Moving shock waves: Difference between revisions

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[[Category:Compressible flow]]
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=== Moving Normal Shock Waves ===
=== Moving Normal Shock Waves ===


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


<math display="block">
{{NumEqn|<math>
\rho_1 u_1 = \rho_2 u_2
\rho_1 u_1 = \rho_2 u_2
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
\rho_1 u_1^2+p_1 = \rho_2 u_2^2 + p_2
\rho_1 u_1^2+p_1 = \rho_2 u_2^2 + p_2
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
h_1 + \frac{1}{2}u_1^2 = h_2 + \frac{1}{2}u_2^2
h_1 + \frac{1}{2}u_1^2 = h_2 + \frac{1}{2}u_2^2
</math>
</math>}}


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 <math>u_1=W</math>, and the gas behind the shock moves to the right with the velocity <math>u_2=W-u_p</math>. Now, let's insert <math>u_1</math> and <math>u_2</math> in the stationary shock relations \ref{eq:stationary:cont} - \ref{eq:stationary:energy}.
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 <math>u_1=W</math>, and the gas behind the shock moves to the right with the velocity <math>u_2=W-u_p</math>. Now, let's insert <math>u_1</math> and <math>u_2</math> in the stationary shock relations \ref{eq:stationary:cont} - \ref{eq:stationary:energy}.


<math display="block">
{{NumEqn|<math>
\rho_1 W = \rho_2 (W-u_p)
\rho_1 W = \rho_2 (W-u_p)
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
\rho_1 W^2+p_1 = \rho_2 (W-u_p)^2 + p_2
\rho_1 W^2+p_1 = \rho_2 (W-u_p)^2 + p_2
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
h_1 + \frac{1}{2}W^2 = h_2 + \frac{1}{2}(W-u_p)^2
h_1 + \frac{1}{2}W^2 = h_2 + \frac{1}{2}(W-u_p)^2
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
(W-u_p) = W \frac{\rho_1}{\rho_2}
(W-u_p) = W \frac{\rho_1}{\rho_2}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
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_1W^2\left(1-\frac{\rho_1}{\rho_2}\right)
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_1W^2\left(1-\frac{\rho_1}{\rho_2}\right)
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
W^2=\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_2}{\rho_1}\right)
W^2=\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_2}{\rho_1}\right)
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
W = (W-u_p) \left(\frac{\rho_2}{\rho_1}\right)
W = (W-u_p) \left(\frac{\rho_2}{\rho_1}\right)
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
(W-u_p)^2=\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_1}{\rho_2}\right)
(W-u_p)^2=\frac{p_2-p_1}{\rho_2-\rho_1}\left(\frac{\rho_1}{\rho_2}\right)
</math>
</math>}}


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}).
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}).


<math display="block">
{{NumEqn|<math>
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]
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]
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
h=e+\frac{p}{\rho}
h=e+\frac{p}{\rho}
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
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]
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]
</math>
</math>}}


which can be rewritten as
which can be rewritten as


<math display="block">
{{NumEqn|<math>
e_2-e_1=\frac{p_1+p_2}{2}\left(\frac{1}{\rho_1}-\frac{1}{\rho_2}\right)
e_2-e_1=\frac{p_1+p_2}{2}\left(\frac{1}{\rho_1}-\frac{1}{\rho_2}\right)
</math>
</math>}}


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.
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.
Line 92: Line 100:
For a calorically perfect gas we have <math>e=C_v T</math>. Inserted in the Hugoniot relation above this gives
For a calorically perfect gas we have <math>e=C_v T</math>. Inserted in the Hugoniot relation above this gives


<math display="block">
{{NumEqn|<math>
C_v(T_2-T_1)=\frac{p_1+p_2}{2}\left(\nu_1-\nu_2\right)
C_v(T_2-T_1)=\frac{p_1+p_2}{2}\left(\nu_1-\nu_2\right)
</math>
</math>}}


where <math>\nu=1/\rho</math>
where <math>\nu=1/\rho</math>
Line 100: Line 108:
Now, using the ideal gas law <math>T=p\nu/R</math> and <math>C_v/R=1/(\gamma-1)</math> gives
Now, using the ideal gas law <math>T=p\nu/R</math> and <math>C_v/R=1/(\gamma-1)</math> gives


<math display="block">
{{NumEqn|<math>
\left(\frac{1}{\gamma-1}\right)(p_2\nu_2-p_1\nu_1)=\frac{p_1+p_2}{2}\left(\nu_1-\nu_2\right)
\left(\frac{1}{\gamma-1}\right)(p_2\nu_2-p_1\nu_1)=\frac{p_1+p_2}{2}\left(\nu_1-\nu_2\right)
</math>
</math><br><br><math>
 
\Leftrightarrow
<math display="block">
</math><br><br><math>
\Leftrightarrow  
</math>
 
<math display="block">
p_2\left(\frac{\nu_2}{\gamma-1}-\frac{\nu_1-\nu_2}{2}\right)=p_1\left(\frac{\nu_1}{\gamma-1}+\frac{\nu_1-\nu_2}{2}\right)
p_2\left(\frac{\nu_2}{\gamma-1}-\frac{\nu_1-\nu_2}{2}\right)=p_1\left(\frac{\nu_1}{\gamma-1}+\frac{\nu_1-\nu_2}{2}\right)
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
\frac{p_2}{p_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{\nu_1}{\nu_2}\right)-1}{\left(\dfrac{\gamma+1}{\gamma-1}\right)-\left(\dfrac{\nu_1}{\nu_2}\right)}
\frac{p_2}{p_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{\nu_1}{\nu_2}\right)-1}{\left(\dfrac{\gamma+1}{\gamma-1}\right)-\left(\dfrac{\nu_1}{\nu_2}\right)}
</math>
</math>}}


<math>\nu=RT/p</math> and thus
<math>\nu=RT/p</math> and thus


<math display="block">
{{NumEqn|<math>
\frac{\nu_1}{\nu_2}=\frac{T_1}{T_2}\frac{p_2}{p_1}
\frac{\nu_1}{\nu_2}=\frac{T_1}{T_2}\frac{p_2}{p_1}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
\frac{p_2}{p_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{T_1}{T_2}\dfrac{p_2}{p_1}\right)-1}{\left(\dfrac{\gamma+1}{\gamma-1}\right)-\left(\dfrac{T_1}{T_2}\dfrac{p_2}{p_1}\right)}
\frac{p_2}{p_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{T_1}{T_2}\dfrac{p_2}{p_1}\right)-1}{\left(\dfrac{\gamma+1}{\gamma-1}\right)-\left(\dfrac{T_1}{T_2}\dfrac{p_2}{p_1}\right)}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
\frac{T_2}{T_1}=\frac{p_2}{p_1}\left[\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right]
\frac{T_2}{T_1}=\frac{p_2}{p_1}\left[\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right]
</math>
</math>}}


Once again using the ideal gas law
Once again using the ideal gas law


<math display="block">
{{NumEqn|<math>
\frac{\rho_2}{\rho_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}
\frac{\rho_2}{\rho_1}=\frac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}
</math>
</math>}}


Going back to the momentum equation
Going back to the momentum equation


<math display="block">
{{NumEqn|<math>
p_2-p_1 = \rho_1W^2\left(1-\frac{\rho_1}{\rho_2}\right)=\left\{W=M_s a_1\right\}=\rho_1M_s^2a_1^2\left(1-\frac{\rho_1}{\rho_2}\right)
p_2-p_1 = \rho_1W^2\left(1-\frac{\rho_1}{\rho_2}\right)=\left\{W=M_s a_1\right\}=\rho_1M_s^2a_1^2\left(1-\frac{\rho_1}{\rho_2}\right)
</math>
</math>}}


with <math>a_1^2=\gamma p_1/\rho_1</math>, we get
with <math>a_1^2=\gamma p_1/\rho_1</math>, we get


<math display="block">
{{NumEqn|<math>
\frac{p_2}{p_1} = \gamma M_s^2\left(1-\frac{\rho_1}{\rho_2}\right)+1
\frac{p_2}{p_1} = \gamma M_s^2\left(1-\frac{\rho_1}{\rho_2}\right)+1
</math>
</math>}}


From the normal shock relations, we have
From the normal shock relations, we have


<math display="block">
{{NumEqn|<math>
\frac{\rho_1}{\rho_2} = \frac{2+(\gamma-1)M_s^2}{(\gamma+1)M_s^2}
\frac{\rho_1}{\rho_2} = \frac{2+(\gamma-1)M_s^2}{(\gamma+1)M_s^2}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
\frac{p_2}{p_1} = 1 + \left(\frac{2\gamma}{\gamma+1}\right)(M_s^2-1)
\frac{p_2}{p_1} = 1 + \left(\frac{2\gamma}{\gamma+1}\right)(M_s^2-1)
</math>
</math>}}


or
or


<math display="block">
{{NumEqn|<math>
M_s=\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}
M_s=\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
W=a_1\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}
W=a_1\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}
</math>
</math>}}


==== Induced Flow Behind Moving Shock ====
==== Induced Flow Behind Moving Shock ====
Line 182: Line 186:
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
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


<math display="block">
{{NumEqn|<math>
\rho_1 W = \rho_2 (W-u_p)
\rho_1 W = \rho_2 (W-u_p)
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
W (\rho_1-\rho_2) = -\rho_2 u_p
W (\rho_1-\rho_2) = -\rho_2 u_p
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
u_p = W \left(1-\frac{\rho_1}{\rho_2}\right)  
u_p = W \left(1-\frac{\rho_1}{\rho_2}\right)  
</math>
</math>}}


From before we have a relation for $W$ as a function of pressure ratio and one for <math>\rho_1/\rho_2</math>, also as a function of pressure ratio.
From before we have a relation for $W$ as a function of pressure ratio and one for <math>\rho_1/\rho_2</math>, also as a function of pressure ratio.
Line 200: Line 204:
Eqn. \ref{eq:unsteady:up:a} togheter with Eqns. \ref{eq:unsteady:W} and \ref{eq:unsteady:density:ratio} gives
Eqn. \ref{eq:unsteady:up:a} togheter with Eqns. \ref{eq:unsteady:W} and \ref{eq:unsteady:density:ratio} gives


<math display="block">
{{NumEqn|<math>
u_p=a_1\underbrace{\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}}_{I}\underbrace{\left[1-\dfrac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right]}_{II}
u_p=a_1\underbrace{\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}}_{I}\underbrace{\left[1-\dfrac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right]}_{II}
</math>
</math>}}


The equation subsets I and II can be rewritten as:
The equation subsets I and II can be rewritten as:
Line 208: Line 212:
Term I:
Term I:


<math display="block">
{{NumEqn|<math>
\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}=\sqrt{\frac{\gamma+1}{2\gamma}\left[\left(\frac{p_2}{p_1}\right)+\left(\frac{\gamma-1}{\gamma+1}\right)\right]}
\sqrt{\left(\frac{\gamma+1}{2\gamma}\right)\left(\frac{p_2}{p_1}-1\right)+1}=\sqrt{\frac{\gamma+1}{2\gamma}\left[\left(\frac{p_2}{p_1}\right)+\left(\frac{\gamma-1}{\gamma+1}\right)\right]}
</math>
</math>}}




Term II:
Term II:


<math display="block">
{{NumEqn|<math>
\left[1-\dfrac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right]=\frac{1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)}
\left[1-\dfrac{\left(\dfrac{\gamma+1}{\gamma-1}\right)+\left(\dfrac{p_2}{p_1}\right)}{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}\right]=\frac{1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
u_p=\frac{a_1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\sqrt{\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)}}
u_p=\frac{a_1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\sqrt{\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)}}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
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}}
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}}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
M_p=\frac{1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\left(\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)}\right)^{1/2}
M_p=\frac{1}{\gamma}\left(\frac{p_2}{p_1}-1\right)\left(\frac{\left(\dfrac{2\gamma}{\gamma+1}\right)}{\left(\dfrac{\gamma-1}{\gamma+1}\right)+\left(\dfrac{p_2}{p_1}\right)}\right)^{1/2}
\left(\frac{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)+\left(\dfrac{p_2}{p_1}\right)^2}\right)^{1/2}
\left(\frac{1+\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)}{\left(\dfrac{\gamma+1}{\gamma-1}\right)\left(\dfrac{p_2}{p_1}\right)+\left(\dfrac{p_2}{p_1}\right)^2}\right)^{1/2}
</math>
</math>}}


There is a theoretical upper limit for the induced Mach number <math>M_p</math>
There is a theoretical upper limit for the induced Mach number <math>M_p</math>


<math display="block">
{{NumEqn|<math>
\lim_{p_2/p_1\rightarrow\infty} M_p\left(\frac{p_2}{p_1}\right)=\sqrt{\frac{2}{\gamma(\gamma-1)}}
\lim_{p_2/p_1\rightarrow\infty} M_p\left(\frac{p_2}{p_1}\right)=\sqrt{\frac{2}{\gamma(\gamma-1)}}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
\lim_{p_2/p_1\rightarrow\infty} M_p\left(\frac{p_2}{p_1}\right)\simeq 1.89
\lim_{p_2/p_1\rightarrow\infty} M_p\left(\frac{p_2}{p_1}\right)\simeq 1.89
</math>
</math>}}


\section{Shock Wave Reflection}
=== 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.
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.
Line 269: Line 273:
The pressure ratio over the incident shock in Fig.~\ref{fig:reflection} can be obtained as
The pressure ratio over the incident shock in Fig.~\ref{fig:reflection} can be obtained as


<math display="block">
{{NumEqn|<math>
\frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}\left(M_s^2-1\right)
\frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}\left(M_s^2-1\right)
</math>
</math>}}


where <math>M_s</math> is the wave Mach number, which is calculated as
where <math>M_s</math> is the wave Mach number, which is calculated as


<math display="block">
{{NumEqn|<math>
M_s=\frac{W}{a_1}
M_s=\frac{W}{a_1}
</math>
</math>}}
 


In Eqn.~\ref{eq:incident:Mach:def}, <math>W</math> is the speed with which the incident shock wave travels into region 1 and <math>a_1</math> is the speed of sound in region 1 (see Fig.~\ref{fig:reflection}).
In Eqn.~\ref{eq:incident:Mach:def}, <math>W</math> is the speed with which the incident shock wave travels into region 1 and <math>a_1</math> is the speed of sound in region 1 (see Fig.~\ref{fig:reflection}).
Line 284: Line 287:
Solving Eqn.~\ref{eq:incident:pr} for <math>M_s</math>, we get
Solving Eqn.~\ref{eq:incident:pr} for <math>M_s</math>, we get


<math display="block">
{{NumEqn|<math>
M_s=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_2}{p_1}-1\right)+1}
M_s=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_2}{p_1}-1\right)+1}
</math>
</math>}}


Anderson derives the relations for calculation of the ratio <math>T_2/T_1</math>
Anderson derives the relations for calculation of the ratio <math>T_2/T_1</math>


<math display="block">
{{NumEqn|<math>
\frac{T_2}{T_1}=\frac{p_2}{p_1}\left(\dfrac{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_2}{p_1}}{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_2}{p_1}}\right)
\frac{T_2}{T_1}=\frac{p_2}{p_1}\left(\dfrac{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_2}{p_1}}{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_2}{p_1}}\right)
</math>
</math>}}


From Eqn.~\ref{eq:incident:tr} it is easy to get the corresponding relation for <math>\rho_2/\rho_1</math>
From Eqn.~\ref{eq:incident:tr} it is easy to get the corresponding relation for <math>\rho_2/\rho_1</math>


<math display="block">
{{NumEqn|<math>
\frac{\rho_2}{\rho_1}=\dfrac{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_2}{p_1}}{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_2}{p_1}}
\frac{\rho_2}{\rho_1}=\dfrac{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_2}{p_1}}{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_2}{p_1}}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
u_p=W\left(1-\frac{\rho_1}{\rho_2}\right)=M_s a_1 \left(1-\frac{\rho_1}{\rho_2}\right)
u_p=W\left(1-\frac{\rho_1}{\rho_2}\right)=M_s a_1 \left(1-\frac{\rho_1}{\rho_2}\right)
</math>
</math>}}


==== The Reflected Shock Wave ====
==== The Reflected Shock Wave ====
Line 310: Line 313:
The pressure ratio over the reflected shock can be obtained from Eqn.~\ref{eq:incident:pr} by analogy
The pressure ratio over the reflected shock can be obtained from Eqn.~\ref{eq:incident:pr} by analogy


<math display="block">
{{NumEqn|<math>
\frac{p_5}{p_2}=1+\frac{2\gamma}{\gamma+1}\left(M_r^2-1\right)
\frac{p_5}{p_2}=1+\frac{2\gamma}{\gamma+1}\left(M_r^2-1\right)
</math>
</math>}}


where <math>M_r</math> is the Mach number of the reflected shock wave defined as
where <math>M_r</math> is the Mach number of the reflected shock wave defined as


<math display="block">
{{NumEqn|<math>
M_r=\frac{W_r+u_p}{a_2}
M_r=\frac{W_r+u_p}{a_2}
</math>
</math>}}


where <math>W_r</math> is the speed of the reflected shock wave and <math>a_2</math> is the speed of sound in region 2 (see Fig.~\ref{fig:reflection}).
where <math>W_r</math> is the speed of the reflected shock wave and <math>a_2</math> is the speed of sound in region 2 (see Fig.~\ref{fig:reflection}).
Line 324: Line 327:
Solving Eqn.~\ref{eq:reflected:pr} for <math>M_r</math> gives
Solving Eqn.~\ref{eq:reflected:pr} for <math>M_r</math> gives


<math display="block">
{{NumEqn|<math>
M_r=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_5}{p_2}-1\right)+1}
M_r=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_5}{p_2}-1\right)+1}
</math>
</math>}}


The ratios <math>T_5/T_2</math> and <math>\rho_5/\rho_2</math> can be obtained from Eqns.~\ref{eq:incident:tr} and \ref{eq:incident:rr} by analogy
The ratios <math>T_5/T_2</math> and <math>\rho_5/\rho_2</math> can be obtained from Eqns.~\ref{eq:incident:tr} and \ref{eq:incident:rr} by analogy


<math display="block">
{{NumEqn|<math>
\frac{T_5}{T_2}=\frac{p_5}{p_2}\left(\dfrac{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_5}{p_2}}{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_5}{p_2}}\right)
\frac{T_5}{T_2}=\frac{p_5}{p_2}\left(\dfrac{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_5}{p_2}}{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_5}{p_2}}\right)
</math>
</math>}}


<math display="block">
{{NumEqn|<math>
\frac{\rho_5}{\rho_2}=\dfrac{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_5}{p_2}}{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_5}{p_2}}
\frac{\rho_5}{\rho_2}=\dfrac{1+\dfrac{\gamma+1}{\gamma-1}\dfrac{p_5}{p_2}}{\dfrac{\gamma+1}{\gamma-1}+\dfrac{p_5}{p_2}}
</math>
</math>}}


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


<math display="block">
{{NumEqn|<math>
u_p=W_r\left(\frac{\rho_5}{\rho_2}-1\right)=M_r a_2 \left(1-\frac{\rho_2}{\rho_5}\right)
u_p=W_r\left(\frac{\rho_5}{\rho_2}-1\right)=M_r a_2 \left(1-\frac{\rho_2}{\rho_5}\right)
</math>
</math>}}


\subsection{Reflected Shock Relation}
==== Reflected Shock Relation ====


\noindent 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}\\
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}


\begin{equation}
{{NumEqn|<math>
M_r a_2 \left(1-\frac{\rho_2}{\rho_5}\right)=M_s a_1 \left(1-\frac{\rho_1}{\rho_2}\right)
M_r a_2 \left(1-\frac{\rho_2}{\rho_5}\right)=M_s a_1 \left(1-\frac{\rho_1}{\rho_2}\right)
\label{eq:relation:a}
</math>}}
\end{equation}\\


\noindent rewriting gives \\
rewriting gives


\begin{equation}
{{NumEqn|<math>
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=M_s \left(1-\frac{\rho_1}{\rho_2}\right) \frac{a_1}{a_2}
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=M_s \left(1-\frac{\rho_1}{\rho_2}\right) \frac{a_1}{a_2}
\label{eq:relation:b}
</math>}}
\end{equation}\\


\noindent Assuming calorically perfect gas gives $a=\sqrt{\gamma RT}$ and thus\\
Assuming calorically perfect gas gives <math>a=\sqrt{\gamma RT}</math> and thus


\begin{equation}
{{NumEqn|<math>
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=M_s \left(1-\frac{\rho_1}{\rho_2}\right) \sqrt{\frac{T_1}{T_2}}
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=M_s \left(1-\frac{\rho_1}{\rho_2}\right) \sqrt{\frac{T_1}{T_2}}
\label{eq:relation:c}
</math>}}
\end{equation}\\


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


\begin{equation*}
{{NumEqn|<math>
M_r \left(1-\frac{\rho_2}{\rho_5}\right)
M_r \left(1-\frac{\rho_2}{\rho_5}\right)
\end{equation*}\\
</math>}}


\noindent 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\\
Using the <math>\rho_5/\rho_2</math> and <math>p_2/p_5</math> from Eqns.~\ref{eq:reflected:rr} and~\ref{eq:reflected:pr} and simplifying gives


\begin{equation}
{{NumEqn|<math>
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_r^2-1}{M_r}\right)
M_r \left(1-\frac{\rho_2}{\rho_5}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_r^2-1}{M_r}\right)
\label{eq:relation:d}
</math>}}
\end{equation}\\


\noindent 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\\
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


\begin{equation}
{{NumEqn|<math>
M_s \left(1-\frac{\rho_1}{\rho_2}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_s^2-1}{M_s}\right)
M_s \left(1-\frac{\rho_1}{\rho_2}\right)=\left(\frac{2}{\gamma+1}\right)\left(\frac{M_s^2-1}{M_s}\right)
\label{eq:relation:e}
</math>}}
\end{equation}\\


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


\begin{equation}
{{NumEqn|<math>
\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}}
\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}}
\label{eq:relation:f}
</math>}}
\end{equation}\\


\noindent Simplifying and inverting gives\\
Simplifying and inverting gives


\begin{equation}
{{NumEqn|<math>
\left(\frac{M_r}{M_r^2-1}\right)=\left(\frac{M_s}{M_s^2-1}\right)\sqrt{\frac{T_2}{T_1}}
\left(\frac{M_r}{M_r^2-1}\right)=\left(\frac{M_s}{M_s^2-1}\right)\sqrt{\frac{T_2}{T_1}}
\label{eq:relation:g}
</math>}}
\end{equation}\\
 
The rightmost term in Eqn.~\ref{eq:relation:g} (<math>\sqrt{T_2/T_1}</math>) needs to be rewritten. Inserting~\ref{eq:incident:pr} in~\ref{eq:incident:tr} and expanding all terms gives
 
{{NumEqn|<math>
\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)^2M_s^2} =
</math>|nonumber=1}}
 
{{NumEqn|<math>
=\frac{2(\gamma+1) + (\gamma+1)(\gamma-1)M_s^2+4\gamma(M_s^2-1)}{(\gamma+1)^2M_s^2}+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\gamma =
</math>|nonumber=1}}


\noindent 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\\
{{NumEqn|<math>
  =\dfrac{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)^2M_s^2}
</math>|nonumber=1}}


\begin{align*}
{{NumEqn|<math>
\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)^2M_s^2} = \\
  \dfrac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\dfrac{1}{M_s^2}\right)
& \\
</math>}}
& =\frac{2(\gamma+1) + (\gamma+1)(\gamma-1)M_s^2+4\gamma(M_s^2-1)}{(\gamma+1)^2M_s^2}+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\gamma = \\
& \\
& =\dfrac{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)^2M_s^2}+\nonumber\\
  & \dfrac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\dfrac{1}{M_s^2}\right)
\end{align*}\\


\noindent Finally we end up with the following relation\\
Finally we end up with the following relation


\begin{equation}
{{NumEqn|<math>
\frac{T_2}{T_1}=1+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\frac{1}{M_s^2}\right)
\frac{T_2}{T_1}=1+\frac{2(\gamma-1)}{(\gamma+1)^2}(M_s^2-1)\left(\gamma+\frac{1}{M_s^2}\right)
\label{eq:relation:tr}
</math>}}
\end{equation}\\


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


\begin{equation}
{{NumEqn|<math>
\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)\left(\gamma+\frac{1}{M_s^2}\right)}
\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)\left(\gamma+\frac{1}{M_s^2}\right)}
\label{eq:relation:final}
</math>}}
\end{equation}\\


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

Latest revision as of 13:37, 1 April 2026

Moving Normal Shock Waves

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

ρ1u1=ρ2u2(Eq. 6.1)
ρ1u12+p1=ρ2u22+p2(Eq. 6.2)
h1+12u12=h2+12u22(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 u1=W, and the gas behind the shock moves to the right with the velocity u2=Wup. Now, let's insert u1 and u2 in the stationary shock relations \ref{eq:stationary:cont} - \ref{eq:stationary:energy}.

ρ1W=ρ2(Wup)(Eq. 6.4)
ρ1W2+p1=ρ2(Wup)2+p2(Eq. 6.5)
h1+12W2=h2+12(Wup)2(Eq. 6.6)

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

(Wup)=Wρ1ρ2(Eq. 6.7)

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

p1+ρ1W2=p2+ρ2W2(ρ1ρ2)2p2p1=ρ1W2(1ρ1ρ2)(Eq. 6.8)
W2=p2p1ρ2ρ1(ρ2ρ1)(Eq. 6.9)

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

W=(Wup)(ρ2ρ1)(Eq. 6.10)

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

(Wup)2=p2p1ρ2ρ1(ρ1ρ2)(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}).

h1+12[p2p1ρ2ρ1(ρ2ρ1)]=h2+12[p2p1ρ2ρ1(ρ1ρ2)](Eq. 6.12)
h=e+pρ(Eq. 6.13)
e1+p1ρ1+12[p2p1ρ2ρ1(ρ2ρ1)]=e2+p2ρ2+12[p2p1ρ2ρ1(ρ1ρ2)](Eq. 6.14)

which can be rewritten as

e2e1=p1+p22(1ρ11ρ2)(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=CvT. Inserted in the Hugoniot relation above this gives

Cv(T2T1)=p1+p22(ν1ν2)(Eq. 6.16)

where ν=1/ρ

Now, using the ideal gas law T=pν/R and Cv/R=1/(γ1) gives

(1γ1)(p2ν2p1ν1)=p1+p22(ν1ν2)



p2(ν2γ1ν1ν22)=p1(ν1γ1+ν1ν22)
(Eq. 6.17)

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

p2p1=(γ+1γ1)(ν1ν2)1(γ+1γ1)(ν1ν2)(Eq. 6.18)

ν=RT/p and thus

ν1ν2=T1T2p2p1(Eq. 6.19)

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

p2p1=(γ+1γ1)(T1T2p2p1)1(γ+1γ1)(T1T2p2p1)(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

T2T1=p2p1[(γ+1γ1)+(p2p1)1+(γ+1γ1)(p2p1)](Eq. 6.21)

Once again using the ideal gas law

ρ2ρ1=(γ+1γ1)+(p2p1)1+(γ+1γ1)(p2p1)(Eq. 6.22)

Going back to the momentum equation

p2p1=ρ1W2(1ρ1ρ2)={W=Msa1}=ρ1Ms2a12(1ρ1ρ2)(Eq. 6.23)

with a12=γp1/ρ1, we get

p2p1=γMs2(1ρ1ρ2)+1(Eq. 6.24)

From the normal shock relations, we have

ρ1ρ2=2+(γ1)Ms2(γ+1)Ms2(Eq. 6.25)

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

p2p1=1+(2γγ+1)(Ms21)(Eq. 6.26)

or

Ms=(γ+12γ)(p2p11)+1(Eq. 6.27)

Eqn. \ref{eq:unsteady:Mach} with Ms=W/a1

W=a1(γ+12γ)(p2p11)+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

ρ1W=ρ2(Wup)(Eq. 6.29)

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

W(ρ1ρ2)=ρ2up(Eq. 6.30)
up=W(1ρ1ρ2)(Eq. 6.31)

From before we have a relation for $W$ as a function of pressure ratio and one for ρ1/ρ2, 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

up=a1(γ+12γ)(p2p11)+1I[1(γ+1γ1)+(p2p1)1+(γ+1γ1)(p2p1)]II(Eq. 6.32)

The equation subsets I and II can be rewritten as:

Term I:

(γ+12γ)(p2p11)+1=γ+12γ[(p2p1)+(γ1γ+1)](Eq. 6.33)


Term II:

[1(γ+1γ1)+(p2p1)1+(γ+1γ1)(p2p1)]=1γ(p2p11)(2γγ+1)(γ1γ+1)+(p2p1)(Eq. 6.34)

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

up=a1γ(p2p11)(2γγ+1)(γ1γ+1)+(p2p1)(Eq. 6.35)

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

Mp=upa2=upa1a1a2=upa1γRT1γRT2=upa1T1T2(Eq. 6.36)

With up/a1 from Eqn. \ref{eq:unsteady:up} and T1/T2 from Eqn. \ref{eq:unsteady:temperature:ratio}

Mp=1γ(p2p11)((2γγ+1)(γ1γ+1)+(p2p1))1/2(1+(γ+1γ1)(p2p1)(γ+1γ1)(p2p1)+(p2p1)2)1/2(Eq. 6.37)

There is a theoretical upper limit for the induced Mach number Mp

limp2/p1Mp(p2p1)=2γ(γ1)(Eq. 6.38)

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

limp2/p1Mp(p2p1)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

p2p1=1+2γγ+1(Ms21)(Eq. 6.40)

where Ms is the wave Mach number, which is calculated as

Ms=Wa1(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 a1 is the speed of sound in region 1 (see Fig.~\ref{fig:reflection}).

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

Ms=γ+12γ(p2p11)+1(Eq. 6.42)

Anderson derives the relations for calculation of the ratio T2/T1

T2T1=p2p1(γ+1γ1+p2p11+γ+1γ1p2p1)(Eq. 6.43)

From Eqn.~\ref{eq:incident:tr} it is easy to get the corresponding relation for ρ2/ρ1

ρ2ρ1=1+γ+1γ1p2p1γ+1γ1+p2p1(Eq. 6.44)

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

up=W(1ρ1ρ2)=Msa1(1ρ1ρ2)(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

p5p2=1+2γγ+1(Mr21)(Eq. 6.46)

where Mr is the Mach number of the reflected shock wave defined as

Mr=Wr+upa2(Eq. 6.47)

where Wr is the speed of the reflected shock wave and a2 is the speed of sound in region 2 (see Fig.~\ref{fig:reflection}).

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

Mr=γ+12γ(p5p21)+1(Eq. 6.48)

The ratios T5/T2 and ρ5/ρ2 can be obtained from Eqns.~\ref{eq:incident:tr} and \ref{eq:incident:rr} by analogy

T5T2=p5p2(γ+1γ1+p5p21+γ+1γ1p5p2)(Eq. 6.49)
ρ5ρ2=1+γ+1γ1p5p2γ+1γ1+p5p2(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

up=Wr(ρ5ρ21)=Mra2(1ρ2ρ5)(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}

Mra2(1ρ2ρ5)=Msa1(1ρ1ρ2)(Eq. 6.52)

rewriting gives

Mr(1ρ2ρ5)=Ms(1ρ1ρ2)a1a2(Eq. 6.53)

Assuming calorically perfect gas gives a=γRT and thus

Mr(1ρ2ρ5)=Ms(1ρ1ρ2)T1T2(Eq. 6.54)

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

Mr(1ρ2ρ5)(Eq. 6.55)

Using the ρ5/ρ2 and p2/p5 from Eqns.~\ref{eq:reflected:rr} and~\ref{eq:reflected:pr} and simplifying gives

Mr(1ρ2ρ5)=(2γ+1)(Mr21Mr)(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

Ms(1ρ1ρ2)=(2γ+1)(Ms21Ms)(Eq. 6.57)

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

(2γ+1)(Mr21Mr)=(2γ+1)(Ms21Ms)T1T2(Eq. 6.58)

Simplifying and inverting gives

(MrMr21)=(MsMs21)T2T1(Eq. 6.59)

The rightmost term in Eqn.~\ref{eq:relation:g} (T2/T1) needs to be rewritten. Inserting~\ref{eq:incident:pr} in~\ref{eq:incident:tr} and expanding all terms gives

T2T1=2(γ+1)+(γ+1)(γ1)Ms2+4γ(Ms21)+2γ(γ1)Ms2(Ms21)(γ+1)2Ms2=
=2(γ+1)+(γ+1)(γ1)Ms2+4γ(Ms21)(γ+1)2Ms2+2(γ1)(γ+1)2(Ms21)γ=
=2(γ+1)+(γ+1)(γ1)Ms2+4γ(Ms21)(2(γ1)(Ms21))(γ+1)2Ms2
2(γ1)(γ+1)2(Ms21)(γ+1Ms2)(Eq. 6.60)

Finally we end up with the following relation

T2T1=1+2(γ1)(γ+1)2(Ms21)(γ+1Ms2)(Eq. 6.61)

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

(MrMr21)=(MsMs21)1+2(γ1)(γ+1)2(Ms21)(γ+1Ms2)(Eq. 6.62)

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