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:
[tex]a_1 = \sqrt{\gamma RT } = \sqrt{1.4* 1716*520 } = 1117.70 \ ft/s[/tex]
γ is the ratio of specific heat, R is the universal gas constant, and T is the initial temperature.
initial mach number
[tex]M_1 = \frac{v_1}{a_1} = \frac{3355}{1117.7} = 3[/tex]
then, [tex]M_n = M_1sin \beta = 3sin(35) = 1.721[/tex]
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.
[tex]M_t_1 = M_1cos \beta = 3 cos(35) = 2.458[/tex]
[tex]w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/s[/tex]
[tex]a_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/s[/tex]
[tex]v_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/s[/tex]
[tex]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[/tex]
