Numerical analysis of hydrogen sulphide conversion to hydrogen during its pyrolysis and partial oxidation

Obtaining of hydrogen during pyrolysis and partial oxidation of hydrogen sulphide is analyzed on the basis of a detailed kinetic model of $\mathrm{H}_2\mathrm{S}$ oxidation. It is shown that the $\mathrm{H}_2$ output in the case of $\mathrm{H}_2\mathrm{S}$ pyrolysis in adiabatic flow reactor with a residence time of $\approx1$ s. Even for the initial temperature of the mixture $T_0=1400$ K, the molar fraction of $\mathrm{H}_2$ is only $12\%$, though the equilibrium value is reached within the reactor. At $T_0<1200$ K, there is no enough time for the chemical equilibrium inside the reactor to be established, and the $\mathrm{H}_2$ concentration is lower than the equilibrium value. At $T_0<1000$ K, there is practically no pyrolysis reaction in the reactor. Addition of a small amount of air to $\mathrm{H}_2\mathrm{S}$ leads to energy release, to an increase in temperature, and, as a consequence, to acceleration of $\mathrm{H}_2\mathrm{S}$ conversion. The normalized output of $\mathrm{H}_2$ can be increased by several times. For each value of $T_0$, there exists an optimal value of the fuel-to-air equivalence ratio $\phi$ that ensures the maximum $\mathrm{H}_2$ output in the $\mathrm{H}_2\mathrm{S}$-air mixture. The process of partial oxidation at high values of $\phi>\phi_b$ and low values of $T_0$ is essentially nonequilibrium; as a result, the $\mathrm{H}_2$ concentration at the exit from a finite-length reactor can be higher than its equilibrium value, e.g., by $30$–$40\%$ at $T_0=800$ K and $\phi=6-10$. The reasons responsible for reaching a “superequilibrium” concentration of $\mathrm{H}_2$ at the flow reactor exit are determined.