Consider an oblique shock wave with a wave angle equal to 35-degrees. Upstream of the wave, P1 = 2000 lb/ft2, T1 = 520-degrees R, and V1 = 3

Question

Consider an oblique shock wave with a wave angle equal to 35-degrees. Upstream of the wave, P1 = 2000 lb/ft2, T1 = 520-degrees R, and V1 = 3355 ft/s. Calculate P2, T2, V2, and the flow deflection angle.

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Doris 3 years 2021-08-23T05:23:01+00:00 1 Answers 123 views 0

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    2021-08-23T05:24:09+00:00

    Answer:

    P2 The pressure is 6570  lbf/ft²

    T2 The temperature is 766 ⁰R

    V2 The velocity is 2746.7 ft/s

    flow deflection angle 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,

    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

    We determine the velocity and deflection angle, first we calculate the mach number.

    .M_t_1 = M_1cos \beta = 3 cos(35) = 2.458w_2 = a_1M_t_1 = 2.458(1117.70) = 2746.7 \ ft/sa_2 = \sqrt{\gamma RT_2} = \sqrt{1.4*1718*766} = 1357.34 \ ft/sv_2 = a_2M_n_2 = 1357.34(0.6355) = 862.59 \ ft/sTan(\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

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