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

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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isoceles right angled triangle