Page 310 - Aerodynamics for Engineering Students

292 Aerodynamics for Engineering Students

giving

Ti = 0.920 T
and

=
a1 = ~(0.920)'/~ 318(0.920)1/2 = 306ms-'
Therefore, Mach number over panel = 338/306 = 1.103.
(b) The alternative method of solution is as follows, with the total pressure of the flow denoted
PO= [1 +;M2]3.5= [1+4
by Po:
(0.85)'
P 3'5
= (1.1445)3.5 = 1.605

Therefore
PO = 46 500 x 1.605 = 74 500 N mP2
As found in method (a)
p1 -p = -11 740NmP2

and
p1 = 34760NmP2

Then

giving
M? = 1.22,Ml = 1.103

which agrees with the result found in method (a).
The total temperature TO is given by
1/3.5
$= e) = 1.1445
Therefore

To = 1.1445 x 248.6 = 284 K
Then

To
_- - (2.15)1/3.5 = 1.244
Tl
giving

7'1 =-- 284 - 228 K
1.244