Question

Air ows over a wall at a supersonic speed. The wall turns towards the ow generating an oblique shock wave. This wave is found to be at an angle of 400 to the initial ow direction. A scratch on the wall upstream of the shock wave is found to generate a very weak wave that is at an angle of 300 to the ow. Find the angle through

Answer 1

Answer:

10.65°

Explanation:

To calculate the initial Mach Number

$Sin \beta _{weak} = \frac{1}{Ma_1}$

$Ma_1 = \frac{1}{sin \beta_weak}$

$Ma_1 = \frac{1}{sin 30}$

$Ma_1 = 2$

Now, to calculate the angle turned by the wall; we have the following expression.

$tan \theta = \frac{2 cot \beta_i (Ma_1^2 sin^2 \beta_i -1 )}{Ma_1^2(k+cos2 \beta_i)+2}$

where:

$\beta_i$ = the angle between the initial flow direction = 40°

K is the ratio of specific heats o fair = 1.4

$tan \theta = \frac{2 (cot 40) (2^2 (sin 40)^2 -1 )}{2^2(1.4+cos2 (40))+2}$

$tan \theta = \frac{1.56}{8.295}$

$tan \theta = 0.188$

$\theta = tan ^{-1}( 0.188)$

$\theta = 10.65^0$