Question #236998
For a normal shock wave in air Mach number is 2. If the atmospheric
pressure and air density are 26.5 kN/m2 and 0.413 kg/m3 respectively, determine the flow condi￾tions before and after the shock wave. Take γ = 1.4.
1
Expert's answer
2021-09-15T11:39:24-0400

P1=26.5  kN/m2M=2e1=0.413  kg/m3P1=e1RT1T1=26.50.413×0.287=223.57  KP_1 = 26.5 \;kN/m^2 \\ M= 2 \\ e_1 = 0.413 \;kg/m^3 \\ P_1 = e_1RT_1 \\ T_1 = \frac{26.5}{0.413 \times 0.287} = 223.57 \;K

From Normal shock table at M1=2M_1 = 2

P2P1=4.5T2T1=1.6875M2=0.577P2=45×26.5=119.25  kN/m2T2=223.57×1.6875=2377.27  KM2=V2γRT20.577=V21.4×287×377.27V2=224.65  m/se2=P2RT2=119.250.287×377.27=1.1013  kg/m3\frac{P_2}{P_1} = 4.5 \\ \frac{T_2}{T_1} = 1.6875 \\ M_2 = 0.577 \\ P_2 = 45 \times 26.5 = 119.25 \;kN/m^2 \\ T_2 = 223.57 \times 1.6875 = 2377.27 \;K \\ M_2 = \frac{V_2}{ \sqrt{γRT_2}} \\ 0.577 = \frac{V_2}{\sqrt{1.4 \times 287 \times 377.27}} \\ V_2 = 224.65 \;m/s \\ e_2 = \frac{P_2}{RT_2} = \frac{119.25}{0.287 \times 377.27} = 1.1013 \;kg/m^3


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