TITLE.PM5

COMPRESSIBLE FLOW 869

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Speed of the aircraft, V :
Refer Fig. 16.3. Let O represent the observer and
A the position of the aircraft just vertically over the
observer. After 4 seconds, the aircraft reaches the posi-
tion represented by the point B. Line AB represents the
wave front and α the Mach angle.
From Fig. 16.3, we have

tan α =

1800

4

450

VV

= ...(i)

But, Mach number, M =

C

V

=

1

sinα

or, V =

C

sinα ...(ii)

Substituting the value of V in eqn. (i), we get

tan α =

450 450

(/sin )

sin

CCα

α

=

or,

sin

cos

α

α =

450 sinα

C

or cos α =

C

450

...(iii)

But C = γRT, where C is the sonic velocity.

R = 287 J/kg K and γ = 1.4 ...(Given)

∴ C = 1 4 287 277.×× = 333.6 m/s

Substituting the value of C in eqn. (ii), we get

cos α =

333 6

450

= 0.7413

∴ sin α = 1 −=−cos^22 α 1 0 7413. = 0.6712

Substituting the value of sin α in eqn. (ii), we get

V =

C

sin

α.

==333 6 ×

0 6712

497 m/s =497 3600

1000 = 1789.2 km/h (Ans.)

16.6. Stagnation Properties

The point on the immersed body where the velocity is zero is called stagnation point. At
this point velocity head is converted into pressure head. The values of pressure (ps), temperature
(Ts) and density (ρs) at stagnation point are called stagnation properties.

16.6.1. Expression for stagnation pressure (ps) in compressible flow

Consider the flow of compressible fluid past an immersed body where the velocity becomes
zero. Consider frictionless adiabatic (isentropic) condition. Let us consider two points, O in the free
stream and the stagnation point S as shown in Fig. 16.4.
Let, p 0 = pressure of compressible fluid at point O,
V 0 = velocity of fluid at O,
ρ 0 = density of fluid at O,
T 0 = temperature of fluid at O,

a

4V

1800 m

O

AB = Vt = 4V

A B

Fig. 16.3