Answer to Question #239569 in Molecular Physics | Thermodynamics for Unknown346307

Question

. A supersonic aircraft flies at an altitude of 1.8 km where temperature is

4°C. Determine the speed of the aircraft if its sound is heard 4 seconds after its passage over the

head of an observer. Take R = 287 J/kg K and γ = 1.4

Answer 1

Altitude of the aircraft = 1.8 km = 1800 m

Temperature, T = 4 +273 = 277 K

Time, t= 4 s

Speed of the aircraft, V :

Refer Fig. Let O represent the observer and A the position of the aircraft just vertically over the observer. After 4 seconds, the aircraft reaches the position represented by the point B. Line AB represents the wave front and α the Mach angle.

From Fig., we have

$tan α = \frac{1800}{4V} = \frac{450}{V}$

But, Mach number,

$M= \frac{C}{V} = \frac{1}{sin α} \\ V=\frac{C}{sin α}$

Substituting the value of V, we get

$tan α = \frac{450}{(C/sin α)} = \frac{450 sin α}{C} \\ \frac{sin α}{cos α} = \frac{450 sin α}{C} \\ cos α = \frac{C}{450} \\ C=\sqrt{γRT},$

where C is the sonic velocity.

$R=287 \;J/kg \;K \\ γ = 1.4 \\ C = \sqrt{1.4 \times 287 \times 277} = 333.6 \;m/s$

Substituting the value of C we get

$cos α = \frac{333.6}{450} = 0.7413 \\ sin α = \sqrt{1 -cos^2 α} = \sqrt{1 -0.7413^2} = 0.6712$

Substituting the value of sin α we get

$V = \frac{C}{sin α} = \frac{333.6}{0.6712} = 497 \;m/s \\ = \frac{497 \times 3600}{1000} = 1789.2 \;km/h$