Answer to Question #236817 in Molecular Physics | Thermodynamics for Unknown346307

Question

. In a duct in which air is flowing, a normal shock wave occurs at a Mach
number of 1.5. The static pressure and temperature upstream of the shock wave are 170 kN/m2
and 23

Answer 1

Let subscripts 1 and 2 represent flow conditions upstream and downstream of the shock wave respectively.

Mach number, $M_1=1.5$

Upstream pressure, $p_1 = 170 \; kN/m^2$

Upstream temperature, $T_1 = 23 + 273 = 296 \;K$

γ = 1.4

(i) Pressure, temperature and Mach number downstream of the shock :

$\frac{p_2}{p_1} = \frac{2γM_1^2 -(γ-1)}{γ+1} \\ = \frac{2 \times 1.4 \times 1.5^2 -(1.4-1)}{1.4+1} \\ = \frac{6.3-0.4}{2.4} \\ = 2.458 \\ p_2 = 170 \times 2.458 = 417.86 \;kN/m^2 \\ \frac{T_2}{T_1} = \frac{[(γ-1)M_1^2 +2][2γM_1^2 -(γ-1)]}{(γ+1)^2M_1^2} \\ = \frac{[(1.4-1) \times 1.5^2 +2][2 \times 1.4 \times 1.5^2 -(1.4-1)]}{(1.4+1)^2 \times 1.5^2} \\ = \frac{2.9 \times 5.9 }{12.96 } \\ = 1.32 \\ T_2 = 296 \times 1.32 = 390.72 \;K = 117.72 \;°C \\ M^2_2 = \frac{(γ-1)M_1^2 +2}{2γM_1^2 -(γ-1)} \\ = \frac{(1.4-1) \times 1.5^2 +2}{2 \times 1.4 \times 1.5^2 -(1.4-1)} \\ = \frac{2.9}{5.9} \\ = 0.49 \\ M_2 = 0.7$

(ii) Strength of shock :

Strength of shock $= \frac{p_2}{p_1} -1 = 2.458 -1=1.458$