Taylor–von Neumann–Sedov blast wave

The gas motion behind the shock wave is governed by Euler equations. For an ideal polytropic gas with spherical symmetry, the equations for the fluid variables such as radial velocity ${\displaystyle v(r,t)}$, density ${\displaystyle \rho (r,t)}$ and pressure ${\displaystyle p(r,t)}$ are given by

${\displaystyle {\begin{aligned}{\frac {\partial v}{\partial t}}+v{\frac {\partial v}{\partial r}}&=-{\frac {1}{\rho }}{\frac {\partial p}{\partial r}},\\{\frac {\partial \rho }{\partial t}}+{\frac {\partial (\rho v)}{\partial r}}&=-{\frac {2\rho v}{r}},\\\left({\frac {\partial }{\partial t}}+v{\frac {\partial }{\partial r}}\right)\ln {\frac {p}{\rho ^{\gamma }}}&=0.\end{aligned}}}$

At ${\displaystyle r=R(t)}$, the solutions should approach the values given by the Rankine-Hugoniot conditions defined in the previous section.

The variable pressure can be replaced by the sound speed ${\displaystyle c(r,t)}$ since pressure can be obtained from the formula ${\displaystyle c^{2}=\gamma p/\rho }$. The following non-dimensional self-similar variables are introduced,[13][14]

${\displaystyle \xi ={\frac {r}{R(t)}},\quad V(\xi )={\frac {5tv}{2r}},\quad G(\xi )={\frac {\rho }{\rho _{0}}},\quad Z(\xi )={\frac {25t^{2}c^{2}}{4r^{2}}}}$.

The conditions at the shock front ${\displaystyle \xi =1}$ becomes

${\displaystyle V(1)={\frac {2}{\gamma +1}},\quad G(1)={\frac {\gamma +1}{\gamma -1}},\quad Z(1)={\frac {2\gamma (\gamma -1)}{(\gamma +1)^{2}}}.}$

Substituting the self-similar variables into the governing equations will lead to three ordinary differential equations. Solving these differential equations analytically is laborious, as shown by Sedov in 1946 and von Neumann in 1947. G. I. Taylor integrated these equations numerically to obtain desired results.

The relation between ${\displaystyle Z}$ and ${\displaystyle V}$ can be deduced directly from energy conservation. Since the energy associated with the undisturbed gas is neglected by setting ${\displaystyle p_{0}=0}$, the total energy of the gas within the shock sphere must be equal to ${\displaystyle E}$. Due to self-similarity, it is clear that not only the total energy within a sphere of radius ${\displaystyle \xi =1}$ is constant, but also the total energy within a sphere of any radius ${\displaystyle \xi <1}$ (in dimensional form, it says that total energy within a sphere of radius ${\displaystyle r}$ that moves outwards with a velocity ${\displaystyle v_{n}=2r/5t}$ must be constant). The amount of energy that leaves the sphere of radius ${\displaystyle r}$ in time ${\displaystyle dt}$ due to the gas velocity ${\displaystyle v}$ is ${\displaystyle 4\pi r^{2}\rho v(h+v^{2}/2)\mathrm {d} t}$, where ${\displaystyle h}$ is the specific enthalpy of the gas. In that time, the radius of the sphere increases with the velocity ${\displaystyle v_{n}}$ and the energy of the gas in this extra increased volume is ${\displaystyle 4\pi r^{2}\rho v_{n}(e+v^{2}/2)\mathrm {d} t}$, where ${\displaystyle e}$ is the specific energy of the gas. Equating these expressions and substituting ${\displaystyle e=c^{2}/\gamma (\gamma -1)}$ and ${\displaystyle h=c^{2}/(\gamma -1)}$ that is valid for ideal polytropic gas leads to

${\displaystyle Z={\frac {\gamma (\gamma -1)(1-V)V^{2}}{2(\gamma V-1)}}.}$

The continuity and energy equation reduce to

${\displaystyle {\begin{aligned}{\frac {\mathrm {d} V}{\mathrm {d} \ln \xi }}-(1-V){\frac {\mathrm {d} \ln G}{\mathrm {d} \ln \xi }}&=-3V\\{\frac {\mathrm {d} \ln Z}{\mathrm {d} \ln \xi }}-(\gamma -1){\frac {\mathrm {d} \ln G}{\mathrm {d} \ln \xi }}&=-{\frac {5-2V}{1-V}}.\end{aligned}}}$

Self-similar solution of Taylor-von Neumann-Sedov blast wave for ${\displaystyle \gamma =7/5}$

Expressing ${\displaystyle \mathrm {d} V/\mathrm {d} \ln \xi }$ and ${\displaystyle \mathrm {d} \ln G/\mathrm {d} V}$ as a function of ${\displaystyle V}$ only using the relation obtained earlier and integrating once yields the solution in implicit form,

${\displaystyle {\begin{aligned}\xi ^{5}&=\left[{\frac {1}{2}}(\gamma +1)V\right]^{-2}\left\{{\frac {\gamma +1}{7-\gamma }}[5-(3\gamma -1)V]\right\}^{\nu _{1}}\left[{\frac {\gamma +1}{\gamma -1}}(\gamma V-1)\right]^{\nu _{2}},\\G&={\frac {\gamma +1}{\gamma -1}}\left[{\frac {\gamma +1}{\gamma -1}}(\gamma V-1)\right]^{\nu _{3}}\left\{{\frac {\gamma +1}{7-\gamma }}[5-(3\gamma -1)V\right\}^{\nu _{4}}\left[{\frac {\gamma +1}{\gamma -1}}(1-V)\right]^{\nu _{5}}\end{aligned}}}$

where

${\displaystyle \nu _{1}=-{\frac {13\gamma ^{2}-7\gamma +12}{(3\gamma -1)(2\gamma +1)}},\quad \nu _{2}={\frac {5(\gamma -1)}{2\gamma +1}},\quad \nu _{3}={\frac {3}{2\gamma +1}},\quad \nu _{4}=-{\frac {\nu _{1}}{2-\gamma }},\quad \nu _{5}=-{\frac {2}{2-\gamma }}.}$

The constant ${\displaystyle \beta }$ that determines the shock location can be determined from the conservation of energy

${\displaystyle E=\int _{0}^{R}\rho [v^{2}/2+c^{2}/\gamma (\gamma -1)]4\pi r^{2}\mathrm {d} r}$

to obtain

${\displaystyle \beta ^{5}{\frac {16\pi }{25}}\int _{0}^{1}G[V^{2}/2+Z/\gamma (\gamma -1)]\xi ^{4}\mathrm {d} \xi =1.}$

For air, ${\displaystyle \gamma =7/5}$ and ${\displaystyle \beta =1.033}$. The solution for ${\displaystyle \gamma =7/5}$ is shown in the figure by graphing the curves of ${\displaystyle \rho /\rho _{1}=G(\gamma -1)/(\gamma +1)}$, ${\displaystyle v/v_{1}=\xi V(\gamma +1)/2}$, ${\displaystyle p/p_{1}=\xi ^{2}GZ(\gamma +1)/(2\gamma )}$ and ${\displaystyle T/T_{1}=\xi ^{2}Z(\gamma +1)^{2}/[2\gamma (\gamma -1)],}$ where ${\displaystyle T}$ is the temperature.

The asymptotic behavior of the central region can be investigated by taking the limit ${\displaystyle \xi \rightarrow 0}$. From the figure, it can be observed that the density falls to zero very rapidly behind the shock wave. The entire mass of the gas which was initially spread out uniformly in a sphere of radius ${\displaystyle R}$ is now contained in a thin layer behind the shock wave, that is to say, all the mass is driven outwards by the acceleration imparted by the shock wave. Thus, most of the region is basically empty. The pressure ratio also drops rapidly to attain the constant value ${\displaystyle p_{c}}$. The temperature ratio follows from the ideal gas law; since density ratio decays to zero and the pressure ratio is constant, the temperature ratio must become infinite. The limiting form for the density is given as follows

${\displaystyle {\frac {\rho }{\rho _{1}}}\sim \xi ^{3/(\gamma -1)},\quad {\frac {p}{p_{1}}}\rightarrow p_{c},\quad {\frac {T}{T_{1}}}\sim \xi ^{-3/(\gamma -1)}\quad {\text{as}}\quad \xi \rightarrow 0.}$

Remember that the density ${\displaystyle \rho _{1}}$ is time-independent whereas ${\displaystyle p_{1}\sim t^{-6/5}}$ which means that the actual pressure is in fact time dependent. It becomes clear if the above forms are rewritten in dimensional units,

${\displaystyle \rho \sim r^{3/(\gamma -1)}t^{-6/5(\gamma -1)},\quad p\rightarrow p_{c}t^{-6/5},\quad T\sim r^{-3/(\gamma -1)}t^{(6/5)(2-\gamma )/(\gamma -1)}\quad {\text{as}}\quad r\rightarrow 0.}$

The velocity ratio has the linear behavior in the central region,

${\displaystyle {\frac {v}{v_{1}}}\sim \xi \quad {\text{as}}\quad \xi \rightarrow 0}$

whereas the behavior of the velocity itself is given by

${\displaystyle v\sim rt^{1/5}\quad {\text{as}}\quad r\rightarrow 0.}$