Page 311 - Aerodynamics for Engineering Students

Compressible flow 293

and the local speed of sound over the panel, al, is
a1 = 20.05(228)'/2 = 305ms-'

Therefore, flow speed over the panel
q = 305 x 1.103 = 338ms-'

which agrees with the answer obtained by method (a).
An interesting feature of this example is that, although the flight speed is subsonic
(M = 0.85), the flow over the panel is supersonic. This fact was used in the 'wing-flow' method
of transonic research. The method dates from about 1940, when transonic wind-tunnels were
unsatisfactory. A small model was mounted on the upper surface of the wing of an aeroplane,
which then dived at near maximum speed. As a result the model experienced a flow that was
supersonic locally. The method, though not very satisfactory, was an improvement on other
methods available at that time.

Example 6.7 A high-speed wind-tunnel consists of a reservoir of compressed air that dis-
charges through a convergent-divergent nozzle. The temperature and pressure in the reservoir
are 200 "C and 2 MNrnp2 gauge respectively. In the test section the Mach number is to be 2.5.
If the test section is to be 125 mm square, what should be the throat area? Calculate also the mass
flow, and the pressure, temperature, speed, dynamic and kinematic viscosity in the test section.

A 1 5+M2 1 5+6.25
-=-- =--
A* M( 6 ) 2.5( 6 )=2.64
( 12q2
Therefore, throat area = - 5920 (mm)*
=
2.64
Since the throat is choked, the mass flow may be calculated from Eqn (6.24), is
(3
massflow=0.0404 - A*

Now the reservoir pressure is 2MNm-2 gauge, or 2.101 MNm-2 absolute, while the
reservoir temperature is 200 "C = 473K. Therefore

mass flow = 0.0404 x 2.101 x lo6 x 5920 x 10-6/(473)''2
= 23.4 kg s-l

In the test section
1 6 25
1 + -M2 = 1 + - 2.25
=
5 5
Therefore
po/pl = (2.25)3.5 = 17.1
Therefore
2.101 x 106
pressure in test section = = 123 kNm-2
17.1
Also
-- - 2.25
To
TI