Add ver-1kc-2

doc/content/verification_and_validation/ver-1kc-2.md

@@ -10,7 +10,7 @@ Two enclosures are separated by a membrane that allows diffusion according to Si
 
 This verification problem is taken from [!cite](ambrosek2008verification).
 
-Unlike the (ver-1kc-1)[ver-1kc-1.md] case, which only considers tritium T$_2$, this setup describes a diffusion system in which tritium T$_2$, dihydrogen H$_2$ and HT are modeled across a one-dimensional domain split into two enclosures. The total system length is $2.5 \times 10^{-4}$ m, divided into 100 segments. The system operates at a constant temperature of 500 Kelvin. Initial tritium T$_2$ and dihydrogen H$_2$ pressures are specified as $10^{5}$ Pa for Enclosure 1 and $10^{-10}$ Pa for Enclosure 2. Initially, there is no HT in either enclosure.
+Unlike the [ver-1kc-1](ver-1kc-1.md) case, which only considers tritium T$_2$, this setup describes a diffusion system in which tritium T$_2$, dihydrogen H$_2$ and HT are modeled across a one-dimensional domain split into two enclosures. The total system length is $2.5 \times 10^{-4}$ m, divided into 100 segments. The system operates at a constant temperature of 500 Kelvin. Initial tritium T$_2$ and dihydrogen H$_2$ pressures are specified as $10^{5}$ Pa for Enclosure 1 and $10^{-10}$ Pa for Enclosure 2. Initially, there is no HT in either enclosure.
 
 The reaction between the species is described as follows
 
@@ -81,11 +81,11 @@ We assume that $K = 10/\sqrt{RT}$, which is expected to result in $C_1 = 10 \sqr
 As illustrated in [ver-1kc-2_comparison_time_k10], similarly to ver-1kc-1, T$_2$ and H$_2$ pressures reach equilibrium in both enclosures. What is new, however, is that HT is produced in both enclosures following the sorption law.
 
 Thus, it is crucial to ensure that the chemical equilibrium between HT, T$_2$ and H$_2$ is achieved. This can be verified in both enclosures by examining the ratio between $P_{\text{HT}}$ and $\sqrt{P_{\text{H}_2} P_{\text{T}_2}}$, which must equal $\eta=2$.
-As shown in [ver-1kc-2_equilibrium_constant_k10], this ratio approaches $\eta=2$ for both enclosures, as observed in TMAP7. However, achieving this balance involves a compromise. On one hand, $K_1$ must be sufficiently large to ensure that the chemical kinetics in Enclosure 1 are significantly faster than other processes, such as diffusion and surface sorption. On the other hand, $K_2$ should not be excessively large, as this could hinder the diffusion of species into Enclosure 2, where no species are initially present.
+As shown in [ver-1kc-2_equilibrium_constant_k10], this ratio approaches $\eta=2$ for both enclosures, as observed in TMAP7. However, achieving this balance involves a compromise to reproduce TMAP7's results. The ratio of $K_1$ and $K_2$ must respect [eq:eta]. The values of $K_1$ and $K_2$ must also be large enough to ensure that the kinetics of chemical reactions are faster than diffusion or surface permeation to be closer to the equilibrium assumption used in TMAP7. Here, the equilibrium in enclosure 1 is achieved rapidly. Increasing $K_1$ and $K_2$ would also enable a quicker attainment of equilibrium in enclosure 2. However, using very high values for $K_1$ and $K_2$ would lead to an unnecessary increase in computational costs.
 
 The concentration ratios for T$_2$, H$_2$, and HT between enclosures 1 and 2, shown in [ver-1kc-2_concentration_ratio_T2_k10], [ver-1kc-2_concentration_ratio_H2_k10], and [ver-1kc-2_concentration_ratio_HT_k10], demonstrate that the results obtained with TMAP8 are consistent with the analytical results derived from the sorption law for $K \sqrt{RT} = 10$.
 
-As shown in [ver-1kc-2_mass_conservation_k10], mass is conserved between the two enclosures over time. The variation in mass is only $0.2$ %. This variation in mass can be further minimized by refining the mesh, i.e., increasing the number of segments in the domain.
+As shown in [ver-1kc-2_mass_conservation_k10], mass is conserved between the two enclosures over time. The variation in mass is only $0.4$ %. This variation in mass can be further minimized by refining the mesh, i.e., increasing the number of segments in the domain.
 
 !media comparison_ver-1kc-2.py
        image_name=ver-1kc-2_comparison_time_k10.png
@@ -97,7 +97,7 @@ As shown in [ver-1kc-2_mass_conservation_k10], mass is conserved between the two
        image_name=ver-1kc-2_equilibrium_constant_k10.png
        style=width:50%;margin-bottom:2%;margin-left:auto;margin-right:auto
        id=ver-1kc-2_equilibrium_constant_k10
-       caption=Equilibrium constant as a function of time.
+       caption=Equilibrium constant as a function of time for $\eta = \sqrt{2K_1/K_2}=2$.
 
 !media comparison_ver-1kc-2.py
        image_name=ver-1kc-2_concentration_ratio_T2_k10.png

test/tests/ver-1kc-2/comparison_ver-1kc-2.py

@@ -142,7 +142,6 @@
 ax.yaxis.set_major_formatter(plt.FuncFormatter(lambda val, pos: '{:.3e}'.format(val)))
 ax.set_xlabel('Time (s)')
 ax.set_ylabel(r"Mass Conservation Sum Encl 1 and 2 (mol/m$^3$)")
-ax.legend(loc="best")
 ax.grid(which='major', color='0.65', linestyle='--', alpha=0.3)
 mass_variation_percentage = (np.max(mass_conservation_sum_encl1_encl2_k10)-np.min(mass_conservation_sum_encl1_encl2_k10))/np.max(mass_conservation_sum_encl1_encl2_k10)*100
 print("Percentage of mass variation: ", mass_variation_percentage)
@@ -154,13 +153,16 @@
 gs = gridspec.GridSpec(1,1)
 ax = fig.add_subplot(gs[0])
 
-ax.plot(TMAP8_time_k10, equilibrium_constant_encl_1, label = r"Enclosure 1", c='tab:blue')
 ax.plot(TMAP8_time_k10, equilibrium_constant_encl_2, label = r"Enclosure 2", c='tab:red')
+ax.plot(TMAP8_time_k10, equilibrium_constant_encl_1, label = r"Enclosure 1", c='tab:blue')
 ax.axhline(y=2, color='tab:green', linestyle='--', label='TMAP7 Equilibrium Constant')
 ax.set_xlabel('Time (s)')
 ax.set_ylabel(r"Equilibrium constant $P_{\text{HT}} / \sqrt{P_{\text{H}_2} P_{\text{T}_2}}$")
+ax.set_ylim(bottom=0)
 ax.legend(loc="best")
 ax.grid(which='major', color='0.65', linestyle='--', alpha=0.3)
+print("Relative variation to equilibrium constant in enclosure 1", abs(equilibrium_constant_encl_1[len(equilibrium_constant_encl_1)-1]-2)/2 * 100)
+print("Relative variation to equilibrium constant in enclosure 2", abs(equilibrium_constant_encl_2[len(equilibrium_constant_encl_2)-1]-2)/2 * 100)
 fig.savefig('ver-1kc-2_equilibrium_constant_k10.png', bbox_inches='tight', dpi=300)