TITLE.PM5

COMPRESSIBLE FLOW 885

dharm

\M-therm\Th16-2.pm5

The quantity within the brackets may be positive or negative depending upon the magni-

tude of Mach number. By integrating eqn. (16.40), we can obtain a relationship between the

critical throat area Ac, where Mach number is unity and the area A at any section where M (^) <> 1
A
Ac =
12 1
1
2
1
21
M
+−M

• L
  N
  M
  M
  O
  Q
  P
  P


• 
  

  ()γ ()−
  γ
  γ
  γ
  ...(16.41)
  Example 16.12. The pressure leads from Pitot-static tube mounted on an aircraft were
  connected to a pressure gauge in the cockpit. The dial of the pressure gauge is calibrated to read
  the aircraft speed in m/s. The calibration is done on the ground by applying a known pressure
  across the gauge and calculating the equivalent velocity using incompressible Bernoulli’s equa-
  tion and assuming that the density is 1.224 kg/m^3.
  The gauge having been calibrated in this way the aircraft is flown at 9200 m, where the
  density is 0.454 kg/m^3 and ambient pressure is 30 kN/m^2. The gauge indicates a velocity of
  152 m/s. What is the true speed of the aircraft? (UPSC)
  Sol. Bernoulli’s equation for an incompressible flow is given by,
  p + ρV
  2
  2
  = constant
  The stagnation pressure (ps) created at Pitot-static tube,
  ps = p 0 + ρ^00
  2
  2
  V (neglecting compressibility effects) ...(i)
  Here p 0 = 30 kN/m^2 , V 0 = 152 m/s, ρ 0 = 1.224 kg/m^3 ...(Given)
  ∴ ps = 30 +
  1 224 152
  2

  
  

. ×^2
×10–3 = 44.139 kN/m^2
Neglecting compressibility effect, the speed of the aircraft when
ρ 0 = 0.454 kg/m^3 is given by [using eqn. (i)],

44.139 × 10^3 = 30 × 10^3 +

0 454

2

0

. ×V^2

or V 02 =

(. )

.

44 139 30 10 2

0 454

−× ×^3

= 62286.34

∴ V 0 = 249.57 m/s

Sonic velocity, C 0 = γRT 0 = γρ

p 0

0

= 1.4

0.454

×30 10×

3

= 304.16 m/s

Mach number, M =

V

C

0

0

249 57

304 16

=.

. = 0.82

Compressibility correction factor =^1

4

0

2

+

F

HG

I

KJ

M

, neglecting the terms containing higher powers

of M 0 (from eqn 16.19).

=^1

082

4

FHG +. IKJ = 1.168

∴ True speed of aircraft =

249 57

1 168

.

.

= 230.9 m/s

Hence true speed of aircraft = 230.9 m/s (Ans.)