The pores in soil are viewed as a network of vertical capillary tubes. The water flow in the capillary can be characterized as a laminar flow with a parabolic flow velocity distribution since the soil water percolates slowly. Without taking into account the slip of water on the pore wall, the flow velocity distribution of water passing through a capillary pore with the radius of R and length of H under pressure gradient Δ P = P 2 − P 1 is v r = 1 4 μ ∞ Δ P H R 2 − r 2 (1)

At the radius of r, the micro ring flow of the width of dr is d q = 2 π r v r . The Hagen-Poiseuille equation can be deduced from the integration of the section from 0 to R in the r direction, and the sectional total flux calculated by this equation is: q = ∫ 0 R 2 π r v r d r = π R 4 8 μ ∞ Δ P H (2) where q is the total flow of water in a capillary tube of radius R; μ ∞ is the viscosity of volumetric water, which is a function of temperature (Thomas and Sansom, 1995) and can be represented as μ ∞ T = 0.6612 T − 229 − 1.562 .

Due to the velocity slip phenomena, as seen in Figure 1, the liquid velocity in the capillary channel close to the wall is not zero, v r ≠ 0 at r = ± R . Additionally, the function μ ∞ T is no longer usable at negative temperatures and narrow capillary tube diameters, therefore effective viscosity is used instead μ d . When the combined water slip length l s , t   a n d   e f f e c t i v e   v i s c o s i t y   μ d on the pore wall are taken into account, the flow velocity distribution and water flow rate are (Holt et al., 2006) v s , t r = 1 4 μ d Δ P L R + l s , t 2 − r 2 (3) q w = ∫ l s , t R + l s , t 2 π r v s , t r d r = π 8 μ d R 4 + 4 R 3 l s , t Δ P L (4)

David M. Huang (Huang et al., 2008) measured the slip length of water flow wall at various contact angles, and found that the real slip length of water l s , t can be described as the function of soil-water contact angle as follows: l s , t = C cos ⁡ θ + 1 2 (5)

At the same time, the critical thickness can be used to compute the effective viscosity μ d of water in frozen soil pores, which is dependent on the area of the interface zone and the bulk water viscosity (Thomas and Mcgaughey, 2008). μ d = μ i A i R A t R + μ ∞ 1 − A i R A t R (6) A i R = π R 2 − R − λ 2 (7) where θ is the soil-water contact angle, θ = 25°; C is a constant, C = .41; μ i is the viscosity of the fluid in the interface area, P a ⋅ s ; μ i / μ ∞ is in a linear relationship with contact angle θ (Wu et al., 2017), μ i / μ ∞ = − 0.018 θ + 3.25 ; A i is the soil-water contact area in the pore; A t is the pore cross-sectional area; and λ is the critical thickness of the pressured fluid.

An enhancement factor is introduced, which is defined as the ratio of the measured water flux to the water flux q when no slip is considered, to examine the impact of the velocity slip phenomena on the water flux. ε = q w q = 1 + 4 l s , t R μ μ d (8)

As the curve shown in Figure 2, The pore radius is inversely related to the enhancement factor, and the smaller the pore, the larger the enhancement factor. When the pore radius is less than 100 nm, the trend of decreasing enhancement factor gradually accelerates, indicating that the effect of velocity slip on water flux gradually increases. Relevant researches indicate that soil pore with a radius of less than 100 nm make up more than half of all soil pores (Sasanian and Newson, 2013), thus it is crucial to account for effective water viscosity and the velocity slip phenomenon when calculating the soil seepage volume.

The fluid in large pore in the soil freezes when the temperature drops below the freezing point, but the water near the pore wall and in the small pore doesn’t freeze because of the surface tension on pore wall (van der Waals force). On this basis, the of pores can be divided into two categories: the frozen pores and unfrozen pores. Water film seepage not only includes the capillary seepage resulting from frozen pore, but also includes the water film seepage induced by frozen pore.

At temperature T, when a capillary tube is frozen with a radius of R, the film flow velocity distribution of unfrozen water calculated by the Navier-Stokes equation yields the following results (Watanabe and Flury, 2008; Ming et al., 2022) v f r = Δ P 4 μ d L R 2 − r 2 + R 2 − r 2 ln R r i ln r R (9)

Besides, in the r direction, the film seepage flux of unfrozen water q w ′ can be represented as the integration from r i to R. q w ′ = ∫ r i R 2 π r v f r d r = Δ P π 8 μ d L R 4 − r i 4 − R 2 − r i 2 2 ln R r i (10)

The following equation can be used to determine the thickness of unfrozen water film d (Dash et al., 1995; Ishizaki et al., 1996) and the connection between temperature T and pore radius R (Mitchell et al., 2008) d T = − A 6 π p i L i T i T i − T 1 3 (11) R T = 2 γ l s ⁡ cos ⁡ θ ρ i L i T i T i − T (12) where, A is the Hamaker constant, A = − 10 − 19.5 ; ρ i is the ice density, ρ i = 918 K g / m 3 ; L i is the latent heat of ice melting, L i = 3.34 × 10 5 J / K g ; T i is the freezing temperature of soil, in K , T is the soil temperature; γ l s is the ice water interface energy, γ l s =4.09×10⁻² J/m²; and r i is the ice nucleus radius, r i = R i − d . In order to visibly respond to the variations in magnitude of each, it is possible to calculate the ratio of water film seepage and capillary seepage. q w q w ′ = R 4 + 4 R 3 l s , t R 4 − r i 4 − R 2 − r i 2 2 ln R r i (13)

A graph of the ratio of q w q ′ w for a capillary tube of radius R before and after freezing is depicted in Figure 3. As can be seen from the figure, as capillary diameter rises, the ratio gradually climbs to five orders of magnitude and still has a tendency to increase. This result indicates that under large tube diameter, the seepage capacity of an unfrozen water film is substantially less than the seepage capacity of unfrozen capillaries. But the seepage of unfrozen water is only 10 times as great when the tube diameter is small. Since small soil pores make up the majority of soil pores, it is important to take into account the film seepage of unfrozen water (Sasanian and Newson, 2013).

As depicted in Figure 4, the pores in soil are simplified as capillary tubes with different radiuses. Using the SFCC curve (unfrozen water content is treated as a function of temperature T), it is assumed that the temperature interval [ T 0 , T n ] is divided into n equal parts, resulting in T 0 > T 1 > T 2 > … > T n − 1 > T n . Then n temperature points are obtained by taking the midpoint temperature of each temperature interval, which are T 01 , T 12 , T 23 , ⋯ , T n − 1 n , which represent the freezing temperature of pores. According to Eq. 12, the pore radius corresponding to each temperature is obtained, i.e. R 1 , R 2 , R 3 , ⋯ , R n , and they are donated as R i ( i = 1,2,3 , ⋯ , n ). Because unfrozen content θ is a function of temperature T, i.e., each temperature correlates to an unfrozen water content, then it is found that θ 0 > θ 1 > θ 2 > … > θ n − 1 > θ n . Therefore, n temperature intervals correspond to n unfrozen water variations of ∆ θ 1 , ∆ θ 2 , ∆ θ 3 , … , ∆ θ n , which is donated as ∆ θ i ( i = 1,2,3 , ⋯ , n ). In saturated soil, water fills all of the pores, and as long as the divided temperature intervals are enough, the unfrozen water variation ∆ θ i is thought to be caused by unfrozen water crystallization in the capillary of radius R i . Then the number n i of capillary tubes with radius R i can be calculated. The seepage amount of unfrozen water Q i and the seepage amount of unfrozen water film Q i ′ before and after freezing of the capillary tube with radius of R i is obtained using Eqs. 4–10. Δ θ i = π ∑ i = 1 n n i R i 2 (14) Q i = n i π 8 μ d R i 4 + 4 R i 3 l s , t Δ P H (15) Q i ′ = n i π 8 μ d R 4 − r i 4 − R 2 − r i 2 2 ln R r i Δ P H (16)

At the meantime, the water migration path is idealized to occur in a curving capillary. Assuming that the frozen soil has a length of L and a height of H (H<L), and combining Eqs. 15, 16, the water flux of frozen soil can be described as: Q w = ρ w g π 8 μ d ∑ i = j + 1 n R i 4 + 4 R i 3 l i + ∑ i = 1 j n i ξ i R 4 − r i 4 − R 2 − r i 2 2 ln R r i (17) where: j=0, 1, 2,..., n-1, and the corresponding j rises by one for each temperature interval of temperature decrease; ξ is the ratio of the soil column length H to the actual path of water seepage L ( ξ =L/H), ξ i = 1 + 0.41 ⁡ ln 1 / θ i (Xiao et al., 2015); Δ P = ρ w g H ; n is the total type number of pore radius in unit area; n i is the number of pores with the radius of R i in unit area.

Assuming that water seepage in frozen soil follows Darcy’s law, Eq. 17 can be combined to determine the permeability coefficient of frozen soil per unit area. K = ρ w g π 8 A μ d ∑ i = j + 1 n R i 4 + 4 R i 3 l i + ∑ i = 1 j n i ξ i R 4 − r i 4 − R 2 − r i 2 2 ln R r i (18)

According to Eq. 18, when j=0, K is the permeability coefficient of the melted soil under the same conditions. The water in the melted soil primarily flows in large pores, and the pore walls almost have no effect on water flow which can be disregarded. The permeability coefficient of melted soil can then be written as: K 0 = ρ w g π 8 A μ ∞ ∑ i = 1 n n i ξ i R i 4 (19)

Consequently, the frozen soil permeability coefficient can also be expressed as: K = K 0 ′ μ ∞ μ d ∑ i = j + 1 n R i 4 + 4 R i 3 l i + ∑ i = 1 j n i ξ i R 4 − r i 4 − R 2 − r i 2 2 ln R r i / ∑ i = 1 n n i ξ i R i 4 (20)

K 0 ′ is the measured value of permeability coefficient of melted soil.

It is required to examine the impact of the amount of salt crystallization on the permeability coefficient in order to adapt the permeability coefficient model of frozen soil to saline frozen soil. The freezing temperature curve of ice, AB, which is the precipitation ice line, and the solubility curve of salt, BC, which is the precipitation salt line, intersect at the eutectic point B, as the phase diagram of the water-salt binary system with three salts, Na₂CO₃, Na₂SO₄, and NaCl shown in Figure 5 The precipitation ice line, precipitation salt line, and eutectic point separate the solution phase diagram into four areas, i.e. the unsaturated salt solution, ice-salt solution, ice-crystalline salt, and crystalline salt-salt solution. The phase diagram can be used to predict the direction of several changes that will take place in the system during the temperature lowering process. It is possible to determine the order of phase change, freezing temperature, eutectic temperature, eutectic concentration, and other important parameters of saline soils during the process of temperature reduction and further analyze the amount of salt crystallization. In Table 1, the values of specific parameters during the crystallization of various salt solutions are given.

Through electron microscopy, Jing Zhao (Zhao and Luo, 2019) discovered that the salt solution in the pores gradually increases in salt crystallization as the temperature decreases. These salt crystals adsorb around the pores, and gradually reduce the pore size. As a result, the permeability coefficient model of impregnated frozen soil only needs to be modified the R i in the permeability coefficient model of frozen soil.

If it is assumed that the salt solution is contained in a capillary tube with a radius of R i i = 1 ⋅ ⋅ ⋅ n , the solution is always saturated after the salt precipitation. If the supersaturation of solution is not taken into account, the following equation is obtained: C ′ = n T C > n T C C ≤ n T (21) where, C is the initial mass concentration of solution, C′ is the subsequent mass concentration of solution after cooling, R i is the initial pore radius, R i ′ is the subsequent pore radius after the salt precipitation due to cooling, and n T is the solubility curve.

Assuming that the salt crystals are evenly deposited on the pore walls after salt precipitation, the quantity of salt crystallization equals the amount of variation in the pore: M c M ρ c π R i 2 C 100 − C − π R i ′ 2 C ′ 100 − C ′ = π R i 2 − π R i ′ 2 (22) M c where M c is the molecular weight of salt after crystallization, M is the molecular weight of salt, and ρ c is the specific gravity of the crystalline salt. For the same salt M c M ρ c is a constant value and let M c M ρ c = α , then Eq. 23 can be used to determine the pore radius R i ′ after salt precipitation: R i ′ = R i α C − 100 − C 100 − C ′ α C ′ − 100 − C ′ 100 − C (23)

Once the temperature is lowered to the eutectic temperature, the pore solution in soil mass gradually freezes. Since the solubility of salt does not significantly change, the concentration of the salt solution in the pore after the soil freezes is regarded as the eutectic concentration. Eq. 24 can be used to obtain the freezing temperature T i ′ for a pore with the size of R i ′ considering the impact of salt (Scherer, 1999): T i ′ = T 0 1 − T 0 R ∗ L i ln ⁡ a w − σ l s ρ i L i T i R i ′ (24) a w where, a w is the water activity, which is used to solve the freezing temperature T i of salt solution, and the water activity can be calculated using the Pitzer model, R* is the universal gas constant, and T 0 is the freezing temperature of pure water.

In conclusion, in this permeability coefficient model of salinized frozen soil, only the SFCC curve of frozen soil and the permeability coefficient of frozen soil need to be precise measured under the same conditions as well as the phase diagram of salt solution. Then the variation relationship between the permeability coefficient of frozen soil and temperature can be theoretically analyzed, as well as the variation relationship between the permeability coefficient of salinized soil.

As seen in Figure 6, the unfrozen water content of the silty clays from the Tibetan plateau and the silt from Oneida fluctuates with temperature (Burt and Williams, 1976; Hu et al., 2016), and the initial volumetric water contents of the two soils are 50% and 48%, respectively. The unfrozen water content decreases with decreasing temperature, and there is a general power function relationship between the two parameters. The measured temperature of data points ranges from 0 to -5°C. In order to improve prediction capacity of the proposed model, the whole SFCC curve can be derived for by fitting the functional relationship between the above two parameters.

As depicted in Figure 7, at the temperatures above −2°C, the difference between the calculated results from the model that takes the influence of an unfrozen water film into account and that does not is relatively minimal. In both cases, the predicted values of the proposed model essentially follow the trend of the test results of soil. When the temperature drops to −5°C, there is a discrepancy between the calculated results of the two models of around one time, and when the temperature drops to −10°C, there is a difference of about nine times. The primary explanation is that, when considering the seepage of unfrozen water film, as temperature decreases, pore water gradually freezes. As a result, the seepage of unfrozen water film eventually occupies the leading position of frozen soil water seepage.

The relevant research (Burt and Williams, 1976; Chen and Zhang, 2020) shows that when the temperature is higher than −2°C, the soil still contains a lot of unfrozen water, the soil permeability coefficient is large, and the water salinity still has a certain seepage capacity; When the temperature is lower than −2°C, the content of unfrozen water is greatly reduced, the permeability coefficient of soil is small, and the water salinity almost does not percolate. Therefore, the influence of unfrozen water film seepage is ignored in the calculation of permeability coefficient model of saline frozen soil. In the salt frost heaving model, when the temperature is lower than −2°C, the calculated result of the model considering the seepage permeability coefficient of unfrozen water film is larger, so the calculated salt frost heaving deformation is larger. This may be conducive to accurate analysis of frost heave deformation of salinized salt.

As noted in Table 1, the eutectic temperature of NaCl solution is −21.1 ° C, and the eutectic concentration is 23.3%, which is substantially higher than the solution concentration in the pores of saline soils. As a result, in the process of temperature decrease, water freezing takes place in the soil before NaCl crystallizes, and NaCl primarily influences the initial freezing point of soil. According to the NaCl phase diagram, the solutions with freezing temperatures of 0, −1, −2, −2, and −3°C are prepared and added into the samples of powdered clay from the Qinghai-Tibet Plateau with the mass concentrations of 1.86%, 3.59%, 5.20%, and 6.64%, respectively, and the salt contents of the prepared soil samples are .59%, 1.16%, 1.71%, and 2.22%, respectively. The calculation results of the permeability coefficient model of NaCl-saline soil is depicted in Figure 8. It is shown that the permeability coefficient gradually lowers as the temperature steadily drops and eventually tends to stabilize. The change in permeability coefficient exhibits a tendency of steep first and then slow change when the soil pores freeze in order from large to small. In terms of numerical magnitude, the initial freezing temperature steadily drops as NaCl content rises, and at the same temperature, the permeability coefficient increases as the amount of unfrozen water increases.

Since Na₂SO₄ will precipitate before the soil freezes and thus has a less substantial impact on the freezing temperature of the soil than NaCl. Table 1 demonstrates that the eutectic temperature of Na₂SO₄ solution is -1.2°C and the eutectic concentration is 4%. The Na₂SO₄ solutions comprising 2%, 4%, 6%, and 8% salt were fabricated, and the mass concentrations of the pore solutions of the powdered clay samples from the Qinghai-Tibet Plateau are 6.01%, 11.34%, 16.11%, and 20.38%, respectively. Due to the presence of supersaturation of the pore solution concentration after freezing, it is difficult to calculate the first freezing temperature using the Pizter theory. Thus, the initial freezing temperature of the soil will be determined using existing studies (Wan et al., 2015; Xiao et al., 2018). Before the soil freezes, salt will precipitate from the Na₂SO₄ solution during the dropping process of temperature, causing the permeability coefficient to start to fall. This is shown in Figure 9 and Figure 10 for the computation of the permeability coefficient model for sodium sulfate-saline soils. The beginning temperature for salt precipitation increases along with the Na₂SO₄ content, as does the volume of precipitated mannite. The permeability coefficient of the soil with an 8% salt concentration steadily drops to 4.25 × 10⁻⁷ at 19.2°C, which was 56% lower than the permeability coefficient of the soil without salt. After soil freezes, the permeability coefficient changes as the temperature progressively drops, and the change is initially steep, then becomes gently, while eventually converges. The permeability coefficient is relatively high, and the initial freezing temperature directly affects it because the amount of unfrozen water steadily increases as the starting freezing temperature decreases.

Na₂CO₃ will crystallize and precipitate before the soil freezes, as indicated in Table 1, where the eutectic temperature of the solution is -1.8°C and the eutectic concentration is 5.8%. The permeability coefficient of Na₂CO₃ saline soils is calculated using the method for Na₂SO₄ saline soils due to the similarity between these two solutions. The initial freezing temperature of the soils is determined from experiment (Hui and Wei, 2011). The powdered clay soils of the Tibetan plateau, with mass concentrations of 6.01%, 11.34%, 16.11%, and 20.38% are manufactured, and pore solutions are prepared using salt concentrations of 2%, 4%, 6%, and 8%. The calculation results of the permeability coefficient of sodium carbonate saline soils are given in Figures 11, 12. The findings show that the soils with 2% salt content do not crystallize before freezing and that the permeability coefficient remains constant. During the decreasing process of temperature, salt gradually precipitates in the pores of soil with salt contents of 2%, 4%, and 6%, and the permeability coefficient of the soil gradually declines. Likewise, the higher the soil’s salt content, the higher the temperature at which the salt begins to precipitate out, and the lower the permeability coefficient at the same temperature. The permeability coefficient of soil with an 8% salt content drops by around 41%, from 9.68 10⁻⁷ cm/s to 5.66 10⁻⁷ cm/s. When the soil freezes, the permeability coefficient is comparatively larger under the same temperature conditions and the amount of unfrozen water increases due to the higher Na₂CO₃ content and lower initial freezing temperature.