Taylor–von Neumann–Sedov blast wave

Taylor–von Neumann–Sedov blast wave (or sometimes referred to as Sedov–von Neumann–Taylor blast wave) refers to a blast wave induced by a strong explosion. The blast wave was described by a self-similar solution independently by G. I. Taylor, John von Neumann and Leonid Sedov during World War II.

History

G. I. Taylor was told by the British Ministry of Home Security that it might be possible to produce a bomb in which a very large amount of energy would be released by nuclear fission and asked to report the effect of such weapons. Taylor presented his results in June 27, 1941. Exactly at the same time, in the United States, John von Neumann was working on the same problem and he presented his results on June 30, 1941. It was said that Leonid Sedov was also working on the problem around the same time in the USSR, although Sedov never confirmed any exact dates.

The complete solution was published first by Sedov in 1946. von Neumann published his results in August 1947 in the Los Alamos scientific laboratory report on , although that report was distributed only in 1958. Taylor got clearance to publish his results in 1949 and he published his works in two papers in 1950. In the second paper, Taylor calculated the energy of the atomic bomb used in the Trinity (nuclear test) using the similarity, just by looking at the time series of blast wave photographs published by Julian E Mack in 1947.

Mathematical description

Consider a strong explosion (such as nuclear bombs) that releases a large amount of energy $E$ in a small volume during a short time interval. This will create a strong spherical shock wave propagating outwards from the explosion center. The self-similar solution tries to describe the flow when the shock wave has moved through a distance that is extremely large when compared to the size of the explosive. At these large distances, the information about the size and duration of the explosion will be forgotten; only the energy released $E$ will have influence on how the shock wave evolves. To a very high degree of accuracy, then it can be assumed that the explosion occurred at a point (say the origin $r=0$) instantaneously at time $t=0$.

The shock wave in the self-similar region is assumed to be still very strong such that the pressure behind the shock wave $p_1$ is very large in comparison with the pressure (atmospheric pressure) in front of the shock wave $p_0$, which can be neglected from the analysis. Although the pressure of the undisturbed gas is negligible, the density of the undisturbed gas $\rho_0$ cannot neglected since the density jump across strong shock waves are finite as a direct consequence of Rankine–Hugoniot conditions. This approximation is equivalent to setting $p_0=0$ and the corresponding sound speed $c_0=0$, but keeping its density non zero, i.e., $\rho_0\neq 0$.

The only parameters available at our disposal are the energy $E$ and the undisturbed gas density $\rho_0$. The properties behind the shock wave such as $p_1,\,\rho_1$ are derivable from those in front of the shock wave. The only non-dimensional combination available from $r,\,t,\,\rho_0$ and $E$ is

$r\left(\frac{\rho_0}{Et^2}\right)^{1/5}$.

The solution is assumed to be a function only of the above variable. The shock wave location $r=R(t)$ will correspond to some value denoted here by $\beta$, i.e.,

$R= \beta \left(\frac{Et^2}{\rho_0}\right)^{1/5}.$

The propagation velocity of the shock wave is

$D=\frac{\mathrm{d}R}{\mathrm{d}t}=\frac{2R}{5t}=\frac{2\beta}{5}\left(\frac{E}{\rho_0t^3}\right)^{1/5}$

With the approximation described above, Rankine–Hugoniot conditions determines the gas velocity immediately behind the shock front $v_1$, $p_1$ and $\rho_1$ for an ideal gas as follows

$v_1 = \frac{2}{\gamma+1}D, \quad p_1 = \frac{2}{\gamma+1}\rho_0 D^2, \quad \rho_1= \rho_0 \frac{\gamma+1}{\gamma-1}$

where $\gamma$ is the specific heat ratio. Since $\rho_0$ is a constant, the density immediately behind the shock wave is not changing with time, whereas $v_1$ and $p_1$ decrease as $t^{-3/5}$ and $t^{-6/5}$, respectively.

Self-similar solution

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 $v(r,t)$, density $\rho(r,t)$ and pressure $p(r,t)$ are given by

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \begin{align} \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{align}

At $r=R(t)$, 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 $c(r,t)$ since pressure can be obtained from the formula $c^2=\gamma p/\rho$. The following non-dimensional self-similar variables are introduced,

$\xi = \frac{r}{R(t)}, \quad V(\xi) = \frac{5tv}{2r}, \quad G(\xi) = \frac{\rho}{\rho_0}, \quad Z(\xi) = \frac{25 t^2 c^2}{4r^2}$.

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

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

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 $Z$ and $V$ can be deduced directly from energy conservation. Since the energy associated with the undisturbed gas is neglected by setting $p_0=0$, the total energy of the gas within the shock sphere must be equal to $E$. Due to self-similarity, it is clear that not only the total energy within a sphere of radius $\xi=1$ is constant, but also the total energy within a sphere of any radius $\xi<1$ (in dimensional form, it says that total energy within a sphere of radius $r$ that moves outwards with a velocity $v_n=2r/5t$ must be constant). The amount of energy that leaves the sphere of radius $r$ in time $dt$ due to the gas velocity $v$ is $4\pi r^2\rho v(h+v^2/2)\mathrm{d}t$, where $h$ is the specific enthalpy of the gas. In that time, the radius of the sphere increases with the velocity $v_n$ and the energy of the gas in this extra increased volume is $4\pi r^2 \rho v_n(e+v^2/2)\mathrm{d}t$, where $e$ is the specific energy of the gas. Equating these expressions and substituting $e=c^2/\gamma(\gamma-1)$ and $h=c^2/(\gamma-1)$ that is valid for ideal polytropic gas leads to

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

The continuity and energy equation reduce to

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \begin{align} \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{align}

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

Failed to parse (Missing <code>texvc</code> executable. Please see math/README to configure.): \begin{align} \xi^5 &= \left[\frac{1}{2}(\gamma+1)\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{align}

where

$\nu_1= -\frac{13\gamma^2-7\gamma+12}{(3\gamma-1)(2\gamma+1)},\quad \nu_2 = \frac{5(\gamma-1)}{2\gamma+1}, \quad \nu_3 = \frac{3}{2\gamma+1},\quad \nu_4 = -\frac{\nu_1}{2-\gamma}, \quad \nu_5 = - \frac{2}{2-\gamma}.$

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

$E=\int_0^R\rho[v^2/2+c^2/\gamma(\gamma-1)4\pi r^2]\mathrm{d}r$

to obtain

$\beta^5 \frac{16\pi}{25}\int_0^1 G[V^2/2+Z/\gamma(\gamma-1)]\xi^4\mathrm{d}\xi = 1.$

For air, $\gamma=7/5$ and $\beta=1.033$.