S. M. Pourmortazavi, K. Farhadi, V. Mirzajani
Mar 18, 2016
Citations
1
Influential Citations
19
Citations
Journal
Journal of Thermal Analysis and Calorimetry
Abstract
This paper reports thermolysis of diaminoglyoxime (DAG) as a ballistic modifier, and its effects on the thermal behaviors, non-isothermal decomposition reaction kinetics, and burning rates of the homogeneous double-base propellant formulations. Thermal analysis studies were performed by thermogravimetric analysis and differential thermogravimetry (TG-DTG) and differential scanning calorimetry techniques. According to the resulted data, it was found that DAG could change the thermal decomposition mechanism function, thermokinetic parameters and kinetic equation of the propellants. Evaluation of DAG as a ballistic modifier in double-base propellant formulations indicated that it brings down the pressure index to 0.068 compared to 0.24 for the control composition in the pressure range 5–7 Mpa. The results showed that the main exothermal decomposition reaction of the propellant sample in the absence of DAG has the mechanism function of $$ f(\alpha ) = \frac{3}{2}\left( {1 - \alpha } \right)^{4/3} [\left( {1 - \alpha } \right)^{ - 1/3} - 1]^{ - 1} $$f(α)=321-α4/3[1-α-1/3-1]-1 and the kinetic equation of $$ {\raise0.7ex\hbox{${{\text{d}}\alpha }$} \!\mathord{\left/ {\vphantom {{{\text{d}}\alpha } {{\text{d}}t}}}\right.\kern-0pt} \!\lower0.7ex\hbox{${{\text{d}}t}$}} = 1.60 \times 10^{17} \left( {1 - \alpha } \right)^{4/3} [\left( {1 - \alpha } \right)^{ - 1/3} - 1]^{ - 1} e^{{ - 2.21 \times 10^{4} /{\text{T}}}} $$dαdt=1.60×10171-α4/3[1-α-1/3-1]-1e-2.21×104/T, while modified propellant with DAG has a different function mechanism and kinetic equation as following: $$ f(\alpha ) = \frac{5}{2}\left( {1 - \alpha } \right)\left[ { - { \ln }\left( {1 - \alpha } \right)} \right]^{3/5} $$f(α)=521-α-ln1-α3/5 and $$ {\raise0.7ex\hbox{${{\text{d}}\alpha }$} \!\mathord{\left/ {\vphantom {{{\text{d}}\alpha } {{\text{d}}t }}}\right.\kern-0pt} \!\lower0.7ex\hbox{${{\text{d}}t }$}} = 9.28 \times 10^{40} \left( {1 - \alpha } \right)\left[ { - { \ln }\left( {1 - \alpha } \right)} \right]^{3/5} e^{{ - 4.66 \times 10^{4} /{\text{T}}}} $$dαdt=9.28×10401-α-ln1-α3/5e-4.66×104/T, respectively.