Unsteady waves

Moving shock waves

Moving Normal Shock Waves

The starting point is the governing equations for stationary normal shocks (repeated here for convenience).

${\rho }_1{u}_1={\rho }_2{u}_2$

(Eq. 6.1)

${\rho }_1{u}_1^2+p_1={\rho }_2{u}_2^2+p_2$

(Eq. 6.2)

$h_1+\frac{1}{2}{u}_1^2=h_2+\frac{1}{2}{u}_2^2$

(Eq. 6.3)

Shock moving to the right with the constant speed $W$ into a gas that is standing still. Moving with the shock, we would see a gas velocity ahead of the shock $u_1=W$, and the gas behind the shock moves to the right with the velocity $u_2=W-u_p$. Now, let's insert $u_1$ and $u_2$ in the stationary shock relations \ref{eq:stationary:cont} - \ref{eq:stationary:energy}.

${\rho }_{1}W={\rho }_2(W-u_p)$

(Eq. 6.4)

${\rho }_1{W}^2+p_1={\rho }_2(W-u_p{)}^2+p_2$

(Eq. 6.5)

$h_1+\frac{1}{2}{W}^2=h_2+\frac{1}{2}(W-u_p{)}^2$

(Eq. 6.6)

Rewriting Eqn. \ref{eq:unsteady:cont}

$(W-u_p)=W\frac{{\rho }_1}{{\rho }_2}$

(Eq. 6.7)

Inserting Eqn. \ref{eq:unsteady:cont:mod} in Eqn. \ref{eq:unsteady:mom} gives

$p_1+{\rho }_1{W}^2=p_2+{\rho }_2{W}^2{\left(\frac{{\rho }_1}{{\rho }_2}\right)}^2\Rightarrow p_2-p_1={\rho }_1{W}^2(1-\frac{{\rho }_1}{{\rho }_2})$

(Eq. 6.8)

$W^2=\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\left(\frac{{\rho }_2}{{\rho }_1}\right)$

(Eq. 6.9)

From the continuity equation \ref{eq:unsteady:cont}, we get

$W=(W-u_p)\left(\frac{{\rho }_2}{{\rho }_1}\right)$

(Eq. 6.10)

Inserting Eqn. \ref{eq:unsteady:cont:modb} in Eqn. \ref{eq:unsteady:mom:mod} gives

$(W-u_p{)}^2=\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\left(\frac{{\rho }_1}{{\rho }_2}\right)$

(Eq. 6.11)

Now, let's insert Eqns. \ref{eq:unsteady:mom:mod} and \ref{eq:unsteady:mom:modb} in the energy equation (Eqn. \ref{eq:unsteady:energy}).

$h_1+\frac{1}{2}\left[\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\left(\frac{{\rho }_2}{{\rho }_1}\right)\right]=h_2+\frac{1}{2}\left[\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\left(\frac{{\rho }_1}{{\rho }_2}\right)\right]$

(Eq. 6.12)

$h=e+\frac{p}{\rho }$

(Eq. 6.13)

$e_1+\frac{p_1}{{\rho }_1}+\frac{1}{2}\left[\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\left(\frac{{\rho }_2}{{\rho }_1}\right)\right]=e_2+\frac{p_2}{{\rho }_2}+\frac{1}{2}\left[\frac{p_2-p_1}{{\rho }_2-{\rho }_1}\left(\frac{{\rho }_1}{{\rho }_2}\right)\right]$

(Eq. 6.14)

which can be rewritten as

$e_2-e_1=\frac{p_1+p_2}{2}(\frac{1}{{\rho }_1}-\frac{1}{{\rho }_2})$

(Eq. 6.15)

Eqn \ref{eq:unsteady:hugonoit} is the same Hugoniot equation as we get for a stationary normal shock. The Hugoniot equation is a relation of thermodynamic properties over a shock. As the shock in the unsteady case is moving with a constant velocity, the frame of reference moving with the shock is an inertial frame and thus the same physical relations apply in the moving shock case as in the stationary shock case. The fact that the Hugoniot relation does not include any velocities or Mach numbers but only thermodynamic properties, the relation will be unchanged for a moving shock.

Moving Shock Relations

For a calorically perfect gas we have $e=C_{v}T$. Inserted in the Hugoniot relation above this gives

$C_v(T_2-T_1)=\frac{p_1+p_2}{2}({\nu }_1-{\nu }_2)$

(Eq. 6.16)

where $\nu =1/\rho$

Now, using the ideal gas law $T=p\nu /R$ and $C_v/R=1/(\gamma -1)$ gives

$\left(\frac{1}{\gamma -1}\right)(p_2{\nu }_2-p_1{\nu }_1)=\frac{p_1+p_2}{2}({\nu }_1-{\nu }_2)$ $\iff$ $p_2(\frac{{\nu }_2}{\gamma -1}-\frac{{\nu }_1-{\nu }_2}{2})=p_1(\frac{{\nu }_1}{\gamma -1}+\frac{{\nu }_1-{\nu }_2}{2})$

(Eq. 6.17)

From this result, we can derive a relation for the pressure ratio over the shock as a function of density ratio

$\frac{p_2}{p_1}=\frac{\left(\frac{\gamma +1}{\gamma -1}\right)\left(\frac{{\nu }_1}{{\nu }_2}\right)-1}{\left(\frac{\gamma +1}{\gamma -1}\right)-\left(\frac{{\nu }_1}{{\nu }_2}\right)}$

(Eq. 6.18)

$\nu =RT/p$ and thus

$\frac{{\nu }_1}{{\nu }_2}=\frac{T_1}{T_2}\frac{p_2}{p_1}$

(Eq. 6.19)

Eqn. \ref{eq:unsteady:density:ratio} in Eqn. \ref{eq:unsteady:hugonoit:c} gives

$\frac{p_2}{p_1}=\frac{\left(\frac{\gamma +1}{\gamma -1}\right)\left(\frac{T_1}{T_2}\frac{p_2}{p_1}\right)-1}{\left(\frac{\gamma +1}{\gamma -1}\right)-\left(\frac{T_1}{T_2}\frac{p_2}{p_1}\right)}$

(Eq. 6.20)

Now, we can get a relation for calculation of the temperature ratio over the moving shock as function of the shock pressure ratio

$\frac{T_2}{T_1}=\frac{p_2}{p_1}\left[\frac{\left(\frac{\gamma +1}{\gamma -1}\right)+\left(\frac{p_2}{p_1}\right)}{1+\left(\frac{\gamma +1}{\gamma -1}\right)\left(\frac{p_2}{p_1}\right)}\right]$

(Eq. 6.21)

Once again using the ideal gas law

$\frac{{\rho }_2}{{\rho }_1}=\frac{\left(\frac{\gamma +1}{\gamma -1}\right)+\left(\frac{p_2}{p_1}\right)}{1+\left(\frac{\gamma +1}{\gamma -1}\right)\left(\frac{p_2}{p_1}\right)}$

(Eq. 6.22)

Going back to the momentum equation

$p_2-p_1={\rho }_1{W}^2(1-\frac{{\rho }_1}{{\rho }_2})=\{W=M_s{a}_1\}={\rho }_1{M}_s^2{a}_1^2(1-\frac{{\rho }_1}{{\rho }_2})$

(Eq. 6.23)

with $a_1^2=\gamma p_1/{\rho }_1$, we get

$\frac{p_2}{p_1}=\gamma M_s^2(1-\frac{{\rho }_1}{{\rho }_2})+1$

(Eq. 6.24)

From the normal shock relations, we have

$\frac{{\rho }_1}{{\rho }_2}=\frac{2+(\gamma -1)M_s^2}{(\gamma +1)M_s^2}$

(Eq. 6.25)

Eqn. \ref{eq:unsteady:Mach:b} in \ref{eq:unsteady:Mach:a} gives

$\frac{p_2}{p_1}=1+\left(\frac{2\gamma }{\gamma +1}\right)(M_s^2-1)$

(Eq. 6.26)

or

$M_s=\sqrt{\left(\frac{\gamma +1}{2\gamma }\right)(\frac{p_2}{p_1}-1)+1}$

(Eq. 6.27)

Eqn. \ref{eq:unsteady:Mach} with $M_s=W/a_1$

$W=a_1\sqrt{\left(\frac{\gamma +1}{2\gamma }\right)(\frac{p_2}{p_1}-1)+1}$

(Eq. 6.28)

Induced Flow Behind Moving Shock

Let's try to find a relation for calculation of the induced velocity behind the moving shock. Once again, the starting point is the continuity equation for moving shocks (Eqn. \ref{eq:unsteady:cont}) repeated here for convenience

${\rho }_{1}W={\rho }_2(W-u_p)$

(Eq. 6.29)

The induced velocity appears on the right side of the continuity equation

$W({\rho }_1-{\rho }_2)=-{\rho }_2{u}_p$

(Eq. 6.30)

$u_p=W(1-\frac{{\rho }_1}{{\rho }_2})$

(Eq. 6.31)

From before we have a relation for $W$ as a function of pressure ratio and one for ${\rho }_1/{\rho }_2$, also as a function of pressure ratio.

Eqn. \ref{eq:unsteady:up:a} togheter with Eqns. \ref{eq:unsteady:W} and \ref{eq:unsteady:density:ratio} gives

$u_p=a_1\underset{I}{\underbrace{\sqrt{\left(\frac{\gamma +1}{2\gamma }\right)(\frac{p_2}{p_1}-1)+1}}}\underset{II}{\underbrace{[1-\frac{\left(\frac{\gamma +1}{\gamma -1}\right)+\left(\frac{p_2}{p_1}\right)}{1+\left(\frac{\gamma +1}{\gamma -1}\right)\left(\frac{p_2}{p_1}\right)}]}}$

(Eq. 6.32)

The equation subsets I and II can be rewritten as:

Term I:

$\sqrt{\left(\frac{\gamma +1}{2\gamma }\right)(\frac{p_2}{p_1}-1)+1}=\sqrt{\frac{\gamma +1}{2\gamma }[\left(\frac{p_2}{p_1}\right)+\left(\frac{\gamma -1}{\gamma +1}\right)]}$

(Eq. 6.33)

Term II:

$[1-\frac{\left(\frac{\gamma +1}{\gamma -1}\right)+\left(\frac{p_2}{p_1}\right)}{1+\left(\frac{\gamma +1}{\gamma -1}\right)\left(\frac{p_2}{p_1}\right)}]=\frac{1}{\gamma }(\frac{p_2}{p_1}-1)\frac{\left(\frac{2\gamma }{\gamma +1}\right)}{\left(\frac{\gamma -1}{\gamma +1}\right)+\left(\frac{p_2}{p_1}\right)}$

(Eq. 6.34)

the rewritten terms I and II implemented, Eqn. \ref{eq:unsteady:up:b} becomes

$u_p=\frac{a_1}{\gamma }(\frac{p_2}{p_1}-1)\sqrt{\frac{\left(\frac{2\gamma }{\gamma +1}\right)}{\left(\frac{\gamma -1}{\gamma +1}\right)+\left(\frac{p_2}{p_1}\right)}}$

(Eq. 6.35)

Since the region behind the moving shock is region 2, the induced flow Mach number is obtained as

$M_p=\frac{u_p}{a_2}=\frac{u_p}{a_1}\frac{a_1}{a_2}=\frac{u_p}{a_1}\sqrt{\frac{\gamma R{T}_1}{\gamma R{T}_2}}=\frac{u_p}{a_1}\sqrt{\frac{T_1}{T_2}}$

(Eq. 6.36)

With $up/a_1$ from Eqn. \ref{eq:unsteady:up} and $T_1/T_2$ from Eqn. \ref{eq:unsteady:temperature:ratio}

$M_p=\frac{1}{\gamma }(\frac{p_2}{p_1}-1){\left(\frac{\left(\frac{2\gamma }{\gamma +1}\right)}{\left(\frac{\gamma -1}{\gamma +1}\right)+\left(\frac{p_2}{p_1}\right)}\right)}^{1/2}{\left(\frac{1+\left(\frac{\gamma +1}{\gamma -1}\right)\left(\frac{p_2}{p_1}\right)}{\left(\frac{\gamma +1}{\gamma -1}\right)\left(\frac{p_2}{p_1}\right)+{\left(\frac{p_2}{p_1}\right)}^2}\right)}^{1/2}$

(Eq. 6.37)

There is a theoretical upper limit for the induced Mach number $M_p$

$\underset{p_2/p_1\to \mathrm{\infty }}{\lim }{M}_p\left(\frac{p_2}{p_1}\right)=\sqrt{\frac{2}{\gamma (\gamma -1)}}$

(Eq. 6.38)

As can be seen, at the upper limit the induced Mach number is a function of $\gamma$ and for air ($\gamma =1.4$) we get

$\underset{p_2/p_1\to \mathrm{\infty }}{\lim }{M}_p\left(\frac{p_2}{p_1}\right)\simeq 1.89$

(Eq. 6.39)

Shock Wave Reflection

When the incident shock wave reaches the wall, a shock propagating in the opposite direction is generated with a shock strength such that the velocity of the induced flow behind the incident shock is reduced to zero. The flow can not go through the wall and thus the velocity must be zero in the vicinity of the wall. The properties of the incident shock wave are directly related to the pressure ratio over the shock wave. Therefore, it would be convenient to have a relation between the reflected shock wave and incident shock wave.

The Incident Shock Wave

The pressure ratio over the incident shock in Fig.~\ref{fig:reflection} can be obtained as

$\frac{p_2}{p_1}=1+\frac{2\gamma }{\gamma +1}(M_s^2-1)$

(Eq. 6.40)

where $M_s$ is the wave Mach number, which is calculated as

$M_s=\frac{W}{a_1}$

(Eq. 6.41)

In Eqn.~\ref{eq:incident:Mach:def}, $W$ is the speed with which the incident shock wave travels into region 1 and $a_1$ is the speed of sound in region 1 (see Fig.~\ref{fig:reflection}).

Solving Eqn.~\ref{eq:incident:pr} for $M_s$, we get

$M_s=\sqrt{\frac{\gamma +1}{2\gamma }(\frac{p_2}{p_1}-1)+1}$

(Eq. 6.42)

Anderson derives the relations for calculation of the ratio $T_2/T_1$

$\frac{T_2}{T_1}=\frac{p_2}{p_1}\left(\frac{\frac{\gamma +1}{\gamma -1}+\frac{p_2}{p_1}}{1+\frac{\gamma +1}{\gamma -1}\frac{p_2}{p_1}}\right)$

(Eq. 6.43)

From Eqn.~\ref{eq:incident:tr} it is easy to get the corresponding relation for ${\rho }_2/{\rho }_1$

$\frac{{\rho }_2}{{\rho }_1}=\frac{1+\frac{\gamma +1}{\gamma -1}\frac{p_2}{p_1}}{\frac{\gamma +1}{\gamma -1}+\frac{p_2}{p_1}}$

(Eq. 6.44)

Anderson also shows how to obtain the induced velocity, $u_p$, behind the incident shock wave, {\emph{i.e.}} the velocity in region 2 (see Fig.~\ref{fig:reflection}).

$u_p=W(1-\frac{{\rho }_1}{{\rho }_2})=M_s{a}_1(1-\frac{{\rho }_1}{{\rho }_2})$

(Eq. 6.45)

The Reflected Shock Wave

The pressure ratio over the reflected shock can be obtained from Eqn.~\ref{eq:incident:pr} by analogy

$\frac{p_5}{p_2}=1+\frac{2\gamma }{\gamma +1}(M_r^2-1)$

(Eq. 6.46)

where $M_r$ is the Mach number of the reflected shock wave defined as

$M_r=\frac{W_r+u_p}{a_2}$

(Eq. 6.47)

where $W_r$ is the speed of the reflected shock wave and $a_2$ is the speed of sound in region 2 (see Fig.~\ref{fig:reflection}).

Solving Eqn.~\ref{eq:reflected:pr} for $M_r$ gives

$M_r=\sqrt{\frac{\gamma +1}{2\gamma }(\frac{p_5}{p_2}-1)+1}$

(Eq. 6.48)

The ratios $T_5/T_2$ and ${\rho }_5/{\rho }_2$ can be obtained from Eqns.~\ref{eq:incident:tr} and \ref{eq:incident:rr} by analogy

$\frac{T_5}{T_2}=\frac{p_5}{p_2}\left(\frac{\frac{\gamma +1}{\gamma -1}+\frac{p_5}{p_2}}{1+\frac{\gamma +1}{\gamma -1}\frac{p_5}{p_2}}\right)$

(Eq. 6.49)

$\frac{{\rho }_5}{{\rho }_2}=\frac{1+\frac{\gamma +1}{\gamma -1}\frac{p_5}{p_2}}{\frac{\gamma +1}{\gamma -1}+\frac{p_5}{p_2}}$

(Eq. 6.50)

The velocity in region 2 which is the same as the induced flow velocity behind the incident shock wave can be obtained as

$u_p=W_r(\frac{{\rho }_5}{{\rho }_2}-1)=M_r{a}_2(1-\frac{{\rho }_2}{{\rho }_5})$

(Eq. 6.51)

Reflected Shock Relation

With the relations for the incident shock wave and reflected shock wave defined, we now have the tools to derive a relation between the incident and reflected shock waves. The induced flow velocity $u_p$ calculated using the relation obtained for the incident shock wave must of course be the same as when calculated using reflected wave properties, {\emph{i.e.}} the result of Eqn.~\ref{eq:incident:up} is identical to that of Eqn.~\ref{eq:reflected:up}

$M_r{a}_2(1-\frac{{\rho }_2}{{\rho }_5})=M_s{a}_1(1-\frac{{\rho }_1}{{\rho }_2})$

(Eq. 6.52)

rewriting gives

$M_r(1-\frac{{\rho }_2}{{\rho }_5})=M_s(1-\frac{{\rho }_1}{{\rho }_2})\frac{a_1}{a_2}$

(Eq. 6.53)

Assuming calorically perfect gas gives $a=\sqrt{\gamma RT}$ and thus

$M_r(1-\frac{{\rho }_2}{{\rho }_5})=M_s(1-\frac{{\rho }_1}{{\rho }_2})\sqrt{\frac{T_1}{T_2}}$

(Eq. 6.54)

Let's first look at the term on the left hand side of Eqn.~\ref{eq:relation:c}

$M_r(1-\frac{{\rho }_2}{{\rho }_5})$

(Eq. 6.55)

Using the ${\rho }_5/{\rho }_2$ and $p_2/p_5$ from Eqns.~\ref{eq:reflected:rr} and~\ref{eq:reflected:pr} and simplifying gives

$M_r(1-\frac{{\rho }_2}{{\rho }_5})=\left(\frac{2}{\gamma +1}\right)\left(\frac{M_r^2-1}{M_r}\right)$

(Eq. 6.56)

Using the same approach on the corresponding term for the incident shock wave on the right hand side of Eqn.~\ref{eq:relation:c} gives

$M_s(1-\frac{{\rho }_1}{{\rho }_2})=\left(\frac{2}{\gamma +1}\right)\left(\frac{M_s^2-1}{M_s}\right)$

(Eq. 6.57)

Now, inserting~\ref{eq:relation:d} and~\ref{eq:relation:e} in Eqn.~\ref{eq:relation:c} gives

$\left(\frac{2}{\gamma +1}\right)\left(\frac{M_r^2-1}{M_r}\right)=\left(\frac{2}{\gamma +1}\right)\left(\frac{M_s^2-1}{M_s}\right)\sqrt{\frac{T_1}{T_2}}$

(Eq. 6.58)

Simplifying and inverting gives

$\left(\frac{M_r}{M_r^2-1}\right)=\left(\frac{M_s}{M_s^2-1}\right)\sqrt{\frac{T_2}{T_1}}$

(Eq. 6.59)

The rightmost term in Eqn.~\ref{eq:relation:g} ($\sqrt{T_2/T_1}$) needs to be rewritten. Inserting~\ref{eq:incident:pr} in~\ref{eq:incident:tr} and expanding all terms gives

$\frac{T_2}{T_1}=\frac{2(\gamma +1)+(\gamma +1)(\gamma -1)M_s^2+4\gamma (M_s^2-1)+2\gamma (\gamma -1)M_s^2(M_s^2-1)}{(\gamma +1{)}^2{M}_s^2}=$

$=\frac{2(\gamma +1)+(\gamma +1)(\gamma -1)M_s^2+4\gamma (M_s^2-1)}{(\gamma +1{)}^2{M}_s^2}+\frac{2(\gamma -1)}{(\gamma +1{)}^2}(M_s^2-1)\gamma =$

$=\frac{2(\gamma +1)+(\gamma +1)(\gamma -1)M_s^2+4\gamma (M_s^2-1)-(2(\gamma -1)(M_s^2-1))}{(\gamma +1{)}^2{M}_s^2}$

$\frac{2(\gamma -1)}{(\gamma +1{)}^2}(M_s^2-1)(\gamma +\frac{1}{M_s^2})$

(Eq. 6.60)

Finally we end up with the following relation

$\frac{T_2}{T_1}=1+\frac{2(\gamma -1)}{(\gamma +1{)}^2}(M_s^2-1)(\gamma +\frac{1}{M_s^2})$

(Eq. 6.61)

The temperature ratio over the incident shock wave is now totally defined by the incident Mach number $M_s$ and the ratio of specific heats $\gamma$. With~\ref{eq:relation:tr} in~\ref{eq:relation:g} we get the sought relation between the reflected and incident Mach numbers.

$\left(\frac{M_r}{M_r^2-1}\right)=\left(\frac{M_s}{M_s^2-1}\right)\sqrt{1+\frac{2(\gamma -1)}{(\gamma +1{)}^2}(M_s^2-1)(\gamma +\frac{1}{M_s^2})}$

(Eq. 6.62)

It should be noted that Eqn.~\ref{eq:relation:final} is valid for calorically perfect gases only.

Acoustic theory

In the following we are going to derive the linear acoustic wave equation starting from the continuity and momentum equations on non-conservation differential form. The equations are repeated here for convenience.

$\frac{D\rho }{Dt}+\rho (\mathrm{\nabla }\cdot 𝐯)=0$

(Eq. 6.63)

$\rho \frac{D𝐯}{Dt}+\mathrm{\nabla }p=0$

(Eq. 6.64)

Remember that $D/Dt$ denotes the substantial derivative operator defined as follows

$\frac{D}{Dt}=\frac{\partial }{\partial t}+𝐯\cdot \mathrm{\nabla }$

(Eq. 6.65)

where $\partial /\partial t$ is the local temporal derivative and $𝐯\cdot \mathrm{\nabla }$ is the convective derivative.

We are going to analyze acoustic waves in one dimension, which means that the equations above reduces to

$\frac{\partial \rho }{\partial t}+u\frac{\partial \rho }{\partial x}+\rho \frac{\partial u}{\partial x}=0$

(Eq. 6.66)

$\rho \frac{\partial u}{\partial t}+\rho u\frac{\partial u}{\partial x}+\frac{\partial p}{\partial x}=0$

(Eq. 6.67)

Pressure is a thermodynamic property and thus it can be expressed as a function of two other thermodynamic properties. Let's express pressure as a function of density ($\rho$) and entropy ($s$).

$p=p(\rho ,s)\Rightarrow dp={\left(\frac{\partial p}{\partial \rho }\right)}_{s}d\rho +{\left(\frac{\partial p}{\partial s}\right)}_{\rho }ds$

(Eq. 6.68)

Since weak acoustic waves are considered, entropy will be constant and thus $ds=0$, which means that

$dp={\left(\frac{\partial p}{\partial \rho }\right)}_{s}d\rho =a^{2}d\rho$

(Eq. 6.69)

$\rho \frac{\partial u}{\partial t}+\rho u\frac{\partial u}{\partial x}+a^2\frac{\partial \rho }{\partial x}=0$

(Eq. 6.70)

The acoustic perturbations can be described as small deviations around a reference state

$\begin{array}{cc}& \rho ={\rho }_{\mathrm{\infty }}+\Delta \rho \\ & p=p_{\mathrm{\infty }}+\Delta p\\ & T=T_{\mathrm{\infty }}+\Delta T\\ & u=u_{\mathrm{\infty }}+\Delta u=\{u_{\mathrm{\infty }}=0\}=\Delta u\\ \end{array}$

Inserted in Eqns.~\ref{eq:unstady:acoustic:wave:cont} and \ref{eq:unstady:acoustic:wave:mom:b} and using the fact that derivatives of the constant reference state flow quantities are zero, we get

$\frac{\partial }{\partial t}(\Delta \rho )+\Delta u\frac{\partial }{\partial x}(\Delta \rho )+({\rho }_{\mathrm{\infty }}+\Delta \rho )\frac{\partial }{\partial x}(\Delta u)=0$

(Eq. 6.71)

$({\rho }_{\mathrm{\infty }}+\Delta \rho )\frac{\partial }{\partial t}(\Delta u)+({\rho }_{\mathrm{\infty }}+\Delta \rho )\Delta u\frac{\partial }{\partial x}(\Delta u)+a^2\frac{\partial }{\partial x}(\Delta \rho )=0$

(Eq. 6.72)

In the same way as pressure, being a thermodynamic variable, can be expressed as a function of two other thermodynamic variables, so can the speed of sound. Once again we will select density and entropy as the two thermodynamic variables

$a^2=a^2(\rho ,s)$

(Eq. 6.73)

and since entropy is constant

$a^2=a^2(\rho )$

(Eq. 6.74)

Taylor expansion of $a^2$ around the reference state $a_{\mathrm{\infty }}$ with $\Delta \rho =\rho -{\rho }_{\mathrm{\infty }}$ gives

$a^2=a_{\mathrm{\infty }}^2+{(\frac{\partial }{\partial \rho }(a^2))}_{\mathrm{\infty }}\Delta \rho +{(\frac{{\partial }^2}{\partial {\rho }^2}(a^2))}_{\mathrm{\infty }}(\Delta \rho {)}^2+\cdots$

(Eq. 6.75)

Inserted in Eqn.~\ref{eq:unstady:acoustic:wave:mom:pert}, we get

$({\rho }_{\mathrm{\infty }}+\Delta \rho )\frac{\partial }{\partial t}(\Delta u)+({\rho }_{\mathrm{\infty }}+\Delta \rho )\Delta u\frac{\partial }{\partial x}(\Delta u)+[a_{\mathrm{\infty }}^2+{(\frac{\partial }{\partial \rho }(a^2))}_{\mathrm{\infty }}\Delta \rho +\cdots ]\frac{\partial }{\partial x}(\Delta \rho )=0$

(Eq. 6.76)

The perturbations $\Delta u$ and $\Delta \rho$ are small, which implies that $\Delta u\ll a_{\mathrm{\infty }}$ and $\Delta \rho \ll {\rho }_{\mathrm{\infty }}$. This means that products of perturbations can be canceled and so can higher-order terms in the Taylor expansion of $a^2$. This means that the continuity and momentum equations reduces to

$\frac{\partial }{\partial t}(\Delta \rho )+{\rho }_{\mathrm{\infty }}\frac{\partial }{\partial x}(\Delta u)=0$

(Eq. 6.77)

${\rho }_{\mathrm{\infty }}\frac{\partial }{\partial t}(\Delta u)+a_{\mathrm{\infty }}^2\frac{\partial }{\partial x}(\Delta \rho )=0$

(Eq. 6.78)

Before making the assumption that the perturbations are small compared to the corresponding reference state flow quantities and thus justifying the cancelation of products of perturbations from the equations, the flow equations were still the exact fully non-linear equations. Eqns.~\ref{eq:unstady:acoustic:wave:cont:linear}. and \ref{eq:unstady:acoustic:wave:mom:linear}, however, are approximations as several terms has been removed. The equations are linear and are good approximations as long as the perturbations are small. The smaller the perturbations, the better the approximation are the linear equations. Eqns.~\ref{eq:unstady:acoustic:wave:cont:linear} and \ref{eq:unstady:acoustic:wave:mom:linear} describes the motion induced in a gas by the passage of a sound wave. By combining the temporal derivative of Eqn.~\ref{eq:unstady:acoustic:wave:cont:linear} with the divergence of Eqn.~\ref{eq:unstady:acoustic:wave:mom:linear}, it is possible to obtain a wave equation describing the propagation of acoustic waves in a quiescent surrounding.

The temporal derivative of the continuity equation:

$\frac{{\partial }^2}{\partial t^2}(\Delta \rho )+{\rho }_{\mathrm{\infty }}\frac{{\partial }^2}{\partial x\partial t}(\Delta u)=0$

(Eq. 6.79)

The divergence of the momentum equation:

${\rho }_{\mathrm{\infty }}\frac{{\partial }^2}{\partial x\partial t}(\Delta u)+a_{\mathrm{\infty }}^2\frac{{\partial }^2}{\partial x^2}(\Delta \rho )=0$

(Eq. 6.80)

The second term in the first equation is the same as the first term in the second equation. Substituting the term, the two equations reduces to one single equation

$\frac{{\partial }^2}{\partial t^2}(\Delta \rho )=a_{\mathrm{\infty }}^2\frac{{\partial }^2}{\partial x^2}(\Delta \rho )$

(Eq. 6.81)

which is a one-dimensional form of the classic wave equation with the general solution

$\Delta \rho =F(x-a_{\mathrm{\infty }}t)+G(x+a_{\mathrm{\infty }}t)$

(Eq. 6.82)

$F$ and $G$ are arbitrary functions. The function $F$ describes the shape of a wave traveling in the positive $x$-direction at the speed of sound of the ambient gas and the function $G$ describes the shape of a wave traveling in the negative $x$-direction at the same speed. In Eqn.~\ref{eq:wave} $\Delta \rho$ appears with second derivatives in space and time. Let's differentiate the proposed solution (Eqn.~\ref{eq:wave:solution}) two times in time and space, respectively, and check that it is actually a valid solution to Eqn.~\ref{eq:wave}.

$\frac{\partial }{\partial t}(\Delta \rho )=\frac{\partial F}{\partial (x-a_{\mathrm{\infty }}t)}\frac{\partial (x-a_{\mathrm{\infty }}t)}{\partial t}+\frac{\partial G}{\partial (x+a_{\mathrm{\infty }}t)}\frac{\partial (x+a_{\mathrm{\infty }}t)}{\partial t}$

(Eq. 6.83)

$\frac{\partial }{\partial t}(\Delta \rho )=-a_{\mathrm{\infty }}{F}^{\prime }+a_{\mathrm{\infty }}{G}^{\prime }$

(Eq. 6.84)

$\frac{{\partial }^2}{\partial t^2}(\Delta \rho )=-a_{\mathrm{\infty }}\frac{\partial F^{\prime }}{\partial (x-a_{\mathrm{\infty }}t)}\frac{\partial (x-a_{\mathrm{\infty }}t)}{\partial t}+a_{\mathrm{\infty }}\frac{\partial G^{\prime }}{\partial (x+a_{\mathrm{\infty }}t)}\frac{\partial (x+a_{\mathrm{\infty }}t)}{\partial t}$

(Eq. 6.85)

$\frac{{\partial }^2}{\partial t^2}(\Delta \rho )=a_{\mathrm{\infty }}^2{F}^{″}+a_{\mathrm{\infty }}^2{G}^{″}$

(Eq. 6.86)

$\frac{\partial }{\partial x}(\Delta \rho )=\frac{\partial F}{\partial (x-a_{\mathrm{\infty }}t)}\frac{\partial (x-a_{\mathrm{\infty }}t)}{\partial x}+\frac{\partial G}{\partial (x+a_{\mathrm{\infty }}t)}\frac{\partial (x+a_{\mathrm{\infty }}t)}{\partial x}$

(Eq. 6.87)

$\frac{\partial }{\partial x}(\Delta \rho )=F^{\prime }+G^{\prime }$

(Eq. 6.88)

$\frac{{\partial }^2}{\partial x^2}(\Delta \rho )=\frac{\partial F^{\prime }}{\partial (x-a_{\mathrm{\infty }}t)}\frac{\partial (x-a_{\mathrm{\infty }}t)}{\partial x}+\frac{\partial G^{\prime }}{\partial (x+a_{\mathrm{\infty }}t)}\frac{\partial (x+a_{\mathrm{\infty }}t)}{\partial x}$

(Eq. 6.89)

$\frac{{\partial }^2}{\partial x^2}(\Delta \rho )=F^{″}+G^{″}$

(Eq. 6.90)

Eqns. \ref{eq:wave:ddt} and \ref{eq:wave:ddx} inserted Eqn. \ref{eq:wave} gives

$a_{\mathrm{\infty }}^2{F}^{″}+a_{\mathrm{\infty }}^2{G}^{″}=a_{\mathrm{\infty }}^2(F^{″}+G^{″})$

(Eq. 6.91)

which shows that Eqn. \ref{eq:wave:solution} is a valid solution to the wave equation.

$F$ and $G$ are arbitrary functions and thus $G=0$ is a valid solution, which gives

$\Delta \rho (x,t)=F(x-a_{\mathrm{\infty }}t)$

(Eq. 6.92)

If $\Delta \rho$ is constant, i.e. a wave with constant amplitude, we see from Eqn.~\ref{eq:wave:solution:F} that $(x-a_{\mathrm{\infty }}t)$ is constant and thus

$x=a_{\mathrm{\infty }}t+c\Rightarrow \frac{dx}{dt}=a_{\mathrm{\infty }}$

(Eq. 6.93)

From Eqn.~\ref{eq:wave:solution:F}, we get

$\frac{\partial }{\partial t}(\Delta \rho )=-a_{\mathrm{\infty }}{F}^{\prime }$

(Eq. 6.94)

$\frac{\partial }{\partial x}(\Delta \rho )=F^{\prime }$

(Eq. 6.95)

and thus

$\frac{\partial }{\partial x}(\Delta \rho )=-\frac{1}{a_{\mathrm{\infty }}}\frac{\partial }{\partial t}(\Delta \rho )$

(Eq. 6.96)

which gives a relation between the temporal derivative of $\Delta \rho$ and the spatial derivative of $\Delta \rho$. With Eqn.~\ref{eq:wave:solution:F:b}, the linearized momentum equation Eqn.~\ref{eq:unstady:acoustic:wave:mom:linear} can be rewritten as follows

$\frac{\partial }{\partial t}(\Delta u)=-\frac{a_{\mathrm{\infty }}^2}{{\rho }_{\mathrm{\infty }}}\frac{\partial }{\partial x}(\Delta \rho )=\{\frac{\partial }{\partial x}(\Delta \rho )=-\frac{1}{a_{\mathrm{\infty }}}\frac{\partial }{\partial t}(\Delta \rho )\}=\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\frac{\partial }{\partial t}(\Delta \rho )\Rightarrow$

(Eq. 6.97)

$\frac{\partial }{\partial t}(\Delta u-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho )=0\Rightarrow \Delta u-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho =const$

(Eq. 6.98)

In an undisturbed gas $\Delta u=\Delta \rho =0$ and thus

$\Delta u-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho =0$

(Eq. 6.99)

or

$\Delta u=\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho$

(Eq. 6.100)

If instead $F$ is set to zero and $G$ is non-zero, we get

$\Delta u=-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho$

(Eq. 6.101)

${\left(\frac{\partial p}{\partial \rho }\right)}_s=a^2\Rightarrow \Delta p=a_{\mathrm{\infty }}^2\Delta \rho$

(Eq. 6.102)

Acoustic wave traveling in the positive $x$-direction:

$\Delta u=\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho =\frac{1}{a_{\mathrm{\infty }}{\rho }_{\mathrm{\infty }}}\Delta p$

(Eq. 6.103)

Acoustic wave traveling in the negative $x$-direction:

$\Delta u=-\frac{a_{\mathrm{\infty }}}{{\rho }_{\mathrm{\infty }}}\Delta \rho =-\frac{1}{a_{\mathrm{\infty }}{\rho }_{\mathrm{\infty }}}\Delta p$

(Eq. 6.104)

Finite non-linear waves

Starting point: the governing flow equations on partial differential form

Continuity equation:

$\frac{\partial \rho }{\partial t}+u\frac{\partial \rho }{\partial x}+\rho \frac{\partial u}{\partial x}=0$

(Eq. 6.105)

Momentum equation:

$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho }\frac{\partial p}{\partial x}=0$

(Eq. 6.106)

Any thermodynamic property can be expressed as a function of two other thermodynamic properties. This means that we can get density as a function of pressure and entropy: $\rho =\rho (p,s)$ and therefore

$d\rho ={\left(\frac{\partial \rho }{\partial p}\right)}_{s}dp+{\left(\frac{\partial \rho }{\partial s}\right)}_{p}ds$

(Eq. 6.107)

Assuming isentropic flow $ds=0$ gives

$d\rho ={\left(\frac{\partial \rho }{\partial p}\right)}_{s}dp$

(Eq. 6.108)

$\begin{array}{cc}& \frac{\partial \rho }{\partial t}={\left(\frac{\partial \rho }{\partial p}\right)}_s\frac{\partial p}{\partial t}=\frac{1}{a^2}\frac{\partial p}{\partial t}\\ & \\ & \frac{\partial \rho }{\partial x}={\left(\frac{\partial \rho }{\partial p}\right)}_s\frac{\partial p}{\partial x}=\frac{1}{a^2}\frac{\partial p}{\partial x}\end{array}$

(Eq. 6.109)

Now, insert \ref{eq:rhotop} in \ref{eq:pde:cont} gives

$\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x}+\rho a^2\frac{\partial u}{\partial x}=0$

(Eq. 6.110)

Dividing \ref{eq:pde:cont:b} by $\rho a$ gives

$\frac{1}{\rho a}(\frac{\partial p}{\partial t}+u\frac{\partial p}{\partial x})+a\frac{\partial u}{\partial x}=0$

(Eq. 6.111)

A slightly modified form of the momentum equation is obtained by multiplying and dividing the last term by $a$

$\frac{\partial u}{\partial t}+u\frac{\partial u}{\partial x}+\frac{1}{\rho a}\left(a\frac{\partial p}{\partial x}\right)=0$

(Eq. 6.112)

If the continuity equation on the form \ref{eq:pde:cont:c} is added to the momentum equation on the form \ref{eq:pde:mom:c}, we get

$[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}]+\frac{1}{\rho a}[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}]=0$

(Eq. 6.113)

If, instead, the continuity equation on the form \ref{eq:pde:cont:c} is subtracted from the momentum equation on the form \ref{eq:pde:mom:c}, we get

$[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}]+\frac{1}{\rho a}[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}]=0$

(Eq. 6.114)

Since $u=u(x,t)$, we have from the definition of a differential

$du=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}dx=\frac{\partial u}{\partial t}dt+\frac{\partial u}{\partial x}\frac{dx}{dt}dt$

(Eq. 6.115)

Now, let $dx/dt=u+a$

$du=\frac{\partial u}{\partial t}dt+(u+a)\frac{\partial u}{\partial x}dt=[\frac{\partial u}{\partial t}+(u+a)\frac{\partial u}{\partial x}]dt$

(Eq. 6.116)

which is the change of $u$ in the direction $dx/dt=u+a$

In the same way

$dp=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}dx=\frac{\partial p}{\partial t}dt+\frac{\partial p}{\partial x}\frac{dx}{dt}dt$

(Eq. 6.117)

and thus, in the direction $dx/dt=u+a$

$dp=\frac{\partial p}{\partial t}dt+(u+a)\frac{\partial p}{\partial x}dt=[\frac{\partial p}{\partial t}+(u+a)\frac{\partial p}{\partial x}]dt$

(Eq. 6.118)

If we go back and examine Eqn. \ref{eq:nonlin:a}, we see that Eqns. \ref{eq:du:b} and \ref{eq:dp:b} appear in the equation and thus it can now be rewritten as follows

$\frac{du}{dt}+\frac{1}{\rho a}\frac{dp}{dt}=0\Rightarrow du+\frac{dp}{\rho a}=0$

(Eq. 6.119)

Eqn. \ref{eq:nonlin:a:ode} applies along a $C^{+}$ characteristic, i.e., a line in the direction $dx/dt=u+a$ in $xt$-space and is called the compatibility equation along the $C^{+}$ characteristic. If we instead chose a $C^{-}$ characteristic, i.e., a line in the direction $dx/dt=u-a$ in $xt$-space, we get

$du=[\frac{\partial u}{\partial t}+(u-a)\frac{\partial u}{\partial x}]dt$

(Eq. 6.120)

$dp=[\frac{\partial p}{\partial t}+(u-a)\frac{\partial p}{\partial x}]dt$

(Eq. 6.121)

which can be identified as subsets of Eqn. \ref{eq:nonlin:b} and thus

$\frac{du}{dt}-\frac{1}{\rho a}\frac{dp}{dt}=0$

(Eq. 6.122)

In order to fulfil the relation above, either $du=dp=0$ or

$du-\frac{dp}{\rho a}=0$

(Eq. 6.123)

Eqn. \ref{eq:nonlin:b:ode} applies along a $C^{-}$ characteristic, i.e., a line in the direction $dx/dt=u-a$ in $xt$-space and is called the compatibility equation along the $C^{-}$ characteristic.

So, what we have done now is that we have have found paths through a point ($x_1$, $t_1$) along which the governing partial differential equations Eqns. \ref{eq:nonlin:a} and \ref{eq:nonlin:b} reduces to the ordinary differential equations \ref{eq:nonlin:a:ode} and \ref{eq:nonlin:b:ode}. The $C^{+}$ and $C^{-}$ characteristic lines are physically the paths of right- and left-running sound waves in the $xt$-plane.

Riemann Invariants

If the compatibility equations are integrated along respective characteristic line, i.e., integrate \ref{eq:nonlin:a:ode} along the $C^{+}$ characteristic and \ref{eq:nonlin:b:ode} along the $C^{-}$ characteristic, we get the Riemann invariants $J^{+}$ and $J^{-}$.

$J^{+}=u+\int \frac{dp}{\rho a}=const$

(Eq. 6.124)

$J^{-}=u-\int \frac{dp}{\rho a}=const$

(Eq. 6.125)

The Riemann invariants are constants along the associated characteristic line.

We have assumed isentropic flow and thus we may use the isentropic relations

$p=C_1{T}^{\gamma /(\gamma -1)}=C_2{a}^{2\gamma /(\gamma -1)}$

(Eq. 6.126)

where $C_1$ and $C_2$ are constants. Differentiating Eqn. \ref{eq:isentropic:a} gives

$dp=C_2\left(\frac{2\gamma }{\gamma -1}\right)a^{[2\gamma /(\gamma -1)-1]}da$

(Eq. 6.127)

Now, if we further assume the gas to be calorically perfect

$a^2=\gamma RT=\frac{\gamma p}{\rho }\Rightarrow \rho =\frac{\gamma p}{a^2}$

(Eq. 6.128)

Eqn. \ref{eq:isentropic:a} in \ref{eq:isentropic:c} gives

$\rho =C_2\gamma a^{[2\gamma /(\gamma -1)-2]}$

(Eq. 6.129)

and thus

$J^{+}=u+\int \frac{C_2\left(\frac{2\gamma }{\gamma -1}\right)a^{[2\gamma /(\gamma -1)-1]}}{C_2\gamma a^{[2\gamma /(\gamma -1)-2]}a}da=u+\left(\frac{2}{\gamma -1}\right)\int da$

(Eq. 6.130)

$J^{+}=u+\frac{2a}{\gamma -1}$

(Eq. 6.131)

$J^{-}=u-\frac{2a}{\gamma -1}$

(Eq. 6.132)

Eqns. \ref{eq:riemann:a:b} and \ref{eq:riemann:b:b} are the Riemann invariants for a calorically perfect gas. The Riemann invariants are constants along $C^{+}$ and $C^{-}$ characteristics and if the situation shown in Fig. \ref{fig:characteristics} appears, that fact can be used to calculate the flow velocity and speed of sound in the location ($x_1$, $t_1$).

$J^{+}+J^{-}=u+\frac{2a}{\gamma -1}+u-\frac{2a}{\gamma -1}=2u\Rightarrow u=\frac{1}{2}(J^{+}+J^{-})$

(Eq. 6.133)

$J^{+}=u+\frac{2a}{\gamma -1}=\frac{1}{2}(J^{+}+J^{-})+\frac{2a}{\gamma -1}\Rightarrow a=\frac{\gamma -1}{4}(J^{+}-J^{-})$

(Eq. 6.134)

Moving expansion waves

Moving Expansion Waves

The expansion wave propagation into the driver section in a shock tube can be described using characteristic lines.

The expansion is propagating into stagnant fluid in region four (the driver section), which means that the flow properties ahead of the expansion wave are constant.

$J_a^{+}=J_b^{+}$

(Eq. 6.135)

$J^{+}$ invariants constant along $C^{+}$ characteristics

$J_a^{+}=J_c^{+}=J_e^{+}$

(Eq. 6.136)

$J_b^{+}=J_d^{+}=J_f^{+}$

(Eq. 6.137)

Since $J_a^{+}=J_b^{+}$ this also implies $J_e^{+}=J_f^{+}$. In fact, since the flow properties ahead of the expansion are constant, all $C^{+}$ lines will have the same $J^{+}$ value.

$J^{-}$ invariants constant along $C^{-}$ characteristics

$J_c^{-}=J_d^{-}$

(Eq. 6.138)

$J_e^{-}=J_f^{-}$

(Eq. 6.139)

$\begin{array}{cc}& u_e=\frac{1}{2}(J_e^{+}+J_e^{-})\\ & u_f=\frac{1}{2}(J_f^{+}+J_f^{-})\\ & J_e^{-}=J_f^{-}\\ & J_e^{+}=J_f^{+}\end{array}\}\Rightarrow u_e=u_f\Rightarrow a_e=a_f$

(Eq. 6.140)

Due to the fact the $J^{+}$ is constant in the entire expansion region, $u$ and $a$ will be constant along each $C^{-}$ line.

The constant $J^{+}$ value can be used to obtain relations for the variation of flow properties through the expansion region. Evaluation of the $J^{+}$ invariant at any position within the expansion region should give the same value as in region 4.

$u+\frac{2a}{\gamma -1}=u_4+\frac{2a_4}{\gamma -1}=0+\frac{2a_4}{\gamma -1}$

(Eq. 6.141)

and thus

$\frac{a}{a_4}=1-\frac{\gamma -1}{2}\left(\frac{u}{a_4}\right)$

(Eq. 6.142)

Eqn. \ref{eq:expansion:a} and $a=\sqrt{\gamma RT}$ gives

$\frac{T}{T_4}={[1-\frac{\gamma -1}{2}\left(\frac{u}{a_4}\right)]}^2$

(Eq. 6.143)

Using isentropic relations, we can get pressure ratio and density ratio

$\frac{p}{p_4}={[1-\frac{\gamma -1}{2}\left(\frac{u}{a_4}\right)]}^{2\gamma /(\gamma -1)}$

(Eq. 6.144)

$\frac{\rho }{{\rho }_4}={[1-\frac{\gamma -1}{2}\left(\frac{u}{a_4}\right)]}^{2/(\gamma -1)}$

(Eq. 6.145)

Shock-tube relations

From the analysis of the incident shock, we have a relation for the induced flow behind the shock

$u_2=u_p=\frac{a_1}{{\gamma }_1}(\frac{p_2}{p_1}-1){\left(\frac{\left(\frac{2{\gamma }_1}{{\gamma }_1+1}\right)}{\left(\frac{{\gamma }_1-1}{{\gamma }_1+1}\right)+\left(\frac{p_2}{p_1}\right)}\right)}^{1/2}$

(Eq. 6.146)

The velocity in region 3 can be obtained from the expansion relations

$\frac{p_3}{p_4}={[1-\frac{{\gamma }_4-1}{2}\left(\frac{u_3}{a_4}\right)]}^{2{\gamma }_4/({\gamma }_4-1)}$

(Eq. 6.147)

Solving for $u_3$ gives

$u_3=\frac{2a_4}{{\gamma }_4-1}[1-{\left(\frac{p_3}{p_4}\right)}^{({\gamma }_4-1)/(2{\gamma }_4)}]$

(Eq. 6.148)

There is no change in pressure or velocity over the contact surface, which means $u_2=u_3$ and $p_2=p_3$.

$u_2=\frac{2a_4}{{\gamma }_4-1}[1-{\left(\frac{p_2}{p_4}\right)}^{({\gamma }_4-1)/(2{\gamma }_4)}]$

(Eq. 6.149)

Now, we have two ways of calculating $u_2$. Setting Eqn. \ref{eq:shocktube:up:a} equal to Eqn. \ref{eq:shocktube:up:d} leads to the shock tube relation

$\frac{p_4}{p_1}=\frac{p_2}{p_1}{\{1-\frac{({\gamma }_4-1)(a_1/a_4)(p_2/p_1-1)}{\sqrt{2{\gamma }_1[2{\gamma }_1+({\gamma }_1+1)(p_2/p_1-1)]}}\}}^{-2{\gamma }_4/({\gamma }_4-1)}$

(Eq. 6.150)