7.3 Shock Wave ReflectionNiklas Andersson
Chalmers University of Technology
Department of Mechanical Engineering
Division of Fluid Dynamics
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.
Figure 7.1: Shock reflection at solid wall (located at x x x
The Incident Shock Wave ¶ The pressure ratio over the incident shock in Fig. Figure 7.1
p 2 p 1 = 1 + 2 γ γ + 1 ( M s 2 − 1 ) \frac{p_2}{p_1}=1+\frac{2\gamma}{\gamma+1}\left(M_s^2-1\right) p 1 p 2 = 1 + γ + 1 2 γ ( M s 2 − 1 ) where M s M_s M s
M s = W a 1 M_s=\frac{W}{a_1} M s = a 1 W In Eqn. (7.43) W W W a 1 a_1 a 1 Figure 7.1
Solving Eqn. (7.42) M s M_s M s
M s = γ + 1 2 γ ( p 2 p 1 − 1 ) + 1 M_s=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_2}{p_1}-1\right)+1} M s = 2 γ γ + 1 ( p 1 p 2 − 1 ) + 1 Anderson derives the relations for calculation of the ratio T 2 / T 1 T_2/T_1 T 2 / T 1
T 2 T 1 = p 2 p 1 ( γ + 1 γ − 1 + p 2 p 1 1 + γ + 1 γ − 1 p 2 p 1 ) \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) T 1 T 2 = p 1 p 2 ⎝ ⎛ 1 + γ − 1 γ + 1 p 1 p 2 γ − 1 γ + 1 + p 1 p 2 ⎠ ⎞ From Eqn. (7.45) ρ 2 / ρ 1 \rho_2/\rho_1 ρ 2 / ρ 1
ρ 2 ρ 1 = 1 + γ + 1 γ − 1 p 2 p 1 γ + 1 γ − 1 + 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}} ρ 1 ρ 2 = γ − 1 γ + 1 + p 1 p 2 1 + γ − 1 γ + 1 p 1 p 2 Anderson also shows how to obtain the induced velocity, u p u_p u p Figure 7.1
u p = W ( 1 − ρ 1 ρ 2 ) = M s a 1 ( 1 − ρ 1 ρ 2 ) 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 ( 1 − ρ 2 ρ 1 ) = M s a 1 ( 1 − ρ 2 ρ 1 ) The Reflected Shock Wave ¶ The pressure ratio over the reflected shock can be obtained from Eqn. (7.42)
p 5 p 2 = 1 + 2 γ γ + 1 ( M r 2 − 1 ) \frac{p_5}{p_2}=1+\frac{2\gamma}{\gamma+1}\left(M_r^2-1\right) p 2 p 5 = 1 + γ + 1 2 γ ( M r 2 − 1 ) where M r M_r M r
M r = W r + u p a 2 M_r=\frac{W_r+u_p}{a_2} M r = a 2 W r + u p where W r W_r W r a 2 a_2 a 2 Figure 7.1
Solving Eqn. (7.48) M r M_r M r
M r = γ + 1 2 γ ( p 5 p 2 − 1 ) + 1 M_r=\sqrt{\frac{\gamma+1}{2\gamma}\left(\frac{p_5}{p_2}-1\right)+1} M r = 2 γ γ + 1 ( p 2 p 5 − 1 ) + 1 The ratios T 5 / T 2 T_5/T_2 T 5 / T 2 ρ 5 / ρ 2 \rho_5/\rho_2 ρ 5 / ρ 2 (7.45) (7.46)
T 5 T 2 = p 5 p 2 ( γ + 1 γ − 1 + p 5 p 2 1 + γ + 1 γ − 1 p 5 p 2 ) \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) T 2 T 5 = p 2 p 5 ⎝ ⎛ 1 + γ − 1 γ + 1 p 2 p 5 γ − 1 γ + 1 + p 2 p 5 ⎠ ⎞ ρ 5 ρ 2 = 1 + γ + 1 γ − 1 p 5 p 2 γ + 1 γ − 1 + 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}} ρ 2 ρ 5 = γ − 1 γ + 1 + p 2 p 5 1 + γ − 1 γ + 1 p 2 p 5 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 ( ρ 5 ρ 2 − 1 ) = M r a 2 ( 1 − ρ 2 ρ 5 ) 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 ( ρ 2 ρ 5 − 1 ) = M r a 2 ( 1 − ρ 5 ρ 2 ) 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 u_p u p (7.47) (7.53)
M r a 2 ( 1 − ρ 2 ρ 5 ) = M s a 1 ( 1 − ρ 1 ρ 2 ) 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 ( 1 − ρ 5 ρ 2 ) = M s a 1 ( 1 − ρ 2 ρ 1 ) rewriting gives
M r ( 1 − ρ 2 ρ 5 ) = M s ( 1 − ρ 1 ρ 2 ) 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} M r ( 1 − ρ 5 ρ 2 ) = M s ( 1 − ρ 2 ρ 1 ) a 2 a 1 Assuming calorically perfect gas gives a = γ R T a=\sqrt{\gamma RT} a = γ RT
M r ( 1 − ρ 2 ρ 5 ) = M s ( 1 − ρ 1 ρ 2 ) 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}} M r ( 1 − ρ 5 ρ 2 ) = M s ( 1 − ρ 2 ρ 1 ) T 2 T 1 Let’s first look at the term on the left hand side of Eqn. (7.56)
M r ( 1 − ρ 2 ρ 5 ) M_r \left(1-\frac{\rho_2}{\rho_5}\right) M r ( 1 − ρ 5 ρ 2 ) Using the ρ 5 / ρ 2 \rho_5/\rho_2 ρ 5 / ρ 2 p 2 / p 5 p_2/p_5 p 2 / p 5 (7.52) (7.48)
M r ( 1 − ρ 2 ρ 5 ) = ( 2 γ + 1 ) ( M r 2 − 1 M r ) 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 ( 1 − ρ 5 ρ 2 ) = ( γ + 1 2 ) ( M r M r 2 − 1 ) Using the same approach on the corresponding term for the incident shock wave on the right hand side of Eqn. (7.56)
M s ( 1 − ρ 1 ρ 2 ) = ( 2 γ + 1 ) ( M s 2 − 1 M s ) 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 ( 1 − ρ 2 ρ 1 ) = ( γ + 1 2 ) ( M s M s 2 − 1 ) Now, inserting (7.58) (7.59) (7.56)
( 2 γ + 1 ) ( M r 2 − 1 M r ) = ( 2 γ + 1 ) ( M s 2 − 1 M s ) 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}} ( γ + 1 2 ) ( M r M r 2 − 1 ) = ( γ + 1 2 ) ( M s M s 2 − 1 ) T 2 T 1 Simplifying and inverting gives
( M r M r 2 − 1 ) = ( M s M s 2 − 1 ) 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}} ( M r 2 − 1 M r ) = ( M s 2 − 1 M s ) T 1 T 2 The rightmost term in Eqn. (7.61) T 2 / T 1 \sqrt{T_2/T_1} T 2 / T 1 (7.42) (7.45)
T 2 T 1 = 2 ( γ + 1 ) + ( γ + 1 ) ( γ − 1 ) M s 2 + 4 γ ( M s 2 − 1 ) + 2 γ ( γ − 1 ) M s 2 ( M s 2 − 1 ) ( γ + 1 ) 2 M s 2 = = 2 ( γ + 1 ) + ( γ + 1 ) ( γ − 1 ) M s 2 + 4 γ ( M s 2 − 1 ) ( γ + 1 ) 2 M s 2 + 2 ( γ − 1 ) ( γ + 1 ) 2 ( M s 2 − 1 ) γ = = 2 ( γ + 1 ) + ( γ + 1 ) ( γ − 1 ) M s 2 + 4 γ ( M s 2 − 1 ) − ( 2 ( γ − 1 ) ( M s 2 − 1 ) ) ( γ + 1 ) 2 M s 2 + 2 ( γ − 1 ) ( γ + 1 ) 2 ( M s 2 − 1 ) ( γ + 1 M s 2 ) \begin{aligned}
\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} = \\
& \\
& =\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{aligned} T 1 T 2 = ( γ + 1 ) 2 M s 2 2 ( γ + 1 ) + ( γ + 1 ) ( γ − 1 ) M s 2 + 4 γ ( M s 2 − 1 ) + 2 γ ( γ − 1 ) M s 2 ( M s 2 − 1 ) = = ( γ + 1 ) 2 M s 2 2 ( γ + 1 ) + ( γ + 1 ) ( γ − 1 ) M s 2 + 4 γ ( M s 2 − 1 ) + ( γ + 1 ) 2 2 ( γ − 1 ) ( M s 2 − 1 ) γ = = ( γ + 1 ) 2 M s 2 2 ( γ + 1 ) + ( γ + 1 ) ( γ − 1 ) M s 2 + 4 γ ( M s 2 − 1 ) − ( 2 ( γ − 1 ) ( M s 2 − 1 )) + ( γ + 1 ) 2 2 ( γ − 1 ) ( M s 2 − 1 ) ( γ + M s 2 1 ) Finally we end up with the following relation
T 2 T 1 = 1 + 2 ( γ − 1 ) ( γ + 1 ) 2 ( M s 2 − 1 ) ( γ + 1 M s 2 ) \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) T 1 T 2 = 1 + ( γ + 1 ) 2 2 ( γ − 1 ) ( M s 2 − 1 ) ( γ + M s 2 1 ) The temperature ratio over the incident shock wave is now totally defined by the incident Mach number M s M_s M s γ \gamma γ (7.63) (7.61)
( M r M r 2 − 1 ) = ( M s M s 2 − 1 ) 1 + 2 ( γ − 1 ) ( γ + 1 ) 2 ( M s 2 − 1 ) ( γ + 1 M s 2 ) \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)} ( M r 2 − 1 M r ) = ( M s 2 − 1 M s ) 1 + ( γ + 1 ) 2 2 ( γ − 1 ) ( M s 2 − 1 ) ( γ + M s 2 1 ) It should be noted that Eqn. (7.64)