A subsonic airplane is flying at a 5000-m altitude where the atmospheric conditions are 54 kPa and 256 K. A Pitot static probe measures the

Question

A subsonic airplane is flying at a 5000-m altitude where the atmospheric conditions are 54 kPa and 256 K. A Pitot static probe measures the difference between the static and stagnation pressure to be 22 kPa. Calculate the speed of the airplane and the flight Mach number

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Trúc Chi 2 months 2021-09-01T07:57:37+00:00 1 Answers 0 views 0

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    2021-09-01T07:58:50+00:00

    Answer:

    A) Mach Number = 0.716

    B) Speed of Aircraft = 229.64 m/s

    Explanation:

    Formula for stagnation pressure is given as;

    Po = P + ΔP = 54 + 22 = 76 KPa

    Also, Po/P = [1 + {((k – 1)m²)/2}]^(k/(k-1))

    Where m is mach number and k is specific heat ratio of air.

    Specific heat ratio of air = 1.4

    Thus we can solve as follows;

    76/54 = [1 + {((1.4 – 1)m²)/2}]^(1.4/(1.4-1))

    1.407 = [1 + (0.2m²)]^(1.4/0.4)

    1.407 = [1 + (0.2m²)]^(7/2)

    Now, raise both sides to the power of 2/7 to obtain ;

    (1.407)^(2/7) = 1 + 0.2m²

    Thus, 1 + 0.2m² = 1.1025

    Subtract 1 from both sides;

    0.2m² = 1.1025 – 1

    0.2m² = 0.1025

    m² = 0.1025/0.2

    m² = 0.5125

    m = 0.716

    Now, the speed of sound in air is given as;

    C = √KRT

    R is ideal gas constant which is 287 J/Kgmol and T is temperature which is 256K

    Thus,

    C = √(1.4 x 287 x 256) = 320.72 m/s

    Thus, Speed of aircraft = Cm

    Speed of aircraft = 320.72 x 0.716 = 229.64 m/s

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