# Find \({P_2},{T_2},\,{\rho _2},\,{M_2},{P_{02}}\) and \({T_{02}}\) downstream of the shock wave.

A flow passes through a normal shock wave has a pressure, temperature and Mach number of \(1\,atm,\,295\,K\) and \(2.8\) upstream of the shock wave. Find \({P_2},{T_2},\,{\rho _2},\,{M_2},{P_{02}}\) and \({T_{02}}\) downstream of the shock wave.

From the table of normal shock properties; for, Mach number of \(2.8\), \(\frac{{{p_2}}}{{{p_1}}} = 8.98\), \(\frac{{{\rho _2}}}{{{\rho _1}}} = 3.664\), \(\frac{{{T_2}}}{{{T_1}}} = 2.451\), \(\frac{{{p_{02}}}}{{{p_1}}} = 10.57\), \({M_2} = 0.4882\).

Therefore,

\[{p_2} = 8.98{p_1} = 8.98\left( {1atm} \right) = 8.98\,atm\]

\[{T_2} = 2.451{T_1} = 2.451\left( {295} \right) = 723.045\,K\]

\[{p_{_1}} = {\rho _1}R{T_1} \Rightarrow {\rho _1} = \left( {\frac{{1 \times \left( {1.01 \times {{10}^5}} \right)}}{{287 \times 295}}} \right) = 1.193\,kg/{m^3}\]

\[{\rho _2} = 3.664\left( {1.193} \right) = 4.371\,kg/{m^3}\]

\[{p_{02}} = \left( {\frac{{{p_{02}}}}{{{p_1}}}} \right){p_1} = 10.57\left( 1 \right) = 10.57\,atm\]

From isentropic flow properties table, for, Mach number of \(2.8\), \[\frac{{{T_{01}}}}{{{T_1}}} = 2.568\]

Since, \({T_{02}} = {T_{01}}\), Total temperature is constant,

Therefore,

\[{T_{02}} = \left( {\frac{{{T_{01}}}}{{{T_1}}}} \right){T_1} = 2.568 \times \left( {295} \right) = 757.56\,K\]