# 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.

Asked on 11th March 2021 in

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,

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