
= 0 , the follow-
and by assuming that γ
By noting that
Vm,H O
g
2
dVm
ing relation is obtained:
dAssm
(
)dT
l
,
*l
*s
= Vm,H 2Od γ ls
*l
l .
Sm,H2O
Sm,H2O
(10)
dVm
By defining capillary pressure (Pa), pc, as
pc = pl ps ,
(11)
it can be seen from eq 8 and 11 that
ls
ls dAs,m
dpc = d γ
l .
(12)
dVm
Therefore, the capillary pressure of liquid water in a frozen porous medium can be
calculated by
*l
*s
pc
Sm,H2O Sm,H2O
T
∫ dpc ≡ pc = ∫
dT
(13)
*l
Vm,H2O
0
T0
where T0 (K) is the melting point of the bulk liquid phase.
To evaluate the righthand side of eq 13, the following properties must be mea-
sured or estimated for all temperatures of interest:
1. The melting point of the liquid phase in bulk
2. Entropy of the liquid phase
3. Entropy of the ice phase
4. Molar volume of the liquid phase.
3
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