Shock Waves

Abstract

The speed of sound is a critical speed (remember what was said in Sect. 2.1 about the time required to restore dynamic equilibrium). In relation to c the fluid motions are classified into subsonic ($v <c$) and supersonic ($v> c$). We use to introduce the Mach number $\textit{Ma} = v / c$, so the motions will be subsonic or supersonic depending on whether $\textit{Ma} <1$ or $\textit{Ma}> 1$.

Appendix

Appendix A: Effects of the Passage of a Shock Wave

We intend to study the effects of the shock on the dynamic and thermodynamic state of the gas, bearing in mind that the direction of the transformation will be determined by the condition that the entropy increases due to the dissipative processes that take place within the shock.

We introduce a reference speed defined by:

$$\begin{aligned} v ^ * = \frac{S}{j} \ . \end{aligned}$$

(5.96)

The Eq. (5.19) can be written in a different form, dividing by $\rho$ and remembering the definition of speed of sound; we have:

$$\begin{aligned} v ^ 2 + \frac{3}{5} c ^ 2- \frac{S}{\rho } = 0 \end{aligned}$$

(5.97)

and with (5.96) we obtain:

$$\begin{aligned} v ^ 2-v ^ * v + \frac{3}{5} c ^ 2 = 0 \ . \end{aligned}$$

(5.98)

The entropy of the unit of mass of a monatomic perfect gas is:

$$\begin{aligned} s = Aln \frac{p}{\rho ^{5/3}} \end{aligned}$$

(5.99)

with $A = 3k _\mathrm{B}/2 \mu m _\mathrm{H}$. It is:

$$\begin{aligned} s' = \exp \left( \frac{s}{A} \right) = \frac{p}{\rho ^{5/3}} \end{aligned}$$

(5.100)

where $s'$ is a monotone function of s and therefore indicates the direction in which s grows.

Let us also say:

$$\begin{aligned} \Sigma = \frac{s'j ^{2/3}}{(v ^ *) ^{8/3}} = \frac{p}{\rho ^{5/3}} \frac{(\rho v) ^{2/3}}{(v ^ *) ^{8/3}} = \frac{v ^ * - v}{v ^ *} \frac{v ^{5/3}}{ (v ^ *) ^{5/3}} \end{aligned}$$

(5.101)

and we introduce a dimensionless variable $\eta$ defined as the ratio between v and the reference speed $v ^ *$:

$$\begin{aligned} \eta = \frac{v}{v ^ *} \ . \end{aligned}$$

(5.102)

Note that $\eta$ is always less than 1, in fact from the definition of $v ^ *$ we have:

$$\begin{aligned} \eta = \frac{vj}{S} = \frac{\rho v ^ 2}{\rho v ^ 2 + p} <1 \ . \end{aligned}$$

(5.103)

It turns out then:

$$\begin{aligned} \Sigma = \eta ^ \frac{5}{3} (1- \eta ) \end{aligned}$$

(5.104)

so $\sigma$ is a dimensionless quantity which is an increasing monotone function of s.

Using the (5.98) the expression of E given by the (5.28) becomes:

$$\begin{aligned} E = \frac{1}{2} v ^ 2 + \frac{5}{2} v ^ * v- \frac{5}{2} v ^ 2 = v \left( \frac{5}{2 } v ^ * - 2v \right) \end{aligned}$$

(5.105)

which allows us to introduce dimensionless quantity

$$\begin{aligned} e _\mathrm{T} = \frac{E}{(v ^ *) ^ 2} = \frac{v \left( \frac{5}{2} v^* - 2v \right) }{v ^ * } = \eta \left( \frac{5}{2} -2 \eta \right) \ . \end{aligned}$$

(5.106)

Two points characterizing the initial and final state of the gas must lie on the curve $e_\mathrm{T}$ and since energy is conserved, these points represent the intersection of the curve $e_\mathrm{T}$ with a straight line parallel to the abscissa axis. This allows once you have chosen the $\eta$ value for a point to determine the value of $\eta$ for the corresponding point. Once can graphically determine the pairs $\eta _A$ and $\eta _B$. Ma which of the two values is the initial one and which is the final one? This can be determined by identifying in a diagram $\sigma - \eta$ the values of $\sigma$ that correspond to $\eta _A$ and $\eta _B$. Since $\sigma$ must increase during the evolution of the system B is the starting point, while A describes the final situation. To preserve the meaning of the symbols, we indicate $\eta _B$ with $\eta _1$ and $\eta _A$ with $\eta _2$.

Having identified the direction in which the gas transformation proceeds, we can study how the characteristic quantities vary.

• (1) The Mach number is given by:

  $$\begin{aligned} \textit{Ma} ^ 2 = \frac{v ^ 2}{c ^ 2} \end{aligned}$$

  (5.107)

  and from (5.98) we get:

  $$\begin{aligned} \textit{Ma} ^ 2 = \frac{3}{5} \frac{v}{v^* - v} = \frac{3}{5} \frac{\eta }{1- \eta } \ . \end{aligned}$$

  (5.108)

  From the behavior of $\textit{Ma} ^ 2$ as a function of $\eta$ we see that the motion is supersonic when $\eta > 0.63$ and otherwise subsonic. The value of $\eta = 0.63$ is the value at which $\sigma$ has the maximum. Thus motion is initially supersonic but the velocity of the gas behind the shock becomes subsonic.

• (2) Based on the definitions of j and $v ^ *$ we get:

  $$\begin{aligned} \frac{S}{\rho } = \frac{S}{j} \frac{j}{\rho } = v ^ * v \ . \end{aligned}$$

  (5.109)

  Since $S/\rho$ has the size of the square of a velocity, let’s say:

  $$\begin{aligned} \lambda = \frac{S}{\rho (v ^ *) ^ 2} = \eta \end{aligned}$$

  (5.110)

  where $\lambda$ is an increasing function of $\eta$ and for $\eta _1> \eta _2$ it results:

  $$\begin{aligned} \lambda (\eta _1)> \lambda (\eta _2) \end{aligned}$$

  (5.111)

  or

  $$\begin{aligned} \frac{S}{\rho (\eta _1) (v ^ *) ^ 2}> \frac{S}{\rho (\eta _2) (v ^ *) ^ 2} \end{aligned}$$

  (5.112)

  from which:

  $$\begin{aligned} \rho _1 = \rho (\eta _1) <\rho (\eta _2) = \rho _2 \end{aligned}$$

  (5.113)

  i.e. the density behind the shock is greater than the density of the unperturbed gas.

• (3) From (5.98) we get

  $$\begin{aligned} \frac{p}{\rho } = v ^ * v-v ^ 2 \end{aligned}$$

  (5.114)

  and dividing by $(v ^ *) ^ 2$:

  $$\begin{aligned} \frac{p}{\rho (v ^ *) ^ 2} = \frac{v ^ * v-v ^ 2}{(v ^ *) ^ 2} \end{aligned}$$

  (5.115)

  waves:

  $$\begin{aligned} p = \rho (v ^ *) ^ 2 \eta (1- \eta ) \ . \end{aligned}$$

  (5.116)

  Dividing by S and remembering (5.110) we get

  $$\begin{aligned} S p = (1- \eta ) \end{aligned}$$

  (5.117)

  so from $\eta _1> \eta _2$ descends $p_1 <p_2$, that is the pressure behind the shock front is greater than the pressure of the unperturbed gas.

• (4) The internal energy of the unit of mass holds

  $$\begin{aligned} \epsilon = \frac{3}{2} \frac{k _\mathrm{B} T}{\mu m _\mathrm{H}} = \frac{3}{2} \frac{p}{\rho } \end{aligned}$$

  (5.118)

  so based on the definition of c and the (5.98) we have:

  $$\begin{aligned} \epsilon = \frac{3}{2} v (v ^ * - v) \ . \end{aligned}$$

  (5.119)

  Dividing by $(v ^ *) ^ 2$ we have:

  $$\begin{aligned} e _\mathrm{I} = \frac{\epsilon }{(v ^ *) ^ 2} = \frac{3}{2} \eta (1- \eta ) \end{aligned}$$

  (5.120)

  which has a maximum of $\eta = 0.5$. From the graph in Figure ?? it results that for each pair of values $\eta _1$ and $\eta _2$ which represents the conditions in front of and behind the shock, i.e. that they correspond to the intersection of the curve $and _\mathrm{T} (\eta )$ with a horizontal line and are ordered in such a way that $\sigma (\eta _1) <\sigma (\eta _2)$, we have:

  $$\begin{aligned} e _\mathrm{I} (\eta _1) <e _\mathrm{I} (\eta _2) \end{aligned}$$

  (5.121)

  and since $e _\mathrm{I}$ is proportional to the temperature, it can be concluded that the temperature increases due to the passage of the shock wave.