Zeldovich–Taylor flow

Consider a spherically outgoing Chapman–Jouguet detonation wave propagating with a constant velocity $D$. By definition, immediately behind the detonation wave, the gas velocity is equal to the local sound speed $c$ with respect to the wave. Let $v(r,t)$ be the radial velocity of the gas behind the wave, in a fixed frame. The detonation is ignited at $t=0$ at $r=0$. For $t>0$, the gas velocity must be zero at the center $r=0$ and should take the value $v=D-c$ at the detonation location $r=Dt$. The fluid motion is governed by the inviscid Euler equationsSedov, L. I., & Volkovets, A. G. (2018). Similarity and dimensional methods in mechanics. CRC press.

:$$

\begin{align}

\frac{\partial \rho}{\partial t} + v\frac{\partial \rho}{\partial r} &= - \rho\left(\frac{\partial v}{\partial r} + \frac{2v}{r}\right),\\

\frac{\partial v}{\partial t} + v \frac{\partial v}{\partial r} &= - \frac{1}{\rho}\frac{\partial p}{\partial r},\\

\frac{\partial s}{\partial t} + v \frac{\partial s}{\partial r} &= 0

\end{align}

where $\backslash rho$ is the density, $p$ is the pressure and $s$ is the entropy. The last equation implies that the flow is isentropic and hence we can write $c^2=d\; p/d\; \backslash rho$.

Since there are no length or time scales involved in the problem, one may look for a self-similar solution of the form $v(r,t)=v(\backslash xi),\; p(r,t)\; =\; p(\backslash xi),\backslash ,\; \backslash rho(r,t)\; =\; \backslash rho(\backslash xi),\backslash ,c(r,t)\; =\; c(\backslash xi)$, where $\backslash xi=r/t$. The first two equations then become

:$$

\begin{align}

(\xi-v)\rho'/\rho &= v' + 2v/\xi,\\

(\xi-v) v' &= p'/\rho = c^2 \rho'/\rho

\end{align}

where prime denotes differentiation with respect to $\backslash xi$. We can eliminate $\backslash rho\text{'}/\backslash rho$ between the two equations to obtain an equation that contains only $v$ and $c$. Because of the isentropic condition, we can express $\backslash rho\; =\; \backslash rho(c),\; \backslash ,\; p=p(c)$, that is to say, we can replace $\backslash rho^\{-1\}d\backslash rho/dx$ with $\backslash rho^\{-1\}c\text{'}d\backslash rho/dc$. This leads to

:$\backslash begin\{align\}$

(\xi-v)\frac{1}{\rho}\frac{d\rho}{dc} c' &= v' + 2v/\xi,\\

\left[\frac{(\xi-v)^2}{c^2}-1\right]v' &= \frac{2v}{\xi}.

\end{align}

For polytropic gases with constant specific heats, we have $\backslash rho^\{-1\}d\backslash rho/dc\; =\; 2/[(\backslash gamma-1)c]$. The above set of equations cannot be solved analytically, but has to be integrated numerically. The solution has to be found for the range $0\backslash leq\; \backslash xi\; \backslash leq\; D$ subjected to the condition $\backslash xi-v=c$ at $\backslash xi=D.$

The function $v(\backslash xi)$ is found to monotonically decrease from its value $v(D)\; =\; c(D)-D$ to zero at a finite value of

From the second equation described above, it follows that when $v=0$, $\backslash xi=c$. More precisely, as $v\backslash rightarrow\; 0$, that equation can be approximated asLandau, L. D., & Lifshitz, E. M. (1987). Fluid Mechanics Pergamon. New York, section 130, pp-496-499.

:$(\backslash ln\; v)\text{'}\; =\; 2c^2/[\backslash xi(\backslash xi^2-c^2)].$

As $v\backslash rightarrow\; 0$, $\backslash ln\; v\backslash rightarrow\; -\backslash infty$ and $(\backslash ln\; v)\text{'}\backslash rightarrow\; \backslash infty$ if $\backslash xi$ decreases as $v\backslash rightarrow\; 0$. The left hand side of the above equation can become positive infinity only if $\backslash xi\backslash rightarrow\; c$. Thus, when $\backslash xi$ decreases to the value $\backslash xi=c\_0$, the gas comes to rest (Here $c\_0$ is the sound speed corresponding to $v=0$). Thus, the rarefaction motion occurs for and there is no fluid motion for $0\backslash leq\; \backslash xi\; \backslash leq\; c\_0$.

Rewrite the second equation as

:$v\backslash frac\{d\backslash xi\}\{dv\}\; =\; \backslash frac\{1\}\{2\}\backslash xi\backslash left[\backslash frac\{(\backslash xi-v)^2\}\{c^2\}-1\backslash right].$

In the neighborhood of the weak discontinuity, the quantities to the first order (such as $v,\backslash ,\backslash xi-c\_0,\backslash ,c-c\_0$) reduces the above equation to

:$v\backslash frac\{d\}\{dv\}(\backslash xi-c\_0)\; =\; (\backslash xi-c\_0)\; -\; (v+c-c\_0).$

At this point, it is worth mentioning that in general, disturbances in gases are propagated with respect to the gas at the local sound speed. In other words, in the fixed frame, the disturbances are propagated at the speed $v+c$ (the other possibility is $v-c$ although it is of no interest here). If the gas is at rest $v=0$, then the disturbance speed is $c\_0$. This is just a normal sound wave propagation. If however $v$ is non-zero but a small quantity, then one find the correction for the disturbance propagation speed as $v+c=c\_0\; +\; \backslash alpha\_0\; v$ obtained using a Taylor series expansion, where $\backslash alpha\_0$ is the Landau derivative (for ideal gas, $\backslash alpha\_0=(\backslash gamma+1)/2$, where $\backslash gamma$ is the specific heat ratio). This means that the above equation can be written as

:$v\backslash frac\{d\}\{dv\}(\backslash xi-c\_0)\; -\; (\backslash xi-c\_0)\; =\; \backslash alpha\_0\; v$

whose solution is

:$\backslash xi-c\_0\; =\; \backslash alpha\_0\; v\backslash ln\; (A/v)$

where $A$ is a constant. This determines $v(\backslash xi)$ implicitly in the neighborhood of the week discontinuity where $v$ is small. This equation shows that at $\backslash xi=c\_0$, $v=0$, $dv/d\backslash xi=0$, but all higher-order derivatives are discontinuous. In the above equation, subtract $v+c-c\_0$ from the left-hand side and $\backslash alpha\_0v$ from the right-hand side to obtain

:$\backslash xi-v-c\; =\; (\backslash xi-c\_0)-(v+c-c\_0)=\backslash alpha\_0\; v[\backslash ln(A/v)-1]$

which implies that $\backslash xi-v>c$ if $v$ is a small quantity. It can be shown that the relation $\backslash xi-v>c$ not only holds for small $v$, but throughout the rarefaction wave.

First let us show that the relation $\backslash xi-v>c$ is not only valid near the weak discontinuity, but throughout the region. If this inequality is not maintained, then there must be a point where $\backslash xi-v=c,\backslash ,\; v\backslash neq\; 0$ between the weak discontinuity and the detonation front. The second governing equation implies that at this point $v\text{'}$ must be infinite or, $d\backslash xi/dv=0$. Let us obtain $d^2\backslash xi/dv^2$ by taking the second derivative of the governing equation. In the resulting equation, impose the condition $\backslash xi-v=c,\backslash ,v\backslash neq\; 0,\backslash ,\; d\backslash xi/dv=0$ to obtain $d^2\backslash xi/dv^2\; =\; -\backslash alpha\_0\; \backslash xi/c\_0v\backslash neq\; 0$. This implies that $\backslash xi(v)$ reaches a maximum at this point which in turn implies that $v(\backslash xi)$ cannot exist for $\backslash xi$ greater than the maximum point considered since otherwise $v(\backslash xi)$ would be multi-valued. The maximum point at most can be corresponded to the outer boundary (detonation front). This means that $\backslash xi-v-c$ can vanish only on the boundary and it is already shown that $\backslash xi-v-c$ is positive near the weak discontinuity, $\backslash xi-v-c$ is positive everywhere in the region except the boundaries where it can vanish.

Note that near the detonation front, we must satisfy the condition $\backslash xi-v=c,\backslash ,\; v\backslash neq\; 0$. The value evaluated at $\backslash xi=D$ for the function $\backslash xi-v$, i.e., $D-v(D)$ is nothing but the velocity of the detonation front with respect to the gas velocity behind it. For a detonation front, the condition $D-v(D)\backslash leq\; c(D)$ must always be met, with the equality sign representing Chapman–Jouguet detonations and the inequalities representing over-driven detonations. The analysis describing the point $\backslash xi-v=c,\backslash ,\; v\backslash neq\; 0$ must correspond to the detonation front.

• Taylor–von Neumann–Sedov blast wave
• Guderley–Landau–Stanyukovich problem

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