Page 309 - Aerodynamics for Engineering Students

Compressible flow 291

i.e.
-0.8 = 1 - (!)’

whence

(!)’= 1.8 giving - = 1.34
4
V
Now speed of sound = 20.05 (273 - 24.6)’j’ = 318m s-l
Therefore, true flight speed = 0.2 x 318 = 63.6m s-l
Therefore, air speed over panel, q = 63.6 x 1.34 = 85.4m s-’
(ii) Here the flow is definitely compressible. As before,
P1 -P
c --
PI - 0.7pMz
and therefore
pi -p = 0.7 x 46 500 x (0.85)2 x (-0.5)
= -11 740Nm-’
Therefore, load on panel = 11 740 x (0.15)’ = 264N
There are two ways of calculating the speed of flow over the panel from Eqn (6.18):
(4

P [ 5 a’
P1 1 q2 - 3 3s

where a is the speed of sound in the free stream, i.e.

Now
p1 -p = -11 740N m-’

and therefore
p1 = 46 500 - 1 1 740 = 34 760 N mP2

Thus substituting in the above equation the known values p = 46 500Nm-2,
PI = 34760Nm-’ and M = 0.85 leads to
(!!)’= 1.124 giving - 4 = 1.06

U
Therefore

q = 1.06~ = 1.06 x 318 = 338ms-’
It is also possible to calculate the Mach number of the flow over the panel, as follows. The
local temperature Tis found from