Q. 3.13

Shock-Wave Induced Velocity

A normal shock wave travels through stagnant air at v = 700 m/s, Assuming $\rm T_{air}$ = 15°C, k = 1.4, and R = 287 m²/(s²K), find the induced velocity $\rm\Delta v =v_2-v_1$.

 Sketch                                          Model Assumptions • Isentropic flow • Steady-state • Normal shock wave

Verified Solution

Using Eq. (3.49) where $\rm M_1=\frac{v_1}{c}= \frac{v_1}{\sqrt{kRT_1}} :=2.06$,

$\rm M_2=\left\{\frac{M_1^2+2/(k-1)}{[2k/(k-1)]M_1^2-1} \right\} ^{1/2}$                           (3.49)

$\mathrm{M}_{2}=\left\lgroup{\frac{\mathrm{M}_{1}^{2}+5}{7\mathrm{M}_{1}^{2}-1}}\right\rgroup^{1/2}:=0.567$

To obtain $\rm v_2$, i.e. $\rm v_2=M_2c_2=M_2\sqrt{kRT_2}$, we need $\rm T_2$. Employing Eq. (3.46)

$\rm T_1\left\lgroup1+\frac{k-1}{2}M_1^2 \right\rgroup =T_2\left\lgroup1+\frac{k-1}{2}M_2^2 \right\rgroup$                     (3.46)

$\rm\mathrm{T}_{2}={\frac{\mathrm{T}_{2}\left({{ M}_{1}}^{2}+5\right)(7{ M}_{1}^{2}-1)}{36{{ M}_{1}}^{2}}}:=500\,{ K}$

We now can calculate

$\rm v_2$ = 254 m/s  and  Δv = 254 − 700 = −446 m/s

Comments: A downstream flow with $\rm v_2$ = 254 m/s is observed from the stationary shock wave where the upstream airflow approaches at $\rm v_1$ = 700m/s. The induced velocity of 446 m/s is in the same direction as the propagating shock wave (see Sketch).