Question

Consider an oblique shock wave with a wave angle of 35 o . Upstream of the wave, the static pressure and temperature are 2,000 lbf/ft2 and 520oR, respectively. The air velocity is 3,355 ft/s. Calculate the pressure, temperature, velocity and flow deflection angle behind the wave.

Answer 1

Answer:

The pressure is 6570  lbf/ft²

The temperature is 766 ⁰R

The velocity is 2746.7 ft/s

deflection angle behind the wave is 17.56⁰

Explanation:

Speed of air at initial condition:

$a_1 = \sqrt{\gamma RT } = \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s$

γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.

initial mach number

$M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7} = 3$

then, $M_n = M_1sin \beta = 3sin(35) = 1.721$

based on the values obtained, read off the following from table;

P₂/P₁ = 3.285

T₂/T₁ = 1.473

Mₙ₂ = 0.6355

Thus;

P₂ = 3.285P₁ = 3.285(2000) = 6570  lbf/ft²

T₂ = 1.473T₁ = 1.473(520⁰R) = 766 ⁰R

Again; to determine the velocity and deflection angle, first we calculate the mach number.

$M_t_1 = M_1cos \beta = 3 cos(35) = 2.458$

$w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s$

$a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s$

$v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s$

$Tan(\beta -\theta) = \frac{v_2}{w_2} = \frac{862.59}{2746.7} \\\\Tan(\beta -\theta) = 0.314\\\\\beta -\theta= 17.44\\\\\theta = \beta - 17.44 = 17.56^o$