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The pressure , velocity and temperature just upstream of normal shock wave in air are 100 kPa(abs), 660 m/s and $-20^\circ$ respectively.

Calculate pressure, velocity and temperature just downstream of the shock wave. [Take ratio of specific heat k=1.4 and gas constant R=287 J/(kg.k)].

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Data:-

Upstream conditions:

$v_1=660m/s, P_1=110kPa=100\times 10^3 N/m^2$

$k=1.4, T_1=-20^\circ C=-20+273=253 K, R=287 J/kgK$

To find:-

$P_2=?,V_2=?,T_2=?$

Solution:-

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Sonic speed at upstream

$c_1=\sqrt{K.R.T_1}=\sqrt{1.9\times 287\times 253}=318.83 m/s$

Upstream mach number

$M_1=\frac{v_1}{c_1}=\frac{660}{318.83}=2.07$

Velocity at downstream, $v_2$

$\frac{v_2}{v_1}=\frac{(K-1)M_1^2+2}{(K+1)M_1^2}$

$\frac{v_2}{660}=\frac{(1.4-1)2.07^2+2}{(1.4+1)2.07^2}$

$\therefore v_2=238.4 m/s$

Pressure at downstream, $P_2$

$\frac{P_2}{P_1}=\frac{2KM_1^2-(K-1)}{K+1}$

$\frac{P_2}{100\times 10^3}=\frac{2\times 1.4\times 2.07^2-(1.4-1)}{1.4+1}$

$P_2=483.238\times 10^3 N/m^2$

Temperature at downstream, $T_2$

$\frac{T_2}{T_1}=\frac{[(K-1)M_1^2+2][2KM_1^2-(K-1)]}{(K+1)^2M_1^2}$

$\frac{T_2}{253}=\frac{[(1.4-1)2.07^2+2][2\times 1.4\times 2.07^2-(1.4-1)]}{(1.4+1)^2\times 2.07^2}$

$T_2=441.53 K$

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