Sources: The only parameter which characterizes the explosion is the
Damköhler number \delta. When \delta is very high, conduction time is longer than the chemical reaction time and the system explodes with high temperature since there is not enough time for conduction to remove the heat. On the other hand, when \delta is very low, heat conduction time is much faster than the chemical reaction time, such that all the heat produced by the chemical reaction is immediately conducted to the wall, thus there is no explosion, it goes to an almost steady state,
Amable Liñán coined this mode as slowly reacting mode. At a critical Damköhler number \delta_c the system goes from slowly reacting mode to explosive mode. Therefore, \delta, the system is in steady state. Instead of solving the full problem to find this \delta_c,
Frank-Kamenetskii solved the steady state problem for various Damköhler number until the critical value, beyond which no steady solution exists. So the problem to be solved is : \frac 1 {\eta^j} \frac d {d \eta} \left(\eta^j \frac{d \theta}{d \eta}\right) = - \delta e^\theta with boundary conditions :\theta(1)=0, \quad \frac{d\theta}{d\eta}_{\eta=0}=0 the second condition is due to the symmetry of the vessel. The above equation is special case of
Liouville–Bratu–Gelfand equation in
mathematics.
Planar vessel For planar vessel, there is an exact solution. Here j=0, then :\frac{d^2\theta}{d\eta^2} = -\delta e^\theta If the transformations \Theta= \theta_m-\theta and \xi^2 = \delta e^{\theta_m} \eta^2, where \theta_m is the maximum temperature which occurs at \eta=0 due to symmetry, are introduced :\frac{d^2\Theta}{d\xi^2} = e^{-\Theta}, \quad \Theta(0)=0, \quad \frac{d\Theta}{d\xi}_{\xi=0}=0 Integrating once and using the second boundary condition, the equation becomes :\frac{d\Theta}{d\xi} = \sqrt{2(1-e^{-\Theta})} and integrating again :e^{(\theta_m-\theta)/2} = \cosh \left(\eta \sqrt{\frac{\delta e^{\theta_m}}{2}} \right) The above equation is the exact solution, but \theta_m maximum temperature is unknown, but we have not used the boundary condition of the wall yet. Thus using the wall boundary condition \theta=0 at \eta=1, the maximum temperature is obtained from an implicit expression, :e^{\theta_m/2} = \cosh \sqrt{\frac{\delta e^{\theta_m}} 2} \quad \text{or} \quad \delta = 2 e^{-\theta_m} \left(\operatorname{arcosh} e^{\theta_m/2}\right)^{\!2} Critical \delta_c is obtained by finding the maximum point of the equation (see figure), i.e., d\delta/d\theta_m=0 at \delta_c. :\frac{d\delta}{d\theta_m}=0, \quad \Rightarrow \quad e^{\theta_{m,c}/2} - \sqrt{e^{\theta_{m,c}}-1}\operatorname{arcosh} e^{\theta_{m,c}/2} =0 :\theta_{m,c}= 1.1868, \quad \Rightarrow \quad \delta_c = 0.8785 So the critical Frank-Kamentskii parameter is \delta_c=0.8785. The system has no steady state (or explodes) for \delta>\delta_c=0.8785 and for \delta, the system goes to a steady state with very slow reaction.
Cylindrical vessel For cylindrical vessel, there is an exact solution. Though Frank-Kamentskii used
numerical integration assuming there is no explicit solution,
Paul L. Chambré provided an exact solution in 1952. H. Lemke also solved provided a solution in a somewhat different form in 1913. Here j=1, then :\frac 1 \eta \frac d {d\eta}\left(\eta\frac{d\theta}{d\eta}\right) = -\delta e^\theta If the transformations \omega= \eta d\theta/d\eta and \chi = e^{\theta} \eta^2 are introduced :\frac{d\omega}{d\chi} = -\frac{\delta}{2+\omega}, \quad \omega(0)=0 The general solution is \omega^2 + 4\omega + C = -2\delta\chi. But C=0 from the symmetry condition at the centre. Writing back in original variable, the equation reads, :\eta^2 \left(\frac{d\theta}{d\eta}\right)^2+ 4\eta \frac{d\theta}{d\eta} = - 2\delta\eta^2 e^\theta But the original equation multiplied by 2\eta^2 is :2\eta^2 \frac{d^2\theta}{d\eta^2} + 2\eta \frac{d\theta}{d\eta} = -2 \delta \eta^2 e^\theta Now subtracting the last two equation from one another leads to :\frac{d^2\theta}{d\eta^2} - \frac{1}{\eta} \frac{d\theta}{d\eta} - \frac{1}{2} \left(\frac{d\theta}{d\eta}\right)^2 =0 This equation is easy to solve because it involves only the derivatives, so letting g(\eta)=d\theta/d\eta transforms the equation :\frac{dg}{d\eta} - \frac{g}{\eta} - \frac{g^2}{2}=0 This is a
Bernoulli differential equation of order 2, a type of
Riccati equation. The solution is :g(\eta) = \frac{d\theta}{d\eta} = -\frac{4\eta}{B'+\eta^2} Integrating once again, we have \theta = A - 2\ln ( B\eta^2 + 1) where B = 1/B'. We have used already one boundary condition, there is one more boundary condition left, but with two constants A, \ B. It turns out A and B are related to each other, which is obtained by substituting the above solution into the starting equation we arrive at A = \ln (8B/\delta) . Therefore, the solution is :\theta = \ln \frac{8B/\delta}{(B\eta^2+1)^2} Now if we use the other boundary condition \theta(1)=0, we get an equation for B as \delta(B+1)^2 - 8B =0. The maximum value of \delta for which solution is possible is when B=1, so the critical Frank-Kamentskii parameter is \delta_c=2. The system has no steady state( or explodes) for \delta>\delta_c=2 and for \delta, the system goes to a steady state with very slow reaction. The maximum temperature \theta_m occurs at \eta=0 :\theta_m= \ln \frac{8B}{\delta} \quad \text{or} \quad \delta= 8B e^{-\theta_m} For each value of \delta, we have two values of \theta_m since B is multi-valued. The maximum critical temperature is \theta_{m,c} = \ln 4.
Spherical vessel For spherical vessel, there is no known explicit solution, so
Frank-Kamenetskii used numerical methods to find the critical value. Here j=2, then :\frac 1 {\eta^2} \frac d {d\eta}\left(\eta^2\frac{d\theta}{d\eta}\right) = -\delta e^\theta If the transformations \Theta= \theta_m-\theta and \xi^2 = \delta e^{\theta_m} \eta^2, where \theta_m is the maximum temperature which occurs at \eta=0 due to symmetry, are introduced : \frac 1 {\xi^2}\frac d {d\xi}\left(\xi^2\frac{d\Theta}{d\xi}\right)= e^{-\Theta}, \quad \Theta(0)=0, \quad \frac{d\Theta}{d\xi}_{\xi=0}=0 The above equation is nothing but
Emden–Chandrasekhar equation, which appears in
astrophysics describing
isothermal gas sphere. Unlike planar and cylindrical case, the spherical vessel has infinitely many solutions for \delta oscillating about the point \delta=2, instead of just two solutions, which was shown by
Israel Gelfand. The lowest branch will be chosen to explain explosive behavior. From numerical solution, it is found that the critical Frank-Kamenetskii parameter is \delta_c=3.3220. The system has no steady state( or explodes) for \delta>\delta_c=3.3220 and for \delta, the system goes to a steady state with very slow reaction. The maximum temperature \theta_m occurs at \eta=0 and maximum critical temperature is \theta_{m,c} = 1.6079. ==Non-symmetric geometries==