Some insights from the second principle of thermodynamics for snowpack modeling
Abstract. Entropy and the second principle of thermodynamics are regularly used as an analysis tool in applied mathematics for physics-based numerical models. In essence, this approach states that the second principle (i.e. the non-destruction of entropy) is closely related to stability. Consequently, numerical models complying with the second principle are expected to be more robust than models that do not. A notable advantage of this method is its straight-forward generalization to nonlinear physics and to systems of coupled equations. The goal of this work is to thus investigate the added-value of such an entropy-based analysis to the case of snowpack modelling. For that, we study the conditions under which the physics describing snowpacks respects the second principle and the numerical schemes that preserve this compliance after temporal and spatial discretization. Specifically, we consider three cases of increasing complexity: (i) a dry snowpack governed by heat conduction only (meant to be an example of the method for unfamiliar readers), (ii) a system composed of a canopy and a snowpack exchanging heat, and (iii) a dry snowpack with heat conduction, vapor diffusion, and ice-vapor phase changes. Overall, the analysis shows that to comply with the second principle, numerical snowpack models should follow three main rules. First, physical variables should be co-localized. In other words, the temperature at a given point (and other intensive variables) should depend on the energy (and other extensive variables) only at that point. This property is naturally met with the finite volume method, but requires adaptation for the finite element method. Second, advected quantities, such as the enthalpy advected with vapor diffusion, should be numerically upstreamed. Finally, thermodynamical fluxes (for instance heat conduction or phase change rate) should be consistent with the end-of-timestep value of the thermodynamical gradients/differences that drive them (for instance temperature gradients or chemical potential differences). This can be achieved by employing a Backward Euler temporal integration and resolving the physical processes in a tightly-coupled manner. While proper compliance to the second principle is a good practice to build robust numerical models, it can however be cumbersome to achieve in practice. Therefore, we suggest to rather see this kind of entropy-based analysis as a tool, helping to diagnostic potential instability issues, rather than a rule to strictly follow.
Competing interests: At least one of the (co-)authors is a member of the editorial board of The Cryosphere.
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Dear Co-Authors
The entropy equation is studied in the paper for the entropy of a snow-pack defined as a mixture of water vapor and ice (Ih), with ice-vapor phase changes.
The entropy of this mixture is assumed to be (p.8, line 217) the sum: S^n = Sum_{k=1}^N (\Delta z_k) s^n_k
The entropy of ice (Ih) depends on a reference value s_i^0, with (p.18, line 446): s_i = s_i^0 + (...)
Similarly the entropy of water vapour depends on a reference value s_v^0, with (p.19, line 479): s_v = s_v^0 + (...)
These reference values are determined in the Section 5.1.3 ("Defining phase equilibrium", p.19-20), with (p.20, lines 491-492): "... we first set the reference temperature to T0=273.15K, without loss of the generality. We then assume (still without loss of generality) that the molar internal energy and entropy of ice vanishes at T0, yielding that u_i^0=0 and s_i^0=0 ..." Accordingly (p.20, lines 499-500): "... s_v^0 = (s_i^0=0) + [ Delta-h(i->v) ] / T0. This naturally enforces the equality of the chemical potentials, and hence phase equilibrium, in the reference conditions."
Indeed, for such a simple system made of two phases of the same substance (water ice and water vapour) the reference entropy can be set to zero for one or another of the phase, with the other computed from the other by adding or removing the quantity [ Delta-h(i->v) ] / T0, where Delta-h(l->v) is the latent heat of sublimation.
However, for a more general mixture made of several substances, we known since Gibbs (1875-1878) that the reference values cannot be determined arbitrarily, and we know since Planck (1914, 1917) that these reference must be determined from the third-law of thermodynamics (with concrete calorimetric and/or statistical values computed by Sackur, Tetrode, Gordon, Barnes, Giauque, ...).
These third-law values should be key features to be determined in next more realistic studies including the Deuterium oxide (D2O-vapour/liq/ice) species, which cannot be linked with the water (H2O) species. These generalisations should be considered after the Section 6 ("Discussion" p.28-20) 6.1 "Extension to other cases"), where (p.28, line 732-736): "Among the missing mechanisms/processes, we can mention the presence of liquid water, thus with phase changes and liquid water percolation under gravity and capillary effects, the mechanical compaction of the snowpack under its own weight, and the presence of water isotopologues. One can thus wonder about the extensibility of the approach and results presented in this work." With the aim (p.28, line 743) to: "Derive the general entropy function of snow by adding the entropy contribution of each phase."
I provide in the following examples of reference entropies for D2O species from recent and old Thermodynamic Tables that I have used to compute and study the absolute entropies for the moist-air atmosphere (since Marquet, QJRMS, 2011) and the sea-salt oceans (since Marquet, 2026-a,b). See my publications in https://sites.google.com/view/pascal-marquet/
Best regards,
Dr.Hab. Pascal Marquet
pascalmarquet@yahoo.com
%=============================================
% "Atkins' Physical chemistry (12th edition)"
% Oxford University Press (2023) 927~Pp.
%---------------------------------------------
% Table 2C.7 (p.892) 25°C (1 atm)
% D2O M(g/mol) S(J/K/mol) Cp(J/K/mol)
% vap 20.028 198.34 34.27
% liq 20.028 75.94 84.35
% D2O M(g/mol) S(J/K/kg) Cp(J/K/kg)
% vap 20.028 9903.1 1711.1
% liq 20.028 3791.7 4211.6
%---------------------------------------------
% Table 2C.7 (p.893) 25°C (1 atm)
% H2O M(g/mol) S(J/K/mol) Cp(J/K/mol)
% vap 18.015 188.83 33.58
% liq 18.015 69.91 75.291
% ice 18.015 37.99
% H2O M(g/mol) S(J/K/kg) Cp(J/K/kg)
% vap 20.028 10482 1864
% liq 20.028 3881 4179.4
% ice 18.015 2109
%=============================================
%=======================================================
% Wagman et al. (1982)
% "The NBS tables of chemical thermodynamic properties"
% J. Phys. Chem. Ref. Data, Vol.11, Supplement Number-2,
% 392-Pp. https://srd.nist.gov/JPCRD/jpcrdS2Vol11.pdf
%-------------------------------------------------------
% Table-2H (p.2-38) 298.15K 25°C and 1bar (0.1MPa)
% D2O M(g/mol) S(J/K/mol) Cp(J/K/mol)
% vap 20.0276 198.339 34.27
% liq 20.0276 75.94 84.35
%-------------------------------------------------------
% Table-2H (p.2-38) 298.15K 25°C and 1bar (0.1MPa)
% H2O M(g/mol) S(J/K/mol) Cp(J/K/mol)
% vap 18.0154 188.825 33.577
% liq 18.0154 69.91 75.291
%=======================================================
%=============================================
% Long and Kemp (1936)
% "The Entropy of Deuterium Oxide and
% the Third Law of Thermodynamics.
% Heat Capacity of Deuterium Oxide
% from 15 to 298°K.
% The Melting Point and Heat of Fusion"
%---------------------------------------------
% Journal of the American Chemical Society
% Vol-58, Number-10, October 9, 1936
%---------------------------------------------
% (1 atm) S0
% D2O gas at 273.10K: 45.89 pm 0.1 cal/K/mol
% D2O liq at 298.10K: 17.27 pm 0.05 cal/K/mol
% D2O ice at 276.92K: 10.38 cal/K/mol
%---------------------------------------------
% (1 atm) S0
% D2O gas at 273.10K: 192.0 pm 0.4 J/K/mol
% D2O liq at 298.10K: 72.3 pm 0.2 J/K/mol
% D2O ice at 276.92K: 43.43 J/K/mol
%=============================================
Heavy Water thermodynamic:
https://inis.iaea.org/records/bgt3d-ng660
Hill_MacMillan_Vee_Heavy_Water_1981.pdf (HMMV-81)
%=============================================
https://webbook.nist.gov/cgi/cbook.cgi?ID=C7789200&Mask=6F
Deuterium oxide (NIST page): D2O
Gas (1 bar) : Svap(1000hPa)=198.34 J/K/mol (Chase 1998)
---------------------------------------------
D2O / Chase (1998, Ideal gas, p.1045):
Mr = 20.020004-g/mol = 0.020020004-kg/mol
R/Mr = 8.3145/0.020020004 = 415.3 J/K/kg
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Chase (1998) D2O Gas (1 bar 25°C) : Svap(1000hPa)=198.339 J/K/mol
Chase (1998) D2O Gas (1 bar 25°C) : Svap(1000hPa)=9907.04 J/K/kg
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HMMV-81 / D2O / 3.82°C / Table-1, p.1 / T-fus = 276.97-K
(HMMV-81, 3.82°C) P-sat = 0.6601-kPa = 660.1 Pa
(HMMV-81, 3.82°C) Delta(S)-(liq/vap) = 8390.2-J/K/kg
(HMMV-81, 3.82°C) Delta(H)-(liq/vap) = 2323.7-kJ/kg
---------------------------------------------
HMMV-81 / D2O / 25°C / Table-1, p.1 :
(HMMV-81, 25°C) P-sat = 2.737-kPa = 2737 Pa = 27.37 hPa
(HMMV-81, 25°C) Delta(S)-(liq/vap) = 7611.6-J/K/kg = Sv-Sl
(HMMV-81, 25°C) Delta(H)-(liq/vap) = 2269.4-kJ/kg = L-vap
(HMMV-81, 25°C) Delta(S)-(liq/vap) = 152.38-J/K/mol = Sv-Sl
(HMMV-81, 25°C) Delta(H)-(liq/vap) = 45433.4-J/mol = L-vap
- - - - - - - - - - - - - - - - - - - - - - - -
0) Delta(H)-(liq/vap) / 298.15 = Delta(S)-(liq/vap)
45433.4 / 298.15 = 152.384 : OK
- - - - - - - - - - - - - - - - - - - - - - - -
1) S-vap(27.37 hPa) = S-vap(1000hPa) - R*ln(27.37/1000)
S-vap(27.37 hPa) = 198.339 - 8.314*ln(27.37/1000)
S-vap(27.37 hPa) = 228.26 J/K/mol
- - - - - - - - - - - - - - - - - - - - - - - -
2) S-liq = S-vap(27.37 hPa) - Delta(S)-(l/v)
= 228.26 - 152.38
S-liq (25°C) = 75.88 J/K/mol
- - - - - - - - - - - - - - - - - - - - - - - -
3) D2O(vap) 100 200 298.15 300
Cp (J/K/kg) 33.299 33.449 34.255 34.278
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(normal) Water (NIST page): H2O
- - - - - - - - - - - - - - - - - - - - - - - -
https://webbook.nist.gov/cgi/cbook.cgi?ID=C7732185&Mask=1E9F
- - - - - - - - - - - - - - - - - - - - - - - -
Gas (1 bar 25°C) : S0=188.885 J/K/mol (Cox, Wagmann et al, 1984)
Gas (1 bar 25°C) : S0=188.84 J/K/mol (Chase 1998)
Gas (1 bar 25°C) : S0=10482.2 J/K/kg (Chase 1998)
---------------------------------------------
H2O / Chase (1998, Ideal gas 25°C, p.1324):
Mr=18.01528 g/mol
R/Mr = 8.3145/0.01801528 = 461.5 J/K/kg
H2O / Gas (1 bar) : S0=188.834 +/-0.042 J/K/mol
H2O / Liquid (1 bar) : S0=69.950 +/-0.079 J/K/mol
- - - - - - - - - - - - - - - - - - - - - - -
25°C P-sat = 3169 Pa = 31.69 hPa
L-vap = 43988 J/mol = 2441.7 kJ/kg
S-vap = 69.950 + 43988/298.15 + 8.314*ln(31.69/1000)
188.79 = 69.950 + 147.536 - 28.70
close to 188.84 indeed
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